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Answer:
hello your question is incomplete below is the missing parts
(a) A\ (A\B) = B\(B\A)
(b) A\ (BA) = B\(A\B)
answer : A\ (A\B) = B\(B\A) = always true
A\ (BA) = B\(A\B) = sometimes true and sometimes false
Step-by-step explanation:
(a) A\ (A\B) = B\(B\A). = ALWAYS TRUE
using de Morgan's law to prove this
A\ (A\B) = A\ ( A ∩ B^c )
= A ∩ ( A^C ∪ B )
= ( A ∩ A^C ) ∪ ( A ∩ B )
= Ф ∪ ( A ∩ B )
= ( A ∩ B )
ALSO : B\(B\A) = attached below is the remaining parts of the solution
B) A\ (BA) = B\(A\B) = Sometimes true and sometimes false
attached below is the prove using De Morgan's law
You would need 7 freezer bags because you have to turn both fractions into improper fractions. to get that you have to multiply 15 by 4 and then add 3 so you get 63/1. Then for the next fraction, you multiply 4 by 2 and then add 1 so you get 9/1.
Now you have to divide the 2 fractions to find out how many freezer bags you need. So it would look like this. 63/1 divided by 9/1.
To divide fractions, you use the rule (keep, change, flip)
So you keep the first fraction 63/1 but you change to division sign to multiplication and lastly you flip the second fraction from 9/1 to 1/9 and now you multiply so it looks like this.
63/1 x 1/9 and you get 63/9 which equals 7 freezer bags.
I hope this helped you?? :)
BC= 30
Reason: 64-8
56/7
8=x
3x=24
24+6
30
Answer:
whats the question
Step-by-step explanation:
