+9466 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.
+9466 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.
Help me please!
