This question has two parts 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 Part A Which absolute value equation could be used to determine the minimum and maximum values of z the circumference of a regulation discus used in the men's Olympic track and held events? [2+3.15|=691.15 vert x+3.15vert =691.15 vert x-3.15vert =691.15 vert x+691.15vert =3.15 vert x Part B Detormine the minumand maximum circumforence in millimeters, of the regulation discus Minumum square mm mm

## Step 1<br />The problem involves the concept of absolute value equations. The absolute value of a number is its distance from zero on the number line, regardless of direction. In this case, the absolute value equation is used to represent the manufacturing tolerance of the discus, which is within 3.15 mm of the ideal circumference of 691.15 mm.<br /><br />## Step 2<br />The absolute value equation that represents this situation is \(|x - 691.15| = 3.15\). This equation means that the circumference of the discus (x) can be 3.15 mm more or less than 691.15 mm.<br /><br />## Step 3<br />To find the minimum and maximum circumference of the discus, we need to solve the absolute value equation. The absolute value equation \(|x - 691.15| = 3.15\) can be split into two separate equations: \(x - 691.15 = 3.15\) and \(x - 691.15 = -3.15\).<br /><br />## Step 4<br />Solving the first equation, \(x - 691.15 = 3.15\), we get \(x = 694.3\) mm. This is the maximum circumference of the discus.<br /><br />## Step 5<br />Solving the second equation, \(x - 691.15 = -3.15\), we get \(x = 688.8\) mm. This is the minimum circumference of the discus.