Chris W. A Level Maths tutor, GCSE Maths tutor, A Level Further Mathe...

Chris W.

Currently unavailable:

Degree: Mathematics (Bachelors) - Durham University

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About me

About you:
  • I have done mentoring previously; from being a one-to-one reading mentor, to tutoring whole class for GCSE Maths.
  • My passion for maths and sciences has always been there; which is why I chose physics modules as well as maths modules here at Durham.
  • I fully understand the A-Level syllabus and exam techniques required to get an A*; as shown by my four A* grades at A-Level.
 
The session:
  • My sessions will focus on me delivering a short and concise run through of the material and then move on to exam questions, slowly offering help as we tactle the problem together.
  • This is the style of tutorials at university and I think it is a fantastic way of improving. Though I can change my style to suit individuals as everyone learns differently.
  • A optional bonus question of greater difficulty will be available, whether this is for homework, or a problem just to think about before next session is completely up to you. I will reveal the answer at the start of the next session.
 
Anything else:
  • Please do not hesitate to contact me about any questions!
 

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr
Science GCSE £18 /hr
-Personal Statements- Mentoring £20 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable:

Questions Chris has answered

How do I find the equation of a line connecting points a(p,q) and b(r,s)?

First we need to find the gradient of the line connecting points a and b: gradient m = (change in y)/(change in x) = (q - s)/(p -r) Now we use the following equation: y - y1 = m(x - x1) substituting suitable values for (x1, y1) (can be points a or b but we'll use point a this time) and m (cal...

First we need to find the gradient of the line connecting points a and b:
gradient m = (change in y)/(change in x) = (q - s)/(p -r)

Now we use the following equation:

y - y1 = m(x - x1)

substituting suitable values for (x1, y1) (can be points a or b but we'll use point a this time) and m (calculated above):

Using point a:

y - q = [(q-s)/(p-r)](x - p)

and so the equation in the form y = f(x) is:
y = [(q-s)/(p-r)]x + [(q-s)/(p-r)](-p) + q

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2 years ago

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