3. The yearly profits of a company is \( \$ 25,000 \). The profits have been decreasing by \( 6 \% \) per year. Write an exponential decay function to represent this \( \begin{array}{ll}\text { uation. } & \text { B. What will be the profits in } 8 \text { years? Round } \\ \text { answer to the nearest dollar. }\end{array} \)

To represent the yearly profits of a company that is decreasing by 6% per year, we can use an exponential decay function. The general form of an exponential decay function is: \[ P(t) = P_0 \times (1 - r)^t \] where: - \( P(t) \) is the profit at time \( t \), - \( P_0 \) is the initial profit, - \( r \) is the rate of decay (in this case, 6% or 0.06), - \( t \) is the time in years. Given that the yearly profits of the company is $25,000 and the profits have been decreasing by 6% per year, we can substitute these values into the function: \[ P(t) = 25000 \times (1 - 0.06)^t \] Now, we need to find the profits in 8 years. We can substitute \( t = 8 \) into the function to calculate the profits at that time: \[ P(8) = 25000 \times (1 - 0.06)^8 \] Let's calculate this value. Calculate the value by following steps: - step0: Calculate: \(25000\left(1-0.06\right)^{8}\) - step1: Subtract the numbers: \(25000\times 0.94^{8}\) - step2: Convert the expressions: \(25000\left(\frac{47}{50}\right)^{8}\) - step3: Simplify: \(25000\times \frac{47^{8}}{50^{8}}\) - step4: Rewrite the expression: \(2500\times 10\times \frac{47^{8}}{50^{8}}\) - step5: Rewrite the expression: \(50^{2}\times 10\times \frac{47^{8}}{50^{8}}\) - step6: Reduce the numbers: \(10\times \frac{47^{8}}{50^{6}}\) - step7: Rewrite the expression: \(10\times \frac{47^{8}}{10^{6}\times 5^{6}}\) - step8: Reduce the numbers: \(1\times \frac{47^{8}}{10^{5}\times 5^{6}}\) - step9: Multiply the fractions: \(\frac{47^{8}}{15625\times 10^{5}}\) The profits in 8 years will be approximately $15,239.22. Rounding this to the nearest dollar, the profits in 8 years will be $15,239.