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Stitz-Zeager_College_Algebra_e-book

# 005 12t 1 10 1005 100000 100512t 11 ln 100512t

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Unformatted text preview: 4 5 5 + = 3. 6+R 6−R Clearing denominators, we get 5(6 − R) + 5(6 + R) = 3(6 + R)(6 − R) which reduces to R2 = 16. We ﬁnd R = ±4, and since R represents the speed of the river, we choose R = 4. On the day in question, the Meander River is ﬂowing at a rate of 4 miles per hour. One of the important lessons to learn from Example 8.7.3 is that speeds, and, more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just ‘distance-rate-time’ scenarios. Example 8.7.4. Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own? Solution. The key relationship between work and time we use in this problem is: amount of work done = rate of work · time spent working We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case, then: 4 The reader...
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