Because eating out is a luxary, not a nessecity
Answer:
A
Step-by-step explanation:
horizontal component=18cos 36.9°≈14.39 m/s≈14.4 m/s
Vertical component=18 sin 36.9°≈10.81 m/s≈10.8 m/s
Answer:

Step-by-step explanation:
The function that could model this periodic phenomenon will be of the form

The tide varies between 3ft and 9ft, which means its amplitude
is

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


giving us

With all of the values of the variables in place, the function modeling the situation now becomes

Answer:
B. because it want to know how far it takes to reach the ground
Step-by-step explanation:
when you drop the ball it take the hight of getting up