This is a strange question, and f(x) may not even exist. Why do I say that? Well..
[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).
[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!
Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.
To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
Ok I know how to do this but what do I need to do?
Answer:
The center is (-6,5) and the radius is 2 sqrt(2)
Step-by-step explanation:
The equation of a circle can be written in the form
(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius
(x+6)^2 + (y-5)^2 = (sqrt(8))^2
rewriting this
(x - -6)^2 + (y-5)^2 = (2 sqrt(2))^2
The center is (-6,5) and the radius is 2 sqrt(2)
Answer: x=-4, y=-7
Problem:
4x-6y = 26
5x-4y=8
You can use substitution or elimination to solve the problem. I’m going to use elimination.
Multiply by -2 => -8x+12y = -52
Multiply by 3 => 15x- 12y = 24
Add those equations together:
7x=-28
x=-4
Plug that into one of the original equations to get y.
4(-4)-6y=26
-16 - 6y=26
-6y=42
y=-7
Check for the other equation to see if this is true. Which in this case it is. So that’s your answer. x=-4, y=-7
