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Solution Manual for Intermediate Algebra, 4th Edition

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Solution Manual for Intermediate Algebra, 4th Edition - Page 1 preview imageSOLUTIONSMANUALCTRIMBLE&ASSOCIATESINTERMEDIATEALGEBRAFOURTHEDITIONMichael Sullivan, IIIJoliet Junior CollegeKatherine R. StruveColumbus State Community Collegewith contributions byJessica BernardsPortland Community CollegeWendy FreshPortland Community College
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Solution Manual for Intermediate Algebra, 4th Edition - Page 2 preview image
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Solution Manual for Intermediate Algebra, 4th Edition - Page 3 preview imageContentsChapter R1Chapter 119Chapter 2120Chapter 3183Chapter 4282Chapter 5374Chapter 6470Chapter 7542Chapter 8675Chapter 9736Chapter 10822Appendix A862
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Solution Manual for Intermediate Algebra, 4th Edition - Page 4 preview image
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Solution Manual for Intermediate Algebra, 4th Edition - Page 5 preview image1Chapter RSection R.1R.1 ExercisesAnswers will vary.Section R.2Section R.2 Quick Checks1.A set is a well-defined collection of objects.2.The objects in a set are called elements.3.LetDrepresent the set of all digits that are lessthan 5.set-builder:{}|is a digit less than 5=Dxxroster:{}0,1, 2,3, 44.LetGrepresent the set of all digits that aregreater than or equal to 6.set-builder:{}|is a digit greater than or equal to 6=Gxxroster:{}6, 7,8,95.The statement is true because the order in whichelements are listed in a set does not matter.6.If every element of a setAis also an element of asetB, then say thatAis a subset ofBand writeA⊆B.7.False. The empty set can be denoted as { } or∅, but not as{}∅.8.The statement⊆BAis true because all theelements that are inBare also inA.9.The statement=BCis false because there areelements inB(aandb) that are not in C. There isalso an element inC, namelyd, that is not inB.10.The statement⊂BDis false. In order forBtobe a proper subset ofD, it must be the case thatall the elements that are inBare also inD. Inaddition, there must be at least one element inDthat is not inB. Since=BD, this is not thecase.11.The statement∅ ⊆Ais true. The empty set is asubset of every set.12.The statement{}50,1, 2,3, 4,5∈is true because5 is an element of the set.13.The statement Michigan∉{Illinois, Indiana,Michigan, Wisconsin} is false because Michiganisan element of the set.14.The statement83∈{|=pxxqwherepandqare digits,0≠q}is true because83is of the formpqwhere8=pand3=q.15.The whole numbers are numbers in the set{0, 1, 2, 3, 4, ...}.16.The numbers in the setwhereandare integers,0,⎧⎫⎪⎪=≠⎨⎬⎪⎪⎩⎭px xpqqqare called rational numbers.17.The set of irrational numbers have decimalrepresentations that neither terminate nor repeat.18.False. If a number is rational, it cannot beirrational and vice-versa.19.False. Decimals that terminate or repeatrepresent rational numbers, but decimals that donot repeat and do not terminate representirrational numbers.20.True. The set of rational numbers is a subset ofthe set of real numbers.21.True. The set of rational numbers and the set ofirrational numbers do not have any numbers incommon.22.10 and1234=are the only natural numbers.23.10,003=, and1234=are the whole numbers.24.9−, 10,003=, and1234=are the integers.
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Solution Manual for Intermediate Algebra, 4th Edition - Page 6 preview imageChapter R:ReviewISM: Intermediate Algebra225.73,9−, 10,4.56,03,47−, and124are therational numbers.26.5.7377377737777… andπand11are theirrational numbers. Note that5.7377377737777… is irrational because thedecimal portion is non-terminating and non-repeating.27.All numbers listed are real numbers.28.a.To truncate, remove all digits after the thirddecimal place. So, 5.694392 truncated tothree decimal places is 5.694.b.To round to three decimal places, examinethe digit in the fourth decimal place and thentruncate. Since the fourth decimal placecontains a 3, do not change the digit in thethird decimal place. Thus, 5.694392rounded to three decimal places is 5.694.29.a.To truncate, remove all digits after thesecond decimal place. So,4.9369102−truncated to two decimal places is−4.93.b.To round to two decimal places, examinethe digit in the third decimal place and thentruncate. Since the third decimal placecontains a 6, add 1 to the digit in the seconddecimal place. Thus,4.9369102−roundedto two decimal places is−4.94.30.{}122, 0,,3,3.5−03−2123.531.36<because 3 lies to the left of 6 on a numberline.32.32−< −because3−lies to the left of−2 on anumber line.33.20.63=and10.52=2132>because23lies to the right of12on anumber line.34.50.7147≈50.77>because57lies to the right of 0.7 on anumber line.35.20.63=and100.615=210315=because the numbers are the same.36.3.14159π ≈3.14π >becauseπlies to the right of 3.14 on anumber line.R.2 Exercises38.{}1, 2,340.{}3,2,1, 0,1, 2,3, 4,5−−−42.∅or{}44.The statement⊆ACis true since all theelements ofAare also elements ofC.46.The statement⊂ACis true because all theelements in setAare also in setCand there areelements in setCthat are not in setA.48.The statement⊂DBis false. In order for setDto be a proper subset of setB, all the elements ofsetDmust be in the setBand there must be atleast one element in setBthat is not in setD.Since=BD, the statement is false.50.The statement∅ ⊂Bis true because the emptyset is a subset of every set, and the setBis notempty.52.{}4.5|is a rational number∈xx54.{}0|is a counting number∉xx56.a.13b.0, 13c.−4.5656..., 0, 2.43, 13d.2
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Solution Manual for Intermediate Algebra, 4th Edition - Page 7 preview imageISM: Intermediate AlgebraChapter R:Review3e.−4.5656..., 0,2,2.43, 1358.a.15b.6 ,151−c.6 , 7.3,151−d.2+ πe.6 ,2, 7.3,151−+ π60.a.To truncate, remove all digits after thesecond decimal place. So,93.432101−truncated to two decimal places is−93.43.b.To round to two decimal places, examinethe digit in the third decimal place and thentruncate. Since the third decimal placecontains a 2, do not change the digit in thesecond decimal place. Thus,93.432101−rounded to two decimal places is−93.43.62.a.To truncate, remove all digits after thesecond decimal place. So,9.9999truncatedto two decimal places is 9.99.b.To round to two decimal places, examinethe digit in the third decimal place and thentruncate. Since the third decimal placecontains a 9, add 1 to the digit in the seconddecimal place. So,9.9999rounded to twodecimal places is 10.00.64.0−545312−2.566.42>since 4 lies to the right of 2 on a realnumber line.68.2235>since23lies to the right of25on a realnumber line.70.8835−< −since83−lies to the left of85−on areal number line.72.Since sea level is an elevation of 0 ft, the DeadSea has an elevation of1349−feet.74.–$0.37Cisco Systems reported a loss of $0.37 in arecent trading day.76.−7The departure from the normal high temperaturewas –7ºF.78.Answers will vary.80.Answers will vary.82.The sum must be irrational because irrationalnumbers have nonterminating/nonrepeatingdecimals, so the sum will have anonterminating/nonrepeating decimal.84.Since there is an infinite number of rationalnumbers, it would not be possible to list all theelements of the set of rational numbers.86.The decimal 0.45 terminates, while the decimal0 45.has the block 45 repeating. Both numbersare rational.88.There is no positive real number that is “closest”to 0.90.Rounded:192.71437≈Truncated:192.71427≈Section R.3Section R.3 Quick Checks1.In the expressiona⋅b, the expressionsaandbare called factors.2.False because |0| = 0, and 0 is neither positivenor negative.3.66=because the distance from 0 to 6 on thereal number line is 6 units.
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Solution Manual for Intermediate Algebra, 4th Edition - Page 8 preview imageChapter R:ReviewISM: Intermediate Algebra44.1010−=because the distance from 0 to10−on the real number line is 10 units.5.()186+ −One number is positive, while the other isnegative. Find the absolute value of eachnumber:1818=and66−=. Subtract thesmaller absolute value from the larger and obtain18612−=. Because the number with the largerabsolute value is positive, the sum will bepositive. So,()18612+ −=.6.2110−+One number is positive, while the other isnegative. Find the absolute value of eachnumber:2121−=and1010=. Subtract thesmaller absolute value from the larger and obtain211011−=. Because the number with the largerabsolute value is negative, the sum will benegative. So,211011−+= −.7.()5.41.2−+ −Both numbers are negative, so first find theirabsolute values:5.45.4−=and1.21.2−=.Now add the absolute values,5.41.26.6+=.Because both numbers to be added are negative,the sum is also negative. So,()5.41.26.6−+ −= −.8.6.54.3−+One number is positive, while the other isnegative. Find the absolute value of eachnumber:6.56.5−=and4.34.3=. Subtractthe smaller absolute value from the larger andobtain6.54.32.2−=. Because the number withthe larger absolute value is negative, the sumwill be negative. So,6.54.32.2−+= −.9.99−+One number is positive, while the other isnegative. Find the absolute value of eachnumber:99−=and99=. Since the absolutevalues are the same, the difference will be 0. So,990−+=.10.For any real numbera, there is a real number−a,called the additive inverse, or opposite, ofasuchthata+ (−a) =−a+a= 0.11.For any real numbera,−(−a) =a.12.The additive inverse of 5 is5−because()550+ −=.13.The additive inverse of45is45−because44055⎛⎞+−=⎜⎟⎝⎠.14.The additive inverse of12−is()1212− −=because12120−+=.15.The additive inverse of53−is5533⎛⎞−−=⎜⎟⎝⎠because55033−+=.16.The additive inverse of 0 is 0 because 0 + 0 = 0.17.()62624−=+ −=18.()4134139−=+ −= −19.()383811−−= −+ −= −20.()12.53.412.53.49.1−=+ −=21.()8.53.48.53.45.1−− −= −+= −22.()6.99.26.99.216.1−−= −+ −= −23.True; the product of two negative real numbersis positive.24.True. The commutative property of additionstates+=+abba, and the commutativeproperty of multiplication states⋅=⋅a bb a.25.−6(8) =−(6⋅8) =−4826.12(−5) =−(12⋅5) =−6027.4 1456⋅=28.−7(−15) = 7⋅15 = 10529.−1.9(−2.7) = 1.9⋅2.7 = 5.13
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Solution Manual for Intermediate Algebra, 4th Edition - Page 9 preview imageISM: Intermediate AlgebraChapter R:Review530.The additive inverse ofa,−a, is also called theopposite ofa. The multiplicative inverse ofa,1 ,ais also called the reciprocal ofa.31.The multiplicative inverse of 10 is110.32.The multiplicative inverse of8−is18−.33.The multiplicative inverse of25is52.34.The multiplicative inverse of15−is551−= −.35.A fraction is written in lowest terms if thenumerator and denominator share no commonfactor other than 1.36.2 622 5⋅=⋅62⋅655=⋅37.333 1111243 88⋅==⋅38.24466204 55⋅−= −= −⋅39.49is in lowest terms.40.5 125 123253 255 3 43 5 55⋅⋅=⋅⋅⋅=⋅⋅=3⋅43⋅5⋅545⋅=41.252 5343 42 53 2 22⋅⎛⎞−= −⎜⎟⋅⎝⎠⋅= −⋅⋅= −532⋅⋅253 256⋅= −⋅= −42.484333384 33 84÷=⋅⋅=⋅=3⋅34⋅212⋅=43.10351210 123510 123 525=⋅⋅=⋅⋅=3⋅43⋅5⋅8=44.3232511111111++==45.8138135515151515−−−=== −35⋅13= −46.3113111427277777−−−⋅−−=== −= −47.8383515555−−===48.The least common denominator is the smallestnumber that each denominator has as a commonmultiple.
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Solution Manual for Intermediate Algebra, 4th Edition - Page 10 preview imageChapter R:ReviewISM: Intermediate Algebra649.202 2 5153 5=⋅⋅=⋅The common factor is 5. The remaining factorsare 2, 2, and 3.2 2 3 560=⋅⋅⋅=LCD3333 392020320 360⋅=⋅==⋅2242 481515415 460⋅=⋅==⋅50.182 3 3453 3 5=⋅⋅=⋅⋅The common factors are 3 and 3. The remainingfactors are 2 and 5.3 3 2 590=⋅⋅⋅=LCD5555 5251818518 590⋅=⋅==⋅1121 224545245 290⋅−= −⋅= −= −⋅51.LCD =2 2 5 360⋅⋅⋅=323324201520315498606098601760+=⋅+⋅=++==52.LCD =2 7 342⋅⋅=()511511142114215311 21432121522424215224271 7426 716−⎛⎞−=+⎜⎟⎝⎠−⎛⎞=⋅+⋅⎜⎟⎝⎠−⎛⎞=+⎜⎟⎝⎠+ −=−− ⋅==⋅= −53.LCD =5 5 6150⋅⋅=()47472530253046752563052435150150243515059150−−⎛⎞−−=+⎜⎟⎝⎠−−⎛⎞=⋅+⋅⎜⎟⎝⎠−−⎛⎞=+⎜⎟⎝⎠−+ −== −54.LCD =2 3 3 590⋅⋅⋅=()515118451845551 218545225290902529027903 910 9310−−⎛⎞−−=+⎜⎟⎝⎠−−⎛⎞=⋅+⋅⎜⎟⎝⎠−−⎛⎞=+⎜⎟⎝⎠−+ −=−=⋅= −⋅= −55.The Distributive Property states thata(b+c) =ab+ac.56.−(a+b) =−a−b57.()5355 3515+=⋅+⋅=+xxx58.()()6166166−+= −⋅+ −⋅= −−xxx59.()()48448432−−= −⋅− −⋅= −+zzz60.()1116969333693323+=⋅+⋅=+=+xxxx61.−(11p+ 8) =−11p+ (−8) =−11p−8
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Solution Manual for Intermediate Algebra, 4th Edition - Page 11 preview imageISM: Intermediate AlgebraChapter R:Review762.−(−9−4t) =−(−9)−(−4t) = 9 + 4tR.3 Exercises64.5566=66.7722−=68.()6106104−+=−+=70.()936+ −=72.()()8.24.58.24.512.7−−=−+ −= −74.−5(−15) = 5⋅15 = 7576.7(−15) =−(7⋅15) =−10578.−10.4(−0.6) = 10.4⋅0.6 = 6.2480.2555405 88⋅5==⋅82.408 58168 2⋅==⋅58⋅522=⋅84.2 152 152 3 52585 85 2 4⋅⋅⋅⋅===⋅⋅⋅35⋅⋅52⋅344=⋅86.7127 127 3 473353353 5 7⋅⋅⎛⎞−−=⋅==⎜⎟⋅⋅⎝⎠3⋅43⋅57⋅⋅45=88.101510213213152 5 3 73 3 525−÷= −⋅⋅⋅⋅= −⋅⋅⋅= −3⋅733⋅⋅5⋅143= −90.1818 1418 1473737 3143 6 7 27 33⋅=⋅=⋅⋅⋅⋅=⋅=67⋅⋅27⋅3⋅12=92.818818105 2555555−⋅−== −= −= −25⋅2= −94.LCD2 2 2 3 372=⋅⋅⋅⋅=373 9748188918427287272272872172−=⋅−⋅=−−== −96.()3533 5315−=⋅−⋅=−yyy98.()()54554520−−= −⋅− −⋅= −+xxx100.()3232262+⋅=⋅+⋅=+xyxyxy102.()22231531533363033210⎛⎞−+= −⋅+−⋅⎜⎟⎝⎠= −−= −−xxxx104.−(11−8k) =−11−(−8k) =−11 + 8k106.41241216−+=+=
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Solution Manual for Intermediate Algebra, 4th Edition - Page 12 preview imageChapter R:ReviewISM: Intermediate Algebra8108.182 92455 95LCD2 2 2 3 5120⋅==⋅=⋅⋅⋅⋅=()182322345245242242355242454811512012048115120163120−−= −−= −⋅−⋅= −−−+ −== −110.2212 1283383 2 4⋅÷=⋅==⋅⋅132⋅⋅1124=⋅112.5.410.55.15.1−+==114.()5 9454545− −⋅= − −= −= −116.48478337377328242121421421−=⋅−⋅=−==118.414215156152658530308530−=⋅−⋅=−−=330333303 10====⋅311010=⋅120.101062110126621211212121101261162121−+= −+⋅= −+−+==122.202045 4 4551554⋅⋅=⋅==4 45⋅⋅16161==124.70−is undefined.126.005=128.()990+ −=Additive Inverse Property130.1,0=≠aaaDivision Property132.()34312−=−xxDistributive Property134.006=Division Property136.83−49 = 83 + (−49) = 34The difference in life expectancy from 1950 to2016 is 34 years.138.()()()400204560105150400150204560105550230550230320−−−−+=+−+++=−=+ −=Paul’s balance at the end of the month was $320.140.()53585358543− −=+=The difference between the highest and lowestelevations in Louisiana is 543 feet.142.a.12351,2,1.5,1.666667,1123813211.6,1.625,1.615385,581334551.619048,1.617647,2134891441.618182,1.617978, ...5589===≈==≈≈≈≈≈
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Solution Manual for Intermediate Algebra, 4th Edition - Page 13 preview imageISM: Intermediate AlgebraChapter R:Review9b.The ratios approach a number that isroughly equivalent to 1.618. The goldenratio is151.61803398875...2+≈c.Answers will vary.144.()(),626288=− −=+==dP Q−260146.()(),1.69.31.69.310.910.9=− −=+==dP Q−1260−9.31.6148.()77657,665515530737375555⎛⎞=−−=+=⋅+⎜⎟⎝⎠=+==dP Q6075−−3150.993311,,9,,,3,,,1424242⎧⎫±±±±±±±±±⎨⎬⎩⎭152.We do not distribute across multiplication.154.No, the order of subtraction is important. Forexample:755722−≠−≠ −156.No, the order of division is important. Forexample:2442122≠≠158.The fraction14is equivalent to28 ,so 2 parts in8 added to 1 part in 8 results in 3 parts in 8.160.162.164.166.168.Section R.4Section R.4 Quick Checks1.In the expression65 ,the number 5 is the baseand 6 is the exponent or power.2.False.()477 7 7 77 7 7 7−= −⋅⋅⋅= −⋅⋅⋅3.344 4 464=⋅⋅=4.()() ()277749−=−⋅ −=5.()() () ()3101010101000−=−⋅ −⋅ −= −6.322228333327⎛⎞=⋅⋅=⎜⎟⎝⎠7.()288 864−= −⋅= −
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Solution Manual for Intermediate Algebra, 4th Edition - Page 14 preview imageChapter R:ReviewISM: Intermediate Algebra108.()() () ()()35555125125− −= −−⋅ −⋅ −= − −=⎡⎤⎣⎦9.5 2610616⋅+=+=10.3 25 663036⋅+⋅=+=11.4(5 + 3) = 4⋅8 = 3212.8(9−3) = 8⋅6 = 4813.(12−4)(18−13) = 8⋅5 = 4014.(4 + 9)(6−4) = 13⋅2 = 2615.37104913+=+16.148 251416522418−+⋅+=−++=−=17.()( )[][ ]25 2 83825 2 5825 10825 250−−=−⎡⎤⎡⎤⎣⎦⎣⎦=−==18.()()358842 942 5105+===−19.The order of operations is (1) parentheses,(2) exponents, (3) multiplication and division,(4) addition and subtraction.20.65 261016+⋅=+=21.()39412 448+⋅=⋅=22.75126410147+==+23.()[]483245 241014+−⋅=+⋅⎡⎤⎣⎦=+=24.4(62)94 491697−−=⋅−=−=25.282 582 2585042−+⋅= −+⋅= −+=26.35 33 25 33 815249⋅−⋅=⋅−⋅=−= −27.225(108)5 25 420−=⋅=⋅=28.223( 2)6 33 43 46 33 16121848304818−+⋅−⋅=⋅+⋅−⋅=+−=−= −29.2464362 322 32436624085++=⋅+⋅++=+==30.()32233 72 853 72 33 492 273 4954353 515−−−= −−⋅= −−⋅= −−= −−= − ⋅= −R.4 Exercises32.()() () ()3444464−=−⋅ −⋅ −= −34.()422 2 2 216−= −⋅⋅⋅= −36.()3( 7)[( 7)( 7)( 7)]343343− −= −−−−= − −=38.4333338144444256⎛⎞⎛⎞−= −⋅⋅⋅= −⎜⎟⎜⎟⎝⎠⎝⎠40.()()22329413−+ −=+=42.5 31215123−⋅+= −+= −44.2 + 5(8−5) = 2 + 5⋅3 = 2 + 15 = 1746.()[]()3 15733 1543 1133−−=−==⎡⎤⎣⎦
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Solution Manual for Intermediate Algebra, 4th Edition - Page 15 preview imageISM: Intermediate AlgebraChapter R:Review1148.12484 242221 2−⋅⋅===−−− ⋅1 2− ⋅4= −50.6 25 3121533⋅−⋅=−= −=52.345154 3129122155⋅+⋅=+=+=54.()()[]()2 252 1042 252 62 25122 1326−−=−⎡⎤⎡⎤⎣⎦⎣⎦=−==56.()()( )()( )()222 5252 352 32562531−−− −= −− −= −−= −−= −58.()()()()()()22 42 352 42 3252 46252 4192 4192 2346−+⋅−= −+⋅−= −+−= −+ −= −+= −= −60.6(3 210)6(610)6( 4)2424⋅−=−=−= −=62.222 32 9181861641641244⋅⋅====−−−36⋅322=⋅64.233(52 )3(54)3 932 272 3 92 3++⋅===⋅⋅⋅⋅9⋅23⋅9⋅12=66.()()()( )[][]()()4 32 864 32 25 142 235 142 54 345 14104 745 4+−+⎡⎤⎡⎤⎣⎦⎣⎦=−+−⎡⎤⎡⎤⎣⎦⎣⎦+=−==754⋅⋅75=68.23359545293 89293 2952924995591⎛⎞⎛⎞⋅=⋅⎜⎟⎜⎟⎜⎟+−⋅⎝⎠−⋅⎝⎠⎛⎞=⋅⎜⎟−⎝⎠=⋅=70.()()522454324516322016121628+−+=+−+=−+=+=72.2225(376 )7 245(3736)7 216546 29546 235(1)1416129545239−⋅−−⋅−+=++⋅−+⋅−−=+−+−=+5323 3915299139=⋅−=−=74.()()224343431615 25(75)41151044( 2)302824 72−−−−−−==−−− −+⋅+423389152103303044303011303043043014⋅−⋅−==−−−⎛⎞⎛⎞==−−⎜⎟⎜⎟−⎝⎠⎝⎠=76.()2354−⋅−=78.() ()356324+⋅−=
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Solution Manual for Intermediate Algebra, 4th Edition - Page 16 preview imageChapter R:ReviewISM: Intermediate Algebra1280.24 3.1416 34 3.1416 912.5664 9113.0976113.10⋅⋅=⋅⋅=⋅=≈The surface area is about 113.10 square centimeters.82.2108100 88003202.52.52.5⋅⋅===The engine has a rating of 320 horsepower.84.()()()()()( )()()()( )2222222222105100801001151009510010510052015554425400225252547001754−+−+−+−+−+ −++ −+=++++===86.Answers will vary. One suggestion: 3 + 2⋅7 is equivalent to 3 + 7 + 7, which equals 17.Since 3 + 2⋅7 = 3 + 14 = 17, we can see that we multiply before adding.88.34−is equivalent to31 41 4 4 464,− ⋅= − ⋅⋅⋅= −whereas3444464()()()().−= −−−= −90.36216375216159335125125125125⎛⎞−=−=−=⎜⎟⎝⎠92.233298113 2167−−==+⋅+94.()24482.33.726.33113⋅−⋅≈ −
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