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__Quiz 23__

**1.)** Graph $2y+3x<4$.

*Solution:* First isolate $y$ to make plotting the line easy:
$$2y < -3x+4,$$
and divide by $2$ to get
$$y < -\dfrac{3}{2}x + 2.$$
Now we will plot the line $y=-\dfrac{3}{2}x+2$ as a dotted line:

We will use the test point $(0,0)$ and plug it into the original inequality to get
$$0 < 0+4,$$
or
$$0<4,$$
which is **true**. Therefore we shade in the half of the plane containing the test point:

**2.)** Graph
$$\left\{ \begin{array}{lll}
y < x +3 & \quad (i) \\
y \geq 2x+1 & \quad (ii).
\end{array} \right.$$
*Solution:* We graph line $(i)$ with a dotted line and graph $(ii)$ with a solid line to get

We use the test point $(0,0)$. Plugging it into $(i)$ yields $0 < 3$ which is **true**, so we will shade the half of the graph below the dotted line. We plug the same test point into $(ii)$ to get $0 \geq 1$ which is **false** so we shade the half of the graph above the solid line:

**3.)** Graph
$$\left\{ \begin{array}{lll}
y > 5 & \quad (i) \\
x \leq 2 & \quad (ii).
\end{array} \right.$$
*Solution:* We plot $(i)$ as a dotted line and $(ii)$ as a solid line:

We use the test point $(0,0)$ and plug it into $(i)$ to get $0 > 5$ which is **false** so we shade the half of the plane above the dotted line. We plug the test point into $(ii)$ to get $0 \leq 2$ which is **true** so we shade in the half of the plane to the left of the solid line:

**4.)** Simplify $\sqrt{50}$.

*Solution:* Since $50=2 \cdot 5^2$ we see
$$\sqrt{50} = \sqrt{2 \cdot 5^2} = \sqrt{2} \sqrt{5^2} = 5\sqrt{2}.$$

**5.)** Simplify $\sqrt{60}$.

*Solution:* Since $60=4 \cdot 15$ we see
$$\sqrt{60} = \sqrt{4 \cdot 15} = \sqrt{4} \sqrt{15} = 2\sqrt{15}.$$