Maths, Addmaths and Statistics: Post your doubts here!

[A star]

Simply draw the graph of the original equation. Now you will see the curve has a minimum point, because the coefficient of x^2 is positive. This minimum point (turning point) if exists below x-axis, ie. in -ve y-value, reflect that part above the x-axis, taking y=o as the reference line for reflection.
This is what modulus does. It always changes the negative y-values to +ve.

yup thts th rule used in sin cos and tan curves. i could not understand what he meant.

 

[HarisLatif]

u do it thn i see it is either 270 clockwise or 90 anti clock wise u try it

Need help in Q16 a Parthttp://www.xtremepapers.com/papers/...Calculator Version) (4024)/4024_s10_qp_11.pdf

 

[asd]

u do it thn i see it is either 270 clockwise or 90 anti clock wise u try it

I did like 3 times, and the result is that its clock wise.
For anti clockwise it should be:
(0 -1)
(1 0)
Try doing this one with a point (2,1) and let me know your results.

 

[A star]

Need help in Q16 a Parthttp://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s10_qp_11.pdf

well one is 7 being highest and 8 being median of three figures mean it was a single value chosen so now put it in mean (7+x+8+x)/3=9 and find the third

 

[Haris Bin Zahid]

Simply draw the graph of the original equation. Now you will see the curve has a minimum point, because the coefficient of x^2 is positive. This minimum point (turning point) if exists below x-axis, ie. in -ve y-value, reflect that part above the x-axis, taking y=o as the reference line for reflection.
This is what modulus does. It always changes the negative y-values to +ve.

O man....thnnnnkkkkkksssssss a lot! Knew u could do it!
Now I got it. Hope u get an A* In Add. Maths.

 

[A star]

I did like 3 times, and the result is that its clock wise.
For anti clockwise it should be:
(0 -1)
(1 0)
Try doing this one with a point (2,1) and let me know your results.

wait by 0 -1 you means (0 top right -1 bottom right and 1 top left and 0 bottom left of the matrix) then u r rite

 

[asd]

wait by 0 -1 you means (0 top right -1 bottom right and 1 top left and 0 bottom left of the matrix) then u r rite

Are you confused between the 'left and right' ?
Read your post again

 

[asd]

wait by 0 -1 you means (0 top right -1 bottom right and 1 top left and 0 bottom left of the matrix) then u r rite

And yes, i meant that. And i understand what you meant to say there (the left - right thing)

 

[syed1995]

Simply draw the graph of the original equation. Now you will see the curve has a minimum point, because the coefficient of x^2 is positive. This minimum point (turning point) if exists below x-axis, ie. in -ve y-value, reflect that part above the x-axis, taking y=o as the reference line for reflection.
This is what modulus does. It always changes the negative y-values to +ve.

haha that's exactly what i thought would be done! .. plot the graph .. reflect all the -ve points in the x-axis (y=0) to make them all positive..

 

[asd]

O man....thnnnnkkkkkksssssss a lot!
Now I got it. Hope u get an A* In Add. Maths.

Inshallah, although my P1 was worst. But i guess my p2 (which alhumdulillah went awesome) can compensate the marks lost in p1 for A*. I HOPE.

 

[A star]

And yes, i meant that. And i understand what you meant to say there (the left - right thing)

yup it ws a misunderstanding thats y i thought how could some one do it three times the same way

 

[A star]

Inshallah, although my P1 was worst. But i guess my p2 (which alhumdulillah went awesome) can compensate the marks lost in p1 for A*. I HOPE.

same here

 

[A star]

HarisLatif got it?

 

[asd]

Oh and if someone is confused between the 2 rotation matrices of 90* clockwise and anti clockwise:
Just remember that clockwise is:
(o 1)
(-1 0)
Write that down an the page and turn your page all the way about 180*. The matrix you will see will be for anti-clockwise.

 

[Haris Bin Zahid]

Simply draw the graph of the original equation. Now you will see the curve has a minimum point, because the coefficient of x^2 is positive. This minimum point (turning point) if exists below x-axis, ie. in -ve y-value, reflect that part above the x-axis, taking y=o as the reference line for reflection.
This is what modulus does. It always changes the negative y-values to +ve.

See page 31 in this http://www.pearsonschoolsandfecolle...ion/Samples/SampleMaterial/Chap02 023-041.pdf
You mean the same....right?

 

[HarisLatif]

And Q 15 PART C!http://www.xtremepapers.com/papers/...(Calculator Version) (4024)/4024_s09_qp_1.pdf

 

[a.abid]

Umm, yes.
For shear along x-axis it is :
(1 k)
(0 1)
And for shear along y-axis it is:
(1 0)
(k 1)
For stretch, parallel to x-axis (y-axis as invariant) it is:
(k 0)
(0 1)
And for stretch parallel to y-axis(x-axis as invariant) it is:
(1 o)
(o k)
For double stretch:
(k1 0)
(o k2)
where k1 is the stretch factor for stretch parallel to x-axis, and k2 for stretch parallel to y-axis
For rotation, i already posted the shortcut methods.

yeah these r the things I hate about transformations they r so difficult to learn Anyways thanx...

 

[asd]

yeah these r the things I hate about transformations they r so difficult to learn Anyways thanx...

It all needs sick practice!

 

[Haris Bin Zahid]

And Q 15 PART C!http://www.xtremepapers.com/papers/CIE/Cambridge International O Level/Mathematics D (Calculator Version) (4024)/4024_s09_qp_1.pdf

highest time taken-lowest time taken

 

[a.abid]

Isn't this for anti-clockwise :O

for 90degrees anticlockwise it is:
(0 -1)
(1 0)