7. the value of a car is \( \$ 26,500 \). It loses \( 13 \% \) of its value every year. Whion of the following functions models this situction A. \( 1=26,500(15)^{+} \) \( B . y=26,500(0.13)^{+} \) \( C .4=26,500(1.13)^{+} \) \( D \cdot y=26,500(0.87)^{+} \)

To model the depreciation of the car's value, we can use an exponential decay function. Here's how to approach the problem: 1. **Initial Value:** The car is initially valued at **\$26,500**. 2. **Depreciation Rate:** The car loses **13%** of its value each year. This means it retains **87%** of its value annually (since 100% - 13% = 87% or 0.87 in decimal form). The general form of an exponential decay function is: \[ y = y_0 \times (1 - r)^t \] where: - \( y_0 \) is the initial value (\$26,500), - \( r \) is the depreciation rate (0.13), - \( t \) is the time in years. Plugging in the values: \[ y = 26,500 \times (0.87)^t \] **Matching with the Given Options:** - **Option D** correctly represents this function: \[ \boxed{y = 26,500(0.87)^t} \] All other options either use incorrect decay factors or do not appropriately model the depreciation. **Answer:** Option D. \( y = 26,\!500 \times (0.87)^t \)