Wajiha I. A Level Maths tutor, GCSE Maths tutor

Wajiha I.

Currently unavailable: until 31/05/2016

Degree: Actuarial Sciences (Masters) - Manchester University

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About me

About Me:

I am a mathematics graduate, currently studying towards an MSc in Actuarial Science at the University of Manchester. I have always enjoyed solving mathematical problems and hope that my tutorials will help you grasp the mathematical concepts I have grown to love.

I am very patient and friendly. I have previously worked as a statistics tutor for a PhD medical student and am currently working as a weekly mathematics tutor for a GCSE student, so I have a lot of experience in teaching people of different ages and backgrounds.

The Sessions:

During the sessions, you will guide me on what we cover. In maths, understanding the concepts is key, so before we do exam questions, we will focus on this.

I will use as many different ways as possible to explain the concepts and use plenty of examples to ensure you understand them. We will then work through relevant mathsy problems, until I am sure you are 100% confident with the concept.

I hope the sessions will be fun! A lot can be achieved in 55mins especially if it is made enjoyable - maths is amazing and hopefully, if you didn’t think that before, you will by the end of the session!

What next?

If you have any questions, send me a 'WebMail' or book a 'Meet the Tutor Session'! (both accessible through this website). Remember to tell me your exam board and what you're struggling with.

I look forward to meeting you!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Maths GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsBachelors Degree1st
MathematicsA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: until

31/05/2016

General Availability

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Please get in touch for more detailed availability

Questions Wajiha has answered

Differentiate y = (3x − 2)^4

We recognise that this is in the form of a function within a function, i.e u= 3x - 2 is within the u^4 function, therefore here we will use the chan rule to differentiate the equation.  The chain rule states that dy/dx = dy/du * du/dx. Here let u = 3x -2, then du/dx = 3. Similarly, y=u^4 so d...

We recognise that this is in the form of a function within a function, i.e u= 3x - 2 is within the u^4 function, therefore here we will use the chan rule to differentiate the equation. 

The chain rule states that dy/dx = dy/du * du/dx.

Here let u = 3x -2, then du/dx = 3. Similarly, y=u^4 so dy/du = 4u^3. Therefore dy/dx = 3 * 4u^3 = 12u^3.

Finally, we substitute u = 3x - 2 into the equation. This therefore gives us, dy/dx = 12(3x - 2)^3.

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9 months ago

259 views

Rearrange the formula to make 'y' the subject: x = (1 - 2y)/(3 +4y)

Ok so we need to get 'y' by itself on the left hand side of equal sign. We will first try to remove the denominator by multiplying 'x' by (3+4y). This gives (3+4y)x = 1 - 2y. We will then expand the brackets so we can get a sum of the single terms i.e this gives 3x + 4yx = 1 - 2y. We will no...

Ok so we need to get 'y' by itself on the left hand side of equal sign. We will first try to remove the denominator by multiplying 'x' by (3+4y). This gives (3+4y)x = 1 - 2y.

We will then expand the brackets so we can get a sum of the single terms i.e this gives 3x + 4yx = 1 - 2y.

We will now gather all terms with 'y'  on the left hand side of the equal sign. This gives 4yx + 2y = 1 - 3x.

Now we factorise to get a single 'y' term, this gives (4x +2)y= 1 - 3x.

We finally try to get 'y' on its own by divinding both sides of the equal sign by (4x +2). This therefore gives us y = (1-3x)/ (4x +2) which is our final answer!

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9 months ago

338 views
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