Use the piecewise function to evaluate points 4-5), 40), and A2). 2x - 1
f(x) = {2,3
3x
x <-3
-3 X>0
(-2x + 3
O A) A-5) = -11, 40) = 0, and 12) = -1
B) (-5) = -15, 10) = 0, and A2) = 6
OC) 4-5) = -11, 10) = -1, and A2) = 6
OD) (-5) = -11, 10) = 0, and A2) = 3

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.