Answer:
2 over 6 as a fraction simplifies as 1 over three.
Step-by-step explanation:
If you are writing as a fraction, you put 2 over 6 and simplify. Your answer is 1 over 3
Answer:
Refer to the attachmenT
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Answer:

Step-by-step explanation:
As the given Augmented matrix is
![\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D9%26-2%260%26-4%26%3A8%5C%5C0%267%26-1%26-1%26%3A9%5C%5C8%2612%26-6%265%26%3A-2%5Cend%7Barray%7D%5Cright%5D)
Step 1 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%267%26-1%26-1%26%3A9%5C%5C8%2612%26-6%265%26%3A-2%5Cend%7Barray%7D%5Cright%5D)
Step 2 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%267%26-1%26-1%26%3A9%5C%5C0%26124%26-54%2677%26%3A-82%5Cend%7Barray%7D%5Cright%5D)
Step 3 :
↔
![\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26-14%266%26-9%26%3A10%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%26124%26-54%2677%26%3A-82%5Cend%7Barray%7D%5Cright%5D)
Step 4 :
↔
,
↔
![\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%264%26-11%26%3A-8%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%260%26-%20%5Cfrac%7B254%7D%7B7%7D%20%26%5Cfrac%7B663%7D%7B7%7D%20%26%3A-%5Cfrac%7B1690%7D%7B7%7D%20%5Cend%7Barray%7D%5Cright%5D)
Step 5 :
↔
![\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%264%26-11%26%3A-8%5C%5C0%261%26-%5Cfrac%7B1%7D%7B7%7D%20%26-%5Cfrac%7B1%7D%7B7%7D%20%26%3A%5Cfrac%7B9%7D%7B7%7D%20%5C%5C0%260%261%26-%5Cfrac%7B663%7D%7B254%7D%20%26%3A-%5Cfrac%7B1690%7D%7B254%7D%20%5Cend%7Barray%7D%5Cright%5D)
Step 6 :
↔
,
↔
![\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%260%260%26-%5Cfrac%7B71%7D%7B127%7D%20%26%3A%5Cfrac%7B176%7D%7B127%7D%20%5C%5C0%261%260%26-%5Cfrac%7B131%7D%7B254%7D%20%26%3A%5Cfrac%7B284%7D%7B127%7D%20%5C%5C0%260%261%26-%5Cfrac%7B663%7D%7B254%7D%20%26%3A%5Cfrac%7B845%7D%7B127%7D%20%5Cend%7Barray%7D%5Cright%5D)
∴ we get

Answer:
Is rational
Step-by-step explanation:
So, adding two rationals is the same as adding two such fractions, which will result in another fraction of this same form since integers are closed under addition and multiplication. Thus, adding two rational numbers produces another rational number.
1) It's best to draw out a picture of a rectangle and label each corner with the coordinates given: Let's say (-5, 2) is point A, (-5, -2 1/3) is point B, (2 1/2, 2) is point C, and (2 1/2, -2 1/3) is point D.
2) That being said, line AB is one side of the rectangle, BC is another, CD is another, and lastly, AD is the fourth side.
3) We can use the distance formula and plug in the coordinates of each line to find how long every side is. Then you just need to solve it.
For example: if I want to find how long side AB is, I would use the point A (-5, 2) and B (2 1/2, 2) and plug them into the distance formula, where (-5, 2) is (x1, x2) and (2 1/2, 2) is (x2, y2) and solve that.
4) Repeat this process with side BC, CD, and AD, and add the results together. This will be your final answer; the perimeter of the rectangle.