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$\noindent We have $x-4\sqrt{x-1}+2=0$ (note that we must have $x\geq 1$ as the expression under the square root needs to be non-negative). Rearranging, we have $x+2 = 4\sqrt{x-1}$ (note that this implies that $x+2\geq 0$, i.e. $x\geq -2$. But we had $x\geq 1$ anyway, so this restriction of...

Yes.

$\noindent Note that $1 \text{ m} = 100\text{ cm}=10^2 \text{ cm}$. Hence (squaring both sides) $1 \text{ m}^2 = 10^4 \text{ cm}^2$ (as $1^2 = 1$ and $\left(10^2 \right)^2 = 10^4$). Now,$ $$\begin{align*}5\text{ ha}&= 5 \times 10^4 \text{ m}^2 \quad (\text{as we are told that }1 \text{ ha} =...

(Btw you did a typo, it should be 3300 in the denominator.) I did do it correctly, but yeah, your method is faster. :)

Re: Flop math question thread Yes

$\noindent We had a number like $x=0.000212121\ldots$, so since it's nicer to work with the decimal with the first non-zero decimal place started right after the decimal point, we rewrote that as $x=10^{-3}\cdot \underbrace{0.212121\ldots}_{\text{call this } y}$ (i.e. shift the decimal place in...

$\noindent This is because$ $$\begin{align*}0.21121212121\ldots &= 0.211 + 0.000212121\ldots \quad (\text{separate out the first three decimal places. It's like how }0.56414 = 0.564 + 0.00014).\end{align*}$$

For these types of Q's, you can do it as I did above (separate out the decimal so it is a finite pattern + simple recurring pattern). An alternate way of converting a simple recurring pattern decimal to a fraction is by using geometric series.

$\noindent Let $\chi=0.211212121\ldots$. Then $\chi=0.211+0.000212121\ldots=0.211+x$, where $x=0.000212121\ldots$.$ $\noindent Note that $x=10^{-3}\cdot y=\frac{1}{1000}\cdot y$, where $y=0.212121\ldots$. Now, $100y=21.21212\ldots$. So $100y-y =\left(21.21212\ldots\right) -...

$\noindent I'm still not sure what the number is. Maybe write it out a few times so we can tell the pattern. Is it $0.211211211211\ldots$?$

$\noindent I don't understand what the number is. Is it: $0.21111...$ (i.e. $0.2\dot{1}$, also written as $0.2\bar{1}$)?$

What I meant by that was that you can "see it" (that the lines have to be parallel because the two points are equal height above a given line). The OP phrased it as though they couldn't see why the lines were parallel, so I provided the intuition for it before providing a proof. Another easy...

$\noindent Well basically, since $\triangle ABD$ and $\triangle ABC$ have a common base $AB$ and have equal area, their height is the same, as you said the answers says. This means that $D$ has to be the same perpendicular height above $AB$ as $C$, and this forces $AB\parallel DC$. This can be...

$\noindent When you come to something like $\sin \theta \times \text{bla}=\sin \theta \times\text{blah}$, instead of going straight to $\text{bla}=\text{blah}$, say something like ``hence $\sin \theta = 0$ or $\text{bla}=\text{blah}$''. (This is due to the fact that in general if $ab=ac$ where...

$\noindent When you divided out the $\sin \theta$ from both sides early on, you implicitly assumed that $\sin \theta \neq 0$ (since we can't divide through by 0). So you missed out the solutions for $\theta$ where $\sin \theta=0$. But clearly if $\sin \theta=0$, the original equation \textit{is}...

Re: Flop math question thread Yeah, they are in 2U.

It is possible I think, but I think you would need to work pretty hard to catch up on maths.

It is a subset of it I presume, which would explain why the proportions for "More than 20 pts..." are always greater than or equal to the proportions for "More than 10 pts..." in one of the figures in the article. Edit: Actually looking at the figure more closely, it doesn't make sense. I guess...

Re: MX2 2016 Integration Marathon Which formula did we prove on the previous page? Can't find any relevant ones there.

Re: HSC 2016 2U Marathon (btw, we should actually say "or" rather than "and", since "and" means they are simultaneously true.)