Prove or disprove. for all sets a, b, and c, (a⋃b)⋂c ⊆a ⋃(b⋂c)

Let x ∈ (A∪B) ∩ C

⇒ x ∈ (A ∪ B) and x ∈ C

⇒ x ∈ A or x ∈ B and x ∈ C

⇒ x ∈ A or x ∈ B∩C

⇒ x ∈ A ∪ (B∩C)

Now, x ∈ A ∪ (B∩C)

⇒ x ∈ A or x ∈ B ∩ C

⇒ x ∈ A or x ∈ B and x ∈ C

⇒ x ∈ (A∪B) and x ∈ C

⇒ x ∈ (A∪B) ∩ C

Since, x shows the an arbitrary element,

⇒ A ∪ (B∩C) = (A∪B) ∩ C

∵ A set always contains itself,

⇒ A ∪ (B∩C) ⊆  A ∪ (B∩C)

⇒ (A⋃B)⋂C ⊆A ⋃(B⋂C)

Hence, proved...

the answer is c.

for continuous functions, they continue in the same pattern and you have to find that pattern kinda like slope

correct answer is infinitely many solutions

step-by-step explanation:

2(2x + 3) = -4 + 2(2x + 5)

solving the above equation:

4x + 6 = -4 +4x +10

combining like terms

4x -4x = -4+10-6

0=0

since both sides are 0=0, we end up into true statement it means it has infinite many solutions.

so correct answer is infinitely many solutions