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If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=

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If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=  [#permalink]

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New post 11 Jun 2015, 13:11
1
2
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

88% (00:55) correct 12% (01:24) wrong based on 124 sessions

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If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=

A. −42
B. −30
C. −6
D. 6
E. 30

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Re: If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=  [#permalink]

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New post 11 Jun 2015, 13:19
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Re: If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=  [#permalink]

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New post 11 Jun 2015, 13:27
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If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=  [#permalink]

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New post 12 Jun 2015, 09:41
1
I have often seen people getting confused by strange operators.

So to avoid confusion, I usually substitute the strange operator with a function.

In this case,

\(xθy\) can be written as \(f(x,y)\)

\(xθy = f(x,y) = x * y * (x - y)\)

So, \((1θ2)θ3\) can be written as \(f(f(1,2),3)\)

Solving this in stages,

\(1θ2 = f(1,2) = 1 * 2 * (1 - 2) = -2\)

and so it now becomes

\(f(-2,3) = -2 * 3 * (-2 - 3) = -2 * 3 * -5 = -6 * -5 = 30\)

OA = E
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Re: If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=  [#permalink]

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New post 29 Dec 2018, 11:37
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Re: If xθy=x·y(x−y) for all x and y, then (1θ2)θ3=   [#permalink] 29 Dec 2018, 11:37
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