Let x={x|x is a whole number less than 15}, y={x|x is a multiple of 3}, z={x|x is a real number greater than or equal to 5.5} what is xny? (1 point) a- {0,3,6,9,12} b- {3,6,9,12,15} c- { -6,-3,0,3,6,9} d- { -6,-3,0,3,6,9,12}

 The correct option is

(A) {0,3,6,9,12}.

Step-by-step explanation:  We are given the following three sets :

X={x|x is a whole number less than 15},

Y={x|x is a multiple of 3},

Z={x|x is a real number greater than or equal to 5.5}

We are to find the set X∩Y.

The elements in the given sets X, Y and Z can be written as follows :

The set X∩Y will contain all the elements that are present in both X and Y.

Therefore, we get

X∩Y = {0,3,6,9,12}.

Thus, (A) is the correct option.

X = { 0, 1, 2, 3, 4, 5, 6, 7, ... , 14 }

Y = { 0, 3, 6, 9, 12, 15, 18, ... }

Z = {x| x ∈ R ∧ x ≥ 5.5 }

X ∩ Y = { 0, 3, 6, 9, 12 }

A

The intersection X ∩ Y, is the set of all things that are members of both X and Y.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3