Answer:
h(t)=0.7sin(πt/6)+1.1
Explanation:
A periodic function that models the height of the tide can be written as

where
A is the amplitude
is the angular frequency
is the central value of the tide
Here we know that:
- the hight tide is 1.8 meters
- the low tide is 0.4 meters
So, the central value of the tide is

Also, the amplitude is the maximum displacement from the equilibrium position (central value of the tide), so:

Now we have to find the angular frequency, which is related to the period T by

Here the high tide occurs at 12 am and 12 pm: this means that the period is 12 hours, so
T = 12

Therefore, the correct equation is
h(t)=0.7sin(πt/6)+1.1