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BloomLocke367

  • one year ago

Hmmm...this is a dumb question, but to get rid of a fraction in an equation, do you multiply everything by the denominator of the fraction, or the reciprocal?

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  1. BloomLocke367
    • one year ago
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    @jdoe0001 lol sorry for bothering you

  2. anonymous
    • one year ago
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    no question is a dumb question

  3. Nnesha
    • one year ago
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    you would multiply `with the reciprocal` of the denominator

  4. BloomLocke367
    • one year ago
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    It is when you're in precal honors and can't remember basic math rules lol XD

  5. BloomLocke367
    • one year ago
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    thank you

  6. Nnesha
    • one year ago
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    \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } }=\frac{ a }{ b } \times \frac{ d }{ c }\]

  7. anonymous
    • one year ago
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    haha

  8. anonymous
    • one year ago
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    im honers to

  9. Nnesha
    • one year ago
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    nice!

  10. anonymous
    • one year ago
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    :3

  11. jdoe0001
    • one year ago
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    there was a fellow the other day here he said he knew physics, but never seen a logarithm :|

  12. jdoe0001
    • one year ago
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    go figure

  13. anonymous
    • one year ago
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    r u Muslim Nnesha

  14. anonymous
    • one year ago
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    i am

  15. BloomLocke367
    • one year ago
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    So I have \(x^2-7x-\frac{3}{4}=0\), and to get rid of that fraction I'd multiply everything} by \(\frac{4}{3}\)?

  16. Nnesha
    • one year ago
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    ahh no!

  17. BloomLocke367
    • one year ago
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    wait...you said reciprocal of the denominator... so 1/4

  18. BloomLocke367
    • one year ago
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    yes, no, maybe so?

  19. BloomLocke367
    • one year ago
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    Oh, and @jdoe0001, I'm taking AP Physics right now XD I love it

  20. Nnesha
    • one year ago
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    yes right

  21. BloomLocke367
    • one year ago
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    That just creates more fractions D:

  22. Nnesha
    • one year ago
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    :3 x^2-7x-\frac{ 3 }{ 4} \times \frac{1}{4} \[x^2-7x-\frac{ 3 }{ 4} \times \frac{1}{4} \] don't see any points to get more messy stuff :P

  23. triciaal
    • one year ago
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    there are no dumb questions!!!!

  24. Nnesha
    • one year ago
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    i didn't noticed it's 1/4 not 4/3

  25. anonymous
    • one year ago
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    thats what i sayed

  26. BloomLocke367
    • one year ago
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    I dunno...

  27. Nnesha
    • one year ago
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    ignore me plz :(

  28. BloomLocke367
    • one year ago
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    ummm... okay. I guess I'll just ask Mrs. Dunn tomorrow

  29. Nnesha
    • one year ago
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    no wait don't close !

  30. anonymous
    • one year ago
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    nesha r u muslim i am

  31. BloomLocke367
    • one year ago
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    Please don't put spam on my question. >.<

  32. jdoe0001
    • one year ago
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    \(\bf x^2-7x-\frac{3}{4}=0\implies 4\left( x^2-7x-\frac{3}{4} \right)=4(0) \\ \quad \\ 4x^2-21x-\cancel{4}\cdot \cfrac{3}{\cancel{4}}=0\implies 4x^2-21x-3=0\)

  33. triciaal
    • one year ago
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    to get rid of a fraction in an equation, do you multiply everything by the denominator of the fraction, or the reciprocal? you multiply everything by the lowest common denominator

  34. jdoe0001
    • one year ago
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    hmmm rather \(\bf x^2-7x-\frac{3}{4}=0\implies 4\left( x^2-7x-\frac{3}{4} \right)=4(0) \\ \quad \\ 4x^2-28x-\cancel{4}\cdot \cfrac{3}{\cancel{4}}=0\implies 4x^2-28x-3=0\)

  35. BloomLocke367
    • one year ago
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    OKAY THANK YOU OMG.

  36. triciaal
    • one year ago
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    remember the denominator of a whole number is 1

  37. triciaal
    • one year ago
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    you are welcome

  38. BloomLocke367
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @triciaal remember the denominator of a whole number is 1 \(\color{#0cbb34}{\text{End of Quote}}\) Yes, I know :D

  39. Nnesha
    • one year ago
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    fraction in the equation i was doing totally different thin'!

  40. Nnesha
    • one year ago
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    srry about that

  41. BloomLocke367
    • one year ago
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    It's okay :) thanks guys

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