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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
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.
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
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.
Answer:
The answer is .
Step-by-step explanation:
Given:
18 2/5 - n = -10 1/3.
Now, to solve the equation:
Converting the mixed fractions to improper:
⇒
Moving the numbers on one side and variable to the other:
⇒
Now, solving the factions:
⇒
⇒
⇒
⇒
Therefore, the answer is .