Answer:
<h3>A) 204m</h3><h3>B) 188m</h3>
Step-by-step explanation:
Given the rocket's height above the surface of the lake given by the function h(t) = -16t^2 + 96t + 60
The velocity of the rocket at its maximum height is zero
v = dh/dt = -32t + 96t
At the maximum height, v = 0
0 = -32t + 96t
32t = 96
t = 96/32
t = 3secs
Substitute t = 3 into the modeled function to get the maximum height
h(3) = -16(3)^2 + 96(3) + 60
h(3) = -16(9)+ 288 + 60
h(3) = -144+ 288 + 60
h(3) = 144 + 60
h(3) = 204
Hence the maximum height reached by the rocket is 204m
Get the height after 2 secs
h(t) = -16t^2 + 96t + 60
when t = 2
h(2) = -16(2)^2 + 96(2) + 60
h(2) = -64+ 192+ 60
h(2) = -4 + 192
h(2) = 188m
Hence the height of the rocket after 2 secs is 188m
Answer:
1*1+4+5+3+6+3+2(6+3+4(+3+5(8+2+8+483+4+8+82+7-74-4-3
16 or 4^2
This is really saying is 1/(1/4^2)
(0,2)
(1,5)
(2,8)
(3,11)
(4,14)
Go to Mathway .com they'll help you out and do it for you
that way you dont have to wait
