math 1160 hw 11.2 & 11.3

# math 1160 hw 11.2 & 11.3 - ‘9 Math 1 16 Lecture...

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Unformatted text preview: ‘9 Math 1 16 Lecture - Homework #8 Answer Key Section 11.2 2. yn = 50 — 2n 6. yzu = 800.02)” 8. y“) = 800.06)” 12a. yu =10, yl = 10, y; = 10, y; =10 Points (0. 10) (l, 10) (2. 10') (3. 10) 12b. yo: 18,371 = l4,y2=12,y3= ll Points (0, 18;) (1, 14) (2, 12) (3, 11.) 12c. y.3=2,y1=6,y2=8.}’3=9 l4. yn = 6 + 4(3)n 16. yn = 2 +1(-.7)n Points (0. 2) (l, 6') (2. 8) (3, 9‘) (a = 3 '9 monotonic, unbounded) As 11 gets larger, yn gets larger. ( a = —.7 9 oscillates, attracted) As 11 gets larger. yII approaches 2. Section 11.3 10. )2. l4. 16. 18. 20. 22. a = .98 (monotonic, attracted to y = 0) 24. a = 1.4 (monotonic, repelled by y = 0') 26. Solve b/(l-a) to get the maximum loan amount. The graph of a loan is monotonic (decreasing balance over time) and is repelled from the line (loan amount maximum). Difference equation 9 yu = 1.02yn-1 A 100,000 a = 1.02 and b = -100,000 So b/(a-l) = -100_.000/(l — 1.02) = -100,000/—.02 = 5,000,000 **The loan amount must be less than \$5,000,000. ...
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