Cecily bought a car that is depreciating at a rate of 18% each year. The value V of her car after t years can be modeled by the function V(t) = 65000(0.82)t. She wants to track the value of her car each year, for 5 years.

Question

What would be a possible range for this situation?
A) 0 < v ≤ 5
B) All real numbers
C) 0 < v ≤ 65000
D) 30 < v ≤ 65000

Answer 1

To analyze the situation, let's consider the function provided:

\[ V(t) = 65000(0.82)^t \]

This function models the value of Cecily's car over time, where:

• \( V(t) \) is the value of the car after \( t \) years,
• 65000 is the initial value of the car,
• \( 0.82 \) represents the depreciation rate (100% - 18% = 82%).

As \( t \) increases, \( V(t) \) will decrease because \( (0.82)^t \) will become smaller over time, approaching but never reaching zero.

Now, let's find the possible range of values for \( V(t) \) over the 5-year period (from \( t = 0 \) to \( t = 5 \)):

1. At \( t = 0 \): \[ V(0) = 65000(0.82)^0 = 65000 \]

2. At \( t = 1 \): \[ V(1) = 65000(0.82)^1 = 65000 \times 0.82 \approx 53300 \]

3. At \( t = 2 \): \[ V(2) = 65000(0.82)^2 = 65000 \times 0.82^2 \approx 43660 \]

4. At \( t = 3 \): \[ V(3) = 65000(0.82)^3 = 65000 \times 0.82^3 \approx 35850 \]

5. At \( t = 4 \): \[ V(4) = 65000(0.82)^4 = 65000 \times 0.82^4 \approx 29480 \]

6. At \( t = 5 \): \[ V(5) = 65000(0.82)^5 = 65000 \times 0.82^5 \approx 24170 \]

Over the first five years, the value of the car decreases from 65,000 down to approximately 24,170.

Thus, the range of values for \( V(t) \) is:

\[ 0 < V(t) \leq 65000 \]

This means the car value is always greater than 0 (the car can’t have a negative value) and less than or equal to 65,000.

The correct option for the possible range for this situation is:

C) 0 < v ≤ 65000