Question 247794
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Let *[tex \Large r] represent the speed of the freighter.  Then *[tex \Large r\ +\ 14] is the speed of the steamboat.  Let *[tex \Large t] represent the common amount of time.


We know that distance equals rate times time, *[tex \LARGE d\ =\ rt], but this can be solved for *[tex \Large t]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{d}{r}]


Now we can describe the freighter's trip like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{49}{r}]


And we can describe the steamboat's trip like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t\ =\ \frac{84}{r\ +\ 14}]


Now we have two expressions that are both equal to *[tex \Large t], so set them equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{49}{r}\ =\ \frac{84}{r\ +\ 14}]


Now all you need to do is cross-multiply and solve for *[tex \Large r].  Let me know what you get for an answer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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