How to graph a two-step inequality

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ACT Math › How to graph a two-step inequality

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1

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

0

0

0

0

Explanation

Inequality_region1

2

Solve and graph the following inequality:

$2x - 7 \geq 6$

$x\geq6.5$

CORRECT

$x\geq13$

0

0

$x\leq6.5$

0

$x\leq 13$

0

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

3

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

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

0

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

0

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

0

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

0

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 .

4

Which of the following lines is perpendicular to the line ?

$y=\frac{-x}{3}+2$

CORRECT

0

$y=\frac{x}{3}+3$

0

$y=\frac{x}{3}+4$

0

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Explanation

The key here is to look for the line whose slope is the negative reciprocal of the original slope.

In this case, $\frac{-1}{3}$ is the negative reciprocal of .

Therefore, the equation of the line which is perpendicular to the original equation is:

$y=\frac{-x}{3}+2$

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