Matt was amazing! I would highly recommend him as a tutor. Eddie has struggled with Maths for years, yet Matt was able to teach him concepts that other teachers had failed to do. He was extremely patient, but above all his sessions were fun. Eddie organised all his own sessions with Matt and was keen to do so. The whole process gave Eddie confidence and self belief, and this was the best outcome, regardless of results.

Nikki, Parent from Cheshire

Very pleased with the way Christopher delivered the lesson, my daughter found it very helpful and came away surprised that she could learn so much so quickly! The lesson was challenging and kept my daughter's focus at all times. Christopher's energetic style of teaching and tangible enthusiasm will hopefully show my daughter that Maths is not a boring subject!

Ayan, Parent

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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.

As you can see from the question there are 6 parts that 270 needs to be divided by. One sixth of 270 is equal to 45 and so one of the parts is equal to 45. To get the ratio, you have to simply multiply the amount of parts needed by 45. So 3 x 45=135 and 2x45=90. This will give you the ratio needed : 135:90:45. To check this is correct, you can add the values up to make sure it totals to 270.

Answered by Sinat L.

Studies chemistry with medical sciences at Southampton

Step 1: Draw a labelled diagram of the triangle, marking the side lengths.
Step 2: Recall Pythagorasâ€™ theorem equation â€˜c2=a2+b2â€™
Step 3: Substitute the values of a and b into the equation. c2=8.652+10.152
Step 4: Solve the equation. c2=117.845
Step 5: Find c by square-rooting both sides of the equation. c=13.3358â€¦cm

Answered by Alisha M.

Studies Psychology at Bristol

I believe that there is no better way to revise for Maths than to continually practice questions (especially those that you find most tricky!) The more you practise, the easier you'll find the questions because the methods will become more routine.

Answered by Alisha M.

Studies Psychology at Bristol

Step 1: Eliminate the â€˜yâ€™ component from the equations by adding the equations together. This gives the new equation â€˜6x=12â€™
Step 2: Find the value of x by dividing both sides of the equation by 6. This gives us the value of x=2.
Step 3: Use this value of x to find the value of y, by substituting x=2 into one of the two original equations. For example, using the first equation: 6+y=7. By subtracting 6 from both sides of the equation, we find that y=1.
Step 4 (optional but recommended): To check that these values of x and y are correct, substitute both these found values into the original equation that you havenâ€™t already used. Using the second equation, we can verify that 3x-y=5 is correct when x=2 and y=1. We can therefore have full confidence that we have the correct answer!

Answered by Alisha M.

Studies Psychology at Bristol

This is a very common IB-level question,
which is solved with using the technique of a "disguised quadratic".
The first thing we want to do in finding the inverse is swapping the places of x and y in the equation y=f(x), to obtain x=f(y), because finding the inverse is equivalent to reflecting the curve in the line y=x, i.e. interchanging the x and y coordinates. Rearranging the equation x=f(y) to obtain y in terms of x will complete the task.
As such, we have: x=(e^y-e^(-y))/2. We want to solve for y to find the inverse. Firstly, multiply both sides by 2 to obtain: 2x=e^x-e^(-x). Now comes the tricky part: this is actually a quadratic in disguise! We can see this firstly from the rules of exponents: e^(-y)=1/(e^y), so we have: 2x=e^y-1/(e^y). Let us write e^y=p for simplicity: this will render it easier to spot the quadratic. Then 2x=p-1/p, so 2xp=p^2-1, which rearranges to p^2-2xp-1=0. It may not seem like we have done anything special, but remember that p=e^y, so if we can get p in terms of x, we will have e^y in terms of x, so taking a natural logarithm of the eventual expression will gives us the inverse.
Now we can use the quadratic formula to solve for p, where a=1, b=-2x, and c=-1.
We have: p= (2x +- sqrt((-2x)^2+4))/2, which is the same as p= (2x+-sqrt( 4x^2+4))/2. Now factor a 4 out of the expression in the square root to obtain: p= (2x+-sqrt( 4(x^2+1)))/2, or bringing the 4 outside the root: p= x+- sqrt( x^2+1).
Recall that p=e^y so we wouldn't want the expression for p to be negative, since we then wouldn't be able to take the logarithm of a negative value. As such, we take the positive square root in the expression for p: p=x + sqrt(x^2+1). So e^y= x + sqrt(x^2+1), or y= ln(x+sqrt(x^2+1)).
And there we have it: the inverse of y=f(x), where f(x)=(e^x-e^(-x))/2, is y= ln(x+sqrt(x^2+1)).

Answered by Edoardo M.

Studies Mathematics at Bath

log2(x^4) = 4log2(x) as the exoponent can be taken in front of the log.
Applying the same logic in the second statement and writing 16 as 2 to the power 4 gives:
log2(16) = log2(2^4) = 4log2(2).
Therefore x = 2.

Answered by Anna W.

Studies Medicine at Edinburgh

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