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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?

 Mar 15, 2019
 #1
avatar+234 
+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!

 Mar 15, 2019
 #2
avatar+102355 
+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

 

 

cool cool cool

 Mar 15, 2019
 #3
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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
 #4
avatar+102355 
+1

At 3 hrs

 

Intake  = 3(3)^2 + 2  = 29 gal/hr

draining = 4(3) - 1  =  11 gal/hr

 

So

 

The rate at which the amount of liquid is changing in the tank = [ 29 - 11]  =  18 gal / hr

 

So....1st answer

 

 

cool cool cool

CPhill  Mar 15, 2019
 #5
avatar+234 
0

Oops. Sorry I couldn't be more help! I misunderstood the question!

 Mar 15, 2019

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