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MAVERICK [17]
4 years ago
11

Miguel is making an obstacle course for field day. At the end of every sixth of the course,there is a tire. At the end of every

third of the course, thre is a cone. At the end of every half of the course, there is a hurdle. At which locations of the course will people need to go through more than one obstacle?
Mathematics
1 answer:
mariarad [96]4 years ago
8 0

The answers are: 2/6, 3/6, 4/6, and 6/6.

There is a tire every 1/6 of the way through the course. This means at locations 1/6, 2/6, 3/6, 4/6, 5/6, and 6/6, there is a tire.

There is a hurdle every 1/2 of the course. This means at locations 1/2 and 2/2 there is a hurdle.

There is a cone every 1/3 of the way through the course. This means at locations 1/3, 2/3, and 3/3 there is a cone.

Firstly we will make all the denominators equal.

Hence, 1/3 = 2/6, and 1/2 = 3/6 and third we have 1/6 already.

Now we will look out for common locations.

1. At 2/6, there are two obstacles - a tire and a cone.

2. At 3/6, there are two obstacles - a tire and a hurdle.

3. At 4/6, there are two obstacles - a tire and a cone.

4. At 6/6, there are three obstacles - a tire, cone, and a hurdle.

Hence, the locations are - 2/6, 3/6, 4/6, and 6/6.



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Step-by-step explanation:

Given

y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1

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In the second equation. Assume that:

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