Mathematics: Post your doubts here!

[Saad the Paki]

M/J 2013 Paper 42 Question 4(d) explain please.

To find the angle of the major arc we'll do 360-(39×2)
Then the formula for sector as u might know is => (X/360)*2*pi*r
Therefore => (360-(39×2))/360*2*pi*8.5

 

[NOneed2speedd]

M/J 2013 Paper 42 Question 4(d) explain please.

Hope u understand

 

[Lana19984]

Guys can anyone of you please tell me how to find nth term of complicated sequences? Some steps please?
For example:
Sequence 1: 2,6,12,20,30
Sequence 2: 1,3,15,61,213

Please tell me some methods guys, I have an exam tomorrow and I don't want to lose marks in this question.

sequence 2 i am pretty sure its coming from a part of question sometimes sequences are solved upon previous sequences

 

[loaie Amgad]

guys a have a small query but Please I need prompt reply if any one pleases
While drawing a tangent to the curve , should we draw the tangent and extend it until it meets both ends of the grid , I mean whether to fill up the grid or no???
and can we use our end points of the tangent to find the slope????
Thanks in advance........

 

[masterex567]

m/j 14 QP 43 question 7! please

 

[AnonymousX9]

Hope u understand

Why is it 4x39????

 

[masterex567]

Why is it 4x39????

since you have to create a line joining a to c to make a major arc between them
they are asking for MAJOR arc AC, which spans from A to C CLOCKWISE. Now if MOB is 39, OBN is also 39. Also AOM is 39 and ONC is 39. So angle AOC will be 360 - 4 * 39. Now calculate arc length using (360-4*39) as the angle.

 

[NOneed2speedd]

What i m doing wrong? Why my answr is wrong?
answr is 1 0
0 -1

 

[masterex567]

What i m doing wrong? Why my answr is wrong?
answr is 1 0
0 -1

dont use that long complicated method, here's a simple one.
First of all the identity matrix is
1 0
0 1
This also represents co-ordinates on a graph as (1,0) (0,1).
Now U onto V is reflection in x axis. If we reflect these points (1,0) and (0,1) by x axis, what will be the new points? (1,0) stays (1,0) and (0,1) reflects to (0, -1).
Now write the new points (1,0) and (0,-1) as a matrix:
1 0
0 -1
Tada!

 

[AnonymousX9]

since you have to create a line joining a to c to make a major arc between them
they are asking for MAJOR arc AC, which spans from A to C CLOCKWISE. Now if MOB is 39, OBN is also 39. Also AOM is 39 and ONC is 39. So angle AOC will be 360 - 4 * 39. Now calculate arc length using (360-4*39) as the angle.

How is AOM and ONC 39? Explain pls..

 

[masterex567]

How is AOM and ONC 39? Explain pls..

they have mentioned that OM is perpendicular to AB, which means that angle AOB is indeed bisected into two equally by line OM into angle AOM and angle MOB.
Hence if MOB is 39, then also AOM is 39, since AOB is bisected equally into these two angles.
same case for onc

 

[AnonymousX9]

OMG thank you so much!!!! BTW, can you tell me how to find center of rotation? :/ Is there a method other than trial and error? And how do we find out if it is 90 or 180 or any other angle?

 

[Saad the Paki]

m/j 14 QP 43 question 7! please

p is simply 180-32=148°
now for q... angle FYA is equal to CBX (32°) becuz of corresponding angles formed by the parallel EY and CB
Angke YFA is 90.. therefore to find angle YAF we'll do 32+90+y=180
y will come out to be 58.. q can then be found by 180-58=122°
t is equal to p cuz of corresponding angles formed by lines ED and AB
x can den be found by adding all angles in the polygon which will be equal to 180*(n-2)

 

[Saad the Paki]

Something like this...

 

[masterex567]

OMG thank you so much!!!! BTW, can you tell me how to find center of rotation? :/ Is there a method other than trial and error? And how do we find out if it is 90 or 180 or any other angle?

there's the long way of joining two corresponding points, then getting perpendicular bisectors of both, and where the two bisectors meet is centre of rotation. then you can measure the angle between centre and image point to centre and object point.
Best way is trial and error with a tracing paper, saves alot of time just use points that you estimate could be the centre,

 

[masterex567]

p is simply 180-32=148°
now for q... angle FYA is equal to CBX (32°) becuz of corresponding angles formed by the parallel EY and CB
Angke YFA is 90.. therefore to find angle YAF we'll do 32+90+y=180
y will come out to be 58.. q can then be found by 180-58=122°
t is equal to p cuz of corresponding angles formed by lines ED and AB
x can den be found by adding all angles in the polygon which will be equal to 180*(n-2)

Thank you so much bro!
One more 10 a in same paper please

 

[NOneed2speedd]

dont use that long complicated method, here's a simple one.
First of all the identity matrix is
1 0
0 1
This also represents co-ordinates on a graph as (1,0) (0,1).
Now U onto V is reflection in x axis. If we reflect these points (1,0) and (0,1) by x axis, what will be the new points? (1,0) stays (1,0) and (0,1) reflects to (0, -1).
Now write the new points (1,0) and (0,-1) as a matrix:
1 0
0 -1
Tada!

So Whenever there is reflection answer will be 1 0 ?
0 -1

 

[masterex567]

So Whenever there is reflection answer will be 1 0 ?
0 -1

reflection in x axis yes that will always be the matrix representing that transformation.
however for refletion in y axis and etc there are different matrices.

 

[AnonymousX9]

M/J 2013 Paper 43, question 2, which region do we shade? The entire sector 6 cm from R?

 

[Kiara P.]

Usually for Enlargement/Reflection questions it get to be a shape on the graph as a rectangle, hexagon..etc. whenever we enlarge an image we usually multiply the vertices of the shape by the enlargement Factor or matrix to get the new set of prime points forming a shape.

OKAY what if this time in the external the shape was a circle...
and the question states to enlarge it using a certain factor or matrix for instance by then we really can't just multiply vertices......because it's rounded what shall we do in this case??

Thank you help is appreciated.