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We can see from the 1957 leaving certificate paper that We can also see from the 2014 HSC Extension 2 exam that These can be combined to produce a new formula for π in terms of binomial coefficients which was discovered in 2007 by J.C. Toloza: as follows:

And in case you missed it, the grim reaper came to the coronation too. At least he was on time. Some think it was Meghan Markle in disguise. Here is a youtube of it: https://www.youtube.com/shorts/aT8XiSEMU8c

And the song sung at the procession was "I was glad" - but maybe it should have been "I was glad but the king was not". The archbishop later said "Helleluia Christ is risen" but maybe he should have said "Halleluia Prince William is finally here".

If you watched the coronation on TV you may not have noticed that Prince William and his family were delayed at the beginning forcing the king and queen to be stranded waiting in their carriage for several minutes before proceeding to enter the abbey. The king was visibly upset: Prince George...

I think it is going to be hard to penalise because some extension 2 texts use cross products despite it not being in the syllabus. They treat it as extension exercises. They don’t really go into it in much detail. Some IB texts have better and more thorough exercises on vectors including cross...

Yeah it’s much shorter that way like \lambda(\vec{i}-\vec{j}+2\vec{k})\times(2\vec{i}+\vec{j}-3\vec{k})=\lambda \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\ 1&-1&2\\ 2&1&-3\end{vmatrix}=\lambda(\vec{i}+7\vec{j}+3\vec{k})\text{ for nonzero real numbers }\lambda but unfortunately that is not in the...

A reference or proof should be given if it is not well known. But for well known theorems it is not necessary. For example one does not need to prove Pythagoras' theorem or give a reference to a proof every time one uses it. One only needs to say "by Pythagoras' theorem, ...." There are 2...

It is a well known theorem. Tchebychev proved it in 1850. Then in 1919 Ramanujan made a shorter proof. Then in 1932 Erdős made a more elementary proof which is the one most commonly used thesedays, for example at https://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate Here is a youtube...

It follows from Bertrand’s Postulate: For all integers n>1 there exists a prime p such that n<p<2n. For n≥3, 2n≤n! and therefore n<p<n!. In any case there are some proofs here: https://math.stackexchange.com/questions/483838/for-all-n2-there-exists-p-prime-npn

Yes you are right. Solution you put there is partly incorrect. It should say z^3=i=e^\frac{\pi i}{2}\therefore z=e^\frac{\pi i}{6} is a solution etc. So correct answer is A.

So this is year 12 Cambridge Advanced 6B Q1u, 4d, 5d, 12a It does seem that the answer in the textbook for this one is wrong. Textbook answer is 4\cos\frac{x}{4} but it should be 3\cos\frac{x}{4}

3 ways. 1. Check with calculator \sin\frac{49\pi}{30}-\sin\frac{41\pi}{30}=0 and \sin\frac{53\pi}{30}-\sin\frac{37\pi}{30}=0 2. Use sum to product...

There have to be only 5 distinct answers. sin(49pi/30) and sin(53pi/30) are included because \sin\frac{49\pi}{30}=\sin\frac{41\pi}{30} and \sin\frac{53}{30}=\sin\frac{37\pi}{30} So it matters not if you list them as...

1 One might think it is undefined and if you try to do it in wolframalpha it can't do it. But if you want properties of matrices to be conserved it will be 1. For example if P=(A,0;0,Q), then |P|=|A||Q| and if A is empty then |P|=|Q| and so |A|=1.

I made a picture if it helps.

Just let \textstyle\frac{k}{n}=x and \textstyle\frac{1}{n}=\delta x As k goes from 1 to n, x goes from \textstyle\frac{1}{n} to \textstyle\frac{n}{n} which as n\rightarrow\infty means x goes from 0 to 1. And of course in the limit \delta x becomes dx

Do it as an integral...