Water is poured into a bucket according to the rate f(t) = (t + 7)/(2 + t), and at the same time empties out through a hole in the bottom at the rate e(t) = ln(t + 4)/(t + 2), with both f(t) and e(t) measured in pints per minute. how much water, to the nearest pint, is in the bucket at time t = 5 minutes. you must show your setup but can use your calculator for all evaluations.

The amount of water in the bucket at t=5, will be estimated as follows:
Assuming that both pipes started running at the same time, the amount of water poured in the bucket at t=5 will be:
F(t) = (t + 7)/(2 + t)
f(5)=(5+7)(2+5)
f(5)=(12)(7)
f(5)=84

Amount of water drawn from the bucket at t=5 will be:
E(t) = ln(t + 4)/(t + 2)
E(5)=ln(5+4)/(5+2)
E(5)=ln(9)/(7)

Thus the amount of water in the bucket at t=5 will be:
84-ln 9/7
~83.74868557
~83.8

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