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If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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Updated on: 05 Mar 2016, 18:43
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If \(xy \neq 0\), what is the value of \(\frac{((x^2)^3 y^4)}{x}\)? (1) \(x^3y^5 = \frac{6y}{x^2}\) (2) \(x = 2y\)
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Originally posted by tonebeeze on 19 Oct 2010, 11:15.
Last edited by ENGRTOMBA2018 on 05 Mar 2016, 18:43, edited 1 time in total.
Reformatted the question



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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19 Oct 2010, 11:23



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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19 Oct 2010, 13:29
Thanks Bunuel. I apologize for the bad formatting. Won't happen again!



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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15 Oct 2013, 09:06
tonebeeze wrote: Thanks Bunuel. I apologize for the bad formatting. Won't happen again! And just to emphasize that we can divide both sides by y ONLY BECAUSE we are explicitly told that x and y are both different from zero as Bunuel clearly explained above. Cheers J



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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05 Mar 2016, 18:38
Bunuel  Can you please add the math formula formatting tag to the OP. For the first minute, I was solving for y as a power of x. :l Thanks!



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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08 Mar 2016, 20:36
jlgdr wrote: tonebeeze wrote: Thanks Bunuel. I apologize for the bad formatting. Won't happen again! And just to emphasize that we can divide both sides by y ONLY BECAUSE we are explicitly told that x and y are both different from zero as Bunuel clearly explained above. Cheers J You are absolutely right... if xy≠0 wasn't mentioned then A would be insufficient as y can be zero too. and hence E would have been the answer.. Thanks
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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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18 Sep 2016, 08:53
A is correct. Here's why:
Manipulate orig eq to the following > (x^5)(y^4) = ?
(1) (x^3)(y^5) = 6y = (x^5)(y^4) = 6
SUFFICIENT
(2) x = 2y > plug into orig eq =(2y)^5[(y^4)] = 32(y^9)
NOT SUFFICIENT



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Re: If xy does not equal 0, what is the value of ((x^2)^3 y^4)/x
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25 Nov 2018, 10:15
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