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Haha thanks, and you're welcome. Congratulations on your HD! :)

$\noindent The answer is the inradius (radius of the incircle). Let the inradius be $r$, the semiperimeter of the triangle be $s$, and the area of the triangle be $A$. Then $A = rs$ (this holds for a general triangle; you can show this result by drawing the perpendicular a from the incentre...

$\noindent It is clear that the locus will be symmetric about the $y$-axis, so if a point $(x,y)$ is on the locus, so is $(-x,y)$. So let's just focus on when $x\geq 0$ (i.e. when $B$ lies on or to the right of the $y$-axis).$ $\noindent Note that the polar equation of the circle in equation is...

$\noindent It's because it's after the angle it makes with the positive horizontal axis. Since the vector in part (b) is in the third quadrant, we add $180^\circ$ to the acute angle it makes with the horizontal. (Like it goes $180^\circ$ from the positive $x$-direction to get to the negative...

$\noindent Q20. The sum of roots is 0, because the roots are equal in magnitude but opposite in sign. From sum of roots formula, setting sum of roots equal to $0$, it follows that $k=-1$.$ $\noindent Q21. The claim follows from the fact that the discriminant of this (real) quadratic polynomial...

What do you mean add 90? If you wanted the direction from the positive horizontal axis, then yes it'd be 90 deg + theta, but is that really what it wants? What type of answer does it want? Some type of compass bearing?

Yeah, English is a lot to memorise if you take the memorisation route (which many people do, I think). It can really help improve speed though if the essays/creative have been properly memorised.

Maybe try memorising an adaptable essay and spend less than five minutes on planning? And maybe try regularly handwriting to improve speed over the coming months.

Well essentially the displacement is the magnitude together with the direction. The magnitude on its own would be the distance of the motorist from the original position. So displacement is basically when we also care about the direction (unlike for distance). E.g. If you walked 1 km North from...

Make sure for the first one to also give a direction, since displacement is a vector :). You can use basic right-angled trig. to provide a direction (kind of like a compass bearing like in HSC maths if you remember the bearing stuff). (For the second one, it's only asking for the magnitude of...

It turns out no. The limit is actually 0. An easy way to show this is with the help of L'Hôpital's rule (which isn't in the syllabus though). Here's the graph of the function: http://wolframalpha.com/input/?i=plot+y+%3D+%28sin+x+-1%29%28tan+x+%2B2%29&x=0&y=0 . Since it approaches 0, we could...

Note x = pi/2 is void because the tan(x) in the LHS is undefined at pi/2 (so pi/2 isn't part of the domain of the LHS). (You may find it interesting to investigate what happens to the LHS as x approaches pi/2.)

$\noindent OK. For non-negative integers $n$ and $k\leq n$, we can define the \textsl{binomial coefficient} $\binom{n}{k}$ (read ``$n$ choose $k$'', with one alternative notation being $^n C_k$, which is why you will find a calculator button like this on some calculators) as the number of ways...

Re: Discrete Maths Sem 2 2016 $\noindent Note for sets $S$ and $T$ we have $T \in P\left(S\right)$ iff $T \subseteq S$. Also for any set $S$, $S$ is in its power set.$ $\noindent Now, suppose $A$ and $B$ are sets with $P\left(A\right) = P(B)$. Since $A \in P(A)$, we have $A \in P(B)$ (as...

You've done your answers assuming there's an order that matters for the cards, but for a hand, the order doesn't matter, so we should take this into account. $\noindent b) Pick the Ace: $4$ ways.$ $\noindent Pick the remaining four cards: We need to pick four cards from the remaining 48...

Re: Discrete Maths Sem 2 2016 $\noindent Well $A^c$ is the set of all things in $U$ (the universal set) that are not in $A$. Let $A \cup B = U$ and $A \cap B = \emptyset$. Suppose $x\in B \subseteq U$. Then since $A\cap B = \emptyset$ (i.e. $A$ and $B$ are disjoint), $x$ is not in $A$, or in...

Re: HSC 2016 3U Marathon $\noindent Yes, $c$ is just $5$. It's probably easier to just get $a$ next, since it is just $2$. This is because $a$ is the amplitude, from the classic S.H.M. equation $v^2 = n^2 \left(A^2 - \left(x-x_0\right) ^2\right)$. So $a$ is just $2$. Now we just need to find...

Re: HSC 2016 3U Marathon Those are correct.

Re: HSC 2016 4U Marathon You can analyse it separately in the intervals x < 0, 0 ≤ x ≤ 2, x > 2; this lets us get rid of absolute value signs in each interval.

I'm guessing there's not much chance he'd get in trouble.