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anonymous

  • 5 years ago

need help with 3 more problems plz help

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  1. anonymous
    • 5 years ago
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    ?

  2. anonymous
    • 5 years ago
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    i have 3 more math problems to do then i am done..lol

  3. anonymous
    • 5 years ago
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    i dont want to keep u up

  4. anonymous
    • 5 years ago
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    it ok I'm free

  5. anonymous
    • 5 years ago
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    bonus #1

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  6. anonymous
    • 5 years ago
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    i did number 25 already

  7. anonymous
    • 5 years ago
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    ok u want do bonus question 1?

  8. anonymous
    • 5 years ago
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    yes plz

  9. anonymous
    • 5 years ago
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    3x^2+5x-1=0 x=(-5+-sqrt(25+4))/6 now simplify

  10. anonymous
    • 5 years ago
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    \[\frac{-5\pm \sqrt{5^{2}-4(3\times (-1)}}{(2\times3)} =\frac{-5\pm \sqrt{25+12}}{6}\]

  11. anonymous
    • 5 years ago
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    OK , bye

  12. anonymous
    • 5 years ago
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    bye

  13. anonymous
    • 5 years ago
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    i need help with 2 more problems

  14. anonymous
    • 5 years ago
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    Which questions do you need help on?

  15. anonymous
    • 5 years ago
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  16. anonymous
    • 5 years ago
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    number 18 on this one

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  17. anonymous
    • 5 years ago
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    Ok opposite sides of a paralleogram are equal in length so make to equations y2 = 49 so y = 7 (answer) Other equation x + 9 = 4x - 27 subtract x from each side 9 = 3x - 27 Add 27 to each side 18 = 3x divide each side by 3 x = 6 *answer()

  18. anonymous
    • 5 years ago
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    can u show me a break down of the first one plz

  19. anonymous
    • 5 years ago
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    The part about paralellograms?

  20. anonymous
    • 5 years ago
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    One side of the parallelogram is y squared and the other is 49 They are equal in length so if we take the sq root of both of them y = 7

  21. anonymous
    • 5 years ago
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    how u got the answer because i dont get it im confused

  22. anonymous
    • 5 years ago
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    oh we are dividing

  23. anonymous
    • 5 years ago
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    No we are finding the square root of 49 and of y squared because they are equal

  24. anonymous
    • 5 years ago
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    wait are we doing number 18 or bonus number 2

  25. anonymous
    • 5 years ago
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    ohh we have to solve twice

  26. anonymous
    • 5 years ago
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    thats where i got confused

  27. anonymous
    • 5 years ago
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    !8 but I can do bonus 2 for you if you wish

  28. anonymous
    • 5 years ago
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    yes please

  29. anonymous
    • 5 years ago
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    I'm on it right now

  30. anonymous
    • 5 years ago
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    thanks a bunch

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