The length of the length will be 5 and the width will be 25.
Length 17 cm. Width 10 cm.
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.
Any length greater than six.
width:10,length;17 17*2+10*2=54
The length of the length will be 5 and the width will be 25.
Length 17 cm. Width 10 cm.
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
(width + 6) feet, of course!
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
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
Any length greater than six.
width:10,length;17 17*2+10*2=54
The length is 44 1/2 feet, and the width is 39 1/2 feet
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.
Area of the rectangle = 36 =Length * Width Assume Width= a Hence Length = 4*a =4a Area = a*4a =36 Therefore, a*a =9 Hence a =3 =Width