MaxWong

\(\text{Let }\dfrac{x}{y}=z.\\ 2 < \dfrac{x - y}{x + y} < 5\\ 2 < \dfrac{z - 1}{z + 1} < 5\\ 2z + 2 \stackrel{\boxed{1}}{<} z - 1 \stackrel{\boxed{2}}{<} 5z + 5\\ \boxed{1}: 2z + 2 < z - 1\\ z < -3\\ \boxed{2}: z - 1 < 5z + 5\\ -6 < 4z\\ z > -\dfrac{3}{2}\\ \text{Combine }\boxed{1} \text{ and }\boxed{2}: -\dfrac{3}{2} < z < -3\\ \text{The only integer solution is } z = -2\\ \therefore \dfrac{x}{y} = -2\)

Domain = all real numbers

Range = all real numbers

Thanks

I got \(\dfrac{1}{8}\), but I am not so sure either. Will wait for someone else to answer.

\(a - b \equiv n \pmod{99}\\ 62 - 75 \equiv n \pmod{99}\\ n\equiv -13 \pmod{99}\\ n \equiv 86 \pmod{99}\\ n \equiv 990 + 86 \pmod{99}\\ n \equiv 1076\pmod{99}\)

So n is 1076

\(\dfrac{2+i}{4+i}\\ = \dfrac{(2+i)(4-i)}{(4+i)(4-i)}\\ = \dfrac{9 + 2i}{4^2 + 1^2}\\ = \dfrac{9}{17} + \dfrac{2}{17}i\)

Slope of (2x + 3y = 4k) = -2/3

Slope of (x - 2ky = 7) = -1/(-2k) = 1/2k

\(\dfrac{1}{2k} = \dfrac{-2}{3}\\ -4k = 3\\ k = \dfrac{-3}{4}\)

Consider the slope:

\(\dfrac{5-3}{t-0} = \dfrac{3-0}{0-(-8)}\\ 16 = 3t\\ t = \dfrac{16}{3}\)

By intercept theorem,

\(\dfrac{\text{AC}}{\text{DA} }= \dfrac{\text{BC}}{\text{BE}}\\ \dfrac{15}{15-10} = \dfrac{8 + \text{BE}}{\text{BE}}\\ 15\text{BE} = 40 + 5\text{BE}\\ 10\text{BE} = 40\\ \text{BE}=4\)

\(\begin{cases} a + ab^2 = 40b \;\;\;\!\;\!--- (1)\\ a-ab^2 = -32b --- (2)\\ \end{cases}\\ (1) + (2) : 2a = 8b\\ b = \dfrac{a}{4}\\ \text{Substitute }b = \dfrac{a}4\text{ into (1),}\\ a + a \cdot \left(\dfrac{a^2}{16}\right) = 40\left(\dfrac{a}{4}\right)\\ a^3 -144a = 0\\ a(a+12)(a-12) = 0\\ a \in\{ 0, 12, -12\}\)