A metal cube contracts when it is cooled. if the edge of the cube is decreasing at a rate of 0.2 cm/hr, how fast is the volume changing when the edge is 60 centimeters?

2160 cm³/hour

Step-by-step explanation:

By default, we know that the volume of a cube is given as s³

Thus, the Volume function, V = s³

When we differentiate with respect to time we have

dV/dt = 3s² (ds/dt), where ds/dt = 0.2

Then we go ahead and substitute all the given parameters

dV/dt = 3 x 60 x 60 x 0.2

dV/dt = 10800 * 0.2

dV/dt = 2160 cm³/hour

This means that the volume decreases by a rate of 2160 cm³/hour at the instant its edge is 60 cm

step-by-step explanation:

b and c

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right