awesome245
New Member
- Joined
- Oct 17, 2023
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- HSC
- 2023
Oh I thought it was a maximum of two students total, not any number of pairs of two students, thanks
To seat the students as one pair and three singles means many more possible arrangements, as it needs to be SS-X-S-X-S-X-S-X-, where X is not a student.Oh I thought it was a maximum of two students total, not any number of pairs of two students, thanks
lmao yeah I think it was badly wordedTo seat the students as one pair and three singles means many more possible arrangements, as it needs to be SS-X-S-X-S-X-S-X-, where X is not a student.
Available to fill the four non-student positions are three teachers, so at least one of the teachers must be sub-divided. Now, a single teacher could be halved but halving vertically (into a left and right half) would be different from halving horizontally into a top and bottom half, two consider just two directions. Then the spacing objects need not be equally sized, so we could use just a fraction of a teacher (say, a leg) for the fourth position... but is a left leg identical to a right leg, or are these yet more possibilities needing to be accounted for? And, do we have to triple the options as deconstructing one teacher is different from another? Perhaps one of the students will volunteer a school bag as a separator and spare the teachers from subdivision... but then, if that student is adjacent to their own bag, does it return to being a single unit, and so the bag only acts as a separator when not adjacent to its owner?
The possibilities under this interpretation are endless... I think the wording needs to be considered as restricting student groupings to a maximum size of two without limiting the number of such groupings.
this question also relates to cross product too. you can only do cross product in R3 but R3 contains R2 so just do a = (a1,a2,0), b=(b1,b2,0) --> Area of parallelogram spanned by a and b = |a x b| = |a1b2 - a2b1| --> triangle area is 1/2 thatView attachment 40843
Ever heard of the shoelace formula?
Some people viewing this would instantly recognise a1b2-a2b1 as being a determinant.
The result in this paper can be generalised to find areas of polygons, not just triangles:
https://en.wikipedia.org/wiki/Shoelace_formula
imo 2023do you guys think 2022 or 2023 was harder? I'm praying on this year's scaling. what do you think is the e4 raw mar cut off?
just checked solutions, should hopefully be 66/70 unless I wrote something wrong and forgot.carrotsss how did u go? 70/70?????/
woooooo noice!!!!just checked solutions, should hopefully be 67/70 unless I wrote something wrong and forgot.
nah he just smartThat must’ve taken some commitment. Nice.
lmao dont snitch on urself then xDnah he just smart
@TankKuno i managed to lose a mark on that projectile collision question due to silly error![]()
mr mai is gonna be so mad bruh we did like 5000 of those questions in class
imo i think they are on par but i failed both exams soo i cant say much more xD2023 was a little easier than 2022 sorry to upset everyone
your signature is so realimo i think they are on par but i failed both exams soo i cant say much more xD
imo i disagree2023 was a little easier than 2022 sorry to upset everyone
There’s supposed to twoHow is it possible for only two students to sit next to each other, as there is no combination that works
YOU GOT 14CII?!!?just checked solutions, should hopefully be 66/70 unless I wrote something wrong and forgot.
nvm 66 just found a mistake
lets just hope the e4 cut off is 20/70 to save all of usI would have (just slightly) preferred 2022, because
I do agree. In 2022 there were quite a few more fragmented questions that you could scoop up marks on, and with enough panning for gold on the harder questions, you could crank out some marks. With q-14 on this year, if you didn’t have an initial suspicion of where to go it became much harder to pick up what was going on, which is not good when it is in 2-3 parts.
I can not be certain though, that’s just how I see it.