Suppose h(t)= -0.2t^2+2t models the height, in feet, of a ball that is kicked into the air where t is given as time in seconds.

after how many seconds does the ball reach its maximum height?

what is the maximum height of the ball?

after how many seconds does the ball reach the ground?

Step-by-step explanation:

Height:  h(t)= -0.2t^2+2t  =  2t(-0.1t + 1)

-0.2t^2+2t  is  a quadratic expression with a = -0.2 and b = 2.

The ball reaches its max height at t = -b/(2a).

Here, t = -2 / (2·[-0.2] ), or  t = 2/0.4 = 5 (sec)

The ball reaches its max height at at 5 sec.

The max ht. of the ball is h(5) = -0.2(5 sec)^2 + 2(5 sec) = (-5 + 10) sec, or 5 ft.

To determine after how many sec the ball reaches the ground, we set h(t) = 0 and solve for t:  

h(t)= -0.2t^2+2t = 0 = 2t(-0.1t + 1).  Thus, t = 0 sec or t = 10 sec

The ball reaches the ground again after 10 sec.

180 square centimeters

step-by-step explanation:

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: