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A new car is purchased for 15600 dollars. The value of the car depreciates at
14.5% per year. What will the value of the car be, to the nearest cent, after 10 years?
Answer:

A new car is purchased for $15600$ dollars. The value of the car depreciates at $14.5 \%$ per year. What will the value of the car be, to the nearest cent, after $10$ years? $\newline$ Answer:

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Exponential growth and decay: word problems

Full solution

Q. A new car is purchased for

15600

dollars. The value of the car depreciates at

14.5 \%

per year. What will the value of the car be, to the nearest cent, after

10

years?

\newline

Answer:

Identify values: Identify the initial value of the car and the annual depreciation rate. $\newline$ The initial value of the car, $P$ , is $\$15,600$ , and the annual depreciation rate, $r$ , is $14.5\%$ .
Convert to decimal: Convert the annual depreciation rate from a percentage to a decimal. $\newline$ To convert a percentage to a decimal, divide by $100$ . $\newline$ $r = \frac{14.5\%}{100} = 0.145$
Determine factor: Determine the depreciation factor. $\newline$ The depreciation factor is what you multiply the car's value by each year to find the next year's value. Since the car is losing value, the factor will be less than $1$ . $\newline$ Depreciation factor $= 1 - r = 1 - 0.145 = 0.855$
Apply over $10$ years: Apply the depreciation factor over $10$ years. $\newline$ The value of the car after $t$ years can be found using the formula $V(t) = P \times (\text{depreciation factor})^t$ . $\newline$ Here, $t = 10$ years. $\newline$ $V(10) = 15600 \times (0.855)^{10}$
Calculate after $10$ years: Calculate the value of the car after $10$ years. $\newline$ Using a calculator, raise $0.855$ to the $10^{\text{th}}$ power and then multiply by $15,600$ . $\newline$ $V(10) = 15600 \times (0.855)^{10} \approx 15600 \times 0.196874 \approx 3071.3344$
Round final value: Round the final value to the nearest cent. $\newline$ The value of the car after $10$ years, rounded to the nearest cent, is approximately $\$3,071.33$ .

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