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anonymous

  • one year ago

Find the height, in feet, of the rock after 5 seconds in the air. A rock is launched from a cannon. It’s height, h(x), can be represented by a quadratic function in terms of time, x, in seconds. After 1 second, the rock is 129 feet in the air; after 2 seconds, it is 236 feet in the air.

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  1. anonymous
    • one year ago
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    @JoKeR0331

  2. anonymous
    • one year ago
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    someone please help

  3. anonymous
    • one year ago
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    pleaseeeee

  4. anonymous
    • one year ago
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    @unknownunknown

  5. unknownunknown
    • one year ago
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    Do we have the angle of the cannon, is it firing directly upward?

  6. anonymous
    • one year ago
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    it dosent say @unknownunknown

  7. unknownunknown
    • one year ago
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    Can you do a screenshot of the question?

  8. anonymous
    • one year ago
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    i cant its a test

  9. anonymous
    • one year ago
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    do you want me to put every detail? @unknownunknown

  10. unknownunknown
    • one year ago
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    We will need every detail.. that's insufficient information as it is.

  11. anonymous
    • one year ago
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    Help please A rock is launched from a cannon. It's height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.After one second, the rock is 129 feet in the air; after 2 seconds, it is 236 feet in the air. Find the height, in feet, of the rock after 5 seconds in the air.

  12. anonymous
    • one year ago
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    @unknownunknown all of it

  13. unknownunknown
    • one year ago
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    Alright well, I'm going to assume a lot, so I can't see if this will be definitely correct, but the question contains incorrect information to answer properly. So we'll just have to proceed with what we have.. Let me confirm the equation to make sure, one sec.

  14. unknownunknown
    • one year ago
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    insufficient information*

  15. anonymous
    • one year ago
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    so you cant solve it at all? @unknownunknown

  16. anonymous
    • one year ago
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    @unknownunknown ?????

  17. unknownunknown
    • one year ago
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    Is there a multiple choice or you must write the answer?

  18. anonymous
    • one year ago
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    must write the answer @unknownunknown

  19. unknownunknown
    • one year ago
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    So, if we let y equal the displacement, y = 9.8*t^2/2 + initial velocity * time So first of all, we'll need to solve for the initial velocity, can you rearrange the equation to solve for v? y=9.8*t^2/2 + v * t

  20. unknownunknown
    • one year ago
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    Actually since it's coming down, we should make 9.8*t^2/2 negative. So y = -9.8*t^2/2 + v*t

  21. anonymous
    • one year ago
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    i think its a large number for feet @unknownunknown

  22. unknownunknown
    • one year ago
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    I'd say the answer is approximately 527, somewhere around there.

  23. anonymous
    • one year ago
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    its not @unknownunknown

  24. unknownunknown
    • one year ago
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    And how do you know that?

  25. anonymous
    • one year ago
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    i just tried

  26. anonymous
    • one year ago
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    @unknownunknown

  27. unknownunknown
    • one year ago
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    Ok wait, let me work it out

  28. unknownunknown
    • one year ago
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    Have you covered gravity in class? g = 9.8?

  29. anonymous
    • one year ago
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    no @unknownunknown

  30. unknownunknown
    • one year ago
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    Okay so. Let's say we have a function.. f(t) = -11*t^2 + 140*t If we plug into this equation t=1, we have -11 + 140 = 129 which satisfies the first part Then if we plug t=2, we have -11*4 + 140*2 = 236 which satisfies the second part

  31. unknownunknown
    • one year ago
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    So then for f(5) = -11*25 + 140*5 = 425

  32. anonymous
    • one year ago
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    right @unknownunknown

  33. unknownunknown
    • one year ago
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    Okay, good.

  34. anonymous
    • one year ago
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    can you help with another one @unknownunknown

  35. anonymous
    • one year ago
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    Widget wonders produces widgets. They have found that the cost, c(x), of making x widgets is a quadratic function in terms of x. The company also discovered that it costs $23 to produce 2 widgets, $103 to produce 4 widgets, and $631 to produce 6 widgets.

  36. anonymous
    • one year ago
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    @unknownunknown

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