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Re: HSC 2013 4U Marathon Alright what about taking the limit to infinity of a partial sum/product? If not, what can I do to fix this?

Re: HSC 2013 4U Marathon I have not defined alpha nor z in the complex plane. Would changing the order of iv and iii change the rigour issue here? EDIT: Or rather how about changing it so we have only a partial sum and a partial product and consider the inequality of the limit instead?

Re: HSC 2013 4U Marathon $ Question Prime: 11 marks$ $For the context of this question, a prime number is a natural number such that it is only divisible by itself and 1. 1 itself is not a prime number$ $The first few prime numbers are$ \ \ \ \ 2, 3, 5, 7, 11, 17 .... $You may assume that...

I don't like the idea of tutoring in general andd hence I don't want to go UMAT tutoring/preperation whatever it might be. I do realise however a lot of doing the UMAT do go to tutoring. Will not going to tutoring effect my chances? Does it have any benefit going to tutoring I mean I see...

Alright thanks for the advice.

Without the word seems, A is making a very big assumption that life is impossible without the given conditions in the question. Which is why I didn't like the answer. It also seems very redundant to base a question off of 'seems', everything can 'seem' to be true if we look at it from a...

True, I guess I didn't really take notice of the 'only' there. Thanks for the help

So I was doing some sample questions for the UMAT, from this page: http://www.umat.net.au/free-umat-sample-questions/ For the first section I got all but 1 right, and that one was: UMAT Sample Question 4 The presence together of carbon, water, and temperature at which water is...

Re: HSC 2013 4U Marathon $Show using integration that$ \frac{\pi}{4} < \sum_{k=1}^{\infty} \frac{1}{1+k^2} < \frac{\pi}{2}

Re: HSC 2013 3U Marathon Thread The easier way, and I will allow it, is to consider (1+x)^n = P(x) Then consider the sum of roots of P(x) 1 at a time 2 at a time 3 at a time and so on.... (all the roots are -1)

Re: HSC 2013 4U Marathon I did part B, but part A I going to attempt later (yes it is the one I am stuck on) though for Spiral's integrals, here is Q2: J=\int e^{3x} \sqrt{1+e^{2x}} \ dx Make the substitution: e^x = \tan u e^x \ dx = \sec^2 u \ du J= \int \sec^3 u \tan^2 u \ du J=...

Re: HSC 2013 4U Marathon Ok, I attempted the first one with my limited integration experience/knowledge and I was able to get it in the end, I probably took the long way but: 1. part i 1. part ii 1. part iii I will attempt the others ones now, then go back to the circle geo one :L

Sure bro, the answers are: C, A, E, B Post some more of these grate questions, they are much apraciated!

Re: HSC 2013 4U Marathon I've had enough of this, this is a marathon thread, I don't want to see your assignment questions again, they are your assignment questions, not ours. I also suggest that if you are going to cheat like this, don't put your real name in your username, makes you very easy...

Re: HSC 2013 4U Marathon Yeah I made the wrong assumption that angle TPC = angle TPE My method is invalid :(

Re: HSC 2013 4U Marathon I got it in the end.... \angle BOA = x \angle BOT = y \therefore \angle DBA = x \ \ \ \ ($due to isosceles triangle BOD and right angles in appropriate places$) \angle CPT = 180 - y \therefore ECB = 180-y \ \ \ \ ($isosceles triangle and right angles$) It...

Re: HSC 2013 4U Marathon :/ Ok, how about this, we know that OPT is collinear, and we know that B and C are collinear. So lets just extend these lines, they are not parallel hence they must meet at some point, lets call it A. By the symmetry argument the same thing must apply to C D, that...

Re: HSC 2013 4U Marathon I got it now relatively simple actually if you look at it what it means to prove something is collinear: Let angle BOA = x , angle BOP = y Now we need to prove x + y = 180 After showing various things like that OTP is collinear already (due to that property of...

Re: HSC 2013 4U Marathon True dat, back to square one.

Re: HSC 2013 4U Marathon Ok, for the first one, I don't know if my proof is allowed, but: Consider the triangles ABO and ACP Now, since radius perpendicular to tangent, \angle ABO = \angle ACP \ \ \ ($radius perpendicular to common tangents$) \angle BAO = \angle CAP \ \ \ ($common angle$)...