[lanky_182]

I never got that then don't see how sin comes into it.

me either

[coca]

I never got that then don't see how sin comes into it.

I used sin to calculate the angle between A and C

[gary]

I don't get that I thought seeing as it said BC was perpendicular then it would be 90-60 for the angle.

[gary]

Share your paper one answers.

[Colin]

I cant remember much from paper1.

3(a) i got F(10,5,0) i think. and (b) AF (2,5,0)

Ithink this is what i got( it may not be though)

[coca]

I don't get that I thought seeing as it said BC was perpendicular then it would be 90-60 for the angle.

Bah. Who knows!

[e-unit]

[right][/right]
[/quote]

Bah. Who knows!
[right][/right]
[/quote]
I do!!!! mwahahaha!!!!!

why did u not just do a.a+a.b+a.c? Or maybe you did and in that case i don't know how sin comes into it. Damn!

[herbeey]

a.b is 0
a.a is a squared.
a.c is essentially a.a, since they are same length.

[e-unit]

a.b is 0
a.a is a squared.
a.c is essentially a.a, since they are same length.

Wrongo! a.a=3.3.cos0=9

a.c=3.3.cos60=3.3.(1/2)=4.5

[coca]

Compare your answers with mine, tell me if I've gone wrong

(Paper 1)
1.

y = 3^1/2 + 2 * 3^1/2

2.

A) Get centres for both circles, P is the midpoint - (0,2)

B) |AB| = 10

3.

A) F = (10, 5, 3)

B) (-2, 5, 3)

[HatingHighers]

Paper 1:

1. y = root3(x+2)

2)a) (0,2)
b) 10

3)a) (10,5,3)
b) -2 5 3 (in component form)

4) (3x-1)^2 + 7
b)i) (1/3, 7)
ii) h greater than or equal to 7

5) 8cosx(1+2sinx)^3

6)a) k = -1/4
b)i) u1 = 3m+5, u2 = 3m^2 + 5m + 5
ii) m = -2

7)a) a = 4, b = 5
b) x greater than 4

8. ----> didnt write answers on sheet, but stationary points were 0 and 7/3. 7/3 was irrelevant to the question so do a nature table for 0 and I found out it was a falling point of inflection. blah blah blah

9) didn't write answer on paper. first i put cos2x = 2cos squared x - 1 and worked out values for cosx. Then put cos2x intot he sin squared x formula

10)a) 2sin(x - pi/3)
b) Graph

11)a) y squared + (x-t) squared = 4
b) root 5

[KhalidBoussouara]

I almost got stuck at that question with the dot product nonsense. However I then noticed the word "equilateral" which made the answer perfectly clear.

[Colin]

I got the same awnser as you on Q3, my earlier post was tosh

[HatingHighers]

bluh, got the maximum wrong for Q8 - 0 was the maximum point. Think I got the minimum right though, so I'll maybe only lose a couple of marks.

[*Suz*]

Im not suprised i got the reccurance relations question wrong on paper 1, was very confused with that one. Also noticed a few marks here and there ive lost....ah wells, hopefully the 8 markers, 7 markers etc that i got right will balance out the odd 1 mark incorrect ones

[gary]

Yes I can't believe I got the 5 mark circle question in paper 1 right. I was arguing with someone who said you didn't use the midpoint.

[jondom]

I think I got a couple of marks for that then since I managed to expand it etc.

[Kirsty]

I didn't think those papers were as hard as some past papers I'd already done - missed out 11 marks in the paper 1 and 11 marks in paper 2 - oh, and btw, my mum works at SQA and wouldn't tell me anything that was in the papers!

[tupacshakur]

btw, my mum works at SQA and wouldn't tell me anything that was in the papers!

is it possible to find out your results BEFORE 10th august by emailing SQA and pleading with them to let you know what you got or by asking someone u know who works there?

someone i know claimed they did that through someone they knew working there. Is it true?

[YIC]

for the log question in paper 2 I wrote 43/15. Is that okay? I never thought we would have to write an approximation.....