Answer:
Infinite number of solutions.
Step-by-step explanation:
We are given system of equations
Firs we find determinant of system of equations
Let a matrix A=![\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%264%265%5C%5C1%261%262%5C%5C2%261%26-1%5Cend%7Barray%7D%5Cright%5D)
and B=![\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C1%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.
We are finding rank of matrix
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C1%261%262%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C1%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%260%26-2%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B9%7D%7B2%7D%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.
Therefore, rank of matrix is equal to rank of B.
Answer:
Five x squared plus two
Step-by-step explanation:
An exponent of two is always pronounced as "squared"
Answer:
Step-by-step explanation:
oh boy
1) 9 < 16
2) -4 < 5
3) 18 > -19
4) -38 < -16
5) 6 > 4
6) -14 > -25
7) -45 > -55
8) -2 > -10
9) -4, -1, 1, 3, 6
10) -15, -13, -5, 8
11) -9, -6, -5, -1, 0
12) -4, -2, 1, 3
13) -10, -9, 7, 8, 11
14) -33, -25, -14, -4, 7
hope this helps <3
Step-by-step explanation:
Formula is:
x=(-b± √b^2-4ac)/2a
a= 1
b= -8
c= 17
x= (-(-8)±√64-4(1)(17))/2
x= (8±√-4)/2
now
x1= (8+√-4)/2
x2=(8-√-4)/2
