IAT SOLUTIONS - C_7.pdf

Chapter 7, continuedCopyright © by McDougal Littell, a division of Houghton Mifflin Company.11. x 1 2 3 4 512.13.14.ln y 2.89 3.58 4.28 4.97 5.66ln y(4, 4.97)(2, 3.58)121x4.97 2 3.58M 5 } ø 0.704 2 2ln y 5 0.70x 1 2.180.70x 1 2.18y 5 ey 5 e 2.18 (e 0.7 ) xy 5 8.85(2.01) xx 1 2 3 4 5ln y 1.19 2.31 3.42 4.53 5.64ln y(4, 4.53)(2, 2.31)121x4.53 2 2.31M 5 } 5 1.114 2 2ln y 5 1.11x 1 0.091.11x 1 0.09y 5 ey 5 e 0.09 (e 1.11 ) xy 5 1.09(3.03) xx 1 2 3 4 5ln y 2.28 2.50 2.72 2.94 3.17ln y(5, 3.17)(2, 2.50)121x3.17 2 2.50M 5 } ø 0.225 2 2ln y 5 0.22x 1 2.060.22x 1 2.06y 5 ey 5 e 2.06 (e 0.22 ) xln y 2 3.58 5 0.70(x 2 2)ln y 2 2.31 5 1.11(x 2 2)ln y 2 2.50 5 0.22(x 2 2)x 1 2 3 4 5ln y 0.34 1.90 3.49 5.08 6.67211ln y(1, 0.34)(4, 5.08)5.08 2 0.34M 5 } 5 1.584 2 1ln y 2 0.34 5 1.58(x 2 1)ln y 5 1.58x 2 1.241.58x 2 1.24y 5 ey 5 e 21.24 (e 1.58 ) xy 5 0.289(4.85) x15. (4, 3): 3 5 a p 4 b → a 5 3 }4 b(8, 15): 15 5 a p 8 b15 5 1 3 }4 b 2 p 8b15 5 3 p 2 b5 5 2 blog 25 5 blog 5}log 2 5 b2.32 ø ba 5 }34 ø 3b } ø 0.122.324So, an equation is y 5 0.12x 2.32 .16. (5, 9): 9 5 a p 5 b → a 5 9 }5 b(8, 34): 34 5 a p 8 b34 5 1 9 }5 b 2 8b34 5 9 p 1 }8 52 b34}9 5 1 }8 52 blog 8/534}9 5 blog }34 9}log }8 55 b2.83 ø ba 5 }95 ø 9b } ø 0.092.835So, an equation is y 5 0.09x 2.83 .xAlgebra 2Worked-Out Solution Key453