So we know that the formula for the area of a rectangle is

.
Now both the length and width of the rectangle increase at 3 km/s, therefore,
![A(t) = (3t+l)*(3t+w). Since the initial length = initial width = 4 km, then the initial area = 16 [tex]km^2](https://tex.z-dn.net/?f=A%28t%29%20%3D%20%283t%2Bl%29%2A%283t%2Bw%29.%20Since%20the%20initial%20length%20%3D%20initial%20width%20%3D%204%20km%2C%20then%20the%20initial%20area%20%3D%2016%20%5Btex%5Dkm%5E2)
. We want to know the time when the area is four times its original area, therefore, our new formula is:

. Plugging in our known
values we have:
![64 [km^2] = (3t + 4 [km])*(3t + 4 [km])](https://tex.z-dn.net/?f=64%20%5Bkm%5E2%5D%20%3D%20%283t%20%2B%204%20%5Bkm%5D%29%2A%283t%20%2B%204%20%5Bkm%5D%29)

The area is four times its original area after <span>\frac{4}{3} s[/tex]</span>.
4x + 9 - (10/(3x)) I favored the x^2 then expanded and simplified each rational term
28 is not a cube, and 63 is not a square. The simplest form is the form you have.
![\sqrt[3]{28}-\sqrt{63}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B28%7D-%5Csqrt%7B63%7D)
(3•b)^10. 3 times b them after you do that, you put that answer to the 10th power
No because a pentagon has more sides than a square does.
to find perimeter you have to add up all the sides.
for the square it would be 9+9+9+9 which is 36
for the pentagon it would be 9+9+9+9+9 which is 42.