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Sree Sujana

Hoskote, Hoskote, India - 562114

Sree Sujana Class 11 Tuition trainer in Hoskote

Sree Sujana

Hoskote, Hoskote, India - 562114.

4.0

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Overview

I am a teacher. I am giving online tuition, since i am certified Masters in Mathematics, Bachelor in Education
Key Skills:
- Good experience in teaching methods.
- Techniques to teach a particular lesson
- Contribution to any special event for the class/entire school.
- Very good exposure to Online Teaching.

. Experience: Having 5.5 years of experience in teaching field as a teacher in Mathematics and Physical Sciences subjects for 6th -12th standards

Languages Spoken

Telugu Mother Tongue (Native)

English Proficient

Education

Sri Krishnadevaraya University 2008

Master of Science (M.Sc.)

Address

Hoskote, Hoskote, India - 562114

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Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

5

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Reviews

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FAQs

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

CBSE and State

2. Have you ever taught in any School or College?

Yes

3. Which classes do you teach?

I teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 8 Tuition and Class 9 Tuition Classes.

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 5 years.

Answers by Sree Sujana (4)

Answered on 26/06/2019 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Given, f(x) = x4 + 4x2 + 5 = (x2)2 + 4x2 + 5 Put x2 = a Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5 = a2 + 4a + 5 To find the zero of the above polynomial, we take a2 + 4a + 5 = 0 The above equation does not have a real root Hence there is no zero for this given... ...more

Given, f(x) = x4 + 4x2 + 5

                 = (x2)2 + 4x2 + 5

Put x= a

Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5

                              = a2 + 4a + 5

To find the zero of the above polynomial, we take a2 + 4a + 5 = 0

The above equation does not have a real root

Hence there is no zero for this given polynomial.

Answers 13 Comments
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Answered on 26/06/2019 Tuition

RHS= (x+y+z)(x2+y2+z2-xy-yz-zx) =x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx) =x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x =x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz =x3+y3+z3-3xyz =LHS Hence proved. ...more

RHS= (x+y+z)(x2+y2+z2-xy-yz-zx)

       =x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx)

       =x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x

        =x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz

        =x3+y3+z3-3xyz

        =LHS

Hence proved.

Answers 278 Comments
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Answered on 26/06/2019 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

...more

Answers 13 Comments
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Answered on 18/06/2019 Tuition

Online tutions for maths 200/- per hour 3 class per week.
Answers 985 Comments
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Class 11 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

5

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 12 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

5

Board

CBSE, State

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 8 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

5

Board

State, ICSE, CBSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry

Experience in School or College

• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class 9 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

6

Board

CBSE, State, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Physics, Mathematics, Chemistry

Experience in School or College

• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.

Taught in School or College

Yes

State Syllabus Subjects taught

Science, Mathematics, EVS

Class 10 Tuition 4.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

6

Board

CBSE, State

CBSE Subjects taught

Mathematics

Experience in School or College

• Oct 2014-2017: Working as a teacher in New Horizon School, Hoskote. • August 2009 – Feb 2011: worked as an online tutor in Tutorvista for 6th to 12th standards in Mathematics.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

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Answers by Sree Sujana (4)

Answered on 26/06/2019 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Given, f(x) = x4 + 4x2 + 5 = (x2)2 + 4x2 + 5 Put x2 = a Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5 = a2 + 4a + 5 To find the zero of the above polynomial, we take a2 + 4a + 5 = 0 The above equation does not have a real root Hence there is no zero for this given... ...more

Given, f(x) = x4 + 4x2 + 5

                 = (x2)2 + 4x2 + 5

Put x= a

Hence x4 + 4x2 + 5 = (x2)2 + 4x2 + 5

                              = a2 + 4a + 5

To find the zero of the above polynomial, we take a2 + 4a + 5 = 0

The above equation does not have a real root

Hence there is no zero for this given polynomial.

Answers 13 Comments
Dislike Bookmark

Answered on 26/06/2019 Tuition

RHS= (x+y+z)(x2+y2+z2-xy-yz-zx) =x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx) =x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x =x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz =x3+y3+z3-3xyz =LHS Hence proved. ...more

RHS= (x+y+z)(x2+y2+z2-xy-yz-zx)

       =x(x2+y2+z2-xy-yz-xz)+y(x2+y2+z2-xy-yz-zx)+z(x2+y2+z2-xy-yz-zx)

       =x3+xy2+xz2-x2y-xyz+x2y+y3+z2y-y2x-y2z-xyz+x2z+y2z+z3-xyz-yz2-z2x

        =x3+y3+z3+(x2y-x2y)+(y2z-y2z)+(y2z-y2z)-3xyz

        =x3+y3+z3-3xyz

        =LHS

Hence proved.

Answers 278 Comments
Dislike Bookmark

Answered on 26/06/2019 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

...more

Answers 13 Comments
Dislike Bookmark

Answered on 18/06/2019 Tuition

Online tutions for maths 200/- per hour 3 class per week.
Answers 985 Comments
Dislike Bookmark

Sree Sujana conducts classes in Class 10 Tuition, Class 11 Tuition and Class 12 Tuition. Sree Sujana is located in Hoskote, Hoskote. Sree Sujana takes Online Classes- via online medium. She has 6 years of teaching experience . Sree Sujana has completed Master of Science (M.Sc.) from Sri Krishnadevaraya University in 2008. She is well versed in English and Telugu.

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