Rom

continuous variables  can be any value at all, such as length, weight, and area

discrete variables must take integer values such as the number of items or events

can you figure it out now?

\(1) ~1 + 5 = 6\\~\\ 2)~\text{The degree of a polynomial $f(x)$ is its highest power of $x$ $\\f(x)+bg(x)$ will be of degree 4}\)

\(miles_1 = M\\ miles_k = miles_1 + 10(k-1),~k=1,2,\dots 5\\ \sum \limits_{k=1}^5 miles_k = \sum \limits_{k=1}^5 M+10(k-1) = 500\\ 5M + 10\sum \limits_{k=0}^4 k = 500\\ 5M + 10\dfrac{(4)(4+1)}{2} = 500\\ 5M + 100 = 500\\ M=80\\ miles_5 = 80 + 10(4) = 120~miles \)

\(\dfrac{du}{dt} = \begin{pmatrix}\dfrac{dx}{dt}\\\dfrac{dy}{dt}\end{pmatrix} = \begin{pmatrix}1&-2\\1&4\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = A u\)

\(\ell = \displaystyle \int \limits_0^\pi \sqrt{\dot{x}^2 + \dot{y}^2}~dt\\ \ell = \displaystyle \int \limits_0^\pi \sqrt{4\cos^2(t) + 4\sin^2(t)}~dt\\ \ell = \displaystyle \int \limits_0^\pi 2~dt = 2\pi\)

\(\text{in case you're not familiar with the notation $\dot{x} = \dfrac{dx}{dt}$}\)

one short syllable followed by one long syllable

I'd say it's B. compare

Warning:  This answer will require you to do significant algebra yourself! 

\(\text{real roots means the discriminant is non-negative}\\ \text{the sum of the squares of the roots of $a x^2 + b x+c=0$ is simply}\\ \text{$\dfrac{D+2ac}{a^2}$ where $D=b^2-4ac$ is the discriminant}\)

\(\text{I'm going to rewrite your polynomial to avoid confusion and call it}\\ p(x) = 4x^2 + 4(\alpha-2)x-8\alpha^2+14\alpha + 31\\ a = 4\\ b= 4(\alpha - 2)\\ c=-8\alpha^2+14\alpha +31\)

\(\text{First we need a non-negative discriminant $D$}\\ D=144(-3-2\alpha+\alpha^2)~\text{(I leave you to show this algebra)}\\ D=144(\alpha+1)(\alpha-3)\\ D\geq 0 \Rightarrow \alpha \in (-\infty,-1]\cup [3,\infty)\)

\(\text{Now you want to minimize $S=\dfrac{D+2ac}{a^2}$}\\ \text{subject to $a \in (-\infty, -1]\cup [3, \infty)$}\\~\\ \text{with more algebra you can show}\\ S= 5 \alpha ^2-11 \alpha -\dfrac{23}{2} \\ \text{The vertex of this parabola lies within the excluded interval}\\ \text{so the minimum must be on this boundary}\\ \text{at either $x=-1$ or $x=3$}\\ \left. S\right|_{\alpha = -1} = \dfrac 9 2 \\ \left. S\right|_{\alpha = 3} = \dfrac 1 2 \\ \text{and so the minimum value $\dfrac 1 2$ occurs at $\alpha = 3$}\\ \text{or translating back to your original problem statement that $a=3$}\)

there's probably a better way to phrase this....

\(1)~\text{rewrite the system as $u^\prime = A u,$ where $u(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}$}\\ 2) ~\text{find the eigenvalues $\lambda_{1,2}$ and corresponding eigenvectors $v_{1,2}$ of $A$}\\ 3) ~u(t) = c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t}v_2\)

\( 924500000=\underline{92450000} \times 10\)

Yeah divide the left side by 10 which just means knocking off one 0 at the end.