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Class 12 Tuition > Class 12 Tuition in Guntur > Naveena K.

Naveena K. Class 12 Tuition trainer in Guntur/>

Naveena K.

locationImg Navabharath Nagar, Guntur
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Online Classes
I am an engineer who loves Mathematics and has a passion for teaching it. I am giving online classes for now.

Through out my life, I taught Math to my fellow undergraduates from different majors, my friends preparing for Entrance exams, my siblings through their high school and also on an online tutoring site as a tutor during my undergrad. As a tutor, I simplified math concepts while coaching students to think critically to solve problems. I appreciated and gave assurance to boost student's self-confidence and self-esteem. I cultivated a fun and interesting learning environment which, encouraged questions and discussions.

I can teach various Math topics like Algebra, Geometry, Mensuration, Trigonometry, Calculus, Differentiation, Integration and many more in Basic/Intermediate/Advanced levels.

P.S: I have also been among the top students in National Olympiad in Math. I was very enthusiastic and learnt ABACUS too.

Languages Spoken

Telugu Mother Tongue (Native)

Hindi Proficient

English Proficient

Education

National Institute of Technology, Calicut 2018

Bachelor of Technology (B.Tech.)

Address

Navabharath Nagar, Guntur, India - 522006

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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, NIOS, CBSE, International Baccalaureate, IGCSE, ISC/ICSE

Subjects taught

Electronics, Geography, Computer Science, Telugu, Sociology, Physical Education, EVS, Social science, Physics, Hindi, Statistics, Environmental Management, Logic, Philosophy, Education, Home Science, English, Mathematics, History

Taught in School or College

No

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Answers by Naveena K.

Answered on 22/06/2020

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<p>5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z</p>

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 = 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... ...more

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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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, NIOS, CBSE, International Baccalaureate, IGCSE, ISC/ICSE

Subjects taught

Electronics, Geography, Computer Science, Telugu, Sociology, Physical Education, EVS, Social science, Physics, Hindi, Statistics, Environmental Management, Logic, Philosophy, Education, Home Science, English, Mathematics, History

Taught in School or College

No

No Reviews yet!

Answers by Naveena K.

Answered on 22/06/2020

Ask a Question

Post a Lesson

<p>5(X^2+Y^2+Z^2)=4(XY+YZ+ZX) then find X:Y:Z</p>

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 = 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... ...more

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.

Answers 353 Comments
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