Rom

\(x^2 + y^2 = r^2 \\ x = \sqrt{r^2 - y^2} \\ m = \dfrac y x \\ m = \dfrac y{ \sqrt{r^2 - y^2} } \\ m = \dfrac {3.144}{\sqrt{7^2 - (3.144)^2}} =0.503\)

You don't.  You can only add matrices that have the same dimensions.

\(2s^2 t^3 + 6s^4t^9 =2s^2t^3(1+3s^2t^6)\)

\(\dfrac {110}{125} = \dfrac {22}{25}\)

\(\mbox{I'm going to assume you mean }\\3x^2 + 2(x^2-20)=16 - 2x^2\)

\(3x^2+2x^2-40 = 16 - 2x^2 \\ \\ 7x^2=56 \\ \\ x^2 = 8 \\ \\ x = \pm \sqrt{8} \\ \\ x = \pm 2\sqrt{2}\)

This isn't quite what you're after but I suspect it can be simply modified to look for the max weight path vs. the longest path.

http://www.geeksforgeeks.org/find-longest-path-directed-acyclic-graph/

\(m = \dfrac{y_1-y_0}{x_1-x_0}=\dfrac{-6 - (-4)}{7-4}=\dfrac{-2}{3}=-\dfrac 2 3\)

\(\dfrac {a}{b} = a \times \dfrac {1}{b}\)

\(\dfrac {2}{9} \div \dfrac {3}{8} = \dfrac {2}{9} \times \dfrac {8}{3} =\dfrac{16}{27}\)

\(\mbox{The maximum volume of the box that is underwater is }11^3 cm^3 = 1331 cm^3\)

\(\mbox{Water weighs }1 \dfrac {g}{{cc}^3}\mbox{ so the maximum weight the box can displace is } 1331 g\)

\(\mbox{The sand has density }3.5 \dfrac {g}{{cc}^3}\)

\(\mbox{and thus a volume of }\dfrac {1331g}{3.5g/cc^3}=380.3 cc^3 \mbox{ of sand will sink the box.}\)

\(f(x)=x^4+2\\ g(x)=x-7\\ h(x)=\sqrt{x}\\ \\ g(h(x))=\sqrt{x}-7\\ f(g(h(x)))=\left(\sqrt{x}-7\right)^4+2\)