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: i need more information 8 apples cost $4, 4???????
Step-by-step explanation:
Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
Negative infinity
Step-by-step explanation:
The line points down infinitely in the graph.
