Expert Answers to Algebra 2 Problems

Modeling simple linear equations
This should be pretty simple but I'm blanking on this. I need to model (graph) how path 1 becomes equally as efficient as path 2 as the distance of path 2 increases.
distance of path 1 (from A to B) $=10$
distance of path 2 (from A to C) $=y$
The path is a path taken by a forklift driver. The vehicle speed is 2.2m/s. The paths are measured in meters.
On path 1, there is an added flat time of 30 seconds (due to something he has to stop and do). On path 2, there is an added flat time of 10 seconds.
So in summary:
$$\frac{x}{2.2}+30=t_{\text{path }1}wherex>0$$
$$\frac{y}{2.2}+10=t_{\text{path }2}wherey>0$$
I know that once the distance of path $2=54$ meters, both routes take the same amount of time (34.545 seconds). But I can't figure out how to put this into equations that I can use to graph.

Modeling IP Problem
Hallco runs a day shift and a night shift. No matter how many units are produced, the only production cost during a shift is a setup cost. It costs $8,000 to run the day shift and $4,500 to run the night shift. Demand for the next two days is as follows: day 1, 2,000; night 1, 3,000; day 2, 2,000; night 2, 3,000. It costs $1 per unit to hold a unit in inventory for a shift. Determine a production schedule that minimizes the sum of setup and inventory costs. All demand must be met on time.
I am confused on how to model these sentences. It costs $1 per unit to hold a unit in inventory for a shift. What does it mean in mathematical language?

Problem involving polynomial function and prime numbers
Let f be a polynomial function, with integer coefficients, strictly increasing on N such that $f(0)=1$. Show that it doesn't exist any arithmetic progression of natural numbers with ratio $r>0$ such that the value of function f in every term of the progression is a prime number. I noticed that the last term of f is 1, due to the fact that $f(0)=1$. I don't know how to continue.

Determine the minimal polynomial of T with respect to ${\left(\begin{array}{ccc}0& 0& 1\end{array}\right)}^t.$.
Let $T:{\mathbb{R}}^3\to <!--\; \to \; -->{\mathbb{R}}^3$ be defined by $T(v)=\left(\begin{array}{ccc}2& -<!--\; -\; -->1& 1\\ -<!--\; -\; -->3& 4& -<!--\; -\; -->5\\ -<!--\; -\; -->3& 3& -<!--\; -\; -->4\end{array}\right)v.$
Determine the minimal polynomial of T with respect to $\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$
$T(v)=\left(\begin{array}{ccc}2& -<!--\; -\; -->1& 1\\ -<!--\; -\; -->3& 4& -<!--\; -\; -->5\\ -<!--\; -\; -->3& 3& -<!--\; -\; -->4\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ -<!--\; -\; -->5\\ -<!--\; -\; -->4\end{array}\right)$
$T^2(v)=\left(\begin{array}{ccc}2& -<!--\; -\; -->1& 1\\ -<!--\; -\; -->3& 4& -<!--\; -\; -->5\\ -<!--\; -\; -->3& 3& -<!--\; -\; -->4\end{array}\right)\left(\begin{array}{c}1\\ -<!--\; -\; -->5\\ -<!--\; -\; -->4\end{array}\right)=\left(\begin{array}{c}3\\ -<!--\; -\; -->3\\ -<!--\; -\; -->2\end{array}\right)$
$T^3(v)=\left(\begin{array}{ccc}2& -<!--\; -\; -->1& 1\\ -<!--\; -\; -->3& 4& -<!--\; -\; -->5\\ -<!--\; -\; -->3& 3& -<!--\; -\; -->4\end{array}\right)\left(\begin{array}{c}3\\ -<!--\; -\; -->3\\ -<!--\; -\; -->2\end{array}\right)=\left(\begin{array}{c}7\\ -<!--\; -\; -->11\\ -<!--\; -\; -->10\end{array}\right)$
The corresponding matrix is
$A=(v,T(v),T^2(v),T^3(v))=\left(\begin{array}{cccc}0& 1& 3& 7\\ 0& -<!--\; -\; -->5& -<!--\; -\; -->3& -<!--\; -\; -->11\\ 1& -<!--\; -\; -->4& -<!--\; -\; -->2& -<!--\; -\; -->10\end{array}\right)$
After row reducing this matrix, we obtain $\left(\begin{array}{cccc}1& 0& 0& -<!--\; -\; -->2\\ 0& 1& 0& 1\\ 0& 0& 1& 2\end{array}\right)$, and so $\text{null}(A)=\text{span}\{\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}2\\ -<!--\; -\; -->1\\ 0\\ 1\end{array}\right)\}.$ From this we obtain polynomials $f(x)=x^2$ and $g(x)=x^3-<!--\; -\; -->x-<!--\; -\; -->2$ and since $\mathrm{deg}<!--\; \; -->f(x)<\mathrm{deg}<!--\; \; -->g(x),$, $x^2$ is the minimal polynomial.
The back of my book says the minimal polynomial is $x^2-<!--\; -\; -->3x+2.$ What am I doing wrong?

Finite field adjunction - isomorphism that maps from the field into a polynomial ring factored over an ideal
A field $(K,+,0,\ast <!--\; \ast \; -->,1)$ is given, and it is a finite field with 25 elements. P is the prime field over K. Further it was given that a polynomial $r(x):=x^2+x+1\in <!--\; \in \; -->P[x]$ is irreducible.
Now, the polynomial $y^2+y+1\in <!--\; \in \; -->K[x]$ has two non-equal zeros in K, say ${\mathrm{\alpha <!--\; \alpha \; -->}}_1\ne <!--\; \ne \; -->{\mathrm{\alpha <!--\; \alpha \; -->}}_2$.
The question is to describe all possible isomorphisms:
$$\mathrm{\phi <!--\; \phi \; -->}:K\to <!--\; \to \; -->P[x]/(r)$$
by giving the values of $\mathrm{\phi <!--\; \phi \; -->}({\mathrm{\alpha <!--\; \alpha \; -->}}_1)$ and $\mathrm{\phi <!--\; \phi \; -->}({\mathrm{\alpha <!--\; \alpha \; -->}}_2)$.
There are a few things that I don't understand about this question.
First of all, what are the properties of K[y] apart from it being a polynomial ring? I think that the additive group $(K,+,0)$ should be isomorph to a cyclic group $C_5\times <!--\; \times \; -->C_5$ - can that help me in any way?
Then - when I try to find the ${\mathrm{\alpha <!--\; \alpha \; -->}}_1$ and ${\mathrm{\alpha <!--\; \alpha \; -->}}_2$ I do my calculations in the Modulo 25 arithmetic - is that right (because I failed to find the zeros)?
And since the $\mathrm{\phi <!--\; \phi \; -->}(k)=k+(r)$ is one of the possibilites to describe this isomorphism, my question is - are there any more of them? Because my textbook doesn't state that this isomorphism is unique.
Any kind of help - or a link to some other question is more than welcome.

Calculating the integral of a logarithmic expression.
The problem I have been working with is
$\int <!--\; \int \; -->\frac{1}{\sqrt{x}(1+\sqrt{x})}dx$
The first step I did to solve this question was to set $u=1+\sqrt{x}$ the set $du=(1/2)x^{-<!--\; -\; -->1/2}$
Then I set ${\displaystyle \frac{1}{2}}\int <!--\; \int \; -->\ln <!--\; \; -->(du/u)$ and finally I think the answer is $\ln <!--\; \; -->{\displaystyle \frac{1}{x}}+c$

${27}^x+3^{x+1}=100$. Find an exact solution in terms of logarithms.
I've tried this in many different ways and can't find any leeway. I also can not find anything to help me online.

Log laws proof using only rational exponents
For all real $a>0$ and rational $b>0$
Show that $\ln <!--\; \; -->(a^b)=b\ln <!--\; \; -->(a)$

Related To Polynomial Division. How to prove the following result. Show how a polynomial with odd number of term will never be divisible by a divisor with $x+1$ as factor for modulo 2 arithmetic.

Continuous Hidden Markov Modeling
I am writing a speaker recognition program in Matlab using Mel Frequency Cepstral Coefficients, and although I have gotten the problem to work using discrete time wrapping, I was interested in try to successfully recognize speakers using Hidden Markov Modeling (HMM). I understand how to implement a HMM for simple examples (like urn and genie), but don't understand how it pertains to my problem exactly. For instance, what are actually the hidden states that the row vectors are trying to encode? And what does it mean to model these states as Gaussian mixtures? For more information on MFCCs, it gives for each speaker a matrix of numbers, where each row corresponds to the power spectrum at one given time.

Different Polynomial Expansions of Natural Logarithm
I was recently Taylor-expanding ln around $(1,0)$. I noticed that this polynomial will have a range of input that converges between $0$ and $2$ regardless of Taylor order. I then found an expansion that did not seem to have this issue, namely:
$\underset{n\to <!--\; \to \; -->+\mathrm{\infty <!--\; \infty \; -->}}{lim}2\cdot <!--\; \cdot \; -->\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\frac{1}{2i+1}\cdot <!--\; \cdot \; -->{(\frac{x-<!--\; -\; -->1}{x+1})}^{2i+1})=\ln <!--\; \; -->(x)$
My question is: how is this formula derived (as in created, not the derivative) from ln?

Use equal products or equal quotients to mentally doe each computation. show your work and explain your reasoning.
$$\begin{gathered} a.250\times <!--\; \times \; -->-<!--\; -\; -->44 \\ b.-<!--\; -\; -->\frac{240}{48} \end{gathered}$$

Solve the logarithmic equation by $x$
${\log }_2<!--\; \; -->(x^2+7)+{\log }_3<!--\; \; -->(x+6)=6$
What I tried: ${\log }_2<!--\; \; -->(x^2+7)=a$ and ${\log }_3<!--\; \; -->(x+6)=b$, then $a+b=6$ and $2^a=3^{2b}-<!--\; -\; -->4\cdot <!--\; \cdot \; -->3^{b+1}+43$. But the problem is $a$ and $b$ don't have to be integers...>br?It is easy to show that for $x>3$ and $3>x\ge <!--\; \ge \; -->0$ there are no soultions. Which leaves us only with the interval $0>x>-<!--\; -\; -->6$

Value of $x$ when $5+\log <!--\; \; -->x=\log <!--\; \; -->\left(x^6\right)$
Find the value of $x$ when
$$5+\log <!--\; \; -->x=\log <!--\; \; -->\left(x^6\right)$$
I've tried many times to solve this, however I can't seem to find a correct (consistent) answer. My solutions range from
$$x=e,x=\sqrt{5},x=\sqrt{6}$$

Continuous-time versus discrete-time stochastic models
When modeling a dynamic phenomenon, (from a general point of view) people can use two type of models: (1) continuous-time models, (2) discrete-time models.
To be more precise, assume that we try to model a phenomenon observed in a time series form via dynamical systems. If the stochasticity involved in the system, we can use two type of models: continuous-time stochastic processes or discrete-time stochastic processes. Since the time-series data is discrete, it seems very natural to model the processes in discrete-time. However, in some application domains, like stochastic finance, continuous-time models are of great importance and interest.
Here are my questions:
1) Is there an advantage of continuous-time modeling? Since the calculus is more difficult (continuous-time models require Ito calculus), what is the computational or modeling advantage of continuous-time models?
2) If there is no analytical solution of a continuous-time model, we discretize it and solve it numerically as far as I know. So, is it so different from discrete-time modeling?

How to prove that: ${\log }_{\frac{1}{2}}<!--\; \; -->(3)+{\log }_3<!--\; \; -->\left(\frac{1}{2}\right)<-<!--\; -\; -->2$
Prove that:
$${\log }_{\frac{1}{2}}<!--\; \; -->(3)+{\log }_3<!--\; \; -->\left(\frac{1}{2}\right)<-<!--\; -\; -->2$$
Please help me solve it.

Determine which of the following differential equations are separable. Do not solve the equations. y'=y

Compare ${\log }_3<!--\; \; -->4$ and $2^{\frac{1}{4}}$
Both ${\log }_3<!--\; \; -->4$ and $2^{\frac{1}{4}}$ are somewhere in between $1$ and $2$. I know I can get approximate values but they are difficult to calculate by hand so I conclude it's not the way I'm supposed to do the exercise.
So how can I compare the two numbers without using a calculator?

How does $\ln <!--\; \; -->\frac{1}{2}$ become $-<!--\; -\; -->\ln <!--\; \; -->2$?
I just saw a video that said you can transfer the two from the denominator to the numerator, but how does that make the 2 negative? Wouldn't that just make the exponent negative?
For example $\frac{1}{2}$= $2^{-<!--\; -\; -->1}$
or am I missing something?

Separating the log of a sum
I know there is no formula to separate the log of a sum, e.g. $\log <!--\; \; -->(X+Y)$ into two parts, but are there any approximation rules that can allow me to achieve this objective?
$E_t(1+r_{t+1}^K)=E_t\left[{\displaystyle \frac{\frac{1}{X_{t+1}}\mathrm{\alpha <!--\; \alpha \; -->}{A}_0\frac{Y_{t+1}}{K_{t+1}}+Q_{t+1}(1-<!--\; -\; -->\mathrm{\delta <!--\; \delta \; -->})}{Q_t}}\right]$
Suppose we ignore the expectations operator for the moment.