mst121 prerequisite revision
TRANSCRIPT
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Using Mathematics
Starting points
MST121
Revision Pack
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ContentsHowtousethispack 2DiagnosticQuiz 3SolutionstoDiagnosticQuiz 13Module1 Numbers 27
1.1 Numbersystemandnotation 271.2 Calculating 301.3 Multiples,factorsandprimenumbers 341.4 Fractions,decimalsandpercentages 371.5 Calculatingwithsignednumbers 451.6 Workingwithpowers,indicesand logarithms 461.7 Workingwithroots 521.8 Ratioandproportion 55
Module2 Measures 592.1 Units 592.2 Inequalities 602.3 Formulas 612.4 Measuringandclassifyingangles 622.5 Statisticalmeasures 66
Module3 Somebasicfigures 713.1 Triangles 713.2 Rectangles 773.3 Circles 793.4 Areas 80
Module4 Coordinatesandlines 834.1 PointsandCartesiancoordinates 834.2 Lines,gradientsand intercepts 84
Module5 Algebra 895.1 Introduction 895.2 Expressions 895.3 Equations 975.4 Inequalities 106
Module6 Trigonometry 1116.1 Trigonometricratios 1116.2 Areaofageneraltriangle 118
Module7 Graphsandfunctions 1217.1 Graphs 1217.2 Functions 125
Module8 Geometry 1418.1 Propertiesofplanefigures 1418.2 Areasandvolumesofsolids 144
SolutionstoExercises 149Index 163
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How to use this packThispackhasbeendesignedtohelpyoutomakethemostefficientuseofwhatevertimeyouhaveforrevision. It is intwomainparts: anauditsectionandarevisionsection. Thereisalsoan index.TheGuide to Preparationcontainsdetailedadviceaboutyourgeneralpreparation forthecourse(s)andforyourmathematicalpreparation inparticular.Youshouldworkthrough theGuide to Preparation,anduse thispackasdirected there.Audit Section
ThissectioncontainstheDiagnosticQuizand itsSolutions.Revision Section
Thissectioncomprisesrevisionnotes,workedexamplesandpracticeexerciseswithsolutions.The index istheretohelpyoufindtopics,butcanalsobeachecklistofmathematicalwordswhichyoushouldunderstand. Ifyoucomeacrosswordsorsymbolswhichareunfamiliar,thenperhapsyoushouldstartyourownmathematicaldictionary.
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Diagnostic Quiz
Thequestionsthatfollowaredesignedtogiveyoua lookatanumberof TheGuide to Preparationmathematicaltechniquesthatwillberequired inMST121(andMS221). explainshowtomakebest
useoftheRevision PackandDont
worry.
This
quiz
is
not
atest
or
an
examination.
Only
you
will
inparticularthisquiz.knowhowwellorhowbadlyyoudid.
Donthurry. Youcantakeaslongasyoulike,anddothequestions inanyorder. Youmaywellfindthatinanyonequestionyoucandoonepartandthengetstuckorslipuponanother.Asyouworkoutyouranswers,keepanoteabouthowconfidentyoufeelaboutthem. Whenyouhavefinishedthequestions,checkyouranswerswiththesolutions,whichstartonpage13. ThesolutionscontainreferencestotheappropriatesectionsoftheRevisionsection,soyoucanthenspendsometimeworkingonthetopicswithwhichyouhaddifficulty.Youmayfindithelpfultoreturntosomequestionsafteryouhaverevisedatopictocheckthatyouarethenabletoanswerthemcorrectly. ThequestionscannotcovereverythingincludedintheRevisionsection,soevenifyougetthemallright,youmaystillfind ithelpfultoreadthroughthatsection.Good luckenjoythechallenge!
1 Numbers
Question 1
Theuseofpowersisverycommon inmathematics,and itis importanttoknowwhattheymean. Seehowmanyofthefollowingyoucando. Ifyougetstuckonone,trythenextone.(a) Evaluateeachofthefollowing,withoutusingyourcalculator.
32, 83, 41, (12
)2, (8)2, (2)3, 32, (3)2, 32.(b) Useyourcalculatortofindeachofthefollowing.
(i) 76 (ii) 3.24 (iii) (3.2)4 (iv) (3.2)4(v) (3.2)4 correcttotwosignificantfigures
Question 2
(a) Aftercarryingoutaseriesofnumericaloperations,acalculatorgivesthefollowinganswer.
1.917150743E5Explainwhatthismeans.
(b) Howwouldyouexpectthesamecalculatortodisplaythefollowingnumber?
32190818670
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REVISION PACK
Question 3
It isveryeasytomakeamistakewhenyouuseacalculator,so it isimportanttocheckthatyouranswerisreasonable. Youcandothisbyestimatingtheanswer.(a) Giveroughestimatesforeachofthefollowingnumbers.
(i) 413 (ii) 2782 (iii) 12.4 (iv) 0.1253 (v) 189025(b) Useyourestimatestoobtainanapproximateanswertoeachofthe
calculationsbelow.12.4
(i) 4132782 (ii) 413189025 (iii)0.1253
27820.1253(iv)
12.4Checktheaccuracyofyouranswersbyusingacalculator.
Question 4
Knowingthecorrectorder inwhichtodealwiththearithmeticoperationsinanexpressionisimportant.Calculateeachofthefollowing.(a) 32+25 + 6 (b) (32 + 2) 5 + 6 (c) 32 + 2(5 + 6) (d) (32+2)52 +6 (e) (32+2)(52 + 6) Question 5
Byfindingtheprimefactorsofeachofthenumbers12,20and45,findthesmallestwholenumberthatcanbedividedexactlybyallthreenumbers.Question 6
Sometimesnumbers lookverydifferent,butareactuallythesame.(a) Useacalculatortowriteeachofthefollowing fractionsasdecimals,
correcttothreedecimalplaces.3 15 13 87
(i) (ii) (iii) 1 (iv) 2140 62 84 93
(b) Withoutusingyourcalculator,evaluateeachofthefollowing.(i) 2
3+27 (ii)
56
23 (iii)
415+
720 (iv) 3
252
78
(v) 381
511 (vi) 1
344
23
Question 7
Inthisquestion,trytouseyourcalculatoronlywhenreallynecessary,andthenwithasfewkeystrokesaspossible.Findthereciprocalofeachofthefollowingnumbers,givinganswersthatarenotexactcorrecttothreesignificantfigures.(a) 10 (b) 5 (c) 101 (d) 72.5 (e) 0.00356
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DIAGNOSTIC QUIZ
Question 8
Percentagesoccur inmanywalksoflife, includingMST121!(a) Increaseeachofthefollowingnumbersbythepercentageshown.
(i) 14by10% (ii) 142by115%(b) Bywhatpercentagehas300beenincreasedordecreasedtoarriveat
thefollowinganswers?(i) 317.5 (ii) 60
Question 9
(a) Evaluateeachofthefollowingwithoutusingyourcalculator. Thenseeifyoucanobtainthesameanswerbyusingyourcalculator.(i) 3+(4)6 (ii) 34 + (6) (iii) 3+(4 + 6) (iv) 3(4 + 6) (v) (6)(5) (vi) 6(35)
(b) Calculateeachofthefollowing,withoutusingyourcalculator. 3(i) 14400 (ii) 0.81 (iii) 12 (iv) 24
1(v)10000
Question 10
Withoutusingyourcalculator,writeeachofthefollowingnumbersinlog10form. Forexample,log10 100=2.
1(a) 102 (b) (c) 0.001
100Question 11
(a) Useyourcalculatortofindeachofthefollowing,givingyouranswerscorrecttofourdecimalplaces.(i) ln3.142 (ii) ln14.16 (iii) ln14658 (iv) ln0.0324
(b) Useyourcalculatortofindthenumbers,correcttothreesignificantfigures,whosenatural logarithmsareasfollows.(i) 0.812 (ii) 1.623 (iii) 2.976 (iv) 0.0356
Question 12
Threepeopledecidetosplitafoodbillbetweenthem,takingintoaccountthenumberofmealseatenathome. Theyagreethat itshouldbesplit intheratio2to3to5. Thebill is70.How much does each person pay? 2 Measures
Question 13
(a) Markthefollowingnumbersonanumberline,andhencearrangetheminascendingorder.
2.9, 4.3, 0.2, 3, 3.5, 1.5.(b) Ineachofthefollowingparts, inserttheappropriate>or
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REVISION PACK
Question 14
Makexthesubjectoftheformula ineachofthefollowingequations.(a) 2x=y3 (b) p=x2 + 3 (c) (x+ 3)2 =t(d) y= 1
2
x (e) m= 3px2
Question 15
Inmathematicsweoftenuseradians,ratherthandegrees,tomeasureangles.(a) Howmanyradiansarethereinonecompleteturn?(b) Sketchanangleof60. Whatisthemeasureofthisangle inradians?
3(c) Sketchanangleof radians. Whatisthemeasureofthisangle in
2degrees?
Question 16
Theword averagehasmorethanonemeaning. Somematchboxesarelabelled Averagecontents50matches.(a) Thecontentsoffiveboxesofmatcheswerecheckedandfoundtobe:
50, 47, 48, 51, 51.Findthemean,medianandmodeofthecontentsofthefiveboxes.
(b) Thecontentsofafurtherfiveboxeswerecheckedandfoundtobe:50, 48, 48, 51, 48.
Combinethesefigureswiththosefrompart(a),andcalculatethemean,medianandmodeforthetenboxes.
3 Some basicfigures
Question 17
Itisimportanttobeabletoobtainanswersfrominformationinadiagram.Intherectangle inFigure0.1,thesideAB isof length6mandthesideBC isof length8m. Youmayfind ithelpfultoinsertthesefigures inthediagram.
A D
B C
Figure 0.1
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DIAGNOSTIC QUIZ
(a) UsePythagorasTheoremtofindthe lengthofsideAC.(b) CalculatetheperimeterofthetriangleABC.(c) Findtheareaoftherectangle,andhencetheareaofthetriangleABC.(d) Confirmthattheareaofthetriangleyouhaveobtainedinpart(c)is
correct,byfindingtheareadirectly.(e) Apathofwidthonemetre isconstructedwhose inneredge isthe
rectangleABCDandwhoseouteredge isa largerrectangle. What istheareaofthepath?
Question 18
InFigure0.2,O isthecentreofacircle,andA,B,C andD lieonthatcircle.The linesegmentsAC andAB arenotequalin length,andAOD isadiameter.
A
B
CO
D
Figure 0.2
(a) ExplainwhyangleACD is90.(b) Identify
(i) threeisoscelestriangles(therearemorethanthree);(ii) twoscalenetriangles(therearemorethantwo);(iii) tworight-angledtriangles.
(c) IfangleDAC is40andangleCBO is15,showthatangleCOD is80andfindtheotherangles inthefigure.
(d) Iftheradiusofthecircle is10cm,find,correcttothenearestwholenumber:
(i) theareaofthetriangleOCD;(ii) the lengthofthearcCD.
(e) What istheareaofthesectorOC Dtothenearestwholenumber?
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4 Coordinates and lines
Question 19
Coordinatesareashorthandwayofdescribingthepositionsofpointsrelativetoafixedpoint(theorigin)andapairofaxes. ThepointsA,BandC areplottedonFigure0.3.
y
3A
2
1
2 1 0 1 2 3x
1
C2
B
Figure 0.3
(a) WritedownthecoordinatesofA,B andC.(b) PlotthepointsD(1,3),E(1,2.5)andF(0.5,2).(c) DrawthestraightlinewhichpassesthroughAandF.
(i) What isthegradientofthis line?(ii)
Where
does
it
cross
the
x-axis?
(iii) Wheredoesitcrossthey-axis?
5 Algebra
Question 20
Ifx=8,findthevalueofeachofthefollowingexpressions.(a) (i) 5x (ii) 5x (iii) 3x2 (iv) 3(x+ 2) (b) (i) x2 + 4 (ii) (x+ 4)2 (iii) (x+24)12 (iv) x2 3x4Question 21
Simplifyeachofthefollowingexpressionsasfaraspossible.(a) 2a+ 3b4a (b) 2a+ 3b4c(a5b)(c) 2a+ 3b4c+ 2(a5b) (d) 2a+ 3b4c2(a5b)(e) 2a+ 3b4c+c(25b)Question 22
Multiplyoutthebracketsineachofthefollowingexpressions,andsimplifytheresultasfaraspossible.(a) (b+ 1)(b+ 2) (b) (c2)(c+ 5) (c) (df)2(d) (3x+4)(2x7) (e) (3x4y)(5y+ 6x)
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DIAGNOSTIC QUIZ
Question 23
Factorisingistheoppositeofmultiplyingout. Youstartwiththesumofanumberoftermsandaimtoexpressitasaproduct. Factoriseeachofthefollowingexpressions.
2(a) ab+a (b) ab+ac (c) ab+ 2ac+a (d) a2 b2(e) a2 + 2ab+b2 (f) 6a2 +a
1
Question 24
Solveeachofthefollowingequations.(a) (i) 2x5=15 (ii) 2(x5)=15 (iii) 2(x5)2 = 32
(iv) 82x=x+ 7 25
(b) (i) p2 + 2p4=0 (ii) 36=v2
(iii) 5t = 26 Inparts(i)and(iii),giveyouranswerscorrecttotwodecimalplaces.
Question 25
Solvethefollowingpairofsimultaneousequations.x+ 2y= 4
2x3y=6Question 26
Solveeachofthefollowing inequalities,andillustrateeachansweronanumberline.(a) 2x35 (b) 2x4 5
6 TrigonometryQuestion 27
Inthetriangle inFigure0.4,angleA is90, angle B is70andthehypotenusehas length15cm.Usingtrigonometricratios,orotherwise,findtheunknownsidesandangleofthetriangle. Givelengthscorrecttothreesignificantfigures.
A
B C
Figure 0.4
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7 Graphs and functions
Question 28
Thegraphonthe leftinFigure0.5showsajourney inwhichChriswalksfromhometothenewsagent,buysanewspaper,andwalksback,stoppingtotalktoafriendontheway. Thespeedofwalking isconstantineachsectionofthejourney. Usethegraphtodeduce:(a) (i) thedistance, inkilometres,fromChrisshometothenewsagent;
(ii) thespeedofwalking ineachsectionofthejourney, inkilometresperhour;(iii) thedistance,inkilometres,fromthenewsagenttothepointatwhichtheconversationwiththefriendtakesplace.
(b) ExplainwhythegraphontherightofFigure0.5cannotrepresentajourney.
distance in kilometres distance from start
time in minutes0
0.2
0.4
0.6
0.8
1.0
10 20 30 40 50
(a) time in minutes (b)
Figure 0.5
Question 29
Inabuildingdevelopmentallplotsarerectangular. Inonesection(A)allplotswillhaveanareaof500m2. Inanothersection(B)theareasofplotswillvary,butthedimensionsofeachplotwillbesuchthatthe lengthisofthebreadth.Letlmetresandbmetresrepresentthe lengthandbreadth,respectively,ofaplot.(a) Forplots insectionA,describe inwordstherelationshipbetweenthe
lengthandbreadth,andrepresentthisrelationship insymbols.(b) Forplots insectionB,representtherelationshipbetweenthe length
andbreadth insymbols.(c) (i) WhichofgraphsI,IIandIIIinFigure0.6representsthe
relationship inpart(a),andwhichgraphrepresentstherelationship inpart(b)?(ii) Describetherelationshiprepresentedinthegraphyoudidnotchoose.
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5
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DIAGNOSTIC QUIZ
l
bgraph I graph II graph III
b
l
b
Area
Figure 0.6
Question 30
Sketchthegraphofeachofthefollowingequations. (Accurateplotsarenotrequired. Nocalculatorsallowed!)
3(a) y=x2 (b) y=x2 2 (c) y= (x2)2 (d) y=xQuestion 31Sketchthegraphofeachofthefollowingfunctions.(a) f(x) = sin xforxbetween2and2(b) g(x) = 1 + cos xforxbetween2and2
x(c) h(x) = e
8 Geometry
Question 32
Tworectangularpiecesofpapereachmeasure31cmby20cm. Oneisusedtomakeanopen-endedtubeofcircularcross-sectionand length20cm.Theotherisusedtoformanopen-endedprism,alsoof length20cm,whosecross-sectionisanequilateraltriangle.Foreachshape,1cm(ofthe31cm)istakenupbytheneedtosecuretheedges.
20cm
31cm
1cm
20cm
20cm
Figure 0.7
(a) Find:(i) theradiusofthecircularcross-sectionandhencetheareaofthecircularholeatthebaseofthecylinder;(ii) theareaofthetriangularholeattheendoftheprism;(iii) thesurfaceareaofeachshape;(iv) theratioofthevolumeofthecylindertothatoftheprismformedbyclosingofftheends.
(b) Canyouthinkofanyeverydayexamplesofsuchcylindersandprisms?11
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Solutions to Diagnostic Quiz
Solution 11 1(a) 32 = 9; 83 =512; 41 =4
; ( 12
)2 =4
; (8)2 =64; (2)3 =8; SeeSection1.1.1
32
= 9 ; (3)2 = 9; 32 =9.(b) (i) 117649 (ii) 104.8576 (iii) 104.8576
(iv) 104.8576 (v) 0.0095(to2s.f.)Solution 2
This isthecalculatorswayofshowingverylargeandverysmallnumbers. SeeSection1.1.(a) 1.917150743E5 means 1.917150743105,which isequalto
0.00001917150743.(b) 32190818670=3.21908186701010,whichthecalculatorwillshow
as3.219081867E10.Solution 3
Youmayhaveroundedthefigurestoothervalues,soyourestimated SeeSection1.2.answerscoulddifferslightlyfromours. The importantthingistomakesurethatyoucanestimateananswer inyourhead.(a) (i) 400 (ii) 3000 (iii) 10 (iv) 1/10(0.1) (v) 200000(b) (i) 4003000=1200000;
calculatedvalue=1148966.(ii) 400200000=80000000;calculatedvalue=78067325.
1
(iii) 10 =100;10calculatedvalue=98.9625(to4d.p.).
1(iv) 3000 10=30;
10calculatedvalue=28.1117(to4d.p.).
Solution 4
(a) 32+10+6=48 (b) 345+6=170+6=176 SeeSection1.2.(c) 32+22=54 (d) 3425+6=850+6=856(e) 34(25+6)=3431=1054Solution 5
12=223,20=225,45=335;sothesmallestnumber SeeSection1.3.divisibleby12,20and45 is22335=180.
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Solution 6
(a) Eachofthefollowinganswersisgivencorrecttothreedecimalplaces.3 15
eeSection1.4. (i) = 0.075 (ii) = 0.24240 62
13 13(iii) = 0.155,so1 = 1.155.
84 8487 87(iv) = 0.935,so21 = 21.935.93 93
14 6 20 5 4 1 16 21 37(b) (i) + = (ii) = (iii) + =
21 21 21 6 6 6 60 60 6017 23 136 115 21 3 16 6
(iv) = = (v) =5 8 40 40 40 8 11 11
7 14 7 3 3(vi) = =
4 3 4 14 8Solution 7
Didyouusethereciprocalbutton(marked1/xorx1)onyourcalculatorforparts(d)and(e)?
eeSection1.4. (a) 0.1 (b) 0.2 (c) 10 (d) 0.0138(to3s.f.)(e) 281(to3s.f.)Solution 8
eeSection1.4. (a) (i) 10%of14is1.4,soa10% increasegives14+1.4 = 15.4.(ii) 15%of142 is21.3,soa115% increaseis142+21.3=163.3. Thetotalistherefore142+163.3=305.3.
(b) (i) Theactualincreaseis17.50;thepercentage increaseis17.5/300%=5.83%(to2d.p.).(ii) Theactualdecreaseis240;thepercentagedecreaseis240/300%=80%.
Solution 9
eeSection1.5. (a) (i) 3+(4)6 = 3 46 = 7(ii) 34 + (6)=346 = 7(iii) 3+(4 + 6) = 3 + 2 = 5 (iv) 3(4 + 6) = 3 2 = 6 (v) (6)(5)=(6)/(5)=6/5 = 1.2
3 5(vi) 6(35)=6 = 6 = 10 5 3Yourcalculatorprobablygavedifferentanswersforsomeofthecalculations.
eeSection1.7. (b) (i) 14400= 122 102 = 12 10=120 81 92 9
(ii) 0.81= = = = 0.9100 102 10
(iii) 12=22 3,so 12= 22 3 = 2 3. 3 3(iv) 24=23 3,so 24= 23 3 = 2 3 3.
1
1 1
(v)
10000
=
100
2
, so
=
= = 0.01.10000 1002 100(Theanswerstoparts(iii)and(iv)havebeen leftin surd form,butcouldbeexpressedindecimalformwiththeaidofacalculator.)
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SOLUTIONS TO DIAGNOSTIC QUIZ
Solution 101
(a) log10 102 =2 (b) log10 =2 (c) log10 0.001=3 SeeSection1.6.100Solution 11
Ifyouhavedifficulty inobtainingtheanswershown,checkyourrounding. SeeSection1.6.(a) (i) 1.1449 (ii) 2.6504 (iii) 9.5927 (iv)
3.4296
(b) (i) 2.25 (ii) 5.07 (iii) 19.6 (iv) 1.04Solution 12
Thetotaloftheratios is2+3+5=10. Onetenthof70is7,sothe SeeSection1.8.firstpersonpays27 = 14,thesecondpays37 = 21andthethirdpays57 = 35.Solution 13
(a)3 0.2 1.5 2.9 4.3
4 532101234 6
3.5
Figure 0.8
So,arranged inascendingorder,thenumbersare SeeSection1.1.3.5, 3, 0.2, 1.5, 2.9, 4.3.
(b) (i) 2.93 SeeSection2.2.Solution 14
(a)(b)
x= y32 orx=
12(y3)
x2 =p3,sox=p3 (provided p3) SeeSection2.3.
(c) x+ 3 = t, so x=3 t(providedt0)(d) 2y=x, so x= (2y)2 = 4y2(e) Multiplythroughbyx2 togive
mx2 = 3p,fromwhich
x2 = 3p.m
Sox=
3pm (providedp/m0).
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Solution 15
eeSection2.4. (a) Onecompleteturn=2radians(=360).(b)
60
Figure 0.9
60 60= 2radians= radians
360 3(c)
3p radians2
Figure 0.10
3 3 360radians= =270
2 2 2Solution 16
eeSection2.5. (a) (50+47+48+51+51)5 = 49.4,sothemean is49.4matches.Themedianisthemiddlevaluewhenthevalueshavebeenarrangedinorderofmagnitude,so is50matches. Themode isthevaluethatoccursmostoftenthat is,51matches.
(b) Themean is49.2matches. Therearenowtenvalues,sothereisnosinglemiddlevalue. Themedian isthemeanof48and50,thetwomiddlevaluesthat is,49matches. Themode is48matches.
Solution 17 eeSections3.1,3.2,3.4 (a) AC= (62 + 82) m = 36+64m= 100m=10m.nd6.2. (b) PerimeteroftriangleABC= 6 m + 8 m + 10 m = 24m.
(c) Areaofrectangle=6m8 m = 48 m2,soareaoftriangleABC is24m2.
1(d) Areaoftriangle= baseheight,so
21
areaoftriangleABC= 8 m 6 m = 24 m2.2
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SOLUTIONS TO DIAGNOSTIC QUIZ
(e)
A D
B C
A
B C
D8 m
10 m
8 m6 m
Figure 0.11
Thediagramshowsthemeasurementsafterthepath,ofwidth lm,hasbeenadded.NewrectangleABCD hasarea10m
8 m = 80 m2.
OriginalrectangleABCDhasarea48m2, from above. Soareaofpath=80m2 48m2 = 32 m2.
Solution 18
(a) AD isadiameterandtheangle inasemicirclesubtendedbya SeeSections3.1,3.3,3.4diameterisalways90. and6.2.
(b) (i) Isoscelestriangleshavetwosidesequal,solookfortrianglesformedbyradiiofthecircle. ExamplesincludetrianglesOBC,OAB,OBD,OCD.(ii) Scalenetriangleshavenosidesequal. ExamplesincludetrianglesABC,BC D,ACD,ABD.(iii) Sincetheangle inasemicircle is90,ACDandABDareright-angledtriangles.
(c) TriangleAOC is isosceles(AO=OC= radius), so angleACO= angle DAC= 40.
HenceangleAOC=180(40+ 40)=100.
AOD isastraight line,soangleCOD =180
100
= 80
.
Theotheranglesareshown inFigure0.12,overleaf.
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SOLUTIONS TO DIAGNOSTIC QUIZ
(c) SeeFigure0.13.(i) Thegradientofthe linethroughAandF is 3(2) = 5 = 2.
2(0.5) 2.5(ii) Thelinecrossesthex-axisat(1
2,0).
(iii) Thelinecrossesthey-axisat(0,1).Solution 20Ifyouranswersarewrong inthissection,donotbedisheartened;justtry SeeSection5.2.toidentifyyourmistakesfromtheworkingshownhere.(a) (i) 5(8)=40
(ii) (5)(8)=40(iii) 3(8)2 = 242 = 26(iv) 3(8 + 2) = 3 (6)=18
(b) (i) (8)(8)+4=64+4=68(ii) (8 + 4)2 = (4)2 = 16
1 1(iii) (8+24) = 16 = 16=+42 2(iv) (8)2 3(8)4 = 64 + 24 4 = 84
Solution 21
(a) 2a4a=2a, so 2a+ 3b4a= 3b2a. SeeSection5.2.(b) 2a+ 3b4c(a5b) = 2a+ 3b4ca+ 5b=a+ 8b4c(c) 2a+ 3b4c+ 2(a5b) = 2a+ 3b4c+ 2a10b= 4a7b4c(d) 2a+ 3b4c2(a5b) = 2a+ 3b4c2a+ 10b= 13b4c(e) 2a+ 3b4c+c(25b) = 2a+ 3b4c+ 2c5bc
= 2a+ 3b
2c5bcSolution 22
(a) b(b+ 2) + 1(b+ 2) = b2 + 2b+b+ 2 = b2 + 3b+ 2 SeeSection5.2.(b) c(c+ 5) 2(c+ 5) = c2 + 5c2c10=c2 + 3c10(c) d(df)f(df) = d2 df f d +f2 =d2 2df +f2(d) 3x(2x7)+4(2x7)=6x2 21x+ 8x28=6x2 13x28(e) 3x(5y+ 6x)4y(5y+ 6x) = 15xy+ 18x2 20y2 24xy
2= 18x2 9xy20ySolution 23(a) a(b+ 1) SeeSection5.2.(b) a(b+c)(c) a(b+ 2c+a)(d) (ab)(a+b)(e) (a+b)2(f) (3a1)(2a+ 1)
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Solution 24
eeSection5.3. (a) (i) 2x= 15 + 5, so 2x= 20 and x=10.(ii) 2x10=15,so2x= 25 and x= 12.5.(iii) 2(x5)2 =32,so(x5)2 =16. Takingsquarerootsgivesx5 = 4,fromwhichx= 5 + 4 = 9 or x= 5 4 = 1. Thesolutionsarex= 1 and x= 9. Alternatively,it ispossibletomultiplyoutandfactorisetheresult,asfollows.
(x5)2 = 16 becomes
x2 10x+ 25 = 16 sothat
x2 10x+ 9 = 0,fromwhich
(x9)(x1)=0.Sox= 9 or x=1. Thefirstmethodiseasier,however.(iv) 8=x+ 7 + 2x, so 8 = 3x+ 7 giving 8 7 = 3x. Hence3x= 1
1andsox= .3
(b) (i) Usingtheformulaforthesolutionofaquadraticequationgives 2 22 4(4)= 2 4 + 16 p=
2 2 2 20 22 5=
2=
2 .
Sop=1 5,thatis,p= 1.24orp=3.24(to2d.p.).
25(ii) Multiplyingbothsidesof36=
v2byv2 gives36v2 =25,from
which25
2 =v36.
Takingsquarerootsgivesv=5 .6
(iii) Becausetheunknowntis intheindexposition, it isnecessarytotakelogsofbothsides,whichgivestlog5=log26. So
log26t= = 2.02 (to2d.p.).
log5(Itdoesntmatterwhetheryouuse logstobase10or ln,providedyouareconsistent: thesameanswerwillbeobtained.)
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Solution 25
There ismorethanonewayofsolvingapairofsimultaneousequations.First,welabeltheequations:
x + 2y = 4, (0.1)2x 3y =6. (0.2)
FromEquation(0.1),x = 4 2y. Thensubstitutingforx inEquation(0.2)gives2(42y)3y =6,
sothat84y 3y =6,
fromwhich7y = 14.
Hencey =2. Substitutingthisvalue inEquation(0.1)givesx + 4
=
4,
sothatx = 0. Thesolution isx = 0, y = 2. Solution 26
(a) 2x 5 + 3, so 2x 8,givingx 4.
x > 4
0 1 2 3 4 5 6
Figure 0.14
(b) 2x < 1 + 4, so 2x < 5,givingx < 2.5.
x < 2.5
2 1 0 1 2 3 4
Figure 0.15
(c) Careisneededwiththisone!3> 5 + x, so 3 5> x,giving2> x, that is, x < 2.
x < 2
5 4 3 2 1 0
Figure 0.16
SeeSection5.4.
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eeSection6.1. Solution 27Thesumoftheanglesofthetriangle is180,sotheremainingangle is
1809070= 20.Inthetriangle,BC isthehypotenuse,so
AB=BCcos70= 15 cm cos70= 5.13cmand
AC=BCsin70= 15 cm sin70= 14.1 cm.(Youcanobtainthesameanswersbyusingdifferentratiosoftheangle20.Havingobtainedthelengthofoneofthesides,youcouldhaveusedPythagorasTheoremtocalculatethe lengthofthethirdside.)Solution 28
eeSection7.1. (a) (i) ThedistancefromChrisshometothenewsagentis0.9km.(ii) Hisspeedinthefirstpartofjourneyis0.9kmin10minutes,whichis5.4km/hour.Attheshop,thespeed is0km/hour.Hisspeedinthefirstpartofjourneyhome is0.5kmin5minutes,whichis6km/hour.Duringtheconversation,thespeed is0km/hour.Hisspeedinthefinalpartofjourneyis0.4kmin5minutes,whichis4.8km/hour.(iii) Conversationtakesplace0.5kmfromthenewsagent.
(b) Thisgraphsuggestsaperiodofnomovement,followedbyaninstantaneouschange indistance,followedbynomovement. Allmovementstakesometimetocomplete,sothemiddlesectionisimpossible.
Solution 29
eeSection7.1. (a) Foreachplot insectionA,thelengthtimesthebreadthequals500m2.Insymbols,
lb=500.(b) Foreachplot insectionB,
l= 85
b.(c) (i) GraphIcorrespondstol= 8
5
b. GraphIIIcorrespondstolb=500.(ii) GraphIIcorrespondstoarelationship inwhichthearea increasesindirectproportiontothebreadth,that is,
area=kb,wherek isaconstant. (Eachplothasconstantlength.)
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Solution 30
Theappropriategraphsaregivenbelow. SeeSection7.2.
(a) y (b) y
25 25
20 20
15 15
10 10
5 5
4 2 2 4x
4 2 2 4x
2
5
y y(c) (d)
25
20
15
10
512 21
6
4
8
2
2
4
6
8
x
4 2 2 4 6x
Figure 0.17
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Solution 31
eeSection7.2. Theappropriategraphsaregivenbelow.(a) f(x) = sin(x)
y
x
1
1
2p 3p/2 p p/2 p/2 p 3p/2 2p
Figure 0.18
(b) g(x) = 1 + cos(x)
y
2
1
2p p p 2px
Figure 0.19
x(c) h(x) = e
y
1
4
2x
21
2
6
8
10
Figure 0.20
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Solution 32
(a) (i) Letrbetheradiusofthecircularhole. Thecircumferenceofthe SeeSections8.1and8.2.circle(=2r) is 30 cm, so
r= 30/2= 4.77cm.(Thisanswerhasbeenroundedtotwodecimalplacesforconvenience,butthecalculatorgivesmoreplaces. Itistheunroundedversionthatshouldbeusedtofindthearea. Alsomakesurethatyouhaveusedthekeyonyourcalculator,andnotanapproximatevaluefor.)Areaofholeisr2 = 71.62cm2 correcttotwodecimalplaces. (Usingtheroundedversionfortheareagives71.48!)(ii) Lengthofeachsideofthetriangle is30cm/3 = 10 cm.
Areaoftriangle= 1 10cm10cmsin602 3
= 50 cm2 = 25 3 cm2.2
Soarea is43.30cm2 (to2d.p.).(iii) Sincethere isnotoporbottomtoeithershape,thesurfaceareaofeachisjusttheareaofthepaperused,minustheareaoftheone-centimetreseam. Sosurfaceareaofeachis
20cm30cm=600cm2.(iv) Volumeofcylinder is
areaofcirclelength=r2 length.The lengthis20cmandr= 30/2frompart(i). Sovolumeofcylinder is
2
30 cm2
20cm=1432.39cm3 (to2d.p.).2
Volumeofprism is
areaoftriangularbaselength=25 3 cm2 20cm=866.03cm3 (to2d.p.).
Soratioofthevolumeofthecylindertothatoftheprism isapproximately
1432.39= 1.65.
866.03(Theratio isanapproximateonesincethenumbersquotedareroundedfigures,notthefullcalculatoraccuracy.)
(b) Therearemanytubes incommonusage;forexample,theinsideoftoiletandkitchenrolls,thetubularpartofaSmartiestube.Theprismhasmore limiteduse,althoughToblerone issold inprism-shapedboxes.
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Module 1 Number s
1.1 Number system and notation
Naming number s
Overthecenturies,ashumanactivitybecamemorecomplex,therewasaneedtodeveloptheconceptofnumberfordifferentpurposes. Initiallytheonlyneedwasforthecountingofnumbers:
one,two,three,. . . . Thesepositivewholenumbersarenowcommonlyreferredtoasthenaturalnumbers. Ascommercial,scientificandmathematicalactivitydeveloped,sodidfractions,negativenumbers,theconceptofzero,andthevarioussymbols,systemsandnotationsforexpressingthem.Thenegativeandpositivewholenumbersand0(zero)arereferredtoastheintegers,andcanbevisualisedas lyingonanumber line.
negative integers zero positive integers
4 3 2 1 0 1 2 3 4
Figure 1.1
Fractions,alsoreferredtoasrationalnumbers,areobtainedbydividingtwointegersandcanbewrittenas barfractions(oftencalledcommonorvulgarfractions),suchas 12 . Rationalnumbersincludetheintegersbecauseeachintegercanberewrittenasabarfraction;forexample,2= 2 . All Also0= 0
1, forexample.
1
rationalnumberscanbewrittenasdecimalswhicheitherterminateforexample, 5
4= 1.25and4
10=0.4 or recurforexample,
1 = 0.333333333 . . . and811
=0.727272727 . . .. (Theshorthandfora3
recurringdecimal isadotabovetherecurringdigitorateachendofaset 0,1,2,3,4,5,6,7,8and9= 0.3, = 0.7 2,and 41 arethedigits.= 0.12 3.)ofrecurringdigits;forexample, 1
3
8
11 333
Figure1.2 illustratesthecompositionofthesetofrationalnumbers.
All natural numbersare integers.
All integersare rationals.
naturalnumbers
integers
rationals
Figure 1.2
Therearealsoinfinitedecimalsthatneitherterminatenorrecur,andthesearecalledirrationalnumbers. Theyincludenumbers like
2 = 1.4142. . ., 5 = 2.2360. . .,= 3.1415. . . ande= 2.7182. . .. istheGreekletterpi. eisthebasefornaturalThesetofnumbersthatincludestherationalsandthe irrationals iscalled logarithms.
thesetofrealnumbers. (Squarerootscanbewrittenaseither 3or 3. Inthistextweuseonlythelatter.)
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Eachrealrationalor irrationalhas itsplaceonthenumber line. SomeexamplesareshowninFigure1.3. Thegapsbetweentheintegersonthenumberlinearecompletelyfilledbyrationalandirrationalnumbers.
negative integers zero positive integers
5 4/10 1/2 2 e p
4 3 2 1 0 1 2 3
Figure 1.3
Aparticularrealnumbermaybewritteninavarietyofforms;forexample,two(word),2(rationalnumber),+2(positive integer),2.0(decimal), 2 or4Powersandrootsare2
(fractions),21 (power)and 4(positivesquareroot)representthesamexplainedinmoredetaillater number. Differentformshavebeendevelopedtosuitparticularpurposes.nthissection.
Experimenttoensurethatyoucaninputthedifferentformsofnumbersonyourparticularcalculator. Forexample,doyouneedtopressfunctionkeysbeforeorafteryouenterdigits?Place value
TheIndianArabic(orHinduArabic)systemmostgenerallyusedinthewestern world has ten numeral symbols (the digits 0, 1, 2, 3, 4, 5, 6, 7, 8,9)whicharecombinedinaplace-valuesystemstructure. Theplace-valuesystemisbasedonpowersoftenandextendstodecimals.Forexample,thenumeral23789.56(twenty-threethousandsevenhundredand
eighty-nine
point
five
six)
has
seven
digits,
the
value
of
each
of
which
is indicated inthefollowingtable.
Ten Thousands Hundreds Tens Units tenths hundredthsthousands
integer/decimal 10000 1000 100 10 1 0.1 0.01powerform 104 103 102 101 100 101 102example 2 3 7 8 9 5 6
1
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Indices
Indexnotation isaconcisewayofsymbolisingtherepeatedmultiplicationofanumberby itself. Forexample,multiplying10stogethersuccessivelygives:
1010=100, 101010=1000, 10101010=10000.Usingtheshorthandnotation,wewrite:
102 =100, 103 =1000, 104 =10000, . . . . Inthisexample,10isthebasenumberandthesuperscriptnumberatthetopright indicateshowmanyofthesebasenumbershavebeenmultipliedtogether. Thissuperscriptnumber isvariouslycalledthepower,indexorexponent. Thisshorthand isextendedasfollows:
1101 = 10, 100 = 1, 101 =
10 (or0.1),1 1
102 =1010 = 100 (or0.01), . . . .
Thenotationalsoextendstofractionalpowers. Forexample:10
11/2 = 10, 101/2 = .10
Ajustificationforassigningthevalue 10tothesymbol101/2 (or100.5) is giveninSection1.6.
Example 1.1
Evaluateeachofthefollowingpowers,firstbyhandandthenusingyourcalculator.(a) 53 (b) 106 (c) 23Solution
(a) 53 = 5 55=1255 [yx]3[=]125(scientificcalculator) Calculatorkeysareindicatedor5 []3[ENTER]125(graphicscalculator) bysquarebrackets []. The
actualkeysusedvaryfrom(b) 106 = 10 1010101010=1000000
onemakeofcalculatorto10 [yx]6[=]1000000 another.or10 []6[ENTER]1000000
1 1(c) 23 =2
2
2
=8
(or0.125)2 [yx] 3 [ +/][=]0.125or2 [] [()]3 [ENTER] .125
Exercise 1.1
(a) Evaluateeachofthefollowing byhand.(i) 104 (ii) 105 (iii) 34
(b) Useyourcalculatortofindeachofthefollowing.(i) 65 (ii) 4.34 (iii) (7.1)3 (iv) (8)7 (v) 124
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cientificcalculatorsfollowhisconventionbutmost
non-scientificonesdonot.
REVISION PACK
Scientific notation
Anyrealnumbercanbeexpressedintheform(numberbetween1and10,butnot including10)(poweroften).
For example:253 = 2.53100 = 2.53102,
25.3 = 2.53
10 = 2.53
101,2.53 = 2.531 = 2.53100,0.253 = 2.530.1 = 2.53101,0.0253 = 2.530.01 = 2.53102.
Thiswayofwritinganumberindicatesitsorderofmagnitudeastheexponent,and isusefulwhencalculatingusingverylargeorverysmallnumbers,andparticularlysowhendealingwithamixtureof largeandsmallnumbers. Itiscommonlyreferredtoasscientificnotationbut isalsoknownasstandard form. Itisthisnotationwhichscientificcalculatorsusetoshowverylargeandverysmallnumberssometimesthepoweris indicatedbytheletterE(forexponent). Forexample,2.53E1means0.253.Exercise 1.2
(a) Expresseachofthefollowingnumbersinscientificnotation.(i) 1427000000 (ii) 8075 (iii) 0.00327 (iv) 0.5672(v) 0.0000004007
(b) Expresseachofthefollowingnumbers infull.(i) 3.298105 (ii) 7.654101 (iii) 1.098103(iv) 3.41010
1.2 Calculating
Order of operations
Imaginethatsomeoneasksyoutodothefollowingcalculations(readthemout loudwith . . . . . . . . . . . . beingadramaticpause).
What is2add3 . . . . . . . . . . . . times4?What is2. . . . . . . . . . . . add3times4?
Thefirst is likelytohaveproducedtheanswer20,thesecond14.Sincetherearenodramaticpauses inwrittencalculations(oroncalculatorsorcomputers),there isauniversallyacceptedconventiontoovercomethisandsimilarpossiblemisunderstandings.
Incalculations,operationsareperformed inthefollowingorder. Brackets Indices DivisionandMultiplication(theorderdoesnotmatter) AdditionandSubtraction(theorderdoesnotmatter)
Onewayofrememberingthis isthemnemonicBIDMAS.
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Exercise 1.3
(a) Insertbracketsineachofthefollowingcalculationswherenecessarytoemphasisetheorderinwhich itmustbeperformedaccordingtotheaboveconvention. Thendothecalculationswithoutusingacalculator.
15+5(i) 3+52 (ii) 103 3 (iii) (iv) 64 + 2
3 + 7 (v)
2
2
+ 3
10
2
(b) Whathappenswhenyouuseyourcalculatorwithoutthebrackets?Process
Foranymathematicalproblemrequiringanextensivecalculationtherearefivestages.1 Establishthecalculationtobedone.2 Make an estimate(particularlyadvisablewhenusingacalculator).3 Dothefullcalculation,usingacalculatororcomputerifappropriate.4 Verifythesolution;firstcompareyourexactansweragainstthe
estimate,toascertainwhetherit isabouttherightsize;thendosomekindofcheck(forexample,puttingthesolutionbackintotheoriginalproblem,orredoingthecalculationadifferentway).
5 Checkthattheanswermakessenseinthecontextoftheoriginalproblem. Roundasnecessary.
Example 1.2
Abuscompanyownsanumberof42-seatercoachesandhas185passengerswishingtogotoLondon. Howmanycoachesareneeded?Solution
Stage1 18542Stage2 Approximateto easynumberstoobtainanestimate:
20040=5Stage3 18542=4.4047. . . (usingacalculator)Stage4 Calculatoranswerisnottoofarfromestimate,and
calculatoranswer42=185(sotherewerenoerrorskeyingin)
Stage5 Fractionsofacoacharenot sensible,sorounduptonearestwholenumber. Answer: 5coaches.
Estimating
Anestimate isaroughanswerproducedbyusingapproximatenumbers.Estimatinganswerstonumericalcalculations is importantas itprovidesawayofcheckingthatthefinalanswer isofthecorrectorderofmagnitude,whetherit isproducedby handorusingacalculator. Theprocessinvolvessubstitutingeasynumbersthatcanbeworkedonmentallyorquicklyusingpaperandpencil.
Roundingtothenearestintegerisnotappropriateinthiscase,sincetheresultwouldleavesomepassengersstranded.
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Thesymbolisreadas ispproximatelyequalto.
Example 1.3
Estimatetheanswerstothefollowingcalculations;thenfindtheexactanswersusingacalculator.(a) 44281+3729 (b) 7002633 (c) 379.420.23(d) 0.0461.5Solution
(a) 4428144000and37294000;soanapproximateanswer is48000.Usingacalculatorgivestheexactanswer: 44281+3729=48010.
(b) 70027000and633600;soanapproximateansweris6400. Usingacalculatorgives7002633=6369.
(c) 379.42400and0.230.2;soanapproximateanswer is80. (Youmighthavesaidthat0.230.25,givinganapproximateanswerof100.) Usingacalculatorgives379.420.23=87.2666.
(d) 0.0461.5 = 0.46150.4515=0.03. Usingacalculatorwhichdisplays10digitsgives0.0461.5 = 0.0306666667.
Exercise 1.4
Carryouteachofthefollowingcalculationsusingyourcalculator,havingfirstmadeanestimateoftheansweryouexpect.(a) 441.75.2 (b) 53.470.922.2 (c) 217.5 + 60.317.7
Notetheuseofbracketsin (d) (1285329)0.023part(d)toindicatethattheubtractionistobeperformed
beforethemultiplication. RoundingIfthefinalanswertoanumericalcalculation involvesanumberwithalargenumberofdigits, it isoftenappropriatetogivearoundedanswer.
nExample1.2,4.4047. . . was Unlessthenatureofaparticularproblemdetermineswhetherasensibleoundedupto5,notdownto answer isobtainedbyroundingdownorup,aconventiononhowtoround. Useoftheconventionis numbers isused.
llustratedinExample1.4.Toroundtoagivennumberofdecimalplaces: lookatthedigitwhich isonemoreplacetotherightofthe
numberof
places;
roundupifthisdigit is5ormore,anddownotherwise.Toroundtoagivennumberofsignificantfigures: lookatthedigitwhichisthatnumberofplacestotherightofthe
firstnon-zerodigit; roundupifthisdigit is5ormore,anddownotherwise.
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Example 1.4
(a) Round0.0306666667tofourdecimalplaces.(b) Round47.1372536tothefollowingnumbersofdecimalplaces.
(i) two(ii) three(iii) four
(c) Roundeachofthefollowingtotwosignificantfigures.(i) 2295673 (ii) 0.027294 (iii) 0.0000405108
Solution
(a) Thedigit inthefifthdecimalplacein0.0306666667is6,which ismorethan5,sothenumber isroundedup,giving0.0306666667=0.0307(to4d.p.). Forconveniencethewords
(b) (i) Thedigit inthethirddecimalplace in47.1372536 is7,so decimalplacesand47.1372536=47.14(to2d.p.). significantfiguresareoftenabbreviatedtod.p. ands.f.(ii) Thedigit inthefourthdecimalplacein47.1372536is2,so respectively.47.1372536=47.137(to3d.p.).(iii) Thedigit inthefifthdecimalplace in47.1372536 is 5, so 47.1372536=47.1373(to4d.p.).
(c) (i) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft, in2295673 is9,so2295673=2300000(to2s.f.).(ii) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft, in0.027294 is2,so0.027294=0.027(to2s.f.).(iii) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft,in0.0000405108is5,so0.0000405108=0.000041(to2s.f.).
Exercise 1.5
(a) Round2.1415996tothefollowingnumbersofdecimalplaces.(i) one (ii) three (iii) four (iv) five
(b) (i) Round63056totwosignificantfigures.(ii) Round0.038toonesignificantfigure.(iii) Round0.04006tothreesignificantfigures.
(c) (i) Round23.009toonedecimalplace.(ii) Round9999tothreesignificantfigures.(iii) Round6080totwosignificantfigures.(iv) Round16.99toonedecimalplace.
Youneedtobecarefulwhenusingroundednumbersincalculations. Ifyouroundanumberthatistheresultofacalculation,andthenusethisroundednumberaspartofafurthercalculation,thentheanswerthatyouobtainforthe latercalculationmaybe inaccurate. Toavoidsuchroundingerrors,itsbesttousethefull-calculator-accuracyversionsofnumbers incalculations. Storingthesenumbers inyourcalculatorsmemorywillhelpyoutodothis.
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Thefirstsixprimenumbersre 2, 3, 5, 7, 11 and 13.
REVISION PACK
1.3 Multiples, factor s and prime number s
Withtheadventofcalculators,theuseofmultiplesandfactorsindealingwithwholenumbersisdeclining,butanunderstandingandfacilitywithmultiplesandfactorsremainsvery important foralgebra.Multiples
Themultiplesofapositivewholenumberarethosepositivewholenumbersintowhichitdividesexactly: forexample,themultiplesof6are6,12,18,24, . . .;thatis,themultiplesofanumberarethatnumbermultipliedby1,2,3,4, . . .. However,findingthecommonmultiplesoftwoormorenumbers ismoredifficult,andtofindthelowestcommonmultiple(LCM)needsaknowledgeoffactors.Factors
Theprocessoffindingallthewholenumbersthatdivideexactlyintoagivenwholenumberiscalled findingthefactors. Factorisingmeanswritinganumberastheproductoftwoormorewholenumbers. Forexample,48=124 and 48 = 8 32.Aprimenumber isapositivewholenumber,otherthan1,that isdivisibleonlybyitselfand1. Factorisinganumbersothateachfactorisaprimenumberisoftenveryusefulthisprocessiscalledfindingtheprimefactors. Forexample,thefactorsof48whichareprimeare2and3;theprimefactorsof48are
48=22223 (= 24 3).Theprocedureforfindingtheprime factorsofanumberaareasfollows.It is illustrated inExample1.5(b)fora=60.To find the prime factor s ofa
Divideaby2(thefirstprimenumber)repeatedlyuntiltheresult isnotdivisibleby2. Ifthenumberofsuchdivisions isN, then 2N isafactorofa.Iftheresultisaprimenumber,theprocessiscomplete. Ifnot,repeattheprocessforeachsuccessiveprimenumber,replacing2by3,5,7,. . . ,asnecessary.
Example 1.5
(a) Findthefactorsof60.(b) Findtheprimefactorsof60.
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Solution
(a) Thequestionisaskingwhatpairsofwholenumberswhenmultipliedtogethermake60. Theyare
160, 230, 320, 415, 512, 610.Thusthefactorsof60are1,2,3,4,5,6,10,12,15,20,30and60.
(b) Divide60by2: 60=2
30. 30 isdivisibleby2.Divide30by2: 30=215. 15 isnotdivisibleby2.Divide15by3: 15=35. 5 isprime.
60 divide by 2
2 30 divide by 2
2 15 divide by 3
3 5prime,so stop
Figure 1.4
Sotheprime factorsof60are2,2,3,5. Thus60=2235 = 22 35.
Exercise 1.6(a) Findthefactorsof36.(b) Findtheprimefactorsof36.Highest common factor
Thehighestcommon factor(HCF)ofanytwopositivewholenumbers isthe largestnumberwhichisafactorofboth.Orputanotherway,theHCF istheproductofthelowestpowerofeachprimefactorcommontoboth.
Thefollowingexample illustrateshowtofindtheHCF.
Example 1.6
With small numbers, you may beablejusttowritedowntheprimefactors,withoutusingtheprocedure.
Thisdefinitionextendseasilytothreeormorenumbers.
What isthehighestcommonfactorofthenumbers18and30?
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Again,thisdefinitionextendsothreeormorenumbers.
REVISION PACK
Solution
First,writeeachnumberintermsofitsprimefactors. Wehave18=232 and 30=235.
Takingthe lowestpowerofthecommonprimefactorsgivesHCFof18and30=23 = 6.
Exercise 1.7
(a) Findtheprimefactorsof:(i) 8 (ii) 16 (iii) 32.
(b) Whatisthehighestcommonfactorof8,16and32?Lowest common multiple
Thelowestcommonmultiple (LCM)ofanytwopositivewholenumbersisthesmallestpositivewholenumberwhichisamultipleofboth.Orputanotherway,theLCM istheproductofthehighestpowerofeachprimefactoroccurring.
Example 1.7
What isthelowestcommonmultipleofeachofthefollowingsetsofnumbers?(a) 10,25. (b) 8,24,60.Solution
(a) Intermsofprimefactors10canbewrittenas25and25as55 = 52. TheLCMof10and25istheproductofthehighestpowerofeachprimefactor,thatis
252 = 2 25=50.SotheLCMof10and25is50.
(b) 8=23.24=2223 = 23 3.60=2235 = 22 35.SotheLCMof8,24,and60 is23 35=120.
Exercise 1.8
(a) WhatistheLCMof28and36?(b) WhatistheLCMof7,10and14?
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1.4 Fractions, decimals and percentages
Somerealnumberscanbeexpressedasexactfractions,that is,asonewholenumberdividedbyanotherwholenumber,suchas:
3 1 = 9 25
(=3 5); 244
(=9 4);7
(=2 7, orequally2 7.)Thenumberwrittenabovethe lineorbar inafraction iscalledthenumerator,andthenumberwrittenbelowthelineiscalledthedenominator.Converting fractions into decimals
Fractionsareconvertedintodecimalsbydividingthenumeratorbythedenominator.
Example 1.8
Convert 3 , 21 and2 intodecimals.5 4 7
Solution
(a) 35
= 3 5 = 0.6(b) 21 = 9 = 9 4 = 2.25
44
(c) 27
=(2 7)=0.285714285714. . .
In(a)and(b)above,thedivisionprocessterminates.In(c)above,thedivisionprocessdoesnotterminatebutgoes intoarepeatingloopthatgeneratesthedigits285714overandoveragain. The
issaidtoberecurring,andcanbewrittenas0.28571 4,decimalfor 27
wherethetwodotsabovethedigits2and4 indicatetheblockofdigitsthatisrepeated.Iftheresultofacalculation isafinalanswertoaquestion, itcanberoundedbymakingastatementlike: 2
7=0.2857tofourdecimalplaces.
Numberswhicharegoingtobeused insubsequentcalculationsshouldnotberounded.Exercise 1.9Converteachofthefollowingfractions intodecimals.(a) 1
5(b) 1
3(c) 33
4(d) 11
8(e) 33
7
Equivalent fractions
Whenthenumeratoranddenominatorofafractionarebothmultipliedordividedbythesamewholenumber(otherthan0),thenewfractionobtained isequaltotheoriginalone,andthetwofractionsarecalledequivalent fractions. Theyarerepresentedbythesamepointonthenumberline.
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Example 1.9
Writedownthefractionsequivalentto 35
whichareobtainedbymultiplyingitsnumeratoranddenominatorbyeachofthefollowingnumbers.(a) 2 (b) 3 (c) 10 (d) 20 (e)
1
Solution
(a) 3 25 2 =
610 (b)
3 35 3 =
915 (c)
3 105 10 =
3050
(d) 3 205 20 =
60100 (e)
3 15 1 =
35
Inacompletesetofequivalentfractions,thefractionwiththesmallestpossiblepositivedenominator issaidtobethefraction inits lowest terms. Thisisfoundfromanyoneoftheequivalentfractionsbycancelling,thatis,bysuccessivedivisionofthenumeratoranddenominatorbyeachoftheircommonprime factors.(ThisisequivalenttodivisionbytheHCFofthenumeratoranddenominator.)
Anywholenumberthatdividesintoeachoftwonumbersiscalledacommonactorofthosenumbers.
Example 1.10
28Express(a)42
and(b) 360 asfractions intheirlowestterms.240
Solution
(a) We have 28 = 2 2 7 and 42 = 2 3 7,sothecommonprimefactorsare2and7. Hence
28 =42
4
6
2
3
(dividingby7)(dividingby2).=
360 = 36 (dividingby10=2 5)(b)240 24
= 32
(dividingby12=2 2 3)Inpractice,spottinglargecommonfactors,ashere,isanefficientwaytoproceed.
Calculations involving fractions
Adding and subtracting fractions
Twofractionsareadded(orsubtracted)byconvertingeachofthemintoequivalentfractionswiththesamedenominator,adding(orsubtracting)theseequivalentfractionsand,whereappropriate,convertingtheresulttoafraction in its lowestterms.
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Example 1.11
Evaluateeachofthefollowingandexpresstheresult in itslowestterms.
Inthissolutiontheequivalentfractiondenominatorused isthelowestcommonmultipleofthedenominatorsoftheoriginal fractions.However,thesamefinalresultwouldbeobtainedbyusingtheproductofthedenominators:
Thereasonforchoosingtouselowestcommondenominators istokeepthenumbers involvedincalculatingtheequivalentfractionssmallenoughtobedonementallyquicklyandeasily.
Multiplying fractions
Theproductoftwofractions isobtainedbymultiplyingtheirnumeratorsandmultiplyingtheirdenominators,togiveanewnumeratoranddenominatorrespectively,then,whereappropriate,convertingtheresulttoafraction in its lowestterms.
Example 1.12
Evaluateeachofthefollowingandexpresstheresult in itslowestterms.(a) Solution
(c) +
Thenextexampleshowshowtodealwithdivision involvingfractions.
Example 1.13
Noticethatinthiscasethedenominatoroftheequivalentfractions istheproductofthedenominatorsoftheoriginal fractions.
.=
= =
10 +15 = 2550 50
(a) +Solution
+
+
(b)
(b)
(a)
(c)
(a)
3
10
5
12=3
128
12=12
43
1
2
5
10
3
10+2
10=3
10
1
5
1= 250310+15
11
54
5=22
3(b)11
32
+3=11632
5
8
66
34
1=2=213632
15=533284
(b)2132
Evaluateeachofthefollowingandexpresstheresult in itslowestterms.(a) 4 (c) (d) (b)1
2
1
2
1
4
2
3
1
3
4
5
2
3
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Solution
(a) 412
Thisquestionisthesameas howmanyhalvesarethere in4?. Theansweris8asthereare2halvesin1,sotherewillbe4times2halvesin4.
.
Analternativewayofsimplifyingaproductoffractionsistoemploycrosscancellation. Forexample,
(here2hasbeendivided into4and2)
Dividing fractions
Todivideanumberbyafraction,multiplythenumberbythefractionturnedupsidedown. Anothernameforthis upsidedownnumber is reciprocal,atermwhichisexplainedmorefully inthenextsubsection.
Mixed number s
Amixednumber ismadeupofawholenumberandafraction. Forexample,thenumber2 isamixednumber. Itcanbeconvertedintotheequivalent improperortop-heavyfraction(wherethenumerator is largerthanthedenominator)asfollows:
2 .
is the same as 4
= 8.
.
That is,4
TheanswerinExample1.13(d)wasintheformofanimproperfraction:
Inwords,thisquestion isaskinghowmanyquartersinonehalf?.Theansweris2.
= 2.
An improperfractioncanbeconvertedintoamixednumberbydividingthetopbythebottomtofindthenumberofwholeones;theremaindergivesthenumberoffractionparts leftover.For example, (75=1,remainder2).Themethodsforcalculatingwithmixednumbersareexactlythesameasthoseforthemethodsgiven inthepreviousexamplesforproperfractions(inwhichthenumerator issmallerthanthedenominator)providedthatyouchangethemixedfractiontoan improperfractionfirst.
(b)
(c)(d)
65
21
12
5=122
4=4
6=12=35102
21
= 2.
1
2
6=331
3
23
1
+
2
5
4=122
.65
=53
1=124
2=4
=2
= 1
5
7
5
12=12
2
11
42
333
4=24535
45
= 2 + 12
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Exercise 1.10
(a) Evaluateeachofthefollowingandexpresstheresultin its lowestterms.
2 1 3 3 18 2 2 3(i) (ii) 1 (iii) (iv) 4 1+ + 3 6 4 8 25 5 7 5
(b) Evaluateeachofthefollowingandexpresstheresult in its lowestterms.
4 2 4 2 (iii) 1225 389 (iv) 42 2(i) (ii) 1 5 7 5 7 7 32 114
(v) 1 13
Reciprocals
Thewordreciprocal isthemathematicaltermforhowmanyofa Thetermappliestorealnumbers,butinitiallyithelps1
2istherefore2asthereare2particularnumberarein1. Thereciprocalof
tothinkintermsofpositivehalvesin1.wholenumbers.
Thereciprocalofanumberis1dividedbythatnumber. Forexample,thereciprocalof10 is 1
10. Ineffectthenumberis inverted,because10when
10
1.writtenasarationalnumberinlowestterms is
1
10or inpowerSothereciprocalof10canbewritten infractionformas
formas101. Asadecimal, itsvalue is0.1.There isakeyforfindingreciprocalsonyourcalculatorthoughtheanswerisgivenasadecimalratherthanafraction. Trythisonyourcalculatornowforthenumber10byputtingin10andthenpressingthereciprocalkey,which ismarked [1/x] or [x1]. Withsomecalculatorstheanswer0.1willbedisplayedimmediately,withothersyoumayneedtopress[ENTER].Leavethe0.1 inthedisplayonyourcalculatorandthenpressthereciprocalkeyagain. Youwillseethatthereciprocalof0.1(1 ) is10.
10
The inverseoperationof findingareciprocal isthesameoperation, i.e.findingareciprocal;this isanunusualresult. Itmeansthatanumbermultipliedby itsreciprocalalwaysequals1.
1
10= 1. Inthecaseof10: 10
Inthecaseof 110
: 110
10=1.1
2= 1. Inthecaseof2: 2
1Ingeneral: n = 1.
nExample 1.14
Writedownthereciprocalofeachofthefollowingnumbers,givingtheanswerinbothfractionanddecimalform.
1
4(a) 5 (b) (c) 100 (d) 3Solution
3(a) 15
= 0.2 (b) 41
= 4 (c) 1100 = 1100 =0.01 (d) 1 = 0.3
Exercise 1.11
Writedownthereciprocalofeachofthefollowingnumbers,givingtheanswerinbothfractionanddecimalform.(a) 8 (b) 10 (c) 1
5(d) 25
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Percentages
Insomecontexts,andespeciallywhenmakingcomparisons, it isconventionaltoexpressafractionasapercentage. This is a way of expressingfractionswithdenominator100;forexample, 3
5= 60 iswritten
100
%isreadpercentwhich as60%.meansperhundred.
Afraction
is
converted
into
percentage
form
by
multiplying
it
by
100
andwriting%aftertheresult.
Example 1.15
2 27 5Expresseachofthefollowingfractionsaspercentages: ,20
, .5 6
100%= 20027 100%= 2700%=135%20 20
5
2
5%=40%
5
6 100%= 500% = 83.33%(to2d.p.)6
Exercise 1.12
Expresseachofthefollowingfractionsasapercentage.1 1 3 1 3(a)
Tofindagivenpercentageofanumber,changethepercentagetoafractionoradecimalanduseittomultiplythenumber.
(b) (c) 3 (d) 1 (e) 35 3 4 8 7
Example 1.16
Find6%of340.Solution
6% is 6100
or0.06.So6%of340 is0.06340=20.40.
Exercise 1.13
Findeachofthefollowing.isthesymbolforgram. (a) 40%of250g (b) 4%of250g (c) 104%of250g
2
1(d) 17Tofindonequantityasapercentageofanother,dividethefirstquantitybythesecond(toformafraction)andthenexpressthefractionasapercentage.
% of 150 (e) 200%of50
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Example 1.17
Express25gasapercentageof125g.Solution
25
(25
125)
100%
=
125
100%
=
20%.
Exercise 1.14
(a) Express32.50asapercentageof50.00(b) Express23asapercentageof115.(c) Express55asapercentageof50.Percentage increase
Findinganewpricewhentheoriginalpricehasgoneupby10%canbedoneintwostages: byfirstfindingthe10% increaseandthenadding ittotheoriginalprice. Sometimesit isusefultofindthenewpriceusingjustonestage.
Example 1.18
Thepriceofan80CDplayergoesupby10%. What isthenewprice?Solution
Firstmethod: 1010%of80 is 80=0.180=8,sothe increaseis8.100
Thenewprice is80+8 = 88.Secondmethod:Theoriginalcostof80canberepresentedby100%,so increasingthisby10%to110%isthesameasworkingout110%of80.
11080=1.180=88.100
Theadvantageofthisone-stagemethod isthatitcanbeusedtoworkbackwardstofindtheoriginalprice ifthenewpriceandthepercentageincreaseareknown.
Example 1.19
Thenewpriceofatelevisionis225andthis is25%morethantheoriginalprice. Whatwastheoriginalprice?
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Solution
An increasefrom100%to125%needsamultiplierof125/100=1.25. Sooriginalprice1.25=newprice.
Taking25%offthenewprice Sothenewpricedividedby1.25givestheoriginalprice:doesnotgivethecorrect
nswer. (Trythisifyouare 2251.25=180.notconvinced.)
Exercise 1.15
Fill inthegapsinthetablebelow.Originalprice() Percentageincrease Multiplier Newprice()100 30% 1.3(130/100) 130
45 3% 1.03(103/100) 46.35230 15%120 150%
20% 1.2 43.2076 17.5% 1.175
17.5% 30
Percentage decrease
Theone-stagemethodcanalsobeusedforfindingreducedprices.
Example 1.20
Oneofthe80CDplayersisshop-soiled,soitisreducedby15%. Whatisthenewprice?Solution
Theoriginalcostof80canberepresentedby100%,sodecreasingthisby15%isthesameasworkingout(10015)%or85%of80. Thenewpriceis
85 80=0.8580=68.100
Exercise 1.16
Fill inthegapsinthetablebelow.Originalprice() Percentagedecrease Multiplier Newprice()100 20% 0.8 80200 25%
45 5%50% 9030% 154
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1.5 Calculating with signed number s
Oftenacalculation involvesamixtureofpositiveandnegativenumbers.Belowisasummaryoftheruleswhichgovernthearithmeticofsigned(positiveornegative)numbers.
Addinganegativenumberisequivalenttosubtractingthecorrespondingpositivenumber,e.g.4+(2)=42.Subtractinganegativenumberisequivalenttoaddingthe Thisisoftendescribedascorrespondingpositivenumber,e.g.4(2)=4+2. twominusesmakeaplus.Multiplyingordividingtwonumberswiththesamesigngivesapositiveanswer,e.g.(4)(2)=42,(4)(2)=42.Multiplyingordividingtwonumberswithoppositesignsgivesanegativeanswer,e.g.(+4)(2)=(42)=8,(+4)(2)=(42)=4
2
=2.
Example 1.21
Evaluateeachofthefollowing.(a) 5+(8)(b) (5)+(8)(c) 5(8)(d) (5)(8)(e) (7)(3)(f) (7)3(g) (12)4(h) (12)(4)Solution
(a) 5+(8)=58 = 3(b) (5)+(8)=(5)8 = 13(c) 5(8) = 5 + 8 = 13 (d) (5)(8)=(5)+8=3(e) (
7)
(
3)=+(7
3)=7
3 = 21 (f) (7)3 = (73)=21(g) (12)4 = (124)=3(h) (12)(4)=+(124)=3.
Exercise 1.17
Evaluateeachofthefollowing.(a) (i) (2)+(7) (ii) (5)+8 (iii) 3+(5)(b) (i) 4
(
2) (ii) (
3)
(
5) (iii) (
3)
(
3)(c) (i) 4(3) (ii) (2)(7) (iii) 3(9)(d) (i) 24(6) (ii) (40)(8) (iii) (45)15
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1.6 Working with powers, indices and logarithms
Inanumberoftheforman,aiscalledthebaseandnthepower(indexorexponent).Calculations involvingdifferentbasesraisedtothesamepowercanbemadesimplerbyfactorisingfirst.
Example 1.22
Rewriteeachofthefollowing infactorisedformandhencesimplify itmentally.(a) 22 (b) 513
42 173
Solution 22 12 = 1(a) 22 =
4=
42 2 4
2
2
2 1(Infulldetail: 22 = 22 =
42 = = 2
2
=1
2
=4
.)42 44 4 4 22 2513(b) 513 =
17= 33 = 27
173
Exercise 1.18
Factoriseandsimplifyeachofthefollowingmentally. Thencheckyourresultusingacalculator.(a) 1002 (b) 44 (c) 93
252 24 33
Thereareseveralrulesforcombiningpowers.Combiningpowers,andnegativepowersTomultiplytwopowerswiththesamebase,addtheindices:
m m+n 32 = 34+2a an =a e.g.34 = 36.Todividetwopowerswiththesamebase,subtractthe indices:
ma an =amn e.g.34 32 = 342 = 32.Tofindthepowerofapower,multiplytheindices:
m)n(a =amn e.g.(34)2 = 342 = 38.Anumberraisedtoanegativepower isthereciprocalofthenumberraisedtothecorrespondingpositivepower.
ma = 1/am e.g.32 = 1/32 and32 = 1/32.
Fractional powers
Theaboverulesapplytofractionalpowersaswellas integerpowers. Forexample,
101/2 101/2 = 101/2+1/2 = 101 = 10.
3 3
10issaidasthecuberoot Thus(101/2)2 =10,andso101/2 = 10. Similarly101/3 = 10.f10.
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Example 1.23
Evaluateeachofthefollowing,firstasapower,thenasanumber.(a) (i) 72 73
(ii) 712
712(iii) 75 73(iv) 75 75(v) 73 75
Solution
(a) (i) 72 73 = 72+3 = 75 =16807Check: (77)(777)=(77777).
Thisconfirmsthat712 = 7.(ii) 712 712 = 7 1 1 = 71 = 7+2 2
7
(iii) 75 73 = 753 = 72 = 49 5 77777Check: =
73 7777
(iv) 75 75 = 755 = 70 = 1 5 77777
Check: =75 77777
1(v) 73 75 = 735 = 72 =
72
73 777Check: =
75 77777
77= .1
1= = 1.
1= 0.0204(to4d.p.)
1= .
77
Infact,a0 = 1 for any real numbera.
Exercise 1.19
(a) Evaluateeachofthefollowing,firstasapower,thenasanumber.(i) 102 103(ii) 103 101(iii) (10)3 (10)2(iv) 1013 1013 1013(v) 102 101
(b) Evaluateeachofthefollowing,firstasapower,thenasanumber.(i) 107 104(ii) 104 107(iii) (10)7 (10)4(iv) 1032 1012(v) 102 101
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ndex,powerandexponentllmeanthesame.
00=102.
Logarithmswereoriginallydevised in the seventeenth
enturytosimplifyomplicatedcalculationshroughtheuseoftables.
Theywerealsothebasisofliderules,whichhavebeeneplacedbyelectronicalculators.
REVISION PACK
Logarithms: common and natural
Beforestartingthissectionmakesureyouarefamiliarwithpower(index)notationandcanconfidentlymanipulatenumberswritteninthisform.Thenumber100canbewritteninpowernotationas:
102,where
the
base
is
10
and
the
power
is
2.
In
this
example
we
say
that
the
logarithmtothebase10ofthenumber100is2,andwrite
log10 100=2.Logarithmswhichusethebase10arecalledcommon logarithmsandarewrittenas log10 orsometimessimplyas log.Anypositivenumbersuchas2,3or5canbeusedasabaseforalogarithm.Natural logarithmsusethebasee, where eisan irrationalnumberwithavalueofapproximately2.71828. Therearetwokindsofshorthandfornatural logarithms: loge and ln. Inthistextweuseln.Youneedtobeabletochangefrompowerformtothecorrespondinglogarithmic formandbackagainwithconfidence,byhandforsimplecasesandusingacalculatorforawkwardnumbers.
Example 1.24
(a) Writethenumber16asapowerof4(powerform)andthenwritedownthecorrespondinglogarithmic form.
(b) Changelog5 625=4topowerformandthenwritethepowerformasasimplenumber.
Solution(a) Powernotation: 16=42.
Logarithmicform: log4 16=2. The latterisreadas logtothebase4 of 16 equals 2. Inboththeseforms,16 isthenumber,4isthebaseand2 isthe indexorlogarithm.
(b) Thebase is5andtheindex is4,sothepowerform is54 andthesimplenumber is625.
Exercise 1.20
Fill inthegapsinthetablebelow.Power form81=341 = 60
0.2 = 515 = 5
1
x= 24
Logarithmic formlog2 4 = 2 log5 125=3
logx8 = 2
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Herearesomemoreexamplesphrased inadifferentway.
Example 1.25
Findthelogarithmof8tothebase2.SolutionWrite8asapowerof2: 8=23; so log2 8 = 3.
Exercise 1.21
(a) Findthe logarithmtothebase2ofeachofthefollowing.(i) 16(ii) 2(iii) 0.5
(b) Withoutusingacalculator,findthelogarithmtothebase10ofeachofthefollowing.(i) 1000(ii) 10 Youwillhavenoticedthat(iii) 0.1 log10 10andlog2 2 have the
value1. Thisresultisgenerallytrue,i.e.logaa= 1.
The inverse of a logarithm
Ifyouknowthe logarithmofanumberusingaknownbase,howdoyoufindthenumberitself? Forexample,ifthe logarithmtobase10ofsomenumber
x, say, is
4,
what is
x? We
have
log10 x= 4,
andthecorrespondingpowerformgivesx:x= 104 =10000.
10000 issaidtobetheantilogarithmof4tobase10.
Example 1.26
Findthe
antilogarithm
of
5to
the
base
2.
Solution
Here,ifxistheantilogarithm,5 = log2 x,
sothatx= 25 = 32.
Sotheantilogarithmof5tothebase2 is32.
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Exercise 1.22
(a) Withoutusingacalculatorworkouttheantilogarithmtothebase10ofeachofthefollowing.(i) 5 (ii) 0 (iii) 2
(b) Findtheantilogarithmtothebase3ofeachofthefollowing.(i) 2 (ii) 1 (iii)
1
Thelogarithmtoagivenbaseofanumberisthepowertowhichthebase israisedtocalculatethenumber.
Mostoftheindicesencounteredsofarhavebeenintegersbutthesystemofpowernotationextendstofractional indices. Thenumber1canbewrittenas100 andthenumber10as101. Thenumbersbetween1and10canalsobewritteninpowernotationusingthebase10andthecorrespondingindices liebetween0and1. Oneofthese,theindex0.5,hasalreadybeenusedasanalternativenotationforasquareroot: 10=100.5 = 10 12 .Thevalueofthesquarerootof10 is3.162(to3d.p.),that is
10=100.5 = 3.162(to3d.p.).
Sothelogarithmtothebase10of 10=3.162(to3d.p.) is0.5:
log10 3.162log10 10=0.5.Notethat,althoughthe index0.5isexactlyhalfwaybetween0and1theactualnumber(3.162) is lessthanhalfwaybetween1and10.
Calculatorsmayhave Logarithmicvaluesofnumberscanbefoundaslists in logtablesbutnowupersededlogarithmsasa theyaremoreusuallyobtainedbypressingtheappropriatecalculatoralculatingtoolbuttwowell keys. Scientificandgraphicscalculatorshavethefacilitytodealwithbothnownmeasurementscales commonandnatural logarithms. Forcommon logarithms,tothebase10,
uselogarithms. Oneisthe thekeytouseisgenerallymarked[LOG]andtheinverse(orRichterscaleforearthquakes antilogarithm) isthesecondfunctionkeymarked [10x].
ndtheotheristhebelequivalentto10decibels) Fornaturallogarithms,tothebasee,thekeytouseisgenerallymarkedystemformeasuringthe [LN]andthe inverse(orantilogarithm) isthesecondfunctionkeymarkedntensityofsound. [ex].
Findthesekeysonyourcalculatorandestablishhowtheywork: formostscientificcalculators,youenterthenumberthenkey(s);formostgraphicscalculators,youpressthekey(s)thenthenumberand [ENTER].Exercise 1.23(a) Useyourcalculatortofindeachofthefollowing,givingyouranswerto
fourdecimalplaces.(i) log10 1.5489(ii) log10 15.1149(iii) log10 734.256
(b) Useyourcalculatortofindeachofthefollowing,givingyouranswertofourdecimalplaces.(i) ln1.5489(ii) ln15.1149(iii) ln734.256
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Exercise 1.24
(a) Useyourcalculatortofindthenumberswhose logarithmstothebase10arethefollowing,givingyouranswerstofourdecimalplaces.(i) 0.5267 (ii) 0.0023 (iii) 2.4593
(b) Useyourcalculatortofindthenumberswhosenatural logarithmsarethefollowing.(i) 0.5267 (ii) 0.0023 (iii) 2.4593Combining logarithmsSince logarithmsareindices,therulesformultiplyinganddividingpowerscanbeadaptedforcalculationswith logarithms.Rule 1
Thelogofaproduct isthesumofthe logs. Forexample,forthenumbersaandb:
log10(a
b) = log10 a+ log10 b,ln(ab) = ln a+ ln b.
Rule 2
Thelogofaquotientisthedifferencebetweenthe logs. Forexample,fortwonumbersaandb:
alog10 b = log10 alog10 b,aln = ln alnb.b
Rule 3
Thelogofanumberraisedtoapoweristheproductofthepowerandthe logofthenumber. Forexample,forthenumberaandpowern:
n nlog10 a =nlog10 a and lna =nlna.
This lastresultwillbeusedlaterforsolvingexponentialequations(seeModule5).
Example 1.27
Simplifyeachofthefollowing.(a) log10 6 + log10 5 (b) 3 ln 2 ln4Solution
(a) log10 6 + log10 5 = log10(65) (byRule1)= log10 30
(b) 3 ln 2 ln4=ln23 ln4 (byRule3)= ln 8 ln4
8= ln (byRule2)
4= ln 2
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Exercise 1.25
Simplifyeachofthefollowing.(a) (i) log10 12+log10 3 (ii) log10 12 log10 6 (iii) 3log10 5(b) (i) ln15+ln7 (ii) ln35 ln14 (iii) 4ln3
1.7 Working with roots Rootsareoftenbestconsideredasfractionalpowers;forexample,
3writteninpowerformis21/2, 27is271/3. However,therearesomepointsworthunderstandingwhenusingtherootform. Eachpositivenumberhastwosquareroots,onepositiveandone
ometimes iswritten in negative;forexample,thetwosquarerootsof9are3(3 3=9)andplaceof, and adenotes 3 (3 3=9). Thesymbol denotesthepositivesquareroot,so bothsquarerootsofa. thetwosquarerootsof9are 9 = 3 and 9 = 3. Calculatorsgive
onlypositivesquareroots,soyoumustrememberaboutthenegativeones.
Anumberdividedbyoneof itssquarerootsgivesthatsquareroot;forexample,3 = 3 because 3 3 = 3.
3 Thepositivesquarerootofapositivenumbercanberewrittenasa
productofthepositivesquarerootsoftheprime factorsofthat number;forexample, 6 = 2 3. (Thenumbers 2 = 1.414. . . and
3 = 1.732. . . are irrationalnumberswritteninsurdform. It isconventiontoremoveasurdinthedenominatorofafractionbymultiplyingthenumeratoranddenominatorbythatsurd.)
Sometimesit ismoreappropriatetosimplifyanexpressioninvolvingsquarerootsusingfactorsratherthancalculateanumericalsolutionusingacalculatorparticularly iftheresult istobeusedinfurthercalculations.TosimplifyapositivesquarerootRewritethesquarerootastheproductofthepositivesquarerootsofthenumbersprimefactors.
e.g. 175= 5 5 7 = 5 5 7.Multiplytogetherpairsof identicalsquareroots.
e.g. 175= 5 5 7 = 5 7.Theaboveprocesscanbeshortenedifyoucanspothowtowriteasquare
Theperfectsquaresare1(12),(22),9(32), . . . . rootastheproductofaperfectsquareandanotherfactor. Forexample,
108=9 12=912 (9 isaperfectsquare) = 3 3 4 = 3 3 2 = 6 3,
or
108= 36
3 = 36
3 (36isaperfectsquare)= 6 3.
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Example 1.28
(a) Evaluate 2 8withoutusingacalculator.(b) Simplify 60.Solution (a) 2 8 = 28 = 16=4(b) Write60asaproductof itsprimefactors: 2235. Then
60= 2235 = 2 2 3 5 = 2 3 5 = 2 15.
Exercise 1.26
(a) Evaluateeachofthefollowingwithoutacalculator. 9 18(i) 100 (ii)
(iii)
9 2(b) Simplifyeachofthefollowingbyfactorisingfirst. 15
(i) 200 (ii) 112 (iii) 256 (iv) 3
Exercise 1.27
(a) Asquarehasedgelengtha. What isthelengthofitsdiagonal?(b) The longestedgeofanisoscelesright-angledtriangle isa. What isthe
lengthoftheshorteredges? A-sizepaperhasedgesof lengths1and 2.(c) Whatisthe lengthofthediagonalofA-sizepaper?(d) ShowthatcuttingA-sizepaper inhalf,paralleltotheshorteredge,
createstworectanglesthataresimilartotheoriginalrectangle.(e) ThelargestpossiblesquareiscutfromanA-sizerectangle. Whatare
thedimensionsofthewastepiece?(f) Twoequalsquares(as largeaspossible),sidebyside,arecutfrom
A-sizepaper,leavingarectangle. Writedownthedimensionandtheareaofeachpiece. Checkyouransweragainsttheareaoftheoriginalrectangle.
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Surd form
In some of the exercises above we have left an answer in surd form(e.g 2and 5)wheretherewasnoexactnumericalformforthenumber. Ifyouinput 2intoyourcalculator itwilldisplaysomething like1.414213562.This is,ofcourse,averygoodapproximationto 2,moreaccuratethanweshouldneedformostpurposes,butit isanapproximation. In fact, there isnowaytowritedowntheexactvalueof 2exceptbyusingthissymbol. So,inmanycircumstances,ratherthanwritedownsomeapproximation like1.4or1.4142weleavetheresult insurdformas
2. Thisensuresthatouranswer isabsolutelycorrect,andenablesustoensurethatsubsequentcalculationsareaccurate. Forexample,suppose thattheexactanswertoaquestion is 2,andwewrite itdownas 2ratherthan1.4or1.4142. If,later,weneedtosquarethisanswerthenit willbecalculatedas 2 2=2ratherthan1.41.4 = 1.96or1.41421.4142=1.99996164.YoucanusePythagorasTheoremtocalculatetheheightofanequilateraltriangle. Forexample,forthetriangleshowninthefigure,
h2
+ ( 1
2)2
= 12
= 1.
1 1h
1
212
3So, ifwehaveh2 =4
, then h= 3/4 = 3/2. Frequentlywewritethisas
3 . Here 3 isasurd,standingfortheexactpositivesquarerootof3. To2
threedecimalplaces, 3 = 1.732,but(1.732/2)2 + ( 1
2)2 = (0.866)2 + 0.25=0.99956,while
3( 3/2)2 + ( 14
+12
)2 =4
= 1 exactly. Surdscanbemanipulatedexactlyasyouwouldmanipulateothernumbers.Someexamplesaregivenbelow.
Example 1.29
Withoutusingacalculator,simplifythefollowing. (a) 2 6 (b) 4/ 2 (c) 18Solution (a) 2 6 = 2 2 3 = 2 3(Usuallywrittenmoresimplyas
2 3.) (b) 4/ 2 = 2 2/ 2 = 2 2 2/ 2 = 2 2
(c) 18= 92 = 9 2 = 3 2
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Example 1.30
Twopeoplecontributetoabudgetintheratio10to30. Putthisratioinits lowesttermsandwriteeachcontributionasafractionofthewholebudget.Solution
Divideeachquantity intheratio10:30bytheirHCF,inthiscase10.Theratiobecomes1:3.Thereare4partsaltogether,soonepersoncontributes 1
4ofthewhole
budgetandtheothercontributes 34
.
Exercise 1.29
Completethetable.Ratio
(comparingpartwithpart) Ratio(in lowestterms) Asfractions(partsofthewhole)4 : 16 10:56 : 9 1 : 7 10:20:3030:25
Example 1.31
600istobesharedbetween2people intheratio2:3. Howmuchwilleachpersonget?Solution
Thereare5parts,soonepersonwillget 25
of600,whichis240. Theotherpersonwillget 3
5of600,which is360.
Exercise 1.30
Shareeachofthefollowingamountsofmoney intheratioshown.(a) 400between2peopleintheratio3:5(b) 250between3peopleintheratio1:2:2(c) 1000between2people intheratio3:2Map scales
Mapsandplansaredrawntoscaletorepresentsomething inthephysicalworld. Inascaledrawing,all lengthsaremultipliedbythesamescalefactoroftendenotedbythe letterk.
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Example 1.32
Amaphasascaleof1:25000. Thismeansthatthescalefactoris25000.(a) Afootpathalongsideacanalmeasures12.5centimetresonthemap.
Howfar, inkilometres,would itbetowalkalongthispath?(b) Thedistancebetweentwoplacesbyroad is8km. What lengthwould
thisberepresentedbyonthemap?Solution
(a) 12.5cmonthemaprepresentsadistanceof12.525000cm.12.525000cm=312500cm.Thereare100cm inametre,sothelengthofthepath is
312500cm=3125mor3.125km.
100(b) 8 km = 8 1000m
= 8 1000100cm=800000cm
800 000Onthemapthislengthwouldberepresentedby25 000
cm=32cm.
Exercise 1.31
Amaphasascale1:50000.(a) What istheactual lengthofaroadshownas16.5cm?(b) Whatwouldbethelengthonthemapofapathof length15km?Sometimesratiosaregivenassinglenumbers;forexample,theratioofthecircumferenceofacircleto itsdiameter isafixednumbercalledpi, the symbolforwhich is. = 3.14159. . ..Enlargement
Intheenlargementofaphotograph,thephotohastostay inproportioninordertoavoiddistortion. Thatis,theproportionate increase inheightofthephoto isthesameastheproportionate increase inwidth.
Example 1.33
Aphotohasaheightof15cmandawidth12cm. Whatwillbethenewwidthiftheheight is increasedto20cm?Solution
20The lengthmultiplier is15
. Thewidthwillgoup inthesameproportion,sothesamemultiplier isused. Thenewwidthis
2012cm=16cm.15
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Exercise 1.32
Completethetableshowingthedifferentenlargementorreductionsizesforthephotographabove.Height 15 20 30 10Width 12 6 40
Proportion
AlanandBethdecidethattheyneedtoputmoremoney intotheirbudgetbuttheyagreetokeeptheamountsinthesameproportion, i.e.BethwillstillpaytwiceasmuchasAlan. IfAlancontributes110,thenBethwillput in220. Theratio is220:110=2:1,i.e. itstaysthesame.Exercise 1.33
Intheabovesituation,ifBethpays450,howmuchwillAlancontribute?Direct and inverse proportion
Iftwoquantitiesaredirectlyproportionaltoeachother,multiplyingonequantitybyanumbermeansthattheotherquantity ismultipliedbythesamenumber. Anexampleofdirectproportion is: thegreateryourrunningspeed,thegreaterthedistancetravelled inafixedtime. Wesaythatspeed isdirectlyproportionaltodistancetravelled.Iftwoquantitiesareinverselyproportional,multiplyingonequantitybyanumbermeansthattheotherquantityisdividedbythesamenumber(ormultipliedby itsreciprocal). Anexampleofinverseproportion is: thefasterthespeedofajourney,theshorterthetimetaken. Inthiscase,speed
is
inversely
proportional
to
time
taken.
Weshallreturntotheseideas inModule7.
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Module 2 MeasuresMeasuressuchasacountof10000carsandaspeedof10kmperhourillustratetheneedforquantitiestobe labelledandforunitstobespecified,sothattheycanbecomparedandusedtoestablishrelationships.
2.1 Units
Systeme Inter national dUnit es (The inter national system of units)
This internationallyagreedsystem,referredtoasSIunits, isbasedonthemetricsystem. Itcomprisesa limitednumberofbaseunitswhichhaveassociatedsymbols. Examplesaremetre(m)for length,second(s)fortime, litre(l)forcapacity. Forlargeorsmallquantitiesthesebaseunitsarecombinedwithprefixesto indicatemagnitude inpowersoften;forexample,
kilometre
(km)
and
millisecond
(ms).
The
table
shows
some
of
theseprefixes.Table 2.1
Prefix Symbol Figures Words Powersof10mega M 1000000 million 106kilo k 1000 thousand 103
1 one 100centi c 0.01 hundredth 102milli m 0.001 thousandth 103micro 0.000001 millionth 106
Baseunitscanbecombinedtoproducederivedunits. Here are some examples.
area m2 (squaremetres)volume m3 (cubicmetres)velocity ms1 (metrespersecond)acceleration ms2 (metrespersecondpersecond
ormetrespersecondsquared)Notethatinaderivedunit,afinespaceisplacedbetweenbaseunits.
(mu) isaGreekletter,pronouncedmew. ThefullGreekalphabetisgivenintheGuide to Preparation.
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Thepictorialrepresentationfinequalitiessuchas
P > 150000arelookedatinModule5.
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2.2 Inequalities
The=signindicatesthattwothingshaveexactlythesamevalue. Ineverydaylifeyouare likelytomeetsituationswhichinvolvearangeofvaluesandusephrasessuchasmorethanorlessthanortheirequivalentinexpressionslike allpricesare lessthan1and thepopulationofMiltonKeynesisgreaterthan150000.Theseexpressionsareexamplesofinequalities. Themathematicalsymbolsusedtoexpressrelationshipsof inequalityareasfollows.
Symbol Meaning isgreaterthan is greater than or equal to
Remember: ifyouthinkofan inequalitysymbolasanarrow itpointstothesmallervalue,asFigure2.1shows.
is greater than is less than
larger > smaller smaller < larger
Figure 2.1
Examplesoftruestatementsinvolvingthe inequalitysymbols< and> are5< 7, 7> 5, 1< 0, 5> 7.
Truestatements,involvingandare11 and 1 1.
Thesesymbols
can
also
be
used
with
unknowns.
For
example,
if
P is
the
populationofMiltonKeynesthenthesecondintroductoryexamplemaybewritten
P > 150000.
Exercise 2.1
(a) Writeeachofthefollowingstatementsinwords.(i) 16> 12(ii) 11< 9
(b) Writeeachofthefollowingstatementsinsymbols.(i) 9 isgreaterthan11.(ii) Thenumberofstudentsintheclassroom isat least20.
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2.3 Formulas
Anexpressionoftherelationshipbetweenquantitiesasanequality iscalledaformula;forexample:
circumferenceofacircle=2timesradius.Formulasareoftenexpressedinsymbols;forexample:
C= 2r,whereC isthecircumferenceandr istheradius.Anotherexampleis:
areaofatriangle = 12
baseheightor,usingtheobvioussymbolsbforbaseandhforheight(seeFigure2.2),
A= 12
bh.
base
height
Figure 2.2
Thetwoformulasabove involvesymbolswhichincludeunits. Forexample,ifr=6cmthen
C= 2
6 cm = 12cm.Sometimesthesymbolsinaformularepresentjustnumericalqualities. Anexample is as follows. ToconvertfromdegreesFahrenheittodegreesCelsius,taketheFahrenheitmeasurementandsubtract32;multiplytheresultby 5
9. Usingtheobvious
symbolsforthenumbersinvolvedgivestheformulaC= 5
9(F32).
Formulasare ineffectarecipefordoingacalculationwhenallbutoneofthequantities isknown. Substitutinginformulas isstraightforwardprovidedconsiderationisgiventotheunits,whereappropriate,andtheformula is therightwayround. Sometimesaformulaneedstoberearrangedor undone. Onewayofdoingthisis illustratedbelow. Ifyouareconfidentinan
alternativemethod,usethat.
Example 2.1
UndotheformulaC= 59
(F32)sothatitreads F =something.
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Solution
WerepresenttheformulaC= 59
(F32)asadiagramtoshowtheorderoftheoperations,from lefttoright.
F 32 F32 59 59
(F32)(=C) (2.1)Toundotheformulatoobtain
F = something,weundoeachoftheaboveoperations inreverseorder, i.e.startingattheright,atC.
C 59 CC+ 32 +32(F =)9 95 5
ThisgivestheformulaF = 9
5C+ 32.
The importantfeaturetonotice inproducingthesediagrams isthatyouneedtostart(Equation2.1)withthesymbol(F) that you wish to make the subjectoftheformula.
Exercise 2.2
Rearrangeeachofthefollowingformulasasindicated. Youshouldusethedoingandundoingmethod inpart(a).(a) Theareaofatriangleformula,A= 1
2bh, to give a formula for h, the
height,intermsofAandb.distance
(b) speed= togiveaformulafortime
(i) distance, (ii) time.(c) Thecircumferenceofacircleformula,C= 2r, to give a formula for r,
theradius.
2.4 Measuring and classifying angles
Anangle isanamountofrotationorturning,normallytakenaspositivewhenmeasured intheanticlockwisesenseanglesmeasuredclockwisearetakentobenegative;seeFigure2.3.
B B
Oanticlockwise rotation
A O Aclockwise rotation
Figure 2.3
Anglescanbemeasuredinvariousunitsforexample,degrees,radiansandgradients,butonlythefirsttwoareusuallyused inmathematics.
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TheBabyloniansdividedthecircumferenceofacircleinto12equalsectionsanddividedeachofthosesectionsinto30equalparts. Theresulting partcorrespondstothepresent-daydegree,becausethereare360degrees inafullrevolution. Figure2.4showshowtheBabylonianandthepresent-daysystemsarerelated.
Figure 2.4
Alessarbitrarywayofdividingthecircumferenceofacircle(2r) is to divide itbytheradius(r). Thisgives 2r = 2partsforacomplete
rrevolution,foranyradius.Itisthisobservationwhich leadstothedefinitionofaradian.
Theanglesubtendedatthecentreofacirclebyanarcequalinlength Anarcofacircleisanypartofthecirclejoiningtwototheradius isdefinedtobeoneradian;seeFigure2.5. Therearepointsonthecircle.2radians inafullrevolution.
1 radian
r
r
Figure 2.5
Exercise 2.3
Drawroughdiagramstoindicateeachofthefollowingangles. Labeltheanglesinbothdegreesandradians. (Thesymbolforthedegree is; for example,360degrees iswritten360.)(a)(b)
1
4
1
3
revolutionrevolution
(c) 16
revolution
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Converting angle measures
Since2radians=360,360 2
1 radian = and1= radians.radian=57.30(to2d.p.); 2 360
= 0.017radians(to3d.p.). Theseequivalencesleadtotherulesbelowforconversionfromradianstodegrees,andviceversa.
360 2=2 360xradians= x and y y radians.
Example 2.2
Convert113.5toradians.Solution
2113.5= 113.5 radians360
= 1.98radians(to2d.p.)
Exercise 2.4
Unlessdegreesarespecified, (a) Converteachofthefollowingtoradians,givingyouranswertotwooushouldassumeanangle decimalplaces.
measurementisgivenin(i) 54 (ii) 125 (iii) 667.18
adians.(b) Converteachofthefollowingtodegrees,givingyouranswerstoone
decimalplace.(i) 2/7radians (ii) 1radian (iii) 0.5 radians
Calculatorscanbesettowork indegreesorradians(andsometimesgradientsaswell). Makesureyouknowhowtowork indegreesandradiansonyourcalculator.Somecalculatorscanbeusedtoconvertbetweendegreesandradiansautomatically. Ifthisfunctionisavailableonyourcalculator,repeattheaboveexerciseusing it.
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Classifying angles
Severaldifferenttypesof(positive)anglearenamed,asfollows.Anacuteangle is lessthanaquarterrevolutionthat is, lessthan90, or /2 radians.
acute angle
Figure 2.6
Arightangle isaquarterrevolutionthatis,exactly90, or /2 radians.
Notetheuseofthemarktodenotearightangle.
Figure 2.7
Anobtuseangle isbetween90and180(butnotequaltoeither)that is,between/2 and radians.
obtuse angle
Figure 2.8
Areflexangle isbetween180and360that is,betweenand2radians.
reflex angles
Figure 2.9
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2.5 Statistical measur es
Statistics involvescollecting,analysing, interpretingandcommunicatingnumericaldata. Oneaspectofanalysingasetofdataistocompare itwithanotherset,sothereneedstobeagreementaboutformsofmeasure.Onewayofprovidingasummaryofabatchofdata isanaverage,andtherearethreetypesofaverage:mean,modeandmedian. Which is mostappropriatetousedependsonwhattheaverage istobeusedfor; insomecasesmorethanonemaybeneeded.Mean
Themean iswhatmostpeopleandthemediausewhentalkingaboutanaverage. Itisfoundbydividingthesumofallthe itemvalues inabatchbythenumberof items(thebatchsize). Thisgivestheformula
sumofvaluesmean= .
batchsizeThemathematicalsymbolfor sumofistheGreeklettercapitalsigma.Ifthevaluesaredenotedbyx,thebatchsizebyn, and the mean by
x(readasxbar),thentheformulaforthemeanmaybewritten insymbolsas
xx= (readasxbarequalssigmaxovern).
nSomescientificcalculatorshavestatisticalfunctionswhichusethesesymbolstodenotethekeystouse. Thevaluesareenteredbyusingaspecialkey;onsomecalculatorsthisismarked [+].
Example 2.3
Theheightsinmetres(measuredtothenearestcentimetre)ofagroupofsevenpeoplearegivenbelow.
1.52 1.72 1.66 1.81 1.69 1.59 1.77What isthemeanheightofthegroup?Solution
Thenumberofvaluesinthebatch(thebatchsize)is7. Hence1.52+1.72+1.66+1.81+1.69+1.59+1.77
meanheight= m711.76
= m7
= 1.68m.
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MODULE 2 MEASURES
Exercise 2.5
Findthemeanofeachofthefollowingbatchesofdata.(a) 23 21 26 26 24 (b) 101 107 98 92 115 102Dataareusually listeddifferentlywhenseveralvaluesoccurmorethanonceandcareneedstobetakentoincludeallthedata. Inthesecases,itismoreappropriatetouseanadaptedformulaforthemean,asshown inthefollowingexample.
Example 2.4
Anumberofpeoplewereaskedwhatsizeofhouseholdtheylived inthatis,howmanypeoplelived intheirhousehold. TheygavetheresponsesshowninTable2.2. What isthemeannumberofpeopleperhousehold?
Table 2.2Sizeofhousehold Numberofresponses
1 32 23 14 35 1
Solution
Letthevaluesofsizeofhouseholdbedenotedbyx,andthenumberofeachsuchhouseholdbyf (forfrequency). Theformulaforthemeanmaythenbewrittenas
xff
where means sum of. The calculation is as follows. Totalnumberofpeople inthehouseholds=(13)+(22)+(31)
+ (4 3)+(51)=27(this is xf).
Total number ofhouseholds = 3 + 2 + 1 + 3 + 1 =10(thisis f).
Meannumberofpeopleperhousehold =
xff =
2710 = 2.7.
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Exercise 2.6
Table2.3belowgives informationonthenumberofbrothersandsistersthatchildrenhaveinaparticularschoolclass. Findthemeannumberofsiblingsofthechildren,correcttoonedecimalplace.
Table 2.3
Numberofsiblings Frequency0 71 18 2 53 24 05 1
Limitations!
One limitationofmeans isthatextremevaluescancausedistortionsandmake
the
summary
meaningless.
(For
example,
the
mean
of
101,
1,
1,
1
and1 is21.) Tomakeameanamore informativesummary,theran