To determine the appropriate range for the value of Cecily's car based on the given function \( V(t) = 65000(0.82)^t \), we can analyze the value of the car over the specified time frame (0 to 5 years).
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Initial Value: At \( t = 0 \) (the year of purchase), the value of the car is: \[ V(0) = 65000(0.82)^0 = 65000 \]
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Value over Time: As time progresses, the value decreases due to depreciation:
- After 1 year: \[ V(1) = 65000(0.82)^1 = 65000 \times 0.82 \approx 53300 \]
- After 2 years: \[ V(2) = 65000(0.82)^2 \approx 43660 \]
- After 3 years: \[ V(3) = 65000(0.82)^3 \approx 35843 \]
- After 4 years: \[ V(4) = 65000(0.82)^4 \approx 29395 \]
- After 5 years: \[ V(5) = 65000(0.82)^5 \approx 24084 \]
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Range: The value of the car starts at \( 65000 \) and drops to a positive value each year, but it will never reach zero. Since the value continues to decrease but does not hit zero, the possible values will fit between:
\[ 0 < V(t) \leq 65000 \]
Based on this analysis, the most suitable response regarding the range of values for Cecily's car after \( t \) years (for 0 ≤ t ≤ 5) is:
0 < V ≤ 65000.
Please note that among the provided options, the closest and most correct choice based on the analysis is: 0 < v ≤ 65000.
However, if this option is not present and we were to consider the potential values including zero we can use a similar option, but keep in mind that the car value never actually reaches zero. Thus, it would be more accurate to state that the possible range is \( 0 < v ≤ 65000 \).