Felix S. IB Maths tutor, 11 Plus Maths tutor, GCSE Maths tutor, 13 pl...

Felix S.

Currently unavailable: for new students

Degree: Mathematics (Masters) - Warwick University

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About me

About me:

Hi, I'm Felix and I'm currently studying for a Mathematics degree at Warwick University. I am very keen to show people not just how to pass maths exams, but how to actually enjoy the subject too! I got A* in GCSE Maths and the maximum mark of 7 in IB HL Maths and HL Further Maths, and I am often commended for my presentation style and clarity. It was while volunteering in a local school that I discovered just how important one to one tuition can be in helping build confidence and understanding, two key aspects to anyone's learning.

Session outline:

Maths is definitely best learnt through examples and applications of the theory, alongside a solid theoretical understanding. To this extent, the sessions will start by ensuring a basic understanding of the topic before moving on to tackling some exam questions together. Sessions with me will focus upon building up mathematical knowledge and encouraging you to apply this knowledge to the questions that you bring. It will be a joint effort between us to reach those wonderful moments of understanding, and I hope that I can make it an enjoyable journey along the way.

How do I get in touch?

Just send me a message through MyTutorWeb and include your exam board and specific questions that you have so that I can make the sessions as useful as possible.

Hope to meet you soon!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Further Mathematics IB £20 /hr
Maths IB £20 /hr
Maths 13 Plus £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
HL MathematicsBaccalaureate7
HL Further MathematicsBaccalaureate7
HL HistoryBaccalaureate6
SL EnglishBaccalaureate6
SL PhysicsBaccalaureate7
SL Ancient GreekBaccalaureate6
Disclosure and Barring Service

CRB/DBS Standard

29/12/2014

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Questions Felix has answered

Express 0.545454... as a fraction in its simplest form.

First, we are going to give our decimal a name, let's call it x. We notice that x repeats after two digits, so how about we look at 100x. 100x = 54.54545454... But, the decimal part of this number is what we called x, so let's subtract it. 100x - x = 54.5454545454... - 0.54545454 Working out...

First, we are going to give our decimal a name, let's call it x. We notice that x repeats after two digits, so how about we look at 100x.

100x = 54.54545454...

But, the decimal part of this number is what we called x, so let's subtract it.

100x - x = 54.5454545454... - 0.54545454

Working out both sides, we get

99x = 54

So, x = 54/99. However, we are not finished as the question said "in its simplest form". Either notice that 99 and 54 are both divisible by 9, or that they're divisible by 3 twice and we get the final answer.

x = 6/11

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5 months ago

202 views

Given a^2 < 4 and a+2b = 8. Work out the range of possible values of b. Give your answer as an inequality.

For this question, we want to start by finding the possible values of a from the first equation and then using that to give us information about b from the second equation. Now, the condition that a^2 < 4 actually gives us two pieces of information (a way to remember this is that for inequali...

For this question, we want to start by finding the possible values of a from the first equation and then using that to give us information about b from the second equation.

Now, the condition that a^2 < 4 actually gives us two pieces of information (a way to remember this is that for inequalities, squared gives us two things). We have that a < 2 but also that a >  -2 since for example, (-3)^2 = 9 which is larger than 4. Therefore, we have -2 < a < 2.

Rearranging the second equation for b, we get b = (8-a)/2. Therefore, putting in our values for a, we get that the maximum value of b is (8-(-2))/2 = 10/2 = 5 and the minimum value of b is (8-2)/2 = 6/2 = 3. Since the inequality for a does not include 2 and -2, we do not include 3 and 5 for b and so we get the answer 3 < b < 5.

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5 months ago

208 views

Prove that the function f:ZxZ -> ZxZ defined by f(x,y) = (2x+y,x+y) is a bijetion.

Here, we must remember the definition of a bijection. To be bijective, a function must be both injective and surjective. For a function to be injective, we must have that f(a) = f(b) implies that a = b. For a function to be surjective, we must have that for all elements of the range (right han...

Here, we must remember the definition of a bijection. To be bijective, a function must be both injective and surjective. For a function to be injective, we must have that f(a) = f(b) implies that a = b. For a function to be surjective, we must have that for all elements of the range (right hand side), there is an element of the domain (left hand side) that is sent to it by the function.

So, let's start by showing injectivity. We take two pairs (p,q) and (r,s) such that f(p,q) = f(r,s).

We have 2p + q = 2r + s (1) and p + q = r + s (2). I have labelled them (1) and (2) for convenience. Now we look at (1) - (2), and get p = r. Applying this to equation (2), q = s. Therefore, (p,q) = (r,s), our function is injective.

Next, we must consider surjectivity. Let us take a point (a,b) in Z x Z. If 2x + y = a and x + y = b, then as before we can find that x = a - b and y = 2b - a. Therefore, for any a and b in Z x Z, we have found values of x and y that map to it. Since a and b are integers, x and y must also be integers. Therefore our function is surjective as required.

Combining the two results, we have that the function is bijective.

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5 months ago

301 views
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