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Firejay5

Simplify. 30. 2x^2(6y^3)(2x^2y) 32. −5x3y3z420x3y−4z4

  • one year ago
  • one year ago

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  1. Firejay5
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    \[\frac{ -5x^3y^3z^4 }{ 20x^3y^7z^4 }\]

    • one year ago
  2. Firejay5
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    Is #32

    • one year ago
  3. Sheng
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    |dw:1350606375165:dw|

    • one year ago
  4. Sheng
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    y^4 * 128 = -1 y^4 = -1/128 take 4th root for answer

    • one year ago
  5. Firejay5
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    what number are you doing 30 or 32

    • one year ago
  6. Firejay5
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    This is monomials

    • one year ago
  7. tjones89
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    \[-\frac{ 5x^2y^3z^4 }{ 20x^3y^7z^4 } = -\frac{ 5 }{ 20 }\times \frac{ x^2 }{ x^3 }\times \frac{ y^3 }{ y^7 }\times \frac{ z^4 }{ z^4}\]

    • one year ago
  8. tjones89
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    \[\frac{ x^2 }{ x^3 } = x^2x ^{-3} = x ^{2-3}\]

    • one year ago
  9. tjones89
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    \[ = x ^{-1} = \frac{ 1 }{ x }\]

    • one year ago
  10. tjones89
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    use these steps for each one of those terms: \[\frac{ y^3 }{ y^7 }\] \[\frac{ z^4 }{ z^4 }\] and then put them all together

    • one year ago
  11. Firejay5
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    I got \[\frac{ - 1 }{ 4 }\] x^1y^4z^1

    • one year ago
  12. Firejay5
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    is that it

    • one year ago
  13. Sheng
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    oh i thought you mean whole equation = 32... then answer is just -1/4y^4

    • one year ago
  14. Sheng
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    all the x terms, z terms cancel

    • one year ago
  15. tjones89
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    not quite, the exponents on x and y should be negative, or you should put them in the denominator of your final answer like this \[-\frac{ 1 }{ 4xy^4 }\] or \[-\frac{ x ^{-1}y ^{-4} }{ 4 }\]

    • one year ago
  16. tjones89
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    and yes \[\frac{ z ^{4} }{ z^4 } = 1 \] so it cancels

    • one year ago
  17. Firejay5
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    you can divide the exponents or subtract the exponents or both whenever you divide a fraction

    • one year ago
  18. tjones89
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    another way to look at dividing variables with exponents...|dw:1350609282625:dw|

    • one year ago
  19. Sheng
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    the x exponents cancel, why are you leaving them in? it's not a 2 on top, it's x^3

    • one year ago
  20. Firejay5
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    @tjones89 Actually both x's are 3s just looks like its a 2

    • one year ago
  21. tjones89
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    hah you're right, its really small print...

    • one year ago
  22. Firejay5
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    yes so x's would be x^1 or x^0

    • one year ago
  23. tjones89
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    x^2/x^2 = x^2-2 = x^0 = 1

    • one year ago
  24. Firejay5
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    x^3/x^3 = x^1 or X^3 - x^3 = 0

    • one year ago
  25. Firejay5
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    x^0

    • one year ago
  26. tjones89
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    did you mean to put divide instead of subtract for x^3 - x^3?

    • one year ago
  27. Firejay5
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    you can divide or subtract depending on the number

    • one year ago
  28. Firejay5
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    like if it's the same you divide and if its different you subtract the big from little( 8-5)

    • one year ago
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