![\bf \int [11\ ln\left( \sqrt[3]{x} \right)]dx\qquad \begin{cases} ln\left( \sqrt[3]{x} \right)\implies ln\left( x^{\frac{1}{3}} \right) \\ \quad \\ \frac{1}{3}ln(x) \end{cases}\qquad thus \\ \quad \\\\ \quad \\ \int [11\ ln\left( \sqrt[3]{x} \right)]dx\implies \int[11\cdot \frac{1}{3}ln(x)]dx\impliedby \textit{scalars to the left} \\ \quad \\ 11\cdot \frac{1}{3}\int[ln(x)]dx\implies \cfrac{11}{3}\int[ln(x)]dx](https://tex.z-dn.net/?f=%5Cbf%20%5Cint%20%5B11%5C%20ln%5Cleft%28%20%5Csqrt%5B3%5D%7Bx%7D%20%5Cright%29%5Ddx%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Aln%5Cleft%28%20%5Csqrt%5B3%5D%7Bx%7D%20%5Cright%29%5Cimplies%20ln%5Cleft%28%20x%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Cright%29%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Cfrac%7B1%7D%7B3%7Dln%28x%29%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%20%5Cquad%20%5C%5C%5C%5C%20%5Cquad%20%5C%5C%0A%5Cint%20%5B11%5C%20ln%5Cleft%28%20%5Csqrt%5B3%5D%7Bx%7D%20%5Cright%29%5Ddx%5Cimplies%20%5Cint%5B11%5Ccdot%20%5Cfrac%7B1%7D%7B3%7Dln%28x%29%5Ddx%5Cimpliedby%20%5Ctextit%7Bscalars%20to%20the%20left%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A11%5Ccdot%20%5Cfrac%7B1%7D%7B3%7D%5Cint%5Bln%28x%29%5Ddx%5Cimplies%20%5Ccfrac%7B11%7D%7B3%7D%5Cint%5Bln%28x%29%5Ddx)
and I'm pretty sure you know what that is
B because you just work it out
Surface area is the sum of the areas of each side of a three dimensional object.
There are two sides that are 11x9, two sides that are 9x3, and two sides that are 11x3.
So the surface area is:
2(11)(8)+2(9)(3)+2(11)(3) = 296 ft^2
Answer:
8
Step-by-step explanation:
2 4 6 8
The answer to your question is 41