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# ​Help me please!

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Jun 11, 2019

#1
+8724
+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.

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
+8724
+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.

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