IAT SOLUTIONS - C_7.pdf

Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.46.y47.y54. Let m 5 }ny 5 2.5e 20.5x 1 2r , so n 5 rm and }r n 5 }1 m .y 5 0.6e x9( 0,2 )A 5 P(1, 1.63)11 1 }n2 r 5 P 11 1 }m2 1 5 P F 11 1 }m2 1 G rt(1, 3.52)As n approaches 1`, m approaches 1`, and3(3, 1.63)5 (1, 1.52)( 0,2 )( 0,5 )11 1 }213 xm2 1 approximates e.1 y 5 2.5e 20.5x21x( 2,5 )21So, A 5 P 11 1 }y 5 0.6e n2 r approximates A 5 Pe rt as nx 2 230f (x)}g (x) 5 }2e23xe 5 24x 2e23x 2 (24x) 5 2e x 2010Domain: all real numbers Domain: all real numbers approaches 1`.Range: y > 2 Range: y > 055. When x 5 2002 2 1997 5 5: y 5 1.28e 1.31x48.1f(x) Domain: all real numbers y 5 1.28e 1.31(5) 5 1.28e 6.55 ø 895f(x) 5 e x 1 3 2 22Range: y > 22About 895 million camera phones were shipped globally1f(x) 5 e xin 2002.2256. When t 5 1999 2 1989 5 10: y 5 738e 0.345t(1, 1.36)y 5 738e 0.345(10) 5 738e 3.45 ø 23,2471 3 x( 0,2 )About 23,247 termites were collected in 1999.3( 23, 22 ) (22, 20.64)57. When P 5 2000, r 5 0.04, and t 5 5: A 5 Pe rtA 5 2000e (0.04 p 5) 5 2000e 0.2 ø 2442.8149.g(x) Domain: all real numbersThe balance after 5 years is $2442.81.(2, 4.62)Range: y > 1(1, 3.62)58. When P 5 800, r 5 0.0265 and t 5 12.5: A 5 Pe rtA 5 800e4g(x) 5 e x 2 1 1 1 73( 1,3 )5 800e 0.33125 ø 1114.17The balance after 12.5 years is $1114.17.44 3 x59. a. When k 5 20.02: L(x) 5 100eg(x) 5 e x 213 ( 0,3 )→ L(x) 5 100e 20.02xL(x)1009050.h(x)Domain: all real numbers80570Range: y > 2360(1, 0.14)50(0, 1)40h(x) 5 e 22x30(21, 22)(0, 22.86)005 x201020 40 60 80 xh(x) 5 e 22(x 1 1) 2 3Depth below water surface (m)51. Using the table feature, you can notice that incrementingsmall values of n increases the function very slowly.b. Using the graph, you can estimate that the percent ofsurface light is about 45% at a depth of 40 meters.Incrementing n by powers of 10 gives values that are onec. Using the graph, you can estimate that the submarinedigit closer to the actual value of e each time. Whenn 5 10 10 can descend about 35 meters before only 50% of, the value of the function is 2.718281828,surface light is available.which is the value of e correct to 9 decimal places.60. a. P(t) 5 P52. No, e cannot be expressed as a ratio of two integers0e 0.116t ; when P 05 30: P(t) 5 30e 0.116tbecause it is an irrational number; it is a decimal thatb. P(t)90neither terminates nor repeats.807053. Choose a, b, r, and q, such that a > 0, b > 0, r < 0, q < 0,60and r 2 q > 0.50Sample answer: f (x) 5 2e 23x , g (x) 5 e 24x4000 2 4 6 8 tHours after 1:00 P.M.Percent of lightPopulationc. Using the graph, you can estimate that the bacteriapopulation is about 48 at 5:00 p.m.Algebra 2Worked-Out Solution Key429