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Firejay5

  • 2 years ago

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

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

  2. Firejay5
    • 2 years ago
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    Is #32

  3. Sheng
    • 2 years ago
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    |dw:1350606375165:dw|

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

  5. Firejay5
    • 2 years ago
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    what number are you doing 30 or 32

  6. Firejay5
    • 2 years ago
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    This is monomials

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

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

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

  10. tjones89
    • 2 years ago
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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

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

  12. Firejay5
    • 2 years ago
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    is that it

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

  14. Sheng
    • 2 years ago
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    all the x terms, z terms cancel

  15. tjones89
    • 2 years ago
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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 \[-\frac{ x ^{-1}y ^{-4} }{ 4 }\]

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

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

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

  19. Sheng
    • 2 years ago
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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

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

  21. tjones89
    • 2 years ago
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    hah you're right, its really small print...

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

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

  24. Firejay5
    • 2 years ago
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    x^3/x^3 = x^1 or X^3 - x^3 = 0

  25. Firejay5
    • 2 years ago
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    x^0

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

  27. Firejay5
    • 2 years ago
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    you can divide or subtract depending on the number

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

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