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For all real numbers v, the operation v* is defined by the

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For all real numbers v, the operation v* is defined by the [#permalink] New post 08 Jun 2003, 11:18
. For all real numbers v, the operation v* is defined by the equation v* = v - v/3 . If (v*)* = 8, t hen v =

?
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Skoper [#permalink] New post 08 Jun 2003, 13:23
Show me the work like your math teacher would.
:lol:

(v*)* = 8
[(2/3)v]* = 8
(2/3)(2/3)v = 8
(4/9)v = 8
(4/9)v x (9/4) = 8 x (9/4)
v = 18
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 [#permalink] New post 09 Jun 2003, 10:22
I have a slight difference

For me it is -6

V-V/3 - (V-V/3)/3=8 I think you forgot to divide into 3

2V/3 - 2V = 8

(2V-6V)/3=8

V= -6
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Re: Skoper [#permalink] New post 09 Jan 2006, 10:26
Curly05 wrote:
Show me the work like your math teacher would.
:lol:

(v*)* = 8
[(2/3)v]* = 8
(2/3)(2/3)v = 8
(4/9)v = 8
(4/9)v x (9/4) = 8 x (9/4)
v = 18



In the 3rd line of your working, shdnt it be (2v/3) (2v/3)
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Re: Skoper [#permalink] New post 09 Jan 2006, 12:45
vizzz wrote:
Curly05 wrote:
Show me the work like your math teacher would.
:lol:

(v*)* = 8
[(2/3)v]* = 8
(2/3)(2/3)v = 8
(4/9)v = 8
(4/9)v x (9/4) = 8 x (9/4)
v = 18



In the 3rd line of your working, shdnt it be (2v/3) (2v/3)


I agree. This would then give v=9
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 [#permalink] New post 09 Jan 2006, 18:26
v* = v - v/3

So (v*)* = v*-V*/3 = (v-v/3) - (v-v/3)/3
= 2v/3 - 2v/9
= 4v/9 = 8

v = 9*8/4 = 18
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 [#permalink] New post 09 Jan 2006, 21:18
How come i am getting -6

v*= v-v/3=2v/3

so (v*)*= (2v/3)*= 2v/3- 2v/3/3 = -4v/3
-4v/3=8 => v=-6
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 [#permalink] New post 09 Jan 2006, 22:02
andy_gr8 wrote:
How come i am getting -6

v*= v-v/3=2v/3

so (v*)*= (2v/3)*= 2v/3- 2v/3/3 = -4v/3
-4v/3=8 => v=-6


2v/3 - 2v/3/3 = 2v/3 - 2v/9 = 4v/9
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 [#permalink] New post 09 Jan 2006, 22:09
andy_gr8 wrote:
How come i am getting -6

v*= v-v/3=2v/3

so (v*)*= (2v/3)*= 2v/3- 2v/3/3 = -4v/3
-4v/3=8 => v=-6


andy,

2v/3 - (2v/3)/3 = 2v/3 - 2v/9
= 4v/9

so 4v/9 = 8 or v = 18 should be it.
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  [#permalink] 09 Jan 2006, 22:09
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For all real numbers v, the operation v* is defined by the

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