Question

brandon is deep - sea diving with irma, who is to the south of him. a fish emerges from a hole to the east of irma. the fish swims 6.1 meters to irma and then swims 7.8 meters to brandon. the fish then swims back to its hole and disappears. how far did the fish travel in all? if necessary, round to the nearest tenth. meters

Explanation:

Step1: Find the distance from the hole to Brandon

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 6.1\) (distance from hole to Irma) and \(b\) is the distance from Irma to Brandon (\(7.8\)). Let \(c\) be the distance from the hole to Brandon.

$$c=\sqrt{6.1^{2}+7.8^{2}}=\sqrt{37.21 + 60.84}=\sqrt{98.05}\approx9.9$$

Step2: Calculate the total distance

The fish travels \(6.1\) meters (hole - Irma), \(7.8\) meters (Irma - Brandon), and then back to the hole (\(9.9\) meters).
Total distance \(d=6.1 + 7.8+9.9\)

$$d=23.8$$

Answer:

\(23.8\)