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Comments
| Anonymous #937463 5 months ago |
????????????????? |
| Anonymous #937468 5 months ago |
Well... technically My Little Pony is infinite(ly growing), so the whole thing would equal to zero. |
| benger #937482 5 months ago |
would only work if it was 1/ (d/dx)mlp:fim |
| benger #937488 5 months ago |
^sorry, 1/ (d/dt)mlp:fim |
| Anonymous #937489 5 months ago |
I would, in all sincerity, love to see this equation answered.
Please, make the descriptions as simple as possible. This is so incomprehensible to me, it might as well be written in Linear A. |
| NightJack #937491 5 months ago |
Actually anon458, that would make MLP infinite not zero. |
| PowerToole #937501 5 months ago |
I... don't really know what that's supposed to be... |
| Anonymous #937502 5 months ago |
1 / (INFINITY) = 0 |
| benger #937504 5 months ago |
^it would make 1/mlp:fim zero because 1/infinity = 0 |
| Anonymous #937507 5 months ago |
pika-pie-brator? |
| Anonymous #937510 5 months ago |
my mind |
| benger #937511 5 months ago |
its made to be solvable, u need to use differentail equations to solve ti though |
| Littlecabbit #937589 5 months ago |
That reminds me, .....I saw a hello kitty dildo on ebay once, ....why did you have to remind me? |
| Irony #937726 5 months ago |
Wait. Am I supposed to take (the integral of pikachu) times the vibrator or the integral of (pikachu times the vibrator).
It makes a difference. |
| benger #938041 5 months ago |
its the integral of (pikachu x purple vibrator),
the color matters |
| Wyrm #938398 5 months ago |
@502
Only if integral ≠ infinity. |
| Anonymous #938858 5 months ago |
if pikachu and vibrator are variables and only pikachu is in the integral:
((.5pikachu^2)(1/(2MLPFiM))*vibrator) on the interval of p(t)-p(pinkiepie) where function p equals pikachu as a function. |
| benger #939128 5 months ago |
no, pikachu and vibrator are both functions of t |
| Anonymous #939308 5 months ago |
 |
| Anonymous #940076 5 months ago |
Okay, so we have 1/MLP:FiM multiplied by the integral of Pikachu * dildo, between the limits of Pinkie Pie and T.
The integral of pikachu*dildo would be, well, I'm not sure, since it's not given what we should integrate with respect to. But assuming that dildo is a constant number to multiply by and we're integrating with respect to Pikachu, it'd work out to be: 1/MLP:FiM[(0.5pikachu^2)*vibrator] between the limits of Pinkie Pie and T. That'd work out, overall, to be: (1/MLP:FiM)*[(0.5(T*pikachu^2)*vibrator)-(0.5(PinkPie*pikachu^2)*vibrator)] Sweet jesus, what am I doing with my time? |
| AspenOnFire #940249 5 months ago |
It needs a d(pikachu) or a d(dildo) or else it really doesn't make sense. |
| Anonymous #940592 5 months ago |
My thoughts exactly, AspenOnFire. Which is why I assumed that there was a d(pikachu) present; or that we were integrating with respect to pikachu.
There'd also have to be a way to relate pikachu to dildo, either implicitly or through an equivalence equation. |
| Anonymous #1275786 2 months ago |
you fucking nerds |