Question

The revenue of an online retailer grew at the rate ofR' (t) = 9.05e^0.025tmillion dollars per year (with t = 0 corresponding to 2016). The revenue in 2016 was $134 million. Find a function to represent theretailer's revenue t years after 2016.OR (1) = 362e0.0254 + 134OR (t) = 36.220.025€ + 97.8R (1) = 36.220.0251 - 134R (1) = 36220.0251 - 228

Answer 1

The rate at which the revenue grows is given as :

`R^{^(\prime)}(t)=9.05e^(0.025t)`

To obtain the retailer's revenue as a function of time, we integrate the given equation with respect to time (t)

`\begin{gathered} \int R^(\prime)(t)\text{ dt = }\int 9.05e^(0.025t)\text{ dt} \\ R(t)\text{ = }(9.05)/(0.025)e^(0.025t)+R_0 \\ \text{Given that R}_0\text{ = 134} \\ R(t)=362e^(0.025t)\text{ + 134} \end{gathered}`

The correct option is option A