Cssa 2004 (1 Viewer)

[withoutaface]

~50% or possibly less I should think for CSSA.

 

[Enigmatic Eve]

Wow!! Are you serious?!!
Ooh.. then hopefully I can get above the state average!!
Very slight chance though
Thanks heaps!!

 

[withoutaface]

we didn't get ANY amendments at all!!
and we should've gotten at least half of the mechanics stuff taken out...

Everyone will be in the same boat, so I wouldn't be too concerned, just consider those marks a write off.

 

[withoutaface]

I have included the alternative question 8b from my paper, excuse my paint skills

 

[Rorix]

hrmhrm that volumes q is pretty easy

 

[coca cola]

Yeh.. I managed to figure that question out... did u get the values of a to be +-sqrt of c?

can it be + root c? i think its only - root c because if its positive then there is no error. right?

 

[coca cola]

hrmhrm that volumes q is pretty easy

but the indcuction question and circle geometry question for Q8 in that trial is even easier. that is by far the shortest question the whole exam.

 

[tempco]

But can't errors be both +ve or -ve depending on the gradient of the function at the point of estimation?

 

[coca cola]

umm i guess, yea i think your right.

i think i put down both + or - root c in the exam, but when i came back i thought it couldn't be zero error, but now i guess it could be that the error is infact zero if first and second estimation is the same.

i guess i got confused by the word "error".

edit: ahh crap, damn this, just did the Q again, seems i messed up the absolute value in the exam.

the answer should be - root c and (root c)/3.

oh well, another mark gone to mistake.

 

[coca cola]

hey how do you guys do the last part of Q3 complex numbers?

i couldn't figure out a gemetric way of finding the area so i integrated one part of it and times it by 4. is there any geometrical way that can be used?

 

[boneyfish]

LOL, all these people freaking out about the paper... our school is apart of the CEO, but we don't do the CSSA ext2 math paper... rather, our teacher makes his own. Does anyone have copy, cause i'd love to see it (i can't access the link on this thread!)

Hope you all go well...!

 

[UrDsteR]

Hmmz the CSSA paper

I honestly think it wasn't as hard as previous years. But I still am aiming for a pass. I reckon I left 30% out... but most of my answers were quite good and not really THAT bullshitty... Lol the volumes question.. i think i made up some equations just to make it show... The conics was good, same as graphing and complex numbers... Really, there wasn't any questions that were dead hard but it wasn't easy too. A lot of repeats from previous years. Omg my friend got that induction... lol i cheated by doing the first 5 lines and the last 5 lines... it kinda looked like... and somehow... this equals this!

Aiming for 50%
Aiming for 80 UAI

 

[Jase]

hey how do you guys do the last part of Q3 complex numbers?

i couldn't figure out a gemetric way of finding the area so i integrated one part of it and times it by 4. is there any geometrical way that can be used?

hmm, well i think if you can find the angle that the chord subtends at the centre of the circle, then you can find the area of each segment. -> A = 1/2 pi r^2 (@ - sin@)
add the two segments and you get the enclosed area.

 

[withoutaface]

umm i guess, yea i think your right.

i think i put down both + or - root c in the exam, but when i came back i thought it couldn't be zero error, but now i guess it could be that the error is infact zero if first and second estimation is the same.

i guess i got confused by the word "error".

edit: ahh crap, damn this, just did the Q again, seems i messed up the absolute value in the exam.

the answer should be - root c and (root c)/3.

oh well, another mark gone to mistake.

Woohooo thats what I got!

 

[McLake]

... speak to me lazarus. speak to me. your silence is killing me. killing me ...

For the moment hold back on posting it. Thank you.

 

[CrashOveride]

hmm, well i think if you can find the angle that the chord subtends at the centre of the circle, then you can find the area of each segment. -> A = 1/2 pi r^2 (@ - sin@)
add the two segments and you get the enclosed area.

You get two equilateral triangles and @ = pi/3. Yeah an almost identical problem was posted in 3unit maths forum....before the exam i would think....hmmmm

*looks at withoutaface* =p

 

[Crikket]

What?
how is that possible?

 

[Crikket]

oopsies, that last message was to tywebb having the paper a week before

 

[velox]

Tywebb, sorry if this sounds a little rude. But i saw your name as the author of some article...and was wondering who you were? Pardon me if its really obvious..

 

[McLake]

i think boredofstudies could make their requirements a bit clearer in future so that such difficulties do not arise again.

Yes, we should have. Unfortunatly we had not decided what our policy should be before the first batch of exams were held. In the future we will be clearer on our intent.