SOLUTION: At what rate is the surface area of a cube increasing if its edges are 2 inches and are increasing at a rate of 3 inches per minute?
A=6s^2
dA/dt=12s (ds/dt)
dA/dt=12(2)(3
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-> SOLUTION: At what rate is the surface area of a cube increasing if its edges are 2 inches and are increasing at a rate of 3 inches per minute?
A=6s^2
dA/dt=12s (ds/dt)
dA/dt=12(2)(3
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Question 368966: At what rate is the surface area of a cube increasing if its edges are 2 inches and are increasing at a rate of 3 inches per minute?
A=6s^2
dA/dt=12s (ds/dt)
dA/dt=12(2)(3)
dA/dt=72 in/min
Is this correct? Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! The value is correct but your units are not.
It should be ()()=
Remember surface area is measured in square units.
