2004 CSSA Solutions? (1 Viewer)
- Thread starter sHin
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Candypants
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Yeh but on sheets of paper... and I ain't typing them up!
withoutaface
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- 2004
Well if you need answers to specific questions myself or others in this forum can help. If I had a scanner I would do the questions myself and put them up, but as it stands I don't:S
withoutaface
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Is it possible for you just to type the final answers without working, because I think that most people just want them to check their own papers. It's cool if you don't want to, just an ideaCandypants said:
Heres some of the Answers for the CSSA Trail paper 2004
If you need more , just ask
I can't write all the working out because of those little symbols but i'll try my best =D cheers guys!
Question 1
a) 5.49
b) 0.3 +0.33333 = 3/10 +1/3 = 19/30
c) tan alpha = 3 therefore alpha = 72 degree / 252
d) 2b / a+b
e) x =5/3
f) a = 2 and b = 1
Question 2
a)i) 7(3x + 7) to the power of 6 . 3 = 21(3x +4) to the power of 6
ii) x to the power of 2 . e to the power of x ( x + 3 )
iii) 5x sec square 5x - tan 5x / 5x square
b)i) e to the power of 3x / 3 + 2 / 3 . x to the power of 3/2 + C
ii) 3 . 3/4 + ln 2
iii) dy/dx = 2x - sin x therefore y = x square + cosx + c
when y = 2, x = 0 therefore c = 1 therefore y = x square + cosx + 1
Question 3
a) a > 5/2
b) k sqyare + 2k - 15 = 0 therfore (k+5)(k-3) = 0 therfore K = 3 or K = -5
c)i) AB = 5 root 2
ii) BC = 5 root 2 therefore Triangle ABC = isosceles
iii) Gradient of AB = 5-0 / 1-6 = -1
therefore equation of line AB is y-o = -1(x-6) therefore x + y = 6(1)
iv) substitute y=7 into equation (1) therefore x = -1 therefore P is (-1,7)
v) Area of Triangle PAC = 1/2 . 6 . 2 = 6units square
If you need more , just ask
I can't write all the working out because of those little symbols but i'll try my best =D cheers guys!
Question 1
a) 5.49
b) 0.3 +0.33333 = 3/10 +1/3 = 19/30
c) tan alpha = 3 therefore alpha = 72 degree / 252
d) 2b / a+b
e) x =5/3
f) a = 2 and b = 1
Question 2
a)i) 7(3x + 7) to the power of 6 . 3 = 21(3x +4) to the power of 6
ii) x to the power of 2 . e to the power of x ( x + 3 )
iii) 5x sec square 5x - tan 5x / 5x square
b)i) e to the power of 3x / 3 + 2 / 3 . x to the power of 3/2 + C
ii) 3 . 3/4 + ln 2
iii) dy/dx = 2x - sin x therefore y = x square + cosx + c
when y = 2, x = 0 therefore c = 1 therefore y = x square + cosx + 1
Question 3
a) a > 5/2
b) k sqyare + 2k - 15 = 0 therfore (k+5)(k-3) = 0 therfore K = 3 or K = -5
c)i) AB = 5 root 2
ii) BC = 5 root 2 therefore Triangle ABC = isosceles
iii) Gradient of AB = 5-0 / 1-6 = -1
therefore equation of line AB is y-o = -1(x-6) therefore x + y = 6(1)
iv) substitute y=7 into equation (1) therefore x = -1 therefore P is (-1,7)
v) Area of Triangle PAC = 1/2 . 6 . 2 = 6units square
rainonmay
floating around
Can you plz post Question 6 (b) not sure if i did it right or juz post up the answers of the other questions without working out coz i juz wanna c if i got the right answer....
rainonmay
floating around
umm.. that doesnt have the CSSA solutions...Tennille said: