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Both have a threshold that you have to meet to get in, and if you meet them both, you're in. There's no weighting. Edit: More precisely, in terms of ATAR, just need 99+. Then in terms of UMAT, they'd give offers to the X highest UMAT scorers out of those who got 99+ ATAR / OP 1 Equivalent and...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Because $(n+1)^2 = n^2 + 2n+1$ and the $n^2$ here is greater than $10n +7>10n$ by the inductive hypothesis. I realise I forgot that the inductive hypothesis was $n^2 > 10n+7$ and was thinking $n^2 > 10n$. So you can rewrite the...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $I assume you can do the base case and conclusion for each question, so I'll just show the induction step.$ $1) Assume $n^2 > 10n + 7$ for some integer $n\geq 11$. We need to show that $(n+1)^2 > 10 (n+1)+7$.$ $We have$...

I think these kinds of divergent sums (and integrals for that matter) come up all in the time in quantum physics. For example, the 1+2+3+4+... one comes up in calculating the Casimir Force in 1D in Quantum Electrodynamics. The 13 + 23 + 33 + ... one comes up in the 3D version of this calculation...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent You can factorise it, but you'd need to justify that the term in the brackets is an integer. The only term in the brackets that you'd need to talk about for this justification is $\frac{k^2}{2}$ (you'd need to explain...

I like the 1 + 2 + 3 + 4 + … = -1/12 result, and find it also quite amazing that this is used in physics and gives some experimentally verifiable results. There's a lot of 'weird' stuff like this in this series of lectures on Mathematical Physics by Carl Bender that can be found on YouTube...

Here it is: http://community.boredofstudies.org/1094/general-discussion-2015-hsc/344201/will-you-stay-bos-after-your-hsc.html

Try rationalising the denominator for each one.

$\noindent I think they want you to assume that $\sqrt{x}$ is irrational. Then the fact that $a$ and $b$ are rational means that $a$ has to be $\frac{2}{3}$ and $b$ has to be $3$. This is based on the fact that if $\alpha$ is an irrational number and $a,a^\prime ,b,b^\prime$ are rational...

I guess I just got used to it haha.

$\noindent (a) We have$ $$\begin{align*}\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}&= \frac{1}{\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}} \\ &= \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left[ \left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5} \right] \left[ \left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5} \right] } \quad...

Re: HSC 2016 4U Marathon - Advanced Level $Tangent at $P$ has equation $\frac{x_1x}{c^2}+\frac{y_1 y}{d^2}=1$, which has slope $m_P = -\frac{\frac{x_1}{c^2}}{\frac{y_1}{d^2}}=-\frac{x_1 d^2}{c^2 y_1}$ for $y_1\neq 0$. Similarly, the slope of the tangent at $Q$ is $m_Q = -\frac{x_2 d^2}{c^2...

Re: HSC 2016 3U Marathon $\noindent \textbf{NEW QUESTION}$ $\noindent Explain the flaw (if any) in the following reasoning.$ $\noindent For a particle moving in a straight line, we know its acceleration is given by $\ddot{x} = v \frac{\mathrm{d}v}{\mathrm{d}x}$. Hence if $v =0$, then...

Re: HSC 2016 4U Marathon There is a "first year university maths thread", which seems to be a bit like these marathons, but yeah, it's a dead thread. Here it is for anyone curious: http://community.boredofstudies.org/1003/maths/321004/1st-year-university-mathematics-thread.html

Re: HSC 2016 4U Marathon Can't it also be done using the Leibniz rule for differentiating under the integral sign and also multivariable chain rule? Or is this assuming too many rules for what you had in mind?

Simply put, internal performance affects how your internal mark will be moderated ("scaled").

No, everyone's exam mark is their own exam mark. So if you got a higher exam mark than your friend, it means you did better than him in the external HSC Exam, and this is unaffected by internal ranks.

$Suppose by way of contradiction that $2+\sqrt{3}$ is rational. Then by definition of ``rational'', we have $2+\sqrt{3} =\frac{a}{b}$ for some integers $a,b$. Thus $\sqrt{3} = \frac{a}{b}-2 =\frac{a-2b}{b}$. The numerator and denominator here are both integers, so $\sqrt{3}$ is rational, which...

It's actually you who did the higher scaling subjects!

$\noindent One way to do it would be to use the fact that $R$ is the point of intersection of the line through $P$ parallel to $OQ$ and the line through $Q$ parallel to $OP$.$