Hannah M. A Level Maths tutor, GCSE Maths tutor, 11 Plus Maths tutor

Hannah M.

£24 - £26 /hr

Currently unavailable: for new students

Studying: Mathematics (Bachelors) - Durham University

5.0
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30 reviews| 126 completed tutorials

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About me

Hi, I'm Hannah and I am studying Maths in my second year at Durham University. I've really loved maths all the way through school (and so far into university!) and have signed up to be a tutor so I can try to share that love with others! I have a younger sister who is currently taking A level maths who asks me for help regularly, so I have some experience explaining this level of maths to others. I can also help with GCSE level maths, and possibly at even younger ages if it was needed. I am very patient and I will always try to come up with different ways of looking at concepts if they don't make sense at first. I also hope to make my tutorials enjoyable, as well as educational. Please don't hesitate to contact me if you would like to organise a free 15 minute "meet the tutor" session to get to know me. I've posted the hours I'm usually free below, however my timetable is quite flexible, so I am willing to try and work around you. I hope to be meeting you soon!

Hi, I'm Hannah and I am studying Maths in my second year at Durham University. I've really loved maths all the way through school (and so far into university!) and have signed up to be a tutor so I can try to share that love with others! I have a younger sister who is currently taking A level maths who asks me for help regularly, so I have some experience explaining this level of maths to others. I can also help with GCSE level maths, and possibly at even younger ages if it was needed. I am very patient and I will always try to come up with different ways of looking at concepts if they don't make sense at first. I also hope to make my tutorials enjoyable, as well as educational. Please don't hesitate to contact me if you would like to organise a free 15 minute "meet the tutor" session to get to know me. I've posted the hours I'm usually free below, however my timetable is quite flexible, so I am willing to try and work around you. I hope to be meeting you soon!

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Ratings & Reviews

5from 30 customer reviews
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Roisin (Student)

May 4 2016

Really helpfull and clear on the solutions to questions.

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Denisse (Parent)

April 16 2016

Thank you Hannah - Natalia enjoyed her tutorial with you

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Roisin (Student)

April 4 2016

Very nice and taught in logical steps which was helpful when I got stuck on parts of the question.

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Kerrie (Parent)

March 9 2017

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
ChemistryA-level (A2)A*
SpanishA-level (A2)A

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further Mathematics A Level£26 /hr
MathsA Level£26 /hr
MathsGCSE£24 /hr
Maths11 Plus£24 /hr

Questions Hannah has answered

Express (2x-1)/(x-1)(2x-3) in partial fractions.

The general form of the partial fractions we want is A/(x-1) + B/(2x-3), where A and B are constants we need to find the values of.

We know that A/(x-1) + B/(2x-3) = (2x-1)/(x-1)(2x-3), so we can conclude that A(2x-3) + B(x-1) = 2x - 1

The best way to solve for A and B now is to equate the coefficients on each side of the equation. On the left, we have 2A + B x's, while we have 2x on the right. This means that 2A + B = 2.

Similarly, we have that -3A - B = -1.

We now have a pair of simultaneous equations to solve in A and B. If we add together the above equations, we get -A = 1, which implies that A = -1. We can then obtain that B = 2. Putting these values into the general equation for the partial fractions gives -1/(x-1) + 2/(2x-3), which is our answer.

The general form of the partial fractions we want is A/(x-1) + B/(2x-3), where A and B are constants we need to find the values of.

We know that A/(x-1) + B/(2x-3) = (2x-1)/(x-1)(2x-3), so we can conclude that A(2x-3) + B(x-1) = 2x - 1

The best way to solve for A and B now is to equate the coefficients on each side of the equation. On the left, we have 2A + B x's, while we have 2x on the right. This means that 2A + B = 2.

Similarly, we have that -3A - B = -1.

We now have a pair of simultaneous equations to solve in A and B. If we add together the above equations, we get -A = 1, which implies that A = -1. We can then obtain that B = 2. Putting these values into the general equation for the partial fractions gives -1/(x-1) + 2/(2x-3), which is our answer.

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2 years ago

2059 views

Prove that 1/(tanx) + tanx = 1/sinxcosx

The key here is to realise that tanx = sinx/cosx. If we write out the left hand side of the equation in terms of sine and cosine we get:

cosx/sinx + sinx/cosx

These two fractions can be put over a common denominator of sinxcosx to give:

(cos2x + sin2x)/sinxcosx

If we then use the well-known identity cos2x + sin2x = 1, we see that the above expression is equivalent to 1/sinxcosx, which is the expression we were required to find.

The key here is to realise that tanx = sinx/cosx. If we write out the left hand side of the equation in terms of sine and cosine we get:

cosx/sinx + sinx/cosx

These two fractions can be put over a common denominator of sinxcosx to give:

(cos2x + sin2x)/sinxcosx

If we then use the well-known identity cos2x + sin2x = 1, we see that the above expression is equivalent to 1/sinxcosx, which is the expression we were required to find.

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2 years ago

4443 views

Solve the equation 5^x = 8, giving your answer to 3 significant figures.

Since there is an x written as a power here, it suggests that this question should be solved using logs (here assume I am using the natural logarithm, which has base e). Taking logs of both sides of the equation gives log5x = log8

Using the rule logab = bloga, the equation becomes xlog5 = log8

If we then divide the equation through by log5, we see that x = log8/log5 = 1.29

Since there is an x written as a power here, it suggests that this question should be solved using logs (here assume I am using the natural logarithm, which has base e). Taking logs of both sides of the equation gives log5x = log8

Using the rule logab = bloga, the equation becomes xlog5 = log8

If we then divide the equation through by log5, we see that x = log8/log5 = 1.29

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2 years ago

1218 views

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