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Fittoniya [83]
3 years ago
7

Jose earns 954 per hour at his job he worked 18 hours last week calculate Jose's pay before taxes​

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
6 0

Answer:

$171.72

Step-by-step explanation:

$9.54 x 18

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The machine bottled 8,000 bottles in 4 hours. if the rate of the machine were doubled, how long would it take the machine to bot
valina [46]
It would take 20 hours since 8,000 times 5 is 40,000 so 4 times 5 is 20.



Logic.
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3 years ago
PLEASE HELP WITH THIS CONDITIONAL PROBABILITY QUESTION AND PLEAS COULD YOU SHOW A STEP BY STEP EXPLANATION.
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5 0
2 years ago
The point nearest to the origin on a line is at (4, -4). Find the standard form of the equation of the line.
MariettaO [177]
The line through that point and the origin has a slope of -1. It is perpendicular to the line you want, so the line you want has slope +1. Its equation can be written as
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5 0
3 years ago
Simplify. Assume that no denominator is equal to zero. <br><br> 4^12/4^5
Georgia [21]

Answer:

16,384 or 2^14

Step-by-step explanation:

4^12= 4*4*4*4*4*4*4*4*4*4*4*4=16,777,216

4^5= 4*4*4*4*4=1024

16,777,216/1024=16,384

16,384=2^14

4 0
2 years ago
Solve the given system of equations using either Gaussian or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTI
cricket20 [7]

Answer:

The system has infinitely many solutions

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

Step-by-step explanation:

Gauss–Jordan elimination is a method of solving a linear system of equations. This is done by transforming the system's augmented matrix into reduced row-echelon form by means of row operations.

An Augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

There are three elementary matrix row operations:

  1. Switch any two rows
  2. Multiply a row by a nonzero constant
  3. Add one row to another

To solve the following system

\begin{array}{ccccc}x_1&-3x_2&-2x_3&=&0\\-x_1&2x_2&x_3&=&0\\2x_1&+3x_2&+5x_3&=&0\end{array}

Step 1: Transform the augmented matrix to the reduced row echelon form

\left[ \begin{array}{cccc} 1 & -3 & -2 & 0 \\\\ -1 & 2 & 1 & 0 \\\\ 2 & 3 & 5 & 0 \end{array} \right]

This matrix can be transformed by a sequence of elementary row operations

Row Operation 1: add 1 times the 1st row to the 2nd row

Row Operation 2: add -2 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1

Row Operation 4: add -9 times the 2nd row to the 3rd row

Row Operation 5: add 3 times the 2nd row to the 1st row

to the matrix

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 1 & 0 \\\\ 0 & 1 & 1 & 0 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

\begin{array}{ccccc}x_1&&-x_3&=&0\\&x_2&+x_3&=&0\\&&0&=&0\end{array}

The system has infinitely many solutions.

\begin{array}{ccc}x_1&=&-x_3\\x_2&=&-x_3\\x_3&=&arbitrary\end{array}

7 0
3 years ago
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