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The operation @ is defined for all non-zero x and y by the

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The operation @ is defined for all non-zero x and y by the [#permalink] New post 16 Mar 2006, 19:07
The operation @ is defined for all non-zero x and y by the equation x@y=x/y. Then the expression (x@y)@z is equal to?

I get as far as:

(x@y)@z

= (x/y)@z

What is next?
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 [#permalink] New post 16 Mar 2006, 22:01
x@y = x/y

(x@y)@z = (x/y)@z = x/yz
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Re: Defined Function Problem [#permalink] New post 16 Mar 2006, 23:20
sideslip wrote:
The operation @ is defined for all non-zero x and y by the equation x@y=x/y. Then the expression (x@y)@z is equal to?
I get as far as:
(x@y)@z
= (x/y)@z
What is next?

=(x@y)@z
=(x/y)@z
=(x/y)/z
=x/yz
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 [#permalink] New post 17 Mar 2006, 07:58
Algebraically, yes: x/y@z---> (x/y)/z --->x/y*1/z ---> x/yz.

Were you given any numerical values for x, y, and z?
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 [#permalink] New post 18 Mar 2006, 15:30
x@y=x/y

x/y@z=(x/y)/z=x/yz
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 [#permalink] New post 18 Mar 2006, 17:47
macca wrote:
x@y=x/y

x/y@z=(x/y)/z=x/yz


i don't understand why @ z becomes /z
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  [#permalink] 18 Mar 2006, 17:47
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