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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.

No

Hindi

Graduate from MJK Collage in 1994.

Bakhtawar Pur, Delhi, India- 110036.

Class 9 Tuition

Board

CBSE

IB Subjects taught

CBSE Subjects taught

Mathematics, Science, English

ICSE Subjects taught

IGCSE Subjects taught

Experience in School or College

Taught in School or College

No

State Syllabus Subjects taught

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

4

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No Reviews yet! Be the first one to Review

"How did the Non-Cooperation movement unfold? Who participated in it?" in 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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| 0

"How did the Non-Cooperation movement unfold? Who participated in it?" in 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.

0

| 0

"If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc "" in 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} = 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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"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." in CBSE/Class 10/Mathematics, Tuition/Class IX-X Tuition

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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Anil Upadhyay address

x Class 9 Tuition

Board

CBSE

IB Subjects taught

CBSE Subjects taught

Mathematics, Science, English

ICSE Subjects taught

IGCSE Subjects taught

Experience in School or College

Taught in School or College

No

State Syllabus Subjects taught

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

4

Class 10 Tuition

Board

CBSE

IB Subjects taught

CBSE Subjects taught

Mathematics, Science, English

ICSE Subjects taught

IGCSE Subjects taught

Experience in School or College

Taught in School or College

No

State Syllabus Subjects taught

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

4

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

No Reviews yet! Be the first one to Review

"How did the Non-Cooperation movement unfold? Who participated in it?" in 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.

0

| 0

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

0

| 0

"If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc "" in 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} = 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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| 0

"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." in CBSE/Class 10/Mathematics, Tuition/Class IX-X Tuition

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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| 0

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