Mathematics: Post your doubts here!

[daredevil]

MATH A2 PAPER 3 2013 OCTOBER

PLEase reply

here i'll just explain the procedure and see if u can do it then.... if not then i'll solve it out for u ... so here goes...

the circle with centre O has the radius r ... however the circle with the center A has the radius AB. so first step will be to find AB (radius of the circle with center A)
AB^2 = r^2 + r^2 - 2(r)(r)Cos(pie - 2thita)

this from the cosine rule.... and triangle BOA. u will then get the expression for the length of AB.
the angle (pie-2theeta) comes from the fact that in triangle BOA, OB and OA are equal so angleOBA=angleOAB=theeta

the shaded region is made up of a sector with (center A and radii AB=AC ) and two equal segments.

find the area of the sector BAC from the formula
(2theeta/2pie)*(pie)(AB)^2

Now u have to find the area of the segments. take one segment first.
calculate the area of sector BOA.
calculate the area of triangle BOA.
(area of sector - area of traingle)=area of segment

Area of shaded region = (area of sector BAC) + 2(area of segment)
[(pie)(R)^2]/2 = xyz + 2abc

(xyz and abc are watever areas u get)



If this whole stuf is too confusing then tell me and i'll solve it and try to send u a pic of that. toodles but it will be better if u work through these steps one by one with ur pen and paper in hand and not just read it and become confused as hell

 

[daredevil]

found those probzz ....
solve em up now ..

ahmed abdulla

 

[ahmed abdulla]

found those probzz ....
solve em up now ..

ahmed abdulla

 

[daredevil]

thanksss i'll be sure to tag u wen i hav more problems

 

[ahmed abdulla]

thanksss i'll be sure to tag u wen i hav more problems

Will be glad to help you when i am free

 

[midha.ch]

A particular solution of the differential equation

3 y^2 dy/dx=4 ( y^3 + 1 )cos^2(x)

is such that y = 2 when x = 0. The diagram shows a sketch of the graph of this solution for 0 ≤ x ≤ 20;
the graph has stationary points at A and B. Find the y-coordinates of A and B, giving each coordinate
correct to 1 decimal place.

october november 2013 paper 33

there you go

 

[midha.ch]

This is the link for oct/nov papers!

i actually also had a doubt in this : http://maxpapers.com/category/quali...l/cambridge/a-level/a-level-mathematics-9709/

I have uploaded the answer
Idk how to tag so couldn't tag you in the post

 

[sitooon]

I have uploaded the answer
Idk how to tag so couldn't tag you in the post

You can tag someone bu putting @ sign and then his name
Thanks for solving. i Found the equation , i only had doubt on how to get the x-values

 

[midha.ch]

You can tag someone bu putting @ sign and then his name
Thanks for solving. i Found the equation , i only had doubt on how to get the x-values

thankss!!
I referred the "Help" site and it showed some weird method which would take me ages to get the hang of!

 

[Manal.A]

may june 09 paper 62 question 1...how do we find the standard deviation from box and whisker diagram??

 

[ahmed abdulla]

Hi . .
I am curremtly doing M2 , and started recently doing pastpapers
i tried to ask my teacher to help me but he is not able to solve them !
maybe because this is the first year of teaching a level so they find it difficult
my request if any one does M2 , plz PM me , so i will be able to tell you my doubts

 

[Talha Irfan]

may june 09 paper 62 question 1...how do we find the standard deviation from box and whisker diagram??

i didn't give the paper a look, but I suppose that the box and whisker plot would be symmetrical (means the box is equally divided by the line of the inter quartile value). This occurrence gives rise to the possibility of applying the normal distribution. Apply it and find the standard deviation accordingly by taking areas such as 0. 75 and 0.25 into consideration

 

[Manal.A]

i didn't give the paper a look, but I suppose that the box and whisker plot would be symmetrical (means the box is equally divided by the line of the inter quartile value). This occurrence gives rise to the possibility of applying the normal distribution. Apply it and find the standard deviation accordingly by taking areas such as 0. 75 and 0.25 into consideration

Yes its the upper quartile minus mean for z score but if its possible can u post a pic or sumthing of how u solve it further? Why is the area 0.75 or 0.25
Btw thanks

 

[rawaaaan]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
in question 4 i
why do we do 360/180=2 to get B
and 9-3 to get A
6+x=9 to get C
PLEASSEEEE ANSWER AS SOON AS POSSILBE

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_1.pdf
in question 4 i
why do we do 360/180=2 to get B
and 9-3 to get A
6+x=9 to get C
PLEASSEEEE ANSWER AS SOON AS POSSILBE

a is the maximum distance from the position of equilibrium so 9-3
b is the no of oscillations ie the no of wave if you see there are 2
and c is the y intercept the point where the curve cuts the y axis here its 3

 

[sitooon]

Any help , ?
for second question tell me the procedure , its a different question and assume my dy/dx = 2x +3

 

[Muskan Achhpilia]

Maths doubt of stem and leaf diagram, please help

Answer is also given in the picture in the smaller font

Thanks a tonne!

 

[Muskan Achhpilia]

The following doubt is of finding median without a cumulative frequency graph, please help!

Answers are given in red font

Thank you very much

 

[Talha Irfan]

The following doubt is of finding median without a cumulative frequency graph, please help!

Answers are given in red font

Thank you very much

The formula to calculate median (50th percentile or any other percentile) without the cumulative frequency curve is :

lower class boundary +
((50n/100 - cf (till prev. class)/ f (of that interval))

where n is the total number of observed items or maximum cumulative frequency. First calculate the corresponding class by checking in which class the n/2th value lies then perform the relative ascribed function

 

[usama321]

Any help , ?
for second question tell me the procedure , its a different question and assume my dy/dx = 2x +3

4(1/2 r^2 [(pie-x)- sin(pie-x)]) = 1/2 pie r^2
since sin 180 - x = sin x
2 r ^2 [(pie-x) - sin x] = 1/2pie r^2

2pie - 2x -2sinx = 1/2 pie
2pie - 1/2pie - 2sinx = 2x
3/2 pie - 2 sinx = 2x
x = 3/4 pie - sinx

For the second question, please send in the whole question and i'll try to solve it