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faltersainse [42]
2 years ago
14

WILL MARK BRAINLIEST! WILL GIVE 57 POINTSSSS! NO FUNNY ANSWERS I WILL REPORT YOU UNTIL YOU GET BANNED.

Mathematics
1 answer:
Step2247 [10]2 years ago
3 0

Answer:

-5

-3

-2

Rule:  Subtract 2 from the input value

Step-by-step explanation:

Examine the pattern of the input and the corresponding output for the two given ordered pairs:

Input -2 → Output -4  

Output - Input =  -4 - (-2) = -2

Input 1 → Output -1

Output - Input = -1 - 1 = -2

Therefore to get the output value (y-value), we subtract 2 from the input value (x-value).

\begin{array}{| c | c |}\cline{1-2} x & y \\\cline{1-2} -3 & -5 \\\cline{1-2} -2 & -4 \\\cline{1-2} -1 & -3 \\\cline{1-2} 0 & -2\\\cline{1-2} 1 & -1 \\\cline{1-2}\end{array}

As the rule for the function is the subtract 2 from the input value to get the output value, we can write this as:

Output = Input - 2

or as an equation:  y=x-2

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Perform the indicated row operations, then write the new matrix.
Studentka2010 [4]

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right]

Operations:

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Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

The first operation:

2R_1 + R_2 ->R_2

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

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2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right]

\left[\begin{array}{ccc}0&5&7|1\end{array}\right]

The second operation:

-3R_1 +R_3 ->R_3

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

Multiply by -3

-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

Add to row 2 elements are: \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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Answer:

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