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I think you're right, but I don't think it would matter for him if rather than med, he might prefer to study something that he's more interested in

free bump

'You must spread some Reputation around before giving it to Trebla again.'

this is very neat wish I could rep

I doubt that's gonna appear in any of your school assessments or the hsc but it's quite intuitive to understand the idea of an improper integral this is an example: in your particular question, we should get(after doing the substitution) :

"P(x) is an even polynomial with real coefficients" yeah, I wasn't really thinking clearly, OP's approach was right (i.e. if z=rcis(theta) is a root, then -z=-rcis(theta) must also be a root since it's even) we have: one complex root z, one real root a, 1< a <2 then zbar is a...

Re: [The Mathematics CORNY Pick Up Lines THREAD] reminds me of this even 'function' , hm

and you can check that the restrictions are satisfied , when you substitue

sorry, my bad they're written in white , you gotta highlight them to read I didn't wanna spoil it, so before you read it, these are some hints: basically you need to compare the inequaity you can obtain from part i) and the one to prove in part ii) Then you'll need to make some...

I found it quite difficult to differentiate 3u from 2u content as well when I was a student

just for the 1st part of ii) compare the given inequality with the one you obtained in part i): y= [ x^k-(c-x)^k ] /2 <= (c/2)^k , we'll need to make the following substitutions : let c=a+b then we can let x=a, so c-x=a+b-a=b and we get a strict inequality since: a=/=b, so x=/= c/2...

btw, this is actually an extension 1 topic so I suppose this post is in the wrong section and this is Year 12 , Chapter 5 in the Cambridge book (binomial probability would be in the last chapter of the Year12 book)

used r instead of k here: n over the r in the brackets is the same as nCr, which is equal to and both of them are read as 'n choose k' they're just different notations that people invented to confuse you

Consider the complex root z=a+bi note: z^2=(a+bi)^2=a^2+b^2+2abi zbar^2=(a-bi)^2=a^2+b^2-2abi So (a+bi)^2=(a-bi)^2 if and only if a=0 or b=0 i.e. z=a+bi is purely real or purely imaginary when z is purely imaginary that is, z=bi , then zbar=-bi but then -z=-bi=zbar, -zbar=bi=z so -z and...

but 'study' and 'mEAT' are equally important !!!

NOTE : (1+x)^8= 8C0 + 8C1 x + 8C2 x^2 +...+ 8C8 x^8 and counting from 1, we have T1, T2, T3, ..., T9 respectively So Tk=8 choose k-1 (*) Tk+1 = 8 choose k the inequality you solved is acutally for Tk+1 > Tk , from which we obtain k<4.5 hence T5>T4>T3>T2>T1 inequality doesn't hold for larger k...

#1 says that there are currently: 34 students , 13 tutors (including 2 who may not be available) it's great that there are gonna be more engLish tutors willing to help on the day BTW, we could post off topic stuff here: http://community.boredofstudies.org/showthread.php?t=280826&page=2865...

would L'Hopital's Rule be included? maybe it's too time consuming