11*5*7
385 breaks down to 7 and 55. Bring the seven down. Break 55 down to 11 and 5. Then bring those down and you get 11*5*7
Answer:
The correct option is;
Yes, the line should be perpendicular to one of the rectangular faces
Step-by-step explanation:
The given information are;
A triangular prism lying on a rectangular base and a line drawn along the slant height
A perpendicular bisector should therefore be perpendicular with reference to the base of the triangular prism such that the cross section will be congruent to the triangular faces
Therefore Marco is correct and the correct option is yes, the line should be perpendicular to one of the rectangular faces (the face the prism is lying on).
the line that is parallel will be y = 1/2x
Answer:
![\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The <em>transpose of a matrix </em>
is one where you swap the column and row index for every entry of some original matrix
. Let's go through our first matrix row by row and swap the indices to construct this new matrix. Note that entries with the same index for row and column will stay fixed. Here I'll use the notation
and
to refer to the entry in the i-th row and the j-th column of the matrices
and
respectively:

Constructing the matrix
from those entries gives us
![P^T=\left[\begin{array}{cc}2&8\\5&1\end{array}\right]](https://tex.z-dn.net/?f=P%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%268%5C%5C5%261%5Cend%7Barray%7D%5Cright%5D)
which is option a. from the list.
Another interesting quality of the transpose is that we can geometrically represent it as a reflection over the line traced out by all of the entries where the row and column index are equal. In this example, reflecting over the line traced from 2 to 1 gives us our transpose. For another example of this, see the attached image!