Math P2 May/June 2001 HELP !!!

[Q80BOY]

IGCSE (CIE) Mathematics Paper 2 May/June 2001

(i have attached the paper)

Page 6, Question 14, Part (b) (Done)

Page 7, Question 15 (Done)

Page 8, Question 19, Part (b) (Done)
 
Page 9, Question 20, Part (b) (Done)

Page 10, Question 21, Angle z (Done)

Page 12, Question 23, Part (c), Section (ii) (Done)

Special thanks to Amber, Aishath, Astarmathsnadphyscis and Sanity_Master

[Christy]

Q14)a) x= 130 (180 - 50)
b) x= 230 and x= 310 ( 180 +50 and 360 - 50 )

Q15) 3* (4x-3/8) - 2*(3x-4/12)

12x-9-6x-8/24 = 6x-17/24

Q20) g inverse = multiply the bracket with a (-) so it can turn positive [x-3/-3]
( x +3/3 )
18+3/3 = 7  hope thats correct

Q21) w = 90
        x = 20
        y = 40 ( 180 - (20+20) = 140 [180 - 140 = 40 ] angles on a st. lne add up to 180 )
        z = 130 ( 180 - (90+40) = 50 [ 180 - 50 = 130 ] opposite angles in a quadrilateral add up to 180 )

Q23) u have to look at the graph ... there will be a point where y=-x thats wat they want u to find and the answer is -.6 draw the curve and u will see that .6 on y axis is equals to -.6 on the x axis .... so the point is (-.6,.6) ... the answer is x = .6  

[Q80BOY]

Q14)a) x= 130 (180 - 50)
b) x= 230 and x= 310 ( 180 +50 and 360 - 50 )

Q15) 3* (4x-3/8) - 2*(3x-4/12)

12x-9-6x-8/24 = 6x-17/24

Q20) g inverse = multiply the bracket with a (-) so it can turn positive [x-3/-3]
( x +3/3 )
18+3/3 = 7  hope thats correct

Q21) w = 90
        x = 20
        y = 40 ( 180 - (20+20) = 140 [180 - 140 = 40 ] angles on a st. lne add up to 180 )
        z = 130 ( 180 - (90+40) = 50 [ 180 - 50 = 130 ] opposite angles in a quadrilateral add up to 180 )

Q23) u have to look at the graph ... there will be a point where y=-x thats wat they want u to find and the answer is -.6 draw the curve and u will see that .6 on y axis is equals to -.6 on the x axis .... so the point is (-.6,.6) ... the answer is x = .6  

14 (a) Thanks !! the ans is correct 
    (b) the m/s says the ans is 210 and 330  im confused !!

15 Thanks for the help, By the way -9--8 = -1, so the ans = 6x-1/24 

20 i dont think it is the m/s says its -5 how did we get -5 !!

21 Thanks a lot !! the ans is correct  (Thanks for the tips)

23 is there a formula or something, cause the graph only goes to -1

Thanks a lot for the help amber. i appreciate it !!!

[Aishath]

oh for Ques 20(b) its:

 y=3-3x
now make y the subject
therefore its: x=(y-3)/-3
then substitute 18 in to y
so its:
x=(18-3)/-3
x=(15)/-3
x=-5
solved the ques!!

hope you understood!!!

[Q80BOY]

oh for Ques 20(b) its:

 y=3-3x
now make y the subject
therefore its: x=(y-3)/-3
then substitute 18 in to y
so its:
x=(18-3)/-3
x=(15)/-3
x=-5
solved the ques!!



hope you understood!!!

ya i understood 100%  Thanks a lot for ur time and help !!!  + Rep

[astarmathsandphysics]

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 

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

[Q80BOY]

@ astar, hey Thanks for the help 

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]

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?

Thanks a lot, how cant i get it, ur explanation is simple and to the point 

[Aishath]

haha yup

ur welcome

[Q80BOY]

I scanned my Q23, what do i do next ??

[shri]

please here too
shrikansal@hotmail.com
Thanks a lot

[Q80BOY]

please here too
shrikansal@hotmail.com
Thanks a lot

sorry ??

[sanity_master]

lol

ppl got used to that when u post anything, it will be old past papers LOL