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If xyz>0, is (xy)^(2z)^3>0 1) y>0 2) x>0

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If xyz>0, is (xy)^(2z)^3>0 1) y>0 2) x>0 [#permalink] New post 27 Apr 2006, 19:17
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A
B
C
D
E

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If xyz>0, is (xy)^(2z)^3>0

1) y>0
2) x>0
Manager
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 [#permalink] New post 27 Apr 2006, 19:51
Quote:
If xyz>0, is (xy)^(2z)^3>0

1) y>0
2) x>0


I would say that neither statements are necessary. Since (xy)^(2z)^3>0 = (xy)^6z>0 = x^6z*y^6z>0. X and Y will always to the power of an even number and thus will be positive.
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Re: DS: Inequality - (III) [#permalink] New post 27 Apr 2006, 20:29
Professor wrote:
If xyz>0, is (xy)^(2z)^3>0

1) y>0
2) x>0



from i

if y>0 then x and z are either both +ve or both -ve

insuff

from ii


if x>0 then y and z are either both +ve or both -ve

insuff


together

z > 0

suff

I'll go with C
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 [#permalink] New post 27 Apr 2006, 20:42
I would go with C.

We can not assume that Z is integer, and 2z^3 is not always even. So we neen x*y to be positive to insure that the whole expression is positive.
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 [#permalink] New post 29 Apr 2006, 11:37
[quote="Charlie45"][quote]If xyz>0, is (xy)^(2z)^3>0

1) y>0
2) x>0[/quote]

I would say that neither statements are necessary. Since (xy)^(2z)^3>0 = (xy)^6z>0 = x^6z*y^6z>0. X and Y will always to the power of an even number and thus will be positive.[/quote]

Have to agree with Charlie - D
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 [#permalink] New post 29 Apr 2006, 14:36
yep it has to be D as charlie explained...
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 [#permalink] New post 29 Apr 2006, 23:57
I think it is C since we dont know if Z is an integer
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 [#permalink] New post 30 Apr 2006, 01:28
shevy wrote:
I think it is C since we dont know if Z is an integer


yep I guess I missed non integer parts!
  [#permalink] 30 Apr 2006, 01:28
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