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If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 =

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Bunuel
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If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   08 May 2015, 06:33
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If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 =

(A) 4/3
(B) 1/3
(C) 1/12
(D) -1/12
(E) -4/3

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C
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funkyleaf
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   08 May 2015, 13:25
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Bunuel wrote:
If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 =

(A) 4/3
(B) 1/3
(C) 1/12
(D) -1/12
(E) -4/3

Kudos for a correct solution.


Working inside to out:
(-1)^2/2= 1/2
gives us: (1/2)#3
(1/2)^2/3 = (1/4)(1/3) = 1/12 answer choice C.
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   09 May 2015, 02:46
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Since you always solve the part in brackets
\(-1#2 = (-1)^2/2= 1/2\)
Thus,
\((1/2)#3 = (1/2)^2/3 = (1/4)(1/3) = 1/12\)

Answer C
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   09 May 2015, 09:33
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Bunuel wrote:
If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 =

(A) 4/3
(B) 1/3
(C) 1/12
(D) -1/12
(E) -4/3

Kudos for a correct solution.


(-1#2)=-1^2/2=1/2
And, (1/2)#3=1/4/3=1/4*1/3=1/12
Answer C
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   08 May 2015, 10:52
(-1)^2 / 2 = 1/2
(1/2)^2 / 3 = 1/12
so C is my answer
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink] New post   04 Sep 2021, 23:14
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Re: If x#y is defined to equal x^2/y for all x and y, then (-1#2)#3 = [#permalink]
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