XICHEN - Maths tutor - Bristol
XICHEN - Maths tutor - Bristol

The profile of XICHEN and their contact details have been verified by our experts

XICHEN

  • Rate 164 GHS
  • Response 4h
  • Students

    Number of students XICHEN has accompanied since arriving at Superprof

    3

    Number of students XICHEN has accompanied since arriving at Superprof

XICHEN - Maths tutor - Bristol

164 GHS/hr

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  • Maths
  • Algebra
  • Geometry
  • Statistics

Mathematics student offering maths (including Geometry, Algebra, Statistics etc.) lessons up to first year university level in Bristol

  • Maths
  • Algebra
  • Geometry
  • Statistics

Lesson location

About XICHEN

I am a second year student in University of Bristol majoring in maths. Two and a half years of university study has given me good mathematical derivation and computing skills. In the summer of 2018, as a maths tutor, in a half-year part-time teaching, I helped a boy with difficulty learning mathematics effectively improve his mathematical ability and made him interested in mathematics. In this teaching, I used some examples from real life to explain abstract mathematical problems (such as the relationship between geometry and algebra), and cooperated with certain homework exercises and encouragement, step by step, finally effectively improved his confidence in learning mathematics.

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About the lesson

  • Primary school
  • Junior high school
  • SHS 1
  • +8
  • levels :

    Primary school

    Junior high school

    SHS 1

    SHS 2

    SHS 3

    BTS

    Adult Education

    Facultate (Licență)

    Master's degree

    Higher national diploma

    Doctor of philosophy

  • English

All languages in which the lesson is available :

English

Feel relaxed about maths! Sometimes you feel puzzled because you think mathematical problems in a bit complicated way. A tricky example (for junior high school students): How to solve [1/(1*2)]+[1/(2*3)]+...+[1/(99*100)]. Of course you can compute it directly but it is wiser to do in this way: Since 1/(1*2)=(1/1)-(1/2), 1/(2*3)=(1/2)-(1/3), ... ,1/(99*100)=(1/99)-(1/100), the previous fomula = (1/1)-(1/2)+(1/2)-(1/3)+...+(1/98)-(1/99)+(1/99)-(1/100) = 1-(1/100)=99/100. See! You have solved it. For mathematics, I will teach you how to find "the most general description" of a maths problem. Please think about this question: how to solve k/[m*(m+n)]+k/[(m+n)*(m+2n)]+...+k/[(m+(t-1)n)*(m+tn)], where t is a positive integer and all k,m,n are rational numbers! (This is what I said "the most general description")

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Rates

Rate

  • 164 GHS

Pack prices

  • 5h: 820 GHS
  • 10h: 1640 GHS

online

  • GHS164/h

Details

I don't need tariffs but if possible, I recommend students to come to my college dormitory since it is better to learn maths face to face.

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