The point (0, 0) is a solution to which of these inequalities? a. y – 4 < 3x – 1 b. y – 1 < 3x – 4 c. y + 4 < 3x – 1 d. y + 4 < 3x + 1

A

Step-by-step explanation:

Replace the point (0,0) in each inequality

A. y - 4 < 3x - 1

B. y - 1 < 3x - 4

C. y + 4 < 3x - 1

D. y + 4 < 3x + 1

A. 0 - 4 < 3(0) - 1

- 4 < - 1

True

B. 0 - 1 < 3(0) - 4

- 1 < - 4

False

C. y + 4 < 3x - 1

0 + 4 < 3(0) - 1

4 < - 1

False

D. y + 4 < 3x + 1

0 + 4 < 3(0) + 1

4 < 1

False

ian gone lie this got me confused

step-by-step explanation:

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.