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yeah literally i looked at it in reading time and did a double take … thought what the heck did he know about it… so glad i did it thank you so much 😆

you can write “vector” CB = e^(ipi/3)CA since it’s equilateral i’m just going to use zA=a, zB=b, zC=c i.e. b-c=(a-c) e^(ipi/3) so b=c(1-e^(ipi/3))+ae^(ipi/3) we know that: b=xb+5i c=xc-5i subbing c and a into above the expression for b, you get an expression for b in terms of xc, and you...

I spent like 20 minutes on that and I finally got it after just staring for so long

yeah it was like a really abstract test that’s a good way of putting it but for the most part just going through with calculations got to the required result so I think it’s ok- predicting a very low E4 cutoff this year though so don’t be too disappointed

that was kinda funny lmfao they wrote some nice creative writings

good luck everyone!! I better see at least one sr from this thread

probably grinding.. I don’t remember where but I saw somewhere that cognitive performance decreases as time you’ve been awake increases (including lower hours… ie waking up closer to exam is better maybe?) getting to see people that do adv having to wake up early is completely worth the pain of...

yeah same- just going to do a few past papers and lazy revision, nothing that will drastically change marks at this point just gotta stay in the groove 😬

lmao i have no idea how you got to that but not the suggested thing… then again i’ve done some stupidly complicated stuff in exams that isn’t required so I can’t really talk 😂

how did you do it with trapeziums though, I’m curious?

yeah that suggested does look like it makes the most sense; I don’t know if I would have thought of it in an exam, my way just seemed the most obvious to me since the concavity implies it increases at a decreasing rate… but suggested sltn is probably more rigorous than my method— although, it...

exponential form gang :) and yeah- I had a go at the q just then and after graphing it becomes clear that |cbrta-cbrt(a-b)| will have a greater magnitude than |cbrt(a)-cbrt(a+b)| due to the concavity? not sure if that’s the right path, but it seems to work; |cbrta-cbrt(a-b)|>|cbrta-cbrt(a+b)|...

whoa, very nice- very similar sum as well -time do to all the 200X HSC papers hahah

yeah - it's not too bad but under exam conditions the notation looks a little scary I guess, especially in like q14-15 (I forget) the other two that threw people off in ours was part (ii) of this (part i is free) and also the later parts of this

lmao literally... i'll just like draw them a nice triangle and leave the question

another inequality q from our trial

yeah same with all the ones I see- considering that 'hard' questions typically like to combine syllabus points ... but no, it's literally just 'hey you remember pythagoreas... yeah, now draw it'

the geometric way is essentially a more convoluted version of the easy algebraic way, you just get to draw a nice triangle instead of immediate substitution lmao edit: in fact, its exactly the same working out as the algebraic way... just imagine the teachers decided to include an extra mark to...

what does \sqrt{y^2+z^2} remind you of

hence why I asked for geometric haha