anonymous
  • anonymous
Help for a medal???
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Charlie needs to simplify the expression below before he substitutes values for x and y. If x ≠ 0 and y ≠ 0, which of the following is a simplified version of the expression below? x^9y^5 x^24y^16 x^6y^3 + x^3y^2 x^15y^8 + x^6y^4
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anonymous
  • anonymous
@zepdrix Can you help?
zepdrix
  • zepdrix
\[\Large\rm \frac{x^{18}y^{12}+x^9y^8}{x^3y^4}\]So um... we have three x's multiplying in the bottom. We'd like to find three x's in the top also, so we can cancel some stuff out. But to do that we need to take the x's out of each term in the top. Do you remember how to factor? :) How do we take three x's out of each term in the top?\[\Large\rm x^{18}y^{12}+x^9y^8=x^3(\quad ?\quad+\quad?\quad)\]

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zepdrix
  • zepdrix
Hannah Bananaaaaa +_+ where you at girrrrrr
anonymous
  • anonymous
Sorry D: Pizza comes first math comes second! Do I add the Xs together?
zepdrix
  • zepdrix
Ooo pizza noice :U no. You're dividing some x's out of each term.|dw:1434569099407:dw|
anonymous
  • anonymous
Ooooh, okay.
anonymous
  • anonymous
So would you divide x^3 by x^9 and x^18?
zepdrix
  • zepdrix
mhm :o
anonymous
  • anonymous
Alright, give me a sec
anonymous
  • anonymous
x^6y^12 + x^3y^8
zepdrix
  • zepdrix
See the first term, you're multiplying 18 x's together. If you pull 3 of them out, then you're only multiplying 15 x's together, ya? I don't think you wanted to divide the 18 by 3 silly :)
anonymous
  • anonymous
Oooooooooooooooooooohhhhhh. Alright, gotcha!
zepdrix
  • zepdrix
So let's see what pulling out the x^3 does for us:\[\Large\rm \frac{x^{18}y^{12}+x^9y^8}{x^3y^4}=\frac{\color{orangered}{x^3}(x^{15}y^{12}+x^3y^{8})}{\color{orangered}{x^3}y^4}\]Ooo notice now that we have some matching x's on top and bottom! So we can cancel out that junk.\[\Large\rm \frac{\cancel{\color{orangered}{x^3}}(x^{15}y^{12}+x^3y^{8})}{\cancel{\color{orangered}{x^3}}y^4}=\frac{x^{15}y^{12}+x^3y^8}{y^4}\]
zepdrix
  • zepdrix
We'll have to do a similar step to deal with the y's.
zepdrix
  • zepdrix
Any confusion on that step Bananz? :o You have pizza in your hand don't you -_- lol
anonymous
  • anonymous
RIGHT
zepdrix
  • zepdrix
prolly cheese everywhere -_-
zepdrix
  • zepdrix
Since we have y^4 in the bottom, we'd like to try and pull out y^4 from each term in top.
anonymous
  • anonymous
Haha pizza is my life :D
anonymous
  • anonymous
Wait, what do you mean pull the y^4 out? Do you just get rid of it? Or do you gotta divide the y^4 by other Y's?
anonymous
  • anonymous
Or is it something else?
zepdrix
  • zepdrix
yes, divide the other y's by y^4.
anonymous
  • anonymous
Alright
anonymous
  • anonymous
\[(x ^{15} y ^{3} + x^3 y^2) = x ^{15} y^{3} + x^3 y^2\]
anonymous
  • anonymous
That took a LONG TIME to type using the equation thing.
zepdrix
  • zepdrix
woops! you did it again silly. when you divide y's like that, you need to `subtract` the exponents.\[\Large\rm \frac{y^{12}}{y^4}=y^{12-4}=y^8\]
anonymous
  • anonymous
OOOOOOOOH
anonymous
  • anonymous
Alright, I'll do it again
anonymous
  • anonymous
\[(x^{15} y^{12} +x^3 y^{4}) = x^{15} y^{8} + x^3 y^4\]
anonymous
  • anonymous
OMG I FORGOT ABOUT THE y^12
anonymous
  • anonymous
This has proved difficult doing while eating pizza
zepdrix
  • zepdrix
careful the way you're writing that +_+ don't forget about the y^4 out front.\[\Large\rm (x^{15} y^{12} +x^3 y^{4}) = \color{orangered}{y^4}(x^{15} y^{8} + x^3 y^4)\]Your work looks correct though :)
zepdrix
  • zepdrix
\[\Large\rm \frac{x^{15}y^{12}+x^3y^8}{y^4}=\frac{\color{orangered}{y^4}(x^{15}y^{8} + x^3y^4)}{\color{orangered}{y^4}}\]
anonymous
  • anonymous
Haha, sorry about that XD
zepdrix
  • zepdrix
Cool, so do you see which option you'll end up with? :)
anonymous
  • anonymous
D?
anonymous
  • anonymous
I dedicate this song to you https://www.youtube.com/watch?v=TjcOJmoJwpk skip to the 0:26 second mark please
zepdrix
  • zepdrix
yay good job \c:/
zepdrix
  • zepdrix
lolol -_-
anonymous
  • anonymous
I know cliche :D But still, thank you!

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