Edit View History Bookmarks Profiles Tob Window Help texs classroom Link ALEKS - Sydney Cano - Know Edgenifity Cowboy Channelt Google Docs Dashboard Google Dive e2020.geniusals. Course Map - Iman. Wrzant Knownedge Check Question 3 Sydney A rowing team rowed 60 miles while going with the current in the same amount of time as it took to row 10 miles going against the current. The rate of the Expatiat current was 5 miles per hour. Find the rate of the rowing team in still water. Rate of the team in still water: \( \square \) miles/hour

To solve this rowing team problem, we can use the formula for speed, which is distance divided by time. When the team rows with the current, their effective speed is the sum of their rowing speed and the current speed. Conversely, when rowing against the current, their effective speed is the difference. The equations can be set up based on these speeds. Let \( r \) be the rowing team's rate in still water. The time spent rowing with the current for 60 miles is \( \frac{60}{r + 5} \), and the time for 10 miles against the current is \( \frac{10}{r - 5} \). Since both times are equal: \[ \frac{60}{r + 5} = \frac{10}{r - 5} \] Cross-multiplying gives \( 60(r - 5) = 10(r + 5) \). Simplifying that will lead you to find the value of \( r \). This problem emphasizes the relationship between rate, distance, and time and is a staple in both classroom math and real-life situations. Fishermen, boats, and even swimming athletes deal with similar timing when currents and waves are involved, showcasing how these concepts weave into daily activities! If you're prone to making mistakes in these types of problems, a common pitfall is missing the signs when dealing with current speeds. Always remember that rowing with the current adds to your speed while rowing against it subtracts. A helpful tip is to clearly label your variables and write down the times upfront to avoid confusion as you work through the calculations!