Answer:
<em> 1,407</em>
<em>x 14</em>
<em>_______</em>
<em> 5,628</em>
<em>+ 1,4070</em>
<em>______</em>
19,698
<em>The two partial products are 5,628 and 14,070</em>
Answer:
The reduced row-echelon form of the linear system is ![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%260%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:
- Interchange two rows
- Multiply one row by a nonzero number
- Add a multiple of one row to a different row
To find the reduced row-echelon form of this augmented matrix
![\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D2%263%26-1%2614%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
You need to follow these steps:
- Divide row 1 by 2

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 from row 2

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 5 from row 3

![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 1

![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 3

![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Multiply row 2 by 2

![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Divide row 3 by −19

![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 16 from row 1

![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 6 to row 2

![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%260%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
Factor:
<span>30n - 40 = 10(3n - 4)
hope it helps</span>
Answer:
The pdf is not showing anything
Step-by-step explanation:I would help you otherwise.I am sorry.Don't let this get you down.I hope you have a good day.
It would make 5 complete frames, and about 8/10 of another frame.
108 / 18.375 = 5.8&*)&^&%*$*
I used a calculator.
You could keep adding 18 and 3/8 over and over again until you get right under 108 if you're good at math like me, but most people would have to add 18 and 3/8 separately.
Thanks, and if you liked my answer, please consider giving me the brainliest answer, because I need 5 of them to get to the next level.