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Lorico [155]
3 years ago
6

I’ll really appreciate it if you help me out with this one .

Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

the answer is d

Step-by-step explanation:

because i said so

but also i figured it out

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Which is true of the slopes of parallel lines?
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They have equal slopes.

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Julissa gave out an equal number of oranges to each of the 6 apartments on her floor. If she gave each apartment 5 oranges, how
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Answer:

30 oranges

Step-by-step explanation:

Total number of apartment in julissa floor = 6

Number of oranges given to each apartment = 5

how many oranges did Julissa give out in all?

Total number of oranges julissa's gave out = Total number of apartment in julissa floor × Number of oranges given to each apartment

= 6 × 5

= 30 oranges

Therefore, total number of oranges julissa gave out = 30

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What is 2 thirds plus 3 fourths
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In order to find this you need to find a common denominator. 2/3 and 3/4 common denominator would be 12 and so what you do to the bottom you must do to the top. 2x4 is 8 and 3x3 is 9. that would make 8/12 and 9/12. 8 plus nine is 17 and 12 stays there so the answer is 17/12 which equals 1 and 5/10.
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Please HELP!<br><br> 1) Enter 6/12 in simplest form.<br><br> 2) Enter 18/21 in simplest form.
scoray [572]
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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:

2R_1 + R_2 ->R_2

-3R_1 +R_3 ->R_3

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]

Multiply by 2

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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3 years ago
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