IGCSE MATHS Doubts

[rahuldevnani93]

Heyy guys,

i needed the formulas used to find the n^th term.
Thanx a lot guys.
if ny files?
Email it to me at thunder.rahul@hotmail.com

[Saladin]

A very good site

If you have any particular questions, please post them here.

And please do not give your e-mails.

[haris94]

Thanks dude........very good site

[The Golden Girl =D]

okay i got this question thati tried so many times to solve but i simply can't get the answer  ..anyway it's nov 02 paper 2 question 12) part  b)..................... plz someone help as soon as u can  i really need to kno the answer 

thx in advance

[Saladin]

Please post the paper.

[Ghost Of Highbury]

IGSCE Math Paper 2 Nov 02, Answer 12)b.

As EF and DC are parallel, ED is the transverse and thus angle EDC is 71 degree.(180-109) (Corresponding angle to DEF)

As ED and BC are parallel , DC is the transverse and thus angle DCB is 109 (180-71) (Corresponding angles rule)

Now, as the polygon has 6 sides, the sum of the angles is (6-2) * 180 = 720 degrees.

They've mentioned that Angle FAB and ABC are equal thus, 720 - 95 - 109 - 71 - 109 = 336 degrees.

336 degrees is the sum of the equal angles FAB and ABC. Therefore , angle FAB = ABC = 336/2 = 168 degrees.

[The Golden Girl =D]

IGSCE Math Paper 2 Nov 02, Answer 12)b.

As EF and DC are parallel, ED is the transverse and thus angle EDC is 71 degree.(180-109) (Corresponding angle to DEF)

As ED and BC are parallel , DC is the transverse and thus angle DCB is 109 (180-71) (Corresponding angles rule)

Now, as the polygon has 6 sides, the sum of the angles is (6-2) * 180 = 720 degrees.

They've mentioned that Angle FAB and ABC are equal thus, 720 - 95 - 109 - 71 - 109 = 336 degrees.

336 degrees is the sum of the equal angles FAB and ABC. Therefore , angle FAB = ABC = 336/2 = 168 degrees.

thx again 

[Ghost Of Highbury]

ur welcome.

[holtadit]

  1.5   0
  0     1

For this matrice isnt the inverse:


                      1  0
                           0 1.5


But the MS says                      1  0
                                                0 1.5

[astarmathsandphysics]

It is the same. multiple each entry in the ms matrrix by 1/1.5 to get your matrix

[Baladya]

Hey guys, i have this really annoying question that i m totally clueless with...
I was solving a question and the latest thing i could get to was: 21^2 = (2x)^2 + x^2 – (2)(2x)(x)(cos120). This is right in the mark scheme. Then what i do is : 441= 3x^2 +2x^2.(because i did cosine 120 times the answer of – (2)(2x)(x) which is -4x^2. And cosine 120 is -0.5... so (-0.5) (-4x^2)= 2x^2) And so its supposed to be 441=3x^2 + 2x^2.
However the asnwer in the mark scheme is 441= 7x^2.
From where the hell did the 7 come from?? plz help

(Answer of mark scheme: M1 212 = (2x)2 + x2 – 2.2x.x.cos120 oe
                                   M1 441 = 7x2                                      )

[The Golden Girl =D]

Hey guys, i have this really annoying question that i m totally clueless with...
I was solving a question and the latest thing i could get to was: 21^2 = (2x)^2 + x^2 – (2)(2x)(x)(cos120). This is right in the mark scheme. Then what i do is : 441= 3x^2 +2x^2.(because i did cosine 120 times the answer of – (2)(2x)(x) which is -4x^2. And cosine 120 is -0.5... so (-0.5) (-4x^2)= 2x^2) And so its supposed to be 441=3x^2 + 2x^2.
However the asnwer in the mark scheme is 441= 7x^2.
From where the hell did the 7 come from?? plz help

(Answer of mark scheme: M1 212 = (2x)2 + x2 – 2.2x.x.cos120 oe
                                   M1 441 = 7x2                                      )

i am confused , u gotta tell  which question it is and where can i find it , so thati can answer

thx

[Ghost Of Highbury]

Hey guys, i have this really annoying question that i m totally clueless with...
I was solving a question and the latest thing i could get to was: 21^2 = (2x)^2 + x^2 – (2)(2x)(x)(cos120). This is right in the mark scheme. Then what i do is : 441= 3x^2 +2x^2.(because i did cosine 120 times the answer of – (2)(2x)(x) which is -4x^2. And cosine 120 is -0.5... so (-0.5) (-4x^2)= 2x^2) And so its supposed to be 441=3x^2 + 2x^2.
However the asnwer in the mark scheme is 441= 7x^2.
From where the hell did the 7 come from?? plz help

(Answer of mark scheme: M1 212 = (2x)2 + x2 – 2.2x.x.cos120 oe
                                   M1 441 = 7x2                                      )

The bold part is correct.

This is what you got incorrect - "21^2 = (2x)^2 + x^2 – (2)(2x)(x)(cos120)."

If you solve the red part u get, (2x)^2 + x^2 = 4x^2 + x^2 = 5x^2. (Notice : it is (2x) the whole squared not 2x^2)

If u carry on, :- 5x^2 + 2x^2 = 7x^2.

Silly mistake, happens. No problem.

[Vin]

hey guys jus wanted to ask what kind of calculators are allowed in the examination??
i use a Texas TI-30XB ..

[Ghost Of Highbury]

We used. Fx - 991 Ms.