最新版商务与经济学完整版课后习题答案

GettingStarted1(a)ThisisshowninFigureSI.1.(e)JustputthecursorovercellC5andtypeinthenewmarkof42.PressingtheEnterkeycausescellsD5andD10tobeautomaticallyupdated.ThenewspreadsheetisshowninFigureSI.3.SolutionstoProblemsFigureSI.1FigureSI.2FigureSI.3(b)TypetheheadingTotalMarkincellD3.Type=B4+C4intocellD4.ClickanddragdowntoD9.(c)TypetheheadingAverage:incellC10.Type=(SUM(D4:D9))/6incellD10andpressEnter.[Note:Excelhaslotsofbuilt-infunctionsforperformingstandardcalculationssuchasthis.Tofindtheaverageyoucouldjusttype=AVERAGE(D4:D9)incellD10.](d)ThisisshowninFigureSI.2.MFE_Z02.qxd16/12/200510:51Page5982(a)14(b)11(c)53(a)4;isthesolutionoftheequation2x−8=0.(b)FigureSI.4showsthegraphof2x−8plottedbetweenx=0and10.2(1)(a)−30;(b)2;(c)−5;(d)5;(e)36;(f)−1.(2)Thekeysequences(workingfromlefttoright)are(a)5×±6=(b)±1×±2=(c)±50÷10=(d)±5÷±1=(e)2×±1×±3×6=(f)sameas(e)followedby÷±2÷3÷6=3(1)(a)−1;(b)−7;(c)5;(d)0;(e)−91;(f)−5.(2)Thekeysequencesare(a)1−2=(b)±3−4=(c)1−±4=(d)±1−±1=(e)±72−19=(f)±53−±48=4SolutionstoProblems599FigureSI.4FigureSI.5Chapter1Section1.11FromFigureS1.1notethatallfivepointslieonastraightline.(c)x2+4x+4;thebracketshavebeen‘multipliedout’intheexpression(x+2)2.(d)7x+4;liketermsintheexpression2x+6+5x−2havebeencollectedtogether.(e)FigureSI.5showsthethree-dimensionalgraphofthesurfacex3−3x+xyplottedbetween−2and2inboththexandydirections.FigureS1.1PointCheck(−1,2)2(−1)+3(2)=−2+6=4✓(−4,4)2(−4)+3(4)=−8+12=4✓(5,−2)2(5)+3(−2)=10−6=4✓(2,0)2(2)+3(0)=4+0=4✓ThegraphissketchedinFigureS1.2(overleaf).MFE_Z02.qxd16/12/200514:38Page599Thegraphshowsthat(3,−1)doesnotlieontheline.Thiscanbeverifiedalgebraically:2(3)+3(−1)=6−3=3≠453x−2y=43(2)−2y=4(substitutex=−2)6−2y=4−2y=−2(subtract6frombothsides)y=1(dividebothsidesby−2)Hence(2,1)liesontheline.3x−2y=43(−2)−2y=4−6−2y=4(substitutex=2)−2y=10(add6tobothsides)y=−5(dividebothsidesby−2)Hence(−2,−5)liesontheline.ThelineissketchedinFigureS1.3.6x−2y=20−2y=2(substitutex=0)−2y=2y=−1(dividebothsidesby−2)Hence(0,−1)liesontheline.x−2y=2x−2(0)=2(substitutey=0)x−0=2x=2Hence(2,0)liesontheline.ThegraphissketchedinFigureS1.4.SolutionstoProblems6007FromFigureS1.5thepointofintersectionis(1,−1/2).FigureS1.4FigureS1.5FigureS1.3FigureS1.2MFE_Z02.qxd16/12/200510:51Page600(b)4x+2y=12y=1−4x(subtract4xfrombothsides)y=1/2−2x(dividebothsidesby2)soa=−2,b=1/2.ThegraphissketchedinFigureS1.7.12(a)0.5;(b)4;(c)−4;(d)−0.5;(e)0;(f)9.13xy086034ThegraphissketchedinFigureS1.9.SolutionstoProblems601FigureS1.6FigureS1.7FigureS1.8FigureS1.98(a)a=1,b=2.ThegraphissketchedinFigureS1.6.9FromFigureS1.8thepointofintersectionis(2,3).10(a)−20;(b)3;(c)−4;(d)1;(e)−1;(f)−3;(g)11;(h)0;(i)18.11(a)1;(b)5;(c)−6;(d)−6;(e)−30;(f)44.14(a)(−2,−2);(b)(2,11/2);(c)(11/2,1);(d)(10,−9).MFE_Z02.qxd16/12/200510:51Page601(b)ThegraphissketchedinFigureS1.11.3x−2y=4x−2y=2−2x=2Step2Theequation2x=2hassolutionx=2/2=1.Step3Ifthisissubstitutedintothefirstequationthen3(1)−2y=43−2y=4−2y=1(subtract3frombothsides)y=−1/2(dividebothsidesby−2)Step4Asacheckthesecondequationgivesx−2y=1−2(−1/2)=1−(−1)=2✓Hencethesolutionisx=1,y=−1/2.Ifyoudecidetoeliminatexthenthecorrespondingstepsareasfollows:Step1Triplethesecondequationandsubtractfromthefirst:3x−2y=43x−6y=6−4y=−2Step2Theequation4y=−2hassolutiony=−2/4=−1/2.Step3Ifthisissubstitutedintothefirstequationthen3x−2(−1/2)=43x+1=43x=3(subtract1frombothsides)x=1(dividebothsidesby3)(b)Step1Itisimmaterialwhichvariableiseliminated.Toeliminatexmultiplythefirstequationby5,multiplythesecondby3andadd:SolutionstoProblems60216(a)16.(b)Presentedwiththecalculation,−42,yourcalculatorusesBIDMAS,sosquaresfirsttoget16andthensubtractsfromzerotogiveafinalanswer,−16.Toobtainthecorrectansweryouneedtousebrackets:(±4)x2=Section1.21(a)Step1Itisprobablyeasiesttoeliminatey.Thiscanbedonebysubtractingthesecondequationfromthefirst:15(a)ThegraphissketchedinFigureS1.10.FigureS1.10FigureS1.11MFE_Z02.qxd16/12/200510:51Page60215x+25y=95−15x+6y=−33+31y=62Step2Theequation31y=62hassolutiony=62/31=2.Step3Ifthisissubstitutedintothefirstequationthen3x+5(2)=193x+10=193x=9(subtract10frombothsides)x=3(dividebothsidesby3)Step4Asacheckthesecondequationgives−5x+2y=−5(3)+2(2)=−15+4=−11✓Hencethesolutionisx=3,y=2.2(a)Step1Toeliminatexmultiplythefirstequationby4,multiplythesecondequationby3andadd:12x−24y=−8−12x+24y=−3+0y=−11Step2Thisisimpossible,sotherearenosolutions.(b)Step1Toeliminatexmultiplythefirstequationby2andaddtothesecond:−10x+2y=810x−2y=−8+0y=0Step2Thisistrueforanyvalueofy,sothereareinfinitelymanysolutions.3Step1Toeliminatexfromthesecondequationmultiplyequation(2)by2andsubtractfromequation(1):2x+2y−5z=−52x−2y+2z=6−4y−7z=−11(4)Toeliminatexfromthethirdequationmultiplyequation(1)by3,multiplyequation(3)by2andadd:6x+6y−15z=−15−6x+2y+4z=−4+8y−11z=−19(5)Thenewsystemis2x+2y−5z=−5(1)4y−7z=−11(4)8y−11z=−19(5)Step2Toeliminateyfromthethirdequationmultiplyequation(4)by2andsubtractequation(5):8y−14z=−228y−11z=−19−3z=−3(6)Thenewsystemis2x+2y−5z=−5(1)4y−7z=−11(4)−3z=−3(6)Step3Equation(6)givesz=−3/−3=1.Ifthisissubstitutedintoequation(4)then4y−7(1)=−114y−7=−114y=−4(add7tobothsides)y=−1(dividebothsidesby4)Finall