"How many ways can 21 people be divided into 3 equal groups?"
Just wondering if the answer should be 21C7 14C7 7C7 or same but divided by 3!, seeing as I don't know if the groups are implied to be 'distinct' or not.
Is it possible that some student results are entered later than others for the same subject? Because mine is a lot of points (like almost 10) over what I'm expecting lol
Re: UNSW chit chat thread 2016
can anyone who's done actl1101 confirm lectures are definitely recorded? would I be missing much just watching them from home
Thanks leehuan and Integrand!! I managed to get the arctan one out.
However for the first one I asked, I just want some clarification. I know 23 < c < 27 so if c < 27 then I can obtain the lower bound. But obviously if I use c>23 then I would require a calculator to show that the 3-23^(1/3)<...
Can anyone prove \frac{4}{27}<3-23^{\frac{1}{3}}<\frac{3}{16} using the mean value theorem? I've been trying to come up with functions that I can apply the MVT to but I've gotten stuck.
Also for \frac{8}{145}<\tan^{-1} \frac{9}{8}-\frac{\pi}{4}<\frac{1}{16}
Let f be continuous in \mathbb{R} with \lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow-\infty}f(x)=0
Showthat if there is a number \xi such that f(\xi)>0 then f attains a maximum value in the reals.
[Note the max min theorem applies to finite closed intervals [a,b] only]...
2) Suppose that for any K>2 the solution to f(x) > K is [tex]x\epsilon(1,\frac{K}{K-2})[\tex]
What is the behaviour of f(x) as x approaches 1+? Carefully explain your answer
This is mainly a problem with my working.
So intuitively I know that the limit must be infinity. the reason being that...
Hey guys,
I don't come on here often but I've seen similar threads so I've decided to make one too :) For your reference I am doing MATH1151 this sem.
1) Find \lim_{x\rightarrow\infty} \sin\frac{1}{x} . I already know it's 0 but they want a solution using the pinching theorem.
You must be doing math1151 too?? haha I got stuck on this for a bit too. I just used a GP summation
1+x+x^2+x^3+x^4=\frac{x^5-1}{x-1}
since we're not considering this when x=1. then the top is negative, and so is the bottom! so when you divide it it's positive.
Of course it would work for...
lmao then you're definitely getting in. the centre only distributes the test to find a class that best suits your learning pace/ability. they take all kids regardless of ability although sometimes they will tell you that 3u isn't for you. in the past they've recommended other centres for 2u but...
Hey guys,
Do any of the actl kids on here know if the 2015/2016 packs will be different? i.e. if they do will they affect me in any way? Someone wants to sell me their pack for cheap. But idk if it's worth the RISK. Should I just buy the 2016 one?