Solution Manual for College Algebra, 12th Edition

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SOLUTIONSMANUALBEVERLYFUSFIELDCOLLEGEALGEBRATWELFTHEDITIONMargaret L. LialAmerican River CollegeJohn HornsbyUniversity of New OrleansDavid I. SchneiderUniversity of MarylandCallie J. DanielsSt. Charles Community College

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1Chapter RREVIEW OF BASIC CONCEPTSSection R.1Sets1.The elements of the set of natural number are{1, 2, 3, 4, …}.2.SetAis a subset of setBif every element ofsetAis also an element of setB.3.The set of all elements of the universal setUthat do not belong to setAis the complementof setA.4.The intersection of setsAandBis made up ofall the elements belonging to both setAandsetB.5.The union of setsAandBis made up of all theelements belonging to setAor setB(or both).6.The set11139271,,,,is infinite.7.Using set notation, the set {x|xis a naturalnumber less than 6} is {1, 2, 3, 4, 5}.8.4, 5A9.16,18, 21, 5015,16,17,1816,1810.16,18, 21, 5015,16,17,1815,16,17,18, 21, 5011.The set {4, 5, 6, …, 15} has a limited numberof elements, so it is a finite set. Yes, 10 is anelement of the set.12.The set {1, 2, 3, 4, 5, …, 75} has a limitednumber of elements, so it is a finite set. Yes,10 is an element of the set.13.The set1112481,,,,has an unlimitednumber of elements, so it is an infinite set. No,10 is not an element of the set.14.The set {4, 5, 6, …} has an unlimited numberof elements, so it is an infinite set. Yes, 10 isan element of the set.15.The set {x|xis a natural number larger than11}, which can also be written as {11, 12, 13,14, …}, has an unlimited number of elements,so it is an infinite set. No, 10 is not an elementof the set.16.The set {x|xis a natural number greater thanor equal to 10}, which can also be written as{10, 11, 12, 13, …}, has an unlimited numberof elements, so it is an infinite set. Yes, 10 isan element of the set.17.There are infinitely many fractions between 1and 2, so {x|xis a fraction between 1 and 2}is an infinite set. No, 10 is not an element ofthe set.18.The set {x|xis an even natural number} hasno largest element. Because it has anunlimited number of elements, this is aninfinite set. Yes, 10 is an element of the set.19.The elements of the set {12, 13, 14, …, 20}are all the natural numbers from 12 to 20inclusive. There are 9 elements in the set, {12,13, 14, 15, 16, 17, 18, 19, 20}.20.The elements of the set {8, 9, 10, …, 17} areall the natural numbers from 8 to 17 inclusive.There are 10 elements in the set, {8, 9, 10, 11,12, 13, 14, 15, 16, 17}.21.Each element of the set11124321,,,,afterthe first is found by multiplying the precedingnumber by12. There are 6 elements in the set,1111124816321,,,,,.22.Each element of the set {3, 9, 27, …, 729}after the first is found by multiplying thepreceding number by 3. There are 6 elementsin the set, {3, 9, 27, 81, 243, 729}.23.To find the elements of the set{17, 22, 27, …, 47}, start with 17 and add 5 tofind the next number. There are 7 elements inthe set, {17, 22, 27, 32, 37, 42, 47}.24.To find the elements of the set{74, 68, 62, …, 38}, start with 74 and subtract6 (or add –6) to find the next number. Thereare 7 elements in the set, {74, 68, 62, 56, 50,44, 38}.25.When you list all elements in the set {allnatural numbers greater than 8 and less than15}, you obtain {9, 10, 11, 12, 13, 14}.

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2Chapter R Review of Basic Concepts26.When you list all elements in the set{all natural numbers not greater than 4} youobtain {1, 2, 3, 4}.27.6 is an element of the set {3, 4, 5, 6}, so wewrite 6{3, 4, 5, 6}.28.9 is an element of the set {2, 3, 5, 9, 8}, so wewrite 9{2, 3, 5, 9, 8}.29.5 is not an element of {4, 6, 8, 10}, so wewrite 5{4, 6, 8, 10}.30.13 is not an element of the set {3, 5, 12, 14},so we write 13{3, 5, 12, 14}.31.0 is an element of {0, 2, 3, 4}, so we write0{0, 2, 3, 4}.32.0 is an element of the set {0, 5, 6, 7, 8, 10}, sowe write 0{5, 6, 7, 8, 10}.33.{3} is a subset of {2, 3, 4, 5}, not an elementof {2, 3, 4, 5}, so we write {3}{2, 3, 4, 5}.34.{5} is a subset of {3, 4, 5, 6, 7}, not anelement of {3, 4, 5, 6, 7}, so we write{5}{3, 4, 5, 6, 7}.35.{0} is a subset of {0, 1, 2, 5}, not an elementof {0, 1, 2, 5}, so we write {0}{0, 1, 2, 5}.36.{2} is a subset of {2, 4, 6, 8}, not an elementof {2, 4, 6, 8}, so we write {2}{2, 4, 6, 8}.37.0 is not an element of,because the emptyset contains no elements. Thus, 0.38.is a subset of,not an element of.The empty set contains no elements. Thus wewrite,. 39.False. 3 is not one of the elements in{2, 5, 6, 8}.40.False. 6 is not one of the elements of{2, 5, 8, 9}.41.True. 1 is one of the elements of{11, 5, 4, 3, 1}.42.True. 12 is one of the elements of{18, 17, 15, 13, 12}.43.True. 9 is not one of the elements of{8, 5, 2, 1}.44.True. 3 is not an element of {7, 6, 5, 4}.45.True. Both sets contain exactly the same fourelements.46.True. Both sets contain exactly the same fiveelements.47.False. These two sets are not equal because{5, 8, 9, 0} contains the element 0, which isnot an element of {5, 8, 9}.48.False. These two sets are not equal because{3, 7, 12, 14, 0} contains the element 0, whichis not an element of {3, 7, 12, 14}.49.True. 1 and 2 are the only natural numbers lessthan 3.50.True. Both sets describe the same elements.For Exercises 51−62,A= {2, 4, 6, 8, 10, 12},B= {2, 4, 8, 10},C= {4, 10, 12},D= {2, 10}, andU= {2, 4, 6, 8, 10, 12, 14}.51.True. This statement says “Ais a subset ofU.”Because every element ofAis also an elementofU, the statement is true.52.True. This statement says “Cis a subset ofU.”Because every element ofCis also an elementofU, the statement is true.53.True. Because both elements ofD, 2 and 10,are also elements ofB,Dis a subset ofB.54.True. Because both elements ofD, 2, and 10,are also elements ofA,Dis a subset ofA.55.False. SetAcontains two elements, 6 and 12,that are not elements ofB. Thus,Ais not asubset ofB.56.False. SetBcontains two elements, 2 and 8,that are not elements ofC. Thus,Bis not asubset ofC.57.True. The empty set is a subset of every set.58.True. The empty set is a subset of every set.59.True. Because 4, 8, and 10 are all elements ofB, {4, 8, 10} is a subset ofB.60.False. Because 0 is not an element ofD, {0, 2}is not a subset ofD.61.False. BecauseBcontains two elements, 4 and8, that are not elements ofD,Bis not a subsetofD.62.False. There are three elements ofA(2, 6, and8) that are not elements ofC, soAis not asubset ofC.63.Every element of {2, 4, 6} is also an elementof {2, 3, 4, 5, 6}, so {2, 4, 6} is a subset of{2, 3, 4, 5, 6}.We write {2, 4, 6}{2, 3, 4, 5, 6}.

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Section R.1 Sets364.Every element of {1, 5} is also an element of{0, 1 2, 3, 5}, so {1, 5} is a subset of the set{0, 1 2, 3, 5}. We write {1, 5}{0, 1 2, 3, 5}.65.Because 0 is an element of {0, 1, 2}, but is notan element of {1, 2, 3, 4, 5}, {0, 1, 2} is not asubset of {1, 2, 3, 4, 5}. We write{0, 1, 2}{1, 2, 3, 4, 5}.66.Because 8 is an element of {5, 6, 7, 8}, but isnot an element of {1, 2, 3, 4, 5, 6, 7}, {5, 6, 7,8} is not a subset of {1, 2, 3, 4, 5, 6, 7}. Wewrite {5, 6, 7, 8}{1, 2, 3, 4, 5, 6, 7}.67.The empty set is a subset of every set, so {1, 4, 6, 8}.68.The empty set is a subset of every set,including itself, so.  69.True. 7 and 9 are the only elements belongingto both sets.70.True. 8 and 11 are the only elements belongingto both sets.71.False. {1, 2, 7}{1, 5, 9}{1, 2, 5, 7, 9},while {1, 2, 7}{1, 5, 9}{1}.72.False.{6, 12, 14, 16}{6, 14, 19}{6, 12, 14, 16,19}, while{6, 12, 14, 16}{6, 14, 19}{6, 14}.73.True. 2 is the only element belonging to bothsets.74.False. The sets {6, 8, 9} and {9, 8, 6} areequal because they contain exactly the samethree elements. Their union contains the sameelements, namely 8, 9, and 6.75.{3, 5, 9, 10}{3, 5, 9, 10}In order to belong to the intersection of twosets, an element must belong to both sets.Because the empty set contains no elements,{3, 5, 9, 10},  so the statement isfalse.76.True. For any setA,.AA 77.True. Because the two sets are equal, theirunion contains the same elements, namely 1,2, and 4.78.False. {1, 2, 4}{1, 2, 4}{1, 2, 4}79.True.80.True.For Exercises 81–110,U= {0, 1, 2 , 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13},M= {0, 2, 4, 6, 8},N= {1, 3, 5, 7, 9, 11, 13},Q= {0, 2, 4, 6, 8, 10, 12}, andR= {0, 1, 2, 3, 4}.81.MRThe only elements belonging to bothMandRare 0, 2, and 4, soMR= {0, 2, 4}.82.MUBecauseMU, the intersection ofMandUwill contain the same elements asM.MU=Mor {0, 2, 4, 6, 8}.83.MNThe union of two sets contains all elementsthat belong to either set or to both sets.MN= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13}84.MRThe unionMandRis made up of elementswhich belong toMor toR(or to both).MR= {0, 1, 2, 3, 4, 6, 8}85.MNThere are no elements which belong to bothMandN, soMN. MandNare disjointsets.86.UNBecauseNU,the elements belonging toUandNare all the elements belonging toN,UN=Nor {1, 3, 5, 7, 9, 11, 13}87.NR= {0, 1, 2, 3, 4, 5, 7, 9, 11, 13}88.MQBecauseMQ, the elements belonging toMorQare all the elements belonging toQ.MQ=Qor {0, 2, 4, 6, 8, 10, 12}89.NThe setNis the complement of setN, whichmeans the set of all elements in the universalsetUthat do not belong toN.N Qor {0, 2, 4, 6, 8, 10, 12}90.QThe setQis the complement of setQ, whichmeans the set of all elements in the universalsetUthat do not belong toQ.Q Nor {1, 3, 5, 7, 9, 11, 13}

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4Chapter R Review of Basic Concepts91.MQFirst formM, the complement ofM.Mcontains all elements ofUthat are notelements ofM. Thus,M{1, 3, 5, 7, 9, 10, 11, 12, 13}. Nowform the intersection ofMandQ. Thus, wehaveMQ{10, 12}.92.QRFirst formR, the complement ofR.Rcontains all elements ofUthe are not elementsofR. Thus,R {5, 6, 7, 8, 9, 10, 11, 12,13}. Now form the intersection ofQandR.Thus, we haveQR {6, 8, 10, 12}93.RBecause the empty set contains no elements,there are no elements belonging to bothandR. Thus,andRare disjoint sets, and.R 94.QBecause the empty set contains no elements,there are no elements belonging to bothandQ. Thus,andQare disjoint sets, and.Q 95.NBecausecontains no elements, the onlyelements belonging toNorare theelements ofN. Thus,andNare disjointsets, andNN or {1, 3, 5, 7, 9, 11, 13}.96.RBecausecontains no elements, the onlyelements belonging toRorare theelements ofR. Thus,andRare disjointsets, andRR or {0, 1, 2, 3, 4}.97.(MN)RFirst, form the intersection ofMandN.BecauseMandNhave no common elements(they are disjoint),MN. Thus,(MN)R.R Now, becausecontains no elements, the only elementsbelonging toRorare the elements ofR.Thus,andRare disjoint sets, andRRor {0, 1, 2, 3, 4}.98.(NR)MFirst form the union ofNandR. We haveNR= {0, 1, 2, 3, 4, 5, 7, 9, 11, 13}. Nowform the intersection of this set withM. Wehave (NR)M= {0, 2, 4}.99.(QM)RFirst form the intersection ofQandM. WehaveQM= {0, 2, 4, 6, 8} =M.Now formthe union of this set withR. We have(QM)R=MR= {0, 1, 2, 3, 4, 6, 8}.100.(RN)MFirst form the union ofRandN. We haveRN= {0, 1, 2, 3, 4, 5, 7, 9, 11, 13}. Nowfind the complement ofM. We haveM= {1, 3, 5, 7, 9, 10, 11, 12, 13}. Now, findthe intersection of these two sets. We have(RN)MNor {1, 3, 5, 7, 9, 11, 13}.101.(MQ)RFirst, findM, the complement ofM. WehaveM{1, 3, 5, 7, 9, 10, 11, 12, 13}. Next,form the union ofMandQ. We haveMQ= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12, 13}.UThus, we have(MQ)RURRor {0, 1, 2, 3, 4}.102.Q(MN)First, form the union ofMandN. We haveMN= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13}.Now form the intersection ofQwith this set.We haveQ(MN)Mor {0, 2, 4, 6, 8}.103.QNUFirst, findQ, the complement ofQ. We haveQ{1, 3, 5, 7, 9, 11, 13}.NNow findN, the complement ofN. We haveN{0, 2, 4, 6, 8, 10, 12}.QNext, formthe intersection ofNandU. We haveNUQUQFinally, we haveQNUQQ Because theintersection ofQandNUis,QandNUare disjoint sets.104.URBecause,U and,UUUwe haveURUURURUor{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.105.|,xxUxMThis means all elements ofUexcept those insetM. This gives the set {1, 3, 5, 7, 9, 10, 11,12, 13}.

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Section R.2 Real Numbers and Their Properties5106.|,xxUxRThis means all elements ofUexcept those insetR. This gives the set {5, 6, 7, 8, 9, 10, 11,12, 13}.107.|andxxMxQThis means all elements that are members ofboth setMand setQ, or.MQNote that setMis a subset of setQbecause all members ofsetMare included in setQ. Thus, the answeris setMor {0, 2, 4, 6, 8}.108.|andxxQxRThis means all elements that are members ofboth setQand setR, or.QRThus, theanswer is {0, 2, 4}.109.|orxxMxQThis means all elements that are eithermembers of setMor setQ, or.MQNotethat setMis a subset of setQbecause allmembers of setMare included in setQ. Thus,the answer is setQor {0, 2, 4, 6, 8, 10, 12}.110.|orxxQxRThis means all elements that are eithermembers of setQor setR, or.QRThus,the answer is {0, 1, 2, 3, 4, 6, 8, 10, 12}.Section R.2Real Numbers and TheirProperties1.{0, 1, 2, 3, …} describes the set of wholenumbers.2.{…, –3, –2, –1, 0, 1, 2, 3, …} describes the setof integers.3.In the expression36 ,6 is the base and 3 is theexponent.4.If the real numberais to the left of the realnumberbon a number line, thena< (is lessthan)b.5.The distance on a number line from a numberto 0 is the absolute value of that number.6.(a)0 is a whole number. Therefore, it is alsoan integer, a rational number, and a realnumber. 0 belongs to B, C, D, F.(b)34 is a natural number. Therefore, it isalso a whole number, an integer, arational number, and a real number. 34belongs to A, B, C, D, F.(c)94is a rational number and a realnumber.94belongs to D, F.(d)366is a natural number. Therefore,it is also a whole number, an integer, arational number, and a real number.36belongs to A, B, C, D, F.(e)13 is an irrational number and a realnumber.13 belongs to E, F.(f)21654100252.16is a rational number anda real number. 2.16 belongs to D, F.7.31010008.2 510210559.4410.74728xyxy For Exercises 11−16,51214846,,,3, 0,, 1, 2, 3,12.A11.1 and 3 are natural numbers.12.0, 1, and 3 are whole numbers.13.–6,124(or –3), 0, 1, and 3 are integers.14.51214846,(or3),, 0,, 1, and 3arerational numbers.15.3, 2and12are irrational numbers.16.All are real numbers.17.422 2 2 216  18.533 3 3 3 3243  19.   42222216   20.622222226421.    5333333243      22.5( 2)( 2)( 2)( 2)( 2)( 2)32 23.42 323 3 3 32 81162   24.34 545 5 54 125500   

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6Chapter R Review of Basic Concepts25.2 5123101231046   26.9 3164271642742327.    334 98724 1724 17847845660          28. 4653265316303163048304818       29.3(42 )( 225)(48)( 25)( 4)(3)12 30.2335316259424 484416  31.215194182895936361818174178253618363636  32.523118521025161511944040101040109162554040408   33.846128241243438276677   34.155 4683 4686582654126828654146655  For Exercises 35−48,p=−4,q= 8, andr= –10.35.22222747 81047 8100167 810016561007210028pqr        36.22242 810162 810161610321042pqr          37.8( 10)218( 4)42qrqp   38.4( 10)1474842prpq  39.353 852451251041045412548252354202020qrp    40.323( 10)23028108101517547145202020rqr   41.55( 10)50232( 4)3( 10)83050252211rpr 42.33 8243234210122102424243122012208qpr 43.810102241123233333( 4)831234848qrpq 44.810102454554882222222222422422qrpq 

Solution Manual for College Algebra, 12th Edition - Page 9 preview image

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Section R.2 Real Numbers and Their Properties745.2222342310228231064310430664302613663prq    46. 22262862444428484484182qpp 47. 333343 4433 48108123 012622ppr 48.3335 82 1452 131035 82275 82377402274054771427qpr  49.distributive50.distributive51.inverse52.inverse53.identity54.identity55.commutative56.commutative57.associative58.associative59.closure60.closure61.No; in general.abbaExamples willvary, i.e. ifa= 15 andb= 0, thena−b= 15−0 = 15, butb−a= 0−15 =−15.62.No; in generalabcabc.Examples will vary, i.e. ifa= 15,b= 0, andc= 3, then150315312,abcbut150315318.abc 63.10102222201111zzz64.3331212129444rrrr 65.565611mmm66.87878715(or15)aaaaa67.31632408927931633234089827893163325898273245+393yzyzyzyz68.1 20832411120832444111208324445+28=528myzmyzmyzmyzmyz   69.814(814)6pppp 70.151015105xxxx71.44444zyzyzy   72.33333mnmnmn   73.The process in your head should be thefollowing:72 1728 17(7228) 17(100) 17170074.The process in your head should be thefollowing:32 8032 2032 802032 100320075.The process in your head should be thefollowing:5151551123123112323182828821(100) 11502   

Solution Manual for College Algebra, 12th Edition - Page 10 preview image

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8Chapter R Review of Basic Concepts76.The process in your head should be thefollowing:232323317141741714454545442171017.4 10517477.This statement is false because6822 and68682. Acorrected statement would be6868or688678.This statement is false because332727 and332727  . A corrected statementwould be3333 or3333 .79.This statement is true because565 630and5 63030 .80.This statement is true because1414722and14772 .81.This statement is false. For example if you let2aand6bthen2644 and26264ab . A correctedstatement isabba, ifb>a> 0.82.This statement is true by the algebraicdefinition of absolute value stated in the text.83.101084.151585.4477 86.7722 87.88  88.1212  89.12011612044dPP ThedPvalue for a woman whose actualsystolic blood pressure is 116 and whosenormal value should be 120 is 4.90.Consider the relation,12017.dPPBecause 103 and 137 both differ from 120 by17, these are the two possible values for thepatient’s systolic blood pressure.For Exercises 91−98,x=−4 andy= 2.91.323( 4)2(2)1241616xy 92.25245 28101818xy  93.343( 4)4(2)1282020xy94.55 241041414yx     95.   232 2344 22 23 441281888yxxy 96.  44444 244 44 2168246444xyx97.88(2)( 4)16( 4)442020544yxx     98.242 242 282444xyx  99.True because2525and2525.100.True because the absolute value of any numberis always greater than or equal to 0.101.False.51388  while51351318. 102.False.81244 while8128124. 103.True because11611 666 and6666.104.True because10552 and10105.22

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Section R.2 Real Numbers and Their Properties9105.,141433dP Q   or,414133dP Q    106.,84841212dP R or,481212dP R  107.,818199d Q R or,1899d Q R  108.,1211313d Q S or,1121313d Q S  109.0xyifxandyhave the same sign.110.20x yifyis positive, because2xis positive for any nonzerox.111.0xyifxandyhave different signs.112.20yxifxis negative, because2yis positive for any nonzeroy.113.Because3xhas the same sign asx,30xyifxandyhave the same sign.114.0xyifxandyhave different signs.115.Because18( 1)1919  and1( 18)1919, the number of strokes between their scores is 19.116.18, 35532241521, 564 yards.No, it is not the same, because the sum of the absolute values is18, 35532241518, 35532241521, 594 The fact that | –15 | = 15 changes the two answers.117.BAC483.20.07519020.0150.031118.BAC364.00.07513530.0150.035119.BAC3203.80.07520040.0150.026If the man’s weight was greater, the BAC would be lower. For example, if the man weighed 250 pounds, hisBAC would be3203.80.07525040.0150.0084.120.BAC26140.07515020.0150.054If the woman consumed the same two glasses of wine over a longer period of time, her BAC would be lower.For example, if she drank the wine over 3 hours, her BAC would be26140.07515030.0150.039.

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10Chapter R Review of Basic ConceptsFor exercises 121−132, we use the formula250100012.56.251250,3CTYIAAAARwhere00.775, 00.11875, 012.5,CTYAAAand00.095.IA121.First, find the value of the numerator.250100012.56.2512503043437059250100012.56.251250339.73435435435435CTYIAAAA339.73113.23R122.First, find the value of the numerator.250100012.56.2512503413843815250100012.56.251250336.56520520520520CTYIAAAA336.56112.23R123.First, find the value of the numerator.250100012.56.2512504083249529250100012.5 66.251250309.95608608408608CTYIAAAA309.95103.33R124.First, find the value of the numerator.250100012.56.25125039539472715250100012.56.251250304.55597597597597CTYIAAAA304.55101.53R125.First, find the value of the numerator.250100012.56.2512503733341099250100012.56.251250292.10582582582582CTYIAAAA292.1097.43R

Solution Manual for College Algebra, 12th Edition - Page 13 preview image

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Section R.2 Real Numbers and Their Properties11126.First, find the value of the numerator.250100012.56.25125045633495217250100012.56.251250291.00659659659659CTYIAAAA291.0097.03R127.First, find the value of the numerator.250100012.56.25125038040476116250100012.56.251250289.55616616616616CTYIAAAA289.5596.53R128.First, find the value of the numerator.250100012.56.2512501411116263250100012.56.251250286.72224224224224CTYIAAAA286.7295.63R129.First, find the value of the numerator.250100012.56.2512501971724838250100012.56.251250286.02312312312312CTYIAAAA286.0295.33R130.First, find the value of the numerator.250100012.56.2512502852034757250100012.56.251250284.87452452452452CTYIAAAA284.8795.03R131.First, find the value of the numerator.250100012.56.25125041528469414250100012.56.251250281.61628628628628CTYIAAAA281.6193.93R

Solution Manual for College Algebra, 12th Edition - Page 14 preview image

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12Chapter R Review of Basic Concepts132.First, find the value of the numerator.250100012.56.25125037931428618250100012.56.251250281.38570570570570CTYIAAAA281.3893.83R133.First, find the value of the numerator.250100012.56.2512503434546436250100012.56.251250367.38502502502502CTYIAAAA367.38122.53R134.First, find the value of the numerator.250100012.56.25125033649455710250100012.56.251250363.32497497497497CTYIAAAA363.32121.13R135.First, find the value of the numerator.250100012.56.2512503243639004250100012.56.251250332.98492492492492CTYIAAAA332.98110.03R136.We use the maximum values for,, and,CTYAAAbut the minimum value forIA(because we want the fewestinterceptions possible.)250100012.56.251250250 0.7751000 0.1187512.5 12.56.251250 0475CTYIAAAA475158.33RSection R.3Polynomials1.The polynomial524xxis a trinomial ofdegree 5.2.A polynomial containing exactly one term is amonomial.3.A polynomial containing exactly two terms isa binomial.4.In the term26,xy–6 is the coefficient (ornumerical coefficient.5.A convenient way to find the product of twobinomials is to use the FOIL method.6.True7.True

Solution Manual for College Algebra, 12th Edition - Page 15 preview image

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Section R.3 Polynomials138.False.326aa9.False.2222xyxxyy10.False.2222xxx11.5252527444 41616xxx xxx  12.434343736361818yyy yyy  13.6464 111nnnnn14.8585 114aaaaa15.35358999916.282810444417.  425425425113643647272mmmm m mmm  18.  3643 6 4364138258258080tttt t ttt  19.23423423145553531515x yx yx xyyxyx y  20.  3223123 134474 72828xyx yxxy yxyx y   21.222212123311882244mnm nmmnnmnm n22.4242411253223535771010m nmnm mnnmnm n  23.252 510(2 )2224.434 312(6 )6625. 33332266666216xxxx 26. 555555252522232xxxx 27.3022320223 20 226026266(4)4 () ()4441416m nmnmnm nmmm       28.0433034330 34 330121212(2)2 () ()228 18x yxyxyx yyy     29.38838 3242232 36()()rrrrssss30.24424 28222()ppppqqqq31.42424488244848424( 4) ()( 4)256mmmmtptptptp32.3434331212223665( 5) ()( 5)125()nnnnrrrr 33.0351x yz 34.02331p qr 35.(a)B;061(b)C;061 (c)B;061(d)C;061  36.(a)D;033 13p(b)E;033 13p  (c)B;031p(d)B;031p37.115xis a polynomial of degree 11. It is amonomial because it has one term.38.54yis a polynomial of degree 5 and is amonomial.39.463xxis a polynomial. It has degree 4because the term43xhas the greatest degree.It is a binomial because there are two terms.

Solution Manual for College Algebra, 12th Edition - Page 16 preview image

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14Chapter R Review of Basic Concepts40.395yyis a polynomial. It has degree 3because the term35yhas the greatest degree.It is a binomial because there are two terms.41.53721zzis a polynomial. It has degree 5because the term57zhas the greatestdegree. It is a trinomial because there are threeterms.42.43987ttis a polynomial. It has degree 4because the term49thas the greatest degree.It is a trinomial because there are three terms.43.23385615121312a ba bbbis apolynomial. It has degree 11 because the term3812a bhas the greatest degree. It is not amonomial, binomial, or trinomial.44.57389101612418xyx yxyxis apolynomial. It has degree 12 because the term5716xyhas the greatest degree. It is not amonomial, binomial, or trinomial.45.523198xxis not a polynomial because oneterm has a variable in the denominator.46.652313ttis not a polynomial because oneterm has a variable in the denominator.47.5 is a polynomial of degree 0. It is amonomial.48.9 is a polynomial of degree 0. It is amonomial.49.222225474355443751122xxxxxxxxxx     50.3232323232334263231461410410mmmmmmmmmm    51. 22222222 12864 3422 122 82 64 3444 224161212168124yyyyyyyyyyyyy52.22222223 855 3243 83 55 3525 424151510209520pppppppppppppp53.423224232243243243263254632546235114162746274mmmmmmmmmmmmmmmmmmmmmmmmmmmm   54.3322332232323283243183243182141331634316344xxxxxxxxxxxxxxxxxxxxx       55.22(41)(72)4 (7 )4 (2)1(7 )1(2)28872282rrrrrrrrrrr56.  22563453546 36 41520182415224mmmmmmmmmmm57.222213533213533122135353333xxxxxxxxxxx      2222224321021539310215339727215153939xxxxxxxxxxxxxx58.333232543112342111123232442316481111664848mmmmmmmmmmmmmmmmmm   

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