Difference between revisions of "User:Memoriesonfilm"
Line 48: | Line 48: | ||
I <3 me too! :p!<br> | I <3 me too! :p!<br> | ||
Your mom <3 him | Your mom <3 him | ||
+ | |||
+ | ====How to determine the height of an infinitely skinny rectangle found on Kat's Facebook profile picture==== | ||
+ | |||
+ | <math>\begin{align} | ||
+ | H(x) | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {\iint_R \ dy\, dx} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} \int_{x+1}^{-x^2+6x-3} \,dy\,dx} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} [y]_{x+1}^{-x^2+6x-3} \,dx} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} (-x^2+5x-4) \,dx} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {[-\frac 1 3 x^3 + \frac 5 2 x^2 - 4x]_x^{x+\Delta x}} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {[-\frac 1 3 (x^3 + 3x^2 \Delta x + 3x(\Delta x)^2 + (\Delta x)^3) + \frac 5 2 (x^2 + 2x | ||
+ | \Delta x + (\Delta x)^2) - (4x + 4 \Delta x)] - [-\frac 1 3 x^3 + \frac 5 2 x^2 - 4x]} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} \frac {-x^2\Delta x - x(\Delta x)^2 -\frac 1 3 (\Delta x)^3 + 5x\Delta x + \frac 5 2 | ||
+ | (\Delta x)^2 - 4\Delta x} {\Delta x} \\ | ||
+ | &{} = \lim_{\Delta x \to 0} (-x^2 - x\Delta x -\frac 1 3 (\Delta x)^2 + 5x + \frac 5 2 \Delta x - 4) \\ | ||
+ | &{} = -x^2 - 0x -\frac 1 3 0^2 + 5x + \frac 5 2 0x - 4 \\ | ||
+ | &{} = -x^2 + 5x - 4 | ||
+ | \end{align}</math> |
Revision as of 19:38, 3 February 2008
This is Max Wang speaking.
I was formerly the official photographer-videographer for LAN.2, currently a onemore. This is summarized by the fact that I always have a camera with me and I'm ready to capture any and all lasting memories on digital film. It is officially true that I had a camcorder and a tripod with me during LAN.07.2.
In 06.2, I'd also taken on the task of expanding circles for American Pie, The Afterdance, and raving songs before the new raving rule was enstated. Similarly, I also attempted to close circles during the slow canon songs, which is quite difficult because there was an overabundance of squirrels at 06.2 who either formed small, three person circles, socialized in the quad, or didn't come at all...which makes me sad. Slightly less similarly, but still similarly, I'd also tried to keep people from standing in aformentioned Pie circles.
In stark contrast, I am now opposed to closing circles because they become absurdly tiny. But sadly, they become not round. Instead, I break up smaller circles and direct them to the bigger one. I'm so bad.
Contents
Sessions
- MIND.A.LAN.05.2
- CHEM.A.LAN.06.2
- ETYM.LAN.07.2 ---- maybe next time, THEO...
- THEO.LAN.08.2 ---- pending
Things that Max is Saying, Apparently
There is no relation between myself and Frank Wang, although Juice has second thoughts about that...
One may refer to me as "A Hallucination," and I promise never to respond. I am also the player two spaces to my own up. Also, anyone else's. Also, I may be Mr. Dictator.
Max is feeling chill. He also likes umbrellas.
I lose both games!
I am SO HAPPY!! to be the Holder of the Duck for 2007-2008 and I love everybody who helped make me duckworthy of it. However, I'm sad to be home.
For various reasons, my last name has vanished from most pages. This may be due to the paranoia monster's very large appetite.
I'm also your local RealCTY admin (the one who's still eligible and who's here a lot).
I would like to apologize to anyone who thinks that I'm a wiki-fascist. I'm only concerned about the quality of the site.
Randomness
- RealCTY Admin: bureaucrat level and re-creator
- Duck 2007-2008
- Videographer (05.2), 06.2, 07.2
Contact Info
- Email: thechaoseternal@gmail.com
- AIM: thechaoseternal
- Facebook! (Network: Great Valley HS)
We <3 you Max Wang.
I <3 me too! :p!
Your mom <3 him
How to determine the height of an infinitely skinny rectangle found on Kat's Facebook profile picture
<math>\begin{align}
H(x) &{} = \lim_{\Delta x \to 0} \frac {\iint_R \ dy\, dx} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} \int_{x+1}^{-x^2+6x-3} \,dy\,dx} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} [y]_{x+1}^{-x^2+6x-3} \,dx} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {\int_x^{x+\Delta x} (-x^2+5x-4) \,dx} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {[-\frac 1 3 x^3 + \frac 5 2 x^2 - 4x]_x^{x+\Delta x}} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {[-\frac 1 3 (x^3 + 3x^2 \Delta x + 3x(\Delta x)^2 + (\Delta x)^3) + \frac 5 2 (x^2 + 2x \Delta x + (\Delta x)^2) - (4x + 4 \Delta x)] - [-\frac 1 3 x^3 + \frac 5 2 x^2 - 4x]} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} \frac {-x^2\Delta x - x(\Delta x)^2 -\frac 1 3 (\Delta x)^3 + 5x\Delta x + \frac 5 2 (\Delta x)^2 - 4\Delta x} {\Delta x} \\ &{} = \lim_{\Delta x \to 0} (-x^2 - x\Delta x -\frac 1 3 (\Delta x)^2 + 5x + \frac 5 2 \Delta x - 4) \\ &{} = -x^2 - 0x -\frac 1 3 0^2 + 5x + \frac 5 2 0x - 4 \\ &{} = -x^2 + 5x - 4
\end{align}</math>