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Must the angles be expressed only in terms of s and t ? The best I could do is s' = \arccos\left( \cos s + \frac{k^2-RX^2}{2\cdot PX \cdot QX}\right), t' = \arccos\left( \cos t + \frac{k^2-PX^2}{2\cdot QX \cdot RX}\right), where k is the side length of the equilateral triangle and...

HSC SDD has very little in common with the introductory programming courses. The latter focuses solely on being able to code properly. There's nothing about ethics, dev approaches or that sort of stuff. 1911 is a cut down version of 1511. 1811 is significantly different to both of those. If...

The first factorisation, (z+1)^8 - z^8 = ((z+1)^4 + z^4)((z+1)^4 - z^4), is straightforward. Realising that you can turn a sum of squares into a difference of squares is not so trivial - that's the part I showed. (actually, this method generalises to any positive even power...)

This is not ideal but one method is to use the difference of squares identity to continuously factorise the LHS of the equation into a product of seven polynomial factors of degree one. You can rewrite a sum of squares as a difference of squares by using i^2 = -1 . e.g. (z+1)^4+z^4 =...

Let x = \sqrt{a+\sqrt b} - \sqrt{a-\sqrt b}. Now, \begin{aligned} x^2 &= (a + \sqrt b) - 2\sqrt{ (a + \sqrt b) (a - \sqrt b)} + (a - \sqrt b) \\ &= 2(a - \sqrt{a^2-b}).\end{aligned} Clearly x \ge 0 , so x = \sqrt{2(a - \sqrt{a^2-b})}. Set a = 11, \, b = 2 and you're done.

I'm not familiar with the new syllabus but I can tell you how it was in the older syllabus. In the older syllabus, circle geometry problems were pretty limited in scope - some diagram involving circles is given and you have to bash out those circle geo-specific theorems you memorised to prove a...

How about something a little bit different? \text{Let } I = \int_0^{\pi/2} e^{\sin x}\, \mathrm dx. Without numerically evaluating I, show that \text{a) } \frac{\pi}2 + 1 \leq I \leq \frac{\pi e}2, \text{b) } \mathrm{(\bold {Harder}.) \,\,} e \leq I \leq \frac{\pi e}2 - \frac{5\pi}8 +1.

Choose the one car that gets the manually operated gate, then choose which gate. \binom{3}{1}\cdot \binom 21 = 6. The remaining two cars must go through the auto gates (of which there are three). \binom 31 \cdot \binom 31 = 9. Finally, 6 \cdot 9 = 54 . A gate can be used by more than one...

Trimesters are fine. Especially if you haven't experienced semesters.

COMP2521 - Data Structures and Algorithms Ease: 8/10. (kind of). Conceptually some of the content gets very tricky to properly understand, but in terms of assessment you won't actually need to be that familiar with them. Midsem is really really easy. Assignments are fairly large. Content: 7/10...

T3 COMP2521 Data Structures and Algorithms...98 HD T3 MATH1081 Discrete Mathematics.............90 HD T3 MATH2701 Algebra and Analysis.............89 HD pretty happy with this

IMO the problem was less about the trimester system and moreso how the transitioning was handled by the staff/course convenors etc. For the most part it was fine to me, but you could definitely tell the course convenors were experimenting with the change and not all of these experiments worked...

MATH1081 ~92 COMP2521 ~92 MATH2701 ~70 😬

Integration is hard - unlike differentiation it's not a mechanical process: there isn't a set sequence of steps you can take that always gives you the answer. In fact, most functions don't even have antiderivatives. The best way to get better is to just simply practice. You'll develop an...

For the proof to be mathematically valid, you need to show that these steps actually are reversible. Then the proof is correct. But I don't know how an HSC marker would react to this - this sort of logic wasn't touched on at all in the old HSC. But it is covered in first year uni math so I...

Whether or not you can work backwards depends on if the steps you take are invertible/reversible, or in other words, the implications you use are of the "if and only if" type. For example, \begin{aligned} x+ 5 &= 0\\ \iff x &= -5 \end{aligned} is an example of such an invertible operation...

One way is by similar triangles. We suppose that A,B,C,O are distinct points (otherwise the question isn't very well defined). In particular, z \ne 0, 1. Then, \frac{OC}{OA} = \frac{|1|}{|z|}, \frac{BC}{CA} = \frac{|1-1/z|}{|z-1|} =\frac{|1/z||z-1|}{|z-1|}, \frac{OB}{OC} =...

It's not negligible at all - it's the biggest defect in the "inverse" trig functions and it can completely throw off your calculations if you're not careful. If instead we had z_2 = -1 then this method would've given a very wrong answer. Suppose you're developing a navigation software with...

The argument of a complex number satisfies \tan(\arg z) = \frac{\text{Im } z}{\text{Re } z}. This is not the same as \arg z = \tan^{-1}\left(\frac{\text{Im } z}{\text{Re } z}\right) because \tan^{-1} is not an inverse function of \tan . With z = \exp({3\pi}/4 \cdot i) it's clear that...

I use it to take notes for uni, but I'm not aware of any handwriting-to-text conversion that it has.