High School

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At a seaport, the depth of the water, \( h \) meters, at time \( t \) hours during a certain day is given by:

\[ h = 6 \cos \left( \frac{2\pi(t-8)}{12} \right) + 14 \]

What is the depth of the water at time \( t = 4 \) hours?

Answer :

Final answer:

The depth of the water at time t = 4 hours is -3 meters.

Explanation:

To find the depth of the water at time t = 4 hours, we can substitute t = 4 into the equation h = 6 cos 2π(t-8)/12.4. This gives us h = 6 cos 2π(4-8)/12.4.

First, calculate the value inside the cosine function: 2π(4-8)/12.4 = -π/3.

Now, substitute this value into the cosine function: h = 6 cos(-π/3).

Finally, evaluate the cosine function: h = 6 * (-0.5) = -3 meters.

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