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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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Shreya G. - Biotechnologist

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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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Teaches

BSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

BSc Bio-Chemistry Subjects

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

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

BSc Micro- Biology Subjects

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

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

Biology, MicroBiology, Biotechnology

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

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

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

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

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

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

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

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

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

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Hindi, Social studies, EVS, Science, Social Science

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 Ask a Question

Post a Lesson 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
Answer Answers 8 Comments Comments
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Answered on 09/12/2017 Learn CBSE - Class 9/Science +1 Tuition/Class IX-X Tuition

Ask a Question Ask a Question

Post a Lesson 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.
Answer Answers 6 Comments Comments
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Teaches

BSc Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

BSc Bio-Chemistry Subjects

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

Type of class

Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

BSc Micro- Biology Subjects

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

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

Biology, MicroBiology, Biotechnology

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

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

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

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

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

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

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

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

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

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Hindi, Social studies, EVS, Science, Social Science

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 Ask a Question

Post a Lesson 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
Answer Answers 8 Comments Comments
Dislike Bookmark

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

Ask a Question Ask a Question

Post a Lesson 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.
Answer Answers 6 Comments Comments
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