Rom

\(\dfrac{8~in}{0.4~in} = 20\)

you mean with compass and straight edge?

#3

\(|z-3|=|z+i|=|z-3i|\\ \text{let $z=x+i y$}\\ (x-3)^2 +y^2 = x^2 + (y-3)^2\\ x^2 - 6x+9 +y^2 = x^2 + y^2 - 6y+9\\ -6x=-6y\\ x = y\)

\(|x-3+xi| = |x + (1+x)i|\\ (x-3)^2 + x^2 = x^2 + (x+1)^2\\ x^2-6x+9 = x^2 + 2x + 1\\ 8=8x\\ x=1\\ z = 1+i\)

#2

\(z^2 = \bar{z}\\ \text{it should be intuitively clear that $|z| = 1$ so let $z = e^{i\theta}$}\\ e^{i2\theta} = e^{-i\theta + 2\pi k},~k \in \mathbb{Z}\\ 2\theta = 2\pi - \theta\\ \theta = \dfrac{2\pi}{3}\\ z = e^{i2\pi/3} = -\dfrac 1 2 + i \dfrac{\sqrt{3}}{2}\)

you can set up a ratio such that both numerator and denominator both go to infinity 

yet the ratio has any finite value you like.

well your answer agrees with sim!

what's the extra factor of 4! for?

oh nm, it's because you are having order matter.

(sorry I edited accidentally, I think it is back to how it was now)

Above is wrong.  Simulation is showing p is approximately 0.14

Still working on this.

\(y = 2x-13\\ y = -3x+92\\ 2x-13 = -3x + 92\\ 5x = 105\\ x = 21\)

\(3x+4y=7\\ 6x+4y=c-4y\\~\\ 6x+8y=c\\ \text{letting $c=14$ makes equation 2 twice equation 1 and thus the system has infinite solutions}\)

\(ab^4 = 12\\ a^5 b^5 = 7776\\ a^5b^{20} = 12^5\\ b^{15}=\dfrac{12^5}{7776} = 32\\ b^3 = 2\\ b= 2^{1/3}\\ 2^{4/3} a= 12\\ a = 3 \cdot 2^{2/3}\)

\(\text{Add the two equations}\\ 2a = 8b\\ b=\dfrac a 4\\ a + a\left(\dfrac a 4\right)^2 = 40\left(\dfrac a 4\right)\\ \dfrac{a^3}{16} =9a\\ a^2 = 144\\ a= \pm 12\\ b=\pm 3\\ \text{$a=0,b=0$ is also a solution} \)