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Bananas1234

  • one year ago

Divide and simplify completely. Assume that no denominator equals zero. d^2-1/d^2-d divided by d - 1/d

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  1. Bananas1234
    • one year ago
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    @UnkleRhaukus

  2. UnkleRhaukus
    • one year ago
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    \[\frac{d^2-1}{d^2-d}\div\frac{d-1}d\\ =\frac{d^2-1}{d^2-d}\times\frac d{d-1}\\ =\]

  3. UnkleRhaukus
    • one year ago
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    factor the denominator of the first fraction . . .

  4. Bananas1234
    • one year ago
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    d^2 @UnkleRhaukus

  5. UnkleRhaukus
    • one year ago
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    huh?

  6. Bananas1234
    • one year ago
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    d^2 - 1/d^2

  7. UnkleRhaukus
    • one year ago
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    what is that susposed to be?

  8. Bananas1234
    • one year ago
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    the answer

  9. UnkleRhaukus
    • one year ago
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    its not right

  10. Bananas1234
    • one year ago
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    ok

  11. UnkleRhaukus
    • one year ago
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    factor the numerator and denominator of \(\dfrac{d^2-1}{d^2-d}\)

  12. Bananas1234
    • one year ago
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    wouldn't the top be the same and the bottom be d^2?

  13. Bananas1234
    • one year ago
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    @UnkleRhaukus

  14. UnkleRhaukus
    • one year ago
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    nope

  15. Bananas1234
    • one year ago
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    hmmm well then i am not sure :/ @UnkleRhaukus

  16. UnkleRhaukus
    • one year ago
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    can you factor d^2 - d ?

  17. Bananas1234
    • one year ago
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    (d^2 -d) (d^2 + d)

  18. UnkleRhaukus
    • one year ago
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    d^2 - d = d*d - d = d(d-1)

  19. Bananas1234
    • one year ago
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    oh

  20. UnkleRhaukus
    • one year ago
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    d^2-1 = (d-1)(d+1)

  21. Bananas1234
    • one year ago
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    what do i do next?

  22. Bananas1234
    • one year ago
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    @UnkleRhaukus

  23. UnkleRhaukus
    • one year ago
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    \[\frac{d^2-1}{d^2-d}\div\frac{d-1}d\\ =\frac{d^2-1}{d^2-d}\times\frac d{d-1}\\ =\frac{(d-1)(d+1)}{d(d-1)}\times\frac{d}{d-1}\\ =\]

  24. Bananas1234
    • one year ago
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    d + 1 / d - 1?

  25. UnkleRhaukus
    • one year ago
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    yep

  26. Bananas1234
    • one year ago
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    Thank you

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