Let u = {q, r, s, t, u, v, w, x, y, z} a = {q, s, u, w, y} b = {q, s, y, z} c = {v, w, x, y, z} determine the following. (a' ∪ c) ∩ b'

U = { q, r, s, t, u, v, w, x, y, z }

A = { q, s, u, w, y } → A' = { r, t, v, x , z }

B = { q, s, y, z } → B' = { r, t, u, v, w, x }

C = { v, w, x, y, z }

(A' ∪ C) ∩ B'

A' ∪ C = { r, t, v, x , z } ∪ { v, w, x, y, z } = { r, t, v, w, x, y, z }

(A' ∪ C) ∩ B' = { r, t, v, w, x, y, z } ∩ { r, t, u, v, w, x } = { r, t, v, w, x }

A' = U - A, B' = U - B

The union A ∪ B, is the set of all things that are members of either A or B.

The intersection A ∩ B, is the set of all things that are members of both A and B

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

idk

step-by-step explanation: