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MATHMATICS

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elevator was in b ,
3 min and 20 sec,
can you tell me the right answer for the matrix
 
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Giraffe195 said:
The max speed question, I think you have to find the height of the triangle. You find the distance which is 130 and then you do .5 * 16(the time) * h = 130.
The midpoint question you have to do the two X coordinates and divide by two to get the X and then add the two y coordinates and divide by 2 to get the Y coordinates. I think the answer was (4,2) or something like that.
No Locus questions came.
I believe the answers for the symmetry was one line of symmetry and 2 for the rotational. But don't take my word for it. I don't remember exactly.



no rational of symmetry was 1 because the star will not fit in the 2
 
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the answer for the maximum speed was 16.25! i was so nervious that i didnt knew how to do the question about the two circles and the one of the fences that was to use limit (32x1.995) now i hope that paper 4 will be easier!
 
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That was one hard exam. I'm usually good at maths and I thought I'd do really well but ..... oh well. Just everyone keep your hopes up and do well il paper 4 :) I'm gonna atudy my ass off for paper 4, I don't even care about paper 6 chem. and bio :p
 
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ZaidOwns said:
Please theres only 3 hours till exam all tell me, what are the most common questions to come?
wat are the answers for the venn diagrams?
and themaximum speed?
icant stop thinking about it
 
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mariomaged said:
Giraffe195 said:
The max speed question, I think you have to find the height of the triangle. You find the distance which is 130 and then you do .5 * 16(the time) * h = 130.
The midpoint question you have to do the two X coordinates and divide by two to get the X and then add the two y coordinates and divide by 2 to get the Y coordinates. I think the answer was (4,2) or something like that.
No Locus questions came.
I believe the answers for the symmetry was one line of symmetry and 2 for the rotational. But don't take my word for it. I don't remember exactly.



no rational of symmetry was 1 because the star will not fit in the 2
OMG I GOT THE ANSWER 17.5!!!!!
cu i calculated the distance rong 140!!
would i get marks for the working out???!!!
how many mrks was it anyway???
 
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kalazooni said:
mariomaged said:
Giraffe195 said:
The max speed question, I think you have to find the height of the triangle. You find the distance which is 130 and then you do .5 * 16(the time) * h = 130.
The midpoint question you have to do the two X coordinates and divide by two to get the X and then add the two y coordinates and divide by 2 to get the Y coordinates. I think the answer was (4,2) or something like that.
No Locus questions came.
I believe the answers for the symmetry was one line of symmetry and 2 for the rotational. But don't take my word for it. I don't remember exactly.



no rational of symmetry was 1 because the star will not fit in the 2
OMG I GOT THE ANSWER 17.5!!!!!
cu i calculated the distance rong 140!!
would i get marks for the working out???!!!
how many mrks was it anyway???

It was five marks. If it's a ecf then you may get 4 marks.
 
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many friends of mine are very upset and they hope that paper 4 will be a nicer one..beacuse i think that this paper had been a nasty one and even my math teacher was surprised.
 
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THAT WAS ONE AWSOME EXAM, it was very hard, i did do from 1998 to 2009, this was hardest, but think about it now the curve is gonna get low, i lost from 6 to 10 marks, anyway, thats just the beginning if math and even paper 6 physics were hard, how about biology and chemistry =D
 
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Seeing as u did so well what was thr answet for the one about tht guys speed

The the hw lng ws pt...please explain how u got it

Also sinx - cosx = 0.5

And did u do the changing of subject of fornular
 
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the speed of guy i got 10 i asked my teacher he said true

PT was hardest i spent about 30 mins looking at it, i was like =0

sinx - cox = -.5 was 66

yes i did change the subject, it is X = 2/p
 
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the question about the max speed was 16.25 not 10..! and the one about putting x as the subject was 3/p-1!!!!
 
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max speed answer:
the total time is 16
distance = 130m
take time till max speed v to be t
thus time after max speed = 16-t
eq: (0.5*v*t) + [0.5*v*(16-t)]=130, using area of a graph is distance
taking l.c.m 2 for the 2 fractions, (vt)+[v(16-t)] /2 =130
taking v common in the eq., v(t+16-t) /2 = 130
thus v*16/2 = 130
v*8 = 130
v = 130/8= 16.25

hope this helped :)
 
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