The table of the value of y 0 to y 6 if the difference equation is y n = − 1.05 y n − 1 + 1.25 with initial value y 0 = 10 .

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Answer 1

Question

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

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The position of a particle that moves along the x-axis is defined by x = - 3t^2 + 12^t - 6 f, where t is in seconds. For the time interval t = 0 to t = 3 s, (1) plot the position, velocity, and acceleration as functions of time; (2) calculate the distance traveled; and (3) determine the displacement of the particleshow the graph and write the solution with a pen

Answer 2

Question

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

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Answer 3

Question

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

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Answer 4

Question

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

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Answer 5

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

Answer 6

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

Answer 7

Chapter 12.1, Problem 34E

(a)

To determine

The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

Answer 8

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

Answer 9

(b)

To determine

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

Answer 10

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

Answer 11

(c)

To determine

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

Answer 12

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Answer 13

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Answer 14

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Answer 15

Ch. 12.1 - Consider the difference equation...Ch. 12.1 - Prob. 2CYU Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E Ch. 12.1 - Prob. 4E Ch. 12.1 - Prob. 5E Ch. 12.1 - For each of the difference equations in Exercises...Ch. 12.1 - Prob. 7E Ch. 12.1 - Prob. 8E

The table of the value of y 0 to y 6 if the difference equation is y n = − 1.05 y n − 1 + 1.25 with initial value y 0 = 10 .

Solution Summary: The author explains how to calculate to the value of y_0to

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The table of the value of y0 to y6 if the difference equation is yn=−1.05yn−1+1.25 with initial value y0=10.

To graph: The terms of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10.

The value of y6 with the help of graph or table of the difference equation yn=−1.05yn−1+1.25 with initial value y0=10 and also determine the value of n when yn approximately 14.4.

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Chapter 12 Solutions