Answer:
C
Step-by-step explanation:
We have the system of equations:

And an ordered pair (10, 5).
In order for an ordered pair to satisfy any system of equations, the ordered pair must satisfy both equations.
So, we can eliminate choices A and B. Satisfying only one of the equations does not satisfy the system of equations.
Let’s test the ordered pair. Substituting the values into the first equation, we acquire:

Evaluate:

Evaluate:

So, our ordered pair satisfies the first equation.
Now, we must test it for the second equation. Substituting gives:

Evaluate:

So, the ordered pair does not satisfy the second equation.
Since it does not satisfy both of the equations, the ordered pair is not a solution to the system because it makes at least one of the equations false.
Therefore, our answer is C.
Answer:
mathematics questions and then you solve your questions

so the slope of that line above is really -4/3, now

Answer:
Part A: Both functions' slopes increase at intervals of four in a positive manner, so they will rise to the right.
Part B: g(x) has a greater y intercept as it is y=3 and f(x) is y=-1
Step-by-step explanation:
Y - y1 = m(x - x1)
slope(m) = -56
(8,-4)...x1 = 8 and y1 = -4
now we sub...pay close attention to ur signs
y - (-4) = -56(x - 8)....not finished yet
y + 4 = -56(x - 8) <===