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.

Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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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

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

17

Board

State, CBSE, ICSE

CBSE Subjects taught

Mathematics, Science, English, Hindi, Computers

ICSE Subjects taught

Science, Hindi, English, Computer science, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

English, Mathematics, Science, Hindi, Computer Science

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

State, CBSE and ICSE

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 24/06/2020 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

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

17

Board

State, CBSE, ICSE

CBSE Subjects taught

Mathematics, Science, English, Hindi, Computers

ICSE Subjects taught

Science, Hindi, English, Computer science, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

English, Mathematics, Science, Hindi, Computer 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 24/06/2020 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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