Answer:
Is a consistent independent system
Step-by-step explanation:
we know that
If a system has at least one solution, it is said to be consistent .
If a consistent system has exactly one solution, it is independent .
If a consistent system has an infinite number of solutions, it is dependent
If a system not have solution, it is said to be inconsistent
we have
----> equation A
----> equation B
----> equation C
substitute equation A and equation B in equation C

Solve for z

The solution is the point (5,6,11)
The system has only one solution
therefore
Is a consistent independent system
A. 1.9 is an irrational number
Answer:n-6=168
Step-by-step explanation:
The statement starts with the variable first.
Answer:
Step-by-step explanation:
HERE ARE THE ANSWERS
<h3>
ax² + bx + c = 0</h3>
<em>Let's write -9 where we see A</em><em>:</em>
<h3>
-9x² + bx + c = 0</h3>
<em>Let's</em><em> </em><em>write</em><em> </em><em>0</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>B</em><em>:</em>
<h3>
-9x² + 0.x + c = 0</h3>
<em>(</em><em>Since B = 0, when it is multiplied by x, it becomes 0 again</em><em>)</em>
<h3>
-9x² + c = 0</h3>
<em>Let's</em><em> </em><em>write</em><em> </em><em>-2</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>C</em><em>:</em>
<h3>
-9x² + -2 = 0</h3>
<em>Now we can move on to solving our equation</em><em>:</em><em>)</em>
<em>Let's put the known and the unknown on different sides:</em>
<em>(</em><em>-2 goes to the opposite side positively</em><em>)</em>
<h3>
-9x² = 2</h3>
<em>(</em><em>i</em><em>t goes as a division because it is in the case of multiplying -9 across</em><em>)</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>
<h3>
x² = 2/-9</h3>
<em>I could not find the rest of it, but I did not want to delete it for trying very hard. Sorry. It felt like we should take the square root, but I couldn't find it, maybe this can help you a little bit.</em>
<em>Please do not report</em><em>:</em><em>(</em>
<em>I hope I got it right, I'm trying to improve my English a little :)</em>
<h3>
<em>Greetings from Turke</em><em>y</em><em>:</em><em>)</em></h3>
<h3>
<em><u>#XBadeX</u></em></h3>