MMMYYYKKK
- Post #862,941
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- Dec 12, 2013 3:39am Dec 12, 2013 3:39am
- | Joined Apr 2011 | Status: Member | 1,152 Posts
MMMYYYKKK
- Post #862,942
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- Dec 12, 2013 3:40am Dec 12, 2013 3:40am
- Joined Dec 2013 | Status: Member | 2,118 Posts
f(z)=z2+c
- Post #862,943
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- Dec 12, 2013 3:42am Dec 12, 2013 3:42am
- Joined Jun 2010 | Status: s = k log W | 20,603 Posts
- Post #862,944
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- Dec 12, 2013 3:42am Dec 12, 2013 3:42am
- Joined Jun 2010 | Status: s = k log W | 20,603 Posts
- Post #862,945
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- Dec 12, 2013 3:42am Dec 12, 2013 3:42am
- Joined Dec 2013 | Status: Member | 2,118 Posts
f(z)=z2+c
- Post #862,946
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- Dec 12, 2013 3:42am Dec 12, 2013 3:42am
- Joined Aug 2011 | Status: Member | 4,739 Posts
As the lights go by so too do the shadows move
- Post #862,947
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- Dec 12, 2013 3:44am Dec 12, 2013 3:44am
- Joined Jun 2010 | Status: s = k log W | 20,603 Posts
- Post #862,948
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- Dec 12, 2013 3:45am Dec 12, 2013 3:45am
- Joined Dec 2013 | Status: Member | 2,118 Posts
f(z)=z2+c
- Post #862,949
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- Dec 12, 2013 3:48am Dec 12, 2013 3:48am
- Joined Aug 2009 | Status: Member | 12,448 Posts
If you are new to Forex? Say Goodbye to your account!!
- Post #862,950
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- Dec 12, 2013 3:51am Dec 12, 2013 3:51am
- | Membership Revoked | Joined Jul 2013 | 4,068 Posts
Wherever you go, there you are.
- Post #862,952
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- Dec 12, 2013 3:52am Dec 12, 2013 3:52am
- | Membership Revoked | Joined Jul 2013 | 4,068 Posts
Wherever you go, there you are.
- Post #862,953
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- Dec 12, 2013 3:55am Dec 12, 2013 3:55am
- Joined Jun 2010 | Status: Hello,Mr.Market | 1,805 Posts
A chart helps those who can read it.
- Post #862,954
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- Dec 12, 2013 3:55am Dec 12, 2013 3:55am
- | Membership Revoked | Joined Jul 2012 | 4,197 Posts
"It's the Margin, Stupid."
- Post #862,955
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- Dec 12, 2013 3:56am Dec 12, 2013 3:56am
- Joined Aug 2011 | Status: Member | 4,739 Posts
As the lights go by so too do the shadows move
- Post #862,956
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- Dec 12, 2013 3:58am Dec 12, 2013 3:58am
- | Membership Revoked | Joined Jul 2013 | 4,068 Posts
Wherever you go, there you are.
- Post #862,957
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- Dec 12, 2013 3:59am Dec 12, 2013 3:59am
- Joined Dec 2013 | Status: Member | 2,118 Posts
f(z)=z2+c
- Post #862,958
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- Dec 12, 2013 3:59am Dec 12, 2013 3:59am
- | Joined Dec 2009 | Status: Bear | 1,009 Posts
- Post #862,959
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- Dec 12, 2013 4:01am Dec 12, 2013 4:01am
- Joined Dec 2013 | Status: Member | 2,118 Posts
f(z)=z2+c