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BSc Tuition > BSc Tuition in Mumbai > BSc Tuition in Borivali West > Shreya G.

Shreya G. BSc Tuition trainer in Mumbai

Shreya G.

locationImg Borivali West, Mumbai
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Biotechnologist

Student's home
Shreya G. describes herself as Biotechnologist. She conducts classes in BSc Tuition, Class 10 Tuition and Class 6 Tuition. Shreya is located in Borivali West, Mumbai. Shreya takes at students Home. Shreya has completed Master of Science (M.Sc.) from Mumbai University in 2017. She is well versed in English, Marathi and Hindi.

Languages Spoken

English

Marathi

Hindi

Education

Mumbai University 2017

Master of Science (M.Sc.)

Address

Borivali West, Mumbai, India - 400091

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Email Verified

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Teaches

BSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

BSc Bio-Chemistry Subjects

Proteins and Enzymes, Metabolism of Carbohydrates & Lipids, Biomolecules, Biophysics, Recombinant DNA Technology, Cell Biology, Molecular Genetics, Metabolism of Amino acids & Nucleotides, Immunology, Enzymology, Genetics & Genomics

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

BSc Micro- Biology Subjects

Bacteriology, Introduction to Microbial World, Plant Pathology, Food and Dairy Microbiology, Immunology, Microbial Ecology, Recombinant DNA Technology and Biotechnology, Molecular Biology, Medical Microbiology, Genetics & Genomics, Industrial Microbiology, Cell Biology

Taught in School or College

No

BSc Branch

BSc Bio-Chemistry, BSc Micro- Biology

MSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Subject

Biotechnology, MicroBiology, Biology

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

Geography, Social Science, Hindi, Chemistry, Science, EVS, Physics, Biology

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

Biology, Physics, Science, Social Science, Hindi, EVS, Geography, Chemistry

Taught in School or College

No

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Science, EVS, History, Chemistry, Biology, Hindi, Social science, Geography, Physics

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Science, Social science, Physics, History, Hindi, Chemistry, Biology, EVS, Geography

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Chemistry, EVS, Social science, Physics, History, Biology, Science, Hindi, Geography

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

EVS, Social Science, Social studies, Science, Hindi

Taught in School or College

No

Reviews

No Reviews yet!

Answers by Shreya G.

Answered on 09/12/2017 Learn CBSE - Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Ask a Question

Post a Lesson

Solve : 55x + 52y = 217; 52x + 55y = 217

Firstly, from the first equation, find x. From eqn 1: x = 217 - 52y/55. Now, substitute this value of x in the first eqn. i.e. 52 X (217 - 52y)/55 = 217 Now, multiply both sides with the denominator 55. Therefore, the eqn is 52(217-52y) + 3025y=11935 Solve for y. y = 217/107 ... ...more
Firstly, from the first equation, find x. From eqn 1: x = 217 - 52y/55. Now, substitute this value of x in the first eqn. i.e. 52 X (217 - 52y)/55 = 217 Now, multiply both sides with the denominator 55. Therefore, the eqn is 52(217-52y) + 3025y=11935 Solve for y. y = 217/107 Now, substitute this value of y in the x = 217 - 52y/55 which was derived from the original eqn. Therefore x = 217 - 52(217/107) /55 which is x = 217/107. Therefore, x = y = 217/107
Answers 8 Comments
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Answered on 09/12/2017 Learn CBSE - Class 9/Science +1 Tuition/Class IX-X Tuition

Ask a Question

Post a Lesson

What is a colloidal solution ?

A colloidal solution is a solution in which a material (which may be either a solid, liquid or a gas), is evenly dispersed throughout the liquid solvent. These differ from true solutions as the solutes in colloidal suspension do not dissolve completely in the solvent, but are found suspended. Based on... ...more
A colloidal solution is a solution in which a material (which may be either a solid, liquid or a gas), is evenly dispersed throughout the liquid solvent. These differ from true solutions as the solutes in colloidal suspension do not dissolve completely in the solvent, but are found suspended. Based on whether the solute is a solid, liquid or a gas, the colloidal solutions are classified as sols, emulsions or foams respectively.
Answers 6 Comments
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Teaches

BSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

BSc Bio-Chemistry Subjects

Proteins and Enzymes, Metabolism of Carbohydrates & Lipids, Biomolecules, Biophysics, Recombinant DNA Technology, Cell Biology, Molecular Genetics, Metabolism of Amino acids & Nucleotides, Immunology, Enzymology, Genetics & Genomics

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

BSc Micro- Biology Subjects

Bacteriology, Introduction to Microbial World, Plant Pathology, Food and Dairy Microbiology, Immunology, Microbial Ecology, Recombinant DNA Technology and Biotechnology, Molecular Biology, Medical Microbiology, Genetics & Genomics, Industrial Microbiology, Cell Biology

Taught in School or College

No

BSc Branch

BSc Bio-Chemistry, BSc Micro- Biology

MSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Subject

Biotechnology, MicroBiology, Biology

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

Geography, Social Science, Hindi, Chemistry, Science, EVS, Physics, Biology

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

Biology, Physics, Science, Social Science, Hindi, EVS, Geography, Chemistry

Taught in School or College

No

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Science, EVS, History, Chemistry, Biology, Hindi, Social science, Geography, Physics

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Science, Social science, Physics, History, Hindi, Chemistry, Biology, EVS, Geography

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Chemistry, EVS, Social science, Physics, History, Biology, Science, Hindi, Geography

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE, ICSE, State

Subjects taught

EVS, Social Science, Social studies, Science, Hindi

Taught in School or College

No

No Reviews yet!

Answers by Shreya G.

Answered on 09/12/2017 Learn CBSE - Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Ask a Question

Post a Lesson

Solve : 55x + 52y = 217; 52x + 55y = 217

Firstly, from the first equation, find x. From eqn 1: x = 217 - 52y/55. Now, substitute this value of x in the first eqn. i.e. 52 X (217 - 52y)/55 = 217 Now, multiply both sides with the denominator 55. Therefore, the eqn is 52(217-52y) + 3025y=11935 Solve for y. y = 217/107 ... ...more
Firstly, from the first equation, find x. From eqn 1: x = 217 - 52y/55. Now, substitute this value of x in the first eqn. i.e. 52 X (217 - 52y)/55 = 217 Now, multiply both sides with the denominator 55. Therefore, the eqn is 52(217-52y) + 3025y=11935 Solve for y. y = 217/107 Now, substitute this value of y in the x = 217 - 52y/55 which was derived from the original eqn. Therefore x = 217 - 52(217/107) /55 which is x = 217/107. Therefore, x = y = 217/107
Answers 8 Comments
Dislike Bookmark

Answered on 09/12/2017 Learn CBSE - Class 9/Science +1 Tuition/Class IX-X Tuition

Ask a Question

Post a Lesson

What is a colloidal solution ?

A colloidal solution is a solution in which a material (which may be either a solid, liquid or a gas), is evenly dispersed throughout the liquid solvent. These differ from true solutions as the solutes in colloidal suspension do not dissolve completely in the solvent, but are found suspended. Based on... ...more
A colloidal solution is a solution in which a material (which may be either a solid, liquid or a gas), is evenly dispersed throughout the liquid solvent. These differ from true solutions as the solutes in colloidal suspension do not dissolve completely in the solvent, but are found suspended. Based on whether the solute is a solid, liquid or a gas, the colloidal solutions are classified as sols, emulsions or foams respectively.
Answers 6 Comments
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