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SilverWolf1234
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A number x is no more than 9 units away from 8. Write and solve an absolute-value inequality to show the range of possible values of x.


|x − 8| ≤ 9

−1 ≤ x ≤ 17


|x + 8| ≤ 9

−17 ≤ x ≤ 1


|x − 9| ≤ 8

1 ≤ x ≤ 17


|x + 9| ≤ 8

−17 ≤ x ≤ −1

Verified Answer :

kmHart

The answer is a (|x - 8| ≤ 9 and -1 ≤ x ≤ 17) Because the difference between x and 8 can't be greater than 9, but it can be less than or equal to it, and because the range that x could be would be -1 to 17, since 8 + 9 = 17 and 8 - 9 = -1.

Answer:

Option A is right

Step-by-step explanation:

Given that a number x is no more than 9 units away from 8

This implies that number can be either larger than 8 by less than 9 units or smaller than 8 by less than 9 units.

Difference between x and 8 is at most 9.

We represent in absolute value as

\(|x-8|\leq 9\\-9\leq x-8\leq 9\\-1\leq x\leq 17\)

Hence option A is right

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