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Law Incorporation [45]
3 years ago
13

Two clothing stores are having a sale on all their dress shirts. Store A is charging $15 for each dress shirt. Store B is $18 fo

r each shirt but giving customers a coupon for a $9 discount when they check out.
Which of the following statements is true?
A. Helen bought 4 dress shirts at Store A for $75.
B. Henry bought 4 dress shirts at Store B for $72.
C. Hank bought 3 shirts from Store A and paid the same as Holly paid for 3 shirts at Store B.
D. Buying dress shirts from Store A will always be cheaper than buying dress shirts from Store B.
Mathematics
2 answers:
Klio2033 [76]3 years ago
8 0

Answer:

C

Step-by-step explanation:

When the steps are followed. They both pay the same amount

Zepler [3.9K]3 years ago
7 0

Answer:

C

Step-by-step explanation:

A: Helen paid $60 at store A for 4 Dress Shirts

B:Henry bought 4 dress shirts at Store B for $63

D: (1 shirt)Store A with discount is $15 and at Store B (with $9 coupon) $9 , though it is true with one shirt, it is not certain this will happen with a greater amount of shirts.

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Use\ (a\cdot b)^n=a^nb^n\\-----------------\\\\(4ab)^2=4^2a^2b^2=\huge\boxed{16a^2b^2}
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Answer:if you need a question you can say “ how far did the toy car move? “

Step-by-step explanation:

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Use the Distributive Property to multiply.) -4x(x - 7)*
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-4x^2-28x

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(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca
Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

\left[\begin{matrix}4 & -2 \\ 1 & 1 \end{matrix}\right]

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation det(A-\lambdaI)=0 where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6

So the characteristic polynomial is \lambda^2-5\lambda+6=0.

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0

So \lambda=3, \lambda=2

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 \lambda=3 we get the following matrix

\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right].

Since both rows are equal, we have the equation

x-2y=0. 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 \lambda=2, 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 A=PDP^{-1}. 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

P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]

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

P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

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Need help with a math question
irina1246 [14]

Answer: 27

Step-by-step explanation:

No. Of juniors: 8

Total students: 30

8/30 x 100 = 26.66

Rounds to (whole percent): 27

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4 years ago
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