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About Me Member Hack FjorxcMale/Canada Recent Activity Deviant for 3 Years
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182 New Messages? Fnah!

Thu Mar 9, 2006, 9:56 AM
It has been a fairly long thirteen weeks, five days, and change. I didn't realize that so much time had passed by, but there you go.

I haven't really done anything because I've been busy with writing essays and have had no real ideas. Though I do have this one sentence that is going to be a component of something I'm planning to write when this frakking essay is done.

"(name) was realistic enough to know that he wasn't getting out of this one, and man enough to admit that the prospect terrified him."

It will probably be better when I do stuff like, y'know, give people names, and figure out some semblance of a plot.

But that's for the future.

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Thanks for the visit ^¸^
thanks for the fav, dude, muchly appreciated.
soons i get some time, i gotta read your stuff, tho many things get in my way now...

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Beautiful Skies
When are you putting up your next deviation?
Keep me posted.

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Did you know that:
E^2 = p^2c^2)+([<delta>m^2]c^4.
which implies that:
E= [mc^2]/[<sq. root> 1-(v^2/c")].
when p=0 (in eq 1, or 2, whichever you find simpler)
E=mc^2........
(In reality, p can never=0, so E should, realistically not be =mc^2, )
I have to write it first. Recently I've come to the decision to disassociate my work from the shared verse in which they were first created, so at the moment I'm caught up in retooling.

Though, since I know that at least a few people will be reading it here, that does give me incentive to produce something in the near future. Thanks for asking, though.
Well.......... good luck with your retooling.
Keep putting up stuff, and maybe more people here will be reading your stuff.

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Did you know that:
E^2 = p^2c^2)+([<delta>m^2]c^4.
which implies that:
E= [mc^2]/[<sq. root> 1-(v^2/c")].
when p=0 (in eq 1, or 2, whichever you find simpler)
E=mc^2........
(In reality, p can never=0, so E should, realistically not be =mc^2, )
gotta read your new stuff.

--
Did you know that:
E^2 = p^2c^2)+([<delta>m^2]c^4.
which implies that:
E= [mc^2]/[<sq. root> 1-(v^2/c")].
when p=0 (in eq 1, or 2, whichever you find simpler)
E=mc^2........
(In reality, p can never=0, so E should, realistically not be =mc^2, )
you've been tagged by me. Go to my journal. That's it. I'm not really into these things, but might as well carry out this localised tradition that's going around DA.

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Did you know that:
E^2 = p^2c^2)+([<delta>m^2]c^4.
which implies that:
E= [mc^2]/[<sq. root> 1-(v^2/c")].
when p=0 (in eq 1, or 2, whichever you find simpler)
E=mc^2........
(In reality, p can never=0, so E should, realistically not be =mc^2, )
Thanks, whenever you have time - I just like it when people read my stuff.
you're welcome for the fav, and btw, i couldn't help it. great writing; i faved it so i could finish reading it later

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Beautiful Skies

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