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Q: How do you solve A intersecting B union C?

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If: a = b+c+d Then: c = a-b-d

not (b or c) = (not b) and (not c)

No- this is not true in general. Counterexample: Let a = {1,2}, b = {1} and c ={2}. a union c = [1,2} and b union c = {1,2} but a does not equal b. The statement be made true by putting additional restrictions on the sets.

ax - b = c ax = b + c x = (b + c)/a

There is nothing to "solve". You can evaluate the expression when each of a, b and c are TRUE or FALSE. But that is not solving.

exactly one

suppose x is in B. there are two cases you have to consider. 1. x is in A. 2. x is not in A Case 1: x is in A. x is also in B. then x is in A intersection B. Since A intersection B = A intersection C, then this means x is in A intersection C. this implies that x is in C. Case 2: x is not in A. then x is in B. We know that x is in A union B. Since A union B = A union C, this means that x is in A or x is in C. since x is not in A, it follows that x is in C. We have shown that B is a subset of C. To show that C is subset of B, we do the same as above.

If: a = 2b+c Then: a-c = 2b And: b = (a-c)/2

complement of c

c equals b over 8

p=b+3a+c p-3a-c=b+3a-3a+c-c p-3a-c=b b=p-3a-c

to solve ab=c+a for a:Divide both sides by b, so:a= (c+a)/b

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