Mathematics: Post your doubts here!

[My Name]

[URL='https://www.xtremepapers.com/community/members/ahmadumar.11149/']ahmadumar[/URL]
Mai(M4!)
DeadlYxDemon
Evangeline
a_wiserME!!
any other person whocan help ur MOST WELCOME!
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

 

[Layla..]

can someone explain how to find a matrix representing a transformation, pleaaaaaassee
I dont want to memorize them, cause they're they wanna know how did u get it.. so please anyone?

Draw the y and axis and mark the points (0,1) and (1,0). The scale does not have to be perfect, it is just a 'sketch'. Afterwards, think where these two points will shift when there is rotation, reflection etc and then use it to find the matrix

 

[dubi2pro]

ahmadumar
Mai(M4!)
DeadlYxDemon
Evangeline
a_wiserME!!
any other person whocan help ur MOST WELCOME!
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

q3 part b

(9-2x) (7-2x) (x)

they asked for volume which is always length x width x height

 

[Layla..]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

I can help

 

[dubi2pro]

ahmadumar
Mai(M4!)
DeadlYxDemon
Evangeline
a_wiserME!!
any other person whocan help ur MOST WELCOME!
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

once u solve it u will end up with 4x^3-32x^2+63x

 

[dubi2pro]

ahmadumar
Mai(M4!)
DeadlYxDemon
Evangeline
a_wiserME!!
any other person whocan help ur MOST WELCOME!
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

q4 part b

sum of the interior angles formula..

180- ( n-1 ) where n = number of sides

substituting for 5 sides of pentagon

which gives 720

 

[Mai(M4!)]

Ahh Yeah! i forgot the total amount!

Listen same paper, Q5, (e)

to help u save time here is the info of previous parts

angle BAC = 39,.8
angle ABC =115
distance AB= 40 KM
DISTANCE BC = 60 KM

i guess they are askin to find AC?

u know when u told me not to waste time reading the Q I wasted time answering what u said which is wrong
he is asking for the length between the east of A and point C which means u will draw a new triangle (down extension of north till the end of the line is horizontal to C so it will be right angle triangle)
AC=

(40^2+60^2-2*40*60*cos(115))=85
as east is 180 degrees from north so 180-80-39.8=60.2
60.2 is oppesite of the east of a and c
using sin
sin(60.2)x(hyp=AC=85)=73.76

 

[dubi2pro]

u know when u told me not to waste time reading the Q I wasted time answering what u said which is wrong
he is asking for the length between the east of A and point C which means u will draw a new triangle (down extension of north till the end of the line is horizontal to C so it will be right angle triangle)
AC=

(40^2+60^2-2*40*60*cos(115))=85
as east is 180 degrees from north so 180-80-39.8=60.2
60.2 is oppesite of the east of a and c
using sin
sin(60.2)x(hyp=AC=85)=73.76

No the info i gave is correct, but even i dint get the end part :S

 

[dubi2pro]

u know when u told me not to waste time reading the Q I wasted time answering what u said which is wrong
he is asking for the length between the east of A and point C which means u will draw a new triangle (down extension of north till the end of the line is horizontal to C so it will be right angle triangle)
AC=

(40^2+60^2-2*40*60*cos(115))=85
as east is 180 degrees from north so 180-80-39.8=60.2
60.2 is oppesite of the east of a and c
using sin
sin(60.2)x(hyp=AC=85)=73.76

what u mean "as east is 180 degrees from north so 180-80-39.8=60.2"?

 

[dishadishki]

paper 43- Winter 2011

Question 8 e....please help...how should we do it?

 

[Mai(M4!)]

ahmadumar
Mai(M4!)
DeadlYxDemon
Evangeline
a_wiserME!!
any other person whocan help ur MOST WELCOME!
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf
question 1 part c
question 3 part b
question 4 part b and d
question 5 part b ii
question6 part b ii steps

question7 part b iii
question 10 part v

PLEASE AND THANKU!

there is no part c for 1
3)b)volume=width*length*hieght=(9-2x)(7-2x)(x)
x(9*7+9*-2x+-2x*7+-2x*-2x)
63x-18x^2-14x^2+4x^3
=4x^3-32x^2+63x


4)b)pentagon=5 sides
n=5
180(n-2)=sum of interior angles
180*3=540


d)i)sum of interior angles of quadrilateral =360
4x-5+3y-20+2x+5+x+y-10=360
rearrange it to be 4x+2x+x+3y+y=360+5+20-5-+10
=7x+4y=390

(ii)as AD and BC are parallel then 2x+5+3y-20=180
rearrange it to be 2x+3y=180-5+20
2x+3y=195

(iii)2x+3y=195
7x+4y=390
you can use a lot of ways to answer a simultaneous equation but i enjoy the substitution the most
so here is it
y=(195-2x)/3
y=(390-7x)/4
(195-2x)/3=(390-7x)/4
65-2x/3=97.5-7x/4
7x/4-2x/3=97.5-65
x=32.5/(13/12)=(12*32.5)/13=30
y=(195-2*30)/3=45

(iv)use the answer in iii
it should be
65
65
115
115


5)b)ii)both uses the internet=3/5*3/4=9/20
Chaminda uses it=3/5*1/4=3/20
Niluka uses it=3/4*2/5=3/10
9/20+3/20+3/10=9/10


Q 6 no ii is found so i answered b
6)b)Surface area =circumference of biggest semi-circle*height+circumference of smallest semi-circle*height+circumference of semi-circle with diameter 17.5*height +area of cross section (solved in a )
1/2π (17.5+6.5)*35+1/2π(6.5)*35+1/2π (17.5)*35 +329.7=1/2π (17.5+6.5+24)*35+329.7=420π +306.25π +113.75π +329.7=840π +329.7=2968.6 but as it should be 3 s.f. so it is equal 2970


7)b)iii)x=8--4=8+4=12
y=14--4=14+4=18


no 10 v

 

[Mai(M4!)]

No the info i gave is correct, but even i dint get the end part :S

yeah u r right sorry

 

[dubi2pro]

yeah u r right sorry

no its fine.. actually this part is hard!

anyway thanks for helping!

 

[Hello090078601]

Hi, I don't know how this works because I'm new here but I've been stuck on a mathematical question for like the past hour and it's driving me insane.

My Math IGCSE is in three days.
Anywho, this question is from October/November 2013, Paper 43 Question 1b.

The question is:

In a car magazine, 25% of the pages are used for selling second-hand cars, 62.5% of the remaining pages are used for features and the other 36 pages are used for reviews.

Work out the total number of pages in a magazine.

NEED HELP ASAP.

 

[dubi2pro]

this ones!

 

[ZaqZainab]

what bio exam did u have

variant 12

 

[Bilal ashraf]

Hey i need help in this x_X
The ratios of teachers : male students : female students in a school are 2:17:18
The total number of students is 665
find the number of teachers

 

[Mai(M4!)]

no its fine.. actually this part is hard!

anyway thanks for helping!

anytime

 

[Mai(M4!)]

Hi, I don't know how this works because I'm new here but I've been stuck on a mathematical question for like the past hour and it's driving me insane.

My Math IGCSE is in three days.
Anywho, this question is from October/November 2013, Paper 43 Question 1b.

The question is:

In a car magazine, 25% of the pages are used for selling second-hand cars, 62.5% of the remaining pages are used for features and the other 36 pages are used for reviews.

Work out the total number of pages in a magazine.

NEED HELP ASAP.

100%-25%-62.5%=12.5%
12.5% represents 36 pages
so 100% represents (100*36)/12.5=288
288 pages are the total number of the magzine

 

[Mai(M4!)]

this ones!

(a)u need to find the other two angles first
angle BDC=BAC=28
DXC=180-DXA =180-52=128(as they are on a straight line)
sum of interior angles of triangle =180
than 180-128-28=XCD=24

(b)angle QPS=22x/2=11x
as PQRS is a cyclic then sum of opposite angles=180
s0 11x+25x=180
36x=180
x=180/36
x=5