Solution Manual For Algebra And Trigonometry, 9th Edition

Preview (16 of 1481 Pages)

100%

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 1 preview image

Loading page ...

1Chapter RReviewSection R.11.rational2.45 63430331+⋅−=+−=3.Distributive4.()536x+=5.True6.False; The Zero-Product Property states that if aproduct equals 0, then at least one of the factorsmust equal 0.7.False; 6 is the Greatest Common Factor of 12and 18. The Least Common Multiple is thesmallest value that both numbers will divideevenly. The LCM for 12 and 18 is 36.8.True9.{}{}{}1, 3, 4,5, 92, 4, 6, 7,81, 2,3, 4, 5, 6, 7,8, 9AB∪=∪=10.{}{}{}1, 3, 4,5, 91, 3, 4, 61, 3, 4, 5, 6, 9AC∪=∪=11.{}{}{}1, 3, 4,5, 92, 4, 6, 7,84AB∩=∩=12.{}{}{}1, 3, 4,5, 91, 3, 4, 61, 3, 4AC∩=∩=13.{}{}(){}{}{}{}()1, 3, 4,5, 92, 4, 6, 7,81,3, 4, 61, 2,3, 4,5, 6, 7,8,91,3, 4, 61, 3, 4,6ABC∪∩=∪∩=∩=14.{}{}(){}{ }{}{}()1, 3, 4,5, 92, 4, 6, 7,81,3, 4, 641, 3, 4, 61,3, 4, 6ABC∩∪=∩∪=∪=15.{}0, 2, 6, 7, 8A=16.{}0, 2, 5, 7, 8, 9C=17.{}{}{ }{}1, 3, 4, 5, 92, 4, 6, 7, 840, 1, 2, 3, 5, 6, 7, 8, 9AB∩=∩==18.{}{}{}{}2, 4, 6, 7, 81, 3, 4, 61, 2, 3, 4, 6, 7, 80, 5, 9BC∪=∪==19.{}{}{}0, 2, 6, 7, 80, 1, 3, 5, 90, 1, 2, 3, 5, 6, 7, 8, 9AB∪=∪=20.{}{}{}0, 1, 3, 5, 90, 2, 5, 7, 8, 90, 5, 9BC∩=∩=21.a.{}2,5b.{}6, 2,5−c.{}16,,1.333...1.3, 2,52−−= −d.{}πe.{}16,,1.333...1.3,, 2,52π−−= −22.a.{}1b.{}0,1c.{}5 , 2.060606...2.06,1.25, 0,13−=d.{}5e.{}5 , 2.060606...2.06,1.25, 0,1,53−=23.a.{}1b.{}0,1c.{}1 1 10,1,,,2 3 4

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 2 preview image

Loading page ...

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 3 preview image

Loading page ...

Chapter R:Review2d.Nonee.{}1 1 10,1,,,2 3 424.a.Noneb.{}1−c.{}1.3,1.2,1.1,1−−−−d.Nonee.{}1.3,1.2,1.1,1−−−−25.a.Noneb.Nonec.Noned.{}12,,21,2ππ++e.{}12,,21,2ππ++26.a.Noneb.Nonec.{}110.32+d.{}2,2π−+e.{}12,2,10.32π−++27.a.18.953b.18.95228.a.25.861b.25.86129.a.28.653b.28.65330.a.99.052b.99.05231.a.0.063b.0.06232.a.0.054b.0.05333.a.9.999b.9.99834.a.1.001b.1.00035.a.0.429b.0.42836.a.0.556b.0.55537.a.34.733b.34.73338.a.16.200b.16.20039.325+=40.5 210⋅=41.23 4x+=⋅42.322y+=+43.312y=+44.24 6x=⋅45.26x−=46.26y−=47.62x=48.26x=49.942527−+=+=50.643235−+=+=51.64 36126−+⋅= −+=52.84 2880−⋅=−=53.458981+−=−=54.834541−−=−=55.1121134333++==56.14132222−−==57.()( )63 52326152161711−⋅+⋅−=−+⋅=−= −

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 4 preview image

Loading page ...

Section R.1:Real Numbers358.()()[][]2834232836328183210320323⋅−⋅+−=⋅−⋅−=⋅−−=⋅ −−= −−= −59.()()2358 2122161416112111⋅−+⋅−=⋅ −+−= −+−=−=60.()()14 3221122211211−⋅−+=−−+=−= −61.()[][][ ]1062 283210645210252107210144−−⋅+−⋅=−−+⋅=−+⋅=−⋅=−= −62.()()[]25 46342206118618612−⋅−⋅−=−−⋅ −= −− −= −+= −63.()()11532122−==64.()()11549333+==65.48126532+==−66.2421532−−== −−67.3 103 2 535215 3 7⋅⋅⋅==⋅⋅25⋅⋅53⋅277=⋅68.535 359 103 3 5 2⋅⋅==⋅⋅⋅3⋅335⋅⋅162=⋅69.6102 3 5 22325275 5 3 9⋅⋅⋅⋅⋅==⋅⋅⋅5⋅25⋅53⋅⋅4459=⋅70.21 1003 7 4 25325325 3⋅⋅⋅⋅==⋅7 425⋅⋅⋅253⋅28=71.3215823452020++==72.4183113266++==73.59255479653030++==74.81516135151921818++==75.511031318123636++==76.28640461594545++==77.1733532163018909045−−== −= −78.3294514214242−−==79.3298120156060−−==80.631215335147070−−== −81.5185275 9 35918119 2 111127⋅⋅⋅=⋅==⋅⋅39⋅15222 11=⋅⋅82.5215355 7 5572127 3 2235⋅⋅⋅=⋅==⋅⋅57⋅2563 2=⋅⋅83.1373737101251010101010+⋅+=+===

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 5 preview image

Loading page ...

Chapter R:Review484.24122 22222235635 3 235 3 2315252102102123515151515154 34 345 35 35⋅⋅+⋅=+=+=+⋅⋅⋅⋅+=⋅+=+==⋅⋅===⋅⋅85.33233636232 4814848428123123158888⋅+=⋅+=+=⋅++=+==86.513513 513 513 621623 223 225151422222⋅⋅⋅−=⋅−=−=−⋅⋅−=−===87.()64624xx+=+88.()4 2184xx−=−89.()244x xxx−=−90.()243412x xxx+=+91.31312 3222242422 222 32312222xxxxx⋅−=⋅−⋅=−⋅⋅=−=−⋅92.21213 23333363633 23 231233 22xxxxx⋅+=⋅+⋅=+⋅⋅=+=+⋅93.()()222442868xxxxxxx++=+++=++94.()()22515565xxxxxxx++=+++=++95.()()2221222xxxxxxx−+=+−−=−−96.()()22414434xxxxxxx−+=+−−=−−97.()()228228161016xxxxxxx−−=−−+=−+98.()()224224868xxxxxxx−−=−−+=−+99.()()23232355xxxxxxx+=⋅+⋅=+⋅=⋅=100.23 421214+⋅=+=since multiplication comes before addition in theorder of operations for real numbers.()2345 420+⋅=⋅=since operations inside parentheses come beforemultiplication in the order of operations for realnumbers.101.()()2 3 42 1224⋅==() ()() ( )2 32 46848⋅⋅⋅==102.4371257+==+, but434 53 220626132.6251010105⋅+⋅++=====103.Subtraction is not commutative; forexample:231132−= −≠=−.104.Subtraction is not associative; forexample:()()52124521−−=≠=−−.105.Division is not commutative; for example:2332≠.106.Division is not associative; forexample:()1222623÷÷=÷=, but()122212112÷÷=÷=.107.The Symmetric Property implies that if 2 =x,thenx= 2.

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 6 preview image

Loading page ...

Section R.2:Algebra Essentials5108.From theprinciple of substitution,if5x=, then()()( ) ( )222552525530xxxxxxx==+=++=109.There are no real numbers that are both rationaland irrational, since an irrational number, bydefinition, is a number that cannot be expressedas the ratio of two integers; that is, not a rationalnumberEvery real number is either a rational number oran irrational number, since the decimal form of areal number either involves an infinitelyrepeating pattern of digits or an infinite, non-repeating string of digits.110.The sum of an irrational number and a rationalnumber must be irrational. Otherwise, theirrational number would then be the difference oftwo rational numbers, and therefore would haveto be rational.111.Answers will vary.112.Since 1 day = 24 hours, we compute12997541.541624=.Now we only need to consider the decimal partof the answer in terms of a 24 hour day. That is,()()0.54162413≈hours. So it must be 13 hourslater than 12 noon, which makes the time 1 AMCST.113.Answers will vary.Section R.21.variable2.origin3.strict4.base; exponent (or power)5.31.234567810×6.True.7.True8.False; the absolute value of a real number isnonnegative.00=which is not a positivenumber.9.False; a number in scientific notation isexpressed as the product of a number, x,110x≤<or101x−<≤ −, and a power of 10.10.False; to multiply two expressions with the samebase, retain the base andaddthe exponents.11.−2.5−152100.253412.1332−22−1.502313.102>14.56<15.12−> −16.532−< −17.3.14π >18.21.41>19.10.52=20.10.333>21.20.673<22.10.254=23.0x>

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 7 preview image

Loading page ...

Chapter R:Review624.0z<25.2x<26.5y> −27.1x≤28.2x≥29.Graph on the number line:2x≥ −0−230.Graph on the number line:4x<4031.Graph on the number line:1x> −−1032.Graph on the number line:7x≤0733.(,)(0,1)1011d C Dd==−==34.(,)(0,3)3033d C Ad=−=−−=−=35.(,)(1,3)3122d D Ed==−==36.(,)(0,3)3033d C Ed==−==37.(,)( 3,3)3( 3)66d A Ed=−=− −==38.(,)(1,1)1122d D Bd=−=−−=−=39.222 3264xy+= −+⋅= −+=40.33(2)3633xy+=−+= −+= −41.525(2)(3)230228xy+=−+= −+= −42.22(2)(2)(3)462xxy−+= −−+ −=−= −43.2(2)4242355xxy−−===−−−−44.23112355xyxy−++=== −−−−−45.3(2)2(3)66320022355xyy−+−++====++46.2(2)343237333xy−−−−−=== −47.3(2)11xy+=+ −==48.3(2)55xy−=− −==49.32325xy+=+ −=+=50.32321xy−=− −=−=51.33133xx===52.22122yy−=== −−−53.454(3)5(2)12102222xy−=−−=+==54.323(3)2(2)9455xy+=+−=−==55.454(3)5(2)1210121022xy−=−−=− −=−==56.323 3223 32 29413xy+=+−=⋅+⋅=+=57.21xx−Part (c) must be excluded. The value0x=mustbe excluded from the domain because it causesdivision by 0.

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 8 preview image

Loading page ...

Section R.2:Algebra Essentials758.21xx+Part (c) must be excluded. The value0x=mustbe excluded from the domain because it causesdivision by 0.59.2(3)(3)9xxxxx=−+−Part (a) must be excluded. The values3 and3xx= −=must be excluded from thedomain because they cause division by 0.60.29xx+None of the given values are excluded. Thedomain is all real numbers.61.221xx+None of the given values are excluded. Thedomain is all real numbers.62.332(1)(1)1xxxxx=−+−Parts (b) and (d) must be excluded. The values1, and1xx== −must be excluded from thedomain because they cause division by 0.63.223510510(1)(1)xxxxx xxxx+−+−=−+−Parts (b), (c), and (d) must be excluded. Thevalues0,1, and1xxx=== −must be excludedfrom the domain because they cause division by0.64.22329191(1)xxxxxxx x−−+−−+=++Part (c) must be excluded. The value0x=mustbe excluded from the domain because it causesdivision by 0.65.45x−5x=must be exluded because it makes thedenominator equal 0.{}Domain5x x=≠66.64x−+4x= −must be excluded sine it makes thedenominator equal 0.{}Domain4x x=≠ −67.4xx+4x= −must be excluded sine it makes thedenominator equal 0.{}Domain4x x=≠ −68.26xx−−6x=must be excluded sine it makes thedenominator equal 0.{}Domain6x x=≠69.555(32)(3232)(0)0 C999CF=−=−==°70.555(32)(21232)(180)100 C999CF=−=−==°71.555(32)(7732)(45)25 C999CF=−=−==°72.55(32)(432)995 ( 36)920 CCF=−=−−=−= −°73.2(4)(4)(4)16−= −−=74.224(4)16−= −= −75.22114164−==76.22114164−−= −= −77.64642211333393−− +−⋅====78.2323144444−− +⋅===79.()()()121223339−−−−===

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 9 preview image

Loading page ...

Chapter R:Review880.()()()313132228−−−−===81.22555==82.23666==83.()2444−= −=84.()2333−= −=85.()()2232368864xxx==86.()122211444xxx−−== −−87.()()()42222121422xx yxyx yy−−−=⋅==88.()()333113333yxyxyxyx−−−=⋅==89.232 134114x yxxyx yyxy−−−===90.22 11 231231xyxyxyx yx y−− −−−−===91.3424222334 1232 13113(2)()839898989xy zx y zx y zx y zxyzx yzx zy−−−−−−=−=−== −92.21211344241 11621624()428481212xy zxyzx yx yxyzxyzx y z−−−−−− −− −−−−−====93.2221222122233441643439yxxxxxyyyy−−−−====94.()()3332222223326363255666562161255yxxyxyxxyy−−−−====95.()()12 22241xxyy−=== −−96.()()13133322yxyx−−−−−===97.()()222221415xy+=+ −=+=98.() ()2222214 14x y=−=⋅=99.()()()()2222124xy=⋅ −=−=100.()()()( )2222111xy+=+ −==101.222xx===102.()22xx==103.()()222221415xy+=+ −=+=104.2221213xyxy+=+=+ −=+=105.1122yx−==106.()211xy=−=

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 10 preview image

Loading page ...

Section R.2:Algebra Essentials9107.If2,x=323223542 23 25 24161210410xxx−+−=⋅−⋅+⋅−=−+−=If1,x=323223542 13 15 1423540xxx−+−=⋅−⋅+⋅−=−+−=108.If1,x=32324324 13 11243128xxx+−+=⋅+⋅−+=+−+=If2,x=32324324 23 22232122244xxx+−+=⋅+⋅−+=+−+=109.4444(666)666381222(222)===110.()333333331(0.1) (20)2 101012101028=⋅⋅=⋅⋅==111.6(8.2)304, 006.671≈112.5(3.7)693.440≈113.3(6.1)0.004−≈114.5(2.2)0.019−≈115.6(2.8)481.890−≈116.6(2.8)481.890−≈ −117.4(8.11)0.000−−≈118.4(8.11)0.000−−≈ −119.2454.24.54210=×120.132.143.21410=×121.20.0131.310−=×122.30.004214.2110−=×123.432,1553.215510=×124.421, 2102.12110=×125.40.0004234.2310−=×126.20.05145.1410−=×127.46.151061,500×=128.39.7109700×=129.31.214100.001214−×=130.49.88100.000988−×=131.81.110110, 000, 000×=132.24.11210411.2×=133.28.1100.081−×=134.16.453100.6453−×=135.Alw=136.()2Plw=+137.Cdπ=138.12Abh=139.234Ax=140.3Px=141.343Vrπ=142.24Srπ=

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 11 preview image

Loading page ...

Chapter R:Review10143.3Vx=144.26Sx=145.a.If1000,x=4000240002(1000)40002000$6000Cx=+=+=+=The cost of producing 1000 watches is$6000.b.If2000,x=4000240002(2000)40004000$8000Cx=+=+=+=The cost of producing 2000 watches is$8000.146.210801202560325$98+−+−−−=His balance at the end of the month was $98.147.We want the difference betweenxand 4 to be atleast 6 units. Since we don’t care whether thevalue forxis larger or smaller than 4, we takethe absolute value of the difference. We want theinequality to be non-strict since we are dealingwith an ‘at least’ situation. Thus, we have46x−≥148.We want the difference betweenxand 2 to bemore than 5 units. Since we don’t care whetherthe value forxis larger or smaller than 2, wetake the absolute value of the difference. Wewant the inequality to be strict since we aredealing with a ‘more than’ situation. Thus, wehave25x−>149.a.110108110225x−=−=−=≤108 volts is acceptable.b.110104110665x−=−=−=>104 volts isnotacceptable.150.a.220214220668x−=−=−=≤214 volts is acceptable.b.22020922011118x−=−=−=>209 volts isnotacceptable.151.a.32.99930.0010.0010.01x−=−=−=≤A radius of 2.999 centimeters is acceptable.b.32.8930.110.110.01x−=−=−=≤/A radius of 2.89 centimeters isnotacceptable.152.a.98.69798.61.61.61.5x−=−=−=≥97˚F is unhealthy.b.98.610098.61.41.41.5x−=−==<100˚F isnotunhealthy.153.The distance from Earth to the Moon is about8410400, 000, 000×=meters.154.The height of Mt. Everest is about388488.84810=×meters.155.The wavelength of visible light is about75100.0000005−×=meters.156.The diameter of an atom is about101100.0000000001−×=meters.157.The smallest commercial copper wire has adiameter of about40.0005510−=×inches.158.The smallest motor ever made is less than20.05510−=×centimeters wide.

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 12 preview image

Loading page ...

Section R.3:Geometry Essentials11159.()() ()()511221.86106102.4103.6510186, 000 60 60 24 365××××⋅⋅⋅⋅=1012586.5696105.86569610=×=×There are about125.910×miles in one light-year.160.72593, 000, 0009.310510186, 0001.8610500 seconds8 min. 20 sec.×==××=≈It takes about 8 minutes 20 seconds for a beamof light to reach Earth from the Sun.161.10.333333 ...0.3333=>13is larger by approximately0.0003333 ...162.230.666666 ...0.666=>23is larger by approximately 0.0006666 ...163.No. For any positive numbera, the value2aissmaller and therefore closer to 0.164.We are given that2110x<<. This implies that110x<<. Since103.162x<≈and3.142xπ>≈, the number could be 3.15 or 3.16(which are between 1 and 10 as required). Thenumber could also be 3.14 since numbers such as3.146 which lie betweenπand10wouldequal 3.14 when truncated to two decimal places.165.Answers will vary.166.Answers will vary.5 < 8 is a true statement because 5 is further tothe left than 8 on a real number line.Section R.31.right; hypotenuse2.12Abh=3.2Crπ=4.similar5.True.6.True.22268366410010+=+==7.False; the volume of a sphere of radiusris givenby343Vrπ=.8.True. The lengths of the corresponding sides areequal.9.True. Two corresponding angles are equal.10.False. The sides are not proportional.11.222225,12,5122514416913abcabc===+=+=+==12.222226,8,68366410010abcabc===+=+=+==13.2222210,24,102410057667626abcabc===+=+=+==14.222224,3,43169255abcabc===+=+=+==15.222227,24,7244957662525abcabc===+=+=+==

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 13 preview image

Loading page ...

Chapter R:Review1216.2222214,48,14481962304250050abcabc===+=+=+==17.222534259162525=+=+=The given triangle is a right triangle. Thehypotenuse is 5.18.22210681003664100100=+=+=The given triangle is a right triangle. Thehypotenuse is 10.19.2226453616253641 false=+=+=The given triangle is not a right triangle.20.22232294498 false=+=+=The given triangle is not a right triangle.21.2222572462549576625625=+=+=The given triangle is a right triangle. Thehypotenuse is 25.22.222261024676100576676676=+=+=The given triangle is a right triangle. Thehypotenuse is 26.23.222634369163625 false=+=+=The given triangle is not a right triangle.24.2227544925164941 false=+=+=The given triangle is not a right triangle.25.24 28 inAl w=⋅=⋅=26.29 436 cmAl w=⋅=⋅=27.211 (2)(4)4 in22Ab h=⋅==28.211 (4)(9)18 cm22Ab h=⋅==29.222(5)25m22(5)10mArCr= π= π=π=π=π=π30.222(2)4ft22(2)4ftArCr= π= π=π=π=π=π31.38 4 7224 ftVl w h==⋅⋅=( ) ()( )()()()22222 842 872 476411256232 ftSlwlhwh=++=++=++=32.39 4 8288 inVl w h==⋅⋅=()()()( )()( )22222 942 982 487214464280 inSlwlhwh=++=++=++=33.333222442564cm33344464cmVrSr=π=π⋅=π=π=π ⋅=π34.33322244336f3344336ftVrtSr=π=π⋅=π=π=π ⋅=π

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 14 preview image

Loading page ...

Section R.3:Geometry Essentials1335.223(9) (8)648inVr h= π= π=π()()( )2222229298162144306inSrrhπππππππ=+=+=+=36.223(8) (9)576inVr h= π= π=π( )( ) ()2222228289128144272inSrrhπππππππ=+=+=+=37.The diameter of the circle is 2, so its radius is 1.22(1)square unitsAr= π= π= π38.The diameter of the circle is 2, so its radius is 1.222(1)4square unitsA=− π=− π39.The diameter of the circle is the length of thediagonal of the square.2222244882222222dddr=+=+======The area of the circle is:()2222square unitsAr= π= π=π40.The diameter of the circle is the length of thediagonal of the square.2222244882222222dddr=+=+======The area is:()222224 square unitsA= π−=π −41.Since the triangles are similar, the lengths ofcorresponding sides are proportional. Therefore,we get8428 244xxx=⋅==In addition, corresponding angles must have thesame angle measure. Therefore, we have90A=°,60B=°, and30C=°.42.Since the triangles are similar, the lengths ofcorresponding sides are proportional. Therefore,we get612166 16128xxx=⋅==In addition, corresponding angles must have thesame angle measure. Therefore, we have30A=°,75B=°, and75C=°.43.Since the triangles are similar, the lengths ofcorresponding sides are proportional. Therefore,we get30204530 4520135or67.52xxxx=⋅===In addition, corresponding angles must have thesame angle measure. Therefore, we have60A=°,95B=°, and25C=°.44.Since the triangles are similar, the lengths ofcorresponding sides are proportional. Therefore,we get810508 501040xxx=⋅==In addition, corresponding angles must have thesame angle measure. Therefore, we have50A=°,125B=°, and5C=°.

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 15 preview image

Loading page ...

Chapter R:Review1445.The total distance traveled is 4 times thecircumference of the wheel.Total Distance44()41664201.1 inches16.8 feetCd==π=π ⋅=π ≈≈46.The distance traveled in one revolution is thecircumference of the disk4π.The number of revolutions =dist. traveled2051.6 revolutionscircumference4==≈ππ47.Area of the border = area of EFGH – area ofABCD2221061003664 ft=−=−=48.FG = 4 feet; BG = 4 feet and BC = 10 feet, soCG= 6 feet. The area of the triangle CGF is:21 (4)(6)12 ft2A=⋅=49.Area of the window = area of the rectangle +area of the semicircle.221(6)(4)224230.28 ft2A=+⋅ π⋅=+π ≈Perimeter of the window = 2 heights + width +one-half the circumference.12(6)4(4)1242216222.28 feetP=++⋅ π=++π=+π ≈50.Area of the deck = area of the pool and deck –area of the pool.2222(13)(10)16910069ft216.77 ftA= π− π=π −π=π≈The amount of fence is the circumference of thecircle with radius 13 feet.2 (13)26ft81.68 ftC=π=π≈51.We can form similar triangles using the GreatPyramid’s height/shadow and Thales’height/shadow:h126114240{{23This allows us to write224032 2401603hh=⋅==The height of the Great Pyramid is 160 paces.52.Letx= the approximate distance from San Juanto Hamilton andy= the approximate distancefrom Hamilton to Fort Lauderdale. Using similartriangles, we get10465853.51046 53.558964.8xxx=⋅=≈104658571046 57581028.0yyy=⋅=≈The approximate distance between San Juan andHamilton is 965 miles and the approximatedistance between Hamilton and Fort Lauderdaleis 1028 miles.53.Convert 20 feet to miles, and solve thePythagorean Theorem to find the distance:222sq. miles1 mile20 feet20 feet0.003788 miles5280 feet(39600.003788)3960305.477 milesdd=⋅==+−=≈3960396020 ftd54.Convert 6 feet to miles, and solve thePythagorean Theorem to find the distance:222sq. miles1 mile6 feet6 feet0.001136 miles5280 feet(39600.001136)396093 milesdd=⋅==+−=≈396039606 ftd

Solution Manual For Algebra And Trigonometry, 9th Edition - Page 16 preview image

Loading page ...

Section R.3:Geometry Essentials1555.Convert 100 feet to miles, and solve thePythagorean Theorem to find the distance:1 mile100 feet100 feet0.018939 miles5280 feet=⋅=222sq. miles(39600.018939)396015012.2 milesdd=+−≈≈Convert 150 feet to miles, and solve thePythagorean Theorem to find the distance:1 mile150 feet150 feet0.028409 miles5280 feet=⋅=222sq. miles(39600.028409)396022515.0 milesdd=+−≈≈56.Given0,0 andmnmn>>>,if2222,2andamnbmncmn=−==+, then()()22222242242242242242abmnmnmm nnm nmm nn+=−+=−++=++and()222242242cmnmm nn=+=++222,andabca bc∴+=→represent the sidesof a right triangle.57.Letl =length of the rectangleandw= width of the rectangle.Notice that22()()[()()][()()](2 )(2)44lwlwlwlwlwlwlwlwA+−−=++−+−−===2214So[()() ]Alwlw=+−−Since2()0lw−≥, the largest area will occurwhenl – w= 0 orl = w; that is, when therectangle is a square. But1000222()5002250lwlwlwllw=+=+=+===The largest possible area is225062500=sq ft.A circular pool with circumference = 1000 feetyields the equation:50021000rrππ==The area enclosed by the circular pool is:222250050079577.47 ftArππππ===≈Thus, a circular pool will enclose the most area.58.Consider the diagram showing the lighthouse atpoint L, relative to the center of Earth, using theradius of Earth as 3960 miles. Let P refer to thefurthest point on the horizon from which thelight is visible. Note also that362362 feetmiles.5280=Apply the Pythagorean Theorem toCPLΔ:()()2221362396039605280d+=+()()()()()222122136252803625280396039603960396023.30 mi.dd=+−=+−≈Therefore, the light from the lighthouse can beseen at point P on the horizon, where point P isapproximately 23.30 miles away from thelighthouse. Brochure information is slightlyoverstated.Verify the ship information:Let S refer to the ship’s location, and letxequalthe height, in feet, of the ship.We need1240dd+≥.Since123.30 milesd≈we need24023.30=16.70 miles.d≥−

Preview Mode

This document has 1481 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

Test Bank for Algebra And Trigonometry, 11th Edition

Precalculus - Polynomial and Rational Functions

Precalculus - Functions

Precalculus - Exponential and Logarithmic Functions

Math Word Problems - Translating Expressions

Math Word Problems - Equations

Algebra II – Word Problems

Algebra II – Sequences and Series

Algebra II - Segments Lines and Inequalities

Algebra II - Relations and Functions

Company

Explore

Study Tools