Answer:
Length = 18 Width = 11
Step-by-step explanation:
Perimeter = 2(length) + 2(width)
So let's label the unknown length as x. Width would then be (x-7). Then plug it into the perimeter equation.
2(x) + 2(x-7) = 58
Then use PEMDAS which will give us
2x+2x-14 = 58
Combine like terms
4x -14 = 58
Add 14 to both sides
4x = 72
To get x by itself you have to divide both sides by 4.
x = 18
So length is 18 and width is (18-7) which is 11.
no because it don’t happen to everyone
Dimensions are length 20 meter and width 14 meter
<em><u>Solution:</u></em>
Let "a" be the length of rectangle
Let "b" be the width of rectangle
Given that,
<em><u>A rectangle has width that is 6 meters less than the length</u></em>
Width = length - 6
b = a - 6
The area of the rectangle is 280 square meters
<em><u>The area of the rectangle is given by formula:</u></em>

<em><u>Substituting the values we get,</u></em>

<em><u>Solve the above equation by quadratic formula</u></em>



Since, length cannot be negative, ignore a = -14
<em><u>Thus solution of length is a = 20</u></em>
Therefore,
width = length - 6
width = 20 - 6 = 14
Thus dimensions are length 20 meter and width 14 meter
Answer:
±16
Step-by-step explanation:
the square root of 256 is positive or negative 16! (if it only lets you put one answer, put positive 16).