Answer:
System of equations:

Augmented matrix:
![\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C2%261%261%266%5C%5C1%261%263%266%5Cend%7Barray%7D%5Cright%5D)
Reduced Row Echelon matrix:
![\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C0%261%261%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Convert the system into an augmented matrix:
![\left[\begin{array}{cccc}1&2&2&6\\2&1&1&6\\1&1&3&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C2%261%261%266%5C%5C1%261%263%266%5Cend%7Barray%7D%5Cright%5D)
For notation, R_n is the new nth row and r_n the unchanged one.
1. Operations:

Resulting matrix:
![\left[\begin{array}{cccc}1&2&2&6\\0&-3&-3&-6\\0&-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C0%26-3%26-3%26-6%5C%5C0%26-1%261%260%5Cend%7Barray%7D%5Cright%5D)
2. Operations:

Resulting matrix:
![\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C0%261%261%262%5C%5C0%26-1%261%260%5Cend%7Barray%7D%5Cright%5D)
3. Operations:

Resulting matrix:
![\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&2&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C0%261%261%262%5C%5C0%260%262%262%5Cend%7Barray%7D%5Cright%5D)
4. Operations:

Resulting matrix:
![\left[\begin{array}{cccc}1&2&2&6\\0&1&1&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%262%266%5C%5C0%261%261%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Answer:
The expression would be 43x + 128
Step-by-step explanation:
43 dollars is multiplied by however many feet is used. Then, the exception(the gate) is added to the charge that is 43x
Answer:
their is a 7/12 chance that u will get a apple out the bag
Step-by-step explanation:
<h3>
Answer:</h3>
System
Solution
- p = m = 5 — 5 lb peanuts and 5 lb mixture
<h3>
Step-by-step explanation:</h3>
(a) Generally, the equations of interest are one that models the total amount of mixture, and one that models the amount of one of the constituents (or the ratio of constituents). Here, there are two constituents and we are given the desired ratio, so three different equations are possible describing the constituents of the mix.
For the total amount of mix:
... p + m = 10
For the quantity of peanuts in the mix:
... p + 0.2m = 0.6·10
For the quantity of almonds in the mix:
... 0.8m = 0.4·10
For the ratio of peanuts to almonds:
... (p +0.2m)/(0.8m) = 0.60/0.40
Any two (2) of these four (4) equations will serve as a system of equations that can be used to solve for the desired quantities. I like the third one because it is a "one-step" equation.
So, your system of equations could be ...
___
(b) Dividing the second equation by 0.8 gives
... m = 5
Using the first equation to find p, we have ...
... p + 5 = 10
... p = 5
5 lb of peanuts and 5 lb of mixture are required.