Solution Manual for Algebra and Trigonometry, 5th Edition

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’SSOLUTIONSMANUALJUDITHA.PENNAALGEBRA&TRIGONOMETRYFIFTHEDITIONPRECALCULUS:ARIGHTTRIANGLEAPPROACHFIFTHEDITIONJudith A. BeecherJudith A. PennaMarvin L. BittingerIndiana University Purdue University Indianapolis

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ContentsChapter1........................... 1Chapter2.........................57Chapter3......................... 107Chapter4.......................... 163Chapter5......................... 249Chapter6......................... 305Chapter7......................... 357Chapter8......................... 399Chapter9......................... 451Chapter10......................... 551Chapter11......................... 635Just-in-Time Review..................... 677ChapterR .......................... 687

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42244224xy(1, 4)(3,5)(0, 2)(4, 0)(2,2)42244224xy(5, 0)(4,2)(1, 4)(4, 0)(2,4)4224424xy(5, 1)(2, 3)(0, 1)(5, 1)(2,1)42244224xy(5, 2)(5, 0)(1,5)(4, 0)(4,3)Chapter 1Graphs, Functions, and ModelsExercise Set 1.11.Point A is located 5 units to the left of they-axis and4 units up from thex-axis, so its coordinates are (−5,4).Point B is located 2 units to the right of they-axis and2 units down from thex-axis, so its coordinates are (2,−2).Point Cis located 0 units to the right or left of they-axisand 5 units down from thex-axis, so its coordinates are(0,−5).Point D is located 3 units to the right of they-axis and5 units up from thex-axis, so its coordinates are (3,5).Point E is located 5 units to the left of they-axis and4unitsdownfromthex-axis,soitscoordinatesare(−5,−4).Point F is located 3 units to the right of they-axis and0 units up or down from thex-axis, so its coordinates are(3,0).2.G: (2,1); H: (0,0); I: (4,−3); J: (−4,0); K: (−2,3);L: (0,5)3.To graph (4,0) we move from the origin 4 units to the rightof they-axis. Since the second coordinate is 0, we do notmove up or down from thex-axis.To graph (−3,−5) we move from the origin 3 units to theleft of they-axis.Then we move 5 units down from thex-axis.To graph (−1,4) we move from the origin 1 unit to the leftof they-axis. Then we move 4 units up from thex-axis.To graph (0,2) we do not move to the right or the left ofthey-axis since the first coordinate is 0. From the originwe move 2 units up.To graph (2,−2) we move from the origin 2 units to theright of they-axis. Then we move 2 units down from thex-axis.4.5.To graph (−5,1) we move from the origin 5 units to theleft of they-axis. Then we move 1 unit up from thex-axis.To graph (5,1) we move from the origin 5 units to the rightof they-axis. Then we move 1 unit up from thex-axis.To graph (2,3) we move from the origin 2 units to the rightof they-axis. Then we move 3 units up from thex-axis.To graph (2,−1) we move from the origin 2 units to theright of they-axis.Then we move 1 unit down from thex-axis.To graph (0,1) we do not move to the right or the left ofthey-axis since the first coordinate is 0. From the originwe move 1 unit up.6.7.The first coordinate represents the year and the second co-ordinate represents the number of Sprint Cup Series racesin which Tony Stewart finished in the top five.The or-dered pairs are (2008, 10), (2009, 15), (2010, 9), (2011, 9),(2012, 12), and (2013, 5).8.The first coordinate represents the year and the secondcoordinaterepresentsthepercentofMarineswhoarewomen.The ordered pairs are (1960, 1%), (1970, 0.9%),(1980, 3.6%), (1990, 4.9%), (2000, 6.1%), and (2011, 6.8%).

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2Chapter 1:Graphs, Functions, and Models9.To determine whether (−1,−9) is a solution, substitute−1 forxand−9 fory.y= 7x−2−9 ? 7(−1)−2∣∣∣−7−2−9∣∣−9TRUEThe equation−9 =−9 is true, so (−1,−9) is a solution.To determine whether (0,2) is a solution, substitute 0 forxand 2 fory.y= 7x−22 ? 7·0−2∣∣∣0−22∣∣−2FALSEThe equation 2 =−2 is false, so (0,2) is not a solution.10.For(12,8):y=−4x+ 108 ?−4·12 + 10∣∣∣∣∣−2 + 108∣∣8TRUE(12,8)is a solution.For (−1,6):y=−4x+ 106 ?−4(−1) + 10∣∣∣4 + 106∣∣14FALSE(−1,6) is not a solution.11.To determine whether(23,34)is a solution, substitute 23forxand 34 fory.6x−4y= 16·23−4·34? 1∣∣4−3∣∣∣1∣∣1TRUEThe equation 1 = 1 is true, so(23,34)is a solution.To determine whether(1,32)is a solution, substitute 1 forxand 32 fory.6x−4y= 16·1−4·32? 1∣∣6−6∣∣∣0∣∣1FALSEThe equation 0 = 1 is false, so(1,32)is not a solution.12.For (1.5,2.6):x2+y2= 9(1.5)2+ (2.6)2? 92.25 + 6.76∣∣∣9.01∣∣9FALSE(1.5,2.6) is not a solution.For (−3,0):x2+y2= 9(−3)2+ 02? 99 + 0∣∣∣9∣∣9TRUE(−3,0) is a solution.13.To determine whether(−12,−45)is a solution, substitute−12 foraand−45 forb.2a+ 5b= 32(−12)+ 5(−45)? 3∣∣−1−4∣∣∣−5∣∣3FALSEThe equation−5 = 3 is false, so(−12,−45)is not a solu-tion.To determine whether(0,35)is a solution, substitute 0 foraand 35 forb.2a+ 5b= 32·0 + 5·35? 3∣∣0 + 3∣∣∣3∣∣3TRUEThe equation 3 = 3 is true, so(0,35)is a solution.14.For(0,32):3m+ 4n= 63·0 + 4·32? 6∣∣0 + 6∣∣∣6∣∣6TRUE(0,32)is a solution.For(23,1):3m+ 4n= 63·23 + 4·1 ? 6∣∣2 + 4∣∣∣6∣∣6TRUEThe equation 6 = 6 is true, so(23,1)is a solution.

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yx422442245x3y 15(3, 0)(0, 5)yx422442242x4y8(4, 0)(0,2)(0, 4)yx422442242xy4(2, 0)yx42242246(0, 6)3xy6(2, 0)Exercise Set 1.1315.To determine whether (−0.75,2.75) is a solution, substi-tute−0.75 forxand 2.75 fory.x2−y2= 3(−0.75)2−(2.75)2? 30.5625−7.5625∣∣∣−7∣∣3FALSEThe equation−7 = 3 is false, so (−0.75,2.75) is not asolution.To determine whether (2,−1) is a solution, substitute 2forxand−1 fory.x2−y2= 322−(−1)2? 34−1∣∣∣3∣∣3TRUEThe equation 3 = 3 is true, so (2,−1) is a solution.16.For (2,−4):5x+ 2y2= 705·2 + 2(−4)2? 7010 + 2·16∣∣∣10 + 32∣∣42∣∣70FALSE(2,−4) is not a solution.For (4,−5):5x+ 2y2= 705·4 + 2(−5)2? 7020 + 2·25∣∣∣20 + 50∣∣70∣∣70TRUE(4,−5) is a solution.17.Graph 5x−3y=−15.To find thex-intercept we replaceywith 0 and solve forx.5x−3·0 =−155x=−15x=−3Thex-intercept is (−3,0).To find they-intercept we replacexwith 0 and solve fory.5·0−3y=−15−3y=−15y= 5They-intercept is (0,5).We plot the intercepts and draw the line that containsthem.We could find a third point as a check that theintercepts were found correctly.18.19.Graph 2x+y= 4.To find thex-intercept we replaceywith 0 and solve forx.2x+ 0 = 42x= 4x= 2Thex-intercept is (2,0).To find they-intercept we replacexwith 0 and solve fory.2·0 +y= 4y= 4They-intercept is (0,4).We plot the intercepts and draw the line that containsthem.We could find a third point as a check that theintercepts were found correctly.20.

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yx422442244y3x12(4, 0)(0, 3)yx422442243y2x 6(3, 0)(0,2)622424xyy3x5yx42244224y 2x142244224xyxy3yx22462246xy44Chapter 1:Graphs, Functions, and Models21.Graph 4y−3x= 12.To find thex-intercept we replaceywith 0 and solve forx.4·0−3x= 12−3x= 12x=−4Thex-intercept is (−4,0).To find they-intercept we replacexwith 0 and solve fory.4y−3·0 = 124y= 12y= 3They-intercept is (0,3).We plot the intercepts and draw the line that containsthem.We could find a third point as a check that theintercepts were found correctly.22.23.Graphy= 3x+ 5.We choose some values forxand find the correspondingy-values.Whenx=−3,y= 3x+ 5 = 3(−3) + 5 =−9 + 5 =−4.Whenx=−1,y= 3x+ 5 = 3(−1) + 5 =−3 + 5 = 2.Whenx= 0,y= 3x+ 5 = 3·0 + 5 = 0 + 5 = 5We list these points in a table, plot them, and draw thegraph.xy(x, y)−3−4(−3,−4)−12(−1,2)05(0,5)24.25.Graphx−y= 3.Make a table of values, plot the points in the table, anddraw the graph.xy(x, y)−2−5(−2,−5)0−3(0,−3)30(3,0)26.27.Graphy=−34x+ 3.By choosing multiples of 4 forx, we can avoid fractionvalues fory. Make a table of values, plot the points in thetable, and draw the graph.xy(x, y)−46(−4,6)03(0,3)40(4,0)

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yx4224422434yx3yx422442243y2x3yx422442245x2y8yx4224422443y2x4224642xyx4y5yx422442246xy4424226xy2x5y10yx42244682244x3y12Exercise Set 1.1528.29.Graph 5x−2y= 8.We could solve foryfirst.5x−2y= 8−2y=−5x+ 8Subtracting 5xon both sidesy= 52x−4Multiplyingby−12onbothsidesBy choosing multiples of 2 forxwe can avoid fractionvalues fory. Make a table of values, plot the points in thetable, and draw the graph.xy(x, y)0−4(0,−4)21(2,1)46(4,6)30.31.Graphx−4y= 5.Make a table of values, plot the points in the table, anddraw the graph.xy(x, y)−3−2(−3,−2)1−1(1,−1)50(5,0)32.33.Graph 2x+ 5y=−10.In this case, it is convenient to find the intercepts alongwith a third point on the graph. Make a table of values,plot the points in the table, and draw the graph.xy(x, y)−50(−5,0)0−2(0,−2)5−4(5,−4)34.

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28644224xyyx2yx2yx4224246864224224xyyx23y4x2yx42244224yx844881244y x22x3yx22x1yx422442246Chapter 1:Graphs, Functions, and Models35.Graphy=−x2.Make a table of values, plot the points in the table, anddraw the graph.xy(x, y)−2−4(−2,−4)−1−1(−1,−1)00(0,0)1−1(1,−1)2−4(2,−4)36.37.Graphy=x2−3.Make a table of values, plot the points in the table, anddraw the graph.xy(x, y)−36(−3,6)−1−2(−1,−2)0−3(0,−3)1−2(1,−2)36(3,6)38.39.Graphy=−x2+ 2x+ 3.Make a table of values, plot the points in the table, anddraw the graph.xy(x, y)−2−5(−2,−5)−10(−1,0)03(0,3)14(1,4)23(2,3)30(3,0)4−5(4,−5)40.41.Either point can be considered as (x1, y1).d=√(4−5)2+ (6−9)2=√(−1)2+ (−3)2=√10≈3.16242.d=√(−3−2)2+ (7−11)2=√41≈6.40343.Either point can be considered as (x1, y1).d=√(−13−(−8))2+ (1−(−11))2=√(−5)2+ 122=√169 = 1344.d=√(−20−(−60))2+ (35−5)2=√2500 = 50

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Exercise Set 1.1745.Either point can be considered as (x1, y1).d=√(6−9)2+ (−1−5)2=√(−3)2+ (−6)2=√45≈6.70846.d=√(−4−(−1))2+ (−7−3)2=√109≈10.44047.Either point can be considered as (x1, y1).d=√(−8−8)2+(711−711)2=√(−16)2+ 02= 1648.d=√(12−12)2+(−425−(−1325))2=√(925)2= 92549.d=√[−35−(−35)]2+(−4−23)2=√02+(−143)2= 14350.d=√(−113−13)2+(−12−52)2=√16 + 9 =√25 = 551.Either point can be considered as (x1, y1).d=√(−4.2−2.1)2+ [3−(−6.4)]2=√(−6.3)2+ (9.4)2=√128.05≈11.31652.d=√[0.6−(−8.1)]2+ [−1.5−(−1.5)]2=√(8.7)2= 8.753.Either point can be considered as (x1, y1).d=√(0−a)2+ (0−b)2=√a2+b254.d=√[r−(−r)]2+ [s−(−s)]2=√4r2+ 4s2=2√r2+s255.First we find the length of the diameter:d=√(−3−9)2+ (−1−4)2=√(−12)2+ (−5)2=√169 = 13The length of the radius is one-half the length of the di-ameter, or 12 (13), or 6.5.56.Radius=√(−3−0)2+ (5−1)2=√25 = 5Diameter= 2·5 = 1057.First we find the distance between each pair of points.For (−4,5) and (6,1):d=√(−4−6)2+ (5−1)2=√(−10)2+ 42=√116For (−4,5) and (−8,−5):d=√(−4−(−8))2+ (5−(−5))2=√42+ 102=√116For (6,1) and (−8,−5):d=√(6−(−8))2+ (1−(−5))2=√142+ 62=√232Since (√116)2+ (√116)2= (√232)2, the points could bethe vertices of a right triangle.58.For (−3,1) and (2,−1):d=√(−3−2)2+ (1−(−1))2=√29For (−3,1) and (6,9):d=√(−3−6)2+ (1−9)2=√145For (2,−1) and (6,9):d=√(2−6)2+ (−1−9)2=√116Since (√29)2+ (√116)2= (√145)2, the points could bethe vertices of a right triangle.59.First we find the distance between each pair of points.For (−4,3) and (0,5):d=√(−4−0)2+ (3−5)2=√(−4)2+ (−2)2=√20For (−4,3) and (3,−4):d=√(−4−3)2+ [3−(−4)]2=√(−7)2+ 72=√98For (0,5) and (3,−4):d=√(0−3)2+ [5−(−4)]2=√(−3)2+ 92=√90The greatest distance is√98, so if the points are the ver-tices of a right triangle, then it is the hypotenuse.But(√20)2+ (√90)2= (√98)2, so the points are not the ver-tices of a right triangle.60.See the graph of this rectangle in Exercise 71.The segments with endpoints (−3,4), (2,−1) and (5,2),(0,7) are one pair of opposite sides. We find the length ofeach of these sides.For (−3,4), (2,−1):d=√(−3−2)2+ (4−(−1))2=√50For (5,2), (0,7):d=√(5−0)2+ (2−7)2=√50The segments with endpoints (2,−1), (5,2) and (0,7),(−3,4) are the second pair of opposite sides. We find theirlengths.For (2,−1), (5,2):d=√(2−5)2+ (−1−2)2=√18For (0,7), (−3,4):d=√(0−(−3))2+ (7−4)2=√18The endpoints of the diagonals are (−3,4),(5,2) and(2,−1), (0,7). We find the length of each.For (−3,4), (5,2):d=√(−3−5)2+ (4−2)2=√68

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864224624xy1284x12844y8Chapter 1:Graphs, Functions, and ModelsFor (2,−1), (0,7):d=√(2−0)2+ (−1−7)2=√68The opposite sides of the quadrilateral are the same lengthand the diagonals are the same length, so the quadrilateralis a rectangle.61.We use the midpoint formula.(4 + (−12)2,−9 + (−3)2)=(−82,−122)= (−4,−6)62.(7 + 92,−2 + 52)=(8,32)63.We use the midpoint formula.(0 +(−25)2,12−02)=(−252,122)=(−15,14)64.(0 +(−713)2,0 + 272)=(−726,17)65.We use the midpoint formula.(6.1 + 3.82,−3.8 + (−6.1)2)=(9.92,−9.92)=(4.95,−4.95)66.(−0.5 + 4.82,−2.7 + (−0.3)2)= (2.15,−1.5)67.We use the midpoint formula.(−6 + (−6)2,5 + 82)=(−122,132)=(−6,132)68.(1 + (−1)2,−2 + 22)= (0,0)69.We use the midpoint formula.(−16 +(−23)2,−35 + 542)=(−562,13202)=(−512,1340)70.(29 +(−25)2,13 + 452)=(−445,1730)71.For the side with vertices (−3,4) and (2,−1):(−3 + 22,4 + (−1)2)=(−12,32)For the side with vertices (2,−1) and (5,2):(2 + 52,−1 + 22)=(72,12)For the side with vertices (5,2) and (0,7):(5 + 02,2 + 72)=(52,92)For the side with vertices (0,7) and (−3,4):(0 + (−3)2,7 + 42)=(−32,112)For the quadrilateral whose vertices are the points foundabove, the diagonals have endpoints(−12,32),(52,92)and(72,12),(−32,112).We find the length of each of these diagonals.For(−12,32),(52,92):d=√(−12−52)2+(32−92)2=√(−3)2+ (−3)2=√18For(72,12),(−32,112):d=√(72−(−32))2+(12−112)2=√52+ (−5)2=√50Since the diagonals do not have the same lengths, the mid-points are not vertices of a rectangle.72.For the side with vertices (−5,−1) and (7,−6):(−5 + 72,−1 + (−6)2)=(1,−72)For the side with vertices (7,−6) and (12,6):(7 + 122,−6 + 62)=(192,0)For the side with vertices (12,6) and (0,11):(12 + 02,6 + 112)=(6,172)For the side with vertices (0,11) and (−5,−1):(0 + (−5)2,11 + (−1)2)=(−52,5)For the quadrilateral whose vertices are the points foundabove, one pair of opposite sides has endpoints(1,−72),(192,0)and(6,172),(−52,5).The length of each of

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yx24242442x2y24yx48484884x2y281Exercise Set 1.19these sides is√3382. The other pair of opposite sides hasendpoints(192,0),(6,172)and(−52,5),(1,−72).The length of each of these sides is also√3382. The end-points of the diagonals of the quadrilateral are(1,−72),(6,172)and(192,0),(−52,5). The length of each di-agonal is 13. Since the four sides of the quadrilateral arethe same length and the diagonals are the same length, themidpoints are vertices of a square.73.We use the midpoint formula.(√7 +√22,−4 + 32)=(√7 +√22,−12)74.(−3 + 12,√5 +√22)=(−1,√5 +√22)75.(x−h)2+ (y−k)2=r2(x−2)2+ (y−3)2=(53)2Substituting(x−2)2+ (y−3)2= 25976.(x−4)2+ (y−5)2= (4.1)2(x−4)2+ (y−5)2= 16.8177.The length of a radius is the distance between (−1,4) and(3,7):r=√(−1−3)2+ (4−7)2=√(−4)2+ (−3)2=√25 = 5(x−h)2+ (y−k)2=r2[x−(−1)]2+ (y−4)2= 52(x+ 1)2+ (y−4)2= 2578.Find the length of a radius:r=√(6−1)2+ (−5−7)2=√169 = 13(x−6)2+ [y−(−5)]2= 132(x−6)2+ (y+ 5)2= 16979.The center is the midpoint of the diameter:(7 + (−3)2,13 + (−11)2)= (2,1)Use the center and either endpoint of the diameter to findthe length of a radius. We use the point (7,13):r=√(7−2)2+ (13−1)2=√52+ 122=√169 = 13(x−h)2+ (y−k)2=r2(x−2)2+ (y−1)2= 132(x−2)2+ (y−1)2= 16980.The points (−9,4) and (−1,−2) are opposite vertices ofthe square and hence endpoints of a diameter of the circle.We use these points to find the center and radius.Center:(−9 + (−1)2,4 + (−2)2)= (−5,1)Radius:12√(−9−(−1))2+(4−(−2))2= 12·10 = 5[x−(−5)]2+ (y−1)2= 52(x+ 5)2+ (y−1)2= 2581.Since the center is 2 units to the left of they-axis and thecircle is tangent to they-axis, the length of a radius is 2.(x−h)2+ (y−k)2=r2[x−(−2)]2+ (y−3)2= 22(x+ 2)2+ (y−3)2= 482.Since the center is 5 units below thex-axis and the circleis tangent to thex-axis, the length of a radius is 5.(x−4)2+ [y−(−5)]2= 52(x−4)2+ (y+ 5)2= 2583.x2+y2= 4(x−0)2+ (y−0)2= 22Center:(0,0); radius:284.x2+y2= 81(x−0)2+ (y−0)2= 92Center:(0,0); radius:9

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yx242244268x2(y3)216(x2)2y2100yx4848488412(x1)2(y5)236yx484488412(x7)2(y2)225yx484481248(x4)2(y5)29yx224242668(x1)2(y2)264yx484488481210Chapter 1:Graphs, Functions, and Models85.x2+ (y−3)2= 16(x−0)2+ (y−3)2= 42Center:(0,3); radius:486.(x+ 2)2+y2= 100[x−(−2)]2+ (y−0)2= 102Center:(−2,0); radius:1087.(x−1)2+ (y−5)2= 36(x−1)2+ (y−5)2= 62Center:(1,5); radius:688.(x−7)2+ (y+ 2)2= 25(x−7)2+ [y−(−2)]2= 52Center:(7,−2); radius:589.(x+ 4)2+ (y+ 5)2= 9[x−(−4)]2+ [y−(−5)]2= 32Center:(−4,−5); radius:390.(x+ 1)2+ (y−2)2= 64[x−(−1)]2+ (y−2)2= 82Center:(−1,2); radius:891.From the graph we see that the center of the circle is(−2,1) and the radius is 3.The equation of the circleis [x−(−2)]2+ (y−1)2= 32, or (x+ 2)2+ (y−1)2= 32.92.Center:(3,−5), radius:4Equation:(x−3)2+ [y−(−5)]2= 42, or(x−3)2+ (y+ 5)2= 4293.From the graph we see that the center of the circle is(5,−5) and the radius is 15.The equation of the circleis (x−5)2+ [y−(−5)]2= 152, or (x−5)2+ (y+ 5)2= 152.

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Solution Manual for Algebra and Trigonometry, 5th Edition - Page 14 preview image

Exercise Set 1.11194.Center:(−8,2), radius:4Equation:[x−(−8)]2+ (y−2)2= 42, or(x+ 8)2+ (y−2)2= 4295.If the point (p, q) is in the fourth quadrant, thenp >0andq <0. Ifp >0, then−p <0 so both coordinates ofthe point (q,−p) are negative and (q,−p) is in the thirdquadrant.96.Use the distance formula:d=√(a+h−a)2+(1a+h−1a)2=√h2+(−ha(a+h))2=√h2+h2a2(a+h)2=√h2a2(a+h)2+h2a2(a+h)2=√h2(a2(a+h)2+ 1)a2(a+h)2=∣∣∣∣∣ha(a+h)∣∣∣∣∣√a2(a+h)2+ 1Find the midpoint:(a+a+h2,1a+1a+h2)=(2a+h2,2a+h2a(a+h))97.Use the distance formula. Either point can be consideredas (x1, y1).d=√(a+h−a)2+ (√a+h− √a)2=√h2+a+h−2√a2+ah+a=√h2+ 2a+h−2√a2+ahNext we use the midpoint formula.(a+a+h2,√a+√a+h2)=(2a+h2,√a+√a+h2)98.C= 2πr10π= 2πr5 =rThen [x−(−5)]2+(y−8)2= 52, or (x+5)2+(y−8)2= 25.99.First use the formula for the area of a circle to findr2:A=πr236π=πr236 =r2Then we have:(x−h)2+ (y−k)2=r2(x−2)2+ [y−(−7)]2= 36(x−2)2+ (y+ 7)2= 36100.Let the point be (x,0). We set the distance from (−4,−3)to (x,0) equal to the distance from (−1,5) to (x,0) andsolve forx.√(−4−x)2+ (−3−0)2=√(−1−x)2+ (5−0)2√16 + 8x+x2+ 9 =√1 + 2x+x2+ 25√x2+ 8x+ 25 =√x2+ 2x+ 26x2+ 8x+ 25 =x2+ 2x+ 26Squaring both sides8x+ 25 = 2x+ 266x= 1x= 16The point is(16,0).101.Let (0, y) be the required point. We set the distance from(−2,0) to (0, y) equal to the distance from (4,6) to (0, y)and solve fory.√[0−(−2)]2+ (y−0)2=√(0−4)2+ (y−6)2√4 +y2=√16 +y2−12y+ 364 +y2= 16 +y2−12y+ 36Squaring both sides−48 =−12y4 =yThe point is (0,4).102.We first find the distance between each pair of points.For (−1,−3) and (−4,−9):d1=√[−1−(−4)]2+ [−3−(−9)]2=√32+ 62=√9 + 36=√45 = 3√5For (−1,−3) and (2,3):d2=√(−1−2)2+ (−3−3)2=√(−3)2+ (−6)2=√9 + 36=√45 = 3√5For (−4,−9) and (2,3):d3=√(−4−2)2+ (−9−3)2=√(−6)2+ (−12)2=√36 + 144=√180 = 6√5Sinced1+d2=d3, the points are collinear.103.a) When the circle is positioned on a coordinate systemas shown in the text, the center lies on they-axisand is equidistant from (−4,0) and (0,2).Let (0, y) be the coordinates of the center.√(−4−0)2+(0−y)2=√(0−0)2+(2−y)242+y2= (2−y)216 +y2= 4−4y+y212 =−4y−3 =yThe center of the circle is (0,−3).

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Solution Manual for Algebra and Trigonometry, 5th Edition - Page 15 preview image

12Chapter 1:Graphs, Functions, and Modelsb) Use the point (−4,0) and the center (0,−3) to findthe radius.(−4−0)2+ [0−(−3)]2=r225 =r25 =rThe radius is 5 ft.104.The coordinates ofPare(b2, h2)by the midpoint formula.By the distance formula, each of the distances fromPto(0, h), fromPto (0,0), and fromPto (b,0) is√b2+h22.105.x2+y2= 1(√32)2+(−12)2? 1∣∣34 + 14∣∣∣∣∣1∣∣1TRUE(√32,−12)lies on the unit circle.106.x2+y2= 102+ (−1)2? 11∣∣1TRUE(0,−1) lies on the unit circle.107.x2+y2= 1(−√22)2+(√22)2? 1∣∣24 + 24∣∣∣∣∣1∣∣1TRUE(−√22,√22)lies on the unit circle.108.x2+y2= 1(12)2+(−√32)2? 1∣∣14 + 34∣∣∣∣∣1∣∣1TRUE(12,−√32)lies on the unit circle.109.a), b)See the answer section in the text.Exercise Set 1.21.This correspondence is a function, because each memberof the domain corresponds to exactly one member of therange.2.This correspondence is a function, because each memberof the domain corresponds to exactly one member of therange.3.This correspondence is a function, because each memberof the domain corresponds to exactly one member of therange.4.This correspondence is not a function, because there is amember of the domain (1) that corresponds to more thanone member of the range (4 and 6).5.This correspondence is not a function, because there is amember of the domain (m) that corresponds to more thanone member of the range (A and B).6.This correspondence is a function, because each memberof the domain corresponds to exactly one member of therange.7.This correspondence is a function, because each memberof the domain corresponds to exactly one member of therange.8.This correspondence is not a function, because there is amember of the domain that corresponds to more than onemember of the range. In fact, Sean Connery, Roger Moore,and Pierce Brosnan all correspond to two members of therange.9.This correspondence is a function, because each car hasexactly one license number.10.This correspondence is not a function, because we cansafely assume that at least one person uses more than onedoctor.11.This correspondence is a function, because each integerless than 9 corresponds to exactly one multiple of 5.12.This correspondence is not a function, because we cansafely assume that at least one band member plays morethan one instrument.13.This correspondence is not a function, because at least onestudent will have more than one neighboring seat occupiedby another student.14.This correspondence is a function, because each bag hasexactly one weight.15.The relation is a function, because no two ordered pairshave the same first coordinate and different second coor-dinates.The domain is the set of all first coordinates:{2,3,4}.The range is the set of all second coordinates:{10,15,20}.16.The relation is a function, because no two ordered pairshave the same first coordinate and different second coor-dinates.Domain:{3,5,7}Range:{1}

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Solution Manual for Algebra and Trigonometry, 5th Edition - Page 16 preview image

Exercise Set 1.21317.The relation is not a function, because the ordered pairs(−2,1) and (−2,4) have the same first coordinate and dif-ferent second coordinates.The domain is the set of all first coordinates:{−7,−2,0}.The range is the set of all second coordinates:{3,1,4,7}.18.The relation is not a function, because each of the orderedpairs has the same first coordinate and different secondcoordinates.Domain:{1}Range:{3,5,7,9}19.The relation is a function, because no two ordered pairshave the same first coordinate and different second coor-dinates.The domain is the set of all first coordinates:{−2,0,2,4,−3}.The range is the set of all second coordinates:{1}.20.The relation is not a function, because the ordered pairs(5,0) and (5,−1) have the same first coordinates and dif-ferent second coordinates.This is also true of the pairs(3,−1) and (3,−2).Domain:{5,3,0}Range:{0,−1,−2}21.g(x) = 3x2−2x+ 1a)g(0) = 3·02−2·0 + 1 = 1b)g(−1) = 3(−1)2−2(−1) + 1 = 6c)g(3) = 3·32−2·3 + 1 = 22d)g(−x) = 3(−x)2−2(−x) + 1 = 3x2+ 2x+ 1e)g(1−t) = 3(1−t)2−2(1−t) + 1 =3(1−2t+t2)−2(1−t)+1 = 3−6t+3t2−2+2t+1 =3t2−4t+ 222.f(x) = 5x2+ 4xa)f(0) = 5·02+ 4·0 = 0 + 0 = 0b)f(−1) = 5(−1)2+ 4(−1) = 5−4 = 1c)f(3) = 5·32+ 4·3 = 45 + 12 = 57d)f(t) = 5t2+ 4te)f(t−1) = 5(t−1)2+ 4(t−1) = 5t2−6t+ 123.g(x) =x3a)g(2) = 23= 8b)g(−2) = (−2)3=−8c)g(−x) = (−x)3=−x3d)g(3y) = (3y)3= 27y3e)g(2 +h) = (2 +h)3= 8 + 12h+ 6h2+h324.f(x) = 2|x|+ 3xa)f(1) = 2|1|+ 3·1 = 2 + 3 = 5b)f(−2) = 2| −2|+ 3(−2) = 4−6 =−2c)f(−x) = 2| −x|+ 3(−x) = 2|x| −3xd)f(2y) = 2|2y|+ 3·2y= 4|y|+ 6ye)f(2−h) = 2|2−h|+ 3(2−h) =2|2−h|+ 6−3h25.g(x) =x−4x+ 3a)g(5) = 5−45 + 3 = 18b)g(4) = 4−44 + 7 = 0c)g(−3) =−3−4−3 + 3 =−70Since division by 0 is not defined,g(−3) does notexist.d)g(−16.25) =−16.25−4−16.25 + 3 =−20.25−13.25 = 8153≈1.5283e)g(x+h) =x+h−4x+h+ 326.f(x) =x2−xa)f(2) =22−2 = 20Since division by 0 is not defined,f(2) does notexist.b)f(1) =12−1 = 1c)f(−16) =−162−(−16) =−1618=−89d)f(−x) =−x2−(−x) =−x2 +xe)f(−23)=−232−(−23)=−2383=−1427.g(x) =x√1−x2g(0) =0√1−02=0√1 = 01 = 0g(−1) =−1√1−(−1)2=−1√1−1 =−1√0 =−10Since division by 0 is not defined,g(−1) does not exist.g(5) =5√1−52=5√1−25 =5√−24Since√−24 is not defined as a real number,g(5) does notexist as a real number.

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