Rebecca M. GCSE Maths tutor, A Level Maths tutor, GCSE Further Mathem...

Rebecca M.

Currently unavailable: for regular students

Studying: Mathematics (Masters) - Warwick University

5.0
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5 reviews| 5 completed tutorials

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About me

About Me: I am a Warwick University maths student who has also dabbled in philosophy and mechanics modules. My Aim: A lot of the people I have helped who struggle, do so because they either lack understanding or lack passion for the subject. I believe that by tackling these individually we can work towards an improvement. I believe that there's multiple ways of looking at the same thing, and those that are struggling to understand just haven't found the right way of looking at it. So by understanding the student (a meet the tutor session should help this) I can try and tailor my teaching to the student individually.  What next? By booking a meet the tutor session, we can work towards making a difference in you or your child's learning. About Me: I am a Warwick University maths student who has also dabbled in philosophy and mechanics modules. My Aim: A lot of the people I have helped who struggle, do so because they either lack understanding or lack passion for the subject. I believe that by tackling these individually we can work towards an improvement. I believe that there's multiple ways of looking at the same thing, and those that are struggling to understand just haven't found the right way of looking at it. So by understanding the student (a meet the tutor session should help this) I can try and tailor my teaching to the student individually.  What next? By booking a meet the tutor session, we can work towards making a difference in you or your child's learning. 

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Ratings & Reviews

5from 5 customer reviews
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Y.S (Parent)

March 5 2017

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Y.S (Parent)

February 26 2017

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Y.S (Parent)

February 19 2017

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Y.S (Parent)

February 12 2017

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further mathematicsA-level (A2)A*
English LiteratureA-level (A2)A*
PsychologyA-level (A2)A*

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
English LiteratureGCSE£18 /hr
MathsGCSE£18 /hr
MathsGCSE£18 /hr
PsychologyGCSE£18 /hr
.STEP.Uni Admissions Test£25 /hr

Questions Rebecca has answered

Given 6x+2y=4 and 5x+y=8, solve the simultaneous equations to find x and y.

The key here is to eliminate one of the variables; it doesn't matter whether we start by trying to get rid of x or y we will arrive at the same solution. For us to eliminate either x or y it is easiest to find a way of making the co-efficient (number before the x or y) the same in both equations by multiplying through by a number as follows. 

If we take 5x+y=8 and multiply through by 2 (don't forget to multiply both sides), we get

10x+2y=16. 

So we now have 10x+2y=16 and 6x+2y=4

We can rearrange both equations to give 2y=10x-16=6x-4.

We have now eliminated y leaving us with 10x-16=6x-4

Rearranging gives 10x-6x=16-4   so 4x=12. Therefore x=3. 

If we substitute this back into one of our original equations such as 5x+y=8, we get 15+y=8.

So y=8-15=-7

Therefore, x=3 and y=-7. 

The key here is to eliminate one of the variables; it doesn't matter whether we start by trying to get rid of x or y we will arrive at the same solution. For us to eliminate either x or y it is easiest to find a way of making the co-efficient (number before the x or y) the same in both equations by multiplying through by a number as follows. 

If we take 5x+y=8 and multiply through by 2 (don't forget to multiply both sides), we get

10x+2y=16. 

So we now have 10x+2y=16 and 6x+2y=4

We can rearrange both equations to give 2y=10x-16=6x-4.

We have now eliminated y leaving us with 10x-16=6x-4

Rearranging gives 10x-6x=16-4   so 4x=12. Therefore x=3. 

If we substitute this back into one of our original equations such as 5x+y=8, we get 15+y=8.

So y=8-15=-7

Therefore, x=3 and y=-7. 

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1 year ago

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Given y=(1+x^3)^0.5, find dy/dx.

In order to solve this question, we need to use the chain rule when differentiating. 

The chain rule formula is dy/dx= (dy/du)*(du/dx). 

Let u=1+x3

Differentiating with respect to x gives du/dx=3x2

We now have y=u0.5

Differentiating with respect to u gives dy/du=0.5u-0.5=0.5*(1+x3)-0.5

Therefore dy/dx= (dy/du)*(du/dx)= 0.5*(1+x3)-0.5*(3x2)= 1.5x2*(1+x3)-0.5

In order to solve this question, we need to use the chain rule when differentiating. 

The chain rule formula is dy/dx= (dy/du)*(du/dx). 

Let u=1+x3

Differentiating with respect to x gives du/dx=3x2

We now have y=u0.5

Differentiating with respect to u gives dy/du=0.5u-0.5=0.5*(1+x3)-0.5

Therefore dy/dx= (dy/du)*(du/dx)= 0.5*(1+x3)-0.5*(3x2)= 1.5x2*(1+x3)-0.5

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1 year ago

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