Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
let's recall that d = rt, distance = rate * time.
he went upstream to a distance "d", got tired and came back to his starting point, so he rowed back a distance "d" exactly.
we know the rates, we also know the trip took 2 hours, let's say on the way over he took "t" hours to get there, on the way back he lasted then "2 - t" hours.
[tex]\begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Upstream&d&2&t\\ Downstream&d&8&2-t \end{array}\qquad \qquad \begin{cases} d=2t\\\\ d=8(2-t) \end{cases}[/tex]
[tex]\stackrel{\textit{substituting on the 2nd equation}}{2t=8(2-t)\implies 2t=16-8t}\implies 10t=16\implies t=\cfrac{16}{10}\implies t=\cfrac{8}{5} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{we know that}}{d=2t}\implies d=2\left( \cfrac{8}{5} \right)\implies \stackrel{\textit{3 miles and 1056 feet}}{d=\cfrac{16}{5}\implies d=3\frac{1}{5}}[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
