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:disco: :disco: :read: :hammer: :cool: :uhhuh: :shy: :angry: :spam: :cool: :uhhuh: :cool: yeah iv dun all the papers back to 1832 and iv never made a mistake:mad: :uhoh: :giddy: :drink: lolololololololololololololololololl!!!!!!!!!!!!11111

:shy: :cool: :uhhuh: :angry: :rolleyes: :uhoh: :spam: the tery lee one kills all others nd all u cambrige losers suck me cos i got it aswell and half the answers are wrong nd for most theres only like 10 questions nd the book is a little bitch:hammer:

yeah i was bored so i did it and cos u get x^[(a-b)(a+b-c)+(b-c)(b+c-a)+(c-a)(c+a-b)] and expanding [(a-b)(a+b-c)+(b-c)(b+c-a)+(c-a)(c+a-b)] gives: a^2 -ac -b^2 +bc +b^2 +b^2 -ab -c^2 -bc -a^2 +ab and all of these cancel out to give 0 and x^0=1 so its 1

fuckin pe

yeah guys best op is take the logarythm then using the Goldstone boson principle we integrate in terms of VALPHI using the equation: {d\over dt} Q_A = {d\over dt} \int_x e^{-x^2\over 2A^2} J^0(x) = -\int_x e^{-x^2\over 2A^2} \nabla \cdot J = \int_x \nabla(e^{-x^2\over A^2}) \cdot J peace out...

Re: CSSA 4 unit paper all u had to do was makes p(alpha)=0 and p'(alpha)=0 then eliminate alpha by making it sugject of the P'(x) one and subbing in to the p(x) one then manipulating

HEY HOTTIES LAWL omg its john r here lawl and i just wanna say i saw the cutest asian boy today he was so cute i just wanted to gobble him up haha ^__^ i asked him out and he said yes but he hasnt called back :( maybe hes playing hard to get lol ;p i cant wait to spank his little bottom teehee...

u take it up tha b really hard get a train like 60 bucks

:eek: :uhoh: :rolleyes: :santa: :santa: :santa: :wave: i did a chick in tha butt:mad: :burn: :headbang: :idea:

ye bra i did the same theres no marks allocated to it apparently you draw the diagram to help you better understand it so its your loss by not drawing it but i swear you didnt need to draw it anyway anyway yeah i dont think you lose marks for it cos thatd be fuckin shit