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

Chris W.

Currently unavailable:

Degree: Mathematics (Bachelors) - Durham University

Contact Chris

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

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

see more

3 years ago

698 views
Send a message

All contact details will be kept confidential.

To give you a few options, we can ask three similar tutors to get in touch. More info.

Contact Chris

Still comparing tutors?

How do we connect with a tutor?

Where are they based?

How much does tuition cost?

How do tutorials work?

We use cookies to improve your site experience. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok