Solution Manual for Algebra and Trigonometry, 6th Edition

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INSTRUCTOR’SSOLUTIONSMANUALDANIELS.MILLERNiagara County Community CollegeALGEBRAANDTRIGONOMETRYSIXTHEDITIONRobert BlitzerMiami Dade College

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iiiTABLE OF CONTENTSALGEBRA ANDTRIGONOMETRY6EChapter PFundamental Concepts of Algebra............................................................................. 1Chapter 1Equations and Inequalities ....................................................................................... 71Chapter 2Functions and Graphs............................................................................................. 201Chapter 3Polynomial and Rational Functions ....................................................................... 325Chapter 4Exponential and Logarithmic Functions ................................................................ 467Chapter 5Trigonometric Functions........................................................................................ 543Chapter 6Analytic Trigonometry........................................................................................... 715Chapter 7Additional Topics in Trigonometry ....................................................................... 829Chapter 8Systems of Equations and Inequalities................................................................... 977Chapter 9Matrices and Determinants .................................................................................. 1097Chapter 10Conic Sections and Analytic Geometry............................................................... 1201Chapter 11Sequences, Induction, and Probability................................................................. 1321

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Chapter PFundamental Concepts of Algebra1Section P.1Check Point Exercises1.2286(3)86(133)x+−=+−286(10)86(100)8600608=+=+=+=2.a.Since 2014 is 14 years after 2000, substitute 14forx.22433033104(14)330(14)33108714Txx=++=++=The average cost of tuition and fees at publicU.S. colleges for the school year ending in2014 was $8714.b.The formula underestimates the actual answerby $179.3.The elements common to {3, 4, 5, 6, 7} and{3, 7, 8, 9} are 3 and 7.{3, 4,5, 6, 7}{3,7,8,9}{3, 7}∩=4.The union is the set containing all the elements ofeither set.{3, 4,5, 6, 7}{3, 7,8,9}{3, 4,5, 6, 7,8,9}∪=5.9,1.3, 0, 0.3,,9,102π−−a.Natural numbers:9because93=b.Whole numbers: 0,9c.Integers:9, 0,9−d.Rational numbers:9,1.3, 0, 0.3,9−−e.Irrational numbers:,102πf.Real numbers:9,1.3, 0, 0.3,,9,102π−−6.a.12−Because21.4,≈the number inside theabsolute value bars is negative. The absolutevalue ofxwhenx< 0 is –x.Thus,()121221−= −−=−b.3π−Because3.14,π ≈the number inside theabsolute value bars is positive. The absolutevalue of a positive number is the number itself.Thus,33.ππ−=−c.xxBecause0,.xxx>=Thus,1xxxx==7.4(5)99−−= −=The distance between –4 and 5 is 9.8.227(43 )2(5)xxxx+++222227(43 )2(5)28211023823xxxxxxxxxx=+++=+++=+9.64[7(2)]x+−−64[72)]64[9]6364424xxxx=+−+=+−=+−=−Concept and Vocabulary Check P.11.expression2.bto thenth power; base; exponent3.formula; modeling; models4.intersection;AB∩5.union;AB∪

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Chapter PPrerequisites: Fundamental Concepts of Algebra26.natural7.whole8.integers9.rational10.irrational11.rational; irrational12.absolute value;x,x13.ba+;ba14.()abc++;()ab c15.abac+16.0; inverse; 0; identity17.inverse; 1; identity18.simplified19.aExercise Set P.11.75(10)75057+=+=2.()86 583038+=+=3.6(3)818810−=−=4.( )8 3424420−=−=5.283(8)642488+=+=6.()265 6363066+=+=7.276(7)3494237310−+=−+=+=8.( )287 84645648412−+=−+=+=9.3345(97)45(2)45(8)44044+−=+=+=+=10.()()( )3365 8665 265 864046+−=+=+=+=11.283(82)643(6)641846−−=−=−=12.()( )284 83644 5642044−−=−=−=13.5(2)5(102)2142(10)145(12)65 210xx++=−−==⋅=14.7(3)7(93)7(6)7 3212162(9)162xx−−===⋅=−−15.23;2,41xyxyx+= −=+()()223 4412882111−+−+==== −−+−−16.2;2 and42xyxyxyx+= −=−()()()()22444002422844−+−+===−−−−+17.5 (5032)9C=−5 (18)109==50°F is equivalent to 10°C.18.555(32)(8632)(54)30999CF=−=−==86°F is equivalent to 30°C.19.2246016460(2)16(2)412016(4)4120641246460htt=+−=+−=+−=+−=−=Two seconds after it is kicked, the ball’s height is60 feet.

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Section P.1Algebraic Expressions, Mathematical Models, and Real Numbers320.2246016460(3)16(3)418016(9)418014418414440htt=+−=+−=+−=+−=−=Three seconds after it is kicked, the ball’s height is40 feet.21.{}{}{}1, 2,3, 42, 4,52, 4=22.{}{}{}1,3, 72,3,83=23.{}{}{}, ,, ,, ,s e tt e ss e t=24.{}{}{}, ,,, ,,, ,,r e a ll e a rr e a l=25.{}{}{}1,3,5, 72, 4, 6,8,10=The empty set is also denoted by.∅26.{}{}{}1,3,5, 75,3,1−−−=or∅27.{},, ,a b c d∅ = ∅28.{},,w y z∅ = ∅29.{}{}{}1, 2,3, 42, 4,51, 2,3, 4,5=30.{}{}{}1,3, 7,82,3,81, 2,3, 7,8=31.{}{}{}1,3,5, 72, 4, 6,8,101, 2,3, 4,5, 6, 7,8,10=32.{}{}{}0,1,3,52, 4, 60,1, 2,3, 4,5, 6=33.{}{}, , ,,, , ,,a e i o ua e i o u∅ =34.{}{},,, ,,,, ,e m p t ye m p t y∅ =35.a.100b.0,100c.9, 0,100−d.49,, 0, 0.25, 9.2,1005−−e.3f.49,, 0, 0.25,3,9.2,1005−−36.a.49b.0,49c.7, 0,49−d.7,0.6, 0,49−−e.50f.7,0.6, 0,49,50−−37.a.64b.0,64c.11, 0,64−d.511,, 0, 0.75,646−−e.5,πf.511,, 0, 0.75,5,,646π−−38.a.4b.0,4c.5, 0,4−d.5,0.3, 0,4−−e.2f.5,0.3, 0,2,4−−39.040.Answers will vary. An example is1 .241.Answers will vary. An example is2.42.Answers will vary. An example is2.−43.true; –13 is to the left of –2 on the number line.44.false; –6 is to the left of 2 on the number line.45.true; 4 is to the right of –7 on the number line.

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Chapter PPrerequisites: Fundamental Concepts of Algebra446.true; –13 is to the left of –5 on the number line.47.true;−π = −π48.true; –3 is to the right of –13 on the number line.49.true; 0 is to the right of –6 on the number line.50.true; 0 is to the right of –13 on the number line.51.300300=52.203203−=53.1212− π =− π54.77− π =− π55.2552−=−56.513135−=−57.33133−−== −−58.77177−−== −−59.373744−− −=−= −=60.51351388−− −=−= −=61.2( 5)33xy+=+ −= −=62.2( 5)77xy−=− −==63.25257xy+=+ −=+=64.25253xy−=− −=−= −65.55155yy−−=== −−66.25251( 1)02525xyxy−+=+=+=+ −=−−67.The distance is2171515−= −=.68.The distance is4151111−= −=.69.The distance is2577−−= −=.70.The distance is681414−−= −=.71.The distance is19( 4)1941515−− −= −+= −=.72.The distance is26( 3)2632323−− −= −+= −=.73.The distance is3.6( 1.4)3.61.42.22.2−− −= −+= −=.74.The distance is5.4( 1.2)5.41.24.24.2−− −= −+= −=.75.6 + (–4) = (–4) + 6;commutative property of addition76.11 (74)11 711 4;⋅+=⋅+⋅distributive property of multiplication over addition77.6 + (2 + 7) = (6 + 2) + 7;associative property of addition78.6 (2 3)6 (3 2);⋅⋅=⋅⋅commutative property of multiplication79.(2 + 3) + (4 + 5) = (4 + 5) + (2 + 3);commutative property of addition80.7 (11 8)(11 8) 7;⋅⋅=⋅⋅commutative property of multiplication81.2 (–8 + 6) = –16 + 12;distributive property of multiplication over addition82.8(311)24( 88)−+= −+ −;distributive property of multiplication over addition83.()131;33xxx+=≠ −+,inverse property of multiplication84.()()440;xx++−+=inverse property of addition85.5(34)45 35 44152041516xxxx+−=⋅+⋅−=+−=+

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Section P.1Algebraic Expressions, Mathematical Models, and Real Numbers586.2(54)32 52 431083105xxxx+−=⋅+⋅−=+−=+87.5(32)125 35 2121510122710xxxxxxx−+=⋅−⋅+=−+=−88.2(51)142 52 11410214242xxxxxxx−+=⋅−⋅+=−+=−89.7(35)2(43)7 37 52 42 32135862929yyyyyyy−++=⋅−⋅+⋅+⋅=−++=−90.4(26)3(510)4 24 63 53 108241530236yyyyyyy−++=⋅−⋅+⋅+⋅=−++=+91.()()5 3272151072812yyyyy−−+=−−−=−92.4(53)(63)2012631415yyyyy−−+=−−−=−93.()[]74 34574 345yy−−−=−−+[]74 84732161625yyy=−−=−+=−94.[][]65[8(24)]65 82465 122660101054yyyyy−−−=−−+=−−=−+=−95.()()2222222222221846251846125184671846718647186111211xxxxxxxxxxxx+−−+=+−−+=+−−=+−+=−++=−+=+96.()()222222222222145724145714414571014571014751014715715xxxxxxxxxxxx+−−+=+−−+=+−−=+−+=−++=−+=+97.–(–14x) = 14x98.()1717yy− −=99.–(2x– 3y– 6) = –2x+ 3y+ 6100.()51315131xyxy−−−= −++101.[]1 (3 )(4 )( 4 )03xyyxx++ −=+=102.()()127702yxxyy+−+=+=103.636363−−>Since63,63 .>−> −104.205020502050−−<Since2050,2050 .<−< −105.30.650.60.60.60.60.60.6−−=Since30.60.6,0.6 .5== −

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Chapter PPrerequisites: Fundamental Concepts of Algebra6106.52.522.52.5.52.52.52.52−−=Since52.52.5,2.5 .2== −107.30314 1540415 143030144040−⋅−1515⋅140101<Since30314 1501,40415 14<−<⋅.108.171850518 1760617⋅−1818⋅17505060601010−>Since171850510,>18 17606>⋅−.109.881131381311381111÷−⋅=Since8811,=1 .1313=÷−110.442171741721742121−÷⋅>Since4421,2 >.1717>−÷111.2281624364164 43644 436416348345−÷⋅−=−÷⋅−=−⋅−=−−=−=112.221010052310010025 231004 231008392389−÷⋅−=−÷⋅−=−⋅−=−−=−=113.2222225 235 29[3( 2)][9( 2)]109[92]109111121⋅−⋅−=− −− −−=+−==114.2221023 4512(123 2)(126)1761736÷+⋅+=−⋅−==115.83[ 2(25)4(86)]83[ 2( 3)4(2)]83[68]83[ 2]8614−−−−−=−−−−=−−=−−=+=116.83[ 2(57)5(42)]83[ 2( 2)5(2)]83[410]83[ 6]81826−−−−−=−−−−=−−=−−=+=117.2( 2)4( 3)4125838383−−−−+=−−= −= −

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Section P.1Algebraic Expressions, Mathematical Models, and Real Numbers7118.6( 4)5( 3)24159101919−−−−+=−−−= −=119.222(56)2 37( 1)24893 25893 512(4)8975181471412−−−−−−=−⋅−⋅−=−−=−== −120.222123 5 23123 5 4973367364 5 13103620(13)262602610÷⋅+÷⋅+=+−+−⋅=−=−= −= −121.()444xxxx−+=−−= −122.()8828xxxxx−−=−+=−123.()6530xx−= −124.()10440xx−= −125.523xxx−=126.()62628xxxxx− −=+=127.()83683656xxxxx−+=−−=−128.()8368318310xxx−+=−−= −−129.a.()()()72201072202010720010140HaH=−=−==The lower limit of the heart rate for a 20-year-old with this exercise goal is 140 beats perminute.b.()()()42205422020542005160HaH=−=−==The upper limit of the heart rate for a 20-year-old with this exercise goal is 160 beats perminute.130.a.()()()1220212203021 190295HaH=−=−==The lower limit of the heart rate for a 30-year-old with this exercise goal is 95 beats perminute.b.()()()3 22053 2203053 1905114HaH=−=−==The upper limit of the heart rate for a 30-year-old with this exercise goal is 114 beats perminute.131.a.222186215,55221(14)862(14)15,55231, 736Txx=++=++=The formula estimates the cost to have been$31,736 in 2014.

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Chapter PPrerequisites: Fundamental Concepts of Algebra8b.This overestimates the value in the graph by$35.c.222186215,55221(20)862(20)15,55241,192Txx=++=++=The formula projects the cost to be $41,192 in2020.132.a.222186215,55221(12)862(12)15,55228,920Txx=++=++=The formula estimates the cost to have been$28,920 in 2012.b.This underestimates the value in the graph by$136.c.222186215,55221(22)862(22)15,55244, 680Txx=++=++=The formula projects the cost to be $44,680 in2022.133.a.()0.050.12 10,0000.0512000.1212000.07xxxxx+−=+−=−b.12000.0712000.07(6000)$780x−=−=134.a.0.060.5(50)0.06250.5250.44ttttt+−=+−=−b.0.06(20)0.5(5020)1.20.5(30)1.21516.2 miles+−=+=+=135. – 143.Answers will vary.144.does not make sense; Explanations will vary.Sample explanation: Models do not alwaysaccurately predict future values.145.does not make sense; Explanations will vary.Sample explanation: To use the model, substitute 0forx.146.makes sense147.does not make sense; Explanations will vary.Sample explanation: The commutative propertychanges order and the associative property changesgroupings.148.false; Changes to make the statement true will vary.A sample change is: Some rational numbers are notintegers.149.false; Changes to make the statement true will vary.A sample change is: All whole numbers are integers.150.true151.false; Changes to make the statement true will vary.A sample change is: Some irrational numbers arenegative.152.false; Changes to make the statement true will vary.A sample change is: The termxhas a coefficientof 1.153.false; Changes to make the statement true will vary.A sample change is:53(4)531237.xxx+−=+−=−154.false; Changes to make the statement true will vary.A sample change is:2 .xxx−−= −155.true156.21.41.41.521.5≈<<157.3.5π−> −158.3.141.5721.57121.571.5713.1422−= −π−≈ −−> −π−> −159.a.437()()bbb b b bb b bb⋅=⋅⋅⋅⋅⋅=b.5510()()bbb b b b bb b b b bb⋅=⋅⋅⋅⋅⋅⋅⋅⋅=c.add the exponents

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Section P.2Exponents and Scientific Notation9160.a.743bb b b b b b bbb b bb⋅⋅⋅⋅⋅⋅==⋅⋅b.862bb b b b b b b bbb bb⋅⋅⋅⋅⋅⋅⋅==⋅c.subtract the exponents161.36.2106.21010106200×=×××=It moves the decimal point 3 places to the right.Section P.2Check Point Exercises1.a.323253333or 243+⋅==b.()()342632464104 10x yx yxxyy=⋅⋅⋅⋅⋅32465104040xyx y++=⋅=2.a.633( 3)( 3)27( 3)−= −= −−b.14814814385113353527279933xyxyxyxyx yxy−−=⋅⋅==3.a.22115255−==b.33111( 3)2727( 3)−−=== −−−c.222211414161144−==⋅==d.4644661333yxyyxx−=⋅⋅=4.a.()233 26333or 729⋅==b.()277(2)14141yyyy−−−===c.()433( 4)12bbb−−−−==5.3333( 4 )( 4) ( )64xxx−= −= −6.a.()55552232yyy−−−==b.()33551533273xxx==7.a.()()() ()4444363612242216x yxyxy==b.()()()25325338636318x yxyxxyyx y−=−⋅⋅⋅⋅⋅= −c.1221221641641001002020xyxyxyxy−− =   12 162(4)4664555xyxyyx−− −−===d.( )()()22242455xxyy−−−−=( )()()22242288228255525xyxyyxyx−−−−−−====8.a.92.6102, 600, 000, 000−×= −b.63.017100.000003017−×=9.a.95, 210, 000, 0005.2110=×b.80.000000068936.89310−−= −×

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Chapter PPrerequisites: Fundamental Concepts of Algebra1010.()727410104.11010×=××()2794.110104.110=××=×11a.()()577.110510−××()()572121217.1 5101035.5103.5510103.5510103.5510−−−−−=⋅×⋅=×=××=××=×b.66331.2101.210331010−−×=⋅×6(3)980.4100.410410− −=×=×=×12.1010105554.08104.08104.08 10680, 0006.86.810100.61060, 000××==⋅×=×=The average salary was $60,000 per U.S. policeofficer.Concept and Vocabulary Check P.21.mnb+; add2.mnb−; subtract3.14.1nb5.false6.nb7.true8.a number greater than or equal to 1 and less than 10;integer9.true10.falseExercise Set P.21.252(5 5) 225 250⋅=⋅⋅=⋅=2.262(6 6) 236 272⋅=⋅⋅=⋅=3.6( 2)( 2)( 2)( 2)( 2)( 2)( 2)64−= −−−−−−=4.4( 2)( 2)( 2)( 2)( 2)16−=−−−−=5.622 2 2 2 2 264−= −⋅⋅⋅⋅⋅= −6.422 2 2 216−= −⋅⋅⋅= −7.0( 3)1−=8.0( 9)1−=9.031−= −10.091−= −11.3311144 4 4644−===⋅⋅12.6611122 2 2 2 2 2642−===⋅⋅⋅⋅⋅13.2323522222 2 2 2 232+⋅===⋅⋅⋅⋅=14.3232533333 3 3 3 3243+⋅===⋅⋅⋅⋅=15.232 36(2 )222 2 2 2 2 264⋅===⋅⋅⋅⋅⋅=16.323 26(3 )333 3 3 3 3 3729⋅===⋅⋅⋅⋅⋅=17.884442222 2 2 2162−===⋅⋅⋅=18.884443333 3 3 3813−===⋅⋅⋅=19.33 12211133333 393−− +−⋅=====⋅20.33 12211122222 242−− +−⋅=====⋅21.3374742111222 2 2 21622−−=====⋅⋅⋅

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Section P.2Exponents and Scientific Notation1122.4473733111333 3 32733−−=====⋅⋅23.2221yxyyxx−=⋅=24.3331xxyxyy−=⋅=25.05551x yyy=⋅=26.70771xyxx⋅=⋅=27.373710xxxx+⋅==28.11511 516xxxx+⋅==29.5105 105xxxx−− +⋅==30.6126 126xxxx−− +⋅==31.373 721()xxx⋅==32.11 511 555()xxx⋅==33.535 315151()xxxx−− ⋅−===34.646 424241()xxxx−− ⋅−===35.1414777xxxx−==36.3030 102010xxxx−==37.1414(7)147217xxxxx− −+−===38.3030(10)30 104010xxxxx− −+−===39.3223223 26(8)8 ()864xxxx⋅===40.4224224 28(6)(6) ()636xxxx⋅===41.33334( 4)64xxx−−== −42.33336( 6)216yyy−−== −43.252222522 25 2410( 3)( 3) ()()99x yxyxyx y⋅⋅−= −⋅==44.463343634 36 31218( 3)( 3) () ()2727x yxyxyxy⋅⋅−= −= −= −45.47474711(3)(2)3 266xxxxxx+=⋅⋅==46.5125125 1217(11)(9)11 99999xxx xxx+=⋅==47.364364361 495( 9)( 2)( 9)( 2)1818x yx yx x yyxyx y++−−= −−==48.47114711471 111112( 5)( 6)( 5)( 6)3030x yx yx x yyxyxy++−−= −−==49.20202041644884422xxxxxx−===  50.242424618662020221010xxxxxx−===51.1341342323132431125255555abababababab−−⋅=−−⋅= −= −52.1461467373147637335357755ababa bababa b−−=−−= −= −53.777 14714147141422277bbbbbbb−−==⋅==

Solution Manual for Algebra and Trigonometry, 6th Edition - Page 15 preview image

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Chapter PPrerequisites: Fundamental Concepts of Algebra1254.101020201020101020201010222bbbbbbb−−====55.3223226266(4)(4)()414116xxxxx−−−−−====56.2332 (3)36366(10)101011011000xxxxx−−⋅ −−−====57.353 75(9)794141442434323434xyxyx yxyyx−− −−−⋅===58.494 129(3)123812128101330133x yxyxyxyyx−− −−−===59.23262655225xxyyxy−−−==−60.3344334 331233327xyyxyxyx−⋅===61.()33234210310435615181535327a bbabababa− −−−−−=−=−=62.()338214817217 14310309303103327abbabababa− −−−−−=−=−=63.052343112aba b−−=64.053354112aba b−−=65.23.810380×=66.29.210920×=67.46100.0006−×=68.57100.00007−×=69.67.16107,160, 000−×= −70.68.17108,170, 000−×= −71.17.9100.79−×=72.16.8100.68−×=

Solution Manual for Algebra and Trigonometry, 6th Edition - Page 16 preview image

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Section P.2Exponents and Scientific Notation1373.34.15100.00415−−×= −74.33.14100.00314−−×= −75.106.000011060, 000,100,000−×= −76.107.000011070, 000,100,000−×= −77.432, 0003.210=×78.464, 0006.410=×79.17638, 000, 000, 000, 000, 0006.3810=×80.17579, 000, 000, 000, 000, 0005.7910=×81.357165.71610−= −×82.338293.82910−= −×83.30.00272.710−=×84.30.00838.310−=×85.90.000000005045.0410−−= −×86.90.000000004054.0510−−= −×87.()()()()43434373102.11032.110106.3106.310+××=××=×=×88.()()4372104.1108.210××=×89.()()()()1511151115( 11)41.6104101.6410106.4106.410−−+ −××=××=×=×90.()()()()1511151115(11)41.4103101.4310104.2104.210−−+ −××=××=×=×91.() ()()()84848(4)12116.1102106.12101012.21012.2101.2210−−−−− + −−−××=××=×=×=×92.()()8412115.11031015.3101.5310−−−−××=×=×93.()()()()8484841213134.3106.2104.36.2101026.661026.66102.666102.6710+××=××=×=×=×≈×94.()()84841213138.2104.61037.721037.72103.772103.7710+××=×=×=×≈×95.88558 538.4108.4104410102.1102.110−×=××=×=×96.88 5356.9102.3102.310310−×=×=××97.44224(2)653.6103.6109910100.4100.410410−−− −×=××=×=×=×98.()44(2)621651.2100.6100.61021061010610− −−−×=×=××=××=×99.22662684.8104.8102.42.41010210210−−− −−×=××=×=×100.226867.5103103102.510−− −−×=×=××101.22662( 6)432.4102.4104.84.810100.5100.510510−−−−− − −×=××=×=×=×

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