Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
6. Absolute Value Equations and Inequalities
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Exercise 39 Page 46

Absolute values can be interpreted as the distance from a midpoint.

|C-27.5|≤ 0.25

Practice makes perfect

Absolute values can be interpreted as the distance away from a midpoint. For one-variable absolute value inequalities, this distance can be represented by two points on a number line. These are the endpoints of the given compound inequality.

Because our inequality needs a distance from a midpoint, we should find the halfway point between the endpoints. We can do this by calculating their mean.

Mean=27.25+ 27.75/2= 27.5 Now we need to find the distance between this midpoint and each of the endpoints. To do this, we will find the difference between each of the endpoints and the midpoint.

We see that both 27.25 and 27.75 are 0.25 units away from 27.5. Notice that the given inequality is an and inequality, and the symbols used can be read as less than or equal to. 27.25≤ C ≤ 27.75 27.25 is less than or equal toC and C is less than or equal to 27.75 To write the given compound inequality as an absolute value inequality, we can show that the difference between a number C and the midpoint is less than or equal to the distance we found above. |C- midpoint|≤ distance |C- 27.5|≤ 0.25