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:
The slope of the line is: 1
The y-intercept is: -3
You can graph the line using the slope and y-intercept, or two points.
It's about 20 units
Hope this helped :)
Answer:
2c(c - 3)(c - 8)
Step-by-step explanation:
Hello!
Factor:
- 2c³ - 22c² + 48c
- 2c(c² - 11c + 24)
Think: What two numbers multiply to 24, and add up to -11?
Answer: -8 and -3
Split up the -11c into -8c and -3c:
- 2c(c² - 8c - 3c + 24) (factor each pair)
- 2c( c(c - 8) - 3(c - 8))
- 2c(c - 3)(c - 8)
