Answer choice 2 would be the correct one! If the sides were all curved they would be cylinders.
Answer:
if the depth is based off of the original spot where the penguin is, then 13 ft.
if not, 10 ft.
Step-by-step explanation:
Answer:


z = 2
Step-by-step explanation:
Given equations are
x - 2y - z = 2
x + 3y - 2z = 4
-x + 2y + 3z = 2
from the given equations the augmented matrix can be written as
![\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C1%263%26-2%3A4%5C%5C-1%262%263%3A2%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C0%265%26-1%3A2%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-1%3A2%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26-1-%5Cdfrac%7B2%7D%7B5%7D%3A2%2B%5Cdfrac%7B4%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)
![=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26%5Cdfrac%7B-7%7D%7B5%7D%3A%5Cdfrac%7B14%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%262%3A4%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26%5Cdfrac%7B-7%7D%7B5%7D%3A%5Cdfrac%7B14%7D%7B5%7D%5C%5C%5C%5C0%261%26%5Cdfrac%7B-1%7D%7B5%7D%3A%5Cdfrac%7B2%7D%7B5%7D%5C%5C%5C%5C0%260%261%3A2%5Cend%7Barray%7D%5Cright%5D)

![=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right]](https://tex.z-dn.net/?f=%3D%5C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%3A%5Cdfrac%7B14%7D%7B5%7D%2B%5Cdfrac%7B7%7D%7B5%7D%5C%5C%5C%5C0%261%260%3A%5Cdfrac%7B2%7D%7B5%7D%2B%5Cdfrac%7B1%7D%7B5%7D%5C%5C%5C%5C0%260%261%3A2%5Cend%7Barray%7D%5Cright%5D)
So, from the above augmented matrix, we can write


z = 2
y+z=10
x+z=8
x+y=12
(y+z)-(x+z)=10-8
y-x=2
y-x=2
y+x=12
solve simultaneously
x=5, y=7
then substitute to get z=3
hence xyz=573