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Math help:
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And don't take this post down, I'm tryna get help here and ya'll keep taking it down, then banning me
Shadow you don't need to write a whole essay lol. THE ANTICIPATION IS KILLING MEEEE!!!
ima eat a sandwhich. I'll be back
First, simplify the coes 12/21 = 2/3 then cancel out x ^8 and subtract the exponents of y: y ^6-2= y=^4 giving the final answer 2y^4/3
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First simplify the coes 14/21* Mb
@sllo wrote:
First, simplify the coes 12/21 = 2/3 then cancel out x ^8 and subtract the exponents of y: y ^6-2= y=^4 giving the final answer 2y^4/3
@gdubb08 wrote:
@sllo wrote:
First, simplify the coes 12/21 = 2/3 then cancel out x ^8 and subtract the exponents of y: y ^6-2= y=^4 giving the final answer 2y^4/3
@sllo wrote:
@gdubb08 wrote:
@sllo wrote:
First, simplify the coes 12/21 = 2/3 then cancel out x ^8 and subtract the exponents of y: y ^6-2= y=^4 giving the final answer 2y^4/3
Yeah that's what I got as well. I used the wrong coefficients at the start, silly me (: But let me know if you don't understand how you get there.
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@shadow wrote:
Yeah that's what I got as well. I used the wrong coefficients at the start, silly me (:
But let me know if you don't understand how you get there.
\[\frac{ 3 }{ x y^8}\] so ur saying that's the end result, for that new expression?
@shadow wrote:
\[\frac{ 3 }{ x y^8}\] so ur saying that's the end result, for that new expression?
\[9x^5y^8/3x^1y^1\] I got \[3x^4y^7\]
yup that's correct.
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@shadow wrote:
yup that's correct.
@shadow wrote:
yup that's correct.
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