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Mathematics
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\[\frac{12x^4y^3z^2w}{2xy^2}\]
all we need to do is use division for the numbers only and use one of the exponent rules for the variables. \[\frac{x^a}{x^b} \rightarrow x^{a-b}\] where x = variable a and b = numbers so for the numbers only we have 12/2, so what's 12 divided by 2
6

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Other answers:

yes.. so for the variables only we noticed that only x and y are present on the numerator and denominator, so we use the exponent law to figure this part out ... so if we let a = 4 and b = 1 from earlier \[\frac{x^4}{x^1} \rightarrow x^{4-1}\] we just have to subtract.. so what's 4-1 ?
3
alright so we have \[x^3\]
we do the same thing for the y variable because that exists on the numerator and denominator \[\frac{y^3}{y^2} \rightarrow y^{3-2}\] so all we need to do is subtract.. what is 3 -2 ?
1
yes so now we have just \[y^1 \] or you can write it as y only because it's to the first power... those 3 arithmetics we've just done are only necessary for this problem because z and w aren't present in the denominator, so we just leave them alone.
\[6x^3yz^2w\]
and that's our final answer.
tyvm<3

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