Mathematics, 28.07.2019 19:30 sese47

The speed of the current in a river is 6 mph. a ferry operator who works that part of the river is looking to buy a new boat for his business. every day, his route takes him 22.5 miles against the current and back to his dock, and he needs to make this trip in a total of 9 hours. he has a boat in mind, but he can only test it on a lake where there is no current. how fast must the boat go on the lake in order for it to serve the ferry operator’s needs?

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Lest organize and label the information the problem is giving us first:
We know that the total time of the trip is 9 hours, so $t_{t}=9$ . We also know that the total distance of the trip is 22.5 miles, so $d_{t}=22.5$ . And we also know that the speed of the current is 6 mph, so $S_{c}=6$ .
Now, lets organize and label the things we don't know:
We don't know the speed of the boat in the lake, so $S_{l}=?$ . We don't know the time of the trip against the current, so $t_{ac}=?$ . We don't know the time of the trip with the current, so $t_{wc}=?$

The next thing we are going to do is set up equations to relate the things we don'k know with the things we actually know.
If the distance of the whole trip is 22.5 miles, the distance of the trip against the current is $\frac{22.5}{2}=11.25$ , so $d_{ac}=11.25$ . Similarly, the distance with the current is 11.25, so $d_{wc}=11.25$ .

Now lets use the equation $speed= \frac{distance}{time}$ to solve our problem:
Speed of the boat against the current:
$S_{l}-6= \frac{11.25}{t_{ac} }$ equation (1)
Speed of the boat with the current:
$S_{l}+6= \frac{11.25}{t_{wc}}$ equation (2)
Notice that we don't know the values of $t_{ac}$ and $t_{wc}$ , but we cant take advantage of the fact that the total time of the trip is 9 hours, so $9=t_{ac}+t_{wc}$ . Lets solve this equation for $t_{wc}$ :
$t_{wc}=9-t_{ac}$ equation (3)

Now we can replace equation (3) in equation (2) to express our equations with only tow variables:
$S_{l}+6= \frac{11.25}{9-t_{ac} }$ equation (4)
Next, lets solve for $t_{ac}$ in equation (4):
$S_{l}+6= \frac{11.25}{9-t_{ac}}$
$9-t_{ac}= \frac{11.25}{S_{l}+6}$
$-t_{ac}= \frac{11.25}{S_{l}+6} -9$
$-t_{ac}= \frac{11.25-9S_{l}-54}{S_{l}+6}$
$-t_{ac}= \frac{-42.75-9S_{l}}{S_{l}+6}$
$t_{ac}= \frac{42.75+9S_{l}}{S_{l}+6}$ equation (5)

Replace equation (5) in equation (1):
$S_{l}-6= \frac{11.25}{t_{ac}}$
$S_{l}-6= \frac{11.25}{ \frac{42.75+9S_{l}}{S_{l}+6} }$ equation (6)

Finally, we can solve equation (6) to find the speed of the boat in the lake:
$(S_{l}-6)(42.75+9S_{l})=11.25(S_{l}+6)$
$42.75S_{l}+9S_{l}^{2} -256.5-54S_{l}=11.25S_{l}+67.5$
$9S_{l}^{2}-11.25S_{l}-256.5=11.25S_{l}+67.5$
$9S_{l}^{2}-22.5S_{l}-324=0$
Using the quadratic formula to solve our quadratic equation, we get that S_{l}= 7.38 or $S_{l}=-4.88$ . Since speed cannot be negative, the solution of our equation is $S_{l}=7.38$ .

We can conclude that the speed of the boat on the lake must be 7.38 mph in order for it to serve the ferry operator’s needs.

Answer from: Quest

this shows a reflection over the x-axis and is translated one unit down and 6 units to the right (x + 6, y - 1)

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Answer from: Quest

answer: yard ramps with a level off platform are typically 36 feet long. truck, rather than at an angle, enabling access to those pallets at the very back of the truck. if you are using your yard ramp to convert a loading dock into a drive-in door, we even at 60 inches (5 feet) high and a 30 foot ramp, the slope of the ramp will be less

step-by-step explanation:

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