Question

dean is playing hide - and - seek with janice and emmet. janice is hiding 3.3 meters south of dean, and emmet is hiding due east of janice. if dean is 7.9 meters from emmet, how far apart are janice and emmet? if necessary, round to the nearest tenth.
meters

Explanation:

Step1: Apply Pythagorean theorem

Let the distance between Janice and Emmet be \(d\). We know that one leg \(a = 3.3\) (distance between Janice and Dean) and the hypotenuse \(c=7.9\) (distance between Dean and Emmet). According to the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), we can find \(b\) (distance between Janice and Emmet) as \(b=\sqrt{c^{2}-a^{2}}\).

Step2: Substitute values

Substitute \(a = 3.3\) and \(c = 7.9\) into the formula: \(b=\sqrt{7.9^{2}-3.3^{2}}=\sqrt{(7.9 + 3.3)(7.9-3.3)}=\sqrt{11.2\times4.6}=\sqrt{51.52}\).

Step3: Calculate the square - root

\(\sqrt{51.52}\approx7.2\)

Answer:

\(7.2\)