The length of a rectangle is three more than twice the width. Determine the dimensions that will give a total area of 27m^ {2}. What is the minimum area that this rectangle can have.

3m and 9m

Step-by-step explanation:

length of the rectangle = 2w+3

width of rectangle = w

area = L*W

if area = 27m²

then (2w+3)w=27

=2w²+3w-27=0

using factorization w= 3 or -4.5

w=3m

L=9m

The length is 10metres and the width is 3 1/2metres or 3.5metres

Step-by-step explanation:

If the length of the rectangle is 3m more than ×2 the width, the dimensions (length & width) that will give an area of 27m^2 =?

Firstly, the area of a rectangle = (L+W)×2

Since the length is 3 more than ×2 the width,

The length, L us therefore = (2×w)+3= 2w+3

Recall that the area is 27m^2

Now, substitute L for 2w+3

[(2w+3)+w]×2 = 27

(2w+3+w)×2= 27

(3w+3)×2 = 27

6w+6=27

6w= 27-6

6w = 21

w = 21/6

w= 3 1/2m

If the width = 3 1/2, and area = (L+w)×2, we substitute "w" for 3 1/2 in the equation below:

(L+ 3 1/2)×2 = 27

2L + 7 = 27

2L =27-7

2L = 20

L = 10m

Therefore, the length = 10m and the width

= 3 1/2m

maybe i can

answer:

24pi or letter c