3XY+2XX = ?SEX (= 3SEX)
3X + 3Y + 2X + 2X = 7X + 3Y
7X + 3Y = ?SEX
6X + 3Y = ?SE
? = (6X + 3Y) / (SE)
(SEXY) = Constant
? = (4XY + 1XX) + EX
:: MMMF with EX
:: Your f(ex) is a bimbo
The swinger theorem complicates the matter, as you need to know the number of individuals in reference to their relationships (R1 + R2 +... Rn = ?SEX) rather than on their own. You run into the issue of combinations and fucktorials. It's classic group sex theory. Using the already stated participants, there are already 8 different combinations for the types of relationships with at minimum 2 members and at most 4, as there needs to be at minimum one individual outside of the original relationship for it to be valid for the theorem. E.g.Code:3XY+2XX = ?SEX (= 3SEX) 3X + 3Y + 2X + 2X = 7X + 3Y 7X + 3Y = ?SEX 6X + 3Y = ?SE ? = (6X + 3Y) / (SE) (SEXY) = Constant ? = (4XY + 1XX) + EX :: MMMF with EX :: Your f(ex) is a bimbo
I'm rather upset by this.
There are 1 equation and 1 unknown, therefore unsolveable.
Unless the equation collapses into a single proof.
Also, how do you collect the terms.
It's implied in a doggy-style gang session proof.
But not the more general case swinger theorem!![]()
1XY + 1XX, 2XY, 2XX,... 2XY + 2XX, 3XY + 1XXIf we consider XY and XX as chromosomes, XY -> male, XX -> female, then 3 XY is not 3X + 3Y, just simply 3 [variable-name].Code:3XY+2XX = ?SEX (= 3SEX) 3X + 3Y + 2X + 2X = 7X + 3Y 7X + 3Y = ?SEX 6X + 3Y = ?SE ? = (6X + 3Y) / (SE) (SEXY) = Constant ? = (4XY + 1XX) + EX :: MMMF with EX :: Your f(ex) is a bimbo
I'm rather upset by this.
There are 1 equation and 1 unknown, therefore unsolveable.
Unless the equation collapses into a single proof.
Also, how do you collect the terms.
It's implied in a doggy-style gang session proof.
But not the more general case swinger theorem!![]()