Question

the circumference of a regulation discus used in the men’s olympic track and field events should be 691.15 millimeters (mm). the manufacturing tolerance is within 3.15 mm. write an absolute value equation that could be used to determine the maximum and minimum value of x, the circumference of a regulation discus used in the men’s olympic track and field events.

Explanation:

Step1: Define the variable and tolerance

Let \( x \) be the circumference of the discus. The target circumference is \( 691.15 \) mm, and the tolerance is \( 3.15 \) mm. This means the difference between \( x \) and \( 691.15 \) should be at most \( 3.15 \).

Step2: Write the absolute value equation

The absolute value equation representing the maximum and minimum values of \( x \) is \( |x - 691.15| = 3.15 \). This equation states that the distance between \( x \) and \( 691.15 \) on the number line is equal to \( 3.15 \), which gives the maximum (\( 691.15 + 3.15 \)) and minimum (\( 691.15 - 3.15 \)) values of \( x \).

Answer:

\( |x - 691.15| = 3.15 \)