Bakhtawar Pur, Delhi, India - 110036.
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Hindi Mother Tongue (Native)
MJK Collage 1994
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Bakhtawar Pur, Delhi, India - 110036
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
4
Board
CBSE
CBSE Subjects taught
English, Mathematics, Science
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
4
Board
CBSE
CBSE Subjects taught
English, Mathematics, Science
Taught in School or College
No
1. Which school boards of Class 10 do you teach for?
CBSE
2. Do you have any prior teaching experience?
No
3. Which classes do you teach?
I teach Class 10 Tuition and Class 9 Tuition Classes.
4. Do you provide a demo class?
Yes, I provide a free demo class.
5. How many years of experience do you have?
I have been teaching for 4 years.
Answered on 22/01/2018 Learn CBSE/Class 10/Social Studies
Answered on 22/01/2018 Learn CBSE/Class 10/Social Studies
Answered on 22/01/2018 Learn CBSE/Class 10/Mathematics
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
∴ D = 0
⇒ b 2 – 4ac = 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 – 2a 2 bc) – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0
⇒ a 4 + b 2 c 2 – 2a 2 bc – b 2 c 2 + ac 3 + ab 3 – a 2 bc = 0
⇒ a 4 + ab 3 + ac 3 –3a 2 bc = 0
⇒ a [a 3 + b 3 + c 3 – 3abc] = 0
⇒a = 0 or a 3 + b 3 + c 3 = 3abc
Answered on 22/01/2018 Learn CBSE/Class 10/Mathematics
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
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
4
Board
CBSE
CBSE Subjects taught
English, Mathematics, Science
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
4
Board
CBSE
CBSE Subjects taught
English, Mathematics, Science
Taught in School or College
No
Answered on 22/01/2018 Learn CBSE/Class 10/Social Studies
Answered on 22/01/2018 Learn CBSE/Class 10/Social Studies
Answered on 22/01/2018 Learn CBSE/Class 10/Mathematics
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
∴ D = 0
⇒ b 2 – 4ac = 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 – 2a 2 bc) – 4 (c 2 b 2 – ac 3 – ab 3 + a 2 bc) = 0
⇒ a 4 + b 2 c 2 – 2a 2 bc – b 2 c 2 + ac 3 + ab 3 – a 2 bc = 0
⇒ a 4 + ab 3 + ac 3 –3a 2 bc = 0
⇒ a [a 3 + b 3 + c 3 – 3abc] = 0
⇒a = 0 or a 3 + b 3 + c 3 = 3abc
Answered on 22/01/2018 Learn CBSE/Class 10/Mathematics
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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