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3 years ago

I need help with this problem

Mathematics

2 answers:

pychu [463]3 years ago

8 0

Step-by-step explanation:

Solution,

Given:-

Length=x+4

Width=x-9

Now,

By applying the formula for area of rectangle, we get

Area=length×width

Area=(x+4)(x-9)

Area=x(x-9)+4(x-9)

Area=x²-9x+4x-36

Area=x²-5x-36

Zarrin [17]3 years ago

5 0

L=x+4
B=x-9

$\\ \sf\longmapsto Area=LB$

$\\ \sf\longmapsto Area=(x+4)(x-9)$

$\\ \sf\longmapsto Area=x(x-9)+4(x-9)$

$\\ \sf\longmapsto Area=x^2-9x+4x-36$

$\\ \sf\longmapsto Area=x^2-5x-36$

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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:

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

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$2R_1 + R_2 ->R_2$

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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:

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

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

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$-3R_1 +R_3 ->R_3$

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

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

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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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