Answer:
for the 1st year she earned 160 so for 3 years she will earn $480
Answer: 38
Step-by-step explanation:
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)
Answer:
If we want the greatest portion of pie, then you must choose the section with the greatest angle. Therefore, we must choose Section 2. But if we want the smallest portion of pie, then we must choose Section 1.
Step-by-step explanation:
From statement, we know that measure of the angle ABC is equal to the sum of measures of angles ABD (<em>section 1</em>) and DBC (<em>section 2</em>), that is to say:
(1)
If we know that
,
and
, then the value of
is:




Then, we check the angles of each section:
Section 1


Section 2


If we want the greatest portion of pie, then you must choose the section with the greatest angle. Therefore, we must choose Section 2. But if we want the smallest portion of pie, then we must choose Section 1.