URGENT!! The length of a rectangle is five more than double the width. Find the dimensions if the
perimeter is 88m.

13, 31

Step-by-step explanation:

Set the width as x.

Length: 2x + 5

Perimeter:

(x) + (2x + 5) + (x) + (2x + 5) = 88

Simplify.

6x + 10 = 88

Solve.

6x = 78

x = 13

Substitute this into the previous equations

Width: 13

Length: 31

The length is 31m and the width is 13m.

Step-by-step explanation:

Start by translating the word problem into an equation. 5 more than double the width is written as 2w + 5. Then substitute that into the equation for the perimeter. It should look something like this:

88 = 2(2w +5) + 2w

Then simplify by combining like terms and solve for the width (w).

88 = 6w + 10

78 = 6w

13 = w

Now that we have the width, we can plug it back into the equation we made for the length.

l = 2w + 5

l = 2(13) + 5

l = 26 + 5

l = 31

I hope this could help!

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!