Anil Upadhyay

Bakhtawar Pur, Delhi, India - 110036

Bakhtawar Pur, Delhi, India- 110036.

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I worked in this profession for 10 years. I am feeling proud to be as Teacher. I am one who have the responsibility of the making one's future.This is my promise that I will not let you down.

Hindi Mother Tongue (Native)

MJK Collage 1994

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Bakhtawar Pur, Delhi, India - 110036

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Class 9 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

4

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

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Answered on 22/01/2018 CBSE/Class 10/Social Studies Tuition/Class IX-X Tuition

No corporation to the British which was rule to the India. Many Indian was stop British job. Mahatama Ghandhi led this movement.

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Answers 3 Comments Answered on 22/01/2018 CBSE/Class 10/Social Studies Tuition/Class IX-X Tuition

No corporation with the British who rule India. Many Indian stopped going to their British jobs. Mahatama Ghandhi led this movement.

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Answers 3 Comments Answered on 22/01/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Given equation is: (c 2 – ab) x 2 – 2 (a 2 – bc) x + (b 2 – ac) = 0 To prove: a = 0 or a 3 + b 3 + c 3 = 3abc Proof: From the given equation, we have a = (c2 – ab) b = –2 (a 2 – bc) c = (b 2 – ac) It is being given that equation has real and equal roots ∴... ...more

Given equation is:

(*c* ^{2} – *ab*) *x* ^{2} – 2 (*a* ^{2} – *bc*) *x* + (*b* ^{2 }– *ac*) = 0

To prove: *a* = 0 or *a* ^{3} + *b* ^{3} + *c* ^{3} = 3*abc*

Proof: From the given equation, we have

*a* = (c^{2} – *ab*)

*b* = –2 (*a* ^{2} – *bc*)

*c* = (*b* ^{2} –* **ac*)

It is being given that equation has real and equal roots

∴ *D* = 0

⇒ *b* ^{2} – 4*ac** *= 0

On substituting respective values of *a*, *b* and *c* in above equation, we get

[–2 (*a* ^{2} – *bc*)]^{2} – 4 (*c* ^{2} – *ab*) (*b* ^{2} – *ac*) = 0

4 (*a* ^{2} – *bc*)^{2} – 4 (*c* ^{2} *b* ^{2} – *ac* ^{3} – *ab* ^{3} + *a* ^{2} *bc*) = 0

4 (*a* ^{4} + *b* ^{2} *c* ^{2} – 2*a* ^{2} *bc*) – 4 (*c* ^{2} *b* ^{2} – *ac* ^{3} – *ab* ^{3} + *a* ^{2} *bc*) = 0

⇒ *a* ^{4} + *b* ^{2} *c* ^{2} – 2*a* ^{2} *bc* – *b* ^{2} *c* ^{2} + *ac* ^{3} + *ab* ^{3} – *a* ^{2} *bc* = 0

⇒ *a* ^{4} + *ab* ^{3} + *ac* ^{3} –3*a* ^{2} *bc* = 0

⇒ *a* [*a* ^{3} + *b* ^{3} + *c* ^{3} – 3*abc*] = 0

⇒*a* = 0 or *a* ^{3} + *b* ^{3} + *c* ^{3} = 3*abc*

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Answers 6 Comments Answered on 22/01/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Let original speed of train = x km/h We know, Time = distance/speed First case: Time taken by train = 360/x hour Second case: Time taken by train its speed increase 5 km/h = 360/(x + 5) Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour 360/x - 360/(x +5) = 48/60... ...more

Let original speed of train = x km/h

We know,

Time = distance/speed

First case:

Time taken by train = 360/x hour

Second case:

Time taken by train its speed increase 5 km/h = 360/(x + 5)

Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour

360/x - 360/(x +5) = 48/60 = 4/5

360 {1/x - 1/(x +5)} = 4/5

360 ×5/4 {5/(x²+5x)}=1

450 x 5 = x² + 5x

x²+5x - 2250 = 0

x = {-5±√(25+9000)}/2

= (-5 ±√(9025))/2

=(-5 ± 95)/2

= -50, 45

But x ≠ -50 because speed doesn't negative,

So, x = 45 km/h

Hence, original speed of train = 45 km/h

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Answers 2 Comments Anil UpadhyayDirections

x Class 9 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

4

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

Class 10 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

4

Board

CBSE

CBSE Subjects taught

Mathematics, Science, English

Taught in School or College

No

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No Reviews yet! Be the first one to Review

Answered on 22/01/2018 CBSE/Class 10/Social Studies Tuition/Class IX-X Tuition

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Answers 3 Comments Answered on 22/01/2018 CBSE/Class 10/Social Studies Tuition/Class IX-X Tuition

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Answers 3 Comments Answered on 22/01/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

Given equation is: (c 2 – ab) x 2 – 2 (a 2 – bc) x + (b 2 – ac) = 0 To prove: a = 0 or a 3 + b 3 + c 3 = 3abc Proof: From the given equation, we have a = (c2 – ab) b = –2 (a 2 – bc) c = (b 2 – ac) It is being given that equation has real and equal roots ∴... ...more

Given equation is:

(*c* ^{2} – *ab*) *x* ^{2} – 2 (*a* ^{2} – *bc*) *x* + (*b* ^{2 }– *ac*) = 0

To prove: *a* = 0 or *a* ^{3} + *b* ^{3} + *c* ^{3} = 3*abc*

Proof: From the given equation, we have

*a* = (c^{2} – *ab*)

*b* = –2 (*a* ^{2} – *bc*)

*c* = (*b* ^{2} –* **ac*)

It is being given that equation has real and equal roots

∴ *D* = 0

⇒ *b* ^{2} – 4*ac** *= 0

On substituting respective values of *a*, *b* and *c* in above equation, we get

[–2 (*a* ^{2} – *bc*)]^{2} – 4 (*c* ^{2} – *ab*) (*b* ^{2} – *ac*) = 0

4 (*a* ^{2} – *bc*)^{2} – 4 (*c* ^{2} *b* ^{2} – *ac* ^{3} – *ab* ^{3} + *a* ^{2} *bc*) = 0

4 (*a* ^{4} + *b* ^{2} *c* ^{2} – 2*a* ^{2} *bc*) – 4 (*c* ^{2} *b* ^{2} – *ac* ^{3} – *ab* ^{3} + *a* ^{2} *bc*) = 0

⇒ *a* ^{4} + *b* ^{2} *c* ^{2} – 2*a* ^{2} *bc* – *b* ^{2} *c* ^{2} + *ac* ^{3} + *ab* ^{3} – *a* ^{2} *bc* = 0

⇒ *a* ^{4} + *ab* ^{3} + *ac* ^{3} –3*a* ^{2} *bc* = 0

⇒ *a* [*a* ^{3} + *b* ^{3} + *c* ^{3} – 3*abc*] = 0

⇒*a* = 0 or *a* ^{3} + *b* ^{3} + *c* ^{3} = 3*abc*

Like 0

Answers 6 Comments Answered on 22/01/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Let original speed of train = x km/h We know, Time = distance/speed First case: Time taken by train = 360/x hour Second case: Time taken by train its speed increase 5 km/h = 360/(x + 5) Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour 360/x - 360/(x +5) = 48/60... ...more

Let original speed of train = x km/h

We know,

Time = distance/speed

First case:

Time taken by train = 360/x hour

Second case:

Time taken by train its speed increase 5 km/h = 360/(x + 5)

Time taken by train in first - time taken by train in 2nd case = 48 min = 48/60 hour

360/x - 360/(x +5) = 48/60 = 4/5

360 {1/x - 1/(x +5)} = 4/5

360 ×5/4 {5/(x²+5x)}=1

450 x 5 = x² + 5x

x²+5x - 2250 = 0

x = {-5±√(25+9000)}/2

= (-5 ±√(9025))/2

=(-5 ± 95)/2

= -50, 45

But x ≠ -50 because speed doesn't negative,

So, x = 45 km/h

Hence, original speed of train = 45 km/h

Like 0

Answers 2 Comments X

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