Hello! My name is Matt and I'm a 4th year Maths student at the University of Warwick. I achieved all A*s at GCSE and 4 A*s at A-Level in Maths, Further Maths, Physics, and Chemistry. I am offering tutoring for A-Level Maths, Further Maths and Physics, as well as GSCE Maths and Science.
At school I informally tutored other students and used to volunteer after school to help younger pupils with their homework. I then worked for a year as a teaching assistant with children ranging from year 7 up to year 13. During this time I discovered my passion for teaching and gained a huge amount of experience working with students of all ability levels and ages. I am extremely patient, easy going, and friendly.
The tutorials will provide you with an opportunity to go through topics that you find particularly challenging, and it will be up to you to decide which areas to focus on. My main focus is to build a solid understanding of the simple concepts and then slowly apply that knowledge to more complex situations.
I also hope to build the confidence of the student, so that eventually they feel comfortable explaining the idea to someone else (nothing tests understanding like explaining it yourself).
I believe that the key to productive learning is through a relaxed, unpressured environment. Therefore, I will try very hard to make the sessions as fun and engaging as possible, so that you actually look forward to learning Maths and Science, rather than dread it.
Feel free to send me a "WebMail" or book a "Meet the tutor session" if you think I might be the tutor for you!
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Step II||Uni Admissions Test||3|
|Step III||Uni Admissions Test||2|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Colette (Parent) March 14 2016
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Colette (Parent) February 22 2016
Suppose you are asked by your teacher to sum all the integers from 1 up to 1000. You might be thinking they must hold some kind of grudge against you. If you try to calculate the sum by adding on each integer one at a time, you will have to perform 999 separate additions, some of which will be quite long and tedious. This would take far longer than anyone can be bothered to spend adding up numbers.
There is, however, a quicker way. Let's give our sum a letter to represent its unknown value. Let's use "S" for "sum". Then:
S = 1 + 2 + 3 + ... + 999 + 1000
But we can rewrite this sum in reverse order. Starting with 1000 and ending with 1:
S = 1000 + 999 + 998 + ... + 2 + 1
We can now add together these two equations to give us:
S + S = (1+1000) + (2+999) + (3+998) + ... + (999+2) + (1000+1)
Simplifying both sides gives us:
2S = 1001 + 1001 + 1001 + ... + 1001 + 1001
The right hand side has 1000 separate terms, since our original sum contained 1000 numbers. So:
2S = 1000 x 1001 = 1001000
Dividing both sides by two we find that S = 500500. Therefore:
1 + 2 + 3 + ... + 999 + 1000 = 500500
This calculation is an example of a more general concept called an arithmetic series, where you sum a sequence of numbers which differ by adding on a fixed amount with each step.see more