Question 1081002
the formula is:


i(t) = 86000 - 65000 * e^-.003t


since e^-a = 1/e^a, this formula becomes:


i(t) = 86000 - 65000 / e^.003t


you can see that, as t gets extraordinarily large, the value of i(t) gets closer and closer to 86000, because the value of 65000 / e^.003t gets closer and closer to zero.


to find the income in the 50th and 100th month, just replace t with 50 and 100 respectively to get:


when t = 50, monthly income = 86000 - 65000 / e^.003*50 = 30053.98153


when t = 60, monthly income = 86000 - 65000 / e^.003*100 = 37846.81566


this can be shown graphically below:


<img src = "http://theo.x10hosting.com/2017/051302.jpg" alt="$$$" </>


as t approaches a higher and higher number, the value of y approaches 8600 as shown in the following graph:

<img src = "http://theo.x10hosting.com/2017/051303.jpg" alt="$$$" </>