Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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?

Sagot :

SchwarzschildRadius
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}
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.