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About me

About me:

I am a third year undergraduate at Warwick studying maths. In my first and second years I achieved a first (with an average score of 86% and 88% respectively).  In my Maths and Further Maths A levels I scored an average of 95%. I have always wanted to study Maths from a young age. I have had a year of experience tutoring students on mytutor and in person. I am currently also working for the Warwick Mathematics Institute teaching first year undergraduate maths students. 

The sessions

I hope you be able to instil an appreciation of the beauty of the Maths whilst simultaneously preparing your child to effectively deal with exam questions.

Every child will learn best in a slightly different way. We will work to find the style of learning that suits your child best by adjusting the pace and depth in which we cover topics, the number and type of examples, the use of visual aids, and so on.

Whilst I hope to teach your child to love Maths, the primary aim of tutoring will probably be to achieve the highest grades in exams. My A level scores show that I know how to execute an exam perfectly.

According to your child’s needs we can:

-supplement school teaching through more and varied examples,

- approach a topic from fresh,

- revise a topic by covering the main ideas and then by doing questions,

- work on the nitty gritty aspects of exam technique and how to approach problems,

- approach a tricky topic from a different perspective to help with understanding.

I will make sure that parents are fully informed about their child’s progress and which areas we will be working on. 

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £24 /hr
Maths A Level £24 /hr
Maths GCSE £22 /hr
Maths IB £24 /hr
Maths 13 Plus £22 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
EconomicsA-LevelA*
STEP IUni Admissions Test1
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Ratings and reviews

5from 23 customer reviews

Amanda (Parent) October 19 2016

My son really values the sessions - many thanks

Callum (Student) June 1 2016

I don't have time to write a long review, but to rate the tutorial 5 stars - as Josh has worked at - I have to write something, so I just want to say Josh is an excellent maths tutor. Thanks for all the help in the run before the exams!

Callum (Student) February 17 2016

Unfortunately I started late. Despite this Josh went overtime to compensate for the missed time - which was really helpful :) Thanks for the excellent tutorials that are really clear and helpful. :)

Callum (Student) October 18 2015

Josh gives really helpful tutorials; he is patient, friendly and is able to help me understand things quickly. Thanks for the helpful sessions.
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Questions Josh has answered

What is the perpendicular bisector of the point (0,2) and (1,0)?

First things first it will definately be useful to make a quick sketch of the x-y axes, the two points, and roughtly what the perpendicular bisector will look like. This kind of question can be made much more painless if you make the following observation: for any point (x,y) on the perpendic...

First things first it will definately be useful to make a quick sketch of the x-y axes, the two points, and roughtly what the perpendicular bisector will look like.

This kind of question can be made much more painless if you make the following observation: for any point (x,y) on the perpendicular bisector (lets call this P) the distance from (x,y) to (0,2) and the distance from (x,y) to (1,0) are equal. 

Let (x,y) lie on P. Using Pythagoras' Theorem the distance (x,y) to (0,2) is  sqrt[(x - 0)2 +  (y - 2)2]. Similairly the distance from (x,y) to (1,0) is sqrt[(x - 1)2 + (y - 0)2]. These two distances must then be equal so:

 sqrt[(x - 0)2 +  (y - 2)2] = sqrt[(x - 1)2 + (y - 0)2] , now squaring gives:

(x - 0)2 +  (y - 2)2 = (x - 1)2 + (y - 0)2 , then simplifying gives:

x2 + (y - 2)2 = (x - 1)2 + y2 , expanding the brackets gives:

x2 + y2 - 4y +4 = x- 2x +1 + y2 , then cancelling the x2 and y2 terms:

-4y + 4 = -2x + 1 , which rearranges to give the line:

y = x/2 + 3/4 

There was nothing special here about finding the perpendicular bisector of (0,2) and (1,0), the exact some method can be used for any two points in the plane. The only thing that must be altered is our two expressions for the distances from (x,y) to the first point and the distance form (x,y) to the second. 

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1 year ago

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