Mathematics: Post your doubts here!

[Ram97]

View attachment 53248 "Within 1 standard Deviation of mean" mtlb :/

It means that the data deviates +1 and -1 of mean
So if the mean is 9 so data would deviate from 8 to 10

 

[asadalam]

It means that the data deviates +1 and -1 of mean
So if the mean is 9 so data would deviate from 8 to 10

Not +1 -1 but +1 -1 of the STANDARD DEVIATION. if mean is 9 and sd is 2 then our range os 7-11 not 8-10

 

[Aina]

Not +1 -1 but +1 -1 of the STANDARD DEVIATION. if mean is 9 and sd is 2 then our range os 7-11 not 8-10

can u plz solve that question part ii only

 

[Turki AbdulAziz]

Q.P: http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_42.pdf
M.S: http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_ms_42.pdf
Can someone explain question 7 (iv)??
I mean i get the -T(A)-2=0.5a but i really cant get the other equation?

 

[BhaiArshad]

Q.P: http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_qp_42.pdf
M.S: http://maxpapers.com/wp-content/uploads/2012/11/9709_s14_ms_42.pdf
Can someone explain question 7 (iv)??
I mean i get the -T(A)-2=0.5a but i really cant get the other equation?

 

[nehaoscar]

How to do this?? Explanation with diagrams would be helpful if possible

 

[TheJDOG]

View attachment 53288

How to do this?? Explanation with diagrams would be helpful if possible

There you go, if you have any more doubts just tell me
https://www.xtremepapers.com/community/attachments/image-jpg.52316/

 

[nehaoscar]

Part (ii) please
TheJDOG

 

[nehaoscar]

There you go, if you have any more doubts just tell me
https://www.xtremepapers.com/community/attachments/image-jpg.52316/

Thanks! but how do you know that the tension in the strings is equal to the weight of A and B respectively?

 

[ZaqZainab]

Hey thanks, I get the working but I want an explanation :/ how do we know that the vertices of the equilateral triangle on the circle on the Argand diagram are the complex numbers z, zw and z/w ? How do we know this?

First get the equations of z*w and z/w in a+bi form and then just plot the "a" on the real axis and "b" on the imaginary axis if you have trouble getting the z*w or the z/w in a a+bi from for a question let me know i might be able to help you out

 

[TheJDOG]

First get the equations of z*w and z/w in a+bi form and then just plot the "a" on the real axis and "b" on the imaginary axis if you have trouble getting the z*w or the z/w in a a+bi from for a question let me know i might be able to help you out

Oh ok! That's amazing, I get it now, thank you

 

[TheJDOG]

View attachment 53289
Part (ii) please
TheJDOG

There you go, part (ii) in the rectangle
https://www.xtremepapers.com/community/attachments/image-jpg.52317/

 

[TheJDOG]

Thanks! but how do you know that the tension in the strings is equal to the weight of A and B respectively?

You mean A and C? It says the whole system is in equilibrium , so for A and C to stay in equilibrium , the tension in the strings they are attached to must be equal to their weights respectively, thus first string has a tension of 5.5 N and the other string has a tension of W N
In other words, just resolve at A and resolve at C

 

[papajohn]

Please anyone solve

 

[Fate0]

9709/43/O/N/11


Q7. iii) The examiner in the marking scheme used the formula: Work done by engine = Work done against resistance + KE gain.
I am wondering why he didnt included work done by driving force since the body does have a driving force causing acceleration.
Thanks!

 

[Ahmed Aqdam]

View attachment 53318
Please anyone solve

W=6N
cos a=5/13 and sin a=12/13
Angle between P and plane is a so component of P perpendicular to the plane will be 12/13 P.
Resolving perpendicular: R=6*5/13 + 12/13 P
F=0.4R Block is on the point of slipping down the plane so Friction will be upwards.
Resolving parallel: 0.4R=6*12/13 - 5/13 P
Solve these equations to get P= 6.12

9709/43/O/N/11

View attachment 53319
Q7. iii) The examiner in the marking scheme used the formula: Work done by engine = Work done against resistance + KE gain.
I am wondering why he didnt included work done by driving force since the body does have a driving force causing acceleration.
Thanks!

Work done by engine is the same as work done by driving force. The driving force is provided by the engine.

 

[papajohn]

W=6N
cos a=5/13 and sin a=12/13
Angle between P and plane is a so component of P perpendicular to the plane will be 12/13 P.
Resolving perpendicular: R=6*5/13 + 12/13 P
F=0.4R Block is on the point of slipping down the plane so Friction will be upwards.
Resolving parallel: 0.4R=6*12/13 - 5/13 P
Solve these equations to get P= 6.12

Work done by engine is the same as work done by driving force. The driving force is provided by the engine.

Please can you draw me the diagram??

 

[papajohn]

Is this correct??

 

[Ahmed Aqdam]

Please can you draw me the diagram??

 

[ashcull14]