Best Answer

The answer depends on whether you mean

A intersecting (B union C) or

(A intersecting B) union C.

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

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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.

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.

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

All angles are formed by two intersecting lines. The pairs of angles opposite each other are called vertical angles. If two angles are vertical, they measure the exact same. Say you name the angles formed as A, B, C, and D. A and C are vertical and B and D are vertical. The angles next to each other formed by intersecting lines are supplementary, and add up to 180 degrees. That means A + B = 180 (since they are next to each other) and B+C = 180. Subtracting the first equation by the second equation gives us A + B - B - C = 180 - 180, which simplifies to A - C = 0, which further simplifies to A = C. The same can be said about B and D being equal.

To solve this you need more information about the relationship between a, b and c.

Related questions

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