Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like
(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}
√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then
(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}
A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that
(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}
I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then
(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.
N is a subset of B, so
(iv.b) N - B = N = {1, 2, 3, ...}
Answer:
-13
Step-by-step explanation:
-4-8+3+6-10
Answer:
1 or x (the equation is y=x+1)
Step-by-step explanation:
Answer:
A
Step-by-step explanation:
Because you have to use the Pythagorean theorem to solve for this.
The answer is 8 because:
24 divided by 3 = y and y is 8
