Joseph P. GCSE Maths tutor, A Level Maths tutor, GCSE French tutor, U...

Joseph P.

Currently unavailable: for regular students

Degree: Mathematics (Bachelors) - Warwick University

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About me

About Me

I am a 20 year old mathematics student at the University of Warwick, about to enter my final year. Though I have always enjoyed a range of subjects (from Latin to Music) I have a particular fondness for maths, and have much experience in teaching it from GCSE right through to STEP, having tutored for many years now and understanding better and better the common problems of secondary school students with maths. 

In case you're also interested in improving your French, I have just spent one year in France and am now close to fluency (level C1). So if you need some help in understanding either the grammar, writing, or simply how to speak colloquially, I would be very happy to pass on what I've learnt. 

The Sessions

Most especially for maths, understanding the subject is key. Through the hour session we can talk about a certain subject in maths (say integration), understand what it means and how to use it, before finally tackling questions from past papers. Having spent much time running through dozens of past papers, I can safely say that practice makes perfect, and that once you have the base of understanding, you must you it until your maths ability becomes second nature. You'll soon be solving complex equations on auto pilot!

What Next

Feel free to ask whatever questions you might have through the 'WebMail' on the site, and try to tell me the main areas what you're struggling with. Look forward to hearing from you soon!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
French GCSE £18 /hr
Maths GCSE £18 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
LatinA-LevelA
R.SA-LevelA
MusicA-LevelA
MathematicsBaccalaureate1st (predicted)
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

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Ratings and reviews

5from 3 customer reviews

Martina (Parent) May 13 2016

Very good lesson. Thanks

Akshay (Student) April 9 2016

Great tutor, clearly explained how to answer hypothesis testing and sampling distribution questions, would definitely recommend.

Jassim (Student) May 3 2016

Questions Joseph has answered

What is exactly differentiation?

This is obviously a very important but somewhat difficult to explain question of maths. Let's try to define these terms for normal functions between R (real numbers) and R.  Now, if we consider the derivative of f at a certain point (let's say x), you can think of it as looking at the gradien...

This is obviously a very important but somewhat difficult to explain question of maths. Let's try to define these terms for normal functions between R (real numbers) and R. 

Now, if we consider the derivative of f at a certain point (let's say x), you can think of it as looking at the gradient of f at that point. So, if you're function f is constant, then we have a flat line, and so we have that it's gradient everywhere is 0, and therefore it's derivative is zero. (i.e. f'(x) = 0 for all x). If, however we have a linear function such as f(x) = 2x + 1, if we look at the graph we see that it's gradient is 2 (using the simple gradient formula), hence f'(x) = 2. 

Of course we don't have to use the gradient formula every time, sometimes we won't even be able to (when the function isn't linear, that is), and there is a very helpful rule for functions of the type f(x) = x^n. 

That is, f'(x) = nx^(n-1). 

But it is still good to understand what the derivative really is, once you understand it's relationship with the gradient, you will already be ahead of most other A-level mathematicians. Indeed, have you ever wondered why f has a minimum at x if f'(x) = 0? Draw a picture of f and what it looks like when the gradient is 0 at a certain point, and it will all become natural. 

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9 months ago

216 views

What is the difference between bon and bien in french?

Though you could say this problem is a fairly basic one, this was always a question I had difficulty with during school, and I felt that I never got a clear answer. Having now spent a year in France, I will do my best to explain. Fundementally, 'bon' is an adjective and 'bien' is an adverb. Th...

Though you could say this problem is a fairly basic one, this was always a question I had difficulty with during school, and I felt that I never got a clear answer. Having now spent a year in France, I will do my best to explain.

Fundementally, 'bon' is an adjective and 'bien' is an adverb. That is, if you want to say a 'good book' you would say 'bon livre' or a 'bonne glace' for 'good ice cream' (here we use bonne as glace is a feminine word). 

On the other hand, you might say 'tu cuisines bien' (you cook well) 'il a bien joué' (he played well). Here bien is the adverb of the sentence, and can be translated as 'well'. 

One difficult more specific difference between the two words is comparing 'c'est bon' and 'c'est bien'. If you go to France you will probably hear these two expressions a lot, and it's important to understand the difference between them. 

'C'est bien' is easier to directly translate into English, it simply means 'that's good'. So, for example, you might after someone gets a good grade, you'll tell them 'c'est bien'. Or if you've just enjoyed a nice holiday and somebody asks you how it was, you might say 'yeah it was good', i.e. 'oui c'était bien' (c'est bien in the imperfect). 

'C'est bon' on the other hand can be roughly translated as 'it's fine' or 'that's ok'. For example, when offered help from someone, you would say 'c'est bon' as in 'that's ok (don't worry)'. Or if you're at a shop and the shopkeeper is giving you a quantatiy of fruit, you would say 'c'est bon' ('that's fine/enough') to tell him to stop. Bon can also be associated with physical senses, so if you want to compliment a meal, you would say 'c'est très bon' for 'it's very tasty'. 

As I said, the difference between 'bon' and 'bien' is very difficult to define clearly, but I hope I've given you a better understanding of it. 

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9 months ago

226 views

What is the highest common factor of 24 and 90?

To find out the highest common factor of two different numbers, first of all we write each number as a product of its primes.  Here we have 24 = 2 x 2 x 2 x 3 = 2^3 x 3^1 180 = 2 x 2 x 3 x 3 x 5 = 2^2 x 3^2 x 5^1 Now comes the important part. We compare the the prime factors of each number, ...

To find out the highest common factor of two different numbers, first of all we write each number as a product of its primes. 

Here we have

24 = 2 x 2 x 2 x 3 = 2^3 x 3^1

180 = 2 x 2 x 3 x 3 x 5 = 2^2 x 3^2 x 5^1

Now comes the important part. We compare the the prime factors of each number, and take the lower power of the two. So for example, we take out the 2^2 (2 squared) from the 180, because that power is smaller that the one for 24 (where we have 2^3). 

Similarly, we notice that 3^1 is the smaller power of the two cases (we have 3^2 for 180), and so we take out 3^1 (i.e. 3).

Finally, we see that the two numbers share no other prime factors, so now we simply multiply what we have taken out, and that it the highest common factor. i.e.

hcf(24, 180) = 2^2 x 3 = 4 x 3 = 12

And there's our answer. 

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9 months ago

336 views
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