Mathematics: Post your doubts here!

[lonelyperson]

assalamualikum warahmatullahi wabarakatu.how r u freinds?
i have doubt in one question if someone can solve it for me.
(a woman stands on the bank of frozen lake with a dog by her side .she skims a bone across the ice at a speed of 3m/s . the bone slows down with deceleration 0.4m/second square ,and the dog chases it with an acceleration 0.6m/second square.how farout from the bank does the dog catch up with the bone?)pls it will be better if u can scan and attach the solution.

thanks in advance

 

[ffaadyy]

assalamualikum warahmatullahi wabarakatu.how r u freinds?
i have doubt in one question if someone can solve it for me.
(a woman stands on the bank of frozen lake with a dog by her side .she skims a bone across the ice at a speed of 3m/s . the bone slows down with deceleration 0.4m/second square ,and the dog chases it with an acceleration 0.6m/second square.how farout from the bank does the dog catch up with the bone?)pls it will be better if u can scan and attach the solution.

thanks in advance

What's the answer to this question?

 

[lonelyperson]

What's the answer to this question?

The answer to this question is 10.8 metres
But how we get this answer i dont know.

 

[abcde]

assalamualikum warahmatullahi wabarakatu.how r u freinds?
i have doubt in one question if someone can solve it for me.
(a woman stands on the bank of frozen lake with a dog by her side .she skims a bone across the ice at a speed of 3m/s . the bone slows down with deceleration 0.4m/second square ,and the dog chases it with an acceleration 0.6m/second square.how farout from the bank does the dog catch up with the bone?)pls it will be better if u can scan and attach the solution.

thanks in advance

W.S!
Dog: a= 0.6 ms^-2 u = 0 ms^-1
Bone: a = - 0.4 ms^-2 u = 3ms^-1
s = ut + 1/2 at^2
Dog's displacement from the bank = 0 + 1/2 (0.6)t^2
Bone's displacement from the bank = 3t + 1/2 (-0.4) t^2
When they catch up, displacements are equal. So:
=> 0 + 1/2 (0.6)t^2 = 3t + 1/2 (-0.4) t^2
=> 0.5t^2 - 3t = 0
=> t^2 - 6t = 0
=> t(t-6) = 0
=> t = 0 or t = 6s
They catch up at t=6s.
=> Displacement = 0 + 1/2 (0.6) (6)^2 = 10.8 m.

 

[rz123]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s07_qp_1.pdf
Q.6 ii how to calculate the point D (any other way apart from doing it by the vector way from ms, if not then explain this vector method..) Thanks a lot!

 

[smzimran]

http://www.xtremepapers.com/CIE/International A And AS Level/9709 - Mathematics/9709_s07_qp_1.pdf
Q.6 ii how to calculate the point D (any other way apart from doing it by the vector way from ms, if not then explain this vector method..) Thanks a lot!

Simple:
Equation of AD and CD may be found using (y-y1) = m(x-x1)
In case of AD, use coordinates of A as (x1,y1) and the gradient of BC as 'm'
Similarly for CD, use coordinates of C as (x1,y1) and the gradient of AB as 'm'
Simultaneously solve the equations of AD and CD to get the point of intersection, D.

 

[Ahmed Ali Akbar]

help me in oct/nov 09 M1 paper 41 question 6 part c..

 

[Unicorn]

In the first part, you'll obtain the answer '64 - 192x + 240x^2'. For the second part, you'll multiply '64 - 192x + 240x^2' with '1 + kx' and equate the coefficients of 'x^2' to 0. Like this:

(1 + kx) (64 - 192x + 240x^2)
64 - 192x + 240x^2 + 64kx - 192kx^2 + 240kx^3

Take the coefficients of 'x^2' and equate them to 0:

240 - 192k = 0
240 = 192k
1.25 = k

Thanks you that was really helpful

 

[lonelyperson]

W.S!
Dog: a= 0.6 ms^-2 u = 0 ms^-1
Bone: a = - 0.4 ms^-2 u = 3ms^-1
s = ut + 1/2 at^2
Dog's displacement from the bank = 0 + 1/2 (0.6)t^2
Bone's displacement from the bank = 3t + 1/2 (-0.4) t^2
When they catch up, displacements are equal. So:
=> 0 + 1/2 (0.6)t^2 = 3t + 1/2 (-0.4) t^2
=> 0.5t^2 - 3t = 0
=> t^2 - 6t = 0
=> t(t-6) = 0
=> t = 0 or t = 6s
They catch up at t=6s.
=> Displacement = 0 + 1/2 (0.6) (6)^2 = 10.8 m.

Thank youvery much for the solution
may Allah give barakah in ur knowledge.
wasalaam.

 

[ffaadyy]

help me in oct/nov 09 M1 paper 41 question 6 part c..

In the second part of the question, we have calculated the height at which the particles are and the speed with which they've moved as soon as the string is cut. Using these details and the fact that both the particles are moving vertically under gravity, we'll find the time taken for each of the 2 particles to hit the ground using the formula
's = ut + (1/2)at^2'.

For P:

3 = 2t + (1/2)(10)(t^2)
3 = 2t + 5t^2
5t^2 + 2t - 3 = 0
5t^2 + 5t - 3t - 3 = 0
(t + 1) ( 5t - 3) = 0
t = -1 and t = 0.6

Since 't' can't be negative, therefore the time taken for particle P to hit the ground is 0.6s.

For Q:

7 = -2t + 5t^2 (we've taken '-2' because this speed was in the upward direction while as now it's moving downwards)
5t^2 - 2t - 7 = 0
5t^2 + 5t - 7t - 7 = 0
(t + 1) (5t - 7)
t = -1 and t = 1.4

Again as 't' can't be '-1', therefore the time taken for particle Q to hit the ground is 1.4s.

Now we'll subtract both the times:

Time taken for particle Q to hit the ground - Time taken for particle P to hit the ground
1.4 - 0.6
0.8s

Hence, 'Q' reaches the ground 0.8s after 'P'.

 

[XPFMember]

Assalamoalaikum wr wb!
I have finished the whole of P3 but there are two chapters which I feel difficult. Vectors and complex numbers. Vectors I have tried much to do it but for complex numbers I have much problem with the argand diagrams. Can anybody please help me with that..
Jazakallah khairen

 

[unique840]

Assalamoalaikum wr wb!
I have finished the whole of P3 but there are two chapters which I feel difficult. Vectors and complex numbers. Vectors I have tried much to do it but for complex numbers I have much problem with the argand diagrams. Can anybody please help me with that..
Jazakallah khairen

post the ques in which u hav probs cox evry ques is of different type

 

[Doctor Nemo]

Assalamoalaikum wr wb!
I have finished the whole of P3 but there are two chapters which I feel difficult. Vectors and complex numbers. Vectors I have tried much to do it but for complex numbers I have much problem with the argand diagrams. Can anybody please help me with that..
Jazakallah khairen

Perhaps this explanation of Argand diagrams might be helpful.

Complex numbers can be represented as a vector in a diagram called an Argand diagram. The x-axis is the real number part and the y-axis is the imaginary part. The length of the vector or line representing the complex number is called the modulus. The modulus of the complex number z = a + b i = |z | = √ (a2 + b2) The argument of a complex number in an Argand diagram is the angle between the line and the x-axis and therefore arg z = tan-1(b/a). A complex number can also be written in terms of the modulus and argument giving z = R (cos ø+ i sin ø) where R is the modulus and øis the argument.

Multiplying a complex number z1 by a second complex number z2 results in the complex number z1z2 whose modulus is the product of the moduluses of z1 and z2 and whose argument is rotated by ø in a counter clockwise direction. The modulus is R1R2 and the arg (z1z2) = arg z1 + arg z2
R1(cos ø 1+ i sin ø 1) X R2(cos ø2+ i sin ø2) = R1R2 (cós(ø1 + ø2)+ i sin(ø1 + ø2))
If a complex number is mulitplied by i then this hás the effect of rotating the complex number by 90º because the argument of i is 90º Division has the opposite effect.

The modulus is R1/ R2 and ø 2 is negative in the above expression so that arg (z1/z2) = arg z1 - arg z2.
General expressions or equations for complex numbers can be written in the terms of the modulus of the complex number, for example |z | = 2. This means that x2 + y2 = 4. This is the equation of a circle of radius 2 with the centre on the origin. Remember that this is a graph with y-axis with imaginary numbers. It would include the complex numbers 2i and -2i.

Since these expressions often result in equations for circles it is useful to remember that in a general equation for a circle
(x-a)2 + (y-b)2 = r2
the circle has a radius of r and is centred on points a, b.

This also means that

|z -2| = 2 This means subtract 2 from the real part and the resultant modulus will equal 2.
This will be a circle centred on 2,0 with radius of 2. (x-2)2 + 2 = 22

|z +2 | = 2 will be a circle centred on -2,0 with radius of 2 (x+2)2 + 2 = 22

|z - i | = 2 will be a circle centred on 0, i with radius of 2 (x)2 + (y-1)2 = 22

 

[XPFMember]

post the ques in which u hav probs cox evry ques is of different type

Assalamoalaikum wr wb!!

ermm....actually there were a few which i found difficult...i know the basics for it...

I'll see if I can find those particular questions....
and btw are we just to make them approximately...or do we do it on a graph paper? using compass and protractor?

 

[XPFMember]

Perhaps this explanation of Argand diagrams might be helpful.

Complex numbers can be represented as a vector in a diagram called an Argand diagram. The x-axis is the real number part and the y-axis is the imaginary part. The length of the vector or line representing the complex number is called the modulus. The modulus of the complex number z = a + b i = |z | = √ (a2 + b2) The argument of a complex number in an Argand diagram is the angle between the line and the x-axis and therefore arg z = tan-1(b/a). A complex number can also be written in terms of the modulus and argument giving z = R (cos ø+ i sin ø) where R is the modulus and øis the argument.

Multiplying a complex number z1 by a second complex number z2 results in the complex number z1z2 whose modulus is the product of the moduluses of z1 and z2 and whose argument is rotated by ø in a counter clockwise direction. The modulus is R1R2 and the arg (z1z2) = arg z1 + arg z2
R1(cos ø 1+ i sin ø 1) X R2(cos ø2+ i sin ø2) = R1R2 (cós(ø1 + ø2)+ i sin(ø1 + ø2))
If a complex number is mulitplied by ithen this hás the effect of rotating the complex number by 90º because the argument of i is 90º Division has the opposite effect.

The modulus is R1/ R2 and ø 2 is negative in the above expression so that arg (z1/z2) = arg z1 - arg z2.
General expressions or equations for complex numbers can be written in the terms of the modulus of the complex number, for example |z | = 2. This means that x2 + y2 = 4. This is the equation of a circle of radius 2 with the centre on the origin. Remember that this is a graph with y-axis with imaginary numbers. It would include the complex numbers 2i and -2i.

Since these expressions often result in equations for circles it is useful to remember that in a general equation for a circle
(x-a)2 + (y-b)2 = r2
the circle has a radius of r and is centred on points a, b.

This also means that

|z -2| = 2 This means subtract 2 from the real part and the resultant modulus will equal 2.
This will be a circle centred on 2,0 with radius of 2. (x-2)2 + 2 = 22

|z +2 | = 2 will be a circle centred on -2,0 with radius of 2 (x+2)2 + 2 = 22

|z - i | = 2 will be a circle centred on 0, i with radius of 2 (x)2 + (y-1)2 = 22

It was helpful...

 

[fahadii92]

Assalam O Alaikum! can someone do Q3 (vector one) june 11/paper 31 plz part 2 is confusing

 

[Silent Hunter]

Asalamoalikum

I wanted to ask a question. It may be simple but i dont get it. Where do we have to use mode?where median and where mean? as averages?

Thank You

 

[unique840]

Assalamoalaikum wr wb!!

ermm....actually there were a few which i found difficult...i know the basics for it...

I'll see if I can find those particular questions....
and btw are we just to make them approximately...or do we do it on a graph paper? using compass and protractor?

walaikum salam
its better if we draw on a graph paper with compass and protractor cox it will give the exact diagram which we hav to use in the succeeding parts of the ques. if there is even a little variation in the diagram, it will give wrong answers.

 

[Mason]

Hello guys! I'm new to the XtremePapers forum as well as to CIE. I was just some self-studying on "representation of data" and I encountered the Histograms topic. Do we have to draw Histograms (the bar chart sort) for the paper (63, Probability and Statistics 1)? Or do we just have to calculate the frequency, height, class width?


-IGNORE THIS IF HISTOGRAMS ARE NOT MEANT TO BE DRAWN IN THE EXAM- If so, for the data of speed (45-60, 60-75, 75-90, 90-105, 105-120, over 120), Class Boundaries - 44.5 </ m < 60.5 , 59.5 </ m < 75.5 etc...
How would be represent the data? With the overlap of class boundaries?

Also, what diagrams will we be meant to draw for the paper? (as in, probability tree, stem-and-leaf diagram etc?)

Thanks!

 

[ezam]

Can someone help me with this 1) solve to correct 3 siginificant figures e^x + e^(2x) = e^(3x)

2) Solve ln(x+2) = 2+Ln X