Answer:
<h3>444m</h3>
Step-by-step explanation:
Given the height of the rocket modelled by the equation;
y=-16x^2+152x+83
The rocket reaches its maximum height when the velocity of the rocket is zero.
Since velocity is the change in displacement with respect to time;
velocity v = dy/dx
dy/dx = -32x + 152
at dy/dx = 0
-32x + 152 = 0
-32x = -152
x = -152/-32
x = 4.75s
To get maximum height reached by the ball, we will substitute x = 4.75 into the modeled function as shown;
y(4.75) = -16(4.75)^2+152(4.75)+83
y(4.75) = -361+722+83
y(4.75) = 444m
Hence the maximum height reached by the ball is 444m
It’s just 324 x 8 because base x height
2592cm^3
5/9-2/12=?
9x8=72
5x8=40
12x6=72
2x6=12
40/72 - 12/72 = 28/72
28/4=7
72/4=18
28/72=7/18
Answer:A
Step-by-step explanation:
