It will still be 6 units high. Even if the width is reduced, that does not effect the height of the farm.
Answer:
Me for sure!
Step-by-step explanation:
(:)
Answer:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given




Required
Find the standard matrix
The standard matrix (A) is given by

Where
![T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=T%28x%29%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
becomes
![Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=Ax%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
The x on both sides cancel out; and, we're left with:
![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
Recall that:



In matrix:
is represented as: ![\left[\begin{array}{c}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D)
So:
![T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]](https://tex.z-dn.net/?f=T%28e_1%29%20%3D%20%281%2C2%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
![T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_2%29%20%3D%20%28-4%2C6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C6%5Cend%7Barray%7D%5Cright%5D)
![T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_3%29%20%3D%20%282%2C-6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-6%5Cend%7Barray%7D%5Cright%5D)
Substitute the above expressions in ![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Hence, the standard of the matrix A is:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Answer: you will have 120 cubes
V = lwh
You're given length, width, and height of the prism, so:
V = (5/3) * (4/3) * (2)
V = 40/9 cm³
Now, you need to find the volume of each cube that'll be filling the prism. Since it's a cube, the length, width, and height are the same.
Therefore the volume is simply:
V = (1/3)³
V = 1/27cm³
To find how many cubes will fill the prism now, divide (40/9) by (1/27).
This results in 120 cubes being able to fill the prism.
Step-by-step explanation:
Answer:
maybe its not the right explanation cuz it doesnt make sense to me
Step-by-step explanation: