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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread We can immediately say the answer to the very last part is just the initial separation between the bulldozers (in this case 36 m) doing essentially no calculation. Can you see why? Also by symmetry, for c) (i), we conclude with...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Note that the fifth year is $\color{blue}{4}$ years after the first year. Let the common ratio be $r$, then since the fifth year's sales are \color{red}{half} \color{black}{that} of the first year, we must have...

$\noindent Recall that the slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$, because the product of slopes of perpendicular lines is $-1$.$

Re: Extracurricular Integration Marathon What techniques of integration do they learn in IB? Do they learn Complex Analysis methods? I'm guessing they'd do double integrals?

Re: Extracurricular Integration Marathon So is this integral from an IB maths textbook?

Re: Extracurricular Integration Marathon Are you absolutely (lol) sure about this?

Also Biology is generally not very mathematical.

USyd is planning on making 2U Maths a prerequisite for several degrees.

You have to keep in mind that you start counting at 0. That's why only memorising the general formulas will probably get you into trouble. Imo it's better to just know the results for a single parabola/interval, and then if multiple ones are used, you should be able to see you're just adding up...

It's one of the classic HSC Physics tricks. :)

Lol, is it because the ring has a gap (isn't a full closed circle)?

It's the place where on either side when you increment x by some amount, the y-values are the same. For example for the first one the y-values are seen to be symmetrical about the x = 0 column, which implies the axis of symmetry is x = 0. (In fact, just having one pair of symmetric values at...

$\noindent For instance for the third one (part (c)), the $y$-values are symmetrical about $x=-1$, and here $y=0$, so the quadratic is $y = A\left(x+1\right)^2 + 0$, i.e. $y= A\left(x+1\right)^2$, for some constant $A$. Since $y=1$ at $x=0$, we have $1 = A\times 1^2 \Rightarrow A = 1$. So the...

$\noindent One way is to let the quadratic be $y=ax^2 + bx + c$, sub. in three given $(x,y)$ pairs, and solve the resulting three simultaneous equations in the three unknowns $a,b,c$. If we know it's a quadratic like this, we only need three points to do it.$ $\noindent However, in each of...

Re: HSC 2016 3U Marathon Yeah, there are congruency tests for higher-sided polygons, it's just that they're going to be more complicated (or less practical to apply) than the standard triangle ones (like for quadrilaterals, the Cambridge Year 10 book is reducing it to the case of triangles...

Re: HSC 2016 3U Marathon No, congruency for shapes with more than 3 sides becomes more complicated. Like a simple example, a square has the same side lengths as a rhombus formed with equal side lengths, but these shapes are not congruent, despite an 'SSSS'. (Similar things occur for shapes...

Has your school released any data on what ATARs its students received in recent years or median/top ATAR etc.? If so, you might be able to gauge roughly what sort of ranks you'd need based on that.

Re: International Baccalaureate Marathon 2016 Pretty much. And the area of a circle of radius 0 is 0, which makes C = 0. Not sure how rigorous you are required to be though with regards to explaining the integral. Here's a paper discussing when the perimeter of a shape will be the derivative...

You should probs just do heaps of practice Q's where they make you do them with multiple sub-intervals, and you'll probably end up just being able to do them without really needing to memorise a formula (like the process would just stick to your head). :) The trouble with memorising formulas...

Re: International Baccalaureate Marathon 2016 Yeah that's essentially one way to do it. The typical HSC method would be to do something like integrate sqrt(r^2 - x^2) between 0 and r (this getting the area of the quarter-disk), and then multiplying by 4. The integration would be carried out...