goss = gay 

when a fellow compadre, say a bretherin, grows up with you and comes along all the adventures that u encounter. Then, one day he stops going on those adventures and is afraid to say the word " bretherin," becasue he is too mature. Also, he likes girls.

Man, when did sam billings become a goss= gay. " i think it was when he stuck his fingers up his ass.

goss = gay by garretmain March 10, 2005

Bag = Secured

When a woman gets pregnant with a man with a lot of money, her finances are secured because the man is now obligated to give his money to her and for the child. This is usually said when the man is wealthy and the girl is not, and she gets pregnant soon into the relationship (assumed to be on purpose) for the reasons stated above.

Random Person: "Yo John! You see Becky got pregnant with that rich dude Jarod? Damn man! They have only seen each other for 2 months!"
John: "Bag, Secured" or "Bad equals Secured" or (by text) "Bag = Secured"

Bag = Secured by Babycakes Lovin May 21, 2018

dy/dx z^omg = f (sub) j00 * dpwn/dx 

reads dy dx of z to the omg equals f sub j00 times d pwn dx.

The new derivative formula. Bish.

please take dy/dx of 563210983x^32 * ln(sin(32x*y*z) * tan^-1(q)

The answer is 42. Bish.
I cite dy/dx z^omg = f (sub) j00 * dpwn/dx

dy/dx z^omg = f (sub) j00 * dpwn/dx by ff8 November 17, 2005

y = g

young equals good. the younger a girl is, the better she feels.

Joe: "What are you talking about? You're way too old for her!"
John: "Wrong!!!!!! Y = G bro!!!"

y = g by John Huckabee December 31, 2007

2 + 2 = 5 

Maths nerds making a joke out of some line from George Orwell's 1984

"Well *actually* 2 + 2 could = 5 !"

*pathetic nerd eagerly waits for confused stares to reinstate his superiority

2 + 2 = 5 by What March 25, 2005

(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}. 

Some good meth for yo head doode.

Person 1: "Yo, do you know that 2*2 =4 and 2+2= 4."
Person 2: "(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}."

(x+y)^{n}=\sum _{k=0}^{n}{n \choose k}x^{n-k}y^{k}=\sum _{k=0}^{n}{n \choose k}x^{k}y^{n-k}. by Lurlei139See December 3, 2018

<(="_"=<) 

*Huggles fer Yew*
X3! X3! X3!

*glomps yew 2 death* <(="_"=<)

<(="_"=<) by Nami February 18, 2004