A rocket is launched from a tower the height of the rocket why in feet is related to the time after launch X in seconds by the given equation use this equation find the maximum height reached by the rocket to the nearest 10th of a foot y=-16x^2+152x+83

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

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

stops on an even number.