A trough is 10 feet long and the ends have the shape of isosceles triangles that measure 3 feet across at the top and have a height of 1 foot. If the trough is being filled with water at a rate of 12 ft3 per minute, how fast is the water level rising when the water is 6-inches deep

0.8 ft³/min

Step-by-step explanation:

Assuming that the cross section formed by the water at a depth of x is b. Then, if we use the property of similar triangles we have...

See the attachment for the workings.

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Continuation of the workings

From the question, we're told that the water is 6 inches deep. Converting this inch to ft, we have

6 ft = 0.5 ft.

So we use 0.5 as our x, and then use 12 ft³/min as our dV/dt. Doing this, we have

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12 = 30 * 0.5 dx/dt

12 = 15 dx/dt

dx/dt = 12/15

dx/dt= 0.8 ft³/min

e) none of the above because the measure of angle cba is 101°