- #1

karush

Gold Member

MHB

- 3,267

- 4

Water is pumped into a tank at a rate of $r(t) = 30(1-e^{e-0.16t})$ gallons per minute,

where t is the number of minutes since the pump was turned on.

If the tank contained 800 gallons of water when the pump was turned on,

how much water, to the nearest gallon, is in the tank after 20 minutes?

\begin{array}{ll}

a. &380 \textit{ gal}\\

b. &420\textit{ gal}\\

c. &829\textit{ gal}\\

d. &1220\textit{ gal}\\

e. &1376\textit{ gal}

\end{array}

so starting take the integral

why is e in the d/dt

$$\displaystyle800-\int_0^{20} 30(1- e^{-0.16t})\ dt $$

where t is the number of minutes since the pump was turned on.

If the tank contained 800 gallons of water when the pump was turned on,

how much water, to the nearest gallon, is in the tank after 20 minutes?

\begin{array}{ll}

a. &380 \textit{ gal}\\

b. &420\textit{ gal}\\

c. &829\textit{ gal}\\

d. &1220\textit{ gal}\\

e. &1376\textit{ gal}

\end{array}

so starting take the integral

why is e in the d/dt

$$\displaystyle800-\int_0^{20} 30(1- e^{-0.16t})\ dt $$

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