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

Question

. She wants to track the value of her car each year, for 5 years.

What would be a possible range for this situation?

(2 points)
Responses

30<v≤65000
30<v≤65000

0<v≤5
0<v≤5

All real numbers
All real numbers

0<v≤65000

Answer 1

To determine a possible range for the value of Cecily's car after t years, we can analyze the function \( V(t) = 65000(0.82)^t \).

1. At \( t = 0 \) (the time of purchase), the value \( V(0) = 65000(0.82)^0 = 65000 \).
2. As \( t \) increases (up to 5 years), the value \( V(t) \) will decrease because \( 0.82 \) is less than 1, meaning the car’s value depreciates.

After 5 years (at \( t = 5 \)): \[ V(5) = 65000(0.82)^5 \approx 65000 \times 0.36049 \approx 23432.85 \]

Thus, after 5 years, the car's value will be approximately $23,432.85.

From this, we can see:

• The value of the car will never drop below 0.
• The maximum value at purchase is $65,000.

Thus, the range of values for the car over the next 5 years will be:

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

So the correct response is 0 < V ≤ 65000.

However, as none of the provided options exactly states that, the closest and most representative answer for this situation is:

0 < V ≤ 65000.