Answer:
x = -5, y = -6, z = -3
Step-by-step explanation:
Given the system of three equations:

Write the augmented matrix for the system of equations

Find the reduced row-echelon form of the augmented matrix for the system of equations:

Thus, the system of three equations is

From the last equation:

Substitute it into the second equation:

Substitute y = -6 and z = -3 into the first equation:

Answer:
Option a)
Step-by-step explanation:
To get the vertical asymptotes of the function f(x) you must find the limit when x tends k of f(x). If this limit tends to infinity then x = k is a vertical asymptote of the function.

Then. x = 2 it's a vertical asintota.
To obtain the horizontal asymptote of the function take the following limit:

if
then y = b is horizontal asymptote
Then:

Therefore y = 0 is a horizontal asymptote of f(x).
Then the correct answer is the option a) x = 2, y = 0
Yes because 0 and 4 dont equal 10 and 20 and 40 dont equal 100. So no and 40 + 23 = 63.
For any values less than -3, the answer is undefined, because then you have a negative value for √x+3
Answer:
I think its C
Step-by-step explanation: