The question is asking what v_final is, given that v_initial is at 300 feet. and v_initial is at 0 feet.
We know there will be a constant downward acceleration of 32.15 ft/s^2.
Use the following equation:
v_final^2 = v_initial^2 + 2ah
v_final^2 = (160 ft/s)^2 + 2(-32.15 ft/s^2)(300 ft) = 6310 ft^2/s^2
v_final = (6310 ft^2/s^2)^1/2 = 79.4 ft/s.

5 0

3 years ago

Find the derivative dy, dx for y equals the quotient of the quantity x cubed minus 5 times x and the quantity x squared minus 1.

Elza [17]

Using the quotient rule

dy/dx = (x^-1) * (3x^2 -5 ) - (x^3 - 5x) * 2x
----------------------------------------------
(x^2 - 1)^2

You are correct

7 0

3 years ago

Jim Tree sells trees. The mean length of the trees purchased was 68 inches with a standard deviation of 10 inches. Jim wants to

Zepler [3.9K]

Step-by-step explanation:

We know that,

$Z=\dfrac{X-\mu}{\sigma}$