Rom

\(-2-2i = 2\sqrt{2}e^{i\frac{5\pi}{4}} = (\sqrt{2})^3e^{i\frac{5\pi}{4}} \\ (-2-2i)^{1/3} = \sqrt{2} e^{i\frac{\left(\frac{5\pi}{4}+2k\pi\right)}{3}},~k=0,1,2 = \\ \sqrt{2} e^{\frac{5\pi}{12}},~\sqrt{2}e^{i\frac{13\pi}{12}},~\sqrt{2}e^{i \frac{7\pi}{4}}\\ \text{of these only }\sqrt{2}e^{i \frac{7\pi}{4}} \text{ lies in the 4th quadrant}\)

\(\sqrt{2}e^{i \frac{7\pi}{4}} = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} - i \dfrac{1}{\sqrt{2}}\right) = 1-i\)

A bit of a different take on it that doesn't require you actually find the numbers.

\(\text{We are looking for }n \pmod{10}\\ m=10a+7\\ 21^6 = (2\cdot 10+1)^6 = \sum \limits_{k=0}^6 \dbinom{6}{k}(2\cdot 10)^k = 1 + 10\cdot (\text{some big number})\\ 21^6 \pmod{10} = 1\\ \)

\(mn \pmod{10} = (m \pmod{10})(n \pmod{10})\pmod{10} = 21^6 \pmod{10} = 1\\ m\pmod{10}=7\\ 7(n \pmod{10}) \pmod{10}=1\\ n\pmod{10} = 3 \)

1-9 gets you up to the 9th

100-999 gets you another 900

10000-10090 - gets you the rest

10090 is the 1000th

\(\text{1 real root indicates that the Discriminate }D = b^2 - 4ac \text{ is equal to }0\\ D=(c+1)^2 - 4(1)(c) = \\ c^2 + 2c + 1 - 4c =\\ c^2 - 2c+1 = 0\)

\((c-1)^2 = 0\\ c=1\)

121, 484, 676

217=216+1 = (36)6 + 1

45 = 42+3 = (7)6 + 3

217 x 45 = (216+1)(42+3) = (216)(42)+(216)(3)+42+3 mod 6 = 3

3 is the units digit of 217d x 45d written in base 6

\(\text{I get the following strings} \\ \left( \begin{array}{cccccc} \text{A} & \text{B} & \text{D} & \text{G} & \text{V} & \text{Z} \\ \text{A} & \text{B} & \text{D} & \text{K} & \text{N} & \text{Z} \\ \text{A} & \text{B} & \text{D} & \text{M} & \text{N} & \text{V} \\ \text{A} & \text{B} & \text{G} & \text{H} & \text{K} & \text{Z} \\ \text{A} & \text{B} & \text{G} & \text{H} & \text{M} & \text{V} \\ \text{A} & \text{B} & \text{G} & \text{K} & \text{M} & \text{P} \\ \text{A} & \text{B} & \text{H} & \text{K} & \text{M} & \text{N} \\ \text{A} & \text{D} & \text{G} & \text{H} & \text{K} & \text{M} \\ \end{array} \right)\)

xxxxx

\(\text{What the sum of these elements is describing is a 5 digit binary number that includes 0}\\ \text{There are }2^5=32 \text{ of these, 0-31}\)

you're gonna have to translate this to complete sentences

\(\text{let }N \text{ be the total number of kids}\\ R \text{ is the set of kids that bring raincoats}\\ U \text{ is the set of kids that bring umbrellas}\\ |U|=2|R|\\ |U\cap R|=0.08N\)

\(N = |U \cup R| = |U| + |R| - |U \cap R|\\ N = |U|+\dfrac{|U|}{2} - 0.08 N\\ 0.92N = \dfrac 3 2 |U|\\ |U| = \dfrac 2 3 0.92 N = \dfrac{184}{300}N\)

\(|\text{kids that only bring umbrellas}| = |U| - |U\cap R| = \\ \dfrac{184}{300}N - \dfrac{8}{100}N = \dfrac{160}{300}N = \dfrac{8}{15}N\\ \dfrac{8}{15} \approx 53.3\%\)