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+
180x+63

To find the maximum of a function in ax^2 + bx + c form, you can use  the formula: maximum = c - (b^2 / 4a).

In this question, a = -16, b = 180, and c = 63.

Maximum = 63 - (180^2/4(-16))

Maximum  = 569.25

Finally, rounding this to the nearest tenth of a foot, the final answer is 569.3 ft.

Let me know if this helps!

1/4 and 8/3 because:

the reciprocal if 4 is 1/4 because 4 is used as 4/1 and the opposite of that is 1/4. the opposite of 3/8 is 8/3 so that is the answer.