Rom

\(\text{An $m\times k$ matrix may multiply a $k \times n$ matrix to produce a $m \times n$ matrix}\)

The example you gave is undefined.

\(\text{Applying Vieta's equations we have}\\ -7 = -(r_1+r_2+r_3)\\ \text{As we are told the roots are distinct positive integers the only possible solution is}\\ r_1=1,~r_2=2,~r_3=4,~\text{or some permutation thereof}\\ \text{Again using Vieta's equations (which return the same answer given all permutations)}\\ k = 1\cdot 2 + 1\cdot 4 + 2\cdot 4 = 14\\ m = 1\cdot 2\cdot 4 = 8\\ k+m=22\)

\(\text{first place the red marbles, then the blue, the green go into the remaining slots}\\ N=\dbinom{13}{6}\dbinom{7}{4} = 60060\)

\(a \cos(\theta) + b \sin(\theta) = \\ \sqrt{a^2 + b^2}\sin(\theta + \phi),~\phi = \tan^{-1}\left(b,a\right)\\ -1 \leq \sin(\theta+\phi) \leq 1\\ \text{The max value is thus $\sqrt{a^2+b^2}$}\)

\(63N^3 - 550 = \dfrac{73}{26}\\ 63N^3 = \dfrac{14373}{26}\\ N^3 = \dfrac{1597}{182}\\ N= \left(\dfrac{1597}{182}\right)^{1/3}\)

\(2)~\text{I assume you mean that $z^*$ is the conjugate of $z$}\\ \text{There's a problem here. If $z^* = 6+8i$ then $|z| = \sqrt{6^2+8^2} = 10$}\\ \text{Yet we are told that $|z|=5$}\\ \text{Is it supposed to be that $z^* w=6+8i$ ?}\)

\(1) ~|zw|^2 = |z|^2 |w|^2 = 16\cdot 4 = 64\\ \left|\dfrac z w \right|^2 = \dfrac{|z|^2}{|w|^2} = \dfrac{16}{4} = 4\\~\\ \text{The remaining two cannot be solved without the angle between $z$ and $w$}\)

\(f(xy)=xf(y)\\ f(79) = f(79 \cdot 1) = \\ 79 f(1) =79 \cdot 25 = 1975\)

hocus pocus indeed.

If a series doesn't have a finite limit then you can't use arithmetic on it.

the addition of two different sized matrices isn't defined.