L = W + 6
If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.If you assume that the width is not greater than the length, it can have any value if the range (0, 9] yards.
yes
The length of the length will be 5 and the width will be 25.
(width + 6) feet, of course!
The width can be any number greater than zero and less than the square root of 35, and the length can be any number greater than the square root of 35, subject to the constraint that the product of the length and the width must be 35.
Length 17 cm. Width 10 cm.
Any length greater than six.
Since the length is greater than the width, any value greater than sqrt(43.7) which is approx 6.61 metres. For example, length = 40m, width = 43.7/40 m or length = 100m, width = 43.7/100 m or length = 1000000m, width = 43.7/1000000 m
There will be Length/Width widths in 1 length. This will normally be a number that is greater than 1.
The perimeter of a rectangle is 2 x length + 2 x width. If the width is 16 then 78 = 2 x length + 2 x 16 2 x length = 78 - 32 = 46 length = 23. For the perimeter to be greater than 78 cm, the length must be greater than 23 cm
width:10,length;17 17*2+10*2=54
Rectangular perimeter = 2(Length + Width) Width = a Length = a+5 Then, 84 = 2(a+a+5) 42 = 2a+5 (42-5)/2 = a = 18.5m = Width Length = a+5 = 23.5m