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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.

To show something is a subgroup we need to show that it satisfies the group axioms. Therefore we need to show that if g and h are in Z(G) then gh is in Z(G), g^-1 is in Z(G), the identity e is in Z(G). As eg = g = ge for all elements g in G we can see e is in Z(G).
Then suppose we have g and h in Z(G). Then for all elements j in G we have ghj = gjh as h is in Z(G) = jgh as g is in Z(G). Therefore Z(G) is closed under the group operation. Also we have g^-1 j = g^-1 j e as e is the identity = g^-1 j g g^-1 by definition of inverses = g^-1 g j g^-1 as g is in Z(G) = e j g^-1 = j g^-1 and so g^-1 is in Z(G) and so Z(G) is closed under inverses and is therefore a subgroup of G

Answered by Alex R.

Studies Mathematics and Computer Science at Oxford, St John's College

Polar coordinates are expressed in the form (r,Î¸), where r is the distance of the point, P, from the origin, and Î¸ (usually expressed in radians) is the angle between the line joining the point to the origin, and the positive x-axis (moving anti-clockwise from the x-axis).
r can be seen as the hypotenuse of a right-angled triangle, where the base of the triangle has a length equal to the value of the x-coordinate of P, and the height of the triangle a length equal to the value of the y-coordinate of P. Therefore r can be calculated using Pythagoras' Theorem (r=(x^2+y^2)^1/2). Î¸ can also be calculated using this right-angled triangle, since tanÎ¸ will be equal to y/x for a point in the top right quadrant (for a point in the top-left quadrant or bottom-left quadrant this value will need to be subtracted from pi, and in the case of the bottom-left quadrant made negative since the smallest angle will be going clockwise from the x-axis, and for a point in the bottom-right quadrant this value of Î¸ will also be also made negative).
(Diagrams would be used throughout explanation)

Answered by Gwen W.

Studies Chemistry and Physics at St. Andrews

We first find the complementary function by guessing y=e^(kx).
Substituting this into the equation d^2y/dx^2 + (3/2)dy/dx + y = 0.
we find k^2 + (3/2)k + 1 = 0
which factorises into (k+2)(k+1/2).
So our complementary function is y= Ae^(-2x) + Be^(-x/2).
Now we find any particular integral by guessing y = Le^(-4x).
Substituting this in to the equation d^2y/dx^2 + (3/2)dy/dx + y = 22e^(-4x)
we find that L(16e^(-4x) - 4e^(-4x) + e^(-4x)) = 22e^(-4x)
and L=2.
So the solution to the differential equation is
y= Ae^(-2x) + Be^(-x/2) + 2e^(-4x) //

Answered by Nathan E.

Studies Mathematics at Cambridge

The first step is to write sinhx in its exponential form and set it equal to y, this will make rearranging easier. Then multiply everything by e^x and rearrange to form a quadratic, in terms of e^x. Express e^x using the quadratic formula and from here rearrange to find x in terms of y. This can then be used to find arcsinhx in terms of x, as sinhx = y and x = arcsinhy.

Finding the cross product of two 3D vectors uses a very similar method to that of finding the determinant of a 3x3 matrix.
So, to find the cross product, just follow these simple steps:
1) To begin, create a 3x3 matrix with the unit vectors i, j and k filling the top row.
2) Take the first vector and put the x component under the i, the y component under the j and the z component under the k. Repeat this for the second vector by putting the x, y and z components in their correct columns in the bottom row of the matrix.
3) Next, cover the first row and the first column (the row and column containing the i unit vector). This leaves a 2x2 matrix still showing. You should find the determinant of this matrix (remember from FP1: determinant = ad-bc). This determinant then becomes the x component of the final vector we are trying to find.
4) Now uncover the first row and column and cover the row and column containing the j unit vector. Find the determinant of the resulting 2x2 matrix and then multiply the determinant by -1. This becomes the y component of the final vector.
5) Repeat this for the k unit vector, but without multiplying by -1, to find the z component of the final vector.
6) You now have the final vector - this is the vector product of your original vectors!
Remember: You only have to multiply the y component (when you cover the j) by -1!

Answered by Luke M.

Studies MMath Mathematics at Warwick

Let M be a 3x3 matrix s.t.
M=
|a b c|
|g h i|
|d e f|
Then Det(M)= a(Det(e,f,h,i))-b(Det(d,f,g,i))+c(Det(d,e,g,h).
Given that the determinant of a 2x2 matrix such as (e,f,h,i) is = ei-fh.
The solution is; Det(M)=a(ei-fh)-b(di-fg)+c(dh-eg).
Since the inverse of a matrix, M^-1 = 1/Det(M) * Adj(M), the inverse does not exist when Det(M)=0.

Answered by Oskar D.

Studies Economics and Mathematics (MA) at Edinburgh

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OurÂ FurtherÂ Maths A Level TutorsÂ will give you that focused, individual attention which could make all the difference to your grade. Gaps in your learning, careless mistakes and ineffective time-management during the exams can all be overcome with the expert guidance of aÂ FurtherÂ Maths A-Level Tutor.Â YourÂ Maths A Level TutorÂ will also ensure you have plenty of past papers to complete, as practice, repetition and excellent exam technique are key to Maths A Level success.