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Matt6288

  • 4 years ago

14ab^3/7a^-2b^-1

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  1. Mertsj
    • 4 years ago
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    \[\frac{14ab ^{3}}{7a ^{-2}b ^{-1}}\]

  2. Matt6288
    • 4 years ago
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    yes

  3. Mertsj
    • 4 years ago
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    Hang on a sec

  4. Matt6288
    • 4 years ago
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    k

  5. Mertsj
    • 4 years ago
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    \[2a ^{3}b ^{4}\]

  6. Matt6288
    • 4 years ago
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    thx

  7. Mertsj
    • 4 years ago
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    Do you understand?

  8. Matt6288
    • 4 years ago
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    can you explain?

  9. Mertsj
    • 4 years ago
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    The factors that are in the denominator have negative exponents (except for the 7) When you move them to the numerator, you change the exponents to positive.

  10. Mertsj
    • 4 years ago
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    \[\frac{1}{a ^{-n}}=a ^{n}\]

  11. Mertsj
    • 4 years ago
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    \[\frac{a}{a ^{-2}}=a(a ^{2})=a ^{3}\]

  12. Matt6288
    • 4 years ago
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    a^3b^3c/1^-3b^-3c^-1

  13. Matt6288
    • 4 years ago
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    would it be a^6b^6c^2?

  14. Mertsj
    • 4 years ago
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    And\[\frac{b ^{3}}{b ^{-1}}=b ^{3}b ^{1}=b ^{4}\]

  15. Mertsj
    • 4 years ago
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    \[\frac{a ^{3}b ^{3}c}{1^{-3}b ^{-3}c ^{-1}}\]

  16. Mertsj
    • 4 years ago
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    Is that the problem?

  17. Mertsj
    • 4 years ago
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    or is there a typo?

  18. Matt6288
    • 4 years ago
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    ya

  19. Matt6288
    • 4 years ago
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    thats the problem

  20. Mertsj
    • 4 years ago
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    So the denominator contains 1^-3?

  21. Matt6288
    • 4 years ago
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    and my answer is A^6b^6c^2

  22. Matt6288
    • 4 years ago
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    ya

  23. Mertsj
    • 4 years ago
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    yes. If that is a^-3 in the denominator instead of 1^-3

  24. Matt6288
    • 4 years ago
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    thx :) i get it now

  25. Mertsj
    • 4 years ago
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    yw

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