Topic: Who would win in a fight between Action Bastard and Bert?

+Action Bastard !AlbLh2dwTI — 13.3 years ago #23,110

In one corner, a fictional cartoon character.

In the other, a drunken old man, desperately looking for someone to blow in order to receive more drug money.

Poll option	Votes	Percentage
Action Bastard	-	0%
Bert	-	0%
Boners	-	0%
Sprite	-	0%

+Anonymous B — 13.3 years ago, 4 minutes later[T] [B] #276,631

Waaaait a minute here...YOU'RE another of The Doctor's accounts too???

+FuckAlms !vX8K53rFBI — 13.3 years ago, 3 minutes later, 8 minutes after the original post[T] [B] #276,634

Action Bastard would arrive fashionably late only to find bert passed out in the middle of the ring from booze consumption.

(Edited 1 minute later.)

+Anonymous D — 13.3 years ago, 56 seconds later, 9 minutes after the original post[T] [B] #276,635

Bert wins.

Fictional 2d cartoon characters has zero inertia and cannot even blow air to ward off a butterfly.

Thinking of one line in the cartoon, as an geometrical object.

y = mx + b.
ax + by = c,
y = (-a/b)x + c/b,

2x + 3 y = 4
4x + 6y = 8
-x - (3/2) y = -2
(1/2)x + (3/4)y = 1

ax + by = c(ka)x + (kb)y = kc.

(a/c)x + (b/c)y = 1.

+Anonymous E — 13.3 years ago, 3 minutes later, 12 minutes after the original post[T] [B] #276,636

@276,631 (B)
The Socktor.

·Action Bastard !AlbLh2dwTI (OP) — 13.3 years ago, 2 hours later, 2 hours after the original post[T] [B] #276,694

@276,635 (D)

> y = mx + b.
> ax + by = c,
> y = (-a/b)x + c/b,
>
> 2x + 3 y = 4
> 4x + 6y = 8
> -x - (3/2) y = -2
> (1/2)x + (3/4)y = 1
>
> ax + by = c(ka)x + (kb)y = kc.
>
> (a/c)x + (b/c)y = 1.

·FuckAlms !vX8K53rFBI — 13.3 years ago, 25 minutes later, 3 hours after the original post[T] [B] #276,700

@276,635 (D)
I believe you have failed to account for the hexagonal motion establishment matrix.

·Anonymous D — 13.3 years ago, 8 minutes later, 3 hours after the original post[T] [B] #276,706

@previous (FuckAlms !vX8K53rFBI)
Its only relevant if you subscribe to the illuminati MATRIX.

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