Mathematics: Post your doubts here!

[Serial_Ripper]

CAN ANYONE TELL ME HOW TO POST A SCREENSHOT? I tried copying and pasting and failed.

 

[awesomaholic101]

62/o/n/14 - q6.iii .. For calculating midvalues, should we always take the limits given or must take the limits we get after making the plots continuous??

 

[The Sarcastic Retard]

View attachment 57230 Plz guyz anyone solve this problem

7(ii)
Take x to be 7.2 and 9.2 and solve it..

 

[Mahreen Mustafa]

Which variant?

 

[Maayee]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w14_qp_31.pdf
how do i solve 8 iii)

 

[eliyeap]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w14_qp_31.pdf
how do i solve 8 iii)

Use sin3theta= 3/4
When they say hence means that there is a connection to the previous question.

 

[musiclover gurl]

2 * |3^x - 1| = 3^x

Square both sides to get rid of modulus (cuz its ugly)

4*(3^2x - 2* 3^x + 1) = 3^2x
Expand bracket:
4•3^2x - 8•3^x + 4 = 3^2x
Move terms:
3•3^2x - 8•3^x + 4 = 0
Make u=3^x :
3u^2 - 8u + 4 = 0
Factorise:
3u^2 - 6u - 2u + 4 = 0
3u(u-2) -2(u-2) = 0
(3u-2)(u-2) = 0
u=2, u=2/3
3^x = 2, 3^x = 2/3
x = lg2/lg3 , x=lg(2/3) / lg3
sorry don't have a calculator

Thanks a lot qwertypoiu ! I just get confused when I multiply numbers with powers, I guess. Thanks

 

[Farhad221]

may june 2015 33 question 8 part 3

 

[RadzMau]

can someone please help out?
June 14 Paper 31 Question 5(ii)

 

[musiclover gurl]

Help please?

 

[The Sarcastic Retard]

Help please?
View attachment 57251

As level ?

 

[Nushad Nahue]

HAS ANYONE SOLVED 9709_s02 paper 3 Q8? I can't seem to match the answer. Please help me.

 

[eliyeap]

Please help with Q8 part ii and iii
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_31.pdf
Thank you.

 

[@cbjamboii]

Greetings. Can you please help me with question 6 part 1?

 

[Farhad221]

Greetings. Can you please help me with question 6 part 1?

 

[Serial_Ripper]

PLEASE HELP ME!!!

 

[Serial_Ripper]

HELP!! IN PART B

 

[qwertypoiu]

View attachment 57262PLEASE HELP ME!!!View attachment 57262

Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBAC
Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))
QUADRITLATERAL OBAC = r²*sin(π-2€)

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

Shaded = 1/2*circle
πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€

 

[Serial_Ripper]

Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBAC
Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))
QUADRITLATERAL OBAC = r²*sin(π-2€)

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

Shaded = 1/2*circle
πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€

THANKS!

 

[manya]

someone tell me how to write equation of plane in the form 3x+2y-3z+1=0 to the form r.n=a.n