Qualification > Math
IGCSE MATHS Doubts
Baladya:
--- Quote from: A@di on May 04, 2010, 07:03:01 am ---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.
--- End quote ---
omg.... I dunno what will i do in the IGCSE if i do these mistakes :S
Anyway thx man :D
By the way about the calculators, The best are fx-570 ES(which i use) and 991 ES. They are so amazing u can do anything especially i love that u can put fractions :D xD
Quicki for dummies :D: Make n the subject in y=m^2-4n^2, Why can't i just do this: minus m^2 from both sides so it will go to the left. It will be like this y-m^2=-4n^2. Then divide by -4 both sides and square root. What is wrong in my method?? The right answer is n= SQUARE ROOT of m^2-y/4
Now a long one :P: http://www.freeexampapers.us/IGCSE/Maths/CIE/2008%20Jun/0580_s08_qp_4.pdf
http://www.freeexampapers.us/IGCSE/Maths/CIE/2008%20Jun/0580_s08_ms_4.pdf
Question Number 7. As the Mark scheme says, this is a cyclic quadrilateral. Is there any new rules about this?? Or is it normal and m just missing smth. Plz answer the questions too.
I know m asking a lot but plz help cuz only a week for exam and somewhy i m confused tho i get A* always :S :(
Vin:
whoaaaa @ haris94 comon every person in the world noes tht igcse allows calci !!
the thing there are specific rules abt which calci u use..
commin to calcis of casio...ny model abv fx-100 ms is NOT ALLOWED !!
thts wat i wantd 2 clear..!! :)
Vin:
@baladya
y= m^2 - 4 n^2
y + 4 n^2 = m^2
4 n^2 = m^2 - y
n^2 = m^2 - y (whole)/4
n = root m2 - y (whole) /4
Ghost Of Highbury:
--- Quote from: Baladya on May 04, 2010, 09:56:56 am ---omg.... I dunno what will i do in the IGCSE if i do these mistakes :S
Anyway thx man :D
By the way about the calculators, The best are fx-570 ES(which i use) and 991 ES. They are so amazing u can do anything especially i love that u can put fractions :D xD
Quicki for dummies :D: Make n the subject in y=m^2-4n^2, Why can't i just do this: minus m^2 from both sides so it will go to the left. It will be like this y-m^2=-4n^2. Then divide by -4 both sides and square root. What is wrong in my method?? The right answer is n= SQUARE ROOT of m^2-y/4
Now a long one :P: http://www.freeexampapers.us/IGCSE/Maths/CIE/2008%20Jun/0580_s08_qp_4.pdf
http://www.freeexampapers.us/IGCSE/Maths/CIE/2008%20Jun/0580_s08_ms_4.pdf
Question Number 7. As the Mark scheme says, this is a cyclic quadrilateral. Is there any new rules about this?? Or is it normal and m just missing smth. Plz answer the questions too.
I know m asking a lot but plz help cuz only a week for exam and somewhy i m confused tho i get A* always :S :(
--- End quote ---
Ok thats the right method..
still ill solve it for ya.
y = m2 - 4n2
y - m2 = -4n2
Dividing both by -4
=
=
therefore
n =
long q, next up
Ghost Of Highbury:
The opposite angles of a cyclic quadrilateral add up to 180.
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