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, there 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?

So,

We have three fractions with different denominators: sixths, thirds, and halves.  The first step is to make all the denominators equal.  In this case we want sixths.

 1/3 = 2/6, and 1/2 = 3/6.  Now we can start solving.

1. There are six tires at the following: 1/6, 2/6, 3/6, 4/6, 5/6, and 6/6.
2. There are three cones at the following (G.C.F.): 2/6 (or 1/3), 4/6 (or 2/3), and 6/6 (or 3/3).
3. There are two hurdles at the following (G.C.F.): 3/6 (or 1/2) and 6/6 (or 2/2).

We simply look for common numbers.

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. (hard!)

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

S = {2/6, 3/6, 4/6, 6/6}

This is the graph x=1

29933333.33

step-by-step explanation:

6×10=604000000÷60=66666.6666766666.6667×449=29933333.33