Airline passengers pay $439 to fly to California. For this price, customers may check 2 pieces of luggage. There is a fee of $25
for each additional piece of luggage a passenger wants to check. Which function can be used to find the amount in dollars a passenger has to pay to fly with p pieces of luggage, where p >2
1 answer:
Answer:
G.
Step-by-step explanation:
Amount paid by customer to fly to California if carrying 2 pieces or less than 2 pieces of luggage for which they won't be charged for = $439
Fee for additional luggage = $25
Fee for additional luggage if an individual carries p luggages = 25(p - 2)
Total amount (C) if p pieces of luggage were carried by a customer travelling = fee for additional p pieces of luggage + $439
The function for this would be:
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Answer:
Step-by-step explanation:
Since this question is lacking the matrix A, we will solve the question with the matrix
so we can illustrate how to solve the problem step by step.
a) The characteristic polynomial is defined by the equation where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,
So the characteristic polynomial is .
b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that
So
c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.
If we get the following matrix
.
Since both rows are equal, we have the equation
. Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)
For the case , using the same process, we get the vector (1,1).
d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are
This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is
which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors