So in your given pattern, you need to find first the derivatives and observe the patter that occurs in the given functions. So with this kind of pattern, every fourth one is the same; that makes the 114th derivative is the same as the second derivative. It is known since 114/4 has a remainder of two
length = x
width = 2x
x(2x) = 1682
2x^2 = 1682
x^2 = 841
x = sqrt(841) = 29
length = 29 meters
width = 29 x 2 = 58 meters
Answer:
perimeter of a square = 4 * side
4 m^2 = 4 * side
side = 4/4
side = 1m
![\bf \textit{volume of a cube}\\\\ V=s^3~~ \begin{cases} s=\stackrel{\textit{length of}}{\textit{a side}}\\[-0.5em] \hrulefill\\ V=7 \end{cases}\implies 7=s^3\implies \sqrt[3]{7}=s](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cube%7D%5C%5C%5C%5C%20V%3Ds%5E3~~%20%5Cbegin%7Bcases%7D%20s%3D%5Cstackrel%7B%5Ctextit%7Blength%20of%7D%7D%7B%5Ctextit%7Ba%20side%7D%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20V%3D7%20%5Cend%7Bcases%7D%5Cimplies%207%3Ds%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B7%7D%3Ds)
now, Check the picture below, since it's really just 6 square faces,
![\bf \textit{surface area of a cube}\\\\ SA=6s^2~~ \begin{cases} s=\stackrel{\textit{length of}}{\textit{a side}}\\[-0.5em] \hrulefill\\ s = \sqrt[3]{7} \end{cases}\implies SA=6\sqrt[3]{7}\implies SA\approx 11.477587](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bsurface%20area%20of%20a%20cube%7D%5C%5C%5C%5C%20SA%3D6s%5E2~~%20%5Cbegin%7Bcases%7D%20s%3D%5Cstackrel%7B%5Ctextit%7Blength%20of%7D%7D%7B%5Ctextit%7Ba%20side%7D%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20s%20%3D%20%5Csqrt%5B3%5D%7B7%7D%20%5Cend%7Bcases%7D%5Cimplies%20SA%3D6%5Csqrt%5B3%5D%7B7%7D%5Cimplies%20SA%5Capprox%2011.477587)