Author Topic: IGCSE MATHS DOUBTS HERE !!!!  (Read 39760 times)

elemis

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #75 on: October 31, 2010, 01:16:22 pm »
Parallel lines ALWAYS have the same gradient.

Thus, m = 2

Next we find the point at which y = 2x +8 crosses the X AXIS. This is the same as point A as marked on the diagram.

Hence, at this point the y coordinate is zero. Thus we get :

2x +8 = 0

x = -4

Since the distance between A and B is 9 we add 9 to -4 to get the corresponding X INTERCEPT for the other line. In this case the x intercept is (5,0)

We input the coordinate (5,0) into y = 2x +c

We get 0 = 2*5 + c

c = -10

Offline ~aanchal~

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #76 on: October 31, 2010, 01:42:44 pm »
Parallel lines ALWAYS have the same gradient.
thanx alott

another question
m/j 2009 q 17(B)

elemis

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #77 on: October 31, 2010, 01:45:05 pm »
Gimme some time.

elemis

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #78 on: October 31, 2010, 01:49:23 pm »
You are allowed to use tracing paper.

So Place your tracing paper over the page and trace the grid and triangle.

With your tracing paper still on top of the page place your pencil on the point 4,4.

Rotate your TRACING paper 90 degrees to the right i.e. if the original position of your tracing paper was portrait it should now be landscape.

Make a note of the vertices of the triangle as in its new position.

Draw in the triangle.

Offline ~aanchal~

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #79 on: October 31, 2010, 04:14:53 pm »
m/j 2003 p2 q 17
FAST

Offline ~aanchal~

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #80 on: October 31, 2010, 04:23:16 pm »
You are allowed to use tracing paper.

So Place your tracing paper over the page and trace the grid and triangle.

With your tracing paper still on top of the page place your pencil on the point 4,4.

Rotate your TRACING paper 90 degrees to the right i.e. if the original position of your tracing paper was portrait it should now be landscape.

Make a note of the vertices of the triangle as in its new position.

Draw in the triangle.
pls answer

Offline Vin

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #81 on: October 31, 2010, 06:16:51 pm »
m/j 2003 p2 q 17
FAST

Q.17 a) [Attachment]

b) [Attachment]

c)

Matrix of Object * A = Matrix of Image.

|O| * |A| = |I|

(0  2)  * (a   b)   =  (0  6)
(2  2)     (c   d)       (2  6)

You can eliminate the coordinates of the bas as it is common for both - the image as well as the object.

Now, you multiply both the matrix as you normally do. You get:

(0 + 2c  0 + 2d)
(2a+2c  2b+2d)

Therefore,
2c = 0
2a + 2c = 2


2d = 6
2b + 2d = 6


^^Look at the colour combinations. The values corresponding are equal.

Solve the eqns.
2c = 0
c = 0

2a + 2 * 0 = 2
a = 1


2d = 6
d = 3

2b + 2*3 = 6
2b = 0
b = 0


Arrange the values of a b c and d to |A|

So, |A| =
(1  0)
(0  3)

Offline Arthur Bon Zavi

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #82 on: November 04, 2010, 05:05:55 am »
Maths papers are all over !

Continuous efforts matter more than the outcome.
- NU

Offline Galleria

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #83 on: November 05, 2010, 09:41:03 am »
loool how do we get an A* in math ?
like if you get an A in paper 2 and 4 both is that A* ?

Offline Arthur Bon Zavi

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #84 on: November 05, 2010, 03:07:59 pm »
loool how do we get an A* in math ?
like if you get an A in paper 2 and 4 both is that A* ?

Well, what you said is one criteria !

Another criteria is that that also have many other criteria s with which they grade candidates.

You will come to know when you have earned your results. ;)

Continuous efforts matter more than the outcome.
- NU

**RoRo**

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #85 on: December 11, 2010, 12:00:50 pm »
Paper 2, June 2001.

The capacity of a jug is 3.5 litres correct to the nearest 0.1 litre.
The capacity of a glass is 0.25 litres correct to the nearest 0.01 litre.

b) Calculate the greatest number of glasses which you can be sure to fill from a full jug.

Can someone please explain to me how do you arrive to this answer? Thanks!

elemis

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #86 on: December 11, 2010, 12:08:31 pm »
One moment.

elemis

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #87 on: December 11, 2010, 12:11:06 pm »
Paper 2, June 2001.

The capacity of a jug is 3.5 litres correct to the nearest 0.1 litre.
The capacity of a glass is 0.25 litres correct to the nearest 0.01 litre.

b) Calculate the greatest number of glasses which you can be sure to fill from a full jug.

Can someone please explain to me how do you arrive to this answer? Thanks!

We need to find the largest possible capacity of the jug and the SMALLEST possible capacity of the glass.

LARGEST capacity of jug = 3.55

SMALLEST capacity of glass = 0.255

Hence, 3.55/0.255 = 13 glasses (dont round up)

Offline SauD~

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #88 on: January 20, 2011, 08:17:35 am »
I want some help, in indices and graph of functions. I only need the method of solving the indices without calculator and in the graphs i want to know how to find solution. here are some eg.
Indices:-   9^1/2 x 3^5/2                        
               --------------                                                       
               3^2/3 x 3^-1/6                                             
                   
 Graphs of function:- a) Plot a graph of function equation y=2x^2-5x-5 for -2 =<x=<5.
b) Use the graph to solve the 2x^2-5x-5=0.                      
c) Show you method clearly, use the graph to solve the equation 2x^2-3x=10.

part c) with the method is necessary :)

if you need the answer then the answer for indices is 27 and the answer for graph thing of part c) is x= -1.6 and 3.1

x --> multiplication sign   &   x --> X, the unknown value..
Thanks, :)

Offline Amii

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Re: IGCSE MATHS DOUBTS HERE !!!!
« Reply #89 on: January 20, 2011, 09:10:32 am »
Quote
Indices:-   9^1/2 x 3^5/2                        
               --------------                                                      
               3^2/3 x 3^-1/6


9^1/2 x 3^5/2

9^1/2 = Square root of 9 = 3 which is same as 3^1

so 3^1 X 3^5/2

1/1 + 5/2 = 3.5

1x2  +  5x1  = 2 + 5 = 7   = 3.5
----    ----    -----    --
1x2      2x1       2       2

3^3.5 or 3^7/2

3^2/3 x 3^-1/6

2/3 -1/6 = 3/6

2x2 - 1x1   = 4-1 = 3   
----  ----     ----   --
3x2    6x1      6      6

so 3^3/6

(3^7/2 ) / (3^3/6)

(7/2) - (3/6) = 3

7x3 - 3x1 =  21-3  = 18    = 3
---    ---     -----    ---
2x3   6x1       6        6

so 3^3 = 27

Okay saud?  :D
« Last Edit: January 20, 2011, 09:26:40 am by Amii »

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