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https://www.urbanpro.com/bangalore/kapil-garg/24677483 # Kapil Garg

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

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## Answers by Kapil Garg (1)

## Answers by Kapil Garg (1)

Kapil Garg

Dooravaninagar, Bangalore, India - 560016

Dooravaninagar, Bangalore, India - 560016.

5.0

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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

Hindi Mother Tongue (Native)

English Proficient

Punjabi Basic

Punjab Engineering College 2000

Bachelor of Engineering (B.E.)

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

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No Reviews yet! Be the first one to Review

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.

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

Like 0

Answers 353 Comments 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

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

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

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