Graham R. A Level Maths tutor, GCSE Maths tutor, A Level Physics tuto...
£18 - £20 /hr

Graham R.

Degree: Geophysics with Geology (Bachelors) - Durham University

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About me

About Me

I am a 3rd year Geophysics and Geology student at Durham University. Ever since I was small understanding the way things work, whether it be how mountains formed, the way planets move or solving a differential equation, has always been a passion for me. My aim is that, through my tutorials, you will also have the same love too. This interest and ability came in use when I was asked to be an academic mentor during my time at 6th form to help students lower in the school who needed additional help. 

I have been a competitive swimmer from the age of 7 and used these skills to provide one to one swim coaching last summer for a 6 year old child. I also have over 200 hours sports coaching volunteering. 

The Sessions

The content of each session is entirely guided by you and what you're needing help with at the time. I hope to make learning enjoyable and easy for you with a variety of different techniques (diagrams, exam questions, analogies) so that you can explain the concepts back to me just as well as I did for you! 

Everybody learns differently and I will personalise the tutorials in order to help you learn to the best of your ability. I firmly believe that I can make each session useful and, more importantly, enjoyable! (even in the grueling exam time). 

If you want to get into contact with me, send me a WebMail or book a Meet The Tutor Session. Just remember to tell me what subject you're studying and what exam board.

Cant wait for our first session! 

Subjects offered

SubjectLevelMy prices
Geography A Level £20 /hr
Physics A Level £20 /hr
Geography GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA
PhysicsA-LevelA
GeographyA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

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Ratings and reviews

5from 1 customer review

Fatima (Student) October 5 2016

Questions Graham has answered

Using logarithms solve 8^(2x+1) = 24 (to 3dp)

Using the laws of logs you can see that if you log both sides of the equation you get:  (2x+1)*log(8) = log(24)  Dividing both sides of the equation by log(8) you get:  2x+1 = log(24)/log(8) Then it is a simple case of solving for x:  x = 0.5*(((log(24)/log(8))-1) x = 0.264

Using the laws of logs you can see that if you log both sides of the equation you get: 

(2x+1)*log(8) = log(24) 

Dividing both sides of the equation by log(8) you get: 

2x+1 = log(24)/log(8)

Then it is a simple case of solving for x: 

x = 0.5*(((log(24)/log(8))-1)

x = 0.264

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3 months ago

122 views

y = x*(x-2)^-1/2. Prove dy\dx = (x-4)/2*(x-2)^3/2

Firstly, when approaching a differentiation question you need to work out what method you need to use to solve it. As you can see there are two terms multiplied by one another (the 'x' term and the '(x-2)-1/2' term), therefore the product rule must be used.  Making u = x and v = (x-2)-1/2 du...

Firstly, when approaching a differentiation question you need to work out what method you need to use to solve it. As you can see there are two terms multiplied by one another (the 'x' term and the '(x-2)-1/2' term), therefore the product rule must be used. 

Making u = x and v = (x-2)-1/2

du/dx = 1 dv/dx = -1/2*(x-2)-3/2

Substituing these things into the Product Rule equation we get: 

dy/dx = -x/2*(x-2)-3/2 + (x-2)-1/2

Now we need to focus on manipulating this equation to match the one given in the question. To start with we will take out a factor of (x-2)-3/2 giving: 

dy/dx = (x-2)-3/2*(-x/2 + x-2) 

Simplyfying : 

dy/dx = (x-2)-3/2* (x/2 -2)

Multiplying by 2 : 

dy/dx = (x-4)/2*(x-2)3/2

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3 months ago

112 views
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