Mathematics: Post your doubts here!

[Chooi1993]

can anyone help tEach me hpw to do Q6?
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w03_qp_7.pdf

Q2(ii) ?
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w07_qp_7.pdf

 

[smzimran]

The first inequality actually makes a vertical line at x = 1 right? I think in your diagram you considered 2 to be the y coordinate by mistake. Also, in your diagram you started the arg angle from the origin. I thought I should start the angle line from ( -2, 1 ) and focus it towards the right till it intersects my vertical x line? That triangle would represent my inequality right?

I've attached a sort of rough sketch. Its quite terrible, I know. Is that how its suppose to look and I assume that I should join the coordinate from u till it touches my vertical line to make my base aswell so it can be a complete triangle which represents the shaded area?

Yes, i misread it, you are right

​

 

[johnsth]

hey could u help me by telling me how to find normal contact force(R) in mechanics...like when a ring is being pulled at some angle from a rod

 

[anonymous123]

Show that sin(-theta) = -sin theta

 

[ffaadyy]

hey could u help me by telling me how to find normal contact force(R) in mechanics...like when a ring is being pulled at some angle from a rod

 

[johnsth]

View attachment 6277

thnx man but still not very clear with the triangle...like what andle does F(N) make with the rod?by the way could u be clear of your location?

 

[johnsth]

Show that sin(-theta) = -sin theta

okk.....could you give me and example?

 

[anonymous123]

okk.....could you give me and example?

imran solved it for me sin(-45) = -√2/2 and - sin (45) = -√2/2

 

[johnsth]

imran solved it for me sin(-45) = -√2/2 and - sin (45) = -√2/2

so where is the normal contact force (R)?

 

[anonymous123]

so where is the normal contact force (R)?

wat force...

 

[ffaadyy]

thnx man but still not very clear with the triangle...like what andle does F(N) make with the rod?by the way could u be clear of your location?

That angle is usually mentioned in the question. Is there any specific mechanics question about this rod-ring thing?

 

[johnsth]

wat force...

i was talking about normal contact force (R) of mechanics..what were u talking about?

 

[johnsth]

That angle is usually mentioned in the question. Is there any specific mechanics question about this rod-ring thing?

yeah like in oct/nov 09 . i.e. 9709 w09 qp 42 question no 5...i did look at the mark scheme and got it...but i need to understand it for other similar questions like in w06 42 question no 2

 

[johnsth]

hey guys could you help me with understanding the sketching the argand diagram like explain what |z-1-i√3|≤ 1 and arg z ≤ pi/3

 

[ffaadyy]

hey guys could you help me with understanding the sketching the argand diagram like explain what |z-1-i√3|≤ 1 and arg z ≤ pi/3

To represent the inequality |z-1-i√3|≤ 1, draw a circle of radius 1 with center at '1 , √3'.
To represent the inequality arg z ≤ pi/3, make an angle of 60° with the x-axis by placing your protractor at (0,0).

This will help you: http://www.xtremepapers.com/community/attachments/sketching-argand-diagrams-pdf.5787/

 

[Chooi1993]

can anyone help tEach me hpw to do Q6?
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w03_qp_7.pdf

Q2(ii) ?
http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_w07_qp_7.pdf

 

[ffaadyy]

yeah like in oct/nov 09 . i.e. 9709 w09 qp 42 question no 5...i did look at the mark scheme and got it...but i need to understand it for other similar questions like in w06 42 question no 2

This's how we'll do Q2, P4 Nov. 06.

Part (i):



We'll be needing to resolve the forces vertically in order to find the magnitude of the frictional force. Note that the vertical component of the 5N force and the frictional force act in the same direction (upwards) while as the weight component acts downwards.

Vertical component of the 5N force + Frictional Force = Weight Component
5 cos 30 + F = 6
F = 6 - 5 cos 30
F = 1.67N

Therefore, Frictional Force = 1.67N.

Part (ii):



To find the coefficient of friction between the ring and the rod, we'll use the formula F=uR. We've already calculated the 'F' in first part of this question and in order to calculate the 'Reaction Force (R)', we'll resolve the forces horizontally.

R = 5 sin 30
R = 2.5

F=uR
1.67 = 2.5u
0.668 = u

Therefore, the coefficient of friction between the ring and the rod is '0.668'.

 

[johnsth]

This's how we'll do Q2, P4 Nov. 06.

Part (i):

View attachment 6287

We'll be needing to resolve the forces vertically in order to find the magnitude of the frictional force. Note that the vertical component of the 5N force and the frictional force act in the same direction (upwards) while as the weight component acts downwards.

Vertical component of the 5N force + Frictional Force = Weight Component
5 cos 30 + F = 6
F = 6 - 5 cos 30
F = 1.67N

Therefore, Frictional Force = 1.67N.

Part (ii):

View attachment 6286

To find the coefficient of friction between the ring and the rod, we'll use the formula F=uR. We've already calculated the 'F' in first part of this question and in order to calculate the 'Reaction Force (R)', we'll resolve the forces horizontally.

R = 5 sin 30
R = 2.5

F=uR
1.67 = 2.5u
0.668 = u

Therefore, the coefficient of friction between the ring and the rod is '0.668'.

thanks a lot man!!really!!!

 

[johnsth]

To represent the inequality |z-1-i√3|≤ 1, draw a circle of radius 1 with center at '1 , √3'.
To represent the inequality arg z ≤ pi/3, make an angle of 60° with the x-axis by placing your protractor at (0,0).

This will help you: http://www.xtremepapers.com/community/attachments/sketching-argand-diagrams-pdf.5787/

dude! u really have helped me

 

[anonymous123]

Q. For the GP 4,6,9.... find the value of n if the nth term is the first term greater than 100.
Help me plzz