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Mathematics: Post your doubts here!

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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?
 
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11208984_10205037046620490_2093874573_n.jpg "Within 1 standard Deviation of mean" mtlb :/
 
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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
 
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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?
IMG_35841.jpg
 
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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
 
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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 :)
 
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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
 
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9709/43/O/N/11

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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!
 
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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

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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.
 
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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??
 
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