Certainly! Let's go through the problem step-by-step.
Deepak bought a car for $25,000. The value of the car decreases by 9% each year.
1. Understanding the decrease:
- If the value decreases by 9%, it means it retains [tex]\(100\% - 9\% = 91\%\)[/tex] of its value each year.
- In decimal form, 91% is 0.91.
2. Formulating the exponential decay model:
- Let [tex]\( v \)[/tex] be the value of the car after [tex]\( t \)[/tex] years.
- The value of the car each year can be modeled as an exponential decay function, where each year the car's value is multiplied by 0.91.
- The general form of an exponential decay model is [tex]\( v = A \cdot (r)^t \)[/tex], where [tex]\( A \)[/tex] is the initial value, [tex]\( r \)[/tex] is the rate of decay (retention rate), and [tex]\( t \)[/tex] is time in years.
3. Defining the variables:
- Initial value, [tex]\( A = 25,000 \)[/tex] dollars.
- Retention rate per year, [tex]\( r = 0.91 \)[/tex].
4. Putting it all together:
- The exponential function that models the value [tex]\( v \)[/tex] of the car after [tex]\( t \)[/tex] years is:
[tex]\[ v = 25,000 \cdot (0.91)^t \][/tex]
Given the options:
- A. [tex]\( v=25,000(1.09)^t \)[/tex] - Incorrect, because 1.09 would imply the value is increasing by 9%, not decreasing.
- B. [tex]\( v=25,000(0.91)^t \)[/tex] - Correct, as it matches our derived model.
- C. [tex]\( v=25,000(1.9)^t \)[/tex] - Incorrect, because 1.9 would imply an increase of 90%, which is not relevant here.
- D. [tex]\( v=25,000(0.1)^t \)[/tex] - Incorrect, because 0.1 would imply the value decreases extremely rapidly, retaining only 10% each year.
Therefore, the exponential function that models the value [tex]\( v \)[/tex] of the car after [tex]\( t \)[/tex] years is:
[tex]\[ \boxed{v=25,000(0.91)^t} \][/tex] Option B is the correct answer.
