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$\noindent Now $a + b = -m$ and $ab = n$. Using the quadratic formula, \\ \begin{align*} x &= \frac{m^2 - 2n \pm \sqrt{(2n-m^2)^2-4n^2}}{2n} \\ &= \frac{(a+b)^2 - 2ab \pm \sqrt{(2n-m^2-2n)(2n-m^2+2n)}}{2ab} \\ &= \frac{a^2+b^2 \pm \sqrt{-m^2(4n-m^2)}}{2ab} \\ &= \frac{a^2 + b^2 \pm...

$\noindent Now $a + b = -8$ and $ab =-5$. The roots $\left(\frac{a}{b}\right)$ and $\left(\frac{b}{a}\right)$ have a sum $\left(\frac{a}{b}\right) + \left(\frac{b}{a}\right) = \frac{a^2+b^2}{ab} = \frac{(a+b)^2-2ab}{ab} = \frac{(-8)^2-2(-5)}{-5} = \frac{-74}{5}$. The roots have a product $\left(...

Re: HSC 2017 Maths (Advanced) Marathon $\noindent $\begin{align*} S_n &= T_1 + T_2 +T_3 + ... + T_{n-1} + T_n \\ S_{n-1} &= T_1 + T_2 + T_3 + ... + T_{n-1} \end{align*} \\ $As can be seen, $S_n - S_{n-1} = T_n

Re: HSC 2017 Maths (Advanced) Marathon $\noindent $T_n = S_n - S_{n-1}.$ Your answer is $T_{16} = 249 - 233 = 16

Re: HSC 2017 Maths (Advanced) Marathon $\noindent \textbf{QUESTION 1} \\ $ \begin{align*} 2\ln{x} &= \ln{(5+4x)} \\ \ln{x^2} &= \ln{(5+4x)} \\ x^2 &= 5 + 4x \\ x^2 - 4x - 5 &= 0 \\ (x-5)(x+1) &= 0 \\ x &= -1, 5 \\ x \neq -1 \textrm{ } (x > 0), \textrm{ } x &= 5 \end{align*} \\\\\\...

$\noindent There may be a shorter method but this is what I have so far; \\ (choosing base $\sqrt{2}$ also works) \\ \begin{align*} \log_a{16} + \log_{\sqrt{2}}a &= 9 \\ \frac{\log_{16}{16}}{\log_{16}{a}} + \frac{\log_{16}{a}}{\log_{16}{\sqrt{2}}} &= 9 \\ \frac{1}{\log_{16}{a}} +...

But if you get first in the internal assessments (even with a terrible mark) then get 100% raw HSC mark its most likely a SR isnt it?

I think you must also be ranked first in your internal assessments amongst your school

I think the syllabus has all of them so if you scroll through they identify all the ones you need. eg on page 22 they have the wave equation.

When is everyone's mathematics half-yearly?

$\noindent \begin{align*} V &= \pi \int_0^1 (x^2)^2dx + \pi \int_1^2 [(x-2)^2]^2dx \\ &= 2 \pi \int_0^1 (x^2)^2dx \\ &= 2\pi \left[\frac{x^5}{5} \right]^1_0 \\ &= \frac{2}{5} \pi \text{ cubic units} \end{align*} \\ The transition from the first step to the second step is due to the fact that the...

$\noindent Simply state that $y' = -3 < 0$ $\forall x$.

Re: HSC 2017 Maths (Advanced) Marathon $ \noindent $\overline{x}_1 = \alpha_1x_1+\alpha_2x_2 + ... + \alpha_nx_n\\ \overline{x}_2 = \beta_1x_1 + \beta_2x_2 + ... + \beta_nx_n \\ \ldots \\ \overline{x}_K = \gamma_1x_1+\gamma_2x_2+ ... + \gamma_nx_n \\\\ $Now the weighted average of...

Re: HSC 2017 Maths (Advanced) Marathon $\noindent Now, the weighted average of $\left\{\overline{x}_{1},\overline{x}_{2},\ldots, \overline{x}_{K}\right\}$ is given by \\\\ $\sum_{i = 1}^{K}(\alpha ')_i\overline{x}_{i}=(\alpha')_1\overline{x}_{1} + (\alpha')_2\overline{x}_{2} + ... +...

I got the same results, thanks

Does anyone have the MC solutions? The paper can be found on dan's THSC

$\noindent $(0.5)^{x} = \left (\frac{1}{2}\right)^{x} = (2^{-1})^{x} = 2^{-x}$, as required. \\\\Now, we have $y = 2^{-x}$. This is not a rising curve as it is simply the reflection of the curve $y = 2^{x}$ about the $y$-axis. \\\\This result can also be obtained as follows: let $y = f(x).$...

$\noindent For $x^{2} + y^{2} = 5$ it becomes obvious that the only two numbers $n_{1}, n_{2}$ such that $n_{1}, n_{2}\in\mathbb{Z}$ and $(n_{1})^{2} + (n_{2})^{2} = 5$ are $n_{1} = \pm1, \pm2$ and $n_{2} = \pm2, \pm1$ respectively \\\\ From this, we obtain the set of coordinates \\$\left...

The answer you attached does not have axis y = 0 nor does the point (3,6) belong to the curve. $\noindent We know that the curve is of the form $y^{2}=4ax.$ Note that I did not write $\pm 4a$, as a parabola with vertex the origin cannot pass through a point in the first quadrant if it's...