At time t is greater than or equal to zero, a cube has volume V(t) and edges of length x(t). If the volume of the cube decreases at a rate proportional to its surface area, which of the following differential equations could describe the rate at which the volume of the cube decreases?

A) dV/dt=-1.2x^2
B) dV/dt=-1.2x^3
C) dV/dt=-1.2x^2(t)
D) dV/dt=-1.2t^2
E) fav/dt=-1.2V^2

  A)  dV/dt=-1.2x^2

Step-by-step explanation:

The rate of change of volume is given by dV/dt. Surface area is proportional to x^2. Since the volume is decreasing, the constant of proportionality between surface area and rate of volume change will be negative. Hence a possible equation might be ...

  dV/dt = -1.2x^2

C

Step-by-step explanation:

V(t) = [x(t)]³

A(t) = 6[x(t)]²

dV/dt = k × 6[x(t)]²

Where k < 0

From the options,

taking k = -0.2

dV/dt = -1.2[x(t)]²

details

step-by-step explanation:

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3