We need to solve two problems: Question 4 and Question 5. Question 4 concerns the depreciation of a laptop, where $V_n$ is the value of the laptop after $n$ years. We are given that $V_0 = 2800$ and $V_{n+1} = 0.91V_n$. Part (a) asks for the rate of depreciation each year. Part (b) asks after how many years the laptop will be worth less than $1000. Question 5 concerns the depreciation of a machine. The machine is purchased for $36,000 and depreciates at a rate of $0.16 for each unit it produces. The machine produces 24,000 units per year on average. Part (a) asks for the annual depreciation amount. Part (b) asks us to use the rule $V_n = 36000 - 0.16n$ to write out a recurrence relation. Part (c) asks for the value of the machine after six years of use.

1. Problem Description

We need to solve two problems: Question 4 and Question

5.

Question 4 concerns the depreciation of a laptop, where $V_n$ is the value of the laptop after $n$ years. We are given that $V_0 = 2800$ and $V_{n+1} = 0.91V_n$. Part (a) asks for the rate of depreciation each year. Part (b) asks after how many years the laptop will be worth less than $

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Question 5 concerns the depreciation of a machine. The machine is purchased for $36,000 and depreciates at a rate of$0.16 for each unit it produces. The machine produces 24,000 units per year on average. Part (a) asks for the annual depreciation amount. Part (b) asks us to use the rule $V_n = 36000 - 0.16n$ to write out a recurrence relation. Part (c) asks for the value of the machine after six years of use.

2. Solution Steps

Question 4:

(a) The recurrence relation $V_{n+1} = 0.91V_n$ means that the value in year $n+1$ is 91% of the value in year $n$. This means that the depreciation rate is $1 - 0.91 = 0.09$, or 9%.

(b) We want to find $n$ such that $V_n < 1000$. We know $V_0 = 2800$ and $V_n = 2800(0.91)^n$.

So we want to solve $2800(0.91)^n < 1000$.

Dividing both sides by 2800, we get $(0.91)^n < \frac{1000}{2800} = \frac{10}{28} = \frac{5}{14}$.

Taking the natural logarithm of both sides, we get $n \ln(0.91) < \ln(\frac{5}{14})$.

Since $\ln(0.91) < 0$, we divide both sides by $\ln(0.91)$ and reverse the inequality:

$n > \frac{\ln(\frac{5}{14})}{\ln(0.91)} \approx \frac{-1.0296}{-0.0943} \approx 10.91$.

Since $n$ must be an integer, the smallest such $n$ is

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1. Therefore, after 11 years, the laptop will be worth less than $

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Question 5:

(a) The machine depreciates at a rate of $0.16 per unit, and produces 24,000 units per year. So the annual depreciation amount is$0.16 \times 24000 = 3840$.

(b) Let $V_n$ be the value of the machine after $n$ units produced. Then $V_0 = 36000$. The recurrence relation is given by $V_{n+1} = V_n - 0.16$.

(c) The value of the machine after six years of use. Since the machine produces 24,000 units per year, after six years it will have produced $6 \times 24000 = 144000$ units. We are given $V_n = 36000 - 0.16n$, so $V_{144000} = 36000 - 0.16 \times 144000 = 36000 - 23040 = 12960$.

3. Final Answer

Question 4:

(a) The rate of depreciation is 9%.

(b) After 11 years, the laptop will be worth less than $

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Question 5:

(a) The annual depreciation amount is $

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8

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0. (b) The recurrence relation is $V_{n+1} = V_n - 0.16$, where $V_0 = 36000$.

(c) The value of the machine after six years is $

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