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Luden [163]
3 years ago
11

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
Mathematics
1 answer:
True [87]3 years ago
5 0

Answer:

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

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1 year ago
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3 years ago
PLEASE HELP
lions [1.4K]
This question is easy if you think about it. First get the area of the rectangular part of the shape and then find area of the two semi-circles then add them together.

First, let's solve for the rectangle because it's easier.

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Now for the semi-circle. First use the equation for finding the area of a circle which is A = Pi * r^2. We'll use 3.14 for Pi.

We know the diameter is 23. And to get the radius we just half it because the radius is half of the diameter.

23 / 2 = 11.5

A = 3.14 * 11.5^2
A = 3.14 * <span>132.25
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Now we have the total area for both semi-circles. Because if we half 415.265 then add the other semi-circle's area it will be the same and adding them up will result in 415.265.

Now add up 851 and 415.265.

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