Ask a New Question
Search
Questions and answers by
sigfig
Answers (2)
when t=2 y= 39690, so 39690=3600e^k(2) 11.025=e^k(2) ln(11.025)=k(2)ln(e) 2.400=k(2) 1.200=k y=36000e^1.2t
Start with what you know. Continuous compound interest tells you that you will be using e^ starting with 1 million dollars. Now just plug in the time(t) at 20yrs and 40yrs into the formula Money received(time)=1million(e^(.10)t) You should have two diff.
