A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 27 mi/hr. Another ship leaves the same port at 1:30 P.M. and tr
avels S20°W at 18 mi/hr. Approximately how far apart are the ships at 3:00 P.M.? (Round your answer to the nearest mile.)
1 answer:
To solve this problem you must apply the proccedure shown below:
1. You must apply the Law of Cosines, as you can see in the figure attached. Then:
- The first ship travels at in for two hours. Therefore, the side
is:
- The second ship travels at for
. Therefore, the side
is:
- Now, you can calculate :
The answer is:
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To be a function, it has to pass the vertical line test, which means a vertical line drawn anywhere in the graph should cut the graph only once, not more.
To be a one-to-one function, it has to pass the vertical line test as well as the horizontal line test, which means that a horizontal line drawn anywhere in the graph should cut the graph only once, not more.
<em>Of course, it passes the vertical line test, so its a function. </em>
<em>Does it pass the horizontal line test? </em><em>NO! </em><em>As shown in the attached picture, the blue line (horizontal line) cuts the graph at 3 places. We therefore eliminate choices A and B. We can eliminate choice C because nowhere is it required a function to pass through the origin for it to be one-to-one. Hence, the answer is D (blue line shows this).</em>
<em>ANSWER: D</em>
The required number is :
Solution is in attachment ~
