Rom

\(i) ~\dbinom{6}{2} = 15\\ ii)~52\cdot 51\cdot 50\cdot 49\cdot 48 = ‭311875200‬\)

wat?

We don't use (x,y,z) for spherical coordinates.  It just causes confusion.

\(\text{I'm going to assume that }(x,y,z) \equiv (\rho, \theta, \phi)\\ \begin{align*} -3 &= \rho \sin(\theta)\cos(\phi)\\ 4 &= \rho\sin(\theta)\sin(\phi)\\ -12 &= \rho\cos(\theta) \end{align*}\)

\(\theta= \arccos\left(\dfrac{-12}{\sqrt{(-3)^2+4^2+(-12)^2}}\right) = \arccos\left(\dfrac{-12}{13}\right)\\ \tan(\theta)=-\dfrac{\sqrt{13^2-12^2}}{12} = -\dfrac{5}{12}\)

\(\phi= \arctan\left(\dfrac{4}{-3}\right)\\ \tan(\phi)= -\dfrac 4 3\)

\(\text{let }a = \dfrac{1}{2+\dfrac{1}{2+\dfrac{1}{2+\ddots}}}\)

\(a = \dfrac{1}{2+a}\\ a^2+2a-1=0\\ a = \dfrac{-2\pm \sqrt{4+4}}{2} = -1 \pm \sqrt{2}\\ \text{we can rule out the negative solution since everything is positive}\\ a = -1 + \sqrt{2}\)

\(x= 1+a = \sqrt{2}\)

\(\text{Let }\Phi(x) \text{ be the CDF of the standard normal distribution.}\\ \text{It's the table you probably have seen somewhere}\\ z = \dfrac{20555-20500}{55}=1\\ \Phi(1) \approx 0.8413 = \%84.13\)

\(\text{probably easiest to un parameterise them and solve}\\ x = 3-2t \Rightarrow t = \dfrac{3-x}{2}\\ y = -2+5\cdot \dfrac{3-x}{2} = -2+\dfrac{15}{2}-\dfrac 5 2 x = -\dfrac 5 2 x + \dfrac{11}{2}\)

\(x = -1+3s \Rightarrow s = \dfrac{x+1}{3}\\ y = 5 - 9\cdot \dfrac{x+1}{3} = 5-3x-3= -3x+2\)

\(-\dfrac 5 2 x + \dfrac{11}{2}=-3x+2\\ \dfrac 1 2 x = -\dfrac 7 2\\ x = -7\\ y = 23\\ (x,y)=(-7,23)\)

what you describe would be a hyperbola

\(\text{let }\sqrt{b+\sqrt{b+\sqrt{b+\dots}}} = \beta\\ \sqrt{b+\beta}=\beta\\ b+\beta=\beta^2\\ \beta^2-\beta-b=0\\ \beta=\dfrac{1\pm \sqrt{1+4b}}{2}\\ \beta>0 \text{ as it's a square root so }\\ \beta = \dfrac{1+\sqrt{1+b^2}}{2}\\ a\bowtie b = a + \dfrac 1 2 (1+\sqrt{1+b^2})\)

\(7\bowtie 9 = 9\\ 7+\dfrac 1 2 (1+\sqrt{1+g^2})=9\\ \sqrt{1+g^2}=3\\ 1+g^2=9\\ g=\pm\sqrt{8} = \pm 2\sqrt{2}\)

14 is correct