Mathematics: Post your doubts here!

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lol

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im having trouble with may june 2010 41 question 7ci and may june 2010 42 question n 3d
please explain with full method as im ckueless, thankyou in advance!
 
Kareem Gamal Omar

Kareem Gamal Omar

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WHAT IS THE PERCENTAGE INCREASE FORMULA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
 
C

Cute-princesa

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Guys plz help me
I cant solve mensuration & probability problems
+ can any1 tell me the A marks for nov2011 math & june plzzzz
 
C

Cute-princesa

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The percentage increase = (new value-old value)divided by old value then multiplied by 100
:) hope you get it ^_^

Ex: 200sr is the new 100sr is the old
So (200-100)divide by 100 then X by 100
 
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anooshraja

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May June 2008 question 10/8/20​
May June 2006 question 22 AND 23​
Oct/nov 2006 question 5/13​
Oct/nov 2007 question 21/16​
May June 2009 question 3/6and 8/11 part b​
May June 2011 question 1.3,4 15 part b 16 part a,17 part c,20 part b[varient21]​

plz kindly some one help​
anooshraja said: ↑
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_qp_21.pdf​
 
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anooshraja

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May June 2008 question 10/8/20​
May June 2006 question 22 AND 23​
Oct/nov 2006 question 5/13​
Oct/nov 2007 question 21/16​
May June 2009 question 3/6and 8/11 part b​
May June 2011 question 1.3,4 15 part b 16 part a,17 part c,20 part b[varient21]​

plz kindly some one help​
anooshraja said: ↑
http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_
 
Yours Truly

Yours Truly

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slayer said:
guys please help me with Q 11 and Q 21 ON M/J 2008 http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s08_qp_2.pdf
plzzz!!
Click to expand...

Q11:
(x)^2 + (2x)^2 - (2 * x * 2x * Cos 120) = 21^2
x^2 + 4x^2 - (4x^2 * Cos 120) = 441
5x^2 - ( -2x^2) = 441
7x^2 = 441
x^2 = 63
x = 7. something :p

Q21:
(a)
AC^2 = 800^2 + 600^2
AC = 1000

Tan X = O/A
Tan X = 200/1000
X = Tan inverse 0.2
X = 11.3 degrees :p

(b)
Tan Y = O/A
Tan Y = 800/600
Y = Tan inverse 800/600
Y = 53.13

Ans = 53.13 + 180
bearing = 233.13 degrees
 
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anooshraja

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May June 2008 question 10/8/20​
May June 2006 question 22 AND 23​
Oct/nov 2006 question 5/13​
Oct/nov 2007 question 21/16​
May June 2009 question 3/6and 8/11 part b​
May June 2011 question 1.3,4 15 part b 16 part a,17 part c,20 part b[varient21]​

plz kindly some one help​

anooshraja said: ↑

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_​
 
A

anooshraja

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May June 2008 question 10/8/20​
May June 2006 question 22 AND 23​
Oct/nov 2006 question 5/13​
Oct/nov 2007 question 21/16​
May June 2009 question 3/6and 8/11 part b​
May June 2011 question 1.3,4 15 part b 16 part a,17 part c,20 part b[varient21]​

plz kindly some one help​

anooshraja said: ↑

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_​
 
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