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It did :p You put in k, not n lol. a_{n}=2^{n-1}+1 \; $for$ \; n=1,2....k-1,k

Re: The Official Cricket Thread luv dat f33l

Fair enough, I had a lot of fun anyways :p. I went past the roundhouse again at like 11 and there was no one there, what time did they close up?

http://press.web.cern.ch/press/PressReleases/Releases2011/PR19.11E.html It appears that it may have been just experimental error..

Bad luck :( Next you you'll win. I showed up to the roundhouse at ~5:30 but I didn't recognise anyone/ no one called out to me, so I left and ended up drink with friends :p

Did you get in?

Well this is what I did. x^2 + y^2 = R^2 \\ y=R-\frac{1}{2}gt^2, x = v_it \\ $Subbing them in, v_i^2t^2 + R^2 - Rgt^2 + \frac{1}{4}g^2t^4=R^2 \\ v_i^2 - Rg + \frac{1}{4}g^2t^2 = 0 \\ v_i^2 = Rg - \frac{1}{4}g^2t^2 \\ $At $ t =0, v_i^2 = Rg, \therefore $ minimum value for $ v_i $ is $ v_i =...

We can always play roll the spiral :D

Do you have friends yet?

Last time, 4am lol. It doesn't make a difference for me, I just go to bed when I'm tired.

A physicist. Probably end up working at a university doing research/ lecturing.

Yeah that is a better way.

Blasphemy! OT: People are people. Deal with it.

lol

Just talk to people. Even if it's a simple hi, it can do wonders.

Whoaaa swearing. Now now no need for that.

$Using $ x+t^2y=2ct, $ we have the tangent at $ P $ having eqn $ x+t_1^2y=2ct_1 $ and similarly tangent at T has eqn $ x+t_2^2y=2ct_2. \\ $Now find where they intersect$.$

Getting friends at uni is easy, just talk to people.

Living on campus is gr8, it comes with many advantages ;)

6"2-6"3 and red hair.