Answer: The answer is = 2x^2 -11x-40
Step-by-step explanation: a= 2x, b=5, c= x, d =8
= 2xx + 2x (-8) + 5x +5(-8)
+ (-a)=-a
then simplify =2xx-2 -8x+5x.8
2xx=2x^2
2.8x=16x
5.8=40
=2x - 16x + 5x- 40
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, ...}
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