CAN SOMEONE PLEASE TELL ME THE FORMULAS FOR A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. these type of problems i cant find anything and ive been up for 4 hrs its like 6 am

9.72 seconds ( approx )

Step-by-step explanation:

Since, the equation that shows the height of the rocket from the ground,

y=-16x^2+143x+121y=−16x

2

+143x+121

Where,

x = time after launch, in seconds,

When the rocket hits the ground,

y = 0,

i.e.

-16x^2 + 143x+12=0−16x

2

+143x+12=0

By the quadratic formula,

x = \frac{-143\pm \sqrt{(143)^2 - 4\times -16\times 12}}{-32}x=

−32

−143±

(143)

2

−4×−16×12

x = \frac{-143\pm \sqrt{20449 + 7744}}{-32}x=

−32

−143±

20449+7744

\implies x = -0.778\text{ or }x=9.716⟹x=−0.778 or x=9.716

∵ Time can not be negative,

So, the time taken to hit the ground = 9.716 seconds ≈ 9.72 seconds.

Step-by-step explanation:

그것이 당신에게 도움이되기를 바랍니다 :)

the answer is (2x+8)²= 16

the answer is two. hope this !