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xlane:
sum1 explain how to solve da 2nd n 3rd part of oct/nov 2002 p1..12 either..please

Ghost Of Highbury:

--- Quote from: xlane on June 09, 2010, 12:12:25 pm ---sum1 explain how to solve da 2nd n 3rd part of oct/nov 2002 p1..12 either..please

--- End quote ---

ii) when theta = 30

R = r(1+1/sin 30) = 3r

Area of circle = pi*r^2

Area of sector = 1/2 R^2(pi/3) = (3pir^2)/2

fraction : pir^2/(3pir^2)/2 = 2/3

iii) tan 30 = r/d (R = 15 , 15 = 3r , r = 5)

    tan 30 = 5/d

d = 5/tan 30 = 8.66

arc length = r*a = 5(2pi/3) = 10.47

perimeter = 10.47 + 2(8.66) = 27.8

xlane:
Thanks ghost o highbury..bt u r hardly a ghost..u can type n ur usin a modern technology :P

adrian1993:
Hmm, can someone who did variant 1 tell us what came out in their paper 1?

Yeah, the permutations question was so easy. I remember the last answer was 100 for that question.

I think relative velocity would probably come out tomorrow...

jellybeans:
OH, & for variant 2, the combination question for choosing 7 pieces from 6 classical & 4 modern pieces..
i got 10C7 = 120ways for i) and 100 ways for ii) :) :) did anyone get the same answers?

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