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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 the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+115x+130 y=−16x 2 +115x+130

Respuesta :

Answer:3.59 s

Step-by-step explanation:

Given

height of rocket is given by

[tex]y=-16x^2+115x+130[/tex]

Rocket will be at it maximum height when [tex]\frac{\mathrm{d} y}{\mathrm{d} x}=0[/tex]

therefore

[tex]\frac{\mathrm{d} y}{\mathrm{d} x}=-32x+115=0[/tex]

[tex]32x=115[/tex]

[tex]x=\frac{115}{32}[/tex]

[tex]x=3.593\ s[/tex]

At [tex]x=3.59\ s[/tex] rocket will be at its maximum height

     

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