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

Naveena K. Class 12 Tuition trainer in Guntur/>

Naveena K.

Navabharath Nagar, Guntur, India - 522006.

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

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 (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

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

Subjects taught

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

Taught in School or College

No

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

ISC/ICSE, NIOS, CBSE 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.

Contact

Answers by Naveena K. (1)

Answered on 22/06/2020 Learn Tuition

<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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Contact

Teaches

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

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

Subjects taught

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

Taught in School or College

No

No Reviews yet!

Answers by Naveena K. (1)

Answered on 22/06/2020 Learn Tuition

<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
Dislike Bookmark

Contact

Naveena K. conducts classes in Class 12 Tuition. Naveena is located in Navabharath Nagar, Guntur. Naveena takes Online Classes- via online medium. Naveena has completed Bachelor of Technology (B.Tech.) from National Institute of Technology, Calicut in 2018. She is well versed in Hindi, English and Telugu.

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NCERT Class 12 Solutions CBSE Syllabus for Class 12 Chapter 1 - Electric Charges and Fields Chapter 2 - Electrostatic Potential and Capacitance Chapter 3 - Current Electricity Chapter 4 - Moving Charges and Magnetism Chapter 5 - Magnetism and Matter Chapter 6 - Electromagnetic Induction Chapter 7 - Alternating Current Chapter 8 - Electromagnetic Waves

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