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For dy/dx, just do it using a fraction. Code to type a fraction in LaTeX is shown below. [tex.] \frac{a}{b} [/tex.] produces \frac{a}{b}. i.e. you write "\frac{a}{b}". So to get dy/dx, the code is \frac{dy}{dx}. For an integral sign: [tex.] \int f(x) dx [/tex.] produces...

Using LaTeX here (on BOS) is a bit different to using it on MiKTeX/TeXMaker etc. The biggest difference is that to use LaTeX here, you need to enclose the code within "TeX tags". These are like this: [tex.] <insert code here> [/tex.], except exclude those red full stops. (I only put the red full...

Assuming a circular clock with hands undergoing uniform circulation motion, the speeds of the hands in revolutions per hour (rph) are: • second hand: 60 rph (since it undergoes one revolution per minute); • minute hand: 1 rph; • hour hand: 1/12 rph (since it takes 12 hours for it to undergo a...

Basically the convention with the "√" symbol for complex numbers is to take the root that has its principal argument in the range (-pi/2, pi/2]. (For any non-zero complex number z, there is a unique square root of z with principal argument in this range. This square root is called the principal...

$\noindent Recall the formula for the area of a segment: $A = \frac{1}{2}r^{2}(\theta -\sin \theta)$. Using this, for 19(b), we have $1 = \frac{1}{2}\times 1^{2} (\theta -\sin \theta)$. Now use your work from part (a). Make sure to express the answer in degrees as the question asks for.$

$\noindent Well here's some hints to at least evaluate that sum in your question. Once you evaluate the integral, you should see that you effectively need to compute a sum of the form$ $$S = \sum_{k=1}^{n}k v^{k-1}.$$ $\noindent There are a few ways to evaluate such a sum. Here's a hint for...

$\noindent \textbf{Hints.} Let $p = m$ (sorry, I started writing with $m$ instead of $p$, and forgot it was $p$ until I'd written up a fair amount). Note that we can break up $(mn)!$ into $n$ chunks of $m$ consecutive integers:$ $$(mn)! = (\color{red}{1}\color{black}m\times (m-1)\times \cdots...

$\noindent Here's some steps/hints for a possible method:$ $\noindent 0. Show in general that if $u,v,w\in \mathbb{C}$ are collinear, then $A u, A v, A w $ are collinear for any given complex number $A$.$ $\noindent 1. Show that the result is true if the circle is centred at the point $(R,0)$...

$\noindent \textbf{Hint.} If you substitute $u = 2^{x}$, then you get a quadratic inequation to solve in $u$ (noting that $u > 0$ also because $u = 2^{x}$). So you can do the question if you've learnt how to solve quadratic inequations. $

This is not correct. You are counting things as though order matters for the total number of ways to pick four candies from the 24 in the box. But when you are counting the number of ways for each of the cases (e.g. RRYB), you have counted the ordering as though the reds must come first, then...

Yes, they have. Even if you don't know anyone personally who has, it turns out to be quite easy to find the answer to this question with a bit of googling (verification of this left as an exercise to the reader :P).

$\noindent \textbf{Hint.} You can do it by breaking it up into cases. Find the number of ways to pick four candies so that your colours picked are (where order of picking is unimportant): (i) RRYB; (ii) RYYB; (iii) RYBB. (Those three cases are the only ways to have all three colours picked in...

$\noindent Are you able first of all to find the general solution to$ $$\cos \theta = \frac{1}{2}?$$ $\noindent If you can do this, you should be able to easily do the questions you wrote. But if you \emph{can't} do this, you should probably work on this first.$

Yes, I do.

Answer to your last question: yes.

$\noindent Well the graph shown is not the graph of $y= 1 -\frac{1}{2x}$ (can you explain why?). Based on that graph, the desired equation may have been $y = 1 -\frac{2}{x}$ instead.$

For 22b), you have made an error in the squaring. Double check the exponential term (and your index laws if need be).

For 7a), that's not the right integral for the area of the given region. Your integral finds the area under the exponential curve, but you want the area between a horizontal line and the exponential curve. A hint to get the right answer for 7a) without doing any more integrating: the desired...

Re: HSC 2018 MX2 Marathon $\noindent \textbf{Hints.}$ $\noindent $\bullet$ A 3U method: you already know a formula for cos-squared:$ $$\cos^{2}\theta = \frac{1}{2}\left(1 + \cos 2 \theta\right).$$ $\noindent (This is known as a \textsl{half-angle formula}.) So to get $\cos^{4}x$...

That is not the right antiderivative. We can't just use a "power rule" like that because there is an x2 term present. It is easier to obtain the answer if you use a well-known formula for the area of a circle.