bloodhound

rui costa missed as well. its now 2-2 in penalties.

oh no. beckham has missed a penalty. noo....

oh no. its penalty shootout. i am too nervous to watch it. so ill just follow it up on the internet

its being shown on the normal terrestial channel

Either i didnt notice it. or portugal has scored before england or after. cos now its 2-2 and still a draw. and penalties looming in sight

Engaldn have sCOREEEEEDDDD . they have SCORREEEEEDDDDDDDDDDD. PLS god . let them defend the lead. PLEASEEe

not out yet. but they should be in semi-s by now. They are playing the second half of the extra time. if its still a draw. then penatlies will be taken. and lets admit it, england dont have a good history in those

what a Rubbish REF, He has made the worst decisions i have ever seen. England should be through to Semi's. the FA should ban that ref from ever being in 1 km of a football stadium.

in the town of Haywood Jablowme, which is named after the great explorer of beautiful

wow, u had a whole module on differential equations!!!. we didnt do much on calculus this year. well for me not much cos i did further maths in a level.

i did quite well in my linear maths. love the vector spaces. Its absolutely mindblowing when u think about them for long enough. and the inner products, and gram-schimd orthogonalisation. it cool.

This is for the second part.

Instead of finding f(n) explicitly we find g(n) for which [math]f(n) \le g(n)[/math] for all [math]n \ge 3[/math]

Here [math]f,g:N \mapsto N[/math] and that [math]f,g \ge 0[/math]

Finding a function g(n)

The number of right angled triangles formed by n points will be always less than or equal to the total number of triangles formed by those n points.

The total number of traingles formed by n points is the number of ways of choosing 3 points out of the n points. i.e [math]g(n)=^{n}C_3[/math]

We have

[math]0 \le \frac{{f(n)}}{{n\^3}} \le \frac{{g(n)}}{{n\^3}}[/math]

It can be shown that [math]\frac{{g(n)}}{{n\^3}}[/math] converges to 0 as n tends to infinity.

Therefore by sandwiching (pinching) [math]\frac{{f(n)}}{{n\^3}}[/math] also must tend to 0

or Maple

then he had sex with a horse when all of

i think most of the laptops operate at 60Hz . I can't seem to notice anything wrong with the graphics.

cubase vst used to ship with dongle cos they used to have it in my school.

neber used deamon tools. heard its good. download most of my games from bit torrent. also use alcohol to copy protected cds. takes a while to skip through all the error blocks in a safedisd 2/3 disc. takes about 2 hours for me. somehow i cant seem to make alcohol fast skip the error blocks.

yeah, its better to work in its native resolution. most laptops have a res of 1024 768

if u get a TFT screen, they may have some strong blur filters to make the graphics acceptable at other than its native resolution

does anyone know how keygens work???? i am astounded by them.

Does .9999999 = 1 ?

can't believe that this question is still being asked, with all of the posts and the proofs posted surely everyone can see that it is indeed true.

i sort of did this kind of stuff in our schoo, but luckily the network admin was too incompetent. like deleting users and adding user names, changing the admin pass. it was quite fun watching his dumb face being terrorised.

i was always rubbish at trig.

what does DGO stand for anyway

never tried mathmatica, will have a go at it afterwards.

ill just stick to maple

so do many other functions. and compared to them this is a pretty nice function

limit of y as x tends to 0 does exist

as

0<= x exp(sin(1/x)) <= ex

0 - > and ex -> 0 as x -> 0

there fore by sandwiching y tends to 0 as well

ive never seen that equation before' date=' or maybe i have with different letters or something, or maybe i learnt it a while ago, but anyway;

 

[/quote']

thats just the "theorem of total probability" which just states that

 

Let [math]E_1,E_2,...[/math] be a partition of [math]\Omega[/math] and let F be the proper subset of [math]\Omega[/math].

 

Then [math]P(F)=\sum_i{P(F|E_i)P(E_i)}[/math]