Wajiha I. A Level Maths tutor, GCSE Maths tutor

Wajiha I.

Unavailable

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!

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!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Qualifications

SubjectQualificationGrade
MathematicsDegree (Bachelors)1st
MathematicsA-level (A2)A

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

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 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.

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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2 years ago

1069 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 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!

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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2 years ago

2888 views

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