Currently unavailable: for regular students
Degree: Maths and Physics (Masters) - Durham University
Hi! I’m Tom and I’m currently studying at Durham, reading Maths and Philosophy. I really enjoy my subjects, particularly the lightbulb moments you get when suddenly you understand a derivation or a proof you’ve been staring at for ages. I also love helping other people towards these moments. You might not empathise with this passion in the slightest – if so, then my aim is to share some of my enthusiasm with you.
In 2014 I scored 44 points out of 45 on the IB, and at uni I’m currently achieving a first. These grades are mainly a result of strong dedication and passion for my subject, and I am just as dedicated and passionate about getting you to do the best you possibly can in your exams.
From my experience, the main obstacle a lot of students face with Maths is a lack of confidence. This often comes from school teachers allowing students to progress without basic conceptual knowledge by teaching them to memorise methods rather than to understand how they work. My aim is to show you that once you understand the fundamental concepts (which anyone can understand if taught properly) Maths is not that hard.
These are of course your sessions so it’s up to you how we spend the time, but my general idea is that either you will come to me with specific questions to answer or exercises to work through together, or you can let me know of a topic that you’d like to work on in advance and I can prepare a lesson on it and some examples that we can work through.
I realise how difficult it must be to choose out of all of the tutors just from these profiles so don’t hesitate to book a free ‘meet your tutor’ session or send me a WebMail and then make your decision after that.
I hope to meet you soon!
|Maths||A Level||£20 /hr|
|Physics||A Level||£20 /hr|
|Physics (Higher Level)||Baccalaureate||7|
|Chemistry (Higher Level)||Baccalaureate||7|
|Maths (Higher Level)||Baccalaureate||6|
|French B (Higher Level)||Baccalaureate||7|
|English A (Standard Level)||Baccalaureate||7|
|Psychology (Standard Level)||Baccalaureate||7|
Bilal (Parent) December 26 2015
Bilal (Student) December 26 2015
Sheila (Parent) November 5 2015
The electromotive force or emf of a cell in a circuit is, for a start, not a force at all: it has the unit of volts. A good definition is ‘the power supplied by the cell per unit current through the cell’ or, equivalently, ‘the energy supplied by the cell per unit charge through the cell’ (IB examiners love to ask for this by the way). When the internal resistance of a cell is zero (or negligible), the emf is exactly the same as the potential difference (or voltage) between the terminals of the cell. But this changes when the internal resistance is significant – this is the resistance inside the actual cell which will ‘use up’ some of the emf before the current even leaves the cell. Since V = IR, you can find the potential difference across this internal resistance by multiplying the resistance by the current through it. Taking away this value from the emf will give you the remaining potential difference between terminals of the cell.see more
When do you use n choose k?
Say you have n objects/elements, and each one can either be selected or not selected. ‘n choose k’ gives you the number of different ways in which you can end up with k selected objects. So for example, if you have a deck of n different cards and you select k of them to take into your hand, the number of distinct hands you could end up with is n choose k. Note that the order of the cards in the hand doesn't matter - only the combination of cards distinguishes one hand from another.
How do you get the formula?
When picking your hand out of the deck of n cards, the first card you pick could have n different values. The second one could have n - 1 values since there is now a card missing from the deck, so the total number of possible combinations for the first two cards is n x (n - 1). Continuing in this way until you have picked k cards, there will be n x (n - 1) x … x (n - k + 1) combinations for the hand, which is written as n!/(n - k)! (it’s like n factorial but with all the terms from n - k downwards cancelled out by the denominator).
But wait! By counting in this way, we have counted hands of the same cards picked in a different order as distinct combinations. To adjust for this, we need to divide our result by however many differently-ordered arrangements there are of each hand. We count in a similar way as before: for a hand of k cards, we could put any of the k cards in the first slot, and then any of the k - 1 remaining ones in the second slot, and so on… so there are k! arrangements. Hence, our final answer is n! / [(n - k)! x k!] (looks much simpler when properly written out…).see more