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$\noindent It is probably easier to prove uniqueness (and get the actual answer) by just solving the given pair of simultaneous equations (fairly easy, just start by taking the log base $a$ of both sides of each equation $\ddot{\smile}$.).$

Correct!

$\textbf{NEW QUESTION}$ $\noindent Let $a$ and $b$ be real numbers with $a>1$ and $b\neq 0$. Suppose that $\frac{a}{b}=a^{3b}$ and $ab=a^{b}$. Find $b^{-a}$.$

$\noindent Note that the L.H.S. factorises into $(x+2y)(x+4y+3)$. This can be motivated by writing the L.H.S. as $x^2 + (6y+3)x+ \left(8y^2 + 6y\right)$, and factorising this quadratic in $x$. We want to write it as $(x+\alpha)(x+\beta)$, where $\alpha \beta = 8y^2 + 6y$ and $\alpha + \beta =...

He probably wanted to devote a thread to it.

I'm guessing the resistance is linear vs. temperature when the temperature is near the reference temperature in that formula? But far away from this, it could be a steeper curve?

Yeah, in fact, nothing in (real) physics can be understood properly without mathematics. Btw, resistance generally increases linearly with temperature, see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html

This is why I was quite surprised that the HSC is well-regarded – at least for sciences, it shouldn't be, since the HSC Science subjects have been essentially reduced to memorisation contests and contain more history and social science in them than actual science (although the syllabi for HSC...

Answer is pretty much "yes" to all the yes/no questions.

Why is the HSC well-regarded?

$\underline{QUESTION 4}$ $\noindent Let $\Upsilon \equiv a^2 + \frac{1}{bc} + b^2 + \frac{1}{ca} +c^2 + \frac{1}{ab}$ (I'm assuming that's what you meant, as there were no brackets). Assume that none of $a,b,c$ are 0. Note that $\frac{1}{bc}=\frac{a}{abc},\frac{1}{ca}=\frac{b}{abc}$ and...

$\underline{QUESTION 3}$ $\noindent Given:$ $$\left\{\begin{matrix}bcd=\frac{a}{9}\quad (1)\\acd=b\quad (2)\\abd = \frac{9c}{4} \quad (3)\\abc = 4d \quad (4)\\ \end{matrix}\right.$.$ $\noindent From (2), we have $cd=\frac{b}{a}$. So subbing in (1), we have$ $$\begin{align*}b\cdot \frac{b}{a}...

That's the cheap way out. :p (Still valid.)

$\underline{QUESTION 2}$ $\noindent The answer is 1. We have $x=\sqrt[3]{\sqrt{2}+1}-\sqrt[3]{\sqrt{2}-1}$. Note that $\left( \sqrt{2}+1\right)\left( \sqrt{2}-1\right)=2-1^2 = 1 \Rightarrow \sqrt{2}+1 = \frac{1}{\sqrt{2}-1}$. Let $\iota=\sqrt{2}+1$, so $\frac{1}{\iota}=\sqrt{2}-1$. Then we...

$\underline{QUESTION 1}$ $\noindent The answer is $-3$.$ $\noindent Let $\Omega \equiv x^5 - 5x$. Note that we have $x^2 + x = 1 \Rightarrow x^5 + x^4 = x^3\Rightarrow x^5 = -x^4 +x^3$. So we have $\Omega \equiv x^5 - 5x = -x^4 + x^3 - 5x$ (replacing $x^5$ with $-x^4 +x^3$).$ $\noindent...

$\noindent He basically expanded it normally and added and subtracted the term $abc$ in order to allow it to be factorised nicely. It is valid to add and subtract $abc$ like this because it is equivalent to adding $0$.$

$\noindent For Question 1, we start off doing it as though it's a usual quadratic in $x$, so treat $y$ as a constant. Calling the expression $E$, we have$ $$\begin{align*}E\equiv x^2+4y^2+x+2y-5xy-2 &= x^2 +\left(1-5y\right)x + \left(4y^2 + 2y-2 \right).\end{align*}$$ $\noindent First we'll...

The classic "add and subtract".

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent This is due to the following trig. identity: $\cos^2 \theta = \frac{1}{2}\left(1+\cos 2\theta \right)$.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Yes, you essentially end up deriving the physics formulae. Like when the acceleration is a constant 2 (all units in SI units), then $\ddot{x}=2$. Integrating gives $\dot{x} = 2t + c_1$. When $t=0, \dot{x}$ is given...