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Not impossible at all.

$\noindent \textbf{Hint.} Use the conjugate root theorem, and the fact that $\overline{p^{k}} = p^{7-k}$ for any integer $k$ (make sure you know why this is true!).$

For the first one, we have 4 different lollipop types and want to buy a total of 10 lollipops. So we use 3 bars and 10 stars. Something like this: *** | ** | **** | *. That above pattern would represent buying 3 of the first lollipop type, and 2, 4, 1 of the second, third and fourth...

The question's wording is such that it wants us to assume we have already picked 4 boys and 4 girls.

The desired line has equation 3x + 2y - 12 + k(5x - y - 7) = 0 for some value of k. All you need to do is find what k is. Hint: To do this, note that the desired line should be of the form ''x = something'', because it should be perpendicular to y = 5,

I doubt it. If it is, it would be from a long time ago and would be coincidence (I didn't pick that question out of a past HSC paper). I included it since it's a nice application of Stars and Bars.

$\noindent 1) There are 4 different types of lollipops available at a lolly store. Alice wants to buy 10 lollipops. How many different selections can she make?$ $\noindent 2) Let $r$ be a positive integer and $n$ a non-negative integer. Find the number of ordered $r$-tuples of non-negative...

$\noindent By the way, the functions $\mathcal{C}$ and $\mathcal{S}$ above are known as the \textit{hyperbolic functions} $\cosh$ and $\sinh$. They have a number of interesting properties (for example, each is the derivative of the other) and are related to the trigonometric functions $\cos$ and...

$\noindent I'll give you an outline of how to compute the \emph{in}definite integral $I = \int \left(\frac{e^{x} - e^{-x}}{2}\right)^{3} \left(\frac{e^{x} + e^{-x}}{2}\right)^{11} \, dx$. If you can do this, you should be able to do your question.$ $\noindent Write $\color{blue}\mathcal{S} =...

$\noindent The mistake is in this line:$ $\noindent Putting $t = 0$ there, we have $\sin \alpha = 0$. This does not imply that $\alpha = 0$, but only that $\alpha = 0$ or $\alpha = \pi$ (note it suffices to take $\alpha$ between $0$ and $2\pi$ due to periodicity). The correct one to choose...

$\noindent Well basically the acceleration of a particle at any moment in time is independent of its ``initial location''. So it suffices to only consider displacement from the ``middle'' of the arc when deducing the acceleration expression.$ $\noindent The reason I said to integrate from...

$\noindent It's because the displacement along the arc is given by $s = L\theta$, so $\ddot{s} = a = L\ddot{\theta}$. (Acceleration being the second derivative wrt time of displacement.)$

$\noindent Ah yes, I meant $-g/L$, not $g/L$, sorry.

$\noindent Here are some hints.$

$\noindent Imagine repeatedly inserting \$1 coins and consider a sequence of random variables $X_{1}, X_{2}, X_{3},\ldots$ defined by $X_{j} = I(A_{j})$, for $j=1,2,3,\ldots$, where $A_{j}$ is the event that the player wins on the $j$-th play. ($I$ represents an indicator function.)$ $\noindent...

"Inspection" :p

$\noindent Similar rectangles implies that $\frac{z}{y} = \frac{y}{x}$, so $xz = y^{2}$. \color{blue}{But $x = z/2$, since we are cutting the paper in half. }\color{black} Substituting $x = z/2$, we have $xz = y^{2} \Rightarrow \frac{z^{2}}{2} = y^{2} \Rightarrow \frac{z}{y} = \sqrt{2}$. That...

Re: HSC 2018 MX2 Marathon Hint: conjugate root theorem.

The basic reasons why are the following: The partial sums of an even number of terms is positive always because an even partial sum is the previous even partial sum, plus a number, minus a number (e.g. +1/3, -1/4); and the term you're subtracting is smaller than the term you're adding, so the...

$\noindent Perhaps this will make it easier: just redo the question by rewriting the condition in the equivalent form $x^{-1}gx = g$. (And of course, make sure you know / can explain why this is equivalent to the given condition $gx = xg$.) I.e. use $C_{g} = \left\{x\in G : x^{-1}gx = g\right\}$.$