The dollar value \( v(t) \) of a certain car model that is \( t \) years old is given by the following exponential function. \[ v(t)=26,000(0.86)^{t} \] Evaluating an exponential function that models a real-world situation the value of the car after 5 years and after 12 years. Round your answers to the nearest dollar as necessary. Value after 5 years: Value after 12 years: T】]

To determine the value of the car after 5 years and after 12 years, we'll use the given exponential depreciation function: \[ v(t) = 26,000 \times (0.86)^t \] ### **Value After 5 Years:** \[ v(5) = 26,000 \times (0.86)^5 \] First, calculate \( (0.86)^5 \): \[ (0.86)^5 \approx 0.470 \quad (\text{rounded to three decimal places}) \] Now, multiply by 26,000: \[ v(5) = 26,000 \times 0.470 \approx 12,220 \] **Rounded to the nearest dollar:** \[ \boxed{\$12,\!220} \] ### **Value After 12 Years:** \[ v(12) = 26,000 \times (0.86)^{12} \] First, calculate \( (0.86)^{12} \): \[ (0.86)^{12} \approx 0.164 \quad (\text{rounded to three decimal places}) \] Now, multiply by 26,000: \[ v(12) = 26,000 \times 0.164 \approx 4,264 \] **Rounded to the nearest dollar:** \[ \boxed{\$4,\!264} \] ### **Summary:** - **Value after 5 years:** \$12,220 - **Value after 12 years:** \$4,264