The time an object is in free fall is given by the formula t=n/2g. where v is the final velocity, g is the gravitational constant of 32 ft/sec sq, and t is the time of the fall. a skydiver jumps out of a plane at 6000 feet. how long did it take to reach the final velocity of -96 feet per second.

Given formula t= v/2g

Where v is the final velocity, g is the gravitational constant of 32 ft/sec^2, t is the time in seconds of the fall.

For the given problem, we have

g=-32 ft/sec^2 ( we took it negative because it's free fall and for free fall the value of g is a negative value).

v=-96 ft/sec.

Plugging values of v and g in given formula, we get

t= -96/2(-32)

t= -96/-64

Dividing -96 by -64, we get

t=1.5 seconds.

Therefore, it take 1.5 seconds to reach the final velocity of -96 feet per second.

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

f(x) = y

f(-3) = -5

step-by-step explanation: