Search results

fucking hell ..effed up my english yearly

Consider Darren and Samantha as a single element. In this element you can arrange the two people 2! ways. Place the element somewhere around the circular table to begin. Consider Fred and Ethel as another element. You can arrange Fred and Ethel 2! ways if they must sit next to one another. You...

yeah

oh yeah true lol..

the lower rectangles start at x=1 though?

Don't you integrate from x=1 to n?

Re: The Nonsense Thread Naa it's okay I was joking lol. We have to base it on our AOS "Changing Self" and I've got an idea of what I'm going to roughly be writing about.

no worries man :)

Say you want to consider all four bases (i.e. 2,3,5,7), the number of factors are 3*2*4*1=24 Say you to consider a particular 3-member combination of these four, the number of factors will be 3*2*4+2*4*1+3*4*1+3*2*1=50 Say you want to consider a particular 2-member combination of these four...

wait yeah this guy is right.

yeah I realised.

Well, by the quadratic formula: z^2=[-b+/-sqrt[b^2-4d]]/2=-b/2. Now z^2>=0 if the polynomial is to have real roots i.e. -b/2>=0. So P(z) has real roots if b<0 since d=/=0. Final answer: b<0.

K thanks carrot.

I can participate in next holidays mock HSC exam right, even though I'm not doing the HSC this year?

lol hahahahahahah. Never knew you had a dirty mind Carrot.

lol fuck no.

hey fawn whudup?

Don't you mean e^x > 1+x+x^2/2 lol?

Re: MX2 Integration Marathon Well I haven't even attempted that part but it doesn't look too difficult. I thought you were talking about UMADBRO's integral.

Re: MX2 Integration Marathon which question are you referring to?