Which ordered pair makes both inequalities true? y > –3x + 3 y > 2x – 2 (1,0) (–1,1) (2,2) (0,3)

All options are wrong.

Step-by-step explanation:

We have y > –3x + 3 and y > 2x – 2

Option A - (1,0)

          y > –3 x 1 + 3 = 0

         We have y = 0, which is not greater than 0.

         Option A is not correct.        

Option B - (–1,1)

          y > –3 x -1 + 3 = 6

         We have y = 1, which is not greater than 6.

         Option B is not correct.

Option C - (2,2)

          y > –3 x 2 + 3 = -3

         We have y = 2, which is greater than -3.

          y > 2 x 2 – 2  = 2

         We have y = 2, which is not greater than 2.

         Option C is not correct.  

Option D - (0,3)

          y > –3 x 0 + 3 = 3

         We have y = 3, which is not greater than 3.

         Option D is not correct.    

All options are wrong.

idk

step-by-step explanation:

use that smart brain

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°