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Im also not sure why you are multiplying by two in the final answer? And where does the \log\left(\frac{3}{2}\right) go?

remember it's \log\left(\frac{1+\sqrt{5}}{2}\right) not \frac{\log(1+\sqrt{5})}{2}

First

Try rechecking your last two lines, can you explain your thought process?

He used the phrase 'goes into' implying its a whole new world.

lol I gave him that yesterday, thats why he created this thread in the first place.

A1La5 posted two other questions?

Over the real numbers ofcourse, more generally x=2,1,-\frac{1}{2}\pm\frac{\sqrt{3}i}{2},-1\pm\sqrt{3}i

Half the total ways are when S is to the left of H and half are when S is to the right of A. So simply work out the total number of ways of arranging the letters in general then divide by 2 (what Bianca did).

Literally subsititute in the point so: 2\sqrt{3}=2sin\left(2 \times \frac{\pi}{6}\right)+C and solve for C

Hey man don't let that thread get to you, you are clearly capable but I think being humble goes a long way!

Breaking bad

Give an example.

Whilst I agree there is a lot of variation most of perms and combs is formulaic just like the rest of the syllabus (by most I mean like 80% whereas 20% is unknown or has a trick). You will find a lot of questions use the same trick whereas a few require you to try something new.

Yes lol, it's based of ppl submitting their raw marks they gained from nesa.

https://rawmarks.info/ check this site for prev years (Note this is for your HSC exam, there is no set mark for internals, as internals are based of ranks).

Now our second way of solving the equation x+2y+4z=10 is to rewrite it as x+2y=10-4z. y can vary such that 0 \leq y \leq5-2z and the RHS must obv be non-negative and an integer so 0 \leq z \leq 2. Now we could use the 'stars and bars method,' but Ill go for a more intuitive approach. Let...

Ahh I missed (2,2,1)

The problem can be rewritten as: 5x+10y+20z=50 or x+2y+4z=10 where x= no of 5c coins, y=no of 10c coins z=no of 20c coins. From here we could just list out the possibilities of of the triplet (x,y,z) given that x,y,z are non-negative integers. So: (10,0,0) (8,1,0) (6,2,0) (6,0,1) (4,3,0) (4,1,1)...

No there is still an interview just no ucat.