5 average from 72,576 reviews

Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your Online Lessons are recorded. Make the most out of your live session, then play it back after

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 Online Lessons. In my Online Lessons, 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 Online 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.

z^{3} = 4 ( sqrt(3) + i ) = 4 (2 * exp(iπ( 1/6 + 2*n*))), where *n* is an integer

z = (8 * exp(iπ (1/6 + 2*n *)))^{1/3} = 2 exp(1/3 (iπ(1/6 + 2*n*))) = 2 exp( iπ (1 + 12*n*) / 18 )

** z = 2 exp ( iπ / 18) ; 2 exp ( 13 iπ / 18) ; 2 exp ( 25 iπ / 18 )**

z = (8 * exp(iπ (1/6 + 2

To solve a linear second order differential equation we first find the complementary function and then the particular integral. To find the complementary function we must find the roots of the auxiliary equation m^{2}+ 5m + 6 = 0 which are m = -3 and m = -2 and which can be found by factorising. This means that the complementary function of the differential equation is x_{c} = A e^{-3t}+B e^{-2t} where A and B are constants.As 4 e^{-t} is to the right of the equals sign, the particular integral must be of the form x_{p} =C e^{-t} where C is a constant. We then differentiate the particular integral twice and substitute into the original differential equation. Differentiating gives dx_{p}/dt = -Ce^{-t} and d^{2}x_{p}/dt^{2} = Ce^{-t}. Substituting these into the differential equation gives 2C e^{-t}= 4e^{-t}, therefore C =2 and so x_{p}=2e^{-t}. Finally we add the complementary function to the particular integral to reach the general solution x = A e^{-3t}+B e^{-2t}+ 2e^{-t}.

Answered by Thomas M.

Studies Engineering Science at Oxford, University College

z=(−b± sqrt(b2−4ac))/2awith a = 1, b = −10 and c = 29. Substituting these coefficients givesz = (10±sqrt((−10)2 −(4×1×29)))/ 2×1= 5 ± 2i

As graphs get more complicated I like to follow a few steps to make sure my graph looks as accurate as possible.

Are there x values that make the equation (y value) get**Big**?* If for example I have 1/(x-2), as x gets closer to 2, y will get bigger. *Where does the graph cross the **Axis**? *If I substitute in x=0, what is my y value? Is there an x value that makes y=0?*What happens to y when x tends to +/- **Infinity**? E.g*. If I have y=1/x, I know y tends to 0 as x tends to +/- infinty.*Where are the **Turning** **points**? *Can i find the stationary points of the equation? Are they maximums or minimums?*

*As you can see the bold words are the important parts to remember.* You can use the acronym **BAIT **too.

Are there x values that make the equation (y value) get

Answered by Jasmin D.

Studies Mathematics at Imperial College London

Say you want to find the square root of the complex number 3+2i.

We can assume that the answer we want will be in the form a+bi.

It follows then, that you can also write 3+2i as (a+bi)^{2}.

Expanding this gives us 3+2i = a^{2}+2abi-b^{2}

Then all we need to do is compare the coefficients of the imaginary and real parts: i.e. 3 = a^{2}-b^{2} and 2 = 2ab.

Solve these 2 simultaneous equations to get a =1.8 and b = 0.56 (ignore any imaginary solutions for a and b - they have to be real).

Therefore the square root of 3+2i is 1.8+0.56i. You can check this by squaring our solution and you'll get back to 3+2i (or near enough due to rounding).

We can assume that the answer we want will be in the form a+bi.

It follows then, that you can also write 3+2i as (a+bi)

Expanding this gives us 3+2i = a

Then all we need to do is compare the coefficients of the imaginary and real parts: i.e. 3 = a

Solve these 2 simultaneous equations to get a =1.8 and b = 0.56 (ignore any imaginary solutions for a and b - they have to be real).

Therefore the square root of 3+2i is 1.8+0.56i. You can check this by squaring our solution and you'll get back to 3+2i (or near enough due to rounding).

Answered by Dan C.

Studies Mechanical Engineering at University College London

A minimum point will have a gradient of 0 (although so will a maximum point or a point of inflection). dy/dx = 3x^{2}-6x. We can substitute x = 2 into this equation to give 0 (alternatively solve 3x^{2}-6x = 0 to give x = 2 as a root). Thus the gradient at x = 2 is 0 so it is either a minimum point, maximum point or point of inflection. To prove it is in fact a minimum point, we can differentiate the original equation of the curve once more to give d^{2}y/dx^{2}= 6x - 6 (the second derivative). Substituting x = 2 into this equation gives a positive value. Positive value = minimum point; negative value = maximum point; zero value requires further investigation by finding gradients of x values slightly smaller and larger than 2 (this is also a possible alternative method to determine which type of stationary point it is to begin with).

Answered by Tom C.

Studies Medicine at Cambridge

mtw:mercury1:status:ok

Our Further Maths A Level Tutors 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 a Further Maths A-Level Tutor. Your Maths A Level Tutor 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.