Solutions Manual for Precalculus Mathematics for Calculus International Metric Edition 8th Edition b

CHAPTER 1FUNDAMENTALS

1.1 RealNumbers1

1.2 ExponentsandRadicals6

1.3 AlgebraicExpressions13

1.4 RationalExpressions19

1.5 Equations27

1.6 ComplexNumbers36

1.7 ModelingwithEquations39

1.8 Inequalities50

1.9 TheCoordinatePlane;GraphsofEquations;Circles72

1.10 Lines91

1.11 SolvingEquationsandInequalitiesGraphically101

1.12 ModelingVariation111 Chapter1Review115 Chapter1Test132

¥ FOCUSONMODELING: FittingLinestoData138 ErratainExercisesandAnswersinFirstPrinting141

1

1.1

FUNDAMENTALS

REALNUMBERS

1.(a) Thenaturalnumbersare 1 2 3.

(b) Thenumbers  3 2 1 0 areintegersbutnotnaturalnumbers.

(c) Anyirreduciblefraction p q with q  1isrationalbutisnotaninteger.Examples: 3 2 , 5 12 , 1729 23

(d) Anynumberwhichcannotbeexpressedasaratio p q oftwointegersisirrational.Examplesare 2, 3,  ,and e

2.(a) ab  ba ;CommutativePropertyofMultiplication

(b) a  b  c   a  b   c ;AssociativePropertyofAddition

(c) a b  c   ab  ac;DistributiveProperty

3.(a) Inset­buildernotation: x 3  x  5

(b) Inintervalnotation:  3 5

(c) Asagraph: _35

4. Thesymbol  x  standsforthe absolutevalue ofthenumber x .If x isnot0,thenthesignof  x  isalways positive.

5. Thedistancebetween a and b onthereallineis d a  b   b a .Sothedistancebetween 5and2is 2

 7.

6.(a) If a  b ,thenanyintervalbetween a and b (whetherornotitcontainseitherendpoint)containsinfinitelymany numbers—including,forexample a  b a 2n foreverypositive n .(Ifanintervalextendstoinfinityineitherorboth directions,thenitobviouslycontainsinfinitelymanynumbers.)

(b) No,because 5 6 doesnotinclude5.

7.(a) No: a b  b a   b a ingeneral.

(b) No;bytheDistributiveProperty, 2 a 5 2

10.

8.(a) Yes,absolutevalues(suchasthedistancebetweentwodifferentnumbers) arealwayspositive.

(b) Yes, b a   a b

9.(a) Naturalnumber:100

(b) Integers:0,100, 8

(c) Rationalnumbers: 1 5,0, 5 2 ,2 71,3 14,100, 8

(d) Irrationalnumbers: 7, 

10.(a) Naturalnumbers:2, 9 

(b) Integers:2, 100

,10

(c) Rationalnumbers:4 5  9 2 , 1 3 ,1 6666

(d) Irrationalnumbers: 2, 3 14

35.(a) False (b) True

37.(a) True (b) False

True (b) True 39.(a) x  0 (b) t  4 (c)

83.(a) a isnegativebecause a ispositive.

(b) bc ispositivebecausetheproductoftwonegativenumbersispositive.

(c) a b  a   b ispositivebecauseitisthesumoftwopositivenumbers.

(d) ab  ac isnegative:Eachsummandistheproductofapositivenumberandanegative number,andthesumoftwo negativenumbersisnegative.

84.(a) b ispositivebecause b isnegative.

(b) a  bc ispositivebecauseitisthesumoftwopositivenumbers.

(c) c a  c   a  isnegativebecause c and a arebothnegative.

(d) ab2 ispositivebecauseboth a and b 2 arepositive.

85. DistributiveProperty

86.(a) When L  150, x  20,and y  15,wehave

220.Because 220  274,thepostofficewillacceptthispackage. When L  120, x  60,and y  60,wehave

360,andsince 360  274,thepostofficewill not acceptthispackage.

(b) If x  y  22 5,then L

184cm  184m.

87. Let x  m 1 n 1 and y  m 2 n 2 berationalnumbers.Then x 

, x y  m 1



,and

.Thisshowsthatthesum,difference,andproduct oftworationalnumbersareagainrationalnumbers.Howevertheproductof twoirrationalnumbersisnotnecessarily irrational;forexample, 2  2  2,whichisrational.Also,thesumoftwoirrationalnumbersisnotnecessarilyirrational; forexample, 2   2  0whichisrational.

88. 1 2  2isirrational.Ifitwererational,thenbyExercise6(a),thesum  1 2  2   1 2   2wouldberational,but thisisnotthecase. Similarly, 1 2 2isirrational. (a) Followingthehint,supposethat r  t  q ,arationalnumber.ThenbyExercise6(a),thesumofthetworational numbers r  t and r isrational.But r  t    r   t ,whichweknowtobeirrational.Thisisacontradiction,and henceouroriginalpremise—that r  t isrational—wasfalse.

89.

(b) r isanonzerorationalnumber,so r  a b forsomenonzerointegers a and b .Letusassumethat rt  q ,arational number.Thenbydefinition, q  c d forsomeintegers c and d .Butthen rt  q  a b t  c d ,whence t  bc ad ,implying that t isrational.Onceagainwehavearrivedatacontradiction,andweconclude thattheproductofarationalnumber andanirrationalnumberisirrational.

As x getslarge,thefraction1 x getssmall.Mathematically,wesaythat1 x goestozero.

As x getssmall,thefraction1 x getslarge.Mathematically,wesaythat1 x goestoinfinity.

90. Wecanconstructthenumber 2onthenumberlineby transferringthelengthofthe hypotenuseofarighttriangle withlegsoflength1and1.

Similarly,tolocate 5,weconstructarighttrianglewithlegs oflength1and2.BythePythagoreanTheorem,thelength ofthehypotenuseis 12  22  5.Thentransferthe lengthofthehypotenusetothenumberline.

Thesquarerootofanyrationalnumbercanbelocatedona numberlineinthisfashion.

Thecircleinthesecondfigureinthetexthascircumference  ,soifwerollitalonganumberlineonefullrotation,wehave found  onthenumberline.Similarly,anyrationalmultipleof  canbefoundthisway.

91.(a) Supposethat a  b ,somax a  b

Ontheotherhand,if b

If a  b ,then a b

(b) If a  b ,thenmin

 0andtheresultistrivial.

and

Similarly,if b  a ,then

92. Answerswillvary.

;andif a  b ,theresultistrivial.

93.(a) Subtractionisnotcommutative.Forexample,5 1  1 5.

(b) Divisionisnotcommutative.Forexample,5  1  1  5.

(c) Puttingonyoursocksandputtingonyourshoesarenotcommutative.Ifyouputonyoursocksfirst,thenyourshoes, theresultisnotthesameasifyouproceedtheotherwayaround.

(d) Puttingonyourhatandputtingonyourcoatarecommutative.Theycanbedoneineitherorder,withthesameresult.

(e) Washinglaundryanddryingitarenotcommutative.

94.(a)

Ineachcase,

andtheTriangleInequalityissatisfied.

(b) Case0: Ifeither x or y is0,theresultisequality,trivially.

Case1: If x and y havethesamesign,then  x  y

y if x and y arepositive

and y arenegative

Case2: If x and y haveoppositesigns,thensupposewithoutlossofgeneralitythat x  0and y  0.Then

1.2 EXPONENTSANDRADICALS

1.(a) Usingexponentialnotationwecanwritetheproduct5  5  5  5  5  5as56

(b) Intheexpression34 ,thenumber3iscalledthe base andthenumber4iscalledthe exponent

2.(a) Whenwemultiplytwopowerswiththesamebase,we add theexponents.So34  35  39

(b) Whenwedividetwopowerswiththesamebase,we subtract theexponents.So 35 32  33

3. Tomoveanumberraisedtoapowerfromnumeratortodenominatororfromdenominatortonumeratorchangethesignof the

4.(a) Usingexponentialnotationwecanwrite 3 5as513

(b) Usingradicalswecanwrite51

(c) No.

5.

6. Becausethedenominatorisoftheform a ,wemultiplynumeratoranddenominatorby

7. 513 523 

8.(a)

(d) No;if a isnegative,then 4a 2

13.(a)

21.(a)

(c) x 16 x 10  x 16 10  x 6

23.(a) a 9 a 2 a  a 9 2 1  a 6 (b) a 2 a 4 3  a 24 3 

22.(a) y 2 y 5  y 2 5  y 3  1 y 3 (b) z 5 z 3 z 4  z 5 3 4  z 2  1 z 2 (c) y 7 y 0 y 10  y 70 10  y 3  1 y 3

24.(a)

28.(a)

30.(a)

75.(a) 1 5 x  1 5 x  5 x 5 x  5 x 5 x (b)  x 5   x 5  5 5  5 x 5 (c) 5  1 x 3  1 x 35 x 25 x 25

79.(a)

81.(a) 69,300,000  6 93  107 (b) 7,200,000,000,000  72  1012

(c) 0000028536  28536  10 5 (d) 0 0001213  1 213  10 4

83.(a) 3 19  105  319,000

(b) 2 721  108  272,100,000

(c) 2 670  10 8  0 00000002670 (d) 9 999  10 9  0 000000009999

85.(a) 9,460,000,000,000km  95  1012 km (b) 00000000000004cm  4  10 13 cm (c) 33billionbillionmolecules

86.(a) 150,000,000km  1 5  108 km

(b) 0 000000000000000000000053g  5 3  10 23 g (c) 5,970,000,000,000,000,000,000,000kg 

87. 7 2  10 9 1 806  10 12  

82.(a) 129,540,000  1 2954  108 (b) 7,259,000,000  7259  109

(c) 00000000014  14  10 9 (d) 0 0007029  7 029  10 4

84.(a) 7 1  1014  710,000,000,000,000

(b) 6  1012  6,000,000,000,000 (c) 8 55  10 3  0 00855 (d) 6 257  10 10  0 0000000006257

93.(a) b 5 isnegativesinceanegativenumberraisedtoanoddpowerisnegative.

(b) b 10 ispositivesinceanegativenumberraisedtoanevenpowerispositive.

(c) ab2 c3 wehave positive

whichisnegative.

(d) Since b a isnegative, b a 3  negative3 whichisnegative.

(e) Since b a isnegative, b a 4  negative4 whichispositive.

(f) a 3 c3 b 6 c6

94.(a) Since 1 2

(b)

(c) Wefindacommonroot:71

95. Sinceonelightyearis9

98. Eachperson’sshareisequalto

99. First,weestimatethetotalmassofthestarsintheobservableuniverse:

whichisnegative.

Thus,thenumberofhydrogenatomsintheobservableuniverseis

100. Firstconvert346meterstokilometers.Thisgives346m  346  1kilometer 1000meters  0 346km.Thusthedistanceyoucansee isgivenby D 

2kilometers.

101.(a) Using f  0 4andsubstituting d  65,weobtain s  30 fd  30  0 4  65

28km/h. (b) Using f

102. Since1day  86,400s,365 25days  31,557,600s.Substituting,weobtain d 

103. Since106  103 103 itwouldtake1000days  2 74yearstospendthemilliondollars. Since109  103 106 itwouldtake106  1,000,000days  2739 72yearstospendthebilliondollars.

104.(a)

Sowhen n getslarge,21 n decreasesto1.

105.(a)

Sowhen n getslarge,

factors .Because m  n ,wecancancel n factorsof a fromnumeratoranddenominatorandareleftwith m n factorsof a inthenumerator.Thus, a m a n  a m n . (b)  a b n 

1.3 ALGEBRAICEXPRESSIONS

1. Thegreatestcommonfactorintheexpression18 x 3  30 x is6 x ,andtheexpressionfactorsas6 x 3 x 2  5

2.(a) Thepolynomial2 x 3  3 x 2  10 x hasthreeterms:2 x 3 ,3 x 2 ,and10 x (b) Thefactor x iscommontoeachterm,so2 x

3. Tofactorthetrinomial x 2  8 x  12welookfortwointegerswhoseproductis12andwhosesumis8.Theseintegersare6 and2,sothetrinomialfactorsas  x  6 x  2

4. TheSpecialProductFormulaforthe“squareofasum”is

2 x  32 

5. TheSpecialProductFormulaforthe“productofthesumanddifferenceofterms”is

6. TheSpecialFactoringFormulaforthe“differenceofsquares”is

7. TheSpecialFactoringFormulafora“perfectsquare”is

8.(a) No;

(b) Yes;

Yes;byaSpecialProductFormula,

(d) No,  x  a  x a   x 2 a 2 ,byaSpecialProductFormula.

9. Type:binomial.Terms:5 x 3 and6.Degree:3.

10. Type:trinomial.Terms: 2 x 2 ,5 x ,and 3.Degree:2.

11. Type:monomial.Term: 8.Degree:0.

12. Type:monomial.Terms: 1 2 x 7 .Degree:7.

13. Type:four­termpolynomial.Terms:

,and

14. Type:binomial.Terms: 2 x and 3.Degree:1.

15.

95. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis, x

96. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis, x

97. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis, x

98. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis, x 13

Thus, x 53  x 23 

99. Startbyfactoringoutthepowerof x 2  1withthesmallestexponent,thatis,

127. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis,

(differenceofsquares)

(eachfactorisadifferenceofsquares)

143. Thevolumeoftheshellisthedifferencebetweenthevolumesoftheoutside cylinder(withradius R )andtheinsidecylinder (withradius r ).Thus

.The averageradiusis R  r 2 and2

2 istheaveragecircumference(lengthoftherectangularbox), h istheheight,and R r isthethicknessoftherectangularbox.Thus

height

144.(a) Movedportion  field habitat (b) Usingthedifferenceofsquares,weget

145.(a) Thedegreeoftheproductisthesumofthedegrees.

averageradius

(b) Thedegreeofasumisatmostthelargestofthedegrees—itcouldbesmallerthaneither.Forexample,thedegreeof

146.(a)

(b) Basedonthepatterninpart(a),wesuspectthat

(b)

1.4

RATIONALEXPRESSIONS

1. Arationalexpressionhastheform P  x  Q  x  ,where P and Q arepolynomials.

(a) 3 x x 2 1 isarationalexpression.

(b)  x  1 2 x  3 isnotarationalexpression.Arationalexpressionmustbeapolynomialdividedbyapolynomial,andthe numeratoroftheexpressionis  x  1,whichisnotapolynomial.

(c) x  x 2 1 x  3  x 3 x x  3 isarationalexpression.

2. Tosimplifyarationalexpressionwecancelfactorsthatarecommontothe numerator and denominator.So,theexpression  x  1 x  2  x  3 x  2 simplifiesto x  1 x  3

3. Tomultiplytworationalexpressionswemultiplytheir numerators togetherandmultiplytheir denominators together.So 2 x  1 x x  3 isthesameas

4.(a) 1 x 2  x  1 x  x  12 hasthreeterms.

(b) Theleastcommondenominatorofallthetermsis x  x  12

5.(a) Yes.Cancelling

(b) No;

6.(a) Yes, 3  a 3

(b) No.Wecannot“separate”thedenominatorinthisway;onlythenumerator,asinpart(a).(SeealsoExercise101.)

7. Thedomainof4

Since x 3  0wehave x  3.Domain:

x

22. x 2 x 12 x 2  5 x  6   x 4

23. y 2  y y 2 1  y  y 

numeratoranddenominatorbythecommondenominatorofboththenumerator anddenominator,inthiscase x 2 y

76. Incalculusitisnecessarytoeliminatethe h inthedenominator,andwedothisbyrationalizingthenumerator:

100.(a) Theaveragecost A 

numberofshirts

(b)

Fromthetable,weseethattheexpression x 2 9 x 3 approaches6as x approaches3.Wesimplifytheexpression: x 2 9 x 3   x 3 x  3 x 3  x  3, x  3.Clearlyas x approaches3, x  3approaches6.Thisexplainstheresultinthe table.

102. No,squaring 2  x changesitsvaluebyafactorof 2  x

103. Answerswillvary.

,sothestatementistrue.

(b) Thisstatementisfalse.Forexample,take

(c) Thisstatementisfalse.Forexample,take

(d) Thisstatementisfalse.Forexample,take

(e) Thisstatementistrue:

(f) Thisstatementistrue: 1  x  x 2 x  1 x  x x 

105.(a)

Itappearsthatthesmallestpossiblevalueof x  1 x is2.

(b) Because x  0,wecanmultiplybothsidesby x andpreservetheinequality:

x  0,andbecauseeachstepis reversible,wehaveshownthat x  1 x  2forall x  0.

1.5

EQUATIONS

1.(a) Yes.If a  b ,then a  x  b  x ,andviceversa.

(b) Yes.If a  b ,then ma  mb for m  0,andviceversa.

(c) No.Forexample, 5  5,but 

2.(a) Takepositiveandnegativesquarerootsofbothsides:

(b) Subtract5frombothsides:

(c) Subtract2frombothsides:

3.(a) Tosolvetheequation x 2 6 x

(b) Tosolvebycompletingthesquare,wewrite x

(c) TosolveusingtheQuadraticFormula,wewrite x 

or x 2.

4.(a) TheZero­ProductPropertysaysthatif a b  0theneither a or b mustbe0.

(b) Thesolutionsoftheequation x 2  x 4  0are x  0and x  4.

(c) Tosolvetheequation x 3 4 x 2  0we factor theleft­handside: x 2  x 4  0,asabove.

5.(a) Isolatingtheradicalin 2 x  x  0,weobtain 2 x  x

(b) Nowsquarebothsides: 2 x 2   x 2  2 x  x 2

(c) Solvingtheresultingquadraticequation,wefind2 x  x 2  x 2 2 x  x  x 2  0,sothesolutionsare x  0and x  2.

(d) Wesubstitutethesepossiblesolutionsintotheoriginalequation: 2  0  0  0,so x  0isasolution,but 2 2  2  4  0,so x  2isnotasolution.Theonlyrealsolutionis x  0.

6. Theequation  x  12 5  x  1  6  0isof quadratic type.Tosolvetheequationweset W  x  1.Theresulting quadraticequationis W 2 5

x  1or x  2.Youcanverifythatthesearebothsolutionstotheoriginalequation.

7. Toeliminatethedenominatorsintheequation 3 x  5 x  2  2,multiplyeachsidebythelowestcommondenominator x  x  2 togettheequivalentequation3 

8. Toeliminatethesquarerootintheequation2x  1   x  1, square eachsidetogettheequation 2

1

2  x  1.(But don’tforgetthatsquaringsometimesintroducesextraneoussolutions.)

9.(a) When x 2,LHS  4  2  7 8  7 

21.SinceLHS  RHS, x 2isnotasolution.

(b) When x  2,LHS  4

3

15.SinceLHS  RHS, x  2isa solution.

10.(a) When x  2,LHS  1

0.Since LHS  RHS, x  2isasolution.

(b) When x  4LHS  1

SinceLHS  RHS, x  4isnotasolution.

11.(a) When x  2,LHS  1 2

 RHS, x  2isasolution.

(b) When x  4theexpression 1 4 4 isnotdefined,so x  4isnotasolution.

12.(a) When

(b) When

4isasolution.

46.

47. 2 x 2  8  x 2  4  x

thereisnorealsolution. 72. z 

D isnegative,this equationhasnorealsolution.

75. D  b 2 4ac  2

Potentialsolutionsare x

4.Theseareonlypotentialsolutionssincesquaringisnotareversible operation.Wemustcheckeachpotentialsolutionintheoriginalequation.Checking

0isfalse.Checking

4istrue.Thus,theonlysolutionis x  4.

0  4 x 2 33 x  63  4 x 21 x 3.Potentialsolutionsare x  21 4 and x  3.Substitutingeachofthesesolutions intotheoriginalequation,weseethat x  3isasolution,but x  21 4 isnot.Thus3istheonlysolution.

0becomes

2 13  40  0

5

8  0. So  5  0    5,and  8  0    8.When   5,wehave x 2  5  x 5.When   8,wehave

are x 4, 3, 1,and0.

94. 2 x 4  4 x 2  1  0.TheLHSisthesumoftwononnegativenumbersandapositivenumber,so2 x 4  4 x

Thisequationhasnorealsolution.

95. Let u  x 23 .Then0

u 3  0,then x 2

Thesolutionsare

1,0,1,or2.If W  0,then W   x 

x 32 neverequals0,andnosolutioncanbenegative,becausewecannottakethesquarerootofanegativenumber.Thus2 istheonlysolution.

Theonlysolutionis256.

usingtheQuadraticFormula,weget

Thusthesolutionsare

.Potentialsolutionsare x  20and x  31.Wemustcheckeachpotentialsolutionintheoriginalequation.

Checking x  20: 

5,whichistrue,andhence x  20isa solution.

Checking x  31:

5,whichisfalse,andhence x  31isnota solution.Theonlyrealsolutionis x  20.

122. 3

solutionsare0and2.

123. x

If  x  312  0,then

124. Let u  11 x 2 .Bydefinitionof u werequireittobenonnegative.Now 11 x 2 2 11 x 2  1  u 2 u

mustbenonnegative,weonlyhave

7.Thesolutionsare 7.

125. 0  x 4 5ax 2  4a 2  a x 2 4a x 2 

 x a 2  36.Checkingtheseanswers,weseethat x a 2  36isnotasolution(forexample,trysubstituting a  8),but x  a 2  36isasolution.

128. Let   x 16 .Then x 13   2 and x 12   3 ,andso

0   3 a  2  b

a  6  x  x  a 6 isonesolution.Settingthefirstfactorequaltozero,wehave

However,theoriginalequationincludestheterm b 6  x ,andwecannottakethesixthrootofanegativenumber,sothisisnot asolution.Theonlysolutionis x  a 6 .

129. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

Using h 0  88,wesolve0

Thusittakes4 28secondsfortheballthehittheground.

130.(a) Using h 0 29,halfthedistanceis14 5,sowesolvetheequation14 5 

t 2  14 5 4

 3  t 3.Since t  0,ittakes 3  1732s.

(b) Theballhitsthegroundwhen h  0,sowesolvetheequation0 4

Since t  0,ittakes 6  2 449s.

131. Wearegiventhat  0  12m/s.

(a) Setting h  7 2,wehave7

2 4 2t 3t 1  0  t  1or t  1 1

Therefore,theballreaches7 2metersin1second(ascending)andagain after1 1 2 seconds(descending).

(b) Setting h  14

4,wehave14

.However,sincethediscriminant D  0,thereisnorealsolution,andhencetheball neverreachesaheightof14 4meters.

(c) Thegreatestheight h isreachedonlyonce.So

(d) Setting h

0hasonlyonesolution.Thus

5.Sothegreatestheightreachedbytheballis7 5meters.

.Thustheballreachesthehighest pointofitspathafter1 1 4 seconds.

(e) Setting h  0(groundlevel),wehave0

Sotheballhitsthegroundin2 1 2 s.

132. Ifthemaximumheightis30meters,thenthediscriminantoftheequation4

30  0mustequalzero.So

24doesnotmakesense,wemusthave  0  24m/s.

133.(a) Theshrinkagefactorwhen 

 0 00055.Sothebeamshrinks 0 00055  12 025  0 007m,sowhenitdriesitwillbe12 025 0 007  12 018mlong. (b) Substituting S  0 00050weget0 00050  0 032

  7 5 0 032  234 375.Sothewatercontentshouldbe234 375kg/m3

134. Let d bethedistancefromthelenstotheobject.Thenthedistancefromthelenstotheimageis d 4.Sosubstituting F  4 8, x  d ,and y  d 4,andthensolvingfor x ,wehave

d 4 .Nowwemultiplybythe LCD,4 8d 

,toget

d  13 6  10 4 2 .So d  1 6or d  12.Since d 4mustalsobepositive,theobjectis12cmfromthelens.



136. Setting P  1250andsolvingfor x ,wehave1250

quadraticformula,

200,heshouldmake50ovensperweek.

137. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x bethedistancefromthecenteroftheearthtothedeadspot(inthousandsof kilometers).Nowsetting F  0,wehave 0  K x 2

0 988 x 2 764 x  145,924  0.UsingtheQuadraticFormula,weobtain



x  386 64 42 35  344.Since429,000isgreaterthanthedistancefromthe earthtothemoon,werejectthefirstroot;thus x  344,000kilometers.

138. Sincethetotaltimeis3s,wehave3

so d  40 46.Thewellis40 5mdeep.

(b)

1.Sincebothsidesofthisequationareequal, x  2isasolutionforeveryvalueof k .Thatis, x  2isasolutiontoeverymemberofthisfamilyofequations.

140. Whenwemultipliedby x ,weintroduced x  0asasolution.Whenwedividedby x 1,wearereallydividingby0,since x  1  x 1  0.

141.(a) x 2 9 x  20  0   x 4 x

4or x

5.Theproductofthesolutionsis4 5  20,theconstant termintheoriginalequation;andtheirsumis4  5  9,thenegativeofthecoefficientof x intheoriginalequation. (b) Ingeneral,theequation x 2  bx  c  0hassolutions r 1  b 

and

142.(a) Wemakethesubstitution

possiblesolutionsare4and1.Checkingwillresultinthesamesolution.

(b) Method1:

Thesolutionsare

Method2: MultiplyingbytheLCD,

1.6 COMPLEXNUMBERS

1. Theimaginarynumber i hasthepropertythat i 2 1.

2. Forthecomplexnumber3  4i therealpartis3andtheimaginarypartis4.

3.(a) Thecomplexconjugateof3  4

(b) 3  4i 

3  4i   32  42  25

2  13.

4. If3  4i isasolutionofaquadraticequationwithrealcoefficients,then 3  4i  3 4i isalsoasolutionoftheequation.

5. Yes,everyrealnumber a isacomplexnumberoftheform a  0i

6. Yes.Foranycomplexnumber z

7. 3 8i :realpart3,imaginarypart 8. 8.

9. 2 5i 3  2 3 5 3 i :realpart 2 3 ,imaginarypart

 2a ,whichisarealnumber.

5 i

5  i :realpart 5,imaginarypart1.

:realpart2,imaginarypart 7 2

11. 3:realpart3,imaginarypart0. 12. 1 2 :realpart 1 2 ,imaginarypart0.

13. 2 3 i :realpart0,imaginarypart 2 3 14. 3i :realpart0,imaginarypart 3.

RHS,thisprovesthestatement.

SinceLHS

SinceLHS

RHS,thisprovesthestatement.

84. Suppose z  z .Thenwehave

,whichisapureimaginarynumber.

,whichisarealnumber.

0,so z isreal.Nowif z isreal, then z  a  0i (where a isreal).Since z  a 0i ,wehave z  z

85. UsingtheQuadraticFormula,thesolutionstotheequationare x  b

b 2 4ac 2a .Sincebothsolutionsarenonreal,we have b2 4ac  0  4ac b 2  0,sothesolutionsare x 

2a  4ac b2 2a i ,where 4ac b2 isarealnumber. Thusthesolutionsarecomplexconjugatesofeachother.

86. i  i , i 5  i 4  i  i , i 9  i 8

Because i 4  1,wehave i n  i r ,where r istheremainderwhen n isdividedby4,thatis, n  4 k  r ,where k isan integerand0  r  4.Since4446  4

1.7 MODELINGWITHEQUATIONS

1. Anequationmodelingareal­worldsituationcanbeusedtohelpusunderstandareal­worldproblemusingmathematical methods.Wetranslatereal­worldideasintothelanguageofalgebratoconstructourmodel,andtranslateourmathematical resultsbackintoreal­worldideasinordertointerpretourfindings.

2. Intheformula I  Prt forsimpleinterest, P standsfor principal, r for interestrate,and t for time(inyears)

3.(a) Asquareofside x hasarea A  x 2 . (b) Arectangleoflength l andwidth  hasarea A  l  (c) Acircleofradius r hasarea A  r 2

4. Balsamicvinegarcontains5%aceticacid,soa960mLbottleofbalsamicvinegarcontains 960 5%  960 5 100  48millilitersofaceticacid.

5. Apainterpaintsawallin x hours,sothefractionofthewallshepaintsinonehouris 1wall x hours  1 x

6. Solving d  rt for r ,wefind d t 

7. If n isthefirstinteger,then n  1isthemiddleinteger,and n  2isthethirdinteger.Sothesumofthethreeconsecutive integersis n 

3.

8. If n isthemiddleinteger,then n 1isthefirstinteger,and n  1isthethirdinteger.Sothesumofthethreeconsecutive integersis n 1  n 

1

 3n

9. If n isthefirsteveninteger,then n  2isthesecondevenintegerand n  4isthethird.Sothesumofthreeconsecutive evenintegersis n

6.

10. If n isthefirstinteger,thenthenextintegeris n  1.Thesumoftheirsquaresis

11. If s isthethirdtestscore,thensincetheothertestscoresare78and82,theaverageofthethreetestscoresis

12. If q isthefourthquizscore,thensincetheotherquizscoresare8,8,and8,the averageofthefourquizscoresis

13. If x dollarsareinvestedat2 1 2 %simpleinterest,thenthefirstyearyouwillreceive0 025 x dollarsininterest.

14. If n isthenumberofmonthstheapartmentisrented,andeachmonththerentis$945,thenthetotalrentpaidis945n

15. Since  isthewidthoftherectangle,thelengthisfourtimesthewidth,or4 .Then area  length  width 

2 m2 .

16. Since  isthewidthoftherectangle,thelengthis   6.Theperimeteris

17. If d isthegivendistance,inkilometers,anddistance  rate  time,wehavetime  distance rate  d 55

18. Sincedistance  rate timewehavedistance  s  45min

19. If x isthequantityofpurewateradded,themixturewillcontain750gofsaltand3  x litersofwater.Thusthe concentrationis 750 3  x

20. If p isthenumberofpenniesinthepurse,thenthenumberofnickelsis2 p ,thenumberofdimesis4  2 p ,and thenumberofquartersis 2 p   4 

4.Thusthevalue(incents)ofthechangeinthepurseis 1

21. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. If d isthenumberofdaysand m thenumberofkilometers,thenthecostofarentalis C

65d  0 40m .Inthiscase, d  3 and C  283,sowesolvefor m :283

220.Thus, thetruckwasdriven220kilometers.

22. Thecostofspeakingfor m minutesonthisplanis

Inthiscase m  250 and C  120 50,sowesolve100 

m

250  82  332.Therefore,thetouristused332minutesoftalkinthemonthofJune.

23. If x isthestudent’sscoreontheirfinalexam,thenbecausethefinalcountstwiceasmuchaseachmidterm,theaverage scoreis

 86.Sothestudentscored86%ontheirfinalexam.

24. Sixstudentsscored100andthreestudentsscored60.Let x betheaveragescoreoftheremaining25 6 3  16students. Becausetheoverallaverageis84% 

 16 x  1320  x  1320 16  82 5.Thus,theremaining16students’averagescorewas82 5%.

25. Let m betheamountinvestedat2 1 2 %.Then12,000 m istheamountinvestedat3%. Sincethetotalinterestisequaltotheinterestearnedat2 5%plustheinterestearnedat3%,wehave 318

8400.Thus, $8400isinvestedat2 1 2 %and12,000 8400  $3600isinvestedat3%.

26. Let m betheamountinvestedat5%.Then8000  m isthetotalamountinvested.Thus 4%ofthetotalinvestment  interestearnedat3 1 2 %  interestearnedat5%

4000.Thus,$4000mustbe investedat5%.

27. Usingtheformula I  Prt andsolvingfor r ,weget262

5%.

28. If$3000isinvestedataninterestrate a %,then$5000isinvestedat a  1 2  %,so,rememberingthat a isexpressedasa percentage,thetotalinterestis I  3000 a 100 1  5000 a  1

25.Sincethetotalinterest is$265,wehave265  80a  25  80a  240  a  3.Thus,$3000isinvestedat3%interest.

29. Let x bethemonthlysalary.Sinceannualsalary  12  monthlysalary  Christmasbonus,wehave 180,100  12 x  7300  172,800  12 x  x  14,400.Themonthlysalaryis$14,400.

30. Let s betheassistant’sannualsalary.Thentheforeman’sannualsalaryis1 15s .Theirtotalincomeisthesumoftheir salaries,so s  115s  113,305  215s  113,305  s  52,700.Thus,theassistant’sannualsalaryis$52,700.

31. Let x betheovertimehoursworked.Sincegrosspay  regularsalary  overtimepay,weobtaintheequation

6.Thus,thelab technicianworked6hoursofovertime.

32. Let x bethenumberofhoursworkedbytheplumber.Thenthecabinetmakerworksfor9 x hours.Thetotallaborchargeis thesumoftheircharges,so2610 

3.Thus,theplumberworks for3hoursandthecabinetmakerworksfor3 9  27hours.

33. Allagesareintermsofthedaughter’sage7yearsago.Let y beageofthedaughter7yearsago.Then11 y istheageofthe moviestar7yearsago.Today,thedaughteris y  7,andthemoviestaris11 y  7.Butthemoviestarisalso4timeshis daughter’sagetoday.So4

3.Thus,todaythemoviestaris 11 3  7  40yearsold.

34. Let h benumberofhomerunsBabeRuthhit.Then h  41isthenumberofhomerunsthatHankAaronhit.So 1469  h  h  41  1428  2h  h  714.ThusBabeRuthhit714homeruns.

35. Let n bethenumberofnickels.Thentherearealso n dimesand n quarters.Thetotalvalueofthecoinsinthepurseisthe sumofthevaluesofnickels,dimes,andquarters,so2

pursecontains7nickels,7dimes,and7quarters.

36. Let q bethenumberofquarters.Then2q isthenumberofdimes,and2q  5isthenumberofnickels.Thus300  value ofnickels valueofdimes valueofquarters,so 3 00  0 05 2

Thusyouhave5quarters,2 5  10dimes,and2 5  5  15nickels.

37. Let l bethelengthofthegarden.Sincearea  width  length,weobtaintheequation1125

thegardenis45meterslong.

38. Let  bethewidthofthepasture.Thenthelengthofthepastureis2 .Sincearea  length  width,wehave 115,200  

39. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let  bethewidthofthebuildinglot.Thenthelengthofthebuildinglotis5 .Sinceahalf­hectareis 1 2 10,000  5000 andareaislengthtimeswidth,wehave5000

31 6.Thusthewidthofthe buildinglotis31 6metersandthelengthofthebuildinglotis5 31 6  158meters.

40. Let x bethelengthofasideofthesquareplot.Asshowninthefigure, areaoftheplot  areaofthebuilding  areaoftheparkinglot.Thus,

landmeasures120metersby120meters. x

41. Let  bethewidthofthegardeninmeters.Thenthelengthis   10.Thus875     10  

0,then

35,whichisimpossible.Therefore

25

0,andso

25. Thegardenis25meterswideand35meterslong.

42. Let  bethewidthofthebedroom.Thenitslengthis   4.Sinceareaislengthtimeswidth,wehave 21 

7or   3.Sincethewidthmust bepositive,thewidthis3meters.

43. Let  bethewidthofthegardeninmeters.Weusetheperimetertoexpressthelength l ofthegardenintermsof width.Sincetheperimeteristwicethewidthplustwicethelength,wehave

100  .Usingtheformulaforarea,wehave2400

40.Sothelengthis60metersandthewidthis40meters.

44. Let  bethewidthofthelotinmeters.Thenthelengthis   6.UsingthePythagoreanTheorem,wehave

0.Sincewidthsmustbepositive,thewidthis120metersandthelengthis126meters.

45. Let l bethelengthofthelotinmeters.Thenthelengthofthediagonalis l  10. WeapplythePythagoreanTheoremwiththehypotenuseasthediagonal.So l 2

Thusthelengthofthelotis120meters.

46. Let r betheradiusoftherunningtrack.Therunningtrackconsistsoftwosemicirclesandtwostraightsections110meters long,sowegettheequation2r

35 03.Thustheradiusofthesemicircleis about35meters.

47.(a) Firstwewriteaformulafortheareaofthefigureintermsof x .Region A has dimensions10cmand x cmandregion B hasdimensions6cmand x cm.Sothe shadedregionhasarea 10 x

6 x

 16 x cm2 .Wearegiventhatthisisequal to144cm2 ,so144  16 x  x  144 16  9cm.

(b) Firstwewriteaformulafortheareaofthefigureintermsof x .Region A has dimensions14cmand x cmandregion B hasdimensions 13  x  cmand x cm.

Sotheareaofthefigureis

48.(a) Theshadedareaisthesumoftheareaofasquareandtheareaofatriangle.So

Wearegiventhattheareais120cm2

(b) Theshadedareaisthesumoftheareaofarectangleandtheareaofatriangle.So

Wearegiventhattheareais1200cm2

y  48cm.

49. Let x bethewidthofthestrip.Thenthelengthofthematis50  2 x ,andthewidthofthematis38  2 x .Nowthe perimeteristwicethelengthplustwicethewidth,so256  2

256  176  8 x  80  8 x  x  10.Thusthestripofmatis10centimeterswide.

50. Let x bethewidthofthestrip.Thenthewidthoftheposteris100  2 x anditslengthis140  2 x .Theperimeterofthe printedareais2 100 

480,andtheperimeteroftheposteris2

.Nowweusethe factthattheperimeteroftheposteris1 1 2 timestheperimeteroftheprintedarea:2 

480

30.Theblankstripisthus30cmwide.

51. Let h betheheighttheladderreaches(inmeters).UsingthePythagoreanTheoremwehave

54meters.

52. Let h betheheightoftheflagpole,inmeters.Thenthelengthofeachguywireis h  1 5.Sincethe distancebetweenthepointswherethewiresarefixedtothegroundisequaltooneguywire,thetriangleis equilateral,andtheflagpoleistheperpendicularbisectorofthebase.ThusfromthePythagoreanTheorem,we get

53. Let x bethelengthoftheperson’sshadow,inmeters.Usingsimilartriangles,

 x  5.Thustheperson’sshadowis5meterslong.

54. Let x betheheightofthetalltree.Hereweusethepropertythatcorresponding sidesinsimilartrianglesareproportional.Thebaseofthesimilartrianglesstartsat eyelevelofthewoodcutter,1 5meters.Thusweobtaintheproportion

5meterstall.

55. Let x betheamount(inmL)of60%acidsolutiontobeused.Then300 x mLof30%solutionwouldhavetobeusedto yieldatotalof300mLofsolution.

Thusthetotalamountofpureacidusedis0

So200mLof60%acidsolutionmustbemixedwith100mLof30%solutiontoget300mLof50%acidsolution.

56. Theamountofpureacidintheoriginalsolutionis300

50%

150.Let x bethenumberofmLofpureacidadded.Then thefinalvolumeofsolutionis300  x .Becauseitsconcentrationistobe60%,wemusthave

75.Thus,75mLofpureacidmustbe added.

57. Let x bethenumberofgramsofsilveradded.Theweightoftheringsis5  18g  90g. 5rings Puresilver Mixture

mustbeaddedtogettherequiredmixture.

58. Let x bethenumberoflitersofwatertobeboiledoff.Theresultwillcontain6 x liters.

59. Let x bethenumberoflitersofcoolantremovedandreplacedbywater.

mustberemovedandreplacedbywater.

60. Let x bethenumberoflitersof2%bleachremovedfromthetank.Thisisalsothenumberoflitersofpurebleachaddedto makethe5%mixture.

Purebleach 5%mixture

06litersneedtoremovedand replacedwithpurebleach.

61. Let c betheconcentrationoffruitjuiceinthecheaperbrand.Thenewmixtureconsistsof650mLoftheoriginalfruitpunch and100mLofthecheaperfruitpunch. OriginalFruitPunch CheaperFruitPunch Mixture

35.Thusthecheaperbrandisonly 35%fruitjuice.

62. Let x bethenumberofpacketsof$3 00packetteaThen80 x isthenumberofpacketsof$2 75packettea.

48.Themixtureuses 48packetsof$3 00packetteaand80 48  32packetsof$2 75packettea.

63. Let t bethetimeinminutesitwouldtaketowashthecarifthefriendsworkedtogether.Friend1washes 1 25 ofthecarper minute,whileFriend2washes 1 35 ofthecarperminute.Thesumofthese fractionsisequaltothefractionofthejobthey candoworkingtogether,sowehave

t  14 35 60 minutes,or14minutes35seconds

64. Let t bethetime,inminutes,ittakesthelandscapertomowthelawn.Sincetheassistantishalfasfast,itwouldtakethem 2t minutestomowthelawnalone.Thus,

assistant2 15  30minutestomowthelawnalone.

65. Let t bethenumberofhoursitwouldtakeyourfriendtopaintahousealone.Thenworkingtogether,ittakes 2 3 t hours. Becauseittakesyou7hours,wehave

.Thus,it wouldtakeyourfriend3 5htopaintahousealone.

66. Let h bethetime,inhours,tofilltheswimmingpoolusingthesmallerhosealone. Sincethelargerhosetakes20%less time,ittakes0 8h tofillthepoolalone.Thus16

h  28 8 0 8  36.Thus,thesmallerhosetakes36hourstofillthepoolalone,andthelargerhosetakes0 8 36  28 8hours.

67. Let t bethetimeinhoursthatittakesyoutowashallthewindows.Thenittakesyourroommate t  3 2 hoursto washallthewindows,andthesumofthefractionsofthejobyoucandoindividuallyperhourequalsthefraction ofthejobyoucandotogether.Since1hour48minutes

or t  21  39 20  3.Since t  0isimpossible,youcanwashthewindowsalonein3hours,andittakesyourroommate

68. Let t bethetime,inhours,ittakesyourmanagertodeliveralltheflyersalone.Thenittakestheassistant t

1 hourstodeliveralltheflyersalone,andittakesthegroup0 4t hourstodoittogether.Thus

2isimpossible, ittakesyourmanager3hourstodeliveralltheflyersalone.

69. Let t bethetimeinhoursthatthecommuterspentonthetrain.Then 11 2 t isthetimeinhoursthatcommuterspentonthe bus.Weconstructatable:

Thetotaldistancetraveledisthesumofthedistancestraveledbybusandbytrain,so480

70. Let r bethespeedoftheslowercyclist,inkm/h.Thenthespeedofthefastercyclistis2r

Whentheymeet,theywillhavetraveledacombinedtotalof90kilometers,so2

speedoftheslowercyclistis15km/h,whilethespeedofthefastercyclist is2

71. Let r bethespeedoftheplanefromMontrealtoLosAngeles.Then

15

30km/h.

isthespeedoftheplanefromLos AngelestoMontreal.

ataspeedof800km/honthetripfromMontrealtoLosAngeles.

72. Let x bethespeedofthecarinkm/h.Sinceakilometercontains1000mandanhourcontains3600s, 1km/h  1000 m 3600 s  5 18 m/s.Thetruckistravelingat80 5 18  200 9 m/s.Soin6seconds,thetrucktravels 6 200 9  133meters.Thusthebackendofthecarmusttravelthelengthofthecar,the lengthofthetruck,and the133metersin6seconds,soitsspeedmustbe

m/s.Convertingtokm/h,thespeedofthecaris 147 6 18 5  882km/h.

73. Let x betherate,inkm/h,atwhichthesalespersondrovebetweenAjaxandBarrington.

Wehaveusedtheequationtime  distance rate tofillinthe“Time”columnofthetable.Sincethesecondpartofthetrip took6minutes(or 1 10 hour)morethanthefirst,wecanusethetimecolumntogettheequation

salesmandroveeither80km/hor384km/hbetweenAjaxandBarrington.(The formerseemslikelier.)

74. Let x betherate,inkm/h,atwhichthetruckerdrovefromTortulatoCactus.

Wehaveusedtime  distance rate tofillinthetimecolumnofthetable.Wearegiventhat thesumofthetimesis11hours.Thuswegettheequation

Hence,thetruckerdroveeither 7 27km/h(impossible)or80km/h betweenTortulaandCactus.

75. Let r betherowingrateinkm/hofthecrewinstillwater.Thentheirrateupstreamwas r 3km/h,andtheirrate downstreamwas r  3km/h.

Sincethetimetorowupstreamplusthetimetorowdownstreamwas2hours40minutes  8 3 hour,wegettheequation

76. Let r bethespeedofthesouthboundboat.Then r  3isthespeedoftheeastboundboat.Intwohoursthesouthbound boathastraveled2r kilometersandtheeastboundboathastraveled2

r  3

 2r  6kilometers.Sincetheyare travelingisdirectionswithare90 apart,wecanusethePythagoreanTheoremtoget

or r  9.Sincespeedispositive,thespeedofthesouthboundboatis9km/h.

77. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x bethedistancefromthefulcrumtowherethe56­kgfriendsits.Thensubstitutingtheknownvaluesintotheformula given,wehave45

6.Sothe56­kgfriendshouldsit16metersfromthefulcrum.

78. Let  bethelargestweightthatcanbehung.Inthisexercise,theedgeofthebuildingactsasthefulcrum,sothe110­kg manissitting7 5metersfromthefulcrum.ThensubstitutingtheknownvaluesintotheformulagiveninExercise77,we have110

550.Therefore,550kilogramsisthelargestweightthatcanbehung.

79. Thevolumeis180m3 ,so

6istheonlypositive solution.Sotheboxis2metersby6metersby15meters.

80. Let r betheradiusofthelargersphere,inmm.Equatingthevolumes,wehave

63mm.

81. Let x bethelengthofonesideofthecardboard,sowestartwithapieceofcardboard x by x .When4centimetersare removedfromeachside,thebaseoftheboxis x 8by x 8.Sincethevolumeis100cm3 ,weget4  x 82  100  x 2 16 x  64  25  x 2 16 x  39  0   x 3 x 13  0 So x  3or x  13.But x  3isnotpossible,since thenthelengthofthebasewouldbe3 8 5 andalllengthsmustbepositive.Thus x  13,andthepieceofcardboard is13centimetersby13centimeters.

82. Let r betheradiusofthecan.Nowusingtheformula V  r 2 h with V  40 cm3 and h  10,wesolvefor r .Thus 40  r 2 10  4  r 2  r 2.Since r representsradius, r  0.Thus r  2,andthediameteris4cm.

83. Let r betheradiusofthetank,inmeters.Thevolumeofthesphericaltankis 4 3 r 3 andisalso2840  0001  284.So 4 3 r 3  2 84  r 3  0 678  r  0 88meters.

84. Let x bethelengthofthehypotenuseofthetriangle,inmeters.Thenoneofthe othersideshaslength x 7meters,andsincetheperimeteris392meters,the remainingsidemusthavelength392 x  x 7  399 2 x .Fromthe PythagoreanTheorem,weget

4 x 2 1610 x  159,250  0.UsingtheQuadraticFormula,weget x-7x

x  227 5,then thesideoflength x 7combinedwiththehypotenusealreadyexceedstheperimeterof392meters,andsowemusthave x  175.Thustheothersideshavelength175

49.Thelothassidesoflength49meters, 168meters,and175meters.

85. Let x bethelength,inkilometers,oftheabandonedroadtobeused.Thenthelengthoftheabandonedroadnotusedis 40 x ,andthelengthofthenewroadis 102  40 x 2 kilometers,bythePythagoreanTheorem.Sincethecostof theroadiscostperkilometer  numberofkilometers,wehave100,000 x



 2 x 2 80 x  1700  68 x Squaringbothsides,weget4 x

islongerthanthe existingroad,16kilometersoftheabandonedroadshouldbeused.Acompletelynewroadwouldhavelength 102  402 (let x  0)andwouldcost 1700  200,000  8 3milliondollars.Sono,itwouldnotbecheaper.

86. Let x bethedistance,inmeters,thatyougoontheboardwalkbeforeveeringoffontothesand.Thedistancealongthe boardwalkfromwhereyoustartedtothepointontheboardwalkclosesttotheumbrellais 2292 642  220m.Thusthe distanceyouwalkonthesandis

Since4minutes45seconds  285seconds,weequatethetimeittakestowalkalongtheboardwalkandacrossthesandto thetotaltimetoget285

 x  213 3or145 4.Checking x  213 3:Thedistanceacrossthesandis64m,so

284 4s.Checking x  145 4:Thedistanceacrossthesandis

 285 0s.Sinceboth solutionsarelessthanorequalto220m, wehavetwosolutions:Eitheryouwalk145 4metersdowntheboardwalkand thenheadtowardstheumbrella,oryouwalk213 3metersdowntheboardwalkandthenheadtowardstheumbrella.

87. Let x betheheightofthepileinmeters.Thenthediameteris3 x andtheradiusis 3 2 x meters.Sincethevolumeofthecone is28m3 ,wehave  3  3 x 2

2

88. Let h betheheightofthescreensincentimeters.Thewidthofthesmallerscreen is h  17 5centimeters, andthewidthofthebiggerscreenis1 8h centimeters.Thediagonalmeasureofthesmallerscreenis

,andthediagonalmeasureofthelargerscreenis

5.Squaringbothsidesgives

5.ApplyingtheQuadraticFormula, weobtain

screensareapproximately34

2centimetershigh.

89. Let h betheheightinmetersofthestructure.Thestructureiscomposedofarightcylinderwithradius3andheight 2 3 h and aconewithbaseradius3andheight 1 3 h .Usingtheformulasforthevolumeofa cylinderandthatofacone,weobtain theequation40

structureis5 7meters.

90. Let y bethecircumferenceofthecircle,so360 y istheperimeterofthesquare.Usethecircumferencetofindthe radius, r ,intermsof y

.Thustheareaofthecircleis

.Nowifthe perimeterofthesquareis360 y ,thelengthofeachsideis 1 4 360 y   andtheareaofthesquareis

Settingtheseareasequal,weobtain y

169 1.Thusonewireis169 1cmlongandtheotheris 190 9cmlong.

91. Let h betheheightofthebreak,inmeters.Thentheportionofthebambooabove thebreakis10 h .ApplyingthePythagoreanTheorem,weobtain

h  91 20  4 55.Thusthebreakis4 55mabovetheground. 10-h h 3

92. Answerswillvary.

93. Let x equaltheoriginallengthofthereedincubits.Then x 1isthepiecethatfits60timesalongthelength ofthefield,thatis,thelengthis60  x 1.Thewidthis30 x .Thenconvertingcubitstoninda,wehave 375  60  x 1

 6or x 5.Since x mustbepositive,theoriginallengthofthereedis6cubits.

1.8 INEQUALITIES

1.(a) If x  5,then x 3  5 3  x 3  2.

(b) If x  5,then3 x  3 5  3 x  15.

(c) If x  2,then 3 x 3 2 3 x 6.

(d) If x  2,then x  2.

2. Tosolvethenonlinearinequality x  1 x 2  0we

firstobservethatthenumbers 1and2arezeros ofthenumeratoranddenominator.These numbersdividethereallineintothethree intervals

Theendpoint 1satisfiestheinequality,because

defined.

Signof x  1

Signof x 2

Thus,referringtothetable,weseethatthesolutionoftheinequalityis[ 1 2

3.(a) Thesolutionoftheinequality  x   3istheinterval [ 3 3].Anynumber thatliesinsidethisintervalsatisfiestheinequality. _33

(b) Thesolutionoftheinequality  x   3istheunionofintervals

 3]  [3 .Anynumberthatlieinsideoneoftheseintervals satisfiestheinequality. _33

4.(a) Thesetofallpointsonthereallinewhosedistancefromzeroislessthan3canbedescribedbytheabsolutevalue inequality  x   3.

(b) Thesetofallpointsonthereallinewhosedistancefromzeroisgreaterthan3canbedescribedbytheabsolutevalue inequality  x   3.

5.(a) No.Forexample,if x 2,then

(b) No.Forexample,if

6.(a) Tosolve3 x  7,startbydividingbothsidesoftheinequalityby3.

(b) Tosolve5 x 2  1,startbyadding2tobothsidesoftheinequality.

(c) Tosolve 3 x  2  8,startbyusingProperty2ofAbsoluteValueInequalitiestorewritetheinequalityas

5 6 ,1, 5,3,and5satisfytheinequality.

1,and0satisfytheinequality.

x 1  2 x 4  7

5 1  14  7;no 1 1  6  7;no 0 1  4  7;no

2 3 1  8 3  7;no

5 6 1  7 3  7;no 1 1  2  7;no

5 1  047  7;no 3 1  2  7;yes 5 1  6  7;yes

Theelements3and5satisfytheinequality.

x 1 x  1 2 5 1 5  1 2 ;yes 1 1

2  3 x  2

2  8  2;no

2  4  2;no

Theelements 5, 1, 5,3,and5satisfytheinequality.

Theelements 5,3,and5satisfytheinequality.

x x 2

2  4

27  4;no 1 3  4;yes

2  4;yes 2 3 22 9  4;yes 5 6 97 36  4;yes 1 3  4;yes

5 7  4;no 3 11  4;no 5 27  4;no

Theelements 1,0, 2 3 , 5 6 ,and1satisfytheinequality.

37.  x  2 x 3  0.Theexpressionontheleftoftheinequalitychangessignwhere x 2andwhere x  3.Thuswe mustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph: _23

38.  x 5 x  4

 0.Theexpressionontheleftoftheinequalitychangessignwhen x  5and x 4.Thuswemust checktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x  x 4or5  x 

Signof x 5

Interval: 

Graph: _45

39. x 2 x  7  0.Theexpressionontheleftoftheinequalitychangessignwhere x  0andwhere x  7 2 .Thuswemust checktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x  x  7 2 or0  x 

Signof x

Signof2 x  7

Signof

Interval:

Graph: 70 2

40. x 2 3 x   0.Theexpressionontheleftoftheinequalitychangessignwhen x  0and x  2 3 .Thuswemustcheckthe intervalsinthefollowingtable.

Fromthetable,thesolutionsetis  x  x  0or 2 3  x 

Signof x

Signof x 2 3 x  

41. x 2 3 x 18  0

3

x

Interval: 

Graph: 2 3 0

x 6  0.Theexpressionontheleftoftheinequalitychangessignwhere x  6andwhere x 3.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x 3  x  6.Interval: [ 3 6]

Signof x  3

Signof x 6

Signof

x

Graph: _36

42. x 2 8 x

0.Theexpressionontheleftoftheinequalitychangessignwhere x  7andwhere x  1.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph: 17

.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

0.Theexpressionontheleftofthe inequalitychangessignwhen x 1and x  2.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis x 1  x  2.Interval:

0.Theexpressionontheleftoftheinequalitychangessign where x 1andwhere x  4.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x 1  x  4.Interval:  1

Graph: _14

0.Theexpressionontheleftoftheinequalitychanges signwhen x  1 2 and x 2.Thuswemustchecktheintervalsinthefollowingtable. Interval

Fromthetable,thesolutionsetis

x  x 2or 1 2  x 

Graph: 1 2 _2

Fromthetable,thesolutionsetis

0.Theexpressionontheleftoftheinequalitychangessignwhen x 3and x  1.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x  x 

Interval:

Graph: _31

0.Theexpressionontheleftoftheinequalitychangessignwhere x 2and where x  2.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph:

0.Theexpressionontheleftoftheinequalitychangessignwhen x

3and x  3.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph: _33

Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph: _23 1

52.  x 5 x 2 x  1  0.Theexpressionontheleftoftheinequalitychangessignwhen x  5, x  2,and x 1. Thuswemustchecktheintervalsinthefollowingtable.

Signof x 5 

Signof x 2  

Signof x  1

Signof  x 5

Fromthetable,thesolutionsetis

.Graph: _15 2 53.

x 4

x

2

0forall x

2,sotheexpressionontheleftoftheoriginalinequalitychanges signonlywhen x  4.Wechecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x

2and x  4.Weexcludethe endpoint 2sincetheoriginalexpressioncannot be0.Interval:

Graph: _24

 x 4 x  22  0.Notethat  x

2  0forall x 2,sotheexpressionontheleftoftheoriginalinequalitychanges signonlyat x  4.Wechecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis x  x  4 (Thepoint 2isalreadyexcluded.)

Signof x 4 

Signof  x  22 

Interval: 4

Graph: 4

55.  x  32  x 2 x  5  0.Theleft­handsideis0when x

Signof  x 

Signof x  5

Signof x 2

Signof  x 2

When x 3,theleft­handsideisequalto0andtheinequalityissatisfied.Thus,the solutionsetis

x  x 5, x 3,or x  2.Interval:  5]   3  [2



.Graph: _52 _3

56. 4 x 2  x 2 9  0  4 x 2  x  3 x 3  0.Theexpressionontheleftofthe inequalitychangessignwhen x 3and x  0.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

.(Theendpoint0isincludedsincetheoriginalexpressionisallowedto be0.)Interval:[ 3 3].Graph: _33 57. x 3 4 x  0 

0.Theexpressionontheleftoftheinequalitychangessignwhere x  0,

Fromthetable,thesolutionsetis

when x 3, x  0,and x  3.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x 4  x

0.Theexpressionontheleftoftheinequality changessignwhere x  0,where x  1,andwhere x 1.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Sincethesearenotrealsolutions.Theexpression x 2  x  1doesnotchangessigns,sowemustchecktheintervalsinthe followingtable.

Fromthetable,thesolutionsetis 

61. x 3 x  1  0.Theexpressionontheleftoftheinequalitychangessignwhere x 1andwhere x  3.Thuswemustcheck theintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Signof x 

Signof x 3

Signof x 3 x 

x  x  1or x  3.Sincethedenominator cannotequal0wemusthave x 1.

Interval: 

1

Graph: _13

62. 2 x  6 x 2  0.Theexpressionontheleftoftheinequalitychangessignwhen x 3and x  2.Thuswemustcheckthe intervalsinthefollowingtable.

Fromthetable,thesolutionsetis x 3  x  2.Interval:  3

Signof2 x  6

Signof x 2

Signof 2 x  6 x 2  

Graph: _32

oftheinequalitychangessignwhen

.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

changessignwhen x 1and x 

Fromthetable,thesolutionsetis

5  0.Theexpressionontheleftoftheinequality changessignwhere x  16andwhere x  5.Thuswemustchecktheintervalsinthefollowingtable.

Signof x

Signof x 5

Signof x  16 x 5

Fromthetable,thesolutionsetis x  x  5or x  16.Sincethedenominator cannotequal0,wemusthave x  5.

Interval: 

Graph: 516

0.Theexpressionontheleftoftheinequalitychanges signwhen x  0and x  3.Thuswemustchecktheintervalsinthefollowingtable.

Sincethedenominatorcannotequal0,wemust have x  3.Thesolutionsetis x

Signof3 x

Signof2 x 

Signof 2 x 3 x 

Interval: [0 3

Graph: 03

table.

Fromthetable,thesolutionsetis

theleftoftheinequalitychangessignwhen

table.

2,where x

1,where x  0,andwhere x  1.Thuswemustchecktheintervalsinthefollowingtable.

Signof

Signof x

Since x 1and x  0yieldundefinedexpressions,wecannotincludetheminthesolution.From thetable,thesolution

Theexpressionontheleftoftheinequalitychangessignwhen

x  0,and x  1.

Thuswemustchecktheintervalsinthefollowingtable.

Signof2 x

Signof2

Signof x

Signof x

Signof

Since x  0and x  1giveundefinedexpressions,wecannotincludetheminthesolution.Fromthetable,thesolutionset is x 2  x  0or1  x  2.Interval: [

1 2].Graph: _22 01

Fromthetable,thesolutionsetis

expressionontheleftoftheinequalitychangessignwhen

2.Thuswemustchecktheintervals inthefollowingtable.

Signof

Fromthetable,thesolutionsetis

1are excludedfromthesolutionbecausetheexpressionisundefinedatthosevalues.Interval:

Graph: _44

Graph: _22

Graph:

Graph: _20

Graph: _3_1

Sincetheabsolutevalueisalwaysnonnegative,the inequalityistrueforallrealnumbers.Inintervalnotation, thisis

Graph:

Graph: 28

.Theexpressionontheleftoftheinequalitychangessign when

Fromthetable,thesolutionsetis

expressionontheleftoftheinequalitychangessignwhere

mustchecktheintervalsinthefollowingtable.

Signof

Fromthetable,thesolutionsetis

1areexcludedfromthe solutionsetbecausetheymakethedenominatorzero.Interval:

_23 01

x

 0.Theexpressionontheleftoftheinequalitychangessignwhen x 3and x 2.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Graph: _3_2

3,sotheexpressionontheleftoftheoriginalinequalitychanges signonlywhen x 1.Wechecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis x  x  1

(Theendpoint 3isalreadyexcluded.)

Interval:  1

Graph: _1

changessignwhen x

Fromthetable,thesolutionsetis

109. For  x 2 9tobedefinedasarealnumberwemusthave x 2 9  0   x  3 x 3  0.Theexpressioninthe

inequalitychangessignat x 3and x  3. Interval

Signof x  3

Signof x 3

Thus x 3or x  3.

110. For  x 2 5 x 50tobedefinedasarealnumber wemusthave x

0.The expressionontheleftoftheinequalitychangessignwhen x 5and x  10.Thuswemustchecktheintervalsinthe followingtable.

Signof x  5

Signof x 10

Thus x 5or x  10.

111. For

expressioninthelastinequalitychangessignat x 2and x  5.

0.The

Thus x  2or x  5,andthesolutionsetis

112. For 4  1 x 2  x tobedefinedasarealnumberwemusthave 1 x 2  x  0 Theexpressionontheleftoftheinequalitychanges signwhen x  1and x 2.Thuswemustchecktheintervalsinthefollowingtable.

Signof1 x  

Signof2  x  

Signof 1 x 2  x 

Thus 2  x  1andthesolutionsetis  2 1].Notethat x 2hasbeenexcludedfromthesolutionsetbecausethe expressionisundefinedatthatvalue.

118. Insertingtherelationship

119. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x betheaveragenumberofkilometersdrivenperday.EachdaythecostofPlan Ais95  0 40 x ,andthecostofPlanB is135.PlanBsavesmoneywhen135  95 

x .SoPlanBsavesmoneywhenyouaverage morethan100kilometersaday.

120. Let m bethenumberofminutesofinternationalcallsplacedpermonth.ThenunderPlanA,thecostwillbe25  0 05m ,and underPlanB,thecostwillbe5  012m .TodeterminewhenPlanBisadvantageous,wemustsolve25

12

 20  0 07m  285 7  m .SoPlanBisadvantageousifapersonplacesfewerthan286minutesofinternationalcalls permonth.

121. Weneedtosolve6400

12,000  m  14,000.Sheplansondrivingbetween12,000and14,000miles.

122.(a) T  20 h 100 ,where T isthetemperaturein  C,and h istheheightinmeters. (b) Solvingtheexpressioninpart(a)for h ,weget h  100 20 T .So0

30.Thustherangeoftemperatureisfrom20 Cdownto 30 C.

123.(a) Let x bethenumberof$3increases.Thenthenumberofseatssoldis120 x .So P  200

.Substitutingfor x wehavethatthenumberofseatssoldis

(b)

290  P  215.Puttingthisintostandardorder,wehave215  P  290.Sotheticketpricesarebetween$215and $290.

124. Ifthecustomerbuys x kilogramsofcoffeeat$6 50perkilograms,thentheircost c willbe6 50 x .Thus x  c 6 5 .Since thescale’saccuracyis 0 03kg,andthescaleshows3kg,wehave3 0

695.Sincethecustomerpaid$19 50,hecouldhavebeenover­or underchargedbyasmuchas195cents.

125. 00004  4,000,000 d 2  001.Since d 2  0and d  0,wecanmultiplyeachexpressionby d 2 toobtain

00004d 2  4,000,000  001d 2 .Solvingeachpair,wehave00004d 2  4,000,000  d 2  10,000,000,000  d  100,000(recallthat d representsdistance,soitisalwaysnonnegative).Solving4,000,000  0 01d 2  400,000,000  d 2  20,000  d .Puttingthesetogether,wehave20,000  d  100,000.

126. 600,000 x 2  300  500  600,000  500  x 2  300 (Notethat x 2  300  300  0,sowecanmultiplybothsidesbythe denominatorandnotworrythatwemightbemultiplyingbothsidesbyanegativenumberorbyzero.)1200  x 2  300 

0  x 2 900  0   x 30 x  30.Theexpressionintheinequalitychangessignat x  30and x 30.However, since x representsdistance,wemusthave x  0.

Signof x 30 

Signof x  30

Signof  x 30 x  30

So x  30andyoumuststandatleast30metersfromthecenterofthefire.

127. Note:Inthefirstprintingofthetext,theanswergivenforpart(a)ofthisexerciseisincorrect. (a) Wesubstitute d  402mand t  10sintothegivenformulaandfindthevalueof a thatresultsinaquarter­miletime ofexactly10s:402  1 2 a 102  402  50a  a  402 50  8 04ms2 .Thus,thequarter­miletimewillbelessthan 10sif a  8 04ms2

(b) Wesubstitute a 

andsolvefor

thatthequarter­miletimefora(downward)quarter­mileunderEarth’sgravityis

Solve13

Thus,theconditionissatisfiedwhenthecarisdrivenbetweenroughly33and67km/h.

and

70.However,since  representsthespeed,wemusthave   0.

Soyoumustdrivebetween0and35km/h.

130. Solve2400



0 0025 x 1 x 4400  0.Theexpressionontheleftoftheinequalitychangessignwhen x  400and x  4400. Sincethemanufacturercanonlysellpositiveunits,wechecktheintervalsinthefollowingtable.

Signof0 0025 x 1

Signof x 4400

Signof 0 0025 x 1 x 4400

Sothemanufacturermustsellbetween400and4400unitstoenjoyaprofitofatleast$2400.

131. Let x bethelengthofthegardenand  itswidth.Usingthefactthattheperimeteris120m,wemusthave2 x  2  120

   60 x .Nowsincetheareamustbeatleast800m2 ,wehave800 

x 2

x

0.Theexpressionintheinequalitychangessignat x  20and x  40.

However,since x representslength,wemusthave x  0. Interval

Signof

Thelengthofthegardenshouldbebetween20and40meters.

132.(a) Let x bethethicknessofthelaminate.Then

 0 0075. (b)

x 005

h 170

0075

0075

arebetween156cmand185cm.

134. Case1:a  b  0Wehave a  a

isodd.

Case2: 0  a  b Wehave a a  a b,since a  0,and b a  b b ,since b  0.So a 2  a b  b2 .Thus0  a  b  a 2  b2 .Likewise, a

,forallpositive

integers n .

Case3:a  0  b If n isodd,then a n  bn ,because a n isnegativeand b n ispositive.If n iseven,thenwecouldhave either a n  b n or a n  b n .Forexample, 1  2and  12  22 ,but 3  2and

135. Therulewewanttoapplyhereis“a  b  ac  bc if c  0and a  b  ac  bc if c  0”.Thuswecannotsimply multiplyby x ,sincewedon’tyetknowif x ispositiveornegative,soinsolving1  3 x ,wemustconsidertwocases.

Case1:x  0Multiplyingbothsidesby x ,wehave x  3.Togetherwithourinitialcondition,wehave0  x  3.

Case2:x  0Multiplyingbothsidesby x ,wehave x  3.But x  0and x  3havenoelementsincommon,sothis givesnoadditionalsolution.Hence,theonlysolutionsare0  x  3.

136.  x 1 isthedistancebetween x and1;  x 3 isthedistancebetween x and3.So  x 1   x 3 representsthose pointscloserto1thanto3,andthesolutionis x  2,since2isthepointhalfwaybetween1and3.If a  b ,thenthe solutionto  x a    x b  is x  a  b 2

137. a  b ,sobyRule1, a  c  b  c.UsingRule1again, b  c  b  d ,andsobytransitivity, a  c  b  d

138. a b  c d ,sobyRule3, d a b  d c d  ad b  c .Adding a tobothsides,wehave ad b  a  c  a .Rewritingtheleft­hand

sideas ad b  ab b  a b  d  b anddividingbothsidesby b  d gives a b  a  c b

Similarly, a  c  cb d  c  c

d ,so a  c b  d  c d

139.(a) Because x isnonnegative, x  y  x 2  xy ,andbecause y isnonnegative, x  y  xy  y 2 .Thus,bytransitivity, x 2  y 2 .

(b) Bypart(a),  xy  x

2 4 .Expanding,thisbecomes4 xy

 x 2  y 2 2 xy  0   x y 

0.Thisistrueforany x and y ,sotheoriginalinequalityistrueforallnonnegative x and y

1.9 THECOORDINATEPLANE;GRAPHSOFEQUATIONS;CIRCLES

1.(a) Thepointthatis3unitstotherightofthe y ­axisand5unitsbelowthe x ­axishascoordinates 3 5 (b) Thepoint 2 7 is2unitstotherightofthe y ­axisand7unitsabovethe x ­axis,soitisclosertothe y ­axis.

2. Thedistancebetweenthepoints a  b  and c  d  is c a 2

3. Thepointmidwaybetween a  b and c 

.Sothedistancebetween

.Sothepointmidwaybetween

and

and

is

4. Ifthepoint 2 3 isonthegraphofanequationin x and y ,thentheequationissatisfiedwhenwereplace x by2and y by3. Wecheckwhether2 3

3.Thisisfalse,sothepoint

Tocompletethetable,weexpress y intermsof x :2

5.(a) Tofindthe x ­intercept(s)ofthegraphofanequationweset y equalto0intheequationandsolvefor x :2 0  x  1  x 1,sothe x ­interceptof2 y  x  1is 1.

(b) Tofindthe y ­intercept(s)ofthegraphofanequationweset x equalto0intheequationandsolvefor y

y  1 2 ,sothe y ­interceptof2 y  x  1is 1 2

6.(a) Thegraphoftheequation  x 12   y 22  9isacirclewithcenter 1 2 andradius 9  3.

(b) Inordertojusttouchthe y ­axis,acirclecenteredat 3

touchesthe y ­axisatthepoint 0 4

musthaveradius3andequation

7.(a) Ifagraphissymmetricwithrespecttothe x ­axisand a  b  isonthegraph,then a  b isalsoonthegraph.

(b) Ifagraphissymmetricwithrespecttothe y ­axisand a  b  isonthegraph,then  a  b isalsoonthegraph.

(c) Ifagraphissymmetricabouttheoriginand a  b  isonthegraph,then  a  b  isalsoonthegraph.

8.(a) The x ­interceptsare 5and3,andthe y ­interceptsare 2.

(b) Thegraphissymmetricaboutthe x ­axis.

9. Yes.If a  b  isonthegraph,thenbysymmetryaboutthe x ­axis,thepoint a  b  isonthegraph.Thenbysymmetry aboutthe y ­axis,thepoint  a  b isonthegraph.Thus,thegraphissymmetricwithrespecttotheorigin.

10. No,thegraphisnotnecessarilysymmetricwithrespecttoeitheraxis.For example,thegraphof y  x 3 issymmetricwith respecttotheorigin,butnotwithrespecttoeitheraxis.

11. Thepointshavecoordinates

12. A and B lieinQuadrantIand E and G lieinQuadrantIII.

21. Thetwopointsare

22. Thetwopointsare

23. Thetwopointsare

24. Thetwopointsare

(b)

27.(a)

Midpoint:

Midpoint:

32. Theareaofaparallelogramisitsbasetimesitsheight. Sincetwosidesareparalleltothe x ­axis,weusethelength ofoneoftheseasthebase.Thus,thebaseis

4.The heightisthechangeinthe y coordinates,thus,theheight is6 2  4.Sotheareaoftheparallelogramis base  height  4  4  16. y 1x 1

33. Fromthegraph,thequadrilateral ABCD hasapairof parallelsides,so ABCD isatrapezoid.Theareais  b1  b2 2  h .Fromthegraphweseethat b

h isthedifferencein y ­coordinatesis

34. Thepoint S mustbelocatedat 0 4.Tofindthearea, wefindthelengthofonesideandsquareit.Thisgives

38.(a)

theorigin.

(b) Thedistancefrom

totheoriginis

39. Sincewedonotknowwhichpairareisosceles,wefindthelengthofallthreesides.

40. Sincetheside AB isparalleltothe x ­axis,weusethisasthebaseintheformulaarea  1 2 base  height.Theheightisthe changeinthe y ­coordinates.Thus,thebaseis  2 4

 6andtheheightis

41.(a) Herewehave

(b) Theareaofthetriangleis

,weconcludethatthetriangleisarighttriangle.Theareais

43. Weshowthatallsidesarethesamelength(itsarhombus)andthenshowthatthediagonalsareequal.Herewehave

therhombusisasquare.

andthepointsarecollinear.

45. Let P  

y  besuchapoint.Settingthedistancesequalweget

.Check:

46. Themidpointof

47. AsindicatedbyExample3,wemustfindapoint

suchthatthemidpoints of PR andof QS arethesame.Thus

Settingthe y ­coordinatesequal,weget

48. Wesolvetheequation6 

tofindthe x coordinateof B .Thisgives6

(b) Themidpointof

(c) Sincethetheyhavethesamemidpoint,weconcludethatthe diagonalsbisecteachother.

areallpointsonthegraphofthisequation.

3,sothe x ­interceptis3,and x  0 

2

6,sothe y ­interceptis 6. x ­axissymmetry:2 x  y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe x ­axis.

y ­axissymmetry: 2 x y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe y ­axis.

Originsymmetry: 2 x  y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe origin.

x

1,sothe x ­interceptis1, and x  0  y  2

2,sothe y ­interceptis2.

x ­axissymmetry: y  2  x 12 ,whichisnotthesameas

y  2  x 12 ,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y  2  x 12 ,whichisnotthesameas y  2  x 12 ,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: y  2  x 12  y 2  x  12 , whichisnotthesameas y  2  x 12 ,sothegraphisnot symmetricwithrespecttotheorigin. 1 1 y x

62.(a) Solvefor y : x 4 y  8  y  1 4 x 2. x y 2 5 2 0 2 2 3 2 4

x ­interceptis8,and

2,sothe y ­interceptis 2.

x ­axissymmetry: y  2 x 6,whichisnotthesameas y  2 x 6,sothegraphisnotsymmetricwithrespecttothe x ­axis.

y ­axissymmetry: y  2  x  6  y 2 x 6,whichis notthesameas y  2 x 6,sothegraphisnotsymmetric withrespecttothe y ­axis.

Originsymmetry: y

2 x  6,whichis notthesameas y  2 x 6,sothegraphisnotsymmetric withrespecttotheorigin. y 1x

y

2,sothe x ­interceptsare

2,and x  0

y

4  4,sothe y ­interceptis4. x ­axissymmetry: y  x 2  4,whichisnotthesameas y  x 2  4,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y 

x 2  4,sothegraph issymmetricwithrespecttothe y ­axis.

Originsymmetry: y  x 2  4,whichisnotthesameas

y  x 2  4,sothegraphisnotsymmetricwithrespectto theorigin. y 1x 1l 0

y  0  0   x  2,sothereisno x ­intercept,and x  0

 y  0  2  2,sothe y ­interceptis2.

x ­axissymmetry: y   x  2,whichisnotthesameas

y   x  2,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y   x  2,whichisnotthesameas

y   x  2,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: y   x  2  y  x 2, whichisnotthesameas y   x  2,sothegraphisnot symmetricwithrespecttotheorigin.

64.(a) y   x 4

x

 0  x  0,sothe x ­interceptis0,and x  0

y

0

 0,sothe y ­interceptis0.

x ­axissymmetry: y 

x   y   x ,whichisnotthe sameas y 

x ,sothegraphisnotsymmetricwith respecttothe x ­axis.

y ­axissymmetry: y 

x

x ,sothegraphis symmetricwithrespecttothe y ­axis.

Originsymmetry: y 

x ,whichisnotthe sameas y   x ,sothegraphisnotsymmetricwith respecttotheorigin. y 5x 5 0

4,sothe x ­interceptis4,and x  0  y  0 4,sothereisno y ­intercept.

x ­axissymmetry: y   x 4,whichisnotthesameas

y   x 4,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y   x 4,whichisnotthesameas

y   x 4,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: y   x 4  y  x 4, whichisnotthesameas y   x 4,sothegraphisnot symmetricwithrespecttotheorigin. y 1l 01x (b) x   y .Here y isnotafunctionof x

y  0  x  0  0,sothe x ­interceptis0,and x  0 

 y   0  y  0,sothe y ­interceptis0.

x ­axissymmetry: x   y    y ,whichisthesameas

x   y ,sothegraphissymmetricwithrespecttothe x ­axis. y ­axissymmetry: x   y ,whichisnotthesame,sothe graphisnotsymmetricwithrespecttothe y ­axis.

Originsymmetry: x   y    y ,whichisnotthesame, sothegraphisnotsymmetricwithrespecttotheorigin.

65.(a) y  4 x 2 x y 2 0 1 3 0 2 1 3 2 0

y  0  4 x 2  0  x 2,sothe x ­interceptsare 2,and x  0  y  4 02  2,sothe y ­interceptis2.

x ­axissymmetry: y  4 x 2 ,whichisnotthesameas y  4 x 2 ,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y  4  x 2  4 x 2 ,sothegraph issymmetricwithrespecttothe y ­axis.

Originsymmetry: y

y 4 x 2 ,whichisnotthesameas y  4 x 2 ,so thegraphisnotsymmetricwithrespecttotheorigin.

the x ­interceptsare0and 2,and x  0  y  03 4 0

 0,sothe y ­interceptis0. x ­axissymmetry:

isnotthesameas y  x 3 4 x ,sothegraphisnotsymmetric withrespecttothe x ­axis.

y ­axissymmetry: y   x 3 4  x   x 3 

isnotthesameas x  x 3 4 x ,sothegraphisnotsymmetric withrespecttothe y ­axis.

y  x 3 4 x ,so thegraphissymmetricwithrespecttotheorigin.

66.(a) y 4 x 2 x y 2 0 1 3 0 2 1 3 2 0

2,sothe x ­interceptsare 2,and x 

y

4 2,sothe y ­interceptis 2. x ­axissymmetry: y 4 x 2  y  4 x 2 ,which isnotthesameas y 4 x 2 ,sothegraphisnot symmetricwithrespecttothe x ­axis. y ­axissymmetry: y 4  x 2 4 x 2 ,sothe graphissymmetricwithrespecttothe y ­axis.

Originsymmetry: y 4  x 2  y  4 x 2 , whichisnotthesameas y 4 x 2 ,sothegraphisnot symmetricwithrespecttotheorigin.

(b) x  y 3  y  3  x y x 8 2 4 3 4 1 1

0

3 4 8 2 y  0  x  03  0,sothe x ­interceptis0,and x  0  0  y 3  y  0,sothe y ­interceptis0.

x ­axissymmetry: x   y 3  y 3 ,whichisnotthesame as x  y 3 ,sothegraphisnotsymmetricwithrespecttothe x ­axis.

y ­axissymmetry: x  y 3  x  y 3 ,whichisnotthe sameas x  y 3 ,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: x   y 3  x  y 3 ,sothegraphis symmetricwithrespecttotheorigin. y 10x 1l 0

67.(a) Tofind x ­intercepts,set y  0.Thisgives0  x  6  x 6,sothe x ­interceptis 6.Tofind y ­intercepts,set x  0.Thisgives y  0  6  6,sothe y ­interceptis6. (b) Tofind x ­intercepts,set y  0.Thisgives0  x 2 5  x 2  5  x 5,sothe x ­interceptsare 5.Tofind y ­intercepts,set x  0.Thisgives y  02 5 5,sothe y ­interceptis 5.

68.(a) Tofind x ­intercepts,set y  0.Thisgives4 x 2  25 0

x 5,sothe x ­interceptsare 5.To find y ­intercepts,set x  0.Thisgives4 

2,sothe y ­interceptsare 2.

(b) Tofind x ­intercepts,set y  0.Thisgives x

x 1,sothe x ­interceptsare 1.To find y ­intercepts,set x  0.Thisgives02

69.(a) Tofind x ­intercepts,set y  0.Thisgives9

2,sothe x ­intercepts are 2.Tofind y ­intercepts,set x 

(b) Tofind x ­intercepts,set y  0.Thisgives0 2 x 

9,sothereisno y ­intercept.

,sothe x ­interceptis 1 4 .Tofind y ­intercepts, set x  0.Thisgives y 2 0 y  4 0  1,sothe y ­interceptis1.

70.(a) Tofind x ­intercepts,set y  0.Thisgives0   x 2 16  x 2  16  x 4,sothe x ­interceptsare 4.Tofind y ­intercepts,set x  0.Thisgives y  02 16,sothereisno y ­intercept. (b) Tofind x ­intercepts,set y  0.Thisgives0  64 x 3  x 3  64  x  4,sothe x ­interceptis4.Tofind y ­intercepts,set x  0.Thisgives y  64 03  y  8,sothe y ­interceptis8.

71. Tofind x ­intercepts,set y  0.Thisgives0

4,sothe x ­interceptsare0and 4.Tofind y ­intercepts,set x  0.Thisgives

0,sothe y ­interceptis0. 72. Tofind x ­intercepts,set y

y ­intercepts,set x

75. x 2  y 2  9hascenter 0 0 andradius3. y 1x 1l

82. Using

83. Theequationofacirclecenteredattheoriginis x 2 

84. Using h 1and k  5,weget

,wesolvefor r 2 .Thisgives

x 

.Usingthepoint

wesolvefor r 2 .Thisgives

.Next,usingthepoint

.Thus,anequationofthecircleis

85. Thecenterisatthemidpointofthelinesegment,whichis

86. Thecenterisatthemidpointofthelinesegment,whichis

87. Sincethecircleistangenttothe x ­axis,itmustcontainthepoint

0

,sotheradiusisthechangeinthe y ­coordinates. Thatis, r   3 0

3.Sotheequationofthecircleis

88. Sincethecirclewith r  5liesinthefirstquadrantandistangenttoboththe x ­axisandthe y ­axis,thecenterofthecircle isat

5

5.Therefore,theequationofthecircleis

89. Fromthefigure,thecenterofthecircleisat

Thustheequationofthecircleis

90. Fromthefigure,thecenterofthecircleisat

91. Completingthesquaregives

92.

95. Completingthesquaregives2

Thus,thecirclehascenter

96. Completingthesquaregives3

97. x ­axissymmetry: 

.Theradiusisthechangeinthe y ­coordinates,so

.Theradiusisthedistancefromthecentertothepoint

.Thus

,whichisnotthesameas y

x 2 ,sothegraphisnotsymmetric withrespecttothe x ­axis. y ­axissymmetry:

x

,sothegraphissymmetricwithrespecttothe y ­axis.

Originsymmetry:

,whichisnotthesameas y  x 4  x 2 ,sothegraphisnot symmetricwithrespecttotheorigin.

98. x ­axissymmetry: x   y 4  y

2  y 4 y 2 ,sothegraphissymmetricwithrespecttothe x ­axis. y ­axissymmetry:  x   y 4 y 2 ,whichisnotthesameas x  y 4 y 2 ,sothegraphisnotsymmetricwithrespecttothe y ­axis.

Originsymmetry:

,whichisnotthesameas x  y 4 y 2 ,sothegraphisnot symmetricwithrespecttotheorigin.

99. x ­axissymmetry:

2,sothegraphissymmetricwithrespecttothe x ­axis. y ­axissymmetry:

Originsymmetry:

y ­axis.

2,sothegraphissymmetricwithrespecttothe origin.

(Notethatifagraphissymmetricwithrespecttoeachcoordinateaxis,itissymmetricwithrespecttotheorigin.The converseisnottrue,asshowninthenextexercise.) 100. x ­axissymmetry:

1,sothegraphisnotsymmetric withrespecttothe x ­axis.

y ­axissymmetry:

 1,sothegraphisnotsymmetric withrespecttothe y ­axis.

Originsymmetry:

101. x ­axissymmetry:  y   x 3  10 x  y  x 3 10 x ,whichisnotthesameas y  x 3  10 x ,sothegraphisnot symmetricwithrespecttothe x ­axis.

y ­axissymmetry: y   x 3  10  x   y  x 3 10 x ,whichisnotthesameas y  x 3  10 x ,sothegraphisnot symmetricwithrespecttothe y ­axis.

Originsymmetry:  y    x 3  10  x   y  x 3 10 x  y  x 3  10 x ,sothegraphissymmetricwithrespect totheorigin.

102. x ­axissymmetry:  y   x 2   x   y  x 2  x ,whichisnotthesameas y  x 2 

x ,sothegraphisnotsymmetric withrespecttothe x ­axis. y ­axissymmetry: y   x

,sothegraphissymmetricwithrespecttothe y ­axis.Notethat

2

Originsymmetry:

x ,whichisnotthesameas y  x 2   x ,so thegraphisnotsymmetricwithrespecttotheorigin.

103. Symmetricwithrespecttothe y ­axis. x y 0

104. Symmetricwithrespecttothe x ­axis. x y 0

105. Symmetricwithrespecttotheorigin. x y 0

107.  x  y 

(andon)thecircle x

.Thisisthesetofpointsinside

106. Symmetricwithrespecttotheorigin. x y 0

.Thisisthesetofpointsoutside

109. Completingthesquaregives

,andthe radiusis4.Sothecircle x 2

y 2  4,withcenter

andradius2 sitscompletelyinsidethelargercircle.Thus, theareais  42  22  16 4

110. Thisisthetopquarterofthecircleofradius3.Thus,the areais 1

y x 3 3

111.(a) Thepoint 5 3

isshiftedto

(b) Thepoint a

b  isshiftedto

(c) Let  x  y  bethepointthatisshiftedto

y ­coordinatesequal,weget

isreflectedtothepoint

(b) Thepoint

(c) Sincethepoint  a

isthereflectionof

(d) A 

3

113.(a) Symmetricaboutthe x ­axis: x y 1 1 (b) Symmetricaboutthe y ­axis: x y 1 1 (c) Symmetricabouttheorigin: x y 1 1

114.(a) Fromthegraph,itappearsthattheclosestthesatellitegetstothecenter ofthemoonisapproximately5Mmand farthestapproximately70Mm.

Setting y  0intheequation(whichobviouslycorrespondstothedesireddistances), wehave

x 32

2.Thus,thesmallestandlargestdistances are52Mmand702Mm.

(b) Setting y  10,wehave

115.(a) d  A

95.Thedistancesofthesepointsfromthecenterofthemoonare

(b) Wewantthedistancesfrom C  4 2 to D  11 26.Thewalkingdistanceis

24  31blocks.Straight­linedistanceis

4 112  2 26

2  72  242  625  25blocks.

(c) Thetwopointsareonthesameavenueorthesamestreet.

116. Lettheareaofthetrianglebe A.Ashinted,wedrawarectangle circumscribingthetriangle.Thisrectangleis7 1  6unitslongand 5 1  4unitshigh,soitsareais A R  6 4  24.Theareasofthethree righttrianglesinsidethe rectanglebutoutsidetheoriginaltriangleare

A1  1 2 7 15 2 

A3  1 2 7 45 1  6.Sincetheareaoftherectangleisthesumof theareasoftherighttrianglesandtheoriginaltriangle,wehave

117. Completingthesquaregives

Whentheequationrepresentsacircle,thecenteris

118. Followingthehint,weconsideranequilateraltrianglewithsidelength1 anywhereintheplane.Eachofitsverticesis eitherredorblue,andeachis1unitawayfromtheothertwo.Soatleasttwohavethesamecolorandareexactly1unit apart.

1.10 LINES

1. Wefindthe“steepness”orslopeofalinepassingthroughtwopointsbydividingthedifferenceinthe y ­coordinatesofthese pointsbythedifferenceinthe x ­coordinates.Sothelinepassingthroughthepoints

2.(a) Thelinewithequation y  3 x  2hasslope3.

(b) Anylineparalleltothislinehasslope3.

(c) Anylineperpendiculartothislinehasslope 1 3

3. Thepoint­slopeformoftheequationofthelinewithslope3passingthroughthepoint

4. Forthelinearequation2 x  3 y 12  0,the x ­interceptis6andthe y ­interceptis4.Theequationinslope­interceptform is y  2 3 x  4.Theslopeofthegraphofthisequationis 2 3

5. Theslopeofahorizontallineis0.Theequationofthehorizontallinepassingthrough

6. Theslopeofaverticallineisundefined.Theequationoftheverticallinepassingthrough

7.(a) Yes,thegraphof y 3isahorizontalline3unitsbelowthe x ­axis.

(b) Yes,thegraphof x 3isaverticalline3unitstotheleftofthe y ­axis.

(c) No,alineperpendiculartoahorizontallineisverticalandhasundefinedslope.

(d) Yes,alineperpendiculartoaverticallineishorizontalandhasslope0.

8. y 5x 5 0 y=_3 x=_3 Yes,thegraphsof y 3and x 3areperpendicularlines.

thatlieontheline.Thustheslopeof

19. Firstwefindtwopoints

theequationofthelineis

20. Wefindtwopointsonthegraph,

equationofthelineis

21. Wechoosethetwointerceptsaspoints,

theequationofthelineis

22. Wechoosethetwointercepts,

.Sotheslopeis

25.

28. Usingtheequation y

29. Firstwefindtheslope,whichis

30. Firstwefindtheslope,whichis

31. Wearegiventwopoints,

32. Wearegiventwopoints,

y1  m  x x 1 ,weget y 7 

33. Wearegiventwopoints,

34. Wearegiventwopoints,

.Thus,theslopeis

35. Sincetheequationofalinewithslope0passingthrough a  b  is y  b ,theequationofthislineis y  3.

36. Sincetheequationofalinewithundefinedslopepassingthrough

37. Sincetheequationofalinewithundefinedslopepassingthrough

38. Sincetheequationofalinewithslope0passingthrough

a

b

a  b

a

b

is x  a ,theequationofthislineis x  3.

is x  a ,theequationofthislineis x  2.

is y  b ,theequationofthislineis y  1.

39. Anylineparallelto y  2 x  8hasslope2.Thedesiredlinepassesthrough

, weget y 4

40. Anylineperpendicularto y  1 2 x  7hasslope 1

,sosubstituting into y y1  m

x x 1

,weget y 2

41. Sincetheequationofahorizontallinepassingthrough a  b 

b ,theequationofthehorizontallinepassingthrough 4 5

is y  5.

42. Anylineparalleltothe y ­axishasundefinedslopeandanequationoftheform x  a .Sincethegraphofthelinepasses throughthepoint

,theequationofthelineis

43. Since3 x  2 y  4 

2,theslopeofthislineis 3 2 .Thus,thelineweseekisgivenby y

44. Since2

and b  6intotheslope­interceptformula,thelineweseekisgivenby y  2

45. Anylineparallelto x  5hasundefinedslopeandanequationoftheform x  a .Thus,anequationofthelineis x 1.

46. Anylineperpendicularto y  1hasundefinedslopeandanequationoftheform x  a .Sincethegraphofthelinepasses throughthepoint 2 6,anequationofthelineis x  2.

47. Firstfindtheslopeof3 x

.Thus,theslopeofanylineperpendicular to3 x  4 y  7  0is m

48. Firstfindtheslopeoftheline4

.Sotheslopeofthe linethatisperpendicularto4

49. Firstfindtheslopeofthelinepassingthrough

ofthelineweseekis

50. Firstfindtheslopeofthelinepassingthrough

,andsotheslopeoftheline thatisperpendicularis

51.(a) y 1x 1 (_2,l 1)

53.

b=6 b=3 b=_1 b=_3 b=_6 b=1 b=0 y 2 x  b , b  0, 1, 3, 6.Theyhavethesame slope,sotheyareparallel.

3 2 .Eachofthelines containsthepoint 0 3 becausethepoint 0 3 satisfieseachequation y  mx 3.Since 0 3 ison the y ­axis,theyallhavethesame y ­intercept.

,

eachequation

57. y  x 4.Sotheslopeis1andthe y ­interceptis 4. y 1x 1 58. y  1 2 x 1.Sotheslopeis 1 2 andthe y ­interceptis 1. y 1x 1

59. 2 x  y  7  y  2 x  7.Sotheslopeis2andthe y ­interceptis7. y 1x 1

2

x .Sotheslopeis 2 5 andthe y ­interceptis0. y 1x 1l 61. 4 x

theslopeis

theslopeis 3 4 andthe y ­interceptis 3. y 1x 1l

63. y  4canalsobeexpressedas y  0 x  4.Sotheslopeis 0andthe y ­interceptis4. y 1x 1l

64. x 5cannotbeexpressedintheform y  mx  b .So theslopeisundefined,andthereisno y ­intercept.Thisisa verticalline.

y 1x 1l

65. x  3cannotbeexpressedintheform y  mx  b .Sothe slopeisundefined,andthereisno y ­intercept.Thisisa verticalline. y 1x 1

66. y 2canalsobeexpressedas y  0 x 2.Sotheslope is0andthe y ­interceptis 2. y 1x 1l

67. 3 x 2 y 6  0.Tofind x ­intercepts,weset y  0and solvefor x :3 x 2 0 6  0  3 x  6  x  2,so the x ­interceptis2.

Tofind y ­intercepts,weset x  0andsolvefor y :

3 0 2 y 6  0  2 y 6  y 3,sothe y ­interceptis 3. y 1x 1

68. 6 x 7 y 42  0.Tofind x ­intercepts,weset y  0and solvefor x :6 x 7 0 42  0  6 x  42  x  7,so the x ­interceptis7.

Tofind y ­intercepts,weset x  0andsolvefor y :

6 0 7 y 42  0  7 y 42  y 6,sothe y ­interceptis 6. y 2x 2

69. 1 2 x 1 3 y  1  0.Tofind x ­intercepts,weset y  0and

solvefor x : 1 2 x 1 3 0  1  0  1 2 x 1  x 2, sothe x ­interceptis 2.

Tofind y ­intercepts,weset x  0andsolvefor y : 1 2 0 1 3 y  1  0  1 3 y  1  y  3,sothe y ­interceptis3. y 1x 1

70. 1 3 x 1 5 y 2  0.Tofind x ­intercepts,weset y  0and

solvefor x : 1 3 x 1 5 0 2  0  1 3 x  2  x  6,so the x ­interceptis6.

1

Tofind y ­intercepts,weset x  0andsolvefor y : 1 3

2

y 10,sothe y ­interceptis 10. y 2x 2

71. y  6 x  4.Tofind x ­intercepts,weset y  0andsolve

for x :0  6 x  4  6 x 4  x  2 3 ,sothe

x ­interceptis 2 3

Tofind y ­intercepts,weset x  0andsolvefor y : y  6 0  4  4,sothe y ­interceptis4. y 1x 1

72. y 4 x 10.Tofind x ­intercepts,weset y  0and solvefor x :0 4 x 10  4 x 10  x  5 2 ,so the x ­interceptis 5 2

Tofind y ­intercepts,weset x  0andsolvefor y : y 4 0 10 10,sothe y ­interceptis 10. y x 2 1

73. Todetermineifthelinesareparallel orperpendicular,wefindtheirslopes.Thelinewithequation y  2 x  3hasslope2. Thelinewithequation2 y 4 x 5 

5 2 alsohasslope2,andsothelinesareparallel.

74. Todetermineifthelinesareparallel orperpendicular,wefindtheirslopes.Thelinewithequation y  1 2 x  4hasslope 1 2

Thelinewithequation2 x  4 y  1  4 y 2 x 1  y  1 2 x 1 4 hasslope 1 2  1 12 ,andsothelinesareneither parallelnorperpendicular.

75. Todetermineifthelinesareparallelorperpendicular,wefindtheirslopes.Thelinewithequation2x 5 y  8  5 y  2 x 8  y

.Thelinewithequation10

4 y

1

y  5 2 x  1 4 has slope 5 2  1 25 ,andsothelinesareperpendicular.

x

76. Todetermineifthelinesareparallel orperpendicular,wefindtheirslopes.Thelinewithequation15 x 9 y  2  9 y  15 x 2  y  5 3 x 2 9 hasslope 5 3 .Thelinewithequation3 y 5 x  5  3 y  5 x  5  y  5 3 x  5 3 hasslope 5 3 ,andsothelinesareparallel.

77. Todetermineifthelinesareparallelorperpendicular,wefindtheirslopes.Thelinewithequation7x 3 y  2  3 y  7 x 2  y  7 3 x 2 3 hasslope 7 3 .Thelinewithequation9 y  21

slope 7 3  1 73 ,andsothelinesareneitherparallelnorperpendicular.

78. Todetermineifthelinesareparallelorperpendicular,wefindtheirslopes.Thelinewithequation6 y 2 x  5  6 y  2 x  5

.Thelinewithequation2

slope 3  1 13 ,andsothelinesareperpendicular.

79. Wefirstplotthepointstofindthepairsofpointsthatdetermineeachside.Nextwe findtheslopesofoppositesides.Theslopeof AB is 4 1 7 1  3 6  1 2 ,andthe slopeof DC is 10 7 5  1

 3 6  1 2 .Sincetheseslopeareequal,thesetwosides areparallel.Theslopeof AD is 7 1 1 1  6 2 3,andtheslopeof BC is 10 4 5 7  6 2 3.Sincetheseslopeareequal, thesetwosidesareparallel. Hence ABCD isaparallelogram.

80. Wefirstplotthepointstodeterminetheperpendicularsides.Nextfindtheslopesof thesides.Theslopeof AB

,andtheslopeof AC is 8  1 9  3  9 6  3

thesidesareperpendicular, and ABC isarighttriangle.

81. Wefirstplotthepointstofindthepairsofpointsthatdetermineeachside.Nextwe findtheslopesofoppositesides.Theslopeof AB

andthe slopeof DC is 6

.Sincetheseslopeareequal,thesetwosides areparallel.Slopeof AD is 6 1 0 1  5 1 5,andtheslopeof BC is 3 8 11 10  5 1 5.Sincetheseslopeareequal,thesetwosidesareparallel. Since slopeof AB   slopeof AD   1 5   5 1,thefirsttwosidesare eachperpendiculartothesecondtwosides.Sothesidesformarectangle.

82.(a) Theslopeofthelinepassingthrough 1 1 and 3 9 is 9 1 3 1  8 2  4.Theslopeofthelinepassingthrough 1 1 and 6 21 is 21 1 6 1  20 5  4.Sincetheslopesareequal,thepointsarecollinear.

(b) Theslopeofthelinepassingthrough

2.Theslopeofthelinepassingthrough

.Sincetheslopesarenotequal,thepointsarenotcollinear.

83. Weneedtheslopeandthemidpointoftheline AB .Themidpointof AB

1.Theslopeoftheperpendicularbisectorwillhaveslope

,andtheslopeof

1.Usingthe point­slopeform,theequationof theperpendicularbisectoris

84. Wefindtheintercepts(thelengthofthesides).When x  0,wehave2 y

3,andwhen

85.(a) Westartwiththetwopoints a  0 and 0 b .Theslopeofthelinethatcontainsthemis b 0 0 a  b a .Sotheequation ofthelinecontainingthemis y  b a x  b (usingtheslope­interceptform).Dividingby b (since b  0)gives y b  x a  1  x a  y b

1. (b) Setting a  6and b 8,weget x 6  y 8 

86.(a) Thelinetangentat 3 4 willbeperpendiculartothelinepassingthroughthepoints

.Theslopeof thislineis

.Thus,theslopeofthetangentlinewillbe

.Thentheequationofthetangent

(b) Sincediametricallyoppositepointsonthecirclehaveparalleltangentlines,theotherpointis

87. Usingthediagramprovided,wecalculatethecoordinatesofthemidpoints: m 1 hascoordinates

Thelinejoiningthemidpointsthushasslope

0,andsoitisparalleltothethirdside(asegmentofthe x ­axis).

Thelinejoiningthemidpointshaslength

,halfthelengthofthethird sideofthetriangle;andsowehaveverifiedbothconclusionsofthetheorem

88.(a) Thesloperepresentstheincreaseintheaveragesurfacetemperaturein  Cperyear.The T ­interceptistheaverage surfacetemperaturein1950,or15 C.

(b) In2050, t  2050 1950  100,so T  0 02

17degreesCelsius.

89.(a) Theslopeis0 0417 D  0 0417 200  8 34.Itrepresentstheincreaseindosageforeachone­yearincreaseinthe child’sage.

(b) When a  0, c  8 34 0  1  8 34mg.

90.(a) y 20x 100l 200l 406080100

91.(a) y x

l 10,000l 5001000

92.(a)

(b) Theslope, 4,representsthedeclineinnumberofspacessoldfor each$1increaseinrent.The y ­interceptisthenumberofspacesat thefleamarket,200,andthe x ­interceptisthecostperspacewhenthe managerrentsnospaces,$50.

(b) Theslopeisthecostpertoasteroven,$6.The y ­intercept,$3000,is themonthlyfixedcost—thecostthatisincurrednomatterhowmany toasterovensareproduced.

(b) Substituting a forbothFandC,wehave

.Thusbothscalesagreeat

93. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

(a) Using n inplaceof x and t inplaceof y ,wefindthattheslopeis

equationis

(b) When n  150,thetemperatureisapproximatelygivenby

94.(a) Using t inplaceof x and V inplaceof y ,wefindtheslopeoftheline usingthepoints

4000

and

4

.Thus,theslopeis m

200 4000 4 0  3800

950.Usingthe V ­intercept,the linearequationis V 950t  4000.

(c) Thesloperepresentsadecreaseinthevalueofthecomputerof$950 eachyear.The V ­interceptistheoriginalpriceofthecomputer.

(d) When t  3,thevalueofthecomputerisgivenby V 950 3  4000  1150. (b) y 2x

95. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

(a) Wearegiven changeinpressure 3­meterchangeindepth  29 9 3  997.Using P for pressureand d fordepth,andusingthepoint P  103when d  0, wehave P 103 

103.

(c) Thesloperepresentstheincreaseinpressurepermeterofdescent. The y ­interceptrepresentsthepressureatthesurface.

(d) When P  690,wehave690 

d  58 9m.Thusthepressureis690kPaatadepthofapproximately 59m. (b) y x

96. Slopeistherateofchangeofonevariableperunitchangeinanothervariable.Soiftheslopeispositive,thenthe temperatureisrising.Likewise,iftheslopeisnegativethenthetemperatureisdecreasing.Iftheslopeis0,thenthe temperatureisnotchanging.

97. Welabelthethreepoints A, B ,and C .Iftheslopeofthelinesegment AB isequaltotheslopeofthelinesegment BC , thenthepoints A, B ,and C arecollinear.Usingthedistanceformula,wefindthedistancebetween A and B ,between B and C ,andbetween A and C .Ifthesumofthetwosmallerdistancesequalsthelargestdistance,thepoints A, B ,and C are collinear.

Anothermethod: Findanequationforthelinethrough A and B .Thencheckif C satisfiestheequation.Ifso,thepointsare collinear.

1.11

SOLVINGEQUATIONSANDINEQUALITIESGRAPHICALLY

1. Thesolutionsoftheequation x 2 2 x 3  0arethe x ­interceptsofthegraphof y  x 2 2 x 3.

2. Thesolutionsoftheinequality x 2 2 x 3  0arethe x ­coordinatesofthepointsonthegraphof y  x 2 2 x 3thatlie above the x ­axis.

3.(a) Fromthegraph,itappearsthatthegraphof y  x 4 3 x 3 x 2  3 x has x ­intercepts 1,0,1,and3,sothesolutions totheequation x 4 3 x 3 x 2  3 x  0are x 1, x  0, x  1,and x  3.

(b) Fromthegraph,weseethatwhere 1  x  0or1  x  3,thegraphliesbelowthe x ­axis.Thus,theinequality x 4 3 x 3 x 2  3 x  0issatisfiedfor x 1  x  0or1  x  3  [ 1 0]  [1 3]

4.(a) Thegraphsof y  5 x x 2 and y  4intersectat x  1andat x  4,sotheequation5 x x 2  4hassolutions x  1 and x  4.

(b) Thegraphof y  5 x x 2 liesstrictlyabovethegraphof y  4when1  x  4,sotheinequality5x x 2  4is satisfiedforthosevaluesof x ,thatis,for x  1  x  4  1 4

5.(a)

y  x 3 x 2 ; [ 2

2] by [ 1

1] (b) x ­intercepts0,1; y ­intercept0. y  x 3 x 2  x 2  x 1,so y  0  x  0or x  1and x  0  y  0. (c) Thegraphisnotsymmetricwithrespecttoeitheraxisortheorigin.

6.(a)

7.(a)

1 1 2 3

2 y  x 4 2 x 3 ;[ 2 3]by[ 3 3] (b) x ­intercepts0,2; y ­intercept0. y  x 4 2 x

(c) Thegraphisnotsymmetricwithrespecttoeitheraxisortheorigin.

2 2 4

(c) Thegraphissymmetricwithrespecttothe y ­axis:

(b) x ­intercepts 1, y ­intercept1. y

(c) Thegraphissymmetricwithrespecttothe y ­axis:

9. Althoughthegraphsof y 3 x 2  6 x 1 2 and y  7 7 12 x 2 appeartointersectintheviewing rectangle [ 4 4] by [ 1 3],thereisnopointof intersection.Youcanverifythisbyzoomingin.

10. Althoughthegraphsof y  49 x 2 and y  1 5 41 3 x  appeartointersectintheviewing rectangle [ 8 8] by [ 1 8],thereisnopointof intersection.Youcanverifythisbyzoomingin. ­8 ­6 ­4 ­2 2 4 6 8 2 4 6 8

11. Thegraphsof y  6 4 x x 2 and y  3 x  18appearto havetwopointsofintersectionintheviewingrectangle [ 6 2] by [ 5 20].Youcanverifythat x 4and x 3areexactsolutions.

12. Thegraphsof y  x 3 4 x and y  x  5appeartohave onepointofintersectionintheviewingrectangle[ 4 4] by [ 15 15].Thesolutionis x  2 627. ­4 ­2 2 4 ­10 10

13. Algebraically:3 x  2  5 x 4  6  2 x  x  3.

Graphically:Wegraphthetwoequations y1  3 x  2and y2  5 x 4intheviewingrectangle [1 4] by [ 1 13] Zoomingin,weseethatthesolutionis x  3.

14. Algebraically: 2 3 x  11  1 x  5 3 x 10  x 6.

Graphically:Wegraphthetwoequations y1  2 3 x  11 and y2  1 x intheviewingrectangle [ 12 2] by [ 2 8].Zoomingin,weseethatthesolutionis x 6.

Algebraically:

and y2  7intheviewingrectangle

Zoomingin,weseethatthesolutionis

Graphically:Wegraphthetwoequations

rectangle[ 5 5]by[ 10 10].Zoomingin,weseethat thereisonlyonesolutionat x 4.

17. Algebraically:4 x 2 8  0  x 2  2  x 2.

Graphically:Wegraphtheequation y1  4 x 2 8and determinewherethiscurveintersectsthe x ­axis.Weuse theviewingrectangle [ 2 2] by [ 4 4].Zoomingin,we seethatsolutionsare x  1 41and x 1 41. ­2 ­1 1 2 ­4 ­2 2 4

18. Algebraically: x 3  10 x 2  0  x 2  x  10  0  x  0or x 10.

Graphically:Wegraphtheequation y  x 3  10 x 2 and determinewherethiscurveintersectsthe x ­axis.Weuse theviewingrectangle [ 12 2] by [ 5 5].Weseethatthe solutionsare x  0and x 10. ­12 ­10 ­8 ­6 ­4 ­2

19. Algebraically: x 2  9  0  x 2 9,whichhasnoreal solution.

Graphically:Wegraphtheequation y  x 2  9andsee thatthiscurvedoesnotintersectthe x ­axis.Weusethe viewingrectangle [ 5 5] by [ 5 30]

20. Algebraically: x 2  3  2 x 

Becausethediscriminantisnegative,thereisnoreal solution.

Graphically:Wegraphthetwoequations y1  x 2  3and y2  2 x intheviewingrectangle [ 4 6] by [ 6 12],and seethatthetwocurvesdonotintersect.

2 2 4 6 ­5 5 10

21. Algebraically:81 x 4 

Graphically:Wegraphthetwoequations y1  81 x 4 and y2  256intheviewingrectangle[ 2 2]by[250 260].

Zoomingin,weseethatsolutionsare x 133.

22. Algebraically:2

Graphically:Wegraphtheequation y  2 x 5 243and determinewherethiscurveintersectsthe x ­axis.Weuse theviewingrectangle [ 5 10] by [ 5 5].

Zoomingin,weseethatthesolutionis x  2 61.

23. Algebraically:  x 54 80  0   x 54  80  x 5  4 80 2 4 5  x  5

Graphically:Wegraphtheequation y1   x 54 80 anddeterminewherethiscurveintersectsthe x ­axis.We usetheviewingrectangle [ 1 9] by [ 5 5].Zoomingin, weseethatsolutionsare x  2 01and x  7 99. 2 4 6 8 ­4 ­2 2 4

25. Wegraph y  x 2 11 x  30intheviewingrectangle [2 8] by [ 0 1 0 1].Thesolutionsappeartobeexactly

x  5and x  6.[Infact

x 2 11 x  30   x 5 x 6.] 4 6 8

0.1

0.1 0.0

27. Wegraph y  x 3 6 x 2  11 x 6intheviewing

rectangle[ 1 4]by[ 0 1 0 1].Thesolutionsare

x  1 00, x  2 00,and x  3 00. ­1 1 2 3 4 ­0.1 0.1

24. Algebraically:3  x

Graphically:Wegraphthetwoequations y

3 and y2  72intheviewingrectangle[ 3 1]by [65 78].Zoomingin,weseethatthesolutionis x 2 12

26. Wegraph y  x 2 075 x  0125intheviewing

rectangle [ 2 2] by [ 0 1 0 1].Thesolutionsare

x  0 25and x  0 50. ­2 ­1 1 2

0.1 0.1

28. Since16 x 3  16 x 2  x  1  16 x 3  16 x 2 x 1  0, wegraph y  16 x 3  16 x 2 x 1intheviewing

rectangle[ 2 2]by[ 0 1 0 1].Thesolutionsare:

x 100, x 025,and x  025. ­2 ­1 1 2

29. Wefirstgraph y  x  x  1intheviewingrectangle [ 1 5] by [ 0 1 0 1] and findthatthesolutionisnear1 6.Zoomingin,weseethatsolutionsis x  1 62. ­1 1 2 3 4 5 ­0.1 0.1

30. 1   x  1  x 2  1   x 1  x 2  0 Since  x isonlydefined for x  0,westartwiththeviewingrectangle

[ 1 5] by [ 1 1].Inthisrectangle,there appearstobeanexactsolutionat x  0and anothersolutionbetween x  2and x  2 5.We thenusetheviewingrectangle[2 3 2 35]by [ 0 01 0 01],andisolatethesecondsolutionas x  2 314.Thusthesolutionsare x  0and x  231.

31. Wegraph y  x 13 x intheviewingrectangle [ 3 3] by [ 1 1].Thesolutions are x 1, x  0,and x  1,ascanbeverifiedbysubstitution.

32. Since x 12 isdefinedonlyfor x  0,westartby graphing y  x 12  x 13 x intheviewing rectangle [ 1 5] by [ 1 1] Weseeasolutionat x  0andanotheronebetween x  3and x  35.Wethenusetheviewingrectangle

[3 3 3 4] by [ 0 01 0 01],andisolatethesecond solutionas x  3 31.Thus,thesolutionsare x  0and x  3 31.

33. Wegraph y  2 x 1and y  3 x 5intheviewingrectangle [0 9] by [0 5] andseethattheonlysolutiontotheequation 2 x 1  3 x 5is x  4,which canbeverifiedbysubstitution.

34. Wegraph y  3  x and y   x 2  1intheviewingrectangle [ 4 4] by [0 4] andseethattheequation 3  x   x 2  1hassolutions x 1and x  2, whichcanbeverifiedbysubstitution.

35. Wegraph y  2 x  1  1and y  x intheviewingrectangle [ 1 6] by [0 6] andseethattheonlysolutiontotheequation 2 x  1  1  x is x  4,whichcan beverifiedbysubstitution.

36. Wegraph y  2 x   x  1and y  8intheviewingrectangle [ 2 6] by [ 2 12] andseethattheonlysolutiontotheequation2x   x  1  8is x  3, whichcanbeverifiedbysubstitution.

37. x 3 2 x 2 x 1  0,sowestartbygraphing

thefunction y  x 3 2 x 2 x 1intheviewing rectangle [ 10 10] by [ 100 100].There appeartobetwosolutions,onenear x  0and anotheronebetween x  2and x  3.Wethen usetheviewingrectangle[ 1 5]by[ 1 1]and zoominontheonlysolution, x  255.

38. x 4 8 x 2  2  0.Westartbygraphingthe

function y  x 4 8 x 2  2intheviewing

rectangle [ 10 10] by [ 10 10].Thereappear tobefoursolutionsbetween x 3and x  3. Wethenusetheviewingrectangle [ 5 5] by [ 1 1],andzoomtofindthefoursolutions x 2 78, x 0 51, x  0 51,and x  2 78.

39. x  x 1 x  2  1 6 x  x  x 1 x  2 1 6 x  0.Westartbygraphing

thefunction y  x  x 1 x  2 1 6 x inthe

viewingrectangle [ 5 5] by [ 10 10].There appeartobethreesolutions.Wethenusethe viewingrectangle[ 2 5 2 5]by[ 1 1]and zoomintothesolutionsat x 205, x  000, and x  1 05.

40. x 4  16 x 3 .Westartbygraphingthefunctions y1  x 4 and y2  16 x 3 intheviewingrectangle [ 10 10] by [ 5 40].Thereappearstobetwosolutions,onenear x 2andanotheronenear x  2.Wethenusetheviewing rectangle [ 2 4 2 2] by [27 29],andzoomintofindthesolutionat x 2 31.Wethenusetheviewingrectangle [1 7 1 8] by [9 5 10 5],andzoomintofindthesolutionat x  1 79.

41. Wegraph y  x 2 and y  3 x  10intheviewingrectangle [ 4 7] by [ 5 30] Thesolutiontotheinequalityis [ 2 5].

42. Since05 x 2  0875 x  025  05 x 2  0875 x 0

25  0,wegraph

y  0 5 x 2  0 875 x 0 25intheviewingrectangle [ 3 1] by [ 5 5].Thusthe solutiontotheinequalityis[ 2 0 25].

43. Since x 3  11 x  6 x 2  6  x 3 6 x 2  11 x 6  0,wegraph

y  x 3 6 x 2  11 x 6intheviewingrectangle [0 5] by [ 5 5].Thesolution setis  100]  [2

44. Since16 x 3  24 x 2  9 x 1  16 x 3  24 x 2  9 x  1  0,wegraph y  16 x 3  24 x 2  9 x  1intheviewing rectangle [ 3 1] by [ 5 5].Fromthisrectangle, weseethat x 1isan x ­intercept,butitis unclearwhatisoccurringbetween x 

x  0.Wethenusetheviewingrectangle[ 1 0]by[ 0 01 0 01].Itshows y  0at x 0 25.Thusinintervalnotation, thesolutionis  1 0 25   0 25 

45. Since x 13  x  x 13 x  0,wegraph y  x 13 x intheviewingrectangle[ 3 3]by[ 1 1].Fromthis,we findthatthesolutionsetis 

AnotherMethod: AsinExample7,wegraph y1  x 13 and y2  x inthesameviewingrectangle,andseethat x 13  x for 1  x  0andfor1  x

47. Since

wegraph y   x  1

2  x 12 intheviewing rectangle [ 2 2] by [ 5 5].Thesolutionsetis 

46. Since

graph y  05 x 2  1 2  x  intheviewingrectangle [ 1 1] by [ 1 1].Welocatethe x ­interceptsat x 0 535.Thusinintervalnotation,thesolutionis approximately  0 535]

Since

x

intheviewingrectangle [ 4 4] by [ 1 1].The x ­interceptiscloseto x  2.Usingatrace function,weobtain x  2 148.Thusthesolutionis [2 148 

49. Wegraphtheequations y  3 x 2 3 x and y  2 x 2  4intheviewingrectangle [ 2 6]by[ 10 50].Weseethatthetwocurvesintersectat x 1andat x  4, andthatthefirstcurveislowerthanthesecondfor 1  x  4.Thus,weseethat theinequality3 x 2 3 x  2 x 2  4hassolutionset

2 4

50. Wegraphtheequations y  5 x 2  3 x and y  3 x 2  2intheviewingrectangle

[ 3 2] by [ 5 25].Weseethatthetwocurvesintersectat x 2andat x  1 2 , whichcanbeverifiedbysubstitution.Thefirstcurveishigherthanthesecondfor x  2andfor x  1 2 ,sothesolutionsetoftheinequality5 x 2  3 x  3 x 2  2is

 2]   1 2  .

51. Wegraphtheequation y   x  32  x 2 x  5 intheviewingrectangle

[ 7 3] by [ 120 20] andseethattheinequality  x  32  x 2 x  5  0has thesolutionset  5]

52. Wegraphtheequation y  4 x 2  x 2 9 intheviewingrectangle [ 5 5] by [ 100 100] andseethattheinequality4 x 2  x 2 9  0hasthesolutionset

[ 3 3]

53. Tosolve5 3 x  8 x 20bydrawingthegraphofasingleequation,weisolate alltermsontheleft­handside:5 3 x  8 x 20 

5 3 x 8 x  20  8 x 20 8 x  20 11 x  25  0or11 x 25  0. Wegraph y  11 x 25,andseethatthesolutionis x  227,asinExample2.

54. Graphing y  x 3 6 x 2  9 x and y   x intheviewingrectangle [ 0 01 0 02] by [ 0 05 0 2],weseethat x  0and x  0 01aresolutionsoftheequation

x 3 6 x 2  9 x 

55.(a) Wegraphtheequation

y  10 x  0 5 x 2 0 001 x 3 5000intheviewing rectangle [0 450] by [ 5000 20000].

100 200 300 400 0 10000 20000

(b) Fromthegraphitappearsthat

0  10 x  005 x 2 0001 x 3 5000for 100  x  500,andso101cooktopsmustbeproduced to begin tomakeaprofit.

(c) Wegraphtheequations y  15,000and y  10 x  0 5 x 2 0 001 x 3 5000intheviewing rectangle [250 450] by [11000 17000].Weuseazoom ortracefunctiononagraphingcalculator,andfindthat thecompany’sprofitsaregreaterthan$15,000for 279  x  400.

56. Note:Onpage123ofthefirstprintingofthetext,thedisplayedequationin thisexerciseshouldread y  2 4 x   x 1000 2 andpart(a)oftheexerciseshouldreadasfollows: (a) Graphtheequationfor0  x  160. (a)

(b) Usingazoomortracefunction,wefindthat y  16for x  1067.We couldestimatethissinceif x  160,then  x 1000 2  0 0256.Sofor x  160wehave 2 4 x 

57. Answerswillvary.

58. Calculatorsperformoperationsinthefollowingorder:Exponentsareappliedbeforedivision,anddivisionisappliedbefore addition.Therefore, Y_1=x^1/3 isinterpretedas y  x 1 3  x 3 ,whichistheequationofaline.Likewise, Y_2=x/x+4 is interpretedas y  x x  4  1  4  5.Instead,enterthefollowing: Y_1=x^(1/3), Y_2=x/(x+4).

1.12 MODELINGVARIATION

1. Ifthequantities x and y arerelatedbytheequation y  5 x thenwesaythat y is directlyproportional to x ,andtheconstant of proportionality is5.

2. Ifthequantities x and y arerelatedbytheequation y  5 x thenwesaythat y is inverselyproportional to x ,andtheconstant of proportionality is5.

3. Ifthequantities x , y ,and z arerelatedbytheequation z  5 x y thenwesaythat z is directlyproportional to x and inversely proportional to y

4. Because z isjointlyproportionalto x and y ,wemusthave z  kxy .Substitutingthegivenvalues,weget 10  k 45  20k  k  1 2 .Thus, x , y ,and z arerelatedbytheequation z  1 2 xy.

5.(a) Intheequation y  3 x , y isdirectlyproportionalto x (b) Intheequation y  3 x  1, y isnotproportionalto x

6.(a) Intheequation y  3 x  1 , y isnotproportionalto x (b) Intheequation y  3 x , y isinverselyproportionalto x

7. T  kx ,where k isconstant.

9.   k z ,where k isconstant.

11. y  ks t ,where k isconstant.

13. z  k  y ,where k isconstant.

8. P  k  ,where k isconstant.

10.   kmn ,where k isconstant.

12. P  k T ,where k isconstant.

14. A  kx 2 t 3 ,where k isconstant.

15. V  kl  h ,where k isconstant.

17. R  kP 2 t 2 b3 ,where k isconstant.

16. S  kr 2  2 ,where k isconstant.

18. A  k  xy ,where k isconstant.

19. Since y isdirectlyproportionalto x , y  kx .Since y  32when x  8,wehave32  k 8  k  4.So y  4 x

20.  isinverselyproportionalto t ,so   k t .Since   3when t  8,wehave3  k 8  k  24,so   24 t .

21. A variesinverselyas r ,so A  k r .Since A  15when r  5,wehave15  k 5  k  75.So A  75 r .

22. P isdirectlyproportionalto T ,so P  kT .Since P  60when T  72,wehave60  k 72

5 6 T

23. Since A isdirectlyproportionalto x andinverselyproportionalto t , A  kx t .Since A  42when x  7and t  3,we

have42  k 7 3  k  18.Therefore, A  18 x t

24. S  kpq .Since S  350when p  7and q  20,wehave350  k 720  350  140k  k  5 2 .So S  5 2 pq

25. Since W isinverselyproportionaltothesquareof r , W  k r 2 .Since W  24when r  3,wehave24  k

So W  216 r 2 .

26. t  k xy r .Since t  125when x  10, y  15,and r  12,wehave125  k 10 15 12  k  10.So t  10 xy r

27. Since C isjointlyproportionalto l ,  ,and h ,wehave C  kl  h .Since C  128when l 

h  2,wehave 128  k 222  128  8k  k  16.Therefore, C  16l  h .

28. H  kl 2  2 .Since H  36when l  2and   1 3 ,wehave36 

29. R  k  x .Since R  2 5when x 

30. M  k abc d .Since

31.(a) z  k x 3 y 2

(b) Ifwereplace x with3 x and y with2 y ,then

32.(a) z  k x 2 y 4

(b) Ifwereplace x with3 x and y with2 y ,then

33.(a) z  kx 3 y 5

(b) Ifwereplace x with3 x and y with2 y ,then

34.(a) z  k x 2 y 3

(b) Ifwereplace x with3 x and y with2 y ,then z

35.(a) Theforce F neededis F  kx .

(b) Since F  30Nwhen x  9cmandthespring’snaturallengthis5cm,wehave30  k 9 5  k  7 5Ncm.

(c) Frompart(b),wehave F  7 5 x .Substituting x  11 5  6into

216.

36.(a) C  kpm

(b) Since C  60,000when p  120and m  4000,weget60,000  k 1204000  k  1 8 .So C  1 8 pm .

(c) Substituting p  92and m  5000,weget C  1 8 925000  $57,500.

37.(a) P  ks 3

(b) Since P  96when s  20,weget96  k 203  k  0 012.So P  0 012s 3

(c) Substituting x  30,weget P  0012 303  324watts.

38. LettheamountofsolubleCO2 be x andthetemperatureofthewaterbe T .Becausethesequantitiesareinversely proportional, x  k T .Wearegiventhatwhen T  273K, x  3g,so3  k 273  k  819.Thus,if T  298,then x  819 298  2 75g.

39. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

D  ks 2 .Since D  45when s  60,wehave45  k 602 ,so k  00125.Thus, D  00125s 2 .If D  60,then 60  0 0125s 2  s 2  4800,so s  69km/h(forsafetyreasonswerounddown).

40. L  ks 2 A.Since L  5950when s  80and A  46,wehave5950  k 802  46  k  0 02021.Thus L  002021s 2 A .When A  56and s  64,theliftis L  002021 642  56  46357N.

41. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. F  kAs 2 .Since F  980when A  4and s  8,wecansolvefor k :980  k 482  980  256k  k  3 83.Now when A  25and F  780weget780 

42.(a) T 2  kd 3

(b) Substituting T  365and d  149  106 ,weget3652  k

(c) T 2  4

5 98  104 days  164years.

43.(a) P  kT V

4 03 

.HencetheperiodofNeptuneis

(b) Substituting P  332, T  400,and V  100,weget332  k 400 100  k  83.Thus k  83andtheequationis P  8 3 T V

(c) Substituting T  500and V  80,wehave P  8

500

80  51875kPa.Hencethepressureofthesampleofgasis about51 9kPa.

44.(a) F  k  s 2 r (b) Forthefirstcarwehave 1  768and s1  100andforthesecondcarwehave 2  1200.Sincetheforcesareequal wehave k 768 1002 r  k 1200 s 2 2 r  16 1002 25  s 2 2 ,so s2  80km/h.

45.(a) Theloudness L isinverselyproportionaltothesquareofthedistance d ,so L  k d 2 . (b) Substituting d  10and L  70,wehave70  k 102  k  7000.

(c) Substituting2d for d ,wehave L 

(d) Substituting 1 2 d for d ,wehave L 

,sotheloudnessischangedbyafactorof 1 4

,sotheloudnessischangedbyafactorof4.

46.(a) Thepower P isjointlyproportionaltothearea A andthecubeofthevelocity  ,so P  kA  3

(b) Substituting2 for  and 1 2 A for A,wehave P 

(c) Substituting 1 2  for

47.(a) R  kL d 2

and3 A for A,wehave

(b) Since R  140when L  1 2and d  0 005,weget140

(c) Substituting L  3and d  0 008,wehave R  7 2400 

(d) Ifwesubstitute2d for d and3 L for L ,then R 

3 ,sothepowerischangedbyafactorof4.

,sothepowerischangedbyafactorof 3 8 .

0 002916.

2400

 137ohms.

,sotheresistanceischangedbyafactorof 3 4

48. Let S bethefinalsizeofthecabbage,inkilograms,let N betheamountofnutrientsitreceives,ingrams,andlet c bethe numberofothercabbagesaroundit.Then S  k N c .When N  600and c  12,wehave S  15,sosubstituting,wehave 15  k 600 12  k  0 3.Thus S  0 3 N c .When N  300and c  5,thefinalsizeis S  0 3  300 5   18kg.

49. Note:Inthefirstprintingofthetext,theanswergivenforpart(b)ofthisexerciseisincorrect.

(a) Forthesun, E S  k 60004 andforearth, E E  k 3004 .Thus E S E E  k 60004 k 3004   6000 300 4  204  160,000.Sothesun produces160,000timestheradiationenergyperunitareathantheEarth.

(b) Thesurfaceareaofthesunis4 696,0002 andthesurfaceareaoftheEarthis4 63402 .Sothesunhas

4 696,0002 4 63402   696,000 6340 2 timesthesurfaceareaoftheEarth.Thusthetotalradiationemittedbythe sunis 160,000   696,000 6340 2  1,928,234,931timesthetotalradiationemittedbytheEarth.

50.(a) Let T and l betheperiodandthelengthofthependulum,respectively.Then T  k l (b) T  k l  T 2  k 2 l  l  T 2 k 2 .Iftheperiodisdoubled,thenewlengthis 2 T 2 k 2  4 T 2 k 2  4l .Sowewould quadruplethelength l todoubletheperiod T

51.(a) Since f isinverselyproportionalto L ,wehave f  k L ,where k isapositiveconstant.

(b) Ifwereplace L by2 L wehave k 2 L  1 2 k L  1 2 f .Sothefrequencyofthevibrationiscutinhalf.

52.(a) Since r isjointlyproportionalto x and P x ,wehave r  kx  P x ,where k isapositiveconstant.

(b) When10peopleareinfectedtherateis r  k 10 5000 10  49,900k .When1000peopleareinfectedtherateis r 

  4,000,000k .Sotherateismuchhigherwhen1000peopleareinfected.Comparing theserates,wefindthat 1000peopleinfected 10peopleinfected  4,000,000k 49,900k  80.Sotheinfectionratewhen1000peopleareinfected isabout80timesaslargeaswhen10peopleareinfected.

(c) Whentheentirepopulationisinfectedtherateis r  k 50005000 5000  0.Thismakessensesincethereareno morepeoplewhocanbeinfected.

53. Wesubstitute e  0 2kWhkmand   100km/hintothesecondgivenequation:0

Now,becausetherangeis R  500kmforthesevaluesof e and  ,wecanusethefirstequationtocalculatethevalueof C :

500  C

02  C  100,soafullchargeforthisvehicleis100kWh.

Substituting e  k  2 intothefirstgivenequation,wehave R  C k

2

Soat130km

54.(a) Wearegiventhatthemassflowrateis m  A 0  0

.Because m isconstant,wecanequatetheflowratesforthewider andnarrowersectionsofpipe: A

A .Thevelocityofthefluidisindeedinverselyproportional tothecross­sectionalareaofthepipe.

(b) Wesubstitutethegivenvaluesintoourexpressionfor

s.Thus, thefluidflowsthroughtheconstrictedsectionofpipeat45m/s.

55. Using B  k L d 2 with k  0 080, L  2 5 

Thestar’sapparentbrightnessisabout3 47  10 14 Wm2

56. First,wesolve B  k L d 2 for d : d 2  k L B  d  k L B because d ispositive.Substituting

,sothestarisapproximately2

57. Examplesincluderadioactivedecayandexponentialgrowthinbiology.

CHAPTER1REVIEW

1. CommutativePropertyforaddition. 2. CommutativePropertyformultiplication. 3. DistributiveProperty.

DistributiveProperty.

mfromearth.

21. 78,250,000,000  7 825  1010

22.

23. ab c 

24.

 x  2or x 5.However,since x  2makestheexpressionundefined,werejectthissolution.Hencetheonly solutionis x 5.

68. x 4 8 x 2 9  0   x 2 9

x 3,however x 2  1  0hasnorealsolution.Thesolutionsare x 3.

 x 3

83. Let x bethenumberofkilogramsofraisins.Thenthenumberofkilogramsofnutsis50 x

20.Thusthemixtureuses 20kilogramsofraisinsand50 20  30kilogramsofnuts.

84. Let t bethenumberofhoursthatthedistrictsupervisordrives.Thenthestoremanagerdrivesfor t 1 4 hours.

Whentheypasseachother,theywillhavetraveledatotalof256kilometers.So72

2.Sincethesupervisorleavesat2:00 P M .andtravelsfor2hours,theypass eachotherat4:00 P. M .

85. Let r betheathlete’srunningspeed,inmi/h.Thentheycycleat r  8km/h.

Sincethetotaltimeoftheworkoutis1hour,wehave

speedis

86. Let x bethelengthofonesideincm.Then28 x isthelengthoftheotherside.UsingthePythagoreanTheorem,we have

2  x 12 x 16  0.So x  12or x  16.If x  12,thentheothersideis28 12  16.Similarly,if x  16,then theothersideis12.Thesidesare12cmand16cm.

87. Let t bethetimeitwouldtaketheinteriordecoratortopaintalivingroomiftheyworkalone.Itwouldtaketheassistant 2t hoursalone,anditwouldtaketheapprentice3t hoursalone.Thus,thedecoratordoes 1 t ofthejobperhour,theassistant does 1 2t ofthejobperhour,andtheapprenticedoes

t ofthejobperhour.So

6t  11  t  11 6 .Thus,itwouldtakethedecorator1hour50minutestopaintthelivingroom alone.

88. Let l belengthofeachgardenplot.Thewidthofeachplotisthen 80 l andthetotalamountoffencingmaterialis

4 l 10l 12  0.So l  10or l  12.If l  10m,thenthewidthofeachplotis 80 10  8m.If l  12m,thenthe widthofeachplotis 80 12  667m.Bothsolutionsarepossible. 89. 3 x 2

Interval: [ 2 0]

Graph: -3

Graph: _20 91. x 2 7 x 8  0   x 8 x  1  0.Theexpressionontheleftoftheinequalitychangessignwhere x  8andwhere x 1.Thuswemustchecktheintervalsinthefollowingtable.

Graph: _18

92. x 2  1  x 2 1  0 

x 1 x

1  0.Theexpressionontheleftoftheinequalitychangessignwhen x 1and x  1.Thuswemustchecktheintervalsinthefollowingtable.

Interval: [ 1

Graph: _11

93. x 4 x 2 4

0  x 4

x 2

0.Theexpressionontheleftoftheinequalitychangessignwhere x

2,where x  2, andwhere x  4.Thuswemustchecktheintervalsinthefollowingtable.

Sincetheexpressionisnotdefinedwhen x 2

Graph: _224

weexcludethesevaluesandthesolutionis

expressionontheleftoftheinequalitychangessignwhen 2

and2.Thuswemustchecktheintervalsinthefollowing table.

Graph: _212

Themidpointis

Graph: 3.984.02 97.(a) y x P Q 1 1 (b) Thedistancefrom P to Q is

Thelinehasslope

Theradiusofthiscirclewasfoundinpart(b).Itis

(c) Themidpointis

98.(a) y x P Q 1 1 (b) Thedistancefrom P to Q is

(d) Thelinehasslope

(e) Theradiusofthiscirclewasfoundinpart(b).Itis

104. Themidpointofsegment PQ

,andtheradiusis 1 2 ofthedistancefrom P to Q ,or

Sincetheleftsideofthisequationmustbegreaterthanorequaltozero,thisequationhasnograph.

(b) Thisistheequationofthepoint

115. x  16 y 2

(a) x ­axissymmetry:replacing y by y gives x  16  y 2  16 y 2 ,whichisthesameastheoriginalequation,so thegraphissymmetricaboutthe x ­axis. y ­axissymmetry:replacing x by x gives x  16 y 2  x  y 2 16,whichisnotthesameastheoriginal equation,sothegraphisnotsymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by y gives x  16  y 2  x 16 y 2 ,whichisnotthesameas theoriginalequation,sothegraphisnotsymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor x : x  16 02  16,sothe x ­interceptis16.

Tofind y ­intercepts,weset x  0andsolvefor y :0  16 y 2  y 4,sothe y ­interceptare4and 4.

116. x 2  4 y 2  9

(a) x ­axissymmetry:replacing y by y gives x

9,whichisthesameastheoriginal equation,sothegraphissymmetricaboutthe x ­axis.

y ­axissymmetry:replacing x by

gives

9,whichisthesameastheoriginal equation,sothegraphissymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by y gives

9,thesameasthe originalequation,sothegraphissymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor

Tofind y ­intercepts,weset x  0andsolvefor

x 2 9 y  9

(a) x ­axissymmetry:replacing y by y gives

3,sothe x ­interceptsare3and 3.

3

9,sothegraphisnotsymmetricaboutthe x ­axis.

y ­axissymmetry:replacing x by x gives

Originsymmetry:replacing x by x and y by

y ­axis.

9 y  9,sothegraphisnot symmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor

3,sothe x ­interceptsare3and 3. Tofind y ­intercepts,weset x  0andsolvefor y

1,sothe y ­interceptis 1.

 x  12  y 2  4

(a) x ­axissymmetry:replacing y by y gives

4,sothegraphissymmetric aboutthe x ­axis.

y ­axissymmetry:replacing x by x gives

aboutthe y ­axis.

Originsymmetry:replacing x by x and y

4,sothegraphis notsymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor

3or1,sothe x ­interceptsare 3and1.

Tofind y ­intercepts,weset x  0andsolvefor y

y ­interceptsare 3and 3.

119. x 2  4 xy  y 2  1

(a) x ­axissymmetry:replacing y by y gives x

1,whichisdifferentfromtheoriginalequation,so thegraphisnotsymmetricaboutthe x ­axis. y ­axissymmetry:replacing x by x gives

1,whichisdifferentfromtheoriginalequation, sothegraphisnotsymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by

1,so thegraphissymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor x

1and1.

Tofind y ­intercepts,weset x  0andsolvefor y

1,sothe y ­interceptsare 1and1.

120. x 3  xy 2  5

(a) x ­axissymmetry:replacing y by y gives x 3  x  y 2  5  x 3  xy 2  5,sothegraphissymmetricaboutthe x ­axis.

y ­axissymmetry:replacing x by x gives  x

y 2  5,whichisdifferentfromtheoriginalequation,sothe graphisnotsymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by y gives  x 3   x  y 2  5,whichisdifferentfromtheoriginal equation,sothegraphisnotsymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor x

5,sothe x ­interceptis 3 5.

Tofind y ­intercepts,weset x  0andsolvefor y :03  0 y 2  5hasnosolution,sothereisno y ­intercept.

121.(a) Wegraph y  x 2 6 x intheviewingrectangle

[ 10 10] by [ 10 10]. ­10 ­5 5 10

(b) Fromthegraph,weseethatthe x ­interceptsare0 and6andthe y ­interceptis0.

123.(a) Wegraph y  x 3 4 x 2 5 x intheviewing rectangle[ 4 8]by[ 30 20]. ­4 ­2 2 4 6 8 ­20 20

(b) Fromthegraph,weseethatthe x ­interceptsare 1, 0,and5andthe y ­interceptis0.

122.(a) Wegraph y  5 x intheviewingrectangle [ 10 6] by [ 1 5]. ­10 ­5 5 2 4

(b) Fromthegraph,weseethatthe x ­interceptis5and the y ­interceptisapproximately2 24.

124.(a) Wegraph x 2 4  y 2  1  y 2  1 x 2 4  y 1 x 2 4 intheviewingrectangle [ 3 3] by [ 2 2] ­3 ­2 ­1 1 2 3 ­2 ­1 1 2

(b) Fromthegraph,weseethatthe x ­interceptsare 2 and2andthe y ­interceptsare 1and1.

125.(a) Thelinethathasslope2and y ­intercept6hastheslope­interceptequation y  2 x  6.

(b) Anequationofthelineingeneralformis2x y  6  0.

126.(a) Thelinethathasslope 1 2 andpassesthroughthepoint

127.(a) Thelinethatpassesthroughthepoints

128.(a) Thelinethathas x ­intercept4and y ­intercept12passesthroughthepoints

129.(a) Theverticallinethatpassesthroughthepoint 3 2 hasequation x  3.

(b) x  3  x 3  0. (c) y 1x 1

130.(a) Thehorizontallinewith y ­intercept5hasequation y  5.

(b) y  5  y 5  0.

y 1x 1

131.(a) Thelinecontaining 2

4

and 4

4 hasslope m

4 4 4 2  8 2 4,andthelinepassingthroughtheoriginwith thisslopehasequation y 4 x (b) y 4 x  4

132.(a) Thelinewithequation x 4

has slope 1 4 ,soanylineperpendiculartoithasslope

4.Sincethe desiredlinepassesthrough 2 3,ithasequation y 3 4  x 2  y 4 x  11. (b) y 4 x

133.(a) Theslope,0 3,representstheincreaseinlengthofthespringforeachunitincreasein weight  .The s ­interceptisthe restingornaturallengthofthespring. (b) When   5, s  0 3 5  2 5  1 5  2 5  4 0centimeters.

134.(a) Weusetheinformationtofindtwopoints, 0 60000 and 3 70500.Thentheslopeis m  70,500 60,000 3 0  10,500 3  3,500.So S  3,500t  60,000.

(b) Thesloperepresentstheaccountant’sannualincreaseinsalary,$3500,andthe S ­interceptrepresentstheaccountant’s initialsalary,$60,000.

(c) When t  12,theaccountant’ssalarywillbe S  3500 12  60,000  42,000  60,000  $102,000.

135. Fromthegraph,weseethatthegraphsof y  x 2 4 x and y  x  6intersectat x 1and x  6,sothesearethe solutionsoftheequation x 2 4 x  x  6.

136. Fromthegraph,weseethatthegraphof y  x 2 4 x crossesthe x ­axisat x  0and x  4,sothesearethesolutionsof theequation x 2 4 x  0.

137. Fromthegraph,weseethatthegraphof y  x 2 4 x liesbelowthegraphof y  x  6for 1  x  6,sotheinequality x 2 4 x  x  6issatisfiedontheinterval[ 1 6].

138. Fromthegraph,weseethatthegraphof y  x 2 4 x liesabovethegraphof y  x  6for   x  1and6  x  , sotheinequality x 2 4 x  x  6issatisfiedontheintervals  1] and [6 

139. Fromthegraph,weseethatthegraphof y  x 2 4 x liesabovethe x ­axisfor x  0andfor x  4,sotheinequality x 2 4 x  0issatisfiedontheintervals  0] and [4 

140. Fromthegraph,weseethatthegraphof y  x 2 4 x liesbelowthe x ­axisfor0  x  4,sotheinequality x 2 4 x  0is satisfiedontheinterval [0 4]

141. x 2 4 x  2 x  7.Wegraphtheequations y1  x 2 4 x and y2  2 x  7intheviewingrectangle [ 10 10] by [ 5 25].Usingazoomortracefunction,wegetthe solutions x 1and x  7.

10 ­5 5

143. x 4 9 x 2  x 9.Wegraphtheequations y1  x 4 9 x 2 and y2  x 9intheviewingrectangle[ 5 5]by [ 25 10].Usingazoomortracefunction,wegetthe solutions x 2 72, x 1 15, x  1 00,and x  2 87. ­4 ­2 2 4 ­20 ­10 10

142.  x  4  x 2 5.Wegraphtheequations y1   x  4 and y2  x 2 5intheviewingrectangle [ 4 5] by [0 10].Usingazoomortracefunction,wegetthe solutions x 250and x  276. ­4 ­2 0 2 4 5 10

144.  x  3 5  2.Wegraphtheequations y1   x  3 5 and y2  2intheviewingrectangle [ 20 20] by [0 10].UsingZoomand/orTrace,wegetthe solutions x 10, x 6, x  0,and x  4. ­20 ­10 0 10 20 5 10

145. x 2  12 4 x .Wegraphtheequations y1  x 2 and y2  12 4 x intheviewingrectangle[ 8 4]by[0 40].

Usingazoomortracefunction,wefindthepointsof

intersectionareat x 6and x  2.Sincewewant

x 2  12 4 x ,thesolutionistheunionofintervals

147. x 4 4 x 2  1 2 x 1.Wegraphtheequations

y1  x 4 4 x 2 and y2  1 2 x 1intheviewingrectangle

[ 5 5]by[ 5 5].Wefindthepointsofintersectionare

at x 1 85, x 0 60, x  0 45,and x  2 00.Since

wewant x 4 4 x 2  1 2 x 1,thesolutionis

 1 85 0 60

146. x 3 4 x 2 5 x  2.Wegraphtheequations

y1  x 3 4 x 2 5 x and y2  2intheviewingrectangle [ 10 10] by [ 5 5].Wefindthatthepointofintersection isat x  5 07.Sincewewant x 3 4 x 2 5 x  2,the solutionistheinterval 507 . ­10 ­5 5

148.  

x 2 16

x 2 16

10  0.Wegraphtheequation

10intheviewingrectangle[ 10 10]by [ 10 10].Usingazoomortracefunction,wefindthatthe x ­interceptsare x 5 10and x 2 45.Sincewe want

x 2 16

10  0,thesolutionisapproximately

149. Herethecenterisat 0 0,andthecirclepassesthroughthepoint  5 12,sotheradiusis r 

13.Theequationofthecircleis x 2  y 2  132  x 2  y 2  169.Thelineshownisthetangentthatpassesthroughthepoint  5 12,soitisperpendiculartotheline throughthepoints 0

.Theslopeofthelineweseekis m 2  1 m 1  1 125  5 12 .Thus,anequationofthetangentlineis y 12  5 12  x  5  y 12  5 12 x 

150. Becausethecircleistangenttothe x ­axisatthepoint 5 0 andtangenttothe y ­axisatthepoint

0

5,thecenterisat 5 5 andtheradiusis5.Thusanequationis

,soanequationofthelineweseekis

25.Theslopeof thelinepassingthroughthepoints

151. Since M variesdirectlyas z wehave M  kz .Substituting M  120when z  15,wefind120  k 15  k  8. Therefore, M  8z

152. Since z isinverselyproportionalto y ,wehave z  k y .Substituting z  12when y  16,wefind12  k 16  k  192. Therefore z  192 y

153.(a) Theintensity I variesinverselyasthesquareofthedistance d ,so I  k d 2

(b) Substituting I  1000when d  8,weget1000  k 82  k  64,000.

(c) Fromparts(a)and(b),wehave I  64,000 d 2 .Substituting d  20,weget I  64,000 202  160candles.

154. Let f bethefrequencyofthestringand l bethelengthofthestring.Sincethefrequencyisinverselyproportional tothe length,wehave f  k l .Substituting l  12when k  440,wefind440  k 12  k  5280.Therefore f  5280 l .For f  660,wemusthave660  5280 l  l  5280 660  8.Sothestringneedstobeshortenedto8centimeters.

155. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let  betheterminalvelocityoftheparachutistinkm/hand  betheirweightinkilograms.Sincetheterminalvelocityis directlyproportionaltothesquarerootoftheweight,wehave   k  .Substituting   14when   70,wesolve for k .Thisgives14  k 70  k  14 70  1 673.Thus   1 673 .When   105,theterminalvelocityis   1673105  17km/h.

156. Let r bethemaximumrangeofthebaseballand  bethevelocityofthebaseball.Sincethemaximumrangeisdirectly proportionaltothesquareofthevelocity,wehave r  l  2 .Substituting   96and r  74,wefind74  k 962  k  0 00803.If   112,thenwehaveamaximumrangeof r  0 00803 1122  100 7meters.

157. Thespeed  isinverselyproportionaltothesquarerootofthedensity d ,so   k d .Infreshwaterwithdensity d1  1gcm3 ,thespeedis  1  1480ms,so1480  k 1  k  1480.Thus,inseawaterwithdensity10273gcm3 ,we have   1480 10273  1460ms.

158.(a) Because   1 22  D ,forfixed D theangulardistance  isdirectlyproportionaltothewavelength .Thus,theangular distanceissmallerforshorterwavelengths. (b) Forfixed ,ifwesubstitute D1  2 D intheformula   1 22  D thentheangulardistancebecomes

 1  1 22  D1  1 22  2 D  1 2 1 22  D   1 2  .Thus,forafixedwavelength,ifthediameterofthemirrorisdoubled, theangulardistanceishalved.

159. Wesolvethefirstequationfor

ByHubble’sLaw,   H0 D  D   H0 ,sosubstituting  fromaboveand H0  208Mlykms,wehave D

2 4  105 208  11,538megalight­years.

160.(a) y  2  x  3isthegraphoftheabsolutevaluefunction y   x ,stretchedverticallybyafactorof2andshifted downward3units.IthasgraphIII

(b) 2 y 3 x 2  y  3 2 x 1representsalinewithslope 3 2 , x ­intercept 2 3 ,and y ­intercept 1.IthasgraphV.

(c) y  x 4 has y 0  0and y  0elsewhere.Itisanevenfunction,andhasgraphII.

(d)  x  12

y  1

2  9istheequationofacirclewithcenter  1 1 andradius3.IthasgraphIV.

(e) 6 y  x  3  y  1 6 x  1 2 representsalinewithslope 1 6 , x ­intercept3,and y ­intercept 1 2 .IthasgraphI.

(f) y  6 x 1  x 4 isanoddfunctionwith y 0  0andhorizontalasymptotethe x ­axis.IthasgraphVII.

(g) x  y 3 isanoddfunctionwhosegraphcontainsthepoints 1 1 and  1 1.IthasgraphVIII.

(h) x 2 2 x  y 2 4 y  1  0

4istheequationofacirclewithcenter 1 2 and radius2.IthasgraphVI.

CHAPTER1TEST

1.(a)

2.(a)

3.(a)

4.(a)

5.(a)

5.(Notethatthisisimpossible,sotherecanbenosolution.) Squaringbothsidesagain,weget1

4.Butthisdoesnotsatisfytheoriginalequation,sothereisno solution.(Youmustalwayscheckyourfinalanswersifyouhavesquaredboth sideswhensolvinganequation,since extraneousanswersmaybeintroduced,ashere.)

(f) x 4 3 x

11. UsingtheQuadraticFormula,2

12. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

Let t bethetime(inhours)ittookthetruckertodrivefromAmitytoBelleville. Then4 4 t isthetimeittookthetruckerto drivefromBellevilletoAmity.SincethedistancefromAmitytoBellevilleequalsthedistancefromBellevilletoAmity,we have80

13. Let  bethewidthoftheparcelofland.Then   70isthelengthoftheparcelofland.Then 

So   50or  120.Since 

Interval: [ 4 3.Graph: _43 (b) x  x 1 x  2  0.Theexpressionontheleftoftheinequalitychangessignwhen x  0, x  1,and x 2.Thus wemustchecktheintervalsinthefollowingtable.

Signof x 1

Fromthetable,thesolutionsetis

Graph: _201

 x 4

 3isequivalentto

50.Thusthemedicineistobestoredatatemperaturebetween 41 Fand50 F.

16. For 6 x x 2 tobedefinedasarealnumber6 x x 2  0  x 6 x 

0.Theexpressionontheleftoftheinequality changessignwhen x  0and x  6.Thuswemustchecktheintervalsinthefollowingtable.

Signof x  

Signof6 x  

Signof x 6 x  

Fromthetable,weseethat 6 x x 2 isdefinedwhen0  x  6.

17.(a) y 1x 1 P Q R S

Thereareseveralwaystodeterminethecoordinatesof S .Thediagonalsofa squarehaveequallengthandareperpendicular.Thediagonal PR ishorizontal andhaslengthis6units,sothediagonal QS isverticalandalsohaslength6. Thus,thecoordinatesof S are 3 6 (b) Thelengthof PQ is

2.Sotheareaof PQRS is 322  18.

18.(a) 1 1 y x (b) The x ­interceptsoccurwhen y  0,so0  4 x 2  x 2  4  x 2.The y ­interceptoccurswhen x  0,so y  4. (c) x ­axissymmetry:  y   4 x 2  y  x 2 4,whichisnotthesameasthe originalequation,sothegraphisnotsymmetricwithrespecttothe x ­axis. y ­axissymmetry: y  4  x 2  4 x 2 ,whichisthesameastheoriginal equation,sothegraphissymmetricwithrespecttothe y ­axis.

Originsymmetry:  y   4  x 2  y 4 x 2 ,whichisnotthesame astheoriginalequation,sothegraphisnotsymmetricwithrespecttotheorigin.

19.(a) y 1x P1

(b) Thedistancebetween P and Q is

(c) Themidpointis

(d) Thecenterofthecircleisthemidpoint, 1

2 ,andthelengthoftheradiusis

89.Thustheequationofthecirclewhosediameteris PQ is

20.(a) x 2  y 2  5hascenter 0

and radius 5.

andthe y ­interceptis 5. y 1x 1

22.(a) Thelinethrough

y 5  0. (b) 3 x  y 10  0

(c) Usingtheinterceptformweget x 6  y 4  1  2 x

3.Usingthepoint­slopeform,

23.(a) When x  100wehave T  0 08 100 4  8 4  4,sothe temperatureatonemeteris4 C.

(c) Thesloperepresentstheraiseintemperatureasthedepthincrease. The T ­interceptisthesurfacetemperatureofthesoilandthe x ­interceptrepresentsthedepthofthe“frostline”,wherethesoil belowisnotfrozen.

24.(a) x 3 9 x 1  0.Wegraphtheequation

y  x 3 9 x 1intheviewingrectangle[ 5 5] by [ 12 12].Wefindthatthepointsof intersectionoccurat x 2 94, 0 11,3 05.

(b) T 20x 6080100120 _5 5 40

(b) x 2 1  x  1.Wegraphtheequations

y1  x 2 1and y2  x  1 intheviewing rectangle [ 5 5] by [ 5 10].Wefindthatthe pointsofintersectionoccurat x 1and x  2. Sincewewant x 2 1  x  1,thesolutionis theinterval[ 1 2].

25. Note:Inthefirstprintingofthetext,theanswersgivenforparts(b)and(c)ofthisexerciseareincorrect.

(a) M  k  h 2 L

(b) Substituting   10, h  15, L  30,and M  21,000,wehave21,000

 280  h 2 L

280.Thus

(c) Nowif L  25,   7 5,and h  25,then M  280 7 5 252  25  52,500.Sothebeamcansupport52,500newtons.

FOCUSONMODELINGFittingLinestoData

1.(a) Usingagraphingcalculator,weobtaintheregression line y  1 8807 x  82 65.

(b) Using x  58intheequation y  1 8807 x  82 65, weget y  18807 58  8265  1917cm.

2.(a) Usingagraphingcalculator,weobtaintheregression line y  017565 x  18265.

(b) ForapercapitaGDPof$80,000,themodelpredicts carbonemissionsof y  0 17565 80  1 8265  15 9tonspercapita. For$32,000,itpredicts y  017565 32  18265  74tonspercapita.

(c) Alinearmodelisreasonableforthesedata.One limitationisthattherearenodatapointsbeyond per­capitaincomeabove$51,000,sothemodel cannotreliablymakepredictionsforGDPsmuch largerthan$51,000.

3. Note:Inthefirstprintingofthetext,theanswergivenforpart(a)ofthisexerciseisincorrect.

(a) Usingagraphingcalculator,weobtaintheregression

(b) Using x  45intheequation y  2 579 x 0 1783, weget y  2579 45 01783  116years.

4.(a) Usingagraphingcalculator,weobtaintheregression line y  8278 x 5393. Temperature (°C) 0 100 200 1015202535x y 30

(b) Using x  38 Cintheequation y  8278 x 5393,weget y  261chirpsper minute.

5.(a) Usingagraphingcalculator,weobtaintheregression line y 013198 x  72514

Years since 1994 y x 2 4 6 8 01020

6.(a) Usingagraphingcalculator,weobtaintheregression line y 0 168 x  19 89. Flow rate (%)

(b) Using x  25intheregressionlineequation,weget y 013198 25  72514  395millionkm2 . Thisisroughly10%lessthantheactualfigureof 44millionkm2 .

(c) Despitefluctuationsoverbriefperiods,themodel seemsfairlyaccurate.Ifexternalcircumstances change(reducedorincreasedCO2 emissions,for example),itmaybecomelessreliable.Itisunlikely tobeaccuratefarintothefuture.

(b) Usingtheregressionlineequation y 0 168 x  19 89,weget y  8 13%when x  70%.

7.(a) Usingagraphingcalculator,weobtain y 39018 x  4197. Noise level (dB) 0 50 100 8090100110x y

(b) Thecorrelationcoefficientis r 0 98,solinear modelisappropriatefor x between80dBand 104dB.

(c) Substituting x  94intotheregressionequation,we get y 3 9018 94  419 7  53.Sothe intelligibilityisabout53%.

8. Studentsshouldfindafairlystrongcorrelationbetweenshoesizeandheight.

9. Resultswilldependonstudentsurveysineachclass.

1 ERRATAinExercisesandAnswersinFirstPrinting

Page1231.11.56Displayedequationshouldread y  2 4 x   x 1000 2

Page1231.11.56(a) Graphtheequationfor0  x  160.

PageA21.5.129(Answer)  424s

PageA21.5.137(Answer) 344 000km

PageA31.7.21(Answer) 220km

PageA31.7.39(Answer) 31 6mby158m

PageA31.7.77(Answer) 1 6mfromthefulcrum

PageA41.8.119(Answer) Morethan100km

PageA41.8.127(Answer)(a) Accelerationgreaterthan8 04ms2

PageA71.10.93(Answer)(a) t  1 8 n  6 (b) 25 C

PageA71.10.95(Answer)(a) P  9 97d  103,where P is pressureinkPaand d isdepthinmeters (c) Thesloperepresentsanincreaseof9 97kPainpressure foreachonemeterincreaseindepth,andthe d ­interceptis theairpressureatthesurface. (d)  59m (b) y x

PageA71.12.39(Answer)  69kmh

PageA71.12.41(Answer)  9 03kmh

PageA71.12.49(Answer)(b) 1 928 234 931

PageA81.Review.155(Answer) 17kmh

PageA91.Test.12(Answer) 192km

PageA91.Test.25(Answer)(b) 280 (c) 52,500N

PageA101.Focus.3(Answer)(a) y  2 579 x 0 1783

CHAPTER 1FUNDAMENTALS1

1.1 RealNumbers1

1.2 ExponentsandRadicals3

1.3 AlgebraicExpressions7

1.4 RationalExpressions10

1.5 Equations14

1.6 ComplexNumbers20

1.7 ModelingwithEquations21

1.8 Inequalities27

1.9 TheCoordinatePlane;GraphsofEquations;Circles37

1.10 Lines47

1.11 SolvingEquationsandInequalitiesGraphically52

1.12 ModelingVariation57 Chapter1Review59 Chapter1Test68

¥ FOCUSONMODELING: FittingLinestoData70 ErratainExercisesandAnswersinFirstPrinting72

1

1.1

FUNDAMENTALS

REALNUMBERS

1.(a) Thenaturalnumbersare 1 2 3

(b) Thenumbers  3

areintegersbutnotnaturalnumbers.

(c) Anyirreduciblefraction p q with q  1isrationalbutisnotaninteger.Examples: 3 2 , 5 12 , 1729 23 .

(d) Anynumberwhichcannotbeexpressedasaratio p q oftwointegersisirrational.Examplesare 2, 3,  ,and e

3.(a) Inset­buildernotation: x 3  x  5

(b) Inintervalnotation:  3 5

5. Thedistancebetween a and b onthereallineis d

7.(a) No: a b  b a   b a ingeneral.

(b) No;bytheDistributiveProperty, 2

9.(a) Naturalnumber:100

(b) Integers:0,100, 8

(c) Rationalnumbers: 1 5,0, 5 2 ,2 71,3 14,100, 8

(d) Irrationalnumbers: 7, 

11. CommutativePropertyofaddition

(c) Asagraph: _35

.Sothedistancebetween 5and2is 2

13. AssociativePropertyofaddition

CommutativePropertyofmultiplication

83.(a) a isnegativebecause a ispositive.

(b) bc ispositivebecausetheproductoftwonegativenumbersispositive.

(c) a b  a   b ispositivebecauseitisthesumoftwopositivenumbers.

(d) ab  ac isnegative:Eachsummandistheproductofapositivenumberandanegative number,andthesumoftwo negativenumbersisnegative.

85. DistributiveProperty

87. Let x  m 1 n 1 and y  m 2 n 2 berationalnumbers.Then x  y  m 1 n 1 

x y  m 1

.Thisshowsthatthesum,difference,andproduct oftworationalnumbersareagainrationalnumbers.Howevertheproductof twoirrationalnumbersisnotnecessarily irrational;forexample, 2  2  2,whichisrational.Also,thesumoftwoirrationalnumbersisnotnecessarilyirrational; forexample, 2   2  0whichisrational.

89.

As x getslarge,thefraction1 x getssmall.Mathematically,wesaythat1 x goestozero.

As x getssmall,thefraction1 x getslarge.Mathematically,wesaythat1 x goestoinfinity.

91.(a) Supposethat a  b ,somax a  b

Ontheotherhand,if b

,thenmax

If a  b ,then a b   0andtheresultistrivial.

(b) If a  b ,thenmin a  b   a and a b

Similarly,if b  a ,then

.Inthiscase,

 b a .Inthiscase a

;andif a  b ,theresultistrivial.

93.(a) Subtractionisnotcommutative.Forexample,5 1  1 5.

(b) Divisionisnotcommutative.Forexample,5  1  1  5.

(c) Puttingonyoursocksandputtingonyourshoesarenotcommutative.Ifyouputonyoursocksfirst,thenyourshoes, theresultisnotthesameasifyouproceedtheotherwayaround.

(d) Puttingonyourhatandputtingonyourcoatarecommutative.Theycanbedoneineitherorder,withthesameresult.

(e) Washinglaundryanddryingitarenotcommutative.

1.2 EXPONENTSANDRADICALS

1.(a) Usingexponentialnotationwecanwritetheproduct5 5 5 5 5 5as56

(b) Intheexpression34 ,thenumber3iscalledthe base andthenumber4iscalledthe exponent

3. Tomoveanumberraisedtoapowerfromnumeratortodenominatororfromdenominatortonumeratorchangethesignof the exponent.So

9.(a)

77.(a) 4  1 4 4  1 64  4  1 4 43  1 4 (b) 5 40   5 40   1 8 

79.(a)  y 4  y 2  y 1

81.(a) 69,300,000  6 93  107

(b) 7,200,000,000,000  7 2  1012

(c) 0 000028536  2 8536  10 5

z .Since  0and z  0,thisisequivalentto 9 z .

(d) 0 0001213  1 213  10 4 83.(a) 3 19  105  319,000 (b) 2 721  108  272,100,000 (c) 2 670  10 8  0 00000002670 (d) 9 999  10 9  0 000000009999

85.(a) 9,460,000,000,000km  9 5  1012 km (b) 0 0000000000004cm

(c) 33billionbillionmolecules

93.(a) b 5 isnegativesinceanegativenumberraisedtoanoddpowerisnegative.

(b) b 10 ispositivesinceanegativenumberraisedtoanevenpowerispositive.

(c) ab2 c3 wehave positive

(d) Since b a isnegative, b a 3 

(e) Since b a isnegative, b a

(f) a 3 c3 b 6 c6 

4 

negative3 whichisnegative.

negative

4 whichispositive.

whichisnegative.

whichisnegative.

95. Sinceonelightyearis9 5  1012 km,Centauriisabout4 3  9 5  1012  4 06  1013 km,or40,600,000,000,000kilometers away.

99. First,weestimatethetotalmassofthestarsintheobservableuniverse:

Thus,thenumberofhydrogenatomsintheobservableuniverseis

101.(a) Using f  0 4andsubstituting d  65,weobtain s  30 fd  30  0 4  65  28km/h. (b) Using f  0 5andsubstituting s  50,wefind d .Thisgives s  30 fd  50  30 0 5 d  50  15d  2500  15d  d  500 3  167m.

103. Since106  103  103 itwouldtake1000days  2 74yearstospendthemilliondollars. Since109  103  106 itwouldtake106  1,000,000days  2739 72yearstospendthebilliondollars.

105.(a) 185 95 

1.3 ALGEBRAICEXPRESSIONS

1. Thegreatestcommonfactorintheexpression18 x 3  30 x is6 x ,andtheexpressionfactorsas6 x 3 x 2

3. Tofactorthetrinomial x 2  8 x  12welookfortwointegerswhoseproductis12andwhosesumis8.Theseintegersare6 and2,sothetrinomialfactorsas

5. TheSpecialProductFormulaforthe“productofthesumanddifferenceofterms”is

6  x 6 x  

7. TheSpecialFactoringFormulafora“perfectsquare”is

9. Type:binomial.Terms:5 x 3 and6.Degree:3.

11. Type:monomial.Term: 8.Degree:0.

13. Type:four­termpolynomial.Terms: x

99. Startbyfactoringoutthepowerof

127. Startbyfactoringoutthepowerof x withthesmallestexponent,thatis,

Startbyfactoring y

143. Thevolumeoftheshellisthedifferencebetweenthevolumesoftheoutside cylinder(withradius R )andtheinsidecylinder (withradius r ).Thus V   R 2

.The averageradiusis R  r 2 and2 R  r 2 istheaveragecircumference(lengthoftherectangularbox), h istheheight,and R r isthethicknessoftherectangularbox.Thus V

145.(a) Thedegreeoftheproductisthesumofthedegrees.

(b) Thedegreeofasumisatmostthelargestofthedegrees—itcouldbesmallerthaneither.Forexample,thedegreeof

x 3

147.(a)

x 3  x  

(b) Basedonthepatterninpart(a),wesuspectthat

Thegeneralpatternis

1.4 RATIONALEXPRESSIONS

1. Arationalexpressionhastheform P  x

(a) 3 x x 2 1 isarationalexpression.

Q  x  ,where P and Q arepolynomials.

,where

isapositiveinteger.

(b)  x  1 2 x  3 isnotarationalexpression.Arationalexpressionmustbeapolynomialdividedbyapolynomial,andthe numeratoroftheexpressionis  x  1,whichisnotapolynomial.

(c) x  x 2 1 x  3  x 3 x x  3 isarationalexpression.

3. Tomultiplytworationalexpressionswemultiplytheir numerators togetherandmultiplytheir denominators together.So 2 x  1 x x  3 isthesameas 2 x  x  1  x  3  2 x x 2  4 x  3

5.(a) Yes.Cancelling x

(b) No;

7. Thedomainof4

11. Since x

.Analternativemethodistomultiplythe

numeratoranddenominatorbythecommondenominatorofboththenumerator anddenominator,inthiscase x 2 y 2 : x y y

101.

(b) Substituting R1  10ohmsand R2  20ohmsgives

Fromthetable,weseethattheexpression x 2 9 x 3 approaches6as x approaches3.Wesimplifytheexpression: x 2 9 x 3   x 3 x  3 x 3  x  3, x  3.Clearlyas x approaches3, x  3approaches6.Thisexplainstheresultinthe table.

103. Answerswillvary.

105.(a)

Itappearsthatthesmallestpossiblevalueof x  1 x is2.

(b) Because x  0,wecanmultiplybothsidesby x andpreservetheinequality:

0.Thelaststatementistrueforall x  0,andbecauseeachstepis reversible,wehaveshownthat x  1 x  2forall x  0.

1.5 EQUATIONS

1.(a) Yes.If a  b ,then a  x  b  x ,andviceversa.

(b) Yes.If a  b ,then ma  mb for m  0,andviceversa.

(c) No.Forexample, 5  5,but  52  52  25.

3.(a) Tosolvetheequation x 2 6 x 16  0byfactoring,wewrite x 2 6 x 16 

(b) Tosolvebycompletingthesquare,wewrite x 2

 x 3

(c) TosolveusingtheQuadraticFormula,wewrite x 

5.(a) Isolatingtheradicalin

(b) Nowsquarebothsides:

(c) Solvingtheresultingquadraticequation,wefind2

0and x  2.

(d) Wesubstitutethesepossiblesolutionsintotheoriginalequation: 2 0  0  0,so x  0isasolution,but

2isnotasolution.Theonlyrealsolutionis

7. Toeliminatethedenominatorsintheequation 3 x  5 x  2  2,multiplyeachsidebythelowestcommondenominator x  x  2 togettheequivalentequation3

9.(a) When x

21.SinceLHS  RHS, x 2isnotasolution.

(b) When x  2,LHS

 RHS, x  2isa solution.

11.(a)

(b) When

thereisnorealsolution.

32.Since D ispositive,thisequationhastworealsolutions.

 0.Since D  0,thisequationhasonerealsolution.

77. D  b 2 4ac  52 4 4  13 8   25 26 1.Since D isnegative,thisequationhasnorealsolution.

79. x 2 x  100  50  x 2  50  x  100

or x 

Potentialsolutionsare x  0and x 4  x  4.Theseareonlypotentialsolutionssincesquaringisnotareversible operation.Wemustcheckeachpotentialsolutionintheoriginalequation.Checking

0isfalse.Checking

4istrue.Thus,theonlysolutionis x  4. 91. Let

x 32 neverequals0,andnosolutioncanbenegative,becausewecannottakethesquarerootofanegativenumber.Thus2 istheonlysolution.

Theonlysolutionis256.

hasnorealsolution).

121.  x  5  x  5.Squaringbothsides,weget

x .Squaringbothsidesagain,we get x  5 

x 31.Potentialsolutionsare x  20and x  31.Wemustcheckeachpotentialsolutionintheoriginalequation.

Checking x  20: 20 

5,whichistrue,andhence x  20isa solution.

Checking x  31: 

5,whichisfalse,andhence x  31isnota solution.Theonlyrealsolutionis x 

127.  x  a   x a  2 x  6.Squaringbothsides,wehave x  a  2

x

x  a

6.Squaringbothsidesagainwehave

 x a 2  36.Checkingtheseanswers,weseethat x 

a 2  36isnotasolution(forexample,trysubstituting a  8),but x  a 2  36isasolution.

129. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

Using h 0  88,wesolve0 48t 2  88,for

Thusittakes4 28secondsfortheballthehittheground.

131. Wearegiventhat  0  12m/s.

(a) Setting h  7 2,wehave7 2 

Therefore,theballreaches72metersin1second(ascending)andagain after1 1 2 seconds(descending).

(b) Setting h 

4 .However,sincethediscriminant D  0,thereisnorealsolution,andhencetheball neverreachesaheightof14 4meters.

(c) Thegreatestheight h isreachedonlyonce.So

(d) Setting h  7 5,wehave7 5 

12t 

0hasonlyonesolution.Thus

5meters.

.Thustheballreachesthehighest pointofitspathafter1 1 4 seconds.

(e) Setting h  0(groundlevel),wehave0

Sotheballhitsthegroundin2

Theshrinkagefactorwhen

(b) Substituting S  0

018mlong.

137. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x bethedistancefromthecenteroftheearthtothedeadspot(inthousandsof kilometers).Nowsetting F  0,wehave

0 988 x 2 764 x

145,924  0.UsingtheQuadraticFormula,weobtain

344.Since429,000isgreaterthanthedistancefromthe earthtothemoon,werejectthefirstroot;thus x  344,000kilometers.

(c)

1.Sincebothsidesofthisequationareequal, x  2isasolutionforeveryvalueof k .Thatis, x  2isasolutiontoeverymemberofthisfamilyofequations. 141.(a) x 2 9 x 

5.Theproductofthesolutionsis4

5  20,theconstant termintheoriginalequation;andtheirsumis4  5  9,thenegativeofthecoefficientof x intheoriginalequation.

(b) Ingeneral,theequation x 2  bx  c  0hassolutions r 1 

b2 4c 2 and r 2  b b2

Method1: Let

1.Checking

onlysolutionis4. Method2:x  x

possiblesolutionsare4and1.Checkingwillresultinthesamesolution.

(b) Method1: Let

thequadraticformula,wehave

COMPLEXNUMBERS

1. Theimaginarynumber i hasthepropertythat i 2 1.

3.(a) Thecomplexconjugateof3

5. Yes,everyrealnumber a isacomplexnumberoftheform

11. 3:realpart3,imaginarypart0.

SinceLHS  RHS,thisprovesthestatement.

81.

SinceLHS  RHS,thisprovesthestatement.

85. UsingtheQuadraticFormula,thesolutionstotheequationare

,whichisarealnumber.

.Sincebothsolutionsarenonreal,we have

isarealnumber. Thusthesolutionsarecomplexconjugatesofeachother.

1.7 MODELINGWITHEQUATIONS

1. Anequationmodelingareal­worldsituationcanbeusedtohelpusunderstandareal­worldproblemusingmathematical methods.Wetranslatereal­worldideasintothelanguageofalgebratoconstructourmodel,andtranslateourmathematical resultsbackintoreal­worldideasinordertointerpretourfindings.

3.(a) Asquareofside x hasarea A  x 2

(b) Arectangleoflength l andwidth  hasarea A  l  .

(c) Acircleofradius r hasarea A  r 2

5. Apainterpaintsawallin x hours,sothefractionofthewallshepaintsinonehouris 1wall x hours  1 x

7. If n isthefirstinteger,then n  1isthemiddleinteger,and n  2isthethirdinteger.Sothesumofthethreeconsecutive integersis n 

n  1

3.

9. If n isthefirsteveninteger,then n  2isthesecondevenintegerand n  4isthethird.Sothesumofthreeconsecutive evenintegersis n 

11. If s isthethirdtestscore,thensincetheothertestscoresare78and82,theaverageofthethreetestscoresis 78  82  s 3  160  s 3

13. If x dollarsareinvestedat2 1 2 %simpleinterest,thenthefirstyearyouwillreceive0 025 x dollarsininterest.

15. Since  isthewidthoftherectangle,thelengthisfourtimesthewidth,or4 .Then area  length  width  4    4 2 m2

17. If d isthegivendistance,inkilometers,anddistance  rate  time,wehavetime  distance rate  d 55 .

19. If x isthequantityofpurewateradded,themixturewillcontain750gofsaltand3  x litersofwater.Thusthe concentrationis 750 3  x

21. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. If d isthenumberofdaysand m thenumberofkilometers,thenthecostofarentalis C

.Inthiscase, d  3 and C  283,sowesolvefor

220.Thus, thetruckwasdriven220kilometers.

23. If x isthestudent’sscoreontheirfinalexam,thenbecausethefinalcountstwiceasmuchaseachmidterm,theaverage scoreis

86.Sothestudentscored86%ontheirfinalexam.

25. Let m betheamountinvestedat2 1 2 %.Then12,000 m istheamountinvestedat3%. Sincethetotalinterestisequaltotheinterestearnedat2 5%plustheinterestearnedat3%,wehave 318 

8400.Thus, $8400isinvestedat2 1 2 %and12,000 8400  $3600isinvestedat3%.

27. Usingtheformula I  Prt andsolvingfor r ,weget262 50  3500 r 1  r 

5 3500  0 075or7 5%.

29. Let x bethemonthlysalary.Sinceannualsalary  12  monthlysalary  Christmasbonus,wehave 180,100  12 x  7300  172,800  12 x  x  14,400.Themonthlysalaryis$14,400.

31. Let x betheovertimehoursworked.Sincegrosspay  regularsalary  overtimepay,weobtaintheequation 814  18 50 35  18 50 1 5 x  814

6.Thus,thelab technicianworked6hoursofovertime.

33. Allagesareintermsofthedaughter’sage7yearsago.Let y beageofthedaughter7yearsago.Then11 y istheageofthe moviestar7yearsago.Today,thedaughteris y  7,andthemoviestaris11 y  7.Butthemoviestarisalso4timeshis daughter’sagetoday.So4

3.Thus,todaythemoviestaris 11 3  7  40yearsold.

35. Let n bethenumberofnickels.Thentherearealso n dimesand n quarters.Thetotalvalueofthecoinsinthepurseisthe sumofthevaluesofnickels,dimes,andquarters,so2

7.Sothe pursecontains7nickels,7dimes,and7quarters.

37. Let l bethelengthofthegarden.Sincearea  width length,weobtaintheequation1125  25

45m.So thegardenis45meterslong.

39. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let  bethewidthofthebuildinglot.Thenthelengthofthebuildinglotis5 .Sinceahalf­hectareis 1 2 10,000  5000 andareaislengthtimeswidth,wehave5000   5 

31 6.Thusthewidthofthe buildinglotis316metersandthelengthofthebuildinglotis5 316  158meters.

41. Let  bethewidthofthegardeninmeters.Thenthelengthis   10.Thus875

 35 25  0.If   35  0,then

35,whichisimpossible.Therefore  25  0,andso   25. Thegardenis25meterswideand35meterslong.

43. Let  bethewidthofthegardeninmeters.Weusetheperimetertoexpressthelength l ofthegardenintermsof width.Sincetheperimeteristwicethewidthplustwicethelength,wehave

100  .Usingtheformulaforarea,wehave2400

40.Sothelengthis60metersandthewidthis40meters.

45. Let l bethelengthofthelotinmeters.Thenthelengthofthediagonalis l  10. WeapplythePythagoreanTheoremwiththehypotenuseasthediagonal.So l 2

Thusthelengthofthelotis120meters.

47.(a) Firstwewriteaformulafortheareaofthefigureintermsof x .Region A has dimensions10cmand x cmandregion B hasdimensions6cmand x cm.Sothe shadedregionhasarea 10 x   6 x   16 x cm2 .Wearegiventhatthisisequal to144cm2 ,so144  16 x  x  144 16  9cm.

(b) Firstwewriteaformulafortheareaofthefigureintermsof x .Region A has dimensions14cmand x cmandregion B hasdimensions 13  x  cmand x cm. Sotheareaofthefigureis

Wearegiventhatthisisequalto160cm2 ,so160  x 2  27 x  x 2  27 x 160  0   x  32

49. Let x bethewidthofthestrip.Thenthelengthofthematis50  2 x ,andthewidthofthematis38  2 x .Nowthe perimeteristwicethelengthplustwicethewidth,so256  2 50  2 x   2

8

 x  10.Thusthestripofmatis10centimeterswide.

51. Let h betheheighttheladderreaches(inmeters).UsingthePythagoreanTheoremwehave

54meters.

53. Let x bethelengthoftheperson’sshadow,inmeters.Usingsimilartriangles,

20  x  5.Thustheperson’sshadowis5meterslong.

55. Let x betheamount(inmL)of60%acidsolutiontobeused.Then300 x mLof30%solutionwouldhavetobeusedto yieldatotalof300mLofsolution.

So200mLof60%acidsolutionmustbemixedwith100mLof30%solutiontoget300mLof50%acidsolution.

57. Let x bethenumberofgramsofsilveradded.Theweightoftheringsis5  18g  90g.

5rings Puresilver Mixture

mustbeaddedtogettherequiredmixture.

59. Let x bethenumberoflitersofcoolantremovedandreplacedbywater.

Water Mixture

mustberemovedandreplacedbywater.

61. Let c betheconcentrationoffruitjuiceinthecheaperbrand.Thenewmixtureconsistsof650mLoftheoriginalfruitpunch and100mLofthecheaperfruitpunch.

35%fruitjuice.

63. Let t bethetimeinminutesitwouldtaketowashthecarifthefriendsworkedtogether.Friend1washes 1 25 ofthecarper minute,whileFriend2washes 1 35 ofthecarperminute.Thesumofthese fractionsisequaltothefractionofthejobthey candoworkingtogether,sowehave

65. Let t bethenumberofhoursitwouldtakeyourfriendtopaintahousealone.Thenworkingtogether,ittakes 2 3 t hours. Becauseittakesyou7hours,wehave

wouldtakeyourfriend3 5htopaintahousealone.

67. Let t bethetimeinhoursthatittakesyoutowashallthewindows.Thenittakesyourroommate t  3 2 hoursto washallthewindows,andthesumofthefractionsofthejobyoucandoindividuallyperhourequalsthefraction ofthejobyoucandotogether.Since1hour48minutes

0isimpossible,youcanwashthewindowsalonein3hours,andittakesyourroommate

69. Let t bethetimeinhoursthatthecommuterspentonthetrain.Then 11 2 t isthetimeinhoursthatcommuterspentonthe bus.Weconstructatable:

Thetotaldistancetraveledisthesumofthedistancestraveledbybusandbytrain,so480 

71. Let r bethespeedoftheplanefromMontrealtoLosAngeles.Then r  0 20r  1 20r isthespeedoftheplanefromLos AngelestoMontreal. Rate Time Distance MontrealtoL.A. r

Thetotaltimeisthesumofthetimeseachway,so9

ataspeedof800km/honthetripfromMontrealtoLosAngeles.

73. Let x betherate,inkm/h,atwhichthesalespersondrovebetweenAjaxandBarrington.

Cities Distance Rate Time

Ajax  Barrington

x Barrington  Collins

x  16

x  16

Wehaveusedtheequationtime  distance rate tofillinthe“Time”columnofthetable.Sincethesecondpartofthetrip took6minutes(or 1 10 hour)morethanthefirst,wecanusethetimecolumntogettheequation

salesmandroveeither80km/hor384km/hbetweenAjaxandBarrington.(The formerseemslikelier.)

75. Let r betherowingrateinkm/hofthecrewinstillwater.Thentheirrateupstreamwas r 3km/h,andtheirrate downstreamwas r  3km/h.

Sincethetimetorowupstreamplusthetimetorowdownstreamwas2hours40minutes  8 3 hour,wegettheequation

r 6  0  r  6.Sotherateoftherowingcrewinstillwateris6km/h.

77. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x bethedistancefromthefulcrumtowherethe56­kgfriendsits.Thensubstitutingtheknownvaluesintotheformula given,wehave45 2  56

x  1 6.Sothe56­kgfriendshouldsit1 6metersfromthefulcrum.

79. Thevolumeis180m3 ,so

solution.Sotheboxis2metersby6metersby15meters.

81. Let x bethelengthofonesideofthecardboard,sowestartwithapieceofcardboard x by x .When4centimetersare removedfromeachside,thebaseoftheboxis x 8by x 8.Sincethevolumeis100cm3 ,weget4  x 82 

 0 So x  3or x  13.But x  3isnotpossible,since thenthelengthofthebasewouldbe3 8 5 andalllengthsmustbepositive.Thus x  13,andthepieceofcardboard is13centimetersby13centimeters.

83. Let r betheradiusofthetank,inmeters.Thevolumeofthesphericaltankis 4 3 r 3 andisalso2840  0001  284.So 4 3 r 3  2 84  r 3  0 678  r  0 88meters.

85. Let x bethelength,inkilometers,oftheabandonedroadtobeused.Thenthelengthoftheabandonedroadnotusedis 40 x ,andthelengthofthenewroadis 102  40 x 2 kilometers,bythePythagoreanTheorem.Sincethecostof theroadiscostperkilometer  numberofkilometers,wehave100,000 x  200,000 x 2 80 x  1700  6,800,000  2 x 2 80 x  1700  68 x Squaringbothsides,weget4 x 2 320 x  6800  4624 136 x  x 2  3 x 2 184 x

16.Since45 1 3 islongerthanthe existingroad,16kilometersoftheabandonedroadshouldbeused.Acompletelynewroadwouldhavelength 102  402 (let x  0)andwouldcost 1700  200,000  8 3milliondollars.Sono,itwouldnotbecheaper.

87. Let x betheheightofthepileinmeters.Thenthediameteris3 x andtheradiusis 3 2 x meters.Sincethevolumeofthecone is28m3 ,wehave 

89. Let h betheheightinmetersofthestructure.Thestructureiscomposedofarightcylinderwithradius3andheight 2 3 h and aconewithbaseradius3andheight 1 3 h .Usingtheformulasforthevolumeofa cylinderandthatofacone,weobtain theequation40

7.Thustheheightofthe structureis5 7meters.

91. Let h betheheightofthebreak,inmeters.Thentheportionofthebambooabove thebreakis10 h .ApplyingthePythagoreanTheorem,weobtain

h

55mabovetheground.

93. Let x equaltheoriginallengthofthereedincubits.Then x 1isthepiecethatfits60timesalongthelength ofthefield,thatis,thelengthis60  x 1.Thewidthis30 x .Thenconvertingcubitstoninda,wehave 375

x 5.Since x mustbepositive,theoriginallengthofthereedis6cubits.

1.8 INEQUALITIES

1.(a) If x  5,then x 3  5 3  x 3  2.

(b) If x  5,then3 x  3 5  3 x  15.

(c) If x  2,then 3 x 3 2 3 x 6.

(d) If x  2,then x  2.

3.(a) Thesolutionoftheinequality  x   3istheinterval [ 3 3].Anynumber thatliesinsidethisintervalsatisfiestheinequality. _33

(b) Thesolutionoftheinequality  x   3istheunionofintervals  3]  [3 .Anynumberthatlieinsideoneoftheseintervals satisfiestheinequality. _33

5.(a) No.Forexample,if x 2,then x  x  1 2  1  2  0,but x  0.

(b) No.Forexample,if x  2,then x  x  1  2 3  6  5,but x  5.

1 0 isundefined;no

Theelements3and5satisfytheinequality.

Interval: [ 3 1.Graph: _3_1

0.Theexpressionontheleftoftheinequalitychangessignwhere x 2andwhere x  3.Thuswe mustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x 2  x  3.Interval:

Graph: _23

39. x 2 x  7  0.Theexpressionontheleftoftheinequalitychangessignwhere x  0andwhere x  7 2 .Thuswemust checktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis  x  x  7 2 or0  x .

Signof x

Signof2 x

Signof x

Graph: 70 2 41. x 2 3 x

0.Theexpressionontheleftoftheinequalitychangessignwhere x  6andwhere x 3.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis x 

Graph: _36

0.Theexpressionontheleftoftheinequalitychangessign where x 2andwhere x  1 3 .Thuswemustchecktheintervalsinthefollowingtable. Interval

Signof3 x

Signof

Fromthetable,thesolutionsetis  x  x 2or x  1 3 .

Interval:  2]   1 3  .

Graph: 1 3 _2

Fromthetable,thesolutionsetis

Graph: _14

0.Theexpressionontheleftoftheinequalitychangessign where x  6andwhere x 3.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x  x  3or6  x .

Graph: _36

0.Theexpressionontheleftoftheinequalitychangessignwhere x 2and where x  2.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

.Interval:

Signof x  2 

Signof x 2

Signof

Graph: _22

51.  x  2 x 1 x 3  0.Theexpressionontheleftoftheinequalitychangessignwhen x

3. Thuswemustchecktheintervalsinthefollowingtable.

Signof x  2

Signof x 1  

Signof x 3 

Fromthetable,thesolutionsetis

Graph: _23 1

53.  x 4 x  22

2,sotheexpressionontheleftoftheoriginalinequalitychanges signonlywhen x  4.Wechecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Signof x 4

Signof

x  x 2and x  4.Weexcludethe endpoint 2sincetheoriginalexpressioncannot be0.Interval:

Graph: _24

55.  x  32  x 2 x  5  0.Theleft­handsideis0when x 5, 3,or2.Wechecktheintervalsinthefollowingtable.

Signof x 

Signof x 2

When x 3,theleft­handsideisequalto0andtheinequalityissatisfied.Thus,the solutionsetis

0.Theexpressionontheleftoftheinequalitychangessignwhere x  0, x 2andwhere x  4.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

0.Theexpressionontheleftoftheinequality changessignwhere x

Signof x

Signof x

Signof x 

Signof x 2

Fromthetable,thesolutionsetis

.Graph: _11

61. x 3 x  1  0.Theexpressionontheleftoftheinequalitychangessignwhere x 1andwhere x  3.Thuswemustcheck theintervalsinthefollowingtable. Interval

Signof x

Signof x 3

Signof x 3 x 

Fromthetable,thesolutionsetis

x  x  1or x  3.Sincethedenominator cannotequal0wemusthave x 1.

Interval: 

Graph: _13

oftheinequalitychangessignwhen x  2and x  5 2 .Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

0.Theexpressionontheleftoftheinequality changessignwhere x  16andwhere x  5.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

x  x  5or x  16.Sincethedenominator cannotequal0,wemusthave x  5.

Signof x 5

Signof x  16 x 5

Interval:

Graph: 516

x  0.Theexpressionontheleftofthe inequalitychangessignwhere x  0,where x 2,andwhere x  2.Thuswemustchecktheintervalsinthefollowing table.

Signof2

Signof x

Signof2 x

Signof

Fromthetable,thesolutionsetis x 2  x  0or2  x .Interval: 

.Graph: _22 0

Signof

Signof

Since x 1and x  0yieldundefinedexpressions,wecannotincludetheminthesolution.From thetable,thesolution

0.Theexpressionontheleftoftheinequality changessignwhere

Fromthetable,thesolutionsetis

Graph: _44

Graph: _713

Graph:

.Theexpressionontheleftoftheinequalitychangessign when x 4, x  0,and x  4.Thuswemustchecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis

Fromthetable,thesolutionsetis

Graph: _3_2

0forall x

3,sotheexpressionontheleftoftheoriginalinequalitychanges signonlywhen x 1.Wechecktheintervalsinthefollowingtable.

Fromthetable,thesolutionsetis x

(Theendpoint 3isalreadyexcluded.)

Interval:

inequalitychangessignat

Thus x 3or x  3.

111. For  1 x 2 3 x 10 

expressioninthelastinequalitychangessignat x 2and x  5.

117. Insertingtherelationship

68  F  86.

119. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let x betheaveragenumberofkilometersdrivenperday.EachdaythecostofPlan Ais95  0 40 x ,andthecostofPlanB is135.PlanBsavesmoneywhen135 



.SoPlanBsavesmoneywhenyouaverage morethan100kilometersaday.

121. Weneedtosolve6400  0

 4900  12,000  m  14,000.Sheplansondrivingbetween12,000and14,000miles.

123.(a) Let x bethenumberof$3increases.Thenthenumberofseatssoldis120 x .So P  200  3 x  3 x  P 200  x  1 3 

(b)

200.Substitutingfor x wehavethatthenumberofseatssoldis

290  P  215.Puttingthisintostandardorder,wehave215  P  290.Sotheticketpricesarebetween$215and $290.

125. 0 0004  4,000,000 d 2  0 01.Since d 2  0and d  0,wecanmultiplyeachexpressionby d 2 toobtain

0 0004d 2  4,000,000  0 01d 2 .Solvingeachpair,wehave0 0004d 2  4,000,000  d 2  10,000,000,000  d  100,000(recallthat d representsdistance,soitisalwaysnonnegative).Solving4,000,000  0 01d 2 

400,000,000  d 2  20,000  d .Puttingthesetogether,wehave20,000  d  100,000.

127. Note:Inthefirstprintingofthetext,theanswergivenforpart(a)ofthisexerciseisincorrect.

(a) Wesubstitute d  402mand t  10sintothegivenformulaandfindthevalueof a thatresultsinaquarter­miletime ofexactly10s:402  1 2

s2 .Thus,thequarter­miletimewillbelessthan 10sif a  8 04ms2

(b) Wesubstitute a

8 .Takingthepositiveroot,wefind thatthequarter­miletimefora(downward)quarter­mileunderEarth’sgravityis

129. 70 

and  70.However,since  representsthespeed,wemusthave   0.

Soyoumustdrivebetween0and35km/h.

131. Let x bethelengthofthegardenand  itswidth.Usingthefactthattheperimeteris120m,wemusthave2 x 

   60 x .Nowsincetheareamustbeatleast800m2 ,wehave800

x 2 60 x 

0.Theexpressionintheinequalitychangessignat x  20and x  40. However,since x representslength,wemusthave x  0.

Signof x 20

Signof x 40

Signof  x 20

x 40

Thelengthofthegardenshouldbebetween20and40meters.

arebetween156cmand185cm.

135. Therulewewanttoapplyhereis“a  b  ac  bc if c  0and a  b  ac  bc if c  0”.Thuswecannotsimply multiplyby x ,sincewedon’tyetknowif x ispositiveornegative,soinsolving1  3 x ,wemustconsidertwocases.

Case1:x  0Multiplyingbothsidesby x ,wehave x  3.Togetherwithourinitialcondition,wehave0  x  3.

Case2:x  0Multiplyingbothsidesby x ,wehave x  3.But x  0and x  3havenoelementsincommon,sothis givesnoadditionalsolution.Hence,theonlysolutionsare0  x  3.

137. a  b ,sobyRule1, a  c  b  c.UsingRule1again, b  c  b  d ,andsobytransitivity, a  c  b  d

139.(a) Because x isnonnegative, x  y  x 2  xy ,andbecause y isnonnegative, x  y  xy  y 2 .Thus,bytransitivity, x 2  y 2 .

(b) Bypart(a),  xy  x  y 2  xy   x  y 2 4 .Expanding,thisbecomes4 xy   x  y 2  x 2  y 2  2 xy  x 2  y 2 2 xy  0   x

 0.Thisistrueforany x and y ,sotheoriginalinequalityistrueforallnonnegative x and y

1.9 THECOORDINATEPLANE;GRAPHSOFEQUATIONS;CIRCLES

1.(a) Thepointthatis3unitstotherightofthe y ­axisand5unitsbelowthe x ­axishascoordinates 3 5

(b) Thepoint 2 7 is2unitstotherightofthe y ­axisand7unitsabovethe x ­axis,soitisclosertothe y ­axis.

3. Thepointmidwaybetween

.Sothepointmidwaybetween

and

5.(a) Tofindthe x ­intercept(s)ofthegraphofanequationweset y equalto0intheequationandsolvefor x :2 0

 x  1  x 1,sothe x ­interceptof2 y  x  1is 1.

(b) Tofindthe y ­intercept(s)ofthegraphofanequationweset x equalto0intheequationandsolvefor y :2 y  0  1  y  1 2 ,sothe y ­interceptof2 y  x  1is 1 2

7.(a) Ifagraphissymmetricwithrespecttothe x ­axisand a  b  isonthegraph,then a  b isalsoonthegraph.

(b) Ifagraphissymmetricwithrespecttothe y ­axisand a  b  isonthegraph,then  a  b isalsoonthegraph.

(c) Ifagraphissymmetricabouttheoriginand a  b  isonthegraph,then  a  b  isalsoonthegraph.

9. Yes.If a  b  isonthegraph,thenbysymmetryaboutthe x ­axis,thepoint a  b  isonthegraph.Thenbysymmetry aboutthe y ­axis,thepoint  a  b isonthegraph.Thus,thegraphissymmetricwithrespecttotheorigin.

11. Thepointshavecoordinates

Thetwopointsare

23. Thetwopointsare

21.

33. Fromthegraph,thequadrilateral ABCD hasapairofparallelsides,so ABCD is atrapezoid.Theareais

h .Fromthegraphweseethat

39. Sincewedonotknowwhichpairareisosceles,wefindthelengthofallthreesides.

41.(a) Herewehave

,weconcludethatthetriangleisarighttriangle.

(b) Theareaofthetriangleis

43. Weshowthatallsidesarethesamelength(itsarhombus)andthenshowthatthediagonalsareequal.Herewehave

rhombus.Also

therhombusisasquare.

45. Let P  0 y  besuchapoint.Settingthedistancesequalweget

4.Thus,thepoint is P

.Check:

47. AsindicatedbyExample3,wemustfindapoint S  x 1  y1  suchthatthemidpoints of PR andof QS arethesame.Thus

coordinatesequal,weget

49.(a) y

(b) Themidpointof

(c) Sincethetheyhavethesamemidpoint,weconcludethatthe diagonalsbisecteachother.

arepointsonthegraphofthisequation.

interceptis3,and

6 6,sothe y ­interceptis 6. x ­axissymmetry:2 x  y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe x ­axis.

y ­axissymmetry: 2 x y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe y ­axis.

Originsymmetry: 2 x  y  6,whichisnotthesameas

2 x y  6,sothegraphisnotsymmetricwithrespecttothe origin.

and

y ­interceptis2.

x ­axissymmetry: y  2  x 12 ,whichisnotthesameas

y  2  x 12 ,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y  2  x 12 ,whichisnotthesameas y  2  x 12 ,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: y  2  x 12  y 2  x  12 , whichisnotthesameas y  2  x 12 ,sothegraphisnot symmetricwithrespecttotheorigin. 1 1 y x

63.(a) y   x  2

9 5 y  0  0   x  2,sothereisno x ­intercept,and x  0

1

 y  0  2  2,sothe y ­interceptis2.

x ­axissymmetry: y   x  2,whichisnotthesameas y   x  2,sothegraphisnotsymmetricwithrespectto the x ­axis.

y ­axissymmetry: y   x  2,whichisnotthesameas

y   x  2,sothegraphisnotsymmetricwithrespectto the y ­axis.

Originsymmetry: y   x  2  y  x 2, whichisnotthesameas y   x  2,sothegraphisnot symmetricwithrespecttotheorigin. y 1l

x  0,sothe x ­interceptis0,and x  0  y 

 0,sothe y ­interceptis0.

x ­axissymmetry: y

x

,whichisnotthe sameas y   x ,sothegraphisnotsymmetricwith respecttothe x ­axis.

y ­axissymmetry: y 

x    x ,sothegraphis symmetricwithrespecttothe y ­axis.

Originsymmetry: y 

x

x

,whichisnotthe sameas y   x ,sothegraphisnotsymmetricwith respecttotheorigin.

65.(a) y  4 x 2 x y 2 0 1 3 0 2 1 3 2 0

y

y ­interceptis2.

x ­axissymmetry: y  4 x 2 ,whichisnotthesameas y  4 x 2 ,sothegraphisnotsymmetricwithrespectto the x ­axis. y ­axissymmetry: y 

2

4 x 2 ,sothegraph issymmetricwithrespecttothe y ­axis.

Originsymmetry: y  4  x 2  4 x 2 

y 4 x 2 ,whichisnotthesameas y  4 x 2 ,so thegraphisnotsymmetricwithrespecttotheorigin.

(b) y  x 3 4 x y x 3 15 2

,so the x ­interceptsare0and 2,and x  0 

y  03 4 0  0,sothe y ­interceptis0.

x ­axissymmetry:  y   x 3 4 x  y  x 3  4 x ,which isnotthesameas y  x 3 4 x ,sothegraphisnotsymmetric withrespecttothe x ­axis.

y ­axissymmetry:

4

,which isnotthesameas x  x 3 4 x ,sothegraphisnotsymmetric withrespecttothe y ­axis.

Originsymmetry: y 

 y  x 3 4 x ,so thegraphissymmetricwithrespecttotheorigin. y x 1 10l 0

x 3 4  x

67.(a) Tofind x ­intercepts,set y  0.Thisgives0  x  6  x 6,sothe x ­interceptis 6.Tofind y ­intercepts,set x  0.Thisgives y  0  6  6,sothe y ­interceptis6.

(b) Tofind x ­intercepts,set y  0.Thisgives0  x 2 5  x 2  5  x 5,sothe x ­interceptsare 5.Tofind y ­intercepts,set x  0.Thisgives y  02 5 5,sothe y ­interceptis 5.

69.(a) Tofind x ­intercepts,set y  0.Thisgives9 x 2 4 02  36  9 x 2 

x 2  4  x 2,sothe x ­intercepts are 2.Tofind y ­intercepts,set x  0.Thisgives9 0

2 4 y 2  36  y 2 9,sothereisno y ­intercept. (b) Tofind x ­intercepts,set y  0.Thisgives0 2 x 

4 x

1 4 ,sothe x ­interceptis 1 4 .Tofind y ­intercepts, set x  0.Thisgives y 2 0

x

71. Tofind x ­intercepts,set y  0.Thisgives0  4 x x 2 

0  x or x  4,sothe x ­interceptsare0and 4.Tofind y ­intercepts,set x  0.Thisgives y  4 0 02  y  0,sothe y ­interceptis0.

73. Tofind x ­intercepts,set y  0.Thisgives x 4  02 x 0  16  x 4  16  x 2.Sothe x ­interceptsare 2and2. Tofind y ­intercepts,set x  0.Thisgives04  y 2

4.Sothe y ­interceptsare 4and4.

75. x 2  y 2  9hascenter 0 0 andradius3. y 1x 1l 77.  x 22  y 2  9hascenter 2 0 andradius3. x y 1 1l 0

79.  x  32   y 42  25hascenter  3 4 andradius5. y 1x 1

81. Using

83. Theequationofacirclecenteredattheoriginis x 2  y 2  r 2 .Usingthepoint

.Thus,theequationofthecircleis

85. Thecenterisatthemidpointofthelinesegment,whichis

wesolvefor r 2 .Thisgives

.Theradiusisonehalfthe

87. Sincethecircleistangenttothe x ­axis,itmustcontainthepoint 7

,sotheradiusisthechangeinthe y ­coordinates. Thatis, r 

 3.Sotheequationofthecircleis

89. Fromthefigure,thecenterofthecircleisat  2

Thustheequationofthecircleis

91. Completingthesquaregives

.Theradiusisthechangeinthe y ­coordinates,so

Completingthesquaregives2

Thus,thecirclehascenter

97. x ­axissymmetry:  y   x 4

,whichisnotthesameas

x

,sothegraphisnotsymmetric withrespecttothe x ­axis. y ­axissymmetry: y

,whichisnotthesameas y  x 4  x 2 ,sothegraphisnot symmetricwithrespecttotheorigin.

99. x ­axissymmetry:

y ­axissymmetry:

Originsymmetry:

2,sothegraphissymmetricwithrespecttothe origin.

(Notethatifagraphissymmetricwithrespecttoeachcoordinateaxis,itissymmetricwithrespecttotheorigin.The converseisnottrue,asshowninthenextexercise.)

101. x ­axissymmetry:  y   x 3  10 x  y  x 3 10 x ,whichisnotthesameas y  x 3  10 x ,sothegraphisnot symmetricwithrespecttothe x ­axis. y ­axissymmetry: y   x 3  10  x

 y  x 3 10 x ,whichisnotthesameas y  x 3  10 x ,sothegraphisnot symmetricwithrespecttothe y ­axis. Originsymmetry:  y    x 3  10

x

y  x 3 10 x  y  x 3  10 x ,sothegraphissymmetricwithrespect totheorigin.

103. Symmetricwithrespecttothe y ­axis. x y 0 105. Symmetricwithrespecttotheorigin. x y 0

107.  x  y   x 2  y 2  1.Thisisthesetofpointsinside (andon)thecircle

109. Completingthesquaregives x 2  y 2 4

,andthe radiusis4.Sothecircle x 2  y 2  4,withcenter 0

0 andradius2 sitscompletelyinsidethelargercircle.Thus, theareais  42

Thepoint

(b) Thepoint

(c) Let

x

isshiftedto

bethepointthatisshiftedto

113.(a) Symmetricaboutthe x ­axis: x y 1

(b) Symmetricaboutthe y ­axis: x y 1 1 (c) Symmetricabouttheorigin: x y 1 1

(b) Wewantthedistancesfrom

(c) Thetwopointsareonthesameavenueorthesamestreet.

117. Completingthesquaregives

equationrepresentsapointwhen

Whentheequationrepresentsacircle,thecenteris

1.10 LINES

,andtheradiusis

1. Wefindthe“steepness”orslopeofalinepassingthroughtwopointsbydividingthedifferenceinthe y ­coordinatesofthese pointsbythedifferenceinthe x ­coordinates.Sothelinepassingthroughthepoints 

3. Thepoint­slopeformoftheequationofthelinewithslope3passingthroughthepoint

5. Theslopeofahorizontallineis0.Theequationofthehorizontallinepassingthrough

7.(a) Yes,thegraphof y 3isahorizontalline3unitsbelowthe x ­axis.

(b) Yes,thegraphof x 3isaverticalline3unitstotheleftofthe y ­axis.

(c) No,alineperpendiculartoahorizontallineisverticalandhasundefinedslope.

(d) Yes,alineperpendiculartoaverticallineishorizontalandhasslope0.

9. m  y2 y

17. For 1 ,wefindtwopoints,

and

thatlieontheline.Thustheslopeof

19. Firstwefindtwopoints 0 4 and 4 0

theequationofthelineis

21. Wechoosethetwointerceptsaspoints,

theequationofthelineis

23. Using

25. Usingtheequation y y

27. Usingtheequation y

29. Firstwefindtheslope,whichis

thatlieontheline.Sotheslopeis

31. Wearegiventwopoints, 2 5 and 5 1.Thus,theslopeis m

33. Wearegiventwopoints, 1

and 0 3.Thus,theslopeis

35. Sincetheequationofalinewithslope0passingthrough a

37. Sincetheequationofalinewithundefinedslopepassingthrough a  b is x  a ,theequationofthislineis x  2.

39. Anylineparallelto y  2 x  8hasslope2.Thedesiredlinepassesthrough  1 4,sosubstitutinginto y y1

, weget y 4

41. Sincetheequationofahorizontallinepassingthrough a

b  is y  b ,theequationofthehorizontallinepassingthrough 4 5 is y

43. Since3 x 

2,theslopeofthislineis 3 2 .Thus,thelineweseekisgivenby

45. Anylineparallelto x  5hasundefinedslopeandanequationoftheform x  a .Thus,anequationofthelineis x 1.

47. Firstfindtheslopeof3

.Thus,theslopeofanylineperpendicular to3 x 

49. Firstfindtheslopeofthelinepassingthrough

_4 _2 0 2 4

m=0 m= 3 2 m= 3 4 m=_ 3 4 m= 1 4 m=_ 3 2 m=_ 1 4 y  m  x 3, m  0,  1 4 ,  3 4 ,  3 2 .Eachofthelines

containsthepoint 3 0 becausethepoint 3 0 satisfies

eachequation y  m  x 3.Since 3 0 isonthe x ­axis, wecouldalsosaythattheyallhavethesame x ­intercept.

59. 2 x  y  7  y  2 x  7.Sotheslopeis2andthe y ­interceptis7.

57. y  x 4.Sotheslopeis1andthe y ­interceptis 4. y 1x 1

2.So theslopeis 4 5 andthe y ­interceptis2. y 1x 1

63. y  4canalsobeexpressedas y  0 x  4.Sotheslopeis 0andthe y ­interceptis4. y 1x 1l

65. x  3cannotbeexpressedintheform y  mx  b.Sothe slopeisundefined,andthereisno y ­intercept.Thisisa verticalline. y 1x 1l

67. 3 x 2 y 6  0.Tofind x ­intercepts,weset y  0and solvefor x :3 x 2 0 6  0  3 x  6  x  2,so the x ­interceptis2.

Tofind y ­intercepts,weset x  0andsolvefor y :

3 0 2 y 6  0  2 y 6  y 3,sothe y ­interceptis 3. y 1x 1

69. 1 2 x 1 3 y  1  0.Tofind x ­intercepts,weset y  0and solvefor x : 1 2 x 1 3 0  1  0  1 2 x 1  x 2, sothe x ­interceptis 2.

Tofind y ­intercepts,weset x  0andsolvefor y : 1 2 0

y ­interceptis3. y 1x 1

71. y  6 x  4.Tofind x ­intercepts,weset y  0andsolve for x :0  6 x  4  6 x 4  x  2 3 ,sothe x ­interceptis 2 3 .

Tofind y ­intercepts,weset x  0andsolvefor y : y  6 0  4  4,sothe y ­interceptis4. y 1x 1

73. Todetermineifthelinesareparallel orperpendicular,wefindtheirslopes.Thelinewithequation y  2 x  3hasslope2. Thelinewithequation2 y 4 x 5  0

alsohasslope2,andsothelinesareparallel.

75. Todetermineifthelinesareparallelorperpendicular,wefindtheirslopes.Thelinewithequation2x 5 y  8  5 y  2 x 8  y  2 5 x 8 5 hasslope 2 5 .Thelinewithequation10 x  4 y

slope 5 2  1 25 ,andsothelinesareperpendicular.

77. Todetermineifthelinesareparallelorperpendicular,wefindtheirslopes.Thelinewithequation7x 3 y  2  3 y  7 x 2  y  7 3 x 2 3 hasslope 7 3 .Thelinewithequation9 y  21 x

slope 7 3  1 73 ,andsothelinesareneitherparallelnorperpendicular.

79. Wefirstplotthepointstofindthepairsofpointsthatdetermineeachside.Nextwe findtheslopesofoppositesides.Theslopeof AB is 4 1 7 1  3 6  1 2 ,andthe slopeof DC is 10 7 5  1  3 6  1 2 .Sincetheseslopeareequal,thesetwosides areparallel.Theslopeof AD is 7 1 1 1  6 2 3,andtheslopeof BC is 10 4 5 7  6 2 3.Sincetheseslopeareequal, thesetwosidesareparallel. Hence ABCD isaparallelogram. y 1x 1lA

81. Wefirstplotthepointstofindthepairsofpointsthatdetermineeachside.Nextwe findtheslopesofoppositesides.Theslopeof AB is 3 1 11 1  2 10  1 5 andthe slopeof DC is 6 8 0 10  2 10  1 5 .Sincetheseslopeareequal,thesetwosides areparallel.Slopeof AD is 6 1 0 1  5 1 5,andtheslopeof BC is 3 8 11 10  5 1 5.Sincetheseslopeareequal,thesetwosidesareparallel. Since slopeof AB   slopeof AD   1 5 

1,thefirsttwosidesare eachperpendiculartothesecondtwosides.Sothesidesformarectangle.

83. Weneedtheslopeandthemidpointoftheline AB .Themidpointof AB is

,andtheslopeof AB is m  2 4 7 1  6 6 1.Theslopeoftheperpendicularbisectorwillhaveslope 1 m  1 1  1.Usingthe point­slopeform,theequationof theperpendicularbisectoris y 1  1  x 4 or x

85.(a) Westartwiththetwopoints a

.Theslopeofthelinethatcontainsthemis b 0 0 a  b a .Sotheequation ofthelinecontainingthemis y  b a x  b (usingtheslope­interceptform).Dividingby b (since b  0)gives y b 

and

b

1. (b) Setting a  6and b 8,weget x 6  y 8  1  4

87. Usingthediagramprovided,wecalculatethecoordinatesofthemidpoints: m 1 hascoordinates  0 

Thelinejoiningthemidpointsthushasslope

 0,andsoitisparalleltothethirdside(asegmentofthe x ­axis).

Thelinejoiningthemidpointshaslength

,halfthelengthofthethird sideofthetriangle;andsowehaveverifiedbothconclusionsofthetheorem.

89.(a) Theslopeis0 0417 D  0 0417 200  8 34.Itrepresentstheincreaseindosageforeachone­yearincreaseinthe child’sage.

(b) When a  0, c  8 34 0  1  8 34mg.

91.(a) y x 5000l 10,000l 5001000

(b) Theslopeisthecostpertoasteroven,$6.The y ­intercept,$3000,is themonthlyfixedcost—thecostthatisincurrednomatterhowmany toasterovensareproduced.

93. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

(a) Using n inplaceof x and t inplaceof y ,wefindthattheslopeis t2 t1 n 2 n

(b) When n  150,thetemperatureisapproximatelygivenby t  1 8 150  6  24 75 C  25 C.

95. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

(a) Wearegiven changeinpressure 3­meterchangeindepth  29 9 3  9 97.Using P for pressureand d fordepth,andusingthepoint P  103when d  0, wehave P 103  9 97 d 0  P  9 97d  103.

(c) Thesloperepresentstheincreaseinpressurepermeterofdescent. The y ­interceptrepresentsthepressureatthesurface.

(d) When P  690,wehave690  9 97d  103  9 97d  587  d  58 9m.Thusthepressureis690kPaatadepthofapproximately 59m.

(b) y x

l

l

l

l

l 204080120 60100

97. Welabelthethreepoints A, B ,and C .Iftheslopeofthelinesegment AB isequaltotheslopeofthelinesegment BC , thenthepoints A, B ,and C arecollinear.Usingthedistanceformula,wefindthedistancebetween A and B ,between B and C ,andbetween A and C .Ifthesumofthetwosmallerdistancesequalsthelargestdistance,thepoints A, B ,and C are collinear.

Anothermethod: Findanequationforthelinethrough A and B .Thencheckif C satisfiestheequation.Ifso,thepointsare collinear.

1.11

SOLVINGEQUATIONSANDINEQUALITIESGRAPHICALLY

1. Thesolutionsoftheequation x 2 2 x 3  0arethe x ­interceptsofthegraphof y  x 2 2 x 3.

3.(a) Fromthegraph,itappearsthatthegraphof y  x 4 3 x 3 x 2  3 x has x ­intercepts 1,0,1,and3,sothesolutions totheequation x 4 3 x 3 x 2  3 x  0are x 1, x 

(b) Fromthegraph,weseethatwhere 1  x  0or1  x  3,thegraphliesbelowthe x ­axis.Thus,theinequality x 4 3 x 3 x 2  3 x  0issatisfiedfor

5.(a)

x

y 

(c) Thegraphisnotsymmetricwithrespecttoeitheraxisortheorigin.

1 y  2 x 2  1 ; [ 5 5] by [ 3 1]

(b) No x ­intercept; y ­intercept 2. y  0hasnosolution; x  0 

(c) Thegraphissymmetricwithrespecttothe y ­axis:

9. Althoughthegraphsof y 3 x 2  6 x 1 2 and y  7 7 12 x 2 appeartointersectintheviewing rectangle [ 4 4] by [ 1 3],thereisnopointof intersection.Youcanverifythisbyzoomingin.

20 13. Algebraically:3 x  2  5 x 4  6

2 4 ­1 1 2 3 11. Thegraphsof y  6 4 x x 2 and y  3 x  18appearto havetwopointsofintersectionintheviewingrectangle [ 6 2]by[ 5 20].Youcanverifythat x 4and x 3areexactsolutions.

4 ­2 2

Graphically:Wegraphthetwoequations y1  3 x  2and y2  5 x 4intheviewingrectangle [1 4] by [ 1 13].

Zoomingin,weseethatthesolutionis x  3. ­1 1 2 3

15. Algebraically:

Graphically:Wegraphthetwoequations y1  2 x  1 2 x and y2  7intheviewingrectangle[ 2 2]by[ 2 8].

Zoomingin,weseethatthesolutionis x  0 36.

17. Algebraically:4 x 2 8  0  x 2  2  x 2.

Graphically:Wegraphtheequation y1  4 x 2 8and determinewherethiscurveintersectsthe x ­axis.Weuse theviewingrectangle [ 2 2] by [ 4 4].Zoomingin,we seethatsolutionsare x  1

21. Algebraically:81 x 4  256

Graphically:Wegraphthetwoequations y1  81 x 4 and y2  256intheviewingrectangle [ 2 2] by [250 260]. Zoomingin,weseethatsolutionsare x 1 33.

19. Algebraically: x 2  9  0  x 2 9,whichhasnoreal solution.

Graphically:Wegraphtheequation y  x 2  9andsee thatthiscurvedoesnotintersectthe x ­axis.Weusethe viewingrectangle [ 5 5] by [ 5 30]

1 0 1 2

25. Wegraph y  x 2 11 x  30intheviewingrectangle

[2 8]by[ 0 1 0 1].Thesolutionsappeartobeexactly x  5and x  6.[Infact x 2 11

Algebraically:

Graphically:Wegraphtheequation y1   x 54 80 anddeterminewherethiscurveintersectsthe x ­axis.We usetheviewingrectangle [ 1 9] by [ 5 5].Zoomingin, weseethatsolutionsare x  2 01and x  7 99.

27. Wegraph y  x 3 6 x 2  11 x 6intheviewing rectangle[ 1 4]by[ 0 1 0 1].Thesolutionsare x  100, x  200,and x  300.

29. Wefirstgraph y  x  x  1intheviewingrectangle [ 1 5]by[ 01 01]andfindthatthesolutionisnear 1 6.Zoomingin,weseethatsolutionsis x  1 62.

31. Wegraph y  x 13 x intheviewingrectangle [ 3 3] by[ 1 1].Thesolutionsare x 1, x  0,and x  1, ascanbeverifiedbysubstitution.

33. Wegraph y  2 x 1and y  3 x 5intheviewingrectangle [0 9] by [0 5] andseethattheonlysolutiontotheequation 2 x 1  3 x 5is x  4,which canbeverifiedbysubstitution.

35. Wegraph y  2 x  1  1and y  x intheviewingrectangle [ 1 6] by [0 6] andseethattheonlysolutiontotheequation 2 x  1  1  x is x  4,whichcan beverifiedbysubstitution.

37. x 3 2 x 2 x 1  0,sowestartbygraphing

thefunction y  x 3 2 x 2 x 1intheviewing rectangle [ 10 10] by [ 100 100].There appeartobetwosolutions,onenear x  0and anotheronebetween x  2and x  3.Wethen usetheviewingrectangle[ 1 5]by[ 1 1]and zoominontheonlysolution, x  2 55.

39. x  x 1 x  2  1 6 x  x  x 1 x  2 1 6 x  0.Westartbygraphing

thefunction y  x  x 1 x  2 1 6 x inthe

viewingrectangle [ 5 5] by [ 10 10].There appeartobethreesolutions.Wethenusethe

viewingrectangle[ 2 5 2 5]by[ 1 1]and zoomintothesolutionsat x 205, x  000, and x  1 05.

41. Wegraph y  x 2 and y  3 x  10intheviewing rectangle[ 4 7]by[ 5 30].Thesolutiontothe inequalityis [ 2 5]

43. Since x 3  11 x  6 x

x

wegraph y  x 3 6 x 2  11 x 6intheviewing rectangle[0 5]by[ 5 5].Thesolutionsetis

 1 00]  [2 00 3 00] 2 4

45. Since x 13  x  x 13 x  0,wegraph y  x 13 x intheviewingrectangle [ 3 3] by [ 1 1].Fromthis,we findthatthesolutionsetis

1

1

AnotherMethod: AsinExample7,wegraph y1  x 13 and y2  x inthesameviewingrectangle,andseethat x 13  x for 1  x  0andfor1  x

47. Since  x  12   x 12   x 

wegraph y 

2 intheviewing rectangle [ 2 2] by [ 5

49. Wegraphtheequations y  3 x 2 3 x and y  2 x 2

4intheviewingrectangle [ 2 6] by [ 10 50].Weseethatthetwocurvesintersectat x 1andat x  4, andthatthefirstcurveislowerthanthesecondfor 1  x  4.Thus,weseethat theinequality3 x 2 3 x  2 x 2

4hassolutionset

51. Wegraphtheequation y 

intheviewingrectangle [ 7 3] by [ 120 20] andseethattheinequality

thesolutionset

53. Tosolve5 3 x  8 x 20bydrawingthegraphofasingleequation,weisolate alltermsontheleft­handside:5 3 x  8 x 20 

5 3 x 8 x  20  8 x 20 8 x  20 11 x  25  0or11 x 25  0. Wegraph y  11 x 25,andseethatthesolutionis x  2 27,asinExample2.

55.(a) Wegraphtheequation y  10 x  0 5 x 2 0 001 x 3 5000intheviewing rectangle [0 450] by [ 5000 20000] 100 200 300 400 0 10000 20000

(b) Fromthegraphitappearsthat

0  10 x  005 x 2 0001 x 3 5000for 100  x  500,andso101cooktopsmustbeproduced to begin tomakeaprofit.

57. Answerswillvary.

1.12 MODELINGVARIATION

(c) Wegraphtheequations y  15,000and y  10 x  0 5 x 2 0 001 x 3 5000intheviewing rectangle [250 450] by [11000 17000].Weuseazoom ortracefunctiononagraphingcalculator,andfindthat thecompany’sprofitsaregreaterthan$15,000for 279  x  400.

1. Ifthequantities x and y arerelatedbytheequation y  5 x thenwesaythat y is directlyproportional to x ,andtheconstant of proportionality is5.

3. Ifthequantities x , y ,and z arerelatedbytheequation z  5 x y thenwesaythat z is directlyproportional to x and inversely proportional to y

5.(a) Intheequation y  3 x , y isdirectlyproportionalto x (b) Intheequation y  3 x  1, y isnotproportionalto x

7. T  kx ,where k isconstant.

9.   k z ,where k isconstant.

11. y  ks t ,where k isconstant. 13. z  k  y ,where k isconstant.

15. V  kl  h ,where k isconstant.

17. R  kP 2 t 2 b 3 ,where k isconstant.

19. Since y isdirectlyproportionalto x , y  kx .Since y  32when x  8,wehave32  k 8  k  4.So y  4 x

21. A variesinverselyas r ,so A  k r .Since A  15when r  5,wehave15  k 5  k  75.So A  75 r

23. Since A isdirectlyproportionalto x andinverselyproportionalto t , A  kx t .Since A  42when x  7and t  3,we have42  k 7 3  k  18.Therefore, A  18 x t

25. Since W isinverselyproportionaltothesquareof r , W  k r 2 .Since W  24when r  3,wehave24  k 32  k  216. So W  216 r 2 .

27. Since C isjointlyproportionalto l ,  ,and h ,wehave C  kl  h .Since C  128when l

29. R  k  x .Since R  2 5when x

31.(a) z  k x 3 y 2

(b) Ifwereplace x with3 x and y with2 y ,then

33.(a) z  kx 3 y 5

(b) Ifwereplace x with3 x and y with2 y ,then z

35.(a) Theforce F neededis F  kx

2,wehave

(b) Since F  30Nwhen x  9cmandthespring’snaturallengthis5cm,wehave30  k 9 5  k  7 5Ncm.

(c) Frompart(b),wehave F  75 x .Substituting x  11 5  6into F  7

5 x gives F  75 6  45N.

37.(a) P  ks 3

(b) Since P  96when s  20,weget96  k 203  k  0 012.So P  0 012s 3

(c) Substituting x  30,weget P  0 012 303  324watts.

39. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

D  ks 2 .Since D  45when s  60,wehave45  k 602 ,so k  0 0125.Thus, D  0 0125s 2 .If D  60,then 60  0 0125s 2  s 2  4800,so s  69km/h(forsafetyreasonswerounddown).

41. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect.

F  kAs 2 .Since F  980when A  4and s  8,wecansolvefor k :980  k 482  980  256k  k  3 83.Now when A  2 5and F  780weget780  3 83 2 5 s 2  81 4621  s 2 ,so s  81 4621  9 03km/h.

43.(a) P  kT V

(b) Substituting P  33 2, T  400,and V  100,weget33 2  k 400 100  k  8 3.Thus k  8 3andtheequationis P  8 3 T V

(c) Substituting T  500and V  80,wehave P 

51 875kPa.Hencethepressureofthesampleofgasis about51 9kPa.

45.(a) Theloudness L isinverselyproportionaltothesquareofthedistance d ,so L  k d 2

(b) Substituting d  10and L  70,wehave70  k 102  k  7000.

(c) Substituting2d for d ,wehave L  k

,sotheloudnessischangedbyafactorof 1 4 .

(d) Substituting 1 2 d for d ,wehave L  k  1 2 d 2  4  k d 2 ,sotheloudnessischangedbyafactorof4.

47.(a) R  kL d 2

(b) Since R  140when L  1 2and d  0 005,weget140  k 1 2

(c) Substituting L  3and d  0008,wehave R 

(d) Ifwesubstitute2d for d and3 L for L ,then R  k

3 L 

2  k  7 2400  0 002916.

 4375 32  137ohms.

2  3 4 kL d 2 ,sotheresistanceischangedbyafactorof 3 4

49. Note:Inthefirstprintingofthetext,theanswergivenforpart(b)ofthisexerciseisincorrect.

(a) Forthesun, E S  k 60004 andforearth, E E  k 3004 .Thus E S E E  k 60004 k 3004   6000 300 4  204  160,000.Sothesun produces160,000timestheradiationenergyperunitareathantheEarth.

(b) Thesurfaceareaofthesunis4 696,0002 andthesurfaceareaoftheEarthis4 63402 .Sothesunhas

4 696,0002 4 63402   696,000 6340 2 timesthesurfaceareaoftheEarth.Thusthetotalradiationemittedbythe sunis

160,000   696,000 6340 2  1,928,234,931timesthetotalradiationemittedbytheEarth.

51.(a) Since f isinverselyproportionalto L ,wehave f  k L ,where k isapositiveconstant.

(b) Ifwereplace L by2 L wehave k 2 L  1 2 k L  1 2 f .Sothefrequencyofthevibrationiscutinhalf.

53. Wesubstitute e  0 2kWhkmand   100km/hintothesecondgivenequation:0 2 

Now,becausetherangeis R  500kmforthesevaluesof e and  ,wecanusethefirstequationtocalculatethevalueof C :

500  C 02  C  100,soafullchargeforthisvehicleis100kWh.

Substituting e  k  2 intothefirstgivenequation,wehave R  C k  2

Soat130kmh,itsrangeis R 

55. Using B  k L d 2 with k  0080,

Thestar’sapparentbrightnessisabout3

2  296km;andat80km

57. Examplesincluderadioactivedecayandexponentialgrowthinbiology.

CHAPTER1REVIEW

1. CommutativePropertyforaddition. 3. DistributiveProperty.

19.(a)

solutiontotheoriginalequation.

2makestheexpressionundefined,werejectthissolution.Hencetheonly solutionis x

83. Let x bethenumberofkilogramsofraisins.Thenthenumberofkilogramsofnutsis50 x Raisins Nuts Mixture

20.Thusthemixtureuses 20kilogramsofraisinsand50 20  30kilogramsofnuts.

85. Let r betheathlete’srunningspeed,inmi/h.Thentheycycleat

8km/h.

Sincethetotaltimeoftheworkoutis1hour,wehave

speedis

87. Let t bethetimeitwouldtaketheinteriordecoratortopaintalivingroomiftheyworkalone.Itwouldtaketheassistant 2t hoursalone,anditwouldtaketheapprentice3t hoursalone.Thus,thedecoratordoes 1 t ofthejobperhour,theassistant does 1 2t ofthejobperhour,andtheapprenticedoes 1 3t ofthejobperhour.So

6t  11  t  11 6 .Thus,itwouldtakethedecorator1hour50minutestopaintthelivingroom alone.

89. 3 x 2  11

Interval:

Graph: -3

91. x 2 7 x 8

x 8

0.Theexpressionontheleftoftheinequalitychangessignwhere x  8andwhere x 1.Thuswemustchecktheintervalsinthefollowingtable.

Interval:

Graph: _18

93. x 4 x 2 4  0  x 4  x 2

0.Theexpressionontheleftoftheinequalitychangessignwhere x

2,where x  2, andwhere x  4.Thuswemustchecktheintervalsinthefollowingtable.

Signof x 4

Signof x 2

Signof x 4

Sincetheexpressionisnotdefinedwhen

Graph: _224

 x 5

Interval:[2 8]

Graph: 28

weexcludethesevaluesandthesolutionis

97.(a) y

Q

(d) Thelinehasslope

Thecenteris

(b) Thedistancefrom P to Q is

(c) Themidpointis

(e) Theradiusofthiscirclewasfoundinpart(b).Itis

x

4,anequationofacircle. (b) Thecirclehascenter

and radius2. 1 1 y x

Sincetheleftsideofthisequationmustbegreaterthanorequaltozero,thisequationhasnograph.

x  16 y 2 (a) x ­axissymmetry:replacing y by y gives x  16  y 2  16 y 2 ,whichisthesameastheoriginalequation,so thegraphissymmetricaboutthe x ­axis. y ­axissymmetry:replacing x by x gives x  16 y 2  x  y 2 16,whichisnotthesameastheoriginal equation,sothegraphisnotsymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by y gives x  16  y 2  x 16 y 2 ,whichisnotthesameas theoriginalequation,sothegraphisnotsymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor x : x  16 02  16,sothe x ­interceptis16.

Tofind y ­intercepts,weset x  0andsolvefor y :0  16 y 2  y 4,sothe y ­interceptare4and 4.

117. x 2 9 y  9

(a) x ­axissymmetry:replacing y by

9,sothegraphisnotsymmetricaboutthe x ­axis.

y ­axissymmetry:replacing x by x gives 

Originsymmetry:replacing x by x and y by

9,sothegraphissymmetricaboutthe y ­axis.

9 y

9,sothegraphisnot symmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor

Tofind y ­intercepts,weset x  0andsolvefor

119. x 2  4 xy  y 2  1

(a) x ­axissymmetry:replacing y by

gives

3,sothe x ­interceptsare3and 3.

1,sothe y ­interceptis 1.

1,whichisdifferentfromtheoriginalequation,so thegraphisnotsymmetricaboutthe x ­axis. y ­axissymmetry:replacing x by x gives

1,whichisdifferentfromtheoriginalequation, sothegraphisnotsymmetricaboutthe y ­axis.

Originsymmetry:replacing x by x and y by y gives

thegraphissymmetricabouttheorigin.

(b) Tofind x ­intercepts,weset y  0andsolvefor

x ­interceptsare 1and1.

Tofind y ­intercepts,weset x  0andsolvefor y

1,sothe y ­interceptsare 1and1.

121.(a) Wegraph y  x 2 6 x intheviewingrectangle [ 10 10] by [ 10 10] ­10 ­5 5 10 ­10 10

(b) Fromthegraph,weseethatthe x ­interceptsare0 and6andthe y ­interceptis0.

123.(a) Wegraph y  x 3 4 x 2 5 x intheviewing rectangle [

(b) Fromthegraph,weseethatthe x ­interceptsare 1, 0,and5andthe y ­interceptis0.

125.(a) Thelinethathasslope2and y ­intercept6hastheslope­interceptequation y  2 x  6.

(b) Anequationofthelineingeneralformis2x y  6  0. (c) y 1x

127.(a) Thelinethatpassesthroughthepoints 

 3 2 ,so,usingthesecondpointforconvenience,an

129.(a) Theverticallinethatpassesthroughthepoint

131.(a) Thelinecontaining

and

4,andthelinepassingthroughtheoriginwith thisslopehasequation y 4

(b) y 4 x  4 x  y  0.

y 1x 1

133.(a) Theslope,0 3,representstheincreaseinlengthofthespringforeachunitincreasein weight  .The s ­interceptisthe restingornaturallengthofthespring. (b) When   5, s  0 3

 4 0centimeters.

135. Fromthegraph,weseethatthegraphsof y  x 2 4 x and y  x  6intersectat x 1and x  6,sothesearethe solutionsoftheequation x 2 4 x  x  6.

137. Fromthegraph,weseethatthegraphof y  x 2 4 x liesbelowthegraphof y  x  6for 1  x  6,sotheinequality x 2 4 x  x  6issatisfiedontheinterval [ 1 6].

139. Fromthegraph,weseethatthegraphof y  x 2 4 x liesabovethe x ­axisfor x  0andfor x  4,sotheinequality x 2 4 x  0issatisfiedontheintervals  0] and [4 

141. x 2 4 x  2 x  7.Wegraphtheequations y1  x 2 4 x and y2  2 x  7intheviewingrectangle[ 10 10]by [ 5 25].Usingazoomortracefunction,wegetthe solutions x 1and x  7.

143. x 4 9 x 2  x 9.Wegraphtheequations y1  x 4 9 x 2 and y2  x 9intheviewingrectangle[ 5 5]by [ 25 10].Usingazoomortracefunction,wegetthe solutions x 2 72, x 1 15, x  1 00,and x  2 87.

145. x 2  12 4 x .Wegraphtheequations y1  x 2 and y2  12 4 x intheviewingrectangle [ 8 4] by [0 40].

Usingazoomortracefunction,wefindthepointsof intersectionareat x 6and x  2.Sincewewant x 2  12 4 x ,thesolutionistheunionofintervals

 6

2

147. x 4 4 x 2  1 2 x 1.Wegraphtheequations y1  x 4 4 x 2 and y2  1 2 x 1intheviewingrectangle [ 5 5] by [ 5 5].Wefindthepointsofintersectionare at x 1 85, x 0 60, x  0 45,and x  2 00.Since wewant

149. Herethecenterisat 0 0,andthecirclepassesthroughthepoint  5 12,sotheradiusis r  

13.Theequationofthecircleis x 2  y 2  132  x 2  y 2  169.Thelineshownisthetangentthatpassesthroughthepoint  5 12,soitisperpendiculartotheline throughthepoints 0 0 and  5 12.Thislinehasslope m 1  12 0 5

 12 5 .Theslopeofthelineweseekis m 2  1 m 1  1 125  5 12 .Thus,anequationofthetangentlineis y 12  5 12  x  5  y 12  5 12 x  25 12  y  5 12 x  169 12  5 x 12 y  169  0.

151. Since M variesdirectlyas z wehave M  kz .Substituting M  120when z  15,wefind120  k 15  k  8. Therefore, M  8z

153.(a) Theintensity I variesinverselyasthesquareofthedistance d ,so I  k d 2 (b) Substituting I  1000when d  8,weget1000  k 82  k  64,000.

(c) Fromparts(a)and(b),wehave I  64,000 d 2 .Substituting d  20,weget I  64,000 202  160candles.

155. Note:Inthefirstprintingofthetext,theanswergivenforthisexerciseis incorrect. Let  betheterminalvelocityoftheparachutistinkm/hand  betheirweightinkilograms.Sincetheterminalvelocityis directlyproportionaltothesquarerootoftheweight,wehave   k  .Substituting   14when   70,wesolve for k .Thisgives14 

673 .When   105,theterminalvelocityis   1 673105  17km/h.

157. Thespeed  isinverselyproportionaltothesquarerootofthedensity d ,so

 k d .Infreshwaterwithdensity d1  1gcm3 ,thespeedis

1480.Thus,inseawaterwithdensity10273gcm3 ,we have   1480 10273

159. Wesolvethefirstequationfor

ByHubble’sLaw,

CHAPTER1TEST

1.(a)

3.(a)

5.(Notethatthisisimpossible,sotherecanbenosolution.) Squaringbothsidesagain,weget1

4.Butthisdoesnotsatisfytheoriginalequation,sothereisno solution.(Youmustalwayscheckyourfinalanswersifyouhavesquaredboth sideswhensolvinganequation,since extraneousanswersmaybeintroduced,ashere.) (f) x

11. UsingtheQuadraticFormula,2

13. Let  bethewidthoftheparcelofland.Then   70isthelengthoftheparcelofland.Then

15. 5  5

41 Fand50 F.

17.(a) y 1x 1

S Thereareseveralwaystodeterminethecoordinatesof S .Thediagonalsofa squarehaveequallengthandareperpendicular.Thediagonal PR ishorizontal andhaslengthis6units,sothediagonal QS isverticalandalsohaslength6. Thus,thecoordinatesof S are

(b) Thelengthof PQ is 0

2.Sotheareaof

19.(a) y 1x P1 Q

(b) Thedistancebetween P and Q is

(c) Themidpointis

(d) Thecenterofthecircleisthemidpoint,

,andthelengthoftheradiusis

PQ is

21. 2 x 3 y  15 3 y 2 x  15  y  2 3 x 5.Theslopeis 2 3 andthe y ­interceptis 5. y 1x 1

23.(a) When x  100wehave T  0 08

 4  8 4  4,sothe temperatureatonemeteris4 C.

(c) Thesloperepresentstheraiseintemperatureasthedepthincrease. The T ­interceptisthesurfacetemperatureofthesoilandthe x ­interceptrepresentsthedepthofthe“frostline”,wherethesoil belowisnotfrozen.

(b) T 20x 6080100120 _5 5 40

25. Note:Inthefirstprintingofthetext,theanswersgivenforparts(b)and(c)ofthisexerciseareincorrect.

(a) M  k  h 2 L

(b) Substituting   10, h  15, L  30,and M  21,000,wehave21,000

(c) Nowif L

FOCUSONMODELINGFittingLinestoData

1.(a) Usingagraphingcalculator,weobtaintheregression

(b) Using x  58intheequation y  1 8807 x  82 65, weget y  1

3. Note:Inthefirstprintingofthetext,theanswergivenforpart(a)ofthisexerciseisincorrect.

(a) Usingagraphingcalculator,weobtaintheregression line y  2 579 x 0 1783.

(b) Using x  45intheequation y  2 579 x 0 1783, weget y  2 579 45 0 1783  116years.

5.(a) Usingagraphingcalculator,weobtaintheregression line y 0 13198 x  7 2514 Years since 1994 y x 2 4 6 8 01020

7.(a) Usingagraphingcalculator,weobtain

(b) Using x  25intheregressionlineequation,weget y 0 13198 25  7 2514  3 95millionkm2 Thisisroughly10%lessthantheactualfigureof 4 4millionkm2

(c) Despitefluctuationsoverbriefperiods,themodel seemsfairlyaccurate.Ifexternalcircumstances change(reducedorincreasedCO2 emissions,for example),itmaybecomelessreliable.Itisunlikely tobeaccuratefarintothefuture.

9. Resultswilldependonstudentsurveysineachclass.

(b) Thecorrelationcoefficientis r 0 98,solinear modelisappropriatefor x between80dBand 104dB.

(c) Substituting x  94intotheregressionequation,we get y 3 9018 94  419 7  53.Sothe intelligibilityisabout53%.

1 ERRATAinExercisesandAnswersinFirstPrinting

Page1231.11.56Displayedequationshouldread y  2 4 x   x 1000 2

Page1231.11.56(a) Graphtheequationfor0  x  160.

PageA21.5.129(Answer)  424s

PageA21.5.137(Answer) 344 000km

PageA31.7.21(Answer) 220km

PageA31.7.39(Answer) 31 6mby158m

PageA31.7.77(Answer) 1 6mfromthefulcrum

PageA41.8.119(Answer) Morethan100km

PageA41.8.127(Answer)(a) Accelerationgreaterthan8 04ms2

PageA71.10.93(Answer)(a) t  1 8 n  6 (b) 25 C

PageA71.10.95(Answer)(a) P  9 97d  103,where P is pressureinkPaand d isdepthinmeters (c) Thesloperepresentsanincreaseof9 97kPainpressure foreachonemeterincreaseindepth,andthe d ­interceptis theairpressureatthesurface. (d)  59m (b) y x

PageA71.12.39(Answer)  69kmh

PageA71.12.41(Answer)  9 03kmh

PageA71.12.49(Answer)(b) 1 928 234 931

PageA81.Review.155(Answer) 17kmh

PageA91.Test.12(Answer) 192km

PageA91.Test.25(Answer)(b) 280 (c) 52,500N

PageA101.Focus.3(Answer)(a) y  2 579 x 0 1783