# Thread: Chk Anwser - Draining tank of water

1. ## Chk Anwser - Draining tank of water

Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is (t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?

A. $\displaystyle y(t) = .25t^2 -30t +c.$
$\displaystyle C = 2000.$ Substitute $\displaystyle t = 25.$
We get $\displaystyle y(25) = .25(25)^2 -30t + 2000 = 1406.25$ Gal remaining.

2. Originally Posted by Yogi_Bear_79
Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is (t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?

A. $\displaystyle y(t) = .25t^2 -30t +c.$
$\displaystyle C = 2000.$ Substitute $\displaystyle t = 25.$
We get $\displaystyle y(25) = .25(25)^2 -30t + 2000 = 1406.25$ Gal remaining.
It looks good to me.

-Dan

3. Originally Posted by Yogi_Bear_79
Q. Suppose that a tank initially contains 2000 gal of water and the rate of change of its volume after the tank drains for t min is `(t) = (0.5)t -30 (in gallons per minute). How much water does the tank contain after it has been draining for 25 minutes?

A. $\displaystyle y(t) = .25t^2 -30t +c.$
$\displaystyle C = 2000.$ Substitute $\displaystyle t = 25.$
We get $\displaystyle y(25) = .25(25)^2 -30t + 2000 = 1406.25$ Gal remaining.
I got the same answer. The only mistake is that,
$\displaystyle y(25)=.25(25)^2-30(25)+2000$ but you just probably forgot to write that.

4. Thanks, this isn't easy for me, so the confirmations and examples this forumn provides are priceless!