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Ask QuestionPosted by Rudrakshi Khetan 6 years, 3 months ago
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Shreya Raj 6 years, 3 months ago
Posted by Tukaram Yedle 6 years, 3 months ago
- 1 answers
Posted by Harshit Agarwal 6 years, 3 months ago
- 2 answers
Yogita Ingle 6 years, 3 months ago
First ten odd natural numbers are
1 ,3 ,5 ,7 ,9 ,11 ,13 ,15 ,17 ,19
mean = sum of all the observations / Total number of observations
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 / 10
= 100 / 10
= 10
So the mean is 10
Posted by Manya Sharma 6 years, 3 months ago
- 1 answers
Shreyas Soni 6 years, 3 months ago
Posted by Tirthesh Shah 6 years, 3 months ago
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Sia ? 6 years, 3 months ago
First of all you should be know that
formula
a³ + b³ + c³ -3abc = ( a + b + c)(a² + b² + c² - ab - bc - ca).
LHS = a³ + b³ + c³ - 3abc
= ( a³ + b³ ) + c³ - 3abc
= ( a + b)³ - 3ab( a + b) +c³ - 3abc
= ( a + b)³ - 3a²b - 3ab² + c³ - 3abc
= {(a + b)³ + c³ } -3a²b - 3ab² - 3abc
= ( a + b + c)³ - 3c(a + b) ( a + b + c ) -3ab(a + b+ c)
= ( a + b + c){ ( a + b + c)² -3c(a +b) - 3ab }
= ( a + b + c){ a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab }
= ( a + b + c)( a² + b² + c² - ab - bc -ca) = RHS
LHS = RHS
Hence, Proved
Posted by Dhananjay Kumar 6 years, 3 months ago
- 1 answers
Posted by Hemavati Gadaraddi 6 years, 3 months ago
- 0 answers
Posted by Satyam Singh 6 years, 3 months ago
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Posted by Salvinder Thind 6 years, 3 months ago
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Shreya Raj 6 years, 3 months ago
Posted by Kessang Tamang Kessang Tamang 6 years, 3 months ago
- 1 answers
Gaurav Seth 6 years, 3 months ago
Triangles on the same base and between same parallels are equal in area.
Given
: Two triangles ABC and DBC are on the same base BC and between same parallels EF and BC.
To Prove : ar(ΔABC) = ar(ΔDBC)
Construction : Through B, draw BE || AC intersecting line AD in E and through C draw CF || BD intersecting the line DA in F.
Proof : EACB and DFCB are parallelograms (Since two pairs of opposite sides are parallel)
Also ||gm EACB and ||gm DFCB are on the same base BC and between same parallels EF and BC.
∴ ar(||gm EACB)= ar(||gm DFCB)...(i)
Now AB is the diagonal of ||gm EACB
∴ ar(ΔEAB) = ar(ΔABC)
∴ ar(ΔABC) = (1/2)ar(||gm EACB) .....(ii)
Similarly,
ar(ΔDBC) = (1/2) ar(||gm DFCB)...(iii)
From equations (i), (ii) and (iii), we get
ar(Δ ABC) = ar(Δ DBC)
Posted by Raghuraj Thakur 6 years, 3 months ago
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Posted by Deepak Sharma 6 years, 3 months ago
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Posted by M Kavi Varshini 6 years, 3 months ago
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Posted by Sanjay Kumar 6 years, 3 months ago
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Posted by Anjali Mehlawat 6 years, 3 months ago
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Posted by Jyoti Ahirwar 6 years, 3 months ago
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Posted by Atinder Kaur 6 years, 3 months ago
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Posted by Novesh No 6 years, 3 months ago
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Posted by Prashant Kumar 6 years, 3 months ago
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Hitesh Bhakuni 6 years, 3 months ago
Posted by Sangeet Kumar Achari 6 years, 3 months ago
- 1 answers
Gaurav Seth 6 years, 3 months ago
In the given triangleABC,
Angle B = 90°
Hence Angle A + Angle C = 180°-90°.....[Sum of all angles of the triangle are 180°.]
Therefore Angle B is the greater angle than Angle A and Angle C. [Angle A>Angle C ]
We know that.. Side opposite to greater angle is always longer.
Therefore.. Side opposite to angle B is AC
[AC > BC and AC > AB]
That is AC is the longest side..
[AC is the hypotenuse]
Posted by Shweta Mishra 6 years, 3 months ago
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Posted by Binu Rooban 6 years, 3 months ago
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Posted by Dharmendra Tiwari 6 years, 3 months ago
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Harshit Maurya 6 years, 3 months ago
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Kanishka Kamboj 6 years, 3 months ago
1Thank You