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Help me please!

 Jun 11, 2019

Best Answer 

 #1
avatar+9488 
+3

1.

 

1x2+2y5x6y2 = 1x2+2y5x6y2 x2y2x2y2 = y2+2x2y5xy26x2     and   y ≠ 0

 

 

2.

 

To graph  y12x+212 ,  graph the line  y=12x+212  then lightly shade the portion above the line.

 

To graph  y<15x+6 ,  graph the line  y=15x+6   but make it a dotted line. Then lightly shade the portion below it.

 

The solution to the system of inequalities is the portion of the graph that is shaded twice.

 

See here: https://www.desmos.com/calculator/hzox2qggig

 

According to the graph, the point  (0, 4)  should satisfy both inequalities.

 

Is it true that  412(0)+212  ?  Yes it is true that  4212

 

Is it true that  4<15(0)+6  ?  Yes it is true that  4<6

 

So  (0, 4)  does satisfy both inequalities.

 Jun 11, 2019
 #1
avatar+9488 
+3
Best Answer

1.

 

1x2+2y5x6y2 = 1x2+2y5x6y2 x2y2x2y2 = y2+2x2y5xy26x2     and   y ≠ 0

 

 

2.

 

To graph  y12x+212 ,  graph the line  y=12x+212  then lightly shade the portion above the line.

 

To graph  y<15x+6 ,  graph the line  y=15x+6   but make it a dotted line. Then lightly shade the portion below it.

 

The solution to the system of inequalities is the portion of the graph that is shaded twice.

 

See here: https://www.desmos.com/calculator/hzox2qggig

 

According to the graph, the point  (0, 4)  should satisfy both inequalities.

 

Is it true that  412(0)+212  ?  Yes it is true that  4212

 

Is it true that  4<15(0)+6  ?  Yes it is true that  4<6

 

So  (0, 4)  does satisfy both inequalities.

hectictar Jun 11, 2019

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