the length of the rectangle is 3 ft less than double the width, the area of the rectangle is 44 ft squared. what is the length and the width?

Respuesta :

A=wl is the area of the rectangle but here’s a picture to show u this might help u with the length and width have a wonderful and nice day
Ver imagen CamaroNeedsHelp
We need to find the length, l, and the width, w. Let's set up the two equations with the information found in the question:

l = 2w - 3
w = w
A = 44[tex] ft^{2} [/tex]

We know that area is equal to length * width ([tex] A=l*w[/tex]), so let's substitute those values

[tex] A=l*w[/tex]
[tex]A=(2w-3)*w[/tex]
[tex]A=2 w^{2}-3w [/tex]

And we know that the area is equal to [tex]44 ft^{2} [/tex]:

[tex]44=2 w^{2}-3w [/tex]

Now we set equal to 0 and factor in order to get w:

[tex]0=2 w^{2}-3w-44 [/tex]
[tex]0=(2w-11)(w+4)[/tex]

Then we set each factor equal to zero and solve for w. We get the two values:

[tex] \frac{11}{2} [/tex] and [tex]-4[/tex] for w.

We know that the width cannot be negative, so we take the value that is positive, [tex] \frac{11}{2} [/tex]

So now we know that w = [tex] \frac{11}{2} [/tex], so we plug in this value into the equation for the length:

[tex]l=2w-3[/tex]
[tex]l=2( \frac{11}{2})-3 [/tex]
[tex]l= \frac{22}{2}-3 [/tex]
[tex]l=11-3=8[/tex] 

So we now know that the length is 8, and the width is [tex] \frac{11}{2} [/tex].