A tank of liquid has both an inlet pipe allowing liquid to be added to the tank and a drain allowing liquid to be drained from the tank.

The rate at which liquid is entering the tank through the inlet pipe is modeled by the function i(x)=3x^2+2, where the rate is measured in gallons per hour. The rate at which liquid is being drained from the tank is modeled by the function d(x)=4x−1 where the rate is measured in gallons per hour.

What does (i−d)(3) mean in this situation?

Guest Mar 15, 2019

#1**+1 **

Since i is how much liquid is being put in, and d is how much is being let out, i - d is how much liquid is in the container at any given time. So (i-d)(3) is how much liquid is in the container in gallons after three hours.

Hope this helps!

LagTho Mar 15, 2019

#2**+1 **

(i - d) is the rate the tank is being filled after x hours

For instance....after 2 hours the tank is being filled at 3(2)^2 + 2 = 14 gal/hr

And it is being drained at 4(2) - 1 = 7 gal/hr

So.....the net is that it is being filled at 2 hours = 14 - 7 = 7 gals/hr

So

(i - d)(3) is the rate that the tank is being filled at 3 hrs

CPhill Mar 15, 2019

#3**0 **

The rate at which the amount of liquid in the tank is changing at t = 3 hours is 18 gallons per hour.

The rate at which the amount of liquid in the tank is changing at t = 3 hours is 40 gallons per hour.

There are 40 gallons of liquid in the tank at t = 3 hours.

There are 18 gallons of liquid in the tank at t = 3 hours.

those are my answer choices

Guest Mar 15, 2019