the session was extremely useful as everything was explained to me in a detailed and organised way. I was provided with different methods to do some problem solving which made things a lot easier and I was given some good advise that will definitely help in the future. Another important point, is that the lesson was very fun which made things much nicer as well.

Laleh, Parent from Buckinghamshire

Jonny is a 5 star tutor and has gone out of his way to help us at times when things haven't gone quite to plan. His knowledge, patience and delivery of problem solving is excellent. I totally recommend Jonny from our experience of him a my son's math's tutor. Thanks Jonny!

Emma, Parent from Cornwall

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

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 calculate the value of a number raised to the power of a fraction, you first find the root of the number according to the value of the denominator (bottom number of the fraction).
In the example given in the question, 9 is the number provided, and the denominator of the fraction is 2.
Therefore, you would, firstly, find the square root of 9, which is 3.
Â²âˆš9 = 3
Then, you would raise the answer you get to the power of the value of the numerator (top number in a fraction). In this example, the numerator is 3, so you would raise 3 to the power of 3, which gives you 27.
3Â³ = 27 [final answer]

Answered by Elsa A.

Studies Communication and Media at Leeds

The ways to differentiate a function depend on the function itself. If there is only one x value in the function you can differentiate as normal (hard to explain on a computer), yet if the function contains more than one x value is a separate place, you will need to use another approach such as the chain, product or quotient rule

Answered by Barnaby N.

Studies Economics at Bath

Step 1: Write 125 as an exponent. Is there any number you can multiply by itself a few times to give you 125? The answer is 5 because 5 x 5 x 5= 5 x 25 = 125.
Step 2: Since we know that 125=5^3, we can replace 125 in the equation by 5^3; (5^3)^(-2/3).
Step 3: Use the power rule (a^b)^c = a^(b*c). Applying this rule to our equation we obtain 5^(3*-2/3).
Step 4: Inside the brackets we have 3*(-2/3). We can break this down further by cancelling out the 3 in the numerator with the 3 in the denominator and we will be left with -2.
Step 5: Now we can simplify 5^(3*-2/3) to 5^(-2)
Step 6: Apply the negative exponent rule a^(-b)= 1/(a^b)---> 5^(-2) = 1/(5^2) = 1/25

Answered by Araba S.

Studies Business & Marketing at Warwick

Step 1: Multiply both sides by x to obtain xy = 3x + 5.
Step 2: Place all the 'x' terms on the same side of the equation to obtain xy-3x=5
Step 3: Factor out x from the left side of the equation: x (y-3) = 5.
Step 4: Now you can single out x by dividing both sides of the equation by (y-3) so that x = 5/(y-3)

Answered by Araba S.

Studies Business & Marketing at Warwick

The objective of this question is to solve for x and y. Since we cannot directly eliminate x or y, find the lowest common multiple for x in both equations by comparing 2x in equation 1 with 6x in equation 2. In this case it is 6 and x can be eliminated by multiplying equation 1 by 3. This gives 6x-9y=72. Now subtract equation 2 from the new equation 1 which you just worked out. Working from left to right, you will have 6x -6x=0; -9y-2y= -11y; and 72-(-5) = 72+5=77. Make sure you watch out for the negative signs. The resulting equation is -11y =77. Divide both sides of the equation by -11 to obtain y= 77/-11= -7. Label the equation y=-7 as (3) and substitute equation 3 into equation 1 as follows: 2x-3*(-7)= 24 --> 2x - (-21) = 24 -->2x + 21=24. Subtract 21 from both sides of the equation to obtain 2x = 24-21 --> 2x=3 and x=3/2. We have now solved the equation for x (x=3/2) and y( y=-7).

Answered by Araba S.

Studies Business & Marketing at Warwick

In order to get from f(x) to f''(x) we need to differentiate the function f(x) with respect to x and then differentiate the resulting function with respect to x again. When differentiating f(x), split the equation f(x) into smaller parts. These parts should be separated by + and - signs. In this case 5x^3 will be one part, -6x^(4/3) will be another then +2x and finally -3.
Next step would be to differentiate each part separately, then in the end add them together. Following the rules of differentiation, 5x^3 will become 15x^2, as the constant - number that is not x (in this case 5) - is multiplied by the power of x (which is 3) and then power is reduced by 1 to give us 15x^2. Next part, which is -6x^(4/3) will differentiate to -8x^(1/3). Following the rules of differentiation: -6*4/3 = -8; 4/3 - 1 = 1/3
Third part, which is +2x, will differentiate into +2. That's because +2x can be written as +2x^1. So by following the rules we get +2x^0. Since any number to the power of 0 equals to 1, +2x^0 can be written as +2*1 which is +2.
Next part would have been -3, but because it is a constant (it does not contain x) when differentiating it will become 0.
to finish the first step, we need to add all differentiated parts together, giving 15x^2 - (9/2)x^(-1/4) + 2. This is now f'(x). To get to the f''(x), the f'(x) function must be differentiated again with respect to x. Just like before break the equation into parts, separated by + or - sign and differente each of them separately. 15x^2 will become 30x
-(8)x^(1/3) will become -(8/3)x^(-2/3). Be careful with the signs when differentiating, as -8 multiplied by 1/3 will give -8/3 and (1/3)-1 will give -2/3.
Next part which is +2 disappears when differentiating as now it is just a constant, so just like -3 in original equation it will become 0.
Finally, add all differentiated parts together to get 30x - 8/3 x^(-2/3).
With questions like that it is very easy to get signs wrong when differentiating. The best way to make sure this does not happen is to practice these types of questions and not rush it.

Answered by Alexey B.

Studies Economics at Durham

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