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\[\frac{ x^-3y^5z^0 }{ x^2y^8 }^2\]

()^2

x^6

y^-15

\[\frac{ x ^{a} }{ x ^{b} }=x ^{a-b}\]

\[\left(\begin{matrix}x^-3y^5z^0 \\ x^2y^8\end{matrix}\right)^2\]

thats the problem

i'm new at this

yeah later on you squared them after simplifying the terms

oh

i'm so lost with these

but first simlipfying the division of powers are adding and subtracting on not multiplying .. :D

so x^1y^-3z^0

\[\frac{ z }{ x^2y^6 }\]

\[\frac{ y ^{-c} }{ y ^{b} }=y ^{-c-b}\]

look i dont understand

\[\frac{ v ^{-c} }{ y ^{b} }=y ^{-c}*y ^{-b}=y ^{-c-b}\]

i get the x^-1*2 y^-3*2

for example only
\[\frac{ y ^{-5} }{ y ^{7} }=y ^{-5}*y ^{-7}=y ^{-5-7}=y ^{-12}\]

oh

when you move the denominator to the numerator the sign of its power is change

ok

\[\left(\begin{matrix}x^0 \\ x^1y^3\end{matrix}\right)^2\]

thats what i got

\[\left(\begin{matrix}z \\ x^1y^3\end{matrix}\right)^2\]

that top is z

hm the idea is to move all the denominator upward to the numerator first then do add or subtract

so x^-3-2y^5-8z^0

yes correct and z^0=1 right?

i guess

so x^-1y-3z^1

any number or letter raised to the power of zero is =1

oh right i have been working on homework for the last 5 hours my mind is fried =)

hmm what is
-2-3=?
5-8=?

your 5-8= -3 is correct :D

-5

yesss :D

yess =)

so whats the answer now?

so x^-5y^-3x^1

than that times 2

so x^-5*2=-10 y^-3*2=-6 x^1*2=2

\[\left(\begin{matrix}z^2 \\ x^10y^6\end{matrix}\right)\]

now that z^0=1
\[(x ^{-5}y ^{-3})^{2}=\]?

x^-10y^-6

\[(x ^{b})^{a}=x ^{ab}\]
multiply now the power here

x^-5*2

=10

yess correct do that to y also

y^-3*2=-6

yess but dont forget about the sign :D

so it's z^2/x^10y^6

oh so it's 1/x^10y^6

yesssssssssssss :D

its right!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

thanks

YW have fun now and enjoy solving good luck ......... :D

would u mind helping me with the other one i have on line

ok yesss where is it? @hannalafave

it did not work but i got one for u now 12a^3-3a^2-5a+10/4a+3

ok the easy one to do for this one is by long division can you try that?

yea can u go to link i sent u?

ok