Mathematics: 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 the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+119x+57

A rocket is launched from a tower. The height of

riri95

17.07.2021

Time required to hit the ground is 7.9 s.

Step-by-step explanation:

The height of the rocket is given by

y =- 16 x^2 + 119 x + 57

For the time to hit the ground, put y = 0

$-16x^2+119x+57=0\\\\16 x^2 - 119 x - 57 = 0 \\\\x = \frac{119\pm\sqrt{14161+3648}}{32}\\\\x = \frac{119\pm133.45}{32}\\\\t = - 0.45 s, 7.9 s$

Time cannot be negative, so time to hit the ground is 7.9 s.

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Nerdylearner8639

30.03.2022

When you use 2 GB

Step-by-step explanation:

Data Plan a Formula 35(12p)

P = per GB

Data Plan B 25(17p)

Set each on equal to each other

Please Check my work, but I think this is on the right track.

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