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

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\[\frac{ -5x^3y^3z^4 }{ 20x^3y^7z^4 }\]
Is #32

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Other answers:

y^4 * 128 = -1 y^4 = -1/128 take 4th root for answer
what number are you doing 30 or 32
This is monomials
\[-\frac{ 5x^2y^3z^4 }{ 20x^3y^7z^4 } = -\frac{ 5 }{ 20 }\times \frac{ x^2 }{ x^3 }\times \frac{ y^3 }{ y^7 }\times \frac{ z^4 }{ z^4}\]
\[\frac{ x^2 }{ x^3 } = x^2x ^{-3} = x ^{2-3}\]
\[ = x ^{-1} = \frac{ 1 }{ x }\]
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
I got \[\frac{ - 1 }{ 4 }\] x^1y^4z^1
is that it
oh i thought you mean whole equation = 32... then answer is just -1/4y^4
all the x terms, z terms cancel
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 \[-\frac{ x ^{-1}y ^{-4} }{ 4 }\]
and yes \[\frac{ z ^{4} }{ z^4 } = 1 \] so it cancels
you can divide the exponents or subtract the exponents or both whenever you divide a fraction
another way to look at dividing variables with exponents...|dw:1350609282625:dw|
the x exponents cancel, why are you leaving them in? it's not a 2 on top, it's x^3
@tjones89 Actually both x's are 3s just looks like its a 2
hah you're right, its really small print...
yes so x's would be x^1 or x^0
x^2/x^2 = x^2-2 = x^0 = 1
x^3/x^3 = x^1 or X^3 - x^3 = 0
did you mean to put divide instead of subtract for x^3 - x^3?
you can divide or subtract depending on the number
like if it's the same you divide and if its different you subtract the big from little( 8-5)

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