Pete H. A Level Maths tutor, A Level Electronics tutor, GCSE Maths tu...

Pete H.

Currently unavailable: for new students

Degree: Electronic Engineering with Artificial Intelligence (Masters) - Southampton University

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About me

Important stuff:

Hi, and thanks for checking out my tutor profile! Hi, I'm Pete, I'm happy to help out with any Maths or electronics exam boards up to A-level, or physics up to GCSE. I really enjoy the logical process required for maths and science, which can be broken down into simple steps to provide usually a single definitive answer.

Previously I have worked as an academic peer tutor in my previous sixth form, helping other students out with their then A-level studies, so I already have experience in tutoring to that level (as well as, of course, having already been through the pain of school exams).


Me, myself and I

Here's just a little bit about me so I'm not a complete stranger on a screen. I'm 21 years old and am currently studying an integrated master’s degree in Electronics and Artificial Intelligence at the University of Southampton. This combines large elements of both maths and electronics theories, but also includes a heavy dose of physics at times too. My current interests for my 3rd year project include agent modelling within a computer program to attempt to simulate swarm behaviour within real life biological systems (such as ants building a nest).

In my spare time I enjoy a wide range of activities, particularly outdoor adventure sports such as rock climbing and mountain biking. Outside of the adrenaline junkie field, I also enjoy other hobbies such as dance, yoga and martial arts.


My Method

Ultimately, the pace and topic of each of our sessions will be determined by you, as it's only fair that you get to decide what happens with your tutor session. However, I suggest that the way to get the most out of each session is for you to message me with topics you'd like to go over or specific questions you'd like to go through before the session itself. This way, I can prepare useful information for you and we'll be able to get a lot more done within the hour slot.

If you have any questions or would like to arrange a free meet the tutor request, please do not hesitate to get in touch. I will try my best to reply to inquiries on the same day, however if that is not possible, it will be next working day at the latest.

Hope to hear from you soon!

 

 

Subjects offered

SubjectLevelMy prices
Computing A Level £20 /hr
Electronics A Level £20 /hr
ICT A Level £20 /hr
Maths A Level £20 /hr
Computing GCSE £18 /hr
Electronics GCSE £18 /hr
ICT GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA
Further MathsA-LevelA
PhysicsA-LevelA
ElectronicsA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5from 4 customer reviews

Geeta (Parent) May 29 2016

i wont change the rating as this could be justified however Pete was't present unfortunately not present.

Neil (Parent) November 3 2015

Really helpful tutor who is very good at explaining maths, very pleasant manner.

Paula (Parent) October 7 2015

Thanks Pete, A very helpful session!

Trevor (Student) April 20 2016

Questions Pete has answered

Perform the 8-bit two's complement binary sum 17 - 24

The first step is to view the sum as an addition of a positive and a negative number, rather than a subtraction: 17 + (-24)   From there we can work out the binary equivalents of the 2 numbers:           b7   b6   b6   b4   b3   b2   b1   b0 17      0     0     0     1     0     0     0    ...

The first step is to view the sum as an addition of a positive and a negative number, rather than a subtraction:

17 + (-24)

 

From there we can work out the binary equivalents of the 2 numbers:

          b7   b6   b6   b4   b3   b2   b1   b0

17      0     0     0     1     0     0     0     1

24      0     0     0     1     1     0     0     0

 

24 then has to be converted into it's equivalent negative number in two's complement. This is done in 2 steps:

1. Flip all the bits in the number

2. Peform binary addition of new number and 1.

 

So first we flip all the bits:

b7   b6   b6   b4   b3   b2   b1   b0

1     1     1     0     0     1     1     1

 

Then we add 1:

b7   b6   b6   b4   b3   b2   b1   b0

1     1     1     0     0     1     1     1

0     0     0     0     0     0     0     1

1     1     1     0     1     0     0     0

This gives us the two's complement representation of -24.

 

Then we simply perform the additon of 17 and -24 in binary:

b7   b6   b6   b4   b3   b2   b1   b0

0     0     0     1     0     0     0     1

1     1     1     0     1     0     0     0

1     1     1     1     1     0     0     1

 

Which, as expected, is the binary representation of -7 (you can check this by performing the same 2 stage method to convert the number back to positive 7, then back into decimal from there).

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2 years ago

2127 views

What is the quotient rule and how is it applied?

The quotient rule is a formula used to differentiate fractions that contain terms of x in both the numerator and the denominator (we usually refer to this as f(x)/g(x)). As it isn't possible to differentiate this by normal methods straight away, we use the quotient rule to allow us to do so. ...

The quotient rule is a formula used to differentiate fractions that contain terms of x in both the numerator and the denominator (we usually refer to this as f(x)/g(x)).

As it isn't possible to differentiate this by normal methods straight away, we use the quotient rule to allow us to do so. For A-level maths, you only need to know how to use the quotient rule, deriving how it is formed is not necessary. The quotient rule takes the following form:

dy/dx = ((g(x)f'(x) - f(x)g'(x))/g(x)^2

Where f'(x), g'(x) are dy/dx of f(x), g(x) respectively. Therefore differentiating the numerator and the denominator seperately, and then plugging these back into the formula, will yield the result of differentiating a fraction of the form f(x)/g(x).

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2 years ago

306 views
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