Rom

It looks like it maps to the strip

\(\{z: 0 \leq Re[z] \leq 1,~Im[z] \in \mathbb{R}\}\)

If you separate the 20 switches into groups of 4 and fully connect within a group you can pull this off in 30 connections.

I'm seeing 28

\(\dbinom{6+3-1}{3-1} = \dbinom{8}{2} = 28\)

\(\left( \begin{array}{ccc} 6 & 0 & 0 \\ 5 & 1 & 0 \\ 5 & 0 & 1 \\ 4 & 2 & 0 \\ 4 & 1 & 1 \\ 4 & 0 & 2 \\ 3 & 3 & 0 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \\ 3 & 0 & 3 \\ 2 & 4 & 0 \\ 2 & 3 & 1 \\ 2 & 2 & 2 \\ 2 & 1 & 3 \\ 2 & 0 & 4 \\ 1 & 5 & 0 \\ 1 & 4 & 1 \\ 1 & 3 & 2 \\ 1 & 2 & 3 \\ 1 & 1 & 4 \\ 1 & 0 & 5 \\ 0 & 6 & 0 \\ 0 & 5 & 1 \\ 0 & 4 & 2 \\ 0 & 3 & 3 \\ 0 & 2 & 4 \\ 0 & 1 & 5 \\ 0 & 0 & 6 \\ \end{array} \right)\)

The ratios will be the equal

\(\dfrac{16}{\frac{11}{2}} = \dfrac{6}{\ell}\\ 16 \ell = 33\\ \ell = \dfrac{33}{16} = 2 ~ft \text{ to the nearest foot}\)

it will travel 20 circumferences

\(C = 2 \pi r = 2 \pi \left(\dfrac D 2\right) = \pi D\\ C = 2\pi~inch\\ 20C = 40 \pi ~inch\)

\(tlc=mm+3\\ 7 mm + 4 tlc = 78\\ 7mm+4(mm+3) = 78\\ 11mm+12=78\\ 11mm=66\\ mm=\$6/lb\)

total up the frequencies, looks to be 90.

divide each of the frequencies by 90.

I'll let you do that.

eh b****y h**l I missed it.

It goes in the denominator so the denominator becomes 18 rather than 9.

\(x^4 -x^3 - 2x - 4 = (x^4-4)-(x^3+2x) = \\ (x^2-2)(x^2+2) - x(x^2+2) = \\ (x^2+2)(x^2-x-2)=\\ (x+\sqrt{2}i)(x-\sqrt{2}i)(x-2)(x+1)\)

2) A fifth degree polynomial will have 5 complex zeros

3) \(x^4+2x^3-16x^2-2x+15 = \\ (x^4-16x^2+15) +(2x^3-2x) =\\ (x^2-15)(x^2-1)+2x(x^2-1) = \\ (x^2+2x-15)(x^2-1) = \\ (x-3)(x+5)(x-1)(x+1) \\ \\ \text{and the zeros can be read off as }\\ \\ x=3,~-5,~1,~-1\)

\(\dfrac{\sqrt[3]{4r^3}}{\sqrt[3]{9r}} = \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \\ \\ \text{This is really as far as it can be simplified}\\ \text{but it's common to want the denominator to be free of radicals}\\ \dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} =\dfrac{r\sqrt[3]{4}}{\sqrt[3]{9r}} \dfrac{\sqrt[3]{9^2r^2}}{\sqrt[3]{9^2r^2}} = \dfrac{r\sqrt[3]{324r^2}}{\sqrt[3]{9^3r^3}}=\dfrac{r\sqrt[3]{324r^2}}{9r}= \dfrac{\sqrt[3]{324r^2}}{9}\)