Mathematics: Post your doubts here!

[Annabelle_xx]

It's really important

 

[hellodjfos;s'ff]

I really need P11 Feb/2018? Who has it or has a link to where I can get it?

There's only one variant for feb/march papers...lol

 

[Mts.99]

how to do part iii

 

[Annabelle_xx]

There's only one variant for feb/march papers...lol

Oh thanks

 

[amina1300]

Anyone tell me the hardest exam papers for S1 and P3.

 

[Nushad Nahue]

M2, s04, Q4 part 1.
According to the figure I can tell that the force in the rod is 0.7m from the hinge and has to act upward for equilibrium.
But mark scheme says ,

Distance of the rod from the hinge is (0.7) 2.4/2.5 or 0.7cos16.26° (=0.672). How did come to be? Please help!

 

[Holmes]

no

yup no.
Quoted this expression mistakenly

 

[Holmes]

Help me out anyone........
Zaki ali asghar

 

[Holmes]

Can anyone tell me how to shade the region?
and how to represent Re z >2
Help me Maths nerds.

Zaki ali asghar

 

[DudeWithoutName]

View attachment 63626
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Help me out anyone........
Zaki ali asghar

Hey! First question, to find the angle between the two planes we can find the angle between the two normals. That is equal to the angle between the two planes.
The first normal is i+j+3k and the second is 2i-2j+k. Using the dot product between these two we have: (1)(2)+(1)(-2)+(3)(1)=sqrt(1^2+1^2+3^2)*sqrt(2^2+2^2+1^2)cos(theta)
theta=72.5

Second question, before i proceed, do you know how use the vector product to find common perpendiculars or do you use another method?

 

[DudeWithoutName]

View attachment 63629
View attachment 63630
View attachment 63631
Can anyone tell me how to shade the region?
and how to represent Re z >2
Help me Maths nerds.

Zaki ali asghar

Modulus of (z-u) being less than 2 is basically a circle with center u of radius 2 and the wanted region is whats inside the circle because its less than.
Re z>2 means real part must be bigger than 2 so the x value of the complex number on the argand diagram must be greater than 2 so draw the line x=2 and the final shaded region should be

The region inside the circle that is to the right of line x=2

 

[Krypton36]

here is paper 3, 4 and 6. feb march 2018

Got the MS for 62 ?

 

[cellular_automata]

Got the MS for 62 ?

 

[amina1300]

https://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_qp_33.pdf
Q4 part ii) Can anyone give me the working? Can the formula be sqrt(2p+3) -p rather than this in ms --> p-sqrt(2p+3) ?

 

[amina1300]

https://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_qp_33.pdf
part ii) Do we only use sinx for the angle between line and plane?

part iii) The point C lies on the line and is such that the distance between C and B is twice the distance between A and B. Find the coordinates of each of the possible positions of the point C.
(can anyone solve this?)

 

[amina1300]

https://papers.gceguide.com/A Levels/Mathematics (9709)/9709_w15_qp_33.pdf
Q9 part (b)

 

[Holmes]

Hey! First question, to find the angle between the two planes we can find the angle between the two normals. That is equal to the angle between the two planes.
The first normal is i+j+3k and the second is 2i-2j+k. Using the dot product between these two we have: (1)(2)+(1)(-2)+(3)(1)=sqrt(1^2+1^2+3^2)*sqrt(2^2+2^2+1^2)cos(theta)
theta=72.5

Second question, before i proceed, do you know how use the vector product to find common perpendiculars or do you use another method?

Common perpendiculars??? finding the intersection line of these two planes(other method)

 

[Holmes]

Modulus of (z-u) being less than 2 is basically a circle with center u of radius 2 and the wanted region is whats inside the circle because its less than.
Re z>2 means real part must be bigger than 2 so the x value of the complex number on the argand diagram must be greater than 2 so draw the line x=2 and the final shaded region should be

The region inside the circle that is to the right of line x=2

Thanks for replying thanks a lot.

 

[amina1300]

How do we draw |z-5| =|z|
where |z|=5 is a circle with radius 5 and center at the origin.
I know that it's going to be a perpendicular bisector of the |z-5| line but is |z-5| a horizontal line along the x-axis with magnitude 5 and the perpendicular bisector at 2.5?

 

[Hunzlah Malik]

How do we draw |z-5| =|z|
where |z|=5 is a circle with radius 5 and center at the origin.
I know that it's going to be a perpendicular bisector of the |z-5| line but is |z-5| a horizontal line along the x-axis with magnitude 5 and the perpendicular bisector at 2.5?

For this you have to plot 2 points
(5,0) and (0,0)
draw perpendicular bisector of these two points
Yes, it is at x=2.5