Acylindrical tank holds 20,000 gallons of water, which can be drained from the bottom of the tank in 20 minutes. the function gives the volume v of water (in gallons) in the tank at time t in minutes, where time t = 0 corresponds to the instant the tank starts to drain.

at time t = 3 minutes, at what rate is the volume of water in the tank changing?
a. –2200 gallons/minute
b. 1700 gallons/minute
c. 2200 gallons/minute
d. –1800 gallons/minute
e. 1800 gallons/minute

You didn't include the function.

If the function is, V = 20,000 (1 - t/20)^2, as you can see, for t = 20, V = 0

Then I am going to show you the work with that function, and you can use the same procedure with your own fucntion.

The rate of change of the volume is the derivative of the function with respect to the time.

[dV/dt] = 20,000 *2 [1 -t/20] * [-1/20] = - 2000 [1 - t/20] = 2000[t/20 - 1]

At t = 3

dV/dt = 2000[3/20 - 1] = -1700 gallons / min

The negative sign mean decreasing.

The answer is - 1700 gallons / min

Now, for the same function, if the time is t = 2, you would obtain one of the results of the list of options.

dV/dt = 20,000 [2/20 - 1] = -1800 gallons/minute, which is the option D of the list.

ay! I finally reached Ace, and to celebrate... I'm giving out

19 points

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divided then multipy then substract

Step-by-step explanation: