Kayode - Mathematics Tutor - Port Harcourt
1st lesson free
Kayode - Mathematics Tutor - Port Harcourt

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Kayode

  • Rate US$4
  • Response 24h
  • Students

    Number of students Kayode has accompanied since arriving at Superprof

    1

    Number of students Kayode has accompanied since arriving at Superprof

Kayode - Mathematics Tutor - Port Harcourt
  • 5 (2 reviews)

US$4/hr

1st lesson free

Contact

1st lesson free

1st lesson free

  • Mathematics
  • Trigonometry
  • Arithmetic
  • Geometry
  • Logic

A dynamic and proficient Math tutor, incorporating sound knowledge of Engineering and Science to solving your mathematical problems. I teach the "understanding" of Math to aid your strength in not j

  • Mathematics
  • Trigonometry
  • Arithmetic
  • Geometry
  • Logic

Lesson location

Recommended

Kayode is a respected member of our tutor community. He is highly recommended for his commitment and the quality of his lessons. An excellent choice to progress with confidence.

About Kayode

I am a Civil Engineer but I've been passionate about teaching. I presently take home lessons and online tutorials in some other platforms. I engage the teaching of basic principle of mathematics and engineering to expose students to the concept behind each Mathematical problem and hence stir up your interest in not just the questions but the topic as a whole. Once you get the understanding of Mathematics, you can easily solve any question even beyond the scope of the particular question.

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About the lesson

  • Compulsory School
  • Secondary School
  • Adult Education
  • +4
  • levels :

    Compulsory School

    Secondary School

    Adult Education

    Master

    Doctorate

    MBA

    Pre-School

  • English

All languages in which the lesson is available :

English

Find the largest of three consecutive odd numbers whose sum is 57. Solution: Let the largest be x Consecutive odd numbers increase/decrease by difference of 2. This implies, other two numbers are (x-2) and (x-4) So, the sum; x + (x-2) + (x-4) = 57 Hence, solving, we have; 3x - 6= 57 Add 6 to both sides, 3x = 63 Dividing both sides by 3 X = 21. This implies that, the largest number is 21. Other two numbers are; X-2 => 21-2 = 19 X-4 => 21-4 = 17 Check: The sum = 21+19+17= 57

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Rates

Rate

  • US$4

Pack prices

  • 5h: US$19
  • 10h: US$38

free lessons

The first free lesson with Kayode will allow you to get to know each other and clearly specify your needs for your next lessons.

  • 30mins

online

  • US$0/h

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