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Dais M. Class 10 trainer in Kochi/>

Dais M.

Labour Colony Road Thammanam, Kochi, India - 682032.

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

A passionate teacher with experience in teaching IGCSE, A level and CBSE syllabi in Physics. Join me to learn Physics interactively through blended learning techniques.

My approach of teaching is to thorough the basics before approaching the problems and exercises.

Classes will be primarily focused at online learners with frequent tests and quizzes to keep them focused. However other modes of interactive learning and personal training are available for those who need a more personalised approach.

Grading will be done and results are analysed and will be made available intermittently. Analysis of the results can help the learners assess where they stand and the level of improvement shown in the period of time.

Candidates showing exceptional learning ability will be challenged with a more fast-track route by emphasising more on practise questions and exercises to double down on higher performance.

Languages Spoken

Malayalam Mother Tongue (Native)

English Proficient

Hindi Basic

Education

VIT University 2013

Master of Engineering - Master of Technology (M.E./M.Tech.)

Address

Labour Colony Road Thammanam, Kochi, India - 682032

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Teaches

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

3

Board

IGCSE

IGCSE Subjects taught

Physics, Geography

Taught in School or College

Yes

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 10 do you teach for?

IGCSE

2. Do you have any prior teaching experience?

Yes

3. Which classes do you teach?

I teach Class 10 Tuition Class.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 3 years.

Contact

Answers by Dais M. (2)

Answered on 08/07/2020 Learn Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx ==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx ==> + + = 0 ==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0 This is possible only when each component is zero. ==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0 ==> 2x =y ; 2y = z; & 2z = x ==> x/y = 1/2... ...more

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx

==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx

==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0

==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0

This is possible only when each component is zero.

==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0

==> 2x =y ; 2y = z; & 2z = x

==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2

==> x = y/2 = z/4; y = z/2

x:y:z = 1:2:4

Answers 353 Comments
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Answered on 08/07/2020 Learn Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx ==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx ==> + + = 0 ==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0 This is possible only when each component is zero. ==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0 ==> 2x =y ; 2y = z; & 2z = x ==> x/y = 1/2... ...more

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx

==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx

==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0

==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0

This is possible only when each component is zero.

==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0

==> 2x =y ; 2y = z; & 2z = x

==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2

==> x = y/2 = z/4; y = z/2

x:y:z = 1:2:4

Answers 353 Comments
Dislike Bookmark

Contact

Teaches

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

3

Board

IGCSE

IGCSE Subjects taught

Physics, Geography

Taught in School or College

Yes

No Reviews yet!

Answers by Dais M. (2)

Answered on 08/07/2020 Learn Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx ==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx ==> + + = 0 ==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0 This is possible only when each component is zero. ==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0 ==> 2x =y ; 2y = z; & 2z = x ==> x/y = 1/2... ...more

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx

==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx

==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0

==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0

This is possible only when each component is zero.

==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0

==> 2x =y ; 2y = z; & 2z = x

==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2

==> x = y/2 = z/4; y = z/2

x:y:z = 1:2:4

Answers 353 Comments
Dislike Bookmark

Answered on 08/07/2020 Learn Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx ==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx ==> + + = 0 ==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0 This is possible only when each component is zero. ==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0 ==> 2x =y ; 2y = z; & 2z = x ==> x/y = 1/2... ...more

5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx

==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx

==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0

==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0

This is possible only when each component is zero.

==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0

==> 2x =y ; 2y = z; & 2z = x

==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2

==> x = y/2 = z/4; y = z/2

x:y:z = 1:2:4

Answers 353 Comments
Dislike Bookmark

Contact

Dais M. conducts classes in Class 10 Tuition. Dais is located in Labour Colony Road Thammanam, Kochi. Dais takes Online Classes- via online medium. She has 3 years of teaching experience . Dais has completed Master of Engineering - Master of Technology (M.E./M.Tech.) from VIT University in 2013. She is well versed in English, Malayalam and Hindi.

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