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We are given the exponential function for the car's value $ v(t) = 26,000(0.80)^t $, where $ t $ is the number of years after the car's purchase.

### 1. **Initial Value of the Car** (at $ t = 0 $):
To find the initial value, substitute $ t = 0 $ into the equation:
\[
v(0) = 26,000(0.80)^0 = 26,000(1) = 26,000
\]
So, the initial value of the car is **\$26,000**.

### 2. **Value of the Car after 12 Years** (at $ t = 12 $):
Substitute $ t = 12 $ into the equation:
\[
v(12) = 26,000(0.80)^{12}
\]
We can calculate $ (0.80)^{12} $ using a calculator:
\[
(0.80)^{12} \approx 0.0687195
\]
Now, calculate:
\[
v(12) = 26,000 \times 0.0687195 \approx 1,787
\]
So, the value of the car after 12 years is approximately **\$1,787**.

Would you like any more details or have any questions?

### Related Questions:
1. How would the value change if the annual depreciation rate was 25% instead of 20%?
2. What is the car's value after 5 years using the same formula?
3. How long will it take for the car's value to fall below \$1,000?
4. What would the car’s value be if the initial value was \$30,000 instead of \$26,000?
5. How would the function change if the car depreciated by 15% per year instead?

### Tip:
In exponential decay problems, the base of the exponent represents the percentage of value remaining each year. For example, 0.80 represents 80% remaining, meaning a 20% depreciation rate.

The dollar value v(t) of a certain car model that is t years old is given by the following exponential function: v(t) = 26,000(0.80)^t. Find the initial value of the car and the value after 12 years. Round your answers to the nearest dollar as necessary.

Learn how to calculate the value of a car after 12 years using exponential decay. The car starts at $26,000 and depreciates at 20% annually. This problem is suitable for students in Grades 9-12.

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