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$\noindent What have been your attempts thus far? To find $y''$ in terms of $t$, first find $y'$ in terms of $t$ using $y'=\frac{dy/dt}{dx/dt}$. Then, we will have $y'' = \frac{dy'}{dx}= \frac{dy'}{dt}\frac{dt}{dx}$. (And make sure you remember how to find something like $\frac{dt}{dx}$!)$...

$\noindent I don't know what the HSC marking standard is, but in my opinion we shouldn't need to show the real parts (essentially because they are completely irrelevant to us and we (should!) know how to find imaginary parts of expansions like this without calculating the real part). We simply...

If you expand by only keeping the imaginary part, then at least you essentially only have to write four terms for the expansion part. $\noindent (Also for brevity, it would be wise to do something like define $c\equiv \cos \theta$ and $s \equiv \sin \theta$.)$ Edit: just had a proper look at...

Isn't that written on the paper?

It looks like only the odd-numbered pages are there.

$\noindent This is a question of \textbf{binomial probability} (though I'm not sure the 2U syllabus calls it that). The probability of any single coin being faulty is $p = 0.03$ (i.e. $3\%$).$ $\noindent To have at least 2 coins in the sample of 3 coins be faulty, this means either exactly 2 or...

$\noindent To get an expression for $P(T)$ and $P(H)$ in terms of $q$, note that $P(T) = 1-P(H)$. Substitute this into $\frac{P(T)}{P(H)} = q$ and solve for $P(H)$. However, we don't actually need to know exactly what $P(H)$ is in terms of $q$ to do this question. It is enough to know that...

$\noindent You can do both of them by using the ``auxiliary angle method'' to convert something of the form$ $$a\cos \theta + b\sin \theta$$ $\noindent into the form$ $$R\cos \left(\theta-\phi\right).$$ $\noindent (You can check your textbook for this if you haven't learnt it yet or want to...

$\noindent Incidentally, you should end up with the right answer (also to fan96's question) if you blindly apply log laws like $\ln a - \ln{b} = \ln\left(a/b\right)$, without worrying about absolute values or negative arguments in the log.$

Have you tried using the discriminant? What do you know about what the discriminant looks like based on whether the quadratic has no real roots, or has a double root?

If you don't want to graph stuff, you can also do it algebraically in cases. The cases to consider are x ≤ -2, -2 < x < 3, and x ≥ 3 (because these are the places where the inputs to the absolute value change sign). E.g. if x ≤ -2, then x + 2 ≤ 0, so |x+2| = -(x+2) = -x-2, and x - 3 ≤ -5 < 0...

$\noindent If given a velocity function $v(t)$, the answer in general is $\color{blue}\int_{t_{1}}^{t_{2}}|v(t)|\, dt$. That is, to get the distance travelled over a given time interval, integrate the \textbf{absolute value} of $v(t)$ over the time interval in question.$ $\noindent (By the way...

$\noindent This is true in general (for any (smooth) flight path, at any point on the curve). If $\beta$ is the angle made by the curve to the positive $x$-axis at some point in time, then $\tan \beta = \frac{v_{y}}{v_{x}}$ at that time (where $v_{x} \equiv \dot{x}$ and $v_{y}\equiv \dot{y}$)...

Did you copy almost all of the answers, or just a small number? When you copied some answers, did you stop and think for any of them whether the copied answer seemed correct for the given question? For example, if most of the time it seemed incorrect, it could be the case the answer orders were...

It's because the number of ways to choose 2 objects is exactly the same as the number of ways to choose 8 objects (from a set of 10 objects). Why? Because for each way of choosing 2 objects, you are also actually choosing 8 objects (and vice versa)! Namely, by choosing 2 objects, you are...

Thanks. The main reason I wrote it was to show students the idea of writing the polynomial as a polynomial in (z+ 1/z). In fact, the polynomial P(z) is an even degree palindromic polynomial, and any even degree palindromic polynomial of degree 2N (in z) can be written as z^N times a polynomial...

$\noindent Oh yes, I was assuming that. (The axis of symmetry for $q(t)$ is at $t = -5/4$.)$

$\noindent Well essentially we have $P(z) = z^2 \left( 2z^2 + 5z + 7 + 5z^{-1} + 2 z^{-2}\right)$. Hence $P(z) = z^2 \left( 2 \left( z + z^{-1}\right)^{2} + 5\left(z + z^{-1}\right) + 3\right)$.$ $\noindent So if $x$ is a real root of $P$, then $x \neq 0$ and $q(u):= 2u^2 + 5u + 3 = 0$, where...

$\noindent By the way, an easier way is to just say the denominator is greater than or equal to $x^2$, since $2y^4 \geq 0$. Then carry on similarly.$

$\noindent \textbf{Hint.} If the probability that the sum of the results of two die is $6$ (respectively 8) is $a$ (respectively $b$), the probability $6$ occurs first is $\frac{a}{a+b}$ (make sure you can show this!).$