5 average from 72,576 reviews

Fantastic first session. My son came away very positive and thought it helped alot. Alexander was very dedicated to finding the best way to help my son who has been struggling in a level maths. thank you so much for a brilliant experience, great service, great value. Will be booking another session.

Carolyn, Parent from Wiltshire

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.

Using rules of logarithms: 3log(x^2)+4log(y^3)=log((x^2)^3)+log((y^3)^4) =log(x^6)+log(y^12) =log((x^6)(y^12)

Answered by Alexander C.

Studies Mathematics and Statistics at Warwick

1 - Multiply each part of equation by denominators to multiply out the fractions. {(6(x-2))/x-2} - {(6(x-2))/x+1} = 1(x-2). This would simply to 6 - {(6(x-2))/x+1} = x - 2. We then do the same for the other denominator. {6(x+1)} - {(6(x-2))/x+1)x+1} = (x-2)(x+1). This would simplify to 6x +6 - 6x + 12 = x^2 - x - 2. 2 - Rearrange quadratic equation to solve for x: x^2 - x - 20. 3 - Factorise. This is done by finding 2 numbers which multiply to equal -20 and add to equal -1. In this case the 2 numbers are -5 and 4. We put these in the brackets like so. (x - 5) (x + 4). An additional step could be to multiply the brackets out to check the equations match, but this would take up time. 4 - Finally, solve for x by making each bracket equal to 0: x - 5 = 0, therefore x = 5 and x+4 = 0, therefore x = -4. Final solutions are are x = 5 and -4.

Answered by Aman K.

Studies MBBS Medicine at Kings, London

I would get the student to draw a diagram of the situation.. so draw the curve y=x^2, the tangent at (2,4) and then the normal. Then I would highlight to them that the equation we want is the equation of a straight line.. with the form y=mx+c. Then we would need m and c. So, m first, let's use the fact that the tangent gradient multiplied by the normal gradient must be minus 1. So we can find the gradient of the tangent to the curve using differentiation, and then utilise that result to find m for the normal. Now we need c... so we know a point the curve passes through (2,4).. now all we need to do is to plug those x and y values that satisfy the equation we want to find and plug in the m value from before into the general equation y=mx+c. From which we can find c. So c and m have both been found. Hence we have found the equation we are looking for.

Answered by Mr M.

Studies Aeronautical Engineering at Imperial College London

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Answered by Kelan T.

Studies Mathematics at Durham

Differentiation allows you to find the rate of change of a variable relative to another, which is basically the gradient. So if you had a speed-time graph, finding the amount that speed changes over a given time will get you acceleration.

Answered by Kourosh S.

Studies Physics at Bristol

First, what is a tangent? A line which touches a circle (or ellipse) at just one point. In our case (-3,2)

As it touches the line at one point, it is perpendicular the radius at (-3,2). Which implies that the gradient of the tangent is the negative reciprocal of the gradient of the the radius at (-3,2).

To work out the gradient of the radius at (-3,2) we need to know the centre point of the circle.

Rearrange the equation of the circle into square form yielding, centre as (-5,-1)

With two points on the same line we can work out the gradient of the radius at (-3,2). It's 3/2

So gradient of the tangent is the negative reciprocal of 3/2 which is -2/3.

We now have a point on the tangent line (-3,2) and the gradient of the line, hence we can work out the equation of the tangent. Solving yields 2x+3y=0

**NOTE: I will draw a diagram on the whiteboard and do the arithmetic in the session also**

As it touches the line at one point, it is perpendicular the radius at (-3,2). Which implies that the gradient of the tangent is the negative reciprocal of the gradient of the the radius at (-3,2).

To work out the gradient of the radius at (-3,2) we need to know the centre point of the circle.

Rearrange the equation of the circle into square form yielding, centre as (-5,-1)

With two points on the same line we can work out the gradient of the radius at (-3,2). It's 3/2

So gradient of the tangent is the negative reciprocal of 3/2 which is -2/3.

We now have a point on the tangent line (-3,2) and the gradient of the line, hence we can work out the equation of the tangent. Solving yields 2x+3y=0

Answered by Patrick D.

Studies Economics and Management at Jesus College, Oxford

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