area= l*w right so then l=6 w=6
6*6=36 again think buddy! ;)
100 x 100
15 yd by 15 yd square (educated guess???)
22 ft x 11 ft
To find the possible dimensions of a rectangular garden using 40 feet of fencing, we can use the formula for the perimeter of a rectangle: ( P = 2 \times (length + width) ). Setting the perimeter to 40 feet gives us the equation ( length + width = 20 ). Therefore, for any integer value of the length (from 1 to 19 feet), the width can be calculated as ( width = 20 - length ). Possible pairs of dimensions include (1, 19), (2, 18), (3, 17), and so on, up to (19, 1).
To find the dimensions of a rectangle with the largest perimeter using 100 feet of fencing, we can express the perimeter ( P ) of a rectangle in terms of its length ( l ) and width ( w ) as ( P = 2l + 2w ). Since the total amount of fencing is 100 feet, we set up the inequality ( 2l + 2w \leq 100 ). Simplifying this gives ( l + w \leq 50 ). The dimensions that maximize the area (which is a related concept) would be when ( l = w = 25 ) feet, creating a square shape.
100 x 100
A circle with a diameter of 11.459 feet would give the most area. (About 103 square feet)
I got no clue.
15 yd by 15 yd square (educated guess???)
Heras fencing has been known to supply railings, sports fencing, mesh fencing, and gates. Acquiring Heras fencing may not be possible outside of Doncaster, South Yorkshire, England.
50' x 50'and its spelled dimension not demension
4 x 4
22 ft x 11 ft
fencing
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To find the dimensions of a rectangle with the largest perimeter using 100 feet of fencing, we can express the perimeter ( P ) of a rectangle in terms of its length ( l ) and width ( w ) as ( P = 2l + 2w ). Since the total amount of fencing is 100 feet, we set up the inequality ( 2l + 2w \leq 100 ). Simplifying this gives ( l + w \leq 50 ). The dimensions that maximize the area (which is a related concept) would be when ( l = w = 25 ) feet, creating a square shape.
625 square feet.. the area would be a square rectangle with 25 feet of fencing on each side.