Maths, Addmaths and Statistics: Post your doubts here!

[SalmanPakRocks]

b(i) a
Consider Tri-QO^N
OQ is the perpendicular bi-sector. And therefore angle O will be 45^degrees. If we consider tri-MO^N then we know than angle N^ is also = 45^ and therefore the Tri-QO^N is an icosceles. Therefore we can say that OQ=QN.
therefore QN = x. Now as OQ is perpendicular bisector so MN = 2QN. there
MN= 2x

 

[SalmanPakRocks]

(ii)
Area of triangle = 1/2 * b * h
=1/2 * 2x * x
=x^2
Area of the sector of circle = 2pieR
2pie8
16pie.
okie now for the shaded region just subtract it.
16pie - x^2

 

[bogus]

Plz explain 20 b) of the attached paper.

 

[scouserlfc]

Plz explain 20 b) of the attached paper.

Easy dude both triangles use the same height BX which means if u use the formulae 0.5 b * h the height and 0.5 will get cut and u will be left with the bases which is 2 is to 5 the ratio

 

[TheEconomist]

which subject?

Sorry i forgot to write it, Math....

 

[multixamza01]

Sorry i forgot to write it, Math....

Guess not

 

[bogus]

Easy dude both triangles use the same height BX which means if u use the formulae 0.5 b * h the height and 0.5 will get cut and u will be left with the bases which is 2 is to 5 the ratio

Thanx...and that was hard.wayyyy beyond my imagination..unlike u i'm not exactly by a mathemaniac so yup thanks anyway.

 

[zeshan huq]

http://www.xtremepapers.com/CIE/Cambridge O Levels/4024 - Mathematics/4024_s10_qp_11.pdf hello im stuck in question number 9)B please help

 

[Narcotic]

http://www.xtremepapers.com/CIE/Cambridge O Levels/4024 - Mathematics/4024_s10_qp_11.pdf hello im stuck in question number 9)B please help

the diameter of the pencils is CORRECT TO NEAREST MM. so first of all we will take the upper bound IN CASE the pencils are more than 7 mm. the new diameter becomes 7.5 and multiply it with the number of pencils, 8. 7.5 into 8=60mm. the answer is to be given in cm and as 1cm=10mm, divide 60 by 10. u get "6cm" which is the answer

 

[scouserlfc]

Thanx...and that was hard.wayyyy beyond my imagination..unlike u i'm not exactly by a mathemaniac so yup thanks anyway.

no problem !

 

[hredoymohammad]

i have a problem can anyone help me...O/N 2011 p-21 i have not understood the question 4(ii) http://www.mediafire.com/download.php?fsglx7320qqfgd5

 

[hredoymohammad]

and also question 5 b.....is there any special way to find out the answers...or we have to carry out trial and error method?

 

[Narcotic]

i have a problem can anyone help me...O/N 2011 p-21 i have not understood the question 4(ii)

in answer to part (i) did u get k=2?
so if k=2 im gonna try to solve below.
integral of 2x/((x^2)+3)^2 =1/((x^2) +3)
so this implies that 3 into integral will equal 6x/....
we can write it as 3(integral of 2x/..) which will equal 3/((x^2)+3)
now put the vales of x n the answer i got is -1/2
is it right? if not thn sm1 plz tell me too how to solve it,,
p.s. i gave this paper in oct/nov, i mean final o lvl. thn it was easy to solve such Qs bt i m out of practice now, so plz sorry if i cant help u..

 

[waztaz123]

and also question 5 b.....is there any special way to find out the answers...or we have to carry out trial and error method?

Yes we have to use just trial and method. We're going to try different functions like kk(x) and hh(x) and kh(x) and hk(x) and see if anyone give us the answer. The last one is a bit tricky......as we have to use the inverse function, taking help from our previous answers.

 

[waztaz123]

in answer to part (i) did u get k=2?
so if k=2 im gonna try to solve below.
integral of 2x/((x^2)+3)^2 =1/((x^2) +3)
so this implies that 3 into integral will equal 6x/....
we can write it as 3(integral of 2x/..) which will equal 3/((x^2)+3)
now put the vales of x n the answer i got is -1/2
is it right? if not thn sm1 plz tell me too how to solve it,,
p.s. i gave this paper in oct/nov, i mean final o lvl. thn it was easy to solve such Qs bt i m out of practice now, so plz sorry if i cant help u..

I think ur value of k is incorrect but otherwise the rest of solution is right. I got k=-2 and answer of 4(ii) 1/2. Haven't checked with the marking scheme though.

 

[beeloooo]

june 2010 p11 4024 maths q.9 part b ? plz can any 1 explain ??? file attached !~!

 

[scouserlfc]

june 2010 p11 4024 maths q.9 part b ? plz can any 1 explain ??? file attached !~!

Its prretty easy concentrate when doing these arithmetic questions
Each pencil has a diameter of 7 correct to the nearest millimeter so the lowest possible diameter is 6.5 mm right so if the box has to have 8 pencils side by side it should have the width equal to the diameter of the pencils,think abt this closely by looking at the diagram !
Since it says the lowest possible width we take the lowest possible diameter and multiply by 8 and we have the answer

 

[bogus]

this is completely of the topic...anyone know the topic for this yrs paper 3 computer studies..it was supposed to come on 1st march

 

[bogus]

explain q11 part b (II)a of the attached paper.Plzzz

 

[Gémeaux]

explain q11 part b (II)a of the attached paper.Plzzz

whenever there's a question abt centre of enlargement, its better if you join the original and the enlarged point with a line. as i did in the image above, once drawn, u can easily divide it into 4 equal parts as the scale factor was 4. if u start from (-2,3), u can consider 3 boxes as one unit, so the black dots represent the other units, in total making up 4.
hope it helps.