Mathematics: Post your doubts here!

[Phoenix Blood]

Need help with mensuration if anyone has notes on that topic or anything that will help my cause please post.

You basically need to know the formulae. Then substitute all the way
https://www.xtremepapers.com/community/attachments/mathematics-0580-formula-sheet-2011-pdf.2387/

 

[Ganey786]

Thanks a lot for the formula sheet, i'm sure it will help me in my final exam in a few months time.

 

[karimalaaa]

I had prob with surname in statement of entry then I changed it but bio paper 1 was printed with the old one elsamdoni instead of el samadoni. I told the supervisor and the invigilators and they told me no problem as centre and candidate numbers and signature are written correctly but iam really unsure about what they told me. Do you think there is a problem?

 

[jseaklfjskldfj]

hey a bit of a random question but I would seriously seriously appreciate an answer :/ So for anyone who did IGCSE Additional Math (CIE) papers 12 and 22, how did you find them?, I thought they were a bit harder than last years paper. What do you think the grade boundaries for IGCSE Additional math (CIE) papers 12 and 22 will be this year? Considering last years was 151 and the grade boundaries every year seem to hover around 150.

 

[Phoenix Blood]

madcap_97

 

[jseaklfjskldfj]

Hi, to get a "TOP OF THE WORLD" award for IGCSE in mathematics, you require to get 100% in the exam, is this correct?

 

[Phoenix Blood]

Hi, to get a "TOP OF THE WORLD" award for IGCSE in mathematics, you require to get 100% in the exam, is this correct?

Pretty much. You just have to get the heighest, be it 100% or 99% or 98%

 

[Phoenix Blood]

Hi, to get a "TOP OF THE WORLD" award for IGCSE in mathematics, you require to get 100% in the exam, is this correct?

And beat others with a gap of much % difference b/w you and the other guys.

 

[jseaklfjskldfj]

so basically, it will have to be 100, you can't get the award with say 99 (this is the UMS mark cie gives)

 

[masterex567]

Can someone please explain how no. 4 c) is done in this question paper: s13_qp_43

Thanks

 

[ahmadumar]

Can someone please explain how no. 4 c) is done in this question paper: s13_qp_43

Thanks

You need to find the angle <FJE. Find the length EJ. Now you now EJ, FJ and FE. All these 3 together make a right angled triangle. The Right-angled is at <FEJ . Now u got all the lengths and the Right angle. now you use the Sine Rule to get the angle ^_^ FJ/sin(<FEJ) = FE/sin(<FJE)
hooraaayyy!

 

[masterex567]

Thanks!

You need to find the angle <FJE. Find the length EJ. Now you now EJ, FJ and FE. All these 3 together make a right angled triangle. The Right-angled is at <FEJ . Now u got all the lengths and the Right angle. now you use the Sine Rule to get the angle ^_^ FJ/sin(<FEJ) = FE/sin(<FJE)
hooraaayyy!

Can you help me out here as well?
No. 7 on w13_qp_23

 

[Atharva Joshi]

Please solve this question for me (Pythagoras Theorem)

12^2 + x^2 = (x+8)^2

(x+8) is the hypotenuse
The answer given is 5 cm

 

[Phoenix Blood]

Please solve this question for me (Pythagoras Theorem)

12^2 + x^2 = (x+8)^2

(x+8) is the hypotenuse
The answer given is 5 cm

12^2 + x^2 = (x+8)^2
12^2 + x^2 = x(x+8)+8(x+8) ...... opening the brackets and finding the products
12^2 + x^2 = x^2+8x+8x+64

x^2+8x+8x+64 = 12^2 + x^2
x^2- x^2+ 8x+8x+ 64- 12^2 = 0
16x-80=0
16x=80
x=5

 

[mohamed kamal]

In paper 23 November 2013 for math (CIE) - Question no. 16 a
Area in triangle in mark scheme is multiplied by pie !
It should be = 0.5 x 3 x 3 x sin 120 ... only !

is that wrong in mark scheme or what

Kindly any one can clarify

 

[Phoenix Blood]

In paper 23 November 2013 for math (CIE) - Question no. 16 a
Area in triangle in mark scheme is multiplied by pie !
It should be = 0.5 x 3 x 3 x sin 120 ... only !

is that wrong in mark scheme or what

Kindly any one can clarify

Solve it by your own method, its correct.

Area of triangle= 0.5 x 3 x 3 x sin 120
Area of triangle= 3.897

Area of sector = Ѳ/360 * π* r^2
=30/360* π* 3^2
= 2.356

Area of both sectors = 2.356* 2
= 4.712

Area of shaded region = area of both sectors + area of triangle
= 4.712+3.897
=8.609
=8.61, as in the mark scheme.

 

[Ionic Blade]

My session has just started , My question is :
About rounding to 3 significant integers for any answer.
I will give some numbers as an example , how should I type it in the solution?
64821.2675 = ?
21.5631
1.5872
312.9859

 

[Phoenix Blood]

My session has just started , My question is :
About rounding to 3 significant integers for any answer.
I will give some numbers as an example , how should I type it in the solution?
64821.2675 = ?
21.5631
1.5872
312.9859

All of these numbers are integers, so you'll be rounding the third integer-
64821.2675
Thus, you will round off the 8 in the above solution.
= 648oo.000
In other words, it is like rounding off to the nearest tens, if to two significant integers, to the nearest hundreds, if stated to round off to three significant figures and so on.

They may also ask you to round off to significant decimals.

 

[M.Omar]

All of these numbers are integers, so you'll be rounding the third integer-
64821.2675
Thus, you will round off the 8 in the above solution.
= 648oo.000
In other words, it is like rounding off to the nearest tens, if to two significant integers, to the nearest hundreds, if stated to round off to three significant figures and so on.

They may also ask you to round off to significant decimals.

wont it simply be 64800 not 64800.000 ?

 

[Phoenix Blood]

wont it simply be 64800 not 64800.000 ?

Same thing. I was just showing Ionic Blade how all the integers after 8, after being rounded off, turn into zeroes.