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YanaSidlinskiy

  • one year ago

Help? I'm stuck. Applications of Quadratic Applications.

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  1. YanaSidlinskiy
    • one year ago
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    \[S=-16t^2+11t\] \[(-1)0= (-1)(-16t^2+112 t)\] \[0=16t^2-112t\] That is what I got. I've tried solving further but I'm confused...

  2. hihi67
    • one year ago
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    what r u tryin to figure out

  3. YanaSidlinskiy
    • one year ago
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    Well, now I'm supposed to be factoring the last equation...

  4. anonymous
    • one year ago
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    try completing the square

  5. Astrophysics
    • one year ago
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    You could use the quadratic equation if you like, c = 0, or complete the square.

  6. hihi67
    • one year ago
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    itll be in the form (a-b)(a+b)

  7. YanaSidlinskiy
    • one year ago
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    Yes, I do know that I just do not know how to go from what I've put up last.

  8. hihi67
    • one year ago
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    just take any random number, plug it in and see what happens i guess lol

  9. Astrophysics
    • one year ago
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    You could just do (11-16t)t=0 that gives your factors

  10. Astrophysics
    • one year ago
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    factoring out a t

  11. YanaSidlinskiy
    • one year ago
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    Nevermind I got it haha. Thanks guys!

  12. Astrophysics
    • one year ago
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    Ok :)

  13. anonymous
    • one year ago
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    \[16t^2-112t+(\frac{ 112 }{ 2 })^2= (\frac{ 112 }{ 2 })^2\]

  14. anonymous
    • one year ago
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    lol ok

  15. Astrophysics
    • one year ago
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    S = -16t^2+11t is different from 16t^2-112t, I'm assuming there was just a mistake there, but you can do the same as above for it in any case, I guess you realized that :P

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