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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??
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7(ii)View attachment 57230 Plz guyz anyone solve this problem
Use sin3theta= 3/4
Thanks a lot qwertypoiu ! I just get confused when I multiply numbers with powers, I guess. Thanks2 * |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
Greetings. Can you please help me with question 6 part 1?
Shaded Area = SECTOR OBAC + SECTOR ABC - QUADRITLATERAL OBACView attachment 57262PLEASE HELP ME!!!View attachment 57262
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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