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
NIOS Subjects taught
Physics, Hindi, Mathematics, Physical Education, Computer Science, Home Science, English
Board
CBSE, IGCSE, State, International Baccalaureate, NIOS, ISC/ICSE
IB Subjects taught
Hindi, Physics, English, EVS, Mathematics
ISC/ICSE Subjects taught
Hindi, History, Mathematics, Geography, Computer Science, Physics, English, EVS, Physical Education
CBSE Subjects taught
Sociology, English, Physical Education, Home Science, Physics, Hindi, Computer Science, Philosophy, Mathematics
IGCSE Subjects taught
Hindi, Environmental Management, Social science, Physics, English, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Telugu, Mathematics, Sociology, Statistics, Computer Science, English, Home Science, Education, Hindi, Physics, Electronics, Logic
1. Which school boards of Class 12 do you teach for?
CBSE, IGCSE, State 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
NIOS Subjects taught
Physics, Hindi, Mathematics, Physical Education, Computer Science, Home Science, English
Board
CBSE, IGCSE, State, International Baccalaureate, NIOS, ISC/ICSE
IB Subjects taught
Hindi, Physics, English, EVS, Mathematics
ISC/ICSE Subjects taught
Hindi, History, Mathematics, Geography, Computer Science, Physics, English, EVS, Physical Education
CBSE Subjects taught
Sociology, English, Physical Education, Home Science, Physics, Hindi, Computer Science, Philosophy, Mathematics
IGCSE Subjects taught
Hindi, Environmental Management, Social science, Physics, English, Mathematics
Taught in School or College
No
State Syllabus Subjects taught
Telugu, Mathematics, Sociology, Statistics, Computer Science, English, Home Science, Education, Hindi, Physics, Electronics, Logic
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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