$\noindent $\int f(\sin x)\, d(\sin x)$ essentially just means $\int f(u)\, du$ where $u=\sin x$. If you don't like using $d(\sin x)$, just write it using a $u$-substitution, integrate with respect to $u$, and then write the answer in terms of $x$.$
$\noindent For example, to find $\int e^{\sin...
$\noindent \textbf{Hints:} From the dot product, you should be able to find $\left\|\mathbf{b}\right\|$ (remember the formula $\boxed{\mathbf{a}\cdot\mathbf{b}=\left\|\mathbf{a}\right\|\left\|\mathbf{b}\right\|\cos\theta}$). You also know / can work out the angle $\mathbf{a}$ makes to the...
$\noindent \textbf{Hint:} Recall that we have $(\mathbf{u}\cdot\mathbf{v})^2 = \left\| \mathbf{u}\right\|^2\left\| \mathbf{v}\right\|^2\cos^2 \theta$, where $\theta$ is the angle between the two vectors. You should get a quadratic equation in $a$ from this, which you should know how to solve...
$\noindent You could in theory add the $\color{blue}\Leftrightarrow$ (or even $\color{blue}\Leftarrow$) symbol before each line after the first line to make the proof valid. However, I'm not sure if the HSC markers would accept it.$
You can show that if there are 7 stamps placed, then there must be a row of 3 stamps as follows:
Suppose 7 stamps are placed, then consider the "blank" squares (squares that don't have a stamp in them). There are 2 blank squares (because there are 7 stamps placed in 9 squares). Since there are...
$\noindent Note that it is not generally true that if $\cos A = \cos B$, then $\cos \left(\frac{\pi }{2}-A\right) = \cos\left(\frac{\pi}{2}-B\right)$.$
$\noinent Do you mean whether you needed to express your answer in radians? I would be very surprised if they penalised you for expressing the answer in degrees.$
$\noindent As long as you can show that $\alpha_{k}$ is a root for all $k =1,\ldots, m$ and can explain why the $\alpha_{k}$ are all different (i.e. if $k\neq j$, then $\alpha_{k} \neq \alpha_{j}$), then you are done, no need to use $p'$.$
Why is it obviously 42C9?
$\noindent This answer follows from noting that we need to insert at least one 0 between each 1. If there are $K$ $1$'s and $N-K$ $0$'s, there are $K-1$ gaps between $1$'s, so after putting one $0$ in each gap, there are now $N-K-(K-1)=N-2K+1$ $0$'s left to place...
I got 42C9 using the method I posted above, so I assumed there was a simpler method.
For the equivalence between the method your teacher used and the original question, it is as follows:
The bit string of length 50 corresponds to your choice of whether you pick each number from 1-50 or not. A...