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Bas_tet [7]
3 years ago
10

The depth of the water at the end of a pier changes periodically along with the movement of tides. On a particular day, low tide

s occur at 12:00 am and 12:30 pm, with a depth of 2.5 m, while high tides occur at 6:15 am and 6:45 pm, with a depth of 5.5 m. Let t = 0 be 12:00 am. Write a cosine model, d = acos(bt) + k, for the depth as a function of time. This amplitude is meters. a =

Mathematics
2 answers:
Valentin [98]3 years ago
8 0
Draw a diagram to illustrate the problem as shown in the figure below.

The minimum depth of 2.5 m occurs at 12:00 am and at 12:30 pm.
Therefore the period i0s T= 12.5 hours.
The maximum depth of 5.5 m occurs at 6:15 am and at 6:45 pm. Therefore the period of T = 12.5 hours is confirmed.

The double amplitude is 5.5 - 2.5 = 3 m, therefore the amplitude is a = 1.5 m.
The mean depth is k = (2.5 + 5.5)/2 = 4.0 m

The model for tide depth is
d=-1.5cos( \frac{2 \pi t}{12.5})+4

That is,
d = -1.5 cos(0.5027t) + 4
where 
d = depth, m
t = time, hours

A plot of the function confirms that the model is correct.

pentagon [3]3 years ago
4 0

Answer:

answers are 1.5 and -1.5, then 12.5 and 4pi/25, then 4, and finally D

Step-by-step explanation:

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Answer:

The amount in the account after six years is $2,288.98

Step-by-step explanation:

In this question, we are asked to calculate the amount that will be in an account that has a principal that is compounded quarterly.

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hope this helps, God bless!
8 0
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