- #1

alllove

- 2

- 0

Select one:

A. 35 ≤ w ≤ 60

B. 10 ≤ w ≤ 35

C. 10 ≤ w ≤ 60

D. 0 ≤ w ≤ 10

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D.In summary, the width of the rectangular enclosure must lie within the range of 0 to 10 yards in order for the area to be at least 600 square yards while using 140 yards of fencing, and the width cannot exceed the length.

- #1

alllove

- 2

- 0

Select one:

A. 35 ≤ w ≤ 60

B. 10 ≤ w ≤ 35

C. 10 ≤ w ≤ 60

D. 0 ≤ w ≤ 10

Mathematics news on Phys.org

- #2

skeeter

- 1,103

- 1

length, $L$ and width $(70-L)$

constraints ...

$70-L \le L$

$L(70-L) \ge 600$

see what you can do from here

constraints ...

$70-L \le L$

$L(70-L) \ge 600$

see what you can do from here

Last edited by a moderator:

- #3

HOI

- 921

- 2

Of course that's exactly the same as the inequality skeeter gave for L. There will be two solutions. It is the condition that "the width cannot exceed the length" that allows you to distinguish between them.

- #4

kaliprasad

Gold Member

MHB

- 1,335

- 0

To find the width of a rectangle, you can use the formula: width = (area/fencing) * 2. In this case, the width would be (600/140) * 2 = 8.57 yards.

Yes, you can also use the formula: width = (area/fencing) / 2. In this case, the width would be (600/140) / 2 = 2.14 yards.

To convert yards to feet, you can use the conversion factor: 1 yard = 3 feet. So, 1 yard is equal to 3 feet.

In order to find the width, you will need to convert both the area and fencing to the same unit of measurement. Once they are in the same unit, you can use the formula mentioned in the first question.

Yes, this formula can be used for any rectangle as long as you have the area and the measurement of one side (in this case, the fencing).

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