Do you do it?
In math class, I found a way to turn 1 and 2 step addition into multiplication.
Quite simple, and quite useless, but I did it anyway.
The tiny fraction in the corner is 1 and N[1] over X
What do you guys think of when you're bored?
Last edited by Nexstudent (2011-11-09 19:52:26)

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How bored I am.

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I'm bored already. Math
Is
So
BORING
However: Steps to a hot-air essay.
First: write the structer
Second: use really long adjective phrases
Example
And the bearer of golden rings road across the solid air to a meeting of mail coats and sword tips.
And the original...
And the king sailed across the sea to fight a battle.

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P.E. = mgh
g = ((Gm)/(d^2) )
So, if all else is constant, the force of gravity is inversely proportional to distance squared, and potential energy is directly proportional to distance.
P.E. = d * g
P.E. = d / (d * d)
P.E. = 1/d
So Potential energy is inversely proportional to distance. So the closer you get to something, the greater potential energy you have (so if you lift a ball one meter high, it has greater potential energy than if you lift it two meters high).
In reality, "P.E. = mgh" doens't work on that large a scale. Too bad, I thought I was onto something![]()
Last edited by MoreGamesNow (2011-11-09 21:51:35)
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the allegory in the cave with the greek guy
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Sometimes I write division problems for myself. Sometimes I visualize some code I'm debugging and try to fix it in my head. Too bad I can't afford to get an interpreter in my head. Maybe I'll have enough money for brain surgery after Christmas
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MoreGamesNow wrote:
P.E. = mgh
g = ((Gm)/(d^2) )
So, if all else is constant, the force of gravity is inversely proportional to distance squared, and potential energy is directly proportional to distance.
P.E. = d * g
P.E. = d / (d * d)
P.E. = 1/d
So Potential energy is inversely proportional to distance. So the closer you get to something, the greater potential energy you have (so if you lift a ball one meter high, it has greater potential energy than if you lift it two meters high).
In reality, "P.E. = mgh" doens't work on that large a scale. Too bad, I thought I was onto something![]()
Oh. For a second I thought you meant P.E. as in the school-type-thing.
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MoreGamesNow wrote:
P.E. = mgh
g = ((Gm)/(d^2) )
So, if all else is constant, the force of gravity is inversely proportional to distance squared, and potential energy is directly proportional to distance.
P.E. = d * g
P.E. = d / (d * d)
P.E. = 1/d
So Potential energy is inversely proportional to distance. So the closer you get to something, the greater potential energy you have (so if you lift a ball one meter high, it has greater potential energy than if you lift it two meters high).
In reality, "P.E. = mgh" doens't work on that large a scale. Too bad, I thought I was onto something![]()
Nope. Sorry. E = MC^2.
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