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.