c=2πr
=2*3.14*12
=75.36 m
it's circumference is 75.36 m
Answer:
The solution of the system of linear equations is 
Step-by-step explanation:
We have the system of linear equations:

Gauss-Jordan elimination method is the process of performing row operations to transform any matrix into reduced row-echelon form.
The first step is to transform the system of linear equations into the matrix form. A system of linear equations can be represented in matrix form (Ax=b) using a coefficient matrix (A), a variable matrix (x), and a constant matrix(b).
From the system of linear equations that we have, the coefficient matrix is
![\left[\begin{array}{ccc}2&3&-6\\1&-2&3\\3&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%26-6%5C%5C1%26-2%263%5C%5C3%261%260%5Cend%7Barray%7D%5Cright%5D)
the variable matrix is
![\left[\begin{array}{c}x&y&z\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%26y%26z%5Cend%7Barray%7D%5Cright%5D)
and the constant matrix is
![\left[\begin{array}{c}12&-2&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D12%26-2%2613%5Cend%7Barray%7D%5Cright%5D)
We also need the augmented matrix, this matrix is the result of joining the columns of the coefficient matrix and the constant matrix divided by a vertical bar, so
![\left[\begin{array}{ccc|c}2&3&-6&12\\1&-2&3&-2\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%263%26-6%2612%5C%5C1%26-2%263%26-2%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
To transform the augmented matrix to reduced row-echelon form we need to follow these row operations:
- multiply the 1st row by 1/2
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\1&-2&3&-2\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C1%26-2%263%26-2%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
- add -1 times the 1st row to the 2nd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\3&1&0&13\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%26-7%2F2%266%26-8%5C%5C3%261%260%2613%5Cend%7Barray%7D%5Cright%5D)
- add -3 times the 1st row to the 3rd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&-7/2&6&-8\\0&-7/2&9&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%26-7%2F2%266%26-8%5C%5C0%26-7%2F2%269%26-5%5Cend%7Barray%7D%5Cright%5D)
- multiply the 2nd row by -2/7
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&-7/2&9&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%26-7%2F2%269%26-5%5Cend%7Barray%7D%5Cright%5D)
- add 7/2 times the 2nd row to the 3rd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&3&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%260%263%263%5Cend%7Barray%7D%5Cright%5D)
- multiply the 3rd row by 1/3
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&-12/7&16/7\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%26-12%2F7%2616%2F7%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add 12/7 times the 3rd row to the 2nd row
![\left[\begin{array}{ccc|c}1&3/2&-3&6\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%26-3%266%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add 3 times the 3rd row to the 1st row
![\left[\begin{array}{ccc|c}1&3/2&0&9\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%263%2F2%260%269%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
- add -3/2 times the 2nd row to the 1st row
![\left[\begin{array}{ccc|c}1&0&0&3\\0&1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%263%5C%5C0%261%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
From the reduced row echelon form we have that

Since every column in the coefficient part of the matrix has a leading entry that means our system has a unique solution.
Answer:
The term that represent the amount of money she will spend on brownies is 2b
Step-by-step explanation:
Let
c ----> the number of cones
b ----> the number of brownies
we know that
The cost of desserts can be found using the expression

where
4c ----> is the term that represent the amount of money spend in cones
2b ----> is the term that represent the amount of money spend in brownies
5 ----> is the term that represent the amount of money spend in water
therefore
The term that represent the amount of money she will spend on brownies is 2b
That means ---> the number of brownies (b) multiplied by the cost of each brownie ($2)
48+27=75
27/75 x 100 = 36% regular soda