No, Because P ≠ 2(24) + 2w
A is correct.
The length of a rectangle is 24 units.
The perimeter of rectangle is P = 60 unit
If width of rectangle is 11 units (w)
Perimeter of rectangle (P) = 2(L+W)
where, L=24 and W=11
P = 2(24+11)
Therefore, P ≠ 2(24) + 2w
No, The perimeter can not be 60 unit of this rectangle.
P = 2L + 2W
P = 2x24 + 2x11
P = 70
So your answer is No, because P ≠ 2(24) + 2w
We are given length of the rectangle = 24 units.
Width of the rectangle = 11 units.
We know, the formula for periemter of the rectangle as following
P = 2 * length + 2* width.
Let us plug values length = 24 units and width 11 units. We get
P = 2*24 + 2*11
P = 48 +22
P = 70 units.
We got perimeter of 70 units for length 24 units and width 11 units of the rectangle.
So, when its width w is 11 units with the length 24 units, the perimeter P of the rectangle can not be 60 units.
Hope this helps :)