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Recent content by biscuitbiscuit

  1. biscuitbiscuit

    Mathematics: Post your doubts here!

    How do we know if the function is 1 to 1. I know it comes for only 1 mark but still someone please help out
  2. biscuitbiscuit

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    Find dy/dx at B Take the reciprocal of the answer and change the sign (This is the gradient of normal) Use the formula, y-y1=mx-x1, where m= gradient of normal To find co-ordinates of B put y= 0 in the equation of the curve :)
  3. biscuitbiscuit

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    This question someone!!!
  4. biscuitbiscuit

    Post your AS-Level Mathematics (P1 and M1) doubts here.

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w05_qp_1.pdf Q5(i), i know it's about similar triangles but still someone explain this
  5. biscuitbiscuit

    Chemistry: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w10_qp_23.pdf Q3(c)(i) and (ii)
  6. biscuitbiscuit

    Physics: Post your doubts here!

    Yeahhh!!, thanks :)
  7. biscuitbiscuit

    Physics: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_s12_qp_21.pdf 'Q5(iii)', why 1.03V?????
  8. biscuitbiscuit

    Physics: Post your doubts here!

    Ohhhhh! formula mey current ko square karna bhol gaya :LOL: aur tum to na kaho phy weak hay. Ghar mey sab nay tang kar rakha hai, chor do phy ko nahi hoti tum sey:p
  9. biscuitbiscuit

    Physics: Post your doubts here!

    well the on the graph y-axis is labelled in ohms!
  10. biscuitbiscuit

    Physics: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_23.pdf "Q8(b)" someone please
  11. biscuitbiscuit

    Physics: Post your doubts here!

    Okay, this is how i did this : Power=V^2/R, therefore resistance of a heater is, R= V^2/P R= 230^2/1000= 52.9 ohms (i) Since the heaters are in parallel therefore voltage across them would be 230V Total resistance of the circuit= 52.9/2= 26.45 Ohms therefore power diss...= 230^2/26.45= 2000W=...
  12. biscuitbiscuit

    Physics: Post your doubts here!

    I didn't get this thing, did u mean 1.5 to 3.5 ohms if not how did you get volts?
  13. biscuitbiscuit

    Physics: Post your doubts here!

    Still in As, maybe next year ;)
  14. biscuitbiscuit

    Physics: Post your doubts here!

    Hey can you please explain Q6(c) also http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_22.pdf
  15. biscuitbiscuit

    Physics: Post your doubts here!

    http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w10_qp_21.pdf "Q6(ii) " sketch would be appreciated
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