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anonymous

  • one year ago

If a football player passes a football from 4 feet off the ground with an initial velocity of 36 feet per second, how long will it take the football to hit the ground? Use the equation h = −16t2 + 36t + 4. quantity of 9 plus or minus square root of 65 all over 8 9 plus or minus square root of 65 end root over 8 quantity of 9 plus or minus square root of 97 all over 8 9 plus or minus square root of 97 end root over 8

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  1. anonymous
    • one year ago
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    You need to find time \(t\) when the ball is at height \(h(t) = 0\), i.e. solve the quadratic equation\[-16t^2+36t+4 = 0\]Can you do that?

  2. anonymous
    • one year ago
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    Notice that there is a common factor of 4. First thing I would do is factor out the 4. Then apply the quadratic formula to get your answer.

  3. anonymous
    • one year ago
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    is it a?

  4. anonymous
    • one year ago
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    How did you get a?

  5. anonymous
    • one year ago
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    sim-36+-square root of (36)^2 -4(16)(4)/2(16)

  6. anonymous
    • one year ago
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    simplify 36/32

  7. anonymous
    • one year ago
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    OK but then it will need to be simplified by factoring out common factors. It is easier to do this beforehand. simplify\[-16x^2+36x+4=0\]by factoring out a common factor of 4

  8. anonymous
    • one year ago
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    Hello?

  9. anonymous
    • one year ago
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    ok so the answer is 9 square root 65 all over 8

  10. anonymous
    • one year ago
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    Not quite. I assume you've factored out the 4 to get\[-4x^2 + 9x + 1 = 0\]Now apply the quadratic formula with a=-4, b=9, and c=1.

  11. anonymous
    • one year ago
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    \[x=\frac{ -9 \pm \sqrt{9^2-4(-4)(1)} }{ 2(-4) } = ?\]

  12. anonymous
    • one year ago
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    is it the whole thing over 8?

  13. anonymous
    • one year ago
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    -9+-square root of 97/

  14. anonymous
    • one year ago
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    -8

  15. anonymous
    • one year ago
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    @ospreytriple

  16. anonymous
    • one year ago
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    Very good. Now you a negative divide by a negative. So it can be simplified a tiny bit more to get your final answer.

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