length = 2 x width Perimeter = 2 x (length x width) =2 x (2 x width + width) = 2 x (3 x width) = 6 x width → 6 x width = 48 units → width = 48 units ÷ 6 = 8 units → length = 2 x width = 2 x 8 units = 16 units Ares = length x width = 16 units x 8 units = 128 square units
the formula for the area of a rectangle is base*width. plug in 2 for the base and 6 for the width. 2*6=12 the answer is 12cm
Ratio of new length to old length = 12/8 = 3/2 So ratio of new width to old width must be 3/2 New width/Old width = 3/2 so New width = Old Width * 3/2 = 6 inches*3/2 = 9 inches
Area= length * width length= width + 5 so.... Area= width * (width + 5) Area= width^2 + 5width (going to use w for width.) 66= w^2 + 5w w^2 + 5w - 66= 0 Now factor! (w - 6)(w + 11) Solve for w, and since area can't be negative, you get.. w=6 length= 6+5= 11
18 - 6 plus 3 times 2 how don't you know that ?
perimeter = (2*length) + (2*width) length = 2*width so perimeter = (2*2*width) + 2*width = 6*width perimeter = 48 so you can figure out the width and length
length = 2 x width Perimeter = 2 x (length x width) =2 x (2 x width + width) = 2 x (3 x width) = 6 x width → 6 x width = 48 units → width = 48 units ÷ 6 = 8 units → length = 2 x width = 2 x 8 units = 16 units Ares = length x width = 16 units x 8 units = 128 square units
the formula for the area of a rectangle is base*width. plug in 2 for the base and 6 for the width. 2*6=12 the answer is 12cm
perimeter = 2 x (width + length) area = width x length If area = perimeter then: width x length = 2 x width + 2 x length width x length - 2 x length = 2 x width length x (width - 2) = 2 x width length = 2 x width / (width - 2) So any rectangle with length and width that satifies the above will work, for example: width = 2.5, length = 10, area: 2.5 x 10 = 25; perimeter: 2x(2.5+10) = 2x12.5 = 25 width = 3, length = 6, area: 3x6 = 18; perimeter: 2x(3+6) = 2x9 = 18 width = 4, length = 4, area: 4x4 = 16; perimeter: 2x(4+4) = 2x8 = 16 By making the length dependent on the width, the width must be at least 2. Once the 4x4 square has been reached, the width becomes larger than the length, but the pair will also be given if the width is the smaller dimension, for example, a width of 6 gives a length of 2 x 6 / (6 - 2) = 12 / 4 = 3, which matches the length of 6 being given by a width of 3.
Ratio of new length to old length = 12/8 = 3/2 So ratio of new width to old width must be 3/2 New width/Old width = 3/2 so New width = Old Width * 3/2 = 6 inches*3/2 = 9 inches
It depends on the gauge of the steel used in the stud and the width of the stud. For example, a 2-1/2" 18 Ga. Steel Stud made from Galvanized Steel could be figured simply from the properties of the steel. The stud is 2-1/2" wide with a 1-1/4" flange and a 1/4" return (on each side). This adds up to 5-1/2" of steel. Galvanized steel weighs 2.156 lbs/sq ft. Now it's just math. 5-1/2" = 0.45833 ft. 2.156 lbs/sq ft * 0.45833 ft = 0.9882 lbs/ft. So, a 2-1/2" 18 Ga. Galvanized Steel Stud weighs approximately 1 lb/ft.
Area= length * width length= width + 5 so.... Area= width * (width + 5) Area= width^2 + 5width (going to use w for width.) 66= w^2 + 5w w^2 + 5w - 66= 0 Now factor! (w - 6)(w + 11) Solve for w, and since area can't be negative, you get.. w=6 length= 6+5= 11
18 - 6 plus 3 times 2 how don't you know that ?
Length: 6 miles width: 2 miles
A rectangle with length L and width W has perimeter = 2*(L+W) So, 2(6 + W) = 220 or 6 + W = 110 so that W = 104 Usually, Length > Width but clearly not in this case.
8 and 6
38'' length 14''heigth 5''width