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area of shaded region = area of medium circle - area of smallest circlehttp://clickpapers.net/past papers/math-/4024_s07_qp_1.pdf
q.14 last part !!
q.15 last part !!
help would be appreciated with explaination
area of shaded region = area of medium circle - area of smallest circle
= 4*pie*x^2 - pie*x^2 = 3(pie)x^2
let perpendicular distance from A to DE be x cm. then perpendicular distance from A to CB is x+4
use proportionality for similar triangles so that 9 : 12 :: x : x+4
3x=36
x=12
x+4 = 16
its pretty simple try solving it and then see for yourself...Thanks
i got q.15 but still struggling to understand q.14 :O
Funny how people think PI's spelling as Pie, even I used to think that way.area of shaded region = area of medium circle - area of smallest circle
= 4*pie*x^2 - pie*x^2 = 3(pie)x^2
let perpendicular distance from A to DE be x cm. then perpendicular distance from A to CB is x+4
use proportionality for similar triangles so that 9 : 12 :: x : x+4
3x=36
x=12
x+4 = 16
area of shaded region = area of medium circle - area of smallest circle
= 4*pie*x^2 - pie*x^2 = 3(pie)x^2
let perpendicular distance from A to DE be x cm. then perpendicular distance from A to CB is x+4
use proportionality for similar triangles so that 9 : 12 :: x : x+4
3x=36
x=12
x+4 = 16
acha acha, bus ziyada nahin !Funny how people think PI's spelling as Pie, even I used to think that way.
Hey everyone....how r u guys preparing for Maths?As in how many past papers r u doing?
gud job....I didnt do P2 yet but I am doing P1 randomly...not good at p2 luv p1 and doing dem only from 02-11 !! im done with the june ones u ?
(a) It is Figure 3. Bcz y-intercept seems to be 2, discriminant (b^2-4ac)<0 which means there are no real roots and the curve does not cut the x-axis. Moreover, the coefficient of x^2 is positive, means it has a minimum point.http://clickpapers.net/past papers/math-/4024_s09_qp_1.pdf
here u go another one q.13 ??? i have no idea about these graphs if anyone could guide me plz
thanks in advance !!
U have to make the equationcmon guys !! do this last part of this paper
http://clickpapers.net/past papers/math-/4024_w10_qp_12.pdf
In transormation first u have to make a line joining corresponding points to find the invariant line...whenu will join thm u will find x= -1 as the invariant line now find the distance of any point of image from invariant line and divide it by the distance of the corresponding point of object to find stretch factor which will be 2co linear points :
Please help me !!! And tell me about all the properties of colinear points !
And also the last two parts of transformation !
In transormation first u have to make a line joining corresponding points to find the invariant line...whenu will join thm u will find x= -1 as the invariant line now find the distance of any point of image from invariant line and divide it by the distance of the corresponding point of object to find stretch factor which will be 2
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