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anonymous

  • 5 years ago

7 over 5x squared -20x+20 plus 1 over 20x squared - 80

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  1. anonymous
    • 5 years ago
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    lol, the question is not clear >_<

  2. anonymous
    • 5 years ago
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    7 over 5x squared minus 20x plus 20 add to 1 over 20x squared minus 80

  3. anonymous
    • 5 years ago
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    i agree with sstarica because 7over 5x or 7over5 then an x ?

  4. anonymous
    • 5 years ago
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    \[\frac{7}{5x^2} - 20x + 20 + \frac{1}{20x^2} - 80\]?

  5. anonymous
    • 5 years ago
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    ______7_____ 5x squared -20x+20 added to 1 over 20x squared - 80

  6. anonymous
    • 5 years ago
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    so is sstarica right?

  7. anonymous
    • 5 years ago
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    sorry for the confusion Ihavent quited figured out how to work this program yet

  8. anonymous
    • 5 years ago
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    \[\frac{7}{5x^2-20x+20} + \frac{1}{20x-80}\]

  9. anonymous
    • 5 years ago
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    yes

  10. anonymous
    • 5 years ago
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    alright lol :)

  11. anonymous
    • 5 years ago
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    how do you get horizontal bars for the fractions with the maths editor

  12. anonymous
    • 5 years ago
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    frac{a}{b} <-- ^_^

  13. anonymous
    • 5 years ago
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    \[\frac{7}{5(x-4)^2} + \frac{1}{20(x-4)}\]\[\frac{7}{5(x^2-4x+4)} + \frac{1}{20(x-4)}\]

  14. anonymous
    • 5 years ago
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    wrong order lol

  15. anonymous
    • 5 years ago
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    lol, I noticed

  16. anonymous
    • 5 years ago
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    multiply top and bottom of second by x-4

  17. anonymous
    • 5 years ago
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    it's actually (x-2)^2 :)

  18. anonymous
    • 5 years ago
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    nvm LOL, I wrote the wrong question

  19. anonymous
    • 5 years ago
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    yeh something like that , im going to go to sleep soon

  20. anonymous
    • 5 years ago
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    no, wait, I wrote the right question >_<

  21. anonymous
    • 5 years ago
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    (x-2)^2 ^_^" instead of (x-4)^2

  22. anonymous
    • 5 years ago
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    its 1.40 in the morning where I am , I am too generous staying up this late to help ppls lol

  23. anonymous
    • 5 years ago
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    lol, go to bed

  24. anonymous
    • 5 years ago
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    \[= \frac{7(4)(x-4) + (x-2)(x-2)}{20(x-2)^2(x-4)}\] \[= \frac{28x - 112 +x^2 - 4x -4}{20(x-2)^2(x-4)}\] \[= \frac{x^2 + 24x -116}{20(x-2)^2(x-4)}\] hmm, there's something wrong

  25. anonymous
    • 5 years ago
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    I'm not sure

  26. anonymous
    • 5 years ago
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    did you understand what I just did?

  27. anonymous
    • 5 years ago
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    not really i'm not very smart with algebra

  28. anonymous
    • 5 years ago
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    I factored, then put both fractions with the same denominator to have them in one :)

  29. anonymous
    • 5 years ago
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    write down the steps on paper, and you'll figure out what I did ^_^

  30. anonymous
    • 5 years ago
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    wait a min, there's a mistake in the last 2 steps, hold on dear ^_^

  31. anonymous
    • 5 years ago
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    I'm lost because under the 1 in 20x squared minues 80

  32. anonymous
    • 5 years ago
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    \[= \frac{28x-112+x2-4x+4}{20(x-2)^2(x-4)}\]<-- for the second line

  33. anonymous
    • 5 years ago
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    I factored, I pulled out 20 as a common factor and was left with (x-4) use the distributive law to check ^_^

  34. anonymous
    • 5 years ago
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    \[= \frac{x^2+24x-108}{20(x-2)^2(x-4)}\]

  35. anonymous
    • 5 years ago
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    sorry about the mess, but are you following now :)?

  36. anonymous
    • 5 years ago
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    i think so, thank you so much

  37. anonymous
    • 5 years ago
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    np ^_^, but there's still something fishy about it lol, I can't seem to factor the numerator

  38. anonymous
    • 5 years ago
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    does the question want you to add only? or simplify?

  39. anonymous
    • 5 years ago
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    simplify

  40. anonymous
    • 5 years ago
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    hmm

  41. anonymous
    • 5 years ago
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    I'm sure of the following steps I just made, but looks it can't be factored

  42. anonymous
    • 5 years ago
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    thank you

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