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Kapil Garg

Dooravaninagar, Bangalore, India - 560016

Kapil Garg Class I-V Tuition trainer in Bangalore

Kapil Garg

IT Professinal with 17 years of experience, tech blogger,...

Dooravaninagar, Bangalore, India - 560016.

5.0

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Education

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Overview

I have professional's degree in Computer Science and Engineering, have got 17 years of industry experience and have taken tuitions for students upto 10th class in Maths, English, Hindi and basic computer science

Languages Spoken

Hindi Mother Tongue (Native)

English Proficient

Punjabi Basic

Education

Punjab Engineering College 2000

Bachelor of Engineering (B.E.)

Address

Dooravaninagar, Bangalore, India - 560016

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Class I-V Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class I-V Tuition

17

Board

ICSE, State, CBSE

CBSE Subjects taught

Mathematics, Computers, English, Hindi, Science

ICSE Subjects taught

Hindi, Mathematics, Computer science, English, Science

Taught in School or College

No

State Syllabus Subjects taught

Mathematics, Computer Science, Hindi, English, Science

Reviews

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FAQs

1. Which school boards of Class 1-5 do you teach for?

ICSE, State and CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class I-V 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 17 years.

Answers by Kapil Garg (1)

Answered on 24 Jun Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX). (X^2+Y^2+Z^2) is always positive since it is a sum of squares.=> left-hand side term is still positive, so by equality; the right-hand side term is also positive:: The first conclusion:- XY+YZ+ZX is always positive. Now it can be re-written as 4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2)... ...more

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX).

(X^2+Y^2+Z^2) is always positive since it is a sum of squares.
=> left-hand side term is still positive, so by equality; the right-hand side term is also positive
:: The first conclusion:- XY+YZ+ZX is always positive.

Now it can be re-written as

4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2) = 4(XY+YZ+ZX)
=> 4(X^2+Y^2+Z^2) - 4(XY+YZ+ZX) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
=> 4((X^2+Y^2+Z^2) - (XY+YZ+ZX)) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0

assuming (X^2+Y^2+Z^2) to be a positiver term A and
(XY+YZ+ZX) to be a positive term B
=> 4(A - B) + A = 0
=> the above relation can only be true when both positive terms A & B are 0.
=> (X^2+Y^2+Z^2) = 0
=> the above can only be true when each of X, Y & Z are 0, since squares of each will always be positive.

Thus X = Y = Z = 0 => X:Y:Z = 1:1:1

Answers 353 Comments
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Class I-V Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class I-V Tuition

17

Board

ICSE, State, CBSE

CBSE Subjects taught

Mathematics, Computers, English, Hindi, Science

ICSE Subjects taught

Hindi, Mathematics, Computer science, English, Science

Taught in School or College

No

State Syllabus Subjects taught

Mathematics, Computer Science, Hindi, English, Science

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No Reviews yet! Be the first one to Review

Answers by Kapil Garg (1)

Answered on 24 Jun Tuition

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX). (X^2+Y^2+Z^2) is always positive since it is a sum of squares.=> left-hand side term is still positive, so by equality; the right-hand side term is also positive:: The first conclusion:- XY+YZ+ZX is always positive. Now it can be re-written as 4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2)... ...more

5(X^2+Y^2+Z^2)=4(XY+YZ+ZX).

(X^2+Y^2+Z^2) is always positive since it is a sum of squares.
=> left-hand side term is still positive, so by equality; the right-hand side term is also positive
:: The first conclusion:- XY+YZ+ZX is always positive.

Now it can be re-written as

4(X^2+Y^2+Z^2) + (X^2+Y^2+Z^2) = 4(XY+YZ+ZX)
=> 4(X^2+Y^2+Z^2) - 4(XY+YZ+ZX) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0
=> 4((X^2+Y^2+Z^2) - (XY+YZ+ZX)) + X^2+Y^2+Z^2) - 4(XY+YZ+ZX) = 0

assuming (X^2+Y^2+Z^2) to be a positiver term A and
(XY+YZ+ZX) to be a positive term B
=> 4(A - B) + A = 0
=> the above relation can only be true when both positive terms A & B are 0.
=> (X^2+Y^2+Z^2) = 0
=> the above can only be true when each of X, Y & Z are 0, since squares of each will always be positive.

Thus X = Y = Z = 0 => X:Y:Z = 1:1:1

Answers 353 Comments
Dislike Bookmark

Kapil Garg describes himself as IT Professinal with 17 years of experience, tech blogger, Coach, Counsellor, Tutor. He conducts classes in Class I-V Tuition. Kapil is located in Dooravaninagar, Bangalore. Kapil takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 17 years of teaching experience . Kapil has completed Bachelor of Engineering (B.E.) from Punjab Engineering College in 2000. HeĀ is well versed in English, Hindi and Punjabi.

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