Solved on Nov 13, 2023

Find the numbers with absolute values of 33, 53\frac{5}{3}, and 9129 \frac{1}{2}.

STEP 1

Assumptions1. Absolute value of a number is its distance from zero on the number line, regardless of the direction. . Therefore, the absolute value of a number is always non-negative.
3. Both positive and negative values of a number have the same absolute value.

STEP 2

The absolute value of a number is the number itself if it is positive or zero, and it is the negation of the number if it is negative. So, for each given number, there are two numbers that have that absolute value the number itself and its negation.
For the number, the numbers with an absolute value of are and -.
=,=|| =, |-| =

STEP 3

For the number 53\frac{5}{3}, the numbers with an absolute value of 53\frac{5}{3} are 53\frac{5}{3} and 53-\frac{5}{3}.
53=53,53=53|\frac{5}{3}| = \frac{5}{3}, |-\frac{5}{3}| = \frac{5}{3}

STEP 4

For the number 9129\frac{1}{2}, the numbers with an absolute value of 9129\frac{1}{2} are 9129\frac{1}{2} and 912-9\frac{1}{2}.
912=912,912=912|9\frac{1}{2}| =9\frac{1}{2}, |-9\frac{1}{2}| =9\frac{1}{2}So, the numbers that have an absolute value of3 are3 and -3, the numbers that have an absolute value of 3\frac{}{3} are 3\frac{}{3} and 3-\frac{}{3}, and the numbers that have an absolute value of 9129\frac{1}{2} are 9129\frac{1}{2} and 912-9\frac{1}{2}.

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