Rom

\(a + b = 260\\ b+c = 245 \\ c+d = 270\\~\\ a+d = (a+b) +(c+d) - (b+c) = \\ 260+270-245 = 285\)

ordinals are able to be sorted

without getting too smart-arsed I'd say D is the only ordinal variable

\(x = n + 4\\ n = x - 4\\ x + n \leq 14\\ x+(x-4) \leq 14\\ 2x - 4 \leq 14\\ 2x \leq 18\\ x \leq 9\)

well for one you've got 25 in the matrix where it should be -25.

\(\begin{pmatrix} 1&-1&-2&|&-6\\ 3&2&0&|&-25\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ -4&1&-1&|&12 \end{pmatrix}\\ \begin{pmatrix} 1&-1&-2&|&-6\\ 0&5&6&|&-7\\ 0&-3&-9&|&-12 \end{pmatrix}\\\)

Now exchange rows 2 and 3 and normalize row 2

\(\begin{pmatrix} 1 & -1 & -2 & -6 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 5 & 6 & -7 \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & -9 & -27 \end{pmatrix}\\ \)

Normalize row 3 and continue

\(\begin{pmatrix} 1 & 0 & 1 & -2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \begin{pmatrix} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 3 \\ \end{pmatrix}\\ \)

\(\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-5\\-5\\3\end{pmatrix}\)

\(\left(7-\dfrac 2 x\right) = \sqrt{\dfrac 7 x}\\ \left(7-\dfrac 2 x\right)^2 = \dfrac{7}{|x |} \\ 49-\dfrac{28}{x} + \dfrac{4}{x^2} = \dfrac{7}{|x|}\)

\(\text{let $x>0$}\\ 49-\dfrac{28}{x} + \dfrac{4}{x^2} = \dfrac{7}{x}\\ 49x^2 - 28x + 4 = 7x\\ 49x^2 - 35x+ 4 = 0\\ x = \dfrac{35\pm \sqrt{(35)^2 - (4)(49)(4)}}{98} = \left(\dfrac 4 7,~\dfrac 1 7\right)\\ \text{We do however notice that $x = \dfrac 1 7$ is not a solution of the original equation $\\$ so it is discarded leaving only $x = \dfrac 4 7$}\)

\(\text{let $x\leq0$}\\ 49-\dfrac{28}{x} + \dfrac{4}{x^2} = -\dfrac 7 x\\ 49x^2-21x+4 = 0\\~\\ \text{We note that this equation has no real solutions and thus would not satisfy the original equation}\)

\(\text{So the only solution is $x= \dfrac 4 7$}\)

\(\text{If the dependence is linear, i.e. $x = \alpha y,~\alpha \in \mathbb{C}$, you can write $x \sim y$}\\ \text{Otherwise I would just write $x=f(y).\\$ I don't know of any single symbol that expresses it.}\)

\(\text{There are 360 people in my school. 15 take calculus, physics, and chemistry, and 15 don't take any of them.$~\\$ 180 take calculus. Twice as many students take chemistry as take physics. 75 take both calculus and chemistry, and $~\\$ 75 take both physics and chemistry. Only 30 take both physics and calculus. How many students take physics? }\)

\(p = \dfrac{\dbinom{4}{3}\dbinom{5}{0}\dbinom{6}{0}+\dbinom{4}{0}\dbinom{5}{3}\dbinom{6}{0}+ \dbinom{4}{0}\dbinom{5}{0}\dbinom{6}{3}}{\dbinom{15}{3}} = \dfrac{34}{455}\)

\(z = r e^{i \theta}\\ z^n = r^n e^{i n \theta}\\ z^n = 1 \Rightarrow r = 1\\ z = e^{i \theta}\)

\(z + \dfrac 1 z = e^{i\theta}+ e^{-i\theta} = 2\cos(\theta)\\~\\ \left(z + \dfrac 1 z \right)^n = 2^n \cos^n(\theta) = 1\\ \cos^n(\theta) = \dfrac{1}{2^n}\\ \cos(\theta) = \dfrac 1 2\\ \theta = \pm \dfrac{\pi}{3}\)

\(z = e^{\pm i \pi/3}\)

You can't convert from volume to weight without having a density.

A cubic meter of feathers weighs a lot less than a cubic meter of neutronium.

You don't mention a density.  Do you have one?

(for the pedantic:  technically density gets you from volume to mass, and then you have to apply

the gravitational acceleration of whatever planet you happen to be on to get the weight.  As most

of us spend 100% of our time on Earth it's fairly common to see densities expressed as a

weight per unit volume, just including Earth's gravitational acceleration in the constant.)