Which system of linear inequalities is graphed? {y< 3x−1x+3y> 6 {y> 3x−1x+3y< 6 {y≤3x−1x+3y≥6 {y≥3x−1x+3y≤6

Option B

Step-by-step explanation:

To check the system of inequalities we will take a point in the shaded area of the graph.

Then we will plug in the coordinates in the inequalities to check the validity of the inequality.

Let a point in the shaded area is ( 0,1)

Option A ⇒  y < 3x - 1

                    1 < 3 × 0 - 1

                    1 < -1

                    False. So this graph is not for this system.

Option B ⇒ y > 3x - 1

                    1 > 3 ×0 - 1

                    1 > -1

                   x + 3y < 6

                   0 + 3 × 1 < 6

                   3 < 6

                  Valid for both the inequalities so the system graphed is the set                       of y > 3x - 1 and x + 3y < 6

Option C ⇒ y ≤ 3x - 1

                    1 ≤ 3 × 0 - 1

                    1 ≤ -1

                    Not valid

Option D ⇒ y ≥ 3x - 1

                     1 ≥ 3 × 0 -1

                     1 ≥ -1

                    Since 1 ≠ -1

So this point is not in the shaded area.

Therefore, Option B is the answer.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116