jocelynnjohnso90
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please help for brainliest
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 maximum height reached by the rocket, to the nearest tenth of a foot.
y=-16x^2+212x+139

Respuesta :

altavistard

Answer:

841.6 ft

Step-by-step explanation:

Solve this using the quadratic formula.

y=-16x^2+212x+139 becomes -16x^2 + 212x + 139.  The x-value of the axis of symmetry is

       -212

x = ----------- = 6.625  (This time measurement is in seconds.)

       2(-16)

The maximum height reached by the rocket is the y value when x = 6.625 and is y(6.625) = 841.6 ft

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