The formula for Area is A = L x W. To solve the above problem, let X represent the width and 2X represent the length. Use the formula to find the value of X: A = 2X x X. Since the "certain area" is known, use that number as A (we'll use 200 sq ft as an example): 200 = 2X x X; rewrite as 200 = 2X-squared; next, divide both sides of the equation by 2: 100 = X-squared; find the square root of 100 to find the value of X; now the equation is 10 = X. Remember X represents the width, so the width is 10. The length is twice the width, so 10 x 2 = 20.
The length of a rectangle is twice its width. If the perimeter of the rectangle is , find its area.
Yes, the diameter of a circle is twice the length of the radius.
The diameter
The diameter
The width is 6 cm and the length is 12 cm.
Length = (Perimeter - twice width) / 2
C4H6. C2H3 gives a molecular mass of 27, 54/27 gives 2. Therefore the molecular formula is twice the empirical formula.
For a rectangle, calculate twice the length, plus twice the width.For a rectangle, calculate twice the length, plus twice the width.For a rectangle, calculate twice the length, plus twice the width.For a rectangle, calculate twice the length, plus twice the width.
If the width is 9' and the length is twice that, the length is 18'; the perimeter is twice the length plus twice the width, which is 36' + 18' = 54 feet.
The width is half the length: The perimeter is twice the length plus twice the width. If the perimeter is 3 times the length, twice the width must be the length.
You just multipy the length (no matter what it is) by the width (no matter what that is). So... L x W = A Where L is Length, W is Width and A is Area. For your question about the area of a rectangle where the length is twice the width but no numbers are given then I guess a formula could be W x 2W = A. Or send me the exact question and I'll see if I can think of a better answer.
About twice the length as a normal wolf.
The circumference of an ellipse is given by an integral, though it can be approximated through some workable formulae. One such formula gives an approximate length of 72.66 ft (if the lengths given are diameters; if they are semi-axes, it gives a length of approx 145.33 ft - twice as long as the ellipse is twice as big).
l = 2*w - 3
The length of the diameter is always twice the length of the radius.
12 meters
The duration of a minim is as twice as it of a crotchet.