Solution Manual for Finite Mathematics with Applications In the Management, Natural, and Social Sciences, 12th Edition

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SOLUTIONSMANUALSALSCIANDRANiagara County Community CollegeMATHEMATICS WITHAPPLICATIONSANDFINITEMATHEMATICSWITHAPPLICATIONSIN THEMANAGEMENT,NATURAL,ANDSOCIALSCIENCESTWELFTHEDITIONMargaret L. LialAmerican River CollegeThomas HungerfordSaint Louis UniversityJohn HolcombCleveland State UniversityBernadette MullinsBirmingham-Southern College

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ContentsChapter 1Algebra and Equations1Chapter 2Graphs, Lines, and Inequalities47Chapter 3Functions and Graphs94Chapter 4Exponential and Logarithmic Functions160Chapter 5Mathematics of Finance198Chapter 6Systems of Linear Equations and Matrices234Chapter 7Linear Programming316Chapter 8Sets and Probability422Chapter 9Counting, Probability Distributions, and Further Topics in Probability453Chapter 10Introduction to Statistics490Chapter 11Differential Calculus521Chapter 12Applications of the Derivative603Chapter 13Integral Calculus677Chapter 14Multivariate Calculus751

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1Chapter 1: Algebra and EquationsSection 1.1 The Real Numbers1.True. This statement is true, since every integercan be written as the ratio of the integer and 1.For example,551.2.False. For example, 5 is a real number, and1052which is not an irrational number.3.Answers vary with the calculator, but2, 508, 429, 787798, 458, 000is the best.4.0( 7)( 7)0 This illustrates the commutative property ofaddition.5.6(4)66 4ttThis illustrates the distributive property.6.3 + (–3) = (–3) + 3This illustrates the commutative property ofaddition.7.(–5) + 0 = –5This illustrates the identity property of addition.8.14( 4) ()1This illustrates the multiplicative inverseproperty.9.8 + (12 + 6) = (8 + 12) + 6This illustrates the associative property ofaddition.10.1 ( 20)20 This illustrates the identity property ofmultiplication.11.Answers vary. One possible answer: The sum ofa number and its additive inverse is the additiveidentity. The product of a number and itsmultiplicative inverse is the multiplicativeidentity.12.Answers vary. One possible answer: When usingthe commutative property, the order of theaddends or multipliers are changed, while thegrouping of the addends or multipliers ischanged when using the associative property.For Exercises 13–16, letp= –2,q= 3 andr= –5.13.()[][]()35325(3)32153 1339pq−+= −−+= −−+= −= −14.()()( )22 352 816qr−=+==15.qrqp3( 5)223( 2)1+ −−=== −+ −16.33(3)99323( 2)2( 5)6104qpr===−−−−−+17.Letr= 3.8.APR1212(3.8)45.6%r18.Letr= 0.8.APR1212(0.8)9.6%r19.LetAPR= 11.1211121112.9167%APRrrrr20.LetAPR= 13.2.1213.21213.2121.1%APRrrrr21.34 55320517512  22.28( 4)( 12)  Take powers first.8 – 16 – (–12)Then add and subtract in order from left to right.8 – 16 + 12 = –8 + 12 = 423.(45) 661 66660   

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2CHAPTER 1 ALGEBRA AND EQUATIONS24.2(37)4(8)4( 3)( 3)( 2) Work above and below fraction bar. Domultiplications and work inside parentheses.2( 4)328322441261266 25.284( 12) Take powers first.8 – 16 – (–12)Then add and subtract in order from left to right.8 – 16 + 12 = –8 + 12 = 426.2(35)2313Take powers first.–(3 – 5) – [2 – (9 – 13)]Work inside brackets and parentheses.– (–2) – [2 – (–4)] = 2 – [2 + 4]= 2 – 6 = –427.32( 2)162( 3)641Work above and below fraction bar. Take roots.32(2)(4)2( 3)81Do multiplications and divisions.3122681Add and subtract.312114222271777 28.226325613Take powers and roots.363(5)3615213736134929.20401894587,,27,, 6.735,475233769130.187385, 2.9884,85 ,,10,6311731.12 is less than 18.5.12 < 18.532.–2 is greater than –20.–2 > –2033.xis greater than or equal to 5.7.5.7x34.yis less than or equal to –5.5y 35.zis at most 7.5.7.5z36.wis negative.0w37.62 38.3 4.7539.3.1440.1 3.3341.alies to the right ofbor is equal tob.42.b+c=a43.c<a<b44.alies to the right of 045.(–8, –1)This represents all real numbers between –8 and–1, not including –8 and –1. Draw parentheses at–8 and –1 and a heavy line between them. Theparentheses at –8 and –1 show that neither ofthese points belongs to the graph.46.[–1, 10]This represents all real numbers between –1 and10, including –1 and 10. Draw brackets at–1 and 10 and a heavy line between them.47.2, 3All real numbersxsuch that –2 <x≤3Draw a heavy line from –2 to 3. Use aparenthesis at –2 since it is not part of the graph.Use a bracket at 3 since it is part of the graph.

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SECTION 1. 1 THE REAL NUMBERS348.[–2, 2)This represents all real numbers between –2 and2, including –2, but not including 2.Draw a bracket at –2, a parenthesis at 2, and aheavy line between them.49.2,All real numbersxsuch thatx> –2Start at –2 and draw a heavy line to the right.Use a parenthesis at –2 since it is not part of thegraph.50.(–∞, –2]This represents all real numbers less than orequal to –2. Draw a bracket at –2 and a heavyline to the left.51.9129(12)3  52.848(4)4 53. 4114(4)1541519      54.6124(6)166(16)22     55.555 __ 55556.444444 57.10331077777758.6( 4)46101010101010 59.282866666660.353( 5) 3–5153 515151561.35352352222 62.5151451464663.Whena< 7,a– 7 is negative.So7(7)7aaa .64.Whenb≥c,b–cis positive.Sobcbc.Answers will vary for exercises 65–67. Sample answersare given.65.No, it is not always true that.ababForexample, leta= 1 andb= –1. Then,1( 1)00ab , but1( 1)112ab.66.Yes, ifaandbare any two real numbers, it isalways true that.abbaIn general,a–b= –(b–a). When we take the absolutevalue of each side, we get()abbaba .67.22bbonly whenb= 0. Then each sideof the equation is equal to 2. Ifbis any othervalue, subtracting it from 2 and adding it to 2will produce two different values.68.For females:|63.5 |8.4x; for males:|68.9 |9.3x

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4CHAPTER 1 ALGEBRA AND EQUATIONS69.1; 3006200770.8; 2008, 2009, 2010, 2011, 2012, 2013, 2014,201571.9; 2006, 2008, 2009, 2010, 2011, 2012, 2013,2014, 201572.4; 2008, 2010, 2013, 201573.4; 2006, 2007, 2009, 201174.8; 2006, 2007, 2009, 2010, 2011, 2012, 2013,2014, 201575.|10.614.9 | |4.3 |4.3−= −=76.| 63.1( 8.0) | | 71.1|71.1− −==77.|1.063.1| |64.1|64.1−−= −=78.|10.6( 5.7) | |16.3 |16.3− −==79.|5.7( 8.0) | | 2.3 |2.3−− −==80.|1.0( 5.7) | | 4.7 |4.7−− −==81.3; 30062010, 2015, 201682.7; 2010, 2011, 2012, 2013, 2014, 2015, 201683.6; 2010, 2011, 2012, 2013, 2014, 201584.3; 2014, 2015, 2016Section 1.2 Polynomials1.611.21, 973,822.6852.11( 6.54)936,171,103.1 3.618289.099133974.75.016333996795.23is negative, whereas2( 3)is positive. Both33and3( 3)are negative.6.To multiply34and54, add the exponents sincethe bases are the same. The product of34and43cannot be found in the same way since thebases are different. To evaluate the product, firstdo the powers, and then multiply the results.7.2323544448.4646104444   9.25257( 6)( 6)( 6)( 6) 10.565611(2 )(2 )(2 )(2 )zzzz11.744 728555uuu12.435320236666(6 )yyyyy13.degree 4; coefficients: 6.2, –5, 4, –3, 3.7;constant term 3.7.14.degree 7; coefficients: 6, 4, 0, 0, –1, 0, 1, 0;constant term 0.15.Since the highest power ofxis 3, the degreeis 3.16.Since the highest power ofxis 5, the degreeis 5.17.323232548xxxxxx 3322342( 58 )xxxxxx 3213xxx 18.32257482pppp 3224( 58)(72)pppp  322439ppp 

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SECTION 1.2 POLYNOMIALS519.22438262yyyy22438262yyyy  22438262yyyy 2242( 36 )(82)yyyy  2636yy 20.22725326bbbb22725326bbbb 22732256bbbb 241b21.32322243281xxxxx32322243281xxxxx 32322243281xxxxx33222228(4 )( 31)xxxxx 2644xx 22.322391184106yyyyy 322391184106yyyyy322394( 1110 )(86)yyyyy 323132114yyy23.29261mmm2( 9) 29(6)( 9)( 1)mmmmm  3218549mmm 24.22468aaa2242 ( 6 )2 (8)aaaaa3281216aaa25.2(35) 421zzz2232232(3 ) 421(5) 421126320105121475zzzzzzzzzzzzz322122061035zzzzz32121475zzz26.32(23) 43kkkk32322433 43kkkkkkk432328621293kkkkkk4328673kkkk27.(6k– 1)(2k+ 3)(6 )(23)( 1)(23)kkk 2121823kkk212163kk28.(8r+ 3)(r– 1)Use FOIL.28833rrr2853rr29.(3y+ 5)(2y+1)Use FOIL.263105yyy26135yy30.(5r– 3s)(5r– 4s)2225201512rrsrss22253512rrss31.(9k+q)(2k–q)221892kkqkqq22187kkqq32.(.012x– .17)(.3x+ .54)= (.012x)(.3x) + (.012x)(.54)+ (–.17)(.3x) + (–.17)(.54)2.0036.00648.051.0918xxx2.0036.04452.0918xx33.(6.2m– 3.4)(.7m+ 1.3)24.348.062.384.42mmm24.345.684.42mm34.2p–3[4p– (8p+ 1)]= 2p– 3(4p– 8p– 1)= 2p– 3(– 4p– 1)= 2p+ 12p+ 3= 14p+ 335.5k– [k+ (–3 + 5k)]= 5k– [6k– 3]= 5k– 6k+ 3= –k+ 3

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6CHAPTER 1 ALGEBRA AND EQUATIONS36.2(31)(2)(25)xxx2235242025xxxx2235242025xxxx2234(520 )(–225)xxxx21527xx 37.R= 5 (1000x) = 5000xC= 200,000 + 1800xP= (5000x) – (200,000 + 1800x)= 3200x– 200,00038.R= 8.50(1000x) = 8500xC= 225,000 + 4200xP= (8500x) – (225,000 + 4200x)= 4300x– 225,00039.R=9.75(1000x) = 9750x22260, 000( 33480325)33480259, 675Cxxxx  22(9750 )( 33480259, 675)36270259, 675Pxxxxx 40.R= 23.50(1000x) = 23,500x22145, 000( 4.23220425)4.23220144,575Cxxxx  22(23,500 )( 4.23220144,575)4.220, 280144,575.Pxxxxx 41.a.According to the bar graph, the net earningsin 2007 were $673 million.b.Letx= 7.( )( )( )32324.79122.5110428634.79 7122.5 71104 72863505.47xxx−+−=−+−=According to the polynomial, the net earningsin 2007 were approximately $505 million.42.a.According to the bar graph, the net earningsin 2015 were $2757 million.b.Letx= 15.()()()32324.79122.5110428634.79 15122.5 151104 1528632300.75xxx−+−=−+−=According to the polynomial, the netearnings in 2015 were approximately $2301million.43.a.According to the bar graph, the net earningsin 2012 were $1384 million.b.Letx= 12.()()()32324.79122.5110428634.79 12122.5 121104 1228631022.12xxx−+−=−+−=According to the polynomial, the netearnings in 2012 were approximately $1022million.44.a.According to the bar graph, the net earningsin 2013 were $8 million.b.Letx= 13.()()()32324.79122.5110428634.79 13122.5 131104 1328631310.13xxx−+−=−+−=According to the polynomial, the netearnings in 2013 were approximately $1,310million.45.Letx= 17.()()()32324.79122.5110428634.79 17122.5 171104 1728634035.77xxx−+−=−+−=According to the polynomial, the net earnings in2017 will be approximately $4036 million.

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SECTION 1.2 POLYNOMIALS746.Letx= 18.()()()32324.79122.5110428634.79 18122.5 181104 1828635254.28xxx−+−=−+−=According to the polynomial, the net earnings in2018 will be approximately $5254 million.47.Letx= 19.()()()32324.79122.5110428634.79 19122.5 191104 1928636745.11xxx−+−=−+−=According to the polynomial, the net earnings in2019 will be approximately $6745 million.48.The figures for 2013 – 2015 seem high, butplausible. To see how accurate these conclusionsare, search Starbucks.com for later annualreports.For exercises 49–52, we use the polynomial329.5401.6612225,598.xxx49.Letx= 10.329.5(10)401.6(10)6122(10)25,5984962Thus, the costs were approximately $4962million in 2010. The statement is false.50.Letx= 15.329.5(15)401.6(15)6122(15)25,5987934.5Thus, the costs were approximately $7934.5million in 2015. The statement is true.51.Letx= 12.329.5(12)401.6(12)6122(12)25,5986451.6Letx= 15.329.5(15)401.6(15)6122(15)25,5987934.5Thus, the costs were $6451.6 million in 2012and $7934.5 million in 2015. The statement isfalse.52.Letx= 11.329.5(11)401.6(11)6122(11)25,5985794.9Letx= 16.329.5(16)401.6(16)6122(16)25,5988456.4Thus, the costs were $5794.9 million in 2011and $8456.4 million in 2016. The statement istrue.For exercises 53–58, we use the polynomial3272.85208216,53259,357.xxx−+−+53.Letx= 7.323272.85208216,53259,35772.85(7)2082(7)16,532(7)59,35720, 663.45xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2007 wasapproximately $20,663 million.54.Letx= 10.323272.85208216,53259,35772.85(10)2082(10)16,532(10)59,35729,387xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2010 was$29,387 million.55.Letx= 12.323272.85208216,53259,35772.85(12)2082(12)16,532(12)59,35734,896.2xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2012 wasapproximately $34,896 million.56.Letx= 15.323272.85208216,53259,35772.85(15)2082(15)16,532(15)59,35733,958.25xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2015 wasapproximately $33,958 million.

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8CHAPTER 1 ALGEBRA AND EQUATIONS57.Letx= 13.323272.85208216,53259,35772.85(13)2082(13)16,532(13)59,35736, 247.55xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2013 wasapproximately $36,248 million.Letx= 9.323272.85208216,53259,35772.85(9)2082(9)16,532(9)59,35726,103.35xxx−+−+= −+−+=Thus, the profit for PepsiCo Inc in 2009 wasapproximately $26,104 million. Therefore, theprofit was higher in 2013.58.By comparing the answers to problems 55 and56, the profit was higher in 2012.59.27.25005230, 000Pxx. Here is part ofthe screen capture.For 25,000, the loss will be $100,375;For 60,000, there profit will be $96,220.There is a loss at the beginning because oflarge fixed costs. When more items aremade, these costs become a smaller portionof the total costs.60.In order for the company to make a profit,27.25005230, 0000PxxSet the profit function equal to 0 and solveforx.284450215, 0000xxBy the quadratic formula,x≈45 orx≈–601.Sincexrepresents a positive number,x= 45.Therefore, between 40,000 and 45,000calculators must be sold for the company tomake a profit.61.Letx= 100 (in thousands)27.2(100)5005(100)230, 000342,500d.The profit for selling 100,000 calculators is$342,500.62.Letx= 150 (in thousands)27.2(150)5005(150)230, 000682, 750d.The profit for selling 150,000calculators is $682,750.Section 1.3 Factoring1.212241212212 (2)xxxxxx x2.5655 (1)5 (13 )5 (113 )yxyyyxyx3. 322255151rrrr rrrrr rr4.322238(3 )(8)38tttt ttttt tt

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SECTION 1.3 FACTORING95.3261218zzz266 (2 )6 (3)z zzzz2623z zz6.3255510xxx255 (11 )5 (2)xxxxx25112x xx7.233(21)7(21)yy22(21) (3)(21)7(21)yyy2(21) [37(21)]yy2(21) (3147)yy22(21) (144)2 2172yyyy8.53(37)4(37)xx323(37) (37)(37) (4)xxx32(37)(37)4xx32(37)942494xxx32(37)94245xxx9.463(5)(5)xx442(5)3(5) (5)xxx24(5)35xx42(5)31025xxx42(5)1028xxx10.243(6)6(6)xx222222222223(6) (1)3(6)2(6)3(6)12(6)36121236361224723622473xxxxxxxxxxxxxx11.254(1)(4)xxxx12.276(1)(6)uuuu13.2712(3)(4)xxxx14.2812(2)(6)yyyy15.2632xxxx16.24551xxxx17.22331xxxx18.21243yyyy19.23414xxxx20.22824uuuu21.291427zzzz22.261628wwww23.21024(4)(6)zzzz24.21660(6)(10)rrrr

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10CHAPTER 1 ALGEBRA AND EQUATIONS25.2294(21)(4)xxxx26.2384(32)(2)wwww27.215234(34)(51)pppp28.28143(41)(23)xxxx29.241615(25)(23)zzzz30.2122915(35)(43)yyyy31.2654(21)(34)xxxx32.2121(41)(31)zzzz33.2102110(52)(25)yyyy34.21544(52)(32)uuuu35.2654(21)(34)xxxx36.212710(32)(45)yyyy37.2325351aaaa38.2264812068206(10)(2)aaaaaa39.22281(9)(9)(9)xxxx40.221772(8 )(9 )xxyyxyxy.41.22229124(3)2(3)(2)2(32)ppppp42.232(32)(1)rrrr.43.22310(2 )(5 )rrttrtrt.44.2226(23 )(2 ).aabbabab45.22222816()2()(4 )(4 )(4 )mmnnmmnnmn46.22816102 4852(21)(25)kkkkkk47.22412923uuu48.222916343434pppp49.225104ppThis polynomial cannot be factored50.2101735123xxxx51.22492323rvrvrv52.2232874xxyyxyxy53.222442xxyyxy54.221612182 8692 4323uuuuuu55.231330(35)(6)aaaa.56.2328(34)(2)kkkk57.2221132(72 )(3)mmnnmnmn58.281100(910)(910)yyy59.22421(7 )(3 )yyzzyzyz60.2499aThis polynomial cannot be factored.61.212164(118)(118)xxx62.22222224144942( )(7)(7)4561964(7 )zzyyzzyyzzyyzy63.333264(4)(4)416aaaaa64.33322166(6)636bbbbb

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SECTION 1.3 FACTORING1165.33332222827(2 )(3 )(23 )2233(23 ) 469rsrsrsrrssrsrrss66.333322100027(10)(3 )(103 ) 100309pqpqpqppqq67.364125m33(4)(5)m  22(45)4455mmm2(45) 162025mmm68.3216343y33(6 )(7)y2(67) 364249yyy69.331000yz33(10 )( )yz22(10) (10 )(10 )( )( )yzyyzz22(10) 10010yzyyzz70.3333221258(5)(2 )(52 ) 25104pqpqpqppqq71.42225623xxxx72.422271025yyyy73.42222111bbbbbbb74.422223441221zzzzzzz75.422221243223xxxxxxx76.422224278149923239xxxxxxx77.4422222216814949232349abababababab78.3366222242242222xyxyxyxxyyxyxyxxyyxxyy79.382262232242882224xxxxxxxxxx80.9336366223333333332264642222224224xxxxxxxxxxxxxxxxxx81.42226332133xxxxis not thecorrect complete factorization because233xcontains a common factor of 3. This commonfactor should be factored out as the first step.This will reveal a difference of two squares,which requires further factorization. The correctfactorization is42422226333 213 2113 21 (1)(1)xxxxxxxxx82.The sum of two squares can be factored when theterms have a common factor. An example is2222(3 )3999(1)xxx

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12CHAPTER 1 ALGEBRA AND EQUATIONS83.32232232(2)(2)(2)(2)(44)428486128,xxxxxxxxxxxxxxwhich is not equal to38x. The correctfactorization is328(2)(24).xxxx84.Factoring and multiplication are inverseoperations. If we factor a polynomial and thenmultiply the factors, we get the originalpolynomial back. For example, we can factor26xxto get(3)(2)xx. Then if wemultiply the factors, we get22(3)(2)2366xxxxxxxSection 1.4 Rational Expressions1.288567 87xxxxxx2.32227271812733mmmmmm3.2232255 5573575pppppp4.4222221863342464yyyyyy5.5155(3)54124(3)4mmmm6.1055(21)120105(21) 22zxzx7.4(3)4(3)(6)6wwww8.6(2)6(4)(2)4xxxxor64x9.23223123 (4)93343yyy yyyyyy10.22154515 (3)33935(3)335(3)3kkk kkkkkkkkkk11.2244(2)(2)2(3)(2)36mmmmmmmmmm12.226(3)(2)2(4)(3)412rrrrrrrrrr13.22312331111xxxxxxxxx14.22224422224zzzzzzzz15.2233383836416264 2aaaaa16.2323442102105918889uuuuuuuu17.333271476676631166111411 14xxxyxyyyxxx18.22621622217xyxyx yyyxyxxy19.215(2) 1515534(2)(2) 12124ababcababccc20.224(2)34(2) 338(2)4(2) 22xwwxwwxw x

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SECTION 1.4 RATIONAL EXPRESIONS1321.15310215336361023(51) 33 2 2 (51)3(51) 333(51) 2 24pppppppp22.283122836363122(4)363(4)6(4)6118(4)183kkkkkkkk23.918369(2)3(2)612 15306(2)15(2)27(2)(2)27390(2)(2)9010yyyyyyyyyyyy24.122461212(2)6(2)36368836(1)8(1)23(2)3(1)4(1)24(1)43(1)3(2)9rrrrrrrrrrrrrrrr25.22224129412202102102094(3)(5)(4)2(5)(3)(3)4(3)(5)(4)2(5)(3)(3)2(4)3aaaaaaaaaaaaaaaaaaaaaaaa26.222618121641296246(3)4(34)2(34)4(3)3 3283282(34)2(34)(2)2rrrrrrrrrrrrrrrrr27.22226341223(3)(2)(4)(1)(4)(3)(3)(1)(3)(2)(4)(1)2(4)(3)(3)(1)3kkkkkkkkkkkkkkkkkkkkkkkkkk28.22226928712(3)(2)(3)(3)(4)(2)(3)(4)3334444434nnnnnnnnnnnnnnnnnnnnnnnnnAnswers will vary for exercises 29 and 30. Sampleanswers are given.29.To find the least common denominator for twofractions, factor each denominator into primefactors, multiply all unique prime factors raisingeach factor to the highest frequency it occurred.30.To add three rational expressions, first factoreach denominator completely. Then, find thelowest common denominator and rewrite eachexpression with that denominator. Next, add thenumerators and place over the commondenominator. Finally, simplify the resultingexpression and write it in lowest terms.31.The common denominator is 35z.212 51 71073757557353535zzzzzzz32.The common denominator is 12z.454 45 316151343443121212zzzzzzz33.22(2)(2)33322433rrrrrr34.3131(31)(31)2188884yyyy 35.The common denominator is 5x.414 512020555555xxxxxxxxx

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