9514 1404 393
Answer:
$2038.85
Step-by-step explanation:
The value of the loan at that point is given by ...
A = P(1 +rt) . . . . . Principal P, rate r, time t (years)
A = $1850(1 + 0.1225·(10/12)) = $2038.85
Ricardo will have paid back $2038.85 at the end of the loan period.
_____
<em>Additional comment</em>
We assume that the loan accrues simple interest and that the amount due is the sum of principal and interest at the end of the loan period.
The question is not specific as to whether interest compounds, or whether intermediate (monthly) payments are made. There are many possible ways the loan could be repaid, generally involving different amounts for the different terms.
Answer: y = 15
Step-by-step explanation:
6y - 10 + 6y + 10 = 180
+ 10 - 10 -10 +10
12y = 180
12y/12 = 180/12
y = 15
Answer:
3) 
Step-by-step explanation:
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