hw11-1 - from the line y = 0. 12c. y n = (1/2)(-2) n In...

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Math 1160 – Section 11.1 Answer Key 2. a = -3, b = 16, b/(1-a) = 16/(1 - -3) = 16/4 = 4 4. a = 1/3, b = 4, b/(1-a) = 4/(1 – 1/3) = 4/(2/3) = 6 6. a = .5, b = -4, b/(1-a) = -4/(1 - .5) = -4/.5 = -8 8a. y 0 = 2, y 1 = 6, y 2 = 8, y 3 = 9, y 4 = 9.5 8b. Points (0, 2) (1, 6) (2, 8) (3, 9) (4, 9.5) **The value of n becomes your x coordinate. The value of y n becomes your y coordinate. 8c. y n = 10 + (-8)(.5) n 10a. y 0 = 8, y 1 = 8, y 2 = 8, y 3 = 8, y 4 = 8 10b. Points (0, 8) (1, 8) (2, 8) (3, 8) (4, 8) **The value of n becomes your x coordinate. The value of y n becomes your y coordinate. 10c. y n = 8 (constant graph) 12a. y 0 = 1/2, y 1 = -1, y 2 = 2, y 3 = -4, y 4 = 8 12b. Points (0, 1/2) (1, -1) (2, 2) (3, -4) (4, 8) **Notice the points are oscillating and being repelled from
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Unformatted text preview: from the line y = 0. 12c. y n = (1/2)(-2) n In this equation, b/(1-a) = 0 20. y n = 1.05y n-1 + 100 , y =1000 25a. y n = 1.04y n-1 + 250 , y =800 25b. y n = 6250 + 7050(1.04) n 25c. y 7 = 6250 + 7050(1.04) 7 & $3027.32 26. Difference Equation is y n = 1 . 10 y n-1 100 *Each value of n is a new compounding period. In this situation, n represents years. To calculate your new balance on the loan (this year), take the old balance (last year), add the interest (compounding period interest times prior balance), and subtract the payment. y 1 = 1.10(y ) 100 & 1.10(317) 100 & 248.70 y 2 = 1.10(y 1 ) 100 & 1.10(248.70) 100 & 173.57 27a. y n = 0.85y n-1 , y =20000 27b. y n = (20000)(0.85) n 27c. y 5 = (20000)(0.85) 5 & $8874.11...
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This note was uploaded on 05/21/2008 for the course MATH 116 taught by Professor Copeland during the Spring '08 term at Western Michigan.

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hw11-1 - from the line y = 0. 12c. y n = (1/2)(-2) n In...

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