A toy rocket launched into the air has a height (h feet) at any given time (t seconds) as
h = -16t2 + 160t until it hits the ground. At what time(s) is it at a height of 9 feet above the
ground?

t =0.06s and t=9.94 s

Step-by-step explanation:

The equation that models the height of the rocket is

when the rocket is at height of 9 fret above the ground, then

This implies that:

Rewrite in standard quadratic form.

The solution is given by

The rocket reach the height of 9 feet at times t=0.06s and t=9.94 seconds

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1