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Vivi
DevinWell, isn't that just a happy little question! The width of an eraser can vary depending on the brand and type, but typically they are around half an inch to an inch wide. Remember, erasers come in all shapes and sizes, so just choose the one that feels right for you and your creative journey!
The width of an eraser can vary depending on the brand, type, and size of the eraser. On average, a standard pencil eraser has a width of around 0.25 inches (6.35 millimeters). However, larger erasers, such as block erasers or eraser sticks, can have widths ranging from 0.5 inches (12.7 millimeters) to 1 inch (25.4 millimeters) or more. It is important to note that these measurements are approximate and can differ based on the specific eraser being used.
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If the length of a rectangle is 24.7 meters and the area is 550.81 meters what is the width of the rectangle?
length*width = area 24.7*width = 550.81 Divide both sides by 24.7 to find the width: width = 22.3 meters
The perimeter of a certain rectangle is 16 times the width The length is 12 cm more than the width Find the lenghth and width of the rectangle?
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
What is the length and width of a rectangle if the perimeter is 32 feet and the area is 48 square feet?
Length is 12ft, width is 4 ft. 2 x (length + width) = perimeter = 32 ft ⇒ length = 16ft - width [1] length x width = area = 48 sq ft Using [1] above: ⇒ (16ft - width) x width = 48 sq ft multiplying out and rearranging: ⇒ width2 - 16 x width + 48 = 0 factorizing: ⇒ (width - 4)x(width - 12) = 0 Thus width = 4 or 12, so using [1] above: ⇒ width = 4ft, length = 16 - 4 = 12ft or width = 12ft, length = 16 - 12 = 4ft Convention is that the length is larger than the width, thus rectangle is 12ft long by 4 ft wide.
How do you calculate the top dimensions of a rectangle if you know its perimeter and bottom area?
Important to note are these formulae: Perimeter_of_rectangle = 2 x (length + width) Area_of_rectangle = length x width So if the perimeter and area are known, then: 2 x (length + width) = perimeter => length + width = perimeter / 2 => length = perimeter / 2 - width length x width = area => (perimeter / 2 - width) x width = area (substituting for length given above) => perimeter / 2 x width - width2 = area => width2 - perimeter / 2 x width + area = 0 which is a quadratic and can be solved either by factorization or by using the formula: width = (perimeter / 2 +/- sqrt(perimeter2 / 4 - 4 x area)) / 2 = (perimeter +/- sqrt(perimeter2 - 16 x area)) / 4 This will provide two values for the width. However, each of these values is the length for the other, so the larger value is the length and the smaller value is the width. Sometimes only 1 value will be found for the width above. In this case, the rectangle is actually a square which means that the length and width are both the same. Examples: 1. perimeter = 6, area = 2 width2 - perimeter / 2 x width + area = 0 => width2 - 6 / 2 x width + 2 = 0 => width2 - 3 x width + 2 = 0 => (width - 2) x (width - 1) = 0 => width = 2 or 1. So the length is 2 and the width is 1. 2. perimeter = 12, area = 9 width2 - perimeter / 2 x width + area = 0 => width2 - 12 / 2 x width + 9 = 0 => width2 - 6 x width + 9 = 0 => (width - 3)2 = 0 => width = 3 So the rectangle is a square with both length and width of 3.
The perimeter of a rectangle is 24 inches. And its length is 3 times its width. What are the length and the width of the rectangle?
perimeter 24 so length + width half of that ie 12. Length 3 times width must be 9 and width 3.
