+0  
 
0
83
3
avatar+20 

need help please and steps would be nice\

gakinotsukai  Aug 21, 2018
 #1
avatar+91381 
+1

For the first one...let's pick a point in the shaded region...I'll choose  (-1, 6)

This point should satisfy all of the inequalities  in any answer

 

Look at the third  answer....we have that

6 - 6  ≤ -11 (-1)         6  + 3(-1) ≥  -4           -2(-1)  + 3(6)   > 15                       

0  ≤ 11   true             6  - 3  ≥ -4                     2    +  18   >   15

                                     3  ≥  -4   true                20  >  15     true

 

So.....this answer is the correct one

 

cool cool cool

CPhill  Aug 22, 2018
 #2
avatar+91381 
+1

Second one

y  + 3x ≥ 2    ⇒  y  ≥ -3x + 2

 

y + 4 ≥ (1/2)x   ⇒   y ≥ (1/2)x - 4

 

Look at the graph  on the bottom left...note that   (2,0)  is a point in the shaded region

 

So...replacing these into the inequalities, we have that

 

0 ≥  -3(2) + 2              and          0 ≥ (1/2)(2) - 4

0 ≥ -6 + 2                                    0 ≥ 1 - 4

0 ≥ -4   true                                 0 ≥  -3    true

 

So.....this graph is the correct one

 

 

cool cool cool

CPhill  Aug 22, 2018
 #3
avatar+91381 
+1

Last  one

x  ≤ -4   this means  ( -inf, 4 ]

or          

x  ≥  5    this means  [ 5, inf )

 

So  we  have

 

(-inf, 4 ] or [5, inf )

 

 

cool cool cool

CPhill  Aug 22, 2018

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