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=-16x2 + 200x + 82

Step-by-step explanation:

dy/dx = -32x + 200

maximum height when dy/dx = 0

-32x + 200 = 0

x = 6.25 seconds

y = -16·6.25² + 200·6.25 + 82 = 707 feet

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

i would be 2,000 just get the square root of 50 which is 2500 and the divid 4/5 and last thing multiply the answer you get for that by the square root of 50

step-by-step explanation: