\(\sqrt[3]{x} > 0\) ==> \(\text{Domain:} \quad x \in (0, \infty)\)
\(\lfloor \pi + 4 \rfloor = \lfloor 7.14 \rfloor = \boxed{7}\)
\((x-1)(x+2) = x^2 + x - 2\)
\(a + b = 1 - 2 = \boxed{-1}\)
\(\ell(y) = \frac{1}{2y+6}\)
The denominator must not equal 0, so the domain is everything that doesn't equal -3.
\(\boxed{x \in (-\infty, -3) \cup (-3, \infty)}\)
Powers of i cycle in groups of 4:
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
\(i^{11} + i^{16} + i^{21} + i^{28} + i^{31} = -i + 1 + i + 0 - i = \boxed{1-i}\)
15/4 = 3.75
\(\lfloor \frac{15}{4} \rfloor = \boxed{3}\)
The denominator cannot equal 0. The asymptotes are at x = 3 and x = -8/3.
Domain: \(x \in (-\infty, -\frac{8}{3}) \cup (3, \infty)\)
\(\frac{5+12i}{2-8i} \cdot \frac{2+8i}{2+8i} = \frac{10+24i+40i+96i^2}{70} = \boxed{\frac{-43+32i}{35}}\)
from 1 to 3: 1
from 4 to 8: 2
from 9 to 15: 3
from 16 to 24: 4
from 25 to 29: 5
3(1) + 5(2) + 7(3) + 9(4) + 5(5) = 95
\((4-2i) - 2(3+4i) = (4-2i) - (6+8i) = \boxed{-2-10i}\)
