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.
Answer: ALZ is congruent to AIR by reason ASA
Step by Step Explanation: i hate rsm
<span>
the main formula is </span>(cd)(x)=c(x).d(x) <span>
If c(x) = 4x – 2 and d(x) = x2 + 5x
so </span> (cd)(x) = (4x – 2)(x2 + <span>5x)=4x^3+20x²-2x²-10x=4x^3+18x²-10x
so the answer is A: </span><span>4x^3+18x²-10x</span><span>
</span>
Answer:
hmmm...
Step-by-step explanation:
This looks like...
"jnxxmppoihuueitsdehbxexdnjussheyetyyuuhggfffwspiritualdjjdmsmmsmyra."
YES!
(lol)
Answer:
rwrrwrwrr
Step-by-step explanation:
wrrwrrw
