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Board of Directors D
Joined: 01 Sep 2010
Posts: 3405
If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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Difficulty:   5% (low)

Question Stats: 84% (01:06) correct 16% (01:19) wrong based on 152 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 58396
Re: If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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

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Math Expert V
Joined: 02 Sep 2009
Posts: 58396
If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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

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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Board of Directors D
Joined: 01 Sep 2010
Posts: 3405
Re: If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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I didn't know this topic. It is very useful to cover topics not so common.

As soon as I will have time, I will compile an archive with questions and explanations.

Thanks
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Manager  Joined: 26 Feb 2015
Posts: 110
Re: If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58396
If the operation ˆˆ is deﬁned for all x and y by the equatio  [#permalink]

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

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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_________________ 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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