Question 1161018
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The volume of the leakage is the integral of the function  {{{6e^(-0.1t)}}} over the time from t= 0 to t= 90 minutes.


This integral is equal to  {{{(-6/0.1)*(e^(-9)-1)}}} = {{{60*(1-e^(-9))}}} = {{{60*(1-2.71828^(-9))}}} = 60*(1-0.000123) = 60 liters

    (with the accuracy absolutely enough for any engineering purposes).



By knowing the leakage volume, you calculate the amount of fuel left in the reservoir as the difference

    1200 liters - 60 liters = 1140 liters.    <U>ANSWER</U>



Notice that starting leakage rate was  6 liters per minute.


At the time t= 90 minutes, the leakage rate is  {{{6e^(-9)}}} = {{{6*2.71828^(-9)}}} = 0.00074 liters per minute.


So, the change of the leakage rate during 90 minutes is very significant.
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