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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    A firecracker shoots up from a hill 160 feet high, with an initial speed of 90 feet per second. Using the formula H(t) = -16t2 + vt + s, approximately how long will it take the firecracker to hit the ground?

  2. Liv1234
    • one year ago
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    Are there any options with this question?

  3. anonymous
    • one year ago
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    Yeah, the options are Five seconds Six seconds Seven seconds Eight seconds

  4. Liv1234
    • one year ago
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    Okay, I can try and help you(:

  5. Liv1234
    • one year ago
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    By the way, I'm still here I'm just working on solving it so I can help you.

  6. anonymous
    • one year ago
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    Alright

  7. Liv1234
    • one year ago
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    Multiply out everything, then move everything over to the right except for "y". If you get something in the form of y = ax^2 + bx + c, where 'a' is not zero, then it's a quadratic equation. 2) Multiply out the right until you get something in the form of y = ax^2 + bx + c, then look at what 'a' would be. 3) y = ax^2 + bx + c is going to be the graph of a parabola. If a > 0, then it opens upwards and the vertex shows the minimum value for y. If a < 0, then it opens downwards, and the vertex shows the maximum value for y. 4) Subtract 24x from both sides, then factor out 8x. 5) See #4

  8. anonymous
    • one year ago
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    Why delete it? I was in the middle of reading it D:

  9. jim_thompson5910
    • one year ago
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    i made a typo lol

  10. jim_thompson5910
    • one year ago
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    \[\large \text{A firecracker shoots up from a hill } {\color{red}{\text{160 feet high}}},\\ \large\text{with an }{\color{green}{\text{initial speed of 90 feet per second}}}.\\ \large \text{Using the formula } H(t) = -16t^2 + vt + s,\\ \large \text{approximately how long will it take the firecracker to hit the ground?}\] From that info, we can pull out \[\large \text{Initial Height: }{\color{red}{ s = 160}}\] \[\large \text{Initial Velocity: }{\color{green}{ v = 90}}\] So the expression \[\large -16t^2 + {\color{green}{v}}t + {\color{red}{s}}\] turns into \[\large -16t^2 + {\color{green}{90}}t + {\color{red}{160}}\] Hopefully this is making sense so far?

  11. jim_thompson5910
    • one year ago
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    but it's fixed now

  12. jim_thompson5910
    • one year ago
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    From here you need to solve \[\Large -16t^2 + 90t + 160=0\] for t. Use the quadratic formula to do so

  13. anonymous
    • one year ago
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    Alright, I can handle it from here. Sorry for the delay, the pizza man came :D

  14. jim_thompson5910
    • one year ago
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    That's ok. Let me know what solutions for t you get.

  15. anonymous
    • one year ago
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    If I remember too then I will!

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