Maths Ext 2 Predictions/Thoughts (3 Viewers)
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ExtremelyBoredUser
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I just thought abt centripetal force, which would have to be to the centre, and acceleration would have to be in the same direction since its moving in a circular motion.
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part i you can just do |xi +yj + zk|=1<=|xi|+|yj|+|zk| = |x|+|y|+|z| since you can multiply lengths of the unit vectors i j and k which are 1
Paradoxica
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Suppose T(2n) = H(m) for some integer m.
2n(2n+1)/2 = m(2m-1)
2n²+n = 2m²-m
m+n = 2m²-2n² = 2(m-n)(m+n)
Since m and n are positive integers, we can divide.
1 = 2(m-n)
This is impossible, as integers cannot have a difference of ½.
Therefore, the even triangular numbers are never equal to any hexagonal numbers.
85+
edit: probably close to 90 or low
Paradoxica
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Here is the region for which Re(z) ≥ Arg(z)
DrugLord69
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Nah mate i reckon i'm gonna ace tf out the test
u contributed to the god-tier scaling, only thing u aced
Paradoxica
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Meme Solution for 15. (d).
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
i did that in the examMeme Solution for 15. (d).
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
I posted a version of that in the discussion in the MX2 forum... I wonder how markers will evaluate anything like it?Meme Solution for 15. (d).
If the result were true, this would be a non-trivial solution to Fermat's Last Theorem for every integer n>2.
Hence, the result must be false for all n>2.
The result is clearly false for n=2 by manually checking.
QED
did you figure this one out? i also got lost with this lol and ran outta time to look at it properly
i think a bit too much is sketched in there
Paradoxica
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I can't say. I know that for proving the irrationality of the nth root of 2, the proof of fermat's last theorem does in fact require some elementary irrationality results (including this very thing) so it would be circular reasoning. This would need the consultation of an expert in algebraic number theory to resolve.
Paradoxica
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specifying the branches of the function is a pain so i'm leaving that as an exercise to the reader
JesusChriiii
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what you guys reckon a raw of 70-75 will align to?
Alistruggles
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i'm not the person u quoted and idk if this would be accepted but i noted that they want u to prove for every odd number and an odd number can be represented by 2n - 1 so....subbing 2n - 1 into the Triangle equation actually gets u the hexagonal equation, therefore hexagonal number??? seems way too simple lol (lmk if you think this would get any marks)