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Well I have the emulator now and it seems there are really 2 questions here. Firstly as I said before this is the go back one screen button: But there is also an Undo. It's not a button, but a menu item in the tools button: which gives this: So suppose you choose to input 9-5 but...

If you check the photo on the officeworks page you will see that it is the AU.

As I said before it probably was going to be released early and it seems that retailers are already selling it now, for example at https://www.officeworks.com.au/shop/officeworks/p/casio-fx8200-scientific-calculator-black-casfx8200 Maybe don't rush out and get it straight away. If you wait a...

\text{Toloza's formula essentially relates } \pi \text{ to one diagonal of Pascal's triangle via a pairwise \text{alternating sum of reciprocals of binomial coefficients.} \text{But Toloza's formula can be generalised to relate }\pi\text{ to every second diagonal of Pascal's triangle.}...

It is based on CASIO's rrp's: It should be clear now. Retailers may have different prices. When the 8200 is released retailers may choose to drop the prices on the 82 and 100 and bump up the price on the 8200 for example. Alternatively simply delay selling the 8200 until their 82's and...

I finally have the answers to these other questions you have: nPr: nCr: factorial: abs: SD: Go back: I got them from the user guides that have recently been made available: https://casioeducation.com.au/wp-content/uploads/2023/05/fx-8200-AU_User-Guide.pdf and...

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}