L + W = P/2 = 14, so L = 14 - W.
L = 3W - 2 so 3W - 2 = 14 - W ie 4W = 16 so Width is 4" and Length is 10"
The exact answer cannot be given because the number of inches for the length is missing. But if it was k inches, the width would be 375/k inches.
The perimeter actually doesn't tell you anything about the breakdown betweenthe length and width. All we can tell about this one is that the sum of (length+width)has to be 15 inches. That still leaves an infinite number of possible rectangles.If the perimeter is 30 inches and the length has to be the longer dimension, thenit can be anything more than 7.5 inches, with no upper limit.For example:Width = 1 millionth of an inchLength = 1 millionth of an inch less than 15 inchesPerimeter = 30 inches.
To find the length of a diagonal in a rectangle, use the Pythagorean method. Diagonal length = square root(length squared + height squared).
You need to know the area of the portion of floor you wish to cover, and you need to know the size of the carpet tile. To measure area, use the formula for a rectangle, which is the length times the width. If the floor is not a rectangle, divide it up into rectangles. For example, an "L" shaped room can be measured as two rectangles. Be sure to measure in consistent units -- such as yards, feet or inches. As an alternative, find out the size of the carpet tiles in inches. Then measure the length of the rectangle in inches, and divide by the tile size. That is the number of rows in that direction; do likewise for the width. Then multiply together for the total number of tiles in that rectangle.
If the length and width of a rectangle are multiplied by the same number, then . . . -- the perimeter is multiplied by the same number -- the area is multiplied by the square of the numbner
The exact answer cannot be given because the number of inches for the length is missing. But if it was k inches, the width would be 375/k inches.
It is 5!!!! It can be any number
The perimeter actually doesn't tell you anything about the breakdown betweenthe length and width. All we can tell about this one is that the sum of (length+width)has to be 15 inches. That still leaves an infinite number of possible rectangles.If the perimeter is 30 inches and the length has to be the longer dimension, thenit can be anything more than 7.5 inches, with no upper limit.For example:Width = 1 millionth of an inchLength = 1 millionth of an inch less than 15 inchesPerimeter = 30 inches.
The number of feet is not the same as the number of inches. In length, inches = feet * 12
the answer to number 20 is B...12
To find the length of a diagonal in a rectangle, use the Pythagorean method. Diagonal length = square root(length squared + height squared).
You need to know the area of the portion of floor you wish to cover, and you need to know the size of the carpet tile. To measure area, use the formula for a rectangle, which is the length times the width. If the floor is not a rectangle, divide it up into rectangles. For example, an "L" shaped room can be measured as two rectangles. Be sure to measure in consistent units -- such as yards, feet or inches. As an alternative, find out the size of the carpet tiles in inches. Then measure the length of the rectangle in inches, and divide by the tile size. That is the number of rows in that direction; do likewise for the width. Then multiply together for the total number of tiles in that rectangle.
A whole number is a pure number: it is not a measure of length. For example, an object which is a whole number of millimetres in length will mostly not be a whole number of inches.
you can't, unless the area was an odd number you can't, unless the area was an odd number
If the length and width of a rectangle are multiplied by the same number, then . . . -- the perimeter is multiplied by the same number -- the area is multiplied by the square of the numbner
The perimeter actually doesn't tell you anything about the breakdown betweenthe length and width. All we can tell about this one is that the sum of (length+width)has to be 98 inches. That still leaves an infinite number of possible rectangles.If the perimeter is 196 inches and the length has to be the longer dimension, thenit can be anything more than 49 inches, with no upper limit.For example:Width = 1 millionth of an inchLength = 1 millionth of an inch less than 98 inchesPerimeter = 196 inches.
4 times the number of tiles in the length of the rectangle.