Currently unavailable: until 01/11/2015
Degree: Mathematics (Masters) - Oxford, New College University
I'm a first year maths student at New College in Oxford. Maths has always felt like a game, rather than a subject to me, and I really hope I can show why, and get you to feel that way.
As well as the online tutoring, I tutored to help pass a school entrance exam at 11+, as well as helping out in the AS maths classes when I was in year 13. I do a lot of public speaking, and I was a House Captain at my school, so I have a lot of experience motivating people and hopefully you'll agree that I'm a very enthusiastic person!
What sort of tutoring do I get?
I'm really happy to tutor all things maths. GCSE, A level (maths and further maths), the MAT and STEP are all things that I'm happy to cover. Personally as a student, I always found 1 on 1 time really helpful, as it let me ask questions to the teacher, so I'll try to keep lessons student led. As well as being helpful for you, it's really helpful for me if I know what you're struggling with!
Personally, I hated just being told things in maths, so where possible, I will try to provide proofs, and lead from there. I find understanding the proof makes the rest of the maths so much easier.
What A level modules did you take in maths?
I took all the pure maths modules C1,C2,C3,C4,FP1,FP2 and FP3. With the applied modules, I took S1, S2, M1,M2, M3, D1 and D2. I was on Edexcel
I am more than happy to teach people all of the pure maths, S1, S2, M1, M2 and M3. I can teach D1 and D2, but I always felt less confident in those two modules. I'm happy to take anything you can throw at me though!
If you have any other questions, feel free to webmail me through this website.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|.MAT.||Uni Admissions Test||£25 /hr|
|Access to HE-Engineering pathway||A-Level||A*|
|MAT||Uni Admissions Test||Pass|
|STEP I||Uni Admissions Test||1|
Since the diameter is 130cm, the radius is half of this, so it is 65cm
Area is Pi*radius2. This means the area is Pi*(65cm)2
This means the area is Pi*4225cm2
The chain rule works so that if you have y=f(x)g(x), where f(x) and g(x) are functions of x, dy/dx = f'(x)g(x) + f(x)g'(x)
If we consider this like a normal quadratic problem, this becomes easy
x4 < 8x2 + 9
x4 - 8x2 - 9 < 0
(x2-9)(x2+1) < 0
This means there are roots of this expression at x2 = 9 and x2 = -1
Since for all reals, x2 > 0, we know the two roots of this expression are x=+-3
Now, since x4 - 8x2 - 9 is a quartic (ie, it has an x4 expression), we know that given any sufficiently positive or negative x, the quartic will be positive (ie, if x is 10000, or -10000)
Therefore, we know for this to be true, -3