To determine the possible range for the value \( V \) of Cecily's car after \( t \) years using the function \( V(t) = 65000(0.82)^t \), we need to consider how the function behaves over time.
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Initial Value: At \( t = 0 \) (when Cecily first buys the car), \( V(0) = 65000(0.82)^0 = 65000 \).
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Depreciation: The value of the car decreases each year due to depreciation. As \( t \) increases, \( (0.82)^t \) becomes smaller, meaning \( V(t) \) decreases.
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Behavior Over Time: As \( t \) approaches infinity, \( V(t) \) approaches 0, but it will never actually reach 0. Therefore, the value will decrease but remain positive.
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After 5 Years: Let's compute the value at \( t = 5 \): \[ V(5) = 65000(0.82)^5 \approx 65000(0.38667) \approx 25134.55 \]
Given these considerations, the value of the car after any number of years must be greater than 0 but less than or equal to the initial value \( 65000 \).
Thus, the possible range for the value \( V \) of Cecily's car is:
\[ 0 < V \leq 65000 \]
From the options you've listed, "0 < v ≤ 65000" matches the analysis (though it's important to note it should be strictly greater than 0, as it will never be exactly 0). Therefore, the answer is:
0 < V ≤ 65000 (though it appears the option you provided is "0<v≤65000", which would also imply the same range).