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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 the maximum height reached by the rocket, to the nearest tenth of a foot.
y=−16x^2+160x+150

Sagot :

Kittyhacker14idn

Answer:

Answer below

Step-by-step explanation:

Time can not be negative from any perspective. Hence, the correct answer is 10.16 seconds.

The equation of the path of a rocket is given that is  y= x^2 + 153x + 98

We need to determine the time that the rocket will hit the ground.

Now, if the rocket hits the ground after the launching then the overall displacement at the time of hitting the ground will be zero.

Therefore, the value of y is 0.

Thus,  -16x^2 +153x + 98 = 0

Now, the formula for finding the roots of the quadratic equation  

 ax^2 +bx +c = 0 is given as:

 [tex]x^{1} =\frac{-b+\sqrt{b^2 -4ac} }{2a}[/tex]

[tex]x^2 = \frac{-b -\sqrt{b^2 -4ac} }{2a}[/tex]

Comparing the expression with the standard form of the equation.

a = -16b = 153 and c = 98

By applying the formula there are two values are come that is,

x^1 = -0.602555

x^2 = 10.1651  

Time can not be negative from any perspective.

Hence, the correct answer is 10.16 seconds.

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(Hope this helps can I pls have brainlist (crown)☺️)

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