Answer:
-6z
Step-by-step explanation:
(3z)2=6z
Add the negative
-6z
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: 1100
Step-by-step explanation: mark x for the total sum.
0.35x+ 0.20x + 495 = x. 495= x-0.35x-0.20x
495= 0.45x. X= 495/0.45
Answer:
The second answer choice is right
Step-by-step explanation: