The next step of your proof is to subtract (a/b) from both sides.
Then you get, x = (m/n) - (a/b)
Since rationals are closed over addition, (m/n) + (-a/b) is a rational number.
Therefore, x (an irrational number) = a rational number <em>This is a false statement which is a contradiction. So, the assumption was incorrect.</em>
Thus, the sum of a rational and irrational number is an irrational number. QED
Answer: f(-16) = -26
Step-by-step explanation: Let's substitute:
f(w) = 2w + 6 --> f(-16) = -32 + 6 =
f(-16) = -26
Hope this helps, please consider making me Brainliest.
Replace x with -4 so it would be:
f(-4)= 2(-4)=-8
-8 is your answer
So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.
![f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%28%5Comega%28x%29%29%5C%5C%5C%5Cf%28%5Cpi%5En%29%3D%5Csin%28%5Comega%28%5Cpi%5En%29%29%3D0%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5C%5C%5C%5C%5Csin%20x%3D0%20%5Cimplies%20x%3Dk%5Cpi%2Ck%5Cin%5Cmathbb%7BZ%7D%5C%5C%5CDownarrow%5C%5C%5Comega%28%5Cpi%5En%29%3Dk%5Cpi%5C%5C%5C%5C%5Cboxed%7B%5Comega%28x%29%3Dk%5Csqrt%5B%5Clog_%7B%5Cpi%7D%20x%5D%7Bx%7D%2Ck%5Cin%5Cmathbb%7BZ%7D%7D)
