Two points of a triangle are located at (0, 0) (4, 0) . Which two of the following correctly describe where a third point can be placed so that the area of the triangle is 4 square units?

A. anywhere along the graph of x = 2
B. anywhere along the graph of x = 4
C. anywhere along the graph of y = - 2
D. anywhere along the graph of y = 2
E. anywhere along the x-axis F. anywhere along the y- axis

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.

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation: