Saheli B.

Airoli Railway Station, Mumbai, India - 400708

Airoli Railway Station, Mumbai, India - 400708.

5.0

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More than 2 years I am attached with ICSE School as Computer and Math teacher.

English Proficient

Wbut 2016

Master of Engineering - Master of Technology (M.E./M.Tech.)

Airoli Railway Station, Mumbai, India - 400708

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Class I-V Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

CBSE Subjects taught

Mathematics

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Answered on 09/12/2014 Tuition/Class XI-XII Tuition (PUC)

just put the value like x=1 ,2,3...similerly for y..answer will be x=4, y=9... OR , IF YOU NEED JUSTIFICATION THEN... ?x+y=11 .(i) ?y+x=7 (ii) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) ?x=k or,x=k^2 So, x?perfect square integer and ?7 Now perfect square integer less than 7... ...more

just put the value like x=1 ,2,3...similerly for y..answer will be x=4, y=9... OR , IF YOU NEED JUSTIFICATION THEN... ?x+y=11…………………….(i) ?y+x=7………………………(ii) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) ?x=k or,x=k^2 So, x?perfect square integer and ?7 Now perfect square integer less than 7 is 4 and 1 and 7 is not a perfect square Thus, x=1 or 4 let, x = 1 then x?-?y (mod 7) 1?-6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x?-?y (mod 7) 4?-3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9

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Answers 85 Comments Saheli B.Directions

x Class I-V Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

CBSE Subjects taught

Mathematics

Class 6 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 6 Tuition

6

Board

International Baccalaureate, CBSE, ICSE

IB Subjects taught

Mathematics, Computers, Science

CBSE Subjects taught

Mathematics, Science, Computers

ICSE Subjects taught

Physics, Mathematics

Taught in School or College

Yes

Class 7 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 7 Tuition

6

Board

International Baccalaureate, CBSE, ICSE

IB Subjects taught

Mathematics, Computers, Science

CBSE Subjects taught

Mathematics, Science, Computers

ICSE Subjects taught

Physics, Mathematics

Taught in School or College

Yes

Class 8 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 8 Tuition

6

Board

International Baccalaureate, CBSE, ICSE

IB Subjects taught

Mathematics, Computers, Science

CBSE Subjects taught

Mathematics, Science, Computers

ICSE Subjects taught

Physics, Mathematics

Taught in School or College

Yes

Class 10 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

6

Board

CBSE, ICSE, International Baccalaureate

IB Subjects taught

Mathematics, Computers

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Computer Application, Physics

Experience in School or College

6 years experience in icse school

Taught in School or College

Yes

Class 9 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

6

Board

CBSE, ICSE, International Baccalaureate

IB Subjects taught

Mathematics, Physics, Computers

CBSE Subjects taught

Bengali, Science, Mathematics

ICSE Subjects taught

Mathematics, Physics

Experience in School or College

6 years teaching experience in this field

Taught in School or College

Yes

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No Reviews yet! Be the first one to Review

Answered on 09/12/2014 Tuition/Class XI-XII Tuition (PUC)

just put the value like x=1 ,2,3...similerly for y..answer will be x=4, y=9... OR , IF YOU NEED JUSTIFICATION THEN... ?x+y=11 .(i) ?y+x=7 (ii) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) ?x=k or,x=k^2 So, x?perfect square integer and ?7 Now perfect square integer less than 7... ...more

just put the value like x=1 ,2,3...similerly for y..answer will be x=4, y=9... OR , IF YOU NEED JUSTIFICATION THEN... ?x+y=11…………………….(i) ?y+x=7………………………(ii) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) ?x=k or,x=k^2 So, x?perfect square integer and ?7 Now perfect square integer less than 7 is 4 and 1 and 7 is not a perfect square Thus, x=1 or 4 let, x = 1 then x?-?y (mod 7) 1?-6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x?-?y (mod 7) 4?-3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9

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Answers 85 Comments X

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