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#97622 THE EXAM!

Posted by Da King on 15 May 2007 - 03:51 PM in Mathematics

gradient was -1 bt cant rememba da equation



#97618 THE EXAM!

Posted by Da King on 15 May 2007 - 03:47 PM in Mathematics

1/24



#97615 THE EXAM!

Posted by Da King on 15 May 2007 - 03:42 PM in Mathematics

Anybody got a link for the solutions/answers???



#80447 How It Go?

Posted by Da King on 20 May 2006 - 12:19 PM in Physics

Does anybody have a link to the solutions or answers for this paper?



#80202 Higher Maths 2006

Posted by Da King on 19 May 2006 - 03:49 PM in Mathematics

factorise!!! take sinx as common factor

anyone got a link to the solutions for this paper or higher physics paper?



#80199 Higher Maths 2006

Posted by Da King on 19 May 2006 - 03:46 PM in Mathematics

QUOTE(loved-up-loon @ May 19 2006, 04:41 PM) View Post

For question 7 i got the 0,180,360 bit but not the other!


Question 7 Paper 1

sinx - sin2x = 0 (sin2x=2sinxcosx)
sinx - 2sinxcosx=0
sinx(1-2cosx)=0

sinx=0
x=0, 180, 360

OR

1-2cosx=0
cosx=1/2
x=60, 300

anyone got a link to the solutions for this paper or higher physics paper?



#80193 Higher Maths 2006

Posted by Da King on 19 May 2006 - 03:40 PM in Mathematics

QUOTE(cfesq909 @ May 19 2006, 04:25 PM) View Post

Paper II - Q.10

Is this correct?

a. k sin(x-a)
k sin x cos a - k coz x sin a
k cos a = 7
k sin a = 24

tan a = 24/7
a = 1.29 radians

k=√(7²+24²)
= 25

therefore,
25 sin (x-1.29)


b. differential sin x
= cos x

cos x=1
x=0

sin(x-1.29)=0
x-1.29=0
x=1.29 radians



Paper II Q10

a) is correct answer was 25sin(x-1.29)

b) dy/dx = gradient of curve

dy/dx = 25cos(x-1.29) (m=1 from question) so

25cos(x-1.29)=1
cos(x-1.29)=1/25

a=cos(inverse)(1/25)
=1.53

cos is +ve so quadrants 1 and 4

x-1.29=1.53, 4.75 (4.75 out of range) so

x= 2.82 radians



#80188 Higher Maths 2006

Posted by Da King on 19 May 2006 - 03:36 PM in Mathematics

Paper II Q10

a) is correct answer was 25sin(x-1.29)

b) dy/dx = gradient of curve

dy/dx = 25cos(x-1.29) (m=1 from question) so

25cos(x-1.29)=1
cos(x-1.29)=1/25

a=cos(inverse)(1/25)
=1.53

cos is +ve so quadrants 1 and 4

x-1.29=1.53, 4.75 (4.75 out of range) so

x= 2.82 radians



#80131 Higher Maths 2006

Posted by Da King on 19 May 2006 - 02:54 PM in Mathematics

yeah thats right. exactly what i got!!!



#80118 Higher Maths 2006

Posted by Da King on 19 May 2006 - 02:46 PM in Mathematics

how you do 6c in paper 2



#80045 Higher Maths 2006

Posted by Da King on 19 May 2006 - 01:21 PM in Mathematics

so you get x=2.82 radians



#80042 Higher Maths 2006

Posted by Da King on 19 May 2006 - 01:20 PM in Mathematics

derivitive is the gradient of the curve. so i derivied 25sin(x-1.29). which i got to be 25cos(x-1.29). then i equaled it to 1 and solved the equation for x-1.29. then added on 1.29 to my answers to get x



#80040 Higher Maths 2006

Posted by Da King on 19 May 2006 - 01:17 PM in Mathematics

what about 6c in paper 2



#80037 Higher Maths 2006

Posted by Da King on 19 May 2006 - 01:15 PM in Mathematics

what is the derivative of 25sin(x-1.29)



#80033 Higher Maths 2006

Posted by Da King on 19 May 2006 - 01:11 PM in Mathematics

what was the answer to Q10b in paper 2?