Hannah D. A Level Maths tutor, GCSE Maths tutor, A Level Further Math...

Hannah D.

Currently unavailable: for regular students

Degree: Mathematics (Masters) - Bristol University

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About me

A bit about me:

Hi! I'm Hannah and I have just received my Masters degree in Maths at Bristol University, graduating with a first. I have been a high achiever throughout my time in education and I now hope to help the next generation of students do the same, working in a friendly, relaxed atmosphere.

I had always gotten high grades in maths at school, but I had never really enjoyed it until I went to sixth form college. There I realised how satisfying maths can be and began to consider studying it at university. However at the time I wasn't taking further maths, so to catch up with the rest of my peers I learnt both maths and further maths modules independently, and was able to finish college with a further maths A level at grade A*. I know the techniques needed to understand these modules so am well equipped to explain them to you!

The sessions:

What we cover in sessions is up to you. We will spend time on those areas of maths you find most challenging, and that are relevant to your specific exam board.

I believe that, particularly in maths, once the key concepts have been covered, the best way to improve your understanding is practice practice practice! Sometimes a topic doesn't click into place until you look at questions based on it. Therefore, during our tutor sessions I would aim to go through example questions in detail, until you feel confident enough to explain them to me.

To help you get the best out of your sessions I will make them as engaging as possible. Actively taking part in learning gives a greater appreciation for the subject (and achieves much better results!) than listening to someone ramble on.  

Any questions?...

Feel free to contact me by sending a 'Webmail' or booking a 'Meet the tutor' session. We can discuss the areas you would like to work on and the best approach to your sessions, should you choose me as your tutor. I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
BiologyA-LevelA*
MathematicsMasters Degree1st
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

General Availability

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Please get in touch for more detailed availability

Questions Hannah has answered

How would you differentiate ln(sin(3x))?

To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx) To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x) Firstly, using a standard result we have dy/du=1/u Secondly, we must work out du/dx. Another standard result is that ...

To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx)

To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x)

Firstly, using a standard result we have dy/du=1/u

Secondly, we must work out du/dx. Another standard result is that d/dx(sin(ax))=a*cos(ax) for any constant number a. This means du/dx=3*cos(3x)

Putting the two parts together, we find that our answer, given by dy/dx, is equal to

3*cos(3x)*1/u=(3*cos(3x))/(ln(sin(3x)))

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5 months ago

195 views

Solve x^2-5*x+6=0

There are 2 methods we can use here. Method 1 - Factorisation If we can rewrite the equation so that it looks like (x+a)*(x+b)=0, where 'a' and 'b' are numbers, then we can easily find a solution. If one or both of the terms inside the brackets equals 0, then their product is also 0. This me...

There are 2 methods we can use here.

Method 1 - Factorisation

If we can rewrite the equation so that it looks like (x+a)*(x+b)=0, where 'a' and 'b' are numbers, then we can easily find a solution. If one or both of the terms inside the brackets equals 0, then their product is also 0. This means the solution(s) are the values of x such that x+a=0 or x+b=0.

Multiplying out the brackets of (x+a)*(x+b)=0 gives x2+(a+b)*x+a*b=0. If we match up the terms with the equation we want to solve (called comparing coefficients), we see that we want two numbers 'a' and 'b' so that a+b=-5, and a*b=6. The two numbers which satisfy this are -2 and -3, so these are our values for 'a' and 'b'.

Now we can write our equation as (x-2)(x-3)=0 (you can check yourself this can be multiplied out to give the original equation!)

If x-3=0 then x=3, and if x-2=0 then x=2, so these are our two solutions.

Method 2 - Quadratic Formula

This method can be used when you can't easily 'see' the numbers to use to factorise the equation.

The quadratic formula states that for an equation of the form a*x2+b*x+c=0, where 'a', 'b', and 'c' are numbers, then

x=(-b+-sqrt(b2-4*a*c)/(2*a)

For our equation a=1, b=-5 and c=6, which can then be substituted into the formula. Again you can check yourself that it simplifies to give the same answers as method 1

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5 months ago

215 views
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