Answer:
Never
Never
Never
Step-by-step explanation:
The equations given are
2x1−6x2−4x3 = 6 ....... (1)
−x1+ax2+4x3 = −1 ........(2)
2x1−5x2−2x3 = 9 ..........(3)
the values of a for which the system of linear equations has no solutions
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
Since X2 and X3 can't be cancelled out, we conclude that the value of a is never.
a unique solution,
Let first add equation 1 and 2. Also equation 2 and 3. This will result to
X1 + (a X2 - 6X2) - 0 = 5
And
X1 + (aX2-5X2) + 2X3 = 8
The value of a = never
infinitely many solutions.
Divide equation 1 by 2 we will get
X1 - 3X2 - 2X3 =3
Add the above equation with equation 3. This will result to
3X1 - 8X2 - 4X3 = 12
Everything ought to be the same. Since they're not.
Value of a = never.
Answer:
Answer B ![(-\infty,1)U(1,2]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C1%29U%281%2C2%5D)
Step-by-step explanation:
Notice that the quotient of f(x)/g(x) is:

therefore, this new function imposes conditions due to the fact that it has a square root in the numerator and a binomial in the denominator both with the variable x. Then, in order for the root in the numerator to be defined, the argument inside the root must be larger than or equal to zero. That is:

So, this condition must be satisfied by the x-values of the domain.
Then we have the binomial in the denominator, which in order to be defined needs to be different from zero. Notice that the only x-value that could cause problems (render zero) is:

Then, 
So we have to eliminate the number 1 from the previous subset that required x smaller than or equal to 2.
The way to represent this Domain is then: ![(-\infty,1)U(1,2]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C1%29U%281%2C2%5D)
1. 4n + 28
2. 30-10x
3. 3m+3n
4. 8+2j