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Sreelakshmi S. Class 12 Tuition trainer in Kochi

Sreelakshmi S.

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experience 2 yrs of Exp
students 2 students
locationImg Edapally, Kochi
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Biotechnology graduate and Medical Coder, experienced tutor for Science subjects and Mathematics.

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I am a graduate of Bsc botany and biotechnology (dual core). I am taking tuitions for school students upto 10th standard and also science tuitions for students upto 12th standards.Especially I can give classes for Genetics and Biotechnology portions of class 10 and 12 as well.I have a strong base for science subjects and EVS.I have also core strength in the subject ,based on competitive exams like NEET. My key methodology is explaining the topic word by word and detailed teaching of the content instead of by-hearting.

Languages Spoken

Malayalam Mother Tongue (Native)

Tamil Proficient

English Proficient

Hindi Basic

Education

SIBBR&D-SCMS, MG University 2019

Bachelor of Science (B.Sc.)

Address

Edapally, Kochi, India - 682024

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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

2

Board

CBSE

Subjects taught

Biology, Biotechnology, Physics, Chemistry

Taught in School or College

No

Class 11 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

2

Board

CBSE

Subjects taught

Physics, Biotechnology, Chemistry, Biology

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

2

Board

CBSE

Subjects taught

Malayalam, Science, Mathematics

Taught in School or College

No

Class I-V Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

2

Fees

₹ 375.0 per hour

Board

CBSE

Subjects taught

Science, Mathematics, EVS

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

2

Board

CBSE

Subjects taught

Science, Mathematics, Social Science

Taught in School or College

No

NEET-UG Coaching Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in NEET-UG Coaching Classes

2

NEET Subjects

Biology

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

2

Board

CBSE

Subjects taught

Science, Mathematics, EVS

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

2

Board

CBSE

Subjects taught

EVS, Mathematics, Science

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

2

Board

ICSE, CBSE

Subjects taught

Chemistry, Science, Biology, EVS, Mathematics, Physics

Taught in School or College

No

Reviews

No Reviews yet!

Answers by Sreelakshmi S.

Answered on 04/07/2020 Learn CBSE - Class 11/Physics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

The position vector is the vector which is measured from a reference point. But displacement vector is the difference between the final vector and initial vector.
Answer Answers 145 Comments Comments
Dislike Bookmark

Answered on 04/07/2020 Learn CBSE - Class 11/Physics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

The position vector is the vector which is measured from a reference point. But displacement vector is the difference between the final vector and initial vector.
Answer Answers 145 Comments Comments
Dislike Bookmark

Answered on 18/06/2020 Learn CBSE - Class 11

Ask a Question Ask a Question

Post a Lesson Post a Lesson

∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant. ...more

∫log x dx = x log x -x

Proof : Using integration by parts,

∫udv = uv - ∫vdu

In ∫ log x dx,

take, u=logx => du= (1/x) . dx

∫dv=∫dx => v=x

Now substituting,

∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C

where C is constant.

Answer Answers 218 Comments Comments
Dislike Bookmark

Answered on 14/06/2020

Ask a Question Ask a Question

Post a Lesson Post a Lesson

Infact online classes have made educating techniques to the next level of smart learning. From whiteboard classes of numerous children for a single teacher, now online classes have taken academics a hi-tech arena with engaging presentations and effective evaluations as well as individual attention a... ...more

Infact online classes have made educating techniques to the next level of smart learning. From whiteboard classes of numerous children for a single teacher, now online classes have taken academics a hi-tech arena with engaging presentations and effective evaluations as well as individual attention a student with adequate communication space.

Answer Answers 2388 Comments Comments
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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

2

Board

CBSE

Subjects taught

Biology, Biotechnology, Physics, Chemistry

Taught in School or College

No

Class 11 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

2

Board

CBSE

Subjects taught

Physics, Biotechnology, Chemistry, Biology

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

2

Board

CBSE

Subjects taught

Malayalam, Science, Mathematics

Taught in School or College

No

Class I-V Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

2

Fees

₹ 375.0 per hour

Board

CBSE

Subjects taught

Science, Mathematics, EVS

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

2

Board

CBSE

Subjects taught

Science, Mathematics, Social Science

Taught in School or College

No

NEET-UG Coaching Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in NEET-UG Coaching Classes

2

NEET Subjects

Biology

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

2

Board

CBSE

Subjects taught

Science, Mathematics, EVS

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

2

Board

CBSE

Subjects taught

EVS, Mathematics, Science

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

2

Board

ICSE, CBSE

Subjects taught

Chemistry, Science, Biology, EVS, Mathematics, Physics

Taught in School or College

No

No Reviews yet!

Answers by Sreelakshmi S.

Answered on 04/07/2020 Learn CBSE - Class 11/Physics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

The position vector is the vector which is measured from a reference point. But displacement vector is the difference between the final vector and initial vector.
Answer Answers 145 Comments Comments
Dislike Bookmark

Answered on 04/07/2020 Learn CBSE - Class 11/Physics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

The position vector is the vector which is measured from a reference point. But displacement vector is the difference between the final vector and initial vector.
Answer Answers 145 Comments Comments
Dislike Bookmark

Answered on 18/06/2020 Learn CBSE - Class 11

Ask a Question Ask a Question

Post a Lesson Post a Lesson

∫log x dx = x log x -x Proof : Using integration by parts, ∫udv = uv - ∫vdu In ∫ log x dx, take, u=logx => du= (1/x) . dx ∫dv=∫dx => v=x Now substituting, ∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C where C is constant. ...more

∫log x dx = x log x -x

Proof : Using integration by parts,

∫udv = uv - ∫vdu

In ∫ log x dx,

take, u=logx => du= (1/x) . dx

∫dv=∫dx => v=x

Now substituting,

∫log x = logx (x) - ∫x . 1/x . dx = logx (x) - ∫dx= x log x -x + C

where C is constant.

Answer Answers 218 Comments Comments
Dislike Bookmark

Answered on 14/06/2020

Ask a Question Ask a Question

Post a Lesson Post a Lesson

Infact online classes have made educating techniques to the next level of smart learning. From whiteboard classes of numerous children for a single teacher, now online classes have taken academics a hi-tech arena with engaging presentations and effective evaluations as well as individual attention a... ...more

Infact online classes have made educating techniques to the next level of smart learning. From whiteboard classes of numerous children for a single teacher, now online classes have taken academics a hi-tech arena with engaging presentations and effective evaluations as well as individual attention a student with adequate communication space.

Answer Answers 2388 Comments Comments
Dislike Bookmark
x

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