MaxWong

The first one is actually an odd function.

That is a straight line with unknown slope that passes through the origin. We can say that its equation is f(x) = mx, where m is the slope.

Now f(-x) = -mx = -f(x), which means it should be an odd function instead.

\(\dfrac{1}{a} +\dfrac{1}{b}=\dfrac{1}c\\ \dfrac{1}a = \dfrac{1}c-\dfrac{1}b\\ \dfrac{1}a = \dfrac{b-c}{bc}\\ a = \dfrac{bc}{b-c}\)

That means {\(\dfrac{a}{1-r} = 2\\ \dfrac{a^2}{1-r^2} = 3\).

\(\dfrac{2a}{1+r} = 3\\ \dfrac{2(2-2r)}{1+r} = 3\\ 4-4r = 3+3r\\ 7r = 1\\ r = \dfrac{1}{7}\\ \dfrac{2a}{\frac{8}{7}} = 3\\ 7a = 12\\ a =\dfrac{12}{7}\\\)

So \((r,a) = \left(\dfrac{1}{7},\dfrac{12}{7}\right)\)

10% of 60% = 0.1 * 0.6 = 0.06

20% of 60% = 2(10% of 60%) = 0.12

25% of 60% = 0.06 * 5 / 2 = 0.15

50% of 60% = 2*(25% of 60%) = 0.3

https://web2.0calc.com/questions/given-that-f-x-x-k-where-k-gt-0-what-is-the-range#r2

\(\begin{align}&306\\&---\\15&|4598\\&\;45\\&---\\&\quad\;98\\&\quad\;90\\&---\\&\quad\quad\! 8\end{align}\)

And here I am, attempting the problem :P

I used a graphing calculator, looked at the graph and then found the area of R xD

That's half of a circle with radius 4 units, which is \(\dfrac{\pi\cdot4^2}{2} = 8\pi\).

I can't find a proper way of doing this. 

This is presumably a geometry question. But the figure is missing, and we don't know the exact location of point F. 

If no figure is given, the problem cannot be solved.

This looks like a fun problem. I will try later, perhaps tomorrow if I remember .