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Tcecarenko [31]
2 years ago
8

A triangular course for a canoe race is marked with buoys. The first leg is 3/10 mi, the second leg is 1/2 mi, and the third leg

is 2/5 mi. How long is the race?
Mathematics
1 answer:
balu736 [363]2 years ago
7 0
<h2>The race is 1.2 miles long</h2>

Step-by-step explanation:

The first leg is 3/10 mi, the second leg is 1/2 mi, and the third leg is 2/5 mi.

\texttt{Length of first leg = }\frac{3}{10}miles\\\\\texttt{Length of second leg = }\frac{1}{2}miles\\\\\texttt{Length of third leg = }\frac{2}{5}miles

Now we need to find length of race

       \texttt{Total length = }\frac{3}{10}+\frac{1}{2}+\frac{2}{5}\\\\\texttt{Total length = }\frac{3}{10}+\frac{5}{10}+\frac{4}{10}\\\\\texttt{Total length = }\frac{3+5+4}{10}\\\\\texttt{Total length = }\frac{12}{10}\\\\\texttt{Total length = }1.2miles

The race is 1.2 miles long

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algol13

Answer:

One watering system requires 36 minutes to complete the job alone while another watering system requires 12 minutes to complete the job alone.

Step-by-step explanation:

Given:

Both the system can complete the job = 9 minutes

We need find the time required by each system to do the job.

Solution:

Let the time required by another watering system to complete the job be 'x' mins.

Now given:

One watering system needs about three times as long to complete a job as another watering system.

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Rate to complete the job by One watering system = \frac{1}{3x}\ job/min

Rate at which both can complete the job = \frac{1}{9}\ job/min

So we can say that;

Rate at which both can complete the job is equal to sum of Rate to complete the job by another watering system and Rate to complete the job by One watering system.

framing in equation form we get;

\frac{1}{x}+\frac{1}{3x}=\frac19

Now taking LCM to make the denominator common we get;

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