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Class 9 Tuition > Class 9 Tuition in Chennai > Class 9 Tuition in Perungudi > Visalakshi V.

Visalakshi V. Class 9 Tuition trainer in Chennai

Visalakshi V.

locationImg Perungudi, Chennai
4 yrs of Exp
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Details verified of Visalakshi V.✕

Identity

Education

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Identity is verified based on matching the details uploaded by the Tutor with government databases.

Math Tutor

Online Classes
Tutor's home
I have been tutoring for the past 4 years. I teach Mathematics for classes (IX- XII) at my house. I have taken online tutoring for US students. Also, been a regular Online Tutor for a registered US Based Tutoring Company.
Trained students for Board exams(India) also, for achieving good score.

Languages Spoken

Tamil

Telugu

English

Education

IGNOU Pursuing

Master of Business Administration (M.B.A.)

JNTU, Hyderabad 2010

Bachelor of Technology (B.Tech.)

Address

Perungudi, Chennai, India - 600113

Verified Info

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Email Verified

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Teaches

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

State, ICSE, CBSE

Subjects taught

Mathematics, Science

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

State, ICSE, CBSE

Subjects taught

Mathematics, Science

Taught in School or College

No

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

EVS, English, Science, Biology, Physics, Mathematics, Chemistry

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

Science, EVS, Mathematics, Biology, English, Physics, Chemistry

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

Biology, Chemistry, Mathematics, EVS, English, Science, Physics

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics

Taught in School or College

No

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics

Taught in School or College

No

Reviews

No Reviews yet!

Answers by Visalakshi V.

Answered on 11/01/2015

Ask a Question

Post a Lesson

Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.

Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b... ...more
Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b = - 6 b = - 6/a a + b = 1 a - 6/a = 1 a^2 - 6 = a a^2 - a - 6 = 0 a^2 - 3a + 2a - 6 = 0 a(a - 3) + 2( a - 3) = 0 (a-3) ( a + 2) = 0 a = 3 or a = - 2 If a = 3, b would be -6/a = -2 If a = -2, then b would be -6/a = 3 Hence equation of line would be either of : x/ -2 + y/3 = 1 x/ 3 + y / -2 = 1
Answers 21 Comments
Dislike Bookmark

Answered on 11/01/2015

Ask a Question

Post a Lesson

4 cards are drawn from a well shuffled deck of 52 cards. what is the probability of obtaining 3 diamonds and one spade ?

There would be total of 13 diamonds. So, drawing 3 from 13 would be written as a selection of 13C3. For drawing one spade out of 13 spades, would be a selection of 13C1 So, total four cards from 52 cards. Probability P(E) = (13C4 * 13C1)/ 52C4 = 858/62475
Answers 43 Comments
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Teaches

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

4

Board

State, ICSE, CBSE

Subjects taught

Mathematics, Science

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

State, ICSE, CBSE

Subjects taught

Mathematics, Science

Taught in School or College

No

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

EVS, English, Science, Biology, Physics, Mathematics, Chemistry

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

Science, EVS, Mathematics, Biology, English, Physics, Chemistry

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

ICSE, International Baccalaureate, State, CBSE

Subjects taught

Biology, Chemistry, Mathematics, EVS, English, Science, Physics

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics

Taught in School or College

No

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

4

Board

State, CBSE

Subjects taught

Mathematics

Taught in School or College

No

No Reviews yet!

Answers by Visalakshi V.

Answered on 11/01/2015

Ask a Question

Post a Lesson

Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.

Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b... ...more
Equation of straight line in intercept form: x/a + y/b = 1 Here a and b are called the intercepts of x and y respectively. Sum of Intercepts : a+ b = 1 Product of the intercepts = a* b = - 6 Let us try to solve a and b, by writing into a quadratic equation. a + b = 1 a * b = - 6 b = - 6/a a + b = 1 a - 6/a = 1 a^2 - 6 = a a^2 - a - 6 = 0 a^2 - 3a + 2a - 6 = 0 a(a - 3) + 2( a - 3) = 0 (a-3) ( a + 2) = 0 a = 3 or a = - 2 If a = 3, b would be -6/a = -2 If a = -2, then b would be -6/a = 3 Hence equation of line would be either of : x/ -2 + y/3 = 1 x/ 3 + y / -2 = 1
Answers 21 Comments
Dislike Bookmark

Answered on 11/01/2015

Ask a Question

Post a Lesson

4 cards are drawn from a well shuffled deck of 52 cards. what is the probability of obtaining 3 diamonds and one spade ?

There would be total of 13 diamonds. So, drawing 3 from 13 would be written as a selection of 13C3. For drawing one spade out of 13 spades, would be a selection of 13C1 So, total four cards from 52 cards. Probability P(E) = (13C4 * 13C1)/ 52C4 = 858/62475
Answers 43 Comments
Dislike Bookmark
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