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Recent content by 19islandprincess96

  1. 1

    Mathematics: Post your doubts here!

    Good Luck for tomorrow everyone! Insha'Allah we'll all do good!
  2. 1

    Mathematics: Post your doubts here!

    When its an even number take n/2 and when its an odd number take n+1/2. I think.:whistle:
  3. 1

    Mathematics: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_y13_sy.pdf Go to page 34, apply the first formula.
  4. 1

    Mathematics: Post your doubts here!

    Please?
  5. 1

    Mathematics: Post your doubts here!

    P(z<a) = direct table P(z>a) = 1- table value P(z<-a) = 1 - table value P(z>-a) = direct table value P(a<z<b) = value of (b) from table - value of (a) from table P(-a<z<b) = [value of (b) from table + value of (a) from table] -1 P(-a<z<-b) = Value of (a) from table - Value of (b) from table
  6. 1

    Mathematics: Post your doubts here!

    Somebody help me with 3b, please. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf
  7. 1

    Mathematics: Post your doubts here!

    When they ask for an Argand diagram, do we draw it on the given graph paper or on the answer booklet? If anyone's awake, please answer...
  8. 1

    Mathematics: Post your doubts here!

    dy/dx = 3cosx - 12cos^2x sinx 3cosx = 12cos^2x sinx 3/12 = cosx sinx 1/4 = 1/2 sin2x 1/2 = sin2x 2x = pi/6 , 5pi/6 x = pi/12 , 5pi/12 Sorry, I cant get 1/2 pi like in the mark scheme...
  9. 1

    Mathematics: Post your doubts here!

    Q 5i) dy/dx = 8 * (1/2)cos(1/2)x - (1/2)sec^2(1/2)x dy/dx is zero at maximum point. so 0 = 8 * (1/2)cos(1/2)x - (1/2)sec^2(1/2)x (1/2)sec^2(1/2)x = 4cos(1/2)x sec(x) = 1/cosx so sec^2(1/2)x = 1/cos^2(1/2)x 1/cos^2(1/2)x = 8 cos(1/2)x 1 = 8 cos^3(1/2)x cos^3(1/2)x = 1/8...
  10. 1

    Mathematics: Post your doubts here!

    1/x + 1/y* dy/dx = derivative of ln (xy) -3y^2 * dy/dx = derivative of y^3 so 1/x + 1/y* dy/dx - 3y^2 * dy/dx = 0 1/x + dy/dx (1/y - 3y^2) = 0 dy/dx (1/y - 3y^2) = -1/x dy/dx [(1-3y^3)/y] = -1/x dy/dx = y/-x(1-3y^3) dy/dx = y/x(3y^3 - 1) Hope this is clear!
  11. 1

    Mathematics: Post your doubts here!

    Thank you so much! I got it now! :)
  12. 1

    Mathematics: Post your doubts here!

    Okay! How do we find the normal of OAB?
  13. 1

    Mathematics: Post your doubts here!

    nooo... but I'm figuring it out...
  14. 1

    Mathematics: Post your doubts here!

    Thanks! :)
  15. 1

    Mathematics: Post your doubts here!

    Yep! I got it 100%, thanks to u. Can u solve another question if u r in the mood? :P Its Q 9 part ii) of this paper http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s07_qp_3.pdf
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