Mathematics: Post your doubts here!

[Stefan Salvatore]

Take the co-ordinates of both the object and image and place them into two matrices, and let x be the required matrix which will take the form
(a b)
(c d)
Solve the matrices to get the required matrix Hope it helps.
Cheers

 

[Arbaz Ahmed]

Can someone attach formula sheet for Additional Mathematics..

 

[QatarOwnz]

Take the co-ordinates of both the object and image and place them into two matrices, and let x be the required matrix which will take the form
(a b)
(c d)
Solve the matrices to get the required matrix Hope it helps.
Cheers

Can you do one example please, so I can fully get it and able to do the rest .

 

[Stefan Salvatore]

yeah sure, tell me which part you wanna know?

 

[QatarOwnz]

yeah sure, tell me which part you wanna know?

just do 3a "B To C"
Thanks a lot

 

[Stefan Salvatore]

Triangle B Co-ordinates:
B= {2 5 5}
{1 1 3}
Triangle C Co-ordinates:
C= { 1 1 3}
{-2 -5 -5}

Let the required transformation matrix be:
T= {a b}
{c d}

Now lets apply the transformation:

TA = C
{a b} * {2 5 5} = { 1 1 3}
{c d} {1 1 3} {-2 -5 -5}

{a*2+b*1 a*5+b*1 a*5+b*3} = { 1 1 3}
{c*2+d*1 c*5+d*1 c*5+d*3} {-2 -5 -5}

{2a+b 5a+b 5a+3b} = { 1 1 3}
{2c+d 5c+d 5c+3d} {-2 -5 -5}

Now solve first two simultaneous equations:

2a+b =1 (1)
5a+b =1 (2)
- - -
_________
-3a=0
a=0

Substitute the value of a in equation (1)
2a+b=1
2(o)+b=1
b = 1

Now solve two other simultaneous equations:

2c+d = -2 (3)
5c+d = -5 (4)
- - +
_________
-3c=3
c=-1

Substitute the value of c in equation (3)

2c+d = -2
2(-1)+d = -2
-2+d= -2
d = 0

So the required transformation matrix is:
{a b}
{c d}

= {0 1}
{-1 0}

 

[QatarOwnz]

Triangle B Co-ordinates:
B= {2 5 5}
{1 1 3}
Triangle C Co-ordinates:
C= { 1 1 3}
{-2 -5 -5}

Let the required transformation matrix be:
T= {a b}
{c d}

Now let apply transformation:

TA = C
{a b} * {2 5 5} = { 1 1 3}
{c d} {1 1 3} {-2 -5 -5}

{a*2+b*1 a*5+b*1 a*5+b*3} = { 1 1 3}
{c*2+d*1 c*5+d*1 c*5+d*3} {-2 -5 -5}

{2a+b 5a+b 5a+3b} = { 1 1 3}
{2c+d 5c+d 5c+3d} {-2 -5 -5}

Now solve first two simultaneous equations:

2a+b =1 (1)
5a+b =1 (2)
- - -
_________
-3a=0
a=0

Substitute the value of a in equation (1)
2a+b=1
2(o)+b=1
b = 1

Now solve two other simultaneous equations:

2c+d = -2 (3)
5c+d = -5 (4)
- - +
_________
-3c=3
c=-1

Substitute the value of c in equation (3)

2c+d = -2
2(-1)+d = -2
-2+d= -2
d = 0

So the required transformation matrix is:
{a b}
{c d}

= {0 1}
{-1 0}

Thanks a lot for your time

 

[Stefan Salvatore]

I don't know whats happening but the brackets format starts to change when I paste the solution over here so i am attaching the solution file, download and check from there.

 

[ahmed faraz]

I am really stuck in matrics, I don't know how to find the position vectors like [1 0 0 -1], is there a easy way?

if we look at B to C, the answer is [0 -1, 1 0] but why?

Special thanks to mr.Stefan Salvatore for clearing up the doubts Explicitly..


And Yeah one thing more which book is this..?

 

[Stefan Salvatore]

Cambridge IGCSE Mathematics Core and Extended Practice Book (Cambridge Igcse Practice Book)
by Ric Pimentel (Author), Terry Wall (Author)

 

[ahmed faraz]

Cambridge IGCSE Mathematics Core and Extended Practice Book (Cambridge Igcse Practice Book)
by Ric Pimentel (Author), Terry Wall (Author)

Thanks

 

[acecie]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s15_qp_22.pdf

qs:21 (a)
can someone please explain how to do this

TIA

 

[Stefan Salvatore]

f(x) = x^2 + 4x - 6 (1)

The question gives you the hint that the function of x can be converted into algebraic identity form by completing the square method and then by comparison you can deduce the values
of m and n.

By looking closely you can easily see that it can be converted into algebraic identity (a + b)^2

Take the co-efficient of x i.e 4 If you remember a(x)^2 + bx +c​

f(x) = x^2 + 4x - 6 Divide it by 2 ----> 2
= x^2 + 4x - 6 Square The Number ----> 4
= x^2 + 4x - 6 + 4 - 4 Then add and subtract it at the same time to make the equation balance
= x^2 + 4x + 4 - 6 - 4
= (x)^2 + (2 * x * 2 ) + (2)^2 - 6 - 4 Rearranging the equation in the form (a)^2 + (2*a*b) + (b)^2 gives:
= (x + 2)^2 - 6 - 4
= (x + 2)^2 - 10 (2)

Finally by comparison of 1 from 2 the values of m and n can be deduced:
m = 2 and n = -10

Cheers

 

[Danish Sardar]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s15_qp_42.pdf Q3 plz help

 

[Stefan Salvatore]

The Rate Of Fuel for (x+20) kilometers is given by the same formula just replacing x with x + 20 as the distance has changed:
(i) This gives Rate Of Fuel = 600/(x+20)

(ii) In part two as clearly stated that the rate of fuel has been decreased by 1.5 liters per kilometer

This means in the second part of journey the rate of fuel was 1.5 less than the rate of fuel for first part of journey. This helps us in constructing the equation:

This means:
(600/x) - (600/x+20) = 1.5
Multiply each term by x(x+20)
600x + 12000 - 600x = 1.5 (x^2 + 20x)
12000 = 1.5(x)^2 +30x
1.5(x)^2 + 30x - 12000 = 0
Divide each term by 1.5
(x)^2 + 20x - 8000 = 0
Hence proved.

(iii) Now what you gotta do is solve the equation by using quadratic formula.
(x)^2 + 20x - 8000 = 0 a = 1 , b = 20 , c = -8000

x = -20 (+ or -) √ (-20)^2 - 4 * (1) * (-8000) / 2

This gives the following roots: 80 , -100

(iv) As we need to find the rate of fuel used by Alan’s car for the complete journey we need to add both the distances for each part of journey:

Also before computing the distances we need to choose one value of x. As the value of x (i.e the distance) can never be negative so we choose the positive root (i.e 80) and substitute it into each of the distance equations giving:

First Distance = x
First Distance = 80

Second Distance = x+ 20
Second Distance = 100

Total Distance = 80 + 100 = 180

So rate of fuel for whole journey is calculated by dividing 600 with the total distance giving answer in liters per 100 kilometer :

Rate Of Fuel For Complete Journey = 600/Total Distance
= 600/180
= 10/3
=3.33 liters per 100 km

Cant add the radical sign in the quadratic formula part (iii). Hope you get it.

 

[Danish Sardar]

The Rate Of Fuel for (x+20) kilometers is given by the same formula just replacing x with x + 20 as the distance has changed:
(i) This gives Rate Of Fuel = 600/(x+20)

(ii) In part two as clearly stated that the rate of fuel has been decreased by 1.5 liters per kilometer

This means in the second part of journey the rate of fuel was 1.5 less than the rate of fuel for first part of journey. This helps us in constructing the equation:

This means:
(600/x) - (600/x+20) = 1.5
Multiply each term by x(x+20)
600x + 12000 - 600x = 1.5 (x^2 + 20x)
12000 = 1.5(x)^2 +30x
1.5(x)^2 + 30x - 12000 = 0
Divide each term by 1.5
(x)^2 + 20x - 8000 = 0
Hence proved.

(iii) Now what you gotta do is solve the equation by using quadratic formula.
(x)^2 + 20x - 8000 = 0 a = 1 , b = 20 , c = -8000

x = -20 (+ or -) √ (-20)^2 - 4 * (1) * (-8000) / 2

This gives the following roots: 80 , -100

(iv) As we need to find the rate of fuel used by Alan’s car for the complete journey we need to add both the distances for each part of journey:

Also before computing the distances we need to choose one value of x. As the value of x (i.e the distance) can never be negative so we choose the positive root (i.e 80) and substitute it into each of the distance equations giving:

First Distance = x
First Distance = 80

Second Distance = x+ 20
Second Distance = 100

Total Distance = 80 + 100 = 180

So rate of fuel for whole journey is calculated by dividing 600 with the total distance giving answer in liters per 100 kilometer :

Rate Of Fuel For Complete Journey = 600/Total Distance
= 600/180
= 10/3
=3.33 liters per 100 km

Cant add the radical sign in the quadratic formula part (iii). Hope you get it.

THNX MAN BUT a(ii) is still very confusing btw thnx for ur time

 

[Anum96]

Can someone attach formula sheet for Additional Mathematics..

http://www.mei.org.uk/files/pdf/6993 Formulae Sheet.pdf

 

[Stefan Salvatore]

You should brush up your facts about rate and ratio. Then, I hope you will get everything.

 

[Danish Sardar]

Question: 25% of a magazine is used to sell 2nd hand cars . 62.5% of the remaining pages were used to show other things and 36 pages of them were used for advertisement ..HOW MANY PAGES ARE THERE TOTALLY??????? TIP : ONLY 17% IN THIS WHOLE WORLD GAVE THE ANSWER WRITE AS MY TEACHER SAID .IF U WANT TO BE THE 18 TH % TTTRRRRYYYYY THHHIISSSSS

 

[Danish Sardar]

Question: 25% of a magazine is used to sell 2nd hand cars . 62.5% of the remaining pages were used to show other things and 36 pages of them were used for advertisement ..HOW MANY PAGES ARE THERE TOTALLY??????? TIP : ONLY 17% IN THIS WHOLE WORLD GAVE THE ANSWER WRITE AS MY TEACHER SAID .IF U WANT TO BE THE 18 TH % TTTRRRRYYYYY THHHIISSSSS

OTHER good tip

Answer is 128 find it how