Thank you SO MUCH for the important part you played in helping Adam get the A level grades he needed to get to the course of his choice in Newcastle. To anyone contemplating getting Adams support - just do it :-) - having a helpful, knowledgable, friendly, same sort of age chap to ask about topics you need a bit more time on to get clarity - is invaluable. And Adam is reliable - flexible - friendly and supportive. Thanks and best of luck. Brenda - Mum to the Adam :-) - this Adam tutored.

Brenda, Parent from berks

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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

In order to understand this question we must define what modulus-argument form is. The modulus of a complex number is its distance from the origin (0,0) on the Argand Diagram. It is written as |z|.
The argument of a complex number is the angle subtended anticlockwise between the x axis and a line drawn from the origin to the complex number. This is written where the angle is between -π and π radians (where the angle is below the x axis, it is written as a negative because the angle is measured clockwise around the origin). It is written as arg(z).
If you are unaware of what the Argand diagram is, don’t worry, here is a brief explanation!
The Argand diagram is a really useful visual aid for the use of complex numbers where they are split into their real and imaginary components. The real part is represented by a value on the x axis and the imaginary part is represented by a value on the y axis. This allows us to visualise the “size” of complex numbers, in other words the modulus.
The number can be easily shown on the Argand diagram, with the x (real part) value of 1, and the y (imaginary part) value of √3.
To find the modulus of this number which we will now refer to as z, we must effectively find the length of the line drawn from the origin to z. By creating a right angled triangle with the modulus as the hypotenuse, we can see that the other two lengths are 1 and √3.
Pythagoras proved that a^2 +b^2=c^2 which we can use to find the length of the hypotenuse: the modulus. So the |z|^2=(1)^2 +(√3)^2
|z|^2=1+3=4
|z|=2
So we have found the modulus to be 2.
We have defined that arg(z) is the angle between the x axis and the line from the origin to z.
Using trigonometric properties of the right angled triangle drawn we can use the tangent function to find the argument.
We know that tan(x)=Opposite/Adjacent.
The opposite in this case is √3 and the adjacent is 1. Therefore tan(x)=√3/1=√3
Using the inverse tangent function we find that arg(z)=arctan(√3)= π/3 radians.
And there is our final answer, that |z|=2 and arg(z)= π/3.
Extension: What is the modulus-argument form of z= -1-2i
Answer |z|=√5
arg(z)=-2π/3

Answered by Tom M.

Studies Maths at Durham

First we must establish how to differentiate terms individually. This is done by using the simple method of multiplying the X by the power, and subtracting one away from the power.
To make it easier we will differentiate each term individually and then put the equation back together at the end.
1. x^2
2*x^(2-1) =2x
2. 9x
1*9x^(1-1) = 9x^0
=9*1 = 9
3. 8
0*8^(0-1) = 0
Therefore dy/dx = 2x+9
This would be useful if the gradient needed to be found. To find the gradient at a point all you need to do is substitute in the X value.

Start by moving all the terms to one side of the inequality. In this case it's easiest to move the 6x to the left hand side by subtracting 6x from both sides, so that you are left with x^3 + x^2 - 6x > 0. Then factorise the cubic equation so that you get x(x+3)(x-2) > 0. From this form you can see that x=0 ; x= -3 and x= 2 solve the cubic equation, so these are the points, where the graph of y= x^3 + x^2 - 6x crosses the line y=0 (the x axis). Next sketch the cubic graph and you will be able to see clearly, which values solve the inequality. In this case, since x^3 + x^2 - 6x >0 it will be all the parts of the graph above the x axis, which are -3 < x < 0 and x > 2.

Answered by Miron S.

Studies Physics at Edinburgh

first the definition of the rank of a matrix is "maximal number of linearly independent column vectors in the matrix"

then the question could be rephrased to " how many independent column vectors are there".

so what we want to do is actually to find how many independent column vectors this matrix has.

to find the number of independent columns, use Elementary Row Operations (would be demonstrated with an example matrix in detail in real class) to find the rank.

Something further things to note after covering the main thing above:

1. The above method can be used to find the rank of a matrix, be it square or not.

2. To save the calculation, determininant of a square matrix can be checked beforehand, if it is non zero then the rank is its number of columns (rows).

3. If the matrix is a zero matrix, its rank is 0.

Answered by Yilin S.

Studies Mathematics and Statistics at Oxford, St Anne's College

First, let's think of how many different possibilities we have to seat 6 people at a 'normal' table, i.e. in a straight line. Let's call the people A, B, C, D, E, F. We now have 6 seats:

__ __ __ __ __

When we pick who gets to sit on the first seat, we have 6 options to choose from. For the second seat, there are still 5 people left to choose from, 4 for the third, 3 for the fourth, 2 for the fifth and only one person is left over to sit on the last seat. So we have:

6 x 5 x 4 x 3 x 2 x 1 = 6! = 720 possibilities.

Now if instead the table is round, there is 'no first seat'. If everybody rotates through one seat to the left, each person still has the same people to their left and right and we say the configuration is identical. There are 6 different seats for A to seat on that correspond to identical configurations (simply keep rotating the whole group through) - so compared to the 'normal' table, we have overcounted by a factor of 6. The number of possibilities is now:

6! / 6 = 720 / 6 = 120

Answered by Sjoerd B.

Studies Natural Sciences (Physics and Maths) at Durham

(bx+d)(bx-d)=b^2x^2-d^2

(ax-c)(bx+d)(bx-d)=(ax-c)(b^2x^2-d^2)=ab^2x^3-ad^2x-b^2cx^2+cd^2

ab^2=3

b^2c=2

ad^2=147

-cd^2=98

From equations:

a=3/b^2

c=2/b^2

d^2=49b^2

Since a, b, c, d are positive integers, b must be 1. Then a=3, c=2, d=7

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