Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

The length of a rectangle is 5 feet more than twice the width. The perimeter is 130 feet. Find the dimensions.

Sagot :

SchwarzschildRadius
So,

The problem tells us that the length of the rectangle is 5 feet more than twice the width.
Mathematically:
length = 5 + 2width
l = 5 + 2w


We know that the perimeter is 2l + 2w.
2l + 2w = 130

2(5 + 2w) + 2w = 130

Distribute
10 + 4w + 2w = 130

Collect Like Terms
10 + 6w = 130

Subtract 10 from both sides
6w = 120

Divide both sides by 6
w = 20

The width is 20.


Since
l = 5 + 2w

Substitute
l = 5 + 2(20)

Multiply
l = 5 + 40

Collect Like Terms
l = 45

The length is 45.

Check
2l + 2w = 130

2(45) + 2(20) = 130

90 + 40 = 130

130 = 130

Width: 20 ft.
Length: 45 ft.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.