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fomenos
3 years ago
7

doug earns $10.50 per hour working at a restaurant on Friday he spent 1 3/4 hours cleaning 2 3/4 hours doing paper work and 1 5/

12 hours serving costumers what were Doug's earnings
Mathematics
2 answers:
sweet [91]3 years ago
6 0
Add up the hours.

1 3/4 + 2 3/4 + 1 5/12

Convert to improper fractions:

7/4 + 11/4 + 17/12

Convert 7/4 and 11/4 into denominators of 12:

21/12 + 33/12 + 17/12

Add the numerators:

71/12

Convert to mixed number:

5 11/12

Multiply by 10.5:

5 11/12 * 10.5 = 62.125

So he earned approximately $62.13
Kruka [31]3 years ago
3 0
10.50(1\frac{3}{4}+2\frac{3}{4}+1\frac{5}{12})\\10.5(3\frac{6}{4}+1\frac{5}{12})\\10.5(4\frac{6}{12}+1\frac{5}{12})\\10.5(5\frac{11}{12})\\10.5(\frac{71}{12})\\\frac{745.5}{12}\\62.125

$62.13
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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]

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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

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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].

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For the case \lambda=2, using the same process, we get the vector (1,1).

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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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3 years ago
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Kryger [21]
To solve this, you need to know slope-intercept form;  
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Where mx is the slope and b is the y-intercept form.  
Now, looking at the values you have been provided, you can just plug them into the formula.  
y= \frac{2}{3}x -2  

6 0
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