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<h2>C. 1,004.8 cubic units</h2>Step-by-step explanation:
A sled is moving down a steep hill in a straight line. At time ts, the
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The given function is a variable separable differential equation. Combine like terms, integrate, apply the appropriate limits, and express V in terms of t. This is done as follows:
dV/dt = -3(V)^1/2
dV/-3V^1/2 = dt
m here is the initial V which is 225. Then after integrating,
-2/3 (√V - √225) = t
-2/3 (√V - 15) = t
That is the expression for V at time t. I hope I was able to help. Have a good day.
dV/dt = -3(V)^1/2
dV/-3V^1/2 = dt
m here is the initial V which is 225. Then after integrating,
-2/3 (√V - √225) = t
-2/3 (√V - 15) = t
That is the expression for V at time t. I hope I was able to help. Have a good day.