Solution Manual for College Algebra, 7th Edition

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CONTENTS¥PROLOGUE:Principles of Problem Solving1CHAPTERPPREREQUISITES3P.1Modeling the Real World with Algebra3P.2Real Numbers4P.3Integer Exponents and Scientific Notation9P.4Rational Exponents and Radicals14P.5Algebraic Expressions18P.6Factoring22P.7Rational Expressions27P.8Solving Basic Equations34P.9Modeling with Equations39Chapter P Review45Chapter P Test51¥FOCUS ON MODELING:Making the Best Decisions54CHAPTER1EQUATIONS AND GRAPHS571.1The Coordinate Plane571.2Graphs of Equations in Two Variables; Circles651.3Lines791.4Solving Quadratic Equations901.5Complex Numbers981.6Solving Other Types of Equations1011.7Solving Inequalities1101.8Solving Absolute Value Equations and Inequalities1291.9Solving Equations and Inequalities Graphically1311.10Modeling Variation139Chapter 1 Review143Chapter 1 Test161iii

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ivContents¥FOCUS ON MODELING:Fitting Lines to Data165CHAPTER2FUNCTIONS1692.1Functions1692.2Graphs of Functions1782.3Getting Information from the Graph of a Function1902.4Average Rate of Change of a Function2012.5Linear Functions and Models2062.6Transformations of Functions2122.7Combining Functions2262.8One-to-One Functions and Their Inverses234Chapter 2 Review243Chapter 2 Test255¥FOCUS ON MODELING:Modeling with Functions259CHAPTER3POLYNOMIAL AND RATIONAL FUNCTIONS2673.1Quadratic Functions and Models2673.2Polynomial Functions and Their Graphs2763.3Dividing Polynomials2913.4Real Zeros of Polynomials3013.5Complex Zeros and the Fundamental Theorem of Algebra3343.6Rational Functions344Chapter 3 Review377Chapter 3 Test395¥FOCUS ON MODELING:Fitting Polynomial Curves to Data398CHAPTER4EXPONENTIAL AND LOGARITHMIC FUNCTIONS4014.1Exponential Functions4014.2The Natural Exponential Function4094.3Logarithmic Functions4144.4Laws of Logarithms4224.5Exponential and Logarithmic Equations426

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Contentsv4.6Modeling with Exponential Functions4334.7Logarithmic Scales438Chapter 4 Review440Chapter 4 Test448¥FOCUS ON MODELING:Fitting Exponential and Power Curves to Data450CHAPTER5SYSTEMS OF EQUATIONS AND INEQUALITIES4555.1Systems of Linear Equations in Two Variables4555.2Systems of Linear Equations in Several Variables4625.3Partial Fractions4705.4Systems of Nonlinear Equations4815.5Systems of Inequalities488Chapter 5 Review500Chapter 5 Test508¥FOCUS ON MODELING:Linear Programming511CHAPTER6MATRICES AND DETERMINANTS5196.1Matrices and Systems of Linear Equations5196.2The Algebra of Matrices5306.3Inverses of Matrices and Matrix Equations5386.4Determinants and Cramer’s Rule548Chapter 6 Review562Chapter 6 Test572¥FOCUS ON MODELING:Computer Graphics575CHAPTER7CONIC SECTIONS5797.1Parabolas5797.2Ellipses5847.3Hyperbolas5937.4Shifted Conics600Chapter 7 Review612Chapter 7 Test622¥FOCUS ON MODELING:Conics in Architecture624

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viContentsCHAPTER8SEQUENCES AND SERIES6278.1Sequences and Summation Notation6278.2Arithmetic Sequences6328.3Geometric Sequences6378.4Mathematics of Finance6458.5Mathematical Induction6498.6The Binomial Theorem658Chapter 8 Review662Chapter 8 Test669¥FOCUS ON MODELING:Modeling with Recursive Sequences670CHAPTER9PROBABILITY AND STATISTICS6739.1Counting6739.2Probability6809.3Binomial Probability6889.4Expected Value693Chapter 9 Review695Chapter 9 Test701¥FOCUS ON MODELING:The Monte Carlo Method702APPENDIXES705ACalculations and Significant Figures705BGraphing with a Graphing Calculator705

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PROLOGUE: Principles of Problem Solving1.Letrbe the rate of the descent. We use the formula timedistancerate; the ascent takes115 h, the descent takes 1rh, and thetotal trip should take230115 h. Thus we have1151r1151r0, which is impossible. So the car cannot gofast enough to average 30 mi/h for the 2-mile trip.2.Let us start with a given priceP. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the pricedecreases to 08P, and after another 20% discount, it becomes 0808P064P. Since 06P064P, a 40% discountis better.3.We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Sincethefirst cut produces 4 pieces, we get the formulafn43n1,n1. Sincef14243141427, wesee that 142 parallel cuts produce 427 pieces.4.By placing two amoebas into the vessel, we skip thefirst simple division which took 3 minutes. Thus when we place twoamoebas into the vessel, it will take 60357 minutes for the vessel to be full of amoebas.5.The statement is false. Here is one particular counterexample:Player APlayer BFirst half1 hit in 99 at-bats: average1990 hit in 1 at-bat: average01Second half1 hit in 1 at-bat: average1198 hits in 99 at-bats: average9899Entire season2 hits in 100 at-bats: average210099 hits in 100 at-bats: average991006.Method 1:After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus,any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup.Method 2:Alternatively, look at the drawing of the spoonful of coffee and creammixture being returned to the pitcher of cream. Suppose it is possible to separatethe cream and the coffee, as shown. Then you can see that the coffee going into thecream occupies the same volume as the cream that was left in the coffee.coffeecreamMethod 3 (an algebraic approach):Suppose the cup of coffee hasyspoonfuls of coffee. When one spoonful of creamis added to the coffee cup, the resulting mixture has the following ratios:creammixture1y1 andcoffeemixtureyy1 .So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing1y1 of aspoonful of cream andyy1 spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is11y1yy1 of a spoonful. This is the same as the amount of coffee we added to the cream.7.Letrbe the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radiusby 1 foot, the new radius isr1, so the new circumference is 2r1. Thus you need 2r12r2extrafeet of ribbon.1

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2Principles of Problem Solving8.The north pole is such a point. And there are others: Consider a pointa1near the south pole such that the parallel passingthrougha1forms a circleC1with circumference exactly one mile. Any pointP1exactly one mile north of the circleC1along a meridian is a point satisfying the conditions in the problem: starting atP1she walks one mile south to the pointa1on the circleC1, then one mile east alongC1returning to the pointa1, then north for one mile toP1. That’s not all. If apointa2(ora3,a4,a5,  ) is chosen near the south pole so that the parallel passing through it forms a circleC2(C3,C4,C5,  ) with a circumference of exactly12mile (13mi,14mi,15mi,  ), then the pointP2(P3,P4,P5,  ) one mile northofa2(a3,a4,a5,  ) along a meridian satisfies the conditions of the problem: she walks one mile south fromP2(P3,P4,P5,  ) arriving ata2(a3,a4,a5,  ) along the circleC2(C3,C4,C5,  ), walks east along the circle for one mile thustraversing the circle twice (three times, four times,five times,  ) returning toa2(a3,a4,a5,  ), and then walks north onemile toP2(P3,P4,P5,  ).

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PPREREQUISITESP.1MODELING THE REAL WORLD WITH ALGEBRA1.Using this model, wefind that ifS12,L4S41248. Thus, 12 sheep have 48 legs.2.If each gallon of gas costs $350, thenxgallons of gas costs $35x. Thus,C35x.3.Ifx$120 andT006x, thenT00612072. The sales tax is $720.4.Ifx62,000 andT0005x, thenT000562,000310. The wage tax is $310.5.If70,t35, anddt, thend7035245. The car has traveled 245 miles.6.Vr2h325451414 in37. (a)MNG240830 miles/gallon(b)25175GG175257 gallons8. (a)T700003h7000031500655F(b)64700003h0003h6h2000 ft9. (a)V95S954 km338 km3(b)19 km395SS2 km310.(a)P006s30061231037 hp(b)75006s3s3125 sos5 knots11. (a)Depth (ft)Pressure (lb/in2)00450147147100451014719220045201472373004530147282400454014732750045501473726004560147417(b)We know thatP30 and we want tofindd, so we solve theequation 30147045d153045dd153045340. Thus, if the pressure is 30 lb/in2, the depthis 34 ft.12. (a)PopulationWater use (gal)00100040100040,000200040200080,0003000403000120,0004000404000160,0005000405000200,000(b)We solve the equation 40x120,000x120,000403000. Thus, the population is about 3000.13.The numberNof cents inqquarters isN25q.14.The averageAof two numbers,aandb, isAab2.15.The costCof purchasingxgallons of gas at $350 a gallon isC35x.16.The amountTof a 15% tip on a restaurant bill ofxdollars isT015x.17.The distancedin miles that a car travels inthours at 60 mi/h isd60t.3

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4CHAPTER PPrerequisites18.The speedrof a boat that travelsdmiles in 3 hours isrd3 .19. (a)$123$1$12$3$15(b)The costC, in dollars, of a pizza withntoppings isC12n.(c)Using the modelC12nwithC16, we get 1612nn4. So the pizza has four toppings.20. (a)3302800109028$118(b)The cost isdailyrentaldaysrentedcostper milemilesdriven, soC30n01m.(c)We haveC140 andn3. Substituting, we get 14030301m1409001m5001mm500. So the rental was driven 500 miles.21. (a)(i)For an all-electric car, the energy cost of drivingxmiles isCe004x.(ii)For an average gasoline powered car, the energy cost of drivingxmiles isCg012x.(b)(i)The cost of driving 10,000 miles with an all-electric car isCe00410,000$400.(ii)The cost of driving 10,000 miles with a gasoline powered car isCg01210,000$1200.22. (a)If the width is 20, then the length is 40, so the volume is 20204016,000 in3.(b)In terms of width,Vxx2x2x3.23. (a)The GPA is 4a3b2c1d0fabcdf4a3b2cdabcdf.(b)Usinga236,b4,c339, anddf0 in the formula from part (a), wefind the GPA to be4634296495419284.P.2THE REAL NUMBERS1. (a)The natural numbers are123   .(b)The numbers   3210are integers but not natural numbers.(c)Any irreducible fractionpqwithq1 is rational but is not an integer. Examples:32,512,172923.(d)Any number which cannot be expressed as a ratiopqof two integers is irrational. Examples are2,3,, ande.2. (a)abba; Commutative Property of Multiplication(b)abcabc; Associative Property of Addition(c)abcabac; Distributive Property3.The set of numbers between but not including 2 and 7 can be written as(a)x2x7in interval notation, or(b)27in interval notation.4.The symbolxstands for theabsolute valueof the numberx. Ifxis not 0, then the sign ofxis alwayspositive.5.The distance betweenaandbon the real line isdab ba. So the distance between5 and 2 is25 7.6. (a)Yes, the sum of two rational numbers is rational:abcdadbcbd.(b)No, the sum of two irrational numbers can be irrational (2) or rational (0).7. (a)No:ab babain general.(b)No; by the Distributive Property,2a5 2a 25 2a10 2a10.8. (a)Yes, absolute values (such as the distance between two different numbers) are always positive.(b)Yes,ba  ab.

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SECTION P.2The Real Numbers59. (a)Natural number: 100(b)Integers: 0, 100,8(c)Rational numbers:15, 0,52, 271, 314, 100,8(d)Irrational numbers:7,10. (a)Natural number:164(b)Integers:500,16,205 4(c)Rational numbers: 13, 13333  , 534,500, 123,16,246579,205(d)Irrational number:511.Commutative Property of addition12.Commutative Property of multiplication13.Associative Property of addition14.Distributive Property15.Distributive Property16.Distributive Property17.Commutative Property of multiplication18.Distributive Property19.x33x20.73x73x21.4AB4A4B22.5x5y5xy23.3xy3x3y24.ab88a8b25.42m42m8m26.436y436y 8y27.522x4y 522x524y 5x10y28.3a bc2d3ab3ac6ad29. (a)3104159308301730(b)141552042092030. (a)23351015915115(b)1581624241524424352431. (a)236322362332413(b)31414512414554513415132032. (a)2232322322312313931383(b)25121103152512110152512110151010451293333. (a)236 and 2727, so 372(b)67(c)357234. (a)3232 and 3067201, so23067(b)23067(c)067  06735. (a)False(b)True36. (a)False:317320517325.(b)False37. (a)True(b)False38. (a)True(b)True39. (a)x0(b)t4(c)a(d)5x13(e)p3 540. (a)y0(b)z1(c)b8(d)0 17(e)y 241. (a)AB 12345678(b)AB 24642. (a)BC 24678910(b)BC 8

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6CHAPTER PPrerequisites43. (a)AC 12345678910(b)AC 744. (a)ABC 12345678910(b)ABC∅45. (a)BC xx5(b)BC x 1x446. (a)AC x 1x5(b)AB x 2x447.30 x 3x0_3048.28] x2x82849.[28 x2x82850.612x 6x 12_61_2_51.[2 xx2252.1 xx1153.x1x1]154.1x2x[12]1255.2x1x21]_2156.x 5x[5_557.x1x1_158.5x2x52_5259.(a) [35](b)35]60.(a) [02(b)20]61.201121_2162.201 10_1063.[46][08[06]0664.[46][08[48_4865.44_4466.6]21026]26

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SECTION P.2The Real Numbers767. (a)100 100(b)73 7368. (a)55 555 5, since 55.(b)10 10, since 10.69. (a)6  4  64  2 2(b)1111 170. (a)2 12  212  10 10(b)1 1 1  1 11  1 0  171. (a)26  12 12(b)1315 5 572. (a)6241414(b)71212755 1 173.23  5 574.2515  4 475. (a)172 15(b)213  213  24 24(c)310118124055406740674076. (a)7151214910551055410518351835(b)3857  3857  19 19.(c)2618  2618  08 08.77. (a)Letx0777  . So 10x77777  x07777  9x7. Thus,x79.(b)Letx02888  . So 100x288888  10x28888  90x26. Thus,x26901345.(c)Letx0575757  . So 100x575757  x05757  99x57. Thus,x57991933.78. (a)Letx52323  . So 100x5232323  1x52323  99x518. Thus,x51899.(b)Letx13777  . So 100x1377777  10x137777  90x124. Thus,x124906245.(c)Letx213535  . So 1000x21353535  10x213535  990x2114. Thus,x21149901057495.79.3, so3 3.80.21, so1 2 21.81.ab, soab  abba.82.ab ab abba2b83. (a)ais negative becauseais positive.(b)bcis positive because the product of two negative numbers is positive.(c)ababis positive because it is the sum of two positive numbers.(d)abacis negative: each summand is the product of a positive number and a negative number, and the sum of twonegative numbers is negative.84. (a)bis positive becausebis negative.(b)abcis positive because it is the sum of two positive numbers.(c)cacais negative becausecandaare both negative.(d)ab2is positive because bothaandb2are positive.85.Distributive Property

Solution Manual for College Algebra, 7th Edition - Page 13 preview image

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8CHAPTER PPrerequisites86.DayTOTGTOTGTOTGSunday687799Monday727533Tuesday747400Wednesday807555Thursday776988Friday717011Saturday707111TOTGgives more information because it tells us which city had the higher temperature.87. (a)WhenL60,x8, andy6, we haveL2xy60286602888. Because 88108 thepost office will accept this package.WhenL48,x24, andy24, we haveL2xy48224244896144, and since144108, the post office willnotaccept this package.(b)Ifxy9, thenL299108L36108L72. So the length can be as long as 72 in.6 ft.88.Letxm1n1andym2n2be rational numbers.Thenxym1n1m2n2m1n2m2n1n1n2,xym1n1m2n2m1n2m2n1n1n2, andxym1n1m2n2m1m2n1n2. This shows that the sum, difference, and productof two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarilyirrational; for example,2 22, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;for example,220 which is rational.89.12 2 is irrational. If it were rational, then by Exercise 6(a), the sum12 212 2 would be rational, butthis is not the case.Similarly,12 2 is irrational.(a)Following the hint, suppose thatrtq, a rational number. Then by Exercise 6(a), the sum of the two rationalnumbersrtandris rational. Butrtrt, which we know to be irrational. This is a contradiction, andhence our original premise—thatrtis rational—was false.(b)ris rational, sorabfor some integersaandb. Let us assume thatrtq, a rational number. Then by definition,qcdfor some integerscandd. But thenrtqab tcd, whencetbcad, implying thattis rational. Once againwe have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number isirrational.90.x121010010001x112110110011000Asxgets large, the fraction 1xgets small. Mathematically, we say that 1xgoes to zero.x1050100100011x11052101101001100100011000Asxgets small, the fraction 1xgets large. Mathematically, we say that 1xgoes to infinity.

Solution Manual for College Algebra, 7th Edition - Page 14 preview image

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SECTION P.3Integer Exponents and Scientific Notation991. (a)Construct the number2 on the number line by transferringthe length of the hypotenuse of a right triangle with legs oflength 1 and 1.10_121Ï2Ï23(b)Construct a right triangle with legs of length 1 and 2. By thePythagorean Theorem, the length of the hypotenuse is1222 5. Then transfer the length of the hypotenuseto the number line.10_121Ï53Ï5(c)Construct a right triangle with legs of length2 and 2[construct2 as in part (a)]. By the Pythagorean Theorem,the length of the hypotenuse is2222 6. Thentransfer the length of the hypotenuse to the number line.10_121Ï2Ï2Ï6Ï6Ï2392. (a)Subtraction is not commutative. For example, 5115.(b)Division is not commutative. For example, 5115.(c)Putting on your socks and putting on your shoes are not commutative. If you put on your socksfirst, then your shoes,the result is not the same as if you proceed the other way around.(d)Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result.(e)Washing laundry and drying it are not commutative.(f)Answers will vary.(g)Answers will vary.93.Answers will vary.94. (a)Ifx2 andy3, thenxy  23  5 5 andx  y  2  3 5.Ifx 2 andy 3, thenxy  5 5 andx  y 5.Ifx 2 andy3, thenxy  23 1 andx  y 5.In each case,xy  x  yand the Triangle Inequality is satisfied.(b)Case 0:If eitherxoryis 0, the result is equality, trivially.Case 1:Ifxandyhave the same sign, thenxy xyifxandyare positivexyifxandyare negative x  y.Case 2:Ifxandyhave opposite signs, then suppose without loss of generality thatx0 andy0. Thenxyxy  x  y.P.3INTEGER EXPONENTS AND SCIENTIFIC NOTATION1.Using exponential notation we can write the product 555555 as 56.2.Yes, there is a difference:545 5 5 5625, while54 5555 625.3.In the expression 34, the number 3 is called thebaseand the number 4 is called theexponent.4.When we multiply two powers with the same base, weaddthe exponents. So 343539.5.When we divide two powers with the same base, wesubtractthe exponents. So 353233.6.When we raise a power to a new power, wemultiplythe exponents. So34238.

Solution Manual for College Algebra, 7th Edition - Page 15 preview image

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10CHAPTER PPrerequisites7. (a)2112(b)2318(c)1212(d)1232388.Scientists express very large or very small numbers usingscientificnotation. In scientific notation, 8,300,000 is 83106and 00000327 is 327105.9. (a)No,23232294 .(b)Yes,54625 and54 54 625.10. (a)No,x23x23x6.(b)No,2x4323x438x12.11. (a)26 64(b)2664(c)15233123352 272512. (a)53 125(b)53 125(c)52252522252413. (a)5302112(b)233012318(c)142421614. (a)2320 123 18(b)2320 23 8(c)2333323 27815. (a)53554625(b)3230329(c)223266416. (a)38353131,594,323(b)6066(c)54258390,62517. (a)54525225(b)1071041031000(c)32341321918. (a)333134134181(b)54553125(c)7275173134319. (a)x2x3x23x5(b)x2313x23 x6(c)t3t5t35t220. (a)y5y2y52y7(b)8x282x264x2(c)x4x3x43x21. (a)x5x3x53x21x2(b)24524511(c)y10y0y7y1007y322. (a)y2y5y25y31y3(b)z5z3z4z534z21z2(c)x6x10x6101x423. (a)a9a2aa921a6(b)a2a43a243a63a63a18(c)2x25x622x25x620x2620x824. (a)z2z4z3z1z24z31z6z2z62z4(b)2a3a242a3242a5424a5416a20(c)3z232z333z232z3 54z63 54z925. (a)3x2y 2x332x23y6x5y(b)2a2b1 3a2b223a22b126b

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SECTION P.3Integer Exponents and Scientific Notation11(c)4y2 x4y24y2x42y24x8y224x8y426. (a)4x3y2 7y547x3y2528x3y7(b)9y2z2 3y3z93y23z2127yz3(c)8x7y2 12x3y28x7y212x3y2228x7y2x32y232x7y2x6y232x76y2232x27. (a)2x2y323y22x22y323y12x4y7(b)x2y1x5x25y1x7y1x7y(c)x2y33x23y333x6y32728. (a)5x4y3 8x325x4y382x32582x46y3320x2y3(b)y2z3y1yy2z31yz3(c)a3b2b32a6b4b6a6b1029. (a)x3y311x3y3(b)a2b23a32a23b23a32a6b6a6b6a12(c)x2y222y3x23x22y2223y33x23x4y48y9x68x46y498x10y1330. (a)x2y43y4x23x2y43x6y12(b)y212x3y43y223x33y43x98y14(c)2a1b23b12a2223a13b23b1222a22132ab831. (a)3x2y59x3y2x y33(b)2x3y1y222x3y32y3222x32y64x6(c)y1x213x3y22y11x2132x32y22x4y5932. (a)12a3b42a5b11221a35b4114a2b3a24b3

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