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Saylilbaby

  • one year ago

help wanted will fan and medal....

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  1. saylilbaby
    • one year ago
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  2. saylilbaby
    • one year ago
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  3. saylilbaby
    • one year ago
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  4. saylilbaby
    • one year ago
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    @amistre64

  5. saylilbaby
    • one year ago
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    @Bee_see @mathmate @mathmath333 @kropot72

  6. Bee_see
    • one year ago
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    In the first problem, work on the numerator part first. Make them have the same denominator.

  7. Bee_see
    • one year ago
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    |dw:1441667474814:dw|

  8. Bee_see
    • one year ago
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    what's missing on the left side? What's missing on the right side? You want only y to be on the denominator

  9. Bee_see
    • one year ago
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    do you still need help?

  10. saylilbaby
    • one year ago
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    @radar

  11. Bee_see
    • one year ago
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    For the first problem, make the denominator the same on the numerator...so: |dw:1441672733625:dw|

  12. Bee_see
    • one year ago
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    |dw:1441672797727:dw|

  13. Bee_see
    • one year ago
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    remember that when you are adding or subtracting fractions, the fractions must have the same denominator. something like: 2/7 + 1/7 = 3/7

  14. Bee_see
    • one year ago
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    You have to ask yourself what the one side has that the other doesn't and you add the missing item on the top and bottom

  15. Bee_see
    • one year ago
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    When you are dividing fractions, you take the denominator, place it besides the numerator, flip it, and multiply...so....like this: |dw:1441673291161:dw|

  16. Bee_see
    • one year ago
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    |dw:1441673720939:dw|

  17. anonymous
    • one year ago
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    Invert the denominator, multiply the numerator by the result and then simplify.\[\frac{1}{7} (3 x) \left(\frac{4 x}{y}+x\right)=\frac{12 x^2}{7 y}+\frac{3 x^2}{7}=\frac{3 x^2 (y+4)}{7 y} \]

  18. Bee_see
    • one year ago
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    in the 2nd question., cross multiply.

  19. Bee_see
    • one year ago
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    |dw:1441674540603:dw|

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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