Write a note on Asteroid 2021 AF8 detected by NASA
Answers
Explanation:
hopefully its helpful
pls mark it brainlest
Answer:
so cute dp.my lady
\red{\large \mathfrak {qestion}}qestion
\:
\tt{n(A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C) }n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)
\:
\green{\large \mathfrak {solution}}solution
\:
We know that,
\:
\tt \: n(A \cup B \cup C) = n(A \cup (B \cup C))n(A∪B∪C)=n(A∪(B∪C))
\:
\tt\: = n(A) + n(B \cup C) - n(A \cap (B \cap C))=n(A)+n(B∪C)−n(A∩(B∩C))
\:
\tt∵ n(A \cup B) = n(A) + n(B) - n(A \cap B)∵n(A∪B)=n(A)+n(B)−n(A∩B)
\:
\tt∴ A \cap (B \cup C) = (A \cup B) \cup (A \cap C) \: \: \: \: \: \: \sf \: ..(distributie \: property)∴A∩(B∪C)=(A∪B)∪(A∩C)..(distributieproperty)
\:
\tt= n(A) + n(B \cup C) - n(A \cap (B \cap C))=n(A)+n(B∪C)−n(A∩(B∩C))
\:
\tt= n(A) + [n(B) + n(C) - n(B \cap C)] - n(A \cap (B \cap C)=n(A)+[n(B)+n(C)−n(B∩C)]−n(A∩(B∩C)
\:
We know that,
\:
\tt(B \cup C) = n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n((A \cap B) \cap (A \cap C))](B∪C)=n(A)+n(B)+n(C)−n(B∩C)−[n(A∩B)+n(A∩C)−n((A∩B)∩(A∩C))]
\:
\tt= n(A) + n(B) + n(C) - n(B \cap C) - [n(A \cap B) + n(A \cap C) - n(A \cap B \cap C]=n(A)+n(B)+n(C)−n(B∩C)−[n(A∩B)+n(A∩C)−n(A∩B∩C]
\:
\tt= n(A) + n(B) + n(C) - n(B \cap C) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C) = RHS=n(A)+n(B)+n(C)−n(B∩C)−n(A∩B)−n(A∩C)+n(A∩B∩C)=RHS
\:
\sf∴ \: LHS = RHS∴LHS=RHS
\:
Hence Proved
\:
\pink{\large \mathfrak { \: !! \: hope \: it \: helps \: you \: !! }}!!hopeithelpsyou!!