Refer to unions and intersections of relations. Since relations are subsets of Cartesian products, their unions and intersections can be calculated as for any subsets. Given two relations R and S from A to B, R ∪ S = {(x, y) ∈ A × B | (x, y) ∈ R or (x, y) ∈ S}
R ∩ S = {(x, y) ∈ A × B | (x, y) ∈ R and (x, y) ∈ S}.

Exercise

Let A = { –1, 1, 2, 4} and B = {1, 2} and define relations R and S from A to B as follows: For all (x, y) ∈ A × B,
x R y ⇔ |x | = |y| and
x S y ⇔ x – y is even.

Required:
State explicitly which ordered pairs are in A × B, R, S, R ∪ S, and R ∩ S.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

the answer is -6

step-by-step explanation:

the answer is -6 because its next to the variable m

2 equivalent expressions to represent the area of a rectangle and the top is n-3 and the side is 2

area of the rectangle = length * width

given length = n - 3

width = 2

replace it in the area

so area of the rectangle = (n-3) *2

= 2(n-3)

one of the expression is 2(n-3)

to get another expression we distribute 2 inside the parenthesis

2(n-3) is 2n - 6

so another expression is 2n - 6

two equivalent expression are

2(n - 3)   , 2n - 6