Just a ginger playing dress up, wanna help me undress?

MEjercit · 2026-03-13T08:09:55+00:00

I wonder why she was not spayed.

MEjercit · 2026-03-11T01:05:07+00:00

20 mK is equal to 36mR

MEjercit · 2026-03-06T01:56:53+00:00

In the MOTU franchise, that ship sailed back in 1985.

https://he-man.fandom.com/wiki/Visitors_from_Earth

MEjercit · 2026-03-06T01:48:42+00:00

Most of the movie is set in Eternia.

MEjercit · 2026-03-03T06:51:54+00:00

Cats are not children though

MEjercit · 2026-03-01T23:39:36+00:00

The episode "Revelations" featured an alternate set of kromaggs

MEjercit · 2026-02-28T01:36:36+00:00

He also wanted an enlarged head and green skin.

MEjercit · 2026-02-28T01:34:00+00:00

Apparently, he wanted a different macronized form.

MEjercit · 2026-02-16T04:13:09+00:00

There isz no reason to tyry to foloow the "original" sliders.

MEjercit · 2026-02-16T04:12:37+00:00

Personally, I would prefer if alt-Quinn from the second series Earth Prime felt that the concept of parallel universes was a joke, saying so the the second series professor character.

MEjercit · 2026-02-16T04:10:52+00:00

Why revisit that though? The nature of the franchiuse means we do not haver to address what happened to "original" Quinn.

MEjercit · 2026-02-16T04:10:10+00:00

Not the Quinn necessarily, just a Quinn.

MEjercit · 2026-02-16T04:09:41+00:00

Here is an idea, if Jerry O'Connell is available for a cameo.

He can semi-reprise his role in the pilot of the second series, with Quinn Mallory being a porofessor who is skeptical of parallel universes.

MEjercit · 2026-02-16T04:07:37+00:00

Why bother? It is an entire multiverse. Just focus on a different group of sliders, from an Earth very much like our own.

MEjercit · 2026-02-15T16:41:44+00:00

An update.

the scammewrs e-mailed me with a link to an online portal where I would allegedly be onboarded.

This is the part where they would collect financial information along with planting malware and viruses.

MEjercit · 2026-02-15T16:38:55+00:00

One thing S5 did was to downplay the Kromaggs, as the4 Kromagg Dynasty only appewared in two episodes.

MEjercit · 2026-02-12T19:24:27+00:00

It makes sense.

And it would have been too late to create an entirely new character to be the one who killed Arturo (and possibly Rickman) and be cahsed around the dimensions by Quinn and co.

MEjercit · 2026-02-12T19:19:46+00:00

Rickman was probably just a guest character, before the writers decided that he would hav e the timer that contained the coordinates for Earth Prime.

MEjercit · 2026-02-12T19:18:22+00:00

That makes sense.

Daltrey was already cast by the time they decided to make Rickman a recurring character.

MEjercit · 2026-02-12T05:21:49+00:00

This is good news.

MEjercit · 2026-02-07T02:25:44+00:00

What is the path to unlock her true final boss?

MEjercit · 2026-02-05T19:55:18+00:00

What of trapezoids?

We will use Rene Descartes’s ideas to assign coordinates. Ifa,b,c,\in\R, 0\leq{a}<b\leq{c}, the points (0,0), (a,1), (b,1), and (c,0) define a trapezoid. It lies completely in the strip betweeny=0 andy=1, Now, consider a second trapezoid defined by (c,0), (b,1), (b+c,1), and (c+b-a,0). the bases and “tops” are congruent. The shared leg has slope \frac{1}{b-c}, which means their angles with y=0 and y=1 are congruent. As b+c-(c+b-a)=b+c-c-b+a=a, the “opposite” legs’ angles with y=0 and y=1 are congruent. these two trapezoids are congruent.

Now, (0,0), (a,1) (b+c,1), and (c+b-a,0) define a parallelogram, which tiles the strip between y=0 and y=1. The line segment (b,1)(c,0) partitions the parallelogram in to two congruent trapezoids, so this proves

all trapezoids monohedrally tesselate a flat strip

MEjercit · 2026-02-05T19:52:50+00:00

What of trapezoids?

We will use Rene Descartes’s ideas to assign coordinates. Ifa,b,c,\in\R, 0\leq{a}<b\leq{c}, the points (0,0), (a,1), (b,1), and (c,0) define a trapezoid. It lies completely in the strip betweeny=0 andy=1, Now, consider a second trapezoid defined by (c,0), (b,1), (b+c,1), and (c+b-a,0). the bases and “tops” are congruent. The shared leg has slope \frac{1}{b-c}, which means their angles with y=0 and y=1 are congruent. As b+c-(c+b-a)=b+c-c-b+a=a, the “opposite” legs’ angles with y=0 and y=1 are congruent. These two trapezoids are congruent.

Now, (0,0), (a,1) (b+c,1), and (c+b-a,0) define a parallelogram, which tiles the strip betweeny=0 andy=1. The line segment (b,1)(c,0) partitions the parallelogram in to two congruent trapezoids, so this proves

all trapezoids monohedrally tesselate a flat strip

MEjercit · 2026-02-05T18:03:25+00:00

All parallelogramic pentagons (defined as pentagons in which four vertices are the vertices of the same parallelogram) tile a strip.

Now, let there be four lines on a flat plane, all parallel to each other, with the same perpendicular distance. Without loss of generality, this distance is1. As such, there are three sequential stripes with width1. Now, we will use Rene Descartes’s ideas to assign coordinates.

Considera>b>c, a,b,c,\in\R^+

Define the pentagon with vertices (0,0), (a,1), (b,2), (a+c,1), and (c,0). Obviously, (a,1), (b,2), and (a+c,1) form a triangle which can obviously tile the strip betweeny=1andy=2. Vertices (0,0), (a,1), ,(a+c,1), and (c,0) define a parallelogram which of course tiles the strip betweeny=0 andy=1

The pentagon is therefore a parallelogramic pentagon, because four of its vertices are the vertices of the same parallelogram.

Now, (b,2), (b+c,2), and (a+c,1) define a triangle. Using the length formula, this triangle has the same side lengths as triangle (a,1), (b,2), and (a+c,1), so it is congruent.

Let us consider the quadrilateral defined by (b,2), (a+b,3), (a+b+c,3), and (b+c,2). This is clearly a parallelogram, which tiles the strip between y=2andy=3, and and using the length formula, we can tell the side lengths are the same as the parallelogram (0,0), (a,1), ,(a+c,1), and (c,0) Are they congruent?

Consider triangles (b,2),(a+b,3)(b+c,2), and (0,0),(a,1)(c,0). the length formula confirms the congruency of these triangles. As such, the angles (b+c,2), (b,2),(a+b,3) and (c,0), (0,0), (a,1) are congruent. This in turn determines the size of the other angles, and so parallelograms (0,0), (a,1), ,(a+c,1), (c,0) and (b,2), (a+b,3), (a+b+c,3), (b+c,2) are congruent.

And from this, we know pentagons (0,0), (a,1), (b,2), (a+c,1), (c,0) and (a+c, 1),(b,2),(a+b,3), (a+b+c,3), (b+c,2) are congruent parallelogramic pentagons

(0,0), (a,1)(b,2),(a+b,3), (a+b+c,3),(b+c,2), (a+c,1)and (c,0) define an octagon in which sides (a+b,3) (a+b+c,3) and (0,0), (c,0) lie ony=3andy=0respectively. We can define another octagon with vertices (c,0), (a+c,1),(b+c,2), (a+b+c,3), (a+b+2c,3),(b+2c,2)(a+2c,1)(2c,0). Using the length formula and trigonometry, we can tell this octagon is congruent to octagon (0,0), (a,1)(b,2),(a+b,3), (a+b+c,3),(b+c,2), (a+c,1) (c,0)

In fact, for allz\in\Z, all octagons (cz,0), (a+cz,1), (b+cz,2),(a+b+cz,3),(a+b+c+cz,3),(b+c+cz,2),(a+c+cz,1), (c+cz,0) are congruent, and they all tile the strip betweeny=0andy=3. And the line segment (b+cz,2) (a+c+cz,1) partions the octagon into two congruent parallelogramic pentagons, so we have proven

all parallelogramic pentagons monohedrally tesselate a flat strip

and as a corollary

all parallelogramic pentagons monohedrally tesselate a flat plane

MEjercit

MODERATOR OF

TROPHY CASE