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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)=3x2+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?

 

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.

 

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

 

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

 Mar 14, 2019
 #1
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(i - d)  = [ 3x^2 + 2 ] - [ 4x - 1 ]  =  [3x^2 - 4x + 3]

 

(i - d)(3) =  3(3)^3 - 4(3) + 3  =   27 - 12 + 3    =  18    means that

 

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

 

[At 3 hours.... It is entering at   3(3)^2 + 2  = 29 gallons/hr   and leaving at  4(3) - 1 = 11 gal/hr....so...the rate difference is 29 - 11  =  18 gal/hr ]

 

 

cool cool cool

 Mar 14, 2019

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