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$Take a regular n-sided polygon, each side of which subtends equal angles of $ \theta $ to the centre, with length of r from centre to the vertices. (As shown in the diagram$ i) \ \ $Explain why $ \theta = \frac{2\pi}{n} $ and hence prove that the area of the polygon A is given by...

SOLUTION Very well written question, while people to Spiral's one Ill think of my own

I dont even know why this thread is still open, people are just following onto a good troll by barbenator but for the sake of dignity could everyone please stop?

I dont think the modulus can be i though, modulus is the the distance of z from the origin (polar form) Hence it cant be i lol Answer is wrong (and since you and I got same answers I will assume we are correct)

This is the 2U section, from what I can google: A relationship between two variables in which both variables move in tandem. This means that if one variable increases so does the other, if one variable decreases so does the other. Now applying this knowledge to the question, with a bit of...

Is the answer 1? (If so Ill post solution)

For 2U it has happened loads of times, for 3U to a lesser extent however it has occured multiple times, for 4U only 1 person so far in 1993 managed to get 100% RAW in 4U (beast)

2 days apart iirc, Monday ext 2, wednesday ext 1?

Ok, so basically since this is 2U and we are going to assume we know nothing of inverse functions, we are going to split the co-ordinate plane into four sections. One section is a rectangle formed by 4, and ln 4 (as seen in my diagram) The other section is the little rectange formed by 2 and ln...

So from c i We find the radius to be: l=r \theta \\ 15=r \times \frac{\pi}{5} \\ r=\frac{75}{\pi} Now since pi/5 is less than pi, the sector formed is a minor sector, so to find the major sector, we can either find the reflex angle at the centre and use the area of sector formula, or we can...

An alternate method is to understand that what we can grasp from this inequality is that the MAXIMUM y-value that y can take is 3a. So lets use our expression from part iii: y=a(1-2\cos\theta) \\ \\ y'=2a\sin \theta \ \ \theta=0^{\circ}, 180^{\circ} \ \ \ ($For all theta in the possible...

Because a minority of the state gets above around there, it might sound hard to believe but a lot of schools give easy exams and give a wrong impression on the real difficulty of the HSC. (Although you may just be smart and be acing it) Of course the alignment in terms of raw starts to hit a...

$Let the expression $ 8^n-1=7p \ \ \ p-integer $Step 1: $n=1 \\ 8^1-1=7=7p \ \ \rightarrow p=1 \ \therefore p-integer \\ \\ \therefore $ It is divisible at n=1$ $Step 2: $n=k \\ 8^k-1=7p \ \ p-integer \ \ \ $Assume true for n=k$ $Step 3: $n=k+1 \\ \\ 8^{k+1}-1=7q \ \...

Another quick question (again) Is it possible to prove something for ALL integers with mathematical induction if we somehow prove for the initial statement, that the limit to negative infinity of both sides is equal. (or is this mathematically incorrect) And if so, then it is possible that if...

I look forward to this next year as I will make it to the best of my efforts to come (if it is on next year that is)

My teacher told me it was one of the hardest English papers he has seen

The fact that it is a meme makes it less legit that he said it, good on him for fooling the stupid reporters though.

So, its best to write our series in expanded form in order to get a good gist of it at first (later on you will become stronger with series and hence be able to spare yourself writing it out expanded form) \sum=2^{0.5}+2^{1}+2^{1.5}+...+2^{8.5}+ 2^{9} So its a Geometric Progression with a...

Yes but we dont care about how many fish were in the damn before 50 were added, since we are asked to estimate 'fish in the dam just prior to the sample of thirty being removed.' Note the word prior to thirty removed, so this was when there were indeed 50 tagged fish in the damn. So our ratio...

This question isnt a Geometric Series: \sum_{n=1}^{18}n^{0.5n}=1^{0.5}+2^1+3^{1.5}+4^2+...+17^{8.5}+18^9 And the only way I can see doing this question is evaluating all the whole numbers seperately then all the surds then simplifying all them together to get the answer (very very tedious...