http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_31.pdf

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- Thread starter XPFMember
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http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s15_qp_31.pdf

Thank you.

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

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Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBACView attachment 57262PLEASE HELP ME!!!View attachment 57262

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€

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THANKS!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€

Take r as (x,y,z) and multiply with your normalsomeone tell me how to write equation of plane in the form 3x+2y-3z+1=0 to the form r.n=a.n

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can anyone do questions 5 part ii) and 6 ii) from the Oct/Nov 2014 paper 32

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Once again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?As level ?

View attachment 57252

Ya, AS level

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(i) OB = 6 cmOnce again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?

Ya, AS level

AO = 6 cm

AB = sqrt(36 + 36)

AB = 6sqrt(2)

(ii) tan^-1 (6/6) = pi/4

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Thanks a lot!(i) OB = 6 cm

AO = 6 cm

AB = sqrt(36 + 36)

AB = 6sqrt(2)

(ii) tan^-1 (6/6) = pi/4

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http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_qp_71/

http://maxpapers.com/syllabus-materials/mathematics-9709-a-level/attachment/9709_s14_ms_71/

Can anyone explain why you do 70/69 × 2.70 = 2.73913?

Thanks.

Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?

Someone please reply as soon as possible

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10cos(37) = 7.98 m/sA ball is kicked with an initial speed v0=10 m/s at an angle of θ= 37◦above the horizontal. Its speed in m/s at maximum height is:

I am guessing its 45Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?