The perimeter of a rectangle is 36 and the length is twice the width. what are the dimensions of this rectangle and what is the area?

let l be the length and w be the width.

so 2l + 2w =36

also, l = 2w,

Putting the value for l in the first equation:

2(2w) + 2w = 36

4w + 2w = 26

6w = 36

w = 6

now solving for the length:

l = 2w

l = 2(6)

so the length will be 12, l = 12

So the width is 6. the length is 12.

and the area is lxw, so the area will be 72

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

your answer is the frist one

hop i