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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 time that the rocket will hit the ground, to the nearest 100th of second. -16x^2+129x+119

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Answer:

8.90 seconds

Step-by-step explanation:

When the rocket hits the ground, its height ( y ) will be zero. This means we can write the equation as:

  • 0 = -16x²+129x+119

So we need to solve that equation:

  • a = -16
  • b = 129
  • c = 119
  • x = [tex]\frac{-b\frac{+}{-}\sqrt{b^2-4ac} }{2a}[/tex]

x₁ = -0.84

x₂ = 8.90

We discard x₁, as it is negative.

So the answer is x₂, 8.90 seconds.

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