P = 50e^-0.004t where t is time in days?

The power of a satellite is the radioscope. The power output P, in watts, decreases ata rate proportional to the amount present; P is given by:



P = 50e^-0.004t

where t is time in days.



a) How much power will be available after 375 days?



b) What is the half-life of the Power Supply?



c) The satellite's equipment cannot operate on fewer than 10 watts of power. How long can the Satellite stay in operation?



d) How much Power did the satellite have to begin with?



e) Find the rate of Change of the Power and Interpret its meaning.P = 50e^-0.004t where t is time in days?
a) sub 375 into t



b) 25 = 50e^0.004t

1/2= e^0.004t



c) 10 = 50e^0.004t



d) 50



e) 50 x 0.004e^0.004t





Thats the farest i can get you without a calculator