Qualification > Math
Math P2 May/June 2001 HELP !!!
astarmathsandphysics:
--- Quote from: Q80BOY on May 06, 2009, 07:19:11 am ---IGCSE (CIE) Mathematics Paper 2 May/June 2001
(i have attached the paper)
Page 6, Question 14, Part (b) < -- it seems ill never get these graphs !!
Page 7, Question 15 (Done)
Page 8, Question 19, Part (b) < -- what are prime factors ??
Page 10, Question 21, Angle z (Done)
Page 12, Question 23, Part (c), Section (ii) < -- what do they actually want ??
Total Questions Left: 4
I really hope i get the same help i got on my last post, help me guys ???
--- End quote ---
Page 9, Question 20, Part (b) < -- i found the inverse but the m/s says otherwise, HELP !!
19b)obvious 3^2 and 11 so 9999=3^2*11*101 by dividing the number by 99
Find g-1(x) y=3-3x so x=(3-y)/3 and g-1(x)=(3-x)/3 and solve do g(18)=(3-18)/3=-5
14b)180+50 and 360-50 =230 and 310
23c)Draw the lin y=-x then where the line crosses the curve find the x valure which is your answer.
Aishath:
oh ur welcome nd thanks :D
Q80BOY:
@ astar, hey Thanks for the help ;D
but what does ^ stand for ??
appreciate ur help ;)
Aishath:
prime factors r numbers which can only be divided by 1 or itself. eg. 1, 2, 3, 5, ect
therefore product of prime factors for 99999 is = 3*3*11*101
cos all these numbers cannot b gotten by dividing by other numbers except 1 or itself
did u get it?
Q80BOY:
--- Quote from: Aishath on May 06, 2009, 09:07:33 am ---prime factors r numbers which can only be divided by 1 or itself. eg. 1, 2, 3, 5, ect
therefore product of prime factors for 99999 is = 3*3*11*101
cos all these numbers cannot b gotten by dividing by other numbers except 1 or itself
did u get it?
--- End quote ---
Thanks a lot, how cant i get it, ur explanation is simple and to the point ;)
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version