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Ask QuestionPosted by ꪊꪑꪖꪀᧁ ΡꪖTꫀꪶ 4 years ago
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Posted by Divyanshi Gehlot 4 years ago
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Gaurav Seth 4 years ago
n(A × B) = 5(4) = 20
n[(A × B) ∩ (B × A)] = 9
Step by step explanation:
Solution :
Given:
n(A) = 5 and n(B) = 4
Thus, we have:
n(A × B) = 5(4) = 20
A and B are two sets having 3 elements in common.
Now,
Let:
A = (a, a, a, b, c) and B = (a, a, a, d)
Thus, we have:
(A × B) = {(a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (a, a), (a, a), (a, a), (a, d), (b, a),
(b, a), (b, a), (b, d), (c, a), (c, a), (c, a), (c, d)}
(B × A) = {(a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a), (a, b), (a, c), (a, a), (a, a), (a, a),
(a, b), (a, c), (d, a), (d, a), (d, a), (d, b), (d, c)}
[(A × B) ∩ (B × A)] = {(a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a), (a, a)}
∴ n[(A × B) ∩ (B × A)] = 9
Posted by Rim Jain 4 years, 1 month ago
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Posted by Rim Jain 4 years, 1 month ago
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Posted by Rim Jain 4 years, 1 month ago
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Posted by Rim Jain 4 years, 1 month ago
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Posted by Soumo Maity 4 years, 1 month ago
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Posted by Soumo Maity 4 years, 1 month ago
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Posted by Sachi Jain 4 years, 1 month ago
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Posted by Sachi Jain 4 years, 1 month ago
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Posted by Jatin Agrawal 4 years, 1 month ago
- 2 answers
Aseem Mahajan 4 years, 1 month ago
b² = ac
c² = bd
ad = bc
For a²+b² , b²+c² , c²+d² in GP , we need to prove :
(b²+c²)² = (a²+b²)(c²+d²)
Solving LHS:
(b²+c²)²= b⁴ + c⁴ + 2b²c²
Solving RHS:
(a²+b²)(c²+d²) = a²c²+a²d²+b²c²+b²d²
= a²c² + b²d² + 2b²c² (because ad=bc)
= (ac)² + (bd)² + 2b²c²
= b⁴ + c⁴ + 2b²c² (because ac = b² , bd = c²)
Hence ,
a²+b² , b²+c² , c²+d² are in GP
Posted by Sahil Manchanda 4 years, 1 month ago
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Posted by Abhijeet Singh 4 years, 1 month ago
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Posted by Sachi Jain 4 years, 1 month ago
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Arinan Aggarwal 4 years ago
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Posted by Aman Thakur 4 years, 1 month ago
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Posted by Anuja Dewan 4 years, 1 month ago
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Posted by Khushi Thakur 4 years, 2 months ago
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Posted by Anurag Agarwal 4 years, 2 months ago
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Yogita Ingle 4 years, 2 months ago
A set is a well-defined collection of objects.
There are two methods of representing a set
- Roster or Tabular form In the roster form, we list all the members of the set within braces { } and separate by commas.
- Set-builder form In the set-builder form, we list the property or properties satisfied by all the elements of the sets.
Posted by Gaurav Kumar 4 years, 2 months ago
- 1 answers
Mani Kiran Batchu 4 years, 1 month ago
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Arjun Beniwal 4 years ago
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