Mathematics: Post your doubts here!

[My Name]

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

U MADE IT SEEM SOO EZ how long did it take u to do this!

 

[Layla..]

If you could please help me answer this question?

Mai(M4!)

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_41.pdf
Q10 part V?

 

[Evangeline]

Can someone please answer this , thanks. Question 7

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s07_qp_4.pdf

 

[Ameena Eesa]

View attachment 41882


shw d b part plsss

i got 4,29,46

bt d ms says 4,30,53

Just solve normally..... the fg(x)

(2x+7)^2 + 2x+7 - 3
2x(2x+7) +7(2x+7) +2x+7 - 3
4x^2 + 14x + 14x + 49+ 2x + 4
4x^2 + 30x + 53

compare this with the original equation
4x^2 + 30x + 53 = px^2 + qx + r

in place of p = 4
in place of q= 30
in place of r= 53

 

[Hello090078601]

I'm stuck on this question, from O580/43/O/N/13 (October November 2013, 43, IGCSE)

It's Question 9ci and 9cii

c) The total number of lines in the first n diagram is 1/2n^3 + pn^2 + qn

i) When n=1, show that p+q=8.5


ii) By choosing another value of n and using the equation in part (c)(i), find the values of p and q.

Need help ASAP.

Link to the paper: http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_43.pdf

 

[Ameena Eesa]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w13_qp_41.pdf
Can anyone help me answer Q 10 a part V? Thanks

compare the original equation with this one
in place of n, you have placed (n-8), thus the answer for the nth term will be the thing placed in place of n, and that is (n-8)

 

[shadil.me]

please i have posted link above help me

1 circle=360
DCE is 1/3 of a circle ,so 1/3*360=120
easy beasy

 

[My Name]

can

1 part d) 540(1+6%)^2 = 606.744
They say it was put 2 years ago means compound interest for 2 years has been given thus the formula
Q3 part aiii) Use englargement matrix and multiply it with the coordinates or make x and y lines at (4,5) and mark coordinates of the triangle along this line then simply mutiply coordinates by 1.5

i use tracing paper for enlaregement!

 

[My Name]

Q 6 part c) TSA = curved surface area of the frustum i.e area of the larger cone - area of the smaller cone + pie x 1.5^2 ( area of the base of smaller cone) +pie x 3^2 + 2 x pie x 3 x 12

i dont get it can u do the workings plz leave sum space after one part of working!?

 

[My Name]

The graph can be weird llol, for part f
x(x^2 - x -2) = -1
x^2 -x-2 =-1/x
-1/x = x^2 - x - 2
-1/x - x^2 = -x-2
1/x + x^2 = x+ 2

fii) Draw the graph for y= x + 2

HOW DIDNU GET EQUATION IN F PART I

 

[My Name]

Layla..
question 9

Done with this paper just wait a sec

 

[Layla..]

Layla..
question 9

Ok I'm starting to solve it now

 

[My Name]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_42.pdf
question 4 part HO W DO I SOVLE ANFLS AND CIRCLES ANY TIPS
question 5 part C ii do i always write full answer on line of answer
question 7 part a and biv and c
question 11 part b

Ok I'm starting to solve it now

may Allah swt reward u for ur hard work!

 

[My Name]

Awesome12

 

[Layla..]

For Q 9) a)
6, 9, 12, 15, 18 (Because of adding 3) nth term would be 6+3(n-1) = 6+ 3n -3 so 3+3n
For second one you have the formula so substitute 4 and 5 with n to find their answers
For Number of triangles its the squares, like 2^2, 3^2, so for 4th and 5th term it would be 4^2 and 5^2 and formula would be (1+n)^2 (Simple way is to find the nth term of underoot of these terms i.e 2, 3, 4, 5 and then square the whole term)

 

[Layla..]

In detail
To find number of triangles: it is 2^2, 3^2, 4^2
so the underoots are 2, 3, 4, 5 etc and their nth term is 2+1(n-1) = 1+n
so the nth term would be squared ie (n+1)^2

 

[Layla..]

For part b) Substitute the equation for number of lines = 360
and solve to find n

 

[rida.12.!!]

AS SALAM O ALAIKUM
Hi!!!
SILLY QUESTION :-
HOW DO YOU CALCULATE UPPER AND LOWER BOUNDS???
THANKS IN ADVANCE :d

 

[My Name]

AS SALAM O ALAIKUM
Hi!!!
SILLY QUESTION :-
HOW DO YOU CALCULATE UPPER AND LOWER BOUNDS???
THANKS IN ADVANCE :d

im not good in explainin this
check oter pages sm 1 didi!

 

[Layla..]

Part b Solution
360 x 2/3 = n^2 +3n +2
240= n^2 + 3n +2
so
n^2 + 3n + 2 = 240
n^2 + 3n + 2 -240= 0
n^2 +3n -238 = 0
n^2 +17n -14n - 238 =0
n(n +17)-14(n+17) = 0
(n+17)(n-14)=0
Since n cant be negative so n-14= 0 so n is 14