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+228x+71
Answers
Answer:
883.3
Step-by-step explanation:
thats all.
An equation is formed of two equal expressions. The maximum height reached by a rocket, to the nearest tenth of a foot is 883.3 feet.
What is an equation?An equation is formed when two equal expressions are equated together with the help of an equal sign '='.
To find the maximum height through which the rocket will reach, we need to differentiate the given function, therefore, we can write,
y=-16x²+228x+71
dy/dx = -16(2x)+228
Substitute the value of dy/dx as 0, to get the value of x,
0 = -32x + 228
228 = 32x
x =7.125
Substitute the value of x in the equation to get the maximum height,
y=-16x²+228x+71
y=-16(7.125²)+228(7.125)+71
y=883.3feet
Hence, the maximum height reached by the rocket, to the nearest tenth of a foot is 883.3 feet.
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Step-by-step explanation:
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Step-by-step explanation:
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Answer:
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Answers
Answer:
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Step-by-step explanation:
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Answers
Answer:
B
Step-by-step explanation:
A=4 so its wrong cause the original one is 10-12=-2