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My son says he's very lucky to have Joseph's help. He's an excellent tutor, 100% reliable, very generous with his time and patient and willing to direct help exactly where it's needed. My son's made progress after only a few lessons and is motivated to work much harder on a subject he finds hard. Couldn't recommend him more highly. Thank you Joseph.

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A very patient tutor. Mayur will pre-plan a lesson in advance if required, this way the lesson begins exactly where you need. Response to messages is immediate and efficiency at the very highest. A most enjoyable and informative lesson. I would highly recommend this tutor to students of all levels; this was my third successful lesson.

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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.

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y=3x+2
x^2+y^2=20
x^2+(3x+2)^2=20
x^2+(3x+2)(3x+2)=20
x^2+(9x^2+6x+6x+4)=20
x^2+(9x^2+12x+4)=20
10x^2+12x+4=20
10x^2+12x-16=0
5x^2+6x-8=0
(5x-4)(x+2)=0
5x-4=0, x+2=0
x=4/5=0.8, x=-2
y=3x+2, y=3x+2
y=3(0.8)+2, y=3(-2)+2
y=4.4, y=-4
When x=0.8, y = 4.4
When x=-2, y=-4

Answered by Daniela C.

Studies History at Warwick

By using the quadratic quation, it can be deduced that x = 1.22 or x = -3.55

Answered by Miranda S.

Studies German and Spanish with Management at Exeter

When simplifying surd expressions we want to look for square numbers that are factors of the number inside the square root. If we list the square numbers (which are numbers that are the result of squaring another number) up to 48 we have 1, 4, 9, 16, 25 and 36 (1^2, 2^2, ... 6^2). Now we see that 1, 4 and 16 are all factors of 48. Choosing the highest we know that 16 x 3 = 48 so the surd becomes sqrt(16x3).
Next, we know that the square root of 16 is 4 so we can apply this and take it outside of the square root giving 4*sqrt(3) (read as 4 root 3). This 4 comes from square rooting 16. As 3 cannot be split up into any more square factors, 4 root 3 is the final answer.

Answered by Matthew H.

Studies Computer Science and Maths at Exeter

One of the easiest ways to find a minimum/ maximum point is using differentiation.
The derivative of the function f(x) is usually written as dy/dx or f'(x) and is essentially a function that tells us the gradient of f(x) at any point x.
One of the key characteristics of a minimum/ maximum of a function is the way the gradient behaves at & around it.
For a minimum point (say m), the function is decreasing before and increasing after m. i.e. it as a negative gradient (f'(x) < 0) to the left of m and a positive gradient (f'(x) > 0) to the right of m (This is reversed for a maximum). In order for the sign of the gradient to change, it has to pass through 0, and this is exactly where m is. i.e. f'(x) = 0.
So if differentiate f(x), we get f'(x) = 2x + 2
and we want to set f'(x) = 0 to find the x value of the minimum, m.
f'(x) = 0 = 2x +2
2x + 2 = 0
2x = -2 (subtract 2 on both sides)
x = -1 (divide by 2 on both sides)
And so we have the x-coordinate of the minimum at x = -1; we can check that this is indeed a minimum by taking a point to the left of -1 e.g. -2 and a point to the right e.g. 0. by inputting these points into f'(x) and checking their signs we can confirm the behaviour of f(x) around x = -1.
f'(-2) = -2 < 0 and f'(0) = 2 > 0 - which confirms that the gradient is indeed negative before m and positive after and so m ( x= -1) is the minimum of f(x).

Answered by Sachin M.

Studies Maths, operational research, statistics and economics at Warwick

For this, we use the FOIL method of expanding brackets; start by multiplying the FIRST terms of each bracket together to get x*x=x^2. Next multiply the OUTSIDE elements, ie the first element in the first bracket and the last in the second, giving x*(-5)=-5x. Then comes the INSIDE elements, giving 4*x=4x, and finaly the LAST elements of each bracket, giving 4*(-5)=-20.
Adding all these terms together gives us our answer; x^2+4x-5x-20, which can be simplified to x^2-x-20.

The solution to this question can be obtained algebraically using substitution. As both equations are equal to y, this also means they are equal to each other. So firstly, substitute the simpler equation which is y=2x+2 into the second equation giving 2x+2 = x^2 - 1. Re-arrangement of this gives x^2 - 2x + 3 = 0. Using quadratic equation theory, this then becomes (x - 3)(x + 1)=0. Any equation that equals zero must have another zero value on the other side of the equation. Therefore when y is 0, x is either 3 or -1. Using these two x values, we can work out the y values from the original equations. Using the simpler equation y=2x + 2, if x is 3 then y=2*3 + 2= 6+2 = 8 and if x is -1, y=2(times-1) + 2 = -2+2=0. Just to check these values are correct, you can then plug them in to the second equation and the same x and y values should be obtained

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