Question

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 there is a hurdle. At which locations of the course will people need to go through more than obstacle?
My daughter's homework

Answer 1

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

Answer 2

@islandgurl

Answer 3

explain ansr please

Answer 4

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.

Answer 5

thank you

Answer 6

np :) :)

Answer 7

http://brainly.com/question/195266