Answer:
Whats the question
Step-by-step explanation:
also if you are looking for the unit rate its <u><em>153.846...</em></u>
or if you are looking for how many in 2 years then <u><em>650,000</em></u>
<u><em>Hope this helps </em></u>
<u><em>Have a nice day</em></u> :)
Answer:
![y(t)= 6-3cos(\dfrac{2\pi}{14}t )](https://tex.z-dn.net/?f=y%28t%29%3D%206-3cos%28%5Cdfrac%7B2%5Cpi%7D%7B14%7Dt%20%29)
Step-by-step explanation:
The function that could model this periodic phenomenon will be of the form
![y(t) = y_0+Acos(wt)](https://tex.z-dn.net/?f=y%28t%29%20%3D%20y_0%2BAcos%28wt%29)
The tide varies between 3ft and 9ft, which means its amplitude
is
![A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}](https://tex.z-dn.net/?f=A%20%3D%5Cdfrac%7B%289-3%29ft%7D%7B2%7D%20%5C%5C%5C%5C%5Cboxed%7BA%20%3D%203ft%7D)
and its midline
is
.
Furthermore, since at
the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is
.
The period of the full cycle is 14 hours, which means
![\omega t =2\pi](https://tex.z-dn.net/?f=%5Comega%20t%20%3D2%5Cpi)
![\omega (t+14)= 4\pi](https://tex.z-dn.net/?f=%5Comega%20%28t%2B14%29%3D%204%5Cpi)
giving us
![\boxed{\omega = \dfrac{2\pi}{14}.}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Comega%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B14%7D.%7D)
With all of the values of the variables in place, the function modeling the situation now becomes
![\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}](https://tex.z-dn.net/?f=%5Cboxed%7By%28t%29%3D%206-3cos%28%5Cdfrac%7B2%5Cpi%7D%7B14%7Dt%20%29.%7D)
The prime number will be 69 and 89
we know that s = 3, so plug it in.
P = 6(3)
P = 18
perimeter = 18 km
Answer: −6p−2
Step-by-step explanation: