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The answer is 3 years from now.

\(\sqrt[3]{x} > 0\)  ==> \(\text{Domain:} \quad x \in (0, \infty)\)

\(\lfloor \pi + 4 \rfloor = \lfloor 7.14 \rfloor = \boxed{7}\)

\((x-1)(x+2) = x^2 + x - 2\)

\(a + b = 1 - 2 = \boxed{-1}\)

\(\ell(y) = \frac{1}{2y+6}\)

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: \(x \in (-\infty, -\frac{8}{3}) \cup (3, \infty)\)

\(\frac{5+12i}{2-8i} \cdot \frac{2+8i}{2+8i} = \frac{10+24i+40i+96i^2}{70} = \boxed{\frac{-43+32i}{35}}\)

The answer is 12 years from now.

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

\((4-2i) - 2(3+4i) = (4-2i) - (6+8i) = \boxed{-2-10i}\)

I use vietas formula

First, simplify the terms in the numerator of the fraction.

\(\frac{5}{\sqrt{80}} = \frac{5\sqrt{80}}{80} = \frac{5\cdot4\sqrt{5}}{80}=\frac{\sqrt{5}}{4}\)

\(\frac{\sqrt{845}}{9} = \frac{13\sqrt{5}}{9}\)

\(\sqrt{45} = 3\sqrt{5}\)

The numerator then becomes

\(\frac{\sqrt{5}}{4} + \frac{13\sqrt{5}}{9} + 3\sqrt{5}\)

Since the denominator contains \(\sqrt{5}\), we can just divide that from the numerator:

\(\frac{1}{4} + \frac{13}{9} + 3 = \frac{169}{36}\)

Now, just take the square root of that to get the final answer:

\(\sqrt{\frac{169}{36}} = \frac{\sqrt{169}}{\sqrt{36}} = \boxed{\frac{13}{6}}\)