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our school always moves the classes around so that the 4u class (we only have one!) is in the same class for 2u and 3u with the same teacher. so we worked through most of the 2u hsc in term 4, most of the 3u hsc + complex/graphs in term 1 (there wasn't a 4u half yearly) and then for the last two...

per the rules and procedures (http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/hsc-rules-procedures-09.pdf), 19.iii: "Equipment should bear only the original inscribed information." that said it really depends on what your supervisors are like, they might not be too fussed about it

8. in a centralised wage system, wage rises are essentially forced on all employers. that means that even if their employees haven't become more efficient (=> decreasing costs), they still have to pay them more. this leads to cost push inflation. under a decentralised system the firm is...

the key with psuedocode questions is that you're trying to communicate that you know how to perform the task programatically. it's not hugely important whether your "programming" is using only available language structures and keywords. if you look back through the past hsc papers you'll see...

well i was going to just link it but since you seem to be incapable of a simple google search (it was, in fact, the first result for "hsc writing booklet") and want to blame us for that, i figured i might help you refine your skills. the link is...

according to the latest ABS data the gini for 2007-8 was 0.331 (pg 12). however! they changed their measurements so that value isn't comparable with the earlier values. if you're talking about trends then the value to use for 2007-8 would be 0.317 (pg 64), reflecting a slight increase in...

it appears like this when you put it in your preferences: so i assume that means you choose which second degree you want to do when you enrol - the law ATAR is higher than for any of the other courses that are combined with it (i think), so it's the law cutoff that will decide whether you get...

\frac{x^2}{\sqrt{x^2-a^2}}, \text{ let } x = a\sec\theta \\ = \frac{a^2 \sec^2\theta}{\sqrt{a^2(\sec^2 - 1)}} \\ = a\sec^2\theta \times \left|\cot\theta\right| \\ \text{from here you could go a number of ways...} \\ = a\frac{1}{\cos^2 \theta} \times \left|\frac{\cos\theta}{\sin\theta}\right| \\...

1. (\sec^2 \theta - 1).\tan(90-\theta) \\ = (\tan^2 \theta).\cot\theta \text{, since } 90-\theta \text{ swaps } \sin\theta \text{ and } \cos\theta \\ = \tan\theta 2. 1-\sin^2(180 + \theta) \\ = 1-(\sin(180+\theta))^2 \\ = 1 - (-\sin\theta)^2 \\ = 1 - \sin^2 \theta \\ = \cos^2 \theta 3...

looks like a 1 to me >_>

are you sure it's not I_{n-1}? i checked it with mathematica and n-2 doesn't work, n-1 does. I_n = \int_0^{\pi/4} \tan^{2n} x\,\text{d}x\\ = \int_0^{\pi/4} \tan^{2n-2}x (\sec^2 x - 1)\,\text{d}x \\ = \int_0^{\pi/4} \tan^{2n-2}x .\sec^2 x\,\text{d}x - \int_0^{\pi/4} \tan^{2(n-1)}x\,\text{d}x...

it's because in these motion problems we are integrating with respect to t rather than with respect to x you're right, \int e^{ax+b}\,\text{d}x = \frac{1}{a}e^{ax+b} + c but here we are doing \int e^{at+b}\,\text{d}t and so we treat the t exactly the same way we treated the x in the...

2.12^c just means 2.12 radians, no powers involved or anything so the angle in degrees would be 2.12^c \times \frac{180}{\pi} = 121.47^\circ