It's so cold today ... it's below zero.
Insufficient funds
I owe you ...
hope this helps you with the question
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.
Answer:
b. x³ +2x² -3x -6
Step-by-step explanation:
For an odd-degree polynomial, the constant term is the opposite of the product of the roots. Here that means we're looking for a polynomial with the constant ...
... -(-√3)(√3)(-2) = -6
There is only one choice.
