Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual

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1Chapter 1GraphsSection 1.11.02.()5388− −==3.2234255+==4.22211601213600372161+=+==Since the sum of the squares of two of the sidesof the triangle equals the square of the third side,the triangle is a right triangle.5.x-coordinate, or abscissa; y-coordinate, orordinate.6.quadrants7.midpoint8.False; the distance between two points is nevernegative.9.False; points that lie in Quadrant IV will have apositive x-coordinate and a negative y-coordinate. The point()1, 4−lies in Quadrant II.10.True;1212,22xxyyM++⎛⎞= ⎜⎟⎝⎠11.(a)Quadrant II(b)Positive x-axis(c)Quadrant III(d)Quadrant I(e)Negative y-axis(f)Quadrant IV12.(a)Quadrant I(b)Quadrant III(c)Quadrant II(d)Quadrant I(e)Positive y-axis(f)Negative x-axis13.The points will be on a vertical line that is twounits to the right of the y-axis.14.The points will be on a horizontal line that isthree units above the x-axis.

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Chapter 1:GraphsISM:Precalculus EGU215.()1, 4−; Quadrant II16.(3, 4); Quadrant I17.(3, 1); Quadrant I18.()6,4−−; Quadrant III19.min11max5scl1min3max6scl1XXXYYY= −=== −==20.min3max7scl1min4max9scl1XXXYYY= −=== −==21.min30max50scl10min90max50scl10XXXYYY= −=== −==22.min90max30scl10min50max70scl10XXXYYY= −=== −==23.min10max110scl10min10max160scl10XXXYYY= −=== −==24.min20max110scl10min10max60scl10XXXYYY=−=== −==25.min6max6scl2min4max4scl2XXXYYY=−=== −==26.min3max3scl1min2max2scl1XXXYYY=−=== −==27.min6max6scl2min1max3scl1XXXYYY=−=== −==28.min9max9scl3min12max4scl4XXXYYY=−=== −==29.min3max9scl1min2max10scl2XXXYYY======

Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual - Page 4 preview image

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ISM:Precalculus EGUChapter 1:Graphs330.min22max10scl2min4max8scl1XXXYYY= −= −====31.2212(,)(20)(10)415d P P=−+−=+=32.2212(,)(20)(10)415d P P=−−+−=+=33.2212(,)(21)(21)9110d P P=−−+−=+=34.()2212(,)2( 1)(21)9110d P P=− −+−=+=35.()()( )222212(,)(53)4428464682 17d P P=−+− −=+=+==36.()()()( )222212(,)214034916255d P P=− −+−=+=+==37.()221222(,)6( 3)(02)9(2)81485d P P=− −+−=+ −=+=38.()()222212(,)422( 3)2542529d P P=−+− −=+=+=39.()222212(,)(64)4( 3)2744953d P P=−+− −=+=+=40.()()221222(,)6(4)2( 3)1051002512555d P P=− −+− −=+=+==41.()221222(,)2.3(0.2(1.10.3)(2.5)(0.8)6.250.646.892.625d P P=− −+−=+=+=≈42.()221222(,)0.31.2(1.12.3)( 1.5)( 1.2)2.251.443.691.92d P P=−−+−=−+ −=+=≈43.222212(,)(0)(0)d P Pabab=−+−=+44.2222122(,)(0)(0)22d P Paaaaaa=−+−=+==45.()()121,3 ;5,15PP==()()()()()221222,51153412161441604 10dP P=−+−=+=+==46.()()128,4 ;2,3PP=−−=()()()()()()()221222,283410710049149dP P=− −+− −=+=+=47.()()124, 6 ;4,8PP=−=−()()()()( )()221222,448681464196260265dP P=− −+ −−=+ −=+==48.()()120, 6 ;3,8PP==−()()()( )()221222,30863149196205dP P=−+ −−=+ −=+=

Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual - Page 5 preview image

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Chapter 1:GraphsISM:Precalculus EGU449.(2,5),(1,3),( 1, 0)ABC= −== −()()()222222222222(,)1(2)(35)3(2)9413(,)11(03)(2)(3)4913(,)1(2)(05)1(5)12526d A Bd B Cd A C=− −+−=+ −=+==− −+−=−+ −=+==− − −+−=+ −=+=Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)1313261313262626d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=⋅. In thisproblem,[] []1(,)(,)21131321 13213 square units2Ad A Bd B C=⋅⋅=⋅⋅=⋅=50.(2, 5),(12, 3),(10,11)ABC=−==−()()()222222222222(,)12(2)(35)14(2)1964200102(,)1012( 113)(2)( 14)4196200102(,)10(2)( 115)12(16)14425640020d A Bd B Cd A C=− −+−=+ −=+===−+ −−=−+ −=+===− −+ −−=+ −=+==Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)10210220200200400400400d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)21 102 10221 100 22100 square unitsAd A Bd B C=⋅⋅=⋅⋅=⋅⋅=

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ISM:Precalculus EGUChapter 1:Graphs551.(5,3),(6, 0),(5,5)ABC= −==()()()222222222222(,)6(5)(03)11(3)1219130(,)56(50)(1)512526(,)5(5)(53)1021004104226d A Bd B Cd A C=− −+−=+ −=+==−+−=−+=+==− −+−=+=+==Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)1042613010426130130130d A Cd B Cd A B+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)211042621 2262621 2 26226 square unitsAd A Cd B C=⋅⋅=⋅⋅=⋅⋅=⋅⋅=52.(6, 3),(3,5),( 1, 5)ABC=−=−= −()()()222222222222(,)3(6)( 53)9(8)8164145(,)13(5( 5))(4)1016100116(,)1(6)(53)5225429d A Bd B Cd A C=− −+ −−=+ −=+==− −+− −=−+=+==− − −+−=+=+=Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()()()222222(,)(,)(,)2911614529116145145145d A Cd B Cd A B+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)21291162129 22921 2 29229 square unitsAd A Cd B C=⋅⋅=⋅⋅=⋅⋅=⋅⋅=

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Chapter 1:GraphsISM:Precalculus EGU653.(4,3),(0,3),(4, 2)ABC=−=−=()()()()222222222222(,)(04)3( 3)(4)0160164(,)402( 3)45162541(,)(44)2( 3)05025255d A Bd B Cd A C=−+ −− −=−+=+===−+− −=+=+==−+− −=+=+==Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()222222(,)(,)(,)45411625414141d A Bd A Cd B C+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)21 4 5210 square unitsAd A Bd A C=⋅⋅=⋅⋅=54.(4,3),(4, 1),(2, 1)ABC=−==()()()()222222222222(,)(44)1( 3)04016164(,)2411(2)04042(,)(24)1( 3)(2)44162025d A Bd B Cd A C=−+− −=+=+===−+−=−+=+===−+− −=−+=+==Verifying that∆ABC is a right triangle by thePythagorean Theorem:[][][]()222222(,)(,)(,)4225164202020d A Bd B Cd A C+=+=+==The area of a triangle is12Abh=. In thisproblem,[] []1(,)(,)21 4 224 square unitsAd A Bd B C=⋅⋅=⋅⋅=

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ISM:Precalculus EGUChapter 1:Graphs755.The coordinates of the midpoint are:1212( ,),224435 ,2280,22(4, 0)xxyyx y++⎛⎞= ⎜⎟⎝⎠−++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠=56.The coordinates of the midpoint are:()1212( ,),222204,2204,220, 2xxyyx y++⎛⎞= ⎜⎟⎝⎠−++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠=57.The coordinates of the midpoint are:1212( ,),223620,2232,223 ,12xxyyx y++⎛⎞= ⎜⎟⎝⎠−++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠58.The coordinates of the midpoint are:1212( ,),222432,2261,2213,2xxyyx y++⎛⎞= ⎜⎟⎝⎠+−+⎛⎞= ⎜⎟⎝⎠−⎛⎞= ⎜⎟⎝⎠⎛⎞=−⎜⎟⎝⎠59.The coordinates of the midpoint are:1212( ,),224631,22210 ,22(5,1)xxyyx y++⎛⎞= ⎜⎟⎝⎠+−+⎛⎞= ⎜⎟⎝⎠−⎛⎞= ⎜⎟⎝⎠=−60.The coordinates of the midpoint are:1212( ,),224232,2221,2211,2xxyyx y++⎛⎞= ⎜⎟⎝⎠−+−+⎛⎞= ⎜⎟⎝⎠−−⎛⎞= ⎜⎟⎝⎠⎛⎞=−−⎜⎟⎝⎠61.The coordinates of the midpoint are:1212( ,),220.22.3 0.31.1,222.1 1.4,22(1.05, 0.7)xxyyx y++⎛⎞= ⎜⎟⎝⎠−++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠=62.The coordinates of the midpoint are:()1212( ,),221.2(0.3)2.31.1,220.93.4,220.45, 1.7xxyyx y++⎛⎞= ⎜⎟⎝⎠+ −+⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠=63.The coordinates of the midpoint are:1212( ,),2200,22,22xxyyx yabab++⎛⎞= ⎜⎟⎝⎠++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠

Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual - Page 9 preview image

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Chapter 1:GraphsISM:Precalculus EGU864.The coordinates of the midpoint are:1212( ,),2200,22,22xxyyx yaaaa++⎛⎞= ⎜⎟⎝⎠++⎛⎞= ⎜⎟⎝⎠⎛⎞= ⎜⎟⎝⎠65.Consider points of the form()2,ythat are adistance of 5 units from the point()2,1−−.()()()()()()222121222222221411612217dxxyyyyyyyy=−+−=−−+ − −=−+ − −=+++=++()()()2222225217521725217028042yyyyyyyyyy=++=++=++=+−=+−404yy+== −or202yy−==Thus, the points()2,4−and()2, 2are adistance of 5 units from the point()2,1−−.66.Consider points of the form(),3x−that are adistance of 13 units from the point()1, 2.()()()()()( )2221212222221232152125226dxxyyxxxxxxx=−+−=−+− −=−++=−++=−+()()()22222213226132261692260214301311xxxxxxxxxx=−+=−+=−+=−−=−+13013xx−==or11011xx+== −Thus, the points()13,3−and()11,3−−are adistance of 13 units from the point()1, 2.67.Points on the x-axis have a y-coordinate of 0.Thus, we consider points of the form(), 0xthatare a distance of 5 units from the point()4,3−.()()()()()22212122222243016831689825dxxyyxxxxxxx=−+−=−+ −−=−++ −=−++=−+()()22222258255825258250808xxxxxxxxx x=−+=−+=−+=−=−0x=or808xx−==Thus, the points()0, 0and()8, 0are on the x-axis and a distance of 5 units from the point()4,3−.

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ISM:Precalculus EGUChapter 1:Graphs968.Points on the y-axis have an x-coordinate of 0.Thus, we consider points of the form()0,ythatare a distance of 5 units from the point()4, 4.()()()()222121222222404416816168832dxxyyyyyyyyy=−+−=−+−=+−+=+−+=−+()()()2222225832583225832087071yyyyyyyyyy=−+=−+=−+=−+=−−707yy−==or101yy−==Thus, the points()0, 7and()0,1are on the y-axis and a distance of 5 units from the point()4, 4.69.The midpoint of AB is:()0600,223, 0D++⎛⎞= ⎜⎟⎝⎠=The midpoint of AC is:()0404,222, 2E++⎛⎞= ⎜⎟⎝⎠=The midpoint of BC is:()6404,225, 2F++⎛⎞= ⎜⎟⎝⎠=()2222(,)04(34)(4)(1)16117d C D=−+−=−+ −=+=()2222(,)26(20)(4)21642025d B E=−+−=−+=+==2222(,)(20)(50)2542529d A F=−+−=+=+=70.Let12(0, 0),(0, 4),( ,)PPPx y===()()()221222122222222222,(00)(40)164,(0)(0)416,(0)(4)(4)4(4)16dPPdPPxyxyxydPPxyxyxy=−+−===−+−=+=→+==−+−=+−=→+−=Therefore,()222248168162yyyyyyy=−=−+==which gives2222161223xxx+=== ±Two triangles are possible. The third vertex is()()23, 2or23, 2−.

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Chapter 1:GraphsISM:Precalculus EGU1071.Let()10, 0P=,()20,Ps=,()3, 0Ps=, and()4,Ps s=.(0,s)(s,0)(0,0)(s,s)XYThe points1Pand4Pare endpoints of onediagonal and the points2Pand3Pare theendpoints of the other diagonal.1,400,,2222ssssM++⎛⎞⎛⎞==⎜⎟⎜⎟⎝⎠⎝⎠2,300,,2222ssssM++⎛⎞⎛⎞==⎜⎟⎜⎟⎝⎠⎝⎠The midpoints of the diagonals are the same.Therefore, the diagonals of a square intersect attheir midpoints.72.Let()10, 0P=,()2, 0Pa=, and33,22aaP⎛⎞= ⎜⎟⎜⎟⎝⎠. To show that these verticesform an equilateral triangle, we need to showthat the distance between any pair of points is thesame constant value.()()()()()22122121222,000dP Pxxyyaaa=−+−=−+−==()()()22232121222222,302234444dPPxxyyaaaaaaaa=−+−⎛⎞⎛⎞=−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠=+===()()()22132121222222,3002234444dP Pxxyyaaaaaaa=−+−⎛⎞⎛⎞=−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠=+===Since all three distances have the same constantvalue, the triangle is an equilateral triangle.Now find the midpoints:1 22 31 3000,, 02223330,22,4422300322,,2244P PP PP PaaDMaaaaaEMaaaaFM++⎛⎞⎛⎞===⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞⎜⎟⎛⎞++⎜⎟=== ⎜⎟⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠⎛⎞++⎜⎟⎛⎞⎜⎟=== ⎜⎟⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠()22222233,0424344316162aaadD Eaaaaa⎛⎞⎛⎞=−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞⎛⎞=+ ⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠=+=()2222223,0424344316162aaadD Faaaaa⎛⎞⎛⎞=−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞⎛⎞=−+ ⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠=+=()22222333,44440242aaaadE Faaa⎛⎞⎛⎞=−+−⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎛⎞=+⎜⎟⎝⎠==Since the sides are the same length, the triangleis equilateral.

Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual - Page 12 preview image

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ISM:Precalculus EGUChapter 1:Graphs1173.221222(,)(42)(11)(6)0366d P P=−−+−=−+==()222322(,)4(4)( 31)0(4)164d PP=−− −+ −−=+ −==221322(,)(42)( 31)(6)(4)3616522 13d P P=−−+ −−=−+ −=+==Since[][][]222122313(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.74.()221222(,)6( 1)(24)7(2)49453d P P=− −+−=+ −=+=()222322(,)46( 52)(2)(7)44953d PP=−+ −−=−+ −=+=()221322(,)4( 1)( 54)5(9)2581106d P P=− −+ −−=+ −=+=Since[][][]222122313(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.Since()()1223,,dPPdPP=, the triangle isisosceles.Therefore, the triangle is an isosceles righttriangle.75.()()221222(,)0(2)7( 1)28464682 17d P P=− −+− −=+=+==()222322(,)30(27)3(5)92534d PP=−+−=+ −=+=()()221322(,)3( 2)2( 1)5325934d P P=− −+− −=+=+=Since2313(,)(,)d PPd P P=, the triangle isisosceles.Since[][][]222132312(,)(,)(,)d P Pd PPd P P+=,the triangle is also a right triangle.Therefore, the triangle is an isosceles righttriangle.76.()()221222(,)4702( 11)(2)121412555d P P=−−+−=−+ −=+==()222322(,)4(4)(60)86643610010d PP=− −+−=+=+==()()221322(,)4762( 3)4916255d P P=−+−=−+=+==Since[][][]222132312(,)(,)(,)d P Pd PPd P P+=,the triangle is a right triangle.

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Chapter 1:GraphsISM:Precalculus EGU1277.Using the Pythagorean Theorem:222229090810081001620016200902127.28 feetdddd+=+====≈90909090d78.Using the Pythagorean Theorem:222226060360036007200720060284.85 feetdddd+=+=→===≈60606060d79.a.First: (90, 0), Second: (90, 90)Third: (0, 90)(0,0)(0,90)(90,0)(90,90)XYb.Using the distance formula:2222(31090)(1590)220( 75)54025232.4 feetd=−+−=+ −=≈c.Using the distance formula:2222(3000)(30090)300210134100366.2 feetd=−+−=+=≈80.a.First: (60, 0), Second: (60, 60)Third: (0, 60)(0,0)(0,60)(60,0)(60,60)xyb.Using the distance formula:2222(18060)(2060)120(40)16000126.5 feetd=−+−=+ −=≈c.Using the distance formula:2222(2200)(22060)22016074000272.0 feetd=−+−=+=≈81.The Neon heading east moves a distance30tafterthours. The truck heading south moves adistance40tafterthours. Their distance apartafterthours is:22222(30 )(40 )9001600250050dtttttt=+=+==d40t30t

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ISM:Precalculus EGUChapter 1:Graphs1382.15 miles5280 ft1 hr22 ft/sec1 hr1 mile3600 sec⋅⋅=()2221002210000484dtt=+=+10022tdSection 1.21.()()2317236336xxxx+−= −+= −+= −= −The solution set is{}6−.2.()()229502150xxxx−−=+−=2102112xxx+== −= −or505xx−==The solution set is{}1 ,52−.3.intercepts4.zeros; roots5.y-axis6.47.()3, 4−8.True9.False; we solve foryin terms ofx.10.False; a graph can be symmetric with respect toboth coordinate axes (in such cases it will also besymmetric with respect to the origin).For example:221xy+=11.12.13.14.

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Chapter 1:GraphsISM:Precalculus EGU1415.5−5y5−5a = (5, 2)b = ( 5,2)−−c = ( 5, 2)−(5, −2)16.17.18.19.20.21.4yxx=−400000=−=411110=−≠40( 1)1011= −−−≠−−(0, 0) is on the graph of the equation.22.32yxx=−3002000=−=3112 111=−≠ −3112 111−=−−= −(0, 0) and (1, –1) are on the graph of theequation.23.229yx=+2230999=+=22039018=+≠220( 3)9018= −+≠(0, 3) is on the graph of the equation.24.31yx=+321182=+≠310111=+=301100= − +=(0, 1) and (–1, 0) are on the graph of theequation.

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ISM:Precalculus EGUChapter 1:Graphs1525.224xy+=2202444+==22(2)2484−+=≠2222444+==()(0, 2) and2,2are on the graph of theequation.26.2244xy+=2204 1444+⋅==2224 0444+⋅==()221224454+=≠(0, 1) and (2, 0) are on the graph of the equation.27.a.Intercepts:()1, 0−and()1, 0b.Symmetric with respect to the x-axis, y-axis,and the origin.28.a.Intercepts:()0,1b.Not symmetric to x-axis, y-axis, or origin29.a.Intercepts:()20,π−,()0,1, and()2, 0πb.Symmetric with respect to the y-axis.30.a.Intercepts:()2, 0−,()0,3−, and()2, 0b.Symmetric with respect to the y-axis.31.a.Intercepts:()0, 0b.Symmetric with respect to the x-axis.32.a.Intercepts:()2, 0−,()0, 2,()0,2−, and()2, 0b.Symmetric with respect to the x-axis, y-axis,and the origin.33.a.Intercepts:()2, 0−,()0, 0, and()2, 0b.Symmetric with respect to the origin.34.a.Intercepts:()4, 0−,()0, 0, and()4, 0b.Symmetric with respect to the origin.35.−10yx5−5(0,9)−10(2,5)−( 2,5)−−36.yx5−5(5, 3)(5,3)−( 4, 0)−−55(0,2)−(0, 2)37.−5yx5π−π,22π⎛⎞−−⎜⎟⎝⎠, 22π⎛⎞⎜⎟⎝⎠(π, 0)(−π, 0)38.

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