Mathematics for CSEC® 2nd edition Online suppor t Achieve your Potential English A for CSEC® Many More Titles Available Less Stress, www.oup.com/caribbean/cxc More Success! Mathematics for CSEC® 2nd edition Andrew Marcus Caine Angella Patricia Ava Manning Finlay George Mothersill 3 Great Clarendon Oxford It University furthers and Oxford The published rights in means, Press, or as Enquiries must impose British Data by the No the the of UK and University excellence worldwide. the the Kingdom Oxford in in is certain a of Oxford. research, scholarship, registered other trade mark countries authors by have Thornes Oxford part of Ltd 2012 permitted Press publication transmitted, permission by in law, by any of licence rights outside Department, in may in writing reprographics reproduction Rights in asserted University this or been the 2017 be or Oxford or by any University under terms organization. scope Oxford reproduced, form of the University above Press, at above. not this of United 2017 prior the 6DP, department appropriate to OX2 objective system, expressly sent in Nelson concerning be address You of retrieval with should Press published without agreed a publishing reserved. a is Manning, rights edition stored the by Andrew moral This Press Oxford, University’s University © First All the education Text Street, circulate same Library this work condition Cataloguing on in in any any other form and you must acquirer Publication Data available 978-0-1984-1452-0 1 3 5 7 9 Printed 10 in 8 6 4 2 China Acknowledgements Cover photograph: Mark Lyndersay, Lyndersay Digital, Trinidad www.lyndersaydigital.com Inside photograph: Illustrations: Page iStockphoto Tech-Set make-up: Limited Tech-Set and Limited, Mike Bastin Gateshead Photographs p47: iStockphoto p49: iStockphoto/Jacqui Although we copyright cases. the the this notied, earliest Links and If to for have holders made every before the effort publication publisher will to trace this has rectify and not any contact been errors all possible or in omissions opportunity. third party information materials work. Paterson websites only. contained are provided Oxford in any by disclaims third party Oxford any in good faith responsibility website referenced for in all at of Contents Introduction 1 Module 3 lebra and Module 1 Number and 1.1 1.2 theory, consumer T ypes of Highest 1.3 common natural 1.4 with factor and 1.5 eal 1.6 oney oring 1.8 tandard with real more with or decimals form 1.1 roperties of 1.12 Interest less and patterns 3.2 irected numbers 3.3 ombining 3.4 inary 3.5 panding 3.6 urther 3.7 hanging 3.8 inear 3. uadratic 3.1 imultaneous 3.11 urther 3.12 ariation 12 3.13 Inerse 14 3.14 elations 16 3.15 unctions 18 3.16 raphs 3.17 The 3.18 oling easures 1.14 arning and 1.15 ets 1.16 ombining 86 factorising factorising the subect of a 1: and numbers seuences and and spending operations depreciation money sets 2 euations 4 euations 6 euations simultaneous 8 euations 1 16 ariation 18 2 of linear functions 11 24 euation of a straight line 112 problems using gradient 28 intercept 114 stimating 2.2 erimeter 2.3 ircles 2.4 urface 2.5 olume 2.6 nits area and 3.1 urther 3.2 Ineualities 3.21 Ineualities 3.22 omposite 3.23 uadratic 3.24 ore 3.25 Trael of linear graphs 116 118 4 eam and and uestions statistics scale drawing area 8 4 and two unnowns inerse functions functions graphs of 12 : 122 124 functions 126 graphs odule 42 area with 6 128 ractice eam uestions 1 44 46 Module of properties 2 ractice Measurement 2.1 measurement 4 eometry 2.7 T ime 2.8 easurement distance 2. ata and and trionometry and 48 ectors speed accuracy matrices 4.1 roperties of lines and angles 12 4.2 arallel 4.3 roperties uadrilaterals 16 4.4 roperties 18 4.5 onstructing angles 4.6 onstructing triangles 4.7 imilarity 4.8 ythagoras’ 4. ymmetry 2 lines 14 4 2.1 isplaying information 1 6 2.11 isplaying information 2 8 2.12 easures 2.13 rouped 2.14 hich 2.15 easures of central tendency data of 2.16 umulatie 2.17 ummary of of triangles and polygons 6 14 62 aerage and dispersion 2.18 robability 2.1 robability and congruence freuency theorem 2: 144 146 68 reections and 148 2 theory ombining odule 142 66 statistics polygons 64 rotations 2.2 formula 14 diagrams 88 odule 2 and 26 Module operations 6 and enn 82 84 12 indices appreciation 1.13 1.17 substitution 22 and epressions numbers: ratios Ordering ases 8 1 1. 1.11 numbers numbers: and for 8 … 1.7 ymbols 4 numbers 3.1 lowest fractions 2 real numbers Operations unctions sets multiple Operations relations, arithmetic number common computation, and raphs probabilities ractice 4.1 urther 4.11 Threedimensional transformations 4.12 Trigonometry 1 4 eam shapes 12 6 uestions 14 8 iii Contents 4.13 urther 4.14 The uses 4.15 The sine 4.16 The cosine 4.17 ircle 4.18 T angents 4.1 ectors area of of a trigonometry trigonometry triangle and in dimensions three rule rule 16 4.2 ector 4.21 atrices 18 4.22 atrices 16 4.23 atri 162 theorems geometry and odule 4: chords can be accessed online at wwwofordsecondarycom8184142 i eam uestions 18 eam practice 18 166 168 support ractice 14 16 164 nswers dditional transformations algebra urther and 1 12 182 ntroduction This tudy with the uide has aribbean been deeloped aminations eclusiely • ouncil n your mars actiities eaminationstyle short proide answer sample and essay ® to be candidates used both as in an and additional out of resource school by following type the uestions answers and with eample feedbac from candidate an eaminer to ® aribbean econdary ducation erti¤cate show programme actiities leel It has been where prepared by a team with epertise will and answers build could your con¤dence in be improed understanding answering These sill eamination in uestions ® the The by syllabus contents proiding are teaching designed tools to help and to eamination support you achiee learning your • best est yoursel proide actiities eperience of are speci¤cally multiplechoice designed to eamination ® in it easier aths for you reuirements to your syllabus reuirements This of tudy and to for uide features master the and the the syllabus full ey o guidance eamination is included mae concepts remember on the uestions and sections to reise refer and inside problem helpful the feedbac study guide will so refer that you you to can areas course This uniue and interactie combination of focused syllabus content format accompanied by a support with inaluable eamination support to practice help you will proide reach your you full ® website you in which includes deeloping good electronic actiities eamination to assist potential in aths techniues: ccess your support website at wwwofordsecondarycom8184142 1 1 Number sets and theory, consumer pes RI stnush tpes • of between number nderstand the 4 3 2 1 0 1 2 3 4 5 dfferent Structure number between of arithmetic S 5 • computation, reatonshp dfferent tpes If we count backwards from 5, we don’t have to stop at zero. of 5, 4, 3, 2, 1, 0, −1, −2, −3, −4, −5, etc. number • he countn numbers • ero • he s not Integers whoe • a natura numbers, • numbers often natura Z are Rational a 3, 4, 5, smbo toether b etc. are caed natural N. wth zero make up the whole W the …, natura −, numbers 2, the number. numbers dented numbers 1, ven Q −, are numbers, −5, a −4, the zero −3, and −2, numbers the −1, that neatve 0, can 1, be 2, 3, …. wrtten R as 7 a fracton. he ncude the nteers, termnatn 1 2.7 decmas Q for eampe −3, −1, 0, 1, 4, 2 , 2 6 and recurrn 11 , , 3 0.5, 5.333. Z • Irrational numbers cannot be wrtten eact as a fracton or a W 0 decma, as the never he numbers, recur. ampes ncude π and √ 2. N 4 • real R, consst of the ratona numbers and rratona numbers. he enn daram daram eampes Factors on are and composite weve and hs s because 2 a s bottes bottes = ne 2 × eft n shows a Venn diagram representaton red. multiples, odd and even, prime and numbers 12 the ven coud t nto a rectanuar bo that s 2 bottes wde on. 12 s n the 2 tmes tabe. e sa 12 s a multiple of factor bottes of 12, w because not t 12 eact can nto be a dvded eact rectanuar bo b 2. that s 2 bottes wde. hs s because utpes he ast 2 of 2 ure eampe, 2. . 34 s 2, or or not 4, , a , unts 156. mutpe 10, dt … of of are an 2. caed even even number numbers. s awas even for umbers that numbers. eampe, ne hs bottes s are 1 or caed suare 4 t because umbers are ke , omposite of awas 2 1, have a 3, 5, unts , … dt are that caed s odd odd for 245. a 3 b 3 mutpe a 100 bo. of suare = 10 3. = 3 × 3. bo × 10 numbers s w caed a nto 4, number hs s mutpes n s suare bottes or t × even ese, w that = not numbers because umbers e.. are dd the onl s resut t onl that a are prime n of a 1 mutpn b mutpe mutpes an nteer b tsef. crate. of of 1 1 and and . themseves and nothn numbers numbers are natura numbers wth more than two factors. IS atura numbers 2 hoe Inteers numbers … −3, 1, 0, 2, 1, −2, 3, 2, 4, 3, −1, … 4, 0, … 1, 2, 3, … I 2 4 atona numbers 2, −5, 0., , 0.4 , … Irratona 6 ea utpes √ numbers numbers he , π, … ratona and rratona numbers • 1 is no t • 2 is the n are the resuts of mutpn n b a no t natura and number. are ven dd numbers are the mutpes of are the natura equa l that are not prim e to or factor s factor s smal ler multip les are mutpes equa l of confus e 2. numbers num ber . even multip les; while numbers prim e only num ber . toether. • Do of a to or larger . 2. actors of prme n are the number eact has dvsors eact two of n factors, 1 and tsef. SR SIS the hch of words prme nteer natura IVI In the that 1th ever centur, even odbach number a reater famous than 2 mathematcan coud be made b stated rratona addn app even to suare the number toether tart 4 to and two check 50 as prme ths the numbers. b epressn addton of two a the prme even numbers numbers the a 1 b c 1 d e − f √ between same 5 prme 5 number can be used 11 twce. 2 nd two factors numbers of 20 and that are are prme numbers. 2 ighest and common lowest factor common multiple Factors RI he • st the factors and a factors postve no ompute two or and postve nteers eatve ou −4, In hen factors list num ber facto rs pairs multip ly the see −, same that to of of facto rs ma e starti ng prime he prme the ents and b whch are 3, 4, , the product of a 3.2, and 1, 2, so can be dvded , 12 and 24. the postve factors of 24 and are neatve aso −1, number −2, −3, −24. factors the factors of same of ten the of − coour are 1, −. mutp −1, to , 2, −3, −2, 3. −. number are are mutpn t 2 and ts b an 3. factors he that are multiples prme of a numbers. number are nteer. of are , 14, 21, 2, 35, etc. highest number common that s a factor factor of of two both or or a more of numbers s the them. he factors of 5 are 1, 2, 4, , , 14, he factors of 4 are 1, 2, 3, 4, , , he common 2, 12, 5 14, 21, 2, 42, 4 confus e 2 an so is cann ot the a 24 factors are 1, 2, 4, , 14, 2 . H will origin al of of mutpes and F be is a s the as t s the arest number n both sts. a smal ler num bers. he lowest smaest common number that s multiple a mutpe of of two both or or more a of numbers s them. be smal ler origin al R 2 num bers. he mutpes of he mutpes of 0 s rme the factors numbers number are 20 as can nto 12 t be pars are s 12, 20, the found of 24, 40, 3, 0, smaest from factors, 4, 0, the 2, 100, number factor wth 0, n trees. , 120, ... ... both sts. factor branches 10, tree endn spts wth a prme crced. 42 6 Prime 4, 120, 126 420 4 t 1. the than are the n factors resuts he with I em embe r it 24 nt −12 wa, R he nteers that hhest so n pars he he than the a F H are I listin g Stud of numbers w −, ote num ber that number of as a nteer the more of remander. he • multiples mutpes wth of and S factors of 2 10 7 7 420 63 = 2 × 2 × 3 × 5 × 7 Prime factors of 126 9 = 2 × 3 × 3 × 7 nother method numbers unt of the ndn end prme resut s factors s to keep dvdn b prme 1 1 ) 00 1 3 ) 021 3 5 ) 2 o ) nd 210 3 prme ) 021 03 420 = the nother ) 2 003 105 ) ) 2 × F wa of 2 × and 5 × ndn 3 × b the 2 using and ) prime s 12 = 2 × 3 × × 3 factors to spt the numbers nto factors. HCF st the hen prme seect the R 420 = 2 12 = 2 = factors, puttn factors that × 2 × 2 3 × 3 are × × factors common to n ne vertca. sts. 3 × dentca 5 3 × × × = 42 LCM st the hen prme seect factors, the factors puttn that dentca appear n factors I n ne vertca. IVI R 420 = 2 12 = 2 = × 2 2 × SR 2 4 × 3 × × 3 × 3 × 3 × 3 5 × 5 × × × nd two of and here = nd the c 2 0 0 nd and b 42 d 4 and 0, 4 42 c 20 1 ane, and and ud b 30 and but ever d ohn the 2 rime b 42. a nd ove t to ver pa 20 and factors prme the a ears, paed toether are end resut he found dvdn numbers s to a b number unt the 1. 1, 20 ud and arnva epensve. ever 3 ears n s n 200, n whch ear rndad ohn w the ommon factors that product of the 30. ecause and hest actor of the ever cost, 5 the net are common to sts. ane he owest ommon ears. utpe If pars 1 pas dfferent of 30 and obao, 10. IS repeated 2 a four of of 4 and the are an SIS a an wth nd 120 numbers pa the s factors other or the n product one st or of the both. perations numbers numbers with real natural and decimals Rounding RI S 400 he • ound off to a ven attendance • ound of • decma off and • to a utp ven match was 32. THE s the newspaper 0 WAT CH CRIC KET MA TCH headne. and he newspaper has he attendance rounded the attendance to the nearest thousand. number ures subtract numbers crcket paces sncant dd a number ere of at natura afwa was between s between 3500, so 3000 the and 4000. attendance was coser to 4000. decmas natura numbers o round off, of nterest. awas ook at the net dt to the rht of the ast dt and If t s 5 or more, round up, otherwse round down. decmas ecimal • vde natura numbers decmas ecma I in you paces o round us whether to a answ ers nd ures are of up decma or the decma paces, the pont. net decma pace tes down. sncant ecept for eadn zeros n a decma. mista es to round 0.00 2 decma paces s 4.32. s the 3rd decma pace s , up. 0.00452 are to 2 not sncant sncant ures s 0.0045. he eadn zeros ures. I he no t round after calcu latio ns. we Sign dts rst 4.313 the gures R your are number to Signicant stim ating helps places and ica the nt same gur es as 3rd sncant ure s 2, so we round down. are decim al places . omputation o add ther and pace ddition subtract, t s wo have ppes tota 3. m mportant subtraction to an the dts accordn + t a 2 enth enth of s 3. m and 1.4 m, he 1.4 m 3. m h respectve. dfference − t n 3 . 1 . 4 3 . 1 . enth h a − arr the 0 4 0 to 1 avod mstakes reate ten ten hundredths 5 . 2 tenths over as 2 . 3 4 one 1 6 1 unt s 1.4 m dd + to vaue. R he and tenth from omputation ultiplication Natural hen R digit 42 × of × 4 multip lyin g of one multip lied 23 4 1 I numbers 2 2 3 1 utp 42 utp 42 dd two b the num ber by o ther every every is digit num ber . 3 2 4 4 0 5 b 20, b nsertn a zero and then mutpn 42 b 2 1 1 0 the cacuated nes toether. Decimals R 4.2 × o 42 poston he × the ueston ou 4 2.3 utp o need 23 above. decma 4.2 3 as × × 23 = 105 pont 2.3 paces 42 of has a tota decmas of n 3 the paces of answer decmas. 10.5 an omputation utp natura both dvsor and dvdend b 10 unt the dvsor s reasonn a dvde 4.1 b 0. utp both b 10 41. ÷ IVI 0 ) ames’ ivision rte foow number. R o the ou down a dt number b 4 4 1. rom the eft 4 ÷ = 0 remander 4 repeatn e.. a 3dt number, 2525 0 5 ) 4 4 1. 41 ÷ = 5 remander 1 vde t b vde the vde that . answer b 11. 0 5 .2 ) 4 4 1 1. 1 ÷ = 2 hat 4.1 ÷ 0. = SR ou 2 se o add 4 o mutp b estmaton the roundn to to decma dvde dvsor ne decmas, s or b ben b an paces or sncant ures. estmate. subtract, numbers ever the round or ou b 13. notce 5.2 IS can do answer a up dts poston addn b the the pace 2.5 to a 2 decma paces b 1 sncant ure. vaue. decma number ound SIS of pont n the decma 2 utp vde 2. b 1.. answer paces n 4.2 b 0.4. mutped. decma. utp both numbers b 10 unt nteer. 4 perations numbers ommon RI nderstand euvaent dd and • and how to to subtract number a fracton a fracton s vdn b a whoe caed n ts owest have no fracton both numerator 1 1 ÷ fractons when the numerator and factors. and denomnator b a common factor s = 24 med and 3 = vde terms common cancelling • terms fractons numbers or lowest nd fractons med utp fractions fractions denomnator • real S ancelling • with 1 ÷ 4 numbers Equivalent uvaent fractions fractons are fractons can fractons whch are the same part of a whoe one. uvaent numerator 2 and 2 × 3 5 × 3 = dding and ractons found b b the 2 × 5 5 × 5 = 15 mutpn same or dvdn the amount. 10 = 5 be denomnator = 25 subtracting can on be added or subtracted f the have a common denomnator. If the denomnators make the are dfferent, denomnators then use euvaent fractons to eua. Adding 3 2 + 5 3 he denomnator s 15 because t s the IVI 10 = 3 of + 15 3 and 5, the orna denomnators. 15 the est Indan mutpn rcket b 3, and 15 10 2 1 of = 5 = mutpn b 5. 15 3 3 uad for the 2010 tour of r here 1 were batsmen, fteen fteenths n a whoe one, = 1 15 were are 4 = 1 anka 1 so 15 = a whoe one and 4 more fteenths. 15 5 bowers and the rest were a Subtracting rounders. 3 1 3 hat fracton were arounders − 1 4 hen 1 ow man crcketers do = numbers. − the there were atoether denomnator n 3 21 suad = − fractons common o = 12 ultipling ut convert nd a to med common = 12 and use euvaent fractons. 1 hane back to a med number. 12 and can 12 dividing be mutped or dvded numbers mutp a must fracton b be an chaned nteer, nto wrte 3 denomnator of 1, e.. wrte 3 as . 1 addn, there s no need to have denomnator. med wth subtractn, ke 5 1 12 n or ust 4 thnk addn ou mproper the fractons. nteer as a fracton a 2 1 1 × hen 2 mutpn or dvdn, an med numbers must be made 4 3 nto an mproper fracton. 5 = here are three thrds n a whoe one, so 1 5 3 hen multipling onl , ou can cance the I 5 2 + = 3 3 • 3 = s 3 4 3 3 2 × numerator ou do nd and × 1 no t the to when 4 denomnator 15 3 = = utp the of dfferent numerators fractons. and vde mutp the the and the 3 denomnators. b 3. multip lyin g fteen but 3 4 you or mus t 4 uarters make 3 whoe ones and three more dividi ng when uarters. addi ng 1 3 need or subtra cting . 1 ÷ 1 hane to mproper • ear n fractons. 4 3 with 10 5 = hen dvdn, invert the second fraction the dvsor 4 10 4 to wo r fractio ns on your and ÷ 3 how calcu lato r . multipl = ow ou are multipling , ou can cance the numerator and × 5 3 denomnator 2 of the dfferent fractons. 4 = × In ths case, cance the 10 and 5 b dvdn both b 5. 1 3 IVI 2 = = 2 3 3 1 acom sas that ÷ = 4. 2 was smpf our answers b wrtn them as med numbers ow where approprate, and wrtn the fracton n ts owest that he coud he s convnce hm wron IS chart beow summarses how ddition to cacuate wth fractons. Subtraction hane an med ultiplication numbers nto mproper rte nd a common denomnator an ivision fractons nteers as a nd euvaent fractons wth If as denomnator subtracton performed, cannot break whoe the dd or subtract the one rst and wth a 1 dvsor be down add across fractons f possbe to fracton fractons utp ance hane SR of Invert ance a fracton denomnator ou terms. to f med numerators, mutp denomnators possbe number f possbe SIS acuate 3 1 a 1 1 − b 2 4 2 5 2 × c 4 3 + 3 1 1 2 budn has a heht of m. 2 1 ow man scaffodn bocks of heht 1 m 4 w be needed to reach the top Real numbers onversions RI Fractions • onvert between decmas between real numbers S and and decimals fractons, ractons percentaes are a short wa of wrtn a dvson. 2 • press one fracton or uantt as percentae 2 a hs s wh the dvson sn ooks ke a fracton. 5 of 5 3 means another • ÷ . acuate a fracton percentae • 3 acuate of the a o or whoe from chane a fracton to a decma, dvde the numerator b the denomnator uantt a 0. 3 5 fracton or percentae ) 3 3. 0 3 4 0 0 = 0.35 o chane wth R t 0. 5 a ecimals has a ast dt n 5 0.5 = and are o, to the dts are chane two = chane dts a two of a 100 whoe tmes Fractions o I wrte dt one, as the decma part as a fracton denomnator. whereas ber decma to than a to 0.0 the = percentaes are part of the euvaent percentae, to mutp b 100 b movn eft. a = decma, 5.5 dvde b 100 b movn and 0. 24 = 0.24 to then a to percentae, a rst percentae b chane t to mutpn a b decma 100. means R percen tage can 2 be 5 writt en fractio as n a comm on with deno minat or • lwa ys answ ers give in 0. 5 a of ) 5 5. 5 5 0 … = … = 55. to 1 decma pace 0 1. their lowes t chane a percentae to a fracton, wrte the percentae as s. fracton of 100. R 4 = 2.5 = 100 12 4 55.555 5 0 fractio nal o term 5 the rht. divisi on. • 100. decma. percentages fracton dvdn a 0.55 percentae = to the 40 and chane b a paces paces 0 n ast 20 o fractio fracton, percentages part ercentaes 1 0.4 • a the = 100 to of pace, so vaue the ecmas hundredths decma pace h the 25 2.5 25 = 125 = 100 25 = 1000 5 = 200 = 40 a pressing one uantit as a fraction or percentage of another IVI 2 s we as meann dvson, a fracton aso means ‘out of’, so means 5 2 out of he 5. of 12 12 as a fracton of 30 s percentae, = = 2 ÷ 5 = 0.4 = • calculate a 3 mon of these ve n amaca. 40 5 30 o popuaton 5 bout a a 2 s has mon. = 30 12 40 2 o arbbean about fraction or a percentage of a uantit hat percentae arbbean of the popuaton ves n amaca utpcaton means ‘ots of’. hree ots of 15 = 3 × 15 = 45. • hen dean wth fractons, we ust sa ‘twothrds of 1’ esearch some of ‘twothrds ots of 1’. In ths sense, ‘of’ means and to nd of of 1, other popuaton arbbean of sands mutp. 2 o the nstead nd the out tota what ve percentae on these cacuate 3 sands. 2 × 1 o × 40 of calculate = 12 k the = = 1 0.4 whole 12 × 1 3 2 = 3 nd 1 2 = 1 × 12 from 12 1 = a 4. k. fraction or percentage 2 chae saves of hs waes. e saves 4. 5 1 o of hs waes s 4 ÷ 2 = 32. s waes are 5 × 32 = 10. 5 are o pas 1 = a 5 50 depost ÷ 5 = of 50 130. he for car a car. costs 100 × 130 = 13 000. ap of the arbbean IS SR o chane a fracton to a decma, dvde the numerator to a fracton, dvde the decma b SIS the denomnator. 2 o chane b the a decma rte part a pace vaue of the ast ure, and 30 as owest 32 0.32 = o chane to the eft n ts = 100 fracton terms e.. a smpf, a 25 decma to to mutp a b percentae, 100, e.. move 0.25 = the dts 2 b 0.45 as c 5.5 a percentae paces as a decma and 2.5 as a fracton n ts owest terms. 4 o chane paces to a percentae the rht to to a dvde decma, b 100, move e.. the 3 dts = two 0.3. 2 o chane a fracton to a percentae, chane t to a hch s reater, and b how much decma 3 and then to a 35 percentae. of 45 or of 40 6 o chane a percentae to a fracton, wrte the percentae 45 a fracton of 100 and smpf, e.. 45 = as schoo has 350 students. = 100 25 5 2 of rs hat them are are rs. efthanded. fracton of the rs are efthanded 6 one rot RI S • dscount, acuate , • prot press saes and prot, ta as a saes dscount, percentae • vaue ove probems purchase, nvovn prot, oss or less loss he cost prce pus he cost prce mnus the prot ves the sen prce. rot and oss are the oss epressed ves as the sen percentages prce. of the cost price of If some more ta oss oss, and … hre I bu a car for percentae and 12 000 oss and se t ater for 10 000, m oss s 2000. s 2000 = dscount 0.1 … = 1. to 1 decma pace. 12 000 iscount R hops off arr bus wants to a char se t to for 250 make a sometmes the prce, for offer a dscount eampe n a ths s a percentae that the take sae. and 15 R 2 prot. ver s prot s 15 of there = 0.15 must 250 × se + 250 t = 3.50, so he = s a some 20 shoes n the sae. he norma cost 0, but dscount. he for 3.50 sees 250 2.50 dscount cost 0 − s 20 14 = of 0, or 0.2 saes ta × 0 = 14, so the shoes 5. V an 15 hs an sands on s added tem 2 = ire that 0.15 nd a hre acuate × purchase or the percentae dea from cataoue. addtona ou pa b a to to = bu prce that 4.20, oods start wth, nstaments. wanted 200 foowed pad hs eter e the otherwse 2 ou R newspaper on mht mone month IVI , or consumpton ta of around ou cost so pa 2 the for has tem oods 15 n the added actua costs shop, on. 15 so of 32.20. purchase ometmes some chare oods. 200 to bu b + 24 hre usua a purchase. depost, costs hs means foowed more than b pan a bun number t motorbke month × 35 usn for 00. nstaments = e of pad a depost of 35. 1040 140 o he pad an etra 140, or = 0.15555 = 15. … 00 hre 2 purchase. to 1 decma of outrht. a 24 on caed pace. Reverse percentages ercentae the chanes original R son sees he wants he cannot sae prce, he or prot, oss, dscount, ta are awas based on amount a handba to know work and reases 4 t the that, n a how out f much b 25 sae. t ndn has 25 cost 25 aread has orna. of been been 0, because taken taken off, off the she has 0 s the orna pad prce. 5, 0.5. S o orna prce × 0.5 = 0 A 25 % N O W orkn back, 0 ÷ 0.5 = 0 orna prce was E O N L Y $6 he L O FF ! 0 0. • se ths technue whenever ou have to work back to the orna ll R percen tages prce. on • the Hire a to 100 costs + 3. 15 rna prce o ncudn prce 3. × ÷ ncudn = 1.15 115 = 1.15 prce ÷ = or at 15, are are based price . = deposi t instal men ts pan 1.15 ncudn 1.15 ou origin al purcha se plus If I = orna prce 3.20 IVI essa to rot, oss, dscount and ta are a cacuated as percentaes. osses and dscounts produce a na vaue ower than make a an artce prot nfortunate so 2 bus and ams IS she puts t of she n a 20. cannot sae se wth t, 20 the off. orna hen the prot workn percentae SR effre hat 2 hrs for hs bus an an the freezer back to to resut nd 100, artce n the a hher orna convert for percentae prot, s taes to a na vaue. amount, decma add and or subtract dvde. pan oss of to artce and sen can of ow be and ses t for hs 0 much s hat makes a 1. 0. has e to then add on adds 15 on 20 . prce bouht and on hre month purchase paments for of a 1. pa y w t cost 6 b she prot for then 0 Just a wh 4. Bu y depost her SIS bus s hs hat and 15 s the more than orna the cash orna cash prce. $70 depo sit month ly of no w, late r! plus instalm ents only $19! prce oring with ratios Ratio RI S • two ompare uanttes ratio If ratos a cass 12 1 • vde a uantt n a atos onvert from one a wa has we 12 sa set of can be two or more uanttes. bos ths and as 1 ‘12 to rs, we sa the rato of bos to rs s 1’. smped, ke fractons. vdn both uanttes b = 2 3. another hs • comparn unts 12 1 to of ven rato • s usn onvert usn be tabes, currenc 2 that roups of we can bos put and 3 the cass roups of nto roups of for 4 there woud rs. converson ou make a pneappe nredents 1 oz canned 4 oz orane oz vana 4 oz water ce he can these 4 R means converson pneappe uce and orane smoothe peope wth chunks frozen concentrate ourt cubes rato of pneappe chunks orane uce vana ourt s 1 4 or 4 1 2 onverting o make conversons, R km In a mture cookes, chp to the for chocoate rato cooke of s has 400 of rato w ow she much to nd cooke need of parts s 20 ÷ so o 1 part 5 the s parts 5 chocoate of the chp = ÷ cooke 5 × oth 200 parts of = of 4 b eua to 5 mes. s 5 euvaent of 20 km n mes, 2.5, = we mutp × 2.5 5 20 km = 12.5 × both 2.5 numbers = 2 = 200 . batter ternatve, the rato are we coud mes. the euaton 20 5 b usn euvaent x 5 sove × 2.5 5 × 2.5 20 = 12.5 n 20 12.5 s 1000 . 200. fractons. 2 km mes = mutped euvaent 20 x = = = of mture, 400 of prncpe batter o 2 the chocoate SI 400 app 2 5. 5 chp. appromate he o, nea s we units chp chocoate batter between fractons. the 5 rato b 2.5. R ne arbadan doars he o rato to doar 1 s worth 1.34 astern arbbean 1.34. of nd the 1 vaue of = 1 1.34. 200, we must sove x = usn 1.34 euvaent fractons. 200 utpn both 200 sdes of the euaton b 200 200x = 1.34 IVI 200 14.2533134 = x o o 200 divide a R In 10. he saes 14.25 uantit ovember op = 2010, essah in a 2 decma paces neckace and whte he rato essah at had number two 2, sons whe n the tua amacan was at beads number tota saes he rato 3 1 1 sae of essah of and tua were the two sons means that for tua, and 1 that pe n the rato s, of f put was ever nto 3 4 000 o the he 4 ÷ eua 4 saes saes of SR vde = eua • of pes essah there there rato are 3 pes of s pes hat each pe • were 1 × 3 × 11 500 11 500 = = n the rato he in 34 500. ob s rato mn concrete 2 3 usn to whte beads to bue s the of of whte beads most bead n common the neckace I ratio s order a rato bue in fo rm of ratio their . quan tities is impo rtant . IS atos the 2 beads 11 500. SIS 45 red simpl est • were red the to hch contans s. essah tua so of s coour atoether, of 3 4. beads was tua. 11 500 of bue 2 3. • ive are red, 3 1. 4 000. saes s here of . beads he made ratio he of s beads. 4 avado was to cement, sand and rave n do sze not of te the ou about uanttes. the 1 2 3. 2 ato s about shares, s usefu when or 3 e uses ow 0 000 cm much of cement sand. and parts. rave w he use ne metre ndrew’ s s appromate heht s 2 eua nches. to hat 3 s nches. hs heht ato performn n metres converson cacuatons. Standard form and indices Suares, RI cubes utpn • nd and roots S suares, cubes, a number b tsef s caed suarn the number. suare 2 roots and cube suared he • rte a ratona number wrtten as and means × = 4. cube of a number s the resut of mutpn three of the same n number standard s roots toether. form 3 • nderstand of and use the aws ndces 4 , or he 4 cubed, suare means root of a 4 × 4 × number s 4 = the 4. nverse of the suare. • acuate wth numbers n 2 he standard suare root of 25 s wrtten √ 25 and s eua to 5, as 5 = 25. form 2 √ 25 s aso eatve he eua to numbers cube root s −5, do the as not −5 have nverse of = a 25. rea the suare root. cube. 3 √ 3 512 our R he popuaton of = s as cacuator = 512. has suare, cube, suare root and cube root kes. the omncan Inde epubc , notation 10 00 000. n a ecause 10 00 000 = 1.0 means a s mutped b tsef n tmes. × 5 4 10 × 10 × 10 × 10, 10 × 10 × 10 = 4 × 4 × 4 × 4 × 4 × n we wrte 10 00 000 he form a s caed nde notaton. a s the base and of the same × 4 n s the nde, as of 1.0 × 10 ran of sand power has a he = ÷ 10 = ÷ 10 eponent. dameter 0.00 cm. 0.00 or × 10 = × 10 3 laws here are of rues power paces the return to of 10 dts ther for combnn powers number. 3 × 10 5 he indices tes have how to orna 3 4 4 × = 4 or enera × man move 4 × a 4 × 4 b y × y × × 4 × 4 × 4 = 4 , a+b = y to postons ÷ × × × × 2 4 = = , × 5 3.2 × 10 a or = 320 000 enera 3 5 ts move 5 paces 4 ÷ 3 = b y = 3 5 × = y 5 a y 3 × 5 b y 3 × 5 12 = 5 , a or enera y b ab = y eft 4 y 4 ÷ y × y 3 y 0 4 . 3 ut y 4 ÷ 0 = y y = 1, so 3 = 1, so y = 3 y and y are recprocas. 3 4. × 10 1 3 hs = means that y = ts move a , or y = 3 0.004 a y 3 paces y rht Standard hen form numbers standard form. et ver ost are or cacuators ver do sma, ths. we often tandard n 6 × 10 , where 1 ⩽ A < 10 and n s an nteer. wrte form s them wrtten n as alculating o add or chane in standard subtract them R out form numbers of n standard standard form form, the safest wa s to rst. 2 5 3.4 × 10 I 4 + . × 10 = 340 000 = 40 000 + 000 hen calcu latin g stan dard form in ma e 5 = 3 4 × 10 4.0 × 10 sure − .2 × 10 = 0.004 − = 0.0030 = 3.0 the answ er stan dard 4 0.0002 part is is form be tween still i.e. 1 the and in rst . 3 o mutp or dvde, the decma × parts 10 and the powers of 10 can be R deat wth separate. 3.4 × 10 3 × 3 × 10 IVI = 3.4 × 3 × 3 10 × 10 uba has a popuaton of 1.1 × 10 5 , and an area of 1.1 × 10 = 2 10.2 × 10 , or, n standard km form, 5 arbados has a popuaton of 2. × 10 , and an area of 10 2 4.3 × 10 • In • ow = 2 1.02 × terms of area, man how tmes the man tmes ber popuaton of than arbados arbados s the s uba popuaton R of 10 km 4 uba 2 • opuaton denst s the number of peope for each km . 4 5 nd the popuaton denst of uba and of × 2 10 ÷ = 5 ÷ = 2.5 2 × 10 arbados. 4 2 × 10 2 ÷ 10 SR IS rte a 0.0005 n standard form aws 5 b 1.4 × 10 as an ordnar of ndces a number. • y b × he un s appromate 1.5 × 10 kometres from • arth. y he speed ow on of ht does t s about take for 3 × ht 10 to kometres et from the per • second. un to ÷ y y y a−b = b y ab = y 0 arth • y = 1 1 −n = b a 5 a+b y a 2 10 SIS × • mpf y = n y a 4 3 × 4 2 b 3 5 ÷ 3 1 c × 2 tandard form s wrtten as 5 n A × and 10 n s , where an 1 ⩽ A < 10 nteer. rdering, patterns and seuences rdering RI o a • rder • enerate a seuence ven set of rea term of a order erve a rue a R a ven the as parces put s t them weh n and R t 4 1 . 1 . 4 the order 3 , and , s , = , the = 3 s 5 weht , = fractons for can has smaest as the more 2.4, have the same number of unts hundredths. 1., 1.45 be a put n order common b chann them to euvaent denomnator. integers = 10 , 40 o order nteers, ncrease n o 4, sze draw from or eft mane to a number ne. he numbers rht. smaest 3 , order −3, 1 and −5 4 10 4 3 2 1 0 1 3, 2 3 4 1, 5 4 is the correct order. I order abou t to 1.45 1. 5, co lder unts. fractions wth 5 5 2 , 40 order , th than but arest ractons 25 , 5 o of 40. 5 40 correct rst 1.45 k 10 s 4 40 so and 2 denomnator 30 most arer rdering 5 1. k 10 3 24 vaues. 4 common 5 5 4, pace arest, the rdering 3 h . tenths, rom 3 ann 2.4 k, order 2 has 1. 5, vertca, seuence 2.4 them terms o wrte rue o of decmas, numbers hree • decimals S enerating integ ers tempe rature it is the a term of a seuence given a rule thin ere s a seuence of suares made from ne sements the lower the tempe rature. so ° is co lder is lower than than ° he number e coud addn o the hs of ne sements n each pattern s 4, , 10, 13. . descrbe ths seuence as start wth 4 and then keep 3. net tabe term shows n the the seuence number umber of suares, umber of line of suares s segments, woud l be and 1. the number of 1 2 3 4 4 10 13 ne sements. he rue s 4, connectn s and l s l = 3s + 1. or eampe, when IVI = l = 3 × 4 + 1 = 13. nd o for 10 suares s = 10 we woud need 3 × 10 + 1 = 31 the net number n ths ne pattern sements. 1, 2, 5, 14, 41, 122, … th eriving hs a rule pattern s given made of the terms re and in a whte seuence the n term crces. IVI chae ne sowed da outsde. he number of whte crces s ven n the number, umber of n white circles, w 1 2 3 w nd s the rue ncreasn between the attern connectn b 2. rows and ecause of number, n the of 10 12 w: the n the same a pot. bean da das the pot ater was bean the ust outsde n a he bean 1 cm was pot. pant ta. n he aread 3 cm ta. ths, nsert the 2 tmes tabe ach n tabe n panted another en pant o bean tabe sowed attern he n a 1 2 3 2 4 10 12 evenn hs two n the the chae measured pants. evenn tte bean of the pant net had da rown ×2 2× table another 2 cm so t was 3 cm + ta. umber of white circles, w ach row rown If we tabe, doube we n, et we et the 2 tmes tabe. If we add to the 2 the the rue s he w w = 2n + sa the nth term of 5 cm the seuence , 10, 12, … s 2n + . two how × can 10 use + the = rue 2 to work whte a common 2 se an out that the 10th pattern woud pant a rew at a da. the das same measured were the heht them n when the have denomnator to ow ta were the order SR a number ne to order nteers. ut a eometrca patterns can be 4.2, numbers. b c for members a seuence of a set t can nto be a found pattern. when work for a he 5 rue as = for an a + seuence b, of sze 4.1 , , −4, 12 , 1 −2, −5, rue rte down the rst ve terms the nth term n the numbers. can be where n s the where s 2n − 1. wrtten y 3., order seuence he n a 2 must 4.0, 2 , 3 rue numbers descrbed 1 usn SIS these had IS se to t crces. fractons. 4 amount before. man pants evenn 2 contnued chae e da outsde fter e t the tmes stead o da doube poston nd the rue for the nth term n these n seuences the seuence, consecutve found term b n a s terms sovn the the n an ncrease the euaton seuence. between seuence. usn b a can a 3, , 11, 15, 1, … b , , 11, 13, 15, … c 1, be ven 15, 12, , , … roperties and of numbers operations Identities RI S Identities • se propertes of numbers have operatons n are numbers whch and no effect when a partcuar computatona operaton s used. tasks • nderstand denttes or mutpcaton the dentt and dvson, and s 1 nverses • now and app operatons • to nderstand dstrbutve the order of × 1 = ÷ 1 = cacuatons the and commutatve, assocatve or addton the dentt and s subtracton, 0 rues 5 + 0 = 5 − 0 = Inverses Inverses • ddton so • have − an and 5 = subtracton he the 20 ÷ 5 = additive dentt he he effect. are inverse operations 4 + 5 = , 4. utpcaton so undon and dvson are nverse operatons 4 × 5 = adds to t to 20, 4. inverse of a number s one whch eua 0. addtve addtve nverse nverse s of 4 found s b −4 , because chann the 4 + −4 number = to 0. ts opposte sn. he b multiplicative to eua the inverse dentt of a number s one whch 1 he mutpcatve nverse of 4 s mutpcatve mproper R he fracton mutpcatve nverse and s mutped 1 , because 4 × 4 he t 1. s found then nverse b wrtn nvertn s aso = 1. 4 the number as an t. caed the reciprocal 2 rackets × 3 + 4 − 2 ÷ 3 he order of operations Indces he cacuaton and then 3 4 of 2 + 3 × 4 ves an answer of 20 f ou add 2 + 3 2 = vdn = × and − 2 ÷ 3 × and − 2 ÷ × and then avod add confuson, 42 − 4, or an answer of 14 f ou mutp the answer to 2. we multipl or divide before we add or o chane ths order, use bracets 3 are four staes to the order of operatons, whch are 3 sometmes 2 b Subtractn here = answer subtract. = the utpn o ddn mutp remembered as IS see orked ampe 1. he associative, • operaton, n a b c = R ddton associative b and and distributive • f n a c. 2 laws operaton, b = b R mutpcaton are assocatve ddton , s commutative and mutpcaton + 3 + 2 = 4 + 3 + 2 = 4 + 3 = 3 + 4 = 4 × 3 × 2 = 4 × 3 × 2 = 24 4 × 3 = 3 × 4 = 12 − 12 and 3 ÷ − 4 operaton, dvson 2 ÷ , = 2 1 = s are − 1.5 not 3 12 − ÷ distributive assocatve 2 4 = ÷ over ubtracton 5 2 = f a 4 n s ubtracton • , a commutative and dvson are are commutatve not − 3 = 3 3 − = −3 4 ÷ 2 = 2 2 ÷ 4 = 0.5 commutatve another IVI operaton, △, f a b △ c = a b △ a c. pan R 34 utpcaton s whch aws are ben used n each ne here 4 dstrbutve over addton and × 2 = 34 × 20 + = 34 × 20 + 34 = 20 × 34 + × 34 = 20 × 30 + 4 + = 20 × 30 + 20 subtracton 4 × 5 + 2 = 4 × 5 + 4 × 2 = 2 × 5 − 3 = × 5 − × 3 = 12 = ses of these these cacuatons × 4 30 + 0 + + × 4 30 10 can be made easer b + 13 + 4 24 4 × 1 = × 13 = × 30 = 240 + 1 IS he addtve dentt 2 he addtve nverse he mutpcatve 4 he assocatve he commutatve 6 he dstrbutve = of 0, a the s mutpcatve dentt a and assocatve rues nverse recproca × 4 = 15 × = 0 = 540 SR × 4 × b s acuate 2 rte aw a * b aw a * * c = a * b the * c aw a b b = △ b c * = a a b △ a c SIS 12 − 2 × 3 + ÷ I 2 • Don’ t • hen fo rge t D S. down you 3 a a 1. __ of b = –a __ ommutatve mutpcatve nverse and b the addtve nverse acuate a 5 × b 2 1 × × 1 a calcu lato r provid e 4 − use pro pe rties menta 2 × cann ot of 5 × usn rue × ncude strbutve 15 + rues. ampes × + × rules = enta 00 × a and these laws can shortc ut. 2 ases ur RI number storca, • tate the vaue of a dt n ven ts ove probems n s unts make number housand a and en, so we coud wrte housands on the have ners so the count n panet on each ehts. or our 2 s wh our a undred, ten undreds make vaues as powers of ens 10 nits 1 0 10 10 the 2 of pace the a of n other nches usn the 3 bases. work digit vaue vaue work and value the n in n a omputers base numeral nde 2312 work n 12. given its base notaton. base 4 4 3 to eet 4 1 2 2 3 s worth 3 × 4 = 3 × 1 = 4. fours need are he hundred we 4 2 vaues pace undreds bnar . eampe, t make 10 down 4 ve hs bases nd or ens 2 ometmes rte ners. hand, o pace our moosh base2, her ten 3 ther four on 10. on. 10 reatures counted on theor o R awas based nvovn a concepts have sstem base en • we a number numera sstem S hts nts o nd he vaue the of value 2312 of base a 4 numeral given its base s and 3 2 × 4 = 2 2 + 3 × 1 4 + 1 × 4 0 + 2 × 4 × 4 + 2 tweves mooshan 123ears od wrote, ‘I s not one twentthree. hundred + 2 + 3 × e 1 × = ehtthree. change + 1 × 1 = 12 × a base number to a different base s + 3 × 2 1 paceaptan ams’ wants ths t n e to base wrte space down for shp the traves at 1010 kmh. mooshans, but he knows the pace vaues n base are 2 ve hundred and tweves e starts e subtracts wth tfours hts 1010 512 4 1 × 512 e must . 22 1 and R 4 4 toda’. o e × am nts wrte IVI e subtracts 44 e subtracts 4 because 44 because 4 s × 4 50 s × a hch o eaves 1010 = 1 2 × pacedmra 1 512 or + ones × 2 × 4 ew ear racker bursts nto star wth 4 arms. 1. + × dd t b remander stfours dvdn + b 2 × 1, or 12 base repeated ach arm then bursts nto 4 ) 15 remander ehts remander 2 unts fresh arms. ) 12 ) 1010 e ot the do ou thnk same s resut of 12 an ou see t hch method easer ach 4, dding and subtracting in other of and that and the pace subtractn base to the R 45 ndcates work how 3 man n n one nto an base. pace ust eua remember one n the at down each of the the number rst ve of arms staes. net base base base 1 5 + = eeven, whch 1 eht 1 and s 3 unts, wrtten 45 base 3 base s tep + he 3 same bursts eft. tep + the then on. bases rte ddn these so 13. 2 whch carred. 103 base 4 s + 1 3 + eht carred and 0 1 = unts, eht, wrtten 10. 1 1 R 4 1101 base 2 111 base 2 tep 1 done − 1 1 = wthout 0. hen neatve 0 – 1 cannot numbers so be pr ess from the 1 whch s worth 2 n the 0 net 0 place the coumn. no ta tion valu es to in calcu latio ns ma e easier . 2 101 base 2 tep 2 2 wthout − the we inde borrow I 111 base – 1 = 1. neatve hen 0 – numbers 1 cannot so we be done borrow from the 2 1 whch s worth 2 n the net coumn. 10 0 2 2 tep − 111 base 3 2 1 = 1. hen 0 – 0 = SR 0. SIS 2 hat s the vaue of the dt 3 110 n 3210 2 rte nd base 321 the 4 base base 10 10 n base vaue . of IS 1011 he pace 2 ace vaue vaues mutpn on of the the eft or eft 1st or dt rht dvdn on of rht a the rht dt the are pace s awas found vaue 1. 4 base 2. acuate a 101 + 11 base b 210 − 123 2 b b the base. base 4 2 2 Interest, and Simple RI nderstandn and interest nterest and ou nvest mone n a bank, the usua pa ou interest cacuatn that smpe depreciation S hen • appreciation s, the ve ou some etra mone. he pa ou a percentae compound of what ou nvest. hs percentae s caed the rate of nterest. nterest • pprecaton and he amount ou he amount of nvest s caed the principal deprecaton nterest that ou receve, I, s ven b the formua PRT I = , where P s the prncpa, R s the rate of nterest per ear, 100 R and If ou and nvest receve 2000 10 for 3 T s the tme nterest, use the rate P = the of 2000, formua to = 10 ost and that ou nvest the mone. interest cacuate nterest. I ears ou ompound can n ears T PRT = 3 banks nterest on nterest s pa compound oans, added ncudn to the interest. credt prncpa ou cards. at the aso th end pa compound compound of the nterest, the ear. I = 100 R 2000 × R × 2 3 10 = 10 = 100 If ou rst ou R = borrow 2000 at 3 per annum per ear, at the end of ear ou now owe owe a nterest tota of of 3 of 2000 = 0.03 × 2000 = 0. 200. 3 In the second = 1.0, In the uck addn a we add decma, o to of then ou have are to an a etra debt or was 3 of of 200 = 0.03 × 200 2121.0. 3 of nterest compound that an chared 215.45, cacuate 3 owe 2121.0 of 3.5, 15.45. nterest ntroduced = n s to use the method 1.. amount, we have a tota of 103 or, as a 1.03. 2000 nterest hree ear, ou percentae R If ou tota wa a ear , so thrd vn s × 1.03 woud ve the tota after the rst ear’ s added. ears’ compound nterest coud be cacuated as 3 2000 × 1.03 × 1.03 × 1.03 = 2000 × 1.03 T R enera formua s na amount, F, = P × ( 1 + ) 100 P 24 the 0R = prncpa, R = rate of nterest and T = tme n ears. , where of ppreciation hen an tem vaue s caed tems that and ans n vaue, we appreciation. sa t appreciates. eweer and antues he are ncrease n eampes of apprecate. an tems caed depreciation become R depreciation worth neckace cost apprecates b ess ever ear – for eampe, a 4 200 car. hs s R n 200. ver ear t 5. car costs deprecates 10 000 b n 12 the ever ear 2010. It ear. 4 2010 1.05 t was worth represents power of 4 s a for 200 5 4 × 1.05 ncrease 5 = 243.10 105, and Its the vaue 0. ears. and n 2015 represents the power w a of be 10 000 12 5 s × decrease for 5 0. =52. 100 − 12, ears. IVI arsha bus two rns. he cost I 400 • ou mus t learn each. fo rm ulae he od one apprecates b 4 a fo r sver hch one w doubn deprecates happen ts vaue rst or b the the 4 a od sver inter ear. • Do • n no t confus e rn the the fo rm ulae two. time havn measu red in vaue em embe r mon ths no t and est. rn is ts simpl e ear, compo und the the is that . year s. year s . IS PRT mpe nterest I = , and 100 T R 2 ompound nterest F = P × ( 1 + ) , where I = nterest, 100 P = prncpa, T = tme • n SR I can nvest ar’ s I nvest earn of nterest, F = na amount and and deprecaton work n the same wa as nterest. 2000 beautfu b 350 100 for nterest. apprecates rate SIS compound 2 = ears. pprecaton compound R ears antue at 5 hch ever 4 s vase ear, smpe at 3.2 better cost her how nterest. smpe and b 4000 much ow s nterest how t n 200. worth on or 3 much w t n If t 2012 take to nterest 2 easures he RI S • nvovn robems and mone, echane • he 12 measures sstem n the metrc sstem use the same sstem. measures ength ncudn the rate and metric 24hour cock s based on the metre, capacit ommon and the parts are thousandths common 1000 mm 100 cm = = 10 mm = 1000 m used 1 m s 1000 m 1 m 100 cℓ 1 cm = m, mutpe 10 mℓ 1 km = 1000 ℓ he he the imperial nches feet 10 = and k he = 1 = 100 c 10 m 1 kℓ = 1 1 20 cwt sstem 1 = etric foot 1 2.5 cm 30 cm me 1 m sstem oz 1 pnts = 1 1 pound b ≈ ≈ ≈ 1 1 1 ≈ 30 1 1 c = 1 k imperial nch foot ard 5 mes and 1 cwt k imperial sstem 1 oz ≈ ≈ 50 k 1 b 1 ≈ tonne etric stone 1 cwt ≈ 1 and ton imperial comparisons ´. oz = 1 pnt 30 mℓ aon 0.5 ℓ 4 roblems onvertn ≈ 0.5 k ton ounces = and etric hundredweht imperial ´ud 1 sstem capacit 1 = = 1000 1 he = tonne stone = = c, comparisons ounces = cent, 1000 m 1 ℓ ard ards stone of on comparisons imperial 14 b mass o 1 cℓ mass 1 and hundredths 1000. km of litre 1 ℓ length 12 3 imperial m ko, = = 1000 k of on gram involving between conversion dfferent unts ≈ ≈ tres 1 ´. 1 ≈ pnt 1 between nvoves oz aon units ratos. IVI or n a partcuar rate for astern was 0.3, da , the arbbean1 and eampe, R ths man nd the appromate metrc nformaton, rato of k stone s appromate how o was €1 worth 12.5 26 euvaent €0.2. he sn to echane stone ≈ × 12.5 = 5 k 1 of 12.5 stone ime here are 24hour • In the two 12hour foowed b mdnht • In the rst was of wrtn the tme the 12hour sstem and the cock. are tmes e.. dts cock, between 45 foowed 24hour two cock, a.m. b a.m., mdnht tmes and mdda between are mdda and p.m. tmes ndcatn and are the wrtten hours as and 4dt the ast numbers, two wth ndcatn the the mnutes. ◦ dnht ◦ fter so o or 100 230 n 1430 045 hen 0 wrtten p.m. the to n workn n R as wrtten s and does as mdda not reset as to 1200. 0, but contnues, 1300. wrtten as 230 p.m. 12hour cock cock. the mornn s 45 a.m. 12hour cock cock. wth an 0000 hour afternoon 24hour mnutes the s 24hour uarter or s mdda, tme, remember t s not metrc. here are hour. 2 o nd starts the at enth 052 of and a bus ends at ourne that • rea t 1025 1 2 11 052 to 1000 s rom 1000 to 1025 s 25 curren cy measu re 1 10 rom I conv and ersion s as 2 ratio s. mnutes. 9 3 • em embe r time is mnutes. 8 no t 4 me tri c. 7 h 5 minut es 6 ota tme = + 25 = 33 mnutes. = . . hour • ear n me tri c 1 = , a.m. = the ey and impe rial 1 = , ko = 1000. 100 mornn, SR no t cent 1000 2 s s. equiva lents . IS hour p.m. = afternoon. SIS rte a 420 m n km 2 ppromate a rte b ow i how 1055 man b 33 cℓ man a.m. mnutes mes and s t n ii mℓ s 2.4 k euvaent 113 from c p.m. 1055 n to the a.m. to n 20 km 24hour 113 cock. p.m. 2 4 arning and spending mone Salaries RI ome • ove and wages S probems peope are pad a weel wage others are pad an annual nvovn salar saares and waes ages • ove and probems uttes nvovn taes eope on number n a etra hher or a of week hours hours a wae week worked usua the are have have caed an to hourl work, the overtime, rate basic and are of pa, and usua pad at rate. eampe R essa pad ast er at earns tme week tme vertme a a − week hour for a basc 35hour week. vertme s haf. s 35 a = means hours wae tota of cacuated hours haf = worked 35 41 and per wae week vertme ota and she week asc = as per foows ate ettn 13.50 + hours. hour hours. 315 41 = pad = = 315. 1.5 × 1.5 tmes = the 13.50 usua rate. 1. 1 = 3. Salaries eope work. 12 who he and that eope on bonus at saar e s the ear end s et of pad the a ed month, cannot ever earn amount so the of mone annua saar for s a ear’ s dvded overtme, but the sometmes et ear. 2 manan an the annua drector saar s bonus 1 s for bonus compan month the s of of made 5 000 of a a compan. 1 400. of prot ÷ 1 240 000 ear b month. e has an annua 5 000. s earnns 2 pad the of saar s amount receves ast a saares R eter earn mone 12 = the of = compan’ s prots. 1 240 000. 250. 12 400, makn a wee hs tota a aes IVI eope pa taes so that the overnment can pa for servces such as • the poce he taes Income In and var ta amaca, over up to n s the ta In a ne pa on mone ncome 5200, arbados, 24 200 amount of ou ta at 25 whereas ncome 12 100, ta s and earn. amaca on on there s chared 35 on annua no at 20 ncome • earnns ncome ta on a n the of arbados ncome above • dded In a or onsumption a s ta ou pa on no obao, enera up n to aes, show ncome on ta earnns the pad up n to whch saar of ta n pad the s the same most amaca and arbados bu. rndad caed the pad 30 000. n n ou show ta earnns same amount amount tems to 30 000. n the 424 200. Value raph ncome arbbean. pad peope 441 1 ahamas. raw hosptas. n aue dded consumpton the rn ta a and s s chared 15, at n amaca 1.5, but t s there s Isands. IVI here are man other taes, whch var from sand to sand. au and choose hrstopher between can two dfferent tilities eectrct an uttes ed charge for eampe, eectrct, water and teephone make a he pus a cost per can per eampe, per uarter a teephone compan mht have a ed chare of months pus a cost of 0.0 per mnute of ca has person makn 342 mnutes of cas n a uarter woud 342 × 0.0 = ncreased b aares and and waes are reduced b deductons for aes uses 540 hch are dependent on the hours worked, the hour et up to works at tme works pas ow ow a 35hour and a do unt unts of eectrct hrstopher tarff shoud hch an au tarff shoud choose euaton the and number sove of the 40 hours ncome much much makes the same b unts under week at .50 per hour, wth overtme I haf. ever week. a e ta at the rate of 20. does does she she earn earn a a ear ear before after ta • Do sure ta rados cost that ncudn se at 45 pus 15 . ow you ta ma e all dedu ction s salarie s. no t e.g. in he per tarff. from 0.40 and necess ary 2 20 SIS • of unts. ves he tarff overtme. SR he 420 nd each pad or chare rate, that son 40 unt used. hrstopher t and ed pus uarter, choose nsurance ta. • of per bonuses. • 2 used, a uarter per are chare 0.3 .3. IS aares has eectrct uses pa au ed pus eectrct of + tarff tme. per 40 a uarter whch choose 40 of 3 ether unit whch or tarffs. mi units do lla rs and cents calcu lations . much 2 Sets RI • se the set concepts descrbe or • se nterpret st • epresent of sets a coecton of obects, usua havn a or somethn n common. ampes mht vowes the a be the bos n cass, the factors of 12, or the to n aphabet. members he and s S set n set members of a set are caed the elements of the etter, the set. notaton varous Set notation forms • econse and nterpret set s usua denoted b a capta and membershp s enn ether descrbed or the eements are sted. he descrpton or stn s darams wrtten • pp the prncpes membershp cardnat , unversa, of a nte, eua, nsde cur brackets. of o set, A nnte, euvaent we = mht 2, 4, wrte , , A = even numbers ess than 12, or 10. sets he ∉ smbo means Venn means not an ‘is an eement element of’. e of’. e coud sa coud that wrte 5 ∉ 4 ∈ A A diagrams Venn daram wrtten ‘s ∈ diagram conssts as ℰ, s of the a a set darammatc rectane to contann representaton represent a eements the of sets. universal that are enn set, reevant to the topc. ther he of sets enn are shown daram as crces. beow s an eampe to demonstrate the prncpes sets. ℰ = natura A = factors numbers of 12, B ess = than or prme eua to numbers 14, and C = mutpes of . ℰ 8 9 10 14 A B 1 5 4 C 2 7 6 12 3 11 13 ver C ⊂ A, eement A. oth factors of n of set the 12. C s aso eements n of set C, A. e sa mutpes C of s , a subset are aso of A, or eements of he vocabular of sets et A n(A) has = eements. sets as the ths ual or n to s cardinalit of set A s and wrte s are sets the = set, a to or the and the we same or has set st contan etters h, s, sa that number A of and B are eements. euivalent e null n the e a of use s an one that the ever the word the same D = se ts. ‘these’ = the etters n the or that are contans mutpes smbo ∅ to no of eements. 4 or represent A = etters n B = etters n a nu set. ‘I’ C = etters n ‘I’ D = etters n ‘’ = etters n ‘I’ he set. countabe nnte number number members of an of of eements. eements. nnte set It snce s mpossbe there s • hch two = p, a, r, sets • hch other are eua no set s euvaent natura numbers = 1, 2, 3, 4, 5, , etters n the word ‘paraeoram’ them … hch set , e, o, , m, or D = a, e, , , m, s a subset of s another D ‘’ neatve membershp. eampe set learn eements. • he to of IVI set, nte has a eact to or impo rtant vocabu lary can t. numbers subset innite count , that e, odd set aso is the I B numbers. set end B contan ≡ ‘sheet’ nite n A eampe nu • as sets suare both empt or • of eampe, word • the • t cardnat • sa 6 he wrte e o, p, r, as order set s • hch two sets have no unmportant. common ets do not contan same he members 2 he number ets 4 he can be the of unversa ua sets same of a set are eements nte consdered, more than once. or set, the ℰ, the = natura numbers = 1, , 1, 25 12, 15, 4, B = 3, , , C = 4, , 12, D = 5, 10, 2 hch hat = set s the cardnat. or a nu same the set, ∅ eements contans eements ben no eements. euvaent sets have SIS A escrbe a contans ℰ n eements. nnte. empt contan caed cardnat. SR F eement IS the eements 15, sets two can 1, A, sets ou mutpes B, of reater 21, than 25 24 24 25 C are sa 1, 20, 20, no and D n words. euvaent about = even numbers and 2 6 ombining ombinations RI of sets sets S ℰ • epresent a set n varous 1 9 forms A • econse and nterpret B enn 3 6 darams C 2 • pp the prncpes of 4 compement of a set, 7 subsets, 8 ntersecton, unon of dsont sets, 5 sets 10 In the ets enn ℰ = A = even B = prme C = mutpes B • we he • s • cross A ⋂ B = A ⋂ B s he enn A ⋃ B = A ⋃ B s he A he e 2 A, ths of A B A n A nsde A′ and and B, as s crce A′, 1, that s 3, the crces do have not no common cross. A the 5, , as A set , contann or odd everthn numbers. B, wrtten whch s where ⋂ B, the s the crces set of ntersect daram. even the and or two daram. on shaded means the wrtten case, enn prme number. daram. B, both, 2, 3, 4, shaded wrtten whch as s A ⋃ B, s everthn the n set of ether eements crce on n the use , , the , 10 daram. regions A 5, n of of two sets A and B conssts of the eements n B two B ths sets compement can of the both hs compete the unon I • n In ntersecton • A. of 10 daram. A he C sets. daram, the escribing • enn on or than 4 2, union ether of disoint intersection n reater numbers the shaded eements or are n no numbers crce not above numbers complement s he C n draw that A′ natura and eements. o daram to A and B conssts of the eements n . of a hep set conssts descrbe of a reons the eements usn set n notaton. that set. or eampe, he In shaded set to descrbe part notaton, s the ths s ths reon secton A ⋂ n A not n IVI B B′ A = etters n ‘cucumber’ B = etters n ‘meon’ C = etters n ‘mano’ • hat • essa s hs one s more dfcut to tr to descrbe t usn words eua A ke , and ake ⋂ sas to number’. descrbe. • e s up B ⋂ C that A etters Is she our ⋃ n B ⋃ ‘on C car correct own puzze ke . ths. he n shaded the reon here s everthn n the unon of A and B and not ntersecton. hat s A ⋃ IS subset 2 he of B a ⋂ set s compement everthn he unon, ⋃, of 4 he ntersecton, ⋂ B′ contaned of ecept SR A a the two ⋂, set A s n of set contans contans set. wrtten eements sets the those A′ and contans A anthn eements n n ether set. both sets. SIS ℰ = natura numbers A = 1, 4, , 1, 25 B = 3, , , 12, 15 C = 4, , 12, D = 5, 10, hat s A ⋂ 2 hat s B hat s the ⋃ 1, 15, no reater than 25 20 20, 25 C C cardnat of the set A ⋃ B ⋃ C ⋃ D′ Venn diagrams Subsets RI • st • S subsets nd the of a ven number of If a A s subsets set wth n onstruct wrte enn etermne eements set ths A s unons compements as A = set A ⊂ are aso eements of set B, then we sa B of probems rhtaned tsef a subset of tranes C = s a subset of B = tranes, poons. ⊂ B ⊂ C and sets umber ove B n A ntersectons, • of of darams whch • eements subset eements he • a of e a the set usn of subsets enn • or a • he set A, the nu set, ∅, and A tsef are subsets of A darams set he • he are set has ∅, a, two subsets. a b has four subsets. I he he a null subs e ts se t of and are • he are set ∅, a, a, b, c, b, d a, has b 1 subsets. he b, are d, ∅, c, a, d, a, b, b, c, c, a, d, b, a, d, b, a, a, c, c, d, a, b, d, c, b, d, c, a, b, c, d n enera, for a set onstructing any ma e ques tion sure you alwa ys univ ersal darams number Venn cardnat of n, there are 2 subsets diagrams of can be eements n used a to show the eements n a set or the set. unde rstand o the a I enn n wth construct a enn daram, we often need to determne the se t. eements n an ntersecton. ℰ R 3 If ℰ = natura numbers ess than or eua to 20, 7 9 11 13 17 19 A B 6 A = even numbers, C = mutpes B = factors of 20 and 8 of 5, 12 then 2 1 A = 2, 4, , , 10, 12, B = 1, 2, 4, 5, 10, 20 C = 5, 10, 15, 20 14, 1, 1, 4 14 20 16 10 20 18 5 hen and o A ⋂ B = 2, A ⋂ C = 10, B ⋂ C = 5, B ⋂ C = 10, construct mstakes 4 ⋂ A to the start enn wth 4, 10, 20 20 10, 20 15 20 daram, the C t prevents ntersectons. Solving hs s problems best epaned R In a o of pa nd draw 40 an Venn diagrams eampe. 2 bos, 1 pa footba and 25 pa crcket. nether. out a usn roup bos using how enn man pa both, ℰ daram. 6 C e know and that outsde the cardnat there the o there 1 footbaers ut ths are unon must be + ncudes of bos of 40 25 C − n and = ℰ s C who ⋃ F’ 7 F . 34 crcketers those F 40, n = C ⋃ F . 41. pa ℰ both, twce. 6 C o the ow etra we 41 can − 34 = compete umber pan umber make the enn footba pan up crcket C ⋂ daram 18 on = 1 − = on = 25 − = 1 IVI SR A = factors of 10 B = factors of C = factors of • ace each D = prme = A ⋃ 4 F = D ⋃ set nto the numbers between 4 and 1 C A B (A correct ⋂ B) = oe suare on the rd 15earod s ardinalit 4 ardinalit a ntha beow. s a rhthanded 15earod. the three 14earod. students wth the set. and ⋃ 2 even IS enn daram cardnat If A he ⊂ B, of can the then show the eements of sets or the ever te eement of A s n a B number of subsets of a set wth cardnat n s 2 does ou B A ⋂ B′ A and cass nu set and A are subsets of cardnat of B of − A ⋃ the B) = the cardnat cardnat enn daram B of b 3 f B ⋂ students the cass, C contans 11 have vsted A the cardnat ths about 1rs. he ⋂ ℰ hat sets. n he A′ A 2 B)′ both eements 14earod. rhthanded eements A 4 a contans ontans efthanded s correct odd students, students atch odd SIS efthanded = hp ardinalit n F F of A ⋂ B) of A + the . cass here who raw a have enn nformaton rs have are not 12 vsted daram and vsted bos nd the to out n the the . show how ths man . odule SI rom ractice eam oncalculator these numbers 5, , , , , 10, choose rte a down seuence tart a prme b a suare c a mutpe wth factor 2 nd the utp of of 4.2 to b 3 2., of 42 vn sncant and 5. × 1 ( our answer he the t and add 2. n − 4 a of ths renada n standard operaton ✪ b = 2a that s ✪ × s 40 000 m form. dened as 3b the operaton ✪ s ures. a assocatve b commutatve. ) 5 as area how 5 1 acuate rte of rue 3. doube rte and 3 terms the 2 2 2 4 ve 4 a correct rst foows number a the that number hen d uestions 2 acuate ow 4 st 4 + ÷ 2 − 2 × 3 decma. 6 pencer rato and ark share some mone n man mnutes s t from 055 to 1025 the 4 3. ark receves ow much a the subsets of A = a, b, c. 4. are the sharn 3 2 of the students n a cass are bos. 5 3 4 × the 3 of 4 3 bos are rhthanded. mpf , wrtn our answer n 2 here 3 nde notaton. rte these wth the ow numbers n −3.4 2 sze, man rhthanded students are bos. there n the cass startn 1 3 −3 4 SI of 1 smaest 3 3. order are 3. √ 2 alculator acuate 45 of 5. 4 ew eaand doar s worth 1.54 arbadan doars. 2 arr ara wants arr sas b 12 hat bouht to s can arr’ s sod sod t a for vase bu she month essa he a for the pa percentae at a for vase. 100 nstaments car ow 40. oss depost of 35. man 320 30. nd the 5, 6 hat 1 2 6 dd she pa for t s 1 1400. of hat orna eaand doars woud I et doars foowed prot of ew arbadan nth , term 11, the of 14, the 1, seuence … mutpcatve nverse recproca SI he 2 number hat alculator s the 3215 vaue hares nvests hat ths s s smpe b compound base the 00 worth a n of at 4 after . dt 2 per addn annum. 3 ears’ nterest nterest 4 car, ever ow bouht for 5000, deprecates b are he ompete an s the car worth after three Income ears ths to annua receves compan ear. much up has aso a saar bonus of of 45 000. 2 of the prots. ta s chared 24 200, and at 20 35 on on a ncome ncome beond 24 200. b er 3 books 12. each 5 pencs 0.35 each 2 notepads 1.5 each 4.50tre compan made a prot of 0 000. acuate 2.5 tres pant a her bonus b the amount of ncome ta she pas. ubtota aes ta 15 one earns week. ast 2 hs A, 14 he much and s hour pad worked dd ueston B per vertme week ow or each true he s at for for a tme 42 the statement beow, sa whether t s fase. a prme numbers b 1 ∈ suare c 15 ∈ mutpes d mutpes ⊂ odd numbers 35hour and a hours. three numbers haf. earn about or 6 sets, aror he C of earns pas an of ⋂ 5 ⋂ factors annua ncome ta factors on of saar her of 20 of 100 = ∅ 55 000. earnns as foows A = factors B = mutpes C = odd a hch b hat set of 24 of 0–20 000 numbers s a nte the taed at 23 taed at 35. hat can cardnat ou sa of A ⋂ about B ⋂ aso ear a roup of 3 bos, 20 pa crcket pa footba. ompete the bos enn pa daram oods that whch she spends have had 23 000 15 n the prce. percentae of her saar s taken n ta and ncome 1 cacuates on C hat In over B ncuded and set a c free. 42 000.01 he s ta 20 000.01–42 000 ta and nether. to show ths nformaton. 2 Measurement 2. and Es sle stimate e d dw Es LEarning • statistics e outcomES area of The e The area of a closed plane shape is the amount of surface inside it. irregular 2 plane gures is measured in square units, such as cm . 2 A • se to maps and determine areas scale square with sides of 1 cm has area 1 cm drawings distances and So, area squares T o estimate Count the squares n can the 11 be thought needed the cover area number which shape green to are of of a as shape, than there which the number of 2 unit 1 cm surface. squares more below, squares a of inside half are are trace it onto the inside than centimetre-squared shape, the complete more a and an shape. squares half include grid. ellow included in and the shape. 2 e estimate This method scale a is area ver as useful + 11 for = 1 cm irregular shapes, and is often used in drawings. Sle n the dw scale drawing, a scale is often written as a ratio, for eample 1 1 . This means drawing, the n real 38 the 1 cm real on the world is 1 drawing times represents greater than the 1 cm or 1 m words, eample world. other ‘1 cm that so occasions, represents a scale 1 m’. is represented in for in orE ere is a EamLE map of Cuba, drawn on a centimetre-squared grid. Malanas HN Cardenas Saga Grande Colon Pinar del Guane Cabarién Santa Rio Clara Nueva Cienfuegos Moron Gerona Nuevista Camguey Banes Holguin Moa Baracao Bayamo Guantànamo Santiago de Scale The As 1 cm represents distance each cm from Cuba 1 m Santa represents Clara to 1 m, uatanamo the actual on the distance is map . mared × in 1 m red = is . cm. m. 2 The area of Cuba, found b counting squares shaded ellow, is 11 cm 2 ach square represents an area of 1 m × 1 m = 1 m 2 So the E estimate 3 the area of Cuba is 11 × 1 m 2 or 11 m actiit ointS The area inside 2 of The se of area a a 2-dimensional shape is the amount of space it. can map be scale estimated to b calculate counting distances • ind • Trace areas on a stimate The ind the shape the the island. esearch is area the of this map of island, drawn to a scale of 1 cm and out area of actual our sie to estimates see are. to Eam actual the stimate wor width shape. an length of the island the greatest ti distance e sure you look island. close ly 3 island. centimetre- and the close • Mak across a grid length, our uEStionS 1 m. 2 onto the how it of map. • Summar map squared squares. and a the area of the island. at the scale on a map. • Rem embe r squa side re is the area equa l lengt h to of a the squa red. 3 2.2 eee The LEarning eee Calculate a shape is the e distance around the boundar of the outcomES shape. • of d the perimeter of t is measured in units of length mm, cm, etc.. a The e of a shape is the amount of at 2-dimensional space polgon inside • se the area formulae the shape. for 2 Area quadrilaterals and is measured in square units, such as square 2 square to calculate areas centimetres cm , triangles or millimetres mm 2 , or square metres m . The area is the missing number of unit squares that would ll the shape. sides T o nd the the area of some eedl special quadrilaterals, multipl the se and e reles The e length b the ele or a square is found b multipling the width. 3 cm w The perimeter is cm + cm + cm + cm = 1 cm. llels 5 cm The e llel l or A = l × a s is found b w multipling the base b the 4 cm Area = cm × = 1 cm cm perpendicular height. h 2 A = b × h Area = cm × cm 2 = easure the 12 cm green sides the 3 cm b are So . cm the ae The Eam • long. perimeter area of is cm + . cm + cm + . cm = 1 cm. le a le is half the area of a parallelogram ti 4 cm Don’ t the fo rge co rrec t t to inclu units – de h area b × h is alwa ys squa re measu red A = in 2 units . cm × cm Area = 2 2 5 cm = b 1 cm a ae e 2 cm The b e the e perpendicular a + is found b multipling the average width height. b A = × 4 cm h 2 2 h + Area 2 = × = 1 cm 2 7 cm cd ses b T o nd can the nd Add all area the the of area areas more comple shapes, cut them into shapes ou of. Area together to nd the total of b area. triangle × h × shape can be split into a rectangle, a triangle and a = 2 cm 2 trapeium Area 3 cm 2 = 2 This = of rectangle 3 cm 2 = Area 4 cm (12 cm l × of w = × = cm trapeium 8 cm) a 5 cm + b + = 5 cm × h = 2 × 2 = 1 cm 2 12 cm 2 T otal 8 cm E 10 cm 2 three congruent the same shape and a parallelograms, height Calculate + 1 = cm erimeter is boundar of the of with a distance around the shape. Area is a base of the a amount of 2-dimensional space shape. and 3 paper + ointS inside out 2 5 cm) actiit sie = 5 cm (10 cm Cut area 8 cm e formulae 1 cm Area of a rectangle = length Area of a parallelogram × width 12 cm. the area of the parallelogram. = Area base of a × perpendicular height triangle base × perpendicular height = 2 Area of a sum trapeium of the parallel sides perpendicular 2 ae a single straight cut on one of = the × height 2 parallelograms rearranged • Calculate to so the mae the a area two pieces can be rectangle. of the rectangle. Summar 3 ae a single straight cut on the parallelogram so the rearranged mae two pieces can ind • Calculate the a area the ae a single of straight of a triangle with a base of cm a height of cm. trapeium. the trapeium. 2 A trapeium cm area be and to uEStionS net cut on the and has cm. parallel The area sides of of the length trapeium is third 2 cm parallelogram so rearranged mae • Calculate to the the area two a pieces can triangle. of the triangle. . Calculate the perpendicular height. be 3 A of rectangle cm. has a perimeter Calculate the area of of cm the and a length rectangle. 2.3 cles s LEarning outcomES • the • Calculate The le distance around the outside perimeter of a circle is called the circumference eee of a circle • Calculate the area of a • Calculate the perimeter • A part • A line of of a combination and crossing is a Calculate the area of the circle through the A line from the centre to the circles length of is a ds plural radii. arc • and an dee circumference • is of • polgons circumference and centre area the circle A line from a point on the circumference sector to is another a point d. The on the diameter • An area cut off b a • An area cut off b two Le magine t taes The eact is chord a is radii special a is over the a diameter three number of sector. d around diameters diameters chord. segment. eee wrapping ust circumference to is the reach called circle. around π pi, the and is circumference. approimatel .11 π is an l number see 1.1, and so cannot be written down eactl. Circumference The f a = An The diameter circle 2 is sie × a of π has × is a 2 π × twice of diameter, the radius or fraction the = of radius, 2 cm, fraction C so the approimatel the or on the π D alternativel circumference 12. cm circumference depends = of sie a to 1 C = = 2πr 2πr decimal place. circle. of the angle at the centre. f the angle is °, the arc is of the circumference, as whole 50° f the angle angle is at °. the centre is θ, then θ θ ength of arc = × circumference ae ere is a le circle, d and a = 2πr se square with an r 2 area of radius magine ept the 2 the × radius, squeeing same circle. the area, or r square but t so inside it r the t would loo lie this. e could t ust over three of actiit these in the circle ee’s ort of S Spain’ s largest is open r space, r trafc The • and fact, it would tae eactl π of perimeter Calculate • them. the area of the circle is π × the area of the ellow square, or A = π r is the about area cd ses fo rm ulae Rem embe r diam e ter Some shapes need to be split into parts in order to nd the area . m. inside it. confus e fo r circu mfer ence • largest ti no t the 2 So Do world’ s roundabout. Eam n the and that is area. the doub le the and radius . perimeter. • Whe n speci orE EamLE door 2.2 m T o it The so is cm wide • the a area rectangle diameter the and of radius is perimeter, and the a 2 we Do no t = 22 height − of = the is = of l π in sure to roun d too off soon . cm, rectangle ointS is 12 cm. Circumference circle Area be a cm. E The gien fo r it. answ ers split e semicircle. semicircle ÷ are alu ques tion use and tall. nd into shown c a The you = of a 2πr rectangle × w = 12 × 2 2 Area of 3 ength a circle = πr 2 = Area 1 2 cm of of arc with angle at semicircle θ = area = πr = π of circle ÷ centre 2 θ = 2π r 2 ÷ 2 38 cm 2 × ÷ 2 2 = 22 cm Area of sector 2 to nearest cm with angle at θ centre θ = 2 πr T otal area 2 = 1 2 + 22 = 1 1 cm 220 cm erimeter Summar uEStionS 182 cm = semicircular top = circumference = 2 = cm × π × to + ÷ ÷ 2 2 2 + + nearest sides 2 + × + bottom 12 + ind of a the area circle and with circumference radius cm. cm 2 A circle of has cm. a circumference ind the diameter, 76 cm radius 3 A and semicircle area. has an area of 2 cm . Calculate the perimeter . 3 2. Se A LEarning -dimensional Calculate solids surface cube, clinder, shape, or solid, has a number of faces or surfaces. outcomES or • e area cuboid, right of eample • A cuboid has • A clinder si rectangular faces. prism, pramid has two circular faces and a curved surface. and hen cone. nding the surface area of a solid, it helps to draw a e ss A s is A clinder a is throughout A cube Cones is a and The A a square cube l × their e side = l as ssse it has the throughout same circular are and not a cuboid prisms as is a rectangular the length. cross-section prism. cross-section – changes ss of l cm a solid has si is the faces. sum ach of the has areas an of each face. area 2 cm 2 Therefore, its lengths. e of l same prism, prism pramids se the length. 2 of with circular its throughout Se solid the surface area of that cube = l 2 cm l cm h cm A cuboid of length l cm, width w cm and height h cm also has si faces. w cm 2 l cm The top and bottom each have an area of l × w = lw cm 2 The front and bac each of The two ends each have an The surface an area of l × h = lh cm Back 2 area of w × h = wh cm 2 w cm T op Base l cm area = 2lw + 2lh + 2wh cm w cm h cm Front l cm or a clinder of radius r cm and a height of h cm r cm 2 The top and The curved bottom are each circles with an area of πr h cm surface could be cut and opened into a rectangle. The width is h cm and the length is the circumference of the circle, or 2πr r The area of The surface the curved surface is 2πrh 2 area of the clinder = 2πr + 2πrh 2π r h Se A right The it pramid pramid has The e a in – d has the its h cm, verte diagram perpendicular height, top of has height each d directl a of e y above square base the with sides section triangular . to the area of all of of x the base. cm, and cm. face can be found 2 thagoras centre blue triangle h b y + appling 2 x 2 = 2 The surface area = xh = triangles + area of square base 2 × + x 2 The diagram height of shows a cone with radius r cm, height h cm and a slant s cm s cm h cm 2 The base is a circle with The curved The circumference of the whole circle The circumference of the sector must surface is a area πr sector of a circle r cm base, or is 2π s equal the 2πr the fraction of the circle required is r the the = 2π s So of r So circumference 2πr area of the sector s 2 is π s × = πrs s cm s 2 Surface area Se of cone e – = πr + πrs see 2 The surface E area of a sphere with radius r cm is π r ointS Surface area is the sum of the areas of all Eam surfaces. ti 2 2 Surface area of a clinder = 2π r + 2π rh When 2 3 Surface area of a cone = πr + area π rs calcu latin g it helps to surfac e draw a ne t. 2 Surface area Summar of a sphere = π r uEStionS actiit ind the surface area of a cuboid cm long, cm wide Show and 2 cm have 2 A cone 1 cm height 3 of is radius stuc 12 cm. A clinder A sphere Show that these two prisms high. has has that on top ind a a the cm, of the height radius sphere height a 12 cm clinder surface of of of area cm and and radius of a slant the the same surface area height cm, shape. radius of cm. cm. and the clinder have the same surface area. 2. le le LEarning outcomES • volume The Calculate of le occupies. prism, clinder, of a solid is the amount of -dimensional space it taen up solids cone, t is measured in cubic units. sphere, or cube, eample, a cubic centimetre, or 1 cm , is the space b a cuboid cube The of edge volume that length of a solid can be paced le elow are 1 cm. can into be the thought of as the number of unit cubes solid. s three prisms, cd all split into tl centimetre cubes. s clde 2 Area of blue top = total l × w b × π r h number of blue squares = × = cm 2 2 2 2 × = π × 2 ≈ 12. cm 2 2 = to 2 d.p. 2 2 = The number of ellow and blue cubes is equal to the number cm of blue olume T otal = of the top laer ellow volume volume of cm = top cm × = 12. cm × = the number of to ≈ 12. × ≈ .2 cm 2 d.p. laer × squares 2 cm = cm to laers s le = le The e volume × e ds of a pramid or es a cone V = d is Ah sees given b le = se e × eedl e 3 So • for a square-based 1 V = pramid with base 2 l h • for a cone of radius 1 V = r and 2 π r h • for a sphere of V = πr radius r height h of side l and height h 2 d.p le Compound d shapes can be orE EamLE The shape shown The cuboid has ses broen into simpler shapes. can be split dimensions into a 1 cm cuboid × cm and × a triangular prism. cm 2 so, the area of the top is 1 × = cm 14 cm The volume The prism The area = has area a × height triangular = front base of × × = base cm height the triangle is and of 8 cm cm × height 1 = − = cm. 2 = 1 cm 5 cm 2 2 10 cm The volume of the prism is area × length = 1 × 1 = 1 cm T otal volume Sl ost + 1 = cm les problems orE A = sphere can be solved EamLE of as a T o calculate radius clinder of b setting up an equation. actiit 2 12 cm radius has the same volume 1 cm. A lad from an the height of the clinder, we set holida collects unns in water iver alls up in a plastic in the container equation olume of sphere = πr = shape π × 12 cuboid, of clinder = Area = πr of top × × π × 12 h = 1 1 cm high. tips the water 2 = 1 πh into a with = and πh 2π 2 cm 2 × She long, height wide 2 So a measuring cm olume of × r on amaica clindrical a radius of drum cm 2πh and a height of 1 m. 2π h = = cm • ow man times will 2π she have to container E ointS olume is Summar the amount of - space inside a solid. 2 olume olume the wide of a prism = area of a cone = of base × height 2 A of 1 3 ind the order plastic to ll the drum uEStionS volume and cm clinder cm. ll in has of a cuboid cm long, . cm high. a radius Calculate the of cm volume and of a the height clinder. 2 πr h 3 olume of a sphere = A pramid has a square base of side cm. t πr has the same Calculate the volume height as of a cube the of side cm. pramid. 2. us Si LEarning Convert units s of length, S Sstème time, se S units which for area, be used units for of measurement length, mass and adopt temperature, commonl used prees hundredth and are milli ilo m meaning meaning 1, 1 1 mg milligram is and 1 m ilometre is c 1 a gram, 1 cℓ centilitre is of 1 mm 1 cm 1 cm = 1 m = There are = rarel 1 1 cg = 1 g 1 g other = 1 cg 1 m Additionall, actiit c 1 mg 1 m litre metres. mss = a 1 1 Le S centi of 1 the of thousandth. esearch series time So • a capacit. volume, meaning mass, can speed The • nternational area, prees capacit, esee outcomES The • = 1 mℓ 1 g = 1 cℓ 1 g 1 ℓ 1 cℓ = = 1 ℓ 1 ℓ tonne used divisions and multiples for eample deci sstem. 1 d is 1 ces Converting b 1, or 1 ewee within or the S Si s sstem onl requires multipling or dividing 1. eample 1. m 2 cℓ = = 1. 2 × ÷ 1 m 1 ℓ = = 1 m .2 ℓ te T ime does not have units that seconds = 1 minute minutes = 1 hour 2 hours .2 ae d = 1 das are multiples of 1, 1 or 1. da ≈ 1 ear le 2 A square with sides of 1 cm has an area of 1 cm × 1 cm = 1 cm 2 t is also 1 mm × 1 mm 2 So 1 cm = 1 mm 2 = 1 mm A cube with sides of 1 m has a volume of 1 m × 1 m × 1 m ut or 8 it is also area, the 1 cm length × 1 cm × conversions 1 cm are = 1 cm squared. = 1 m or volume, the or eample length conversions are cubed. Eam 1 m = 1 m 2 So 1 m And 1 m • 2 = 1 = 1 2 Rem embe r = 1 m = 1 m when of States the of world uses m the S smal ler Sstem, but most notabl the do not. uses imperial • The SA the rea unit and rates c • 12 inches = 1 1 foot = feet = 1 ounces 1 o pound 1 lb = 1 ounces to pints = 1 caref ul when with calcu lato r . gallon hae lengt h. wo rk ing pint a time . on hour s a is ards = = uid 1 e diide con ersion ard 1 lb 1 o lum e sstem mss to a unit. differe nt Le to and chan ging nited larger America multip ly chan ging when ost to 2 m ti 1 hundredweight no t cwt hour s mile but 2 cwt ae = 1 eles Le e w mss inch ≈ 2. cm 1 o 1 ard ≈ 1 m 1 lb 1 mile ≈ 1 m 2.2 lb ≈ 1 g miles 1 ≈ g m ≈ ≈ ton 1.1 hour s minut es. sses c 1 ≈ ton minut es g 1 uid g 1 pint 1 gallon ton ≈ 1 ounce ≈ ≈ mℓ . ℓ ≈ ℓ tonne actiit 2 artinique has an area of 112 m or sq. miles. 2 • se this mae information up 1 square to nd out how man m mile. 2 • ow • oes man this miles E m are agree ≈ there with the in 2 square approimate miles equivalent m ointS ilo 2 centi Summar means 1 A recipe 3 milli The the c m means means for soup litres of includes these ingredients water hundredth g salted 22 g breadfruit beef thousandth conversions length uEStionS for area are the squares g of conversions. hat are coco these quantities in imperial units 1 The conversions for volume are the cubes of 2 ow man seconds are there in 2 hours the length conversions. 3 t is important imperial to now equivalents. the approimate A rectangle rite down is 2 cm the area 2 cm long and 1 cm wide. in 2 mm square inches 2. te dse d seed Seed LEarning outcomES Seed • Solve problems involving A distance and car at or is speed s in hour kmh mus kilom e tres in constantl that speed is metres ti distan ce measure ilometres Speed • a of how quicl something is moving. changing speed, but we sa we are travelling at speed ‘ Eam is time, t be and in the ecause time is for usuall per aee the per hour’. an This hour, means we measured in would mh that, travel if we continued travelling ilometres. ilometres per hour or ms second. seed speed calculated tends b to change, dividing the we distance often tal travelled of b ee the time seed. taen, This so s. distance • speed Rem embe r is no t . = minut es hour time s. or A eample bus taes minutes to travel 1 m. 1 m The average speed = actiit The n 1 August 2, sain speed triangle amaica set a world record for the 1 m Calculate his • At this speed, he run in useful wa of remembering the D between time, distance and speed the letters distance, speed and time go in alphabetical order T S sprint. T o • a of for . s is mh hours olt connection of = . speed in calculate the time travelled, cover up the T for ms. D __ time, how far and it shows that ou must calculate would D S • ow far 1 minute would he t run also shows that distance = speed × time, and S in distance that 1 • speed = hour hat time was his speed in mh se–e ournes orE ere is graph of travels calculate write 1 m a the speed time in in in ourne the rst 2 minutes. mh, hours. minutes is of an hour, so his speed for this part of the 25 20 ourne is ecnatsiD 1 2 shown on a distance–time graph. 30 to bac. morf rst ason’ s and emoh ason T o a supermaret often )mk( the EamLE are s 15 10 5 1 1 ÷ = mh. 0 0 10 20 30 40 50 T ime 60 70 (minutes) 80 90 100 110 120 t taes him minutes to travel the 2 m to the supermaret, so his average speed for the ourne is 2 2 ÷ = . mh. e is in the travelling The supermaret towards fastest he or for awa travelled minutes, from was on shown b the horiontal line. t is horiontal because he is not home. the rst part of the ourne bac home. e now this because it is the 1 steepest part. e travelled 2 − = 1 m in 2 minutes, an average speed of = mh 1 e met his average sister speed arsha of in the supermaret. She left ust as he arrived, and she travelled home at an 2 mh. 30 leaving plot point, position closer is an to at home, the graph passed minutes, oining with She 1, we can 2 m later, so . see after over her her is minutes arsha ust b 25 20 ecnatsiD ason 2, later. point the from , hour sing ourne morf she her emoh can )mk( e that about 15 10 5 11 m home. 0 0 10 20 30 40 50 60 T ime Summar 70 travel travel at 2 mh for 1 hour 1 minutes. ow far he grap h ccles 1 m to wor. She goes to wor at an does that of 2 mh. She returns at 2 mh. ow much her ourne to wor than her ourne ason The graph running shows the the 1 m distance travelled uphi was ll t show and s him home trae lling 3 no t quicer down hill. is 120 average trae lling speed 110 ti show adine 100 do • 2 90 uEStionS Eam 80 (minutes) b ames and when • he sprint. from towa rds steep greater 120 er the home home. the line the speed . 100 )m( 80 E ointS ecnatsiD 60 T ime, speed distance and D are 40 connected S T 20 2 e consistent with units. o not multipl 0 0 1 2 3 4 5 6 T ime 7 8 9 10 11 12 a speed in mh b a time in (s) minutes. hat was his average speed over the rst seconds hat was his average speed over the last hat was his average speed over the whole 3 A travel graph shows how seconds race distance from changes over a point time. 2.8 mesee d Es LEarning o • State the error associated given measurement, be wholl whether length, mass, time or other measure, can accurate. measurement The • esee with ever a outcomES etermine the range and possible values for a are affected b human errors, or the limitations of eesight, of the inaccurac of measuring equipment. ie all solids, a ruler given will epand slightl the temperature. when heated, so its length will var according to measurement • Solve measurement A ruler is unliel to have smaller marings than problems millimetres, so it cannot be used to measure to a greater degree of accurac. easurements unit or larger or eample, millimetre half a given a pencil might millimetre cll orE actiit An elevator ‘aimum has a sign weight saing 2 g’. Three width mm The four have people weights g and of g in the g, all to • s in it the particular unit ma actuall be up to half measured anthing more w or plates, . cm to to be less as 12. cm between than the stated d to and the nearest 12. cm long, measurement. les each the long 12. of nearest bolted together, with of 1. cm a bolt 2 g, length the the nearest mm. g. safe time a elevator to nearest be EamLE metal are to smaller. for them elevator at all the to travel same Calculate thread to how will attach be to much showing the nut. Sl ach The plate total and The bolt The greatest plates There ut if ere is are the we is between be amount 1. cm 1. plates are . cm 1. cm possible small might minimum between × and . cm . cm and thic. × . cm, 1. cm. between could thread be thicness 1. cm 2 could be are − and 1. combining a measurement. of and 1. large onl and − 1. cm. thread the = the bolt will is . cm bolt 1. maimum is = be showing large if the 1. cm. showing. small, the amount . cm. measurement with a of or a orE n a of EamLE lemonade 2 mℓ factor, per ach litre ow much 2 second, bottle is lemonade to lled lemonade the for will ows nearest seconds, the through a pipe at a rate 1 mℓ bottles to the nearest second. contain Sl The quantit . s t × of lemonade 2 mℓs could be as = little could be as much as 112. mℓ as . × 1 mℓ = . mℓ Eam ti • ook orE EamLE like oe ran 1 m to the nearest m in 1 seconds to the nearest and second. e ran somewhere said, 1. is 1. ‘ teacher . ÷ Again, we so m = are and run as ou 1. . m 1. far average ‘ut taen 1. minimum have said, have between seconds might seconds, might fo r as was have seconds, 1. m, 1. m speed ma and in a ‘min imum ’ abou t time seconds. so our as quicl run ÷ 1. = . m, speed • as . ms’. and might no t maimum to tions racy measurement with alwa ys um some times it onl Do combi ne be with a accu maim . ms’. combining ques in measu remen t. 1. onl wo rd s ‘ma imum ’ recog nise between e out 3 a a combi ne alu es you need maim to um minim um . a measurement. E ointS measurement actiit A circular to the o is cae nearest has has a and some radius of 1 cm pin or frill icing to to go around go on the 2 cae, the greatest possible area for the uEStionS A pacet of butter weighs 2 g to the nearest pacets are paced in a the maimum and minimum weight of the 2 A bird ies at 2. ms to 1 decimal place for seconds the maimum possible distance the bird has 2 A rectangle length width = ma a + ma b + b = min a + min b − b = ma a − min b − b = min a − ma b × b = ma a × ma b to inimum the × b = min a × min b second. Calculate 3 b aimum a nearest + pacets. 2 smaller. bo. a Calculate can bigger 1 g. 2 unit inimum a a to unit aimum a Summar half given one inimum a icing. be of aimum a top. 3 and accurac alwas it. green A an cm. decorating She a is of has . cm the an to area the rectangle. of cm nearest own. 8 to mm. the nearest Calculate cm the . aimum a 2 ÷ b = ma a ÷ min b ma b The minimum inimum a ÷ b = min a ÷ 3 2. t es LEarning An • ecognise tpes of information plural ungrouped, means. frequenc class band, collect different are reggae, e features interval, for • a the limits, midpoint • mass see ata as can rouped depends EamLE collecting d data cs such was are d uantitative class boundaries, aisie decision-maing tpes not of numerical, calpso, pop, b data, test T allies so scores are she ranie are grouped orE for eample are numerical, such as heights, of d our d a test be can a gaps was the favourite the sets all in mars up a of students frequenc integers, ves heartower row, tae or put whether continuous on an or as cm. in n value on a are to the so mae heights continuous so scale, the with heights rst a the data. such height row second a heartower of tae a or on shoe music or ages. eample, certain values, usuall integers, sie. ed into the categories. data are The discrete method or of grouping continuous. in a test. These are table lie se the data are counting discrete. the t ll ee this | frequencies easier . 1 11–2 21– 11 1– 1– of heartowers in her garden. She as must 1. mae cm or sure 1. there cm are are x no t ll ee | included. 1 <x≤1 the of eactl height 1 cm must be is less recorded than or in the equal 2 1<x≤2 rst row, the height must be greater than to the second row. 1 cm 1.1 cm is recorded in 1 12 <x≤ clss hen or ls we d round eample, nearest sies for <x≤ so or discrete 2<x≤ 1 is 2 recording table, shoe || A ata EamLE heights in d. questionnaire ar. onl score are be can in ed data on can e These called data 1–1 The is surve, etc. t es discrete data discrete continuous etermine There le • orE help often tables groupedungrouped, set to e quantitative steel Construct data used datum. grouped • qualitative, • of continuous other • d data the discrete, outcomES cm, if off the then des continuous students we could in use a data, class this the all appear to measured chart be their discrete height data. to the e t ll ee Eam ti 11–1 • las s 11–1 class boun darie s limits are no t 11–1 necess arily the and same. 11–1 11–1 n this table, the because the lss ls of the rst row are 11 cm and 1 cm. actiit ut 1. cm and data have 1. cm, been rounded, because an the height lss from des 1. cm up are to n but less than 1. cm will be included in this 1, a orld of clss 1. calculate the lower class n above class interval boundar from or the width upper of the class class, subtract the SA set ecord in the 2 m race seconds. eample, the class interval is boundar. 1. − to 1. = 1, timings 1 the Smith el rom T o T ommie categor. 2 decimal orld ecord 2 m places place were instead used given of the previousl . 1 cm. olt’ s ecord all the decimal sain md for set amaica in 2 orld was 1.1 lss seconds. The midpoint is the mean of the class limits. • 11 + n the eample, it hat are the 1 is = limits 1. cm of 2 T ommie All the above 130 131 information 132 133 can be 134 seen 135 on 136 this 137 number 138 line 139 Smith’ s time 140 • 141 hat are the limits of sain olt’ s time Midpoint Class Class E ointS ata can Summar be qualitative 2 limits boundaries quantitative numerical or non-numeric. uantitative data can be uantitative data can be grouped or discrete Class limits stated in are the the upper frequenc and Class boundaries continuous data go The midpoint of a been class about ass fruit parcel Apple –1 anana 11–1 lower these of three Cherr 1–2 ength g l measures the limits rounded. hich when 2 chart cm 1 < l ⩽ 1 1 < l ⩽ 2 2 < l ⩽ 2 is the mean of shows continuous hat are the boundaries are charts. table. beond have tall avourite questions or continuous. the incomplete ungrouped. 3 All uEStionS data class for the width rst qualitative and the categor data class in the ass the chart upper and lower limits. 3 hat the is the ength midpoint of the rst categor in chart 2. sl LEarning s outcomES A • Construct and interpret bar chart has parallel rectangular bars or columns of the same width, bar usuall with a space between them. ach bar shows the quantit of a charts different • • Construct and frequenc polgons Construct and categor of data. interpret interpret ar charts but not are for used for qualitative continuous data or discrete orE A EamLE compan uses man e reams The When paper table draw ing gives in a the res e sed ear. an–ar 2 April–un ul–Sep 1 ct–ec data last remem ber a • Whe n sens ible look need to choose scale. scale. inter pre tin g charts we to the choose ear bar irst charts the of ti for • data, line graphs Eam quantitative data. hen choosing a scale, Bar bar caref ully we choose for each a round number chart showing amount of paper used 450 at scale. square on the 400 ais. 350 The biggest number is , 300 we that need shows a vertical at least smaeR so scale . 250 200 150 actiit A scale of one square to 100 will • Conduct to nd a out surve their of 2 tae squares. friends favourite 50 sport. Then we chart as can draw the 0 bar Jan–Mar • ut the results in a bar chart. A frequenc continuous T o draw a EamLE frequenc Jul–Sep Oct–Dec Quarter ee orE Apr–Jun shown. ls polgon is used to represent quantitative data, usuall data. 2 polgon, plot the points class midpoint, frequenc 7 and oin them. The end points are the midpoints of the intervals on either side. 6 t meets the x-ais at the centres of 5 adacent or this empt bars. t es • the Se ee data points are 1., 2, ., , ., 1–2 2 21– 1– 1– 1–1 ycneuqerF the 4 3 , 2 ., and ., 1 • the net interval on either side would be 0 −1 to points , are and 11 −., to 12. and So the 11., 0 end . 20 40 T est 60 score 80 100 Le s A graph line straight the is lines. drawn b suall, horiontal if EamLE The of ere are is points one a the line and the connecting quantities, it them is with placed on water 3 in a water cooler graph, the is measured te results e As of ais. orE depth plotting time results loo lie a.m. ever 1 1 2 a.m. hours. 12 11 noon 2 this p.m. p.m. 1 Graph p.m. showing depth p.m. 1 of water 2 (cm) 40 A line graph is useful as it sometimes 35 allows the us estimate readings. estimate a.m. was we dashed these eample, the depth of we might water 30 at cm. need to tae section of the it loo points between maes care. line. as oo 25 htpeD owever , the or that values )mc( 11 to at 20 15 oining though 10 it too 2 hours to rell the container , 5 whereas two to it will have replace the taen empt ust one a minute with a or full 0 8 a.m. one. The graph should, in fact, be 10 a.m. 12 noon 2 p.m. 4 p.m. 6 p.m. 8 p.m. vertical T ime at the but moment the graph Summar raw a the does container not tell is us replaced what time this happened, onl that it uEStionS bar chart to was between E show this information from a surve 2 s ar Cricet T ennis 2 charts or gap ootball Athletics on 3 frequenc polgon information ae qualitative data. the ine to show these two sets of about passengers on a the There is bars. polgons of data same graphs to allow be shown aes. allow estimates us to between the data measured showing p.m. for discrete sets mae a are between requenc two raw ee a 2 and ointS data e p.m. values. bus. 1–1 11–2 21– 1– 1– 1– 1 mle ee ele ee 3 hich tpe of chart would ou draw to represent qualitative data 2. sl 2 ss LEarning • outcomES Construct and histograms • Construct and A histogram loos A histogram is similar to a bar chart. interpret interpret pie ecause it is used when continuous, data there are are continuous. no gaps between the bars. charts orE The table ages of EamLE below some shows people the on a ere is the histogram. bus. 7 ae es ee 6 1–1 2–2 – – 5 ycneuqerF 4 3 2 1 The data ou are are continuous, classed as 1 as 0 right 10 up to the da before 15 20 25 Age 2th So e A pie orE A surve EamLE of people’ s 1–1 bar occupies chart is circular graphs charts or are actual in because shape. it does fractions, particularl frequencies music gave these A T o 8 full this turn nd the information is °, angle so for space from 1 right up to 2 A pie not the useful less chart show is different frequencies. from into each each a pie chart, person sector , we nd occupies we 50 on the nstead multipl this the ÷ total 2 result = b to show important the results than s the band of a sample, ap Salsa of the sample 1°. the shows proportions. ee eggae sie it total. results Steel put of are t e T o the 2 favourite 45 chart. proportions, the 40 s other ie 35 (years) birthda. the the 30 our frequenc . + + + = 2 as So the angles t e Steel are s ee band eggae ap Salsa And The but the pie chart it chart does shows ecause pie loos not quite charts lie show show Steel × 1 = ° × 1 = 1° × 1 = 1° × 1 = ° band Reggae Rap this the clearl Salsa ale actual that rap fractions, quantities, was we the can most use popular. fractions to solve problems. Eam orE EamLE ti • efo re 3 chart A pie chart ‘mango’ shows as their that in a favourite surve fruit of were people, those represented b who an chose angle of angl es how man people chose mango. • f mango sector is of the e sure pro tra ctor your Sl The add lled circle. a the number of people choosing mango is of up to pie the . you ° . use your co rrec tly. pie you So the sure °. • Mak Calculate draw ing mak e chart hae mista ke. is no t mad e hec k your calcu latio ns = × check your × then measu ring. 2 = and 1 = people. actiit Summar uEStionS • n a pie chart coconut chose is showing °. f favourite people ice too cream, part in the the angle surve, for how Conduct friends man a to favourite surve nd tpe of out of 2 their music. coconut 2 n what 3 n this wa is a histogram pie of chart test different results, from how all man the other people too charts the • ut the results in a pie • ut the results in a bar • hat are the chart. chart. advantages of a advantages of a test bar chart 8 • hat pie are the chart 6 ycneuqerF 4 E ointS 2 20 istograms data, 0 30 40 50 60 70 80 90 and continuous are no gaps 100 between T est show there the bars. result 2 ie charts rather show than proportions frequencies. 2.2 meses el ede An LEarning of • ee is a single piece of information used to represent a set outcomES etermine the mode, data. There are three commonl used averages. median mde and mean from ungrouped raw data frequenc and tables The A de group ere of are the common counted the piece of number data. of letters in their rst names. results. mode most friends the The is is , as it occurs more than an other number. me The e is orE n a the librar, the ividing The So mean the number b as of could of boos be data equall. on each of ve shelves is boos there the of number boos sharing number total of EamLE The result are is ve boos + + shelves, on arranged a 2 shelf with is + ÷ = + = 2. . . on each sum shelf. of data The mean is calculated as mean = number of items of data med The orE acob The EamLE records results ut in or a the are order ed is the middle piece of data when put in order. 2 number of oranges on each branch on an orange n + tree. 1 list of n numbers, the middle one is in position 2 11 + 1 There are 11 numbers in the list, so the middle one is , or the th number. The median is . 2 f there or the is an data even , , number , , + 1, of pieces 1, there of data, are there is numbers, no so middle the item. middle 1 one is in position = . 2 This is mean halfwa of between those + = 2 . the rd and th numbers, so we nd the aees orE ne • EamLE hundred ere are The the because • 3 children or this largest The les were ased how man das the had been absent from school in the last month. results. de the ee the is dl the most frequenc ed is the e common of of das number, is as ne , it has ds se ee 1 2 2 . middle of 1 the + 1 so we must nd 1 the person in position = . 2 This shows middle that pair, in there is not positions a person and in the middle, but a 1. TTA The rst The • net , so or the students have both the e, an have absence th we an and must absence rate 1st add of rate 1. student together f of we has put an these absence students rate values. mae a There are The t is 2 There total of are have quicest to eros, 1. in line, So the the will median occup is positions which a sum add a total of column to 1. ds se ee Sl which 1 2 2 . ones, twos of all ne 1 1 . . and to so the on. × = × 1 = 2 × 2 = × = 1 × = 12 table So sum of data 1 ean = = number of items of = data 1. 1 TTA 1 1 actiit • Si • Si natural mode of numbers . natural mean of ind have the si numbers . ind the a mean of , a median of and a a of and a numbers. have si a median of , mode numbers. Eam • E he ti mode high est The de is the st the common. no t the The eian is the idle value. • sum of freque ncy alue high est 2 is ointS When that has it is the freque ncy. usin g a data 3 The A freque ncy is number of items of middl e uEStionS of the • Whe n The mean total height height of of the four four students is 12 cm. Calculate the Three numbers have a students. mean of is and a mode of . hat the no t row usin g three ive people and a at a mode -ear-old hat alue a of diide the are no t the numbers num ber 3 the item . table sum freque ncies the the but middl e freque ncy by 2 the data media n Summar table bus of oins happens the to stop . A have a mean -ear-old age leaves of , the a median queue, and of rows . of a queue. the mean, median and mode 2.3 LEarning ged d ged les Sometimes • etermine the modal class and the mean frequenc these cases, from nstead, we The will helpline ‘n not have the nd the the dl group as ed raw data, lss the containing how to estimate motorists This on information a is a so most the median stretch used ee 1– 2 – 1– – 1 1– 12 d we cannot calculate common middle group, value, and an – in of in unit road the 2.1. are recorded discussions in the below. f compan average telephone we within mdl 1 of below. actiit the do lss discover Seed that us e speeds table answer we can ed e states to grouped tables esed computer given average. the A is estimate an of information class, n median ee outcomES lss seconds’. The ere are the particular data for So da. it te t e se 1– modal class or modal group is the one which is most common. a the is modal class is the one with the largest frequenc. n this case – mh. ee f med The + lss median speed of motorists is the speed of one in position 1 –1 = 2., so the median is the speed of the 2th and 2th 2 motorists. 11–2 21 21– 11 1– 1–12 1 There There 11 are are s their • plain claim up to had a speed altogether speeds of with between a 1 1 − speed and m h. between m h, 1 and and m h. 2 had m h. the 2th and 2th motorists are in the − m h group. answer. The actual − 2 motorists with ustied So our motorists motorists speeds • 2 speeds m h is are the unnown median as class the as it data are contains grouped. the median drivers. Ese e cannot nstead T o do we this midwa e nd the nd an we unit Seed eact Then for mean estimate assume speed. previous e that the nding of as we the everone mean ee not have eact data. mean. in calculation the do each is ust from f a group the drives same frequenc mdw as in at the the table. le m f × m 1– 2 1 – 1 1– Eam • he – 1 12 that 2 middl e TTA •n = = onl motorists, As there middle liel estimate instead are be a and as using liel value to E an to be some good we the have middle some not used value motorists travelling more of the actual each travelling slowl, modal speeds group faster the in the than estimated T o nd contains group is median the is ate of the calcu lated no t table. the mean is Summar the the gues sed. of the most common group. Twent the 2 the estimate. ointS The no t the class . estim is is .2 mh. mean This class alu e e the contai ns esed is high est middl e The class the media n class – with freque ncy. 122 modal class • he 1– ti class, median of nd all which the uEStionS friends sent in a compared da. ere the are number the of tets results. class data. ne 3 T o estimate the mean, assume all items in es 1– –1 11–1 1–2 21–2 2 1 the se table and have then the use middle the value for the rule sum of group, ee data mean = number of items of ind the modal 2 Calculate 3 ind class. data the an estimate median of the mean. class. 3 2. Leels LEarning nderstand the levels The into etermine best to use which in a average given are different levels of measurement. of measurement • esee outcomES There • ee l sle is the most basic and means ust putting data categories. is situation amples There ou The is no could 1 might onl order mode is both thin of and a ice red of cream could such or colours come of before doors. or after blue. data. numerical friend nicest 1 slightl each a sle ou from might An nd of but the numbers do not indicate intervals. magine cola avours obvious dl equal ou include to choose is much better. are both least the The to ran different brands of good. same better, ased colas whereas “distance” for positions our friend between 1 1 and might and 2 is 2. ut thin it different ou is for ou. ordinal scale is ordered but the distances between positions is unequal. ou f could instead to ou A ou ver intervals. nd This is is nd sle that ength are median ased satised, could point the called the is ratio to to median an absence scale, to as cm such cola split data. from the 1 most points dissatised into equal sle. and interval of of each tr el mean, an mode score would an similar means a ou and mode scale of but interval has a data. true ero point, a measure. means no length. o T emperature is not a ratio scale, as does not mean no temperature. o The ou temperature could The nd can the of an below mean, ee purpose go s median and mode of such data. es average is to represent a set of data with a single gure. • The we e now calculate is the the the sum. value, it will unaffected. be will onl mean average and the owever, inuence the that uses number if there mean of is ever pieces an whereas piece of outlier the of data. data, an we f can etreme median and mode t • is best The for ed from and large the larger • can as ma can The is median used does be for is nd a for scales from set of a not be if list data eample scales there or are , , no frequenc consisted , , representative onl ecept average depend on mode, or for below all the that adding none representative used table f will all the one be to ratio of , as table, small , it outliers. , will but not numbers , reect , the groups. not than not eas data. and numbers, the mode it more t of be The ordinal grouped then t the of can or at the nominal be scale. used ordering. all, and if for ut the qualitative there sample data, might is be small it data. scales. summarises the strengths and weanesses of each average. Ses me ses Can all the be med data. used calculate the eesses the nuenced to or sum Sles b etreme outliers values. nterval atio of data. as to Alwas calculate. oes central. values not tae into all rdinal account. nterval atio actiit mde Can be used qualitative for ight data. not for data, a one at ere all be sample. more mode, or rdinal esign nterval wages atio Tr than none the friends rst question with 1 carried questions hat include each out a each surve one about E mathematics. one scale our was umber favourite nglish ased our percentage in the last maths The nominal qualitative these and subects as our in least order of preference, 2 favourite The aths istor ordinal ordering not question to 1, where or each 1 ow means much “not do at ou all” 1 maths on means “ a scale love of 1 ordinal, state whether interval or the scale but hich average mean, the equal. t is be mode. an intervals can answers with be mode or it.” collected will be The interval scale has equal intervals. ratio. 2 the usuall onl median. 3 question, b is can Art eno and scale and summarised nominal, question of ointS are ames’ about factor. summarised question as a measurement. eamination arcus’ questionnaire all. Sall’ s to using a in uEStionS school are of particularl small ight Three ominal representative the Summar be median, mode would ou use The ratio scale has a true to ero. summarise the answers given to each question 2. meses meses LEarning nderstand dispersion range, measures range, interquartile semi-interquartile etermine There are measures for raw means how spread a number of out is a set of data. was and ichael both of measuring dispersion. range pla cricet. of atric’ s dispersion dses of atric • dses outcomES ses • scores over ve innings are data , , , ichael’ s , 21, The a both their scores and scores , median ut 1 over and have of a mean are more ve innings are . and scores are . a score mode quite spread of of , . different ichael’ s out. re The e between ooing This of the at a set the data value eample or atric, or ichael, shows of largest that the is difference the smallest. above range the the and is range ichael’ s is − scores − are = . = . much more spread out than atric’ s. The problem values, the orE A She the range and the EamLE 1-metre 11. with largest sprinter seconds, tripped 12. over in but is that it taes into account two has times of 12.2 seconds, 1. one which it onl smallest. race 1. seconds, affects the 1. − 11. = . seconds. 12. − 11. = . seconds. seconds, seconds and accounts range ithout for hugel. the 12. 12.1 the er slow seconds, seconds. slow range race, her time of is range is iele e • The median is • The lwe actiit the middle value of a set of data. Tulane le cuts off the bottom quarter, and so for ere data sets it is the n + value one quarter of the wa from the are 2 her maths tests. scores smallest, 1 in too small Se position ee • The e le cuts off the n + top 2 of data, and so for 2 1 2 2 1 1 small 1 data sets it is in position 11 + 1 or eample, if there are 11 items, the median is the = th 2 item. 11 + 1 The lower quartile will be in position = . 11 + 1 The upper quartile will be in position , or the th item. • The ele upper quartile middle e and of the the measures lower data, the quartile. and so is difference t not shows affected between the range b the of the outliers. • ind her range te seele median and score, the the interquartile the quar tiles e range. The seele suall this quartile E to gives the a e good median, is half estimate and the of of the median the interquartile distance to the from upper range. the lower quartile. ointS Eam The range = largest data item − smallest data • 2 The median is the middle data item. o nd data in 3 The lower from the quartile is the data item one quarter of the The wa • lwa ys smallest. upper quartile is the data item one quarter of the wa as from the mus The gie sing le largest. interquartile range = upper quartile − lower arrang ed the rang e num it rang e bers migh t fo r be quartile. no t interquartile be inter quar tile eam ple t order . and ti item. to . range The semi-interquartile range = 2 Summar ilar uEStionS maes uring a badges 1-da de ne sld sell period, ne to records the number she maes and the number she sells 2 3 8 2 3 2 2 1 2 2 1 1 1 2 2 1 1 21 1 1 1 21 12 1 2 11 1 2 2 2 1 2 2 22 ind the range 2 ind the interquartile 3 hich of she tourists. set to for data each is set range the of data. for more each set spread of data. out 2. cle er LEarning Construct a frequenc table grouped, the cumulative median graphs. large • we have Construct a frequenc graph n we and se a to nd quartiles, range, • a given with a large the range, the value frequenc A median, b quartiles using and the amount mean. of e data. cumulative medians tend can f also the data frequenc to be estimate tables more and useful for sets. from or a cumulative the interquartile ee les sum of frequenc all table frequencies to shows a Le ee given 11– 2 above ere cumulative is pencils a table in an showing the length 1– 1– 1– 1 1–11 11 111–1 11–1 2 of ofce. graph There long, 12 point. range percentage table Le quartiles estimate frequenc semi-interquartile Calculate deal onl cumulative cumulative graph can realit, data cle • to outcomES are • often, ee ee are ve up to two up pencils to up mm mm long. to long eees, be the to m and These le added mm are and can table le e d cle ee 11– 2 . cm 2 1– . cm 1– . cm 12 1– 1 . cm 2 1–11 11 11. cm 111–1 1. cm 2 11–1 2 1. cm cle ee s The 50 last used to two columns draw a of the cumulative table can frequenc be graph. 45 40 e now ycneuqe rf than 35 30 ecause evitalumuC nd 25 the that there 1. mm, the the so sample quartiles pencils b were that is and is no our pencils starting reasonabl median b less point. large, we dividing 20 15 ÷ the = length 11, of so the the lower 11th quartile is pencil. 10 The 5 11 0 0 10 20 30 40 50 60 70 80 90 the upper quartile (mm) pencil 8 is length of the 22nd pencil 2. 100 110 120 130 140 150 The Length median × 11 × . is the length of the rd These values • The median • The the lower can is be read from the red lines on the graph Eam mm quartile interquartile is mm range is and 11 − the upper = quartile is 11 mm, so • he ti cum ulati e table a co lum n fo r maim • The semi-interquartile range is = 1. mm. alu e 2 The graph can also be used to obtain other and eample, to nd the percentage of pencils that are at mm So to that freque ncies alu e. long • The freque ncy all least up blue line there are shows − that 1 = there 2 are 1 pencils pencils mm shorter or than mm long. um ulati e grap hs do longer. freque ncy no t necess arily the origin start 2 The um the cum ulati e information. inclu des or has mm. percentage is × 1 = .1 to 1 decimal from . place. actiit The cumulative blue and girls frequenc in red in graph a shows the heights of bos in school. 70 males 60 females ycneuqe rf 50 40 evitalumuC E ointS cumulative 30 A table shows frequenc the sum of the 20 frequencies up to a given point. 10 2 A cumulative graph alwas left right. frequenc ascends from 0 110 120 130 140 150 Height 160 170 180 10 (cm) 3 • The graphs are different shapes. hat does this difference The tell be ou Summar Andrew did a measured The results raw a median read and from a quartiles can cumulative frequenc graph. seweed uEStionS surve on the beach. Le e to of the the lengths surve cumulative this information. 2 se our 3 hat graph to percentage of are pieces shown frequenc nd of the the in table lower seaweed of seaweed. the and and table. graph upper was over to ee 1–2 2 21– 22 1– 1 1– 1 show 1–1 11–12 121–1 1 quartiles. cm long 2. S Sdd LEarning use standard deviation range sets of spread mae inferences summar interquartile out the data range are both measures of dispersion, or are. data ach • and to how compare de outcomES The • sss has a weaness. from statistics The b range onl uses the largest and smallest values, and so is affected outliers. The interquartile gives no ames range indication and enr of onl how both uses the spread too 11 middle the tests. top ere and are of the bottom their data, 2 scores, and so are. put in order totaL es 11 1 1 1 1 1 1 1 1 1 2 1 e 1 1 1 1 1 1 1 2 2 1 The range This ut for suggests ames’ suggests ust as the d a oth The sets table EamLE of data below 2 – enr’ s ames’ is an = scores is scores average measure eed 11 range of e e. desd orE that mean is that is interquartile This deviation ames . are te more 1 1 are that uses that . is 2 = 1. enr’ s is 1 1 = all the uses data, ever e elw the piece standard of sdd s s data. de el . have a shows total how of 1, much and each so the score is mean above is 1 + or ÷ 11 below = 1. the mean. totaL es e f we So add we 2 these square up, these we 2 2 get scores . to 1 This will mae be them the case +1 +1 + + + +1 +1 + + + + for an set of data. positive. totaL . varied. lle ele range varied. = more dispersion le enr’ s es 2 1 1 1 1 e 1 1 1 1 1 12 The mean ames = inall, squared tae ÷ 11 the difference = for each These The √ = . square gures steps root of the mean are 2. the squared 12 ÷ 11 = 1. enr sdd = √ 1 = difference . de. were ind • Square • ind the mean • ind the square • This tells us = = • the than is enr ames student . difference those us between each piece of the data and mean. differences. of those root that, of squares. the we mean ll of esls the e squares e of the differences. sde, enr’ s scores s comparing sets of data, the central measures mean, mode give measures deviation an overall impression of which set is ointS The standard of give dispersion an range, indication of interquartile which set is range more orE EamLE and standard sues table group a wee. shows of 2 data males of ever about and 2 the number females of spent hours per watching wee that television ele ean 1. 1.2 edian 1. 1 ode 1 1 ange 1 averages and mode 3 in data The is range deviation which spread on as it median average is a than small median which set . . 1. and was 12. both the suggest males. that The the mode females sample it is less reliable suggests than the of higher. dispersion range deviation set of data is tell more out. each. was uEStionS Sarah too obert’ s and his mean standard maths mean standard score was deviation score deviation 2 was and 1. watch otherwise, but ou interquartile ou her more tell standard Sarah’ s and data. generall and tests mean of mean, measures range, obert The piece The Summar Standard is that 2 mle nterquartile dispersion varied. of a deviation higher. 2 The out median measure The spread sss and more ames’. E hen are mean ho do ou thin ow can did better and overall ou tell median. 2 The males have a higher range, interquartile range and hose scores consistent deviation. whereas This the means females’ that the gures male will be results closer are more were more standard spread closer together out ow can ho do ou tell together. 3 highest ou ou thin single test had the score Can tell 2.8 l l LEarning • is the measure of how liel an e is. outcomES nderstand the probabilit • An ssle • Something • All ee that is has e a is probabilit given a of . probabilit of 1. scale • etermine other probabilities lie between and 1. eperimental probabilities of robabilities can l sle be represented b maring the position on a events Impossible Unlikely Even chance actiit Likely Certain 1 • ut • oll a counter in a STAT square. aving a to etting a Flipping birthdays STAT STAT STAT STAT STAT STAT 1 2 year this hen is dice impossible as rolling is are a and unlikely there one is only and ive on it Eeel hen f ou throw a 1, go a square. f ou throw a 2 or a is ou not ip eas even there as other ◦ f T ou go a something throw T Carr on a a , or arbara , off today is likely are and equally likely l to drawing sa pin, which is it might more land point up or point down. liel. EamLE performed an eperiment. square. until our the ipped a drawing pin 1 times. counter t comes very chance, heads tails an eat square. She ◦ a is to , orE go aving t ◦ coin landing heads numbers ◦ a dice. landed point up times and point down 2 times. grid. ◦ f it goes T off ou to the T So or it landed point up times out of 1, or 1 lose. e sa the eperimental probabilit of the pin landing point up ◦ f it goes off the TT, is 1 ou win. hen • eep plaing different • hich the start start game, using nding perform the an eperimental eperiment a probabilit, large number it of is ver important to times. squares. square gives robabilities can be so sa the written as fractions, decimals or percentages, ou we could probabilit in the eperiment above was , 1 the best chance of winning . • Compare our results or . with others. te e l A probabilit of means that the outcome is unliel, as it 1 Eam ti 1 is less than t means 2 • he eperim ental um ber of probab ility to times the = o tal een t of up average, times we out would of epect ever the drawing pin 1. is important to understand that this is merel an epectation trials in 2 point on occurs t num ber land that, the real world, the unepected happens occasionall . us l robabilit help us can be calculate used the to sa number umber how of of liel times times something an the event event will is. t can also occur. occurs robabilit = T otal number of trials So, robabilit × T otal = orE oll T o win, the she to win had to of of trials times the event occurs 2 a cuddl hoo a to at plastic the duc fair. with a winning number on bottom. She watched There a umber EamLE wanted number were winning people 1 plaing. plastic ut ducs. of oll people, wanted to won now a cuddl how man to. had number. She decided that the eperimental probabilit of winning was = 2 So of the ducs might be winners. 2 of 1 = 1. 2 So she thins Summar sobel about 1 winning ducs. uEStionS made coctail are a spinner E out of card and 3 stic. ointS robabilit as a can fraction, be written decimal or 4 a there stic She not quite get the through the centre. recorded the results coctail of spins percentage. 2 5 did 2 She robabilit is alwas between 1 Se 1 2 1 1 1 11 robabilit Calculate can eperiment 2 trials and 1 certain. 3 ee impossible the eperimental probabilit that she scores a the eperimental probabilit that she scores a is if be the found b number sufcientl of large. one. 2 Calculate arsha She ow 3 wors tests a man eorge colour these are has eperiment and a factor of liel bag then , to maing and wor containing he puts taes it a bac. light nds out of some bulbs. that a do bo of counters. counter, maes e this does not e a 1 wor. bulbs tries note times an of its with results ee there are a where cl f in sample ve. are there 1 liel ed lue lac hite 2 1 11 counters to in the bag, how man of each colour be 3 2. l Ell LEarning lel etermine the sample simple eperiments set possible etermine faced with a number of events which are all eample, when we ip a coin, it is ust as liel to land on heads outcomes as • are liel. of or all we space equall for es outcomES Sometimes • e tails. theoretical hen probabilities of we roll a dice, all si outcomes 1, 2, , , and are equall events liel. • nderstand that the sum hen of the probabilities possibilities must of all outcomes are equall liel, the probabilit of an event is all equal number 1 of successful outcomes total number of possible outcomes 1 The probabilit of a coin landing on heads is , as there are two 2 possible A dice outcomes, has si heads possible and tails. outcomes, 1, 2, , , and . 1 So the probabilit of it landing on is 2 The probabilit of it landing on a number greater than is 1 = Sle hen se answering possible ds probabilit outcomes. This questions, helps us to it nd is often the useful number of to list all equall the liel outcomes. magine The possible outcomes e are might tails ¢ ¢ ipping or thin one of a 1¢ there each, coin are so and three the a ¢ coin. possible probabilit outcomes of both both coins heads, landing on both heads 1 is . owever, this is incorrect. h 1 The probabilit of both coins The probabilit of getting landing on heads is t 2 T can be head and a tail is 1 = h T This a A listing A 2-wa of all possible outcomes is called a 2 sle se shown in a 2-wa ¢ or h t t Th Tt table is a quic and accurate wa of constructing a sample table space diagram. eample, here is a le ¢ 2 3 sample space diagram for rolling two de ed 2 3 der The green squares show the was of scoring The probabilit of scoring a total of is a 1 = d t total of . dice te s les actiit The sum of the probabilities of all the possible outcomes of an event A ve-sided spinner and a dice 1. are thrown. multiplied orE ith this EamLE spinner, The scores are together. the probabilit 4 of 5 1 it landing on red is The probabilit of it landing on 3 1 is blue 2 1 is 2 The probabilit 1 or blue is of it landing on • red 1 + hich liel = probabilit of it landing 1 blue or ellow 1 is + will add land up on orE to 1 red, = because or EamLE Charles and each of these it is certain it ellow. 2 avid are having a ti race. • that scores 1. Rem embe r 1 probabilit most probabilit red, The are the 2 blue on Eam en, is 1 + The scores 2 of The two hat en wins how to add is fractio ns 1 by ndi ng a 1 The probabilit that Charles wins comm on is deno minat or . 2 So the 1 probabilit 1 + 1 The 1 + 1 sum of = the or Charles wins is 1 all en = 2 that = 1 probabilities is 1. So, the probabilit that avid wins is 1 − 2 = E ointS hen all Summar the probabilit outcomes of an event are equall liel, the T eri of successful rolls a dice and raw a sample ips a coin. is number uEStionS space diagram to show the outcomes possible total number of possible outcomes 2 2 A sample space diagram outcomes. is a list of all hat is the probabilit of getting an even possible number on the dice and a head on the coin outcomes. 3 3 The sum of the probabilities of all T eri uses green outcomes is either a red dice, a blue dice or a possible dice. The probabilit that she uses a red 1. dice is blue she ., dice uses and is a the .. green probabilit hat is the that she uses probabilit a that dice 2.2 c te LEarning ll se contingenc se the addition rule not inuence se the events other where the outcome of one event event. rolls a fair -sided dice and ips a fair coin. These are events independent • are the for Sophie eclusive ees tables does • le outcomES ideede • les multiplication rule as the coin has an even chance of landing on for heads independent events or tails regardless of the score on the dice. events The sample space diagram shows all possible Se Coin 2 outcomes de 3 eds 1 eads 2 eads eads eads eads eads t ls 1 T ails 2 T ails T ails T ails T ails T ails outcome There are 12 possible equall liel outcomes. 1 The probabilit of her rolling a and ipping a tail = 12 The T o multiplication nd the multipl = probabilit the robabilit of 1 = × sas of probabilities rolling probabilit 1 rule of a rolling 2 or of and a more each independent ipping × events occurring, event. a tail probabilit of ipping a tail 1 = 2 12 orE EamLE 1 The probabilit that rian is late for school tomorrow is 1 1 The probabilit that Sarah is late for school tomorrow is 1 oth The = events are probabilit probabilit independent. that that rian rian is and on Sarah time are both on probabilit time that Sarah is on time 1 1 = ( te 1− 1 same hen a × ( at ) 1 = 12 1 × 21 = 1 1 = 1 2 le else ees are events that cannot both happen at time. Sophie the 1− dd mll the ) rolls same her time. dice The and are ips her mutuall coin, she eclusive. cannot roll a and She can roll mutuall rian T o arriving mutuall The a the occurring, atricia The rule ip for a head school as the at the same time. The are not is table the Sarah might arriving be either of 2 probabilities of each of EamLE maing and and both late for school are not late sas probabilit add orE beads late eclusive addition nd and eclusive. blac shows a eclusive events event. 2 neclace beads. the mutuall She using red chooses probabilit of beads, each blue bead choosing a at beads, white random. red, blue or white bead. l red le e .2 . . l E ointS ultipl probabilities independent The probabilit of the net bead being white or blac of white + probabilit of of to nd = the probabilit events probabilit of both blac occurring. = . + 1 .2 . . = . + .1 = . 2 Add probabilities eclusive the ce A contingenc les table is a events probabilit of to of mutuall nd either occurring. table for showing results in two different classes. Summar orE A EamLE compan produces tests an a uEStionS 3 new drug improvement on in a 1 sin patients to see if it The table and right-handed shows the number students in of a left-handed class. condition. gede The results are shown below mle ele 1 1 gede dedess resl mle ele totaL 2 1 iee Le ded r ded n 2 2 2 ee Two students, selected totaL patient selected at random has a probabilit showing male 2 the probabilit selected are that both right-handed. selected at random has a probabilit hat is the student probabilit selected is that at least one left-handed of showing improvement. patient selected at random showed improvement. 1 probabilit that the patient was female is selected the are probabilit left-handed, that and both use students our = Calculate 1 The Calculate students 2 3 A are improvement. patient = gender, of each 12 = 2 A of of 1 of one random. 1 A at . answer eactl to calculate one student the probabilit selected is that left-handed. mdle The to island 2 below is e drawn to a scale of 1 cm e ais ccles ess 12 miles in 2 hours 1 minutes. m. oughl how man ilometres is 2 stimate the area of the island in m 12 miles ive an speed elissa of approimate in value for ais’ s mh. travels m at an average speed mh. Calculate the time it too in hours and minutes. 8 A water to the The butt contains nearest water minute 1 runs to the 2 gallons of water gallons. out at a nearest rate of gallons per integer. Calculate 2 A trapeium and a triangle have the the maimum the minimum it same The trapeium cm, The and The a has the diagram quarters 1 cm, of and has parallel height triangle Calculate 3 will butt area. a a of of shows circle two of cm the of time for the empt. and cm. height base sides tae to time 12 cm. atch the data descriptions with the data tpes triangle. 1 Shoe sie A ualitative data 2 Shoe colour iscrete oot C Continuous data three- of radius length data radii. Calculate the area the perimeter answers to 1 of the shape, decimal giving ll in a questionnaire. What our place. p to is your ne question sas heigh t cm cm– cm A cone has a radius of cm and a height of cm– cm cm. Calculate its volume. cm– cm A clinder of has a radius of cm and a er height cm 1 cm. Calculate the volume the surface area. height hich hat is 1. cm. categor are the categor 8 do class select boundaries for this The in bar the chart shows families in a the number school of children hich Cop is the modal group class. the table and use the two 12 additional estimate 10 columns for hamsters in the to calculate mean weight ronwen’ s an of the shop. ycneuqerF 8 raw a cumulative frequenc graph to 6 show the weights of the hamsters. 4 d ind the interquartile range of the weights. 2 0 1 2 3 Number Show this information 4 of in 5 children a pie roll chart. a -sided raw a sample possible 2 ichael draws a pie chart to show the age in his out are of hat a people total angle a space -sided diagram dice. to show all outcomes. hat is the probabilit that the two dice street. show There and of people dice of in should the age group the same score – residents. he use for this categor A bag contains red, blue and green counters. 3 ive friends ne friend have a moves mean awa. age The of 1 mean ears. age of The probabilit of choosing a red counter at 1 the random is , and the probabilit of choosing remaining four is 2 1. a blue one is hat is the age of the friend who moved hat is the probabilit of choosing a awa green ronwen The of table owns hamsters e w ses a gives in pet shop. information ronwen’ s about the f the one bag either weights are shop. contains red there or of 1 green, each counters how man that are counters colour ne ses rissa 12 mh. e 2– 1–2 – – 2 ccles At 8 A does what to The the has at ourne return speed cuboid wor an average taes ourne does he him in ccle dimensions of 1 2 speed of minutes. minutes. home cm, cm and 1 cm. ind the same dimensions surface area of as a the cube which has the cuboid. 3 Algebra and functions and Why us symbols symbols and letters are numbers, of umbs? times when we use symbols or letters in place of numbers. it is used as a short hand when we do not know what a operations number • istad umbs to Sometimes represent and fo e There se graphs ymbols LeArnn • relations, is. ariables Translate expressed words For statements algebraically and ice example, into I ersa think Then If we use a much I the + etters a like away − to 4 need before ‘n’ n I add the three. number stand for I I double rst the the thought number, answer . of. we The I subtract answer could write 4. is . this in way: × we out say: number . letter emember it of might take shorter n work we in − to n = put . brackets multiplying algebra by around the n + to show we . represent specic numbers and are called ukows n other matter For occasions, what the example, b use number the × we area of a letter or symbol when it does not is. a triangle, A, is gien by h A = , where The b is the formula etters like vaiabls, is ‘b’ write letters or algebra T o add • T o subtract a aoid and a and of T o • hen whateer ‘h’ in iision h the b we and h is alues the of b represent is allows from way write , multiplication a by b multiplying rst: is m we we a not us or + b write signs a b is write letter multiplied usually written a written as b 80 alue same __ ÷ the algebra their symbols in and multiply written a base xed perpendicular and h happen numbers, and can and height. to are be. called ary. to be we or b − concise. will + be ut we all need a h between letters and between a number as ab and by a 4 a number, is to misunderstood. letter. • • the otatio • e true because Algbaic sing length written fraction the as number 4m, not is always m4. rackets are used to show an operation that must be carried out rst. A • a − b multiply means by subtract b from a rst as it is in brackets, and then The . names gie ote that the next to the bracket means multiply, ust as in to and symbols numbers are we arbitrary, a although for higher there system. numbers In an expression like ab , we carry out the suare before multiplying. So is a • ab means b × b × The a dix French huit’, call which ‘uatre translates ingt to but ‘four twenties, ten and eight’. • ab means a × b × a × b ere Wods obert’ s older to arlene obert. arcia is twice arcia is three younger e do as old times know is Their daughter, not can use the in numbers from rawak. . ba . ian . abun 4. iti . badakabo . batian . ianitian . abunitian . istia years mother as as two arlene. old as her iriam. their ages, but show the • we are algba sister than to algebra to hat the . eidence rawak is ianidakabo there people that counted connections. in If obert arlene arcia is is is r r years + r + r years + old, es rawak old years as old she as is her two age years is akabo word for is an hand. older. twice arlene’ s. iriam is years old, one third of arcia’ s age. Algba to wods x − 4 The expression − looks a little daunting at rst, but it is eA easier to understand if we use IS, which we met in • rackets rst: Start Indices: there iision and ddition e will meet There Subtraction: this Ar ultiplication: and again in with are x and subtract 4. ultiply x − 4 from n by the t many o’clock there ing and diide by . .. coconuts coconuts BIDM AS answer. e on a tree. nother tree has two more coconuts. ow use perfo rm calcu latio ns. en are Alwa ys when none. Subtract .. are were there c on cars etween o’clock and etween o’clock and in the a second car o’clock, tree park. more cars arried. n etters and stand for or any for ab symbols unknown means number a × can numbers, ariable. b a __ ow many cars were in o’clock, the car half park at the cars left. means a ÷ b b o’clock 4 Find the alue of + × 8 ictd umbs ad substitutio ngativ LeArnn any • erform arithmetic children directed Substitute algebraic th umb li numbers symbols learn to add and subtract on a number line. add 3 + 4: start at 3, numbers move • ad operations T o inoling umbs e for 5 − 4 4: move in to the start 4 to at right 5, the left. expressions 0 For uestions the number 5 4 4 like line 3 − − , the 2 = Additio 4 to 1 with left, 1 2 4 negatie to 0 3 2 6 7 answers, introduce 1 5 we numbers 3 4 8 9 need below 5 10 to extend ero. 6 7 8 − ad subtactio with gativ umbs A ll • In the walls adacent below, numbers add to two addition line, number in the and like subtraction the ones uestions can be soled on a number aboe. calculate For the ust brick example: −4 + aboe Start them. at 5 1 3 the 4 part −4 2 1 3 oweer, when negatie, For rst then the you example: 0 − moe 1 second must and 2 in to the 3 number moe to the right 4 be +. 5 added opposite 6 or 7 8 subtracted is direction. −4 4 Start 6 2 at , face left −. negatie 4 9 2 5 −4, 4 moe 3 2 ow time try again, multiply numbers number 4 to as 1 the the number 0 right to be subtracted is 1 instead. 2 3 4 5 6 7 8 3 ultilicatio • ut 5 to in but two adacent calculate the ad divisio with gativ umbs this brick e the aboe know If one × that: 4 = number is negatie, the ÷ = 4 answer is × × × − = negatie: them. − If two three = × −4 are = × − negatie, are × ÷ the − negatie, − = − = −4 answer ÷ the − = is = × − − positie: 4 answer × is − × = negatie: − general: if there is ositiv the 8 − numbers − In 4 numbers − If × an v and answer is if number there is gativ an of negatie odd numbers number of the negatie answer is numbers ubstitutio ubstitutio number, and is where calculate we the replace letters in an expression with a result. a − 2b _______ For example, to nd the alue of if a = , b = − and c c = −4, a the − alue 2b of _______ − × − = eA c −4 − − • Rem embe r = to fo llo w −4 BIDM AS. = −4 = − A ork through this mae. • ou can only moe horiontally • ou can only moe onto a or suare ertically. with a alue of when a = , b = − and c = −. Ar − c b ______ a − c a − b bb a + c a ac + b b − ac a + c b − c c + c a − + ab bc − b a + c c __ a + a + bc a − b c + b a b + c + ac − c a c ______ c − a ab − + a abc c 4a + c a + b a + b c − b − b b a + − a n b _______ b e b ab − c add or number b bc n T o a + + Ar subtract line, directed numbers, remembering to use if the last number is − a c b − a en alculate: moe a backwards c 4 + − b − − negatie. 4 c − − − d T o multiply calculate then if or the there diide directed numerical is an een answer number rst, there the is the an answer odd answer will remains number be of − of −4 and If a = − and positie negatie but IS when numbers ab b hich of whateer negatie. calculating and −, ealuate: c ab a b d ab if these the se = n b negatie a numbers × numbers, can alue neer of −n be negatie, n n substituting. 8 ombiig ollctig LeArnn erform four basic algebraic se laws of are terms containing identical indices • emoe x, x and x are like terms. So are alculate 4ab, −ab ut a the corresponding and a and ba are not, neither are using letters are not ab and the a b because the powers of same. the law Additio • combinations expressions brackets distributie or to manipulate ariables ariables. expressions So • tms operations of with tms e Lik • lik ssios with ad subtactio algebraic nly like terms can be added or subtracted, in much the same way fractions only fractions So a + with a b − terms a × = ab can b = denominator b can be added b − because a ultilicatio ny same or subtracted. a − a = the be + ab ad b = is a the − same a − as ab b + ab = a + ab divisio multiplied or diided. ab 4a × ab = 4 × a × 4a ÷ ab There of are e 4 = raise 4 we × × a multiply = a rules × a when powers a × a × b = a a of using the indices: × a we a another 4 × a = × a same × a number , × a = we add the indices. a power, multiply the powers. a b × n 4 n heneer a = to 4 general: = × a b n diide powers of the same number, we subtract indices. d d × d d 4 general: ny n ut, ÷ n ÷ the = rule = to a for × d 4 d power is . n = n = the + − = a a of diided by itself is . indices = a n same positie = = n number subtraction and general: any a 8 d b the power × In × n as n a a n negatie a d b ÷ raised = using n d × × d = n number d = a In × _____________________ ÷ b a +b n power = × b n × a a In 4 × a = b important general: T o b = a In × idics heneer a a ab Laws × 4a a power = , so a are reciprocals. and a are reciprocals the sing these a rules × 4a allows b 6 ÷ =a × 4a us 4 a complex expressions: eA 4 ÷ simplify b b to a • =a 4 b ÷ a b Rem embe r to b the multip ly num erato r 5 and a 5 =a 5 b or deno minat or 5 by the b same rmovig a oth e a can inside − a b and = −b always the ai of a − hae brackets b by a + − b multiplied pair the a of term + a by − b = a − b . brackets by multiplying each term A outside. • For nt. backts been remoe amou atch the expressions on example the left with the simplied aa − b + c = a − ab + ac ersions on the right. Algbaic 4x factios − x y y e deal with algebraic fractions in exactly the same way as numeric y x + x ___ y × fractions. T o add y or x xy subtract: x a − y − x x − y c ____ ___ − 2bc ab y x ___ xy + First, nd a common denominator. y e can use abc, as bc × a = ab × c = x abc a c a × a − = bc c × c − ab bc × a × × a c abc xy y ___ = abc 4x y _____ multiply or y x ÷ c T o y abc − − − ____ x y c = ab 4x a y x diide: x b a ab ÷ hen diiding, turn the second fraction upside y 4x down c y x _____ _____ + 4cd y x b a 4cd ____ = × ultiply numerators, multiply denominators c ab • ake up your own matching bcd a exercise = nd ab like this one, and test simplify c it on a friend. d a = b c Ar e en n Simplify: T o multiply brackets a by term the by term a bracket, outside. multiply each term in the a − b b The laws of indices: n b × n b = ac n 4b a b a ab = n a a b ÷ n = a = b × 4 n a a+b n − + a a + d c c b n n a n = a n 8 iay e LeArnn hae erform that letters and symbols can be used to represent numbers. e e • seen oatios binary operations can also use symbols to represent oatios by ddition, subtraction, multiplication and diision are all biay substitution oatios e can The dene perimeter written a ∆ by the e a ∆ our of they own a operate binary rectangle on two numbers. operations. with sides of a cm and b cm could be as □ b = emember a because that height could a + to and write b nd then this a b b × as the area diided a binary of by a triangle we multiplied the base . operation, Δ, so that b a Δ b = , where b a is triangle = the has base length base cm and and b is the height perpendicular cm. For this height. triangle a = and . × a Δ b = Δ = = cm h ou associativ rst ere h ou we met are these going to notice ut alues b T o of a and operation check it is true b we this, for a is b, ab b = h The if the Δ be rules = sure, use then apply to our Δ we if a will which need Δ algebra it , b to = b rather be true Δ is = b Δ is numerical whateer if a a Δ b Δ c = a × ab a Δ b Δ c = c abc Δ c = = 4 bc a bc Δ b Δ c = a = Δ is associatie. abc a × = 4 for be all Δ b Δ examples. alues law associatie might true ba 8 is commutatie. operation operations. a than for it it associativ Δ binary suggests know ab So own = Δ to Δ laws law must and distibutiv choose. a So see commutatie we ad .. that commutativ. The in commutativ might commutativ c. of a If and h The idtity identity is , as it does not change the alue of a eA number: • Us e a algebr a Δ a = = whe th er are distibutiv associa tie distibutiv law inoles two operations. we dene the operation J as x J y = x + y, T o a J check Δ x 4 = × whether J y = a Δ Δ + is 4 = J a Δ a Δ x oer a Δ x J y = J, we need to nd + y × x + y = ay ax Δ x J a Δ y = out Δ x J y = a Δ x J ax + ay A ay = ax + + • ay Show a Δ y, so or Δ is by distributie a law is the rule we use to that J is not associatie = commutatie. distributie oer Two J a multiply a single operations ✪ a The is bers. • a num will law ax J So all this the if = a as y. a fo r or . distributie x e whe th er then true utati distrib utie show If de ter mine o pera tions law comm The to a h b = ab, b = a + are dened as and b − . term • bracket: Show that both ✪ and are ✪ and are associatie. ab + c = ab + ac • T o check e if need J to is distributie check if a J oer Δ: Δ y = = a x Show that both commutatie. a J x Δ a J • y. Find the identity for each operation. xy xy a J x Δ y = a J + • Is one operation distributie oer a + xa + the other y a J x Δ a J y = a + x Δ a + y = J is not distributie e n binary two oer Δ Ar operation is a method of combining numbers. These en uestions dened are about The associatie law: The commutatie The distributie The Identity, abc law: ab = = Ia The = a, for inerse aa′ = I. I, all of is the ab◦c number operations a ♣ b x ♥ y = a + b , and abc = x − y. ba law: the by = Find the alue a ♣ 4 b ♥ 4. of ab◦ac for which elements. a, a′, is the element for Show that ♣ is commutatie. Show that ♥ is not which associatie. 8 eadig ad factoisig rmovig LeArnn • pply the factorise distributie or expand expressions, law + b ai of backts reminder: to algebraic e learnt = in . that a − b means × a − b = a − b. e.g. oth xa a e xa + a and –b hae been multiplied by . xb, a + b x + = a + bx = ax lso y + a + by ay + by e + bx + aa can Simplify algebraic • Factorise b expand carefully • − and + c and then = a − simplify ab more + ac complex expressions algebraic simplifying: − 4 − ba − expressions ote nding a common = factor ab − a − ab + ewrite a uadratic expression = 4ab − a + the last term is positie, as b b • expanding expressions ab by by × − = +b b in the form ax completing + the h + k suare. actoisig actoisig we split sing and is the dig inerse factors, example factorise b the is into the to a it by hae a highest of a commo expanding which usually to facto factorise inoles an expression, brackets. aboe, a − b common common we must factor factor of of turn , as the it back a two = into × a a and − b. b = × b terms. So a eA all • factor ising you hae comm on ou can = × = a a − − b × b sing the distributie law in reerse ach term They also etra cted seein g if your to the has a as a factor common − of 4ab +a has three terms. a hae common factors of , and a, but a is the highest factor. chec factor ising epand s such ma e factor s. alwa ys expression a your b n • he n sure − − 4ab + a = a × = a4 4 − a × b + a × a by answ er origin al This is not as complicated as it − b + seems. a The sing numerical the part distributie of each epres sion. term, , 4 and , hae a common factor of , and the letter parts, a, ab and which 88 a can , be hae an a extracted in common, from each so a term. is the common factor law eadig Sometimes x two we + ssios hae x two − pairs of brackets to multiply, for eA example: • o epand remem ber ultiplication is distributie oer addition, + × x − = xx = x = x − epres sions so • x two I . + x − istributie law Rem embe r to simpl ify if possib le. − x + x + x − − xpanding The four terms come from multiplying Simplifying the A irst term uter in each bracket: terms: x × x = x × − x atch = −x = +x left nner terms: + × x the with expressions the on expansions the on the right. Last terms: + × − = − x + x − x x − x + 4 x − 4x − 4 − 4x omltig th sua ot all expressions will factorise. xx − 4 n expression in the form x ax + bx + c can always be written in the + x − + x − x x +x − 4 form ax + h + k This is useful when ompleting the soling suare uadratic uses the x + euations, = x + nx + n = you will see in unit x − 4x + . identity n as − x + nx + n b eplacing n with gies e b (x + ) b = x + bx + or x + bx = (x + + bx + c (x = b ) − ( b ) + − ( + ) n ) b x b , ) So ( se FI to expand two c expressions. Wre eALe ou can factorise extracting a by common factor. T o write x − x + in completed suare form, use the identity x + bx + c (x = + b = −, c = lways T o ( ) + complete x + = + ( b + bx − c = (x + ) ( + ) x x suare, − the use x fully. c . So factorise b ) + b ) + = x − − + b − ( ) + c = Ar x − − en xpand xa xpand and xpand a Factorise − 4x − simplify − 44a x + − 4y − x − y fully: a b + b b xy − y c − c 8 uth actoisig LeArnn Factorise including two algebraic two ais . we found out how to factorise difference e between also saw backts two by nding a common factor. that suares roe of expressions x • ito e In • factoisig + x − = x − x + x − = x + x − algebraic T o expressions to be factorise, we need to turn x + x − back into x + x − . identical The expression x • The x eA the great sign s. care ou multip lyin g but × x − contains three terms: term, x , is the product of the First terms in each bracket, x • • ae x + addi ng Wre to x term, terms, are to The with • The x +x, × − constant is the and term, sum + −, × is of products of the uter and Inner x the product of the ast terms, + × epand simpl ify. eALe T o factorise x − x − 4: x 4 can can only factorise factorise into 4 into × x or × x × , but one factor must be negatie as the constant is −4. So the possibilities products of the that uter yield and x Inner and −4 terms, are listed which T The 0 below. must make n the right of each possibility I + 4x − x × −= −x +4 × x = +4x −x x − 4x + x × += x −4 × x = −4x x x + x − 4 x × −4= −x + × x = +x −x x − x + 4 x × +4= x − × x = −x x x + x − x × −= −x + × x = +x −x x − x + x × += x − × x = −x x factorisation is x − 4x + as the outer the sum S x correct is −x and inner terms add up to + + + + 4x = +x −4x = −x x = −x −x = +x x = −4x −x = +4x + + −x of − h diffc btw two suas eA The expression aboe. This is 4x − because looks there different is no to term the in type x • Do no t confus e the factor ising oweer, it factorises in the same way, and is called difference between x two suares, because + 4x is = of with . and is • Som e tim es sua res It factorises a product to x + x − . The outer terms hae fo r The sum in in x = the of of the −x and these is original the , inner which is terms why multiply there factorises any to expression ax was +x. no = term no t eam ple + differe nce be tw een imm ediatel y two obiou s = . expression. enerally, to is a + bax of − the form a x − b b. A ovig two algbaic ssios a ual The expressions x − x − and lgebraic expressions can often be written in more than one x easiest way to show that they are eual is to simplify the expression until it resembles the simpler eALe that x + look ery but the factorisations uite different. of each Find the expression. Show one. factors Wre + more are complex x way. similar, The − + x − 4 = x − x + . olutio x + + x − 4 = x + x + + x − = x 4x − 4 + x + x + + 4x − x − x + = x − x + = e ou can factorise of between lways two two Factorise by pairs x + extracting of a brackets, common as the en fully: + b − 4 b x − c b c + − 4c + Show x fully: or difference b Factorise a the fully. a factor, including suares. factorise Ar − n product x that x − + x − x + = x + x − . hagig of LeArnn • hange a th subct fomula h subct ost formulae of a fomula e the subect of are written to enable us to calculate one specic a ariable. That ariable is called the subct of a fomula formula bh • hange the subect of For a example, the formula for the area of a triangle, A = , is written formula including roots and in such a way If, instead, that if we know b and h, we can calculate A powers • hange the subect of we formula where appears on the would • Sole word wanted need a to draw formula to a triangle calculate with the a gien height. e area and would base, change subect the both we a subect so that the formula showed how to calculate h sides problems hagig formula T o tells change inerse th us the subct the of a calculations subect, we undo fomula reuired the to work calculations out by a gien applying ariable. the operation. bh The formula A = tells us b that × h ÷ 2 = A A ___ sing inerses, we deduce that b ÷ h 2 A or = b h ote that your socks, T o it is Wre eALe the ‘undo’ and the a the socks. easier perform rearrange of we shoes to take same formula last operation ecause your changes by you put shoes using performing off rst. It’ s your a bit shoes before same formula. a area of a trapeium, A, is gien by the formula A + b = h, where T o a make e and b apply b the are the lengths subect: inerses in in the the of the parallel formula, reerse a A both sides by h: a A __ + + b h ÷ b a A ___ both sides by : + b = h A ___ Simplify: = a + b h A ___ Subtract a: − a = a + h A A ___ Simplify: ___ − h a = b, or b to = h ultiply added = h Simplify: is and order: __ iide b sides = − h a b − a h a, h is the then perpendicular diided by , height. then your to multiplied by h off your socks. terminology, operation The taking after remoing mathematical the like on both we sides Wre The to time, swing eALe T seconds, is gien by that the it takes for a pendulum formula l __ T π = , √g where l is the length in m and g is the acceleration T o make l due the to graity subect, in rst ms diide π: by eA π T l √g π __ = Simplify: π l T __ = √g π • Rem embe r to apply T l Suare both sides: = ) ( g : g h subct Wre Suppose alues eALe we of T o side l we by area = = make First, wanted the rea lw aas e of the rst last . T Simplify: l = g ) ( π g o o pera tion g __ π iners = ) π = ) ( ( __ lg T by Simplify: the l ) ( √ g π ultiply T __ both sids of th fomula to and draw a rectangle perimeter were so that the numerical eual. perimeter l + the w subect: must remoe subtracting l the term from containing both l from the righthand A sides: w ______ lw − l = l + w − • l sing the formula l = , w lw − l = w lw − = w lw − nd T aking a common w factor • 4, = = an w you of nd l if and a w w alue = = , . of l and = − as alue w w l the − iiding w − both sides by w w so that the area of the rectangle is and cm the w l = perimeter w − ake n n en cm e Ar is the subect of this T o change formula, n − the subect of a formula: perform the same S = operations on both sides. ake k the subect of this formula: se inerse operations to r = k + p undo the formula. If new sphere length rite r the r l of has in radius the terms subect. r has same of l. a surface surface This area area means as of the write a 4π r . cube of side the both sphere. formula and make sides remoe it subect of the from is on formula, one side and factorise. 8 Lia Lia LeArnn Sole uatios e • uatios linear euations in lia uatio is a mathematical sentence stating that two one expressions are eual. They are recognisable because they contain an unknown eual • Sole linear euations in one e unknown containing sign. are usually calculate • Sole linear unknown euations containing in Sole a simple linear the The method one subect one Wre eALe sole the 4x The left terms sides to 4x = side in x, an letter euation or this ukow means in the we should euation. soling a a linear formula, euation which you is saw similar in .. to e changing applied to both sides of the euation, use to inerse rearrange side x will so + + of the = sually to is the the to remoe euation, the other all and terms containing remoe all terms the not unknown the containing from the side. to uatios the simplest remoe Wre + with method to backts sole an euation containing brackets them: both −. x is from olvig contain add aim euation undo − of of The unknown − sole the unknown terms. T o to of ineuality operations, on alue one fractions the • asked brackets + eALe 4 − a = a + Simplify: emoe 4x = x + brackets: The right side will contain There terms without x, so subtract both are for − = more the a + 4 unknowns on the right, so we will make x = x + − − a = + a = a + 4 + = a + 4 4 = a + 4 − = a a dd to both sides. Simplied now subtract − nd the alue of x, sides x by nd a, − both sides by : a = = diide : 4 diide T o both − Simplify: T o or r a x = = − eA • Alwa ys aim remo ing than the x Simplify: x this unknowns. sides: 4x a x side from − the a to the so ma e smal ler we the unn own uan tity. remo ed the −a posit ie −a by is by smal ler addi ng. 4 euatios The best by a all the cotaiig techniue common for factios dealing denominator. with The fractions is denominators to multiply will cancel, each term remoing fractions. Wre eALe = x + The − common ultiplying x + x denominator each term − by x is the x x + common + − − x denominator: x Simplify the fractions = by x + − cancelling x A − x = x x = x + xpand dd − + x = x = x to brackets both sides to to make the + the Subtract iide term in from both x both sides by . per = charge irect T axis charge usig represent, and Wre There are stones steps are than choosing formulating eALe Sole an euation distance rst the what the euation. unknown onsider should this hat the second. ow to nd which charge the both same. is the cost of this ourney problem. in three pile. The piles. third The pile second has pile twice as has 4 many Ar en euation more Sole many stones are there in each the stones x as for stones the 4 uatios • important plus . companies The mile. the oblms plus • olvig mile. sides per x T axis positie − = x + pile Sole the euation olutio x − x = et the rst pile contain n − 4 Then the second pile stones. contains n + 4 stones. Sole the euation 4 The third pile contains n + 4 stones. = x There n are + n + stones 4 + altogether, n + + n + 4 + 4 so 4 = e n + n + 4 = emoing 4n + = Simplifying − = n brackets If there are fractions, simplify them. Subtracting 4n + − both 4n = from sides emoe y any brackets. 4 using inerse operations, 4 4n = 4 iiding by 4 create a side for the 4 unknowns n = and a side for the numbers. So the piles contain n = stones, n + 4 = stones and n + 4 = lways keep the unknowns stones. positie. uadatic uadatic LeArnn Sole uatios e uations • uatios uadratic euations by euations by with a suared term are called uadatic uatios factorising olvig • Sole uadratic uadatic uadratic completing the Sole uadratic can sometimes by be factoisig soled by factorising. This suare method • euations uatios euations relies on a special property of the number : by If two numbers hae a product of , then one of the numbers must be . formula Wre eALe For Wre eALe the rectangle euation rectangle greater has than its a aboe, to sole w + 4w = 4, make the length width. ero by subtracting 4: 4 cm The eual w + 4w − 4 = area Then factorise the lefthand side is 4 cm If we . w call the width w cm, ut length is area found is length the w and + 4 cm, by and, as multiplying width, we + − = arrie if these two factors hae a product of , the either the at w = w + − or = w = or w − = euation uadratic ww + 4 = remoing the euations can hae two solutions. 4 In or, w the brackets, this case, original one solution problem, as a does not rectangle make cannot sense hae a in relation width of to the − cm. w + 4w = 4 So, the rectangle olvig ot all is cm uadatic expressions wide and uatios will factorise, so cm by the long. comltig aboe method th will sua not always work. saw in unit . that x b ou + bx + c = (x + b ) ( ) + c x 4x + = cannot be soled by factorising. First, diide by = so the coefcient of x the number of the x term x x + sing the identity aboe, b = – and c = So x x + = becomes x + = x = x = adding to both sides x = + or Suare √ roots can be x = + or + √ x = + √ or x = – x = . √ or x = . to decimal places positie or negatie. is : Wre eALe eA x − 4x + = cannot be soled by factorising. • First, diide term is by so the coefcient of x the number of the three meth ods f or soli ng : uad ratic euat ions as x − x + = the identity + x bx + c = x ( eam ination ues tions b b sing might speci f y wh ich + − ) ( ) + c, b = − meth od to use. and c = • − x x + = becomes ues tion as s for answ ers x − to a − − + = − − = num ber of decim al places or sig nic ant x If an eam ination So ou m ust be able to use al l x gur es th is tel ls yo u that the u adratic will no t − x = adding to both facto rise. sides x − = ± Suare √ roots can be positie or negatie. x = ± + √ e x = + √ n or x = − + √ uadratic euations hae euations hae a term x = . or x = . to decimal places uadratic two olvig uadatic uatios by the suare is really only solutions. fomula ompleting x useful if the coefcient of x is uadratic euations can . be soled by factorising, For the general uadratic euation ax + bx + c = , completing −b √ ± b − 4ac using we use the formula x the suare or by the formula = a −b ± √ b − 4ac x = a A • Sole the euation ◦ Factorising ◦ ompleting ◦ sing the the x − 4x + = by Ar en euation suare formula Sole the x + x − = by factorising. Wre eALe Sole the euation x T o sole the euation x − x + = + 4x − completing a = , b = −, c = = by : the suare. −b ± √ b − 4ac rectangular eld has a Substitute into x = width a of x metres and a −− ± √ − − 4 × × length of x + metres. x = Its × area is 4 m . rite ± √ an − euation and sole it by x = formula to nd the length and + √ x = − or x = .4 or width of the eld. = x √ x = . to decimal places 0 imultaous imultaous LeArnn Sole simultaneous buy in two word and sugarapples for . an aocado costs x and a sugarapple costs y, we can write unknowns algebraically Sole aocadoes linear If euations • uatios e I • uatios x + y = problems ne solution There In If are fact, we This, = its knew too, . single = has y e and the is = = for . example euation aocados has and x an = and innite y = .. number sugarapples cost of solutions. ., then solutions, only for example one solution that two unknowns x = . satises and both y = .. euations .. with can solution. y . euation olvig other there solutions. uniue own, y and = solutions, that + oweer, x x other on x is sole Such a pair of euations simultaous has euations are called an in innite two number unknowns simultaous uatios by of to gie a uatios limiatig a ukow Wre The eALe x + 4y = x − y = rst step unknowns and eA term s to in multip ly the cons don’ t y so by , so the euations coefcient. they both e so that multiply contain × x + y = 44 × x − y = y = y = there is the subtract iide both hae no we the sides know uation coefc ient multip ly euations: by Substitute Subtract the of the or need to eua tions . : iide : alue e can x y x = : + x of : check − y x we = = are 4y y, we = 4 = x = x = , + 4: by Substitute 8 same one of euation the by , x ote: y − −y = can calculate x: t eua tions alrea dy same fo rge tant. Som e tim es will manipulate the ow eua tion • to the Rem embe r all is has euation ow • correct by y = substituting into , y = : − = orrect. euation : y ow consider Wre this example: eALe x + 4y = x − y = e can euate coefcients x + 4y = x − 4y = of y by ecause × we add multiplying the the signs of the euations to 4y by : terms are eliminate opposite, them. eA x = 44 x = 4 • o iide by elim inate unn own uation x + 4y = x = 4 + 4y = the • Subtract 4y = if the If the Sam e sign s − are then by 4 y = − , x = 4 in x − y = aDD the eua tions . heck s Sub tra ct eua tions . the Differe nt iide sign are Substitute an Substitute y = − , x = 4 − − = orrect. 3x + 2 A • Find the alue • suare is cm of x and y in this rectangle. 2x has a length of a + b and width of a − b. Its y 22 cm area . alculate a and b 17 cm e n Simultaneous euations can be soled by eliminating one unknown. T o eliminate each the unknown, Ar en simultaneous Sole a Sole b − = b must be eual in the simultaneous + y = 4 + y = sum e of my years old are euations 4x ow coefcients = x The In the + a the euation. age time, we euations and my both my father father’ s will be age is . exactly twice my age. now uth simultaous uatios In LeArnn unit ., Sometimes • Sole one simultaneous linear, Sole soled pairs of simultaneous euations. one word a better method is to eliminate an unknown by substitution. euations, nonlinear olvig • we e simultaous uatios by substitutio problems This method Wre Suppose b cm The is total useful eALe a and ery cuboid a length of of one of the euations is nonlinear. has height when a suare base of side h cm. its edges is cm, and h its surface There eight area are is cm four ertical horiontal edges edges of of length length b, h and so: b b 4h The four + b = ertical faces hae an area of bh and the two horiontal faces hae an area of b , so: 4bh e + cannot b = eliminate b or h in the same way as before, as is nonlinear. oweer, 4h e can we + can rearrange to make h the b = 4h = − b Subtracting h = − b iiding substitute this into by subect: b 4 : 4bh + b b + b = 4b − = xpand to gie: b − b + b = = = b = = Simplify: b − b earrange: − b + iide by : b Factorise: So b − = or b b = or b = Substituting If b = : ecause pairs 00 of into − b − − + b = − : 4h + b = 4h + = 4h = h = 4, there b was solutions. a If b = uadratic b = : 4h + 4h + euation inoled, b = 4 = 4h = h = , we get b = two idig itsctios Simultaneous graphs can gahs be used to nd the points where two intersect. Wre The euations of eALe diagram shows the graphs of y = T o x − nd we x the sole − and points the of y − x = intersection, simultaneous euations From y = x − y − x x = euation y = x in and : + Substituting − to : y = x − x − − x − − x − − 4x − x + = x Subtract x and x + − x − = x = x = − x − from each side x − + − = or x = or x = Substituting y y , x x − − into + Factorising = − : x = x = : = y = and x y − y −, are the x = x = −: = y = + eA • A can check in pair intersections. : y = x x − y = of = , = y = − : − x − x − x = −, y = + = − r of : oth are r will and hae a non li near will hae two so lutio ns. correct. e Ar linea so lutio ns. linea pairs x of eua tion − pair eua tions • A e n en Sole the euations rectangle is 4 cm x + y = and y = x − 4x + a length of x cm and a width of soled y cm. The If one there the perimeter is euations can substitution. euation is uadratic, area and by has Simultaneous be will be two pairs of cm. solutions. So xy = 4 alculate and the x + length y and = . width of the rectangle. alculate from The graphs one point of y = x x + 4 and y = x the + x − cross unknown euation. at Simultaneous simultaneous euations to nd the coordinates used where they to euations nd the can points of where point second only. be the the linear − Sole the two graphs intersect. cross. 0 aiatio ict LeArnn • epresent direct and hose Sole problems 4 litres of water in minutes. in 4 minutes it will dispense 4 litres of water. and are two uantities here: the time in minutes and the olume inoling of direct dispenses symbolically There • pipe indirect So ariation vaiatio e inerse water in litres. ariation oubling time e by say sing we the , that for write time we the the it e The aries all ates = the onathan number is of of I run mph, s the distance, = or , Substituting So d So running o t ∝ kt, = d d is vais in t. litres e water. by If dictly and read t multiply with for this we as the “ the time. time is in where , so k is 4 a = constant. k, or k = . is litres minute. directly proportional to t, ∝ t and per hour. in I = = into at is In total this minutes a and pay, case, p p, = aries directly aries distance speed, of directly with my d = to a decimal m. mph ary directly, I can write k gies = k, or k = I would dict run d = × = m vaiatio diameter of cm has of cm an area of . cm place. circle to with a decimal diameter 0 has an area of .4 cm place. iamt d with speed. . eALe with = h k mph, coml circle coer d m h. minutes, directly Wre the . kt ariation. eALe at = = worked, run t, I hen of water thing. inole water as ¬ow with paid ∝ 4, same distance d water of t hours Wre If = rate always The write directly mean of of amount amount t”. also hen olume olume to the the algebraically proportional So doubles multiply iamt d Aa A . .4 the oubling the diameter produces a 4fold increase in area. eA In fact the area aries directly A ∝ d with the suare of the • , or A = diameter. If kd a and b ary direc tly a then Substituting A = . and d = is cons , tant. b . = k k = .4 So A = .4d A Wre eALe suare has an area, A cm , The olume radius, of a sphere, cm aries directly as the cube of the and a diagonal of length d cm. r cm. The area aries directly with hen the radius of radius is cm, the olume is . cm . T o nd the the suare of the length of the a sphere with ∝ r a olume of cm : diagonal. , or = . = k k = 4. kr hen the diagonal is cm long, the to decimal area is cm . places • Find the euation connecting area suare = 4.r A and d. So for a sphere with a olume of cm , • Find the of a with a = 4.r = . diagonal of cm. r • Find the length of diagonal r = r = √ . of a suare with an area of Ar h aries en directly omplete h k cm . cm the with table e k. of alues for h and k . n If x x ∝ k is aries y, a directly and x constant, calculated The mass directly side of with a suare the tile suare aries of the for ariation referred tile with side length cm mass Find of the length and gien x is to and a can pair be of y sometimes as proportion. The ariation and might roots, inole for 4 g. mass of a tile with example if with cube x aries directly side the of y, then cm. x then has powers a y, length. with ky alues = ∝ y and x = ky x aries root of hen directly with the cube y x = alculate y , y = when . x = . 0 vs vs LeArnn epresent indirect Sole If problems inerse ., we saw how uantities can ary directly. ariation symbolically • vaiatio e In • vaiatio inoling one uantity Sometimes doubles, uantities so ary does the other. ivsly ariation Imagine If you you had oubling Inerse other If the diided the of means by cat food number ariation is enough cats, the ssuming 4 hae food would cats that to only means if feed one last the 4 cats for food uantity for days. days. lasts is for half multiplied the by n, time. the n cats eat once a day, you actually hae enough food for meals. c is last, the number of cats and d is the number of days the food will then 4 d eA • If a c hen and then = ab b is ary d and c ary inersely, we write iners ely cons tant. k d ∝ or d = c e can say c d Wre y is inersely eALe ourney to work proportional to is km kmh, it takes me 4 minutes. If I trael at kmh, it takes me minutes. I multiply say sing the that for speed hales speed the by speed speed in , and kmh the I diide time and t the vay for or t = w cm, of a and rectangle width, ary Substituting so = , t = 4, that the area is xed at 4 cm . k • • Find some that t. rite the alues of l and = k = w so euation 4 that The connects 0 l and w. relationship is t = we get time by . ivsly time k ∝ with time. A t inersely long. at the aries trael e l cm, d I If length, or If oubling The c, in minutes, we write this as c o s coml with For rest direct example, aries ivs ariation, the time inersely vaiatio inerse it takes with the ariation for a car suare can to root inole trael of the powers m or roots. from acceleration. If it accelerates at 4 ms , it takes seconds. e The can use time, this t, to aries nd inersely t = with it takes the if the suare acceleration root of the is ms acceleration, a ∝ , √ time k t the or t = √ a when a = 4, a so k = √ k = 4 So t = √ a If a = , t = = √ . seconds to decimal place A cuboid • rite • omplete a has a rule suare or the base formula table and a showing showing olume the the of cm connection possible base between lengths the and base length and the height. heights. h as lgth ight h b cm cm 4 . b b • escribe the Ar w aries type of ariation between the en inersely omplete the with table of base length and the height. e p. alues for w and If n x aries p inersely with then x ∝ , and x = y w p k is a constant, alues intensity indirectly distance The of with from intensity light, the the of l, suare light light of the source, is of cm from for ¬ux d at the ariation source the distance where the x and a can pair be of y referred from intensity is to sometimes as proportion. a The ariation might inole source. powers alculate gien aries distance and 4 calculated The y y, k the is and roots, for light example if x aries with cube indirectly ¬ux. the of y, then k x r aries inersely with the cube root of ∝ r = alculate , = when and x = y hen , y . r = . 0 rlatios athmatical LeArnn • latios e ppreciate what makes relation list is simply a relationship between two sets of elements. a of all the students in a class matched to their heights is an relationship example • nderstand • nderstand range, • types the image, epresent a graphically , ordered of of a relation. relationship terms Imagine a number put machine This ‘+’ a set of numbers is fed in on cards, and for each domain, in, there is an output. codomain is a machine. goes in, comes out. relationship algebraically , pairs, arrow set of diagrams e could and y to call this relation represent omai the y = x + , using x to represent the input output. codomai ag ad imag 2 ll IN the case, 7 numbers the that domain can might be be fed all in the are real called the domai. In this numbers. 1 0 OUT 5 The Domain numbers that can be fed out are called the ag Range If maps onto , we say that is the imag of . The relation might numbers. hen T o classed more all domain for if either a The 5 4 4 3 3 2 1 0 0 as the alues could 2 This is 1 a 1 1 2 2 3 3 4 4 a y can be also −4 − − need to eery to this or a the subset element of of In real numbers. as relation real state codomain. not hae domain of real the and suare as the domain numbers, x, = shown x to + on the can e real neer be numbers. domain the the and codomain. domain other codomain. words, must eery map to input so roots, hae then would y = √ nonnegatie y a = not suare √ root. x would be a x is real not only a numbers. If the be dened relation. the the element of the domain can be written . an relation, of represent − for the outputs, output. do adds y member of latios o–o uniue x that output we the domain allowable nonnegatie eual the of a or the the in elements uatio, This e 0 relation an be one dene all hae relation, least rstig 5 is relation, numbers the negatie practice, range a as we was might codomai, elements at negatie relation Then, in will hae x the dening or must s so range be one = the ut negatie, The y dene aow as diagam: eery member of the domain maps range. the relation − − − using a tabl 4 4 to (range) dd to ais represent e can ordered a then pairs such as , , , 4, , , −, can also be relation. 6 represent as y used the shown relation on the by a gah, by plotting 5 the right: 4 Some relations are o–may. This means that members of the 3 domain can map to more than one member of the range. 2 n example is y √ = x 1 ere, 4 maps to and to −. (range) y 0 4 maps to and 3 2 a graph it 1 (domain) 2.5 n x 1 −. looks like 1 this: 2 2 1.5 1 eA 0.5 • 0 he rule fo r x 1 2 3 4 a relat ion 5 mus (domain) t 0.5 be true elem ents of fo r the 1 all domai n. 1.5 2 Ar en 2.5 Some of the −, relations domain Some are are may–o, map to the such same as y = x . member of the relations example This two range: , members is are contains y = the the a may–may. √ In what and examples. n ere, 2 2 1 1 0 0 relation is the y = image 4 b hat type of a x − , of: −4 relation is x . example shown on the arrow y 1 diagram: = x hat − is the range of the 1 relation 2 n a graph, it would look like y = x − 2 this: e y n 5 There are relations 4 four types one–one, many–one and of one–many , many–many . 3 The allowed inputs make 2 up 1 3 2 1 domain, outputs x 4 the The set the of the allowed codomain. all actual outputs 1 is 1 2 called the elations an range. can euation, be a shown set of as ordered 3 pairs or as a graph. 4 5 0 uctios athmatical LeArnn • State the properties function a istinguish relationship se :x and ordered diagrams • and function →, a relation element where of the each element of the domain maps to codomain. one–one and many–one relations are functions, but one–many between and using one function So • is that exactly dene fuctios e x function pairs, = are not. by uctio notation y relations arrow graphs =, many–many s x with relations, ordered e otatio can pairs also we and use can graphs two use to other euations, show forms a arrow diagrams, tables, function. of algebraic notation. The function y = x − x can also x ll three that a we ne way line x, or mean dealing from to written as − vtical ertical eALe x forms are mapping h Wre = be the with one li make : x → x same a − x thing. function, number to The and x the notation : x → reinforces notation suggests another. tst sure that a relation is a function is to use the test. onsider hen e x can ere is x f x the = function −, write a x this table of = as x = x − . . − alues of = . x − −4 − − for x = − x = − − − − to . 4 y ere is the graph of the function: 25 The vtical li tst states that a relation is a 20 function if intersect the In this ust there case, once, graph any so are it no at more ertical is a ertical lines than line will that one cross point. the 15 10 graph function. 5 0 5 4 3 2 5 08 x h ivs of a fuctio A The inerse of a function y = x is x = y . y It reerses the effect of the function nd the inerse of the function y = x − interchange hange the x and y to gie x + = y choose codomain : The First, x is a function if we T o √ = − the domain and carefully. codomain could be dened . subect: as the so it nonero is no real longer a numbers, many–one relation. x + = y , or y √ = x + The function y = x − will hae an inerse of y = √ x + • hat on ecause like a suare root can be positie or negatie, the limits must be placed . graph looks • this: the The the domain domain function y • 3 of codomain hat and a function of the ice would the is inerse ersa. graph of y = x − look like now 2 1 0 4 3 2 x 1 1 2 3 4 5 6 eA 1 • nly one– one func tions 2 will 3 This a fails the ertical line test, is and so although it is a relation, it is hae also a an iners e that func tion. not function. many–one function Ar will hae an inerse that are a one–many relation. en ere is the graphs of y = x and y = x − x e n function is many–one y 30 20 20 The inerse y x = x= 10 the 2 0 x 1 1 2 3 4 3 y , The 30 by then writing changing to y. ertical on a line test graph to can be check relation is a if function. 30 the why function 20 the xplain a 10 20 and subect used of of found 1 hich is x 1 10 or 10 0 3 one–one y 30 4 a relation. two the functions other is a function one–one will not function hae an inerse function. hoose a sensible domain so that : x → √ x + is a function. 0 ahs of lia fuctios What LeArnn • nderstand linear the concept of • lia e function raw and linear functions Find of the linear and • interpret x and graphs the fuctio? yintercepts functions gradient concept of graphically of a fuctio is one that produces a straightline graph. can of recognise three a linear possible function from its euation, as it will hae forms: • y = c, where c is a constant for example, y = 4, y = −, y = • x = c, where c is a constant for example, x = , x = −, x = π • y = mx are constants y = x + − c, where 4, y = m − and x, c y = for example, x straight It line lia of algebraically Find a a one • is e is possible that the euation might be in a different form, but a slope linear For function example, can x − x = = y y = always y + = be can arranged be into rearranged one as of the three forms. follows: y A x • ake y the subect of − these functions to decide x − whether x they are linear or not. nd = can be rearranged to y y x + y = = x = y = y x y x xy = = x x y + awig y x nce If gahs = x • lia = you can, use a graph plotter we linear, know we only a function need to is y identify 5 to check your answers. two points to be able to draw 4 the graph, straight two as line gien there that is only passes one through 3 points. 2 sually, an easy point to identify 1 is the intercept Substituting x on = the yaxis. into the 0 x 1 function tells us the intercept 1 on the yaxis as all points on the 2 yaxis eA hae an xcoordinate of . alue of 3 e • It helps to nd to points in on case you mista e. points the If r ma e the any other x nd a second point on the line. 4 we can nd the intercept 5 on a xaxis by substituting y = . 6 three ma e line choose line the don’ t straig ht can three a chec your For of example, y = x − to draw the graph . calcu latio ns hen So 0 the x = , graph y = × passes − through = −. , −. hen x = , y = × − = −. eA So the graph passes through , −. • he n hen y = , line The = x = x x = . graph h linear e for hs straig h t use a passes through of , ruler . . gadit straightline graph has the same gadit along its B D length. gradient measure one The grap coct entire The − draw ing unit of a the of gradient is straight gradient line of a is a measure line by horiontal distance. calculated as: ertical distance of nding its steepness. the ertical distance C between two points horiontal distance between the same two points 4 A The gradient of = = E This tells us that for eery one unit horiontally, the line climbs eA suares • gradient of = hen The ertically. = grad 4 calcu latin g ient sua res line which slopes downwards from left to right has a by coun ting pay specia l negatie atten tion to the gradient. the scale on aes. − The gradient of = = − = − − The gradient of = The gradient of = = and e are aalll, and so hae the same gradient. n inear or y Ar straightline = mx + c, graphs where m can and c be are written as y = c, x = c Find axes constants. y T o draw a linear Substituting graph, nd three points and oin = gies the intercept on the Substituting x = gies the intercept on the arallel lines hae the same gradient. graph the of . raw = the x + graphs and of y = − x. yaxis. the + on xaxis. y of x intercepts them. y = the en Find the graph in gradient uestion of each h uatio staight h LeArnn uatio raw a graph gien li a staight li the euation y = x − . the x e euation of a straight can Find line the euation from a of construct a table − − − − 4 − − of line alues • a e onsider • of of a for x and y : y straight n graph e y as the left saw for in is the . eery graph how unit to the of y nd line = x the − . gradient moes to the of a right, line. it The climbs gradient units here is , upwards. 5 The graph The table has a gradient of because of the x in , y to dictate the euation. 4 shows that, as x increases by increases by . This is the gradient. 3 The 2 effect passes of the through − , in the −. euation hen x = is , y = × that − the = graph −. This 1 point 0 2 x 1 1 The is the euation 1 graph: 2 straight the intercept yaxis of line at a the straight with , on yaxis. line euation tells y = us mx two + c important has facts gradient m about and the crosses c. 3 awig e 5 saw line staightli in . that gahs we need to nd two points to draw a straight graph. 6 ow we can draw one straight from the euation, using the 7 intercept eA and Wre • ou can your alwa ys eam ple that eALe chec y answ er y = draw the graph of y = − x: 8 In the subs titut e = y T o by subs titut ing. into gradient. when to e = chec = at so know , will with mark a the cross the gradient point , yaxis of −, , 7 and 6 then . , we it plot other points one unit to 5 the right and represent a two units gradient of down to −: 4 A • raw the graphs of: e oin the graph. the points to complete 3 2 y = x y = − y = x + 1 − x 0 x • For each graph, the coefcient check 1 that 1 to the the gradient, constant intercept of on is x is and eual the eual that to yaxis. the 2 idig e can th easily gradient and Wre uatio nd the fom euation of a a gah graph using the concept of intercept. eALe y onsider this graph: 3 The graph crosses the yaxis at −. 2 T o nd the gradient, we take two points on the graph. 1 The gradient, ertical distance between two points 0 3 horiontal distance between − y the same two 2 1 1 points y x 1 2 can be written as gien two points x , y x − x and x , y 3 . 4 The graph passes through , − and 4, , so using these 5 − − points, the gradient = 4 The the euation of a − straight . line is So the gradient is 4 is 4 gien by y = mx + c, where m is the gradient and c is the intercept on yaxis. So the euation is y = x − 4 Ar hat is en the gradient e of: n The gradient measure a y = x − b y = − x c x + y = of of a line is a steepness, − y y int for c: make y the calculated subect as . x − atch the graphs with x y the The graph of y gradient of m = mx + c has 8 correct euation: a and crosses 7 y = x + the yaxis at , c. 6 y = x + x = y + x + y = 5 4 3 2 1 x 0 2 1 1 2 3 4 5 6 rite down through , the 4 euation with a of a gradient straight of line which passes . 8 olvig oblms gadit In LeArnn unit Find line the euation from the of a gradient straight y = Find line two where that m is the the euation gradient of and a c straight is the line is intercept of on the the can use this to nd the euation of a straight line in a number the euation from the of a situations. straight coordinates of o d th uatio of a staight li fom th points Sole problems of parallel perpendicular ad a oit inoling This gradient is best shown through an example. and lines Wre straight T o nd In the the T o nd = eALe line the , c, we = o , + c − = . euation the th , − and passes through the point , . = + c, −x the m + = − the gradient. c euation passes through , , so when is y y = = −x + uatio of here two is into y = −x + c gies . of a staight li fom th oits similar to the one aboe, except rst we −4, must gradient. Wre T o mx y = = techniue nd is x coodiats The = know c d gradient . or y euation y euation: Substituting The has euation So x nd eALe the euation of the line that passes through and : − y y The gradient is gien by − − x = x − = −4 = The euation is y = x + c e can positie use either alues = c = 4 + , point to : c The euation is y = x of point gadit • c, and different • discoered mx+ yaxis. e one we itct e form • . ad usig + 4 nd c it might be easier to take the aalll aalll and ad lis dicula must are hav th lis sam B gadit perpendicular. 4 The line has gradient of The gradient of is − . This is the negatie reciprocal 4 of the gradient of . enerally, C if two lis a dicula ad th gadit of o li is m th oth has gadit − m oblms ivolvig Wre line has line is nother Find eALe euation parallel line the is to y y gadit = = x x − − perpendicular euations A of to these . and y = passes x through and passes , . through , . lines. olutio The parallel line will hae the same gradient as y = x − , A which So is the . euation is y = x + Try c. to answer without It passes through , , so when x = , y = these drawing uestions an accurate . diagram. Substituting, = + c = − c The euation is y = The perpendicular x line − has a gradient of the negatie reciprocal , 4, , − • of , or uadrilateral Find − , the and has , ertices −, euations of and , , . So the euation is y = − • x + xplain how the euations c show It passes through , , so when x = , y = that and are . parallel. Substituting, • = − + c or c = xplain show how that the euations and are perpendicular. The euation is y = − x + Ar line oins • en , and , hat e −. Find the gradient of Find the euation the the ines with the same gradient parallel. line. of is n are shape erpendicular lines hae line. gradients Find the euation perpendicular to of the the line original which passes through , , of m, − . m line. uth lia h LeArnn etermine the length of a line a li sgmt mathematician without thinks end of and a li haing as straight, no continuing in both thickness. li sgmt is part of a line. It has a beginning and an end, and coordinates a • of segment from gahs and directions midpoint of e The • lgth otis Sole graphically two denite The euations in two length. linear graph below shows a line segment that runs from −, − to , 4. ariables hen we ertical were nding lengths. The gradients horiontal in ., length is we x used − x height is y − y So, for e this line ertical segment: the ertical also length height use Wre these of can the be is is 4 − lengths eALe length segment the and . horiontal The and horiontal the can the − − − to = = . nd the length of the line segment. y line found 5 by ythagoras: 4 If the length is l, then: 3 + = l 2 l = + l = l = √ 1 . to 0 d.p. 3 2 x 1 1 1 2 3 eA • Don’ t confu se the h techn iues for ndi ng The lengt h and the midoit a o nd line midpoint the lengt h to b b you a befor e o a nd and b the apply ing a by So add of befor e we The the line need easiest nd line the segment will xcoordinates of hae the an xcoordinate ends of the line exactly segment, exactly halfway between the ycoordinates of the segment. to the nd and way mean to of the number halfway a nd and halfway between the number − between and halfway 4 for − and the ycoordinate. between a and for b the is to mean of b . So the the the ycoordinate of xcoordinate, midpo int you diid ing of between subtra ct ytha goras. • sgmt from ends a li segm ent. and • a midpo int halfway of of the midpoint coordinates. of a line segment is found by nding the x + the midpoint is gien ) − + − this case, the midpoint is In In unit simultaous . we saw how to + 4 , ( , . uatios sole or ) olvig y ( by + y x So a pair of gahically simultaneous euations algebraically. e can also sole Wre T o sole eALe the = x − = x y = − x by drawing − has has and a a a graph. simultaneous y y them y = gradient gradient y euations − of of 10 x and − passes and through passes , through 8 −. , 6 . 4 lotting the two graphs on the same axis gies this result: 2 The the intersection point of where the the graphs two where euations the share lines the cross same shows alues. 0 1 So the solution to the euations is x = ., y = 1 . 4 e A n The length of a line segment oining x , y and x , y aisy’ s is electricity company lets √ x − x her + y − y choose T ariff + of the line segment is + y x midpoint ( : Fixed euations can graphs euations and per ) week, the be soled nding graphically the by coordinates of per k of used. drawing of : Fixed charge of 4 the per point plus T ariff of of electricity Simultaneous charge y , tariffs. x The two between week, plus per k of intersection. electricity T ariff y Ar en = cost x has + and an , x used. is euation where the y is the number total of k used. se a graph and y = to sole the simultaneous euations y = x + T ariff − y uestions to −, and has an euation x are about the line segment which oins , . − • = x lot + the axes. Find the length and Find the euation midpoint of the line is segment. 4. graphs For the what cost the on the same number same for of k each tariff midpoint at right of the angles line to which the passes original through the line. 0 ualitis olvig LeArnn n • lia epresent the solution ineuality ineualities notation, Wre using number There are four sole − x ⩽ x + − − x ⩽ x + x + x ⩽ x + ⩽ 4x + ⩽ 4x + ⩽ euation, except that it tells us that two less • > means ‘is greater • ⩽ means ‘is less than or we than’, so use: − than’, < π so than or eual eual to . . > . to’, so if x ⩽ , x can be any alue ⩾ means than + or ‘is greater eual to than and or less eual than to’, so if 4 > x ⩾ , x is greater 4. x when − are soled multiplying multiplying or the or in exactly diiding diiding direction an of by the a same negatie ineuality an way by a as euations number. negatie e except must number, aoid as this ineuality. 4x 4x that ‘is changes signs means Ineualities − an to eual. < • not • smaller T o similar are set line eALe is of expressions linear iualitis e For example, − < > , but multiplying both sides by − gies ⩽ 4 x −4. ote that the ineuality has been reersed. 4 ⩾ So we must make rstig e saw aboe sure a that the unknown term is always positie. iuality ineualities can be soled in a ery similar way to euations. The of a main the difference solutions. uadratic of ineuality does not euation euation has euations an up has to a hae with two pair a and one and solutions, single ineuality unknown solutions, of an a one solution has pair in the one of each instead, is it format solution, linear unknown. has a range solutions. e can represent Wre Suppose e 8 linear euation simultaneous n between can this in eALe we hae sole this + x + ⩾ x x + ⩾ ⩾ x ⩽ the by a number of ways. ineuality splitting it x into and + ⩾ two x x x x x solutions tell less or than us that eual to x is . any real + separate > and + x The x > − . ineualities: x − > x − > − > −. number x greater than −. sing x ∈ e st : read greater e otatio, −. < this as than could x Wre top The lower So x is an and the arrow element less same for the example aboe is satisfy 8 real eual information x > 6 both line the or on numbers to a such that x is ’. number line. 7 number set of than represents 8 9 solution represents arrow must the the . eALe 9 ut ‘x −. show The ⩽ −. x ⩽ 5 the 4 conditions. 6 5 circle circle 3 representation 7 the is 2 This is looks 4 is lled, 1 true like 3 empty because because 0 1 where x cannot can 2 the x 3 lines eual 4 eual . . 5 6 7 8 9 10 5 6 7 8 9 10 oerlap. this: 2 1 0 1 2 3 4 A eA illie of bought some chocolates er bag illie bag. is put The that strong two bag realised b • Sole She ⩽ this then • rite to that, if of She also bought a to carry • up to and g the in in of • lemonade weighs b g, or ⩽ a the or ine ualit y use on hen is then the > circle the weight. bottle hen < weight. chocolates box g. lemonade the a lemonade. ⩾ an is unl led num ber line. ine ualit y use a lled circle . . ineuality. out The down nd enough bottles took lemonade. of weighed carried illie + bottles the the bag an chocolates ineuality range and put in a third bottle of broke. of to possible Ar en Sole the ineuality x Sole the ineuality x show this weights information, for a bottle of and sole e + 4 < 4x − . it lemonade. n n ineuality written + ⩽ x + < x + in set with one notation unknown or shown can on a be number , line. writing your number answer line in set notation and as a diagram. Filled circles circles If y the < y + possible and integer 4y + alues ⩽ y of y − , nd are are part of the solution unlled not. all ualitis with two ukows Lia LeArnn The • iualitis i two vaiabls e epresent the solution of ineuality contains ineualities using a y < x − y linear two ariables, 6 graph and so is best represented 5 • raw a graph of linear on a graph. 4 ineuality in two ariables First, we draw the graph 3 • se linear programming of y = x − it has an techniues 2 intercept yaxis of and − a on the gradient of 1 . y The line y x shows where 0 3 2 x 1 1 6 = − . 1 5 n one y x side of the line, 4 3 other − side and y on x the − 3 . 4 2 T o 1 decide where y which x − side , is 5 we 6 select 0 3 2 a test point on x 1 1 one side. , is a good point to use if it is not on the line. e 1 substitute true 3 it Required is then false it in the the side then the gien with ineuality. the other If selected side is the the resulting point is reuired on statement the reuired is side. If side. region 4 5 sing , y into = , which y < x is − aboe , the which line, we substitute x = and gies 6 eA < × This is false, The or on • a or the > ine ualit y use a do tte d is the ⩾ ine ualit y use line use a e can is Wre toy They If they For 0 and Sarah takes takes can < < x −. − is on the other side of the line. so lid each make Sarah, x the Thomas total of the region, as y cannot eual for to up and time 2x − , line. the region y < x − by shading. ogammig line. two minutes planes part ogammig is a method of soling problems ineualities. make work not is Lia a eALe plane Thomas or y indicate graphing so dotted Lia Thomas , line grap h. hen ⩽ − • he n < y is different minutes assemble to hours boats, x the + types to a toy. assemble, boat, or total y of ⩽ and and Sarah Sarah spends takes minutes minutes to painting paint minutes. time for Thomas is x + y ⩽ it. it. by so we T o draw the graphs: Thomas x If x = + , y Sarah y = x + y = If + = y = y = + , y = y = 4 4 x , + = x = x + , = x = The rawing the each they toy graphs can shows the number of lines they make. the show work the unwanted y the full maximum number regions to of number hours leae the they aailable. possible 25 25 20 20 15 15 10 10 5 5 0 x 0 5 possible . ne plane prot 15 numbers of these they for sell each x 20 they can ertices makes ertex 0 make are represents them in a the prot, uadrilateral optimum and eery 5 with best boat makes them , planes = , boats at = . T otal prot = planes = , boats at = 4. T otal prot = 4 , : planes = , boats at = . T otal prot = planes = , boats at = . T otal prot = : maximum a prot they make planes and en graph to show n ⩾ a graph, − and show x + region region ⩽ ananas cost , and the . rite y ⩾ x − . where y > number can of and Solid lines are pineapples cost an bananas, ineuality . b, and to are part of the solution dotted not. , . down , n hen linear programming, regions, so the shade clear the part is the I solution hae , boats. e the the y , prot. : se I , is: unwanted at 20 : for x ertices 15 result. lines 10 , Ar 10 , , So get: Thomas 0 The shade we 30 Thomas ery we region, if Sarah 30 , If make y Sarah The can set. show pineapples, p, buy. omosit ad ivs fuctios omosit LeArnn ein • erie fuctios e composite and aria are operating two function machines. functions ein’ s e.g. • g, State the function • erie relationship and the its between inerse inerse aria’ s The , carries out the function x = x + machine output carries from out ein’ s the function machine is fed gx into = x − aria’ s machine. function So machine if ein inputs , his output is g = + = . • aluate a, a, ga, g • se This a the er relationship g = g is fed into output is e hae input So the = used for × oerall x aria’ s as − = . the machine. effect of both machines • In machine. g eA aria’ s gf f calcu lated resul t put into we is rst and g . the write hen is as two can which g x. functions comosit e g x, are combined in this way, they are called fuctios replace the gx = x gx = x composite − , function, g x, with a single function: so − = x = x = x + − + − − gx e can check that g = : g If we = connected ein’ s − the machine, = machines we would so get a that gx = gx aria’ s different output was fed into output: + = x − + = 4x − x − − x x + + gx fg(x) is = ot 4x ual to + gf(x). In general, composite functions are not commutatie. vs In the For So of aboe example, x x will as function fuctios example, not the will = hae x − an is = also many–one be function. . inerse nonnegatie then a real the function unless numbers. The nonnegatie we dene codomain real the of numbers. domain the inerse T o nd the inerse of x, written as x: Similarly, to nd x: g rite x as y = x + gx is The inerse is x = y = y y = represented + g x x r x − is + x = y = y y = x _____ √ as y = x − , so x − − + _____ x So x x − g x + √ = = T o nd the inerse of g x, rst, so g x The inerse of The function a = and its x gx hae to nd effect of the g x x undoes inerse are the original function. commutatie. = uaig If would g function So we = x a − x = x A fuctio , then ggx = gx − = x − − = 4x − • W wit gg(x as g If x gx x = = x x functions their olvig oblms usig ivs − + , on and match the euialent left the with function on fuctios the right: e can sole idtity o problems by using the fact that f fx is ual to th x + x 4 Wre eALe g Suppose x = x − and gx = x + , and we want to nd x 4x + 4x + gx. gx = x + , so, operating on both sides by x gx gx = x + . x x = , the identity, + gx ut so gx = x + . x g x is represented by y = x − , so = y x + − 4x − 4x x is represented by x − , or y = g x x + x = , so = x x x + + x gx + = + − x = x g − x 4 Ar en x − If x = e , nd x. n gx means apply function g followed by function . x If x = − and gx = x + , nd gx. − x is the inerse function hat does your answer tell you about x and of gx x. − x g − x = g − x = x a hat alue b hat is of x must be excluded from the x = x domain special about x LeArnn uadatic awig uadatic function fuctios gahs e • raw graphs of uadratic T o functions • se graphs to nd y • se graphs to nd all gien plot such • se of x gien graphs to nd of the alue function, x a graph, from raw and wish −4 and − x + is not linear. need to calculate a number of alues. to to draw the graph of y = x − x + for alues 4. construct axis of a table, showing each part of the euation on a the line: symmetry x • we eALe we separate of x minimum e euation = Suppose or a y possible y maximum of as x Wre alues such uadratic interpret graphs 4 4 4 of emember uadratic functions to nd x the interal which in the elements domain of the 4 for x cannot range 4 x may be > or < a be negatie gien 4 The + 4 = = −4 4 4 point constant term y = x The x + has alues plot it from will the −4 drop points to a 4, little and oin and y has below them alues when with a from we sum of the although e draw smooth to three parts , the graph. cure. y 25 20 15 10 5 0 5 There is a 4 symmetry 3 to 2 the 1 graph, x 1 which 2 can 3 also 4 be 5 seen in the table. The axis of symmetry is the line x = . eA The minimum x . alue is found on this axis of symmetry, where • u adra tic = cur e s. n x accurate = . alue into the of this minimum = be found by substituting points euation: • y can x − x + = . − . + = . grap no t with uad ratic alwa ys Do a a hs are oin the ruler . grap hs are Ush ape. If the term U radig valus e can use the the range For example, fom graph a shown is is nega tie the inerte d. gah to match elements of the domain x to y. A when corresponding x alue = of ., y, we can which is follow the red line to nd the 4.. raw T o nd elements in the domain which map to in the range, the taking the green arrows. The alues of x are −. and . to graph of y = x − x, follow alues of x from − to 4. decimal n the same set of axes, draw place. the graph of y = x − . This tells us that the solutions to the euation x − x + = are • x = −. and x = . to decimal hat are the xcoordinates of place. the points of intersection o solv th iuality x −x < • earrange the euation First, add euation to of both the sides so that the lefthand side matches the x graph: − x = righthand x − side so = that the . x − x + < • From x = The the .. This graph below e line to solutions shown y the that = by the the is x is − orange region the ineuality to part x + line where < x y < = on between −. the < are = −. uadratic your euation answers to and the points of the graphs. intersection of graph. the these x the compare part of solutions, the so cure the .. n T o draw term is the shows the solution graph, Sole and ou a uadratic separately can use and the graph, then graph to start add with them read a table. alculate each up. intermediate alues and sole ineualities. Ar en raw to the graph of y = x + x − , taking alues of x from − . se your graph to nd the se your graph to sole alue of y when x = .. x − the ineuality x + < . o gahs of fuctios h LeArnn • stimate gadit the gradient at uadratic Find the intercepts of a Sole at a oit graph is cured, changing as we and so moe the along steepness, the or gradient, is graph. uadratic estimate the gradient of a cure at a point, we draw a tagt function at • cuv point T o • a a constantly gien of e that point. tangent is a line that ust touches the cure without euations passing inside the cure. graphically e • Find the axis maximum of a of or minimum uadratic then nd the gradient of the tangent. symmetry, in the alue form ere is the graph y of y = x − x − . 6 ax + h + k T o • Sketch the graph of nd where uadratic in the the gradient 5 a x = −, we 4 form draw the tangent at ax + h + k and determine that the number of point, shown 3 in roots green. 2 • raw and interpret the graphs The of other nonlinear red ( y = show 1 that a lines functions the tangent has a = y y = x ax ) a ertical drop of 0 x x 1 −4 and length a of 1 horiontal , so at 2 x = of − the the gradient cure is 3 −4 = −4. oints called h A where the tuig tangent is horiontal hae a gradient of . These are oits itcts of a cuv with th as The • Sole the linear x − the x − = formula to by hat are y graphs, these = x solutions found the the − x − constant , crosses term the indicates y axis the at , −. intercept on s the with yaxis. The to where y the = uadratic euation x − x = can be . adantages These and of using check solutions. • graph euation disadantages are the x coordinates where the graph crosses the x axis. of soling uadratic graphically euations So, x reading = −.4 from and omltd x the = graph, the solutions to x − x = are .4. sua fom ad sktchig uadatic cuvs riting the euation y = x − x − in completed suare form, we get y = x ompare − this − to . the minimum point of the graph, , −. The minimum and the This allows axis alue of us of the symmetry to sketch euation is the cures line y x = a + drawing up = without x + b is at −a, b, −a a table of alues. y Wre eALe T o sketch The the graph intercept on of the y = x yaxis + is at 4x , + : . The coefcient of x is positie so it has a shape. In completed the suare minimum alue form, is at the −, euation , and is the y = x cure + is + , so 5 symmetrical ( about the line x = 2, 1) −. x So the graph will look like the graph shown opposite: The euation does not x cross + 4x the + = has no solutions, as the graph A xaxis. The will th cuvd In we no gahs • ., drew graphs by constructing a euation not x + factorise, 4x as + = there are solutions. se the completed suare table. form, The same techniue can be applied to draw the graphs x , y = + = what happens , to when you = + of see y x , y try = to sole the euation by x x completing x −4 − − − −4 − − − −. −. −. − can 4 4 . . . you the tell suare. there are ow no solutions x x A . . . . . . • x raw the check = accurately sketches to aboe. y graphs the and y = are not dened when x = . x The x graphs look like this: 1 3 y = x y 1 y = = x y 2 x eA y ear n grap x e x shap Ar gradient of a cure at a point is es of these hs. x n The the estimated from the rite en down the intercept on gradient of the tangent at that point. the yaxis of The graph intersect of the an euation yaxis at , in the form y = ax The graph of an bx + c will c. euation in rite point at −h, form y = ax + h uadratic suared graphs, graphs cubic each x − 4x + . x + k has − 4x suare the + in form, coordinates and of a minimum point. k. = identify the turning y completed the = + y graphs, hae their reciprocal own graphs identiable and reciprocal shape. Sketch the graph of y = all x the − 4x + , intercepts identifying on the axes. avl It LeArnn is often useful raw time • and time interpret and draw graphs of ourneys, as we can read them. istac–tim interpret graphs distance from distance– graphs raw to e information • gahs to arsha speed– isit determine time lies her in aunt, gahs renada. a distance She of traels from Sauteurs to renille to km. speed She cycles for an hour and a half at kmh, and then nishes the acceleration ourney She at kmh. returns emember home the She at a stays with constant triangle from her aunt speed. for The an hour ourney and takes minutes. hours. .: D ere is the information about arsha’ s ourney in a T S table, scitio im . . Start ourney . Finish ourney . Stays . eturn with with hour hours ÷ speed information hour . • A is hour s useful to tae unit minut es no t . hour is hour s. can be fo r time = = kmh km kmh km kmh to a graph: 18 16 14 12 10 8 6 4 2 0 1 2 time sua res − transferred 0 scale is ecnatsiD metr ic . × d km morf a time calculated. km emoh no t that = min Rem embe r times Speed )mk( • and hours The eA distances istac istance aunt missing 3 T ime 4 5 (hours) fo r The hour sua and re is then rst part of minut es. h gadit hen arc from 8 is a eALe skydier . ms to seconds e ms later, he graph has a gradient of = kmh. . of d–tim Wre the each speed is a distactim gah sts th sd gahs constantly changing, a speed–time graph is more useful. umps after opens from a plane seconds. his e at has parachute. an altitude then is of reached speed m. terminal immediately is speed elocity , drops towards and to a falls earth at a constant increases constant ms steadily speed. until he lands. arc’ s speed–time graph looks like this: 70 The rst section of the graph shows a steady 60 increase in speed. This is called acclatio )s/m( cceleration second measured per in ms , read as ‘metres deepS per is second’. arc’ s acceleration is ms , because 50 40 30 eery 20 second his speed increases by ms. 10 This is the gradient of the graph. 0 0 h gadit of a 5 10 15 T ime gah e For know the This is that distance freefall the during section area h So sts of aa the base × the rst th = of a speed the height is stop, a a of × is ms the graph. gah distance × tlls traelled is us the s = th m. distac area of the tavlld green triangle, = 4 m. en graph gets that section the ere time. graph, = Ar × sd–tim stage, 25 (seconds) acclatio yellow ud 20 sd–tim of bus A dward’ s and then ourney walks the to work. nal e walks to the bus • part. se the how )mk( 16 far graph arc emoh morf ◦ second ◦ seconds ◦ seconds ◦ 4 seconds ◦ seconds ◦ seconds ◦ seconds to calculate had fallen after 14 12 10 ecnatsiD 8 6 • se this information to draw 4 a distance–time graph of the 2 rst 0 0 10 20 30 40 T ime 50 60 hat n is the aerage speed of part of the ourney the bus in the did he walk faster to the bus obin is descent. graph n The gradient graph of a distance– shows the speed. bus running in the m race. e accelerates from The gradient ms in seconds. e maintains a speed of ms of a speed– ms time to his or of your kmh time from is 70 which seconds shape (minutes) e hat graph shows the for acceleration. seconds, seconds. Show how this far and then starts slow down, so that he stops in information he to runs on a altogether. speed–time graph, and calculate The area graph under shows a the speed–time distance traelled. odul upert is has sister Their actic n ucinda brother has dward more has than twice as upert. much am atch on the the ustios relations on the left with the types right. as rlatio y ucinda. ow If a much = , b money = − do and they c = hae 4, altogether nd the alue y = x − y = x ne–one + any–one a − b y √ = x − ne–many of c x x The b a Find = x operation ab − is dened by a Find the alue of b Find the alue of c hat xplain a raw b your answers tell you about xx − − of the function why the inerse is not a function. axes for raw from − to 4 for x, and − to y the graphs of y = x − and y xpand inerse x b do = the binary a + + Simplify c y = se x + your graph euations y = to x sole − the and simultaneous y = x + Factorise ake fully x + x − the subect of the formula the passes euation through of the −, straight and , line that −. = + a 8 8 Find Sole the euation x − = x + line goes from −, − to 4, . Find: a the length of b the midpoint the line Sole the euation x − x + = by − 4x − = by of the line. factorisation. raw axes from − to 4 for x, and to 0 Sole the euation completing the x for suare. y ark and Sole the simultaneous a + b a − b = rectangular w m The 0 4 x Find garden is l m long x length + w = x + and gx = x by − a b g− and width are connected by the c gx d x > , y > a raw = x + w the aries hen = length x directly = and width of the , y as = the suare of x when the alues garden. y se of your when x graph x from graph = of y = x − to . to nd − the x, taking alue of y −. . c alculate y = stimate . x 0 contained x and b − wide. l region Find: euations l ⩽ euations = the y = . the gradient at the point where atch the graphs x to their euations: a y = – b y = c y = x − x y y y x ere is a x speed–time graph of ina’ s x race. 10 9 8 )s/m( 7 6 deepS 5 4 3 2 1 0 0 5 10 15 T ime a alculate b Find The the mass coin ina’ s total of with a acceleration distance coin aries diameter ina oer the rst seconds. ran. directly . cm 20 (seconds) has with a the mass suare of of the of y diameter. g. alculate: a the mass of a b the diameter coin of a with coin a diameter with a mass of cm of g. alculate the points of intersection of the graphs = x − and x + y = . 4 Geometry vectors 1 & & trigonometry and matrices Pptis lis ad als Pits LeArnIng ad • A • nderstand the terms pit ray , line segment, intersecting that it does lines, acute, right, plane, solid face, and li angle measure properties opposite, is a angles problems ay • A li using is straight, smt has to no see size. it, so Of course when we this mark a size, but mathematically we assume it a has no height. continues in both directions without end, and apart has is a start part of a point line; but it continues has a to innity. beginning and end, and a length. • Paalll vertically alternate, point, or straight, lis along are their lines that entire are length, the same and so perpendicular never distance meet. cointerior , • Itscti at but able thickness. denite geometric have width no • A accurately • olve be edge, verte • raw position not obtuse, has ree, would curve, • A angle denite lines, length, perpendicular a we parallel point lines, has point, means line, lis ooe complementary lis are lines that meet at a common point. and • Ppdicula supplementary • A cuv is a lis meet smoothly at right owing line angles. with no sharp changes in direction. Als An al Angles is are a measure commonly of the turn measured from in one degrees. direction to degrees another. ° make o a full turn, so ° is a half turn and is a uarter turn or iht al Acute less e ight than ° = ° to use a protractor measure rotractors have two clockwise other t Obtuse greater than °, less than ° is the angles. usually scales, and one the anticlockwise. important correct o measure of the centre zero an zero this line use angle, lines mark he to scale. is scale eactly in place eactly the the protractor coincides on the with on one the arm angle of the 132 than ° is an illustration anticlockwise that one and the verte. is the inner scale, and ee greater so angle, turn. he angle is °. to turn from o measure part n of the he the a ree A full of half is angle, turn, angle at turn sum ree illustration, Als A a full a and the = obtuse − pit °, so instead subtract angle ad angles measure your = is remaining from °. °. °. als that the answer make up a a staiht full turn li have °. turn, or ° angle, makes a straight line. AII e can ook at use these these facts to calculate missing angles. • A diagrams. full turn angles, b is twice three d 130° 110° is a a + the + = a + = b + = ve b = = size the full into and times d, of size the of four so a, of is of two ppsit lines cross, a split in the ratio ind the size of angle. full turn is split ° nine angles in the ratio ind als the are and angle. . hen is ° into tically that c a, size each turn angles • Another a the . each split c size • Another into is b, times four ind b a, the angles opposite each other at the size of each angle. verte eual. hey are called vtically ppsit angles. a Angles a and c are Angles a and b eual, and angles b and d are eual. c called wo add up to supplmtay angles that add °. wo angles that add up to ° are als up to ° are called cmplmtay als Ar e PoIn 1 A point 2 A line eIon x y has has a no denite ends, a location ray has but one no end 48° size. and a line segment has 63° two ends. z 3 arallel lines Angles can never meet, perpendicular lines meet at right angles. 1 and ° Angles to at be or a acute ree point < > sum °, right °, obtuse between to °, angles on a straight line omplementary add up opposite to °. angles x, y and z is a ree c obtuse acute sum 2 ertically of °. °. hich ° angles angles are add a eual. up to °. alculate upplementary angles 3 the x plain the between a size of angle y z difference line and a ray. 133 2 Paalll spdi LeArnIng geometric problems alternate, corresponding; parallel, lines are identied on a diagram with arrows. using A angles als ooe arallel • olve lis tasvsal is a line that crosses a pair of parallel lines. adacent, cointerior; he transversal creates four pairs of cspdi als transversal orresponding angles are pairs of angles in similar, or corresponding positions. ou can imagine orresponding n the are angles diagram marked in on although down back on transversal of are angles of on are to the other. eual. corresponding angles an upside lines the of parallel the make might creates two Alternate lines, transversal. a shape, be back to front stretched. ecause of the rotational symmetry of the IP diagram, no t assu me paralle l ques tion If you to give your sides the lines make be als. between alternate although or parallel altat Alternate • might A on are it pairs pair front. als angles Do to the colour. always Altat pairs • angles of parallel left, same shape, or one on the the orresponding eA sliding are angles on parallel lines are eual. lines unles s tells alternate the iti als you. n the diagram aske d reaso ns answ er , fo r use the a = b alternate b + c = ° angles angles on on parallel a lines straight line a co rrec t mathe matica l o vocabu lary. • If a you are calcu late mus t no t aske d an + angl e, must eual b °. c measu re Angles like a and c inside a pair of you parallel lines and on the same side of it. the are 13 c to transversal called are supplementary, cointerior angles. hey and make a like shape. lvi B plms E n comple n the diagrams, diagram, the we use angle of three ° is letters called to identify angle A an or angle. A. A D magine drawing the angle with a continuous , we movement; you 53° would draw from indicates the A to verte to where so the name angle it is. A . he he rst middle and last G letter pairs of F C letters the A arms Angle and of the need to any angles show reuire calculate you ore the lines which calculate angle correct other eAPLe calculate the surround the angle. hey are H angle. problems you o angles on the choice along of the angle way. fact. t ometimes helps to write diagram. AII 1 in the diagram A above parallelogram pairs A = = ° orresponding angles on parallel of + = ° Angles on a straight = − − = that the angles add the opposite up • rove °. that angles eual. PoIn 1 he 2 orresponding an angle A Alternate ointerior of the make the angle angles angles on angles transversal. a Ar the is on at , between parallel lines are A and eual. . hey make shape. 3 se two °. are e of line to consists lines. lines • rove + parallel parallel are lines inside are the ointerior eual. parallel angles are hey lines make on the a shape. same supplementary. side hey shape. eIon diagram to answer these B uestions. C A 1 hich angle is alternate to 2 hich angle is cointerior 3 f D with E angle a = ° and = °, calculate c angle G F H 13 3 Pptis ad tials uadilatals ials LeArnIng ooe A A • nderstand properties euilateral, of x is properties uadrilaterals; D triangle. etended or pducd to . right, isosceles • nderstand a of triangles; is suare, is e drawn know A = parallel to A. angle A = x x y z y alternate E rectangle, rhombus, parallelogram, kite, angles geometric congruent similar B lines C A = = y corresponding angles on parallel lines problems he using parallel trapezium And • olve on three angles at A, A and show that triangles, x gures o the he + angle z = can of and in any have angles angles ° angles sum acute acute + three riangles two y angles the an a scal different is acute right obtuse Acute-angled • A a triangle, triangle three and on x straight + y + line z = ° °. angles angle angle an a acutald ihtald an tusald Right-angled triangle has all sides of tial, tial or two tial. Obtuse-angled different length and all angles sizes. • An isscls triangle • An uilatal has triangle two has eual three sides eual Scalene and sides two and eual three angles. angles Isosceles of °. Equilateral uadilatals uadilatals uadrilaterals the angle sum can of uadrilaterals hese properties may perpendicular. foursided be all ome diagonals 13 are split two uadrilaterals have include be into shapes. eual, special eual they triangles is × by a = diagonal, and so °. properties. sides may and bisect eual each angles. other, heir and they may be he table below lists those shapes and their hap properties. ids uare All sides Als eual. Opposite sides All four angles = iaals iagonals ° parallel. eual, bisect, perpendicular. ectangle hombus sides eual. Opposite sides parallel. All a sides four angles = iagonals ° eual, bisect. eual. Opposite Opposite sides parallel. angles eual. iagonals bisect, Opposite sides eual. Opposite sides parallel. perpendicular. a a Opposite angles eual. iagonals bisect. iagonals eual. b a b rapezium One pair sides sosceles All b b arallelogram Opposite One a a trapezium of opposite parallel. pair sides of opposite parallel, the wo other pairs Opposite of eual angles angles. supplementary. b pair ite eual. wo pairs adacent a of eual One pair of angles eual. One sides. diagonal bisected, a perpendicular. eA IP AII • • raw each of the uadrilaterals shown in the given • T he the in diagonals the to check the diagon als line table. • raw T he of properties table. • of diag onal s sym me try T he perpen dicul ar are An (of order nd if there there or is A is a vertices . is a line sides. ro tatio nal more . two them triangle angles. has here an angle are two of °. possible ind the size answers; of can PoIn the you 1 both riangles isosceles can or be uadrilateral has three angles of scalene, euilateral; acuteangled, 2 there opposit e o ppos ite if if eIon isosceles other equa l bisect e 1 throu gh throug h diag onal s sym me try Ar are symm etry rightangled or °. obtuseangled. a alculate hat the fourth angle. 2 type of uadrilateral is he special suare, 3 A uadrilateral are eual but has do two not pairs of uadrilaterals are it eual bisect.hat type angles of and the diagonals uadrilateral is it rectangle, parallelogram, isosceles rhombus, trapezium, trapezium and kite. 13 Pptis plys Plys LeArnIng ooe A • nderstand angle ply more of is the general name for a closed shape made of three straight sides. polygons h • olve problems in als i a ply geometry • A heagon triangles o • A the × angles the = they hese an angles sides, a of has have eti and can be add up split into verte. a heagon to °. angles enerally, o has from decagon o of sides, a nsided an polygon angle inside als and decagon sum the make of can split into up to can be split n polygon a be add − are straight × × into triangles n from = °. − a verte. triangles. °. called line with iti the als interior angles. AII • magine on the • tart the – or draw – a triangle oor. halfway sides sides of and the along walk one of along the Interior angle triangle. Exterior • y the time you return starting position, made complete a • epeat with you to will turn, other always because eterior turn have or °. o polygons. you round turn angles, an up to to angle and the interior angle are supplementary °. he the eterior angles of any polygon add up to °. always A add eterior up once, through which angle your add ou ula ply has all sides eual and all angles eual. °. o a regular nams heagon ad num has al sids eual angles pptis of ÷ riangle uadrilateral entagon eagon eptagon Octagon onagon °. plys nam = Al sum ° AII he maimum angles in a • ind the number triangle uadrilateral of is is interior of , right but in × = ° × = ° × = ° × = ° × = ° × = ° × = ° a . maimum pentagon right and in number angles a in a heagon. 13 or properties ecagon lvi he plms interior solve and eterior angle properties of polygons can be used eA to • or eample, make a a set of congruent regular pentagons are connected egu lar equa l to prove that they form a regular decagon at the o eterior the o the angle interior Angles all a and of a angles b are remaining regular are each angle all pentagon − = = ÷ = °. = eter ior °. and °. c − × regul ar centre the he means and all sides angl es equa l. loop. • or o IP problems. = • °. or ÷ the of inter ior eter ior sum inter ior the angl e num ber the − po lygo ns, of or sides, angl e = angl e. the angl es, fo rm ula = learn use the sequ ence. a c b ut the And ut o eterior the this the interior is PoIn 1 A regular 2 he 3 × 1 A 2 his and a each nd A has suare it is of a − angles regular eual an of three decagon in = = ÷ = °. °. the central shape. decagon. sides nsided a polygon and eual polygon angles add up to add up to °. of the ° size each, of one and of the these other two angles. pattern octagons meeting se interior angles alculate regular verte. this at fact angle of to a octagon. regular does the has angles eual. two the = regular eIon are regular 3 shape tessellation shows all angles pentagon angles for a °. eterior Ar of angle polygon interior − he true central e n angle polygon have has int interior nd the angles size of of an °. eterior ow many angle sides rst. 13 stucti here LeArnIng • onstruct °, a number of constructions we can perform without a ruler. ooe hese °, are als angles of °, °, are really used ° are • onstruct parallel perpendicular are to sometimes straightedge draw used called for straight marking rulerandcompasses and lines compasses rather eual than constructions, constructions, to measure. as he the but ruler is compasses distances. and lines stuct An a euilateral al triangle has ° angles of °. o, to construct an angle of C °, we irst, draw Open the B A iagram mark the line, and line vertices and compasses, crossing eeping a the the draw it at mark put , arc to a the and compasses an of an point, point the open cut euilateral the A, on other at the other triangle. near A one and above same arc at end. make the two arcs; one along line. distance, put the point on . B oin A to complete the angle of °. iagram A iscti o bisect the iagram a li line length smt segment AB, open the compasses to more than half AB ith C the epeat compass with iagram he B A iagram a line the point on compass A, draw point on arcs B, so on either the arcs side of AB intersect. oining the iscti a o cut intersections bisects AB al bisect or compasses each arm sing C raw on of and the eactly the the angle. as angle in verte, half A, an and iagram centres, bisector draw from angle, make the point marks at of the and along two A put eual arcs to through cross inside the angle at . . D A bisects A. angle iagram stucti o construct o construct make Opening B is a diagonal of the rhombus a ppdicula t a li a perpendicular a right arcs the B angle and C, at at compasses to a A, line, place eual wider, we simply the compass distances make bisect on eual oin A angle point either two an arcs side on A of A from B C A 1 A and iagram because B A iagram A to cross on the intersection to same side construct of the the line. perpendicular. to the point iagram of of and °. C stucti th als AII • An angle of ° can be constructed by bisecting an angle of °. • An angle of ° can be constructed by bisecting an angle of °. sing straightedge compasses, • An angle of ° can be constructed by constructing a ° and construct angles of angle • ° on a straight line; the obtuse angle is °. • ° • ° stucti a ppdicula m a pit t a li hat o construct a perpendicular from C the other angles could you construct A point A to the line , put the point of the E compasses at ith more on and . and eual A and draw iagram as to arcs crossing centres, arcs two draw intersect at two . eA D A is B • perpendicular to lwa ys is because A is a diagonal of the • C iagram il fo r sharp cons truct ions. tra igh te dge A A. a penc rhombus use . iagram his IP compa sses and cons truct ions E do no t invo lve measu ring. • ak e sure your D F compa sses B enou gh iagram o cstuct construct a a line pai through paalll parallel so are tigh t they do no t slip. lis to A, F rst draw a line se compasses from to A through draw three . arcs of the C same radius F D • entre A, at • entre A, at on A B G on A C E A • ong arc, centre at on A produced. D iagram iagram B E Open an the arc e 1 to compasses cross the to long the arc distance at . . is ut the parallel point to A. PoIn All the on to A draw iagram iagram Ar angle constructions markings. ake symmetry to sure help you you use can symmetrical see remember he parallel copying measure an lines construction angle, and copy and a uses onstruct 2 isect distance. an angle of °. the your angle of °. heck your them. relies the eIon 1 accuracy 2 by measuring with a protractor. on compasses to 3 plain how you could construct an angle of °. 11 stucti ad he LeArnIng tials plys constructions here involve a ruler, protractor and compasses. ooe • onstruct triangles • onstruct regular • onstruct irregular stucti polygons polygons e can tials construct a triangle • he length of all • he length of two of two A; • he • he in between length of stucti o draw • raw • ith and a cm the a and draw pieces of information or and the size of the and the size of an angle between them A; and sides the of rst open to shown open the angle other than the or size of two angles AA. tial side draw connect a between • raw cm, cm and cm put the point on one end of the line blue. cm, arc two cm, cm. in to an ends of to put the point cross the rst ends of the line on arc to the other shown the in end of the green. intersection of the with line stucti to two a construction sides side angle the tial of cm and cm, with an angle ° longest an this of A triangle the • tend a them measure his three ; red. stucti o arc compasses line, side with longest an sides compasses • inally, arcs sides them one triangle draw • ith given sides or length angle three of cm lines A is cm ° at rst, one shown green and then use a protractor to end. in to blue and complete oin the the remaining triangle. tial known as the amiuus cas, as it usually C 1 gives o rise to construct two a different triangle possible A A = with answers. A = raw A ° at cm long, shown ith A to and in = cm and point cross mark an angle of blue. compasses the C cm, ° on the open A, to draw ° line cm an and arc shown in green. 2 he arc crosses labelled at and two . points oining B either of these to A gives a A triangle 12 matching the description. stucti a o triangle construct angle A a = AA ° tial A and with angle A A = = cm, ° C • raw the • easure shown • he an in the ° cm angle of long. ° at A at − angles tend A blue. angle = to line in − a shown the = °, triangle in lines if as add up green. necessary so B they intersect at stucti egular A full o is ° sides by • irst, can be plys constructed inside a circle. °. construct with A ula polygons turn . a regular a draw polygon nonagon, a ÷ = circle divide °. and a eA • rom the radius, ak of ° e sure shown in co rrec t the ends of the radii the regular the in green. line stucti o construct sketch rst, iula irregular and polygons plan the 1 order including in which uadrilaterals, to carry out make the , marke d lengt h no t of from . a . a construction. AII onstruct a with of parallelogram PoIn A triangles usually triangle and one egular polygons have two solutions, obtuseangled one acuteangled triangle. 2 can be drawn inside a circle by dividing sides and an he diagram , 2 from the plys 1 e scale the on turn ing nonagon • ea sure shown by to from construct use scale blue. pro tra ctor • oin you measure the angles IP radius. aint the number of of and cm °. shows incent arbados and ° renada by cm angle . sides. 100 V miles B Ar eIon 60° 1 onstruct a triangle A where A = cm, angle A = ° G and A= °. easure the lengths of A and . sing 2 onstruct a regular 3 onstruct angle the a A length pentagon = of A. a scale of cm to octagon. = A = with °, A = A = = A = = °. = cm, easure miles, drawing. the make an angles an you accurate construct without using a protractor 13 imilaity ad cuc imila LeArnIng f • se tials ooe properties of the three angles of one triangle are eual to the three angles of similar another triangle, then the two triangles are identical in shape even if triangles they • dentify and use are different in size. congruent uch triangles are called simila tials triangles he lengths of each pair ore eAPLe n triangle ABC, n triangle DEF, of corresponding sides are in the same ratio. 1 o angle B = = D o angle D = = 7.8 cm A o angle A = angle F 84º Angle B = angle E Angle C = angle D 15.2 B orresponding cm 41º sides 9.5 cm C are 55º AB and FE, AC and FD, BC and . ED E = = 84º ED 10.0 F cm . . BC o the DF = AB = sides . EF of × ÷ DEF AC . ut = = are . . × . × . ÷ the = . sides . = . of ABC cm cm tials A riangles size he the he are that are called three three the cut angles angles three eactly sides of of of one the one same shape and tials triangle other are eual to triangle. triangle are eual to the D three here to be sides are the four sure . of that . other sets of two he triangle. minimum triangles three reuirements are ides of congruent one triangle D A are of . the A. the are E B 1 C F eual in length other he the three sides triangle. triangles included eual. to have Als two the pairs angles of eual between ides these and two sides . AA pair of . he triangles corresponding have ides two pairs of eual Angles and one eual. D A A D E B . F C r. he two triangles are right angled, have eual ypotenuses B and one other ore riangles are ABC and not so this and sides H is riangles JKL F E 2 are D O angles the eual IHG he between BCHG C ides. C and A G sides A. and A. are eual FDE eual not ABC congruent B and he between riangles of eAPLe congruent. E pair and are eual the angles B BAHI and MON C E F I H eual P M J S Q are o congruent JL = MN . K = O hypotenuses = , and K KL = riangles Angles 1 O JKL and JKL KT PQR and are are UTS congruent are eual, O as are AA orr congruent. LS, but the . he K = Q, L = hypotenuses eual R and JLUS sides KLSU same shape. are JL = are not PR not in which eual are so corresponding corresponding imilar triangles have sides eual are in angles the so same are the eA ongruent proofs not apply. positions. of triangles are congruence the are same , shape A, IP AA and orr size. he and or the cong ruen ce, four angl es . • Ar or the eIon be ABCD is a parallelogram. parallel to A B the the equa l in proof the mus t be tween is does sides. proportion. • EF U PoIn orresponding 2 T R riangles e N L ON be the equa l o rr sides of equa l angl e sides . proof , have to co rres pond ing positio ns. AD E 12 AD FC 1 = = DF = cm cm and cm. rove ABC cm, that are triangles ACD and congruent. 2 rove that 3 alculate triangles the D CEF length of and CAD 3 are cm F 5 cm C similar. EF 1 Pythaas’ Pythaas’ LeArnIng ythagoras’ nd missing angled theorem sides in theorem rightangled theorem alive and around is best bc. e remembered was for a his about triangles. the theorem had been used previously by abylonians and it is believed that ythagoras ythagoras’ theorem states was the rst person to prove problems AII result. rightangled a the the suare on each side of smallest a cm, and the suare the the has middle largest side one − to on each triangle, areas of that side then if of a a the the he red suares the larger area is of suare. triangle rightangled is a triangle. c cm, he b the smaller eual triangle. length drawn triangle. two • raw is rightangled sum a b cm was to suare • f who philosopher, right the of reek and ndians, raw a mathematician triangles ythagoras’ solve was to Although • se thm ooe ythagoras • se thm sum of the areas of a calculate the green suares is eual to b − the area of the yellow suare. a • easure cm clockwise si from each b cm verte suare. of oin the the Pythaas’ marks ore as eAPLe 1 shown. oe • ut into ut out the four out • Arrange eactly to thm middle suare pieces the the along smallest ve cover show and that the cut the lines. ythagoras make e to largest o a it diagonal suare. pieces makes rectangular wooden frame for a door. it suare he wants to put in a piece. draws length rigid, a and sketch, width and of marks the on it the door. 2.1 m was oe correct. realises theorem. to that o he help, can he use colours show the rightangled e does not draw e knows the the area ythagoras’ of the diagram triangle. suares. the bottom suare 0.9 m b a would be . × . = . m 2 he suare on the left would be . × . = . m o the large suare would have an area of . + . = . m he to diagonal the he f nearest longest angle. the t is side shorter 2 1 2 + b 2 = c ., so the diagonal = √ . , or . m cm. called hyptus a = of sides is the the c, of triangle is always opposite the right hyptus a then rightangled ythagoras’ triangle theorem are can a and be b, and written the as lvi oe plms realised sosceles the usi diagonal triangles also Pythaas’ of can a rectangle be split into thm created two rightangled congruent eA triangles. IP rightangled • ytha goras’ triangles. only work s angl ed ore harlotte eAPLe is he is he screws hanging going to 2 a hang large it on a painting wire on the attached wall. to two lwa ys • dd . m apart, and the wire is . m the diagram shows the a diag ram . squa res the ubt ract when hypo tenus e. the squa res long. when he draw ndi ng • are right trian gles. • screws. theorem for arrangement. ndi ng shorter one of the sides . 1.5 m 1.6 m harlotte will wants to know how far down from the screws the wire reach. he draws two a sketch. congruent he triangle rightangled is isosceles, so she divides it into triangles. 0.75 m a 0.8 m sing ythagoras’ a a theorem, + . = . + . = = . . . a − = . a he wire e 1 = he f . will = reach . m . cm the screws. a Ar longest the and b side of a rightangled are hypotenuse, ectangles into two triangle is 1 hypotenuse. perpendicular then and a b isosceles congruent and + A c is the 2 A eIon rightangled and 3 below PoIn called 2 √ cm. suare triangle ind has the a has length diagonal shorter of of the sides of cm hypotenuse. cm. ind the = c length triangles rightangled can be split triangles. 3 An of side of euilateral alculate the the suare. triangle area of has the sides of cm. triangle. 1 ymmty ad tatis rcti LeArnIng symmty ooe A • dentify ctis reection and shape has cti symmty if one half is a mirror image of rotation the other half. symmetry ere • ind the the image obect of given an the obect are two eamples, with the mi li shown as a dotted line. or image AII ome he amaican lines of ag symmetry symmetry of has and order two shapes others have have a display of ags ark any symmetry and write of rotation a school possesses forms eA of either the lines of symmetry symmetry, same. th No lines of symmetry the shape or image can be rotated and it still ow d many matches tati there are as you go around once symmty symmetry . ag or are some eamples. that both symmetry. Order es withou sym me try having sym me try are t 1 Order 2 Order 3 ro tatio n describ ed rcti ro tatio n of order A cti is a cut tasmati . t changes hen we the position reect transformation the 1 while IP • h ap as symmetry, symmty rotation called ere • esign reection the is order of lines looks of line that ith symmetry . one rotation rtati have than . Two • ake more none. ct the a is of shape the the in same original a shape without mirror line, the perpendicular shape, but on affecting ima distance the its the from opposite shape side result the of or of size. the mirror the as mirror . o reect the ag in the mirror line y = x + y 5 • urn the diagram so that the mirror line is vertical. 4 • easure the horizontal distance 3 of each verte from the mirror 2 shown in green, and measure 1 an eual distance the other y x 0 of the mirror in red. oin 3 5 side 2 1 1 2 3 4 1 to create the image 4 vertices 4 the 3 2 2 2 blue. 1 o dene a reection, we must 0 line. eA IP 3 − − 2 2 − mirror − the 1 1 name − • T o reec t the paper a shap e, so that rtati mirr or line is congruent transformation is as the 6 tati. hen we rotate a shape, it is the shape is on a large wheel. e reec tion is then as ho ri onta l. 5 if the verti cal, y Another turn need 4 to know what angle to rotate it through, 3 and the position of the centre of the 2 ‘wheel’. 1 o rotate the ag in the diagram ° x 0 2 anticlockwise • magine at , the wheel, point with the , 1 1 1 centre . • otate and a about the the diagram ag is now ° anticlockwise, horizontal, with the eA top of the agpole • T o directly above the centre of rotation. se tracing paper if it IP units ro tate helps. tracin g a shap e, paper y the paper 6 imag 5 e to or see shou ld use ro tate where the be. 4 3 2 1 1 e PoIn 1 shape x 0 2 1 1 A has symmetry line Ar that escribe fully the splits the is a mirror maps A onto and its shape into an reection. y transformation 2 that reection there eIon obect 1 if . he order of rotation symmetry 5 is the number of matches with B 2 escribe that fully maps the onto transformation the C . original through shape as it rotates °. 3 3 escribe that fully maps A the onto transformation 3 o dene a reection, give the A . euation of the mirror line. 1 0 2 1 x 1 1 2 3 o dene centre of a rotation, rotation give and the the angle 4 and direction of the rotation. 1 1 uth tasmatis aslatis LeArnIng ooe A • epresent translations taslati is a congruent transformation. A translation using changes the position of a shape by sliding it, but does not alter the vectors itati • ecognise positive, scale enlargements fractional and with ranslations negative us how far of the are to shape dened move it by the does a not clum shape reect or vct. horizontally rotate A and it. column vector tells vertically. factors move x units to the right; a negative number means x • ocate the image under a ( combination of y means ) move ( transformations move y units up; a to negative the left ) number means • he translation that moves A to is down − ) ( move ; B to C is ( ) − y elamts 4 An enlargement is a simila tasmati as the image is the 3 same shape as the obect with the same angles, but a different size. B he 2 original dimensions y C are increased by the same 1 6 multiplying 2 or scal act 0 3 factor, 5 x 1 1 2 3 4 5 A 1 Additionally, the distance of 4 each verte from the ct 2 3 lamt by the same is scale increased factor. 2 C n the diagram, is the 1 centre eA of enlargement. IP 0 he • T he centre • or Draw of outsid the centre help • T o of with nd enlargement e can the rays be on, shap e. from the enlarg emen t the the triangle is 2 an to draw ing. centre with a ach side the he scale side of of the of the red distance scale the factor red of the of and rays 1 triangle . blue 2 triangle triangle, from factor as can and the be is each times the verte vertices of of the length the red a given imag e, back to inter section . a of blue the corresponding triangle triangle is times are. y of fraction; a scale factor of enlarg emen t obec t x 1 1 enlarg emen t in blue the draw point produces size of an the distance image obect, from the half half 7 the the centre 5 of enlargement. Although 4 the image is the obect, smaller we still than call it an enlargement. 2 A negative scale factor 1 produces other an side image of the on the centre of 0 3 enlargement. he 2 1 1 shows factor 1 an enlargement, −, centre , . x 1 diagram scale 2 2 3 4 5 6 7 mii hen we another two perform a transformation transformation transformations ore 1 tasmatis otate with eAPLe A on the a on image, single an we obect, can and then sometimes perform replace the one. 1 eA y ° anticlockwise about • lwa ys • If a 6 , , image ranslate is . B don’ t 4 ) , image is a diag ram . asks for a trans form ation , give a stag e answ er . . D draw ques tion singl e C through − ( IP A E 3 ingle from transformation A to 2 otation G ° anticlockwise 1 about −, AII F . 0 3 2 eect A in x = 2 1 x 1 2 3 4 5 raw diagrams these are to show that , 1 image is eect in y = 2 , image is ingle transformation from A • wo . be to about A can a always single translation. , perpendicular reections . − ranslate by ° can 3 translations replaced • wo rotation true. . through be replaced by a rotation. ) ( always , image is . • wo nlarge , scale factor , with centre −, , image is parallel always ingle transformation from A to enlargement, scale reections can . factor be replaced by a , translation. centre , . Ar eIon e PoIn y 1 ranslate triangle A 1 − through ( ranslations column 6 are dened by a vector. ) abel the image 2 5 . nlargements a scale factor are and dened centre by of 4 2 nlarge factor −, of . with , a scale enlargement. centre abel 3 3 A the ome combinations of 2 transformation image replaced 1 3 escribe single fully the to transforms A by a be single transformation. 0 transformation 3 that can . 2 1 x 1 2 3 4 5 1 . 2 11 11 hdimsial shaps lids LeArnIng ooe hdimsial • olve geometric A using • se faces, classes pyramids, edges, of shaps are sometimes called slids problems solids cylinders, twodimensional shape is a at shape, such as a triangle or a circle. vertices prisms, hreedimensional shapes have volume as well as surface area. cones, lasss slid sphere ome slices solids in olids ere the are one of are this type some entire the same direction, are shape we called eamples. shape. idden rise from Square-based solids acs • A he Vertex Edge • wo surface faces • dges yellow edges are meet the ad on meet at a base to pyramid include ds at the a at a their same length. f we take cssscti crosssection shown Cylinder shapes Other get pisms Cuboid Other throughout always a with Triangular point. is constant dotted hese through lines. prism are the T etrahedron pyamids Cone sphere. vtics solid an is called a ac d vt. he plural shown here of verte is vtics Face he dodecahedron has many hidden faces, edges and vertices. he whole ach pentagon ach ÷ ach edge = edge formed 12 shape by is is has made ve formed of sides, where so pentagons, that two makes sides a so it has total coincide, of so faces. × there = must sides. be edges. has two three ends ends, making so there a total must be of ÷ ends. = ach verte vertices. is eonhard rule that uler, for an any thcentury cv wiss mathematician, plyhd, the number of discovered faces, F, the edges, eA E and vertices, , are connected by the • F + V = E + IP formula se the shap n this case, + sym me try of 2 = + es to hidd en help faces, coun t edges and verti ces. A conve interior A polyhedron is a solid with entirely at faces with no ree angles. cylinder is not a conve polyhedron because it has a curved surface. AII lids ad symmty • ount wodimensional shapes have mirror lines, or lines of symmetry, the vertices dimensional shapes have plas symmty. A plane is a number of faces, but and edges for at a cuboid a suarebased a tetrahedron a triangular surface. A cuboid the has shape three were cut planes in of half, symmetry; one half that would is, be a three planes mirror where image of if the pyramid prism. other. • heck he cuboid has its planes of symmetry marked with coloured agree lines. F A cube edges has and nine si planes through of symmetry three through the midpoints + that with = E your answers uler’ s + ule, . of vertices. AII ind the four symmetry Ar eIon 1 has A cuboid through the a suare • a triangular • a suarebased e hole passing 1 centre. many vertices faces, does the olids edges 3 ow uler’ s aw hold for this planes of and uler’ s shape vertices faces corners. ule + vertices symmetry = edges + olids have planes of does symmetry the at where shape 3 many faces edges and faces oes pyramid. have 2 2 prism have surfaces, shape of PoIn meet ow planes of – the surface where have a knife into would two cut the mirrorimage shape pieces. 13 12 LeArnIng ooe • etermine acute imty the angles trig of AII ratios in raw a rightangled triangle with an angle of °. rightangled • easure the length of the side opposite • easure the length of the hypotenuse. the ° has an angle. triangles • se trig ratios in rightangled • hat triangles in realworld do • hat geometry and scale notice happens if the rightangled triangle angle of .° drawing, bearing, heights, angle elevationdepression of you practical distances, imila ihtald All rightangled o if in one triangles such tials with triangle, the an angle side of ° are similar . ppsit AII the A mnemonic such as it ‘ome ° will angle be true is half for all of the ° hyptus, rightangled then triangles. 30° Old airy amels Are airier e say that, for an angle of °, Adjacent han Others Are’ can help opposite remember the Hypotenuse you side side rules. = .. hypotenuse • an you one you t make has will to up be a better something his is called the si of °. remember. Opposite he third because side it is is called adacent, length of the or adact net the to, the adacent side side, angle of °. side he result of is length opposite of the called the csi of °, hypotenuse side and eA ou mus is adacent IP t hese learn these are called the tat usually abbreviated to the = si nd Alan he the looks looks length up up at a top the letters, so we write cos A = opp , tan A = hyp ad timty 1 of the from three , hyp To rst ad opp sin A eAPLe °. three fo rm ulae. ore of side side of a (1) tree. horizontal he is al lvati the angle °. x e is m e uses e draws e calls from the foot of the tree. Opp this information to nd the height of the tree. 35° a sketch, and labels the sketch. 20 m the adacent formula 1 side side, with he and wants wants adacent to to and calculate calculate opposite x. the in is e knows opposite the the side. tangent he formula. Adj opp e writes tan = ad is calculator tells him that tan = ., so he writes x . = x = . o the To nd An height the isosceles e he e × are of know = the length to . m. tree of triangle going dotted is a has nd side two the adacent line side plus the height of his eyes from the ground. (2) angles length perpendicular the . m, of of the creates and ° other two want and a base cm. sides. congruent the of rightangled hypotenuse, so we use triangles with an adacent side of cm. cosine ad cos = hyp Hyp . = x x . x × Opp x = ultiplying both sides by x = = . cm to decimal place . Adj To calculate an 8 cm angle 2.8 m elody measures the end of her shed. t is . m wide. Hyp Opp he roof is . m long. x elody wants to calculate x, the angle of the roof with the horizontal. Adj he knows the adacent and hypotenuse, so she chooses 2.4 m cosine. ad cos x = hyp . cos x = = . . As she knows the cosine and wants the angle, she must use the inverse function of cosine, cos .. er calculator has an key. x = cos . = .° to decimal place. e 1 Ar hese eIon uestions refer to PoIn rigonometry angled triangle is used in triangles. ad opp A. A 2 sin = , cos = hyp 1 A = cm, angle A = right , hyp °. opp alculate . tan = ad 2 A = cm, angle A = °. C B alculate 3 dentify sides 3 A = the known or wanted A. cm, A = cm. alculate angle A. the and angles correct to choose function. 1 13 uth uss timty rigonometry LeArnIng some • se trigonometry in of of triangles elevation • se a number of practical uses. ere we will eplore them. right Als angled has ooe and with depression trigonometry lvati ad dpssi angles ore eAPLe 1 with eter is relaing on his hotel balcony . ohn is standing on the beach. bearings hen he they angle look at each upwards other, from the ohn looks horizontal is up through called the angle al x lvati y eter he b looks angle down through downwards angle from the y horizontal is called the al dpssi he f ohn he x angles can e is eual, m use knows uses are from as they the trigonometry the adacent are hotel, to alternate and the calculate side, and angle the wants angles of height the on parallel elevation lines. is °, b opposite side, so he tangent. opp tan = ad b Hyp . = Opp b 200 m b = . m to d.p. Adj ais e or A can use compass bearings to describe a direction, such as orth, outhast. more Angle precise method bearings always written are is to use measured with three al ais clockwise, digits, so a from bearing orth. of ° earings is written are as °. arbados aint is km incent is due km ast due of aint orth of incent. renada. North o Adj nd the bearing of renada from arbados, we rst 160 km S and label it. B x e calculate e know angle x 92 km Opp the opposite and adacent, Hyp opp tan x = = ad = . G x 1 = tan . = .° so we use tan draw a sketch he bearing ere is a marked more in comple red is − . = .° eample. 114° Puerto he bearing of amana from uerto lata is °, Plata anto Samana omingo lata e is from can on a bearing anto draw a of omingo sketch ° is map on of from a amana, bearing this, and of and uerto °. create rightangled Santo Domingo triangles. 329° he angles of subtracting map. All °, the calculated. °, ° ° other his and and angles would ° have ° in from the allow us been the angles diagram to use calculated can on now by the be trigonometry to solve problems. Puerto amana and anto omingo are both km from uerto Plata lata, 24° so the triangle is isosceles. y x Angle x is Angle y = − − = ° Samana − − = ° Puerto o the o nd triangle the looks distance like Plata 35° this a 59° Santo Domingo opp sin . = 170 km 16.5° hyp a Samana . = a a = . km Santo o to the distance amana = from × anto Domingo AII omingo . km = km to the nearest km. A is on he e Angles down 2 of elevation from earings the are and depression are angles looking up or • ow far orth • ow far ast measured in a clockwise direction from orth. • hat oward of looks depression away he is down of oward boat is km from °. he from the from is is a a cliff to height a of boat the through cliff is an m. and ow far to the from and . km from °. ow is of port, east of on the a bearing port is the of ° from A on a km from on is a know the from right bearing bearing bearing of of A from from A IP roun d end roun ded keep angle. and of a migh t your ind of connection the off until calcu latio n. write distance down version , that A A 140° 11 km A the ever 50° angle is bearing a B you bearing °. how of the a plain A boat 5 km a is far ou bearing boat the of of angle • is the the eA port. A A between eIon 3 ° horizontal. • hat Ar 2 of between km. from 1 bearing PoIn is 1 a distance the full but displa y in calcu lato r . of A. C 1 1 h aa ad timty th rigonometry LeArnIng the a tial i dimsis can be applied to other situations. n this unit we ooe use • ind area of a triangle using trigonometry trigonometry to to nd three the area of a triangle, and also apply dimensions. the formula A = ab sin C h • olve practical involving distances heights in aa a tial problems and e situations can sides use and trigonometry the angle to nd between the area of a triangle, given two them. n the diagram, Area = a opp A sin = hyp o sin = , or = b sin . b c h b ubstituting = b sin into the area formula, we get Area = ab sin e C B a could also write it as D Area = bc sin A, ac sin or Area = he o, formula if a = uses cm, two b = sides and the cm and angle = in between those two sides. °. Area = ab sin Area = × × × sin = . cm imty A e can apply i th trigonometry dimsis in three dimensions. B he maor step is identifying rightangled triangles. 4 cm C D ook at the diagram of the cuboid A. F G e are going to nd out which of these three angles are right angles 7 cm E A H A A 11 cm A good sides of the put in • f A is • f is a a way vertical • ut if nd out is to imagine horizontal. e the then cuboid decide if with the one other of the side can be position. horizontal, then A is horizontal and is vertical. A angle. is right horizontal, then is horizontal and A is vertical. A angle. vertical. 1 is right a to angle o is horizontal, A is not a right is horizontal angle. but A cannot be made ore eAPLe 1 eA n the cuboid A rod rests A, = cm, = cm and = IP cm. • ev er inside the bo from A to the . ou e are going to calculate the angle between the rod and know calculate A = by cm, and we ythagoras’ A can + migh t = keep the = • F + = ou of ten H tan √ = = . cm opp in to d.p. D to use theo rem trigo nom e try in problem s. x displa y have and but calcu lato r . ytha go ra s’ down version , theorem. until write full x off calcu latio n. C a roun ded your of . a e roun d end = = = ad . . h ook x al at the ometimes plane. A = or and tan . t cuboid you will plane o understand is, rotate a .° li A be eample, the = asked we ad at to might the nd be to d.p. a pla start the asked of this angle to unit. between nd the a angle line and a between . which angle F this e G the cuboid so is PoIn B A 1 horizontal. he area of a triangle is given by A = ab sin magine shining a light vertically E H at the 2 shape. D A would cast a shadow at he angle and a between plane is angle between Ar 1 alculate 2 A the pyramid our One and is the angle to the plane. A. A a and of triangle suare the base base bamboo is A A, °. canes where are and A a alculate used to = top the make cm, verte height a . = . cm he of support base the for and has angle edges A of = cm, °. and the angle pyramid. A some beans. end suare area has metre runner line angle eIon between 3 A the . perpendicular he a C of of alculate each side the cane m. angle is he stuck top in the ground ends are tied in the form of a together. 3 m A. B E 1 m D C 1 1 LeArnIng • se a the h si Although sine, functions that cosine ul and tangent A are ooe sine rule to angled calculate triangles, formulae side can for use be we in used can any in right develop triangle. c • se an the sine rule to e calculate use case angle a new letters angle they to notation, label are using sides after b lower the opposite. C B a A h si f draw we two h a perpendicular rightangled I c ul tial A c sin = c sin = b c y using a sin his he is the sine include he steps irst e eAPLe label know are . m the , and perpendicular, choose the b A angle of long. e A, b, and part ., = sine rule both sides by sin ° with are and going , need the is used sin when and lth and to horizontal, to the nd nd the sides could show and two the length opposite of the sine rule containing , b, a the features we know and want to nd angles. sid ith th si ul steps of the them make slide, as a, b an = ° and = angle of ° with the horizontal. x and c and c c °, = sin x = A sin . x c = . . 2.7 m x b x . = . 75° 35° B x 1 both = sides . m by to sin that c sin ultiply × rule. sides th sin = create = sin . b we c A ubstituting sin we b we tial ividing o , 1 vertices at different two idi an = has I b sin a sin slide meet child’ s to = D sin A A b C B ore from b o a line triangles. sin the ° . nearest cm a C idi th si a al usi th si ul AII ecause an be a triangle obtuse aware angle, of the can we y have need values of • se to your calculator to check 2 sin that x 1 when he x is greater graph looks like of y than = sin °. 0 x this x can take between he x = values and ecause an lead of means a ° sin ° = sin ° sin ° = sin ° some that sin that angle uestion sin x and = an having eAPLe sin obtuse more − = values sin to − show x. x. angle than have one the same sine, this eA frame is hen is be ndi ng easier 2 to IP solution. rule triangular x graph it A other 2 • ore sin . the acute to = 1 symmetry about can − ° x • ry in sin to upside an turn angl e, the down . made B from he three base piece and is is aluminium cm welded the third strips. long, at an strip is the angle cm left of x hand ° long c and 11 cm a pivots e from are the going right to end. nd the angle x 56° when the meets e o end the left know we A, of the right strip ust A strip. a and b and want to nd b C 12 cm . use sin A sin = a sin ubstitute A, a and b b sin x = ivide sin ° by sin x . = sin x = . x = sin ut, as sin x Ar 1 se the the size = sin . − x, x = .° could also be − .° = .° eIon sine of rule to angle A. e A calculate 1 PoIn he sine a rule is b sin 2 alculate angle se the sine A = sin sin . b c 3 c = rule to 10 cm calculate 2 nvert the rule to nd an angle. the length of c 3 65° he sine and rule involves sides C B 9 cm a angles. 11 1 h csi ul A he LeArnIng cosine rule can also be used in any ooe triangle. • se the cosine rule to calculate As with the sine rule, lower case c a b side letters • se an the cosine rule to calculate angle label they the are sides after the opposite. angle C B a h C csi sing apply a similar b h diagram ythagoras’ ul = b − c = b = b b to − the one to both theorem x , and we used = a for the rightangled sine rule, triangles to we can obtain − c − x a o a − x = b − x Or B A x D (c x) a a a = = b − x − x + c + c − + c − cx x c ut x 2 o a his = cos t can be • a = • c c with two use a, cs A ul has three different forms, which can be nd e − bc cos A, − ac cos − ba or c by replacing A with , a with the a cosine and a cos rule the sid side b by replacing when included A with , a ith th the uestion gives us all angle. csi ul b in angle the , triangle so A we use ac cos the formula = a + c − = = = . b b + − cos b − . 58° 8 cm 12 b b C B and c b with c and a. know 9 cm by as A o derived or + sides idi 2bc + b x diagram. c = e + a with + − written b = b rule the • b c csi cosine cx A + relabelling − 2 b the 2 is he a = . cm to d.p. with in it three sides or h csi aph AII he graph of y = cos x looks like this • se your calculator to check y that 1 cos ° = cos cos ° = −cos cos ° = cos = −cos x 0 • ry 1 t is a similar translated he cos x = shape to graph the has to the of y = sin x, but it has is − rotational symmetry symmetry angle between there is the idi a sing the map a = b and ith we about ° ambiguity eAPLe ° same al ore about x = , values = cos = −cos to − show x − x. been so eA th , has as a , cos discrete there csi so is with x = −cos value sin of IP cos x, − x. ngl es ° be tw een have a ° nega and tive cosine . so x ul 1 can nd the angle A at ingston. + c − bc cos A B Montego ubstituting, other x ° x. very not cos ° ° left. reection cos graph • here some that ° we Bay 98 get miles a C = + − cos A Port 85 c cos A = + cos A = − − Antonio miles b 23 miles Kingston A − cos A = = −. A = Ar cos −. = .° eIon e A 1 f angle = °, a = cm and c = 1 cm, PoIn he cosine calculate 2 f a = the cm, length b = of b . cm a and c = . cm, 2 c calculate the size of angle b . = he all 3 f a = cm, b = . cm b the is + cosine three and rule c − rule sides bc is or included cos used two A. when sides angle are and given. c = . cm, three angles. calculate all C B a 13 1 icl arlier LeArnIng ◦ theorems Angle angle at at centre are known = twice Angles ◦ angle properties of polygons. number the of cicl facts about angles in a circle. hese facts are thms in the same thms segment al at th ct is tic th al at th cicumc eual Angle in a Opposite of the the semicircle = ° n the diagram, circle, ◦ considered circumference h are a as icl ◦ we ooe here • ircle on thms cyclic and eterior whose A, centre and is are points on the circumference of a O. angles uadrilaterals O is etended O = AO as to they , are and angle radii, so AO = triangle x and AO O is = y isosceles. o C A ngle AO= AO A ngle AO= A ngle AO= x = x y − x Angles in a triangle add up to ° x imilarly, triangle O as is AO + AO isosceles, so = O = Angles y, O = on a straight − y line and O O = y he angle at the centre from A he angle at the circumference and , angle AO, is x + y B from A and , A, is x + y D A o the his D is angle true Als at for i the all th centre is twice the angle at the circumference. circles. sam smt a ual C n this also diagram, angles and on lie A the and A circumference share on the the vertices same side A and of , and A. Angle A = AO Angle at centre = × angle at circumference AO Angle at centre = × angle at circumference and O angle A = so angle A = angle A. B h al i a smicicl is a iht al A C n o this the AO diagram, angle = at AO the is a diameter. centre, °. Angle A = A of the angle at the centre, so O A = ° B 1 yclic uadilatals AII f the a uadrilateral four vertices of hree lie of the four angles a, b, c on B and A the circumference of d are eual. a x circle, it is called a • hich cyclic what uadilatal a A is a O three can are you eual, say and about the fourth b cyclic E uadrilateral. z a y is an Opposite cyclic eterior angles C angle. of uadrilateral a are D b supplementary. f A = x, and = y, c b = x Angle at centre = × angle at circumference a = y Angle at centre = × angle at circumference b = ° Angles y = ° = ° = ° and a so x or + + x + y round a d point eA A + • o the opposite angles of a cyclic uadrilateral are supplementary. hen ti iti a cyclic uadilatal y = ° Opposite y + z = ° Angles x = z the eterior e is angle of a angles on cyclic a of a straight cyclic the t th fo r your give answ er . uadrilateral line uadrilateral is eual to the interior angle. PoIn he ual ns geom e try alwa ys al + opposite 1 ppsit x so o, al so lvin g problem s, reaso h IP angle Ar at the centre = × the angle at A centre Angles 3 he in the a same segment are in a semicircle is a with O. eual. alculate angle A cyclic uadrilateral, circumference. 2 is eIon right angle angle. 1 A 2 O Opposite angles of a cyclic uadrilateral are 160° supplementary. 3 he eterior angle of a cyclic uadrilateral O B is 55° eual to the interior opposite angle. D C 1 1 ats ere LeArnIng we will ◦ the to ◦ theorems tangent the the relationship chds between ooe tangents, • ircle study ad • A is perpendicular tat touches radius tangents radii a entering from a point and is a chords. line circle that ust without it. are • A chd is a line oining two Radius Chord eual points ◦ alternate segment ◦ line centre the circumference. theorem • A from on of adius is a line circumference to midpoint of perpendicular chord to from the circle to the centre. is T angent chord at • he angle contact • rom • he • he is between a tangent and the radius at the point of °. any point tangents hd AII pptis outside from a a circle, point to there are are two a circle the circle to the perpendicular to the tangents. eual in length. pptis line from midpoint of the a centre chord is of A chord. • he perpendicular circle to the chord from the bisects centre the of a chord. C O D h A altat chord splits a smt circle into two thm segments. B he diagram shows a tangent and a chord A at the point of contact. 1 n the diagram, centre of diagram the to alternate O is circle. the se the demonstrate segment the he angle between calculating is in the and altat smt to the angle between a tangent and chord is eual angle the alternate segment. AO A AO AO A 2 epeat angle uestion A = 1, but with C °. a D 3 epeat angle uestion A = 1, but with a x. B 1 angle A, chord. theorem • he by A tangent to the angle in lvi here are plms many theorems. to be he used ith problems theorems together. cicl that from Always can thms be . posed and annotate using those a the presented diagram to eA circle here mark any need • IP an y of the circle are deriv ed angles theo rem s obtained. from the centre ore eAPLe angl e = × 1 at the angl e at circu mfer ence. A is a cyclic • T he A uadrilateral. tang ent, and is , a tangent and is high ligh t a the sym me try circle s. line. • radius pro pe rties at of straight chord T he alter nate segm ent Angles theo rem A = comm only 33° = ° = ak e sure alculate 13° a angle A angle A c angle • 70° rstand remem ber C hen use it three fo rm at, and it. writi ng the tten. you °. unde 1 fo rgo and D most °, B is i.e. an angl e le tter angl e G no t 2 rove that A is a angl e . diameter. luti 1 a A = interior A = angle c 2 = ° opposite in the = eterior ° angle alternate A = A − A = A = ° A o = A A is a + angle of a cyclic uadrilateral = angle diameter between = − angles = tangent and chord = segment angle + in in a = the = ° same segment are eual °. semicircle = ° e 1 PoIn he to Ar tangent the is radius perpendicular at the point of eIon contact. A are points on a circle, A 2 centre O and radius angents from a point are cm. eual. and are tangents to B the circle = such 5 cm that 3 he angle and a between a tangent O cm and = °. the chord angle in is eual the to alternate alculate segment. 1 the length of O 2 the size of angle A 3 the size of angle O. C he a line circle chord 12 cm D from to is the the centre midpoint perpendicular of of to a the chord. 1 1 cts ectors LeArnIng represent movement, and are used in transformation ooe geometry • nderstand the concept of here to are dene three a translation common ways see of .. representing a vector. vectors • he vector from to can be written as a column vector, ( • ombine vectors triangle or › • t parallelogram can and also be written as written as , the arrow indicating the direction law from • Add ) − subtract to . column • t can also be . ecause we cannot easily write in bold vectors type, • nderstand and use when answering uestions we indicate a vector by writing k ∙ vector olumn vectors number represents follow the conventions of coordinates. he rst algebra • press a point a, b as a horizontal direction and the second number a position represents vector a vertical direction, so ( means ) to the right a − • etermine direction the of a magnitude negative and number vector number is owever , y specic indicate signies and left, a and downward vectors have vct epresents B it coordinates lum 6 would negative down because the movement. completely different properties diats movement magnitude or of a as size. no point movement which – remains it signies static, a and 5 k has no size. l C 4 as a denite direction. as no movement therefore, no and, direction. m as no position – there n innite number of 1 as a dened one position. point , here is −. vectors. − 0 2 an only ) ( 1 3 are x 1 2 3 4 5 Psiti vcts 1 Psiti vcts give the vector from the origin to a point. 2 a o the position vector of the point a, b is ( ) b Additi vcts he movement from A to is represented by the vector ) ( , and the movement from to by ( ) − Adding the horizontal components + to = terms, moving + ( ) = moving from from A to A ) ( − , or A. to and then from . › › − = ) ( − ( ) = ( − ) › ubtracting 1 get imilarly A we ) ( directly components › vector vertical n the › › A and › has the same effect as adding . to is identical h tial la ad paalllam y la 6 o add two vectors, we can complete add triangle or a parallelogram. › › o a A, ( 5 and ) , draw ) ( − the two vectors endtoend, C 4 and complete + his › is triangle › › A the = A 3 A. the triangle law for addition of vectors. 1 › › B o add A and A, complete parallelogram A. › › the › 0 A + A = diagonal 3 A. 2 x 1 1 2 3 4 5 1 his is the parallelogram law. 2 y 6 AII C 5 • raw 4 four D 3 your own points • rite A, down A, • Add the A B 0 3 2 1 2 3 4 + A. + • rite your down vectors 2 • hat a the is vct a vector can be found column › › , A. connection the column vector using , the between of A. › A, › size, › + answer. › dicti 5 ad vectors › plain x 1 1 or › › › magnitude, , column he column › , mark . 2 A maitud and and › › h the vectors grid, , › and ythagoras’ theorem. a he vector ( ) has a horizontal component b of a and a vertical component of b, so the eA magnitude is + b . b • he angle the IP √ a vector makes with hen you the squa re a ϴ nega b horizontal, θ, is given by tan answ er is ectors have 2 ectors can magnitude be subtracting added the or and direction, subtracted horizontal but no ed algebraically, components, and position. by the adding Ar eIon vertical components. f a = ) ( , = − 3 ectors can be added graphically using the triangle law of parallelogram position from the = ( vector origin to magnitude of ) and ) , nd law. x he ( − the c the positiv e. PoIn 1 or num ber , a a e tive of x, x, y is ( y ) , and represents a 1 a + 2 c 3 he movement − a y. x he ( y ) is √x magnitude of d, where + y and the angle with the a + + d = c y horizontal is tan x 1 2 ct ultiplyi LeArnIng a vct y a scala subtracted vectors. ooe n • ultiply mty a vector by a . we added ultiplying • nderstand the and scalar meaning two vectors makes little sense, but we can multiply a of vector by a scala or a single number. collinearity • se vectors to solve problems or eample, ( ) , or times the vector ( in ) , = ) ( geometry his is the same as ( ) + ( o the resultant vector , and vector goes in has the + ) ( three same ) times the direction, so magnitude is parallel to of the the original original vector . enerally x n ( y nx ) = ( ) ny AII a x ( y ) and n triangle A, A = of › A and = if ( y ) = ( ( y a = ) ) ( b , and the magnitude b a x and parallel b a x are ) ( › ) is times the magnitude of ) ( b y are the midpoints of A lliaity and A. hree = points, A, and , are said to be cllia if they lie on a straight line. C C y B E 3b b A y F a D 2a E D x x › n ind, in terms of diagram, o › A d show › › A that = a, › › = a, = and = , where is the midpoint A, + A = and a are collinear + + › › A = › › A › of › = › › A › c the y, › a and + = a + a + = a + = a + › › › ow o A o A, = A, can are you tell that A, A and A are parallel and both pass through A. and are collinear. and collinear lvi he key notation e use prove vct to geometry 1 so A solving and the a, vector algebra. eplains general plms , what c A problems labelled we are notation statements is to see diagram them is as a essential, combination and doing. for vectors rather than as we are calculate often eact of algebraic trying values. to ore A he is a eAPLe 1 triangle. position vectors of A and are a and , respectively, and › A = is a a the ind + i ii midpoint the in how of A. position terms that of O, vector a and and of are collinear. › c ind in terms of a and luti tart with a diagram eA › a i O = IP › OA C › + A a + 2b • lwa ys = a + a + = a + A › O = seem › › • ak D ii check answ ers OA + e sure you A direc › › tion into = OA + a A that your realis tic. take accoun t. B › › › = OA + AO + O b = a + −a + O = a + = a + › › O = a + = a + = O O and are are › c O › O = − = a is both passing through O, a + OA O, and + + a › + + A a + eIon trapezium, with e A parallel to and twice the length of . = , and the diagonal A = ind, in terms of a and , y nx ) oints on the A c 3 o A intersects at , so that ( ny are ) collinear same prove that 2 = if straight they lie line. › › ( › a n a 2 1 PoIn x 1 › › A so › = Ar A parallel, collinear. = collinearity, vectors are show multiples › ind of in terms of a and each other and so are parallel and have a common › 3 ind A in terms of a and point. 11 21 atics atics LeArnIng ooe A • nderstand concepts mati hey matrices concept of column, order, • Add have a rectangular array of numbers. graphics in to the real enable a world. or eample, dimensional they image to are be used drawn types, a dimensional subtract matrices ere we show matri is screen. use and • ultiply applications computer on practical matics matri, in row, plural of matrices; by scalar commutativity focus we on use how them to in manipulate matrices, and . will transformations. multiply non of will how A matri A row is described by the number of s and clums it has. matri is horizontal, a column is vertical. multiplication • he determinant and inverse o − − is ) ( a by or × matri. matri he column ati Only vectors additi matrices omponents of in we ad the the considered in . are × matrices. sutacti same same size can position be are added added or or subtracted. subtracted. o − − + ) ( − − − − ) ( = − + + − + − + − ) ( = − − ) ( and − ) ( ultiplyi a A be matri by the can − ) ( ultiplyi the o = t matri by numbers number of a we will resultant − − − − − − − − ) = − − ( ) a scalar by multiplying each component ) ( matics all of he ore involves the components can be only an eAPLe multiply multiplying components a a c a of row column an all a of of a the the the × b atri − ( of rst ) the in a second matri second matri 1 A components column matri. ) in multiply × atri in ( 12 − ( scala by components Algebraically, o y multiplied multiplication rst the the mati = scalar. atri ) ( must row of matri. match matri. by a b × c matri. AII ultiply column the of components matri , and of the add rst the row of matri A by the rst results atri × + × + × = . his goes column irst row, second column × in of + the the × rst row, answer. + × • f A = = econd row, row, Answer h he rst second identity for column × × + + × × + − + − × × = = = A × not ) , , and nd × mati for diagonal eA an of s × n from matri top is left an to n × n bottom A matri of zeros eample, ) − ( ( ) = − IP • at ri multip licat ion is right. no t or ) ( matri a is ) ( idtity ecept column ( econd multiplication commutative. rst comm utativ e. ) ( AII h f ivs the matri ad a b c d ( dtmiat is ) multiplied a 2 d −b −c a 2 mati • how ), ( by × the resultant of ad − bc ), ( ad − or ad − bc ( ) bc , where ( ) is the the inverse of b ( c d ) ad − • rite − bc is called the ) ( . the inverse matri . a dtmiat of the that multiplied ) ( matri. its inverse is ( ) . PoIn • hat 1 determinant bc by e is down • how ad the ) ) −c d identity . −b ( is ( of a o that is An n × matri is a rectangular array the of happens inverse by if you multiply ) ( numbers. 2 atrices or of the subtracted same by corresponding size adding can or be added subtracting Ar A 3 An n × × matri matri to can be multiplied produce an n × by o multiply matri is second, he two multiplied and the identity matri other matrices, for he an containing of ad − bc × from is the = ) ( rst the an s, the n with = − ) ( , ) , = − − ) ( − 1 × , ome of n the a A − d A these or can those be calculated; that + A can, nd others the answer. c A A . d d − cannot. matri of added. n ( matri. from column b ) is row diagonal all ( c where a components inverse a answers a by a = an eIon components. − bc determinant. ind the 3 oes inverse matri of ), −c ad 2 −b ( is a this A + = A + hat do we call property 13 22 atics ad tasmatis atrices LeArnIng can be used to produce transformations. ooe rctis • etermine associated the with × matri he specic green ag has vertices at −, , −, , −, and −, . transformations − As • etermine the representation × of the position these can be written as result − , of as a × matri, P − − − − ) we multiply this by ( − ) and ) ( ) ( transformations f − , ( ) ( or, two vectors, matri matri, , ) ( , we arrive at, P = P′ y − − − − ) ( ( 6 = ) − − − − ) ( 5 he four columns he matri 4 are position produces ) ( the a vectors reection of in the the red line ag. y = x, 3 and ) − 1 he x 3 2 − ( 2 1 1 2 3 4 produces the line y = −x matri produces − a produces ) ( reection in the x a reection in the y ais, and − ) ( ais. 5 1 rtatis 2 o rotate rotation eA • T o and produ ce A these trans form ation matri is writt en ) is matri an − produces ) produces by a ° the a ° clockwise matri anticlockwise rotation. − − ) ( −sinθ produces ) ( sinθ through ( produced cosθ he origin, − rotation we alwa ys prem ultip ly – th at the ° the ( IP trans form ation s, is, about an anticlockwise rotation cosθ angle of θ rst. Ivss f we the reect same a shape mirror reections are line, in a mirror line, we return to slivs; the and the then starting inverse of a reect point. the image his reection is in means the that same reection. elamts e know that the identity matri is ) ( . An enlargement, n the origin, scale factor n is produced by the matri ( miatis n the the centre 1 diagram yais to the on give origin, the the by centre n ) tasmatis net red page, image. scale factor the he to green red ag ag has has produce been then the blue reected been in enlarged, image. AII • how the that origin, the is inverse − of a ) ( − , a ) ( ° ° clockwise anticlockwise rotation about rotation the about origin. y he matri for the position vectors of the green ag is 6 − − − − ) ( 5 he reection is achieved by the operation 4 − − − − − ) ( ( = ) ) ( 3 2 he enlargement is represented by 1 ) ( ( = ) ) ( 0 3 his could be written as − ) ( ( − − − ) ( x 1 1 2 3 4 5 1 ) 2 − 2 ultiplying the two transformation − ) ( ( = ) − matrices, ) ( eA his is the single transformation that maps the green ag to the • − − − − − ) ( ( = ) ) hen ( IP blue combi ning matric es fo r two trans fo rm e the saw in order . is a matri important. transformation put that matri matri after e in multiplication produce front. another it is his is not commutative, transformations is called called by putting premultiplying. ation s, so prem ultip ly, the hen so trans fo rm ation we righ t postmultiplying. of the we the is rst on the second . AII • raw of your the • se • ind own vertices your the own shape as inverse a × on a single matri matri grid, and write the position vectors matri. and to transform check it your reverses shape. the transformation. e 1 PoIn ome can Ar reections, be dened rotations by a × and enlargements 1 matri. raw , . eIon the triangle abel ransform 2 f matri obect using P A performs then AP = the a transformation image P′ can be on f matri A obtained performs can matri be matri a transformation, transforms replaced A vertices raw the , , , and with the matri image, and label − ( it ) . P′ ransform and − ( then it with A. an 2 3 it by the the single image, A and transformation 3 ) ind the escribe . the image raw single fully the new matri this with the image, that matri and label transforms A it to . . transformation. 1 23 ati lvi LeArnIng in matrices to arithmetic, geometry to × simultaus solve algebra linear is a pair of simultaneous − y = euations e he o − y = can write − − these euations x ) ( y the − = ) inverse ( of − − ) ( − − a the in matri the − the your inverse both sides − − of − − ) ( ( answers are each the − euation x ) ( matri. − − by the × = −. inverse matri y ) = − x ) ( y − − ) ( ( ) − ) = − ( ) the he for × is ( same to bottom row heck = ) ( ) ( by ) manipulating using euation − produce matri matri remultiplying a a = ) y determinant as euation x ) ( matri euations up AII ( matics and the ith problems ( • olve uatis ooe ere • se ala lefthand side can be simplied because a matri multiplied inverse euals the x ( y = uppose f the ( = e ) , y th the and a b c d = − mati points ′, matri that are going ) use 7 determinant matri and , get a tasmati transformed by a matri this transformation − is a b c d ), ( then ) ( inverse ) ( = matri. × − × = , so the inverse is 6 of = an y he pducs respectively. produced to that A, − ) ( ( − − ) ( 5 ostmultiplying both sides of the euation by the inverse gives 4 a b c d 2 ) ( ( B 3 a b c d − − ) ) ( ( × ) ( . −. −. . ) ) ( = = − − − ) ( − − − × ) ( . −. −. . ) ( ( ) A 1 A a b c d ) ( 0 1 B 1 1 = ( ) x 1 its matri. − idi A′, identity ) x 2 by method. 3 4 5 e have found the matri that produces the transformation. to lvi nverse hey mati matrices allow us plms are to very useful simplify when matri solving euations as problems. shown in the above eample. ore eAPLe 1 eA a ind the matri such ) c • em embe r a b ) ) ( ( − that d − c = ( ) d − • inverse matri of ) ( is − − or ) ( − matri by ) ( − its we premultiply − − − − both sides − − )( ( of the a b c d a = by − − − − = inverse − )( − ) the • − − a b − − c d − − the the or depend ing positio n matri ) d rem ultip ly on − c = matri . postm ultip ly ) ( e − − ( b ) ( ( euation ) ) ( no t multip lied invers iden tity o matri is comm utativ e. − that multip licat ion luti he IP b ( you of need the to elim inate. ) ( emember nd the that matri a b c d multiplication a b c d − − ) ( such ) ( ( we matri would ) postmultiply a b c d such ) ( a b c d by − commutative, the inverse − − − − − − ) ( ) a b c d )( ) a b c d ) = = = − − − − − ) ) ( ( − − − − ) ( ) ( Ar euations 1 ax + by = , cx + dy = can be a b c d ows rows 3 he written ( can can y a ) ( ( ) = ( ) , nd a and b. b 2 a matri olve euation the x − simultaneous y = y = − euations = ) be or x ) ( − be inverse algebraic f eIon and as x ) ( 2 to matri PoIn he so ) ( 1 not that ( ( e is ) ( that ) ( ( = = multiplied added matri by a together is very geometric constant, or useful and subtracted. when 3 ind , the − matri to , that maps , to −, and −. solving problems. 1 dul 1 alculate diagram the angles below, Pactic marked stating the a, b and c in the am reasons. A regular is heagonal attached heagonal as ustis to a pyramid regular prism shown. 104° a a ind the faces, number edges vertices the of and solid b possesses. c ind the number of 47° planes for 2 A regular ow polygon many sides has has interior the angles of of the symmetry shape. °. polygon A picnic table has a symmetrical side view as shown. 3 An obtuseangled isosceles triangle has an alculate angle of °. alculate the size of the angle a between the leg and the other horizontal. two angles. ithout angle A of he a foot feet a protractor, construct an 1.4 m °. foot against is using ladder leans vertical of the from 1.1 m wall. ladder the a bottom of the wall. 1.9 m ow wall far up does ladder the the reach n triangle A a otate , . eect c hat maps triangle abel is A in the the A the ° line single onto anticlockwise image y = = A, cm angle and A = = °, cm. about a alculate the length of A. alculate the size angle c alculate the area . . abel the transformation image of A. . of the triangle A. that 1 A garage has a base . m long and . m y 6 wide. t tall the at is diagonal . m tall highest at the point. support, as sides, he shown and roof by a . m contains broken line. 5 alculate the angle between the 4 support and the horizontal. 3 A 2 1 1.8 m 2.6 m 0 3 2 1 x 1 2 3 4 5 1 2 1 5.5 m a wooden › › 11 A ship, sailing towards renada, is miles 13 A is a pentagon. est miles of uadeloupe. uadeloupe orth of what bearing is the = , › = a and = a + renada. rove On a, is due = › due A ship that A and are eual and parallel. sailing D Ship 420 miles Guadeloupe a + b 2a E N 280 C miles b Grenada 12 A and is a are point on tangents the to a circle, circumference centre such a O. that › › › angle AO = °. Angle A = 1 °. f A = − and ) ( A = ( ) a alculate your A, giving a reason for A, giving a reason for column 1 f angle as a vector. A = − ) ( and = − ( ) , answer. a c write answer. alculate your angle , plain why AO is a cyclic ind uadrilateral. i A ii A B , the inverse of A iii C , the inverse of iv A , the inverse of A O how that A = A D 75° A 1 Further 1 Lionel type put carried of pizza the out a survey people results in a in bar to his nd class out liked exam what best. practice b alculate c rite the surface area of 1 the cuboid. He down the dimensions of a different chart. cuboid with the same volume. 16 14 5 12 he table minutes shows of the people waiting times catching a in bus. ycneuqerF 10 Waiting time (minutes) Frequency 8 to to to to to 6 4 2 0 Cheese and Ham Pineapple Mushroom tomato Pizza types a a Five children entry for chose pineapple pineapple. that Draw should go the in a histogram show this information. rite down the modal class for these chart. waiting b to the b bar Draw How many more children chose cheese and c tomato than chose times. ne of these passengers is chosen at mushroom? random. c f the same chart information what angle was would be shown used in for a pie hat ham? is the passenger d he probability waited probability that this minutes that a or less? passenger selected 2 Here are the marks of children in a at test. random is male is i selected ii a How at many random males is in the survey? alculate i the range ii the median iii the mode iv the mean. a omplete this x table − for − y o pass the + x you needed to score at − . hat the x test +x least = − x b female? were percentage of the children passed y = + x x − test? b Draw of 3 a c ii x − a ork − b out g = rite + the a + to y = and x y + x from for − values to . ork down value the minimum ii the euation the same of f + g when f = of of the y ais of symmetry. n aes draw the graph of −. out the volume of a cuboid = se x + your cm wide and cm high. . graph to cm solve the simultaneous euations long value b e 180 − i y a from of x d and 4 graph implify i b x the y = x + x and y = x + . 7 Solve a 5y the − equations: 3 = 10 27 a b rite i 5 in o b 11x − 3 = 4x + the 5 a T o 2 in make pastry, our an utter are the ratio T o the to make the hat that , 5 o the oul our is the loest ommon multiple o 5 an 7 a varies Trapeium ontains 5 orret possile ontaine amount in the o is ran on a the area o 5, b ith = the square o b a in b alulate a hen b = 3 c alulate b hen a = 45 an equation onnetin a an b arton 1 m is square trapeium the an out = pastry ri ork inversely a neee 5 least e muh pastry pastry nearest is ho o 12 a ator 7 7 : 4 nearest paket to 9 ommon ators mie hen prime = 11 o 2 in b hihest an 8 prout x __ + 3 a 18 ii x __ c as is is lie a nle entre an o a isoseles on the tanent = irle trianle irle, to the suh an irle at that = , 5° y a alulate anle b alulate anle c alulate anle d A the the raius area o o the the irle shae is 8 m, alulate setor x A B O b otate ° antilokise aout the E oriin ael the imae 56° c eet d esrie in the x ais ael the imae C that ully maps the onto sinle transormation D urther eam pratie an e oun online: oorseonaryom781841452 181 Answers Module 1: Summary 1.12 questions 1 e simple interest is better b 1.45 13 90 14 minutes or ∅, a, b, c, a b, a c, b c, 1.1 320 1 a natural, b irrational integer, 318.55 square 2 4900.17 3 7 15 a b c 36 students 1 c natural, integer, d natural, prime, even, ears 7 square S e integer f none integer 2 2 and 2: alculator 1.13 1 a 4.27 m b 330 ml 1 38.25 2 8 3 2000 4 207.79 5 3n 6 7 72 8 a 9 4021.79 1 3 5 c 2400 g 2 12–13 3 a i miles b 138 1.2 1 a 12 2 a 3 2039 b 60 42 b c 180 6 c d 90 1055 ii 6 d + 2 1313 2 minutes 3 180 1.14 1 1 a 2 4.68 2.57 3 10.65 b 2 15 028 3 51.75 3 1.15 1 672 b 674.92 18 785 1.3 A = square 10 3 boos 12.99 5 penils 0.35 ea = 1.75 2 notepads 1.75 ea = 3.50 2.5 = 11.25 ubtotal ales litres paint ea = 4.50litre 38.97 numbers 1.4 3 1 B = multiples o 3 C = multiples o 4 D = multiples o 5 and 55.47 a 4 ta 15 8.32 1 b 13 2 1 c 5 2 A 3 e 11 637 12 a c t 63.79 D 2 1 2 7 blos ill be needed 8 ( 1 ÷ 1 4 = 6 4 2 are equal and ) 5 1.16 23 a b 45 c 2 3 ∅ 1 4, 16 2 3, 4, 3 12 Cricket 0.575, 10 empt Football 3 1 is 13 1.5 b innite. 40 35 o 45 is 15.75, 0.75 6, 8, 9, 12, 15, 16, 20 more 8 3 tan o 40 8 1 3 1.17 28 out o 196 6 7 1 ilip is A ⋃ B , ntia is A ⋂ B, oel 1.6 is 1 35 2 82.80 3 a 184 b 160 2 A ⋂ B a B is a subset b B ⋂ C is te o A null 14 a 17 800 b 18 350 15 a alse b rue c alse d rue set. 3 16 USA 9610 inome ta + 3000 Girls = 12 610 = 22.9 1.7 1 18 2 30 000 m and 27 7 3 4 13 3 o ement, 90 000 m o Module gravel 2: Summary questions 12 2.1 3 1.846 … m 2 4 girls ave visited te 1.8 1 12 m 2 53 m 3 1200 m 4 1 a 2 500 5.6 3 a × 2 10 b Module 140 000 seonds Practice eam questions 9 3 4 1: b 2.2 2 3 c 8 2 S 1: oncalculator 1.9 1 a 5 c 8 or 7 b 9 d 9 1 17.5 m 2 8 m 3 54 m 2 1 a 3.999, 2 4.08, 4.19, 4.2 5 1 7 3 18 b 9 c 5, 2 3 12.2 4 = 14, = 168 12 2.3 4, 2, 7, 8 2 1 rea = 113.1 m , 5 2 1, 3, 5, 7, 9 irumerene 12 3 a 4n 1 b 2n + 5 c 3n + = 37.7 m 21 5 0.625 6 196 7 3 8 9 2 iameter = 21.0 m, radius = 10.5 m, 2 area = 346.6 m all to 1 d.p. 1.10 7 1 3 25.9 m 0 3 1 5 2 a 3 a 3 3 b 8, 3.687, 3.7, 3 4 2 2.4 5 3 5 √ 3.4, × 19 × 4 = 5 × 4 × 19 = 3, 8, 18, 38, 78 2 1 urae 2 e surae 3 e spere 380 area = 52 m 5 10 b 27 × 10 = 4.7 × 10 2 270 11 a 1.11 a✪b✪c = 12ab =2a × 3c × = 3b✪c = a✪b✪c 1 192 2 501 3 11 4 a 1000 b 21 = 2a × = = and te × 3c 36abc = = area o 144 π m a✪6bc a✪b✪c a✪b = 2a × 3b = b✪a = 2b × 3a= 6ab 2.5 1 84 m 2 175 π 3 10 3 base 2 6ab = a✪b ≈ 1 _ base 4 12 1 m 8 182 linder 2 surae 3 b 210 π m 36abc a✪2b 18bc = 6ab✪c a area 549.8 m . bot ave Answers 2.6 1 2.14 2 gallons 1.1 b 8 o 17 16 or pints 18 o breadruit, o ater, salted 2 o 1 bee, Module all or mean mode or median ames mean arus or median 2 8100 3 a median. 1 375 m 2 5 m 3 a 4 235.6 m 5 a 6 a 7 3 8 72.9 9 1, seonds 2 2 1 17 made, 25 2 20 000 mm c 32 square 2 6 3 e made, 67.1 m 11 3308.1 m b 1244.1 m 20 m b 9 m sold data son b interquartile 1 65 m 2 1 sold 3 as is more spread minute 24 te larger range and minutes 10 1 Maximum b 11 45 minutes minutes, 2, 54.4 minutes 3. against minutes 3 ms ours range. ms length a 171 m–180 m b 170.5 m, umulatie 1 a 2 out, 2.16 3 b ines 2.7 235.6 m sold 3 b eam 2 200 m 25 Practice oo 2.15 2: questions cm 180.5 m freuenc 3 11 1 c 9 ms 20 2 40 24 60 39 80 49 100 55 120 59 140 60 5 11 1 4 2.8 1 5880 g, 2 14.025 m 6120 g 3 4.57 m to 2 3 2 d.p. 12 2.9 1 a engt, b mass avourite 9 14 a 3 12.5 m upper 5 g loer boundar boundar = = 10.5 g 2.10 1 b 31.4 g 25 60 50 40 evitalumuC 29 g–30 g c 5.5 g 10 ycneuqerF 8 6 20 ycneuqerF idt ycneuqer 2 ears ruit 70 72° 13 15 10 5 30 0 27 20 29 31 33 Weight 35 37 (g) 10 4 d 3.5 g 0 2 15 0 20 40 60 80 100 120 140 a 160 1 2 3 4 5 6 0 Maximum cricket football tennis Favourite 2 length (cm) 1 athletics 2 sport oer quartile = 31, pper quartile = 70 2 8 3 13 3 7 = 21.7 60 4 6 ycneuqerF 2.17 4 1 __ 5 _ b 1 obert iger 2 obert smaller 3 ossibl = 24 mean 6 4 1 __ standard 16 deviation a 12 3 ara, but ou annot be sure. b 12 red, 32 blue, 4 green 2 1 a 0 t 10 20 30 40 50 60 bar is a 10 m × 10 m × 10 m ube. 10 70 2 352 3 5 Module Age 18 b 8 3 9 m 3 1 0 17 2.18 1 red, 3 blue, 6 bla, 2 3: Summary questions ite art 3.1 2.19 2.11 eri 1 6 2 t rolls a die and ¤ips a 1 n + 2 oin. c + 12 _______ 1 does sos 3 not so requenies. nstead 2 3 it 2 proportions. 1 2 3 4 5 6 ead, ead, ead, ead, ead, ead, 1 2 3 4 5 6 19 36 ead 3.2 2.12 ail 1 728 m 2 3, 3 e 6, ail, 3 6 2 inreases b 0.6, no is unanged, and 2 ail, 3 ail, 4 ail, 5 ail, 1 tere ill mode. a −1 b −6 d −18 e −1 a 6 b −12 c −18 d 36 c 4 6 1 2 4 te 3 median ail, = 12 mean 1 0.05 2 be 3 n is never negative 2.20 3.3 15 2.13 7 1 2 22 1 e modal lass is 22 1 3 ot letanded . o eatl e estimate o te mean is 7 e median is in te lass a 11–15. 2ac = 22 33 2d 2 19 1 = + ________ 13.5 one 3 − bc 33 2 b 1 16–20 66 7 8b _____ 3 2 9a 4 c 183 Answers 3.4 1 2 3.13 2 a 5 ♣ 4 = 2 5 + 4 = 5 ♥ 4 = 5 4 = √ = 6 9 12 8 24 3 10 1 1 3 y = 2 a ♣ 3 a b ♥ = b 2 a ♥ + c b = 2 = ♥ = b ♥ c b + a = a a + b ♥ ♥ a c = = b ♣ a b p a + 1 b c = a b c 3.20 2 30 m 3 s = 8 c c 3.5 2 1 6ax 2 3x 12x 2 72 3.14 8.94, 2 2 b a ≈ 1, 1 x 2 2 80 = 1 w b engt midpoint 41 1 a 13 b 2 an–one. 3 ll 1 x > 2 x ∈ 3 3.5 R 1 ⩽ x < 7 13 21x 3 te real numbers ⩾ 2 1 0 1 2 3 4 5 6 7 8 9 −3 5y 2 3 12a 4 a bb 7a b y2x y ∈ 4, 3 12 3.15 + 5 3 1 y = x 2 y = x is a one–one 3.21 untion. 3 3 c 9c4c 2x is a man–one untion, 1 y 1 so its inverse tereore ill not a be one–man, 8 and untion. 7 3.6 1 a b b x + a 3c b 5c 8b 3 possible anser is te real numbers 6 3 ⩾ −6. 5 2 2 1x 1 1c = x 1 4 1 3.16 + 45c 4 3 1 2 0, 6 and 2, 0 2 3 x + 2x 3x 2 = + 3 2 y 2 x + 2x + 6x 3x 9 1 12 2 = 3x +3x = 3x 9 2 + x 10 3 0 2 x 1 1 1 2 3.7 6 2S + 3 1 _______ 1 n = 4 3 _____ 2 r 5 p _____ 2 k = √ 6 3 ___ 0 3 x 2 1 2 2 3 4 πr = 6l 7 2 3l ___ 2 , so r = √ 2π 8 4 3.8 6 2 y 1 x = 5 2 x = −3 3 x = −1 3 2, 1 6 3.17 1 5 a 3 4 Required 3.9 b 3 region 1 x = −5 or 2 x = −2 ± 3 or x = x2x x √ 1.32 + 3= = 3 11 , c x = 0.5 2 −5.32 2 3 2 d.p. 4000 y = x + 2 x = y + 2 y = 2x + blue y green = x 2x + y + 1 red = 2 purple 4 2 0 2x or + x 3x = 4000 = 0 x = 43.98 6 −45.48. is 90.96 m 4 × 43.98 m 2 d.p. 1 2 y = 3 y = 3 b + x 1 3p ⩽ 1 2 3 4 5 6 30 2 −2x x + + 9 3.22 4 2 1 2 1 3.10 3 3.18 ield 5 a = 7, b = 2x 1 f + 1 _______ 1 1 x = 3 2 x 3 = 1.5, am 17 y = −1 3.19 ears old, m ater is 39 ears 2 fgx = x fx and gx are inverses y 1 ea old. oter. 8 3 a x b f 7 ≠ 0 1 x = fx fx is selinverse. 3.11 6 1 x 2 e = 3 1, 1, y = 4 or x = 2, y = 3 3.23 5 retangle is 8 m × 5 m y 1 2 4 10 3 8 3.12 2 6 1 h 5 10 25 8 k 7 14 35 112 1 4 0 2 x 1 2 765 g 3 729 1 0 4 2 3 1 2 x = 2, y = 5 4 184 x o Answers 2 y = 3 0.75 2.8 < 20 x < a 7 c fgx b 7 1.8 4.6 x 2x 9 d f 1 _____ 1 = x = 2 3.24 21 1 0, 2 y a 1 2 se sould 3 be angles 11.6 m, o 45° at te 12.3 m entre. y 3 16 2 = x 2 1 minimum point at 14 2, 1 12 2 3 x 4x + 3 = 0 as solutions o x = 1 10 and 1, x = 0 3, and so 3, te interepts are 0, sould be 8.7 m 3, 0 8 4.7 6 3.25 ABCD 1 2 3 o te is a parallelogram. EF is parallel to AD 4 40 m bus 8 m against 1 AD = BC AB = CD, opposite sides 6 m o parallelogram, ongruent 10 0 )s/m( 3 2 AC is ommon. x 1 1 2 FEC = DAC alternate angles, EFC = ADC alternate angles, FCE = DCA ommon 2 8 deepS 6 b e eat c 2 22 a 2 23 a 3.5 ms 24 a 36 g b 1.5 m 4 value is 5.25 b 3 c b 112 m 3 7.5 m 1 2 2 4.8 0 0 5 10 T ime 15 (seconds) 25 e runs Module 115 m 3: 3, 1 and 1, 1 2 3 10 m √ 72 = 8.5 m 3 to 1 d.p. 2 altogeter. Practice 27.7 m to 1 d.p. eam Module 4: Summary questions questions 4.9 1 e¤etion 2 90° 3 e¤etion in y = 3 4.1 1 4n + 6 2 4 7x + 1 a z b y c loise rotation about 1, 3 x 3 in y = 4 x _______ 3 2 a 3 132° b 48° c 297° 6 4 a 9 b line ontinues is not innit in bot 5 4.10 diretions. c to ra starts rom a ed ommutative. point and ontinues to innit in 1 riangle 2 riangle 3 nlargement. one 2 5 2x 6xy diretion. 6 2x + 5x 2 7 s = 8 x = 4.5 10 x = 2 −1.16 2 4.2 2a x = 11 a = 12 12 m, ale ator 2, entre 5, 3 2 u v _______ √ ± 4, b 10 x to = = 9 x 5.16 = 1 or x or = 2 y 1 2 3 a 6 5 132° b 104° c 124° 2 d.p. B −1 4.3 C 8 m 13 x = 43.75 1 55° and 2 a 150° b t 55° or 70° and A 2 40° 2 14 y = x y = 3x 2 + an–one 1 ne–one _____ y = √ x 3 ne–man _____ 15 a y = 16 b t is a b √ a x + 3 n is a 1 ite isoseles 0 trapeium 3 2 x 1 1 2 3 4 5 1 7 one–man relation. y 4.4 1 2 2 114° 10 360 90 4.11 _________ = 135°. 8 2 6 360 1 10 2 es aes, 3 3 10 24 + edges 16 = and 24 + 16 verties 2 __________ 3 = 10 sides. 4 180 144 2 4.12 4.5 0 x 1 2 1 onstrut 2 iset = 60°, son in so = 120°. x 17 y = 18 a 19 = 1.5, −2x y = 2.5 5 10 b 1, 1 28.3 m 2 11.4 m 3 63.6° 4.13 C c red. 3 1 148.2 m to 1 d.p. 2 2.49 m to 3 s.. 3 a e bearing = 50 + o 180 rom = 230° y B 3 onstrut an angle o 90°. daent − = bearing bearing o o rom rom 8 to it onstrut an angle o 60°. iset = 230 − 140 = 90° 6 te rigt angle, so tat an angle o 45° b is 12.1 m rom on a bearing o 4 and 0 60° togeter mae 105°. 115.6° x 1 2 185 Answers 1 4.14 Module 4: Practice b eam 1 1 A = − 1 2 1 1 ) ( 3 2 1 35.2 m 2 15.2 m 3 27.3° questions 1 × 1 a − 2 4.15 b 76° c 29° 1 1 54.7° 2 60.3° 3 9.6 m = 2 12 3 38° 4 onstrution 1 4.16 1 8.2 m 3 = 55.8°, 2 = 51.4°, and 104° = = 72.9° onstrution sould o a inlude blue, rigt olloed b ) + 0 1 4 1 + 0 1 2 ) 1 5 1 1 ) te angle son 1 = in 2 ( 6 103.8° 1 0 1 ( 6 1 ( 65° bisetion o A rigt 4.17 angle 1 80° 2 100° 3 red 45° urter eam ractice 1 4.18 1 13 m 2 64° 3 1 a 26° 16 14 4.19 1 2 ) ( ( 4 3 √ 12 7 ycneuqerF 4 ) 3 68 = 8.25 to 5 18.7 eet to 6 a b c e¤etion 1 d.p. 2 d.p. 4.20 ee diagram. in y = 3 10 8 6 x 4 1 a a b 2a c a y b 2 6 2b 0 2b 5 Cheese Ham Pineapple Mushroom 2 2 a and b ___ ___ › Pizza ___ › 3 = types › 2 tomato 4 3 + = b + 2 a b = 3 3 a b 3 9 A B c 100° a i 5 ii 8.5 4.21 1 1 a 3 2 3 0 8 2 2 ) ( C 2 x 1 iii8 1 1 8 c ( i9.25 ) 4 f 2 2 9 1 10 9 4 ) ( 7 a 13 b b d and 1 2 e 3 3 1 2 annot be 3 es, te edges, 13 3 bot equal 10 a i 2x ii 7a 8 77.2° 9 a to 78.6° to to ) 4 6 1 d.p. 11.1 m b 10 3 4 a 240 m b 248 m c n 1 d.p. 2 1 d.p. atri multipliation matri addition. is distributive c over tree lengts tat multipl 2 5 59.8 m to 1 d.p. to 3b 6 b ( 17 50 verties. 2 24 omputed. ) ( 9 aes, b 10 7.1° 11 123.7° to 12 a 75° alternate b 65° angle 240, or eample 1 d.p. to 1 m × 1 m × 240 m, 5 m × 6 m × 8 m or 1 d.p. segment teorem 5 a 9 4.22 at entre = 2 × angle at 8 1 and 2 on diagram irumerene 7 3 0 c ) ( 0 , an enlargement, sale ator = 90° angle beteen 2, 2 tangent = ycneuqerF 2 entre 0, and radius 0 o y opposite so ___ a li ___ ___ › 6 is angles up to 180°, quadrilateral. 5 4 3 ___ › › add 6 › 2 13 = + + = a = b + b 2a 1 5 ___ C ___ › a ___ › › 0 4 = + 3 = b + a + b 0 2 4 Waiting a 6 time 8 10 (minutes) › › A −2a ___ ___ 2 = = , so and are equal b and 2 to 4 minutes 7 parallel. c 1 12 B 0 d ) ( i 8 3 x 1 5 5 14 1 ii 1 15 a i 1 5 1 1 ) ( 2 1 ii 6 1 1 0 2 ( 4.23 9 a x ) 2 −3 −2 −1 9 4 1 −6 −4 −2 3 0 0 1 2 3 0 1 4 9 0 2 4 6 0 3 8 15 2 1 a = 2, 2 x = 0.5, b = x −1 1 iii y = 1 2 1 1 ) ( 1.5 3 + 3 1 2 0 2 ) ( 1 i 5 1 1 ) ( 6 186 1 2x 2 y = x + 2x −1 Answers 2 y b,d 4 a,c 17 3 a 2x b (2x 2 14 8 x 12 2 3) =4x 12x +9 +3 _____ c 7 2 10 4 6 thgieH 6 )m( 8 5 Further exam practice 3 4 5 1 5×10 2 a 3 2 i 3x(x ii (x x 4 3 1 2 3 7) +4)(x 4) 2 4 b 1 4x 4x 3 2 0 3 a 16 000 b 17 000 2,0)n(1.5,7) 4 0 c i e x 1 = ii −1.7,y = x = −0.5orx 7 a y =6.5 y = x 3 −1 4 5 T ime 6 7 8 9 (seconds) 4 ( 5 20.25 6 trong,ositiecorretion d 5.9secons a 133.7 cm 6 b 2 =1.7, b 1 5 √ √ 3 4.5 5 4 Renmer ✓ ✓ ✓ ✓ ✓ nteger ✓ ✗ ✗ ✓ ✓ trnmer ✓ ✗ ✗ ✓ ✗ rimenmer ✓ ✗ ✗ ✓ ✗ Rtionnmer ✓ ✓ ✗ ✓ ✓ 3 =3 2 b c x 8 a 318 g = 9 a 7 cm 1147.9 cm 12 b 6 7 8 9 49.82 497.5 g 5 2 12 b,c y B a 787.40 b 12.5 a Male Female Youngest 20 18 oweruartile 24 23 Median 27 25 A 7 a 42 b 4.8oss c 55 x C eruartile 33 34 ldest 56 51 nteruartilerange 9 8 2.5< 9 a 11 x i 7 ii 5 ⩽4 iii13 ange 36 33 i19 d Reectionin y = −x b 3 10 a b 2×2×2×7or2 i 14 ii 280. i re ii se 4 ×7 b x a i 1 iiise 5 10 ) ( iocnnotte. 0 180 ____ 11 10 a a = b a =20 a x =17,y b m a 11.6 cm = −3 2 5 b 2 = a ii +3 ) ( 3 11 12 c b a 56° 14 =2 b 34.2° c 15.1 cm iii ) ( 7 2 b 112° c 28° d 14 5.2 cm b 5c b= 4 ) ( =3.5 x 13 ( ) ( 7 12 =3.5a 2 =3 2 62.6 cm Further 1 a 8 b 63 (to3s.f.) exam practice 2 14 a 6 14 1 4 11 a x b (3, c x =3± x = 4.41(to3s.f..) ycneuqerf d =3 35 ) i 1,3,4,5,7,9 ii 3,5,7 iii4,6,8,10 12 b A ⋂(B ⋃ 6 a 81 km 3 b 1325 =1,3,5,7,9 (A ⋂ B)⋃(A ⋂ C)=1,9⋃ 3,5,7=1,3,5,7,9 15 b =88°,=52°n =40°. 16 a i 2b ii 3a iii b evitalumuC 2 2 orx =1.59or y a 1,2,3,4,5,7,9 3 √ C)=1,3,5,7,9 ⋂ 2 c 2) B A 30 25 20 15 10 5 i2a ___ 0 › b x =2a ___ › = ___ › 0 ___ 10 20 30 40 50 60 ___ › › + = b + a Height b=a (cm) ___ › b i 26 cm ii 34 =2 20=14 cm 187 Answers x 13 a +2 _____ 12.1 km b ominmigteR x ⩾0. 12 2 b coominmigteR y ⩾ −2 12 13 4 √ a √ a 40 ≈6.3 cm 89 ≈9.4nits )mk( 2 b 10 b (4.5, 19.1 cm (to1 ..) 1) 14 ecnatsiD 8 5 2iseennrimensote 6 a b 37.8 7 a 2736.95krone b 40.19 8 a (a♣b)♣c y intersectionisnotemt. 6 6 30–40 4 4 2 0 2 0 10 20 30 40 50 T ime 60 70 80 90 =2ab♣c (min) =2×2abc 14 a x 0 2 4 =4abc b 3 2 a♣(b♣c) = c y =2x d y = =2×2abc 1 − x 4 2 2 15 a♣2bc 2 =4abc a,b b a♥b =2a b b♥a =2b a,soitisnot 6 y 6 commttie. c a♣(b♥c) = a♣(2b c) 15 23mintes 16 a i 24 ii 44 4 =2a(2b =4ab c) iii80 2ac 2 2 b (a♣b)♥(a♣c) =2ab♥2ac (a +1) = a 2 (a 1) 2 =2×2ab x 0 4 2ac =4ab 2 2ac 2 101 o♣isistritieoer ♥ 9 a + c a + d =180 (oositengesofccic 17 = 180 (cointeriorngeson 6 c 0 1 ) ( Further 1 2 a i 64° ii 64° b 12 cm a ( reines) 18 0 od = c =180 imir, a otetreimisisoscees. = b = a 180 c. b exam =114°,b practice =43°, c 4 19 =55° a a +1) 1 1 4 3 =4×100=400 ) 2 4 6 4 0 2 ) ( i 72° ii 54° b cisosceestringecnesit crigtngetringesse oecrigtngetringesn intotorigtngetringes. b d 2a 2 99 a ) 4 +1–(a =4a C 2 1 2 +2a 4 cmneigtof4tn54°. c 1 reof ×4×4tn54°=8tn 2 10 223(se6) 54° 2 3 a oreereof x,y = x 4 11 a x +2y b a + rocesstoneeof y.(tis erere10sctringes(toin 12b ecisosceestringe),sototre 5 c mn–oneming.) 12a =10×8tn54° =80tn54° ty = √ x 2 4sesof xfor 2b 2 ____ d icyisnene(e.g. x = = −2), 2 b 2 20 a 14 a 2 3a 3 b 36 c 24 nissoone–mnming (e.g.x =9giesy = soisnotfnction. 188 −1or 7),n Index distributive A addition 168, 6, 8, 20, 76–7, 82, 84, 172 algebra 80–1, 85, law 21, 87 8–9, 20, 82, division 7, domain 106 matrix 84 algebra 176–7 transformations mean 60, 63, 176 64 measurement 91 S 174–5, sample scalars 38–9, 48–9, space 170, diagrams 74 172 scale 64 scale drawing scale factors E ambiguous case 52–3, 142 enlargements angles 132–5, 138, measures 94–101, central 112–15, theorems sine rule rule arcs 88–9 median metric 25 40–1, 48–9, 42–3, 44–5, 158 averages 2, formula, law 50, (axis) factors 106 21, 60, of 86 61, 88, 90, 91, 4–5, 88, 150 changing the subject money 10–11, 18, 85, 42–3 tables 61, 62–3, self-inverse 65 26, semicircles 49 sets 62 12–13, 2, 84, 28–9 18–19 combinations of complements 32 8–9, 20, 76, set 172–3 notation unions Venn 68 sets 32–3 30 intersections 7, 174 30–1 elements 4–5 170, reections 164 sequences 116–17 multiplication 82, frequency sectors 148 60, multiples 8–9, 62, 55, line mode 95 126–7 60–1 66–7 system mirror 96 92–3 fractions 64–5 60, midpoint F diagrams associative axes 76–7 factorising 38, arrow 72, expanding 163 42 area tendency dispersion 161 appreciation 38–9 150 117 164–5 events cosine 26–7 106, 159 circle 64 174 140–1, equations 156, 150, 32 30, 119 32 diagrams 30, 34–5 N functions 108–9, shapes 110–11, nets B 122–3, bar charts SI 124–5 numbers 56 41, 2–3, 16, 42, 82–3, 22–3 number G line units 48 gures 156–7 gradient 111, 114, 115, number v126, systems 20 128, real 129 numbers 2, 6–7, operations 86–7 graphs 50–1, 57, 68–9, 154, 20, 90, 85, 88, 94 107, 110–13, 101, symbols 116–17, for numbers 152–3 80–1 speed 124–9, grouped chords circles 42, circle frequency tables objects 166 theorems segments 42, circumferences 164–7 164, 42, (highest common factor) histograms law 21, 20, parallel 87 lines matrix 134–5, 112–15, 133 34 83, 111, 115, 6, 100 8, 20, 82, 84, 132, 173 141 summary 86 square parallelograms 106 40, surface (reections) patterns 148 area 41, 43, 26, percentages 49 10–11, 12, 80–1 13 symmetry 47 indices (index) 16, perimeter 84–5 70–1 44–5 19 symbols shapes system statistics 169 89, imperial computations 30, 70–1 16–17 172 images compound lines subsets 74 126–7 148–9, 153 40 6–7 inequalities 118–19, 120–1, perpendicular lines 115, 132, T 46 tables between 2, interest planes 126–7 points 24 62–3, 126, 68, 154, 77, 106 166 153, 114, three-dimensional 159 126, 132, 133, shapes 152–3 141 lines polygons 132 56, 138–9, time 143 27, 48, 50–1, 128–9 168 inverses prisms 20 44–5, 46, transformations 152 148, 150–1, 162–3 cross-sections cumulative tangents 58–9 153 intersecting 114, 154, charts 48 polyhedrons coordinates pie 18 units intercepts 26, 61, 77 integers conversion tables 140–1 125 contingency 14–15, form substitution image 45, deviation standard subtraction the 128–9 46 standard straight 150 72, 45, P identity completing cosines 86–7 170 commutative convex 20, 164 identities cones 8–9, 106 collinearity 96–7, 6–7, 18–19 outcomes 58 I codomain ordering orientation 4–5 166 spheres 148 operations H HCF graphs 62–3 49 42–3 128 speed–time O 26, 50, 163 C capacity 176 160–1 10–11 solids brackets 117, 8–9, sines binary 6 equations 22 98–101, BIDMAS 152–3 82 simultaneous bearings 47, 91 signicant bases 43, 44 44, inverse functions 109, inverse matrices inverse operations inverse variation 122–3 probability 174–5, 72–7 176 152 frequency 173, 174 pyramids 45, 46, translations 152 150 graphs Pythagoras’ 20 theorem 146–7 transversal lines 134 68–9 curves 126–7, trapeziums 104–5 132 Q triangles cyclic 41 quadrilaterals 165 quadratic L equations 136, 144–5, 146–7, 96–7 154 laws of indices 16, 84–5 quadratic functions quadratic graphs 124–5 area D LCM data (lowest common multiple) 124–5, 40, 158 126 constructing 54–5 quadrilaterals 4–5 class intervals class limits 40, 136–7, cosine length and 26, quartiles 49 sine terms line graphs information 160 law 169 57 68–9 radii line displaying rule triangle R frequency 162 84 54–5 cumulative rule 67 boundaries like 142–3 165 55 segments 116–17, (radius) 42, trigonometry 166 154–5, 158–9 132, 56–9 range 66–7, turning 106 points 126 140 grouped data 54, 62–3 ratios linear measures of central equations functions linear graphs 132 unknowns 110–11 60–1 reciprocals measures of dispersion 64 U rays linear 14–15, 94–5 tendency 110–11, 80, 94, 98–9, 120–1 20 112–13, 66–7 rectangles V 40 116–17 summary statistics 70–1 reection linear decimals 6, 10–11, inequalities reection 120–1, depreciation relations lines 111, direction 134–5, 140–1, magnitude 50–1, mass 26, matrices 80, variation 102–5 Venn 40 120 150, 168–71 diagrams vertical 149, rotation 169 rounding 49 (matrix) vectors 106–7 16 rotation M graphs variables 132, roots 66 148 2, 30, 159 42 distance–time 128 116, rhombuses 169 dispersion 112–15, 173 133, diameters 174 symmetry 125 25 determinants 148–9, 118–19, 18 172–3 6 volume 174 symmetry 148 line test 46–7, 108 48–9 34–5 Mathematics for CSEC® 2nd edition Less Stress, More Success Developed Guide in will CSEC® Written exclusively provide a team examination, easy-to-use outcomes ● you the with Caribbean additional Examinations support to Council®, maximise your this Study performance Mathematics. by enhance with this of experts Study Guide double-page from your the of Examination-style the the and all Each Mathematics the topic contains subject , practice CSEC® covers format . syllabus study in such questions a essential begins range syllabus information with of and the key features in an learning designed to as: to build confidence ahead of your examination ● Engaging the ● activities help you develop the analytical skills required for examination Examination ● that tips with essential advice on succeeding in your assessments Did You Know? boxes to expand your knowledge and encourage further study This Study Guide multiple-choice answers to and achieve The range questions top marks subjects in and CSEC® Press at by to CSEC® additional to aid revision Mathematics Council produce and a 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(CXC®) series has of worked Study exclusively Guides across a with wide CAPE®. How 1 online examination-style feedback, Examinations University of supported examiner Caribbean Oxford is to get in touch: web www.oup.com/caribbean email schools.enquiries.uk@oup.com tel +44 (0)1536 452620 fax +44 (0)1865 313472