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If the operation ˆˆ is defined for all x and y by the equatio

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New post 21 Jun 2014, 08:19
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If the operation ^ ^ is defined for all x and y by the equation x ^ ^ y = \(\frac{x^2 y}{2}\) , then (2 ^ ^ − 1) ^ ^ (−2 ^ ^1) =

(A) -4
(B) -2
(C) 2
(D) 4
(E) 8

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Re: If the operation ˆˆ is defined for all x and y by the equatio  [#permalink]

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New post 21 Jun 2014, 08:41
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New post 21 Jun 2014, 10:59
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Bunuel wrote:
carcass wrote:
If the operation ^ ^ is defined for all x and y by the equation x ^ ^ y = \(\frac{x^2 y}{2}\), then (2 ^ ^ − 1) ^ ^ (−2 ^ ^1) =

(A) -4
(B) -2
(C) 2
(D) 4
(E) 8


(2 ^ ^ − 1) = \(\frac{2^2*(-1)}{2}=-2\);

(−2 ^ ^1) = \(\frac{(-2)^2*1}{2}=2\);

Thus, -2 ^ ^ 2 = \(\frac{(-2)^2*2}{2}=4\).

Answer: D.


Just compiled questions about various functions in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functios

Hope it helps.
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Re: If the operation ˆˆ is defined for all x and y by the equatio  [#permalink]

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New post 21 Jun 2014, 11:20
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Re: If the operation ˆˆ is defined for all x and y by the equatio  [#permalink]

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New post 05 Mar 2015, 03:52
Bunuel wrote:
Bunuel wrote:
carcass wrote:
If the operation ^ ^ is defined for all x and y by the equation x ^ ^ y = \(\frac{x^2 y}{2}\), then (2 ^ ^ − 1) ^ ^ (−2 ^ ^1) =

(A) -4
(B) -2
(C) 2
(D) 4
(E) 8


(2 ^ ^ − 1) = \(\frac{2^2*(-1)}{2}=-2\);

(−2 ^ ^1) = \(\frac{(-2)^2*1}{2}=2\);

Thus, -2 ^ ^ 2 = \(\frac{(-2)^2*2}{2}=4\).

Answer: D.


Various Functios

Hope it helps.


I thought you were supposed to take those two results you got minus each other, ie -2 - (2) = -4?

I mean, the two things you are solving for, arent those the actual equation?
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If the operation ˆˆ is defined for all x and y by the equatio  [#permalink]

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New post 05 Mar 2015, 06:08
erikvm wrote:
Bunuel wrote:
Bunuel wrote:
If the operation ^ ^ is defined for all x and y by the equation x ^ ^ y = \(\frac{x^2 y}{2}\), then (2 ^ ^ − 1) ^ ^ (−2 ^ ^1) =

(A) -4
(B) -2
(C) 2
(D) 4
(E) 8

(2 ^ ^ − 1) = \(\frac{2^2*(-1)}{2}=-2\);

(−2 ^ ^1) = \(\frac{(-2)^2*1}{2}=2\);

Thus, -2 ^ ^ 2 = \(\frac{(-2)^2*2}{2}=4\).

Answer: D.


Various Functios

Hope it helps.


I thought you were supposed to take those two results you got minus each other, ie -2 - (2) = -4?

I mean, the two things you are solving for, arent those the actual equation?


There is function sign between them: (2 ^ ^ − 1) ^ ^ (−2 ^ ^1)

(2 ^ ^ − 1) = -2 and (−2 ^ ^1) = 2, hence (2 ^ ^ − 1) ^ ^ (−2 ^ ^1) = (-2) ^ ^ 2 = (-2)^2*2/2 = 4.

Hope it's clear.
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Re: If the operation ˆˆ is defined for all x and y by the equatio  [#permalink]

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New post 22 Oct 2018, 23:29
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Re: If the operation ˆˆ is defined for all x and y by the equatio   [#permalink] 22 Oct 2018, 23:29
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