How to graph a two-step inequality

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Geometry › How to graph a two-step inequality

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Let D be the region on the (x,y) coordinate plane that contains the solutions to the following inequalities:

$x\leq k$ , where is a positive constant

$0\leq y\leq 12x$

Which of the following expressions, in terms of ___, is equivalent to the area of D?

CORRECT

Explanation

Inequality_region1

Solve and graph the following inequality:

$2x - 7 \geq 6$

$x\geq6.5$

CORRECT

$x\geq13$

$x\leq6.5$

$x\leq 13$

Explanation

To solve the inequality, the first step is to add to both sides:

$2x - 7 +7 \geq 6+7$

$2x \geq 13$

The second step is to divide both sides by :

$\frac{2x}{2} \geq \frac{13}{2}$

$x \geq 6.5$

To graph the inequality, you draw a straight number line. Fill in the numbers from to infinity. Infinity can be designated by a ray. Be sure to fill in the number , since the equation indicated greater than OR equal to.

The graph should look like:

Number_line

What is the area of the shaded region for this system of inequalities:

$0\leq y \leq \frac{1}{2}x+3$ ; $\frac{1}{2}x \leq y \leq 5$

CORRECT

$\infty$

Explanation

Once graphed, the inequality will look like this:

Inequality b

To find the area, it is easiest to consider it as 2 congruent triangles with base 6 and height 3.

The total area will then be

$2\cdot \frac{1}{2}(6)(3)$ , or just $6\cdot 3 = 18$ .

Which inequality does this graph represent?

Inequality a

$0<y\leq \frac{1}{2}x+1$ ;

$0<y\leq-\frac{3}{5}x+3$

CORRECT

$0<y\leq-\frac{1}{2}x+1$ ;

$0<y\leq \frac{3}{5}x+3$

$-\frac{3}{5}x+3 \geq y \geq \frac{1}{2}x+1$

$0\geq y \geq -\frac{3}{5}x+3$ ;

$0 \geq y \geq \frac{1}{2}+1$

Explanation

The two lines represented are $y=\frac{1}{2}x+1$ and $y=-\frac{3}5{}x+3$ . The shaded region is below both lines but above

Points and lie on a circle. Which of the following could be the equation of that circle?

$(x-3)^{2}+(y-3)^{2}=9$

CORRECT

$(x-3)^{2}+(y-3)^{2}=3$

$(x-3)^{2}+(y+3)^{2}=9$

$(x-3)^{2}+(y+3)^{2}=3$

$(x+3)^{2}+(y+3)^{2}=9$

Explanation

If we plug the points and into each equation, we find that these points work only in the equation $(x-3)^{2}+(y-3)^{2}=9$ . This circle has a radius of and is centered at .