Oliver J. GCSE Maths tutor

Oliver J.

Unavailable

Economics with Industrial Experience (Bachelors) - Exeter University

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About me

I'm Ollie and I've had a years experience in the tutoring business working for the tutoring school 'Kip McGrath West Bridgford'. In terms of grades, Maths-wise I received an 'A* ' at GCSE and an 'A' at A-level helping me achieve A*AA at A-level. I'm a chatty person and will aim to help with as many questions as I can in the hour. 

I'm Ollie and I've had a years experience in the tutoring business working for the tutoring school 'Kip McGrath West Bridgford'. In terms of grades, Maths-wise I received an 'A* ' at GCSE and an 'A' at A-level helping me achieve A*AA at A-level. I'm a chatty person and will aim to help with as many questions as I can in the hour. 

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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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19/12/2013

Qualifications

SubjectQualificationGrade
EconomicsA-level (A2)A*
MathsA-level (A2)A
FrenchA-level (A2)A

Subjects offered

SubjectQualificationPrices
MathsGCSE£18 /hr

Questions Oliver has answered

How do you differentiate x^2?

You bring the power down to multiply by the existing coefficient of x =1 in this case, and then you reduce the existing power by 1 leaving you with dy/dx= 2x. 

You bring the power down to multiply by the existing coefficient of x =1 in this case, and then you reduce the existing power by 1 leaving you with dy/dx= 2x. 

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3 years ago

946 views

How do you find the equation of a straight line on a graph?

Firstly you need to find the gradient and to do this you identify two points on the graph and the gradient is the change in the value of the y values divided by the change in the value of the x values. (y2-y1)/(x2-x1). 

Then to find your 'c' value for the y=mx+c (straight line equation where m=gradient), you either sub in a your x and y values for one point as well as your gradient and find the missing number or you identify where the curve cuts the y axis. At this point you will have your equation of the straight line.

Firstly you need to find the gradient and to do this you identify two points on the graph and the gradient is the change in the value of the y values divided by the change in the value of the x values. (y2-y1)/(x2-x1). 

Then to find your 'c' value for the y=mx+c (straight line equation where m=gradient), you either sub in a your x and y values for one point as well as your gradient and find the missing number or you identify where the curve cuts the y axis. At this point you will have your equation of the straight line.

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3 years ago

1079 views

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