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1. 8y 7 ÷ 4y 5 2. (18x 5 + 6x 4 - 12x 3) ÷ 6x 2 helppppp???

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huh ?
so first take care of the constants, \[\frac{8}{4}=\]

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Other answers:

then you have \[\frac{y^7}{y^5}\] since they have the same base you can use the above property so \[\frac{y^7}{y^5}=y^{7-5}\]
so combine your constant and your variables to get \[2y^{7-5}=2y^2\]
\[ (18x ^{5}+6x ^{4}-12x ^{3})\div6x ^{2} \] is the q btw it didnt post the powers
the second equation you need to split everything up into their own fractions
same for the first one
I have no idea how to do this
so \[\frac{18x^5+6x^4-12x^3}{6x^2}=\frac{18x^5}{6x^2}+\frac{6x^4}{6x^2}-\frac{12x^3}{6x^2}\]
then do what we did in the first one
Still have no idea lol
Thanks guys

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