Sure, or inches.
If you take the standard piece of paper as having a width of 8.5 inches, then the conversion to cm and mm is as follows:8.5 inch x 2.5 cm/inch = 21.25 cm21.25 cm x 10 mm/cm = 212.5 mm
Density = mass / volume.Density = 146 g /[ height(cm) * length(cm) *width(cm)].
milimetres.
192.5 liters or 50.85 gallons
what is the width of a page in mm or cm
The length is 4 cm, the width is 26 cm.
Its width is: 161/14 = 11.5 cm
the width equals 14 cm. and the length equals 31 cm.
Area = length x width ⇒ width = Area ÷ length = 34 cm2 ÷ 9 cm = 37/9 cm ~= 3.78 cm
The area of a rectangle is: length * width= 45 cm * 30 cm = 1350 cm^2
Width would be 14 cm
15cm in width and 10 cm in length
Any area you want greater than 0 cm2 and less than or equal to 16 cm2. perimeter_of_rectangle = 2 x (width + length) 16 cm = 2 x (width + length) ⇒ width + length = 8 cm ⇒ length = 8 cm - width Neither dimension can be negative, so they are both greater than 0 cm and less than 8 cm. Area_of_rectangle = width x length ⇒ area = width x (8 cm - width) = 8 x width - width2 cm2 This has a minimum value when width = 0 cm or 8cm of 0 cm2, so the area of the perimeter must be greater than 0 cm2 (as the width can be neither 0 cm nor 8 cm). This has a maximum value when width = 4 cm: max_area = 8 x 4 - 42 cm2 = 16 cm2. when the width is between 0 cm and 4 cm (and so the length is between 8 cm and 4 cm) the area is between 0 cm2 and 16 cm2 When the width = 4 cm, the length is also 4 cm, that is the rectangle is a square.
Standard width (with no riggings): 24 1/2"
Area = Length*Width so Width = Area/Length = 220/5.5 = 40 cm.
Width of rectangle: 20/5 = 4 cm