Navabharath Nagar, Guntur, India - 522006.
Details verified of Naveena K.✕
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Telugu Mother Tongue (Native)
Hindi Proficient
English Proficient
National Institute of Technology, Calicut 2018
Bachelor of Technology (B.Tech.)
Navabharath Nagar, Guntur, India - 522006
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Board
ISC/ICSE, NIOS, IGCSE, CBSE, International Baccalaureate, State
Subjects taught
Mathematics, Geography, Hindi, Social science, Philosophy, Environmental Management, English, Education, Physical Education, Statistics, Telugu, Logic, Physics, Home Science, Computer Science, Sociology, EVS, History, Electronics
Taught in School or College
No
1. Which school boards of Class 12 do you teach for?
ISC/ICSE, NIOS, IGCSE and others
2. Have you ever taught in any School or College?
No
3. Which classes do you teach?
I teach Class 12 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 less than a year.
Answered on 22/06/2020 Learn Tuition
Rearrange the equation as (x^2 + y^2 + z^2) + 4*(x^2 + y^2 + z^2) - 4*(xy + yz + zx) = 0
Now rearrange this again such that (x^2 + y^2 + z^2) + 2[(x-y)^2 + (y-z)^2 + (z-x)^2 ] = 0
Since a squared number is always positive, the above equation leads us to (x^2 + y^2 + z^2) = 0 and (x-y)^2 + (y-z)^2 + (z-x)^2 = 0, which means x=y=z=0 is the ONLY possible solution.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Board
ISC/ICSE, NIOS, IGCSE, CBSE, International Baccalaureate, State
Subjects taught
Mathematics, Geography, Hindi, Social science, Philosophy, Environmental Management, English, Education, Physical Education, Statistics, Telugu, Logic, Physics, Home Science, Computer Science, Sociology, EVS, History, Electronics
Taught in School or College
No
Answered on 22/06/2020 Learn Tuition
Rearrange the equation as (x^2 + y^2 + z^2) + 4*(x^2 + y^2 + z^2) - 4*(xy + yz + zx) = 0
Now rearrange this again such that (x^2 + y^2 + z^2) + 2[(x-y)^2 + (y-z)^2 + (z-x)^2 ] = 0
Since a squared number is always positive, the above equation leads us to (x^2 + y^2 + z^2) = 0 and (x-y)^2 + (y-z)^2 + (z-x)^2 = 0, which means x=y=z=0 is the ONLY possible solution.
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