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Help me please!

 Jun 11, 2019

Best Answer 

 #1
avatar+8966 
+3

1.

 

\(\dfrac{\frac{1}{x^2}+\frac{2}{y}}{\frac5x-\frac{6}{y^2}}\ =\ \dfrac{\frac{1}{x^2}+\frac{2}{y}}{\frac5x-\frac{6}{y^2}}{\color{blue}\ \cdot\dfrac{x^2y^2}{x^2y^2}}\ =\ \dfrac{y^2+2x^2y}{5xy^2-6x^2}\)     and   y ≠ 0

 

 

2.

 

To graph  \(y\geq-\frac12x+2\frac12\) ,  graph the line  \(y=-\frac12x+2\frac12\)  then lightly shade the portion above the line.

 

To graph  \(y<\frac15x+6\) ,  graph the line  \(y=\frac15x+6\)   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  \(4\geq-\frac12(0)+2\frac12\)  ?  Yes it is true that  \(4\geq2\frac12\)

 

Is it true that  \(4<\frac15(0)+6\)  ?  Yes it is true that  \(4<6\)

 

So  (0, 4)  does satisfy both inequalities.

 Jun 11, 2019
 #1
avatar+8966 
+3
Best Answer

1.

 

\(\dfrac{\frac{1}{x^2}+\frac{2}{y}}{\frac5x-\frac{6}{y^2}}\ =\ \dfrac{\frac{1}{x^2}+\frac{2}{y}}{\frac5x-\frac{6}{y^2}}{\color{blue}\ \cdot\dfrac{x^2y^2}{x^2y^2}}\ =\ \dfrac{y^2+2x^2y}{5xy^2-6x^2}\)     and   y ≠ 0

 

 

2.

 

To graph  \(y\geq-\frac12x+2\frac12\) ,  graph the line  \(y=-\frac12x+2\frac12\)  then lightly shade the portion above the line.

 

To graph  \(y<\frac15x+6\) ,  graph the line  \(y=\frac15x+6\)   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  \(4\geq-\frac12(0)+2\frac12\)  ?  Yes it is true that  \(4\geq2\frac12\)

 

Is it true that  \(4<\frac15(0)+6\)  ?  Yes it is true that  \(4<6\)

 

So  (0, 4)  does satisfy both inequalities.

hectictar Jun 11, 2019

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