The 2004 HSC - Mathematics Paper (1 Viewer)

[Willmaker]

srsly 110/120.. is that 95?

um....I thought you did maths?

 

[coroneos]

lol.. srsly.. u prolly need 97% to get 95 scaled?

 

[slammindas]

ey wat did every1 get for the shaded area of question 8 b

 

[coroneos]

a weirdo number i fuked it.

 

[Managore]

slamm: 45(triangle)-11(part found out before)=34units^2

 

[smallcattle]

ey wat did every1 get for the shaded area of question 8 b

I think i got 27

45-18

dont know if it is right or not

atleast its a nice answer

 

[amy_lou]

argh!! i completely stuffed the simpson's rule question up!! what did everyone else do this part??i went in there confident and came out that disappointed - there goes my prospects of getting a UAI of 85 or above!!!

 

[Willmaker]

ey wat did every1 get for the shaded area of question 8 b

I worked out the length of AB to be the square root of 153. I then worked out perpendicular distance from C to AB to be 30/(root 17). Times the two together and half then take away watever you got in part iv.

(root 153) x 30/(root 17) x 1/2 - 18 = 27

 

[Candypants]

Fair albeit challenging up until question 9 and then it was just crap crap crap.

 

[fr0ggy]

I think i did too. i got something very close but not exact, i think something like 1291/1296

yeah. that's what i got

 

[Sarzzabelle]

I thought it was so totally awesome. I loved it!

 

[jiegu]

yeah i gor 45-18 =27 too!
i use 1/2absinc - the part 4 answer.

 

[amime]

I worked out the length of AB to be the square root of 153. I then worked out perpendicular distance from C to AB to be 30/(root 17). Times the two together and half then take away watever you got in part iv.

(root 153) x 30/(root 17) x 1/2 - 18 = 27

yupz that's wat i got!

 

[acmilan]

yeah. that's what i got

Finally someone agrees with what i got, it think there have been about three different answers so far for this question

I worked out the length of AB to be the square root of 153. I then worked out perpendicular distance from C to AB to be 30/(root 17). Times the two together and half then take away watever you got in part iv.

(root 153) x 30/(root 17) x 1/2 - 18 = 27

I got that too

 

[Bi0mic]

i totalled myself in that area one :|, completely skipped the fact there was a triange and went for 2 integrals ...

the pressure of a test clouds my mind

 

[Managore]

18??? How'd you get that as the area of the sector???

 

[becany]

yeh i got 27 as the shaded area..
as for 9b...isnt part i when t = 2?
and part ii...particle furthest from origin means that x is a stat pt?
that means v = 0?
and since we were given that initially it is at rest at the origin (ie t = 0, v = 0)...i put time is between 0 and 1...but now that i think about it..im not too sure anymore

 

[Jennibeans]

What I did was find the probability of you getting higher that 45 which ended up to be:

(1/6 x 1/6) x (1/6 x 1/6) x 2 - 5x5 then 5x5 or 5x5 then 5x4 is the only way you could get higher than 45 so I then went 1-(that answer) which ended up to be 647/648. Of course I could be wrong but thats the way I did it.

OMG I got that answer too...woo hoo actually got something right...maybe
I did it 1 - P(>45) which was 1/648 so the answer I got was 647/648

 

[becany]

oh one more question...what did ppl get for 6ii?
the ratio of MY:AC?
i got like...1/root3 : 2

 

[Bi0mic]

aww i got 1/4 :s