Answer:
The Answer is 2
Step-by-step explanation:
The rule is 4 or less let it rest. 5 or more raise the score. 6 is greater than 5 so you raise it to two
Answer:

Step-by-step explanation:
It is a result that a matrix
is orthogonally diagonalizable if and only if
is a symmetric matrix. According with the data you provided the matrix should be

We know that its eigenvalues are
, where
has multiplicity two.
So if we calculate the corresponding eigenspaces for each eigenvalue we have
,
.
With this in mind we can form the matrices
that diagonalizes the matrix
so.

and

Observe that the rows of
are the eigenvectors corresponding to the eigen values.
Now you only need to normalize each row of
dividing by its norm, as a row vector.
The matrix you have to obtain is the matrix shown below
Answer:
asdasd
Step-by-step explanation:
Answer:
The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.
Step-by-step explanation:
The question is incomplete. The complete question should be
The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?
Given the initial bearing of a lighthouse from the ship is N 37° E. So,
is 37°. We can see from the diagram that
would be
143°.
Also, the new bearing is N 25°E. So,
would be 25°.
Now we can find
. As the sum of the internal angle of a triangle is 180°.

Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.
And let us assume the distance between the lighthouse and the ship at N 25°E is 
We can apply the sine rule now.

So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.