I cannot praise this tutor enough. Highly skilled in mathematics, teaches at the student's pace, patient and super efficient. A conscientious person who responds immediately to messages. The lesson was tailor made to suit my own needs. Mayur is excellent at pin pointing math problems quickly - taking things back to basic principals where needed. A most enjoyable tutorial - I would highly recommend this very gifted tutor to students of every ability.

Elizabeth, Student

Lloyd is a really good tutor, I would highly recommend him. He has been reliable, available and booked times in to suit us at short notice. He worked hard with my son and helped him achieve a grade higher in his GCSE than in his mocks and made sure he understood a topic before moving on to the next one.

Simone, Parent

Jack has been tutoring my 12 year old son for about 2 months now and is really beginning to improve his confidence and mathematical ability. The lessons are fun and very thorough, often reinforcing a topic covered in classwork or tricky homework, leaving him confident and ready to move on to the next one. Unlike in school, during sessions with Jack my son is never afraid to say when he doesn't fully understand something, ensuring 100% confidence in a topic. Jack has been very flexible with the timings of our sessions and has endless patience. This is a great way to sign up to a tutor and if you are lucky enough to have Jack you will not be disappointed.

Julia, Parent from Hampshire

Parent review - it's a relief to see my son start to understand GCSE maths topics that he has struggled with. His confidence is growing with Chris's help. Chris isn't phased by any questions or topics thrown at him, and is able to clearly explain the methods needed to answer maths questions.

Kelly, Parent from Kent

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 tutorials 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 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 solve the quadratic, we need to factorise it. Consider the coefficient on the x^2, its only factors are 1 and 3 so this tells us that in our factorising we will have something that looks like (x...)(3x...). All the signs are positive so we know that it will also be a case of positive numbers (x+...)(3x+...).
Consider the factors of 6, by looking i can see that a combination on 3 and 2 is going to add up to get to 11 as 3x*3=9 and x*2=2x which totals 11x. Hence we have (x+3)(3x+2)=0. From that, I can show that my solutions will be x=-3, -2/3 after equating each bracket to 0.

Answered by Natalie F.

Studies Mathematics at Warwick

Solving quadratics means we find out at what values of x the equation is equal to 0. This one we can solve by factorising: First take out a common factor of 2, to get: 2(x^{2} - 4x + 3). Now you have to find two numbers that add up to (-4) and also multiply to get 3. In this case the numbers are (-1) and (-2). So now we can re-write it as 2(x - 1)(x - 2). You can check this by multplying out the brackets and you should return to the original equation. (Now, remembering that when we multiply anything by 0 it is equal to 0.) The equation will equal 0 when (x-1) = 0 or when (x-2) = 0. These are two very simple equations to rearrange for x. __ANSWER: x = 1 and x = 2__ You can check this by substituting these numbers back into the original equation.

Answered by Hilde M.

Studies Physics & Music at Edinburgh

The nth term of the sequence is a formula that lets us find any value in the sequence if we know its position. In the given sequence there is a difference of 8 between each term, 13-5 = 8, 21-13=8, 29-21=8. This tells us that the formula for the nth term is going to look like 8n + c for some number c, but we need to find what this number c is. The first term in the sequence (n=1) is 5, so if we plug this into our equation we get 8(1) + c = 5, rearranging this we get c = 5 - 8 = -3. So c is -3.We could have chosen any term to find c. If we had chose use n=2, for example, 8(2) + c = 13, c = 13 - 16 = -3, which is the same as we got before. This is because our formula is correct for every value so it doesn't matter which one we pick but using the first term is usually the easiest. To check our answer, we can check another term, the fourth term is 29 so putting n=4 into our equation we get, 8(4) - 3 = 32 -3 = 29, which is what we expected, so we can be confident that we have the right answer So the formula for the nth term is 8n - 3

(x+3)(x-7)
= (x*x) + (x*-7) + (x*3) + (3*-7)
= x^2 - 7x + 3x - 21
= x^2 - 4x - 21
Key here is to ensure you multiply each term in one bracket by both in the other, and to keep careful track of the signs.

[use of board]
So, Pythagoras' theorem states that a^2+b^2=c^2. Therefore, we need to mark the triangle with a, b &c. a and b do not matter, c must be the longest edge, that creates the diagonal. Put the lengths a and b into the theorem. According to BIDMAS, you must square the numbers first, then add them together afterwards. This will give you a value for c^2. Use a calculator (or your brain) to find the square root of this number... This will leave you with c, the hypotenuse.

the first thing we need to do is rearrange the order of our question to ( 4 x 9 ) x (10^-3 x 10^14).
now (4 x 9) = 36
(10^-3 x 10^14) = 10^(-3 + 14) = 10^11
so our answer is 36 x 10^11 however this is not in standard form. In standard from the first number has to be below 10. To do this we have to rewrite the number 36 in a different way.
36 = 3.6 x 10^1 so our final answer in standard form is 3.6 x 10^(11+1) = 3.6 x 10^12

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With such a variety of questions, GCSE Maths can be a real headache for students. In individual sessions with a GCSE Maths tutor you can focus on exactly the topics you need help with. In our online classroom, tutors use diagrams, graphs and illustrations to enhance their lessons.

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