Transcript for:

hi welcome to the court Maps ultimate GCSE higher revision video in this video we're going to go through the topics on the GCSE higher maths course and the aim is to make sure you're familiar with those topics now if you've got your GCSE Maps exams coming up your GCSE Hammer apps exams coming up this video is going to be absolutely perfect for you also it's useful for anyone who wants to brush up on their mouth skills now this video is quite large I spend two or three minutes going through every single topic so rather than watching it one go I highly recommend you watch it in chunks and you make notes as you go through and watch the video but also there's an accompanying booklet and that booklet has a question every single topic as we go through so that booklet can be useful to do as you're watching the video alternatively you can keep it to the end of the video and then do that booklet now in terms of this video we're going to go through the topics on the GCSE higher revision checklist so this is the Corp Maps GCC higher revision checklist and if you go to the website there's a checklist for lxl aqn OCR so those checklist super spoke for each of the examples and they look something like this now they're pretty much identical with only a couple of subtle differences and as we go through this video I'll point out those subtle differences as we go through that particular topic but there's not many of them an example being a topic such as capture recapture where one exam board covers it and the other ones the other two don't but the different is a very subtle and I'll talk about those as we go through so we're going to go through the topics in red to begin with so the topics in red are the number or the ratio and proportion topics such as standard form limits of accuracy proportion reverse percentages compound interest topics such as those then we're going to go for the topics in green which are the geometry or the ships based the measures topics such as trigonometry Pythagoras your circle firms vectors topics such as those then we're going to go through and blue the algebra topics topics such as simultaneous equations quadratics types of graphs and so on and then finally we're going to go through the topics in Orange which are the statistics or the probability topics now this revision checklist is going to be really useful for you so I've put a link to it in the description below you may want to print it keep it up on your wall bring it with you on the bus it's very important that you have that with this checklist and this checklist has a video tutorial every single topic so if you want a more detailed explanation rather than the two or three minute explanation I'm going to do in every topic in this video if you wanted a five or ten minute explanation on the topic let's do a more detailed video tutorials on each one of these these video numbers and you can go to coopermobs.com and watch those now as I've mentioned I've made an A booklet to accompany this video and it's called the ultimate GCSE higher revision question booklet catchy title I know and there's a link to in the description below and this booklet is going to be really useful as you watch this video as I've said there's a question every single topic as we go through there's a QR code which brings you to this video also there's a QR code which it brings you to the answers as well on the front of the booklet so that book is going to be really useful now as we go through this video sometimes I use some of the information from the chord Mars revision cards and these are going to be really useful for you if you're studying for your GCSE higher exams so I'd highly recommend those and there's a link to them in the description below so they're the corporation cards also whenever you're revising for your GCSE higher exams I'd highly recommend a little and often approach so the chord Mass five a day booklets are going to be really useful for that and we've got three versions that'll be particularly useful for you if you're selling for your higher exams and they will be the yellow Foundation plus booklets the green higher booklets and the blue higher plus booklets and there's five GCSE questions for every single day of the year in each of those books and there's answers at the back of those books as well okay okay so our first topic is adding fractions in this video 133 in corporate maths so whenever I go through this video for every single topic we go through in the top right hand corner there's the video number from quadmavs.com so if you want a more detailed explanation rather than the two or three minute explanation I'm doing in this video if you want the longer 5 10 15 minute explanation if you record mouse.com and go to this video number then that will be the video tutorial that you can watch okay let's get started so to add fractions it's very important that the fractions have the same denominator a common denominator so for instance if we were asked to work out two-fifths plus a quarter as you can see the denominators the numbers on the bottom of the fraction aren't the same we've got a five and a four so what we're going to do is we're going to find the lowest common multiple of five and four well that's going to be 20 because 5 10 15 20 4 8 12 16 20. the lowest common multiple of both of them is 20. and what we're going to do is we're going to find two equivalent fractions now remember to find an equivalent fraction we consider what do you multiply 5 by I to get the 20 well that's four so you multiply the numerator by 4 as well so 4 times 2 is equal to 8. so two-fifths is the same as eight twentieths next our quarter we want to have 20 on the denominator so to get from 4 to 20 you multiply by five so you need to multiply the one by five as well so that's going to be five so one quarter is five twentieths and finally we've got e at twentieths plus five twentieths when we're adding fractions with the same denominator we just add the numerators so if we had eight twentieths plus five twentieths that's going to be thirteen twentieths and that's it so two-fifths plus one quarter is thirteen twentieths okay let's have a look at our next question so our next question is the workout two and seven ninths plus one and two thirds so here we've got some mixed numbers and the mixed numbers where you've got a whole number and a fraction so for instance four and a half and we're going to be dealing with these mixed numbers and whenever we're dealing with mixed numbers whenever add nurse subtracting or multiplying dividend fractions it can be very useful to turn them into top heavy fractions so agitate the whole number which is 2 and times by the denominator so 2 times 9 is 18 and add on the 7 is 25 so that gives me the 25 9. next I take the whole number which is 1 and I do 1 times 3 is 3 and add on the numerator which is 2 so 3 plus 2 is equal to 5 so it'd be Five Thirds so here we've written our mixed numbers as top heavy fractions so we want to have the same denominators so we get ninths and we've got thirds so the lowest common multiple of nine and three would just be nine we could multiply both the numerator and the denominator of this fraction by three and that would give us 9 on the denominator so let's do that so we've got 25 ninths and then we're going to add or if we multiply both of these by three we get well 5 times 3 is 15 and 3 times 3 is equal to 9. so we've got twenty five nines plus 15 nines and that's fantastic because they've both got the same denominator so if we just add the numerators 25 plus 15 is 40 so answer would be 40 ninths I would change this top heavy fraction or answer it into a mixed number as well remember the line in a fraction means divided by so how many 29 Squints of 40 well 4 9 is equal to 36 so it's going to be 4 and 36 where we have 40 so our remainder would be four so our answer would be 4 and 4 9 and that's it our next topic is multiplying fractions so multiplying fractions is really easy you just need to multiply the numerators and multiply the denominators and that's it so if you had two fifths times a quarter if you multiply the numerators two times one that's two and if you multiply the denominators 5 times 4 that's 20. now the answer here is two twentieths I'm going to cancel this down I'm going to simplify it we can divide both of these by two so final answer would be one tenth okay next we've got 1 and 1 7 for multiplied by two-thirds again this is a mixed number so let's make this a top heavy fraction one times seven is seven plus the one is eight so one and one seventh would be eight sevenths and then multiply by two thirds so let's multiply the numerators so 8 times 2 is 16 and 7 times 3 is 21. so our answer would be 16 over 21. 16 and 21 have no common factors apart from one and that's not going to help us so if you want more practice and multiplying fractions if you watch video 142 on corporate maps you can watch the video tutorial on it you can do some practice questions so the textbook exercises but also remember you've got that practice question booklet so you can't do these questions down that booklet and just make sure you're you're happy with this topic okay let's have a look at our next topic our next topic is dividing fractions which is video 134 in corporate Maps so to divide fractions what we do is we multiply by the reciprocal of the number we're dividing by and that's really complicated but what it means is if you've got 7 15 divided by three quarters instead of doing 7 15 divided by three quarters what we can do is we can do 7 15 multiply by and then we can find the reciprocal of three quarters and the reciprocal is a fancy word for flipping over so instead of writing three quarters we could write four thirds Okay so we've at 7 15 multiplied by four thirds now we've got multiply so we can just multiply the numerators and multiply the denominators so 7 times 4 is equal to 28 and 15 times 3 is equal to 45. so answer is 28 40 fifths and apart from one they've got no over common factors so that's it okay next question our next question is to work out two and a half divided by one and three-fifths so we've got mixed numbers here so let's turn them into top heavy fractions so two well we've got two times two is four plus one is five so it's going to be five halves and we're dividing by well if we look we've got one and three-fifths so one times five is five plus three is eight so it's going to be eight fifths now remember we wanna multiply by the reciprocal of the number we're dividing by so we're going to write five over two multiplied by five over eight next we're going to multiply so we've got 5 times 5 is 25 and 2 times 8 is 16. so our answer would be 25 over 16. now because the question is given in mixed numbers I'm going to write this top every fraction as a mixed number so the line means divide by so we've got 25 divided by 16. so there's 116 and 25 and the remainder is 9 so we've got 1 and 9 16. and that's it okay let's have a look at our next topic so next topic is reciprocals and that's video 145 from corporate Maps so reciprocals we talked about that whenever we're dividing by fractions we multiply by the reciprocal so to find the reciprocal of a number we flip it over another way to look at it is a number multiplied by its reciprocal will always give you one so if we've got five five is the same as five wholes five four one if we flip that over we would get one fifth so the reciprocal of five is one-fifth and if we think about it five times one-fifth is equal to one so a number times this reciprocal is always equal to one so the reciprocal of five is one-fifth okay next one find the reciprocal of a third well if we find the reciprocal of a third we flip it over which is three over one which is just three so the reciprocal of one third is free and if you just check it a third times three is equal to one okay our next one the next one says find the reciprocal of seven tenths well flipping it over would be ten over seven alternatively we're gonna write this as a mixed number so how many sevens go into ten one remainder three sevenths so the reciprocal of seven tenths is one and three sevens or ten over seven okay next we've been asked to find the reciprocal of 0.75 well 0.75 is equal to three quarters so when we find this reciprocal we flip it over so it's going to be four thirds so the reciprocal of 0.75 is four thirds or alternatively we could write as a mixed number so it'd be one because one three goes into four remainder one to one and one third and then finally we've been asked to find the reciprocal of two and a quarter now whenever we're finding the reciprocal of a mixed number I like to write as a top every fraction to begin with so two and a quarter well two times four is equal to eight plus one is nine so it's gonna be nine quarters and then if we find the reciprocal that would be flipping over 4 four notes and that's it so the reciprocal of two and a quarter is four ninths so we've now got multiplying decimals and dividing decimals so multiplying decimals now to multiply decimals we can do this in two ways we can either just ignore the decimal point and do 2854 and multiply that by eight and then looking at our question we've got one two three four digits after the decimal point so you put the decimal point in the answer so there's four digits after it alternatively you could multiply this number by a thousand to get two thousand eight hundred and fifty four and you can multiply this number by ten to get eight and then with your final answer you divide it by a thousand and divided by ten and then that would give you your answer as well okay so let's do 2854 multiplied by eight and see what we get so 8 times 4 is equal to 32 so put the two down and carry the three eight times five is forty plus three is forty three so put the three down carry the four eight times eight sixty four plus four sixty eight so put the eight down okay the six and eight times two is equal to sixteen plus six is 22 so we've got 22 832 and to get our final answer we can either consider the decimal places in the question so it was one two three four digits after decimal points in the question so it's going to be four after it in the answer so the answer would be 2.2832 like so alternatively you can take this twenty two thousand eight hundred and thirty two divided by a thousand because you need to multiply this number by a thousand to get a whole number so dividing this by five would be 22.832 and then dividing by another 10 would be 2.2832 and that's it okay and the next question our next question is dividing by decimal so we've got one point divided by 0.04 now remember that if you have something like 6 divided by three that's equal to two and if you multiply above those numbers the number you're dividing and the number you're dividing by above by ten you get 60 divided by 30 and then whenever you divide that you get two and if you had 600 divided by 300 you get two and so on so as long as you multiply both the number you're dividing and the number you're dividing by by either 10 or 100 or a thousand or whatever you'll always get the same answer so if we had 1.2 divided by 0.04 I want to turn this 0.04 into a whole number so I want to multiply it by 10 by 10 again to get 4. I want to multiply by 100. so I'm going to multiply both of these numbers by 100. so 1.2 times 100 is 120 and 0.04 times 100 is equal to 4. so we're going to do 120 divided by 4 and that's equal to 30. so answer would be 30. okay let's have a look at our next topic so our next topic is fractions decimals and percentages and last video is 121 to 129 in corporate Maps so it's going to be very important that you're able to convert between fractions decimals and percentages so let's have a look at these questions that'll help us remember how to do that so let's change 17 20 to decimal and let's do that by dividing the numerator by the denominator so the change from a fraction to a decimal we divide the numerator by the denominator now we know it's going to be decimal number so let's put a decimal point and some zeros and will that'll help us understand the point there so how many 20 please go into one that'd be zero remainder one how many twenties go into Seventeen zero remainder Seventeen how many twenties go into 170 well eight twenties would be 160 so it's going to be eight remainder ten and how many twenties you got 100 that'll be five so the answer is 0.85 so no point eight five so Seventeen twentieths is 0.85 now let's change this decimal to a percentage so the change from a decimal to a percentage would be just times about 100 so 0.85 multiply by 100 would be 85 or 85 and that's it okay now we've got a decimal number we have got 0.315 and we need to change that to a fraction that we need to change that to a percentage so let's change it to a percentage to begin with because that's quite straightforward so times in this about 100 would be 31.5 and then we'll put percent 31.5 and then the change from our decimal to our fraction I just consider what we've got we've got 0.315 so we've got our units or our ones we've then got our tenths we then get our hundredths and then we've got our thousandths so we've got 315 thousands so it's 300 115 thousandths and that's it as a fraction and then we could just cancel this down by dividing both of these by five would give us 63 over 200 so not point three one five would be 63 200 and that's it okay next one we've got 194 so let's write that as a fraction to begin with so it's percent that means out of 100 so that's 194 out of 100 and then we can just cancel this down by half of them both and that would give us that's equal to 97 over 50. so 194 as a fraction would be 97 50ths we could write this as a mixed number so we could say how many 50s go into 97 one remainder 47 over 50. so we could write 97 50 or 1 and 47 50ths okay let's write our percentage as a decimal to divide by 100 would be 1.94 so 194 would be 1.94 and finally we're going to convert 2 11 to the decimal and as a percentage so 2 11's I've chosen this one on purpose because this isn't a terminating decimal these are terminating decimals decimals that stop a recurring decimal is one that goes on forever I think of your third which is 0.33333 and so on so two elevenths so to change this to a decimal we're going to divide so let's do two divided by 11 and see what we get so let's put a decimal point and some zeros and the same with our numerator put our decimal point above how many elevens go into two zero remainder two how many elevens go into Twenty One remain under nine how many elevens go into 90 that's going to be eight which is 88 remainder two how many elevens go into Twenty One remainder nine you can see a pattern here forming how many elevens go into ninety eight remainder two how many elevens go into Twenty One remainder nine and so on how many 11s go into how many elevens go into 90 eight remainder two and so on so you can see we're getting 0.18 18 and so on so we could write that down 0.181818 and so I want to put some dots to say that it just carries on like so or alternative it's a recurring decimal we could put dot above it so we could write naught point and then because it goes one eight one eight one eight we do one it and a DOT above the one and a DOT above the eight and that just says that the one on the eight repeats so it's not point one eight one eight at one eight one eight and so on that's it okay and there's a percentage we want to write this as a percentage now we had 0.181818 and so on so we're going to times this about 100 so it'll be 18.1818 and so on percent and that's it so it's going to be very important that you're able to convert between your fractions your decimals and your percentages okay our next topic okay so next topic is recurring decimals and converting recurring decimals to fractions so let's look and see how we would do that that's video 96 and corporate Maps so we're going to use algebra to write this recurring decimal as a fraction so here's our recurring decimal 0.0262626 and so on it could be written as 0.026 with little dot above the two and a little dot above the six that would also be the same thing and we're gonna look at how to write this as a fraction so first of all what we're going to do is we're going to let this be called X we're going to write x equals 0.0262626 and so on okay so that's X that's our recurring decimal we were given in the question now what we're going to do is we're going to look at our number our 0.026262626 and so on and look at the bit that's recurring forever so that is our 2 6 it goes 2 6 2 6 2 6 and so on so we want to find a number of where we have the decimal point and then we've just got that recurring bit 2 6 2 6 2 6 2 6 and so on now if we multiplied this number by 10 what that would do is move all our digits one column to the left so 10x if we wrote down 10x that would be equal to if we multiply this by 10 we would get naught point and then that would be 2 6 2 6 2 6 and so on and that's going to be really useful for us now what we're going to do is we want to get another number when we get 2 6 2 6 2 6 2 6. so if we go back to our original number if we multiplied this by well if we multiply it by 10 that's 10x if we multiply by 100 we would have 2.626262 well that's not what we want we want two six two six two six two six so let's try multiplying X by a thousand so let's write a thousand X so if we multiply this by a thousands we would get well moving all the digits three places to the left would give us 26.262626 and so on so we multiply that by five so what we've done is we've taken our X and we've multiplied it by 10 to get an 0.26262626 and so on and we've also gone back to our X and we've multiplied about a thousand to get 26.26262626 and so on now what's really fantastic here is if we subtract these from each other are 26.26262626 and we take away 0.26262626 the part after the decimal point will cancel out so let's write that down and see what we get so let's write down our 10x and then equals 0.262626 and so on and when we take these away from each other we get well on our left hand side we've got a thousand X take away 10x well a thousand X take away 10x will be 990x so 990 X and on our right hand side we had 26.262626 and we take my 0.2626 626 and if you ever look thick with six is zero two take away two zero six take away six is zero two take away two zero six take away six is zero so you take away two is zero so we would just have so all those numbers after the decimal point would just cancel out so we'll just be left with 26 take away zero and 26 take away 0 is 26. so whenever you take 26.262626 and you take 0.2626 away from that you'll just be left with 26. so we've got 990x equals 26. now remember we let the number we started with in the question be X so what we want to do is we want to solve this equation and find out what x is so if we divide both sides of this equation by 990 we'll find X so divide by 990 and divide by 990 and that gives us on the left hand side x and on the right hand side we would have 26 over 990 and that's it so we have written our 0.0262626 as a fraction as 26 over 990 and that's it now we could sometimes in the question be asked to cancel it down so if we were asked to cancel this down well we can see first of all the both of these numbers are even so we can divide both of them by two so that would give us well 26 divided by 2 is 13 and if we divided 990 by 2 we get 495. and then finally 13 and 495 don't have any common factors apart from one so that means that that's it we've simplified it as far as we can and that's it so that's how you write a recurring decimal as a fraction so let's have a look and see how we'd round numbers to two significant figures so let's have a look at our first one so we've got 394. so we want two significant figures so for this number that would mean we would want two digits and Then followed by zeros so options would be 390 or 400 because there are two digits so three nine and Then followed by zeros or four hundred that's four zero followed by zeros and we would run around to two significant figures so let's choose the closest one so it's either 390 or 400 so it's going to be 390 because this is closer to 390 than it is to 400. so answer would be 390. or an alternative way to look at it is if we have 394 we wanted to run this to two significant figures where we want to have two numbers Then followed by zeros so we look at the third significant figure which is a four and that means that we round down so it'd be 390 not 4 400 okay next one is 1273 we want to run that to two significant figures so we either want to have one thousand two hundred or one thousand three hundred but we've got two digits Then followed by zeros as you can see one thousand two hundred and seventy three would be closer to one thousand three hundred than it is to one thousand two hundred it's also be one thousand three hundred or again we could look at the very significant figure we've got the first two and then the third significant figure is a seven so we round up so answer would be one thousand three hundred okay our next one our next one is seven thousand nine hundred and sixty one and we want to run that to two significant figures so that means we want to have two digits followed by zeros so we could either have seven thousand nine hundred or eight thousand because eight zero and Then followed by zeros so because this is seven thousand nine hundred and sixty one it's closer to eight thousand than it is the seven thousand nine hundred so that means your answer would be eight thousand or again another way to look at it is if we want to run to this to two significant figures we look at the third significant figure which would be a six and that means we round up so it'd be eight thousand okay let's have a look at our next one our next one is 0.618 so we want to round this to two significant figures now remember with decimal numbers we ignore the note point and the zeros in the front so we're just going to look at our six one eight and we want to run this to two significant figures that means we either want 0.61 or 0.62 and because it's 0.618 we're going to round up so it's going to be 0.62 okay the last number is 20.501 we want to write this to two significant figures so we want two digits Then followed by zeros or if it's just a two digit number we just wanna two there's your number so we've got 20.501 so our choice is either going to be 20 or 21 and because it's 20.501 we're going to round upside down to a B21 or again another way to look at it is because we're 20.501 we look at our significant figures well our first significant figure is a two our second significant figure is a zero and then we look at our third significant figure which is a five that means we round up so answer B21 okay let's have a look at our next topic so the next topic we're going to look at is use of a calculator and that's video 352 in chord Maps that's very important you know where all the buttons are on your calculator that you need to know so for instance making sure you know where your squared is your cubed your power button your square root your cube root your sign your call is your channel and so on making sure you're able to convert between fractions and decimals and so on now you there's different calculators you could have you could have the stepper calculator that your display would look something like this you might have a calculator where the buttons are in slightly different places but all the calculators will have the key buttons you need to know all your scientific calculators you need to have your sign your calls your tan your fraction buttons and making sure that you're able to use them all so you may have a calculator that looks like this so let's just have a look at one of these I'm just going to use this calculator display for the minute and this is our question it says work out five cubes subtract square root of 2 all divided by tan 32. so if I want to type something like this into my calculator the first thing I would do is because it's got a fraction I would press the fraction button so here my fraction button is there and I'd press that button and then my display would look something like this next I would type in 5 cubed subtract the square root of 2 by pressing 5 and then the cubed button now obviously in your calculator it might be in different places so for me I'd press five and then I'd press shift and then the squared button and then that would give me the cube symbol and then take away and then I would press the square root of two and make sure you know you're familiar of your calculator and you know where those buttons are for your calculator then because that's the numerator I then press the down button and then I would press the tan So Tan 32 and it would look something like this now with my calculator remember I type in tan32 it opens the brackets but it doesn't close the brackets so I then make sure that I close my bracket so it would look like this and then I would press equals and whenever I press equals I get the answer 197.7786013 now if that came up as a fraction I wanted this decimal I would then press this button SD standard decimal button and I would then write it as if it was a fraction which converted into a decimal for me and that would be quite useful now if we've got this calculator what I would do is I would press the fraction button and then I would type in my five and then I would press my power button and then type in three and then use my arrows to then go across and then type the square root of two press done to go to the denominator press tan 32 and make sure I close brackets and then press equals and again my answer would be 197.7786013 now I've had a different question amount it came up as a fraction and I wanted the red as a decimal for this calculator press format and then I would press down and then press decimal and then I would write as a decimal for me so instead of having the SD the standard decimal button it has this format button on this type of calculator and that's it okay let's have a look at our next topic so next topic is estimation and sometimes in estimation questions we're given questions like this where we've been asked to work out an estimate for 40.18 multiplied by 6.87 all divided by 0.512 now whenever we're doing a question like this with estimation it can be useful to round our numbers to one significant figure so let's write our numbers to one significant figure so let's start off with our first number 40.18 so we're going to run that to one significant figure so but that means one digit followed by zeros so that means we're even gonna have a choice of 40 or 50 because it's 40.18 I'm going to choose 40. it's much closer to 40 that is the 50. our next number well we've got multiply by and then we've got 6.87 so our choices will be six or seven now 6.87 will be closer to seven than it is the six so we're going to choose seven and then finally we've got divided by and then we've got 0.512 we want to run this to significant figure so our choices will be 0.5 or 0.6 this number is closer to 0.5 because it's 0.51 then it is the 0.6 or we're going to write then 0.5 now whenever I rounded my numbers to one significant figure obviously whenever I carry on the answer is not going to be the exact answer so instead of writing an equal sign here I do this curly equal sign it looks like this and that means it's approximately equal to so whenever you round your numbers whenever you do an estimation question you write your numbers rather than putting an equal sign after you've run them it can be useful to put this approximately equal to symbol it just shows your teacher your examiner or even just remind yourself that you've rounded the numbers so now what we're going to do is we're going to work this out so we're going to work out 40 multiply by 7 well 4 times 7 is 28 so 40 times 7 would be equal to 280 and then we've still got divided by 0.5 so when I just need to work out 280 divided by 0.5 and that will give us our approximate answer to this question our estimate and what you would do is you would just times both these numbers by 10 do 2800 divided by five how many fives are there and two zero remainder two how many fives are there in 28 well that'd be five remainder three how many fives are there in 30 that's gonna be six and how many fives are there and zero zero so the answer is 560. so our estimate for the answer to this question would be 560. that's not going to be exactly 560 unless we're very very lucky um but the at our estimate that's our the proximate answer okay so let's have a look at our next topic so our next topic is Best Buys which is video 210 and corporate Maps so here we've got a question of Best Buy's question that says packets of biscuits are sold in two sizes you've got a regular box of biscuits which has 10 biscuits for 95p or you've got a large box of biscuits and we've got a large box of biscuits which is 15 biscuits and it costs 1.50 and the question asks which packet of biscuits is better value for money now there's two ways which I typically answer questions like this so using the first approach we could divide the total cost the 95p by the 10 biscuits to find the cost per biscuit so I could do 95 P divided by 10 and that would tell me how much it costs per biscuit in the regular box so 95 divided by 10 is 9.5 P so it's 9.5 P per biscuit here per biscuit and in the large box it's 1.50 which is 150 Pence and that's for 15 biscuits so you can divide a 150 for 150 Pence by 15 biscuits and that would tell us that's equal to 10 P per biscuit so 9.5 P per biscuit is cheaper than 10p per biscuit so that means that the regular box is better value for money so that's one approach to divide the total amount by how much you get another approach is to buy the same amount of Biscuits by either buying just the regular boxes or the large boxes so with the regular box you could buy 10 biscuits I could buy 20 biscuits I could buy 30 Biscuits by buying just regular packets of the large box I could buy 15 biscuits or if I bought two packs that'd be 15 plus 15 which is 30 biscuits ah so I could buy three regular boxes and that would be 30 biscuits so that would be three boxes at 95 pH so 3 times 95 P which is equal to 285 P or 2.85 or alternatively if I was looking at just the large boxes to buy 30 biscuits well 30 biscuits would be 2 two boxes of those so that would be two lots of one pound fifty and two lots of 1.50 is three pound so if I wanted to buy 30 biscuits I could buy use the regular boxes which are 95 pH and that would cost me two pound eighty five alternatively I could use the large boxes to buy the very biscuits now cost me three pounds so as you can see it's better value for money to buy the regular boxes so the regular is better value okay let's have a look at our next topic which is exchange rates or currency and that's video 214 and corporate maps and we're told Nicola went to Italy and she changed 800 pound into euros and the exchange rate was one pound is one Euro 40. so in other words for every pound the bank give her one Euro 40. and the question says to change 800 pound into Euros so to change from pounds into Euros or Pounds into whatever currency you're converting into if you've got what one pound is you need to multiply how many pounds there are the 800 by the number in the exchange rate so here we've got one part is one Euro 40. so if we do 800 multiplied by 1.4 or 1.40 well we don't really need the zero there so if we do 800 multiplied by 1.4 we'll see how many years she gets because for every pound she gets one Euro 40. so 800 multiplied by 1.4 equals one thousand one hundred and twenty euros and that's it now while she's there Nicholas Cena watches she likes and it costs 105 euros and you also know how much that is in pounds well if she wants to convert back from euros into pounds we need to divide the 105 by the number and the exchange rate the 1.4 so if we do 105 divided by 1.4 that will tell us how many pounds it would be so if we do 105 divided by 1.4 that is 75. so the watch costs 75 pounds and that's it so our next topic is to look at conversion graphs and here we've got a conversion graph and conversion graphs can be used to maybe compare distances like in this graph here where we've got miles and kilometers conversion graphs can be used to convert money perhaps you could then from currencies and you want to convert between one and the other and the conversion graph can be quite useful for that so here we've got a conversion graph and horizontally we've got miles going naught one two three all the way up to 10 miles and vertically we've got kilometers going naught one two three and so on and let's convert four miles into kilometers using our conversion graphs it's important to be able to know how to use a conversion graph so get a learner pencil and we want to convert four miles so we'll go to four miles on the horizontal axis here for and we'll get our learner pencil and we'll go up to the line and it's just there and then we go across from the line and do sort of learner pencil on the test paper like so so as you see we've gone to the second box above the six so let's find out where each one of those boxes is worth so from naught to one there's one two three four five boxes that must mean we're going up a 0.2 is 0.2 0.4 0.6 0.81 so that would be 6.2 6.4 so 4 miles would be 6.4 kilometers using this conversion graph now we've been asked to change eight kilometers into miles so again getting a roll on a pencil so getting your pencil Ritter and going from eight kilometers across to the line and then down from the line we get to exactly five miles so using this graph eight kilometers would be equal to five miles and that's it now if you were asked to maybe change something like 80 kilometers into miles I would go from eight across and get five and then say well if eight kilometers is five miles eighty kilometers would be 50 miles and so on so you can use conversion graphs to convert what's on the scales but you can also use that information to work out other conversions as well that go beyond the skills topic is index notation and there's video 172 on corporate Maps so if I had five times five times five that is 5 cubed and if you remember your Cube numbers which is multiplying the number by itself and by itself again you may want to write an index notation which means write it as a number with a power so we could write that as five cubed because there's three fives multiplied together here we've got two times two times two times two times two times two so write that in index form because there's one two three four five six twos we would write two to the power of six and if we had y times y times y times y because it's four y it's multiplied together we would say y to the power of four it's very important to be able to write these in the index form particularly for a topical product of primes and I'll talk about that in a moment so here for instance you wanted to work out two to the power of six we would just do two times two which is four times two which is it times two which is 16 times 2 which is 32 times 2 which is 64. so 2 to the power of 6 would be 64. or you can use a calculator to help you so here's a calculator so what I would do is I'd press the two button so I'd press two and then this is the power button and on my calculator it's an x with little white square just above it like so I'd press that button and on your calculator display above the two would be a little rectangular little box appears and then press six and then press equals and then you will get the answer so 2 to the power of 6 is 64. and that's just a quicker way of working out on your calculator rather than writing 2 times 2 times 2 times 2 times 2 times 2. so that's index notation okay let's have a look at our next topic so next topic is laws of indices and there's three very important laws of indices that I would recommend you know and this is video 174 on corporate Maps so our first law well if you're multiplying things with the same base so for instance if you had M cubed to multiply by m to the power of four well that would be M times M times m m cubed multiplied by m times M times M times M and all together the B7 of them so that would be m to the power of seven and a quick way to work that out is because we've got m to the power of 3 and we're multiplying by m to the power of four you can add these Powers you can do three plus four is equal to seven and this is the chord Mouse revision card on laws of indices so if you've got the revision card this revision card will be very useful for you okay next one if we're dividing if you had m to the power of 8 so if you had M times M times M times M times M times M times M times M and you divided that by m squared it has M times M two of the M's would cancel out with two of the m so you'd be left with M times M times M times M times M times M and that'd be m to the power of 6 which is m to the power of six and a quick way to work that out is if you're dividing and you've got the same bases you can take away the powers you can do eight take away two and that's m to the power of six and finally we've got a pi over par so if you've got a power and then another Power you multiply the powers together and let's have a look and see why that works so if you get M cubed squared remember squared means multiplied by itself so with an M cubed multiplied by itself so that's M times M times M multiplied by itself so that's M times M times M and if you multiply all those together you get m to the power of six and a quick way of doing it is to multiply the power so if you go to power to a power you can multiply the two powers of three times two is equal to six so let's have a look at some examples so if I had three to the power of 4 multiply by 3 squared I'd add the powers together because it's multiplied and that would be to the power of six if I had three to the power of 10 divided by 3 squared I would subtract the power so 10 take away 2 is equal to 8 and finally if I had 4 squared cubed I would multiply the powers because it's a power of a power you've got a power and then an over power so you'd multiply the powers together 2 times 3 is equal to 6 so it'd be 4 to the power of 6. and that's it okay let's have a look at our next topic so our next topic is negative indices and that's a number topic because it's in red and this video 175 on corporate maps and let's consider this pattern multiple 5 to the power of three five to the power of two five to the power of one five to the power of zero five to the power of negative one and five to the power of negative two so five cubed well 5 times 5 times 5 is 125. 5 squared plus 25 5 times 5 is 25 5 to the power of one that's just a five so our answer would be five okay next next we've got 5 to the power of zero well five to the power of zero is one okay and next one is five to the power of negative one well if we look at our numbers here we've got 125 25 5 and 1. now there's a pattern here to go down in this pattern we're dividing by five we're dividing by five we're dividing by five we're dividing by five and then if we want to find out our next one we would just divide by five again so if we do one divided by five well that'd be 0.2 or we could just write as a fraction which is one over five if we want to get our next answer we could divide by five again okay so we do one over five or fifth divided by five so that'll be one over twenty five now what's actually quite useful the spot here is because we've got five to the negative two that's the same as putting one over and five squared and five squared is 25. so that's our rule if we have x to the power of negative n that's the same as 1 over x to the N so you just put one over and then just use the positive power so here if we were asked to work out 2 to the power of negative three we could just put one over and then just write two cubed change in our negative three to just three and then two cubed is so answer would be just one over eight or one if so two to the power of negative three is one over eight next 10 to the power of negative 2 where we put 1 over and then just write 10 squared and 10 squared is 100 so answer would be 1 over 100. so if we had 10 to the power of negative 2 our answer would be 1 over 100. that's it so if we want to work out a negative power you just put one over it and then just use the positive power on the denominator okay now let's have a look a fractional indices in this video 173 in corporate Maps so if we have something like X and we've got a power of 1 over n so perhaps it was x to the power of a half or x to the power of a third x to the power of a quarter or something like that that is equal to the nth root of x so for instance if you had x to the power of a half that'll be the square root of x if you had x to the power of a third that'll be the third root or the cube root of x if you had x to the power of a quarter it'll be the fourth root of x and so on so whatever the denominator is on the fraction that's what root you take off the base number of this X and that's how you deal with fractions where you've got a 1 on the numerator if there's another number on the numerator such as x to the power of M over n so what you do is whatever is on the denominator of the power you take that root so you're going to take the M fruit of X and then whatever is on the numerator you then work out that power so whenever you get a fractional indices what you do is whatever is on the denominator you take that root and whatever's on the numerator you then do that power and the rule looks really complicated but it's much more easier whenever we look at numbers so let's have a look at the revision card and let's get started so if we had something such as 25 to the power of a half to the power of a half means square root we don't need to worry about the ones we need to square root it so 25 to the power of a half we just square root the 25 which is 5. if we had a power of a third such as 8 to the power of a third well again because there's a one on the numerator we just need to look at the denominator because it's a 3 we're going to take the cube root and the cube root of eight is two so eight to the power of a third is just 2 because you take the cube root okay now let's have a look at some of the numbers on the numerator so if we had something like 27 to the power of two thirds well we're going to look at the denominator to begin with and it's a 3. so we're going to work out the cube root of 27 and the cube root of 27 is 3 and then you're going to look at the numerator and it's a 2 so you're going to square that answer so 3 squared is 9. so 27 to the power of two thirds would be nine and finally if we had something such as 16 to the power of three quarters again you look at the denominator and it's a 4 so we're going to take the fourth root of 16 well 2 times 2 times 2 times 2 is 16 so it's going to be 2 and then you're going to look at the numerator which is a 3 and you're going to do 2 cubed which is it so whatever is on the denominator you take that root and then whatever's on the numerator you then do that power okay let's have a look at some examples so first question says work out 16 to the power of a quarter so here we've got 16 and we've got a power that is a fraction so this is a fractional indices and what's good about it is it does have a one on the numerator so we just need to work out whatever roots on the denominator so it's going to be the fourth root so we need to work out the fourth root of 16. that's a Forefront of 16. now 16 is equal to two times two times two times two and as you can see whenever you've got four twos and you multiply them all together we get 16. so that means that the fourth root of 16 would be two so that means it's 16 to the power of a quarter is equal to two Okay this time we've got four to the power of 5 halves so again we've got a power that's a fraction and this and we do have something on the numerator but other than one so we are going to need to work out that power but first of all we're going to look at the denominator and it's a 2. so we're going to work out the square root of four so you're going to do the square root of 4 and the square root of 4 is equal to 2. now you take that answer that 2 and we're going to do 2 to the power of 5. so 2 to the power five and then whenever you do that you get 2 times 2 times 2 times we can have two times two times two times two times two and well 2 times 2 times 2 times 2 we know is equal to 16 and the hemisphere 2 again that's going to be 32. and that's it so 4 to the power of 5 halves is equal to 32. okay now next one is 125 to the power of two-thirds so we're going to look at the denominator and it's a 3 so we're going to work out the cube root of 125 so let's work out the cube root of 125 to begin with and the cube root of 125 is 5. now we're going to look at the numerator and that's a 2 so we're going to do 5 squared and 5 squared is equal to 25. so 125 to the power of two-thirds is equal to 25 by taking the cube root and then squaring it okay now we've got our negative fractional indices so we're putting together our negative indices and our fractional indices so we've got 49 to the power of minus a half so first of all because it's a negative power we're going to write 1 over and then 49 to the positive power so that's just going to be to the power of a half so 49 to the negative a half I would write that as 1 over 49 to the power of a half 9 to the power of a half means square root so we're going to square root 49 and the square root of 49 is 7. so answer would be 1 over 7. so if you had 49 to the power of negative a half it'd be equal to one over seven and finally if we had 64 to the power of negative two thirds again I would write it as one over 64 to the power of two thirds getting rid of that negative power and writing it as one over 64 to the power of two thirds then we're going to take the cube root of 64 and then Square it so the cube root of 64 the cube root of 64 is equal to 4 because 4 times 4 times 4 is equal to 64. and then we're going to square it so we're going to take our 4 and we're going to square that and that's equal to 16. so 64 to the power of two thirds is equal to 16. so we had one over that so it's going to be 1 over 16 and that's it so the answer is 1 over 16. Okay so we've looked at fractional indices and we've looked at negative indices let's have a look and see what happens whenever we work out the fractional or a negative indices of a fraction so if we had 4 25 to the power of a half remember to the power of a half means the square root so we're going to square root this fraction now if whenever we Square this fraction we can square root the numerator and square the denominator and that will give us our answer so if we work at the square root of 4 that's 2 and if we work at the square root of 25 that's 5. so 4 25 to the power of a half would be two-fifths and we could just check it two-fifths times two-fifths would be four twenty-fifths and that's it so if you have a fraction to a power of a half you can just square root the numerator and the denominator if you get a fraction to the power of a third you can cube root the numerator and cube root the denominator and if for instance you had a fraction to a power of maybe two-thirds you could cube root the denominator and the numerator and then Square your answers and then that would give you your answer and that's it okay now let's have a look and see what happens when we have a fraction to a negative power so whenever we have a number to a negative power we normally do one over and we're doing one over because we're working out the reciprocal so if we've got a fraction with a negative power to get rid of the negative sign in the power what we can do is we can do the reciprocal of the fraction so if we had 2 4 to the power of negative three to get rid of the negative we can take the reciprocal of two thirds which would be three halves so we then have three halves and then instead of being negative three a b to the power of three and now we've got three halves cubed well we can just Cube the numerator and Cube the denominator so three cubed is 27 and 2 cubed is equal to eight and that's it so if we had two thirds to the power of negative three we would get rid of the negative sign by doing the reciprocal of two thirds to get three halves and then we'd have the power of three so we would just Cube the numerator and the denominator and three cubed is 27 and 2 cubed to zero now here as you can see our answer is atop every fraction so what we can do is we can see how many eights go into 27 so that'll be three because three is twenty four remainder three and then the denominator will still be it so we've got three and three oops okay let's have a look at our last question so this time we've got a fraction to a negative fractional indices so we've got 8 27 so the part of negative two-thirds so first of all what I'm going to do is because it's a negative power I'm going to take the recip circle of the fraction so the reciprocal of 8 27 would be 27 eighths by flipping it over so that gets rid of the negative power so we'd have 27 eighths to the power of two-thirds so that gets rid of the negative power taking the reciprocal now we need to work out what 27 eighths to the power of two thirds would be so let's do the cube root of both of these so the cube root of the numerator would be 3 and the cube root of the denominator would be two so we're going to have three halves and we've done the cube root so we now just need to square it so we've got 3 squared is equal to 9 and 2 squared is equal to 4. so answer would be nine quarters or if two and a quarter and that's it okay let's have a look at our next topic our next topic is called LCM or lowest common multiple and that's video 218 and 219 on corporate maps to find the lowest common multiple of two numbers you consider there are multiples so if we had six the multiples of six would be 6 12 18 24 30 36 and so on the multiples of 15 would be 15 30 45 and so on and the lowest common multiple or the LCM is the first number in both of those lists and as you can see 30 is the lowest common multiple it's the first number in both the multiples of 6 and the multiples of 15. so 30 is the lowest common multiple and that's it so let's have a look at the highest common factor so let's consider 16. so 16 is equal to 1 times 16 2 times 8 and 4 times 4. so the factors of 16 are 1 2 4 8 and 16. and if we consider 20 well 20 is equal to 1 times 20 2 times 10 and 4 times 5. so the factors of 20 are one two four five ten and twenty and let's look at our common factors our common factors are one two and four so there are common factors and the highest common factor but that's going to be 4. so the highest common factor of 16 and 20 is 4. and this topic is particularly useful if you want to cancel fractions and factorize and things like that okay our next topic okay let's have a look at our next topic so our next topic is product of primes and that's video 223 on corporate Maps so every single whole number that is greater than one is either prime or can be written as a product of primes and the word product means to multiply so it e remains that every single number that's greater than one is either a prime number or is equal to prime numbers multiplied together so our first question our example says write 60 as a product of primes this is very important that you know your prime numbers and your prime numbers are 2 3 5 7 11 13 17 19 23 29 31 and so on and it's very important you know those prime numbers so we're going to write 60 as a product of primes and we're going to give our answer in index form so let's start off with 60 and I like to do this using a prime factor tree you can use another approach by using sort of an upside down by shelter whatever approach you prefer is totally fine but I like to use a prime factor tree like so so 60 I think of two numbers that multiplied together to give me 60. so I'm going to go forward 2 times 30. so 2 times 30 is equal to 60. now we look and see if either of these numbers are prime and 2 is prime so I'm going to circle it but 30 is not prime so what I'm going to do is I'm going to think of two numbers at times together to give me 30. so 3 times 10. now 3 is prime so Circle it whereas tens not so I'm now going to think of two numbers that multiply together to give me 10 and that's going to be 2 times 5. I never choose 1 and the number itself because you don't get any further so 10 so 10 would be 2 times 5 so 2 times 5 and let's Circle them and that's it we're finished so 60 is equal to 2 times 3 times 2 times 5. so let's write that out 60 is equal to and let's just do it in order so instead of doing 2 times 3 times 2 times 5 I'm going to do 2 times 2 times 3 times 5. so 2 times 2 times 3 times 5. and let's just check it works 2 times 2 is equal to 4 times 3 is equal to 12 times 5 is equal to sixty and as you can see we've written 60 as a product of primes but the question says they're at an index form as you've just seen earlier on with index notation if you've got something such as 2 times 2 that's 2 squared so we can write 60 is equal to 2 squared multiplied by 3 multiplied by 5. so we have written 60 as a product of primes in index form and that's it and it's very important to be able to do that and video 223 we'll go through that and also if you go to corporatemath.com and you go to videos and worksheets and you scroll down to 223 as well as having the video tutorial there will be some practice questions and textbook exercises on this topic so if you do want extra practice feel free to do those but also remember there's the revision booklet which you can print it's the links in the description below and if you have that there's some questions on this now okay let's have a look at our next topic so our next topic is applying product of primes now it's video 223 a in corporate Maps so let's start off by looking at this question so a number M has been written as a product of primes as 2 multiplied by three squared so this is m 2 multiplied by 3 squared and the first part is the same what number is M so let's work this out now remember we've got our order of operations that is we do any brackets no orders that's another name for Powers so that's squared so we're going to work out the squared first of all so we're going to do 3 squared so 3 squared is 9 and then we're going to multiply by the 2 2 times 9 is equal to 18. so that means m is equal to 18. and Part B is write the number at 10 m as a product of primes so 10m well if we know that m is equal to 18 10 m would be equal to 10 times M so that would be 180 so you could take 180 and do that prime factor tree and find out what 180 is as a product of primes and what's really useful here is we know that 10 is equal to 2 times 5 so 2 prime numbers 2 and 5 multiplied together would give me ten two times five now we know that m is equal to 2 times 3 squared so m is equal to 2 times 3 squared so we multiply this 10 by m which is 2 times 3 squared that would be our answer but we want to write this obviously in index form and as you can see we're 2 times 5 times 2 times 3 squared I would look at the twos first of all you get 2 times 2 which is 2 squared we've then got our 3 squared so multiply by 3 squared and then multiply by 5. so 2 squared times 3 squared times 5 would be 10 m and we can check that because remember we knew that 10 times M was 180 let's check this is 180 2 squared is 4. 3 squared is 9 4 times 9 is 36. I multiply by 5 would be 180 so that means it's 10 m as a product of primes would be 2 squared times 3 squared times 5. okay our next part now we can also use product of primes to find square numbers and Cube numbers so for instance if we knew the 280 was 2 cubed multiplied by 5 multiplied by seven and we were asked what the lowest whole number the 280 would need to be multiplied by to give a square number what we can do is we can consider what a square number is so a square number is a number multiplied by itself and that would give you a square number so if we take what we've been given here so 280 is 2 cubed that's 2 times 2 times 2 times 5 times 7. just write an iron fill if we share these prime numbers out as evenly as possible and then find out what extra ones we need then we can find out what number you can multiply 280 by to get a square number so let's share these numbers out as evenly as we can so we've got two two that's three twos so we're going to have to give one Circle two twos and one Circle one two we've got a five well unfortunately that's just going to have to go in one of the circles as well we can't you know we don't have two fives and we've got a seven so again that's going to have to go in one of the circles now we want the same thing in both of these circles so let's put some extra prime numbers in this circle so we had two twos so we're going to need number two we're going to need a five and we're going to need a seven if we multiply 280 by 2 5 and 7 that should give us a square number so let's see what 2 times 5 times 7 is so 2 times 5 times 7 is equal to well 2 times 5 is 10 times 7 is 70. so if we multiply 280 by 70 we should get a square number so 280 multiplied by 70 is equal to nineteen thousand six hundred and if we work out the square root of that we get that's equal to 140 so it is a square number so if you want to find a square number what you need to do is share out the prime numbers you've been given as evenly as possible and then add in the extra ones you need and there is a bit of a shortcut if we take all the numbers in the circles here are two times two times five times seven times two times two times five times seven that would be 2 times 2 times 2 times 2 which is 2 to the power of 4 times pi and we've got 5 times 5 that's 5 squared and we've got 7 times 7 that's 7 squared if you notice all the powers are even so if you want a square number all you'd need to do it is to make all the powers even so if I was doing this question I would first of all just look and see well is 2 cubed if I'm going to need an extra two I've got five what's five to the power of ones I need an extra five and I've got seven to the power of one so I need an extra sevens so if I multiply this by two and by five and by seven I will get a square number and likewise if I wanted to find a cube number all the powers would have to be multiples of three so for instance if I had 2 to the power of 9 multiplied by 5 to the power of 3 that would be a cube number because this Powers a multiple of three and this Powers multiple of three we use the product of primes to work out the LCM that's the lowest common multiple or the hcf the highest common factor so let's have a look at our first question so let's write some numbers as product of primes and then we'll use that information to find the lowest common multiple and the highest common factor so first part says write 92 is a product of primes so I'm going to do my prime factor tree so 92 that's 2 times 46 let's Circle the 2. and 46 that's 2 times 23 and they're both Prime so let's Circle them so 92 as a product of primes would be 2 times 2 times 23 and in index form that would be 2 squared times 23. now our next question is to write 48 as a product of primes so I'm actually going to use my calculator to write 48 as a product of primes so on my calculator I've got this yellow fact there so what I do is I press 48 then press equals and 48 comes up in my calculator display then I press shift and then where it says fact there and on the display it will show me 2 to the power 4 multiplied by 3. so 48 equals 2 to the power of 4 times 3 or 2 times 2 times 2 times 2 times 3. so we've got 92 as a product of primes and with 48 is a product of primes let's now use that information to work out the lowest common multiple and the highest common factor of those numbers so here we've got our question it says write down the highest common factor of 48.92 and write down the lowest common multiple of 48 92. so let's write down what they were as product or primes again so 48 was equal to 2 to the power of 4 multiplied by 3 or if we wrote it out in full it would be 2 times 2 times 2 times 2 times 3 and 92 92 is equal to 2 squared multiplied by 23 or if full 2 times 2 times 23. now let's put those numbers in our Venn diagram so here we've got a Venn diagram and we've got one Circle for 48 and with one Circle for 92 and we're going to put their prime factors into this Venn diagram so first of all let's see if they share anything so as you can see 48 is 2 times 2 times 2 times 2 times 3 and 98 is 2 times 2 times 23 so they've both got two twos 92 has got two twos and 48 has got two twos so we're going to put two twos in the middle now as you can see here 48 had two more twos so let's put those two twos in the 48 side 48 also had a three but 92 doesn't so let's put the three there as well and finally 92 has a 23 so we'll put that into the 92 side so as you can see in both circles we've got 48 and we've got 2 times 2 times 2 times 2 times 3 and that's 48 and for the 92 Circle we've got 2 times 2 times 23. so you put what they share in the middle and then you put the extras onto each side so the first question says work out the highest common factor of 4892 so to find the highest common factor we multiply the prime numbers in the middle so we're going to do two times two so the highest common factor will equal 2 times 2 and 2 times 2 is equal to four so the highest common factor is four okay our next question says to work out the lowest common multiple of 48.92 so to find the lowest common multiple we multiply all the numbers in the Venn diagram so we're going to do 2 times 2 times 3 times 2 times 2 times 23 or in order the lowest common multiple equals 2 times 2 times 2 times 2 because there's four twos times three times twenty three and when we do that we get that's equal to one thousand one hundred and four so we've worked out the highest common factor and the lowest common multiple of 48.92 really quickly and easily by using this Venn diagram so what you do is you write each number as a product of primes you put the numbers into the Venn diagram and make sure whichever one says share go in the middle and put the extras on each side and that's it and to find the highest common factor you multiply the prime numbers in the middle and to find the lowest common multiple you multiply all the numbers in the Venn diagram okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is called standard form and a number is in standard form if it's in this format we've got a number between one and ten so it's a number bigger than or equal to one but less than ten so for instance 1 or 4.22 or 9.9 multiplied by 10 to a certain power so maybe 10 cubed or 10 to the power of 8 or even 10 to the power of negative 3. and I'm actually going to do this part of the video twice I'm going to use two different techniques here because some teachers will use the technique where they add zeros or move decimal points and some teachers will move the digits and say the decimal points have to stay fixed my job is to help you make sure that I cover the way that you are familiar with and I don't want to confuse you by using just one technique and then you think hold on my teacher talks about it in a slightly different way so to begin with I'm going to talk about adding zeros and moving decimal places and then I'm going to do the questions again by moving the digits rather than the decimal points so first of all we had seven thousand and we wanted to write it in standard form well a number between 1 and 10 to choose when I think second seven would be a good choice and then we're going to multiply it by 10 to a certain power now whenever we have a whole number and we multiply by 10 we add a zero on so 7 times 10 is 7 take better than zero if here we want to add three zeros on so what we actually want to do is multiply seven by a thousand and that would be ten cubed because 10 times 10 times 10 is a thousand now notice that if you have a whole number and you want to add one two three zeros on the power would just be three so that would be seven multiplied by ten cubed if we add a number such as 80 000 which is 8 followed by four zeros we could have eight multiplied by it because this is a whole number we could say it's 10 to the power of 4 because it's one two three four zeros okay let's have a look at our next one so next one is to write two million five hundred thousand in standard form so we're going to choose a number between one and ten and we've got two five so we're going to choose 2.5 here because that's a sensible number to choose between 1 and 10. now what we're going to do is we're going to figure out if we have 2.5 how many times we need to move the decimal point together at the end of the number so we want to move the decimal point from here to here so we would move it one two three four five six times so that means we need to multiply 2.5 by 10 to the power of six because that would move the decimal place one two three four five six places to the right okay next let's look at some small numbers so here we've got 0.4 so that would be four would be our number we would choose between one and ten we need to multiply by 10 to the power of now we've got a small number here which is less than one so this is going to have a negative power and we need to figure out how many times we move the decimal point to get from 4 to be this number so if we had four the decimal point would be here and we would move it one two three four five six seven times so answer would be four times ten to the power of negative seven or shouldn't once actually tell me a bit of a shortcut they said well so there's one two three four five six seven zeros in front so you can just write four times ten to the negative seven and I thought that was quite a neat way to do is just count the number of zeros okay and finally we've got 0.018 and if we want to write that in standard form well it's going to be 1.8 because that's the number we choose between 1 and 10. multiply by ten and now we've got 1.8 so the decimal points here we want to move the decimal point one twice to the left so the power would be negative two alternative you might have spotted the pattern that my student mentioned to me and because it's a small number and no point number and it's got two zeros it'll be to the power of negative two okay now let's have a look at these questions using the other approach so let's get rid of that okay let's have a look at our first question so first question is to write 7 000 in standard form so we want a number between one and ten and a sensible number to choose here would be seven and we're then going to have multiplied by 10 and to a certain power now if we have seven sevens in the units of the ones column and we want to move it one two three columns to the left so we want to move the seven three columns to the left so we would multiply by ten a hundred and a thousand so it's actually a thousand is ten cubed so if we want to move three columns to the left we have a power of three so that would be seven multiplied by ten cubed next we've got two million five hundred thousand so we've got to choose a number between one and ten so a sensible Choice here would be 2.5 and then times ten and to a certain power so let's put 2.5 in so the two would be in the ones column of the units column there would have a decimal point and we'd have our five and we want to move the digits one two three four five six columns to the left so we would have to the power of six so two point five times ten to the power of 6 would be two million five hundred thousand okay let's have a look at some small numbers so we've got 0.4 so that would be well we need a number between one and ten so we're gonna have four times ten now because this is a small number we're going to have a negative power because 10 to the negative 1 is 0.1 so we'll be multiplied by 0.1 10 to the power of negative 2 would be 0.01 and so on so we want to we want a negative power a very small number and if we put our 4 in our units or ones column there we want to move the four one two three four five six seven columns to the right so because we want to move the seven columns to the right the power would be negative seven so answer would be four multiplied by 10 to the power of negative seven and finally 0.018 well a sensible number to choose between 1 and 10 would be 1.8 and we want to multiply it by 10 to the power of negative something because we want it to be smaller and we put one R 1.8 in our columns our one is in our units column are one's column and we want to move it to our hundredths column so we want to move the one one two columns to the right so we want them times it by 1.8 times 10 to the power of negative two okay next now let's look at how we would write numbers that are almost in standard form in standard form so if we have a look at this number 562.8 times 10 to the power of 5. this number is almost in standard form and what I mean by that is we need a number between 1 and 10 and and this number at the front isn't between 1 and 10. so we need to make it between 1 and 10. so if I had the number of 562.8 I would divide that by 100 because if I divide that by 100 I would get 5.628 and that's between 1 and 10. but if I divide one number by 100 so for instance if I had 700 multiplied by 2 the answer is equal to one thousand four hundred if I divide this number by 100 and get seven I would need to multiply this number by 100 to get 200 to still keep the same answer so if I've divided this by 100 I need to multiply this by 100 to make sure answer stays the same so it's going to be 5.628 multiplied by 10. so if I multiply this by 100 I'd multiply by 10 so that'll be 10 to the power of 6 and I'm multiplied by 10 again so that'll be 10 to the power of 7. so answer would be 5.628 times 10 to the power of 7. and that's it okay let's have a look at another one okay this time we've got 0.024 times 10 to the power of 9. now this isn't a standard form because this number needs to be between 1 and 10. so we're going to have to write 2.4 here because that would then make sure that we would have a number between 1 and 10 at the front and then we would do to multiply by 10 to the power of something so we multiply this by we'll move the 2 1 2 3 places to the left so we've multiplied this by a thousand so if we multiply this by five and we need to divide this by a thousands so dividing it by 10 would be 10 to the power of 8 divided by 10 again would be 10 to the power of 7 and dividing it by 10 again would be 10 to the power of 6. whenever to look at it is if we dividing it by a thousand remember that's 10 cubed and then if we take away the powers we get 10 to the power of 6. that's it so just make sure that if you have a number that's almost in standard form if you make the number at the front 10 times larger you need to make the Times by 10 bit 10 times smaller to make sure that the answer is the same okay next okay let's look at how to do our arithmetic numbers in standard form so how to do adding numbers in standard form multiplying numbers and standard form dividing numbers and standard form and so on now with adding numbers in standard form there's no set rule so if I had 5.3 times 10 to the power of 7 plus 7.96 times 10 to the power of 8. the way I would do this question would be to ride them out in full so I'm going to write R 5.3 times 10 to the power of 7 I'm full so that would be 5 3 and then one two three four five six zeros and then with 7.96 times 10 to the power of 8 will be plus seven nine six that'd be 796 and then we would then have another six zeros one two three four five six and then we need to add these numbers together and whenever it happens together I'm just going to use the column method so let's line them up so we've got 7 9 6 and then one two three four five six zeros five three and then six zero so one two three four five six and then five okay and then we're going to add them so let's just add them and see what we get so it would be zero plus zero is zero zero zero zero zero zero we've then got six plus three is nine nine plus five is equal to fourteen so put the four down either one and seven plus one is equal to eight so answer would be 849 million or we might want to write it in standard form again so it'll be eight point four nine times ten to the power of one two three four five six seven eight so beats part of it and that's it okay let's have a look at our next question so next question is multiply numbers in standard form and I like this because we can now put together our laws of indices if we have nine times ten to the power of four multiply by two times ten to the power of five so what we can do is we can multiply the numbers at the front we can do 9 times 2 is equal to eighteen and then we can multiply the ten to the powers so we could do 10 to the power 4 multiply by ten to the power of five and whenever we multiplying things with the same base they're both 10 to the power of something we add the powers so we would do 10 to the power of 4 plus fives to the power of 9. so that's not our answer because that's 18 times 10 to the power of nine remember if four answer to be in standard form we want it to be between 1 and 10. so we're going to do 1.8 times 10. now we've made eighteen ten times smaller so we need to make this 10 times larger so 10 to the power of 9 times my time would be ten to the power of 10. so that would be 1.8 times 10 to the power of 10. now as well as being able to do this in a non-calculator paper it's also important to be able to do in a calculator paper so let's have a look and see how we would type this in on our calculator so on our calculator we've got our standard form button here down at the center there at the bottom we've got our multiply by 10 to the power of X there now whenever I'm doing a question like this I tend to like to use brackets um so I would write it like this so I would open up my brackets to begin with I would press my 9 because that's the number between 1 and 10 we're going to use and then I would press the standard form button there and then you would get it would come up with 9 times 10 and then you press four and that would be how you type in 9 times 10 to the power 4 would look something like this and then close brackets so you press so you open up your brackets you press nine the standard form button and then press four to show that the power is 4 and then close brackets then press times and then for this 2 times 10 to the power of 5 you're going to do two the standard form button and then five and then close brackets and press equals and you should get this as your answer it'd be a good idea to practice that now get your calculator and try typing this in and the answer in your display should look something like this where we had a 1.8 times 10 to the power of 10 and that was our answer and that's it okay let's have a look at our next question so this time we've got a division question and we've got 4.5 multiplied by 10 to the power of 7 divided by or over 9 times 10 to the power of negative 5. so we've got a large number divided by a very small number and whenever we divide by a very small numbers are going to get even bigger so let's have a look at this question see what we're doing so we've got 4.5 times 10 to the power of 7 and we're dividing that by 9 times 10 to the power of negative 5. so we're going to do 4.5 divided by that unless I should do that quickly now I know the answer is 0.5 but I'm just going to show you how 4 divided by 9 is 0 remainder 4 and 45 divided by 9 would be equal to 5. so that'll be equal to 0.5 so 4.5 divided by 9 is 0.5 then we've got our multiply by 10 parts so we've got 10 to the power of 7 divided by 10 to the power of negative 5. remember when we were dividing we take away the power so we'll get 7 minus minus 5. and 7 minus minus 5 we would add on the 5 would be multiplied by 10 to the power of 12 because 7 minus minus 5 would be 12. so we would get 0.5 multiplied by 10 to the power of 12. now this isn't a standard form because this needs to be between 1 and 10. so we're going to multiply that by 10 to get 5 but we need to divide this by 10 so 10 to the power of 12 take out one of those tens would be 10 to the power of 11. so that would be 5 multiplied by 10 to the power of 11. okay let's have a look at our next question so this time we've got a situation that involves standard form and the question says an asteroid travels a 25 kilometers per second and we're asked how far does it travel in one hour give your answer in meters and in standard form and because we've been asked to give our answer in meters less changes 25 kilometers into meters so that'll be 25 000 meters because there's a thousand meters in a kilometer so it's traveling at 25 000 meters per second we're asked to find out how far it travels in one hour so let's work out how many seconds around one hour so the 60 seconds in a minute and then the 60 minutes in an hour so if we do 60 times 60 that's equal to 3600. so there's three thousand six hundred seconds in one hour and we've been asked to work out how far does it travel so remember we've got speed distance and time so distance is equal to speed times time so if we multiply the speed by the time because it travels 25 kilometers every second and it travels for 3600 seconds if we multiply those two together we should get our answer so we've got 25 000 multiplied by three thousand six hundred and we get that sequel to 9 follow by seven zeros or 90 million so 19 million meters now the question says to give our answer in standard form so this is a very large number so we want to give this in standard form so a number between 1 and 10 well let's choose nine multiplied by 10 to the power of one two three four five six seven so the answer would be nine multiplied by 10 to the power of 7 meters okay so our next topic is expressing as a percentage so whenever I'm expressing someone as a percentage I tend to express it as a fraction first of all and then I change it into a percentage so the question says in a box there's 20 counters nine of the countries are blue what percentage of the counters are blue so I'm going to express that as a fraction to begin with I know that nine of the countries are blue so I know there's 20 countries in the box and nine of them are blue so that means that 920 of Sir blue now if I'm doing this on a non-calculator paper I want to get this to a percentage which means that I want to write this as a fraction with 100 on the denominator to get from 20 to 100 we multiply by 5 so 20 times 5 is 100 and then look at my numerator which is nine and I times that by five as well and 9 times 5 is 45 so that means that if I know that nine twentieths of the countries are blue that would be the same as 45 out of 100 being blue which is 45 percent so that's how we do a value calculator but if I was doing this on a calculator what I would tend to do is I would write it as a fraction to begin with which is 9 over 20 like so and then I want to change that into a percentage so I would just do 9 divided by 20 on my calculator and 9 divided by 20 on my calculator is equal to 0.45 so I would just do the Division 9 divided by 20 and that gives us a decimal 0.45 and then I times about 100 so I Times by 100 and that would give me 45 okay our next topic is percentage of amounts and this is a very important topic and quite often we'll be asked to do either with or without a calculator I'm going to do it without a calculator to begin with and then I'm going to show you how to do it on a calculator so this is videos 234 and 235 in corporate Maps now whenever I'm working at percentage of amounts without a calculator I tend to remember these four building blocks you've got 50 to find fifty percent of something you divide by two to find 25 of something you divide by four or you half it and half it again and that will give you 25 so that's divided by four half and half and again then we've got to find 10 you divide by 10 and the found one percent you divide by 100 and with these four building blocks we should be able to work out these questions quite nicely so first question says work out 25 of 60. so to find 25 of something you divide by four so we're going to take our 60 and we're going to divide it by four so 60 half is 30 and half again is 15. so answer would be 15 centimeters okay next question our next question says find 60 of 800. so what I'm going to do is I'm going to find fifty percent I'm going to find 10 and then I'm going to add them together so if we take our 800 and we take 800 and we divide it by 2 we will find fifty percent so 800 divided by 2 is equal to 400. so that's equal to 400 so that's fifty percent now we want to find our 10 so to find 10 we divide by 10 so we take our 800 and we divide it by 10 and 800 divided by 10 is 80 and that's 10 percent now we want to find what 60 is so we would add these two together so if we add our 50 and 10 we will find 60 now 400 plus 80 is 480. so 60 of 800 is equal to 480. okay and our last question our last question says find five percent of 90. now to find five percent I tend to want to find ten percent and then half it so let's find ten percent of 19 to begin with so to find ten percent we divide by 10 so we're going to do 90 divided by 10. and 90 divided by 10 is 9 so that's ten percent so we find 10 is equal to 9 but we want five percent which is half of ten percent so if we do 9 divided by two that's equal to 4.5 so answer would be 4.5 so we never found a percentage of amounts with a calculator there's two common approaches one of them is to find one percent and then multiply by the percent you want so for instance if we were asked to find 19 of 240 what I would do is I would divide 240 by 100 to find one percent so I would do 240 divided by 100. to get one percent and because it's a calculator you can type in 240 divided by 100 and you get 2.4 so 2.4 and then if we want to Define 19 of 240 I would just take our one percent which is 2.4 and I would multiply it by the percent we want which is 19. so then you do 2.4 multiply by 19 and you'll find an answer of 45.6 and that's it so 19 of 240 is 45.6 so to find the percentage of an amount on a calculator I tend to divide by 100 dividend one percent and then multiply by the percent you want and there is another approach and that's by changing the percentage that you want to find of the number into a decimal so changing 19 to a decimal that's 0.19 this is called a multiplier and you multiply the 0.19 by the 240 and that would give you 45.6 as well so if you want to find a percentage of an amount on a calculator you can divide by 100 to find one percent and times where the percent you want or you can change the percentage of the amount you want to find so 19 into decimal and then you can multiply that decimal that multiplier by the number you're finding the percentage off so you could do 0.19 times for 240. okay let's have a look now finding a percentage of an amount where the percentage is greater than 100 so we're going to find 180 of 300 pounds there's lots of different ways to do this but I'm going to show you one way to begin with and then I'll talk about a few other approaches you could use at the end so we're going to find 180 of 300 pound so what I'm going to do is I'm going to find 80 of 300 pound and then add it on to 100 of 300 pound so let's find 80 of 300 pound to begin with so let's find 80 so to find the 80 what I'm going to do is I'm going to find 10 and then times will buy it and that will give me 80 so let's find ten percent off 300 pound so 10 of 300 pounds if I divide 300 pound by 10 I will get 10 which is equal to 30 pound so 30 pound is 10 or 300 pound now we want 80 so I'm going to times that by eight to get eighty percent so I'm gonna find eighty percent off 300 pound so 80 of 300 pound or if I times that by eight well eight times three is twenty four so eight times thirty would be 240 pounds so we've now found eighty percent of three hundred pound which is 240 pound now to find 180 I just need to add that on to 100 so 100 of 300 Pounder with obviously you'd just be 300 pound so if I take our 300 pound or 100 and I add the 80 which is 240 pound that will tell me 180 percent of 300 pound so 300 plus 240 would be equal to 540 pounds so 180 or 300 pounds would be 545 so that's one way to do it there's other approaches that we could use we could take our 300 pound and we could divide it by a hundred to get one percent which would be three pound and then we could just Times by 180 and that would tell us 180 which is equal to 540 pound okay and that's it okay so our next topic is percentage change and this is video 233 and corporate Maps so percentage change is calculated by change divided by original times 100. sometimes in a question you might be asked to find the percentage increase percentage decrease percentage profit or percentage loss and they'll all be calculated same way by using this formula this is a typical question that says the height of a plant increases from 30 centimeters to 46 centimeters work out the percentage increase so the formulas change so as you can see the change is 16 centimeters so it changes 16 divided by the original which would be 30 and then times 100. so if we do 16 divided by 30 times 100 we get we get that's equal to 53.333 and so on percent so the percentage increase would be 53.333 so one percent or we could round this sometimes in the question you ask to give your answer to one or two decimal places so let's round this to two decimal places so answer would be 53.33 and that's it so percentage change is changed divided by original times 100. okay our next topic okay let's have a look at our next topic so our next topic is simple interest so simple interest is where interest is added on to money in a bank account and simple interest is where the cmite is given every single year so the question says 600 Pounders invested for three years at five percent per year simple interest work out the total interest so in this question we're told that it's five percent so let's work out five percent of 600 so let's take our 600 and divide it by 100 so I'll tell us one percent which is six and then take our six and multiply it by the percent we want which is five six times five is equal to 30. so every single year was a simple interest every single year 30 pound of interest is earned so in the first year 30 pounds added on in the second year 30 pounds out at all and so on and we've been asked to work at the tool to limited interest so the money's been invested for three years so that means that three lots of 30 has been earned in interest so 30 multiplied by three is equal to 90. so the question asks us to work at the total interest earned that would be 90 pound and that's it if we were asked how much money would be in the bank account at the end of the three years that 90 would be added on to the 600 so that would be 690 pound but the question just asked us for the total interest that would be 90 pound right so our next topic is compound interest that's video 236 on corporate Maps so compound interest is where the interest is added at the end of every single year so our question says GM's invested eight thousand pound in the bank for two years and it earns compound interest at a rate of five percent per year calculate the total amount of money that James has in the bank at the end of two years there's two different ways to do this question and I'm going to show you both approaches now the first approach is to just treat it like a percentages question so we're going to increase it by five percent to see how much James has at the end of the first year and then to work out another five percent of that and add that on to see what it has at the end of two years so at the beginning he has eight thousand pound and let's work out five percent of that so let's divide that by 100 and Times by five so when we divide by 100 that we get that's equal to 80 and if we do 80 multiplied by five because it's five percent that gives us 400 pound so five percent of eight thousand pound is 400 pound now if we add that on that's how much money James has at the end of the first year so the end of the first year James would have eight thousand plus four hundred that's eight thousand four hundred pounds now to see how much money James has at the end of the second year we now need to work out five percent of this and Aaron to find out what he has at the end of the second year so if we take our eight thousand four hundred pounds because that's how much he has divide that by 100 we get that's equal to 84 pound and then if we take our 84 pound and we times that by five we get five percent so 84 multiplied by 5 is equal to 420 pound so in a second year James earns more interest he earns 420 pound and if you add that onto the 8000 400 we'll see how much money GM says in the bank account at the end of two years so 8400 plus 420 is equal to eight thousand eight hundred and twenty pound so the end of two years GMS has eight thousand eight hundred and twenty pound in the bank so that's one approach that's quite useful approach for a small number of years maybe two years but as we go into three years four years five years it can get quite time consuming to do it that way so there is a quicker way to do that and to do that I'm going to show you another topic called multipliers and then come back to combat interest so we're going to look at multipliers which is video 239 in corporate Maps now increasing by a percentage and decreasing by a percentage now we can use a thing called a multiplier to do that really quickly and easily so if we were asked to increase 500 by 8 we can use a multiplier now if we start off with a hundred percent and we increase by eight percent we'll have 108 and 108 is the same as 1.08 as a decimal number so that would be our multiplier so if we increase by eight percent we'll have 108 and that's 1.08 so if we take our number 500 and we multiply by 1.08 that will tell us our answer straight away it will increase 500 by 8 so that would be if we do 500 multiplied by 1.08 that gives us 540. so if we increase 500 by 8 the answer would be 540 so we can use this multiplier really quickly and easily to do it so next one if we wanted to increase 3500 by 13 well the multiplier well if we had 100 and we increased by 13 that's 113 so multiply is 1.13 that's an increase of 13 so if we take our 3500 and multiply it by 1.13 we get our answer so three thousand five hundred multiplied by 1.13 is equal to 3955. so these multipliers are really great for increasing by percentage really quickly and easily and that'll be useful for compound interest likewise we can decrease using a multiple if we decrease something by two percent we're left with 98 because we would have started with 100 and decreasing by two percent leaves us with 98 so there's a multiplier 98 percent 98 is the same as 0.98 converting the percentage to a decimal so we can just multiply three thousand by 0.98 and that would tell us our answer right away so three thousand multiply by 0.98 is equal to 2940. so you can use a multiplier to decrease by a percentage as well so if we wanted to decrease eight thousand by twenty four percent we would take our 8 000 and we would multiply by well we would have started off with a hundred percent we're decreasing by 24 so 100 take away 2476 so I'll multiply would be 0.76 and if we multiply those we'll get our answers straight away so eight thousand multiply by 0.76 is equal to 6000 and 80 and that's it so this topic of multiplash will be really useful whenever we're looking at compound interest because if we go back to our question we had James invested 8 000 pound in the bank account for two years it earned compound interest of five percent per year so five percent increase would be 1.05 as a multiplier because it's gone from 100 to 105 so we can use this multiplier really quickly and easily to work out our answer so if we want to work out how much money GM's had in the bank account after one year we could take the eight thousand and multiply it by 1.05 if we wanted to work out how much money he had at the end of two years we could then times buy another 1.05 and that would tell us how much money has in the bank after two years or a really quick and easy way to do is to do eight thousand multiply it by because we're multiplying by 1.05 and multiplying by 1.05 we could write 1.05 squared using that index notation so we could do eight thousand multiply it by 1.05 squared and let's see what we get so 8 000 multiplied by 1.05 squared equals eight thousand eight hundred 120 and that's the same answer as before so this is really quick and easy way of doing compound and address questions and it's got a formula it's an initial so the amount of money or whatever started with multiplied by the multiplier to the power of time so this formula will help us work our compound interest questions really easily so our initial is the amount of money that was invested to begin with the 8 000 are multiplier because it's a five percent increase we used 1.05 so that's the multiplier and because it was two years so we use squared if it was three years we would use cubed if it was four years reduced to the power of four and so on and that's it okay let's have a look at our next topic so next topic is growth and Decay and here's a typical question says the population of an island at the beginning of 2018 was fifty thousand and each year the population increased by two percent work out the population of the island at the beginning of 2023 give your answers to the nearest hundred another question like this we can use the compounded interest formula we can use the initial times multiplier to the power of time and that'll be really useful in a situation like this because we're increasing by two percent each year so first of all our initial the population initially was fifty thousand so we're going to use fifty thousand and then we're going to multiply that by the multiplier well it's a two percent increase so that will be as a multiplier 1.02 because you're increasing by two percent you're going from 100 to 102 as a multiplier that would be 1.02 and to the power of time well between 2018 and 2023 that's five years and it's increasing at two percent per year so we've got five years so it's going to be to the part of five so that's it so we're going to work out 50 000 multiplied by 1.2 to the power of 5 and when we do that we get we get an answer of 55 204.04 and so on and the question said give our answer to the nearest hundred so to the nearest hundred are two would be 55 200 and that's it and this compound interest formula can be really useful whenever you're dealing with situations that involve growth so when everything's getting bigger such as this an increase of two percent each year or Decay so it might be whenever things are getting smaller so whenever your multiplied will be a no point something number and that's it so here we've got a finance question and you'll see lots of videos on corporate Maps under Finance so if you go to video 400 there's 400a BC and so on and there's lots of different Finance topics there that will be useful so here's a typical question it says the value of a car decreases by five percent each year and so if we bought a car two years ago for ten thousand pound work out the value now so we can use the same formula initial multiplied by the multiplier to the power of time so initially the car cost ten thousand pounds and then the multiplier well it's decreasing by five percent each year so if we have a hundred percent and we took away five percent we'll be left with 95 so that's 0.95 and we're wanting to find the value of the car after two years so that's going to be to the power of two so we do ten thousand multiplied by 0.95 squared and that gives us nine thousand and twenty five pounds so the value of the car after two years would be nine thousand and twenty five pound and that's it the next topic is a reverse percentages and that's video 240 in corporate maps and remember you can watch that video tutorial video 240 and quarter maps and I will go through reverse percentages in a lot of detail also beside that video 240 incorp Maps you'll find the practice questions which are really great because they will have a lot of curveball questions in there which are really fantastic to do they'll have the textbook exercises and I'll have the answers as well so video 240 and also remember there's reverse percentages questions in that bumper pack of questions which is in the description below so here's a typical reverse percentages question and it says Rebecca is given a 35 pay rise she has now paid 13.77 per hour what was Rebecca's pay before the pay rise so to do this what we need to do is consider the fact that we have increased her Pair by 35 so that means she now has 135 because she had 100 before her eyes and then she's given a 35 increase so she's now got 135 percent and 135 is equal to 13 pounds 77. now to find what her pair was before which we want to find 100 so let's find one percent and then find 100 so if we know what 135 is we can divide by 135 we can divide by 135 and we can divide by 135. and when we do that we get one percent and one percent is equal to 0.102 pounds and then we want to find what 100 is so we'll multiply by 100 so let's multiply by 100 and let's multiply by 100 and when we do that we'll get well one percent times 100 is 100 her pair before the pay rise and whenever we multiply 0.102 pounds by 100 we get that's equal to 10.20 so Rebecca's pair before the pay raise was 10.20 now there is another way to do this I went for a topic called multipliers and to increase by 35 we use a multiplier of we multiply by 1.35 so if you want to go backwards and find what number was increased by 35 we could divide by the multiplier so if you do 1377 divided by 1.35 you'll also find 10.20 and that's it okay next topic okay let's have a look at our next topic so our next topic is simplifying ratios that's video 269 in corporate maps and we're told the ratio of oranges to Apples in a box is four to six and whenever you get those two little dots you read it as the word two so four to six and what that means is for every four oranges the six apples so as you can see here for every four oranges the six apples and sometimes we can cancel the ratios down because here we've got simplify four to six well both of those numbers the divisible by two so we divide both of them by two so four divided by two is two and six divided by two is three so if we had the ratio of four to six we could simplify it two two to three and that would be an equivalent ratio and that's it and we've simplified these as far as we can go okay let's have a look at some more ratios and simplify those and our next one is simplify 35 to 10. so both of these numbers are divisible by five so let's divide both of them by five so 35 divided by five that's seven and ten divided by five is two sounds would be if we have if we we're asked to simplify 35 to 10 our answer would be 7 to 2. okay let's have a look at our next question so the next question says simplify 20 to 40. now both of these numbers are divisible by 4 so we can divide 28 by 4 and we can divide 40 by 4. so 28 divided by 4 is 7 and 40 divided by 4 is 10. so 28 to 40 would be the same as 7 to 10. so if you're asked to simplify 28 to 40 the answer would be 7 to 10 and we can't simplify this any further because the only thing you can only hold number you can divide these by is one and that would actually help okay let's have a look at our next topic so next topic is ratio and writing ratios in the form of 1 to n or n to one and that means one to a certain number or a certain number to one and in certain situations it can be very useful to do that so here we've got a question says the ratio of adults to children on a skill trip is 5 to 24. so that means that for every five adults there's 24 children so there could be five adults and 24 children or it could be 10 adults and 48 children it could be 15 adults and 72 children it could even be 50 adults and 200 140 children and so on I've been asked about this ratio in the form of 1 to n so instead of being 5 to 24 we want in the form 1 to a number so we want the left-hand number and the ratio to be one so to go from five to one we're going to have to divide by five so five divided by five is one so if we divide the left-hand side of the ratio by five we're gonna have to divide the right hand side of the ratio by five so we're going to have to do 24 divided by 5 as well so divided by five and 24 divided by 5 is equal to 4.8 so if we were asked to write the ratio of adults to children in the form 1 to n the answer would be 1 to 4.8 and that means that for every one adult there's 4.8 children now obviously there can't be one adult and 4.8 children but what it does is it lets us know that for every one adult there's just under five children or perhaps there's even a rule it might be there for a school trip there needs to be a ratio of one to four point five so for every one teacher there's 4.5 children and if they're being one to four point eight in this trip that they're actually going past that rule so writing a ratio in the form of one to a certain number or a certain number to one can be very useful and if in this question instead of being asked to write you in the form of one to n we were asked to write it in the form n to one what we would want to do is make the number on the right hand side of the ratio one so we'd divide both of these numbers by 24 and that would give us no point something to one okay let's have a look at our next topic okay let's have a look now at fourman ratios so we're told there's red and purple countries in the bag and there's five times as many purple counters as red counters in the bag write down the ratio of red counters to purple counters in the bag so we don't know the exact number of countries in the bag but we do know this five times as many purple counters as red counters so for instance there could be five purple counters on one red counter there could be ten purple counters and two red counters there could even be 50 purple counters and 10 red counters so we don't know the exact numbers but we do know this five times as many purple counters as red countries in the back and we've been asked to write down the ratio of red counters to purple counters in the back so for instance if there was one red counter there would be five purple counters that means the ratio of rake Hunters the purple counters in the bag would be one to five for every one Ray counter there's five purple countries and that's it now we could have chosen other numbers here we could have said for instance there were two red and 10 purple and then we could simplify the ratio and when we divide both of these numbers by two we would get one to five so the ratio of red counters to purple counters in the bag is one to five so our next topic is ratios and fractions and that's video 269 and Cobra maths and our question says a box contains white beads and gray beads and the ratio of white beads the gray beads is two to three so we don't actually know how many white beads and gray beads there are but what we do know is that for every two white beads there will always be three gray beads like so and we're asked what fraction if the beads are great well if we have a look at this diagram this little sketches will help us so if we work out what fraction of this is great that would be the answer to how many fractions of the beads in the Box will grow so we've got all together five beads here one two three four five so we'll put five on the denominator and for the numerator what fraction are gray well there's three that are gray so that means that three-fifths of these five beads are gray so means of three-fifths of the beads will be gray and that's it so whenever you ask to write a ratio as a fraction a diagram a little sketch like this can be very useful an answer here would be three-fifths and instead of being asked what fraction the beads are great we could have been asked what percentage of the beads would grow so we would then read as a fraction as we've just done three-fifths and then we'll just change the Three-Fifths to a percentage so then that would be sixty percent because three-fifths is sixty percent if we're asked the question what percentage of the beads were white well all together there was five beads and two of thermal weights we would write that as a fraction two out of five two-fifths of the beads are white and then let's change that to a percentage that would be well two fifths is forty percent because one fifth is 20 so two-fifths is forty percent so answer would be forty percent so if you ask to change your ratio to a percentage I would often change it to a fraction to begin with and then change that fraction into a percentage now whenever we're dealing with ratio questions we may encounter questions to look like this almost says there's 91 beads in the Box explain why he must be incorrect so it's very useful to consider what the ratio means so we've got the ratio of white beads to gray beads in the boxes two to three so that means in the Box it could be we could have white beads and gray beads white and gray there could be two because the ratio is two to three there could be two white and three gray and all together the total number of beads would be five there now there may not be five beads in the box or maybe more so there could be another two white and another three great so another two white would be four another three gray would be six so that would be ten beads all together now there may not be ten so there may be more beads so maybe if we add another two white and another three gray that would be six and nine and six plus nine is equal to Fifteen unless you notice the total number of beads is a multiple of five because if we add the two and three together every time we're adding another five beads so Omar says there's 91 beads in the Box well there can't be 91 beads in the Box because if then the ratio of two to three that means it must be a multiple of five there must be a multiple of five number of beads in the box and 91 isn't a multiple of five okay let's have a look at our next topic so our next topic is sharing in a ratio in this video 270 in corporate maths so here's part of the chord miles revision card I'm with Jack and Chloe share 75 pound in the ratio two to three how much money do they each receive it's very important to be able to share a number in a ratio and a share a number and a ratio the first step is to add together the parts in the ratio because jackets two parts and Chloe gets three parts so all together two plus three would be five so there's five parts in total now what we're going to do is we're going to take the grand tool of what they're sharing so in this case it's 75 pound and we're going to do 75 divided by whatever you get whenever you add the ratio together so we're going to divide it by five because two plus three is five so we know there's five parts in total so if we divide 75 by 5 we'll see how much money is in one part so 75 divided by 5 is 15. so that means it's 15 pound in one part now we know the Jack gets two parts and Chloe gets three parts so that means that Jack has two dots of 15 and Chloe gets three lots of 15. so for Jack we're going to do 15 multiplied by his number in the ratio which is two so 15 times 2 is 30 pound and for Chloe she gets three parts so we'll do the 15 pound multiplied by three her number in the ratio so for Chloe we'll do 15 multiply by three and that's equal to 45. and what's great is if we add 30 pound and 45 pounds we get the 75 pound that we started with so if you asked to share something on a ratio the steps would be first of all add the parts together in the ratio then divide the grand total by the total number of parts and then times each of the numbers by how much is in one part and that's it okay let's have a look at our next topic which is whenever we're dealing with ratio questions whenever we're given one quantity and that's video 271 on corporate Maps so here we've got the ratio of lemon sweets to Strawberry sweets in a tub is five to three so in this tub for every five lemon sweets there's three strawberry sweets and we're told there's 120 lemon sweets in the tub so rather than the last time when we're told how many sweets were all together what we're told is how many lemon sweets there are and we're asked how many strawberry sweets will there be in the tub so in a question like this whenever you give them one quantity we know there's 120 lemon sweets now we're told the ratio of lemon to Strawberry is five to three so the lemon number in the ratio is five so if we divide 120 by five we'll find out how many sweets there are in one part so 120 divided by 5 is equal to 24. Sammy says 24 in one part and we find that by dividing how many lemon sweets there were by the lemon number and the ratio of the five now we're asked how many strawberry sweets are in the tub now strawberry is three parts so what we'll do is we'll multiply 24 by 3. so we'll do 24 multiply by three and that's equal to 72. so whenever you give them one quantity so in this case we were told those 120 lemon sweets what you do is you divide that by the lemon part in the ratio which was the five I'm going to redo 120 divided by five that tells you how many sweets are in one part and then you can times divide the other number in the ratio to see how many strawberry sweets there were in this case that's it and if you were asked how many sweets to wear in total you could add together the 120 and the 72 and I'll tell you how many sweets are all together so that's it okay so we're now going to look at what happens whenever we're given two ratios and this is video 271 and corporate Maps so here we told the farmer keeps sheep cows and chickens on a farm and the ratio of sheep to cars has sheep the cows on the farm is four to three so that means that for every four sheep there's three cows and the ratio of coaster chickens is ten to seven so that means it for every 10 cars the seven chickens we've been asked to find the ratio of sheep to cows to chickens so whenever I'm giving a question like this I like to do a little table so I like to write down what we've been given so we've got sheep so we've got sheep cows and chickens and I could have been in the question it may not have told us what we're looking at could have just been the ratio of a to B and B to C and then the headings of my table would be a b and c but in this case we're told what we're looking at sheep cows and chickens so the ratio of sheep to cows is four to three so sheep decodes is four to three so I'm going to write that ratio down four to three and then we're told the ratio of carbs to chickens is ten to seven so close to chickens and I'm just going to write that beneath is ten to seven so if we have a look we've got the ratio of sheep to Coast is four to three and the ratio of cars to chickens is ten to seven so we've been asked to write this as one ratio of sheep to Coast to chickens so the key thing is going to be the cows because if we have like the cows are on both ratios for every four sheep there's three cows and for every 10 cars there's seven chickens so if we can find a common multiple of three and ten that will help us combine these as one single ratio now I'm going to find the lowest common multiple of three and ten with the lowest common multiple of three and ten would be 30 because if you write down the multiples of three three six nine twelve and so on and if you write down the multiples of 10 10 20 30 30 would be the first number before those lists so I'm going to write 30 beneath the 3 and the 10 because that's the lowest common multiple of 3 and 10. now what I'm going to do is I'm going to have a look at this top ratio and I want to instead of being four to three I want it to be something to 30. so to get from 3 to 30 we multiply by 10. so if we multiply the 4 by 10 as well then we will know what number this is in the ratio so if we times about 10 and Times by 10 that would give us 4 times 10 is equal to 40. so 4 to 3 is the same as the ratio of 40 to 30. now if we have a look at our next ratio that the car is the chickens ratio which was 10 to 7 well we instead of having 10 to 7 we want to write 30 to something so if we multiply 10 by 3 we get 30. so if we multiply 7 by 3 we'll find this number in the ratio so 7 times 3 is equal to 21. so we've got that the ratio of course the chickens would be 30 to 21 and that's an equivalent ratio as 10 to 7. so now we've got one ratio we've got the ratio of sheep to college to chickens is 40 to 30 to 21. okay now let's have a look at right now ratios as equations and equations as ratios and that's video 271d on corporate Maps so we've been given the ratio X to Y is equal to one to four and we've been asked to write this as an equation Link in X and Y so here we've got the ratio and we know that the ratio of x to Y is one to four and that means that Y is four times larger than x because for every one of X is four of Y so it could be that X is equal to 1 and Y is equal to four it could be that X is equal to 2 and Y is equal to eight it could be x equal to 10 and Y is equal to 40 and so on but what we know is the value for y is four times larger than the value for x so let's write out as an equation Y is equal to 4 x because 4 times the smaller number x 4 times x is equal to Y because Y is four times larger and that's it and we can show this just look into this diagram here we've got Y and we know the ratio of x to Y is one to four so in other words if we've got y y is four times larger than x if we multiply X by 4 we would get Y is equal to 4X and that's it okay let's have a look at our next question so next question says given the ratio of x to Y is equal to two to three write an equation Link in X and Y so here we've got X and Y and X and Y are two numbers and when we simplify the ratio we get two to three now it could be that X is equal to 2 and Y is equal to 3. it could be that X is equal to 4 and Y is equal to six it could be the x equal to 20 and Y is equal to 30. but what we know is that when we simplify that ratio we go two to three and that means for every two of X is three of Y so Y is going to be bigger than x and I've represented it here as a diagram where I've got that my X is equal to two centimeters or more Y is equal to three centimeters so that's a little diagram here where x equals two centimeters and Y is equal to three centimeters so it represents this ratio now if we were to Double Y and we were to treble X it would be the same as each other if we have a look here if we had two y's and we had three X's they would be the same length and that's it that means we've got an equation that 2y is equal to 3x so that's our equation Link in X and Y now that is an equation Link in X and Y but if I had something like this I would tend to make y the subject or X is subject but if I was making y the subject I would divide both sides by 2 so divide by 2 and divide by two and my left hand side of the equation would just become y because 2y divided by 2 is just Y and if we have 3x divided by 2 we could write 3x over 2 like so so y equals 3x over 2. I would tend to write it like this as 3 over 2 x so y equals 3 over 2 x okay let's have a look at our next question so we're going to write an equation as a ratio so if we've got the equation y equals 2x and we've been asked to write down the equation of x to Y let's have a look at the equation to begin with we know that Y is equal to 2 times x that means that Y is two times larger than x for instance if x is equal to 10 y would be equal to 20. so let's write that as a ratio so for instance if x is equal to 10 then y would be equal to 20 and then you can just divide those by 10 and you get one to two we could have just written one to two to begin with for instance we know that Y is twice as big as X so if we just say x equal to 1 Y is equal to two and that's it so our ratio of x to Y would be one to two and that's video 271 D and code Maps Okay let's have a look at our next topic so our next topic is direct proportion last video 254 on corporate Mouse and here's part of the corporate Mouse revision card and we're told the C is directly proportional to the square of D so let's write that statement down so we could see it's directly proportional so we'll use the proportion symbol um some emotions call it fishy but the proportion symbol to the square of D so that means D Squared because it says the square of D so if we read this it says C is proportional to D Squared C is directly proportional to D Squared sometimes the question might say is varies directly and that means the same is directly proportional and then we've got some information and this information is going to be really useful we're told when C is equal to 200 D is equal to two so let's write this out again and let's get rid of this proportion symbol so we're left right down C and then equals and then we'll put in a K in that case the consonant proportionality so it's a certain value we're going to find multiplied by D Squared so we've got C equals K times D Squared we've got rid of the proportion sign and we'll put in the K which stands for the constant of proportionality now what we're going to do is we're going to substitute in our values we've got C is equal to 200 so we've got 200 equals k multiplied by and with D Squared now D is equal to 2 so that's going to be 2 squared now let's work this out so we've got 2 squared is equal to 4 so we've got 200 equals K multiplied by 4 and if we divide both sides by 4 we get 50 is equal to K so K is equal to 50. and let's put that back into our formula so we've got C equals instead of K times D Squared we're going to have C is equal to 50 times D Squared or 50 D Squared and that's our formula and what I'm going to do is I'm going to put a box around it it's very important and that's what we would use for the rest of this question so that's how you approach directly proportional questions not all finished yet but that's how you approach them so you write down the statement you're given with the proportion symbol you then get rid of that proportion symbol and put in a care for the constant proportionality a certain number we're going to work out in this case it was 50. then there'll be some information that you can put into your formula that has the K to work out what the value of K is and then once you find it then you've got your formula which is really useful okay so the next part of the question says find C whenever D is equal to five so D is equal to five so C is equal to 50 multiplied by D Squared but D is equal to five so we're going to have C is equal to 50 multiply by and instead of D Squared it's going to be 5 squared now remember our order of operations we have to do squaring before we do any multiplication so we're going to do 5 squared is 25 so C is equal to 50 multiplied by 25 and then finally 50 multiplied by 25 is 1250 and that's it and if you look at the chord match revision card there'll be step-by-step guide on how to do that and these revision cards will be really useful for you if you do have a set of the higher revision cards it's really useful and there's a link to it in the description below one other thing that's useful with direct proportions to know what the graphs look like and here are some situations and what their graphs look like if you have wires directly proportional to X that means as X increases so does y and the graph looks something like this this is a nice straight line going up like that okay next one is why it's proportional to x squared and the graph of that will look something like this it's half of the quadratic graph so starts with the origin zero zero and curves upwards like that and finally if we had a wise proportional to the square root of x it looked like the square root of x graph so it would look at sort of zero zero and then go upwards like that and that's just some of the proportionality graphs which might be useful for you okay let's have a look at our next topic and our next topic is inverse proportion and that's video 255 in corporate Maps we're now going to look at inverse proportion and what that means is as one value gets bigger the other value gets smaller so here's our example we've got T is inversely proportional to the cube of Bell so let's write that down we've got T is inversely proportional so we write T is proportional and because it's inversely we put 1 over and then we put whatever it says next on the denominator so it says T is inversely proportional to the cube of L so we do L cubed so here we've got T is proportional to 1 over L cubed so if it says inversely proportional you do the proportional sign and you put one over and then we're given some important information we're told whenever L is equal to 0.2 T is equal to 5. we've been asked to find the formula connecting T and L so whenever we have direct proportion we wanted to get rid of this proportional sign and we put in K so what going to do is we're going to write T and then what we'll do is we'll get rid of this so we'll put in equals and whenever it's inversely proportional we're going to put our K in the numerator okay so we'll have K Over L cubed the reason is we've multiplied this by K and whenever you multiply this by K the K will be on the numerator now we've told some information we're told whenever L is equal to 0.2 so whenever L is equal to 0.2 T is equal to 5. so let's substitute those in so we've got 5 instead of T equals K which is what we want to find out over L cubed so that's going to be 0.2 cubed let's work out with 0.2 cubed is so 0.2 cubed there's no point not not it now we want to get counter tone so we don't want this divided by 0.0 it's what we're going to do is multiply both sides by no point no not eight so we'll multiply both sides by 0.08 multiplied by no point not not it and when we do that we get on the left hand side 5 times 0.08 would be 0.04 on the right hand side we had K divided by 0.08 we Times by no point not not it to get rid of that so we're just F of K so equal to 0.04 so the question says write down a formula connecting T and L so we go back up to here we're told that t is equal to K Over L cubed now we know that K is equal to 0.04 so that means we're going to have 0.04 over L cubed and that's our formula that connects T and L T is equal to 0.04 divided by L cubed next we're asked to work out the value of T whenever L is equal to 0.5 so that means the t is equal to 0.04 divided by 0.5 cubed and when we work that out we get T is equal to 0.32 so we substituted our value for LR 0.5 into our formula and we find that t would be equal to 0.32 okay next next we're asked to find the value for L whenever T is equal to 2. so let's write down our formula again we've got T is equal to 0.04 divided by L cubed and we're told that t is equal to two so we're going to have 2 equals 0.04 divided by L cubed so let's solve this let's multiply both sides by L cubed so we would get 2 L cubed is equal to 0.04 next we want to get the L on its own so let's get rid of the two so let's divide both sides by two so we'll get L cubed equals 0.02 and then finally we want to get the error so we want to get rid of this Cube so let's cube root both sides so L is going to be equal to the cube root of not 0.02 which is equal to 0.27144 and so on but let's just run that to two decimal places that's going to be 0.2722 decimal places and that's it so whenever you do an inverse proportion questions you want to make sure that whenever you put the proportional symbol in because inverse is going to be one over and then whatever comes next in this statement then whenever you get rid of the proportion symbol and put in the equal sign then you put the Canton on the numerator and then you can just solve the questions quite nicely another thing I want to show you then is the proportionality graphs so we've already looked at why it's proportional to X so this graph and we've looked at y's proportional to x squared so this graphs are not the origin of current upwards like the x squared graph and we've also looked at y's proportional to the square root of x so it would start at the origin and then curve like this or whenever you get something such as Y is inversely proportional to X the graph will look something like this where it starts up quite High remember as one value gets bigger the other value gets smaller and it'll curve downwards like so now sometimes whenever we're dealing with proportion we need to have questions that involve time and this is proportion dealing with time and this is video 256a and corporate Maps so here we've got our question it says it takes 20 hours for free pumps to fill a swimming pool how long would it take if four pumps were used so in a question like this we're told that takes 20 hours for free pumps to fill a swimming pool so what I'm going to do is I'm going to consider how long it would take for one pump to fill a swimming pool well if there's three pumps one pump would take three times longer so if there's 20 hours if we multiply that by three that would tell us how long it would take one pump to fill the swimming pool so 20 multiplied by three would be equal to 60. so if there was one pump it would take 60 hours if there was two pumps it would take half of that time it would take 30 hours if there's three pumps you would divide it by three to get 20 hours and so on and the question says how long would it take if those four pumps used so if there's four pumps being used well if it takes one pump 60 hours four pumps would take four times less time we would divided by four so we're going to take 60 and divide it by four and 60 divided by 4 is equal to 15 hours so if there was four pumps used instead of taking 20 hours for three pumps it would take for four pumps 15 hours that's it okay and that's video 256 a in corporate Maps Okay so we looked at rounding now we're going to look at finding what the highest number could have been the lowest number could have been so here we've got the population of wells and that's 12 000 to the nearest thousand and we've been asked to find what was the lowest possible population of Wales and what's the highest possible population of Wales so we're told the population of Wales is 12 000 to the nearest thousand and we've been asked to find the lowest possible population of Wales so let's consider the numbers below 12 000 that would Round Up to be 12 000 to the nearest thousand so the population of Wales could have been 1199 and that number would obviously Round Up to be 12 000 to the nearest thousand it could have been even lower it could mean 11 900 if we rounded that to the nearest thousand it'd be twelve thousand it could be even lower it could have been eleven thousand six hundred if we rounded that to the nearest thousand the answer would be twelve thousand it could even be as low as eleven thousand five hundred if that number would round two twelve thousand to the nearest thousand but it couldn't be eleven thousand four hundred and ninety nine because that would run down to eleven thousand so the lowest possible population of Wales would be eleven thousand five hundred eleven thousand five hundred the lowest possible number that would Round Up to be 12 000 to the nearest thousand okay next question what's the highest possible population of Wales so let's consider the numbers above 12 000 that would round down to be twelve thousand at the nearest thousand so the population of Wales could have been twelve thousand and one obviously if we rounded that to the nearest thousand it would be twelve thousand it could be something even higher it could be twelve thousand two hundred if we round that to the nearest thousand to be twelve thousand not thirteen thousand it could be something even higher twelve thousand four hundred would run down to be twelve thousand but it wouldn't Round Up to be thirteen thousand so it could be twelve thousand four hundred and it could be anything as high as twelve thousand four hundred ninety nine because twelve thousand four hundred and ninety nine would run down to be twelve thousand and not up to thirteen thousand it couldn't be twelve thousand five hundred because twelve thousand five hundred if we render that to the nearest thousand because it's a five in the hundreds column we would round it up to be thirteen thousand so it couldn't be twelve thousand five hundred so they also be twelve thousand four hundred ninety nine so that would be 12 thousand to 499 so the lowest possible population of Wales would be 11 500 and the highest possible population of Wales could have been 12 499. now in this question we've looked at what we call discrete data and that's data that can only take certain values whether it be in population we didn't need to consider decimal numbers later on this video we'll look at what we call continuous data and that's data that can take any value on a given scale and we'll look at that one if we look at a topic called error intervals okay let's have a look at our next topic okay so let's have a look at our next topic so next topic is error intervals and that's video 377 on corporate Maps so we've looked at rounding and we've looked at finding the lowest number or the highest possible number whenever we've looked at data such as the population of Wells now let's have a look at what happens whenever we run numbers that can take any value at all and we often do that whenever we're looking at error intervals so here we've got the mass of a letter is 80 grams to the nearest 10 grams so because the mass of the letter it can take any value it can be whole numbers or could even be decimals and we've been told to write down an error into for the mass of the letter Y so the mass of the letter has been rounded to the nearest 10 grams that means that the master letter may not be 80 grams it could have been 80 grams it could have been 79 grams it could have been 76 grams it could be anything as low as 75 grams because that's the lowest possible value that it could be because 75 grams would Round Up to early but anything below 75 grams so it would be 74 point something and that would run down to be 70 grams to the nearest 10 grams so 75 grams would be the lowest possible mass of the letter now let's consider the masses above 80 grams because as I said the letter could be 80 grams but it could be something that's even higher than 80 grams as long as it rounds to 80 to the nearest 10. so it could have been 81 grams it could have been 82 grams it could have been 84 grams it couldn't be 85 grams because 85 grams would Round Up to be 90 grams but it could be 84.7 grams it could be 84.9 grams it could be 84.99 grams and so so on it could be anything that is up to but not including 85 grams so let's write this as an error interval so let's start off with the mass of the letter which is y so Y and we know that it's bigger than or equal to 75 grams so it could be 75 grams or a numbers bigger than that so we're going to write bigger than or equal to 75 grams but it can go up to but not include 85 grams so we then write this less than 85 grams so this reads y the mass of the letter is bigger than or equal to 75 grams but less than 85 grams and that's our error interval it shows all the possible values for the mass of the letter why okay let's have a look at another question okay let's have a look at our next example so Nigel runs the number X to one decimal place so there's a certain number X and he's rounded to one decimal place and his answer is 7.3 write down an error interval for X so the number could have been less than 7.3 it could have been 7.29 that would round to 7.3 to one decimal place it could have been 7.26 it could be anything as low as 7 1.25 because 7.25 when we round that to one decimal place because it's a five and above in the second decimal place when we round that to one decimal place it'll be 7.3 but it couldn't be anything lower than that so Nigel's number could be as low as 7.25 but it could have been a number bigger than 7.3 it could have been 7.3 it could have been 7.31 it could be 7.34 but it couldn't be 7.35 because then that would Round Up to be 7.4 so a natural number could be n of and up to but not including 7.35 because it could be 7.348 and so on right up to but not including 7.35 so let's write that down X his number could be bigger than or equal to 7.25 but it can be up to but not including 7.35 and that's it so that's the error interval for X okay so let's have a look at our next question so next question says Chloe truncia to number W to one decimal place so Chloe's got a number actually truncated so she's not rounding it she's trunk internet and what that means is if you get a decimal number and you just chop it off so for instance if we had Pi which is 3.14159 and so on and if I rounded this to three decimal places it'll be 3.14 and then because the fourth decimal places if I would round up so be 3.142 so that's rounding if I was to truncate this to three decimal places what that would mean is I just chop off I just ignore anything after the third decimal place so if I truncated this to three decimal places it'll be 3.141 so in this case Chloe has truncated a number W to one decimal place and her answer is 1.4 so she's got 1.4 so Chloe's number it could have been 1.4 it could have been 1.41 if we truncated that to one decimal place that'd be 1.4 it could have been 1.47259 if we truncated that to one decimal place it'll be 1.4 Chloe's number could have been 1.498972 and if we truncated that to one decimal place it'd be 1.4 but it couldn't be 1.5 so it'd be anything up to but not including 1.5 so let's write down the error interval for Chloe's number W so w it could have been 1.4 so it could be bigger than or equal to 1.4 but then it can go up to 1.5 but not be 1.5 so less than 1.5 so that would be the area integral for w that's it so here's a typical question it says the length of a field is 120 meters to the nearest 10 meters and the width of the field is 70 meters to the nearest meter and we've been asked to work at the lower Bound for the area of the field so it's a rectangle field perhaps you should have said that my question it's a rectangular field and we've defined this area we're going to multiply the length from the width together and we want to find the lower bounds we want to find the lowest possible area of the field so let's start off by looking at the length of the field we know it's 120 meters to the nearest 10 meters so it's lower bound the lowest possible value that it could be would be 115 meters that's the lower bound and the upper Bound for the length of the field well the upper bound would be 125 meters in terms of the width of the field well this has been rounded to the nearest meter so the lower Bound for the width of the field would be 69.5 meters and the upper Bound for the width of the field would be 70.5 meters so there are the bones and we've been asked to work out the lower Bound for the area of the field so the lower bound will be the smallest possible area so we would do 115 multiplied by 69.5 so we're going to do 115 meters multiplied by 69.5 and that will give us the lower Bound for the area of the field so at 115 multiplied by 69.5 and that equals 7992.5 meters squared so that's the lower Bound for the area of the field if we were asked for the upper bound if there of the field we would do 125 meters multiply by 17.5 and that will give you the upper Bound for the area of the field so we've got the lengths of time taken by four people to complete a puzzle are listed below and it says each time is given to one decimal place so the four people have taken seconds to one decimal place if 0.4 seconds to one decimal place 35.5 seconds to one decimal place and 19.8 seconds to one decimal place and we've been asked to work out the greatest possible range so remember the range is equal to the largest subtract the smallest so the largest subtract the smallest so here our largest value would be this one 35.5 and our smallest value would be 8.4 but each of these times have been rounded to one decimal place okay so let's find the upper bound and lower Bound for these values so the upper bound so this number has been rounded this 8.4 has been rounded to one decimal place so the upper bound would be 8.45 seconds because it could go up to 8.45 seconds but not including and the lower bound would be well it could go down as far as 8.35 seconds because that's the lowest possible value that would Round Up to 8.4 seconds likewise for 35.5 the upper Bond would be 35.55 seconds and the lower bound would be equal to 35.45 seconds so we find the upper bound and lower Bound for each of those values and we've chose those ones because we're looking at the range which is the largest subtract the smallest so we're going to want to use the upper Bound for the largest value and we're going to want to use the lower Bound for the smallest value because when we take them away we'll get the biggest possible answer so we're going to do 35.55 seconds and we're going to take away 8.35 seconds and whenever you do that you get 27.2 seconds if they wanted the smallest possible range you'd want them to be as close together as possible so you use the lower Bound for the largest value and the upper Bound for the lower value so that they're closer together and they give you the smallest possible range okay next question says Natalie runs 100 meters to the nearest 10 meters so it's 100 meters to the nearest 10 meters that means it could be the lower bound would be 95 meters and the upper bound would be equal to 105 meters there the upper and lower bounds for how far she's ran and it takes her 14 seconds to the nearest Second so the lower boundary so she could do it in 13.5 seconds or the upper Bond the greatest possible amount of time it would take would be 14.5 seconds there the lower band and upper Bound for 14 seconds at the near second so the question says to work out the greatest possible speed so we're working at speed so the speed is found by the distance divided by the time now but we've been asked to find her greatest possible speed so to do that what you want to do is you want to run the largest distance in the smallest possible time because the largest distance in the smallest possible time will give you the greatest speed so the greatest possible distance would be 105 meters that's the greatest possible distance she could run and in the smallest amount of time would be 13.5 seconds because if she ran a large distance in a smaller amount of time that means she's running even faster 105 the upper Bond divided by the lower Bond that's going to give you the greatest possible speed so 105 divided by 13.5 would be equal to 7.778 meters per second to three decimal places that'd be her greatest possible speed if you want to find her slowest possible speed you should do the smallest distance in the greatest amount of time that would mean she's traveling slower and so on okay let's have a look at our next topic and our next topic is swords and this is video 305 to 308 on corporate maps and whenever we're dealing with search there's loads of certs so if you're multiplying two thirds together such as the square root of three multiplied by the square root of five you can just multiply those to get the square root of 15. so the square root of 3 multiplied by the square root of 5 is the square root of 15. so if you've got two thirds you can multiply them together that also works even if you've got numbers in front of them for instance if you had three root 2 multiplied by 2 root 5 you can multiply the numbers at the front so three times two is six and then you can multiply the thirds root 2 times root 5 is root 10 so the answer would be 6 root ten so you can multiply thirds together if they're searched without numbers in the front so if it's just root 3 times root 5 you can multiply them to get root 15 or if you're multiplying Thirds with numbers in front of them such as 3 root 2 multiplied by two root five you can multiply the numbers in front first of all to get six and then you can multiply root 2 by root five root 10. okay next if you've got root a times root a the answer is just a so for instance if you had root 7 times root seven well root seven times root seven and looking at the low above would be root 49 because 7 times 7 is 49 and remember the square root of 49 is just seven so the answer would be seven so so view to root seven multiply by root seven the answer would just be seven root two times root 2 would be two would be ten times root ten would be ten and the reason is root a times root a is equal to a and also we can divide thirds as well so for instance if we had Root 6 divided by root 2 we just do six divided by 2 which is three so answer would be square root of three and also that works with numbers in front as well so for instance if we had 12 root 14 divided by 3 root 2 we can divide the numbers in front to begin with a bit like algebra we can do 12 divided by three which is four so four and then we could do the square root of 14 divided by the square root of 2 which would be the square root of seven so it would be four root seven and that would be it so these are the laws of thirds and they're very important so you can multiply those together you can divide thirds and also if you've got something such as root a times root a it's just a okay let's have a look at our next topic okay we've looked at our rules now let's look at how we can simplify some search so if we had something such as the square root of 200 if you type that into your calculator and press the square root of 200 and press equals you'll find it comes up as 10 root 2 so it doesn't come up as a square root of 200 it comes up as 10 root 2. whereas if you press the square root of 10 and press equals it would just come up as the square root of 10 and that's because some thirds can be simplified and some so it just can't be simplified so here we've got a third that can be simplified the square root of 200. so what you do is you consider what's the biggest square number that you can divide 200 by now the biggest square number that I can divide 200 by is 100 because 200 divided by 100 is 2 I can divide 200 by 100. so I'm going to write down the square root of 100 that biggest square number multiplied by the square root of 2. now what's fantastic is if you've got a square number you can work out the square root of it and the square root of 100 is 10. so we've got 10 times the square root of 2. and then so it's a bit like algebra we tend not to write the multiplication sign so instead of writing 10 times root 2 we just write 10 root 2 and that means 10 lots of root 2. and that's it so if we want to simplify our third you look for the biggest square number you can divide it by and then you split it up into two thirds timesing together so the square root of 100 times the square root of 2 you square root the square number and then you just put them together let's have a look at another one okay let's have a look at our next one so we're going to simplify the square root of 75. so what we're going to do is look for the biggest square number we can divide 75 by so let's write down our square numbers 1 4 9 16 25 36 49 64 81 and so on now I'm not going to go any further because we've gone past 75 and we're looking for the biggest number the biggest square number we can divide 75 back well one what else I can actually help us because we'll end up with the square root of 1 times the square root of 75 that's not actually going to simplify 4 but 4 is not a factor 75 9 is another factor of 75 16 is not a factor 75 25 is because 75 divided by 25 is free 36 49 64 and 81 are not factors so 25 would be the biggest square number that is a factor of 75. so let's write that down the square root of 25 multiplied by the square root of 3 because those two were multiplied together to give us 75. now we can square root of 25 the square root of 25 is 5 so it'll be 5 times the square root of 3 and then again we tend not to write that multiply sign so that would be 5 root 3. and if you've got your calculator and typed in the square root of 75 and press equals you should get 5 root free okay let's have a look at our next topic let's add inserts so whenever we want to add thirds or take thirds away from each other we want to simplify them and then hopefully we can add or take them away so let's have a look at our first example we've got the square root of 18 plus the square root of 8. so let's start off with the square root of 18 so the square root of 18 would be equal to well let's consider the biggest square number that's a factor of 18 and that would be 9 so we'd have root nine multiplied by root 2. and the square root of 9 is 3 so it'll be three times root 2 or 3 root 2. so instead of root 18 we're going to write 3 root 2 because that would simplify to three root 2 and root 8 well the biggest square number this will factor of eight is four so we're going to write root 4 multiplied by root 2 and square rooting the 4 would give us 2 root 2. so root 8 would be two root 2. so we've got 18 plus root 8 we would have three root 2 or 3 lots of root two plus two lots of root two and that's great because we're now dealing with just root twos and if you have three root twos and you add a two root twos all together that would be five lots of root two so answer would be five root two and that's it okay this time we've got root 75 plus 9 root 12. so let's start off with our Route 75 that would be Route 25 that's the biggest square number that you can divide 75 by multiply by root 3 and the square root of 25 is 5 so that'd be 5 root 3. so root 75 is 5 root 3. and then in terms of our 9 root 12 let's deal with our root 12 to begin with so root 12 would be well the biggest square number we can divide 12 by would be four so it'll be root 4 multiplied by root 3 and the square root of 4 is 2 so that'd be 2 root 3. and here we didn't just have our root 12 we had 9 root 12 so that means nine lots of root 12. so we're going to multiply this by nine so we're going to multiply the number two by nine so to be 18 root 3. so we've got five lots of root 3 plus 18 lots of root three so all together that would be 23 lots of root 3. so root 75 plus 9 root 12 would be 23 root 3. so sometimes with surge we're asked to expand and simplify brackets that involve zones so here we've got three minus root 2 close bracket squared and whenever we've got a bracket squared it means they're multiplied by itself so let's write the bracket out beside itself so 3 minus root 2 close brackets three minus root two close brackets and we're going to multiply this by itself so we're going to do 3 times 3 which is nine so that's equal to nine then we're going to do three times minus root two so it's gonna be minus and when we're multiplying the whole number of our third we can just put them together so 3 times root 2 would be three root 2 or in this case it's gonna be minus three root 2. then we've got minus root 2 times 3 again putting them together would be minus three root two and finally we've got minus root two times minus root 2. well negative times a negative is a positive so plus and then you've got root 2 times root 2. now that could be root four but remember the square root of four is two so answer just there would be two so now we need to simplify so let's look at our whole numbers we've got nine and then we've got plus two well nine plus two is eleven so we're just going to write 11. now looking at our thirds we've got minus three root 2 take away another three root two well all together that'd be minus six lots of root two so it'll be minus six root two and that's it okay let's have a look at our next topic okay so let's have a look at our next topic so our next topic is rationalizing denominators now Mavs we tend not to like to have rational numbers on the denominators of fractions so for instance something like 10 over root two we tend not to like to have that because we've got a rational number root 2 on the denominator so whenever we've got a fraction such as 10 over root 2 we are sometimes asked to rationalize the denominator and what that means is get rid of that root 2 on the denominator now whenever we're dealing with fractions we can multiply both the numerator and the denominator by the same thing and we get an equivalent fraction so for instance if we had 10 over root 2 we can multiply these both by four or by nine and you would get an equivalent fraction but what be really great is that whenever we choose our number so whenever we multiply above the numerator and the denominator by it that we can turn this denominator into our rational number now remember if you've got something like root 2 you can do root 2 times root 2 and that would give you 2 because root 2 times root 2 is equal to root 4 which is 2. so if we multiply both the numerator and denominator of this fraction by root 2. what we will do is we will get rid of this root 2 on the denominator so whenever we do that we get 10 times root 2 is equal to 10 root 2 and on the denominator we would have root 2 times root 2 which is root 4 and the square root of 4 is 2. so that we would get 10 lots of root 2 10 root 2 divided by two and that's fantastic because we have rationalized the denominator we've changed our rational number our root 2 into our rational number now if something like this you can actually simplify a little bit forever if you had 10 root 2 and you divided it by 2 you would get 5 root two and that's it okay so if you had something like 4 over root 3 and you were asked to rationalize the denominator what you do is you multiply both the numerator and the denominator by the third on the denominator of this root 3. so multiply by root 3 and multiply by root 3 and let's see what we get four times root 3 is 4 root 3 and on the denominator root 3 times root 3 is root nine which is three and that's it so we'd have 4 root three over three and again this is great because we have rationalized the denominator we have changed our root three our rational number into a rational number so if you ask the rationalize the denominator where you've just got a third on the denominator you can just multiply both the numerator and denominator by that third let's have a look at another one okay our next fraction is five over four root 3. now on this one it's a little bit different because it's not just root 3 on the denominator we've got 4 root 3. now we've got a choice here we could multiply both the numerator and the denominator by four root 3 if you wanted to I personally just multiply it by the third part I'm just going to multiply by the numerator and the denominator by root 3 and let's see what we get so multiply by root three and multiply by root 3. five times root 3 is just five root three and on the denominator we had four root three and we multiply it by root three well if we're four root three and multiply it by root 3 that would give us four times the square root of nine and the square root of nine is three so we would have four times three and four times three is equal to twelve so if you had four root three and you multiply it by root 3 you would get four times three which is twelve and that's it so we have rationalized the denominator and we can't cancel this down because 5 12 can't be canceled then that's it okay we're now gonna look at how to rationalize the denominator for something like this where we've got three over root two plus one and because we've got this root two plus one on the denominator it's where the exam boards will certainly differ from each other AQA only expects you know how to rationalize the denominators whenever you've got a third like root 2 or root 7 on the denominator or even four root seven or two root two but not where you've got this root two plus one but if you're studying for OCR at Excel you're expected to know how to rationalize the denominators like this so if you're studying for OCR or at Excel you going to need to know how to do this if you are studying for AQA feel free to skip on to the next topic alternatively if you could watch it anywhere because it's going to be very useful for you if you're going to go on to a level Maps anyway okay let's get started so if for instance you were given a question such as this rationalize the denominator of 3 over root two plus one now if you're given something like this okay and you want to rationalize the denominator what we've got to do is multiply above the top and the Bottom by what we call the conjugate of the denominator and what that means is because we've got root two plus one if we want to rationalize denominator we need to multiply both the numerator and the denominator by root 2 minus 1 root two minus one and if we multiply both the numerator and denominator of this fraction by the conjugate of root two plus one what that means is change the plus sign to a minus sign and if we multiply both the numerator and denominator by that we will rationalize the denominator let's have a look and see what we get so with the numerator we would just get well 3 times root 2 minus 1 would be three bracket root 2 minus one and then on the denominator we would have root 2 plus 1 bracket root 2 minus one close brackets so let's expand our brackets the numerator expanding that will give us 3 root 2 minus three multiplying both of these by three and in terms of the denominator let's expand these brackets and see what we get so if we had root 2 plus 1 bracket root 2 minus one if we expand these brackets we'll get root 2 times root 2 is 2 root two times minus one would be minus root two 1 times root 2 would be plus root 2. and then finally one times minus one will be minus one and when we simplify that we get two minus one is one then we've got minus root two plus root two well that's zero so the answer would just be one so on our denominator we would have one and that is rationalized ones rational and then finally because we can divide both of these by one our answer would just be three root two minus three and that's it so if you do get a fraction where you've got a third and then a number or a number then a third then what you do is you multiply both the numerator and denominator by the conjugate and what that means is the same expression but if it has a plus sign put a minus sign or if this had a minus sign put a plus sign so the next topic is the product rule of frequency now this video 383 in corporate Maps so the product rule for content is a really useful technique to work out how many possible ways are of doing things so for instance if there's n ways of doing a first task and N ways of doing a second task if we multiply those two together we find that the total possible number is a way of doing those two tasks so for if for instance we went into a restaurant and there was eight different starters and there was 10 different main courses and we wanted to find out how many possible ways of choosing a starter and a main course so one start on one main course we want to listen all the possible options if we do eight times ten and that gives us 80 that would tell us the answer to how many possible ways we can pick a starter and we mean of course there would be 80 possible ways we can choose one star and one main course for instance we could have the first order with each of the 10 possible main courses we could have the second stutter with the 10 different main courses the third stutter with the 10 possible main courses and so on up up until the IP starter and the 10 possible main courses and that would mean there would be 80 possible combinations of a starter on the main course so this product rule for content is a really useful technique to work out how many possible ways you can do things so let's have a look at an example and this is the corporate revision card and we've got Grace picks a four digit number so she's got a padlock and she's going to choose a four digit code for it the first digit is a prime number the third digit is greater than seven and we've been asked how many possible different four digit numbers can she choose so let's look at each of the digits so the first digit is a prime number so that could be two three five seven but it can't be 11 because that's got two digits so she's got four possible options for the first digit she can choose a two a three a five or seven so that means she's got four possible options for the first digit the third digit is greater than seven so that could be an eight or a nine we can't have ten because that's the two digits so the possible digits are zero one two three four five six seven eight nine and eight nine are the only two that are greater than seven and because it's greater than we're not going to include seven itself so there's two possible digits we can choose for the third digit in terms of the second digit and the fourth digit we've not been given any rules for those that means there can be any of the ten digits it could be zero one two three four five six seven eight nine so there's ten different choices for digit two and there's ten different choices for digit four so what that means is if we want to find out how many possible four digit codes you can choose we could multiply these numbers together we could do four times ten times two times ten and four four times ten is forty times two is eighty times ten is it hundred so that means is 800 different possible numbers you could choose and we could try and list them all if you wanted to but with there being 800 I'd rather use this technique this is video of 383 in corporate Maps if you want to try some practice questions on it this practice questions beside those as well and also remember in the description below there's a link to that bumper booklet that accompanies this video and so if you want to do a practice question on this you can click that link and try that practice question okay so let's have a look at our next topic and that's angles and polygons and a polygon is just a straight sided shape so we've got our triangles our quadrilaterals pentagons hexagons heptagons octagons and so on and this is the cool master revision card on angles and polygons so it tells you the angles in a triangle add up to 180 degrees the angles on a quadrilateral add up to 360 Degrees the angles in a pentagon add up to 540 degrees and it's very useful to know that or alternatively know that if a quadrilateral adds up to 360 you can just add on another 180. if we add on another 180 that's 720 so the angles on a hexagon add up to 720 degrees the angles in a heptagon add up to 900 Degrees the angle is an octagon add up to 1080 degrees and so on and this is very useful this is the corporate Master revision card and if you want to get a set of those they're in the description below lower so here we've got a polygon and it's got one two three four five angles so and it's also got one two three four five sides that means it's a pentagon and the angles on a pentagon add up to 540 degrees so if we want to find the size of this missing angle X we can add together the angles we're given the 80 the 140 the 95 and the 130 and we can take that away from 540 to see what's left for X so let's do that so 140 plus 80 plus 95 plus 130 is equal to and I'll just do that my calculator 445 degrees and if we take that away from 540 we can see what x is so 540 take away 445 equals 95 degrees so that means X is 95 degrees and that's it if you were given a hexagon they would give you five angles and you would add those up and take it away from 720 to find the size of the sixth one and so on now there's a formula that's really really useful and that if you want to find out the sum of the interior angles or what the angles in the shape add up to you could use this formula which is n take away 2 so the number of sides take away 2 multiplied by 180 so for instance if I wanted to know what the angles added up to an octagon I could take the number of sides which is 8 take away two that gives me six and then multiply by 100 and 80 and that gives me an answer of 1080. likewise if we knew what the angles in a shape added up to and we needed to find out how many sides there were we could work backwards so we could divide by 180 and then add to that's it so this formula is really useful okay let's have a look at a question based on that so here we've got a question and it says find the sum of the interior angles of a 12-sided polygon so a 12-sided polygon that's a dodecagon I believe and we need to find the sum of the interior angles we need to find out what the Angles and that 12-sided shape will add up to and the formula is to do n take away 2 times 180. so if we take our 12 sides and do 12 take away 2 that's equal to 10 and then do 10 multiplied by 180 that's equal to one thousand eight hundred so in a 12-sided polygon the angles would add up to 1 800 Degrees okay our next example here our next question says if some of the interior angles of a polygon is 4500 degrees so it's got a few more sides than a 12 sided one and the question says how many sides does it have so to find this answer at this 4 500 Degrees we've taken the number of sides we've subtracted two and then we've multiplied by 180 degrees so if we want to work backwards we're going to take our 4500 divided by 180 and then that will give us 25 but remember we've taken away two so because we're going backwards we're going to add two so we're going to do 25 plus 2 is equal to 27. so the angles on the 27-sided polygon will add up to 4 500 Degrees and that's it okay and that's it now this formula is really useful it's really useful for both regular polygons so regular polygons are where all the sides are the same length and the angles are the same size and it's also useful for irregular polygons polygons where the angles aren't all the same size so for instance if we knew that it was irregular we could add up the angles that we're given and then take it away and then find out the size of the missing one but if it's a regular polygon what we can do is so for instance with this 27-sided polygon if we divided 4 500 by 27 and we could find the size of each one of those angles now the sum of the rules with polygons and they are the exterior angles so if you've got the interior angle and you carry on that straight line the angle between the polygon and that straight line that's what we call the exterior angle and I've marked them all on this diagram in Red so all of these red angles are exterior angles so the sum of all those exterior angles all of these red angles will be 360 degrees so so for no matter what polygon you've got if you add up all the exterior angles you will always add up to 360 degrees and then also the interior angle and the exterior angle because they make a straight line there will add up to 180 degrees so there are two very useful rules as well the sum of the exterior angles all those outside angles there all those red angles will add up to 360 degrees and the interior angle and the exterior angle allow up to 180. so let's do some questions based on that now so here we've got a typical question it says work out the size of each exterior angle for a 20-sided regular polygon so this is a regular polygons I mean that all the interior angles are the same size and also all the exterior angles are the same size and it's got 20 sides and we know that all the exterior angles will always add up together to give you 360 degrees so if we divide 360 by 20 we will find the size of each exterior angle so if we do 360 divided by 20 that's 18. so that means that each of the exterior angles will be 18 degrees so we'll have 20 exterior angles and they'll all be 18 degrees each this time we're asked to work out the size of each interior angle of a 20-center regular polygon now we've got two different ways to do this we could use our first Formula where we take away two so that's 18 times that by 180 and that will tell you what the angles add up to in a 20 set a polygon and then divide that answer by 20 to find the size of each one of those angles so that's one approach alternatively if we know that all the exterior angles are 18 degrees each if we take that at 18 degrees away from 180 we will find the size of the interior angle because remember the interior angle and the extra angle will always add up together to give us a straight line 180 degrees so if we do 180 take away 18 that's equal to 162 degrees that's it our next topic is angles and parallel lines now whenever you get two parallel lines and a line that crosses them are straight line that crosses them you'll find that many of the angles are the same um because as you can see you've got lots of vertically opposite angles like for instance here this obtuse angle is vertically opposite to this one so they're the zoom this acute one is equal to that one now as is a straight line that crosses the two parallel lines then the angles of below will be exactly the same you'll also have a 75 75 and so on now we've got some special knee mcu's whenever we're talking about these angles being the same as each other so if you've got two parallel lines and a line that crosses it and you've got this Z shape these are called alternate angles because they're alternate to each other so the 75 is the same as a 75 because there are alternate angles and it's very important to know that that word alternate angles try not to use Z angles use the term alternate angles so this angle is the same as this angle because they're alternate to each other here these angles these green angles are 130 degrees these obtuse angles and they're the same as each other so they're called corresponding because you've got if you look at it you've got the angles at the top and the angles at the bottom they're both in the bottom right corner here and here so they're corresponding to each other at the bottom right angle and the bottom right angle are the same because they're corresponding some people call it an F angle because you've got this F shape and the angles at the bottom below the f are the same but try to learn the word corresponding so as well as our alternate and corresponding angles we've also got co-interior and vertically opposite angles so if we go to two parallel lines this angle and this angle will add to 180 degrees and that's really useful because if we know one of them we can then work out the other one by taking it away from 100 nearly so this angle and this angle will add up to be 180 degrees so they are co-interior angles and also this angle and this angle will add together to be 180 degrees so their core interior angles as well that's really useful because if you know one of them you can then find the other one also we then go vertically opposite angles so remember if you've got two straight lines across each other the opposite angles will be equal to each other so if this angle is 135 degrees the angle opposite it will be 135 degrees also if this angle is 45 degrees the angle opposite is equal to 45 degrees so opposite angles where you've got two lines across each other the opposite angles are equal to each other and again that's really useful for core interior angles so if we could look at this diagram here we know that if this angle is 60 degrees this angle be 60 degrees because they're opposite each other and if this angle is 120 degrees this angle is 120 degrees because they're vertically opposite each other and that's it so here we've got a question just to practice we remember which ones were which so we've got our two parallel lines RS and T U and you've got that straight line that crosses them and we've been asked to find which angle is corresponding to a so we've got a here and there's the top left angle in this little section here so if we look at the angles below at the other line the angle on the top left there is e so A and E are corresponding to each other a b would be corresponding to f d would be corresponding to H and C would be corresponding to C our next question says which angle is alternate to C now when we see alternate we're thinking of that Z angle and you can see here we've got that c here and if you look at my pen here you've got this sort of Z ship here where you've got the c in there and the F there they're alternate to each other so C is alternate to F so C is alternate to f d would be alternate to e and they would be the alternate angle C is I'll turn it to F and D is alternate T and that's it bearings and here's a typical bearings question and bearings this video 26 incorporate math so if you don't have full video talking through bearings if you watch that video 26 and called the maps it'll go through bearings in a bit more detail but as I said the end of this video has been about three or four minutes in each topic to make sure you're familiar with the topics so here we've got two times we've got Antrim and we've got Belfast and the question says write down the three figure bearing off Belfast from Antrim so bearings are a direction of travel and they measure clockwise from North so in this question we've been asked to write down the three figure bearing of Belfast from Antrim so if I was after that the first thing I would do is get a ruler and pencil and join up the two times so I'll join up Antrim and Belfast like so then next we'll draw Northline at Antrim so I just get a ruling pencil and I would draw another flying gun upwards from Antrim like so because that's where we're measuring the bearing from it's the bearing off Belfast from Antrim so there's our Northland and I'm just going to put a little n beside it just to show that it's north next we want to Mark in the angle that we want to measure now bearings are measured clockwise from North so the bearing will be this clockwise angle going around from the north line to the line joining Antrim and Belfast and we're going to get up a tractor so we're going to get our projector and we're just going to rotate it so we've got our projector and we're just going to put it on top of Antrim so we've got the crossover on top of Antrim and we've got the zero going along the north flat and we've got zero on the top now the bearing is measured clockwise from North so we just need to measure this angle so we've got zero at the top so we're looking at these outside angles and we're going to start from 0 10 20 30 40 50 60 70 80 90 100 and then we've got 101 two three four five so that is 105 degrees so if I move my protractor a 105 degree angle and the question said write down the three figure bearing off Belfast from antrum so that would be 105 degrees okay let's have a look at another one so we've got two made up times here we've got Castle town and Milton and we've been asked to write down the three figure bearing of Castle town from Milton so again our first step is to join up the two tones so get your ruler and pencil and join up Milton and Castleton like so so we've joined up our two times next we want to just make sure we know where we're measuring the bearing from so we want to measure the bearing off Castle tone from Milton so we're in Milton so let's draw an offline at Milton and we'll just put a little n at the top to show that it's north next let's get our protractor and we want to rotate it so that we've got the zero at the top so we've rotated our protractor and now let's just double check with angle we want to measure so we want to measure the angle going clockwise from North so mark it in the angle it's clockwise from North so that's the angle we want to measure and we're going to get our protractor and we're going to put the zero on the North Line and the cross on Milton so let's now measure our angle so we start at zero so it's 10 20 30 40. and as you can see we haven't reached 45 so we've actually got 44 degrees so that angle is 44 degrees so let's just move our protractor and Mark it into this 44 degrees so 44 degrees and we've been asked to write down the three figure bearing off Castle town from Milton but remember the bearing has to have three figures so it's going to be zero four four degrees we're just going to put a zero in front of that so zero four four degrees and that's it so the bearing of Castleton from Milton is zero for four degrees that's it okay our next question okay this time we've been asked to write down the three figure bearing off-port Rush from alarm so this time we're going to join up the times as before so we've joined them up now we're going to make sure we know which way North is so North is going upwards and we want to just check where we're starting from so it's writing down the free figure bearing off point Rush from Milan so we're in line and we're going to do our North Planet larne so there's our North Planet let's just mark it with a little n now we've got a mark on the angle we want to measure so remember bearings are measured clockwise from North so we want to go clockwise from North all the way around to the line join in Port rush and learn so we need to measure this angle here it's a reflex angle so it's going to be bigger than 180 degrees but less than 360. so we want to measure that Angle now there's three different ways we can measure this angle the first way is my favorite way it's actually getting a 360 degree protractor so this protractor only goes up to 180 degrees so if you've got a 360 degree protractor that actually really really fantastic because you can put the zero on the North Line and you can just go around and read off even reflex angles like this one and if you imagine this one what you could do is to one approach is you could actually turn it the other way so we can measure this small angle here so we could actually get our protractor and measure the angle anti-clockwise so it's starting at zero on the inside 0 10 20 30. and we're in the middle here between 30 and 40 so that's 35 degrees so one approach is actually to measure this angle which is 35 degrees and because we know the angles on a full turn out up to 360. we could take the 35 away from 360 and do 360 subtract 35 which is equal to 325 degrees so that means that this angle is 325 degrees and then that would be our bearing because as the angle measured clockwise from North all the way around to the line join in the tones so our answer would be 325 degrees so that's one approach another approach is actually is the measure that reflex angle and what we could do is so answer is 325 we can just go back to the beginning and we want to measure this reflex angle so what we can actually do is draw a line straight down like so and we know that a straight line the angles add up to 180 degrees so we do know this is 1 180 degrees on the right hand side so just drawn the line straight down our suppose our South line which would be 180 degrees there and then we could just get our protractor and we could turn it around so that we've got zero at the bottom this time and we want to go from zero around to where the line is and as you can see if we start at zero which is at the bottom here we go 0 10 20 30 all the way around to in between 140 and 150 and it's in exactly the middle so that means that's a 145 degrees so if we wanted to measure this clockwise angle this reflex angle we know that we've got 180 degrees and another 145 degrees and if we add those two together we also get 325 degrees so that's it so the bearing measured clockwise from North from Lauren to Port Rush would be 325 degrees okay okay let's have a look at our next topic and that's actually bearings as well it's working out back bearings so in other words here's an example if we know the bearing off Nottingham from Dublin as 0.98 degrees whilst the bearing off Dublin from Nottingham to go in the other way around so there's two different ways you can actually approach this the first is actually just a little sketch so the bearing off nothing I'm from Dublin so let's do a little X for Dublin so let's just label that Dublin and then we've got Nottingham and it's on a bearing of 0.98 degrees so let's put our Northline in and 098 degrees or 98 Degrees if we started at the top we went around 98 Degrees it would be go past East and just a little bit beyond it so it means that it wouldn't be here it would just be a little bit further so let's put a little cross there and say that's Nottingham so we've got Dublin and we've got Nottingham let's join them up with a straight line now let's mark in the angle so we know that angle is 98 degrees because the bearing is 0.98 so that's 98 degrees there now let's also draw another nor flying at Nottingham so we're drawn our North line with the little land at Nottingham and we want to measure the bearing off Dublin from Nottingham so we want to measure the angle going clockwise from the north line at Nottingham going around to the line joining Dublin and Nottingham like so so we want to find the size of that angle now here we've got two Northlands and they are parallel lines and remember with parallel lines the two interior angles are called co-interior these two angles add up to 180 degrees and actually if we go back to our revision criteria we've got cointerior angles two angles inside our parallel lines add up to 180 degrees so if this angle is 60 that angle is 120. that's quite important so if we go back to our bearings we've got this is 98 degrees so this angle and this angle here will add up together to give us 180 degrees so if we do 180 subtract 98 we get that's equal to 82 degrees that means this angle here is 82 degrees next we want to find the size of this reflex angle now we know the angles on a full tone add up to 360 degrees so if we take 82 away from 360 we can find the size of this angle so we can do 360 take away 82 and that's equal to 278 degrees and that's it so that means that this angle will be 278 degrees and that would be the bearing of Dublin from Nottingham and that's it or another way to do it is to actually seen notice a bit of a pattern so if you're doing back bearings question and the bearing that you're given in the question is less than 180 degrees to get the bearing going the other way you could just add on 180 degrees and that would give you your answer and if the bearing you're given in the question was more than 180 degrees if you take away 180 degrees you can get your answer and that's a bit of a shortcut now the way I do that is I visualize it if I was traveling from Dublin to Nottingham and then I wanted to go the other way I would route him through 180 degrees to go the other way if you want to watch that more detailed you can watch video 27A and chord Mavs and there's questions there on it as well okay let's have a look at our next topic and our next topic is constructions and they are video 78 72 and 79 on corporate maps and if you want to watch me do these constructions where I'm filming myself actually doing the constructions those videos will actually show that but here I'm going to show you step by step and how I would do the constructions so first of all our first question says construct the perpendicular bisector of a b so first of all water is a perpendicular bisector so perpendicular perpendicular means 90 degrees and base sector means to cut in half so if we were asked to construct the perpendicular bisector of a b it would be a straight line that cuts a b in half going through the middle at 90 degrees so we'll be cutting this line a b in half at 90 degrees and to do that by using a pair of compasses and a pencil what you do is first of all you would get your pair of composters and you would put the point of the compass on air and you put the pencil on the compass and you would set it over halfway of the line so I've just chosen this point it's definitely over halfway off the line and I've drawn this semi-circle starting at a and I've drawn the semicircle around so keeping the point here I'm making sure I draw a nice semicircle going around now lifting up your compass and your pencil and making sure it doesn't change size put the point of the compass on B and then what we're going to do is we're going to do the same thing so we're going to do another Arc we're going to put a point at the compass on B and we're going to do an overarch I'm making sure it's the same size as the arc we've done at a and then finally what we're going to do is get a ruling pencil draw a nice straight line through those and this line will be our perpendicular bisector so that line will cross our line a bit 90 degrees so it's perpendicular and it's a bisector because it cuts the line a be exactly in half so if that line a b was 10 centimeters you would have five centimeters to the left here and you'd have five centimeters to the right so that's it okay next we're going to construct the angle bisector so here's an angle a b c and we're going to cut this angle in half by using a compass a pencil and a ruler so the first thing we're going to do is we're going to put our point of our compass on B so we're going to put the point of our Compass here and we're going to do an arc on the line a b here and an arc on the line BC here and it's important that you keep the compass the same size for both of those arcs so you put the point here and you just move it so you've got an arc there on the line a b and Arc here on the line BC okay so next we're going to lift up our compass and we're going to put the point of the compass here so we're going to put the point of the compass where the arc and the line a b meet and we're now going to do an arc in this direction so we're going to get our compass and we're going to do an arc in this direction and it'll look something like this so that's where the point of the compass is and we do this Arc looking like that now we're going to lift it up and we're now going to put the point of the compass here and we're going to do another Arc in that direction and I would look something like this and then finally what we're going to do is get a ruler and a pencil and we're going to join up the center of the angle ABC to where those two arcs meet and that will be our angle by sector that line will cut this angle exactly in half so if this was a 60 degree angle you would now have 30 degrees above it and 30 degrees below so that line is called the angle bisector that's a very important construction so we've constructed the perpendicular bisector and we've now constructed the angle bisector okay our next construction is the Constructor line perpendicular to a b so it's going to be align this at a right angles to a b and it passes through the point C so it's going to be a line that passes through C and there's a right angles to a b and we're going to do that again with our Compass our pencil and our ruler so first of all what we're going to do is get our compass and we're going to put the point of the compass on C and we're going to set it so that it's longer than the distance between C and the line so it's a bit further than that so it's not just going to reach the line it's going to be a bit further but it also needs to make sure that it reaches the line on two different places so if I set it too large so if I set it over here it would cross somewhere over here but it wouldn't actually meet the line so I've set it about here so it's a bit longer than the distance to the line and we're going to do two arcs one here to the left and one Arc to the right and again it's very important that you keep your compost the same size the whole way through next what we're going to do is keeping the compass the same size we're going to now lift it up and we're going to put it on the point here okay so putting the compass here what we're doing is an arc below the line and it's going to be below C so it's going to be an arc in this direction and then we're going to lift it up and we're going to repeat it here so we're going to put our point of our Compass here and we're going to do another Arc below and if you get your ruling pencil join up from C to where those talks meet at the bottom here that line will be perpendicular to the line a b and it will obviously pass through C and that's it so we've constructed the perpendicular to the line a b that passes through the point C we're now going to construct a perpendicular to the line a b that passes through the point C which is on the line so in other words we're going to make a perpendicular line using our ruler our compass and our pencil and it's going to be a perpendicular line to a b so it can be at right angles and it's going to pass through the point C so first step is to put the point of the compass on C and we're going to set the compost so that it's getting close to the distance between C and B so I'm going to put it about this distance here and we're going to do an arc on this side and we're going to keep it the same size and do an arc on the other side so it would look something like this so we've put the point of the compass there and we've done an arc on this side and on that side and it's the same distance okay so now what we're going to do is we're actually going to look at this line going from here to here and we're now going to construct the perpendicular bisector of that so using the same steps as before so first step is to put the point of the compass here and set the pencil his pass C and then we're going to do an arc above C and Below C and keeping the compass the same size we're then going to swap over and we're going to put the point of the compass here and again keep it the same size we're going to do an arc above and below and then finally if we join up these two points there and there that line will pass through C at 90 degrees so there's a line that is perpendicular to a b and passing through C and that's it our last construction that we're going to look at is constructing an equilateral triangle and remember an equilateral triangle all the sides of the same length so this is quite a nice one that if you put your point of your compass here and you put your pencil of your compass on the end of the line that'll be five centimeters and then if you do an arc above it would look something like that so all of these points are five centimeters away from here now keeping the composition size and point putting the point of the compass on this side and putting the pencil here now doing the same thing and then we'd end up with something that looks like this and this point here would be five centimeters away from this end of the line and would also be five centimeters away from this end of the line so if we get a return pencil and join them up like so so there and there that will be an equilateral triangle so it'll be a triangle where the sides are five centimeters so that's five centimeters that's five centimeters and that's five centimeters and also the angles will be 60 degrees so that'll be 60 degrees that'll be 60 degrees that'll be 60 degrees and that's it okay so there were our constructions if you do want to watch full videos on them and be showing you how to do each one of them really clearly using a composter ruler and a pencil watch the videos above there okay let's have a look at our next topic so our next topic is lokai and that's video 75 76 and 77 on corporate maps and my question says to draw the locus of all the possible points that are two centimeters away from the line below so in other words we've been given a rule and we've got to draw on all the possible positions the points can be that follow that rule so we want to draw all the points that are two centimeters away from this line so the points above the line move quite straightforward you could be two centimeters just above the end you could be two centimeters above there you could be two centimeters above here and you could be anywhere there's two centimeters above the line like so alternatively it could be two centimeters below the line so it could be two centimeters there two centimeters there and you get your ruler and pencil to make sure all the points are two centimeters below the line so it looks something like this where you've got all the possible points above the line and all the possible points below the line and then what you do is get your ruler and join and go through the points they're all two centimeters above and below now obviously if you're using a ruler and pencil they would all be in a nice straight line so this point here wouldn't be there it would be actually up there okay and so these points above here and below here are all two centimeters away from the line but what about the end of the line so if we then the line here we could have two centimeters to the right so that would be there and then we could be two centimeters here or two centimeters here or two centimeters here and as you can see we're forming this semi-circle where they're all two centimeters away from the end of the line so it looks something like this and with a compass and a pencil you'd put the point of your compass here and your pencil here you draw a nice semi-circle going around like so and that would join up perfectly and excuse my freehand one but that's just a little Sketcher and likewise if we're in the end of this line here we could have two centimeters here and if we measure the points that are all two centimeters away from the end of the line again it would be another semi-circle and all of these points would be two centimeters away from this end of the line so what we'll do is again get the point of our compass and put it there and then put the pencil two centimeters away and then draw a nice semi-circle going around through those points so that's it okay let's have a look at our next question so this time we've got a diagram and we've got the points A and B and we've got a scale that one centimeter is one mile and we've got land and we've got C and the question says a boat is within eight miles of a so it's within eight miles of air and it's within five miles of B shared the possible positions off the boat and one thing I should have said in this question is the boat's at Sea so it's not on land it's at sea so we know that it's within eight miles of air and we know that one centimeter is one mile so what I would do is I would measure eight centimeters and I would set my compass so that is eight centimeters so I put the point of the compass on air and I would put the point my pencil so it's eight centimeters away so I put my pencil up with a c in the land meat and I would draw a circle and this circle has got a radius of eight centimeters and it would look something like this it's a freehand sketch sorry I'm just going to radius of eight centimeters and that's a circle that looks something like that okay next we're now told that the ball is within five miles of beep So within five miles of B and that's five centimeters we get our Compass we'd measure it so the distance between the point and the pencil is five centimeters and we put the point here and it would be over here somewhere so we put the pencil with us see in the land meat and we draw another Circle and this circle's a bit smaller it's only got a radius of five centimeters this time and it would look something like that and again sorry excuse my diagram so this circle has got a radius of five centimeters and this circle has a radius of 8 centimeters and we were that the boats within eight miles of 80 so in other words it's in here somewhere but it's also within this circle as well so that means the possible positions of the boat would be in this region here and that's it if you want more practice on lokai video 75 76 and 77 will go for more examples also there's some fantastic practice questions that you can do on corporate Maps if you go to corporate maps and go down to worksheets and click on it and go down to those video numbers beside that you'll see practice questions and they'll be ideal to practice as well the next topic is views and elevations so views and elevations are whenever you're looking at 3D shapes looking at from different perspective and considering how the shape will look if you look at it from those angles of those perspectives so here we've got a ship it's a load of multi-link Cubes stuck together and I'm going to look at it from different angles we've been asked to draw what the front elevation would look like so if I was standing here if I was small and I was looking at it from here and this was the front of the ship I would just see this these four blocks this one this one this one and this one so I would see this ship here and I could get my pencil and ruler and draw these really tightly excuse this and that's what I would say I would say a rectangle that was one two three four blocks across but that's what I would say I would see that rectangle so again that's the front again I'm going to throw the side elevation now there's two different side elevations here I could draw it from the left hand side or I could draw it from the right hand side and I'm actually going to draw both so let's start off with the left hand side so if I was here at the side and I was looking at it that way I would again see a rectangle on a b one two three four blocks across so it would be four blocks across like so and I would draw like that alternatively I could drop from the other side so I could be standing over here and I could be looking at it from this side here and again I would see one two three four blocks now they're not all level with each other but that's what I would say I would see the block on the left the Block in the middle the next one back and the back one there so I would say again four blocks some people like to put in the lines as well in between the one on the left hand side that won't be a further forward and then I might see that one a bit further back and that one a bit further back and that one a bit further back but again it would just be a rectangle four blocks across okay and finally we've been asked to draw the plan view the plan view is from above it's the bird's eye view as such so we're going to pretend that we're above the ship and we're looking down on it so if we were looking down at it from above we would see our four blocks at the bottom so our four blocks would have gone along the bottom like so one two three four and I'm looking straight down so then on the left hand side it would go up so it would go like one two three four like so if I'm looking down from the top then it will go across and down across and down across and down and across and down so it would look something like this from above the ship so and that will be the plan view the view from above so whenever you're drawing shapes from different perspectives you've got the front elevation which is the view from the front you've got your side elevations which are the views from the sides and you've got your plan view which is the view from above okay let's have a look at our next topic so our next topic is speed distance and time and that's video 299 on corporate Maps so if we consider speed such as 30 miles per hour we would know that in one hour we would travel 30 miles because 30 miles per hour in two hours we would travel 60 miles in 10 hours we would travel 300 miles and so on now if we divide the distance we travel by the time we will always get that speed 30 divided by one it's 30 60 divided by 2 is 30 300 divided by 10 is equal to 30. so that means that our speed is equal to distance divided by time also if we multiply the speed by the time so 30 times 1 we get 30 30 times 10 we give 300 if we multiply the speed and the time together we'll get the distance traveled and finally if we divided the Distance by the speed we get time so it's very important to be able to know these if speed is equal to distance divided by time distance is equal to speed times time and time is equal to the distance divided by speed so let's have a look at our first question so first question says a car drives 171 miles in 4 hours and 45 minutes calculate the average speed in miles per hour of the car so speed is equal to distance divided by time if we divide the distance the 171 by the time we'll find the speed of the car so speed is equal to distance divided by time the distance the car travels is 171 miles so it's going to be 171 and we're going to divide that by the time now the time is 4 hours 45 minutes now whenever we're dealing with speed this is time we've got to be careful we're finding the speed of this car miles per hour so we want to get this as ours now if we think of 45 minutes 45 minutes is three quarters of an hour so we're going to divide by 4.75 because 45 minutes is three quarters of an hour so it's going to be 4.75 and if we do 171 divided by 4.75 let's see what we get 171 divided by 4.75 is equal to 36 so the average speed of the car is 36 miles per hour and that's it so speed is equal to distance divided by time and just be careful that if you have hours and minutes that you write that as just hours so for instance in this case it was 4.75 if it was 15 minutes that'll be a quarter of an hour so be 4.25 if it 4 hours and 30 minutes that'll be 4.5 because 30 minutes is half an hour if you're not sure what the minutes is if you divide the minutes by 60 because it's 60 minutes in an r that will give you the decimal part so if we done 45 divided by 60 that's 0.75 so we know it'll be 4.75 okay let's have a look at our next question okay next question says Josie drives 200 miles at this average speed of 60 miles per hour how long does the journey take so we're trying to find how long it takes us time and time is equal to the distance divided by the speed if we divide the Distance by the speed we'll see how long the journey takes so the distance is 200 miles and the speed is equal to 60 miles per hour so divided by 60. so if we do 200 divided by 60 that will tell us how long the journey takes now this could be a calculator question or a non-circulated question I'm going to do this question as if it was a non-calculator question so I'm going to do 200 divided by 60 and see what we get so how many 60s go into two zero remainder two how many 60s go into twenty zero remainder 20. how many 60s go into 200 that's going to be 3 which is 180 so 3 and then that's going to leave us with 180 the remainder would be 20. how many 60s go into 203 remainder 20. how many 60s go into 200 three remainder 20 and so on so we have got 3.33333 goes on forever so 3.3 reoccurring so 3.3 reoccurring hours so her journey takes 3.3 very recurring hours now in some questions we might want to change this into hours and minutes so let's do that so it's going to take three hours three hours now we've got 0.33333 and so on 0.3 reoccurring is equal to a third so that's going to be a third of an hour and an hour has 60 minutes and a third of 60 is equal to 20 minutes so the journey will take 3 hours 20 minutes and that's it this time we're told an aircraft flies an average speed of 350 kilometers an hour for two hours and 24 minutes how far does the aircraft fly so if we want to find distance distance is equal to speed times time so if we multiply the speed by the time we'll find how far the aircraft travels so the speed well the speed is equal to 350. and the time we're going to multiply by the time last 2 hours and 24 minutes so I want to multiply by two point something and I'm going to need to change my 24 minutes into a decimal number so it's 24 minutes so it's going to be 24 and if we divide that by 60 we'll give that as a decimal 24 divided by 60 is equal to two-fifths or no point four so it's going to be 2.4 hours so if we do 350 multiply by 2.4 that will tell us how far the aircraft travels so 350 multiplied by 2.4 is equal to 840. now make sure we put the right units on kilometers so the aircraft flies 840 kilometers and that's it let's have a look at our next topic so our next topic is called distance time graphs last video 171 on COBRA maths so here we've got a distance time graph and we've got the distance from home so it starts at zero so that means they're at home and it goes up in two so it goes two kilometers four six eight ten and so on and horizontally we've got the time so it starts at 8 AM and we've got 9 A.M 10 a.m and so on and we're given some information we're told Rosie jogged our friend's house so Rosie starts at home and she jogs to her friend's house and after having a short rest Rosie then jogged home arriving at half past 10 and we were told to complete the distance time graph so let's find out where half past 10 would be so if we look at 8 AM and 9 A.M let's find out what half past would be so half past is one two three four blocks going from 8 AM to 9 A.M and after two of them would be half past year so if we look at 10 AM we've got one two blocks so that would be half past 10 there so that is 10 30. and we've been asked to complete the distance time graph while she jogs home so we now need to draw a line Australia line going from here down to half past 10 and it would look something like this okay Part B so Part B says how long did Rosie rest for so as you can see she rested for one block and then when I say block that's a collection of five of the little small squares one two three four five it's where the line is slightly thicker so she rests from one block so as we spotted earlier we knew the two of them was half past so this should be quarter past let's check so we've got eight o'clock eight AM quarter past eight half past eight quarter to nine nine o'clock so each one of the smaller blocks of those collection of five squares is 15 minutes so Rosie rested for 15 minutes okay and part C says when did Rosie jog at the fastest speed so in other words did she jog to her friend's house at a faster speed than whenever she jogged home if it's steeper on a distance time graph it means it's a faster speed and as you can see the line going to her friend's house is slightly steeper than the line coming back this is a steeper line than this one so the jog to her friend's house is faster than the jog home because the line is steeper and I've just written that down going to a friend's house because the line is steeper okay let's have a look at our next topics our next topic our next compound Majors topic is density and that's video 384 in corporate Maps so here's part of the chord manager vision card and density is equal to mass divided by volume so if you want to find the density of something you do it's mass divided by its volume but we can rearrange this we can multiply both sides of this formula by volume so if you multiply both sides of this formula by volume you'll get density times volume is equal to the mass so if you want to find the mass of something you can multiply its density by the volume and finally if you divide both sides of this formula by density you get mass divided by density is equal to volume so if you want to find the volume of a material you can divide its mass by its density and that will give you its volume so let's have a look at some questions now so first of all we've been given a piece of metal has a volume of 50 centimeters cubed and a mass of 900 grams calculate the density of the metal so we want to find the density of the metal so the density is equal to mass divided by volume so we want to find density we do mass divided by volume so the mass of the metal is 900 grams we've got a new 900 divided by the volume of the metal which is 50 and if we do 900 divided by 50 we'll find the density of the metal so 900 divided by 50 is equal to 18. so it's going to be 18. now we're dividing Mass which is grams by the volume which is centimeters cubed so the units for density will be 18 grams per centimeter cubed that means the density of this metal is 18 grams per centimeter cubed okay let's have a look at our next example okay let's have a look at our next question so next question says a glass paper has got a mass of 420 grams and the density of the glass used is 2.5 grams per centimeter cubed find the volume of the paper weight so the volume is equal to mass divided by density so what we want to do is we want to find the volume of the property so we want to do its mass divided by its density so the volume of is equal to the mass which is equal to 420 divided by the density which is 2.5 and when we do 420 divided by 2.5 we will find the volume of this paper width so 420 divided by 2.5 is equal to 168 and our units will be centimeters cubed and that's it okay let's have a look at one more example so this time we've been told Kylie has a solid glass cube so she's got a solid glass cube the length for each side of the cube is three centimeters so the cubes get a sideline for three centimeters the density of the glass uses 2.5 grams per centimeter cubed and it says find the mass of the cube so mass is equal to density multiplied by volume so we want to find the mass of this Cube so we know it's density as density is equal to 2.5 and we want to multiply that by the volume so to find the mass we're going to do 2.5 the density multiplied by the volume but in this question it's not obvious what the volume of the cube is we're told the side length of the cube is three centimeters but we don't know its volume but we know enough information to work it out the volume of a cube is the length multiplied by the way multiplied by the height and because there's a cube all of the side lengths are the same so it's going to be three multiplied by three multiplied by three and three times three is nine times three is twenty-seven so the volume of the cube is 27 centimeters cubed so to find the mass of the cube we're going to do the mass is equal to the density multiplied by the volume so we're going to do the density 2.5 multiplied by 27. so 2.5 multiplied by 27 equals 67.5 grams and that's it so the mass of this Cube will be 67.5 grams okay let's have a look at our next topic so our next topic is pressure and that's video 385 in Code maps so pressure is equal to force divided by area so this is part of the quart Mass revision card and pressure is equal to force divided by area so if we divide the force by the area we'll find the pressure and if we rearrange this if we multiply both sides of this formula by area we would get the pressure times area equals force so that means that the force is equal to the pressure times the area and if we divided both sides of this equation by the pressure we'll find the force divided by pressure is equal to area so it's very important to remember the pressure is equal to force divided by area so here we've got our question and it says a box exerts a force of 4000 newtons on a table so box has been placed on a table and the exerts a force of 4000 newtons on that table and the area of the base of the box is 250 centimeters squared work out the pressure on the table in Newtons per centimeter squared so you want to find the pressure on the table so pressure is equal to force divided by area pressure is equal to force divided by area so if we divide the force which is four thousand Newtons by the area which is 250 centimeters squared that would tell us the pressure so 4000 divided by 250 is equal to 16. that's measured in Newtons because we had 4 000 Newtons and we're divided by 250 centimeters squared so it's going to be nutrients per centimeter squared and that's it so the pressure on the table is 16 Newtons per centimeter squared and that's it so if you want any extra practice on pressure if you go to that ultimate GCSE practice booklet there's a question out of pressure for you okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is population density in this video 384 iron quarter Maps so population density is calculated by taking the population and dividing it by the area so here we've got a question it says the population of a county is 980 000 and there if the county is 3200 square kilometers work out the population density of the county so to work out the population density is equal to the population which is 980 000 and we'll divide that by the area of the county which is 3200 square kilometers and that gives us an answer of 300 and 6.25 people per square kilometer and that's it so the population density of the county is 306.25 people per square kilometer okay let's have a look at another question so this time we're told the area for county is 5 000 square kilometers and the population density of that county is 315 people per square kilometer work out the population of the county now there's two different ways we can consider this the first approach is by looking at the air for the current day which is 5 000 square kilometers and the population density is 315 people per square kilometer so if we take the 5000 and multiply that by 315 that will tell us the population of the county and that's equal to one million five hundred and seventy five thousand and that's it so if we take the area and multiply that by the population density we'll find the population of the county another way to do is to actually look at the formula population densities equal to population divided by area if we want to find the population well we don't want this divide by area so we'd multiply both sides by the area so it would multiply the left-hand side by area and the right hand Inside by area and that would give us the population density times the area is equal to the population so we get the population density multiplied by the area is equal to the population so we would do the population density which is 315 multiplied by the area which is equal to 5000 and that would give us the population which is one million five hundred and seventy five thousand and that's it so this has been population density okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is parts of the circle so here we've got some parts of the circle that you need to know off by harp we've got the radius and the radius is the distance from the center to the edge of the circle we've got the diameter that's the distance across the middle of the circle so through the center you've got the circumference that's the distance around the outside of the circle and you've got the chord the chords align the joins one part of the circle to another part of the circle now let's look at some more parts of the circle so we've got the arc which is part of the circumference we've got a tangent which is a straight line that touches the circle once and Carries On we've got a sector which is part of the circle this green region here or I like to think about it in terms of specific pizza and then we've got a segment and so a segment is if you've got that chord that goes from one part of the circle to the other part of the circle and the segment is one of those regions such as this okay so it's very important to know these parts of the circles as well and if you've got the chord Master vision card on that that's very useful the next topic we're going to look at is finding the circumference of the circles that the circumference is a fancy name for the perimeter of the circle it's the distance around the outside of the circle and the circumference of a circle is found by working out pi times diameter so if you multiply Pi by the diameter the distance across the circle going through the center that will tell you the circumference of a circle so it's very important to know where Pi is on your calculator and as if you look down here on my calculator I've got a yellow Pi there so because it's in yellow that means I need to press the shift button first of all and then press the button here just beneath the pi symbol and that then will bring up Pan the calculator so in this question we've been asked to find the circumference of a circle that has got a diameter the whole air Cross of 14 centimeters so the circumference of the circle is pi times diameter so it's going to be pi times the distance across the circle which is 14. so that's going to be pi times 14 so we're on our calculator we press shift and then the pi button this one here it'll bring up Pi then you'll do times 14 and then press equals and it will tell us the circumference of a circle now sometimes it will come up like this 14 Pi depending on the mode of your calculator so you'd press the SD button there and then it would tell you the circumference of a circle and here we've got the circumference of the circle would be equal to 43.98229 and so on centimeters and I'm going to round this to two decimal places so I'm going to write 43.98 centimeters so the circumference of the circle would be 43.98 centimeters and always remember your units the circumferences that are perimeter is the distance around the outside of the circle so if the diameter is 14 centimeters the circumference will be measured in centimeters also okay and also we might be given a circle like this where we've got the radius is five centimeters so remember the circumference is pi times diameter so the diameter is the whole way across the circle so if the radius is five that means the diameter of this circle will be 10. so we would do circumferences pi times diameter so we'd do pi times 10 and then you would just work out what that is okay our next topic is the perimeter of a semicircle so here we've got a semicircle you can see it's got a radius the distance from the center of the circle if it was a full circle to the edge is three centimeters so the whole way across would be six centimeters and we've got this Arc at the top and we want to find this distance from here around to here and that will be half of the full circle because it's a semicircle so if I treat this as a full circle to begin with and I do pi times diameter I get the circumference of the whole circle and then if we have and you'll get just this distance here just the length of this Arc at the top so we can then do six plus whatever that is so we're going to need to use the circumference formula so the circumference is equal to Pi times diameter so we're going to do PI on our calculator Times by the diameter of the circle and the diameter is 6 here the whole way across so we're going to do pi times six that's equal to 18.8495592 and so on and it's important not to round this because we're going to be doing more calculations with this um so you may just want to keep it on your calculator display so then we can do the next stage so whenever you do pi times six and press equals keep that on your calculator display so that's the circumference of the whole circle now we're dealing with a semicircle which is half of a circle so if we take that 18.849 and so on I'm just going to do some dots so that I don't have to write the whole thing down and if I divide that by two so in my calculator I've already got that on my calculator if I press divided by 2 so it comes up with answer A and S divided by 2 on my display and press equals I get it says 3 pi to begin with and then I press the SD button and that gives me 9.42477961 and so on so that 9 0.424 so 1 centimeters would be the distance from here all the way around the semicircle to here so that distance from here around to here would be 9.42477 and so on centimeters now we want to find the perimeter of the semicircle so that's going to be the length of all the sides so we've got the base and this Arc so we're going to do 9.4247 and so on and just keep it on your calculator display plus and we've got the base now the base of the shape is six centimeters we've got our three and a number three so plus 6 gives us 15.42477 and so on now I'm going to round this to two decimal places so whenever I round this to two decimal places that would give me 15.42 centimeters as perimeter and we're dealing in centimeter so our units were centimeters and that's why I use centimeters and that's it so the next topic is arc length and the formula to find the Arc Length so the length from here round to here is given by Vita the angle over 360. multiplied by pi multiplied by the diameter and this is video 58 in corporate maps and this is part of the core mods revision card and our question asks is to find the perimeter of this sector so here we've got a sector it's got a radius of 8 so the distance from the center to the edge of the circle is eight centimeters and we want to find its perimeter so we're going to do 8 plus 8 plus whatever this arc length is so the arc length is Vita so the angle is 14 degrees divided by 360. multiplied by pi and now we're going to multiply by the diameter of the whole circle so if we did have a whole circle here the diameter would be 16 the whole way across so multiply by 16. and that gives us 1.9547 and so on centimeters so we've been asked to find the perimeter of the sector so we're going to do 8 plus 8 plus 1.9547 and so on and we're going to get our answer so I've already got this on my calculator display so I'm just going to press plus 8 plus 8 and that gives me 17.9547 and so on centimeters or if I round it to two decimal places the answer would be 17.95 centimeters to two decimal places and that's it okay let's have a look at our next topic so the next topic is area of a trapezium and this is part of the chord Mouse revision card I've actually got into little pieces so you've got the area for a trapezium it's given by the formula a half bracket a plus b so A and B are the two parallel sides of the trapezium so we're going to add those two together divide that answer by two and then times it by the height and that will give you the area of trapezium so here's a trapezium so the area would be equal to a half a plus b they're the two parallel lines so that's seven plus eleven and then we'll get that answer and we'll multiply it by the distance between the parallel lines which is five so when we do that we get a half and then 7 plus 11 is equal to 18. so half of 18 multiplied by five a half of 18 is nine so we've got nine times five that's equal to 45 centimeters squared that's it to define the area for trapezium you add together the two parallel sides you divide by two and then you times but I have the height the distance between the two parallel lines and that's it and then if you want more practice on that that's video 48 in corporate Maps okay next topic is to find the area of compound ships so that's whenever ships have been put together now those shapes could be rectangles triangles semi-circles quarter circles sick for circles parallelograms um trapezia trapeziums um it could be loads of different ships and they've just been put together and you need to find the area of them so if you have a look here we've got the ship it looks like an arrow but if we divide it into two we can see that there's two parallelograms so we've got two parallelograms and each of them has a length of 35 centimeters for the base and the height of them were the distance between the two parallel lines as you can see the whole thing is 52. when you divide that by two we get 26 that means the higher this parallelogram is 26 centimeters and the height of this parallelogram is 26 centimeters so all we need to do is work out the area of this parallelogram and there of this parallelogram and then add them together therefore parallelogram is the base times the height so that's 35 multiplied by 26 and when we do that we get 910 centimeters squared now these two parallelograms are identical or congruent so that means that they both have the same area so it's going to be 910 plus 910 and that's equal to 1820 centimeters squared our next topic is the area of a circle and if you want to revise this topic and called Mouse and watch the field videos video 59 so the area of a circle is given by the formula pi r squared I remember our order of operations you square you do any orders before you do any multiplications so what it means is we're going to square the radius and multiply it by pi so here we've got a circle and we want to find the area of this circle it's got a radius of six centimeters the distance from the centers to the edge is six centimeters so the area is equal to pi r squared so it's going to be area is equal to Pi times the radius 6 squared so what we could do is on our calculator we're just going to type this in pi times 6 squared the great thing is your calculator will know the order of operations whenever you type in pi times 6 squared and I'll work it out for you alternatively you could do 6 squared which is 36 and then do pi times 36 and you get the same answer but I'm just going to type this in I'm going to type in pi so shift and then the pi button here and then that'll bring up pi and then Times by so multiply by six and then squared pressing the square button and then I'm going to press equals my answer would be 113.0973355 so the area for the circle would be 113.973355 I'm going to run this to one decimal place so 113.1 centimeters and it's squared because this area that's it now if we were given the circle like this where we knew the diameter being 20 centimeters well it was pi r squared so it's pi times the radius squared so if you know the diameter you might need to half it so in this case the radius would be ten so do pi times 10 squared okay so let's have a look at finding the area of a semicircle so that's video number 47 on corporate Maps so here we've got a semicircle and using a similar approach to finding the perimeter of a semicircle we're going to consider the whole circle so here's a whole circle um yeah there we go there's a filled Circle and you can see the radius of the whole circle is three centimeters so what we're going to do is we're going to find the area of the whole circle and then divide it by two so we're going to do pi times 3 squared because that'll give us the area of the whole circle and when we do that we get 28.27433 so on and then what we're going to do is we're going to divide there the whole circle by two so we'll take our 28.27433 so on and we'll divide that by two and we find that that's equal to 14.1371661 or 14.1 to 1 decimal place okay let's have a look at our next topic so the next topic is the area of a sector which is video 48 and corporate maps and just remember you do have those practice questions in the link below so if you do want to practice any of these questions as we're going through there's questions there for you to try Okay so for the area of a sector here's part of the corporate Master revision card so because the area of the sector is Vita over 360. so the fraction of the circle Times by High Times by the radius squared so here we've got a sector and again it's got an angle which isn't necessarily a nice angle we've got an angle of 109 degrees so we're going to have 109 over 360 times by pi multiplied by the radius of the circle so if we had a whole circle here the radius would be 0.7 centimeters so no 0.7 squared and when we do that we get that's equal to 0.46609 and so on centimeters squared I'm going to round this to three decimal places so I'm going to write 0.466 centimeters squared to three decimal places and that's it okay so let's have a look at our next topic so our next topic is the volume of a prism so to find the volume of a prism we find the cross-sectional area so whenever you've got a prism it's got that constant cross section so it's got that shape that's the same the whole way through it so for instance this triangular prism has a triangle as the cross section and it just sits the same the whole way through so you get the area of that shape and you multiply that by how long it is so to find the volume of this triangular prism we're going to find the area of the triangle at the front and then just multiply it by four so to do that let's work out the area for the triangle so therefore triangle is half the base times the height so there this triangle would be half the base so half times seven multiplied by the height five so whenever we do that we get 7 times 5 is 35 and multiply by half we're dividing by 2 would give us 17.5 meters squared so the area of this triangle at the front is 17.5 meters squared now we just need to multiply by how long the shear puts on the shape is four meters long so if we do 17.5 multiplied by 4 that gives us the volume of this triangular prism so multiply it by 4 equals 70 meters and the units would be cubed because it's the volume so 70 meters cubed so Define the volume of a prism you get the area of the cross section and then multiply by how long it is or alternatively it stands and upright you multiply by how tall it is our next topic is the volume of a cylinder and as you can see this cylinder standing upright so we're going to find the area of the circle and then multiply by how tall it is so the area of a circle area equals pi r squared so that's equal to High Times the radius so this is the radius it so pi times eight squared and when we do pi times 8 squared we get that's equal to 201.0619298 and so on centimeters squared so that's the area for the top now we're going to multiply that by how tall it is don't round this answer because we're multiplying it by 30. so if you round it that multiply by 30 you can you know that you're somewhere out from the actual answer so we're going to take our 201.0619 and so on if you've got a new calculator display just leave it there and press multiply by 30 and then whenever you do that you get times where 30. 6031.857 and so on and to one decimal place that would be 6031.9 and our units would be centimeters cubed and that's it okay so the next topic is looking at given our answers in terms of Pi now I've previously looked at topics such as circumference and over Circle volume of a cylinder and so on and in all of those questions so far we've been given our answers as decimals now I've actually hinted in previous videos that we could actually give our answer in terms of Pi and that's working exactly so rather than giving a decimal we're giving it an exact answer in terms of Pi so let's have a look at a typical question so we've got a cylinders get a diameter of eight centimeters so we've got a cylinder surfboard magic we've got a cylinder and let's get a diameter of eight centimeters so that means that the width of it is eight centimeters so the whole way across the circle at the top or the bottom is eight centimeters and the height of the cylinder is 15 centimeters so the height is 15 centimeters so we've now got our sketch now what we need to do is work out the volume of the cylinder into terms of Pi and what that means is rather than given our answer is the decimal number we're going to give it as perhaps something like 100 Pi or 75 pi and that means we've got an exact answer so rather than actually having to work it out and then rounding it we can actually just give it in terms of pack so the volume of a cylinder so we will find the area of the cross section so in this case a circle and then multiply by how tall the shape is so let's get the area of the circle so the circle let's get a diameter of eight so the radius of the circle is four centimeters and the area of a circle is pi r squared so pi times 4 squared now 4 squared is 16 so that means we would have pi times 16 and then if we just put them together like an algebra where would maybe write 16y or 16x we're just going to write that SQL to 16 Pi we're just going to put the number in front of Pi so it means the area of the circle is 16 Pi centimeters squared so that's the area of the circle and rather than rounding that as a decimal number and then multiplying it by 15 and so on what we're just going to do is use that 16 pack so to find the volume of a cylinder we'll find the area of the circle now we need to times about how tall or how long the shape is so the height of the cylinder is 15 centimeters so to get the volume we're going to do 16 Pi there if the circle Times by how tall the shape is which is 15. so what we're going to do is we're going to do 16 times 15 and then get our answer and just put Pi after it so 16 times 15 is equal to 240 and then Pi so answer would be 240 Pi centimeters cubed and that's it so we've left our answer in terms of Pi and our answer would be 240 Pi centimeters cubed so we're not actually having to work out with 240 times pi is and then write that as a decimal number we can actually just give it in terms of power and this is particularly useful on non-calculated papers our next topic now is to find the volume of a cone the volume of a cone is one third pi r squared H that means we're going to find a third of Pi times the radius squared multiplied by the height of the cone so in this cone we've got a radius of 20 and we've got a height of 21. so we're going to do one third multiplied by pi multiplied by the radius which is 20 squared and then multiply it by 21 the height of the cone and when we do that we get an answer of 2800 Pi which would be 8796.46 centimeters cubed to two decimal places and that's video 359 according math so the volume of a cone is one third pi r squared hitch here's a pyramid and if you've been given a pyramid and you've asked to find the volume of it again the volume is equal to third the area the base times the height so it's going to be the volume is equal to Third times the area of the base well it's going to be nine times six so nine times six is equal to 54. so if it times 54 times its perpendicular height so that's going to be from the base straight up to the top which is seven centimeters and whenever we work that out we get that's equal to 126 centimeters cubed our next topic is the volume of a sphere so the volume of a sphere four thirds pi R cubed and this is part of the chord modular vision card here and of course Mazda's video 361. and here we've got our sphere and it's got a radius of 14 centimeters so to find its volume we will do four thirds multiply by pi multiply by the radius which is 14 cubed so I'm going to press the fraction button on my calculator type in four thirds I'm then going to press multiply by and then pi times by 14 and then the cubed button eleven thousand four hundred and Ninety Four 0.04 centimeters cubed to two decimal places and that's it okay next we've been asked to find the volume of a frustum our first term sounds complicated but it's just a a cone or a pyramid with the top chopped off so here we've got this is video 360 and covered Mavs and it says shown as a frustrum of a cone so as you can see the Top's chopped off that had a perpendicular height of 20 centimeters so the it did go up to a top point just say there and down again and I did have a height of 20 centimeters in total now the height of this part is 10 so I would have had another 10 centimeters to have a total height of 20 centimeters and we've been asked to find the volume off the thrust room so that's the bit that's left this part here so to do that what we're going to do is we're going to work out the volume of the cone to begin with so the whole big cone we'll then work out the volume of the cone that's been chopped off and then take them away and see what's left so let's do the volume of the big cone so the volume is equal to third times pi times the radius so for the big cone the radius is eight so eight squared times its height and we've been told its height was 20 centimeters okay whenever we work that out we can either get 1 280 over 3 Pi alternatively you can write as a decimal number as 1 340.412866 and so on centimeters cubed that's the volume of the whole big cone to begin with now the top was chopped off and that is a smaller cone and it's similar to the large cone you'll learn about its similarity later which means enlargement but so we'll just work out the volume of this uh so working at the volume of this smaller cone and it's got a volume of a third times pi times the radius of four squared times its height 10. and whenever we work that out we get and that's equal to 160 over 3 Pi or alternatively 167.5516082 centimeters cubed and then finally to work out the volume of the frustrum well we're going to take that away so we're going to do 1 340.412866 take away the 167.5516082 and when we subtract them we get 1272.86 centimeters cubed and I've rounded that to two decimal places so if you asked to find the volume of a frustrum you just work out the volume of whatever shape there was to begin with whether it's a big cone or a big pyramid you work out the volume of what's being chopped off and then you take them away and that'll tell you the volume of what's left this time we've been asked to find the surface area and here's a cuboid and we've been asked to find the surface area of our cuboid and the surface area is the area of all the faces of that particular ship so this is a cuboid so I'll have six faces you've got the top which is green you've then got the bottom which would be the same size it would have the same area you've then got the rectangle on the right hand side here and blue and that would be the same as the rectangle on the left hand side and you've got the rectangle at the front the fifth the front this red one and that'll be the same as the back so what we've got is we've got six rectangles we need to find the area often we need to add them all up now the great thing is that you've got pairs of them that are the same the top and the bottom the right the left and the front the back are the same so if we work out the area of each of these ones so what we can do is then we consider two of them and then add them all up let's call the front one the right hand side two and the top three let's consider the front to begin with so for first one we've got the width of the rectangle is it multiplied by the height of the rectangle which would be five because here we've got a height of five so that means that the height here of this rectangle is five so we've got eight times five and eight times five is equal to 40 centimeters squared so the area of the front is 14. that means the area of the back would also be 40 as well so I'm just going to write it there as well and then 2 we've got the right hand side here if we look at this rectangle it's got a length of seven and a height of five so we're going to do 7 multiply by five and seven times five is 35 so that's 35 centimeters squared so that means the area of this rectangle on the right hand side is 35 and that would be the same as the rectangle on the left hand side so there's another 35 centimeters squared and then the top which is phase three we've got three that has a width of it and a length of seven because you've got it here so that's going to be it there and the 7 here will be there so we've got eight times seven and here times seven is 56 centimeters squared and the top would be the same as the bottom so we've got another 56 centimeters squared so to find the total surface area of this cuboid we're going to do 40 plus 40 plus 35 Plus 35 plus 56 plus 56 and that would tell us the total surface area of this cuboid and when we do that we get the total surface area of 262 centimeters squared so the surface area is in the area of all the faces of that 3D ship okay let's have a look at our next topic so our next topic is the surface area of a cylinder and that's video 315 in corporate Maps now here we've got a cylinder and we've been asked to find the surface area of the cylinder now this cylinder has three faces it's got a circle on the top a circle on the bottom and it's got this curved face going around the outside now there if the circles will be straightforward we'll do pi r squared for the top and then we'll do power squared for the bottom and they'll also be the same area so you could just do power squared for the top and multiply it by two or add up to itself now for this code fee it's going to run the outside if we were to straighten that out that would be a rectangle and the height of the rectangle would be six meters and the length of the rectangle that would be the circumference of the circle so we're going to get the circumference of the circle and multiply it by 6 the head of the cylinder and that would give us the area of the curved face and then we'll add together the two circles and the curved face so let's do that now so let's start off by getting the area of the circle so that would be pi times 5 squared and that will give us the area of the circle which is 78.53981 and so on then we've got the circle at the bottom and that'll have the same area so we'll have a circle beneath and that'll have the same area and then finally we need to find the area of this curved face going around the outside of the cylinder and to find that we're going to find the circumference of the circle so we'll do pi times diameter so we'll do pi times 10 because the diameter of the circle will be 10. so pi times 10 will be 31.4159 and so on and then we'll take the circumference of the circle and multiply it by the height of the cylinder and that will tell us the area of the curved phase so we're going to do 31.4159 and so on and I've got that in my calculator display so I'll just press times six and that will give me 188.4955 and so on so I can add together now the area of the circle on the top the 78.53981 and so on I'll add another Circle to 78.53981 and so on and I'll also add the area of this face the curve face which was 188.4955 and so on and when I do that I get that's equal to 345.575 to three decimal places and then the units because of surface area and its meters will be meters squared and that's it okay let's have a look at our next topic so the next topic is a surface area of a cone so here we've got a cone and to find the surface area of a cone we have to get the area of the two faces and that's the area of the circle at the bottom which will be pi r squared so that's straightforward and then we need to find the curved surface area so the area for the top of the cone and that's given by the formula curved surface area is pi RL where Pi is obviously pi r is the radius of the base of the cone so that will be six and then L is what we call the slant height so this diagonal length here of 11 centimeters so let's find the surface area of this cone so for this cone we would do pi r squared so pi times 6 squared for the base and when we do pi times 6 squared we get that's equal to 113.0973355 and so on and then this area of the base now we need to find the curved surface area so it's going to be pi times the radius 6 times the slant height which is 11. and when we do that we get that's equal to 66 Pi or 207.34 five one one five one and so on and when we have those together we'll find the surface area of this cone which would be 320.44 centimeters squared to two decimal places and that's it okay our next topic our next topic is the surface area of a sphere and so the surface area is 4 pi r squared so here we've got a sphere and we're going to find a surface area so we're going to do 4 multiplied by pi multiply by the radius which is 5 squared so we're now just going to work this out in our calculator so we're going to do 4 multiplied by pi multiply by 5 squared and that gives us 100 Pi which is equal to 314.16 centimeters squared to two decimal places and that's it so our next topic is convert metric units for area and volume so let's start off with converting metric units for area so here we've got a rectangle and it measures three meters by two meters so three meters long and two meters wide and here's the exact same rectangle and instead of writing three meters I've written 300 centimeters because 100 centimeters in a meter so three meters would be 300 centimeters and instead of writing 2 meters I've written 200 centimeters because obviously 2 times 100 would be 200 centimeters now let's find the areas of these rectangles because they're identical the areas will be the same so to find the area of this rectangle we multiply the length from the width together so we're going to do three times two and three times two is equal to 6 meters squared so the area of this rectangle is six meters squared now if we look at this rectangle this rectangle is 300 centimeters long and 200 centimeters wide so to find this area we're going to multiply these together so 300 times 200 about 3 times 2 is 6 and then add one two three four zeros one two three four zeros and then that's measured in centimeters squared so we've got the same rectangle and its area will be six meters squared or sixty thousand centimeters squared so that means that six meters squared is equal to sixty thousand centimeters squared or if we divide both of these by six we get one meter squared is equal to ten thousand centimeters squared so in one square meter there'd be ten thousand square centimeters and that makes sense because if you get a square which was one meter by one meter that would be 100 centimeters by 100 meters that'll be 100 rows of 100 which would be ten thousands little smaller centimeter squares in there that's it so if we want to convert between meter squared and centimeters squared we can multiply by ten thousand or I like to multiply by 100 I multiply by 100 again because it's squared so I know to multiply by 100 twice and that's it okay let's have a look at our next one converting metric units for volume so here we've got two cubes one of them's got a side length of two meters and one's got a side length of 200 centimeters so these cubes are identical so if we know the side length was two meters we can write two meters for the length two meters for the width and two meters for the height so to get the volume of this Cube we're going to do the length times the width times the height so we're going to do two times two times two which is equal to eight meters cubed so the volume of this cube is eight meters cubed so this Cube's identical so it will have a length of 200 centimeters a width of 200 centimeters and a height of 200 centimeters so if we do 200 multiplied by 200 multiply by 200 that will give us the volume of this Cube and the volume of this cube is equal to 2 times 2 times 2 is equal to 8 and then we're going to add on six zeros one two three four five six so eight million centimeters cubed so eight meters cubed would be equal to 8 million centimeters cubed and if we divide both of these by eight we get one meter cubed is equal to one million centimeters cubed and that's it so that's going to be really useful for converting between meters cubed and centimeters so you can multiply by a million to convert between meters cubed and centimeters cubed and likewise to convert back from centimeters cubed to meters cubed so you can divide by a million I like to multiply by 100 and by 100 and by 100 again and divide by 100 100 and 100 again if I want to convert between them and that will give me the same answer so we're now going to look at Imperial units so that's video is 347 and 348 in corporate math so we have our metric units which are centimeters meters grams kilograms milliliters liters and so on and the modern units we've then got our Imperial or older fashioned units um hover not cause any effects anyone there but you've got an older fashioned your imperial units such as Pines ounces inches feet miles and so on and we're going to need to be able to convert between them and here's a typical question where we're told the length of a table is 35 inches so we've been given it in inches which is going to be in Imperial and we're told that one inch is 2.54 centimeters so every inch is 2.54 centimeters and the question says how long is the table in centimeters so we want to convert this 35 inches into centimeters and we're told that every single inch is 2.54 centimeters so in other words we've got 35 lots of 2.54 centimeters so if we do 35 multiply by 2.54 that will tell us then how many centimeters long the table is so we'll do 35 multiplied by 2.54 and when we do that we get that's equal to 88.9 centimeters and that's it and this is a calculator question so you can just type in 35 multiplied by 2.54 and then just make sure we're putting centimeters on the end okay so our next topic is Pythagoras and here's part of the chord miles revision card on pythagorasis firm which is a squared plus b squared equals c squared where n b are the two shorter sides and C is the length for the longest side of the hypotenuse of the right angle triangle so here's a right angle triangle and we've been asked to find X the length of the hypotenuse the longest side so I label the sides first of all so I label shorter side a the next shortest B and the longer side C now in terms of A and B it doesn't actually matter which way around your label those as long as a and b are the two shorter sides and we're going to substitute those values into Pythagoras and sphere so Pythagoras firm is a squared plus b squared equals c squared so if we substitute these values in well instead of a squared we're going to write 5 squared instead of B squared we're going to write 12 squared and instead of c squared we're going to write x squared so we know that 5 squared plus 12 squared equals x squared so now we're going to work out 5 squared and 12 squared so 5 times 5 is 25 and remember this is a calculator paper so you can't just write 5 squared plus 12 squared and press equals on your calculator plus and 12 squared is 144 and that equals x squared now 25 plus 144 is 169 so that gives us 169 equals x squared so we've got 169 equals x squared now obviously X is the length of the side here so we can square root both sides so it gives us the square root of 169 equals x and the square root of 1069 is 13. so 13 equals x that means x equals 13 centimeters and that's it so that means that x equals 13 centimeters okay let's have a look at a right angle triangle where we're trying to find the length of one of the shorter sides so here's a right angle triangle and again Pythagoras is theorem is a squared plus b squared equals c squared so let's label our sides now we don't actually know which ones the shortest here so I'm just going to label the side the X a I'm going to label the five centimeters B and obviously the longer side will be the hypotenuse the side opposite the right angle so that will be C so we write down Pythagoras is 5 and that's a squared plus b squared equals c squared and so a squared about is going to be x squared plus and instead of B squared that will be 5 squared and that equals c squared which is 8 squared so x squared that's x squared plus I'm working out 5 squared well 5 times 5 is 25 so that's 25 equals 8 squared which is 64. now here we've got x squared plus 25 equals 64. so we want to get the x squared on its own so let's take away 25 from both sides so that gives us x squared equals and 64 take away 25 equals 39 so we've got x squared and to find X we're going to square root 39 so we'll work out the square root of 39 and that equals 6.244997998 and so on centimeters so let's round this to two decimal places that means that x equals 6.24 centimeters and that's it so Pythagoras Theorem is really useful for right angle triangles to find the length of missing sides if you know two of them and perfectionism is also useful in situations where you've got shapes that are made up of right angle triangles so for instance if there's a rectangle and it's cut across diagonally you may need to use Pythagoras's Theorem to work out the lengths of the sets there and so on okay let's have a look at our next topic so with Pythagoras we're dealing with right angle triangles and we're using the length of two sides to find the length for the third side with trigonometry we're going to be dealing with right angle triangles again but this time we're going to be involving angles we're either going to be using two sides to work out the size of one of the angles alternatively we would be using an angle and one of the sides to work at the length of another side and it's very important whenever you're dealing with trigonometry to know the trigonometric ratios or the trig ratios and they are sine is equal to opposite divided by hypotenuse the COS is equal to the adjacent divided by hypotenuse and that the tan is equal to the opposite divided by adjacent and some students remember sand such as sock or I like to remember two old Angels skipped over Heaven carrying a harp so it's very important you remember these trig ratios whenever you're answering a trigonometry question so let's have a look at one now so here we've been given a right angle triangle we've been asked to find the size of this angle X and it's a right angle triangle we know that because it's been marked on and we've been given the length of the hypotenuse is 10 centimeters and we've been given the length of this side as eight centimeters and we've been asked to find the size of this angle so the first thing I do on a trigonometry question is to write down the trig ratios and they are two old Angels skipped over Heaven carrying a harp so two old Angels skipped over Heaven carrying a harp or Soca tour and they are the tan is equal to the opposite divided by adjacent sine is equal to the opposite divided by hypotenuse and that the COS is equal to the Json divided by hypotenuse now let's have a look at our triangle and figure out which trig ratio we're going to use in this question and I do that by labeling the sides as the opposite hypotenuse and adjacent so let's start off with the opposite so the opposite is the side opposite the angle we're using or trying to find in the question so we're trying to find this angle so the side opposite that is this one so this is the opposite then let's label the hypotenuse that is the longer side of a right angle triangle and it's the side opposite the right angle so this is the hypotenuse and that means the side that we've got left is the adjacent and it's adjacent to the angle we're trying to find but it's not the hypotenuse of that surgery isn't so that is our opposite hypotenuse and adjacent so we've labeled the three sides of the right angle triangle so now we've labeled the triangle as the opposite hypotenuse and adjacent now we need to figure out which trigger issue we're going to use in the question so here we've been given the hypotenuse is 10 centimeters we've been given the adjacent is 8 centimeters so we're going to use those two and the opposite will not be given and we're not trying to find out so we're just going to cross it off we're not going to be using any trigger issue this involving the opposite so we're not going to be using the tan and we're not going to be using the sign we're going to be using the cause and this question the cos x is equal to the adjacent divided by the hypotenuse that's the trig ratio we're going to use in this question so let's substitute in the values for the adjacent and the hypotenuse into our trig ratio so we've got the cos x is equal to the adjacent which is 8 so 8 divided by the hypotenuse which is 10. so we've got the COS of the angle X is equal to eight tenths so we know that the COS of the angle is equal to 8 over 10 or no point in but this angle obviously isn't 0.8 degrees this angle is much bigger so what we want to do is we want to find out what x is so we want to get rid of this cos so we're going to do the inverse cos so the opposite of cos to both sides of the equation so whenever we do the inverse cos to both sides on the left hand side we'll just be given X the angle we're trying to find and then we want to do the inverse COS of eight temps now on your calculator just above the cause in yellow you've got a COS with a little minus one that's the inverse cos so we would then press shift and then the cause button and it will come up as cos with the little minus one and then you just type in 8 over 10 or 0.8 and then close brackets and press equals and that will give you the size of the angle and we find that X is equal to 36.86 nine eight nine and so on degrees and let's just run that to two decimal places that'll be the x is equal to 36.87 degrees and that's it so in this question we're asked to find the size of this angle X so my steps were first of all I wrote down the trig ratios then what I've done was I looked at my right angle triangle and I labeled the side so I labeled the side opposite the angle as the opposite the hypotenuse as the hypotenuse H and the adjacent as a then we knew we weren't dealing with the opposite in this equation so we crossed off any of the trig ratios involving the opposite so we know it was a cause question we know that the COS of X the COS of this angle is equal to the adjacent divided by the hypotenuse so we found that cos x was equal to e attempts this angle obviously isn't the attempt so we need to do the inverse because of that so we press shift cos and then type in attempts and press equals and we get the size of this angle which is 36.87 degrees okay let's have a look at our next question so this time we're going to find the lengths of one of the sides so we've got this right angle triangle and we know that the hypotenuse of this right angle triangle is four kilometers and the size of this angle is 60 degrees we've been asked to find the size of X so again let's write down our trig ratio so we've got two old Angels skipped over Heaven carrying a harp so here we've got a right angle triangle let's label the sides so we have got the opposite so in the question we've been given the 60 degrees so the side opposite it is X so that's our opposite then we've got our hypothesis which is the side opposite the right angle so our four kilometers is the hypotenuse and the third left is digits and so this side is our adjacent next we cross off any side that has not been given or looked for in the question so in this question we're looking to find the opposite we've been given the hypotenuse and we don't need the adjacent or we're given the adjacent so we cross it off so we then look at our trig ratios and we cross off any trig ratio that involves the adjustments we're not using tan and we're not using the cause so in this question it's a sign question that sine X is equal to the opposite over a hypotenuse so let's substitute in our values so we've got the sine of our angle so that's going to be sine 60. so sine 60 is equal to the opposite which is equal to X and this question divided by the hypotenuse which is 4. so we've got sine 60 is equal to x divided by 4. so we want to find out what x is so we don't want this divided by 4 and this right hand side of this equation so let's multiply both sides of the equation by four so we'll get sine 60 times 4 so that's going to be sine 60. times 4 and that's going to be equal to and on the right hand side of the equation we had x divided by 4 we Times by 4 to get rid of the divided by 4 so we're just going to be left with X so we've got the x is equal to sine 60 times 4. so let's work that out so on our calculator press sine 60. now make sure you close brackets so close brackets multiplied by 4 is equal to 3.4641 and so on kilometers and that's equal to X so X is equal to 3.464 kilometers to three decimal places and that's it so X is equal to 3.464 kilometers to three decimal places okay let's have a look at our next topic so our next topic is exact trig values and it can be useful to remember some of your exact trig values so in other words if you were to work out the sine of zero degrees that's equal to zero the sine of 30 degrees that's equal to half that's one I remember off by heart the sine of 30 degrees is equal to half the sine of 45 degrees that's equal to root 2 the square root of 2 divided by two the sine of 60 degrees is equal to the square root of 3 divided by 2 and the sine of 90 degrees is equal to one so it can be useful to remember those the COS of zero degrees is equal to one the COS of 30 degrees is equal to the square root of three divided by two so the COS of 30 degrees is equal to the sine of 60 degrees they're the same as each other because of 45 degrees is equal to the square root of 2 divided by two the same is the sine of 45 degrees and the COS of 60 degrees is equal to a half and I always remember that one as well always remember the sine of 30 degrees is equal to half and the cause of degrees is equal to half and the cause of 90 degrees is equal to zero and finally tan the tile of zero is equal to zero the tan of 30 degrees is equal to the square root of three divided by three the town of 45 degrees is equal to one and the tan of 60 degrees is equal to the square root of three and the tan of 90 degrees is on the fan there's no answer for that okay so we want to write down the values of these so the sine of 30 degrees would be equal to a half the ton of 45 degrees would be equal to one and the COS of 30 degrees well the COS of 30 degrees is equal to the square root of 3 divided by 2. and that's it the next topic now is to have a look at 3D Pythagoras so this is whenever we're dealing with Pythagoras in three dimensions so here we've got a cuboid and we've been asked to find the length of AEG so the length of a g is from a going up diagonally all the way up to G so let's find the length for this line a g this diagonal going from one corner up to the opposite corner so if we want to find the leg for this line well let's have a look and see what we've been given so we've been given the length of the cuboid is six we've been given the width is two and the height is three so let's start off by looking at the base of this cuboid so the base of the cuboid is a rectangle we've got six and we've got two so if we turn this base into a right angle triangle we can find the length AC and that's going to be very important because we know this is a right angle triangle so if we know what this is and we've got the three we can then find the length of AG so let's find the length of a c to begin with so a c squared this is a right angle triangle here it's a cuboid so so we've got SC squared is equal to 2 squared plus 6 squared so 2 squared plus 6 squared would be equal to AC Squared using Pythagoras so that would mean that a c squared is equal to 2 squared is 4 and 6 squared is 36. so we've got a c squared would be equal to 40. so that means the AC will be equal to the square root of 40. and these questions I tend not to want to square them I like to work with surge because you're going to see in a minute we're going to square this again so that makes it easier if we don't actually use decimal numbers and we use our exact answer so we find the distance from a to c is equal to the square root of 14. so we've got this distance now if we look at this triangle a c g this is the right angle triangle also where this is our right angle there so we know that a g is going to be the hypotenuse it's the longest side and our AC and our c g are our two shorter sides so that means that a g squared a g squared is equal to a c squared plus c g squared so substitute in our values we've got a g squared is equal to a c squared so that's AC which is the square root of 40. squared plus c g squared will be 3 squared now let's work this out so we've got a g squared will be equal to now if you've got the square root of 40 squared that means the square root of 40 times the square root of 40. now considering thirds that will be equal to 40 because root a times root a is a so Roots 40 squared is 40 plus and then 3 squared is 9. so we've got that's equal to 9. so we've got a g squared is equal to 49. well this is grip now we just want to find a g so we want to square root so we've got the square root of 49 which is equal to 7. so that means that from a to G will be equal to seven centimeters and that's it okay let's have a look at our next topic so next topic is 3D trigonometry so as well as the Pythagoras in three dimensions we also know I need to know for me and how to do trigonometry in three dimensions so here we've got a cuboid and we've been asked to find the size of angle s or Ace so we've got a c e so that's going to be the angle if we go from a to c there and then we're going to go from C up to e so that's going to be this way the angle we want to find is this angle in here let's call this angle X so when I find this angle in here now if we want to find this angle we're going to need to know some of the lengths of the sides of this right angle triangle so you can see AC is a right angle triangle and we need to know the lengths of some of the sides to be able to find the size of that angle two sides will be useful so if you notice CG is equal to three the height of the cuboid is three centimeters sum is AE will also be equal to three centimeters that means no we need to even find the length of EC or AC and then and that means we can use trigonometry to find the size of this angle let's find the length of AC so here we've got on the base we've got our ABC and this is a right angle triangle as well where we've got our two shorter sides of six centimeters and two centimeters and this length this diagonal will be the hypotenuse so it'll be the longer side in this right angle triangle so let's write that down a c squared equals and we've got a B squared plus BC squared so it's going to be equals 6 squared plus 2 squared now 6 squared is 36 and 2 squared is 4. so we've got a c squared is equal to 36 plus 4. so that means that a c squared will be equal to 40. and if we square root we get AC is equal to the square root of 40. so we've got the length of this line going diagonally from a across the C will be equal to the square root of 40. now we've got this triangle ace and I tend to like to draw a sketch of it so let's draw a sketch of that triangle and let's label we'll both go up so we know it's a right angle triangle we know the height from a to e is three so we've got that's equal to three centimeters so a e and then C is down there we're trying to find this angle X down there we've got that's equal to three and we'll get a to c being the square root of 40. so that is our triangle we've got a a c a e c we've got three centimeters the square root of 40 and we've got our X now if we have a look at this triangle we're now going to use basic trigonometry to find the size of this angle X so we've got R3 that's our opposite so we're using our two old Angels skipped over Heaven carrying a harp so we've got our opposite we've got our hypotenuse here and we've got the side down here this is our adjacent now we're not actually using that hypotenuse so we can cross that out and we can cross out any cars the hitch in it for hypotenuse so to find the size of this angle here this x we're going to use tan we're using two old Angels So Tan opposite adjacent we want to find the tan of the angle So Tan x is equal to the opposite divided by the adjacent so the opposite is three and the adjacent is the square root of 40. so that's great we now know that the tan of this angle is 3 divided by the square root of 40. so if we do the inverse time we will find the size of the angle so the inverse tan of three over the square root of 40. and we can type that in on our calculator so shift tan the fraction button three and then on the denominator of the square root of 48 and if we press equals we get that's equal to 25.377 degrees to three decimal places that's it so we have used trigonometry in three dimensions to find the size that missing angle our next topic is a sine rule so the sine rules are really useful whenever you want to find out mess the angles or missing sides on non-right angle triangles because with right angle triangles we can use basic trigonometry so the skipper tool then just skipped over Heaven carrying a harp whenever you've got triangles that aren't right angles the sine rule is really useful and here's part of the corporate Master revision card so if you've got a triangle such as this where you've got capital a capital B and capital c b in the Angles and then you've got the opposite sides being a little layer little B and little C the sound rule is a over sine a would be equal to B over sine B which is equal to C over sine C so the sine rule is really useful whenever you know a side and the opposite angle such as this triangle here we've got a side nine and we've got the opposite angle 85 degrees so that's going to be really useful in this question so let's use the sound rule to work out the size of X this inside so the sun rule is a over sine a equals B over sine B so what that means is we've got X over sine 45 so X over sine 45 because they're opposite each other will be equal to 9 over sine 85 so 9 over sine 85. so we've got our x divided by sine 45 and we've got that's equal to 9 divided by sine 85. so if I do 9 divided by 785 in my calculator that gives me X over sine 45 will be equal to 9.0343 and so on and I keep that on my calculator display I'm just going to write it on the page like this now I want to find what x is so I don't want this sine 45 on the denominator so if I want to get X on its own I want to get rid of the divide by sine 45 so I'm going to multiply both sides by sine 45. so multiplying the left hand side by sine 45 we'll just leave me with X on the right hand side I'm going to have my 9.0343 and so on and it's still in my calculator display then I'm going to type in times sine 45. so if I just press times sine 45 my calculator and press equals I get X is equal to 6.388 and so on or I could round this to two decimal places and I could round it to x equals 6.39 to two decimal places and I should put in my units which is centimeters and that's it so the sound rule it's really useful particularly if you've got a side and the opposite angle and that is the a over sine a will be equal to B over sine B and you can use it to find missing sides like this example or we can use it to find Miss angles like our next example so in this case we've got our side and our opposite angle we've got our 35 centimeters and our 80 degrees and we've got our 16 centimeters and our Vita and the sign rule is a over sine a equals B over sine b equals c over sine C now one thing you can do is whenever you didn't miss an angle questions rather than using a percent a you can actually flip it over and write sine a over it little a will be equal to sine B over a little B and that's equal to sine C over little C so if you are finding angles it can be useful to actually flip over the four to write it the other way around so sine a over little a equals sign B over little b equals sine C over little C now if we have a look at our triangle we do have an angle on the opposite side and we're trying to find another angle and we've got its opposite side so we can write that down we could write sine Theta what we're looking for Vita sine Theta over 16 will equal sine 80 over 35 sine 80 over 35. now we can do sine 80 divided by 35 on our calculator so that would give us 0.02813 and so on and the calculator display carries on and I just leave that in my calculator and that gives us a sine beta divided by 16 is equal to 0.02813 and so on now I don't want this divided by 16. I just want the video on its own so let's multiply both sides by 16. so let's multiply by 16 and multiply by 16. and when we do that on the left hand side we'll just be left with sine Theta and on the right hand side we're just going to press on our calculator we've still got this on our calculator to display so we're just going to press times 16 equals and we get 0.45019 and so on now we found with the sound of the angle is the sound of the angle is equal to 0.45019 and so on but we don't want the sign of the angle we want to know what the angle Vita is so we want to get rid of the signs we have to do the inverse so that's going to be the inverse side and when we press shift sign so we'll press we'll write down Vita equals sine to the negative 1 the inverse sine of 0.45019 and so on now we've still got this in our calculator display so if we just press shift and then sign we get our sine to the negative 1 or inverse sine and then just press A and S on our calculator so we'll just type in this sine negative 1 so shift sign and then to press the answer button and then press equals and it will give us the answer of theta equals 26.756 degrees and that's to three decimal places and that's it so we have used the sine rule to find the size of this missing angle so the sine rule is a over sine a equals B over sine b equals c over sine C and that version of the formula I use a lot whenever I'm finding missing sides because the sides on the numerator and if I'm finding them it's an angle I flip it over and write sine a over a equals sine B over b equals sine C over C and the sine rule is particularly useful whenever you've got an angle and a side opposite each other and then you're trying to find a side or an angle Okay so we've looked at the sine rule we're now going to look at the ambiguous case and that's video 334 and corporate Maps now the ambiguous case it's mentioned on two of the exam boards but not mentioned in one of the others um I believe it's mentioned in that XL and AQA but I don't think it's mentioned in OCR and it may be worthwhile to watch anywhere because if you are learning the sound rule there's a good chance you're going to be going on to do a level Maps anyway and it became quite useful to just know this whenever you go into a level Maps anywhere so here we've got our triangle we've got ABC and we're trying to find the angle Theta here and we've been asked to find the two possible sizes of V to calculate the two possible sizes of Vita so here we've got 30 degrees and 8 centimeters and we've got 14 centimeters and Vita so remember with the center of we're finding in the angle we're going to write sine Theta over the opposite side which would be 14 is equal to sine 30 over it so sine 30 over M so we multiply both sides by 14 we're going to get sine Theta is equal to 14 sine 30 divided by 8 and whenever we work this out because sine 30 is equal to half we get 14 times a half which is seven so we're going to get 7 ifs or 0.875 so that means that the sine of this angle is 0.875 and if we do the inverse sine we're going to get the beta is equal to sine the inverse sine of 0.875 and when we do that on our calculator we the Vita is equal to and we get Theta is equal to 61.04497 and so on degrees and I'm just going to round this to three decimal places that means that beta is equal to 61.045 degrees so that means of beta is equal to 61.045 degrees but the question answers for the two possible sizes of Vita now if you actually take a calculator and you do the sign of 80 degrees and you do the sine of 100 degrees let's have a look and see what you get 0.984807753 and if we do the sine of 100 degrees we get that's also equal to 0.984807753 and whenever we do the sine of 100 we can also get that's equal to 0.984807753 so the sine of 80 and the sine of 100 are both the same so if on your calculator you do the inverse sine the sine minus one the inverse sine of 0.984807753 let's have a look and see what we get so shift sign and then answer is equal to 80 degrees so the calculator has just told me 80 degrees it didn't say anything about 100 degrees so in this question here we've got 61.045 but there's going to be another possible answer an obtuse angle that could also work here for Vita and as you can see the question says not drawn accurately so what we're going to do is we're going to look at the graph of y equals sine of X and later on in this video we're going to look at our trigraphs but the graph of y equals sine of x looks something like this this is the graph of y equals sine X and as you can see this graph it goes up from 0 up to 1 and 90 because the sine of 90 is equal to 1 and then it starts coming back down again between naught to 180 degrees it's symmetrical between here and here and there's a line of symmetry here and because there's a line of symmetry if we were to work out the sine of 80 degrees and then we also work out the sign of 100 degrees it would give us the same answer if we would work out the sign of 89 degrees and the sine of 91 degrees it would give us the same answer so that means that whenever we're doing the same rule questions and we're finding the angles there sometimes can be two possible answers now for instance in this question this angle here couldn't be 179 degrees because we if they're already being a 30 degree angle we know that this angle has to be less than 150 degrees but let's figure out what this angle could be so we know that one of our Solutions is 61.045 so that's one of our Solutions so here is equal to 61.045 whenever we work out the sign of that we get that's equal to 70th or no 0.875 and then there must be another angle over here that whenever we do the sign of that we also get 0.875 now to work out what this angle is what I do is I take this angle 61.045 away from 90 to work out how much below 90 it is and then if I add it on to 90 it would then tell me what this angle would be so let's do 90 take away this so 90 take away 61.045 28.955 so that means it's 61.045 is 28.955 below 90. so to find this value what we're going to do is we're going to add 28.955 to 90 and then that would tell us this angle because obviously this is symmetrical so if we do 90 plus 28.955 that would be equal to 118.955 degrees so that's one way we could find the second angle another approach is to take our angle for Vita r60 1.045 and just take that away from 180 degrees and that would also give us 118.9 and 55 degrees so that means that this angle of Vita could either be 61.045 degrees or 118.955 degrees and that's it so sometimes useful to know whenever using the sign rule that there can be two possible solutions Okay so we've looked at the sign rule let's now have a look at the cosine rule and the cosine rule is video 335 and 336 in corporate maps and here's part of the copper Mouse revision card and we've got our cosine rule here which is a squared so we've got this triangle we've got a squared equals B squared plus c squared minus 2 b c cos a so in other words if you have got a side another side and an angle between them you can find the length of the opposite side so here we have got a side and a side and the angle between them and we can find the length of this opposite side so we if we wanted to find this value what we would do is we would do 20 squared plus 15 squared minus 2 times 20 times 15 times the cause of 120 and that will tell us the value of this value squared and then you'd square root it so let's do that now let's find the value of x so we have got the cosine real which is a squared equals B squared plus c squared minus 2 BC cos a so we've got our two sides and we've got the angle in between them so the little layer will be the side opposite the angle so this means that the label will be X so we'll write x squared equals now B and C doesn't actually matter which way around these go so we're just going to write 20 squared plus 15 squared and then we've got minus and then we've got two BC cos and I tend to write this in a bracket so I tend to put a bracket down and do 2 times B which is 20 times C which is 15 times the COS of a and that's the COS of 120 so COS of 120 close brackets okay next let's work this out so we've got x squared equals and then on our calculators we can type in 20 squared which is 400 and then 15 squared which is 225 and then we'll put minus and then let's type this in our calculator and see what we get so 2 times 20 times 15 times the COS of 120. and whenever I do that I get that's equal to negative 300. so our answer to that was negative 300 and let's close brackets around that so we've got x squared is equal to 400 plus 225 minus negative 300. now let's work out what we get so x squared equals so 400 plus 225 minus minus 300 is equal to 925 so this side is not 925 centimeters it wouldn't make sense for this question because that's x squared so we need to square root it to find the value of x so X will be equal to the square root of 925 which is equal to 30.414 centimeters so that will be 30.414 centimeters to three decimal places I've rounded up the three decimal places and that's it so the cosine rule is really useful whenever you've got two sides and the enclosed angle and you're trying to find the length of the other side the one opposite the angle and that's the cosine real formula the cosine rule can also be really useful if you have a triangle and you know the length of the three sides you can work out the size of one of the angles but just remember the little a is opposite capital A so whenever you're trying to find the angle so cos a little a will be the side opposite it and the B and C it doesn't actually matter which area they go so that could be B and that could be C so let's substitute those values into the cosine rule formula and work out the size of this missing angle layer so the cosine rule is a squared so remember the little a and the Capital Area opposite each other so a squared so in this case the a will be 7 so 7 squared is equal to B squared plus c squared so b and c will be 8 squared plus 5 squared minus 2 times B times c times cos a so times so 2 times B which is it times C which is 5. times the COS of a so in this case a is beta so the COS of Vita so let's work this out so we've got 7 squared is 49 equals 8 squared is 64. plus 5 squared which is 25 then we've got minus 2 times 8 times 5 times cos Theta well 2 times 5 is equal to 10 times 8 is 80 and then so it's 80 cos Theta multiplying the numbers in front of the COS beta together and then just put the cosmeter after it so let's simplify this so we've got 64 plus 25 that's going to be 89 so we've got 49 equals 64 plus 25 is 89 minus 80 cos Theta now we want to get the COS Theta on its own so let's get rid of this 89 so let's minus e89 from both sides so minus 89 and minus 89 and 49 take away 89 would be equal to negative 40 is equal to and with the 89 take away 89 is zero so we'll be left with negative 80 cos Vita I've got a couple of options of things I can do here I could divide both sides by negative 80 and that would give me negative 40 divided by negative 8 equals cos Theta well one thing I like to do is if I've got an equation where I've got things above being negative I can multiply both sides by negative 1. 40 times negative 1 is 40 and negative 80 cos beta Times by negative 1 is 80 cos Theta now I'm just going to divide both sides by area to get what cos Theta is so divide by area and divide by area so 40 divided by 80 is not 0.5 and on the right hand side 80 cos Theta divided by 80 just leaves us with cos Theta so that means that the COS of this angle is 0.5 we don't want to know the cause of the angle we want to find the angle so we need to do the opposite so we're going to do the inverse COS of 0.5 and that will give us the size of our angle and when we do that we get that's equal to 60. so that means the beta equals 60 degrees so Vita equals 60 degrees and that's it so the cosine rule can be really useful to find missing angles as well as missing sides okay let's have a look at our next topic so next topic is found in the area of a triangle using trigonometry and that's video 337 on COBRA maths so the area of a triangle is found by a half a b sine C where A and B are two sides and a couple of C's to angle in between them so if you have a triangle such as this one if you want to find the area of this triangle we can do area equals a half times the two sides so 14 times 8 times and then the sine of the angle in between them through the sine of 70. and when we do that on our calculator we get we've got a half multiplied by 14 multiplied by 8 multiplied by the sine of 70. you get that equal to 52.623 centimeters squared to three decimal places so if we're under that to three decimal places and also because it was an area question we've got our units of centimeters squared and that's it so if you want to find the area of a triangle and if you know two sides in the angle in between them you can just do a half a b sine C okay now we've looked at how to find the area of any triangle what we're going to do is we're going to use that to find the area for Segment so here we've got a sector which is opr and we've got a segment which is pqr this segment here and we want to find the area of this segment this green region here so if I wanted to find the area for this green region what I would do is I find the area of the whole sector to begin with so remember to find the area of a sector we use the formula Vita which is the angle which is 20 5 over 360 times by pi times by the radius now if we had the whole circle the radius would be 6 squared so if we do 25 over 360 times by pi times by 6 squared that would give us the area of this whole sector and let's do that and see what we get and whenever I do that I get 5 over 2 pi or 2.5 Pi alternatively we could write as a decimal number and as a decimal number would be 7.853981634 and so on centimeters squared so that's the area of this sector now what I want to do is I want to find the area for the triangle and if I take the area of the triangle away from the area of the sector it will leave me with the area of the segment so let's find the area for the triangle so let's write sector up here that's the area of the sector now let's find the area of the triangle so remember the area of a triangle is a half a b sine C so it's going to be a half now if we look at this triangle we've got two sides in the angle in between them so it's going to be a b and c so half of 6 times 6 times by the sine of 25. that's equal to 7.607128711 and so on centimeters squared so we've now got the area of the triangle and we've got the area of the sector if we take them away we'll find the area of the segment so let's write that down segments and that would be 7.853981634 and so on take away 7.607128711 and so on and that will give us the area of the segment and let's see what we get so whenever we tap this in or use 2.5 Pi take away the answer we get an answer off 0.24685 and so on centimeters squared and that's it we could round this to perhaps three decimal places that'd be 0.24 and then we've got six years that'd be seven centimeters squared that's it see if we want to find the area for segments it can be useful to find the out of the sector then find the area of the triangle and take them away to leave us with the area of the segment that's it okay so our next topic is translations last video 325 in corporate maps and a vector looks something like this we're going to have a number on the top and a number beneath and some brackets around it the number on the top will tell you how to move the shape horizontally and it'll tell you how many squares to the right to move it so for instance if it's an E8 you'll move eight squares to the right if it's a five it'd be five squares to the right if it's a zero you won't move it any Square to the right it'll stay where it is horizontally if it's a negative so for instance if it was negative three you would move it three squares to the left if it was negative nine it'd be nine squares to the left so the top number tells you how to move the shape horizontally and if it's positive it'll be to the right if it's negative is to the left the number beneath that will tell you how to move the shape vertically so if it's positive here it's a one so that's going to move the shape one square upwards if it's a four you move it's four squares upwards and if it's a negative you move it downward so if it's negative three you'd move it three squares downwards so here we've got a typical question and we've got a little shape it's a little T shape and the ship's called C and we've been asked to translate shape c by and we've got minus one five that will mean one square to the left so one left because it's negative it's going to go to the left so one left and then we've got a five that's going to be five up so we're going to move the shape one square left and five squares up and that's it so next topic is rotations so here we've got rotate shape at 90 degrees anti-clockwise about the point two minus one so let's mark on the point two minus one so this is going to be our Center of rotation and we're going to rotate chip a 90 degrees anti-clockwise so going this way anti-clockwise about that point so get your tracing paper and put it on top of the center of rotation and the shape like so and make sure your tracing paper is straight so it's straight and it's not diagonal like something like that so make sure in this case I've put it landscape so going that way and what you're going to do is we're going to draw over the center of rotation and we're going to draw over the shape a so we've drawn over the center of rotation and we've drawn over a shape a so now what we're going to do is we're going to put our pencil on the center of rotation and we're going to rotate the tracing paper 90 degrees anti-clockwise so when we do that we turn it and it's important to keep the center of rotation on that point returning it like so until we've rotated the tracing paper in 90 degrees anti-clockwise and we can now see the position of where a would go after we rotated at 90 degrees anti-clockwise about 2 negative one so now what I would do is I would just draw over this rectangle a couple of times just to make sure that whenever I move my tracing paper there's a sort of a light impression on the paper so it looks something like this and that's it so we've repeated a 90 degrees anti-clockwise about the 0.2 negative one let's have a look at our next question so this time we've been given a triangle and we've got a set of coordinate axes we've got our x-axis and we've got our y-axis and the question says reflect triangle a so we've got this triangle a and we've been asked to reflect it in the y-axis in other words this is where the mirror is Okay so we've got this y axis and this is where the mirror is we want to reflect this triangle across to the other side so let's do each point of the times so this top point of the triangle it's one line away from the y axis so we need to go another one to the other side there the point down here it's one away from the y axis so we need to go one the other way and the bottom right hand point of the triangle it's one two three four away from the mirror line so we need to go another four one two three four so it'll be there and if we get a learner pencil and join those up we'll have reflected triangle a in the y-axis and that's it okay let's have a look at our next question so we've got the same triangle and we've been told this time to reflect triangle a in the x-axis so we've got this x-axis so this is where the mirror line is this time and we're going to reflect the triangle downward so it's going to move down here somewhere so again let's look at each corner of the triangle so let's start off with this corner and it's one two away from the x-axis so we go down to number two one two to here this corner is one two away from the x-axis so we're going over two one two two there and this point up here it's one two three four away from the x axis so we need to go down four one two three four down to there and if we get around a pencil and join those up that'll be answer for reflecting triangle a and the x axis and that's it so we've been asked to reflect this ship the ship C in the line x equals negative one so this line x equals negative one is going to be a line going straight through negative one on the x axis so if we go to the x axis and got a negative one here it's going to be a vertical line going straight through that negative one so that's where our mirror line is going to be so in x equals line will be a vertical line going through whatever number that is on the x-axis now we have to reflect the ship C in that mirror line so let's start off with this corner so this corner is one two to the mirror line so we need to go another two one two so it's going to move to here the bottom right hand corner of C well it's one two three four to the mirror line so we're gonna go one two three four to here and the top of triangle sees here so it's one two three four to the mirror line so we need to go another four one two three four so it's going to be there and we get a ruler and pencil and we just join up that ship and we have a reflected C in the line x equals negative one next we're going to reflects the ship's C in the line Y equals one so let's just draw about what we've done so we've got our shape c again and we're going to reflect it this time in the line Y equals one so what we're going to do is we're going to go to the y axis and the point one so the y axis and the point one and we're going to draw a horizontal line going through that point so going through one on the y axis so if you've got a y equals and a number it'll be a horizontal line passing through that number on the y axis and we have to reflect this shape this triangle in that mirror line so let's start with this bottom left hand corner of C it's one above the mirror line so let's reflect it to one below this point is one above the mirrorland so we're going to go another one downwards so there and finally at this point to get to the mirror line we go one two three down so let's go number three one two three to there and let's join up those points and that's it we need to be able to reflect shapes in the lines Y equals X so the diagonal line going upwards this way and the line Y equals negative X the line going down this way here okay and it would look something like this so it's very important to know what the lines Y equals X and Y equals negative X look like so that whenever you're doing Reflections you can then draw them really quickly and easily now to reflect the ship in that line it's quite easy we just count the diagonals so let's look at each corner so let's start off by looking at this corner the bottom left hand corner of B and as you can see if we go diagonally to our mirror line it's one diagonal so we need to go another diagonal the other way so it'll move to here this point in the bottom right hand corner if we count the diagonals it's one diagonal and then we've got half a diagonal so if we go another one and a half so a half and then one would be here so we've reflected these two points now let's look at the points at the top of the rectangle so let's look at the top left hand corner that would be one diagonal two diagonals so we need to go in over two diagonals that'd be one diagonal two diagonals would be there and finally the top right hand corner of B would be up here so if we turn the diagonalize that with one diagonal two diagonals and a half so we go a half one and two and as you can see we've got a rectangle shift now we just need to get a ruler and pencil draw a nice rectangle and that's it's a reflected shape be a rectangle B in the line Y equals negative X okay so let's have a look at this question so this is a large breast scale factor half using negative five negative five is the center of a large one so let's find negative five negative five so that's our Center of enlargement there and we've got our original shape our object and we're going to enlarge a rescue a factor of half that means all the points in the image will be half the distance away from the center of enlargement so for instance we looked at this point at the top of the kite this point at the top of the kite is one two three four five six squares to the right and one two three four five six seven eight squares upwards so because we're using a scale factor of a half we're going to half those distances so instead of going six to the right we're going to go three to the right one two three and instead of going eight up we're going to go four up so one two three four so the top of our count will move to here okay now let's look at the left hand side of the code here so this point is one two three four squares to the right and one two three four five six squares upwards so we're gonna half those distances so instead of going four squares to the right we're gonna go two squares to the right one two and instead of going six up we're gonna go three up one two three so that means that the left hand side of the kite will move to here the right hand side of the card so we can kind of guess it's going to move to this point here but let's check it so going back to our Center of enlargement it is one two three four five six seven eight squares to the right and one two three four five six squares upwards we're gonna half those so that'll be four to the right and three up so one two three four and one two three so that's the right hand side of the code and let's be careful with the bottom of the cape because we need to make sure we get the right height here so it is whenever we look at our bottom of our code go back to the center of enlargement it's one two three four five six squares to the right and one up so we're gonna have our distances instead of going six to the right we're gonna go three to the right one two three and instead I've got one square we're gonna go half a square up so it's going to be there and then we're gonna relearn our pencil and we'll join those up and that's it so we've enlarged this year by scale factor of a half so it's got smaller using the center for enlargement negative five negative five okay next topic is negative skill factor so we're looking at enlargements and we're looking at how to enlarge things with a negative skill factor so whenever you enlarge by a negative skill factor what happens is the points move in the opposite directions so let's have a look at our question so here we've got a square and it's shown on our grid and we've been asked to enlarge by scale factor negative 4 using negative three negative 1 is the center of enlargement let's mark on that Center of enlargement so that Center of launchment will be here negative 3 negative one and we're not going to enlarge each of these points so let's start off with this point here okay so to get from our Center for enlargement to the point we would move one square to the left so if we're enlarging by negative four what that means is instead of going one to the left we're going to go four to the right so we're gonna go one two three four so that point will move there okay our next point so this point here so this point here was one two to the left from our Center of enlargement so instead of going two to the left we're going to go eight to the right so it's going to be one two three four five six seven eight so it's going to move to here so we've done these two points now let's choose this point here so this point is one to the left and one down so we're going to go four times as far away but in the opposite direction so we're going to go instead of going one to the left and one down we're going to to go 4 to the right and 4 up so one two three four and four up one two three four will be here okay and finally I'll Point down here the bottom left hand point of our Square it's one two to the left and one down so instead of going two to the left and one down we're gonna go four times as far away but in the opposite direction so instead of between two to the left it's going to be eight to the right and instead of being one down it's going to be four up so one two three four five six seven eight and four up one two three four so we've now got our four points we just need to join them up and that's it with the largest Square by scale factor negative four using negative three negative one is the center of enlargement and whenever you enlarge them by negative skill factors it's enlarged but it goes in the opposite direction and that's it and if you want more practice on this this video 100 Newton corporate maps and also remember you've got those practice questions and textbook exercises beside that number and also remember there's a bumper Packer question so if you do look at the booklet in the description below and open it up there'll be a question there on negative skill factors okay let's have a look at our next topic so our next topic is invariant points in this video 392 Accord Maps so an invariant point is whenever we carry out a transformation if a point or any point stay in the exact same location after that transformation is applied then they're what we call in variant points see here we've got a triangle ABC and we've been asked to reflect the triangle ABC and the x-axis so here you can see here's the x-axis and we're going to reflect this triangle in the x-axis so let's reflect the triangle in the x-axis to see where the points move to so the point a well it's 1 2 to the x axis so we're going to go one two up so a is going to move to here so that'll be a dash B is 1 2 to the x axis so we go up another two of one two so the point B will move to here so it's going to be B Dash and the point C well it's on the x-axis when we reflector and the x-axis is going to stay where it is so the point c will stay where it is so let's draw a triangle and the question says are there any invariant points as you can see the point C stay where it is so the answer is yes there are invariant points and is the point C the point see so the point C stays where it was after we carried out the transformation so it's an invariant point and that's it so if you want to find out more invariant points if you watch a video of 392 on called maps that would be the video tutorial and variant points also remember you've got that bumper pack of questions the ultimate GCSE higher revision question booklet I think that's the name of it um but in that booklet you've got a questionnaire and invariant points as well okay our next topic so similar ships is where one ship is an enlargement of another so if you've got one ship and another ship that's similar to it then it's an enlargement of it so whenever you've got shapes that are similar such as this rectangle air and this rectangle B we know that to find the lengths we multiply by a certain skill factor now notice that the sides will become so many times bigger or smaller but the angles will all stay the same because obviously the shape is staying the same shape so here we've got the core management revision card part of it and we've got here are two similar rectangles so we've got this rectangle rectangle a which is six centimeters long and four centimeters wide and we've got this rectangle B which is similar to it so it's a large and it's got a width of 12 centimeters and a length of well we need to find that we've been asked to find the length of rectangle B so let's look at the sides so let's look at the widths of these rectangles we've gone from four centimeters to 12 centimeters so to find the scale factor of enlargement if we divide 12 by 4 we will find the scale factor of enlargement so if we do 12 divided by 4 12 divided by 4 is equal to 3. so that means that this width is three times larger than this width so multiply by three so if we multiply the width of rectangle a by three that means we're going to multiply the length of rectangle a by three as well and that would give us the length of rectangle B so if we do six multiply by three that's going to be equal to 18 centimeters so the length of rectangle B would be 18 centimeters so similar ships are where one chip is an enlargement of another that's it okay let's have a look at our next topic so our next topic is congruent triangles and that's video number 67 on corporate maths so congruent means identical it means the same shape and the same size so it means that if you've got two ships that are congruent it'll mean that the angles are the same and the sides are the same length as well now whenever we're dealing with triangles in a triangle we've got three angles and there's three sides now if you've got two triangles if you want to see if they're congruent to each other you don't actually need to know all six angles in both triangles and the lengths of all six sides what you can actually do is know if they're congruent by just knowing some of that information and these are the conditions to know if triangles are congruent or not so the first condition is what we call side side or SSS and what that means is that the sides are the same size so here we've got five centimeters and five centimeters then we've got seven centimeters and seven centimeters and nine centimeters and nine centimeters so if you've got two triangles that both of the same lengths of the sides so side side and side those two triangles will have to be congruent to each other so they will be the same shape and size okay let's look at the next condition so the next condition is what we call angle side angle or ASA and what that means is if you've got two angles and you know the side in between them for two triangles those two triangles will be congruent to each other so for instance here we've got 70 degrees 30 degrees and in between those four centimeters so if you draw that triangle and then if you go to another triangle where you've got 30 degrees and 70 degrees so the same two angles and four centimeters in between them those two triangles will have to be congruent to each other because there's only one possible triangle that will have 70 degrees 30 degrees and four centimeters in between them so it's called angle side angle or angle side angle okay let's look at the next Second Edition the next condition is SAS or side angle side so that means if you've got two sides and you know the angle in between them and they're the same for two triangles those triangles will be congruent so as you can see here we've got 12 centimeters and 14 centimeters and 30 degrees in between them and then this triangle here we've got 12 centimeters and 14 centimeters and 30 degrees in between them so these two triangles will have to be congruent to each other because there's only one possible triangle you can draw which has 12 centimeters and 14 centimeters and 30 degrees in between them so those two triangles will be congruent okay in another condition is what we call rhs which stands for right angle hypotenuse side so if you've got two right angle triangles and you know the hypotenuse and one of the shorter sides those two triangles will have to be congruent to each other because if you use Pythagoras's film you can find the length for the third side and then it would be side side side so these two triangles will have to be congruent to each other so this is the chord master of vision card and it's very useful to remember the conditions for congruent triangles whether it's side side side angle side angle side angle side or rths and that's it okay let's have a look at our next topic okay so let's have a look at our next topic so our next topic is similar shapes and this video is 292 and 293 equivalent Maps so similar shapes are shapes that are enlargements of each other so if two ships are mathematically similar one is an enlargement of the other if the scale factor of enlargement for the size is n the skill factor for the area is N squared and the scale factor for the volume is n cubed in other words if you really enlarge a ship and the sides are perhaps three times bigger the area will be nine times bigger because 3 squared is nine and the volume if for instance it was Cube and we enlarge a bicycle factor of three the volume wouldn't actually be three times bigger you would do three cubed which is 27 so be 27 times bigger if we enlarge the shape by scale factor five the area would become well 5 times 5 is 25 times bigger and the volume would be 5 cubed which is 125 times bigger that's really useful whenever we're dealing with similar ships so here we've got two similar shapes two trapeziums two trapezia if we have a look at this first trapezium you can see one of the lengths of the parallel lines is five centimeters and for the larger trapezium the corresponding length has a side length of 15 centimeters which if we look at it is three times bigger because 5 times 3 is 15. that means because the sides are three times larger we will square that that's nine so I mean the area is actually nine times bigger and if we were given the area the smaller trapezium which is 20 centimeters squared and we wanted to find the area of the larger trapezium we could do 20 times 9 which is 100 nearly and that's it if we were given the area of the larger trapezium we would divide by 9 and then that would give us the area of the smaller trapezium but if we know the scale factor for enlargement for the sides we can square that to get the scale factor of enlargement for the areas and we can Cube it to find the scale factor for the volumes and that's it so this time we've got two cuboids and we've got a smaller cuboid which is a side length of seven centimeters so the length of is seven centimeters and the length of the larger cuboid is 14 centimeters and as you can see the sides are two times larger so that means the skill factor for larger for the sides is two that means if we want to find the scale factor for the volume we will Cube that 2 cubed is eight that means the volume so eight times bigger so the volume of this cuboid will be eight times larger than the volume of that cuboid so if we do 100 times a that will give us the volume volume of the larger cuboid 100 multiplied by 8 would be 800 centimeters cubed and that's it so we have a look at these prisms prism a and prism B pentagonal prisms pentagonal prism a this prism has got a surface area of 300 centimeters squared whereas the surface area of B is 1 200 centimeters squared well if we divide the surface area of B by the surface area of a we can find the scale factor of enlargements for the areas so we divide 1200 by 300 we get that's equal to four that means the surface area is a four times larger so that means that our scale factor RN squared is equal to four that must mean that our sides where if that's four times bigger that means our sides if we square root this four that means our sides will be two times bigger so for instance if that was 10 centimeters that would be 20 centimeters and to get the scale factor of enlargement for the volume well we Cube this so 2 cubed is 8. so if we know what the scale factor of enlargement is for the areas in this case we know it was four we could square root that which is 2 and we can Cube that then to find the scale factor enlargement for the volumes so we can then find out the volume of this larger Prism by doing 600 multiplied by ear and if we do 600 multiply by 8 that's equal to 4800 centimeters cubed so that's great that means if we know what the surface areas are we can actually divide them and see how many times larger the surface areas for the larger ship and the smaller one we can then square root that and then that'll tell us the scale factor of enlargement for the sides If instead we knew the volumes we could divide the volumes and then in this case that would have been eight and then we could cube root that to find the scale factor of enlargement for the sides and then we could Square it to get the areas and so on and if you want to do more practice on this topic you can watch video 293 be in corporate maps and it will go through more examples of how to deal with questions whenever you're dealing with surface areas and volume and that's it and we're going to start off by looking at Circle firms and these are videos 64 and 65 in corporate maths so here we've got our first Circle theorem so our first Circle firm and this is part of the core masterovision card is the angle of the semicircle is 90 degrees so if you've got the diameter of the circle and you've got two lines that go from the diameter up to a point on Circle or down the angle will always be 90 degrees so here's an example so if you're given this question that's to find the value of x well here we've got our triangle we know that's a right angle so we need to find the X well we know that that's 90 degrees and then we can just use the angles on a triangle to find X so 90 plus 34 190 plus 34 is equal to 124 degrees and if you take that away from 180 we can see what's left for x so 180 take away 124 would be 56. so that means that X is 56 degrees okay let's have a look at our next example now one thing to look out for whenever you're doing Circle theorem questions is looking for isosceles triangles so from the center of the circle to the edge is the radius so here we've got a radius and a radius there to radii that means that the same length as each other so this would have to be an isosceles triangle so if we know that this angle is equal to 98 Degrees we know this angle X and this angle over here would be the same so that's X as well so we can work out what x would be we could do 180 minus 98 which would be equal to 82 and then if you've done 82 divided by 2 that's equal to 41 degrees so X here would have to be 41 degrees so look out for isosceles triangles in circle theorem questions where you've got particularly whenever really I are involved okay next now here again you can see we've got from the center to the circumferences already so that's a radius that's a radius and that's a radius so here we've actually got we've got two isosceles triangles this is an isosceles triangle and this is also an isosceles triangle and also another thing to note is the big triangles a right angle triangle because we've got that diameter as well so that's a right angle at the Top If you had these two angles together so let's look at our X Y to begin with so we know that's the radius and that's the radius that means the Y and 55 of the same as each other y equals 55. knife 55 degrees now if Y is equal to 55 you've got 55 plus 55 or 55 plus 55 is equal to 110 degrees so 180 minus 110 is equal to 70 degrees so that means X is equal to 70 degrees and finally because this is the diameter we know that x and z are in a straight line so we know that if we do 180 minus 70 so that's equal to 110 degrees so Z is equal to 110 degrees and that's it so it is useful to look out for those SLC triangles okay next okay so our next Circle theorem is that the angle of the circumference is half the angle at the center so here we've got our Circle we've got the center of the circle you can see this angle is 120 degrees between the two Radia and the angle at the top here between the two chords be equal to 60 degrees so the angle of the circumference is half the angle of the center so let's have a look at some examples okay with our first example okay so we know the angle of the circumference is half the angle at the center but that means if we double the angle of the circumference we get the angle of the center so if we do 51 multiplied by 2 that's equal to 102 degrees so that means the X here is equal to 102 degrees next we've got the angle at the center and we're trying to find the angle of the circumference so if we divide the 122 by 2 we'll get the angle of the circumference so that would be 61. so that means that x equals 61 degrees for this circle okay so the angle at the circumference is half the angle of the center or the angle of the center is double the angle of the circumference okay let's have a look at one more question so here we've got our Circle this time here we've got a circle and we've got the angle of the circumference but we don't actually have the angle at the center we've got the angle inside of the ship here so what we're going to do is we're going to take the 234 away from 360 to find the angle at the bottom it has to be like this where you've got the angle at the circumference and the angle at the center so if we do 360 subtract 234 we get that's equal to 126 degrees so that's 126 degrees at the center now the angle of the circumference is half of that so if we do 126 divided by 2 we get that's equal to 63 degrees so the angle for X would be 63 degrees now before we move on to the next circum I just want to show you that we can move the angle of the circumference over to the side so it could instead of be looking something like this our diagram might look something like this and the angle of the circumference here 60 degrees is still half of the angle at the center so just be aware that it might not always look like this that the point at the circumference might be over to the side and the angle the circumference is still half the angle at the center and that's it so the next Circle theorem is the angles in the same segment from a common quarter equal so if you look at this chord this blue line This dot or blue line you've got a pair of green lines coming up from either end of that chord and they're meeting at a point on the circumference of the circle and likewise we've got a pair of red lines that are coming off from either end of that chord meeting at a point at the circumference that means at this angle and this angle will be equal to each other now if you have a look at this example we've got a common chord here and we've got all these lines coming off from that chord either end of the chord to the circumference so that means that a b and 37 are all equal to each other so if a is equal to 37 degrees B is equal to 37 degrees as well that's it they're all lines coming off from that same chord okay let's have a look at our next example so we've got this circle we've got our 27 degrees our 88 degrees and our angle X and we've been asked to find this angle X as you can see we've got this chord at the bottom here and we've got our lines coming up here and here and also the lines coming off across here and here so let me set this 27 will be the same as this angle here so that means that that angle there is 27 as well next thing to notice is we've got two straight lines across each other that means the opposite angles are equal so if this angle is 88 degrees that means that that angle is 88 degrees also they're vertically opposite angles and finally you've got this little small triangle here with this x 88 and 27. that means that those three angles will add together to be 180 degrees so if we do 88 plus 27 that's equal to 115 degrees next if we do 180 subtract 115 degrees we get that's equal to 65 degrees so that means the x is 65 degrees now you could have done this question in a state different way where we looked at the 88 and 27 and worked out this missing angle at the bottom and then looked at the chord and said well X is the same as that angle there and worked it that way but that would be 65 as well anyway okay let's have a look at our next Circle theorem so our next Circle theorem is if you've got a cyclic quadrilateral so that's a quadrilateral whether all the corners lie on the circumference of a circle the opposite angles add up to 180 degrees so as you can see here we've got 70 degrees and 110 I'm going 80 and 100 so the opposite angles add up to 180 degrees okay and let's have a look at our example so here you can see we've got a cyclic quadrilateral this point this point this point and this pointer all on the circumference of a circle that means the opposite angles will add up to 180 degrees this 114 is outside but we can actually subtract that from 180 to find the angle inside so 180 subtract 114 degrees is equal to 66 degrees so that means this angle in here is 66 degrees now we can find our X and our y really quickly and easily so because the 66 is opposite the X and the second quadrilateral we know if we take away the 66 away from 180 degrees we can find out what x is so 180 subtract 66 is equal to 114 degrees so that means X is 114 degrees next our y well the opposite angles add up to 180 so if we do 180 subtract 121 that's equal to 59. so we know that Y is equal to 59 degrees so we've worked out what x's and y's by using the fact that the opposite angles and a cyclical quadrilateral always add up to 180 degrees okay next Circle theorem my next Circle theorem is if you have the radius and a tangent they meet at 90 degrees so here you can see a b is the tangent so that means that the tangent and the radius meet at 90 degrees that's a 90 degree angle there now here we have got if this is 154 you can take that away from 180 that would be equal to 26 so that would be 26 degrees in there in a straight line next we've got a triangle we've got 90 degrees and 26 degrees if we add those together we get 116 degrees so 180 minus 116 degrees equals 64 degrees so that's 64 degrees there and then finally we have a straight line here so if we do 180 minus 64 we get that's equal to 116 degrees so that means x equals 100 116 degrees and that's it so the radius and the tangible meter 90 degrees our next Circle firm it's got a name that I always remember that's alternate segment theorem and that is if you have a tangent the angle between the tangent and a chord equals the size of the angle opposite it inside the triangle so this 70 is the same as the angle opposite it in the triangle and the 60 is equal to the 60 the angle opposite it inside the triangle and that's called alternate segment theorem so let's have a look at our example so here we've got our tangent and the angle between the tangent and the chord is 55 so that 55 will be equal to the angle opposite it so X is equal to 55 degrees and here we've got our y That's equal to the angle opposite it's a y would be 60 degrees and that's it now what's interesting is if you go to the videos and worksheet section called Maps there's a video showing you the proof of this so if you want to see why they're equal to each other have a look at chord maps you can see me explaining why okay next okay two more interesting Circle theorems are if you've got a point the two tangents from that point to the circle will be the same length so as you can see here is this point C and you can draw two tangents to the circle the meter is d and e and the lengths of c d and c e will be equal to each other and then finally if you've got a chord the radius that cuts that chord in half will meet the chord at 90 degrees like so okay let's have a look at our next topic so next topic is geometric proof in this video 366 and corporate Maps now if geometric proof I highly recommend you watch that video and try the practice questions because the more practice you do with geometric proof the better and here we've got a diagram and we're told the a PC is a straight line and we're told that CP so CP is the same length as BP so these two lines are the similar if I'm actually just going to put dashes in here and here they show they're the same length as each other and we're told the angle ABP ABP is twice angle so this angle ABP is twice angle CBP so that angle is double the size of that angle and it says angle BCP so b c p is equal to X so let's put X into here now this is actually an isosceles triangle so if this is X this is X up here as well because it's an isosceles triangle and actually we were told that this angle a b p is twice this angle here so if that's X this angle will be 2X and we've been asked to prove that angle BAC is equal to 180 minus 4X so let's actually write down some of the things we've written down so far already so we've said that angle because we were given this angle is equal to X we wrote down the angle CBP was equal to X as well so that's our first step we've wrote that this was equal to X because this is an isosceles triangle because they're the same as each other then we know the angle a b p is equal to 2x because that's double angle CBP and we give that the question so let's write that down okay now we're trying to show that this angle here is equal to 180 degrees minus 4X now if geometric proof questions there can be different approaches we can take now this is actually my second time filming this I filmed it first of all I'm in my first approach what I've done at this point was I looked at this triangle here this triangle CBP and I said that this was a triangle and I knew that the angles would add up to 180 degrees so out of the X and the X together to get 2X and then I took that away from 180 to get this angle to be 180 degrees minus 2x then I said this was a straight line so if this was 100 180 degrees minus two axis would have to be 2x so that adds together to make 180 and then if this was 2X and this was 2x then I know that this triangle the angles add up to 180 degrees so if I add 2X and 2x together that's 4X and then that means that this angle here would have to be 180 degrees minus 4X so I've done that in my first approach and then an edited it I realized hold on a minute there's a much better way to do this question now so we're trying to show that this angle here is 180 degrees minus 4X so let's have a look and see what we've got so if we look at this big triangle a b c this big triangle the angles in this triangle will have to add up together to give us 180 degrees so we said this angle this angle at the top and this angle would have to add together to give us 180 degrees so if we have a look at this angle at the top here we've got 2X and X if we add them together we'll get the sides of this angle altogether which would be 3x so this angle up here would have to be 3x altogether so that means that the angles in this triangle add up together to give us 180 so if we do three exercises this angle plus this angle which is X we'll get 3x plus X and that's equal to 4X so these two angles added together this angle at the top this whole big angle and this angle added together is 4X and then if you take that away from 180 degrees we'll find the size of this angle so 180 degrees minus 4X would have to be the size of the cycle and that's what we're asked to show so this angle is 180 degrees minus 4X and that's it so let's write that down and that's it so I've just said the angle BAC would have to be 180 degrees minus 4X because the angles in a triangle the whole big triangle have to add together to give us 180 degrees there is also that question in the bumper pack of questions that you've got also to try and the more practice you do with geometric proof the better okay let's have a look at our next topic okay let's have a look at our next topic so the next topic is vectors now Vector is something that's good a direction and it has got a size or an over name for size is a magnitude so it's got a Direction so we've got a certain way and has got a magnitude or a size got a certain length so these lines are all vectors they've all got they're all going in a certain direction shown by their arrows and they've all got a certain magnitude or size that's its length and we've already looked at vectors whenever we're dealing with translations because we move the ships we slid the ships and we use the column vector and it showed us how many squares to the left or the right and how many squares up or down and what we're going to do is we're going to represent these vectors as column vectors and we're going to use the same approach where the number on the top will be how many squares left or right and the number on the bottom will be how many squares up or down and if the top number is positive it's to the right and if it's negative it's to the left and the number beneath if it's positive it's up and if it's negative it's done so let's start off with looking at this Vector here so this Vector here we can see the starting point and the arrow shows the switch where we're going and we're starting here and we're going one square to the right so let's write that as a one and then we're going two squares up one two so that would be represented so this Vector will be represented by the column Vector 1 2 1 to the right and two up okay now let's have a look at this column Vector so this column Vector starts here here and it goes in this direction and if we count the squares to the right it goes one two three four five to the right and then one two down so because it's five to the right we're gonna write five and then because it's two down we're gonna write negative two so this Vector will be represented by the column Vector five negative two okay our next Vector is this one and we start here and as you can see from the arrow we're going in this direction here and we're going three squares to the left one two three and one square up so because it's three squares to the left we're gonna write negative three because if we go to the left it's negative and then it's one square up so we then write one so this Vector would be negative three one okay next column Vector is this one we're starting here and we're going up two squares one two so to the left or right well it doesn't go to the left or right so we're gonna write zero and then it goes two up one two so then it'll be two so this is the vector zero two and finally our last Vector of this one we're starting here and we're going down to this point here so we're going one two to the left so it's gonna be negative two and then one two three down so that's negative three so this would be the vector negative two negative three which means two to the left and three done okay so that's how you represent vectors as column vectors now let's have a look and see what happens whenever we add and subtract and multiply these column vectors so here are some column vectors we've got a the vector a is three negative one so the vector a is three to the right and one down and the vector B is 4 6 which means four to the right and six up now these letters A and B they're the vectors and they're in bold which means that they've been typed in a thicker ink obviously now whenever we're writing we can write a and bold we can just sort of write it over and over and over what we could do is to show that a is to show that something is a vector we write it and then put a line underneath it so instead of writing a bold a which is very difficult we can just write a with a line underneath it so our first question says find the vector 2A so we've got the vector a which is three to the right and one down so the vector 2A will be double that so instead of being three to the right it'll be just right 2A is equal to and so instead of being free to the right it's going to be six to the right instead of being one down we're going to double it it's going to be two down so the vector 2A will be double the vector a so it would be 6 negative 2. if we're asked to find the vector 3A we would multiply both these numbers by three if we're asked to find the vector 10A we'll multiply both of these numbers by 10. if we're asked to find the vector negative 4A we would multiply both these numbers by negative four and so on okay let's have a look at our next one so this time we've been asked to find the vector a plus b so here's the vector a which is 3 to the right and one down and here's the vector B which is 4 to the right and six down so to find the vector a b we'll just add these vectors together so the vector a plus b would be equal to and if we add them together 3 plus 4 is equal to 7 and then negative one plus six well negative one plus six is equal to 5. so the vector a plus b would be seven five and that makes sense because if we start off at a certain point and we go three to the right and one down so three to the right one down and then we go four to the right and six up all together we have gone seven to the right and five up and that's and that's what the vector a plus b is okay and finally we've been asked to find the vector B subtract 2A so we've got the vector B which is four six and the vector 2A is 6 negative two so we've got the vector B and we've got the vector 2A so now we just need to do B subtract two out so that's going to be equal to B subtract two a would be equal to well four subtract 6 would be negative two and then six subtract negative two well six subtract negative two would be six plus two which would be it so the vector B subtract 2A if you get B and 2A and you subtract 2A from B we would give you negative two here and that's it okay so that's column vectors now this video 353 a and corporate Maps Okay let's have a look at our next vectors questions so this end we've been given a diagram and it's a triangle and we've got the points o a b and Q and Q is the midpoint of a b and we'll be told that the vector o a is equal to 4A and the vector let's write that down o a is equal a is equal to four a and the vector OB is equal to B so o b is equal to Little B and we've been asked to find the vector ba so we want to get the vector from B to a now we don't know B to a but what we can do is we can go from B to O and then from o to a so let's write that down so b a equal to b o plus o a so if we go from B to O and then from o to a that would be the vector b a so let's write down what that would be so b o well b o is going to be well it's not B it's going to be going the opposite direction so it's going to be negative B or minus B so it's equal to minus B and then we're going to go from o to a and we know that's equal to 4A so plus 4A and that's it so the vector from B to a would be equal to minus B plus 4A or we could write it the way around as four a minus B and that's it okay next would be nice to find the vector BQ so we want to go from B to Q now Q is the midpoint of a b so Q's in the middle so that means that the vector BQ will be half of the vector ba so if we have the vector ba we will get B Q so let's write that down b q is equal to half of ba and that's going to be half off four a minus B and that'll be equal to well half of 4A is 2A and then half of minus B would be minus a half b and that's it so the vector B Cubed would be equal to let's write it down BQ is equal to 2 A minus a half B okay let's have a look at the next part so the next part asks us to show that oq is parallel to the vector 8 a plus two B so we want to show the oq and 8 a plus 2B are parallel to each other now to do that I have two vectors are parallel to each other there are multiples of each other so for instance if we had the vector B and with the vector 2B they're both going in the same direction it's just that 2B is twice as long as the vector B so let's find what oq is so if we want to find oq we're going to have to go from o to B and then from B to Q so let's write that down oq is equal to OB plus b q now OB well OB is equal to B so that's going to be B plus and then B Q is equal to 2A minus a half B so two a minus a half b now let's simplify this we've got b or 1B we're taking away half B so it's going to leave us with semi is the vector oq is equal to a half B plus 2A now we want to show us parallel to eight a plus two B so let's actually read the silver way around let's put the A's at the front so that would give us 2A plus a half b so that means the vector oq is equal to 2A plus a half b and we want to show that's parallel to 8 a plus 2B now if we multiply oq by 4 if we multiply this by 4 we would get 8A plus and four times a half is 2b so if we multiply the vector oq by 4 we get that's equal to the vector we want to show is parallel to so that means oq and this Vector are multiples of each other so let's write that down and that's it so just written down the 40q so if we times this by 4 we get the vector it a plus 2B and the 8 a plus 2B is a multiple of oq so therefore they're parallel to each other and that's it okay let's have a look at our last vector's question so this time we've been given a diagram and we've got o a b and c and we're asked is a b c a straight line so we're looking to see it is this a straight line now vectors questions if I want to see if something has a straight line or not what I would want to do is for this ABC I would want to find the vector at a B I would want to find the vector BC then I'd want to see if there are multiples of each other that would mean they're parallel to each other and if they both pass through b and they're parallel then it must be a straight line because if they pass through the same point and they're parallel to each other then it's a straight line so let's do that let's find the vector a b and let's find the vector BC so a b to get from A to B well we're going to go from a to O and then from o to B let's write that down so AO well AO is equal to four a minus 3B so 4 A minus 3B and the vector OB is equal to 5A plus 2B so plus 5 a plus 2B and when we add these together 4A plus 5a is equal to 9A and then minus 3B plus 2B is equal to minus B so that means the vector a b would be equal to if we write it down nine a minus B and that's it so that's the vector a b now we want to find the vector BC so to get from B to C we're going to go from B to O and then from o to C now in this one we're going to need to be careful because if we look at the arrows they're going in the opposite directions to the way we want to travel so let's be careful with that so we're going to do b c is equal to b o plus OC now b o well B to O would be equal to now we're going the opposite direction so it's going to be minus five a minus 2B so minus five a minus 2B and then going from o to C where we're going the opposite direction so instead of doing B minus 32A we're going to go the opposite way so it's going to be minus B plus 32A so that means if we simplify this let's see what we get so that means the vector from B to C this Vector would be equal to 27 a minus 3B now if we look at these vectors A B and B C we can see that there are multiples of each other that means they're parallel and they both pass through b that means it is a straight line so let's write that down and that's it so I've written down that BC is equal to 3 a b because if we look b c is three times a b that means there are multiples of each other so BC and a b are parallel and they both pass through the point B so a b c is a straight line and that's it okay our next topic is on collecting like terms and so that means we're going to simplify expressions such as this one such as 7x plus y take away X plus five Y and that's video nine in corporate map so if you do want to recap this watch video9 and Maps Okay so we've got 7x plus y take away X Plus 5y so when we're collecting like terms we collect the like terms so let's start off with our X's we've got seven X's so here we've got seven x's and we've got subtract an X so if you get seven x's and you take away an X or 1X you'll be left with six X's so let's write that down 6x now let's deal with our y's we've got plus y so that's positive y or one y and we're going to add another five wise so if we added one y and over five y's all together that's adding six wise so it'd be six x plus six Y and that's it so we've simplified an expression by collecting the like terms the next topic is substitution so that's video 20 equivalent miles and we've been given the question given that W equals nine and Y equals five find the value of 8w minus three y so remember in algebra we don't read the multiplication sign so a w means eight times W or eight times whatever value W is in this case it's 9 and then subtract 3 times y so that's three times whatever value Y is and that's five so let's work out what 8 W is so that's eight times W that's eight times nine and that's equal to 72. so 8w is 72 and we're going to take away whatever three times wise so 3 times 5 is equal to Fifteen so three Y is 15 and then we're going to do 72 to take away 15 that's equal to 57 so 8w subtract 3y if W is equal to 9 and Y is equal to 5 then that would be equal to 57 and that's it sometimes you give them substitution questions that involve odd and even numbers are positive and negative numbers and I want to look at one of these now so here we've got X is a positive number and Y is a negative number and we've been asked to state if the following expressions are positive or negative so we've got 5x so that means 5 times x and x is a positive number so it could be numbers such as 10 and 5 times 10 would be 50. but remember X is a positive number any positive number so 5 is a positive so positive times a positive would always be a positive so 5x would be positive to our next expression X Y that means x times y now X is a positive and Y is a negative well a positive times a negative is a negative for instance if we had 10 and negative 2 10 times negative 2 is negative 20. so this would be a negative and finally we've got y squared so y squared means multiply by itself so Y is a negative number so we're going to multiply the negative number by itself and a negative times a negative is a positive so that means that y squared would be a positive and that's it so sometimes you're asked to do substitution questions where you're asked to find if things are positive or negative or odd uneven and if you have a look at that bumper pack of questions there's another one there for you to practice now okay let's have a look at our next topic so our next topic is laws of indices last video 174 in corporate maps and then what we're going to do is we're going to look at laws of indices whenever we're using algebra as opposed to using numbers but the same rules apply so for instance if we had m to the power of 3 multiplied by m to the power 4 we just add the powers to be m to the power of seven if we had m to the power of 8 divided by m to the power of 2 we take away the powers if we're dividing so be m to the power of six and then finally if we had a power to power so we've got so a power and then a bracket and then a power outside we'll multiply the powers together so if we had M cubed squared we then have m to the power of 6 by multiplying the powers together okay and this is the chord Master revision card on laws of indices so this might be quite useful if you've got the revision cards as well so let's have a look at our examples we've got y to the power of 8 multiplied by y to the power of three so we've got y times y times y times y eight times and then we're multiplying that by another y and times y times y so that also go over to be 11 y's multiplied together so it'll be y to the power of 11. in other words adding the powers together eight plus three is eleven so be y to the power of 11. next we've got y to the power of 15 divided by y to the power of 5 well if we're dividing and we've got the same base so they both got a base of Y we just take away the powers so we've got 15 take away 5 is 10 so be y to the power of 10. and finally if we had y to the power of 6 squared we've got a power over power so we multiply the powers together six times two is twelve so to be y to the power of 12. or another way to look at this one is we've got y to the power of 6 multiplied by itself so we could then add the powers and that would give us 12 as well that's it okay now we're going to look at expanding brackets and X bracket 3x plus 4. so remember to expand our brackets we multiply what's inside by the term outside so we're going to do 3x multiplied by X and we're going to do 4 multiply by X so 3x multiplied by X well that would be 3 x square squared because we're multiplying the X by X would give us the x squared and we've got a 3. so 3x multiplied by X is 3x squared then we've got our plus sign and then we've got 4 multiplied by X so that's just going to be 4X and that's it okay our next question so here we've got our termite side 2y and we're going to expand our brackets y minus 3. so we're going to multiply what's inside by the term outside so we're going to do y multiply by 2y well that would be 2y squared because we've got y times Y is y squared and then we've got the 2 as well so that's 2y squared then we've got our minus sign and then we're going to do 3 multiply by 2y well 3 times 2i would be 6y so answer would be 2y squared minus 6y and if you want more practice in this remember you can go to video 13 on corporate maps you can watch that video beside that this is the practice questions and also the textbook exercises and then also remember there's that bumper booklet of questions and there'll be questions there that involve expand and brackets okay our next topic okay so let's have a look at our next topic so our next topic is expanding two brackets so as you can see here we've got two brackets X plus 6 and x minus two and we've been asked to expand and simplify them so to expand two brackets I use this approach I take my first term my X and I do x times x so x times X is x squared then I do my x times minus 2. and x times minus 2 is equal to minus 2x so 6 times x is 6X so plus 6x and then we've got six times minus two and six times -2 is equal to minus 12. so whenever we expand these brackets we get x squared minus two X plus six x minus 12. as you can see we've been asked to expand and simplify and if you look at these two middle terms we've got minus two X plus six x now minus two X plus six x is 4X so our final answer would be x squared plus 4X minus twelve and that's it and another approach to expand on two brackets is to use a grid so here we've got X plus six we write X and then plus six and then we'll get X subtract two so we write X and negative 2. and what we're going to do is we're going to multiply each of these terms and put the answers inside of our grid so we've got x times x well that's x squared we've got x times 6 or x times positive 6 that's going to be plus six x then we'll get negative 2 times x so it's going to be negative 2X and finally we've got negative 2 times 6 well negative 2 times positive 6 well negative times a positive is a negative and 6 times 2 is equal to 12. so we've got x squared plus 6X subtract 2x subtract 12. so let's write that down x squared plus 6 x subtract 2x subtract 12 and then we need to simplify so let's collect our like terms we've got 6X take away 2x well 6X take away 2x is 4X so that would be x squared plus 4X subtract 12. so that's another approach we can use to expand in two brackets is the user grid okay let's have a look at our next question so our next question says expand and simplify 2x plus 1 X plus four so again we're going to multiply both of these terms by the 2X and then we're going to multiply both of these terms by the one so 2x times x well 2x times x would be 2x squared so 2x squared then we've got 2x times 4 well 2x times 4 would be 8X so plus 8x now we're going to multiply both of these terms by the one so 1 times x is equal to X or 1X so that's X Plus X and then we've got 1 times 4 well 1 times 4 is equal to 4. so what we're going to do now is we're going to simplify these two middle terms ax plus X is equal to 9x so our final answer would be 2x squared plus 9x Plus 4. and that's it and again we could use the grid to expand these brackets so we've got 2x plus 1 and X plus four well x times 2x well x times x is x squared and then we've got Times by 2 so that's two x squared then we've got x times one that's going to be plus 1X or plus X we've then got 2x times 4 well 2x times 4 would be 8X so plus 8X and then we've got 4 times 1 but 4 times 1 is equal to four so plus four and then we can just write it all out to x squared plus X plus eight X plus four and then collecting our like terms X plus eight X is 9x so we've got two x squared plus nine X plus four and that's it okay let's have a look at our next topic so our next topic is expand and three brackets and that's video number 15 in Cobra Maps so here's our question it says expand and simplify x minus two X plus three X plus five so expand these brackets what we're going to do is expand the first two get that answer and then multiply it by X plus five so let's start off with the first two brackets so we could use the grid I'm just going to use this approach x times x is x squared x times 3 well that's gonna be plus 3x so plus 3x minus 2 times X or minus 2 times x would be minus 2X and then minus 2 times 3 well negative 2 times 3 would be negative 6. so we've got x squared plus 3x minus 2x minus 6. and if we simplify the two terms in the middle our three x t equal to X well 3x take away 2x is X or 1X so we've got x squared plus x minus 6. so we've expanded our first two brackets now we need to multiply that by X plus five so we're going to multiply that by X plus 5 and then we'll see what we get now you could expand this using the grid approach but I'm just going to use this approach x squared times x well x squared times x is X cubed x squared times 5 well that'll be plus 5 x squared so we've done the x squared term now let's move on to the X term well x times x would be plus x squared x times 5 would be plus 5X and then finally we've got our negative six negative 6 times x is negative 6X and finally negative 6 times 5 would be negative 30. so we've expanded our brackets now we just need to simplify so we've got X cubed if we have a look at our x squared terms we've got 5x squared plus 1X squared that'd be 6X squared so plus 6 x squared if we look at our X terms we've got 5x take away 6X that's going to be negative 1X or minus 1X or just a minus X and then finally we've got negative 30 so we've got minus 30. that's it so if we were asked to expand and simplify x minus 2 X plus 3 X plus 5 our answer would be X cubed plus 6X squared minus x minus 30 and that's it now one thing to note if we had a question like this where we've got X Plus 1 and then X plus 3 squared what we would do is because we're multiplying the X plus 3 by itself you'd write out the brackets as X plus one and then in Brackets X plus 3 and then another X plus three and then you would just expand those in the same way we've done this by expand on the first two and then just multiplying by X plus three again and that's it factorize W squared plus eight W as you can see both terms have W so let's take W out so W squared divided by W that would just leave U of w and then we've got our plus sign and then 8w divided by W would just be 8. so our answer would be W bracket W plus 8 and you can test it by expanding the brackets so W Times W is W Squared and W Times 8 is 8w okay next one we've been asked to factorize 4y squared plus 6y so we're looking for common factors of 4y squared and 6y in terms of the numbers I can see I can divide 4 and 6 by 2. so I'm going to take a 2 out as a common factor and also in terms of the letters we've got y squared and Y so we can divide both of those by y so we're going to factorize this by taking out 2y so we're going to divide both of these terms by 2y and see what we get well 4y squared divided by 2y well that will leave me with 2y because 2y times 2A is 4y squared and then plus and if we had 6y divided by 2i that would just leave us with 3. so answer would be 2y bracket 2y Plus 3. okay our next topic okay so our next topic is factorizing quadratics in this video 118 on corporate Maps here we're going to be factorizing quadratics that's expressions with an x squared term where it is just an x squared term and whenever we factorize these quadratics what we need to do is figure out what the two brackets were that we expanded to get that so we want to figure out what two brackets we have multiplied together are expanded to get x squared plus 8X plus 15. what I'm actually going to do is I'm actually going to just pick two brackets to begin with okay and I'm going to call it X plus 4 and X Plus 3. and I'm going to expand these brackets so if we had the pair of brackets X plus 4 and X plus 3 and we expanded them we would get x squared plus 7x plus 12. now one thing that animus is that the 4 multiplied by the 3 gives us the number on the end the 12. so the two numbers in the brackets at the end the three and the four were multiplied together to give you the number at the end of the quadratic so here where we've got our x squared plus 8X plus 15. the number here and the number here were times together to give us 15. another thing to notice about these two numbers the four and the three four plus three is equal to seven they add together to give you the term in the middle so whenever we're factorizing this quadratic x squared plus eight X plus 15 well because it's x squared we know there's an X at the front of both brackets because x times x is x squared and in terms of the two numbers that follow the X's they will times together to give you the 15 on the end and they're going to add together to give you the eight in the middle so we want to find two numbers that we'll multiply together to give you 15 and add together to give you it now first of all I'm thinking here is going to be three and five because if you had three and five three times five is equal to 15 and 3 plus 5 is equal to n so let's just check if we had X plus 3 and X plus five if we expanded these brackets we would get x squared plus 5x plus 3x plus 15. and the 5x and the 3x would add together to give you the 8X so that's it so to factorize quadratic where it's just an x squared term at the front you're going to factorize and put your two brackets down you're going to put x's in the front of them and you're going to think of the two numbers that will times together to give you the number at the end and then we'll add together to give you the number in front of the X the coefficient of x okay let's have a look at for example so this time we've been asked to factorize x squared plus x minus 6. so we're going to have two brackets and we're going to put X at the front of both of them and we want to find two numbers in these brackets that will time us together to give you minus six and then we'll add together to give you the number in front of the X and as you can see here it's just X that means it's one so they're going to add together to give you one so we want to find two numbers so we'll times together to give you minus six and add together to give you one so let's think of numbers that times together to give you minus six so minus six we could have minus six times one we could have six times minus one we get a three times minus two or we could have minus three times two so these are all options so which were times together to give you minus six we want the one that we'll add together to give you one because it was plus X or Plus One X so as you can see here we've got minus six plus one well minus six plus one's minus five that's not going to be a right answer we've got six plus minus one that's going to be five that's not going to work you've got minus two plus three well minus two plus three is one so that's going to be our correct option we're gonna have X plus three and x minus two so that means our answer must be X plus three and x minus two okay let's have a look at our next question so our next question says factorize x squared minus eight X plus seven so we've got our quadratic and we're going to start off by putting our brackets down because we know that we're going to have brackets with x's at the front of both of them and we want to find two numbers that will times together to give us seven and then we'll add together to give us negative eight so let's start off by thinking of numbers there were times together to give a seven so it could be one and seven because one times seven is equal to seven but it could also be negative one and negative seven because negative one times negative seven is seven now we want the option where they will add together to give us negative eight so first of all I know it's not going to be the top option because they will add together to give us a year let's check our other option for negative one plus negative seven well negative one going down another seven would be negative eight so this will be our correct option so in the brackets we will have x minus one and x minus seven and that's it factorize five x squared plus thirteen X plus six so it's a quadratic and we're going to factorize it so there's going to be two pairs of brackets like so and we're trying to factorize something that has a 5x squared at the beginning that means that the front above brackets it won't just be an X and an X like it was in this case it's going to be a five x and then X because they will multiply together to give us our five x squared if we had something such as 4x squared it's a bit harder because you're then having to figure out whether it's 4X and X or 2X and 2x but with this one being 5x squared that's quite nice so we've got 5x and NX now then if we were expanding our brackets we would do 5x times x which is our 5x squared then we would do 5x times this number and then x times that number and they would add together to give us our 13x's in the middle and then finally the number here times the number here will give us 6. so a little times together to give us this number on the end so we know that we're looking for two numbers which will times together to be equal to 6 to go at the end of both brackets now they could be one and six they could be six and one it could be two and three or three and two um they it could also in theory be negative one and negative six and so on but they wouldn't work for this question because if we had 5x times negative one and then x times negative six we need to add them together it wouldn't give you a positive term in the middle so I know that because this is positive 13x these are going to be positive numbers here I'm looking at this bottom option here because I know if I use two I can do 5x times 2 which is 10x and then if I put my plus 3 here that will give me x times 3 which is 3x and if you add them together you get 13x so let's just check it 5x times x is 5X squared 5x times 2 is plus 10x 3 times x is plus 3x and 3 times 2 which is plus 6. and then if you add the terms in the middle you get 5x squared plus 13x plus 6. that's it and our answer would be 5x plus 3 X Plus 2. so the main thing with factorizing quadratics is practice the more often the factorize quadratic the better you'll become added so you get to spot how to put those numbers in really quickly and easily okay so if you want to factorize this using the split in the middle technique I'd highly recommend you watch that video in corporate miles but the technique is this first of all you take the number in front of the x squared this 5 the coefficient of x squared the number in front of the x squared the five and you multiply that by the number on the end of the quadratic the six so five times six is equal to Thirty so you'll do five times six is equal to 30. now we're going to want to find numbers which will times together to be 30 but they will add together to give you the term in the middle which is 13. so in terms of numbers which will times together to be 30 and add together to be 13 they would be 3 and 10 because 3 times 10 is 30 and 3 plus 10 is 13. so what we're going to do is we're going to split this middle term up this 13x into a 3X and a 10x and if we do that we get 5 x squared now we could write plus 10x plus 3x or we could write plus 3x and plus 10x and it doesn't actually matter which way around you do that and I'll show you that in a minute so let's do plus 10x and then plus 3x and then we've still got our plus 6 on the end so we've split this middle term up into 10x and a 3X so what we're now going to do is we're going to factorize the first two terms of this expression and we're going to factorize the second two terms of this expression so starting off with the first two terms of this expression we'll go 5x squared plus 10x now you can see that these two terms have 5x is a common factor so we'll write five x and then we'll do a bracket and then factorizing this by taking out the 5x we'd leave us with an X to begin with because 5x squared divided by 5x is just X and then 10x divided by 5x is plus 2. now let's factorize the second two times this expression so we've got 3x plus 6 well we can divide both of these by three so we'll write plus three and then open up our brackets and divide them both of them by 3 will also give us X Plus 2. and that's great because we've got the same bracket this X plus 2 twice we've got five x lots of X plus two and we'll give three lots of X plus two now because we've got five x lots of it and we've got three lots of it all together that would be five X plus three lots of X plus two and that factorizes our five x squared plus 13x plus 6 because as you can see that's what we got whenever we factorize it using inspection okay so that was one approach I just want to show you that if you had written it as plus 3x plus 10x you'll also get the same answer so this time let's write our five x squared plus three X plus ten X so splitting up the 13x into 3x and 10x plus 6. now we're going to factorize the first two terms so we've got our 5x squared and our three x well the common factor there would just be X so we'll take X out as a common factor so dividing both of them by X will give us five X plus three and then our second two terms we've got ten X plus six well if we had 10x Plus 6 the common factor would be 2 so we'd write plus two and then bracket 5 X plus three and what's great is again there we've got the same bracket twice we've got X lots of five X plus three and we've got two lots of five X plus three so all together that would be X plus two lots of five X plus three so as you can see there are two different approaches to factorize this quadratic okay our next question is to factorize seven x squared plus twenty x minus three so if I was using inspection I'd put my brackets down like so and put a 7x and an X now we know the numbers were times together to be negative three so it's going to be one and negative 3 or negative one and three or three and negative one or negative three and one so they would be my options but I know that I want to get 20x whenever I expand so whenever I do 7x times that number and x times that number whenever I add them together they were simplified together to be 20x now straight away I know that if I want 20x it'd be really great at positive 20x out if I put my plus 3 here I would have 7x times 3 which is 21. then if I had a minus 1 here I would then have x times -1 I'll bring me down to 20x so that would be it so my answer would be 7x minus 1 and X Plus 3. alternatively we could use the split in the middle technique and I'll just do that in blue ink now so 7 multiply by minus three is that'd be minus 21 so we want two numbers which will times together to be minus 21 but we'll add together to be 20. and straight away I'm thinking 21 a minus one so 21 times -1 will be minus 21 but also minus 1 plus 21 would be 20. so there would be our two options so we've got our 21x and our X so let's split up the middle term so we could write seven x squared minus X Plus 21 x minus 3. so we've split up our middle term our 20x into a minus X and A 21x now let's factorize the first two terms so if we had 7x squared minus X we could divide both of those by X that would be X bracket 7x and then taking X out here would be -1 I'm going to refactorize these first two terms so we get X bracket 7x minus one and our second two terms we can divide both of those by three so we would write plus three bracket and divide both of these by three would give us seven x minus one and that's great because we've got the same bracket twice so putting our terms in front are X plus three in a bracket we would get X plus three seven x minus one and that's it now this actually isn't really a different topic this is really a special case so this is the difference between two squares and this is video 120 on corporate maths so what I'm going to do is I'm going to expand these brackets where I've got x minus 7 and X plus seven now I've just used these brackets where we've got our x minus and an X Plus and then the same number and I just want to show you what happens whenever you have this this case x times x is x squared x times positive seven will be plus seven X then we've got minus seven times x that's minus seven X and then we've got minus seven times seven well a negative times a positive is a negative and 7 times 7 is 49 so it's gonna be minus 49. and whenever we look at our two middle terms here we've got seven x minus seven X now they will give you zero whenever you do seven X take away seven X that's zero so we would be left here with x squared this and this cancels out and we're just left with minus 49. so if you're expanding brackets where you've got an x and x and then the same number but then one with a plus and one with a minus sign the two middle terms will cancel out now that's very useful because sometimes we can factorize expressions like this and it's called difference between two squareds because whenever we expand these brackets the answer will always have a squared term so x squared and it's maybe a square number and then you'd have a takeaway in the middle so that's your difference you're taking away the difference between two squares okay let's have a look at some factorizing using this then okay so our first question says factorize x squared minus 49 Okay so we've just done that one actually so we know whenever we expanded x minus seven X plus seven we got x squared minus 49. so if we're asked to factorize this so our answer would be x minus seven and X plus seven and the shortcut would be well we had x squared so the square root of x squared is just X so that goes at the front of both of them and then we've got 49 so you're square root 49 that's seven so you know that it's going to be seven at the end of both brackets and then you put one with a plus sign and one with a minus sign and that's it so if I wanted to factorize x squared minus 9x I would put two brackets down I'd square root both terms because they're both squares so square root X where just x and x the square root of nine is three and three and we put one with a plus sign and one with a minus sign and it doesn't matter which way around they go and that's it so if we were to factorize x squared minus 9 the answer would be X plus three x minus three okay next we've been asked to factorize x squared minus 36 so again bracket bracket bracket bracket x x because the square root of x squared is X and then 36 well the square root of 36 would be six so six and six one for a minus animal over plus sign that's it okay now the last one our last one is the factorize 64 minus x squared and this is just to show you that it's the difference between 2 squared so it doesn't have to be x squared at the beginning it could be the other way around so let's factorize and we'd get the square root of 64 is 8 so 8 is going to be at the front of both brackets we've got our two squares so one's going to have a plus sign and One's Gonna Have A minus sign and the square root of x squared is X so it's going to be 8 plus X and 8 minus X and that's it our next question our next question is to factorize fully 20x squared minus 125 y squared and this is video 120 now so this is the difference between two squareds and as you can see here we've got 20 x squared and 125 y squared so we've got the x squared and the Y squared but we don't have square numbers in front of them we've got 20 and 110 25 so let's see if we can actually factorize by taking out any numbers to begin with now because we've got 20x squared minus 125 y squared I can see that both of these the 20 and the 125 have a factor of 5. so let's take 5 out as a common factor and let's divide both of them by five so it would be 4 x squared minus 25 y squared now that's great now this is difference between two squares which you can see you've got four x squared and you get 25 y squared and the 4 and 25 are both square numbers so would be so put our 5 down and now remember the difference between two squares you put your brackets down you square root both of them so square root the 4x squared that'd be 2X and that'll be the front above brackets and square root and 25y squared that'd be 5y and 5y and you put one with a plus sign and one with a minus sign and that's it okay we're not going to look at equations and we're going to deal with equations with letters on both sides so here's an equation with letters on both sides and we've got 5x plus 1 equals 3x plus 19. now whenever I'm solving equations with letters on both sides I want to get rid of the lowest number of x's now if you look at this the lowest number of x's would be the 3x so I'm going to get rid of this 3x remember we do the inverse to both sides so if I want to get rid of 3x I'm going to take away 3x so minus 3x and minus 3x so if we had 5x plus 1 and we take away 3x that leaves us with 2x we've still got our plus one and on the right hand side of the equation we had 3x plus 19 we've taken away the three X's so that just leaves us with 19. so we can solve this by taking away one and taking away one so that would leave us with two x equals 18 and then we can divide by 2 and divide by two and see that X is equal to nine and again we can check our answer 5 times 9 is 45 plus 1 is 46 and 3 times 9 is equal to 27 plus 19 is 46 so that's right okay next question so our equation this time says 4X plus 5 equals twenty subtract X now whenever I'm solving equations with letters on both sides I always want to get rid of the lowest number of x's now minus X is lower than 4X so I'm going to get rid of the minus X now if I want to get rid of minus X I'm going to do the opposite which is ADD X to both sides of the equation so I'm going to add an X and add an X so on the left hand side I had four x's and I add an X or one more X it's going to be 5x plus 5. the numbers will stay the same equals and on the right hand side of the equation I had 20 minus X I added X so the minus x and x will add together to give you zero so they're just going to this appears you're just gonna be left with 20. so we've got 5x plus 5 equals 20. and this is just like the equations you will have seen in the M1 video so we've got a minus five minus five so that will give us five x equals 15. and then finally we're going to divide by 5 and we're going to divide by 5 and that gives us x equals three and that's it okay let's have a look at our next topic so our next topic is forming equations so here we've got some information we're going to form an equation out of it so with Daniel is X years old Daisy is five years older than Daniel so she's five years older than him so he's X so to find her age you would do X plus 5 so we'd write X plus five that's Daisy's age and Chris is twice Daniel's age So Daniel has X and we're going to times it by two to get Chris's age that would be 2X and the sum of their ages is 53 and our first question says a form an equation using the information given so we've got the sum of that ages is 53. remember the word sum means to add up so if we add up our X our X plus 5 and our 2x we will get the sum of their ages and we know that's equal to 53. so X plus X plus 5 plus 2X the sum of their edges equals 53. so let's collect our like terms let's do X Plus X plus two x's so X Plus X is 2x plus another two x's is four x's and we've still got our plus five and that's equal to 53. so that is our equation we've got an equation 4X plus 5 is equal to 53. and part basis solve the equation to find Daniel's age so we're going to solve this equation so let's then get rid of this plus 5 by take away five and take away five so we'll leave us on the left hand side with 4X because we took away 5 to get rid of the plus five on the right hand side of the equation we had 53 take away 5 is equal to 48. now we've got 4 times x is equal to 48 well we don't want this multiplied by 4 so let's divide by 4 and divide by 4. so 4X divided by 4 is just X and on the right hand side we had 48 divided by 4 that's equal to 12. So Daniel was X years old so we know that X is equal to 12 so Daniel is 12. so the next topic is formal equations and that's videos 114 and 115 on corporate Maps so here we've got an isosceles triangle and we've been told the angle at the top is 2X and we've been told the angle at the bottom left is X Plus 20. now from where the sides of the mark the same on the left hand side on the right hand side I can see that this angle is equal to this angle so this is going to be X plus 20 as well now the angles in a triangle add up to 180 degrees so if we add up all these terms for the angles the 2x the X plus 20 and the X plus 20 we'll find that equal to 180 degrees so let's do that so let's let's take our 2X and add our X plus 20 and add another X plus 20 and that will be equal to 180 degrees so let's add up our X's we've got 2x plus X Plus X that's four x's and then we'll put 20 plus 20 so that's going to be plus 40. and that equals 180. now we've got 4X plus 40 equals 180 so let's take away 40 and take away 40 so take away 40 and take away 40 leaves us with 4X equals 140 and then we want to get X on its own so we're going to divide by 4 and divide by 4 and that gives us x equals and 140 divided by 4 would be 35. if the question asks us to find the size of the angles then you would substitute the 35 into these Expressions so 2 times x would be 2 times 35 so this angle would be 70 and if we took our 35 and added 20 this angle would be 55 and this angle be 55 so our angles would be 70 55 and 55. but the question didn't ask is that just asked us to find X and that's what we've done okay so our next topic we're going to look at solving quadratic equations and that's video 266 in corporate Maps so to solve this equation we're going to use the technique we just looked at that factorization so if you have a quadratic equation y equals zero you can try to factorize the left hand side and if it factorizes then you can solve it really quickly and easily so let's factorize x squared plus 14x plus 45 so we'll put our brackets down bracket bracket bracket bracket and we're going to put X's at the front of both of them and we're going to look for two numbers that will multiply together to give us 45 and we'll add together to give us 14. you could do 5 times 9 5 times 9 is 45 and they add together to be 14. fantastic so we've got X plus 5 and then we've got X Plus 9 equals zero so we've factorize this quadratic and we've got in Brackets X plus 5 and then X Plus 9 equals zero now whenever you get two things that multiply together to give you zero one of them has to be equal to zero so that means that either this bracket's equal to zero or this bracket's equal to zero so let's look at this bracket so you have a number plus five X plus five equals zero so X plus five equals zero or X plus nine is equal to zero either that bracket's equal to zero so let's solve this equation if you had X plus five equals zero well you could subtract five from both sides and you'd get x equals negative five and in this case you had X plus nine is equal to zero so if you subtracted nine and subtracted nine you'd get x equals negative nine so if you had this equation x squared plus fourteen X Plus 45 equals zero we've got two possible answers and they are x equals negative five or x equals negative nine and one other thing to note is sometimes I take a little bit of a shortcut whenever I'm doing these questions so if I had X plus five and I'm trying to find when that bracket is equal to zero I just know it's a negative five because negative five plus five is zero so I would often take a bit of a shortcut so whenever I'd be doing these questions I would often just go straight to this point here so I would say well X plus five zero so x equals negative five or and in this case I know that it'd be negative nine plus nine is zero so I just jump to x equals negative nine and that's it okay so let's have a look at our next question so next question is to solve x squared minus four x minus twelve equals zero and that's great because the equation the quadratic equals zero if it's not equal to zero you'd want to make it equal to zero and then let's try and factorize this left hand side so let's put our brackets down and put equals zero so we put our X's at the front of both brackets and we're trying to find two numbers so we'll multiply together to give us negative 12 and then I'll add together to give us negative four and two minus six when negative six plus two would be negative four so that's going to be our option so our brackets will be X plus two and x minus six now we're trying to find out when each of these brackets could be zero we know this bracket would be equal to zero whenever X is equal to negative two because negative two plus two is zero and for this bracket we know it'll be whenever X is equal to six so it would be x equals six and that's it so x equals negative two or x equals six so we've been asked to solve 3x squared minus ten x minus 48 equals zero so what I'm going to do here is I'm going to factorize this left hand side so let's put our brackets down bracket bracket bracket bracket equals zero and we've got three x squared so let's go for a 3X and then X now we know the two numbers were multiplied together to be -48 and that means that one's going to be positive and one's going to be negative and then whenever we expand our brackets we're going to get minus 10 x in the middle plus a minus six well three x times x is three x squared 3x times minus six would be minus 18 x x times eight is plus 8X fantastic and then minus 48 so that would be 3 x squared minus ten X minus 48. so now we've factorized let's then solve it so because we know that these two brackets will multiply together to give a zero that means that the brackets will be equal to zero so either this bracket is zero or this bracket zero now this bracket is quite easy to solve we know that X will equal 6 there because six minus six is zero this bracket whenever I have a 3X it plus 8 equals zero I tend to write three X plus eight equals zero and then just minus C it from both sides so three x equals minus eight and then divide by three x equals minus eight thirds like so so there are two solutions the either x equals minus eight thirds or x equals six and that's it so we have solved a harder quadratic equation by factorizing okay so our next topic is the quadratic formula and that's video 267 on corporate Maps now sometimes we're given a quadratic that can't be factorized so we solve it by using the quadratic formula so if you've got a quadratic in the form ax squared plus BX plus C equals zero where a is a number in front of the x squared so as the coefficient of x squared B is the coefficient of x the number in front of the X and C is the number on the end then you can substitute them into the quadratic formula and it will tell your solution straight away and the quadratic formula is x equals negative B plus or minus the square root of B squared minus 4se all over 2A and this is part of the chord miles revision card and here's our example we've been asked to solve 2x squared minus x minus 9 equals zero now sometimes in a quadratic formula question it will say solve 2x squared minus x minus 9 equals 0 and give your answers to one or two decimal places and that's a clue that you're going to be using the quadratic formula so let's label our a b and c to begin with so a equals b equals and C equals so a is the number in front of the x squared so a is going to be equal to two B is the number in front of the X so it says minus X that means minus 1X so that means that b is equal to minus one and finally season number on the end which is our minus nine so we have got a B and C now we're going to substitute those into the quadratic formula so x equals negative B well B is minus one so we'll get negative B well if it's already negative that's going to make it positive so that's me one plus or minus the square root of B squared well minus 1 squared is going to be minus 1 squared minus 4 times a and I just put it in Brackets so that's 2 times C which is equal to minus 9 like so and that's all divisible by 2A and 2 times a is equal to 4. so we've got x equals 1 plus or minus the square root of minus 1 squared minus 4 times 2 times minus 9 all divided by 4. so now let's simplify what's under the square root sign so we've got x equals 1 plus or minus the square root of we've got minus 1 squared well negative 1 times negative 1 is 1 so it's going to be one and then we've got subtract and whenever I work out this 4AC part I always put my answer in a bracket so we've got 4 times 2 times minus nine and four times two is equal to 8 times minus nine is minus 72. so we've already got our minus sign outside here so we've got 4 times 2 which is eight times minus that which is minus 72 so that goes there and then all divided by 4. next we've got 1 minus minus 72 that's going to be 1 plus 72 which is 73 so we're going to get x equals 1 plus or minus the square root of 73 all divided by 4 and now I'm actually just going to move this down here so just pulling it down here now here we're going to have two solutions we're going to work out one plus the square root of 73 divided by four and we're going to do 1 minus the square root of 73 divided by 4. so let's write those out separately so we're going to have 1 plus the square root of 73 divided by 4 and we're going to have or x equals 1 minus the square root of 73 divided by 4. and so let's just work those out okay let's see what we get so we get x equals 2.386 or x equals negative 1.886 and that's it so we've used the quadratic formula so x equals negative B plus or minus the square root of B squared minus 4AC all divided by 2A the silver quadratic which couldn't be factorized okay let's have a look at our next topic so our next topic is complete in the Square now we've looked at how to solve quadratics by using factorization and by using the quadratic formula and completing the square is a technique that we can use to solve quadratics and also to help us with things such as finding the turning points of quadratic graphs and this videos and corporate maps of videos 10 and 267a so first of all we've been asked to write the quadratic x squared plus 6X plus 15 in the form of X plus a close bracket squared plus b so right in this quadratic in this form is what we call completing the square so to do that what I do is I first of all start if I open up my brackets and put my x inside the brackets because it's x squared I'm just putting X inside the bracket because when I multiply this bracket by itself we'll get the x squared then to get the a what we do is we half the coefficient of x so the coefficient of x is the number in front of the X which is six so we have six that'll be plus three so that's the value of a three if it was plus 10x we would then have plus five if it was x squared minus 8x we would have x minus four halving the coefficient of x then we have r squared now whenever we expand this brackets when we multiply it by itself when we get x squared plus 6X plus 9. what I'm not going to do is I'm going to take away 9. and I'm taking away a 9 to just leave us with the x squared plus 6X part of the quadratic and to get this now and this minus 9 what we do is we take away whatever this number is squared so for instance if it was plus 6 we would take away 36 if it was plus seven we would take away 49 if it was minus 4 well minus 4 squared is 16 square minus 16 is always take away whatever this number is squared and then we just put the plus 15 on the end so this plus 15. so we've got x squared plus 3 close bracket squared minus 9 plus 15. now we we just want plus b so we don't want the minus 9 plus 15 so we're going to do minus 9 plus 15 which is 6. so that would be X plus 3 close bracket squared and then we've got minus 9 plus 15 which would be plus 6 and that's it so we've written this quadratic in the form of X plus a close bracket squared plus b where I've got X plus 3 close brackets squared plus six that's it right so now let's have a look and see how we can use completing the square to solve quadratics and we've got x squared plus 6X plus 2 equals zero so let's solve it using completing the square so let's complete the square for this quadratic to begin with so that would be X and then half of 6 would be three so plus 3 close bracket squared we then take away this number squared so 3 squared is 9 so take away nine and then we've still got our plus two and that equals zero now negative nine plus two well negative nine plus two would be negative seven so we've got X plus three close bracket squared subtract 7 equals zero so we now want to solve this equation so we want the X on its own so let's add 7 to both sides of the equation so that would give us X plus three close bracket squared equals seven so we now want to get rid of the squared so we're going to square root both sides of this equation but just be careful when we're square root them we can have the positive or negative solution so we're going to have on the left hand side X plus three and on the right hand side of the equation we're going to have the square root of seven over root seven but it'll be positive or negative because root seven times root seven is seven or a negative root seven times negative root seven is equal to 7 as well so I'm going to write plus or minus the square root of seven and then finally we want the X on its own so we want to get rid of this three so we're going to take away 3 from both sides of this equation so we're going to take away 3 and take away three on the left hand side that would just leave us with X and on the right hand side whenever I'm taking away 3 here instead of writing plus or minus root 7 take away 3 I tend to put the negative 3 at the front so I then have negative 3 plus or minus the square root of seven and that's it so that means that X is equal to minus 3 plus the square root of seven or X is equal to negative 3 minus the square root of seven and that's it so this is a great way to solve quadratics that can't be factorized instead of using the quadratic formula and that's it okay next topic is changing the subject so here we've got a formula we've got T is equal to a w take away C and at the minute the capital t is the subject because it's on its own the t is on it so and there's so T is the subject but we want to make W the subject so what that means is we want to get the W on its own so we don't want this to take away C and we don't want this to multiply by a so what we're going to do is like an equation and we're going to get rid of this takeaway C and multiply by a so let's write out our formula we've got T equals Aw take away C now we want to forget the W on its own so let's get rid of this take away C to begin with so if I wanted to get rid of take away C I'm going to add C to both sides so add C and add C well t plus C was just t plus C we just write it out and then the right hand side of our formula but we had aw minus C we added C to get rid of the minus C because minus c plus c is zero so we would just be left with aw now we want to make W the subject which means we want to have the W there on its own this is a times W we're multiplying the W by a we don't want to multiply the W by a so let's divide both sides by so let's divide by a and divide it by a on our right hand side we had aw we divided by a to get rid of it so we'd just be left with W on the left hand side of this formula we had t plus C and we're dividing that by a so what we're going to do is write t plus C now remember in algebra we don't write the divided by sound and we do over a and that means we've got t plus C divided by a and that's it so we've got WRX W as a subject so our answer would be W equals t plus C over a and that's it and changing the subject is video seven on corporate map so if you want to watch more examples of continue the subjects starting off with some simple questions and building up to questions like this video seven will be really really useful and also remember there's the practice questions in the textbook exercise there as well if you do have the revision cards changing the subjects is one of the revision codes and as you can see here we've got a typical change in the subject question where it says make W the subject and it goes through step by step explaining how you would make W the subject of that formula so changes very important topic it's video seven on corporate maps and if you do have the revision cards that revision card will be really useful for it okay so let's have a look at this exam so we've got make y the subject of w equals y plus 2 over y so what I'm going to do is I don't want to have this y on the denominator so I'm going to multiply both sides of this by y so multiply by Y and multiply by y w times y will be w y just putting them beside each other and on our right hand side we had y plus 2 divided by y we multiply by y to get rid of the divided by y so we're just going to be left with Y Plus 2. we've now got YW equals y plus 2. now we want to make y the subject which means we want to have so at the end y equals something but here we've got two terms with y's in it we've got wi and we've got the Y so what we want to do is we want to bring them over together on the same side so what I'm actually going to do is take away y from both sides of this formula and what that will do is bring the Y's to the left hand side here so if we're wi and we're going to take away y so we're going to write w y minus y so we've brought all the Y's over to the left hand side and we've got the plus 2 or just 2 on the right hand side so we've got the terms with y's on one side and everything else on the other side now we want to make y the subject now we've got w y minus y we can actually factorize this because they both have y's in them so we can write Y and then brackets and we can factorize this by dividing both of these by y so w y divided by y would just leave us with w and then we've got minus and Y divided by Y is one so if we were to factorize Wy minus y taking the Y out we'd leave us with Y bracket W minus 1 and that equals 2. now we're almost there we've got y times W minus 1 equals two but we don't want the W minus 1 there we just want the Y so if we want to get rid of this W minus 1 what we're going to do is we're going to divide both sides by W minus 1. so dividing the left hand side by W minus one well that would just leave us with Y which is great and on the right hand side we had two and we're going to divide it by W minus one so we'll just write that down we'll write 2. now remember we could donate the divided by sine and algebra we write the big line so over and then we write w minus 1 on the denominator so we've got 2 divided by W minus one so that's our answer we've got y equals 2 over W minus 1. and let's just write that out 2 over W minus one so we've made y this objective and these are videos seven and eight and copper Maps changing the subject and changing the subject Advanced and this particular type of question would be video hit the advanced video so I'd highly recommend watching that video on corporate maps and there's practice questions there I've actually updated those recently and there's actually video solutions to those so whenever you go through those questions you want to see have done each question I've actually made a video talking about every single question in that booklet okay next topic is algebraic fractions and that's videos 22 23 and 24 on corporate Maps so here we've got an algebraic fraction it's a fraction with algebra and we've been asked to simplify that means to cancel it down and whenever we canceling down fractions so we tried to divide the numerator and the denominator by the same thing so they get simpler so as you can see here if we look at the top here we've got x minus 3 times X plus 2 and on the bottom we've got X plus 2 plus X Plus 9. as you'll notice we've got two brackets to the same we've got this X plus 2 and X Plus 2. so if we divide the numerator of this fraction by X plus 2 and the denominator of the fraction by X plus 2 we can cancel it down so let's divide it by X plus 2 and that just means we're canceling it out would give us x minus 3 on the top and X Plus 9 on the bottom okay so next question is to simplify x squared minus 2x over x squared plus 2X minus in so this time we've got to factorize before we can handle it down so let's factorize it so our numerator we can just take X out as a common factor so that's going to be X bracket and then if we divide both of these by X we get x minus 2 close brackets and then the denominator we can factorize it so it's going to have two brackets we can have X's at the front of both of them and we want to find two numbers that will times together to give us minus 8 and add together to give us two now I'm thinking minus two and four because minus two and four they were multiplied together to give us our minus here and they will add together to give us R2 so that means it's going to be x minus 2 and X Plus 4. and that's great because as you can see we've got x minus 2 on the numerator and we've got x minus 2 on the denominator so we can divide both the numerator and denominator by x minus 2 so that means they'll cancel out and then you'll be left with X on the numerator and on the denominator we'll be left with X plus four and we don't write the brackets we just write X plus four and that's it okay our next topic our next topic is algebraic fractions but this time we're going to look at multiplying algebraic fractions so remember to multiply fractions we multiply the numerators and multiply the denominators so we've got simplify X over 7 multiplied by X over three so let's multiply the numerators well x times x is x squared and on the denominator we've got 7 times 3 that's 21. so answer would be x squared over 21. this time we've got 3x over 2 multiplied by 5x over 6. so this time again we're going to move through our numerators 3x times 5x well that's going to be 15 x squared and then multiplying our denominators 2 times 6 is equal to 12. now one thing I notice on this fraction is we've got 15 on the numerator and 12 in the denominator now they're both divisible by 3. so we can actually cancel down this fraction by dividing both the numerator and denominator by three so that will give us if we divide both of them by 3 5x squared over four and that's it so this would be 5x squared over 4. okay let's have a look at our next question so our next question says to simplify 1 over 2x multiplied by x squared over three so again multiplying the numerator 1 times x squared is x squared and then the denominator 2x times 3 would be 6X so we've got x squared on the numerator and we've got 6X on the denominator so we can actually cancel down this fraction so you could divide the numerator and denominator by X so if you did that you'd be left with on the numerator X and on the denominator if you divide this by X you'll be left with six another way to look at it would be you could just divide the x squared by X and that would leave you with X and then again that would just leave you if x over 6. this time we've been asked to simplify x minus 3 divided by X plus 4 multiplied by X plus 4 over X Plus 1. so let's multiply these together so I'm going to write x minus 3 multiply by X plus 4 over X plus 4 multiply by X plus one now one thing I've noticed is we've got the same bracket on the numerator and denominator so we can cancel down this fraction so we can cancel those out and that would leave us with x minus 3 on the numerator and X Plus 1 on the denominator and that's it okay let's have a look at some questions now where we're dividing algebraic fractions so we've got x minus 2 and we've got divided by remember whenever we're dealing with fractions we multiply by the reciprocal of the fraction we're dividing by so we're going to change the divided multiply and we're going to refine the reciprocal which means flip over the second fraction and then we just multiply so if x times 3 is 3x and then on the denominator we've got 2 times 2 which is four so our answer would be 3x over 4. so when we're dividing algebraic fractions we change the divider multiply I find the reciprocal of the fraction we're dividing by okay our next question we've been asked to simplify 3x plus 9 over 2 divided by X plus 3 over 4. so again we're going to keep the first fraction the same 3x plus 9 over 2 times by and finding the reciprocal of the second one we're going to have 4 over X Plus 3. now when we multiply these together we get 4 bracket three X plus nine close brackets over two bracket X plus three just putting them together now one thing I've noticed is we've got a four and a two so we can divide both of the numerator and denominator by 2 and that would give us on the numerator if we divide the numerator by two we get 2 bracket 3 X plus nine and on the denominator we just left with X plus three now another thing is 3x plus 9 will also factorize if we had three X plus nine we could take three out as a common factor they're both divisible by three and it would give us three bracket and divide both of them by three would be X plus three so if you've got three X plus nine that's the same as three bracket X plus three but we had two lots of that so instead of having three lots of X plus three we're going to have six lots of X plus three so it's gonna be six bracket X plus three another way to look at it is if we take the three out we could do two times three is six and that's why there's a six in front there but that's divided by X plus three now another thing to notice is we've got X plus three on the numerator and the denominator so we can cancel those out so answer would just be six and I think that's fantastic we had this complicated algebra and the answer would just be six now I want to sort of show you another approach this question so one thing to notice is also rather than multiplying and then counting down you can actually cancel down at this point here by canceling down the numerator of this fraction in the denominator of this one and the numerator of this fraction in the denominator of this one because when we multiply them they're going to be together on the numerators and the denominators anywhere so a few 2 and 4 you could divide both of those by two that would give you one and two and here if you have 3x plus 9 and X plus 3 you could factorize this to get free bracket X plus three and then if you cancel down these ones when you get divide both of them by X plus three so you'd be left with here three over one multiplied by two over one because obviously you divide this by X plus three you'd be left with one there and the answer would again just be six over one which is six so as long as there's a multiplication you cancel that in that technique where you cancel down the numerator with the denominator another one and the numerator by the other denominator and so on Okay so we've looked at simplified algebraic fractions we've looked at multiplying them and dividing them let's look at and then subtracting them so our question says simplify X plus four over three plus X plus one over two so to add algebraic fractions they need to have the same denominator if we had three and two a common denominator would be six so six and six and to have a six on this denominator we would need to double but if the numerator and the denominator to get from three to six you multiply by two so we need to multiply this by two multiplying both of these terms by 2 would give us two X plus eight and to get from two to six you'd multiply by three so you need to multiply these by three and we'll find these both by three would give us well 3 times x is 3x and 3 times 1 is plus three now we can just add them together so you know we're gonna have six on the denominator and two X plus three would be 5x and 8 plus 3 is equal to 11. so our answer would be five X plus eleven over six okay so our next question so our next question is X plus one over four subtract X plus four over six so if we want to find a common denominator well four and six well I think 12 would be a good choice so 12 and 12 and let's take away now to get from 4 to 12 we multiply by three so we need to multiply both of these by three so three times x is three x and three times one well plus three and to get from 6 to 12 we double it so we need to double the numerator so doubling X and four will be two X plus eight and again we want to subtract so we're going to have equals and we're going to have our denominator of 12. in terms of the numerator we've got 3x take away 2X and 3X take away 2x is 1x or X and then we've got 3 subtract 8 and 3 subtract 8 will be minus five so answer would be x minus 5 over 12. we've been asked to express as a single fraction two over three X plus one plus one over x minus 1. now whenever adding together fractions we want the same denominator and to do that what we're going to do is we're going to multiply above the numerator and denominator of the first fraction by the denominator of the second fraction so we're going to multiply both of these by x minus one and for our second fraction we'll multiply both the numerator and denominator by the denominator of the first fractions so we'll multiply both of these by three X plus one so whenever we do that we get well we'll write 2 and then we're going to multiply our numerator by the denominator of this fraction so 2 bracket x minus 1 like so and then our denominator would be well we would have three X plus one in a bracket and then we're going to put another bracket beside it with x minus one like so so we've multiplied above the numerator and denominator both for x minus one now we're going to multiply both of these the numerator and denominator by three X plus one so when we do that we get we're multiplying the numerator by 3x plus 1 because it was a one we could just write three X plus one because we know that we're just multiplying it by one and then with our denominator of x minus one we're multiplying that by our three X plus one so we've got x minus 1 bracket three X plus one and that's it and as you notice we've got the same denominators for these fractions so we've got three X plus one times x minus one and here we've got three X plus one times x minus one so now that we've got the same denominators we can combine our fractions as a single fraction so we'd have our numerator so we've got two bracket x minus one and then we've got our second numerator so plus and then three X plus one and then we've got our denominator which is just three X plus one bracket x minus one like so now let's simplify our numerator so let's expand that bracket so 2 times x is 2x 2 times minus one that's minus two and then we've still got our plus three X plus one and our denominator is still three X plus one and x minus one and then finally simplifying our numerator well we've got 2x plus 3x that's going to be 5X and then we've got minus two plus one well minus two plus one is minus one and our denominator was three X Plus One X minus one like so and that's it so we've expressed two over three X plus one plus one over x minus one as a single fraction as five x minus one over three X plus one times x minus one like so Okay so we've looked at algebraic fractions and now we're going to look at some equations that involve algebraic fractions so we've got solve x plus five over three plus X plus one over two equals it so whenever you've got an equation like this um I would deal with the left hand side to begin with so I would look at these algebraic fractions and I would simplify them so we're going to give them a common denominator of six and six to get from three to six you multiply by two so let's multiply both of these by two so that would be two X plus ten on the numerator and for this fraction to get from two to six you multiply by three so multiplying both of these by three would give us three X plus three equals it so as you can see we've written both fractions with a common denominator okay so now let's focus on this left hand side and write this as a single fraction so let's just put a big line and six on the denominator and we're going to do 2x plus 3x 2x plus 3x is 5X and 10 plus 3 well 10 plus 3 is 13 so plus 13. and that still equals eight so we have just combined this left hand side as a single fraction now we want to solve it so because we've got this big divided by 6. I want to get rid of that so I'm going to multiply both sides by six I'm going to multiply the left hand side of this equation by 6 and the right hand side of the equation by 6. so I'm multiplying the left hand side of the equation by six to get rid of the big divided by six so it would just leave me with five X Plus 13. and on the right hand side of the equation I had 8 times 6 and 8 times 6 is 48 so that equals 48. no I just want to solve this and so subtract 13 and subtract 13 and that will give me five x equals 35 and divide by 5 and divide by 5 and you get x equals seven and that's it so here's our example it says solve 2 over x minus 3 plus 1 over x minus 4 equals two so as you can see this left hand side we've got two algebraic fractions and we're going to add them together and to combine them as a single fraction so we'll add them together to write them as a single fraction so let's do that so let's multiply above the numerator and the denominator of this fraction by x minus four and then we'll multiply both the numerator and denominator of this fraction by x minus three so doing that would give us two bracket x minus four over x minus 3 times x minus four so multiplying both of these by x minus 4 and then plus and we're going to multiply both of these by x minus 3. now because it's a one I'm just going to write x minus 3 on the numerator if there was a number there I would put it at a bracket and put the number in front then over and then I had x minus 4 and then I'm times that by x minus 3. and although the brackets are in a different order remember a times B and B times a is the same thing that equals 2. okay now we're going to combine them as a single fraction because they've got the same denominator I can write my x minus three x minus four my fraction I went up a bit there so I can write my x minus 3 and x minus 4 and on the numerator I have 2 bracket x minus four plus and then we've got our x minus three and that equals two so I've just written the two fractions as a single fraction but with the same denominator no what we're going to do is we're going to simplify this numerator so let's expand the brackets so 2 times x is 2X and 2 times -4 that's minus 8 plus x minus three over x minus three x minus 4 like so and that equals two so just expanding these brackets now let's simplify the numerator 2x plus X is 3x and then we've got minus eight minus three that's minus 11 over x minus three x minus four and that equals two so now we've got 3x minus 11 over x minus 3 times x minus 4 equals two now I don't want these brackets on the denominator here I want to actually get these off the denominator so what I'm going to do is multiply both sides of the equation by x minus 3 and x minus 4. so multiplying both sides by x minus 3 and x minus four so whenever we do that well we're timesing this side by x minus 3 and x minus 4 to get rid of it on the denominator so we just be left with three x minus eleven and then on the right hand side where we're multiplying two by x minus 3 and x minus 4 so we would have 2 bracket x minus three and x minus four so now we're going to multiply this right hand side so we've got three x minus 11 and then let's expand these brackets so let's put our two down and put it as the big bracket and then expand our brackets so x times x is x squared x times minus 4 is minus 4X minus 3 times x is minus 3x and minus 3 times minus 4 is Plus 12. now let's simplify it so let's simplify these middle terms so we've got three x minus 11 equals 2 bracket x squared minus 7X Plus 12. then expand our brackets so we've got three x minus 11 equals two x squared minus 14x plus 24 and then finally whenever you're solving quadratics we want it to equal zero so let's get rid of this 3x and minus 11. so that's minus 3x from both sides so that would give us minus 11 equals 2x squared minus 17x plus 24 and then we want to get rid of this minus 11 so let's add 11 to both sides so it'll be zero equals two x squared minus 17 X Plus 35. so we've now got a quadratic which we can hopefully factorize and solve so when we do that we get 0 equals so we factorize our quadratic and we put 2X and X now we want two numbers that will multiply together to give us 35 and whenever we expand and add the terms together will give us minus 17x so I'm thinking it's going to be a negative and a negative so let's put a 5 here so that's 2x times -5 3 minus 10x and then put a 7 here so it's minus 7 times x which is minus 17x fantastic and that's a factorized now let's solve it so we know that whenever two brackets multiplied together to give a zero that they have to be zero so that means that for our first bracket we've got two x minus seven equals zero so adding seven to both sides two x equals 7 and dividing by 2 x equals seven over two so X can be equal to seven over two or three point five if you prefer decimals and then I'll write our bracket well we know then the x equals five so our two answers are x equals seven over two or x equals five and that's it so there are two solutions so we've been given a question and it says Circle the identity so here first of all we've got an equation we've got four X plus one equals twenty one and we can solve this we can take one away from both sides and we'll get an answer and if you solve this you would find x equals five now because that's just got an answer that's an equation so it's not an identity now next we've got 3x minus 7 is less than or equal to 50. that's what we call it inequality because we've got this inequality time we've got the less than or equal to sign so if it's got a greater than sign a greater than or equal to a less than or less than or equal to sign they're called inequalities that's an inequality it's not an identity next we've got 6X minus here and that's what we call expression we've got these terms 6X and minus eight so that's just what we call an expression and finally we've got an identity and an identity is something that's always equal to each other so we've got five bracket X plus one close brackets and then we've got this symbol here of three lines that's the equivalent to symbol and then we've got five X plus five and if you expand this set of brackets you get five X plus five so five bracket X plus one is always equal to five X plus five so that is an identity it's always equal to it so an identity is something that's always true and that's it so Okay so we've had a look and see what identity is now let's have a look at a question where we've given an identity and we need to find some more knowns and in this case we need to find the values of A and B so here we've got on the left-hand side of our identity in Brackets three x minus 1 multiplied by x minus five close brackets plus ax plus b is equivalent to or always equal to 3x squared plus 11x minus one so let's have a look at the left hand side of this identity so let's multiply these brackets so let's expand these brackets and see what we get 3x times x is three x squared then we'll go three x times -5 that's going to be minus 15x then we've got minus 1 times x as we minus X and then finally we've got negative 1 times negative five well negative times a negative is a positive and one times five is five and then we've still got our plus ax plus b so that's the left hand side of our identity and that's equivalent to 3x squared plus 11x minus one now let's simplify this because we've got our minus 15x and minus X so minus 15x take away another X would be minus 16x so we've got three x squared minus 16 X plus 5 plus ax plus b is equivalent to 3x squared plus 11x minus 1. Okay so we've expanded our brackets and we've got 3x squared minus 16x plus 5 plus ax plus b and we know that's going to be always equal to 3x squared plus 11x minus 1. now if we have a look at our x squared terms we've got 3x squared on the left hand side of 3x squared on the right hand side so that's fantastic because they're the same as each other now if we have a look at our X terms we've got minus 16x plus ax would have to be equal to 11x now if we've got minus 16x and we're going to add so many x's and we want to get 11 X's that means that a would have to be equal to 27 because minus 16 plus 27 is equal to 11 so minus 16x is plus 27 X's would give us our 11 X's so a would be equal to 27. okay now let's find B so if we have a look at our constants or numbers we've got 5 plus b on the left hand side on the right hand side that's equal to negative 1. so we want to take five and we want to add a number to it to get negative one so that means B is going to be a negative so B will have to be a negative and if we add negative six to five that's going to give us negative 1 so B would have to be equal to negative 6 and that's it so a is equal to 27 and B is equal to negative 6. okay let's have a look at our next topic our next topic is drawn linear graphs so that has drawn a straight line graph and we've been asked to draw the graph of y equals two X plus one so we're going to plot the coordinates where the Y value of the coordinate is twice the x coordinate plus one so to do that sometimes in the question they give you a table but occasionally they haven't given a table so I'm going to show you how to do this without a table but if they give you a table it should look something like this so it starts off with X and Y and then you've got some values of X at the top so it could be something like minus two minus one zero one two and if they haven't run the table for you you would just do something like this and to find what Y is well Y is equal to 2 times X plus one so we're going to substitute in these values of X into this and we're going to find what Y is so we've got 2x plus 1 so we're going to times all the x coordinates all these numbers by 2 and add one so let's start off here with our two two times two is four added one would be five this point is one so two times one is two plus one is three and doing that for the rest of the points we'll get and and that's it so we've got our coordinates now we just need to plot them this is set of coordinates here two five one three zero one negative one negative one and negative two negative three so we'll plot these five points and then draw a nice straight line through them and then get a ruler and a pencil and draw a nice straight line through those points and it looks something like this and that's it so that's how you draw a straight line graph our next topic is to find the midpoint of a line now the midpoint of the line it can be done in two different ways if it's drawn on a grid for you sometimes you can do it just by inspection just by having a look and seeing if you can spot where the midpoint is so here we've got the line a b and if I wanted to find the midpoint of this line I can see straight away it's here that means that the coordinates of the midpoint would be one minus two because it's one across two down so it's one minus two so that would be the midpoint of the line so if you've got on a grid you can just sometimes spot it but sometimes you're given the coordinates so the coordinates of B would be the point three zero the coordinates of the point a here would be minus one minus four minus one minus four and if you're given the coordinates and you're asked to find the midpoint what you do is just add the coordinates together and then divide by two and that will tell you the coordinates of the midpoint so if we have minus one and three well minus one plus three is two and divide by two is one so fantastic and then in terms of the y coordinate if we add the Y coordinates together the minus four and zero or minus four plus zero is minus four divided by two is minus two so if you add the coordinates together and divide by two you'll find the coordinates of the midpoint okay let's have a look at our next topic so our next topic is coordinates and we're going to look at coordinate questions whenever we're dealing with ratio and this video 87a and corporate maths so here we've been given a question we're told that is the point of coordinates negative 20 negative seven and B is the point we'll coordinate 1611 and there's a point on this line somewhere called p and the ratio of a to P to P to B is equal to four to five so it means that the distance a p to PB whenever we simplify we get four to five and we've been asked to work out the coordinates of the point P so the point p is going to be around here somewhere because we know the distance from a to P is smaller than the distance from P to B because we know that the ratio of a p to PB is four to five so I've just put it roughly here on the diagram it's just to the left of the midpoint I've been asked to work out the coordinates of this point so to start off with what I'm going to do is I'm going to look at the points a and the point B and what I'm going to do is I'm going to turn it into a right angle triangle so I'm going to go across and then I'm going to go up and it's a right angle triangle now if we have a look at this triangle if we consider the distance we've gone across we've gone from negative 20 to 16. so we'll go from negative 20 to 16. so let me see we're going 36 across so 36 cross and then if we look at how much we've gone up by so the point here it's got a height of negative seven and the point B is get a height of 11. so we've gone from negative 7 up to 11. so we've gone up to 18. so that means to get from the point a to the point B we would move across 36 and we would go up 18. now if we have a look at the ratio we're told that the point p is a point on this line such as a p so PB is four to five so that means that if we add four and five together so four plus five we get that's equal to 9. so if we divide both of these distances by nine and times them by four that will tell us how we would get from a to p and that means we can work out the coordinates of the point P so let's start off with horizontally so if we take our 36 and we divide it by 9. so 36 divided by 9 is equal to four and then we're going to multiply by 4 because we want to get from a to P so we're going to do 4 times 4 is equal to 16. so that means you go from a to P we would move 16 across so we move 16 across so that means that if we were negative 20 and we add 16 that means the point P will have an x coordinate of negative 4. now let's consider vertically well we're going up 18 altogether so we're going to take our 18 and we're going to divide it by 9 so 18 divided by 9 is equal to 2. now we want to go from a up to P so we're going to multiply that by 4 so we're going to do 2 times 4 is equal to 8. so it means to go from a to P we're going to go across 16 and we're going to go up in and that means that if we take our negative 7 and add on 8 that's going to be one so the point P has coordinates negative four one and that's it okay our next topic is to find the length of the line now if you have a look at this line you can see if we just go across and then go up we can turn it into a right angle triangle and the great thing is then we can use Pythagoras's Theorem to find the length of this line so this is a right angle triangle so let's mark on our length so we've got this point here which is the 0.11 in terms of horizontal you can see we're going across one two three so it's going to be three and in terms of the heights you can see here we're at one and we're going up to five so that means the head of this triangle would be one two three four so we've got the lengths three and four now they're not centimeters they're just three and four units and we want to find the length of the line so remember Pythagoras Theorem is a squared plus b squared equals c squared where A and B are the shorter sides and C is the longest side so A and B well so a is going to be three so three squared plus and B let's let b equal four so four squared equals and then we've got c squared well that's going to be a b so we could call it X if we wanted to and just write x squared then let's work out 3 squared plus 4 squared of 3 squared is 9. so we've got 9 plus and 4 squared is 16. so 16 equals x squared so we've just squared 3 and squared 4. then we're going to add our 9 and 16 together so 9 plus 16 is 25 and that equals x squared so we've got 25 equals x squared or x squared equals 25. now we want to find the length for this line so what we're going to do is we're going to square it the opposite of square square root Define the length of this line so we're going to the square root of 25 and the square root of 25 is 5. so that means the length of this line is 5 units and that's it our next topic is to work out the gradient or the steepness of a line so to work out how steeper line is we use the formula gradient equals rise over run so if we've got a straight line drawn on a grid if we do the rise of the line divided by the run of the line that will tell us the gradient of the steepness of the line and this is video 189 in corporate Maps so let's have a look and see what rise and run are so if you've been given a line drawn on grid such as this one and you've been asked to work out the grain of it choose two points on the line to begin with now in terms of the points here we could choose this point here minus one minus two or zero zero or one two or two four or three six or four eight they would be good points to choose I'm just going to choose this one here one two and I'm going to choose this point here three six and I'm going to turn it into a little right angle triangle so I'm going to score across and then I'm going to go up so the rise is the rise here so it's how far up you've gone so whenever you draw your little right angle triangle it's the Rises how much the line's gone up by so if you can see we've gone up and make sure you know what you're going up in so we're going up once here so we've got about one two three four so the rise here would be four and the run and again check what we're going to cross in we're going to cross them ones so the Run would be well we're going to cross one two so the Run would be two so we've got our rise of four and we've got a run of two so let's work out the gradient the gradient is the rise 4 divided by the run two and that's equal to two so the gradient of this line is two and what that means is for every one unit you go across the line's gone up two and as you see if we started at zero zero if we go across one it goes up two if you go across another one to two you go up another two if you go across another one to three you go up an over two and so on so the gradient of this line is two and the gradient is far better than the rise divided by the run and you just draw a little right angle triangle you just work out what the Run would be and the rise would be and then you do the rise divided by the run so if you had a line going downwards like this and you wanted to find the gradient of this line um again you would draw a right angle triangle you choose two points so I'm going to choose this point and I'm going to choose this point and as you can see we're going across one so we're going from five to six our run is one and in terms of our rise well we're going downwards this time so a rise is negative four because we went down and you would do negative 4 the rise divided by one and that would be negative four and that means for every one you go across the line goes down four if you go across one it goes down four and so on okay let's have a look at another gradient question this time I've changed the scale so the numbers are a bit bigger and we've been asked to work out the gradient of the this line so here we've got a line and we're going to choose two points on them I'm going to choose this point which is the point zero fifty I'm going to choose another point and another point I think looks quite good would be 25 250 here and let's draw our little right angle triangle so we're going to go across and then we're going to go up now in terms of the run we're going from zero here all the way across to 25 so the run here would be 25 we're going across 25 and the rise well we're going from here up to here so we're going from 50 up to 250 so our rise would be 200. and then to find the gradient we're going to do the rise divided by the run so we're going to do the rise 200 divided by 25 and that's equal to Ian so the gradient of this line is it and that what that means is for every one you go across one unit you will cross you're going to be going up here so the green of this line would be it so the next topic is looking at the equation of a line so the equation of a line is in the form y equals MX plus C so for instance something like this y equals 4X plus 3. now we're going to do is we're going to look at this equation so whenever we got in the form y equals MX plus C the number in front of the x is the gradient so if I had the line Y equals 4X plus 3 the gradient of that line is four and you'll remember that means that for every one square you go across a one unit you go across the graph goes up four units so you go across one it goes up four and so on and then R plus C is the Y intercept this shows you where the graph crosses the y axis so for this graph here this graph would have y equals 4X plus 3 has a gradient of four and it crosses the y-axis at zero three it has a y intercept of three now here's part of the chord match revision chord on the equation of a line and you've got y equals MX plus C where m is the gradient so this number is the gradient and C is the Y intercept and if we had this line here if we wanted to find this equation well first of all we would find the Y intercept and crosses the y axis or meets the y axis at one so that means it's going to be plus one on the end in terms of its gradient well we can make a little triangle and we can do rise divided by run and if the rise was 4 and the Run was two but four divided by 2 is 2 that means the grid of this line is 2 for every one you go across it goes up two when you go across goes up two so the Grid in this line is 2 so the equation of this line would be y equals two X plus one so let's find the equation of this line that's been drawn on the grid so we know it's going to be in the format y equals m x plus C we know the M's degree in so let's find the gradient to begin with so let's make a little triangle so let's choose two good points I think that a good point would be this one here at minus one one and another good point would be this one here I you could have also chosen zero four so I'm going to make a little triangle it's going to cross and up and for my triangle it's got a run of two it goes across two units it goes from minus one to one it's going to run a two and the height it goes from one up to seven so it's got a rise of six so for the gradient we've got rise divided by run so m equals rise over run so it's going to be equal to the rice 6 divided by the Run 2 and 6 divided by 2 is 3. so the gradient of this line is three so we know it's going to be y equals three x now in terms of the Y intercept it crosses a 4 positive four so it's going to be plus four if a cross down here minus one you would write minus one and so on if we cross through the origin zero you wouldn't write anything you would just write y equals three x but this graphic cross that had a one step to four so we wrote y equals three X plus four and that's it okay next question says a straight line with gradient minus two passes through the point one five find the equation of the line so we know first of all it's a straight line so we notice in the form y equals MX plus C so we know that the answer is going to be y equals something X plus or minus something and it says it's a straight line with green negative two so we know it's going to be y equals minus 2X and then something after it and it passes through the point one five and it says find the equation of this line so we need to find this y-intercept and that's where the straight line meets the y axis there's two ways which I would typically do this question one is by looking at the gradient the gradient's minus two so that means that this graph is actually going down so we've got a x and y axis it's got a gradient of minus two so it's coming downwards like so and when all goes through the point one five so it goes through the point one five so because we know the gradient minus two that means for every one you go across it comes down two you go across one goes down and two so we for this point here if we want to find where it crosses the Y ax is here so the y axis we know that it's going to cross one and it's come down two and it's now going a height of five that means it must have had a height of seven to begin with because it must have had a higher seven to go across one and come down two until you now have the height of five so that means that a y intercept would be seven so the equation of our line is y equals minus two X plus seven so one approach is to know what the gradient means and to look at the point you've been given and to work out where it would have crossed the y axis now there is another approach and this is actually the person I would typically use and that is whenever we had the whenever we knew it was y equals minus two X plus something I would just write plus C which stands for the Y intercept and I know the point that they've given me is one five so I know that it has an x coordinate of one and a y coordinate of five and we can substitute these in to our equation so we can instead of writing y equals we could write five equals so 5 equals and then next we know we've got minus two x so that's minus two times whatever X is and X is equal to one so minus two times one is equal to minus two and then we've got our plus C and then we want to fines we'll see is our y-intercept well we want the C on its own so we want to get rid of this -2 so we'd add 2 to both sides of this equation and you get 7 equals c so C is equal to seven so you get y equals minus 2x plus 7. okay let's have a look at our next question so this time we've been given a straight line passes through the points three eight and five twenty and we've been asked to find the equation of the line that passes through those points so again we know it's going to be in the form y equals MX plus C so we want to find the gradient of the line that passes through these points and we want to find where it crosses the y axis so first of all let's find the gradients okay so let's do a little sketch so we've got our x and y axis and we've got our first point which is 3 8 so 3 across eight up somewhere like that and then we'll put 520 so a little bit across a much further up like that so we've got 5 20 and we've got three it and we want to find the gradient of that line so the grid of that straight line and so what I do is as before I would draw a little right angle triangle like so and get our rise in our run so I run where we've gone from three across the five so our Runners two and a rise we're going from eight up to twenty so a rise would be 12. so we've got a rise of 12 and a run of two so the m equals rise of a run so rise over run so that's going to be equal to rise 12 over 2 is equal to six so the gradient of this line is six so for every one we go across we go up six and let's just check it if we went across one we would get to four and then if we go up six we'd get to 14 and then if we went across another one we get to five another six we get to twenty so yeah that's right okay so we know it's y equals six x plus something so plus C let's just choose one of our points because we know the line passes through both of these points three eight and five twenty so let's choose one of these points I'm going to choose 520 I could have chosen three here and I know X is equal to 5 and Y is equal to 20. the x coordinate is 5 and the y coordinate is 20. so let's substitute those into our equation so y that's 20 equals 6 times x so that's 6 times 5 6 times 5 is 30 plus C so to get to the other two and I'm going to want to take away 30 from both sides so that will leave me with 20 take away 30 is negative 10 and then we've got 30 plus C take referee just leaves us with C so C is equal to negative 10. so a y intercept is equal to negative 10. so our answer would be y equals six x minus ten and again let's just check it we knew the gradient of the line was equal to six so I meant so for every one we go across we got six so if we're at this point if we go across one we go down six that would then bring us to 2 two two if we go across one and down six that would bring us to one minus four and across one and down six would bring us to zero negative ten so yeah the Y intercepts would be negative ten and that's it okay so let's have a look at our next topic and let's parallel lines which is video 196 on corporate maps and parallel lines their lines that go in the same direction and they never meet each other and because they go in the same direction they have the same gradient of the same steepness so if we had y equals two X plus one and Y equals two x minus three those lines would be parallel to each other because they've got the same gradients and this is the couple miles revision card on that so if you do have the chord Mouse revision cards make sure you know this one and stick it up on your wall or bring it with you on the bus in the morning or whatever make sure you remember the parallel lines have the same gradient and that's it that's a very useful bit of information that parallel lines have the same gradient and you can use that in questions as well okay let's have a look at an example so here we've been told the line a is parallel to the line Y equals three X plus nine and passes through the point seven negative one find the equation of a line F so because the line a is parallel to y equals three X plus nine we know that it's got the same gradient so the line a will have an equation in the form y equals m x plus C and we know the gradient of the line will be three because it's parallel to this line so they must have the same gradient three so we know it's going to be y equals three X plus C now we just need to find C but we know the line a passes through the point with coordinates seven negative ones let's get an x coordinate of seven and a y coordinate of negative one so if we substitute these values into our y equals three X plus C we can find our value for C and then we will know the equation of the line a so we know that X is equal to 7 and Y is equal to negative one so y that's going to be negative 1 is equal to 3 times x now X is equal to seven so three times seven is twenty one plus c now we want to find C so we want to get rid of this twenty one so subtract 21 from both sides of the equation to get rid of this 21 on the right hand side so we know that c is equal to negative 22. so that means the equation of the line a will be y equals three x minus 22 and that's it so let's have a look at our next topic now we're going to look at perpendicular lines and these are lines that meet each other at 90 degrees they cross each other at a right angle now here before the court Mouse revision card and as you can see we've got an example here of a pair of perpendicular lines we've got the line y equals 2x which is going up in this direction and then going down in this direction we've got y equals minus a half X plus one and as you can see these two lines meet at 90 degrees across at 90 degrees now one thing to notice about the ingredients is well here we've got two and here we've got minus a half now the reciprocal of two is a half the negative reciprocal of two would be minus a half so if you've known the gradient of one line if you find the negative reciprocal of it you'll find the gradient of the other line so here if you know the gradient was equal to two if you find the negative reciprocal of it so we take the reciprocal which is a half and then make it negative a half you'll find the gradient of the line that's perpendicular to it another interesting point is if you multiply the gradients of two perpendicular lines you will get negative one so if you have two times negative a half well 2 times negative a half is negative one so the product of the gradients for perpendicular lines is negative one I tend to just use a negative reciprocal so if you know that the gradient of this line is a negative a half well the reciprocal of a half is two and then because it was negative it's not positive so the the gradient of the perpendicular line would be two okay let's have a look at a question based on that so here we've got a question it says the line is drawn that is perpendicular to y equals negative four X plus seven and it passes through the point eight three work out the equation of the line now we know that it's perpendicular to y equals negative four X plus seven so it's perpendicular to this line and this line has a grain of negative four so if we find the negative reciprocal of negative four you will find the gradient of the line that's perpendicular to it so we first of all know this gradient was negative so the grid and fire line is going to be positive so we know it's going to be positive and the reciprocal of four well the reciprocal of four is equal to a quarter so we know that the gradient of our perpendicular line would be a quarter because the negative reciprocal of the gradient that we were given so we know the equation of our line will be y equals a quarter X plus c now we know that it passes through the 0.83 so if this is the x coordinate that's the y coordinate we can substitute those into our equation and we can find RC so we know that our Y is equal to 3 so we've got three equals a quarter X so it's a quarter of X and X is eight so it's a quarter of eight plus c so next we've got three equals a quarter of eight is two so we've got two plus C well C is going to have to equal one here so we've got what C is so that's it so we know that the equation of the line is perpendicular to y equals minus four X plus seven that passes through the point eight three would have to be y equals a quarter X plus one and that's it I've got an example and the question says the graph below shows the cost of higher in a hot tub and this is video 171 a and Corbin Mavs so here we've got a graph that shows uses the cost of hiring a hot tub and you've got the number of days going along horizontally and we've got the cost going up vertically so going from zero up to 350. so the question says the graph intersects the vertical axis at 100 so as you can see here it starts at 100. what does this represent so as you can see here this would be on zero day so this is like a set fee it's like a a set charge that you're charged for higher in the hot tub so no matter what there's 100 pound to begin with and then you're going to be charged so much per day so what this represents so this represents a fixed fee fixed cost of 100 pound so that means no matter how long you hire this hot tub for is a hundred pound fee to begin with if this was the cost of a plumber that might be a call out fee or if this is a taxi that whenever you go into a taxi sometimes you see the meter starts at a particular number like a set fee that you've prepared to begin with so this is a fixed cost this value of 100 pound the next question says find the gradient of the graph so we've got this graph and we want to find its gradient so let's choose two points I'm going to choose the point here zero one hundred I'm going to choose this point up here which is 10 350. and let's make our little right angle triangle so we're going across and going up and let's work out the rise divided by the runs so our gradient M that's the letter for gradient m is Rise divided by run so for our right angle triangle the rise well it's gone from 100 up to 350 so that's a rise of 250 so it will be 250 divided by and the run if you look at how much we've gone across by we've gone from 0 to 10 so the room would be 10. so we're going to do 250 divided by 10 and that's equal to 25. so that means the gradient of this line is equal to 25 and the question says explain what the gradient represents so remember the gradient it means for every one you go across how much the line goes up by now in terms of this context of higher in a hot tub what that would mean is for every one more day you hire the hot tub how much the price increases by so that's what it means it means the hot tub means the cost per day of power in the hot top is 25 pounds so that'd be the cost per day 25 pound so the next topic is simultaneous equations and here we've got a Paris to multinational equations we've got 3x minus y equals 23 and 2x plus 3y is equal to it so whenever we could simultaneous equations what we want to do is cancel out one of the letters we want to add or subtract the equations so that one of the letters disappears so as you can see here we've got minus y and we've got 3y now if we had minus 3y and 3y we could add them together to get zero so what I'm going to do is I'm going to multiply this top equation by 3 to begin with so let's number them one and two so we've got equation one and equation two we're going to do three times equation one so remember if we do three times equation one we're multiplying this by three gives us 3 times 3x is nine X then 3 times minus y was going to be minus 3y and 3 times 23 well 3 times 23 is 69 then we've got equation two which is two x plus 3y equals eight we've got one equation with minus three Y and one equation with 3y so that's fantastic what we can do now is we can add these two equations together and our minus three Y and our three y will add together to give us 0. so we'll just be left with x's and that's fantastic okay so let's add the two equations together so let's write the word add I tend to avoid putting a plus or minus symbol just in case if it was a subtract I put a minus here in front of the two and make it confused and think it's a minus two x so our add or sub now what I'm going to do is I'm going to add these two equations together so 9x plus 2X well that would be 11x minus 3y plus three why well that's zero so they cancel out and finally we've got 69 Plus 8 that's equal to 77. so we've got 11x equals 77. now we can divide by 11 and divide by 11 and we get the x equals seven so x equals seven so that's fantastic so we know X is equal to seven so let's now find y so let's substitute this x equals 7 into one of our equations and I'm going to substitute into equation two just because we have positive 3y and let's write that down to sub x equals 7 into 2. just to show what we're doing so whenever we substitute x equals 7 into this equation we get 2 times 7 so that's 14 plus 3y equals it now we're going to solve this equation so minus 14 and minus 14. well we had 14 plus 3y and we took away the 14 so we're just left with 3y and on the right hand side we've got 8 take away 14 that's going to be equal to negative 6. now we're going to divide both sides by three and whenever we divide both sides by three well three y divided by 3 is just Y and negative 6 divided by 3 is equal to negative 2. so let me see we've got our answers we've got x equals seven and Y equals negative 2. and let's just check our answer now we substitute X into equation two to find our value of y let's now substitute both of them into equation one so let's check in one so check in one well if we put these numbers if we substitute these numbers into equation one we've got three times x so that's 21 3 times 7 is 21. minus y is minus two so we're going to do minus minus two and that's meant to be equal to 23. so let's check it 21 minus minus 2 that's 21 plus 2 and that's 23 so that's right so we know we've got this question right and that's it okay let's have a look at our next question so next question says solve the simultaneous equations 2x plus 3y equals 1 and 7x plus 2A equals negative 22. and I've chosen these numbers on purpose because what we're going to have to do is multiply both equations by certain numbers to get the same number in front of V for the X's or y's so let's look at our equations equation one and equation two now we could cancel out our X's so we could multiply the top equation by seven to get 14x and we could double equation two to get 14x and we could cancel those or we could cancel our y's so we could multiply let's see if we've got three and two let's get six so we can multiply our topic equation by two and our bottom equation by three and that would give us 6y and six way and because they're both six ways we would then take the equations away from each other to get zero let's actually do that let's cancel our y's so let's multiply equation one by two because we want to get six y so we're going to do two times one and when we do two times equation one we get we'll multiply this whole equation by two would be four X plus six y equals two now we've got equation two but we want to get six y so let's now multiply this whole equation by three so let's do three times equation two and when we do that we get multiplying these all by three we get three times seven X is Twenty One X three times two I was going to be plus 6y and three times minus 22 that's going to be minus 66. so this is great because we've now got two equations both with six y's in them now what we can do is we can now take away these equations from each other the only thing is that whenever we take these away we're going to have 4X take away 21x which would be negative 17x and so on so what I'm actually going to do is I'm going to write this top equation again just beneath this other one so I'm going to write 4X plus 6 y equals two just so that whenever I take these equations away from each other I'm going to get 17x rather than minus 17x so now let's put a line beneath them and let's say subtract so we're going to subtract these equations from each other so 21x take away 4X that's 17x 6 YT equals 6 y well that's zero they cancel out and then we've got minus 66 take away two so when we've got minus 66 take away two we're going to go down another two so it's going to be minus 68. so we've now got 17x equals minus 68. so let's divide both sides of this equation by 17 so divide by 17 and divide by 17 and we're going to get that x equals negative 4. so we now know that X is equal to negative 4. we can substitute that into either equation one or equation two to get our y let's substitute it into equation one so equation one so sub x equals negative 4 into 1. and when we do that we get well 2 times x well that's going to be 2 times negative 4 and 2 times negative 4 is negative 8 plus 3y equals 1. now if we want to solve this we want to get rid of our negative 8 so let's add it to both sides so add it and add it so that will give us three y equals and one plus it's nine and divide by 3 and divide by 3 and we get Y is equal to three so our answers are x equals negative 4 and Y equals three and again we can check our answer in equation two check in two and when we do that we get well 7 times 8 X was seven times -4 will be -28 then we've got plus 2y well Y is equal to 3 so 2 times 3 6 we're going to add 6 and that should be equal to negative 22. and let's just check it negative 28 plus 6 would be negative 22. so that is equal to negative 22. so we're right we know that our answers would have to be X is equal to negative 4 and Y is equal to 3. okay let's have a look at our next question a wordy context maybe where you're buying so many of something and so many of something else and you're given the price and so on and you've got to create your own simultaneous equations and solve them so let's have a look at a typical question so we've got five adult tickets and three child tickets cost 58 pound and two adult tickets and eight child tickets cost forty seven pound find the cost of each type of ticket so we've got our two equations we've got five adult tickets but let's let an adult ticket be X pounds unless that a child ticket be y pounds so let's make our equations we've got five x plus 3y is equal to 58 pound so equals 58 and then we've got 2X plus 8y and that's equal to 47 pound so we've got our two equations now what we want to do is either get the same numbers in front of our x's and y's or the same but one being positive one mean negative because if they're both the same we can take them away from each other or if they're both the same but once plus and one's negative we can add the two equations to get zero we could cancel out our X's or we can cancel out our Y's just for fun we're going to cancel out our X's this time so I can actually cancel out our X's so I'm going to multiply our top equation let's equation one by two to get 10x I'm going to multiply the second equation equation two by five to get 10x as well and then what I'm going to do is take the two equations away from each other so let's do 2 times equation one and when we do two times equation one we get well multiplying these all by 2 would get 10x plus multiplying three y by 2 would be 6y equals and multiplying 58 by 2 would be 116. and then multiplying equation two by five because we want to get 10x so 5 times equation 2 would be multiplying these all by five would give us 5 times 2x is 10x 5 times 8y well it's going to be 40y and then multiplying 47 by 5 will be 235. now we want to cancel that our 10 X's so if we've got 10x and 10x we want to take them away to get 0. now because we're taken away rather than 6y take away 40y I'm actually going to write this top equation just beneath the other one again so let's scroll down and let's write 10 X plus 6y equals 116. and let's write sub because we're going to subtract them from each other so 10x take away 10x is 0. 40y take away 6y well that's 34y and then we've got 235 take away 116 and when we do that we get an answer of 119 pounds or 119. so we now know that 34y equals 119 we can divide both sides by 34 to get y equals 119 divided by 34 is equal to 3.5 or 3.50 so we now know what Y is and that's the cost of a child ticket which is 3.5 or 3.50 let's find the cost of an adult ticket so that's sub y equals 3.5 and we can choose either equation one or equation two I'm going to choose equation two so n 2 two when we substitute 3.5 in for the value of y into equation 2 we're going to get we've got 2x plus and then we've got 8 times 3.5 that's equal to 28 and that's equal to 47. so we've got 2x plus 28 is equal to 47 so let's take away 28 from both sides so that gives us 2x equals and 47 take away 28 was equal to 19 and then if we divide that by 2 we get x equals and dividing 19 by 2 is 9.5 so as the question says find the cost of each type of ticket so we've got an adult ticket which is X and that's going to be equal to 9.5 or 9.50 so let's write it as a price 9.50 and the child ticket which was equal to Y is equal to 3.50 and that's it okay let's have a look at our next topic and our next topic is solving simultaneous equations by using graphical approaches in other words drawing the graphs and finding out where the graphs meet and using that to find the answer to simultaneous equations so here we've got a pair of simultaneous equations so that's two equations at the same time we've got y equals three minus x and y equals 2x minus 3. so to solve this what we're going to do is we're going to draw suitable lines in other words we're going to draw the graph for y equals 3 minus X and we're going to draw the graph of y equals 2x minus 3 and where those two straight lines mean will be our answer so let's first of all draw the graph for y equals 3 minus X so to draw this graph I'm going to draw an X Y table so it's X and Y and what I'm going to do is I'm going to choose some values for X with 0 1 2 3 and I'm going to work out the values for y so y equals 3 take away X so if I take away X from 3 I can find the value for y so if x is equal to 0 well 3 take away 0 is 3 so that means the Y is 3. 3 minus X well if x is one it would be three minus one that's two if x is equal to 2 or 3 minus two is one so that'll be one and if x is equal to 3 I'd have 3 minus 3 which is zero so the points in this graph will be 0 3 so 0 3. there'll be one two so one across two up there will be two one so two across one up and there'll be three zeros so three across and zero up like so and if I get my ruler and pencil I can draw a nice straight line through those points and I would look something like this and that's it that's the graph for y equals three minus X now in a typical question I would imagine what they would do is draw one of these two lines for you and you just have to draw the other one and then find out where they meet okay let's have a look at our next one so our next graph is y equals 2x minus three so we'll do an X Y table X and Y and we'll choose some points and again I just tend to use zero one two three and I choose a few of them or in this case I choose four of them just so whenever I plot them I can see that all four they're on a nice straight line if they're not on the straight line then I know I've done something wrong so what I'm going to do to find y we do 2 times x and we take away three so we're going to times all these numbers by two and then take away three so let's start off with zero two times zero is zero take away three would be minus three then we've got one well two times one is two take away three would be minus one now we've got two two times two is four take away three is one and finally we've got three two times three is six take away three is three so we've got our coordinates zero three one minus one two one and three three so let's plot them zero minus three one minus one two one and three three and get our ruler and pencil we'll draw a nice straight line through those points and as you can see here the two straight lines meet at the point two one this is the point of intersection here two one there so that means I mean that the 0.21 now when we're solving some multinous equations what we're trying to do is find the value for x and we're trying to find the value for y so if we look at this point two one we're going 2 across that means the x value is 2 because the coordinates two one the first number is the answer for x and the second number is the answer for y so X is equal to 2 and Y is equal to one and let's just check that 2 across one up the x value is 2 and the Y value is one and that's it okay let's have a look at our next topic so next topic is solving non-linear simultaneous equations and this video 298 in corporate Maps we may have the source simultaneous equations such as this well here we've got a non-linear this is a quadratic now we may be asked to solve some multinase equations and find the values for X and Y we may be asked to find where quadratic or a circle meets or intersects a straight line and so let's solve these simultaneous equations so here we're quite lucky we've got y equals x squared plus x minus 14. and we've also got y equals x minus 5. now if we've got two simultaneous equations whereas to y equals and Y equals we can put them equal to each other the reason is we know Y is equal to this and we also know that Y is equal to this so that means that we could write x squared plus x minus 14 equals x minus five so if they're both in the form y equals and Y equals you can just put them equal to each other like this and then you can solve that and find their points of intersection or solve the simultaneous equations so let's solve this so whenever we're solving a quadratic we want everything brought over to one side and zero on the other side so let's bring everything to the left hand side here so we've got x squared plus x minus 14 equals x minus five so let's get rid of this x so let's take our X from both sides of this equation so that would leave us with x squared x take where X is zero so we've got their minus 14 and on the right hand side we had x minus 5 we've taken away the X we're just going to be left with minus five next well we want this quadratic to equal zero so let's add five to both sides of this equation so let's add five and add five and that will give us x squared and then we've got minus 14 plus 5 5 or minus 14 plus 5 is going to be minus 9 that's equal to well minus 5 plus 5 is 0. so we've got x squared minus 9 equals 0. now whenever you solve an equation like this you could have something like this something which is x squared minus nine so that's going to be difference between two squareds or you may have a quadratic that you may need to factorize or sometimes you might even need to use the quadratic formula and a hint in the question there would be to find your answer to one or two decimal places but in this case here we've got difference between two squared so we've got x squared and minus nine so let's put bracket bracket bracket bracket and then equal zero we'll put X at the front of both brackets and the square root of nine is three so let's put 3 in both brackets one with a plus sign and one with a minus sign and then solving this but we want to make each bracket zero so in this bracket would be x equals negative three or x equals three so it would be our two solutions for X but remember we need to find our solutions for y as well for simultaneous equations so let's substitute these back into one of our equations and I do think this equation is much easier to substitute into the second one so let's sub into y equals x minus five so let's start off with x equals negative three so let's write and x equals negative three we'd have y equals and instead of X we're going to have minus three minus five and minus three minus five would be minus eight so y would equal minus eight so that's one pair of solutions X can be equal to negative three and Y would be equal to negative eight or let's Sub in x equals three and that would give us we'd have y equals and then we've got x minus 5 so it's going to be 3 minus 5. and 3 minus 5 is equal to negative 2 so y would be equal to negative 2. so that's another pair of solutions and that's it so we'll find our Power Solutions so we've got that x equals negative 3 and Y equals negative eight or x equals three and Y equals negative two and that's it so our Solutions are simultaneous equations if you're asking this question to find where the quadratic y equals x squared plus x minus 14 and the straight line y equals x minus 5 intersect then those would be our coordinates we would have the coordinate negative three negative eight and three negative two but this question was just a simultaneous equations question so we've got our pairs of solutions we've got x equals negative three and Y equals negative eight or x equals three and Y equals negative two okay let's have a look at our next topic so the next one we're going to look at is the equation of a circle that's video 12. and here we've got the chord Mouse revision card and the equation of a circle our GCSE level is x squared plus y squared equals r squared where the center of the circle is always the origin a GCSE level that's the point zero zero so the circle will always have a center of zero zero and the radius of the circle is R so this R is equal to the radius so if you have a look at this circle the center of the circle is the origin so that means it'll be in the form x squared plus y squared its equation and it'll be equal to well its radius is equal to four so four squared is 16. so the equation of this circle would be x squared plus y squared equals 16. and if we would look at the revision code that's what this says it says it's got a center of zero zero and since 4 squared is equal to 16 the radius is 4. so that's it so that's the equation of a circle this is the core Mass revision card if you are interested in those then the link below and they're really really useful and this is one of the revision cards okay let's have a look at our next topic it's the equation of a tangent to the circle that's video of 372 in corporate Maps so here we've got a circle and there's a tangent to that Circle and it's a tangent at the 0.26 and the question says the diagram shows the circle x squared plus y squared equals forty so it's a circle with the center of the origin and it's equal to 40. let's just check 2 square squared is 4 6 squared is 36 4 plus 36 is 48 yep that's a point on the circle and we're trying to find the equation of the tangent at that point two six so we've got the 0.26 and this is a tangent and we're going to find the equation of that tangent so let's find the equation of that tangent so first of all we know it's a straight line so we know the tangent will be in the form y equals m x plus C so we need to find M the gradient of this tangent and we also need to find C where this tangent will meet the y axis so this point here and if we know the gradient of this line and if we know where the line crosses the y axis we will be able to find its equation so first of all let's find the gradient of this tangent now here we've got the radius of the circle so this is the radius of the circle I would think back to our Circle theorems a radius and a tangent always meet at 90 degrees so that is a right angle and because that's a round angle that means that the tangent and the radius are perpendicular to each other so that means if we know the gradient of the radius we can then find the negative reciprocal and that will be the gradient of the tangent so let's first of all find the gradient of the radius so here we've got a radius and it's going let's change color of pen to Blue so it just stands out a bit more so here we've got the point zero zero and here we've got the 0.26 so let's find the gradient of this radius here so it's rise of a run so let's turn it into a right angle triangle so we've got rise over run and it's runable we're going from zero across the two so it's run as two and it's rise well we're going from zero up to six so it's rise of six so that means the gradient of this radius will be m equals rise which is 6 divided by run which is 2 which would be equal to 3. so the gradient of this radius is equal to 3. now the tangent is perpendicular to it that means it is gradient is the negative reciprocal so that means that our gradient of our tangent will be the negative reciprocal of three so that means it would be where the reciprocal of three is a third and the negative reciprocal will be minus a third so that means that for our tangent it'll be y equals negative a third X plus C and if we can now just find its Y intercept this point C where it crosses the y axis we can then find out this equation all we need to know is one point that that line passes through and then we can find this plus C what's great because we know that it passes through the 0.26 so we know a point on the line so if we substitute in our values for X and Y so x equals 2 and Y equals 6. if we substitute those into our equation y equals minus a third X plus C we can then find R plus C so let's substitute those in so Y is equal to six so we're going to get 6 equals then we've got negative a third X well X is two so that means negative a third times two and then we've got R plus c now negative a third times two well a third times two is two-thirds negative referred times two would be negative two thirds so we've got six equals negative two-thirds plus C now we want to get C on its own so we don't want this negative two-thirds here so let's add two thirds to both sides so six add two thirds would be six and two thirds and then on the right hand side we added two thirds to get rid of the two thirds and then we'll just be left with C so that's it we've got our intercept this point here is six two thirds or six point six six six six six reoccurring so that's it so we have found our Y intercept and we know the gradient of the tangent is equal to negative a third so that means that the equation of this tangent will be y equals negative a third X plus six and two thirds alternatively this could be written as a top heavy fraction we could have written y equals negative a third X Plus and six times three is equal to eighteen plus two is twenty over three and either of these would be acceptable so this is the equation of the tangent the equation of transition I really like this topic because you put several topics together we use the equation of a circle we could work out the gradients of lines we consider the circle theorem so the radius and the tangent made at 90 degrees we look at the gradient of perpendicular lines being the negative reciprocal and things like that and you just put them all together and it's just a it's a great topic okay so okay let's have a look at our next topic so our next topic is rates of change and last video 390 a in corporate Maps so here we've got a question it says below is the depth of water in a container over six seconds so at zero seconds the depth is just over 12 centimeters at one second it looks like it's about 12 centimeters at two seconds it's just above 11 centimeters and so on and it actually hits six seconds it's actually empty it's the depth of the water at six seconds is zero and the question says estimate the rate of change of the depth of the water at 2.5 seconds so we've got the point of 2.5 seconds so this is the depth of the water at 2.5 seconds which looks like it's about 11 centimeters and we've been asked to find the rate of change of the depth of water at that time so the gradient of this graph at that particular time will be its rate of change but obviously this is a curve so we can't just do rise over run because obviously it's curving so what we're going to do is we're going to draw a tangent at that particular point so let's draw our tangent so we've just run the tangent to the curve at 2.5 seconds and if I find it Greater enough that tangent that's going to be the same as the gradient of the curve at that particular time so that would tell us the rate of change of the depth of water at 2.5 seconds so let's find the gradient of this tangent so let's find two suitable points on that tangent to find the gradient so let's start up with this point here so we've got zero and then if we look at where it meets the y axis it meets the y axis of one square down altogether there's one two three four five little boxes that make up two two divided by 5 is 0.4 so this is going to be 0.4 below 14 so that will be not 13.6 so that's one particular point on the tangent so let's go for another point on the tangent I'm going to go for this one here which is at four seconds at four seconds the tangent passes through the second the first Square below 10 so it's going to be 9.6 so we've got two suitable points two points that are on the tangent so remember the gradient is equal to gradient m is equal to rise over run so let's look at our points let's look at first of all the Run well we're going from this point across all the way across to four seconds so the Run would be four seconds we're going four across and in terms of the rise but we're going down here the graph's actually going down which is going to have a negative rise because it's going downwards and it's going down from 13.6 to 9.6 so it's going down by four so that means our rise is negative four because it's going down by four so the gradient is equal to the rise so that's a negative 4 divided by the run which is 4 and negative 4 divided by 4 would be minus one so that means that the gradient of this tangent is negative one that means the gradient of the curve that particular moment is negative one also and the question wants us to find the rate of change of the depth of water so it's going to be negative one centimeters per second so the rate of change of the depth of water are 2.5 seconds is negative one centimeters per second so in other words the container is decreasing by height at one centimeter every second at that particular moment okay so we've looked at instantaneous rate of change that's found in the rate of change at a particular moment or a particular time now what we're going to do is we're going to find the average rid of change so if we look at this question we've got blows the depth of water in a container over six seconds as you can see the depth of water is decreasing over six seconds and we've been asked to estimate the average rate of change of the depth of water between one second and six seconds so first of all let's have a look at one second and six seconds so if you have a look at one second you can see it one second the depth of water is 12 centimeters so the depth of water is 12 centimeters that's the point 1 12. and then if we have a look at six seconds at six seconds the container is not empty so it's going to be six zero and we've been asked to find the average root of change so what we're going to do is we're going to join those points up we're going to join up the 112 and the six zero so we're just going to join those up and to find the average derivative change what we're going to do is we're going to find the gradient of this chord that joins up those two points and that will tell us the average rate of change so whenever I find an instant initiative change we draw a tangent and we find the gradient of that tangent Define the average derivative change We join up the points that were asked in the question to find the average derivative change between and then what we do is we find the gradient of that chord so we need to find the gradient of this chord and that would tell us the average rate of change so let's do that so we're going to do rise over run so let's make our little right angle triangle so we've got a right angle triangle here so we've got our run so run has come from one second to six seconds so run will be five and then our rise where we're going down and we're going down from 12 down to zero so our rise will be negative 12. so we've got a run which is five and we've got a rise which is negative 12. so to find the gradient we're going to do m is equal to rise over run so it's going to be negative 12 over 5. so we're going to do negative 12 divided by 5 which is negative 2.4 so the gradient of this chord is negative 2.4 now let's make sure we put our units on which would be centimeters per second so the average rate of change of depth of water between one second and six seconds is negative 2.4 centimeters per second so in other words on average every second the depth of water is decreasing by 2.4 centimeters and that's it so this has been average rate of change okay okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is the area under a graph which is video 389 in corporate Maps so here we've got a speed time graph we've got speed on the vertical axis starting at zero meters per second and going up to 14 meters per second I've got time on the horizontal axis starting at zero seconds we're going across the 30 seconds and this is the speed time graph for a bicycle and it shows us how a bicycle travels for 30 seconds so starts at rest it accelerates for the first 10 seconds up to a speed of 12 meters per second and then for the next 20 seconds it travels at a constant speed of 12 meters per second we've been asked to find how far does the bicycle travel over the 30 seconds now if you find the area under a speed time graph that tells you the distance traveled because if you do speed times time you get the distance so if we find the area under this graph it will tell us how far the bicycle has traveled over the 30 seconds now if you have a look at this graph I think there's two regions we have first of all got this triangle on the left hand side let's call that a and then we've got this rectangle here on the right hand side so from 10 10 seconds to 30 seconds and let's call that b so if we find the area of the triangle and the area of the rectangle and add them together that will tell us the distance traveled so let's find the area of a so a so a is a triangle and as you can see it's got a base of 10 and it's got a height of 12. so let's do half the base so half of 10 times its height which is 12 and a half of 10 is 5 times 12 is equal to 60. and our units will be meters because we're traveling meters per second so that means that in section A the bicycle for the first 10 seconds the bicycle has traveled 60 meters now the next 20 seconds is section B so it's a rectangle and to find the area of a rectangle we just do the length multiply by the whip for the length multiplied by the height so the length of the rectangle well that's 20 times the height which is 12 so we're going to do 20 times 12. and 20 times 12 is equal to 240 meters so that means from 10 seconds to 30 seconds at 20 seconds the bicycle travels another 240 meters so it travels 60 meters and then another 240 meters so if we do 60 plus 240 that's equal to 300 meters so that means that the bicycles travels of 300 meters in total over the 30 seconds and that's it so if you find the area underneath a speed times graph you find the distance traveled okay this time we've been given a speed time graph for a bicycle we'll have a look at the graph in a second and we've been asked to find an estimate for the total distance shot by the bicycle in the first 12 seconds I'm going to use three strips of equal width so let's have a look at our speed time graph so this is our speed time graph you've got our speed in our vertical axis and we've got a time and a horizontal axis and we've been asked to find out an estimate for the distance traveled by the bicycle in the first 12 seconds so what I'm going to do is I'm just going to put a dash down here at 12 seconds and we want to find the distance traveled over the first 12 seconds now unlike our last question which was a straight line so that's quite nice to find there underneath that this is a curve so it means that we can only find an estimate for the area underneath this curve and we've been asked to use three strips of equal width so as you can see there's 12 seconds if we divide that by three that's equal to four so we're going to use strips of force seconds so we're going to look at this strip here we're going to use this strip here and we're going to use this strip here so one two three strips and let's label them air so this is air this stripper let's call that b and let's call this strip C and let's find the areas of a b and c now because this is a curve we're not going to be able to find the exact area of these strips at GCSE level so what we're going to do is we're going to find the approximate area of a the approximate area of B and the approximate area of C so what I'm going to do is I'm going to start off with the 0.00 I'm going to join it up here to this point here which is the 0.4 0.8 then I'm going to join that point up to the next Point here which is the 0.82 and then finally I'm going to join the 0.82 oh with this point here which is the point 12 4. and as you can see here we have now got a triangle a trapezium and a trapezium and if we find the area off the triangle the trapezium and the trapezium and add them together that will give us the approximate area underneath this curve that's not going to be the exact area it's actually going to be slightly over because actually if you have a look at cesium the line we've drawn is slightly above that curve which is underneath it so we're actually going to find an overestimate for there underneath this curve but it's going to be it's going to be close enough okay so let's start off with a so if we have a look at a as a triangle it starts at the point zero zero and goes up to this point here actually let's label the coordinates that we have joined up the triangle there if the triangle is half the base times the height the base is four so we're going to do half of four times the height so the height is 0.8 so Times by 0.8 that will give us our estimate for the area of X so half of four is equal to two times where 0.8 is equal to 1.6 and our units will be meters so that means over the first four seconds our estimate for the distance travel by the bicycle is 1.6 meters and if we have a look at babies or trapezium we've got two parallel lines we've got this parallel line and this parallel line and then we've got the distance between them this was the height of the trapezium if we turn on hide the heads outwards so the average medium let's write down B is equal to half brackets a plus b times H which is the distance between them so that means if we know that links to the two parallel sides and the distance between them we can find the area of this trapezium so let's work that out so the two parallel sides well this length will be 0.8 because it's 0.8 high this length will be two so it's going to be 0.8 plus two and then Times by the distance between them which would be the height of the trapezium if we turn our heads outwards which is four so Times by four so if we work out a half of 0.8 plus two and then times that by four that would tell us the area of the trapezium and when we do that we get that's equal to 5.6 meters so that means that the bicycle traveled approximately 5.6 meters over the next four seconds and then if we have a look at the next four seconds from 8 to 12 that's C so if we work at the area of trapezium C then that will tell us approximately how far the bicycle traveled in these four seconds so again it's a trapezium we've got two parallel lines we've got this parallel line with a length of two this parallel line with a length of four because the height of this is two and the header this is four so it's got a length of two a length of four and the distance between them the height of the trapezium if we turn our heads outwards is four so it means we're going to do a half of two plus four the two lengths the two parallel sides here and here and then Times by the distance between them which is four 2 plus 4 is equal to 6 and half that's equal to 3 Times Square 4 is equal to 12. so over the next four seconds the bicycle travels approximately 12 meters so we now know approximately how far the bicycles traveled over the first four seconds the next four seconds in the next four seconds so if we add them together that would tell us approximately how far the bicycles traveled in total over those 12 seconds so if we do 1.6 plus 5.6 plus 12 that will give us our answer and that's equal to 19.2 meters and that's it okay our next topic is functions and functions is one of the few topics where the example is different subtly from each other at Excel and AQA require each another function notation that's what I'm going to go through in this video OCR however don't require you to know the function notation so whenever I do each section I'll just say hi OCR would expect you to approach those topics okay so let's get started okay let's have a look at our next topic so our next topic is composite functions last video 317 corporate Maps now if we have a particular variable let's call it X and we apply a function two let's apply the function G we would get G of X and then if we apply another function let's call that function f we would get f g of x now rather than apply in function G and then function f particularly if you've got this lots of times it can be useful to find what we call a composite function and a composite function would mean that instead of applying function G and then function f we can apply this composite function this FG and I'll bring us straight from X to our end result rfg of X okay so here we've got an example we've got two functions G of X is equal to x squared plus five and we'll get f of x equal to 2x plus 1. and we've been asked to find F G of X in other words there's got to be a variable X the function G is going to be applied and we get G of X and then we're going to apply the function f to that so let's see how we get this composite function so if we have start off with a variable X and we apply the function G to it because G is the ones closest to it so we're going to apply G to it first of all that would give us x squared plus 5. then we're going to apply the function f in other words we're going to substitute this x squared plus 5 in for the X in our function of f of x and then that would give us our composite function so that means that our composite function f g of X would be equal to 2 bracket and instead of X we're going to write x squared plus 5 so x squared plus five we're then going to close brackets and then write plus one now we're going to expand and simplify this so 2 times x squared would be 2x squared and 2 times 5 is 10 so plus 10 and we've still got A plus one and then if we add the 10 and the one together we get that's equal to 2 x squared plus 11. so a composite function f g of X is equal to 2x squared plus 11. okay let's look at our next example so our next example says find g f of x in other words we're going to apply the function f first of all and then apply the function G so if we apply the function f first of all we get 2x plus 1. and then if we apply the function G so let's substitute that in for the value for x and function G and that will give us our g f of x so g f of x and that would be equal to instead of x squared we're going to write in Brackets 2x plus 1 close bracket squared then plus 5. and then if we expand and simplify this we'll find g f of x and let's see what we get so I've written the bracket out twice because we're squaring it we're going to multiply by itself and when we expand these brackets we get four x squared plus two X plus two X plus one and we've still got our plus five and then simplifying our like terms we get 4x squared plus 4X plus 6. so that means that g f of x is equal to 4x squared plus 4X Plus 6. now if you're asked to find something such as gfr4 what you then do is substitute the value for 4 into this and you do 4 times 4 squared plus 4 times 4 plus 6 or if you're asked to find g f of minus seven you would do four times minus seven squared plus four times minus seven plus six and so on and that's it if you're asked to find a composite function such as f f of x then what you would do is get f of x and substitute into f of x so you would write 2 bracket 2x plus 1 close brackets plus one and so on now if you're studying for OC or higher if you're doing composite functions you aren't expected to know that notation of f g of X and g f of x or you'll often see in OCR how is you maybe give them two function machines function M function B and then you have to find a composite function and what you'll do is you'll do one of the function machines you get the result you then put it into the other function machine and that'll give you a composite function okay let's have a look at our next topic so next topic is inverse functions and that's for your 369 corporate Maps so sometimes you might be given a function so is f of x and here we've got f of x is equal to 3x plus 2 and we need to find an inverse function so in other words if we know what the result is of f of x how we would go back to find the value for x and if I want to find an inverse function what I do is I look at my function my f of x equal to 3x plus 2. and I write instead of f of x equals I write y equals three X Plus 2. and as you can see here Y is the subject and what I'm actually going to do is I'm going to make x a subject here and once I make x a subject that will tell me what my inverse function will be so let's make x a subject here so let's take away 2 from both sides so minus 2 and minus two that would give me y minus 2 is equal to 3x and then if I divide by 3 and divide by 3 so divide by 3 and divide by 3 I get y subtract 2 all over 3 is equal to X so if me at X is subject so this tells me my inverse function so my inverse function of X my f minus 1 of X will be equal to and instead of writing y minus 2 over 3 I'm going to write x minus 2 over 3 and that's it let's just have a look at the chord Mouse revision card let's just make it a bit bigger so you've got given f of x equal to 3x plus 2 find F minus 1 of X the inverse function so step one let y equals the function step two make x a subject and step three write F minus one of X in terms of X so replace the Y with X and that's it now if you're studying for OCR how you aren't expected to know that function notation that F minus one of X so what that you will often be given is a function machine or a diagram that shows a function machine and to find the inverse function you'll work backwards to find that inverse function and that's it so that's inverse functions okay let's have a look at our next topic okay let's have a look at our next topic and our next topic is quadratic graphs in this video 264 on corporate Maps so what we're going to do is we're going to draw this quadratic graph and whenever we draw quadratic graphs what you'll find is instead of being straight lines they'll either be this u-ship or a parabola like so this u-ship Parabola or they could be an n-shaped Parabola like so and so whenever you draw a quarter on a graph it will either have this U shape or this N Shape and it depends on whether this x squared is positive or negative so because this is a positive x squared I can tell it's going to be a u ship if it was minus x squared it would actually be an engine okay so the question asked us to complete the table for y equals x squared plus X minus four so we've got this X Y table and we've got lots of points the reason we've got lots of points is it's not a straight line any water curve so we're going to need more points so we can draw a nice curve through them and we've got our X values of minus 2 minus one zero one two and three and what we're going to do is we're going to work out our y values and to find our y values we're going to square X add X and take away 4. so let's start off with three so we're going to do 3 squared plus 3 and then take away 4. so 3 squared is 9 plus 3 is then going to be 12. take away 4 is it so that means that would be it and doing that for the rest of the points we'll get so we've now found our coordinates now what we're going to do is we're going to plot them so so we've now got our points and as you can see they make a curve so what we're going to do is we're going to draw a nice curve through them so we're going to do this on the computer and that can actually be quite tricky on the computer so um please bear with me whenever I'm doing it so it might not be absolutely fantastic but I'll do the best I can hey that's pretty good um so that's it so we've drawn a quadratic graph and we've drawn a nice curve for our points and if you want more practice video 264 Maps will give you more practice on that and remember you've got that bumper pack of questions so um ultimate video practice question booklet and if you go to that in the description below and click on it there'll be questions there on the quadratic graph so I'll give you a chance to draw one yourself okay so let's have a look at our next topic so our next topic is solving quadratics graphically so it could be something like this where using your graph estimate the values of X whenever Y is equal to negative three so if we go back to our graph we need to now draw the graph of y equals negative 3 because we're told to find the values of X whenever Y is equal to negative 3. so if we draw that graph of y equals negative that's going to be all the coordinates with a height of negative 3 such as 0 negative three one negative three two negative three three negative three and so on the graph y equals negative three is a horizontal line that passes through negative three is on the y axis so y equals a number graph will be a horizontal line and it'll be horizontal going through whatever the number is so if it was negative y equals negative three it'll be horizontal line going through negative three on the y axis because all the coordinates on that line have a height of negative three if it was y equals eight it would be horizontal line going through eight and so on okay but in our question we were asked to find an estimate of the values of X whenever Y is equal to negative three so if you have a look at our Parabola and our straight line our y equals negative three you can see they actually meet each over twice they meet here and they meet here and what we're actually going to do is we're going to go up to the x-axis so we're just going up to here there I'm going up from this intersection point the second intersection here we're going to go up again to the x-axis there and we're going to read off what those values would be and there would be our two solutions and it says estimate because remember you we've drawn a curve and we've drawn it freehand so it's not going to be a perfect discretion it's an estimate to answers so let's have a look and see what we've got so let's start off with this one we've got zero we've got one and then the middle is not 0.5 I think that's about 0.6 so I'm going to say x equals 0.6 would be one of our Solutions and this one over here well we've got zero we've got negative one we've got negative two that's negative 1.5 I think that's about negative 1.6 so negative 1.6 so our two estimates would be x equals 0.6 or x equals negative 1.6 so it will be our two estimates so x equals 0.6 or x equals negative 1.6 and just to also point out that the question could be written in a slightly different way rather than saying find estimates of the value of x whenever Y is equal to negative three it could have had the equation of the graph that we drew in the first question this x squared plus x minus for and it could have just written equals negative 3 and then what you do is you look at that value you just draw the graph of y equals negative 3 like we did so we just draw a horizontal line going through negative three and go up to the x axis and find those values and another thing you could be asked whenever you're dealing with quadratic graphs is to find the roots of the quadratic so here we've got a quadratic graph and it's y equals x squared minus x minus two and we've been asked to use this graph to find the roots of x squared minus x minus 2 equals zero so when the quadratic equals zero and that means where the quadratic has a height of zero and that means where it crosses the x-axis so if you have a look at where this quadratic crosses the x-axis it crosses the x-axis at negative one and two so means the roots of this equation will be x equals negative one or x equals two and that's it so to find the roots of the quadratic if you've been given the graph and then they've written the graph equals zero just find where that graph crosses the x-axis and another thing that they sometimes ask you is to write down the coordinates of the turning point and that would be the coordinates of this point here so you just right down the coordinates off that point and that's it okay let's have a look at our next topic and that's sketching quadratics so if we had a quadratic such as y equals x squared minus 7x plus 10 it can be useful to be able to sketch this and whenever you sketch in a quadratic first of all I would consider the shape with the quadratic whether it is a U-shaped Parabola or an anterior Parabola and what determines if it's a u-ship parabola or an n-shaped Parabola is this x squared term if this x squared term is positive so for instance x squared or 2x squared or so on it'll be a U-shaped Parabola if it's negative so if there's a minus x squared or minus 2x squared it will be an entry problem so that's very important whenever we're sketching a quadratic and what's also important is where the quadratic meets the y axis and if and where means the x-axis so let's start off by looking at this quadratic y equals x squared minus seven X plus 10 and let's first of all consider it well it's going to be a u-ship parabola because it's an x squared it's positive x squared okay next let's find where the graph meets the y axis or sometimes it's called the y-intercept so to find the y-intercept or where the graph means the y axis what we do is we consider where x equals zero because we know that the y axis is whenever x equals zero so if we substitute in x equals zero into our equation we'll find this y-intercept quite easily so let x equals zero and if we do that we get y equals zero squared minus seven times zero plus ten so we've got y equals zero squared minus seven times zero plus ten and when we work with that we get y equals zero Square to zero seven times zero is zero so zero take away zero is zero plus ten is ten so whenever x equals zero y equals ten so that would be the coordinate 0 10. summaries are quadratic y equals x squared minus seven X plus ten we'll pass through the point zero ten so point zero ten will be we're going zero across and tenant will be here so that's the point zero ten okay so we now know where our quadratic means the x is let's see if we can find where it meets the x axis so the final word meets the x axis well we know all the points on the x axis have a height of zero so if they've got a height of zero that means that y equals zero so let's let y equal zero and whenever we let y equal zero that will give us 0 equals x squared minus seven X plus ten so we've got a quadratic equation and if we can solve this we can find where it meets the x-axis so let's try and solve this using factorization so 0 equals and bracket bracket bracket bracket let's put X in the front of both brackets and then we're looking for two numbers that multiply together to give us 10 and add together to give us negative seven well it's going to be negative 2 and negative 5 because negative 2 times negative 5 is 10 and negative 2 plus negative 5 would be negative seven so let's find our Solutions well if we know the two things times together to give a zero if that means the either bracket would be zero so that means that here x would be equal to two because two take away two is zero so that means one answer will be x equals two or and then look at the upper bracket that means that X could be equal to five so we've got our two solutions two and five that means our quadratic will pass through the coordinates two zero and five zero and there will be our two points where the quadratic graph crosses the x-axis so let's put those on our graph and then before we do that I'm just going to actually move something I'm actually going to move this to the other side and I'm now going to put on where the media x axis 2 0 and 5 0 so that'll be two zero roughly there and five zero roughly there and if we do a nice quadratic through that nice u-shaped Parabola through those points it would look something like this and that's it and let's label our coordinates two zero and five zero and that's it so we've done a sketch of the quadratic y equals x squared minus seven X plus ten and we've shown the shape of the quadrilateral so you see a parabola we'll find out where met the y axis so the Y intercept which was 0 10 and we found that this quadratic did meet the x axis twice and at the points two zero and five zero sometimes whenever you solve the quadratic you might find there's only one solution and what that would mean is the quadratic touches the x-axis and goes back up and sometimes whenever you try and solve the quadratic you might find actually there's no Solutions and that will mean actually that the quadratic graph the parabola will be above the x-axis okay let's have a look at our next topic so next topic is quadratic graphs and we're going to be looking at videos 267 and 267 D in corporate maps and those videos focus on Drawn quadratic graphs so we're going to draw a quadratic graph here this y equals x squared plus x minus 1 on the set of axes and what we're going to do is we're going to write down the coordinates of where this quadratic meets a straight line graph so first of all let's draw this graph so let's do an X Y table so X and Y and for the axes we've been given we've got the X values of minus two minus one zero one two and three so let's put those in the table minus two minus one zero one two and three so what we're going to do with all these numbers is we're going to square them then we're going to add the number again and then we're going to take away one so looking at three we're going to do 3 squared which is nine then we're going to ad3 which is 12 take away 1 which is 11. 2 squared is four plus two is six take away one is five whenever X is equal to one one squared is one plus one is two take away one is one zero zero squared is zero plus zero zero two where one is minus one next negative one or negative one squared is one plus minus one well one plus minus one is zero take away one is negative one and actually without actually working this one out I can see we're gonna have a value of one in here because with this quadratic we know it's going to be symmetrical and because we've got a height of negative one a negative one the next two will be symmetrical so if this one's one this one should be one and negative three will be five and so on but let's just check it negative two squared is four plus negative two is two take away one is one and that's it now let's pull out our points and now we'll put a lot of points we're going to do a nice Parabola through them and now please bear with me whenever it is on the computer it's quite tricky to actually do it uh nice and accurately okay so that's our Parabola is the best I could do on the computer and we've been asked to write down the coordinates of where this quadratic this Y equals X Gray plus x minus one intersects with this straight line graph y equals three X plus two now I could do an X Y table for this so I could do X Y zero one two three and multiply the numbers by three and add two I'm actually going to draw this graph using a slightly different technique I know it meets the y axis it's got a y intercept of two so I'm going to start at two and then I'm going to consider its gradient which is three so I'm going to go one unit across so one unit across and I'm going to go up three so that'll be five I'm going to go another unit across and go up three that'll be eight and again because the gradient's three I'm gonna go another unit across so going from two to three and go up three so bring me to 11 and so on so I've got these four points I'm going to get my ruler and pencil draw a nice straight line through them and that's it so we've drawn the graph y equals three X plus two and we've got our quadratic and we've been asked to write down the coordinates of where they intersect or where they mean so if we have a look we can see they meet twice they meet here at the coordinate negative one negative one so negative one negative one and they also meet again up here at three eleven and that's it so they're the coordinates of the two points of where they meet okay let's have a look at our next topic so next topic is finding graphical solutions to quadratics and that's video 267 D and these are slightly more complex than the ones we looked at previously so here we've got the graph of y equals x squared plus two x minus three and it says by drawn our previous straight line use your graph to find estimates to the solutions of x squared plus 2X minus three equals X plus two now whenever we looked at simultaneous equations earlier like this one here where we've got y equals and Y equals I mentioned that whenever you solve this whenever you solve that it will give you the coordinate software they intersect each other so if you had y equals something and Y equals something if you then solve the simultaneous equations you find out where they intersect each other so if we go back to our graph as you may notice here we've got our quadratic our x squared plus 2X minus 3 on this left hand side of the equation and on the right hand side of the equation we've got X Plus 2. so if we want to solve this equation what we can do is we can draw the graph of y equals x plus 2 on this grid and then if we find where they intersect each other we can then go to the x-axis and find those values of X and and there'd be estimates because we're using the graph work so let's draw the graph y equals X Plus 2. so we're going to use an X Y tip up and let's choose some values for x or minus 1 0 1 and 2. and to find our y values we're going to add 2. so -1 add 2 is 1 0 add 2 is 2 1 R2 is 3 2 R2 is 4 and so on so let's plot these coordinates now let's get a return pencil and draw a line through those and I straight line through those points and we've just drawn that straight line of Y equals X Plus 2. and if you notice here we've got y equals x squared plus 2X minus three that left hand side of the equation and the right hand side of the equation we've got that X plus 2 that y equals so we've got our Y equals X plus 2 and a y equals x squared plus 2X minus three and we've drawn them on the graph and we can see they intersect each over twice so let's find where they intersect so if we go down to our x axis so we're going down to our x-axis so between one and two we've got 10 little squares and that's the eighth one so it's going to be 1.8 and then if we look at the F1 they intersect here so we've got up to the x-axis that'll be here which is at minus two minus 2.5 that'll be minus 2.8 so our approximations our Solutions will be x equals negative 2.8 or x equals 1.8 and there are estimates to our Solutions and that's it so whenever you've got the quadratic equals something else if you just draw that something else as long as the quadratic you're given in the question drawn on the grid is the same as the one in the equation if you just draw that other line and find where they meet each other and then go to the x-axis you'll then find your approximations okay let's have a look at a more complex one now so we've got the graph y equals x squared minus x minus 2. so drawn force is y equals x squared 3 minus x minus two and the question says by drawing an appropriate straight line use your graph to find estimates to the solution of x squared minus 2x minus 1 equals zero but what we're trying to solve doesn't actually have x squared minus x minus two it's actually slightly different it's got x squared minus 2x minus 1 equals zero but it does equal zero if we were trying to find where straight line intersects a quadratic we would normally put them equal to each other but you notice that this equals zero so that means that whatever the linear was we've taken it away from both sides to then leave us with x squared minus 2x minus 1 equals zero so if we take what we've been given this y equals x squared minus x minus two and we write down what we've been given with the zero at the front so zero equals x squared minus two x minus one if we take these away from each other that will tell us the straight line graph we should draw so let's take those away from each other and see what we get so y take away 0 is y so x squared minus x squared they would cancel out then we've got minus x sub subtract minus 2X or minus x minus minus 2x will mean minus X plus 2X and minus X plus 2X would just be X and then we've got minus 2 minus minus 1. so whenever we do minus 2 minus minus 1 we're going to add one on so minus two add one would be minus one so that tells us we should draw the graph of y equals x minus one and we can check this if we took the quadratic in the question this x squared minus x minus 2 and we put it equal to x minus 1. and then if we take that away from both sides to make it equal to zero hopefully we'll be left with this so if we take our X from both sides we'd have x squared minus two x because if you take away X from both sides that will be minus X take 1 over X B minus 2X and if we added 1 to both sides as well we'd have minus two plus one which would be minus one equals zero and as you can see here then that gives us that quadratic we're trying to solving the question so that's it so if you're given a quadratic graph y equals something and you then ask to solve something which is slightly different and it equals zero if you're at the zero at the front and then take it away from the Y equals it will tell you what straight line to drop so if we draw the line of y equals x minus one and we find where that straight line intersects y equals x squared minus x minus two it will give us our solutions for the quadratic we're given so let's do that so let's do our X Y table X and Y and we'll go for zero one and two and to find our y values we're going to do x minus one so we're going to take one away from each of these so it'll be minus one zero and one and let's plot those points and that's it so we've drawn the graph y equals x minus one and we want to find where that graph intersects the quadratic so let's find the points of intersections that's here and it's also here so we're looking at this point we've got well between two and three so this one two three four five squares that means we're going up in point twos so it's 2.2 2.4 so this will be 2.4 and this point here would be well let's go down in point two so it's a minus 0.2 minus 0.4 so our two solutions would be x equals minus 0.4 or x equals 2.4 and that's it if you do want to watch another video on that topic video 267 deal called Maps we'll go through this in a bit more detail okay let's have a look at our next topic so our next topic is reciprocal graphs and this video 346 in corporate Maps so a reciprocal graph is something in the form of y equals a number and then over X so something like y equals 2 over X and you will have heard of the term reciprocal before whenever you're dividing by fractions because whenever you're dividing by fractions instead of dividing by a fraction you multiply by the reciprocal that's what you get when you flip it over and that's what we mean whenever we've got a graph in this format y equals 2 over X so we're going to draw a graph and we've got an X Y table and we've got quite a lot of points but because it's this curves involves not nice straight lines so we're going to substitute in these values of X into our equation to get our values for y so let's start off with the Positive values so let's use x equals five so two divided by 5 is 0.4 then we've got 2 well 2 divided by 2 is 1. then we've got one two divided by one is two then we've got 0.5 so we've got two divided by 0.5 well there's four halves and two so that would be four now let's move on to our negative numbers so we've got two divided by negative five well a positive divided by a negative is a negative so it's going to be a negative answer and 2 divided by 5 is 0.4 it's going to be negative 0.4 well for the angle 2 divided by negative two well that's going to be well positive divided by a negative would be a negative so negative one now we've got 2 divided by negative one that'll be negative two and finally 2 divided by negative 0.5 would be negative 4. okay so we're not going to have points now let's plot them so we've got five across to 0.4 up so five across would be here 0.4 up but we've got ten little squares for two two divided by 10 is no point two so it'll be two of them upwards or two of them up so be like that somewhere like that then we've got two on So 2 across one up would be there then we've got one across two up so one across two up and then we've got 0.5 across four up so 0.5 across would be halfway in between and four up would be there so we've now got our point now what we're going to do is do a nice curve through them now this is quite tricky on the computer so just bear with me okay this won't be perfect but it's just gonna be a nice curve but looks something like so that's not too bad and as you can see the curve it never actually reaches the y axis because if you've done two divided by zero well that's on the find you can't divide two by zero so this curve this Curve will never actually reach the y axis and as you divide by smaller smaller and smaller decimals so so 2 divided by 0.1 would actually be twenty so that's really how 2 divided by 0.01 would be equal to 200. so what happens is the graph will just shoot off to infinity and it would never actually reach the y axis and similarly whenever we're going across horizontally to the right here and because we're doing two divided by something even if we divided by a million there'll still be a tiny tiny number there would be no point no no no no and so on so the graph will approach the x-axis but never actually reach it they're what we call asymptotes it's quite a fancy word but that's just what they're called okay and these points on our left hand side of the table we've got negative 0.5 and negative four so that would be there then we've got negative one negative two so negative 1 negative 2 would be there negative 2 negative 1 would be there negative 5 negative 0.4 would be there so we've got our points and we're going to draw a nice curve through them so that would be the graph of y equals two over X and if you had something like y equals 4 over X it would have the same shape but it would just be slightly further out so it'd be slightly higher up and slightly further down and so on okay let's have a look at our next topic so the next topic is cubic graphs so here we've got the graph y equals X cubed so cubic is whenever we've got this cubed and here we've got Y equals X cubed so here we've got our X Y table so let's Cube all these values to get our y values two times two times two is eight one times one times one is one zero times zero times zero is zero negative one times negative 1 times negative one well negative 1 times negative one is one and times that by negative 1 is negative one because the negative times a negative times a negative is a negative then we've got negative 2 so negative 2 times negative 2 is 4 times negative 2 is negative eight and finally negative or three times negative three times negative 3 would be negative 27. and now let's pop these points on our grid let's draw our curve so it'll come upwards we'll go through the points and then what will happen is it'll flatten out and come through zero and then curve back upwards and look something like that and excuse my sketch it is free hand on the computer is quite tricky and now let's have a look at our y equals negative X cubed so what this means is we've got a cube our value of x and then we're going to make it negative so we're going to do two cubed so 2 times 2 times 2 is it but we're going to make it negative so it's going to be negative 8. then we've got one times one times one which is one but we're going to make a negative so negative one then we've got zero times zero times zero zero well that's just zero then we've got negative one times negative one times negative one that's negative one but we're gonna make it negative for that means then because already negative means we're going to make a positive it's going to be one and then we've got negative two so negative two times negative two times negative two is negative eight but then we're going to make a negative but it's negative ready so it makes it means make it positive we're going to change the sign so that's it and then finally we've got negative three times negative three times negative three which is negative 27. we're going to make a negative it's already negative so it's going to be 27. and another way to consider this negative signs remember if you've got something like X cubed that means One X cubed or one lot of X cubed if you've got Negative X cubed that's the same as negative One X cubed so really what we're doing is we're finding our value for x cubed and then we're multiplying it by a negative one so here we had two cubed which is here multiply but negative one that's negative eight whenever we had our negative one cubed that's equal to negative one when we times it by negative one that's one and so on and just what it does is just it just changes the science whenever there's a negative sign in front of it now let's pull out our points okay we're now plotted on the points now let's try and draw a curve and that's meant to be a curve and what you'll find is it's the same as the execute graph but it's just flipped around the other way okay let's have a look at a more complex cubic and here's an example of chord major vision card where there's Y equals X cubed minus three x minus one and what we've got is we've got the table of values so we've got our x coordinates our three two one zero minus one minus two minus three and so on and we've worked out that number cubed minus three times that number take away one and we've got our values for y and whenever you plot them it has this shape so there's a simple cubic graphs which will look something like this where it's quite steep flatten out pass through the origin and then curve upwards that's typically whenever you get something such as Y equals X cubed or Y equals X cubed plus one or something like that but if you do have a more complex cubic where you've got x times or X squareds and so on it could look like this where it actually comes up reaches this point this maximum comes down again which is the minimum it comes up again and so on okay let's have a look at our next topic so our next topic is exponential graphs now this video 340 five in corporate Mouse so here's an example of an exponential graph where we've got y equals 2 to the power of X and you may notice that the x is in the power so that's why it's exponential you've got a number positive number and you've got an X is the power so let's have a look at our table of values so if we had an X Y table where we've got the values of negative 2 negative 1 0 1 2 3. let's substitute those values of X in and find out what we were to get so if we had a power of 3 we would have 2 cubed and 2 cubed is it so that means that y would be equal to eight if we had a power of 2 we would have 2 squared and 2 squared is 4. if we had a power of one we'd have 2 to the power of 1 and that's equal to two Okay next one whenever X is equal to zero well that's an important point because that's going to be the y-intercept it's going to be where it crosses the y axis and 2 to the power of 0 is equal to one that's quite important because if it was y equals 5 to the power of X it'd be 5 to the power 0 which would be one if it's Y is equal to 10 to the power of X that would also cross at one even if it was Y is equal to 0.5 to the power of x 0.5 to the power of 0 is also equal to one so that's a very important Point it'll cross through the L across the y axis at 0 1. okay next whenever X is equal to negative one well it's going to be 2 to the power of negative one that means we would have 1 over 2 to the power of one and two to the power of one is just two so the answer would be a half so our next point would be negative one a half and finally if with the value for x negative 2 we'd have 2 to the pi of negative 2. and that would be 1 over 2 squared and 2 squared is four so that'll be a quarter that's it now what's important to notice is if we had another point if we had x equal to 4 well that would be 2 to the power 4 which is 16. if we had five that'll be 2 to the power 5 which is 32. you may notice these values getting very big very quickly and that's why you might have heard of the term exponential growth because as the values for X just increased by one each time you're going to get very large values where it crosses through the points 0 1 the Y intercept it goes through the points one two two four three six four sixteen five thirty two and so on so the right hand side of this graph gets very steep very quickly and on the left hand side well we're going to have minus one a half minus two a quarter and what happens is this graph approaches the x-axis but never reaches it so it goes down like so okay so this is the graph of y equals 2 to the power of x if you had a graph y equals 10 to the power of X it looks similar it would still pass through zero one it would get steeper quicker on the right hand side and it will go just a bit shallower quicker on the left hand side died but it was still just approach the excess has not made it if you had something in the form of y equals and then something else such as 0.5 to the X what would happen for that graph is it would actually be go the other way around okay let's have a look at our next topic so our next topic is the trigonometric graph so the trigraphs and there are videos 338 339 and 340 in corporate Maps Okay so let's have a look at our first graph so our first graph is y equals sine of x so this is the y equals sine of x graph of the sine graph and it passes through the points 0 0 91 180 0 217 negative one 360 degree zero and it carries on 450 one 540 zero and so on and it just carries on on that pattern and it would carry on in the same pattern to the left as well so we could die on it up and so on okay so let's have a look at our next graph so this is the y equals cos x graph so this is the cosine graph and it passes through the point zero one ninety zero 180 degrees negative 1 270 degrees zero 360 Degrees 1 4 450 degrees zero and so on it just carries on on that pattern and to the left it carries on in the same way so it just comes down and so on okay let's have a look at our next graph so next graph is y equals Tan x so the tangent graph so this is the time graph it passes through the point zero zero and then curves up to infinity and just keeps going up here then we've got an asymptote so the graph doesn't have a point at 90 degrees then it starts down here at negative infinity and comes up and passes through the point 180 degrees and zero and then curves up again up to infinity and then again we've got another asymptote for 270 degrees because if you try to work out the tan of 270 degrees try in your calculator you'll get an error message and then it comes up from negative Infinity again and passes through the point 360 degrees and so on and then I'll just cover up to infinity and so on and then have another asymptote about 450 degrees and so on and again if we were to draw the left hand side of this graph it would just curve down and then never asymptote at negative 90 degrees and so on so this is the 10x graph that's important to know what these graphs look like and be able to spot which ones which so the time graph is quite unique so you can use spot the time graph quite easily especially with his asymptotes and then in terms of the sound graph and the cosgraph I just remember that the sine graph passes through the origin whereas the closed graph starts at the point zero one and that helps me recognize which ones which and that's it okay let's have a look at our next topic so our next topic is transforming graphs and this video is 323 and 324 in corporate Maps so GCSE level there's four Transformations that we need to know so if we're given the graph of y equals f of x we need to know how to sketch or how to draw y equals minus f of x y equals F of minus x y equals f of x plus a and Y equals f of x plus a so let's have a look at each of these now I'm going to use the chord manager revision cards to help us so let's start off with our first transformation so our first transformation is y equals minus f of x so if we want to sketch y equals minus f of x we reflect the graph of y equals f of x in the x axis so if we have this curve this black curve and we reflect it it becomes this green curve when we reflect it in the x-axis now the points on the x-axis will stay where they AR and the point above will go below and the points below will go above so let's have a look at a question here so we've got the curve y equals f of x here and we're going to sketch the curve y equals minus f of x so to sketch it we're going to reflect this graph in the x axis so the point minus 6 1 will go to minus six minus 1 so we'll move to here minus six minus one and this curve this part all the points that were above the x axis will go below the x-axis like so and then the points are below will go above so like so and then this side will look the same on the other side so we'll reflect the graph this y equals f of x and the x axis and it would look something like this so what happens is it flips the graph or reflects the graph vertically like so so let's have a look at our next transformation so an X transformation is y equals F of minus X and Y equals F of minus X reflects the graph of y equals f of x in the y axis so as you can see here if we started off with this y equals f of x this black curve if we reflect it in the y axis the points are on the right hand side of the y axis when I move to the left hand side of the y axis and the points are on the left hand side will move to the right hand side and any points that were on the waxes will stay where they are so if we have this curve here y equals f of x and we want to sketch y equals F of minus X we'll reflect it in the y axis so any points are on the y axis we'll stay where they are the coordinates to the maximum Point were negative six one whenever we reflect that in the y axis what will happen is instead of being six to the left it'll be six to the right so it'll move to six one so move to here six one and the points are on the right hand side will move to the left hand side and the points are on the right hand side we move to the left hand side so it'll look something like this and that's it so this would be the graph of y equals F of minus X okay our next transformation our next transformation is if we're given y equals f of x if we want to sketch y equals f of x plus a we shift the graph a squares upwards or if we subtract a we move the graph a squares downwards so if we have this graph this y equals f of x here and we want to sketch the graph of y equals f of x plus two we're going to shift this graph all the points in this graph two squares upwards so let's start off with the maximum point it was negative six one so it's going to move upwards to negative six three like so and then all the points will move two squares upwards so it would look something like this and that's it so if we start off with the graph y equals f of x this black curve here and we added two we had a y equals f of x plus two we shift all the points two squares upwards like so if it was subtract four we would move it all the points four squares downwards and so so on okay and let's look at our last transformation our last transformation is if we have y equals f of x how to sketch y equals f of x plus a or x minus a and this one's it's a bit counter-intuitive if you're adding a instead of moving the graph a squares to the right we actually move it a squares to the left so if we had y equals f of x like so and then we wanted to sketch y equals f of x plus three we move the coordinates all three squares to the left so instead of it being four zero it moves to one zero instead of being minus four zero it's now minus seven zero instead of having a minimum point at zero negative four it's not negative three negative four so if you add it moves to the left and if you subtract inside the brackets it moves all the points a squares to the right okay so let's have a look at this one if we have y equals f of x and we then want to sketch f of x plus two what we're going to do is we're going to move all the points two squares to the left so instead of having a maximum point at negative six one it's now going to move two squares to the left so it's now going to be negative eight one instead of crossing the x axis here it's going to move two squares To the Left To here instead of crossing the x axis here it's going to move two squares To the Left To here and then we just sketch our curve and it looks something like this and that's it so our four Transformations that we need to know GCSE level are y equals negative f of x that's a reflection of the graph in the x axis y equals f of negative X that's a reflection of the graph and the y axis y equals f of x plus a if the plus a is outside of the brackets you just move the graph a squares upwards or if it's take where a you move it a squares downwards and then finally y equals f of x plus F this one's counter-intuitive instead of moving the graph a squares to the right we actually move it a squares to the left and if it's subtract we move it to the right okay that's it if you want more practice of these you can go to videos 323 and 324 in corporate Maps there's some questions in that book on transforming graphs so they'll be ideal to practice now as well and also remember if you have the chord Mouse revision cards you've got those chord Mouse revision cards that include transforming graphs and they'll be quite useful to pin up on your wall and study and learn and memorize as well okay let's have a look at our next topic now what's fantastic is now we know how to transform graphs and also we know the topic complete in the Square we can use that information to help us find the turn and points and lines of symmetry of quadratic graphs so if for instance we were asked to use completing the square to find the coordinates of the permanent points of the graph y equals x squared minus eight X plus one we can use completing the square and transforming graphs to do that so let's consider this quadratic x squared minus eight X plus one and let's do complete the square on this so we would put our bracket then we'll write X and then instead of minus 8x well we're going to half the minus here it's just going to be minus four close brackets squared now minus 4 squared for minus 4 times minus 4 16 so we're then going to take away 16 because we always take away whatever number is in here squared and then we've still got our plus one so we've got y equals this now we've got minus 16 plus 1 that's going to be if we go up one from -16 that'd be minus 15. so we've got y equals and then in Brackets x minus 4 close bracket squared minus 15. now if we consider the graph y equals x squared that's the x squared graph it's a parabola and as you should know it should look something like this it passes through the origin and then back up like so so that's the graph y equals x squared now if we have a look at the graph that we've been given in the question are y equals x squared minus eight X plus one we've been able to write it in the form x minus 4 close bracket squared minus 15. now considering our Transformations as a graph we've got a minus 4 inside the brackets so what that does is it translates the graph it slides the graph four squares to the right remember if we take where it's counterintuitive instead of moving the graph to the left it moves it to the right so it translates it four squares to the right and then so that means that the graph would look something like this where I would go through and touch there and go up like that and then we have a minus 15 outside of the brackets that's then going to move our graph 15 squares down so let's actually translate it move it down 15 squared so it's going to move down to somewhere like that so whenever we have the graph in the form of y equals x minus 4 close bracket squared minus 15. we've got a minus 4 inside the brackets so it moves the graph four squares to the right and then we've got our minus 15 so it moves at 15 squares down so the coordinates of this turning point it was the origin we moved the four squares to the right so it's going to be four and then we moved it 15 squares down so it's going to be minus 15. so the coordinates of that Turning Point would be 4 minus 15. and that's fantastic being able to use completing this square and our Transformations are graphs to find what that graph would look like that quadratic also if we were asked to find the line of symmetry of that quadrant I like what we could do is we just consider if that's the turn and point at four minus 15 the line of symmetry would pass through that as a vertical line like so and that is the line x equals four it's a straight line passing through four in the x-axis so if you were asked for the line of symmetry it'd be x equals 4. okay and that's it okay let's have a look at our next topic so our next topic is solving inequality so that's video 178 in corporate maths so our first question says solve 5x is bigger than 30. now whenever I'm solving inequalities I use a similar approach to I would use for solving equations so here we've got five X is bigger than 30. now we don't want 5x we want just X so we want to get rid of this multiplied by 5. so to get rid of this 5 we're going to do the opposite which is divide by five so we're going to divide both sides by 5. so if we divide both sides by 5 well 5x divided by 5 would be 1X or X and that would be bigger than and if we done 30 divided by 5 that'll be six so answer would be X is bigger than 6. okay this time we've been asked to solve 3x plus 4 is smaller than or equal to 31. so if this was an equation the first thing I would want to do is get rid of this plus 4 so let's take away 4 and take away 4 from both sides of the inequality so we'll go 3x plus 4 we're taking away four so that leaves us with 3x and then we've got our less than or equal to sine and then we're taken away 4 31 take away 4 is 27. so we've got three x is smaller than or equal to 27. now we don't want 3 times x we just want X so let's divide both sides by three so dividing by 3 and dividing by 3 would give us well on our left hand side we're dividing by 3 to get rid of this Times by three so we're just left with X and then we've got a smaller than or equal to and then if we do 27 divided by 3 well 27 divided by 3 is 9. so if we've been given the question 3x plus 4 is smaller than or equal to 31 well we know that X will have to be smaller than or equal to 9. that's it okay let's have a look at our next question so this time we've got an inequality with letters on both sides so remember whenever we're solving equations with letters in both sides we want to get rid of the lowest number of x's so here we've got e at X plus one is smaller than 10x track six so with this inequality I'm going to get rid of the lowest number of x's which is our 8X so let's take away 8X from both sides of this inequality so when we do that we had 8X plus 1 we took away the e at X so we're just going to be left with plus one or one and that's smaller than well we have 10x take away 6 and we're taking away e and X so it's going to leave us with 2x so we've got 2x take away 6. okay now I like an equation I would want to get rid of this take away 6 so let's add 6 to both sides of this inequality so 1 plus 6 is 7. so we've got 7 is less than 2X and we added six to get rid of the minus six so we're just left with 2X and finally we want the X on a tone so we want to get rid of this multiply by two so let's divide by two and divide by two and whenever we divide 7 by 2 we get 3.5 so that's 3.5 is less than x and that's it so we've got 3.5 is less than x now we might want to write this the way around so we've got the X at the front so if we write this the way around just be aware whenever we relative way around we've got X now we've got 3.5 is less than X or if we read the way around we'll get X is bigger than 3.5 so we need to flip that inequality sound right whenever we're writing the way around because X is bigger than 3.5 that's it so whenever three solvent inequalities is similar to solving equations but you just need to be aware of whenever you are wanting to flip it around so in other words whenever you want the X at the front rather than the number at the front whenever you flip around that you need to be aware of that inequality sign okay let's have a look at our next topic so the next topic is inequalities but this time we're dealing with number lines so this is represented inequalities on our number line and it's also looking at how to solve inequalities and then represent the answer on a number line and this would be video 177 on corporate Maps so let's look at how we'd represent some inequalities on a number line so here's a number line and our first inequality is X is greater than negative one so X is bigger than negative one so first of all what we do is we go to negative one and because X is greater than negative one it can't actually be negative one so what we do is we put a Hollow Circle at negative one and because X can be bigger than negative one what we do is we put an arrow to the right like so and we just did an arrow and that tells us that X can be any value that is bigger than negative one and because the circles Hollow it shows us that it can't actually be negative one so it's Any number greater than negative one okay this time our inequality says X is greater than or equal to two now this time we've got X is greater than or equal to two so that means it can actually be two and any number that's bigger than two so instead of doing a Hollow Circle what we do is we actually do a circle and color it in so she added in circle and then a narrow to the right like so and what that means is the X can be 2 or any number are bigger than two so next inequality is X is smaller than negative two so this time what we're going to do is because it's smaller than so it's not equal to we do a Hollow Circle and the arrow points to the left because it's smaller than negative two okay and our next inequality so our next inequality is X is smaller than or equal to three so because the smaller than or equal to instead of doing a Hollow Circle we just shaded in circle up three and we do an arrow to the left like so and that's it so that's how we represent inequalities where we've just got X is bigger than a number or bigger than equal to a number or less than a number or less than or equal to a number and if we had an inequality such as this where X is bigger than one but less than or equal to three and I wanted to represent the inequality of that on the number line because we know that X is bigger than one but we know it's bigger than one so it can be equal to one so we do a Hollow Circle at one and because X is smaller than or equal to 3 because it can be three with a shaded in circle of three and because X can take the values in between those numbers We join them up like so so if we wanted the representative inequality such as X is bigger than one but less than or equal to three we would do a Hollow Circle at one a shaded in circle of three and that represents that inequality on a number line okay let's look at our next question okay so let's have a look at another type of inequalities question so we've been asked to list all the integers that satisfy the inequality 2N is greater than 7 but less than or equal to 14. so we're looking for integers so numbers such as negative two negative one zero one two three and so on this satisfy this inequality now if we have a look at this inequality we've got 7 is less than 2N that is less than or equal to 14. now we've got 2N here what I'm going to do is I'm actually going to divide this inequality by 2. I'm going to divide everything by two and that will give me n instead of 2m so if I divide everything by 2 7 divided by 2 is equal to 3.5 is going to be less than and then 2N divided by 2 would be n and then we've got our less than or equal to symbol and then we've got 14 divided by 2 that's 7. so so we know that n is greater than 3.5 but less than or equal to seven so we need to list all the integers that are greater than 3.5 but less than or equal to 7. so let's list our integers so our first integer that is greater than 3.5 would be four and then we've got five and then we've got six and then it says less than or equal to seven so seven is going to be an integer that works so I integers to satisfy this inequality would be four five six and seven and that's it okay let's have a look at our next topic our next topic is graphical inequalities and I really like this topic in this video 182 in quick maths so here we've got a set of axes we've got our x axis and our y axis and we've been asked to show the region that satisfies X is smaller than or equal to seven Y is bigger than or equal to 3 and Y is less than or equal to half X plus two so what we're actually going to do is we're going to draw these lines to begin with we're going to draw the line off x equals 7. we're going to draw the line of y equals 3 and we're going to draw the line of y equals half X Plus 2. and so we'll draw those lines to begin with okay and then once we draw those we'll then be able to see which region satisfies these inequalities so let's start off with x equals seven let's look at the line of x equals seven so x equals 7 is a vertical line passing through 7 on the x-axis so here we've got our x axis and here we've got seven and the line x equals seven is a vertical line that passes through 7 on the x axis you might have seen it before whenever you've done Reflections when you're reflecting ships on axes so that's the line x equals seven because all of the points on that line have an x coordinate of seven there's seven across one up seven across two up and so on okay all right next time y equals three so we want to draw the line Y equals three and that's going to be a horizontal line passing through three on the y axis so here we've got the y axis here we've got the number three and then I'm y equals 3 will be a horizontal line that passes through that point so let's get our written pencil and let's draw that line so we've drawn the line Y equals three and finally we need to draw the line of y equals a half X plus two so what we've done is I've just taken the inequalities we've been given change all the inequality signs to equals and we're just going to draw those lines to begin with so what we're going to do is we're going to draw this line of y equals a half X plus two so I'm going to do an X Y table X and Y and choose some values for X so 0 1 2 3 and we're going to find the Y value it's half for the x value plus two so half of 0 is 0 plus 2 is 2. half of 1 is 0.5 plus 2 is 2.5 half of 2 is 1 plus 2 is 3 and half of three is one point five plus two is three point five so here we've got our coordinates 0 2 1 2.523 and 3 3.5 so zero two one one point five two three three point five and so on so let's get our ruined pencil and draw a nice straight line through those and it would look something like that okay now one thing to know in this question we have X is less than or equal to seven Y is bigger than or equal to 3 and Y is less than or equal to half X plus two all of them have equals to a bit and that's why we've drawn three solid lines if any of the inequalities were just greater than or less than then you would do a dashed line instead okay so in this question we've been asked to show the region that satisfies these inequalities now we've drawn the line of x equals seven but we've been told in the question that X is less than or equal to seven so that means our region will have to be less than seven so it means that our region will be on the left of this line so it can't be anything on the right hand side of this line because because anything over here would have an x value bigger than seven so what I'm actually going to do is just shade this right hand side to show that it can't be anything on this side of the line because X has to be less than or equal to seven okay let's have a look at our next one so we've got Y is bigger than or equal to 3. so we drew the line of y equals three and if we wanted a y is bigger than or equal to 3 it would be anything above that line anything below it would have a y value of something less than three so that means the elephant below this horizontal line couldn't work so we have to reject n of and below that line so let's just shade in just beneath it like this just to show that we're rejecting anything below that line anything below this line would have a y coordinate of less than three and we want to define the region where the height is bigger than or equal to three okay let's have a look at our last inequality our last inequality is y is less than or equal to half X plus two so here we've got the stagna line we've drawn the line of y equals half X plus two and one of these sides either the top of the bottom we want to keep and one side we want to reject now in terms of figuring out which side I want to keep or which side I want to reject what I do is I look at the inequality Y is less than or equal to half X plus two and what I do is I pick any point that's not on this line I I'm just going to pick the origin this point here the coordinate 0 0. okay so I'm going to choose this coordinate 0 0. and I'm going to switch it in x equals 0 and Y equals 0 into this inequality and I'm going to see if it works or not so let's substitute x equals 0 and Y equals 0 into this inequality so we had a y so instead of Y we're going to write 0 and that's less than or equal to half of X well a half of 0 is 0 plus 2. so we've got zero is less than or equal to zero plus two now zero plus two is two so because zero is less than or equal to two now let's have a look at this inequality we've got zero is less than or equal to 2 where 0 is less than two so that means that this inequality works okay zero is less than or equal to two if we had something like 9 over here and we had nine is less than or equal to two it wouldn't work because nine is not less than or equal to two but here we had zero is less than or equal to two so that works so that means at this point this point the origin zero zero satisfies this inequality it works that means that below the stagnant line is the region we want to keep and above the diagonal line is the region we don't want to keep so what we're going to do is we're just going to Shear in above the diagonal line like so to show that this section of the line This section of the graph anything above this line is rejected and NF and below is accepted and as you can see we're left with this triangle and inside this triangle in this region it satisfies all three of these inequalities so then in the question I might ask you to label it with an R to show that's the region that works it satisfies all three inequalities or sometimes the question could say shared in the region that satisfies each of them and then you shared in that region really carefully to show that it sort of satisfies them but typically correction the ones that I have normally see say label it with an R to show that that's the region that satisfies all three inequalities that's it okay let's have a look at our next topic so our next topic is quadratic inequalities now this video of 378 in corporate Maps so here we've been given an inequality and we've been asked to solve x squared plus 2X minus 24 is less than or equal to zero as you can see it's a quadratic because we've got our x squared term and it's an inequality because it's less than or equal to zero now we want to find the values for X so whenever we substitute them into this quadratic they give us an answer that is either negative or equal to zero because it says less than or equal to zero now I've had a really useful to sketch the quadratic graph for this because if we sketch the quadratic graph we can look at the graph and see when the heights are less than zero in other words when they're negative below the x-axis or equal to zero on the x-axis and that'll help us solve this quadratic inequality so let's sketch this quadratic let's sketch the quadratic y equals x squared plus 2X minus 24. so to sketches quadratic what I'm going to do is I'm going to find where it crosses the x axis to begin with in other words when the height is equal to zero so that means that zero will equal x squared plus two x minus 24. and if I solve that quadratic it will tell me the points where it crosses the x-axis so we've got zero equals and if we factorize this that will give us in our brackets x and x and we've got two numbers that were times together to give us negative 24 and add together to give us two so I'm thinking plus six and minus four so that's our quadratic factorize that means that X will be equal to negative 6 or X will be equal to 4. so they're the points where our quadratic will cross the x axis is going to cross here at negative six and here at 4. so our quadratic would look something like this where it crosses at negative six and four and because this x squared it's going to be a u-ship parabola so it looks something like this so now we've sketched our quadratic we want to find where the quadratic has a height that is less than or equal to zero in other words the values for X so whenever we substitute them into the quadratic will give us an answer that is negative or equal to zero now if we have a look here we can see that any value of x so whenever we substitute into the quadratic that is in between negative six and four so for instance if we had three if we'd on 3 squared plus 2 times 3 minus 24 that will be negative if we try 2 the answer would be negative zero if the answer would be negative 24. if we tried negative 5 the answer would be negative so all the values of X in between negative 6 and 4 would give us a negative answer and also whenever X is equal to negative six and whenever X is equal to 4 we get zero so all of those points in between negative six and four including negative six and four would satisfy this inequality so that means that X will be any value that is bigger than or equal to negative six but any value that is less than or equal to four and that's our answer I'm okay let's round over one okay let's have a look at our next quadratic inequality so in the X quadratic inequality we're going to solve is x squared minus seven X plus ten is bigger than zero so my first step again is to sketch the quadratic and we've got y equals x squared minus seven X plus ten and we want to sketch that quadratic so I'm going to factorize it and let it equal 0 and find where it crosses the x axis so I'm going to write 0 equals x squared minus seven X plus ten and let's factorize it so 0 equals and if we factorize that quadratic we get x minus 5 is negative 2. so that means that X will be equal to 5 or X will be equal to 2. so this is an x squared quadratic it's going to be U shape Parabola and it's going to cross the x axis at 2 and 5. so let's sketch it so it's going to look something like this so we've now sketched the quadratic we now want to find the values for X so whenever we substitute them into the quadratic we will get another is bigger than zero in other words when is the quadratic graph above the x-axis so as you can see here any value for exit is bigger than 5 will give us an answer that is positive so that means that X will be bigger than 5. so that's one possible set of answers any value of x is bigger than 5 would work for instance if we try 10 10 squared is 100 take away 7 times 10 right 70 that would leave us with 30. add 10 is equal to 40 and 40 is bigger than zero and any value of x that is above 5 would work even 5.1 5.01 and so on but five wouldn't work because if we tried five that gives us an answer of zero now the values of X between 2 and 5 they give us negative answers now we want an answer that is positive so we want to find where the graph is above the X Factor so let's have a look at this part of the quadratic whenever the values of X are less than 2. so if we try our value less than two such as negative one well negative 1 squared is one take away 7 times negative one where seven times negative one is negative seven so we've got one take away negative seven by one minus minus seven is eight plus ten is equal to 18. so any value of x that is less than two would work also so answers would be X is less than two or X is greater than 5. so any value of x is less than 2 or any value of x is bigger than 5 would work because that's where the graph is above the x axis and that's it so if we want to solve quadratic inequalities well let's have a look at the chord Mouse revision card so whenever a solving a quadratic inequality I recommend that you sketch the quadratic to begin with if you want to find where the quadratic is less than zero you look and see where the quadratic graph is below the x-axis and you write the values of X whereas the graph is below the x axis so for instance if we look at this quadratic x squared plus X subtract 12 is less than zero you can see that it's below the x axis for any value of x in between negative four and three so we write down the X is bigger than negative four or less than three and if you have a look at this inequality if you want to find out where quadratic is bigger than zero you scared to put radical again and you look at the parts of the graph above the x-axis so for this graph of x squared plus 2X minus 15 is bigger than zero as you can see here any value for exit is bigger than 3 or less than negative 5 would works we would write X is less than negative five or X is bigger than 3 and that's it and if it has less than or equal to zero or bigger than or equal to zero you make sure that you put those signs on your inequalities as well so if it was less than or equal to zero then it would be X is bigger than or equal to negative 4 and less than it equal to three and that's it okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is iteration and that's video 373 in corporate maths so in this video I'm going to go through two different types of iteration question I'm going to go through this one to begin with and then I'm going to look at an iteration question in context and part is to show the equation x squared plus 3x minus 1 equals zero can be a rearranged to give x equals a third minus x squared over three now if we have a look at what we've been given we can see here we've got over three and over three I think we're making this x a subject let's try and make this x a subject and see what we'd get so if we start by adding 1 to both sides of this equation we would get x squared plus 3x is equal to one now we want to make this x a subject so let's take where x squared from both sides of the equation so then that would give us three x is equal to one subtract x squared and then finally we want to make an x a subject so let's divide everything through by three so we get x equals 1 divided by three that's a third subtract x squared over three and that's it so we have rearranged this equation to give x equals a third minus x squared over three okay let's have a look at Part B so Part B says starting with x zero equals zero use the iteration formula x n plus 1 is equal to third minus X N squared over three and we'll have to use the iteration formula three times to find X1 X2 and X3 so whenever we use an iteration formula we need to know what we're starting with so we're starting with x is equal to zero and we're going to substitute that in here and we're going to do a third minus 0 squared over three and we're going to get that answer and that will become our X1 once we know X1 we then substitute that in again and we're going to do a third minus whatever the value for X1 squared is over 3 and then that will give us a value for X2 we'll then get the value for X2 substitute it in again and that will give us our X3 and we can just keep going but in this question we've been asked to use the iteration formula three times so we're told that x 0 is equal to zero so let's write that down x 0 is equal to zero now we want to find X1 so X1 will be equal to so the next term X1 will be equal to a third so third minus and then it would be x 0 squared now x0 is zero so zero squared over three and if we do 0 squared we get 0 and then divided by three that's equal to zero and a third minus zero will be equal to a third so that means that x 1 will be equal to a third to find our value for X2 well we're going to substitute our value X1 into here so we're going to do a third minus whatever X1 is which is a third squared over three and then when we work that out that will give us a value for X2 and then we can just keep going so let's do a third minus now we could do 0.3333 reoccurring squared or we could write a third there over three and then if we do a third minus 0.3 reoccurring squared over three we would get that's equal to 8 27. or 0.296296296 and so on okay or 0.296 reoccurring and then to get our value for X3 we're going to substitute that into here to get our X3 is equal to a third minus X2 squared over three so we'll substitute our value for X2 which is equal to 8 27 so this into here and then that will tell us our value for x free so it's going to be a third minus and we could write it 27 squared or we could write 0.296 reoccurring squared over three and we'll resubstitute that into our calculator we get that's equal to 665 over 2187 or as a decimal number that's equal to 0.3040695016 and so on and that's it so whenever you've been given an iteration formula what you do is you start off with the value you've been given and you substitute that in and you find your value for X1 you get that value you switch it in to get your value for X2 and so on now there's quite a nice way to do this on the calculator because we were told to start with zero what I'm going to do is I'm going to press zero equals on my calculator so I've got 0 up as the answer then I would type into my calculator a third minus and then what I would write is I'd write answer A and S I'll just press answer and then squared over three and I would type that into my calculator so I would type in a third minus and then the fraction button answer squared divided by three and if I press equals that's substitutes in the value for zero in here so it does a third minus zero squared over three and it gives me a third if I then just press equals again it's substituted in that value of a third into here and I get 8 27 so we're not 0.296296 and so on if I just press equals again it gives me 665 over 2187 or my 0.304 and so on so whenever you're using an iteration formula it can be very useful to just use the answer button then if you just keep pressing equals it would just tell you each one of your uh to each one of the next values so X1 X2 X3 I can press equals again and I get X4 press it again I get X5 and so on okay let's have a look at part C part C says explain the relationship between the values of X1 X2 and X3 and the equation x squared plus three x minus one equals zero now this is the equation we started off with from the we rearranged and if we look back at our values we've got 0.3333 and so on no point two nine six two nine six two nine six and then we get 0.304069 and so on and each one of these values are getting closer and closer closer to one of our Solutions one of our roots in that quadratic so let's write that down and I've just written down they're increasingly accurate approximations to a solution or a root of x squared plus 3x minus 1 equals zero and if you use a quadratic formula and solve this you should see that they are getting closer and closer to one of those Solutions okay okay let's have a look at our next question so this time we've got an iteration question that's in a context we've got T N plus 1 is equal to 0.9 TN plus 200 and this formula tells us that if we take the number of season tickets sold and we times about 0.9 and our 200 we will get the number of season tickets sold in the next season so we've been asked to work out how many season tickets are sold in their fourth season and the first season we know that's equal to one thousand so we know there's one thousand tickets sold in the first season in the second season what we could do 0.9 times a thousand which is 900 plus 200 which would be equal to one thousand one hundred then to find how many season tickets were sold in the third season we just need to times this by 0.9 and our 200 so let's do that so let's do 0.9 times 1100 that's equal to and that's equal to 990 and if we add 200 that's equal to 1190 and finally in the fourth season if we times this by our 10190 by 0.9 so multiply by 0.9 that's equal to 0.9 times by 1190 is equal to 1071 and if we add 200 that would be equal to one thousand two hundred and seventy one so that would be how many season tickets were sold in the fourth season and that's it because we're told if we multiply the number of season tickets sold by 0.9 and add 200 we'll find how many season tickets are sold in the next season now there was a shortcut as I said to use your calculator and what you can do is because you're starting off with a thousand you can press 1000 and then press equals and then because we know that we multiply the number of season tickets by 0.9 and now 200. we can then type in this we can write 0.9 times answer plus 200 and then that means that it's going to be doing 0.9 times a thousand plus two hundred and that will give us one thousand one hundred then if you press equals again it will do 0.9 times 1100 plus 200 and that would be equal to 1 ninety and then if you just press equals again it would tell us our answer of 1271. so I like using that answer key whenever I'm doing iteration questions okay let's have a look at our next topic so here we've got a pattern of dots and these are our triangular numbers where we start off with one dot then we add a row of two dots beneath it so one plus two is three then we add a row of three dots beneath that so three plus three is six then we add our four dots well six plus four is ten then we add a row of five dots which would be 15 and so on so to get our triangular numbers we start with one and then we add two add three add four add five add six and so on and there are triangular numbers and there are one three six ten fifteen twenty one twenty eight thirty six and so on okay the number type of sequence that we should know Fabs is our Fibonacci Sequence so Fibonacci sequence is we choose two numbers such as one and one we find the next term of Fibonacci sequence by adding the two previous terms so one plus one is equal to two now we do one plus two is equal to three now we do two plus three is equal to five then we do three plus five is equal to eight five plus eight is equal to thirteen and so on and that's it so if Fibonacci sequence is to find the next number if we add the two previous numbers so here's an example it says find the next three terms of this Fibonacci style sequence so we've got 2 7 9 and 16. so to get the next number we need to add the 9 and 16. so 9 plus 16 is 25. get the next number we're going to add 16 and 25 that's equal to 41 and finally to get our next number we're going to do 25 plus 41 and 25 plus 41 would be equal to 66 and that's it so Fibonacci sequence is where we found the next term by adding the two previous terms okay let's have a look at our next topic so our next topic is the nth term so whenever we've got a sequence there's often a rule or an M term for it which means that we can find out any term in the sequence by substituting which value we're looking for so here we've got our sequence 3 8 13 18 and so on as you can see we're going up on fives because we're adding five adding five adding five and so on and we've been asked to find the nth term of the sequence in other words the rule for the sequence so if we want to find the nth term our first step is to consider the sequence and see what is going up in or going down by and this as the sequence has gone up in fives we're adding five adding five adding five what we're going to do is we're going to write the five times tables beneath the sequence so the five times tables are five ten are the multiples of five five ten fifteen twenty and so on and because we're multiplying the numbers by five we're doing five times one is five five times two is ten five times three is fifteen and so on we're going to write five n because that means five multiplied by n because five times one five times two five times three and so on so if the sequence has gone up in five we write the five times tables or the multiples of 5 beneath it and we write 5 and I'm from now our sequence is not 5 10 15 20. our sequence is actually 3 8 13 and 18. and to get from five to three we take away two to get from 10 to 8 we take away two to go from 15 to 13 we take away two to get from 20 to 18 we take away two so if we took away two from all of these numbers we would get our sequence so that means if we had five n minus two that would be three eight thirteen and eighteen semis are F term is five n minus two five n minus two and let's just check it well the first term in our sequence well if we do five times one is five take away two is three great the second term in our sequence is eight what if we do five times two that's ten take away two is it fantastic our third term five times three fifteen take away two is thirteen and so on let's work out another one so if we get the sequence 7 9 11 13 and so on as this sequence is getting bigger by two each time we write the multiples of two being the sequence that's two four six eight and so on and that's 2N our sequence though isn't two four six eight our sequence is 7 9 11 and 13 so to get from the multiples of two to our secants we'd have to add five add five add five and add five so that means our nth term would be two n plus five so that means the nth term of the sequence would be two n plus five and let's just check it two times one is two plus five is seven two times two is four plus five is nine great two times three is six plus five is eleven and so on so to find the M term of a sequence you consider what the sequence is going up or down by you right there multiples beneath it and then you figure out what you would need to do those multiples to get to the sequence now because I said down by let's just have a look at an example where a sequence is going down instead of up so let's have a look at one of those now so if we had the sequence fifteen twelve nine six and so on and if we look at our sequence we can see we're going down by three each time we're taking over three taking away three take over three and so on so that means that what we're going to do is we're going to write down the negative three times tables so that's going to be minus three minus 6 minus 9 minus 12 and so on so that would be minus 3 n because minus 3 times 1 minus three times two and so on but our sequence is not minus three n it's not minus three minus six minus nine minus twelve our sequence was fifteen twelve nine six to get from -3 to 15 we would have to add eighteen to get from -6 to 12 we would have to add eighteen to get from minus nine to nine we'd have to add eighteen and so on so that means our nth term for our sequence would be minus three n plus 18. now the F term is really really useful because what we can actually do is we can use the M term to work out terms in the sequence if we have the sequence 3 8 13 18 and so on and someone asked me to find the hundredth term in the sequence rather than carrying on the sequence for 100 terms I can use this amp of term so I can do if I want the hundredth term I can do 5 times 100 which is 500 take away two which would be 498. so the hundredth term in this sequence without having to list them all down would be 498 and that's great so it saves us a lot of time and effort and another reason why the Avatar is quite useful is we can use it to work out if the terms in the sequence or not so for instance if somebody came along to me and said is 200 in this sequence well the first one I can tell it's not because our sequence goes 3 8 13 18 and so on and as you can see the numbers end in a three or an eight a three or an eight so because 200 ends in a zero it's not going to be in the sequence but we can actually use the nth term to show it up we can actually write five n minus 2 the nth term and write equals 200. so that's an equation and if we solve this equation and we get a whole number so we get n equals a whole number it would mean that 200 is in the sequence and whatever whole number that is tells us which term in the sequence it is but if whenever we solve this equation we get a decimal number it would mean that the number 200 wouldn't be in the sequence so let's solve our equation and see what happens so if we had 5m minus 2 equals 200 we would add 2 and add two well we add 2 to get rid of the minus two so we're left with five n and on the right hand side we've got 200 add 2 is 202. now we've got 5n we don't want 5 event so let's divide by 5 and divide by 5 so that would give us five n divided by 5 is n and 202 divided by 5 would be 40.4 and as that's a whole number that means that 200 is not in the sequence and because it's 40.4 it means it's not the 40th term and it's not the 41st term in the sequence so that means that 200 is not a term in the sequence okay so our next topic is the quadratic nth term and that's video 388 in quarter Maps well actually I saved video there's actually 388 a 388 B and 388 C and I've actually had to make three different videos there because there's three approaches three popular approaches with teachers with how they find the quadratic term now in this video I'm going to go through one approach so here we've got a quadratic sequence and it goes 5 6 11 20 and 33 and if I want to find the M term for this quadratic nth term I'm going to find what we call the first differences so the first inferences are the difference between these terms and then I'm going to find the second differences which are the differences between the first differences so let's write down the first inferences to begin with so we go from five to six that's a difference of one we then go from six to eleven that's a difference of five we go from eleven to twenty that's a difference of nine now we're going to find our second differences and we'll quadratic term the second emphasis should always be the same number so from one to five the second difference will be four five to nine the second difference will be four and from nine to Thirteen the second Severance will be four so we've got the first inferences and the second differences now we need to consider what the quadratic nth term would look like so the linear nth term is something such as three n plus seven that's the nth term for a linear sequence for a quadratic sequence is something in the form a N squared plus b n plus C so if we consider our quadratic graphs that's something in the form of y equals x squared plus so many X's plus a number and so on here we've got the quadratic n for terminal will be in the form of a N squared plus b n plus C okay so now let's have a look and see how we would find a b and c now if you do watch the micro video on chord Mouse the 388a video it'll show you clearly why this is the case but if we have a look at these numbers this number here this first second difference or actually any of the second differences that's equal to 2A so if we have this number we will find a next if we want to find our B we would look at this number the first first inference and that number would be equal to 3 a plus b and once we know a we can then find out what B would be and finally this term in the sequence the first term in the sequence would be equal to a plus b plus C and that's it now if you do watch that video 388 a I go through and show you why that's the case but let's use this information to find what a B and C are so 2A is equal to four so let's write that down 2A is equal to 4 and if we divide by 2 we get a is equal to two that's fantastic we've got two N squared now we want to find our value for B well if we look at our first first inference the one that's equal to 3A plus b so 3 times a 3 times 2 is equal to six actually let's write that down three a plus b is equal to one and a is equal to two so three times two six so we're gonna have six plus b is equal to one and then if we take away six from both sides we get B is equal to negative five so that means B is equal to negative five or minus five so that means that here we're going to have two N squared minus 5n and then finally we just need to find our C the first term in the sequence is equal to a plus b plus C so let's write that down a plus b plus C is equal to five and we know that a is equal to two so that's going to be two and then B is equal to minus five so plus minus five and then plus C is equal to five and then we've got two plus minus five well two take away fives minus three so we've got minus three plus C is equal to five and then if we add three to both sides we're going to get C is equal to n that's it that means that our c is equal to eight so plus it so our quadratic term for this sequence would be two N squared minus five n plus eight now once you know the quadratican term of a sequence you can also do other things with it so for instance if we want to know what the hundredth term in the sequence would be we would just substitute in N is equal to 100 so we would a hundred squared times two subtract five times a hundred plus eight and so on that's it okay let's have a look at our next topic so our next topic is geometric progressions that's video 375 in corporate Maps so we've looked at arithmetic progressions that's where we're adding our t taken away the same number each time now let's look at geometric progressions where we're multiplying by the same number each time a common ratio so here we've got two geometric progressions and we've been asked to find the next two terms for each one of the following geometric progressions so let's start off by looking at this one we go from 2 to 40. so let's see what we're multiplying by we're multiplying by 20 because if we do 2 times 20 we get that's equal to 40 or we could just do 40 divided by 2 and that's equal to 20. if we're for geometric progression if we divide the next term by the previous term that gives us what we'll multiplying by each time so if we do 800 divided by 40 we will also get that's equal to 20. so here each time we're multiplying by 20. so if we want to find the next two terms we're going to Times by 20 again so multiplying by 20 would give us 16 000 and then if we multiply by 20 again that would give us our next term which would be be 320 000 and that's it our next geometric progression this time we've gone from 3 to 15 to 75 as you can see we're multiplying by five each time multiplying by 5 multiply by five so if we multiply by 5 again so 75 times 5 75 times 5 is equal to 375 and then we're going to take our 375 and times that by 5 and that will tell us our next term which would be 1875 and that's it okay let's have a look at another geometric progressions question so this time we're told that s is a geometric sequence so we've got a geometric sequence so that means that we're multiplying by the same number each time in the sequence and all terms of the sequence are positive integers so they're all positive numbers and the first three terms in the sequence are 8 x and 72 and we've been asked to find the value of x so in this question we don't actually know what we'll multiply by each time because we're trying to find this term and without knowing this term we don't know what we're timesing by but what I do know is with the geometric sequence or geometric progression if I divide the next time by the previous term that will tell me what I'm multiplying by and remember I'm always multiplying by the same thing so that means that if I do x divided by 8 that would be the same as 72 divided by X so let's write that down x divided by 8 the second term divided by the first term will be equal to the third term divided by the second term so that's 72 divided by X now this is just an equation and we can solve it now I want to find what x is so I don't want X on the denominator so I'm going to multiply both sides of this equation by X so times in both sides by X would give us we'll multiply the left hand side by X we would have X over 8 times x that would be x squared over 8 and on the right hand side of the equation we had 72 divided by X we times in by X to get rid of the X that's going to leave us with 72. now we don't want to have this divided by 8 so let's multiply both sides of the equation by 8 that would give us x squared equals and 72 times 8 would be equal to 576 so that means that x squared is equal to 576. now whenever we're solving an equation like this we've got to be careful because there's a positive solution and a negative solution that means X is going to be equal to plus or minus and then we'll just square root 576 and the square root of 576 is 24. so that means that X is equal to plus or minus 24 so that means X could be 24 or X could be negative 24. now in the question said that all terms of X are positive integers that means that it can't be the negative number so it's going to have to be that's equal to 24. so X is equal to 24 and that's it okay let's have a look at our next topic so our next topic is algebraic perfect last video 365 and corporate Maps so here's another very proof question we're going to look at two different types of algebraic proof questions in this video we're going to start off by looking at this type of one where we're trying to show that this is a multiple of 8 and then we're going to look at a question which involves odd and even numbers so here we've been asked to prove n plus six close brackets squared minus n plus 2 close bracket squared is a multiple of it so let's expand this and see what we get so we've got M plus six squared so it's going to be n plus 6 brackets n plus six and we're going to take away n plus 2 multiplied by itself n plus 2. so let's expand this and simplify so n times n is N squared n times 6 would be plus 6 n 6 times n is plus six n and six times six is plus 36 and then we're going to take away now I'm going to expand this but I'm actually just going to put the answer in a bracket just to remember that I'm taking away the whole thing so n times n is N squared n times 2 is plus 2N 2 times n is plus two n and 2 times 2 is Plus 4. now let's simplify we've got N squared plus 6n plus 6n so this could be N squared plus 12 n plus 36 and then we're going to take away and then we've got N squared and we've got 2N plus 2N that's going to be plus 4N and then we've got plus 4. so we've got N squared plus 12 n plus 36 take away N squared plus four n plus 4. so we need to take away the N squared 4N and 4 away from what we have here so N squared take away N squared will be just 0. so we've got 12n take away 4N 12 n take away 4 n would be equal to 8m and finally we've got 36 take away 4 and 36 take away 4 is equal to 32. so whenever we expand this in simplify we get the answer of 8 n plus 32. so if we just factorize this and take our 8 out we get 8 bracket n plus 4. and that means that this would be a multiple of it because we've got eight times something so that means that and that's it so because we have been able to expand it and then factorize and take 8 out that means it's 8 times something so therefore will be a multiple of 8 and that's it and okay let's have a look at our next question so next question involves odd and even numbers now whenever I'm dealing with algebraic proof If I wanted to use an even number I wouldn't just use n because N is a number it could be 1 2 3 4 that could be either odd or even so if I wanted to have an even number I would multiply this number by 2 because if I do two times a number the answer is always even so that would be an even number so 2N is even and if I wanted an odd number now I could minus one and say it's 2 m minus one I like to add one um quite positive person I like to add one so I'm going to do two n plus one because if I know that 2N is even 2N plus 1 would have to be odd so that's what I would use if I was asked to use even numbers or odd numbers I would use 2N for an even number and 2N plus one for an odd number now in this question we've been asked to show that the sum of two consecutive odd numbers is even so it's consecutive odd numbers so let's use our first odd number let's let our first odd number be two n plus one so that's our first odd number now it's consecutive odd numbers so I want to find the next odd number now if I added one to this that would be even because obviously 2N plus 2 would be even so the next odd number would be two n plus three there will be two consecutive odd numbers where you've got two n plus one mean an odd number and then if we add two more we then get to the next odd number so that would be two n plus three so there are two consecutive odd numbers and we've been asked to prove that the sum of these two odd numbers is an even number so let's add them together because there's some if the question said product we would times them together but in this question we're asked to show that the sum of two odd numbers is even if the question said product you'd multiply things together if the question is a difference you would take them away and so on so let's add these together and see what we get 2N plus 2N is equal to 4m and 1 plus 3 is equal to 4. so if we add two consecutive odd numbers two n plus one and two n plus three we get four n plus four now as you can see if you factorize this we want to show it's even so let's take a factor out let's take two out and that's going to be two times and if we divide both of these by two we get two n plus two so that's two times something and two times something is always even so therefore and that's what I've written down the sum of two consecutive odd numbers two n plus one and two n plus three is always even because whenever we added them together we got four n plus four we're able to take two out as a factor so it's two times something so therefore it's even and that's it okay let's have a look at our next topic our next topic is frequency tree so here's a frequency tree and we've got one appointments even an appointment so we've got this 40 appointments all together we know there's 23 in the morning and there's some in the evening and then we've got on time late on time later so we've got 40 appointments all together 23 in the morning so if we do 40 take away 23 we'll find out how many evening appointments there were so 40 take away 23 is 17. so there must have been 17 even appointments because they need to add together to be 40. if we focus on the 23 morning appointments 21 were on time so that means there must be two that were lit because 21 plus this number must be 23 so that must be two and in the evening appointments where there was 17 altogether five were laid so if we do 17 take away five that's 12. so 12 must have been on time and that's it so we've completed the frequency tree and video of 376 and corporate miles we'll give you more information on that okay let's have a look at our next topic so our next topic is two-way tables and if you want to revise two-way tables it's video 319 on corporate Maps so here we've got a two-way table we've got some subjects at the top English and art and then total and then we've got our information about whether they pass the course or failed the course and again we've got total at the bottom so we've been asked to complete this two-way table but we haven't but I'm saying it now we're going to complete this two-way table uh so we've got um first of all let's look at Art we've got 12 students that have failed art of 19 in total so 19 shouldn't study that and 12 of them failed so it must be quite a hard test and we would we can find out how many students passed because we know there's 19 all together and 12 Fields so if we take that 12 away from 19 that leaves us with seven so seven students must have passed the r course next well I'm looking at this top row here for how many students passed English and art and we can find this total we know 25 students passed English and seven students passed art so if we add those two together we can get the total 25 plus 7 is 32. so 32 students passed their courses next I'm looking at these total so we know 32 students passed the course and 13 failed their course so if we add together the 32 and the 13 we can see how many students are in total so 32 add 13 is 45 so let's put that in next let's look at the students that failed their courses we'll all together 13 students filled their course and 12 of them were from art so that means that only one student failed English so that means that we've got that number of one and then finally how many students studied English in total well we've got 25 that passed and one that failed to all together that b26 and let's just check our answers 26 plus 19 is 45. okay let's have a look at our next topic so our next topic is composite bar charts and that's video 148 a on corporate Maps so here's a composite bar chart and it's a bar chart but yeah as you can see each of the bars is split up into different sections so we've got this bar for January and in if we have a look at our key in pink we've got hot drinks and in Gray we've got cool drinks we've then got for February the hot drinks and the cool drinks and for March we've got the hot drinks and the cool drinks and we're told this composite bar chart shows us information about the number of drinks sold over three months so if we look at the bars and we look at the totals in January we can see the 600 drinks sold in February we can see the 650 drink sold and in marches 750 drinks sold but also because it's a composite bar chart we can actually see the breakdown of cool drinks and hot drinks sold as well by you looking at each of these regions so for instance for January there's 450 hot drinks sold and there's 150 equal drinks sold and we can see the same for February and March and the question says in which two months were the same number of hot drinks sold so let's have a look at our key in the question we wanted to find the two months that got the same number of hot drinks sold so that's going to be the pink sections of each bar so if we look at the bar for January and the bar for February they both go up to 450 so that means that in January there's 450 hot drinks sold and in February there's 450 hot drinks sold whereas in March there's not as many hot drinks could sold maybe it's a bit warmer and a March is only actually 300 hot drinks sold so in which two months with the same number of hot drinks sold that will be January and February next topic is pie charts so here we've got our table we've got rugby team and we've got frequency so 90 rugby fans were asked which team they supported England France Ireland Scotland and Wales and we've got 20 supported England five support of France 15 supported Ireland 25 supported Scotland and 25 supported Wheels okay and we're going to draw a pie chart for this information so if you run a pie chart typically they'll be Circle drawn for you the line there going from the center up to top for you and we're going to draw a pie chart for this so to draw a pie chart the first thing you need to do is add up the frequencies well we know there's 90 Ruby fans so if we add them all up we'll find this 90. then we need to divide the whole circle 360 by that number of people so 360 divided by 90 is equal to 4. so that means that each person is given four degrees in the circle so each person is 4 degrees now 20 people supported England now each of those people get four degrees so if we multiply that by four we'll see how what sides of angle they should have in the pie chart 20 times 4 is equal to 80. so the angle we're going to draw for England pounds would be 80 degrees next we're going to multiply 5 by 4 and we're going to multiply all these numbers by 4 because each person gets 4 degrees so we're going to multiply by four multiply by 4 multiply by 4 and multiply by 4. so 5 times 4 is 20 degrees so we'll draw 20 degrees for the French fans 15 times 4 is 60 degrees for the Irish fans that should be much bigger then we've got 25 times 4 25 times 4 is 100 so 100 degrees for the Scottish fans and 25 times 4 25 times 4 is 100 and then we've got all our angles the Highlight to add these up to make sure I get 360. 80 plus 20 is 100 plus 60 is 160 plus the number 100 is 260 plus 1100 is 360 Degrees that's it so whenever you run a pie chart you need to add up the frequencies then divide 360 by the total frequency to find the number of degrees per person or per item or whatever you're looking at and then multiply all the frequencies by that number to find the angle you should draw okay so we're going to draw 80 degrees for the English fan so let's start off by going to our pie chart and we're going to get our protractor and we're going to line it up like so so the cross goes in the in the center of the circle like so and then the zero goes at the top and we're drawing an 80 degree angle so we're going to go around to our 80 degrees is so that's here and we'll do a little Dot now you'll move your protractor and you will draw a nice straight line so let's move our protractor and you draw a nice straight line from the center through that point to the edge of the pie chart like so so that's for the English fans so let's label it for English France for England select so and that's that sector done and it's an 80 degree angle so we can put that in as well if we want to next we're going to draw an angle of 20 degrees for the French fans so we're going to get our protractor we're going to have to turn it so that we put our zero along the line we've just drawn you always put the zero on the line you've just drawn or if there's no lines There Yet the one at the top so we started off with the zero at the Top Line because I was only line now we've drawn this line we're going to put 0 there and we're going to draw an angle of 20 degrees so we go to zero we go around to where 20 degrees is that's here so put a dot and then move our protractor into a nice straight line through there with a pencil so it looks something like that so that's 20 degrees you don't necessarily need to put the angles in but we do need to put it in we do need to label that it's France okay next was Ireland and if we go back and look that was 60 degrees so we're going to get our protractor we're going to turn it so the zero is along the line we've just drawn so we're going to rotate it slightly again the cross has to go in the center of the circle and zero is on the line like so and then we're going to go from zero around to 60 degrees for the Irish fans so to there 16 move our protractor so move our protractor and draw a nice straight line from the center of the circle through that point and to the edge and again what we're going to do is we're going to label it for Ireland and we can put the angle in if we want to it's at 60 degrees so you don't need to do that bit but again label it for Ireland next was the Scottish fan so that was 100 degrees so again take up retractor and rotate it so that the zero is on that new line like so so start off with zero and we're going to go around to 100 degrees so we're going to sign that out go around 10 20 all the way around to 80 90 and 100. so remember we are dealing with the outside numbers here so don't look at this in inside 100 we're dealing with the outside number so we're going zero all the way around to 100 there again move our protractor and draw a nice straight line from the center of the pie chart through that point to the edge and that will be for the Scottish fans so Scotland to 100 degrees and finally the last sector will be for wheels that should be already drawn for us for 100 degrees because that's all that's left but let's check it so let's take our protract and rotate it and when you do that you can clearly see that is 100 degrees and that's it Okay so we've had a look at drum pie charts now let's look at reading them or interpreting pie charts and that's video 164 on corporate Maps so we've been told 180 years students were asked how they travel to school and the pay chart shows their responses so we've got bus some travel to School by bus some walk some travel by car and some cycle and we're asked what fraction the students travel by bus so sometimes when we're given a pie chart we're asked to find a fraction or a percentage so we could be asked what fraction the students walk to school or what percentage of students travel by bus to school and to do that what we're going to do is we're going to look at the whole circle the whole pie chart and figure out what percentage or what fraction is that region so if we have a look at this sector for a bus we can see it's got a right angle that means it's a 90 degree angle and an 90 degree angle will be a quarter of a full circle somebody said this is a quarter of the pie chart so what fractions students travel by bus a quarter off them if you're asked what percentage students traveled by bus that would be 25 if we're asked what fraction students traveled by car we would write 80 out of 360 or 80 over 360 and then just cancel it down so our next question says how many students walk to school so we've got 120 degrees for walk so that's going to be a third so we need to work out a third of 100 nearly we're just going to do 180 divided by three it's equal to 60. so that means to 60 students walk to school okay let's have a look at our next topic and our next topic is scatter graphs and their videos 165 up to 168 in corporate Maps so here's a scatter graph and it shows the cost of 10 plumbing jobs so we've got these 10 plumbing jobs you've got the duration from not up to five hours and the costs from not up to 300 pounds and you've got all the points there and our first question says what type of correlation is showing so because the points are going upwards in other words as the hours are going up the cost is going up that is a positive correlation so positive so if the points were coming downwards in this direction it would be a negative correlation and if the points were just scattered around everywhere that would be no correlation okay our next question our next question says draw a line of best fit and I've just shown Orlando best fit like so I've got my real learner pencil I've drawn a line going straight through as many of the points as possible or as close to the points as possible this point here is an outlier so I've just ignored that one and I've drawn my line of best fit so it's as close to possible to these points so I've tried to sort of minimize the distance between my line and all the points that I've been given our next question says estimate the cost of a job lasts in two and a half hours so if we go down to our horizontal axis here for hours two and a half hours would be here and if we get our ruler and pencil and so use a ruler and pencil whenever you're doing this and go up to the liner best fit and then go across you can clearly see that the cost would be 150 pound so our estimate for the cost of a job lasts in two and a half hours would be 150 pounds next the question says estimate the duration of a job costs them 180 pound so if we got a 180 pound on the vertical axis here and we get our learn pencil and we draw a cross to our line of sfit and then we drill down we can see that's three and a half hours so our answer would be 3.5 hours that's it Okay so we've drawn a line of best fit and we've used that to make some estimations okay let's have a look at the next part the next part says Circle the outlier so here we've got a scatter graph and you can see this point just stands out it just stands out from all the rest so it's what we call an outlier okay another type of question we can encounter whenever we're looking at scatter graphs there's a question like this so here we've got our scatter graph we've got hours along the bottom we've got zero one two three four five so this is a cost of a plumber and we've got the cost of the job going from zero all the way up to 300 pounds so it's going zero 30 60 90 120 and so on and then you see the most expensive job we've got here is this job here which is for five hours and it's just below 240 pounds and the question says here explain why it might not be sensible to use this scatter graph the estimate the cost of a job lasting 18 hours now if you have a look here the hours we've got for the jobs we've been given only go up to five hours now with this job it'll be in 18 hours that's a very long job it might not actually be a one day job or maybe a two or three day job and the plumber may have to add in more traveling costs if he's gone forward and back and forth to do that job so whenever we're dealing with scatter graphs we've got to be very careful whenever we're using the land of best fit so if we were to use this liner best fit for this scatter graph to estimate the cost of a job that lasts in between 0.5 hours and five hours that would be okay because that's what we call interpolation that we're going inside the data that we've been given but if we go beyond the data that we've been given then we've got to be quite careful so just shutting down this beyond the range of the given data it's extrapolation so if we go beyond the data we've been given it may not carry on in the same Trend so it might not be sensible to use lands are best fit to make estimates there that's it okay so our next topic is frequency polygons frequency polygons a bit of difference between the exam boards where Ed Excel says you should know frequency polygon whereas AQA and OCR don't mention frequency polygons so if you're studying for lxl you're going to have to watch this section it should take a minute or two and if you're studying for AQA and OCR you may want to skip on but then again it's only a minute or two so I'd probably watch it anyway just to see what frequency polygons are and they're not that complicated so let's have a look at frequency polygon so that's videos 155 and 156 and corporate Maps so here we've got a table and we've got time not the 20 minutes 20 to 40 minutes 40 to 60 Minutes 60 to 80 minutes and 80 to 100 minutes and we've been given some frequencies for them so there's five times that are between not and 20 minutes 11 times between 20 and 40 and so on so it's a draw frequency polygon what we're going to do is we're going to plot the frequencies in the midpoints of each of the categories so if we've got naught to 20 minutes we're going to go to 10 minutes and then we're going to go up to five so we're going to go 10 minutes and up to five next between 20 and 40 minutes well 30 minutes is in the middle that's the midpoint so we're going to go 30 across and 11 up so 30 across and 11 up now whenever you're dealing with how far up to go make sure you know the scale so if we have a look there's 10 boxes for five that means each little box is not 0.5 here so we're going to go 30 across we're going to go up to 10 and then we're going to got two more boxes to get to 11 so we're going to go to there now our next one we've got between 40 and 60. so it's going to be 50 across and 20 up so 50 across and 20 up next we've got between 60 and 80 so it's going to be 70 across and 15 up so 70 across and 15 up and finally we've got between 80 and 100 that's going to be 90 across a nine up so 90 across and then 9 up would be well we've got 10 here each little box is 0.5 so we want to go down two and then that would be there so we've plotted our points now what we're going to do is going to learn our pencil and we're going to join them up so like so and that's it that's our frequency polygon one thing to note is not to join up the first point and the last Point you're only joining up the consecutive points so you're not the first one to the second the second to the third the third to the fourth and the fourth to the fifth so it's a frequency polygon like so don't join up these points and that's it another thing you could be asked to do with frequency polygons is to compare them so this could be the times for class A we could also have the same grid of number frequency polygon for class B and you can have a look at them and compare them and see you know which class was faster whenever you look at their frequency polygons okay sometimes we're asked to find the mode from a table so here we've got some information and we've got the age of some people and there are five or people are animals or whatever it is with age of something and the ages are five six seven and eight and we've got the frequency so there's two five year olds two six year olds five seven-year-olds and one eight-year-old now whenever you find in the mode from a table it changes the word from often the mode age to the modal age and we're trying to find the Moodle age that just means the most common age and remember we had five of seven-year-olds so that means that that was the most common age so the mode here would be the one with the highest frequency which is seven so the mode of the modal age is seven our next topic is finding the mean from a table so we want to find the mean from this table and remember to find the mean we add up all the values and divide by the number of values so we want to find the ground rule now this if there's two five year olds I add on this column called the FX column and that stands for the frequency multiplied by whatever column this is so if there's two five-year-olds well five plus five is ten another way to do that is just 2 times 5 and 2 times 5 is 10. if there's two six year olds we could do six plus six but two times six is twelve so 12. if there's five seven year olds well 5 times 7 is 35 and 1 8 year old well that's gonna be one times eight is it so we'll find this column called the f x column and it's fired by multiplying the frequency by whatever the values are in the table and then if we add that up we will get the grand total and that's equal to 65. so if you added up all the ages that would be 65 now we need to divide that by how many people there were well if we look at the frequencies and add those up so two plus two is four plus five is nine plus one is ten so all together those 10 people so if we do the total which is 65 divided by 10 that would tell us the mean age so 65 divided by 10 would be 6.5 so the mean age is 6.5 okay our next topic the next topic is to find the median from a table so the median is the middle value so if you look at this table we're trying to find the middle value so one way to do is to arrange them all in order so there we've got all the edges and then we want to find the median in the middle one so we can then just work out the middle one so the medium would be 20. so that's one way to do is to list down all the edges in one list and then just find the middle one so another way to do it though is we can consider the frequencies now all together we've got two people three people another 13 people and one person so if you add those up you'll find there's 19 people all together and if we were to line up 19 people well the median person would be the 10th person so if you had 19 people you could add one which is 20 and divide by two which would be the tenth person and the 10th person if you line these people up an order of age the 10th person wouldn't be in here there's only two 18 year olds the 10th person wouldn't be here because there's only five so far and the tenth person would definitely be in this group of people so the 10th person would be 20 years old okay next topic is to look at the combined mean so to work out the combined mean let's just make sure we know what the mean is so the mean is fine by adding up all the values and dividing by the number of values so it's going to be very useful and what's also useful is if we know the means found by adding up all the numbers and dividing by the number of numbers if we have the mean and multiply it by the number of values that will give us the grand total so that's very useful as well okay so our question says there's 20 students in class air and 10 students in class B and they sit the same test the main test score for class A is 60 and the mean test score for class B is 14 and the question says work out or calculate the mean test score for all 30 students so we're going to do this question by considering if we know what the mean is for class A and we know how many students there are we can find out the total number of marks received by a class out by the students in class A so we know the mean score for class A is 60 and we know there's 20 students so if we do 60 times 20 that gives us 1 200. that means there are 1 200 marks obtained by the students in class there all together as a total number Mark scored and if we divided that by 20 we get the mean score of 60. so that's the total for class A now let's get the total for class B so their class average was 40 and there's 10 students so if we multiply 40 by 10 that's 400 so that means in total the 10 students in class B scored 400 points or 400 marks all together so if we add together the 1 200 and the 400 that tells us how many marks were received by all 30 students all together in this test so 1 200 plus 400 is 1 600. now we're trying to work out the mean score for these 30 students so if we divide the grand total the 1 600 by 30 we will get the mean test score for these for any students so 1 600 divided by 30 is equal to that'll be 53.3333 so on or 53.3 reoccurring or let's just run our answer to two decimal places so 53.33 and that would be the mean test score to two decimal places and that's it so whenever we're working out the combined mean it's very useful to be able to work out the grand total and that can be found by taking the mean and multiplying it by how many people were involved with that mean and that will tell you the total and that'll be very useful and if you want to watch the video on this one called Mouse is 53a and just remember you do have that bumper pack of questions that practice booklet which has loads of questions and there'll be questions on that on the combined mean also okay our next topic okay let's have a look at our next topic our next topic is called the estimated mean and that's video 55 in corporate Maps it's one of my favorite topics and we've been asked to work out an estimate for the main Edge so we've got a group of people and we know that there's nine people from zero up to ten we've got 13 people from 10 up to 20. we've got 16 people from 20 up to 30 and we've got two people from 30 up to 40. and we've been asked to work out the mean age and remember to work out the mean age we add up all the ages and we divide by the number of people let's actually start off by finding the number of people and to do that to allowed up 9 13 16 and 2 and that's equal to 40. so we know there's 40 people now we want to add up their ages and divide by 40 the number of people but unfortunately we're not able to do that in this question because we don't actually know all their ages we've made this group frequency table to make the data a bit more easier to interpret but unfortunately we're not going to be able to work out the exact mean that's why the question says an estimate for the mean age Okay so we've got nine people that have an eh from zero up to 10. we don't actually know their ages there could be four there could be eight there could be they could all be nine rules we don't actually know but what we're going to do is we're going to choose a fairy ager a representative age for these people because they're from not to 10 we're going to use the midpoint which is five years old we're going to add a column on called midpoint and that will be the age that we're going to use for these people so for the not 10 year olds we're going to pretend that they're five years old each we're going to pretend there's nine and five-year-olds we don't actually know their ages that's the fairest thing to do now we've got 13 people with an age between 10 and 20. so let's use the midpoint of 15 and for a 20 to 30 the midpoint would be 25 and for 30 to 40 the midpoint would be 35. what we're going to do is we're going to just do 9 times 5. so we're going to do 9 times 5. 13 times 15 16 times 25 and 2 times 35. so that tells our estimate for the total ages of the nine people between 0 and 10 the 13 people between 10 and 20 and so on so we'll call this column FX and whenever you're doing the estimated mean question you add on a column for midpoint and you'll add on this FX column and your times the midpoint by the frequencies and that will tell you your FX column so 5 times 9 or 9 times 5 is 45 13 times 15 is 195. 16 times 25 is 400 and 2 times 35 is 70. so we've completed our FX column now what we're going to do is we're going to add those up to find our estimate for the grand total so 45 plus 195 plus 400 plus 70 give gives us a total of 710 so our estimate for the total of the edges would be 710 now what we're going to do is we're going to divide that by the total frequency so the ground total divided by the total frequency gives us our mean our estimated mean so 710 divided by 40 is equal to 17.75 so our estimated mean age is 17.75 years old and that's it sometimes we've grouped frequency tables what we need to do is find the modal class interval so remember the word modal is similar to mode and that just means the most common class interval so the most common class interval well we just look at the frequencies so we can see that this category with 16 is the frequency is the most common so that means that this is our modal class interval so D is larger than or equal to 20 but less than 30 and that's it okay let's have a look at our next topic so our next topic is found in the class interval that contains the median so here we've got a crypt frequency table with duration going from 0 up to 10 from 10 up to 20 20 up to 30 and from 30 up to 40. and we've got the frequencies and all together and if we add up these frequencies there's 40 all together we've been asked to find which class interval contains the median so if we knew these 40 durations and if we wrote them all out we're asked which of these categories from naught to 10 from 10 to 20 from 20 up to 30 or from 30 up to 40 would the median lion would be in this category this category this category and this or this category now straight away I can see there's not going to be in the first category because there's 40 altogether and so because there's 40 altogether if we divide 40 by 2 that gives us a rough idea as to where the medium would be so 40 divided by 2 is the 20th so the medium would be roughly the 20th value and because this is grouped it and we don't actually know the actual numbers we may as well just use that 20th value now strictly speaking if we knew all the values and we could write them all out and there's 40 numbers we would do 40 plus 1 which is 41 and we would divide that by 2 41 divided by 2 is equal to the 20.5th value so so if you had 40 people and you'd line them all up and we wanted to find the median the median would really be the 20.5th value but for grouped frequency tables where you don't actually know the values anywhere where it's just nine people in between these values 14 in here and so on we can just divide this frequency by two so that's the 20th and find where the 20th one is so normally in a question like this the 20th and the 20 point fifth value will be in the same category anyway just to make sure that you know whenever they're making the questions that you know you don't have a sort of a difference so that if students you went for this value or if students went for the 20th one they would design the questions that they'd be in the same category anyway and let's just see if that's the case so here we've got nine people in the first category now we're looking for the 20th one it's not going to be in there because that's there's only nine people there now we've got another 13 well 9 plus 13 is 22. so up to the end of this category there's 22 people that means if you were to line up those people someone in this category would be the median or it would be either the 20th or the 20 point for value and as you can see that means that both of them will be in this category so the question says which class interval contains the median well it's going to be this class interval 10 Which is less than or equal to D Which is less than 20. and that's it okay let's look at our next topic so our next topic is the median and find the median from group data now I use this topic of this approach a lot whenever I'm finding the median for all my histogram if I've been given a histogram and I must Define the median or calculate an estimate of the median I use the I use this approach quite a lot also if I was asked to find an estimate of the lower quartile and upper quartile and so on and this is video 52 in corporate Maps so here we've got a group frequency table we've got hours going from zero to five five to ten ten to Fifteen and fifteen to twenty and we've got the frequencies 27 44 21 and here and if you add them together all together there's a hundred and what we're going to do is we're going to find an estimate of the median so if I wanted to find an estimate of the median from this data what I would do is I would use an approach called linear interpolation so what I would do is I would find which group the median is in to begin with so let's do that now because this is group data we're finding the estimate of the median anywhere I'm going to take the frequency and I'm just going to divide it by 2 to find the position of the median of obviously I could add one divided by two but because it's an estimate anyway I just find it easier to just tick the frequency and divided by two so I'm going to take our 100 and I'm going to divide it by 2 so that's equal to the 50th number so the median is the 50th number the 50th person absolute maybe this is the amount of R spent revising for a test and we're looking for the 50th person that's the median so the 50th person it wouldn't be in here there's 27 people here they've revised between naught and five hours now the 50th person then would be in here because we've got another 44 if we line them up and we had 27 and then another 44 the 50th person would be in this group here so the median is in this group so now we know the medians in this group we're going to use interpolation to find out an estimate of the median so the formula for linear interpolation is the lower bound plus the number into the category divided by the number in the category times the class width so in other words now we know the values in this category the lower bound but the lower Bound for this category is five so let's write that down to five and then we're going to add the number into the category well we're looking for 50th person we know that the 27 first 27 people are in this category so we want to go in over 23 people so it's going to be 23 people into this category so we're looking for the 23rd person in this category that would be the 50th person overall divided by the number in the category so that's the 44 people so this gives us the fraction into the class and then Times by the class width and the class width is how wide the class is which is five because from five to ten that's going to whip for five so if we do five plus the 23 44 Times by five that would tell us our estimate of our median and whenever I do that I get an answer of 7.61363 and so on hours and that's it that would be our estimate of the median and that makes sense because if we think about it the person we were looking for was the 23rd person into this group well the 22nd person's halfway and halfway in this group would be 7.5 23 is just over so that'll be 7.61363 and so on hours and that's it this approach can also be used to find to it for the lower quartile and upper quartile even an estimate for the interquartile range and if we wanted to find an estimate for the lower quarter we would find the 25th person so we would look at this group and we would use the same formula if we wanted to find an estimate for the upper quarter we would look for the 75th person and so on and as I say I use this technique a lot whenever I'm dealing with histograms and whenever I'm asked to find an estimate for the median or the lower quartile or upper quartile from a histogram now you could consider the squares and so on I just find this approach can be quite useful in some situations okay let's look at our next topic okay let's have a look at our next topic so our next topic is stem Leaf diagrams so stem Leaf diagrams is one objects which differ really between the exam boards where edxl cover stem Leaf diagrams and frequency polygons whereas OCR and AQA don't if you're studying for OCR or AQA it's only two minutes I would probably watch it because you could practice for the mode and the range in the median anyway but feel free to skip ahead if you do OCR AQA but it's up to you okay our next topic is stem and leaf so stem Leaf is a great way to represent information in order so represent numbers in order and we've got here this would be videos 169 and 117 corporate maps and the question says the following stem and leaf diagram shows the times taken for 15 people to complete the jigsaw so we've got the key and that's very important for a stem and leave so three line one means 31 minutes so we're dealing with minutes here and the question said what does the modal time taken so the motor is the most common so we've got 31 39 48 and so on so if you look here we've got 57 and 57 so the modal time taken will be 57 minutes the next question said find the range of the times taken so the range is the largest takeaway the smallest so the largest amount of time taken was 75 minutes and we're going to take away the shortest time which was 31 minutes and 75 take away 31 will be equal to 44 minutes so the range the difference between the largest and the smallest is 44 minutes and the last question says find the median time taken so the median is the middle value so I like to do this by Crossing off the values and pencil of course across off the smallest and then the largest the next smallest which would be fair 39 then the next largest 66 the next smallest 40 64. the next smallest 43 the next largest 63 the next smallest 46 the next largest 60. the next smallest 51 next smallest 59 and then cross off cross off and we're left with 57. so the median time taken is 57 minutes and make sure that with this seven you remember it's 57 so it's 57 minutes that's it okay let's look at our next topic so our next topic is called quarters and that's video 57 and corporate Maps now we've looked at the medium before and the median is the middle value so if you've got a set of data the median is the middle value if you've got a set of data and you're looking for the quarter value that's called the lower quartile so if you arrange the values in order from the lowest to the highest a quarter of the way into the data is called the lower quartile and three quarters of the way into the theater is called the upper quartile now those are very useful for working at a thing called the interquartile range I'll talk about that in a minute so the median is the middle value the lower quartile is a quarter of the way into the data and the upper quarter is three quarters away into the data now to find the lower quartile I tend to put them in order I find the median and then I find the median of the lower half of the data and for the upper quartile I again range them in order I find the median I work out the median for the upper fifty percent of the data and also I mentioned a concept called the interquartile range and the interquartile range is worked out by working out the difference between the upper quartile and the lower quartile so if you do the upper quartile subtract the lower quartile we get the interquartile range now Mars again you've seen a topic called the range before which is the highest value you subtract the lowest value but sometimes there's outliers and that will affect the range massively if there's one value which is really really big for instance and the rest are much lower whenever you work out the range you get a really large range well if we work out the interquartile range so the difference between the three quarters of the way through the data and a quarter of the way through the data and you subtract those that will tell us how to spread out the middle fifty percent of the data is and that's really useful because it it means that the sort of the effective outliers is reduced or eliminated so let's have a look at our some data so we've got 5 5 6 8 9 10 and 12. and let's find the lower quarter the upper quartile and the interquartile range so if I wanted to work out the lower quartile to begin with well first of all we know that e it's the median so it's the median here and then if we look at the bottom 50 of the data we've got five five and six so the lower quartile would be five and in terms of the upper quartile well that would be ten so the lower quartile would be five the upper quartile will be 10 and the medium would be it and that's the lower quartile and that's the upper quarter and if we wanted to work out the interquartile range we would just do the upper quartile 10 subtract the lower quartile five and that gives us an interquartile range of five okay now these ideas of quartiles lower quartiles and interquartile range are very useful at topical cumulative frequency curves or graphs and I'll talk about those in a second so let's have a look at our next topic okay so our next topic is cumulative frequency so that's videos from 153 and 154 on corporate Maps so I'd highly recommend you watch those videos because this is a very important statistics topic now in terms of the words cumulative frequency so the cumulative frequency is the run on total so here we've got our frequency table and it's a group frequency table and we've got the lengths of some items and we've got zero to five centimeters five to ten ten to Fifteen fifteen to twenty and twenty to twenty five and we're given the frequency so Windows 3 in between naught and five ten between 5 and 10 21 between 10 and 15 and so on and sometimes we're asked to complete a cumulative frequency table so it looks something like this we've got the length but this time instead of going naught to five and five to ten it goes naught to five naught to ten not to Fifteen naught to 20 and not the 25 and we may be asked to fill in the cumulative frequencies so that's the run and total so we notice 3 in between naught and five so that means that the cumulative frequency would be free in terms of not to ten well we know there's three between naught and five and 10 between 5 and 10. so if we add those together three plus ten that means there's 13 altogether between 0 and 10. now I from null to 15 well we know there's a number 21 so if we add 21 on we'll find the accumulative frequency so 13 plus 21 is 34. we know there's another 4 so it's not going to be adding on 4 38 and adding on 2 would be 40. so we've completed the cumulative frequency table now let's draw the cumulative frequency curve now this is sometimes called cumulative currency curve accumulency frequency graph I tend to call it the cumulative frequency curve because it's quite nice to draw a nice curve for it so the plot the points for the cumulative frequency curve we're going to plot each of these cumulative frequencies at the end of each of these groups so we're going to go five across and three up ten across and 13 up 15 across and 34 up 20 across from 38 up and 25 across from 40 up we know these are the cumulative frequencies up to the end of each of these categories so five across and three up ten across and 13 up 15 across and 34 up 20 across on 38 up and finally 25 across and 40 up so points would look something like this now we're going to draw a curve in terms of your curve you're going to start at the the bottom of the first category so it starts at zero so we're going to go start at zero here and we're going to go up and do a curve going through the points like so and this is quite tricky on the computer but like something like that so that's our cumulative frequency curve now sometimes after we draw a cumulative frequency curve we're asked to work out things such as the median the lower quartile the upper quartile the interquartile range and so on so let's have a look at those now so now we're going to find those medians lower quartiles up quartiles and interquartile range for a cumulative frequency curve that I've already drawn for us and it's for pocket money and all together the 600 children and the pocket money goes all the way up to 12 pound okay let's start up a fan in the median so the median's the middle value now this is grouped frequency so we can just divide the frequency by two to find the position of the median so for 600 we're going to divide it by two so that's the third hundredth value so we've got a 300 on the cumulative frequency axis and we're going from 300 across curve and it'll look something like this and then down and make sure you do that with varuna and a pencil and we'll find that we're exactly halfway between this six and this value here that's six point that's eight pounds that's seven pound in the middle and we're exactly in the middle of those so it's going to be six pound fifty so our median amount of pocket money is 6.50 now our lower quarter well all together there were 600 children and the lower quartile will be a quarter of the way into the data so we're going to divide 600 by 4 to find the position of the lower quartile so 600 divided by 4 is equal to the 150th value so we're going to go from 150 across and down and when we do that we get to Five Pound so the lower quartile is five pound the upper quartile was three quarters of the way through the data so we're going to work out three quarters of 600 but we know a quarter is equal to 150 so if we do 3 times 150 that's equal to 450. so we're looking for the 450th so if we go across from 450 and down we get to eight pound so the upper quartile is eight pound and finally we've been asked to work out the interquartile range so that's how spread out the middle fifty percent of the data is so the upper quartile is eight the lower quartile is five so we're going to do eight take away five and that's equal to three pound so the interquartile range is three pound and that's it okay so let's have a look at some more questions that you may be asking cumulative frequency so you're often asked to find the median the lower quartile the upper quartile and the interquartile range but sometimes you asked over a question so let's have a look at another question now so here's a cumulative frequency curve is the same one with pocket money and cumulative frequency we've been asked to estimate how many people earned less than six pound so if we go from six pound up and across we get to 240. so that means that up to six Pi there was 240 children so how many children earned less than six pound well ours would be 240 and this is an estimate because this would have been grouped data and we don't know the exact figures so this is why it's an estimate okay our next question our next question says estimate how many people earned more than three pound so if we go from three pound which is in between two and four so here and up and across so we've gone to three pound we've gone up and across and that tells us that 40 children earn up to three pound pocket money so in terms of an estimate for how many children earn more than three pound that should be the rest of them so there are 600 all together so if we do 600 take away 40 That's equal to 560. so our estimate for how many children are more than three pound pocket money would be 560. okay next question our next question says estimate the amount of Poker money that 100 people receive less than so we're going to go from 100 people across and down and when we do that we get to the second Square across after the four so we know five's there so there's going to be one two three four five so there's five squares for one pound so one pound divided by five is twenty P so each square is 20p and we've gone to the second one there so it's going to be four pound forty so we've got four pound forty so that means that 100 children who receive the least amount of pocket money will receive less than four pound forty and that's our estimate also make sure you use a pencil and a ruler for this I'm doing a freehand on the computer so make sure using a rulona pencil whenever you're using these questions okay now next topic so we've looked at cumulative frequency now cumulative frequency links into another topic called box plots or sometimes you'll see them called box and whisker diagrams but I tend to call them box plots and that's video 149 and 150 in corporate Maps so this is a box plot now box plot has five vertical lines we've got a line for the lowest value a line for the lower quarter a line for the median a line for the upper quarter and a line for the highest value so we've got five vertical lines and the three in the middle the lower put out the median in the upper quartile are joined up into a box like so where a lion will go horizontally at the top and the bottom and then there will be two lines going out to the lowest value and to the highest value like so so this is called a box plot and one of the benefits of box plots is you can see how the data is spread out because between each pair of vertical lines 25 of the data lies so in between the lowest and the lower quarter there'll be 25 percent of the data between the median and the lower quarter number 25 between the media and the upper quartile and over 25 and between the upper quarter and the house value another 25 percent of the data so box spots are really useful ways to represent data and this is the core management revision card so if you do have that revision card make sure you have that and you sort of learn it and bring it to school with you make sure that you and you know how to draw and to read a box plot okay let's compare to box plots now so here we've got two box plots and they showed the reaction times of two groups so we've got group a and Group B and we've got their reaction times the quicker zones reaction times the less time it takes so values down to the left hand side here really good that meant someone's got quick reaction times because it's taken less time to react and values over to the right hand side means they're a bit slower now if I was to compare these box plots for group a and Group B the first thing I notice is the medians are the same so that means that their average the median reaction time is the same it's equal to each other another thing that I notice is that this group a is much it's much wider it means it's much more spread out so it means there's some people with really fast reaction times and then you've got some people with really slow reaction times so they're much more spread out and you can see that by particularly the Box itself because that represents the middle fifty percent of the data and it goes from 0.2 the lower quartile up to 0.35 so the interquartile range here would be 0.15 whereas if we have a look at this box the integral type ranges are much smaller because our lower quartile is 0.24 and our upper quartile is 0.33 and if we take those away that's much smaller so if I was comparing two box plots I would first of all look at the medians and see how they compare so here we can see the medians actually equals each other so it tells you the average is the same but you can see that the data is much more spread out for group a than Group B our next topic is histograms so a histogram is a way of representing group data and here we've got our group data and so we're representing it and in a Fair Way whenever you've got categories that have got different sizes so as you can see here we've got time not the two two to four four to six six to ten ten to 14 and 14 to 20. so as you can see here we've got groups that are quite small so the first group's only got a width for class width of two seconds it's only two seconds wide from null to two but the last group goes from 14 to 20. so it's actually got a width of six so it wouldn't be fair to draw a bar chart because some of the categories are much bigger or much wider than the other categories until you draw a thing called a histogram and the histogram looks at the frequency density and the frequency density is found better than the frequency divided by the class width now just before I go on histograms is a very important topic so if you go to videos 157 to 159 in corporate Mars you'll see me talk about histograms in great detail so what we're going to do is we're going to start off by drawing a histogram for this grouped frequency table so if we want to draw a histogram we're going to work out a thing called frequency density so let's put a column on for frequency density so I'm going to write f d standing for frequency density here so we've added on a column for FDA or frequency density now what we're going to do is we're going to work out the frequency density for each one of these groups or classes so from naught to two seconds well that's going to collapse width of two it's too wide it's got a frequency of 10. so if we do 10 divided by 2 so the frequency divided by the class we have 10 divided by 2 so 10 divided by 2. that gives us a frequency density of 5. so this category has a frequency density of 5. our next category well let's get a frequency of 13 and then it's got a class width of two it's too wide going from four to two four take away two is two so we're going to do 13 divided by 2. so 13 divided by 2 is equal to 6.5 our X group well again it's frequencies 18 and it's class width well it goes from four to six so it's class width again is two so we're going to do 18 divided by 2 and 18 divided by 2 is 9. our next group it's going from six to ten so it's got a class with a four because it's four wide and it's got a frequency of 16 so we're going to do 16 divided by 4 because we do the frequency divided by the class with that's four next the next category is go to frequency of eight and we're going to divide that by the class width where that's equal to 4 because it's four wide so 8 divided by four is two and finally our last group has got a class width of six because it goes from 14 to 20 that's six wide and it's frequency six so we're going to do 6 divided by six which is one so we've got our frequency densities now what we're going to do is we're going to draw a histogram so we're going to draw bars for our each of our groups so our first group is from naught to two so we're going to do a barrel from not to two and having a frequency density of five so it's going to go up to five so we've written on a pencil you'd go draw a line from five across over an opacity draw down and you draw your bar like so so that is your bar going from not the two and with the frequency to enter there five our next bar from two to four it's going to go up to 6.5 so from two to four well I actually need to make our bar a little bit higher so we're going to go now up to 6.5 so that's going to be halfway between six and seven so that's there we're going to go across the four and then down our next bar well from four to six goes up to nine so from four to six we'll look up to nine so again we need to make that bar a bit higher we're going to go up to nine now across and down so we've done from naught to two up to five we've done from two to four up to six point five and we've done from four to six up to nine now we're gonna do from six to ten and with a frequency density of four so from six which is here to ten we're gonna have a bar that goes up to four so we'll draw across the ten and then we're gonna go down like so so that bar from six to ten let's get a height of four next we're gonna go from ten to Fourteen and let's get a height of two so from 10 to 14 I'm gonna go across from two and down and finally the last group goes from 14 to 20 so we're going to go from 14 to 20 and it's got a height of one so we're going to go make sure it's Heights one and we're going to go down like so and that's it we've drawn our histogram so we've drawn our histogram for our group frequency table now sometimes we're given a histogram and we might need to find out the frequency now if we go back frequency density was equal to frequency divided by class width if we wanted to make frequency the subject we would multiply both sides by class width so that would give us the frequency density times class width is equal to frequency so the frequency is equal to the frequency density times the class width so in other words if you multiply the height of each bar the frequency density by its clouds width how wide it is you will find the frequency so let's complete this group frequency table so we've got our categories naught to 40. so as you can see our first bar went from North to 40. now if we look vertically we're going from 0 up to one there's ten little squares that means that each square is 0.1 so this bar has a height of 0.1 our frequency density of 0.1 one and it's got a collapse width from 0 to 40 or 40. well if we do the frequency density 0.1 Times by 40 That's equal to 4. so that means the frequency for this bar is equal to four so as you can see here the frequency would be four next our next bar our next bar goes from 40 to 60 and it's frequency density well it goes up to 2.4 so the frequency density is 2.4 so we're going to do 2.4 multiplied by the class width which is well from 40 to 60 is 20. so multiply by 20 is equal to 48 so that means that this category from 40 to 60 has a frequency of 48 there was 48 people and then our next category from 70 to 80 well the class with is 10 its frequency density well it goes up to 3.9 so we're going to do 3.9 times 10. so 3.9 times 10 is 39 so it would have a frequency of 39. so to draw a histogram we use the formula frequency density is frequency divided by clat's width so if you've been given a histogram and you want to find the frequencies well we would do frequency density times class width okay let's have a look at a typical histogram's question now so the histogram below shows information about the number of hours flown by some pallets and we're told that 45 Pilots flew from more than 400 hours so we know that between 400 and 450 hours there's 45 pallets in here okay so first question says the estimate how many Pilots flew for under 100 hours so if you have a look at our histogram the first category goes from naught to 200 so that's why the question says estimate because whenever we had this group frequency table it would have said not to 200 and it would have the frequency so because of script frequency we don't actually know how many were between no a 100 or 100 and 200 so this is going to be an estimate so if we have a look at this bar let's work out the frequency for this bar it's got a frequency density of 3.5 and the Cloud's width is 200 it goes from naught to 200. so if we do 3.5 multiplied by 200 that would tell us the frequency and that's equal to 700 so that means there's 700 pad it's in here now under 100 hours well that would be half of the pallets well our estimate our guess will be it's half of the pilots so we're going to do 700 divided by 2 and that's equal to 350. so that's our educator guess to how many pallets have flew under 100 we do know the 700 the flu between naught and 200 hours and our estimate is that half of those 350 have flew for under 100. okay next question our next reason says what fraction of pallets flew between 200 and 400 hours so we want to work at the fraction of pallets so we're going to need to know how many pallets there are all together so let's get rid of that dotted line to begin with and we know the 700 in this bar we've been told this 45 in this bar because 50 multiplied by 0.9 is 45 so we know there's 45 in there let's work out how many pallets are in this bar and this bar so for this bar here it has a frequency density of 4.7 so a frequency density of 4.7 multiplied by a class width of 100 would be equal to 470 so this 470 pilots in here and this bar has a height frequency density of 2.5 and again the class move is 100 2.5 times 100 is 250. so we know the 700 Pilots between not 200 470 pallets between 200 and 300 250 pallets between 300 and 400 and 45 pallets between 400 and 450. we've been asked to find out what fraction of pallets flew between 200 and 400 so that's going to be all together so 470 plus 250 that's 720 out of all the pallets all together so it's going to be 700 plus 4 170 plus 250 plus 45 that's equal to one thousand four hundred and sixty five and if I tap that into my calculator I'll simplify it nicely for me to 144 over 293. okay our next question okay let's have a look at our next question so our next question says the histogram shows the height of 2 400 trees so we've got their heights in centimeters from not all the way up to 250 and we've been asked to find an estimate for the mean height we've been asked to find an estimate for the median height so if I was working this out the first thing I'd want to know is the frequencies so we've got the frequency densities of 8 16 20 and 5. and I would want to work out the frequencies for each one of those bars so we've got here this bar 75 this bar would be 125 and yeah okay so for the frequency for this bar from 0 to 75 well I would do the frequency density eight times the class width which is 75 so it has a frequency of 600. next from 75 up to 125 it's going to classify for 50 times by 16 is 800. next from 125 to 150 that's a classroom for 25 Times Square 20 is equal to 500 and finally from 150 up to 250 that's 100 is a class with Times by 5 is equal to 500. and let's just add our frequencies up 600 plus 800 plus 500 plus 500 is equal to 2400 which we're told in the question which is fantastic so the question asks us to work out an estimate of the mean height and an estimate of the median height let's start off with the mean height now I'm actually just going to draw frequency table like so and our frequencies were 600 800 500 and 500 and we've got our height so it's not to 75 75 to 125 125 to 150 and 150 up to 250. now we want to find an estimate the mean so remember that's where we add on our FX column and we do the midpoint times the frequency midpoint times the frequency and so on so the midpoint for our first category from not the 75 well well that's going to be 37.5 so we're going to do 37.5 times 600. that's equal to 22 500. next we'll do the midpoint well the midpoint of 75 125 is 100 times that by the frequency is 80 000. next we'll do the midpoint of 125 150 and that's 137.5 remember we can find the midpoint by just adding the numbers and dividing by Two Times by 500 is equal to 68 750 and finally we're going to do the midpoint of those numbers which is 200 times by 500 which is equal to one hundred thousand next to find the estimated mean we add up our FX column so we find the grand total and when we do that we get 271 250 and then we divide that by the total frequency but we're told in the question that's two thousand four hundred so we could add up the frequencies or we could just use the fact that we told it in the question so when we divide 271 000 of 250 by 2400 the grand total divided by how many there are that would tell us that our estimate for the mean height remember the estimates because we're using the midpoints and we don't actually know the actual values that's equal to 113.02 centimeters and that's it so that's the r estimate of the mean so that's the two decimal places and that's it okay next we will find ask then to find an estimate of the median okay so our next question says find an estimate of the median height so if we go back to our histogram and we look at each bar its area represents its frequency because the frequency density the height of the rectangle multiplied by the class width the width for the rectangle that gives us our frequency or the area of the rectangle that means that frequency is represented by the air of the rectangle so that means that we can find the vertical line that we draw down so the half the area of the scrams on the left and half the area of the histograms on the right that means that we can read off and find out what the median would be so let's find the area for this histogram now I could cut the little tiny squares if I wanted to but I'm actually going to count the bigger blocks because I can actually see in this question that there's lots of these whole blocks and each of the bars sort of is split up into these larger blocks apart from obviously this one where I've then got half blocks so if I count them so there's 48 Big Blocks so if I can find out whether halfway the 24th block is and divide it so there's 24 blocks on the left and 24 blocks on the right we can find out where the medium would be so let's find out where the 24th block would be so we've got one two three four five six seven eight nine ten eleven twelve so we've got 12 blocks in that one so we need number 12. we've got one two three four five six seven eight well twelve plus eight is twenty now we need another four but we've got if we look across here we've got one two three four five six seven eight so we actually only need four blocks that's half of those so if we took our learner pencil and we went halfway down here that would mean that we would have 24 blocks worth on the left and 24 blocks worth on the right and that value would be whereas in between 100 and 125 and exactly in the middle so the median would be 112.5 centimeters and that's it so that's one we're found in the median now you may have been able to guess that that's not actually the approach that I use whenever I work out the median just because I don't like counting the little blocks and trying to figure out where that line is I use approach called linear interpolation so it's video 52 in corporate maps and the formula for it is the lower bound plus number into category divided by number in category times class width it sounds quite complicated but it's not really so it's lower bound plus number into category divided by number in category times clouds width and let me explain what I mean by that so we're looking for the medium value so if this 2400 we divide that by two because remember for group data it's just an estimate anyway so we can do 2400 divided by 2 is equal to the 1 200 value and if I was looking for the 1 200 value I would look at my frequency table and I would say well the 1200 values not in the first category because there's only 600 in there if I look at the next category there's 800 in there so all together if I work out the cumulative frequency there's one thousand four hundred trees all together so the 1200 tree is in this category here so we're going to look at this category and we're going to say what is lower bound at 75 plus the number into the category we're looking for the 1 200th tree so we've gone 600 we need to go another 600 so we need to go in over 600 into this category so we'll write 600 it's 600 into the category divided by the number in the category well that's 800. and then times where the clouds with the cloudswift is 50. so if we do 75 plus 600 over 800 times 50 we should get that median that we got before and the answer is 112.5 centimeters which is the same as before the reason I like that formula is you can then work out the lower quartile upper quartile interquartile range really quickly and easily from histograms and the formula is not actually that complicated okay let's have a look at our last histogram question this is the hardest histogram question we've got here and it says 24 cars traveled faster than 40 miles an hour so we know 24 cars traveled faster than 40 miles an hour and it says estimate the number of cars traveling between 25 and 35 miles per hour so here we've got our histogram and we've got the frequency density gone up vertically with no numbers that's not going to be very helpful and we've got our speed going across horizontally and we've been told that 24 cars so we know there's 24 cars in this bar here and it says a workout an estimate for the total number of cars traveling between 25 and 35 5 miles per hour so we want to work out an estimate for how many cars would be in here now first of all what makes this question a lot harder than the previous ones is we don't know the frequency densities so there's two different approaches you could use for this question the first one because we know that 24 cars travel more than 40 miles per hour we know that the frequency for this bar is 24. so if you do 24 the frequency divided by the class where if we can get the frequency density 24 divided by 10 is 2.4 so we know the frequency density there would be 2.4 so we know the frequency density is 2.4 now we can figure out our scale so we've got 2.4s here well there's one two three lines going up to that so if we divide that by three that's not 0.8 so 0.8 1.6 2.4 3.24 4.8 5.6 so now that we've got our scale for our frequency density we can now work out our estimate for how many cars are traveling between 25 and 35 miles per hour so let's work out the frequency for this 20 to 30. so we would do the class width which is 10 times 4 so that means there's 40 all together in this bar so that means that if we divided it by 2 this should be 20 in here and likewise for this bar this one going from 30 to 40 miles per hour so we would do 10 times 5.6 which is 56 and if we divided that by 2 the be 28 and 20 plus 28 is equal to 20 plus 28 is equal to 48 cars so that's our estimate for how many cars should be traveling between 25 and 35 miles per hour so that's one approach now that question was quite nice because the information we were told is there's 24 cars traveling more than 40 miles per hour so we were told the frequency for one of the bars sometimes in the question you might be given sort of you know so many cars are traveling more than 35 miles per hour it's a bit more complicated or another approach is to consider the frequency and to look at the area of that bar that represents that frequency so as you can see there's one two three four five six seven eight nine ten little squares going across this 5 10 15 of those so there's 150 little squares so 150 squares equals 24 cars you could then divide 150 by 24 and say well 150 divided by 24 is equal to 6.25 so 6.25 squares equals one car then you could look at the 25 to 35 region and see how many little squares there are in those so as you can see we've got 5 10 15 20 25 times 5. so it's 125 little squares there and from 30 to 35 there's 5 10 15 20 25 30 35 times 5. is equal to 175 so all together there would be 125 plus 175 which is 300 squares and if we divide that by 6.25 we can see how many cars would be so 300 divided by 6.25 is also equal to 48 of course and that's it okay let's have a look at the probability question so here's our question it says there are only pink yellow green and blue countries in a bag so there's four different colors of counters in the bag and the table shows the probability of picking each color so the probability of picking a pink counter is 0.5 the probability of picking a green counter is 0.1 the probability of picking a blue counter is 0.2 and the question says find the missing probability so that's the probability of picking a yellow counter because there's only pink yellow green and blue countries in the bag there's no other colors that means it's certain that we're going to pick one of these colors so I mean to these probabilities will add together to give us one so if we add up the properties that we know so if we add those properties together we get 0.5 plus 0.1 it's 0.6 plus 0.2 is 0.8 so that means the probability of picking a pink green or blue counter is not 0.8 so the fact that probably found in the yellow we're going to take this away from one to see what's left tell me it's a probability of picking a yellow counter would be 0.2 and that's it so the next topic is actually the probability of something not happening and that's video 250 in corporate Maps see the probability of something not happening is one take away the probability of it happening and that will tell you the probability of something not happening so here we've got the probably of hammer when in the competition is 0.28 and we've been asked to find out the probability that hammer does not win the competition to do that we just need to take away not point to it from once we just do one take away 0.28 and that would be equal to 0.72 okay let's have a look at our next next topic so our next topic is expectation which is video 248 and Corporal Maps now if you know this probably if something happened and you know how many times an experiment or something has taken place if you multiply those two things together you'll find out how many times you should expect something to happen so let's have a look at an example we've got a factory make 6 000 plates each day so the factory makes 6 000 plates each day and the probability that a plate is faulty is 0.035 so it's quite unlikely it's very unlikely that one's faulty and it's got a probability of 0.035 and we've been asked to find out how many faulty plates would be expected in one day so if we multiply how many plates that are made by the probability of one of them being faulty we'll find out how many faulty plates we'd expect if we do six thousand multiply by 0.035 that will tell us how many faulty plates we would expect and that is 21. so we would expect 21 faulty plates if you're asked to find out how many plates weren't faulty then you could just take the 21 away from 6 000 and that'll tell you how many players would be not faulty okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is called relative frequency or sometimes called experimental probability and that's video 248 in corporate Maps so this time we've been given a table and with two people Susan and Helen and they've done an experiment and Susan's done the experiment 20 times she span a spinner she's spam the spinner 20 times and on that spinner there's different letters and we're told the number of bees that she got whenever she smile at 20 times was it so if we're asked to find the relative frequency of Spin and a b so because you span the spinner 20 times it would be out of 20. and because she got eight B's the number of successes is eight we write 8 out of 20 or you get 20th like so we could write this as our decimal or we could even simplify our fraction simplify our fraction we'll divide in both of these by four would give us two fifths so we could write an answer of eight twentieths or two-fifths if the question doesn't tell you to simplify it we don't have to so we could just write eight twentieths alternatively because Helen's result is a decimal number we could also give it as a decimal number we could write well two-fifths as a decimal will be no point four okay let's look at our next person Helen and Helen spun the spinner more times she span 120 times and this time we don't know how many beats she's got but we know the relative frequency if we're getting a b we know the relative frequency is 0.35 now to find the number of bees that Helen got all we need to do is multiply the relative frequency by her number of spins thinking back to we had a topic called expectation where to find how many times you expect something to happen you just multiply the number of Trials by the probability so if you multiply the relative frequency of the experimental probability by the number of Trials it'll tell you how many successes you had so if we multiply 120 by 0.35 that will tell you how many times Helen spinner landed on a b whenever she spell it so that would be 42. okay let's have a look at our next topic so our next topic is called tree diagrams and it's a probability topic and a tree diagram is a really useful way of showing what can happen Whenever two or more events take place and this video 252 according maths is a very important topic so I highly recommend you watch that video and try the practice questions as well so let's have a look at our questions our question says John and Dan are taken in terms of throw a ball at a Target so we've got two different events John's gonna throw the ball at the Target and then Dan's gonna throw the ball at the Target and what's great is this diagram will show us all the possible outcomes of what can happen and we've been given some probabilities so we've been given the property that John hits the target is 0.6 and they're probably the Dan hits the target is 0.7 now before we get started let's actually label the probabilities on the diagrams now the problem is that John hits the target is 0.6 so let's work out the probably the John misses the Target now it certain that he does one of these two things he's either going to hit it or he misses it that makes these two probabilities need to add together to be equal to one so if we want to work out this missing number we're going to do one take away 0.6 and that would be 0.4 so these two properties will need to add together to give us one okay now with Dan throwing the ball at the Target when he can hit the target which is not 0.7 so the probability that Dam misses the target is 0.3 and below for down again we've got 0.7 is the probability that he hits the target or not 0.3 is the probability that he misses the target Okay so we've now got our tree diagram with our probabilities labeled on it now let's consider the outcomes of John through and the ball and then down from the ball well if we go along the branches we can find out all the possible outcomes and that's the great thing about a tree diagram so let's start here and John throws the ball and he can hit it and then dye can hit it so we've got a hit and a hit hit and a hit so that's a hit and then a hit then we could have John Hits the Target and Dan misses the target so it'll be a hit and a miss so that's a hit and a miss now if we go down this way John can actually missed the Target and then Dan could hit the targets out of a miss and a hit and finally John could miss a Target and dying can miss the target so be a miss and a miss so we've got all the possible outcomes here we've got a hit hit Miss Miss hit and a miss miss and what's fantastic is we can use the properties we've been given in the question to work out the probabilities of a hit hit hit Miss Miss hit a miss miss really easily and all we do is multiply the numbers along the branches so if we wanted to work out the probability of John hitting the Target and Dan hitting the Target because it's and what we do is we multiply the properties together so we're going to do 0.6 multiplied by 0.7 so we go along the branches and we do 0.6 multiply by 0.7 so let's write that down 0.6 multiplied by 0.7 is equal to 0.42 so the probability of John hitting the Target and down hitting the target hit would be 0.42 next let's work out the probability of John hitting and down missing so a hit Miss so I'll be 0.6 multiplied by 0.3 so 0.6 multiplied by 0.3 will be 0.18 next we could have John Missin and Dan hitting so that would be 0.4 multiplied by 0.7 so 0.4 multiplied by 0.7 and that'll be equal to 0.28 and finally we've got a miss miss that's John missing and down missing so that would be 0.4 multiplied by 0.3 so 0.4 multiply by 0.3 would be equal to 0.12 and what's great is if we add these properties together 0.42 0.18 that's 0.6 plus 0.28 that's not 0.88 and then plus 0.12 that gives us one because obviously one of these four things will have to happen over hit hit Miss Miss hit or miss miss okay so let's go to our question so our question says what is the probability that at least one man hits a Target so at least one man hits a Target means either one of them could hit it or both of them can hit it so hit hit that would work hit Miss yep one of them's hitting the Target that would work Miss hit yep one of them's hitting it that would work but miss miss wouldn't work semis were interested in this problem ability this probability and this probability and we want to find the probability that either it's a hit hit or a hit miss or a Miss hip so if we add these three proper leads together we will find the probability that at least one man hits the target so we could do 0.42 plus 0.18 plus 0.28 and when we do that we get 0.88 so the probability that at least one man hits the target would be 0.88 and that's it okay let's have a look at our next topic so the next topic is independent events an independent events is a bit like tree diagrams but we're just not going to be using our tree diagram so then this would be 249 so we've got Naomi has a bar spinner so here we've got a spinner in this bias that means it's not equally likely to land in any of the sections so we've got the probability of the spinner lands in red being 0.1 the probably lands on white is 0.4 the probability that it lands on black is not 0.3 and the probability that lands on Brian is 0.2 and she spins the spinner twice and the question says what's the probability that the spinner lands on black twice now you could draw a tree diagram here and you could do one two three four branches for the first outcome the first Spin and then we could do another four branches for each one of them like so and it'd be quite complicated what would be great is if we could do this question without drawn out the true diagram so we want to find the probability of a black and a black now the probability that the spinner lands on black any time is 0.3 so what we're going to do is if we imagine our branches we're going to do black and black so that would be 0.3 multiplied by 0.3 because that's probably for black multiplied by the probability of a black so 0.3 multiplied by 0.3 would be 0.09 and that's it if we wanted to find out the property had landed on Brown twice we would do 0.2 times 0.2 if we wanted to find out the property it lands on white twice we would do 0.4 times 0.4 red twice 0.1 times 0.1 and so on okay that's it so let's have a look at our next topic so the next topic is tree diagrams but this time we're going to be dealing with tree diagrams whenever objects perhaps are taken out they're not put back in so we know the probabilities will change depending on what happens okay so let's have a look at typical questions so Kieran has seven blue pens so blue pen blue pen blue pen blue pen blue pen blue pen blue pencil we've got seven blue pens and he has got three red pens so he's got three red pens one two three so Karen's going to pick a pen out at random so he's going to pick one of these pens out at random and it says with our replacement so then he doesn't put it back in and then he picks his second pen at random and the question says what's the probability that the pens are different colors so here we've got a tree diagram and we've got the first pen and then the second pen let's deal with the first pen to begin with because we know they just got seven blue and three red to begin with so the proper way from getting a blue would be 7 out of ten so seven temps and the probability if I'm getting a red will be 3 out of ten so three out of ten so if Libra 2 probabilities to begin with now the second branches are going to be a little bit more complicated because we're going to have to think about it a little bit so he could have taken out a blue pen to begin with because he could have taken out a blue pen and now we're going to consider what the properties would be if he's taking out the blue pattern so let's get rid of a blue pen so I've got rid of a blue pen unless I think of what the properties would be then for the second pen if you took out the blue pen to begin with so all together he's now got six blue and three red so the probability of a blue pen well there's nine pens all together so it's gonna be nine and there's one two three four five six blue pens so the probability of the second pan being blue would be six ninths so if Kieran took a blue pen to begin with the probability of the second pen being blue would be six ninths if we took a blue pen to begin with and we want to know the probability of it being red well there's nine pens and one two three of them are red so the probability of the second pair being red if the first pen was blue would be three names okay so these are the probabilities if he took a blue pen to begin with now let's find the probabilities if we took a red pen to be on with so let's put that blue pen back so there's that blue pen back again and if you took a red pen out to begin with so let's get rid of one of the red pants so he's taking that one out and now let's find these probabilities so he's taking out a red pen also probably for blue pen now well again there's still nine pens because he's taking one of them out so we've got one two three four five six seven eight nine so there's nine pens and there's one two three four five six seven he still got seven blue pins in there because he took a red one out so the probably for blue would be seven ninths well the probe for red well if we took a red one out to begin with there'd be nine pens and there'd only be two Reds now so the probably for red would be two ninths so we've now labeled the probabilities on the tree diagram and as you can see the properties change dependent on what he does first now let's consider outcomes so we could have a blue and blue he could have a blue and a red he could have a red and a blue or he could have a red and a red and to find these probabilities remember we multiplied the numbers along the branches so for a blue and a blue we would do seven temps multiplied by six ninths and that would give us 7 times 6 is 42 and 10 times 9 is equal to 90. so the probability for blue and a blue would be 42 90th the probability of a blue and a red well that would be a blue seven tenths to begin with and then we're going to multiply that by three ninths so multiply by three ninths and that would be equal to 7 times 3 is equal to 21 over 90. now the probability of a red and a blue well that would be three Temps multiplied by seven ninths so multiplied by seven ninths three times seven is twenty one ten times nine is 90. so that would also be 21 over 90. and finally the probability of a red and a red well that would be three tenths times two ninths and that would be equal to 3 times 2 6 and 10 times that is ninety so the probability for red and a red would be six ninetieths now the question says find the probability that the pans are different colors so if we have blue and blue that wouldn't work they're both the same colors if we had a blue and a red that would work so that probability is a good probability that would work a red and a blue well there are different colors as well so that's another probability that we want to look up and a red and a red wouldn't work they're both the same color so we want to consider these two probabilities so see if we're going to be a blue and a red or a red and a blue and remember in the OR real if it's one probability or another one we add them together so we're going to do 21 over 90 plus 21 over 90 and that would be equal to 42. either 90. so the probability the pairs of different colors would be 42 out of 90 or 42 90ths now normally with property questions you need to simplify if they ask you to so if we simplify this at the very end the answer would be 7 15. and that's it okay let's have a look at our next topic okay let's have a look at our next topic so our next topic is conditional probability as video 247 of corporate maps and here's our question it says inside a bag there's two apples so an apple and an apple and there's three bananas so there's three bananas so in a bag there's five pieces of fruit there's two apples and three bananas and Olivia's gonna pick out two items from the bag at random without replacement so she's going to take out two atoms you can take the first one out she's not going to put it back in and then she's going to take a second one out and the question says what's the probability of her picking two apples that's the probability of an apple and another Apple so probably of apple and Apple okay so all together Olivia has five pieces of fruit in the bag now we want to find the probability of her picking an apple to begin with so all together there's five pieces of fruit two of them are apples so the probability that first piece of forbidden apple is two-fifths it's also probably the first piece of Rubina Apple two-fifths so now she's picked an apple she's taken it out of the bag and nana leaves her with four pieces of fruit in the bag and she's got an apple and three bananas and we want to find the property with an apple and an apple so remember on that tree diagram we want to find the probability of the first one on Apple and then we're going to multiply by the probably the second one being an apple but she's taken one of those apples out so now all together there's four pieces of fruit in the bag and she's only got one apple so the probably if the second one being an apple would be one quarter now we've got our two probabilities we can just times them together two times one is equal to two and five times four is equal to twenty so the probability of Olivia picking an apple and an apple would be two twentieths or we can counter that down to one tenth and that's it if we ask the probability of her picking two bananas so the probability of banana and banana we would do well to begin with let's put that apple back in the bag so let's put that a back so if we're asked the probability of Olivia picking two bananas well first of all there's five pieces of fruit in the bag and three of them are bananas so to be three-fifths then she would have taken out a banana and then that would leave her with four pieces of fruit and two of them being bananas so that would be the probably of the second would be in a banana would be two fourths or two quarters so multiply by two quarters and then we would just multiply them together three times twos equal to six and five times four is equal to twenty and then we can cancel that down to be three out of ten that's it so whenever you're dealing with conditional probabilities just very important to think about what the probability would be after the first thing has happened so for in other words after she's taken out the first Apple what was the probability of her getting that second Apple likewise with the bananas once she takes out that first banana what would the problem is then be off the second one being a banana and so on so we're gonna look at Venn diagrams now and that would be 380 a and corporate maps and the question says 90 people were asked if they liked free drinks so we've got drink a drink B and drink C and 11 people liked all three drinks so let's put that on the Venn diagram they're like all three drinks so it's gonna be 11 people in the middle if people liked drinks A and B but not C so they like a and b so they're going to be in the middle of A and B so here but not C so it's going to be here eight people like drinks A and B so then there are circles but they're not in the C Circle so that's there next 17 people like drinks A and C so A and C would be here but not B so that means that 17 people will go in there and nine people like drinks B and C so they like drinks B and C but not air so it means that's going to go there we're also told that four people only like drink base with four people only four people like drink B and we're told that 50 people like drinking a so that means that the whole circle here for a will be 50. so if we add up the numbers we know in this circle for a we've got 17 plus 11 that's 28 plus and over 8 is equal to 36 so that means that 14 people must go there and three people did not like any drink so they're gonna go outside and we've been asked to complete the Venn diagram so we've done most of it and there's just one missing number here now we're told that 90 people were surveyed so if we take all these numbers away from 90 we'll see what number must go here and when we take away all these numbers away from 90 we're left with 24 so that means it must be 24 people in here so we've completed our Venn diagram Okay so we've looked at a Venn diagram in question whenever we're dealing with a worthy situation now let's have a look at some notation that you may encounter whenever you're looking at Venn diagrams so here we've got a Venn diagram so we've got a and b and as you can see here's a here's B this is the section where they overlap and this is the section that is neither in air nor B and then this is the universal set which is the whole rectangle and you've got a and a few been asked the sheared a that would be inside of a if we're asked to sheared B that'll be inside of B there next we've got this a with a little Dash above it that means the complement of a and the complement of a means not a it means anything that's not a so if you have a look here whenever we sheared it in it's going to be anything that's not in a so a dash means not a are the complement of a and that means anything that's not a and then here if we've got B Dash that means the complement of b or not B so it's n of an outside of B so that's a b a dash which means the complement of a which is n of in this not a we've got B Dash which is the complement of B which is not B now we've got a and then this symbol B and this is a union B and that means A over B it means anything that's in a or b anything at all and that means it would be this region here and if it's in a or a B because it's a union B A or B now we've got a and then this symbol B and that means a intersect B and that means A and B so it's the region that's in a and it's in B so that's the overlap region in the middle so this is the chord Mass revision card it's very useful and it's very important to know this notation that a is inside of a b is inside a b a dash means the complement of it means not A and B Dash means the complements of B so it means anything that's not in b a union b means n if in an A or B anything tall it's in a or b and a intersect B Means A and B it means anything that's in that middle region that overlap and that's it okay let's have a look at the question now where we're dealing with that notation okay so here's a Venn diagram Square question and we've got the symbol as well and this symbol means a universal set and we've got the universal sets that's all the numbers that we're going to put in the Venn diagram are one two three four five six seven eight nine and ten and we're told that the numbers inside of a are two five seven and eight and the numbers inside of B are two four and nine and we've been asked to complete the Venn diagram so whenever I'm complete an event diagram like this the first thing I would do is look and see what numbers are going to be in the middle so in other words what numbers are in a and in B and if we have a look here two is in both of them A and B so we're going to put a 2 in the middle and then we've got no other numbers that means let's put the rest of the numbers in the a in a inside so it's going to be five seven and eight and if we look at B B has got two four and nine so that's two four and nine so if we look in a we've got 2 5 7 and 8 and then B we've got two four and nine now we're going to put the rest of the numbers outside of A and B because the rest of the numbers are in the Venn diagram but they're not in a or a B so it's going to be one we've done two we've then got three four five and we've got six seven eight nine and ten so we've completed a random grammar let's just check we should have ten numbers in this diagram one two three four five six seven eight nine ten so we've completed the Venn diagram we've done part A and then we're told a number is chosen at random so one of these numbers is going to be chosen at random and we've been asked to find the probability of a intersect B so in other words that's a and b so it's any number that's in A and B so if we look that's going to be the region here so we want to find the probability of choosing a number that's in this region well there's ten numbers all together so we're going to put 10 on the denominator and there's only one number in a intersect B in this region in the middle so the probability of choosing this number would be one out of ten or one tenth okay next question part C part C says find the probability that the number chosen at random is in a union B in other words it's an A or B so that means there's anywhere inside of a or b or both any of these numbers at all so if we have a look we've got one two three four five six numbers in A or B so it's going to be six and then there's ten numbers all together so the probability of choosing a number in a or b in a union B would be six out of ten or six tenths so probability would be six tenths or if you wanted to simplify that would be three-fifths and that's it now samples can be very very useful rather than a surveying or interviewing everybody for instance in a school if you wanted to find information about what color Blazer you would like rather than asking every single student it can be very useful that's called a census asking everybody in a population rather than a census you can do a sample where you ask a smaller number of people but it's very important whenever you do samples they're fair so for instance if I was doing a survey of students about Blazers it's making sure they ask people from different year groups you ask a range of different classes so it's very important that you get a good sample of fair sample to make sure that your sample is representative of the whole population but here we've got a question and it says 480 students attend the school and a teacher asks for 50 students which color Blazer they would like so we've got black 20 students with like a black blazer 15 students want a Navy Blazer nine students want a green Blazer and six students won a maroon Blazer and the question says estimate how many of the 480 students would like a Navy Blazer now altogether we knew there was 50 students all together and we want to work out an estimate with for how many students with like a Navy Blazer now if I look at the sample we know that 15 out of the 50 wanted Navy now if we work out 15 over 50 as a fraction that's equal to three tenths so that means that three tenths of the students asked would want a Navy Blazer now if this was a good sample a representative sample that means that it should be the same fraction of the whole school would want an AV Blazer so if we work out three temps of all the students in the school we'll find out how many should want a Navy Blazer so if we work out three tenths of 480 that would tell us a good estimate so we'll do 480 divided by 10 that's equal to 40 it and then we'll do 48 multiply by three and that's equal to 144. so that's it if we're asked how many of the 480 students would like a black blazer we would have done 20 over 50 which is two-fifths and then work out two-fifths off the 480 and so on okay let's have a look at our next topic so our next topic is capture recapture which is video 391 on corporate Maps so capture recapture is one of the subtle differences in the exam boards at Excel mention capture recapture whereas OCR and AQA don't actually mention capture recapture but they do mention sampling so it may still be useful to spare a minute or two just watching this part anyway but it's up to you okay so capture recaptures the technique that we can use to estimate the total population and this is the formula that we use whenever we're working our capture recapture and that's M the total marked over n the total population is equal to R the total recaptured divided by T the total capture on the second visit that sounds quite complicated but let's have a look at an example now whenever I was in school our biology teacher actually taught us capture recapture by bringing us to the greenhouse and she got us to work out an estimate for how many would likes lived in the greenhouse and what she did was she brought us to the greenhouse on one day perhaps Monday and we captured as many with lights as we could and we marked them with tbx we put a little bit of Techs in each one of their backs we just put a little bit of white tip X in each one of the Woodland back and what we've done was we then released all those wood lights we put them back to where they we found them and we counted how many wood lights we captured to begin with and we captured as many as we could and let's just say for instance we caught 300 red lights so we would put 300 here for the total Mark 300. so then what happened to us we went back at the end of the week went back on Friday we give the woodless chance to mingle and we captured as many as we could we we captured as many as we could on that second visit and let's just say we captured and we looked to see how many of those would like so out of those 80 had white marks on their back how many we recaptured let's just say for instance there were 60 that had setbacks out of the earring then that gave us a formula that we could work out an estimate for the total population because we have got 300 the total that we captured on the first visit and marked over the total population the how many are there in total is equal to how many we recaptured that's a 60 with the white dots divided by the tool we captured in that second visit which was eerie and we could use that information to work out an estimate for how many woodlice lived in the greenhouse and if we just solve this we can work out enesma for the total number of wood likes that lived in the greenhouse so that means if we times both sides of the equation by n we would get 300 is equal to 60 n divided by 80. and then if we multiply both sides of the equation by 80 we get we get 24 000 is equal to 60 n and then if we divide both sides of the equation by 60 we get that's equal to 400 so we get n is equal to 400. so our estimate for the total number of woodlice that lived in the greenhouse would be 400. okay let's just have a look at one more example so we've got some students want to estimate the number of wood likes to live in a greenhouse now it doesn't always have to be woodlice in Greenhouse it could be fish in a pond and the catch a mark 80 would license so in their first visit they capture 80 would like so let me say 80 out of the total population will be equal to and then the next day they go back and they capture 50 with lice and 32 of them are marked so 32 out of the 50 are marked so that means the 80 over the total population is equal to 32 divided by 50. we've been asked to estimate how many would like to live in the greenhouse so let's just solve this for n so let's multiply both sides by n that gives us 80 is equal to 32 n divided by 50. let's multiply both sides by 50 so that'll be 4 000 is equal to 32 M and dividing both sides by 32 will give us 125 is equal to n so n is equal to 125. so our estimate for the total number of fertilized that would live in that Greenhouse would be 125 and that's it and that's it so this has been the ultimate GCSE higher revision video in this video I've spent two or three minutes going through every single one of the topics on the GCSE higher revision checklist some of the topics such as Surge and standard form are spin it's a bit longer than two or three minutes because they're quite important topics and but I've spent roughly two or three minutes on the topics on the GCSE higher revision checklist the video has been quite long roughly 10 hours long so I hope you watched it in chunks and written notes on it and remember you've got that booklet that ultimate GCSE higher revision booklet where there's a question on every single one of the topics as we've gone through for you to practice and this video is absolutely perfect for anyone that's studying for edxl hair AQA higher no CR now remember there are some subtle differences around top of such systematic diagrams frequency polygons ambiguous case and rationalizing denominators and if you watch those bits of the video as I mentioned the difference between them but they're very subtle and that's it so remember there's that ultimate GCSE higher revision question booklet and the link for that's in the description below also as we've gone through this video I've used a lot to the chord Mouse revision card so if you do want to get the code mileage revision cards in the description below there's a link to the higher revision cards and you can find them on the website as well and they're fantastic for you if you're revising for your higher GCSE Maps also on the website we've got the five video books so rather than cramming Eurovision to the very end using a little and often approach is fantastic so if you're revising for GCSE hire I'd highly recommend the yellow Foundation plus books the green higher books and the blue higher plus books and that's it so this has been the chord Maps ultimate GCSE higher revision video it's quite a long video I really hope you found it useful the whole purpose of the video is to make sure you're familiar of all the topics on the GCSE higher revision checklist so I really hope that you're confident with those topics and you do really well in your GCSE haramav's exams so if you have found it useful please like the video please subscribe to the YouTube channel all the very best of luck with your higher Mavs good luck cheers bye