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2 years ago

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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 th

e given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot. Show work. y=-16x^2+200x+82

1 answer:

dolphi86 [110]2 years ago

7 0

Answer:

Step-by-step explanation:

y=-16x^2+200x+82

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EleoNora [17]

Answer:

d = $\frac{11}{12}$

Step-by-step explanation:

1) To solve for d, isolate it in the equation. So, to get rid of the $-\frac{1}{2}$ with the d, add $\frac{1}{2}$ to both sides.

$d-\frac{1}{2} = \frac{5}{12} \\d - \frac{1}{2} +\frac{1}{2} = \frac{5}{12} + \frac{6}{12} \\d = \frac{11}{12}$

Thus, d = $\frac{11}{12}$ .

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