Annu

Chicalim, Mormugao, India - 403711

Chicalim, Mormugao, India - 403711.

3.8

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1 YEAR of teaching experience in Jaipur as a maths teacher and one year of experience in mumbai

English

Hindi

Banasthali University 2015

Master of Science (M.Sc.)

Chicalim, Mormugao, India - 403711

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Class 6 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 6 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics, Science

Taught in School or College

Yes

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Answered on 27/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

Radius and slant height of cone are given ,now first we will find height of cone with the help of pythagores theorem. P2+b2=h2 P2+(4)2=(5)2 P2+16=25 P2=25-16 P2=9 P=sqr root of 9 P=3 Volume of cone =1/3*pi*r2*h Volume =1/3*22/7*16*3 On solving this Volume of cone =16*pi

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Answers 6 Comments Answered on 21/12/2017 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

It will b 0 Since sin45=1/sqr root of 2 And cos45=1/sqr root of 2 Having the same value and if we perform operation Sin45-cos45 it will result into 0.

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Answers 6 Comments Answered on 21/12/2017 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha... ...more

Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha +beta= -b/a Alpha*beta=c/a For the verification of cofficients take both the scenario L.H.S alpha+beta and R.H.S -b/a -3 + - 2 -5/1 -5 -5 Hence L.H.S=R.H.S Now Again check for L.H.S alpha*beta R.H.S c/a -3 * -2 6/1 6 6 Hence L.H.S = R.H.S So the relationship between zeros and cofficients of given polynomial is verified.

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Answers 5 Comments AnnuDirections

x Class 6 Tuition 3.8

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 6 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics, Science

Taught in School or College

Yes

Class 7 Tuition 3.8

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 7 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics, Science

Taught in School or College

Yes

Class 8 Tuition 3.8

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 8 Tuition

2

Board

CBSE

CBSE Subjects taught

Mathematics, Science

Taught in School or College

Yes

Class 9 Tuition 3.8

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

2

Board

CBSE

CBSE Subjects taught

Science, Mathematics

Taught in School or College

Yes

Class 10 Tuition 3.8

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

2

Board

CBSE

CBSE Subjects taught

Science, Mathematics

Taught in School or College

Yes

Computer Course classes 3.3

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

BCA Tuition 3.3

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

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Answered on 27/12/2017 CBSE/Class 9/Mathematics Tuition/Class IX-X Tuition

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Answers 6 Comments Answered on 21/12/2017 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

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Answers 6 Comments Answered on 21/12/2017 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha... ...more

Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha +beta= -b/a Alpha*beta=c/a For the verification of cofficients take both the scenario L.H.S alpha+beta and R.H.S -b/a -3 + - 2 -5/1 -5 -5 Hence L.H.S=R.H.S Now Again check for L.H.S alpha*beta R.H.S c/a -3 * -2 6/1 6 6 Hence L.H.S = R.H.S So the relationship between zeros and cofficients of given polynomial is verified.

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