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About me

About Me
I'm a second year student at Exeter University. I've been passionate about maths from a young age and have carried that love through my school and uni career. I hope that I can generate the same passion for my tutees as well.
I'm a patient and friendly person, and have always enjoyed helping friends and family with their work. I enjoy teaching people about my subject and I will always bring enthusiasm to our sessions.

The Sessions
You guide the topics that we will cover in our sessions. I'll make sure you understand the basics before we tackle the harder topics, once you have basic understanding of a topic, the rest falls into place.
I will do my best to find out what style of learner you are so that I can tailor each tutorial to you, such as including more examples, more diagrams or by finding new ways of explaining ideas. I will not simply talk at you, I will be engaging and make sure that any question you have is answered.

What Next?
If you have any questions, don't hesitate, send me a "WebMail" or organise a Free Trial Session, I look forward to meeting you!

Subjects offered

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

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

4.9from 33 customer reviews

Lisa (Parent) November 19 2016

Very clear and informative

Stephanie (Parent) October 31 2016

Just gets it all spot on :)

Ami (Parent) September 10 2016

very good explained every thing, good understanding

Fergus (Parent) May 18 2016

I can't speak highly enough of James B. He is so patient with my son. He understands him so well. Just one thing, the tutorial did not end at 17.55, it ended at 18.09, because my son kept asking more questions and James B answered them
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Questions James has answered

How do you differentiate X to the power of a?

To differentiate Xa, where a is any real number, you multiply X by a, and subtract 1 from the power. i.e. d(Xa)/dX=aXa-1

To differentiate Xa, where a is any real number, you multiply X by a, and subtract 1 from the power.

i.e. d(Xa)/dX=aXa-1

1 year ago

307 views

How to prove that (from i=0 to n)Σi^2= (n/6)(n+1)(2n+1), by induction.

First you must show that the statement on the right hand side is true for n=1: Σi=0 i2 when n=1, is equal to 12=1 (1/6)(1+1)(1+2)=(1/6)(2)(3)=1 This means that the statement is true for n=1. Next you assume that it is true for 'k', where k is any number, and so you get; Σi=0 i2 when n=k, is ...

First you must show that the statement on the right hand side is true for n=1:

Σi=0 iwhen n=1, is equal to 12=1

(1/6)(1+1)(1+2)=(1/6)(2)(3)=1

This means that the statement is true for n=1.

Next you assume that it is true for 'k', where k is any number, and so you get;

Σi=0 i2 when n=k, is equal to (k/6)(k+1)(2k+2)

You then have to show that the statement is true for n=k+1 which would make;

Σi=0 i2 when n=k+1, is equal to (k+1)/6(k+2)(2k+3) call this 1)

As the left hand side is a sum, it can be written as;

Σi=0 i2 when n=k + (k+1)2

We already know the sum of i2 when n=k and so we can substitute it in;

(k/6)(k+1)(2k+1) + (k+1)2

We then try and reach 1)

We can factorise out (k+1)

(k+1)[(k/6)(2k+1) +k+1]

Next, multiply the inner brackets;

(k+1)[2k2/6+k/6 +k+1]

Take out a factor of 1/6

(k+1)/6(2k2+k+6k+6)= (k+1)/6(2k2+7k+6)

Finally, factorise the inner bracket;

(k+1)/6(k+2)(2k+3)

As this is equal to 1), we have proven that the statement is true for all values of n.

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1 year ago

491 views

What is "Standard Form"?

Standard form is a way of writing large numbers to make them more compact, for example Instead of writing: 197,000,000 We say: 197,000,000= 1.97 x 100,000,000 And so we can write: 1.97 x 108, which is more concise.

Standard form is a way of writing large numbers to make them more compact, for example

Instead of writing:

197,000,000

We say:

197,000,000= 1.97 x 100,000,000

And so we can write:

1.97 x 108, which is more concise.

see more

1 year ago

306 views
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