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23 July, 17:04

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

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Answers (2)
  1. 23 July, 17:53
    0
    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
  2. 23 July, 19:49
    0
    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
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