Expert Answers to Algebra 2 Problems

Sine and Cosine Models
This is a general question about modeling the seasons using sine and cosine functions; I am trying to use sine and cosine to model cyclic behavior in sales due to the seasons (spring, summer, fall and winter); I have the following seasonal factors so to speak:-
$t=0$ Start of year
$t=3$ Spring - $\frac{\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}t)}{12}$
$t=6$ Summer - $\frac{\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}t)}{12}$
$t=9$ Fall - $\frac{\sin <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}t)}{18}$
$t=12$ Winter - $\frac{\cos <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}t)}{6}$
I want to get sales to be max in each period for a given product. For example I want sales for sandals to max in the spring and sales for cardigans to max in the fall.
I Would appreciate advice and guidance as am not sure if these are correct. But looking at the graphs, they seem reasonable.

Solve each equation
$$\begin{gathered} 1)r+6-<!--\; -\; -->5=-<!--\; -\; -->1 \\ a)\{-<!--\; -\; -->11\} \\ b)\{-<!--\; -\; -->15\} \\ c)\{7\} \\ d)\{-<!--\; -\; -->2\} \end{gathered}$$
$$\begin{gathered} 2)-<!--\; -\; -->3n-<!--\; -\; -->3n=-<!--\; -\; -->24 \\ a)\text{ No solution} \\ b)\{-<!--\; -\; -->15\} \\ c)\{11\} \\ d)\{4\} \end{gathered}$$
$$\begin{gathered} 3)-<!--\; -\; -->7b-<!--\; -\; -->2b=18 \\ a)\text{ No solution} \\ b)\{2\} \\ c)\{-<!--\; -\; -->2\} \\ d)\{9\} \end{gathered}$$
$$\begin{gathered} 4)7b+5-<!--\; -\; -->1=-<!--\; -\; -->10 \\ a)\{4\} \\ b)\{6\} \\ c)\{-<!--\; -\; -->3\} \\ d)\{-<!--\; -\; -->2\} \end{gathered}$$
$$\begin{gathered} 5)6(7-<!--\; -\; -->3a)-<!--\; -\; -->4=128 \\ a)\{-<!--\; -\; -->14\} \\ b)\{0\} \\ c)\{-<!--\; -\; -->10\} \\ d)\{-<!--\; -\; -->5\} \end{gathered}$$
$$\begin{gathered} 6)6(8k-<!--\; -\; -->7)=150 \\ a)\{15\} \\ b)\{-<!--\; -\; -->5\} \\ c)\{-<!--\; -\; -->14\} \\ d)\{4\} \end{gathered}$$
$$\begin{gathered} 7)2k-<!--\; -\; -->5(8+6k)=156 \\ a)\{13\} \\ b)\{4\} \\ c)\{-<!--\; -\; -->13\} \\ d)\{-<!--\; -\; -->7\} \end{gathered}$$
$$\begin{gathered} 8)-<!--\; -\; -->5(-<!--\; -\; -->2x+4)=-<!--\; -\; -->100 \\ a)\{-<!--\; -\; -->8\} \\ b)\{3\} \\ c)\{9\} \\ d)\{1\} \end{gathered}$$
$$\begin{gathered} 9)-<!--\; -\; -->2-<!--\; -\; -->5x=-<!--\; -\; -->10-<!--\; -\; -->6x+3x \\ a)\{4\} \\ b)\{-<!--\; -\; -->11\} \\ c)\{15\} \\ d)\{3\} \end{gathered}$$
$$\begin{gathered} 10)3m-<!--\; -\; -->8+3m=16-<!--\; -\; -->m+3m \\ a)\{12\} \\ b)\{1\} \\ c)\{-<!--\; -\; -->5\} \\ d)\{6\} \end{gathered}$$

Arithmetic of Polynomials
if k is a field and f,g,h are polynomials such that $f=g+h$, then a polynomial that divides two of the three will divide the third.
I'm having trouble showing doing the problem as polynomials. I know how to do it is f,g,h are integers.

Separable differential equation modeling salt dissolved in water
I have no idea what to do for this. The equation is supposed to model a solution having salty water dumped into a tank that leaks the solution.
? A tank contains 1000L of pure water. Brine that contains .05 kg of salt per liter of water enters the tank at a ra te of 5L/min. Brine that contains .04 kg of salt per liter of water enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. From Stewart Calculus 7e
I am supposed to find the salt at t minutes and at one hour.
I try to simplify this problem by making it .07kg per lit of water at 5L/m and from here I have no idea how to make an equation for this.

Integration - involving logarithm
I have a slight confusion on how to integrate functions of the form:
$$\int <!--\; \int \; -->\frac{a}{x}dx$$
Suppose we have the following function:
$$\int <!--\; \int \; -->\frac{-<!--\; -\; -->2}{x}dx$$
There are two ways we can proceed to integrate this function. One is to treat the $-<!--\; -\; -->2$ sign as a constant and take it out of the integration function:
$$-<!--\; -\; -->2\int <!--\; \int \; -->\frac{1}{x}=-<!--\; -\; -->2\ln <!--\; \; -->x=\ln <!--\; \; -->\frac{1}{x^2}$$
Another way to do this is to treat the $-<!--\; -\; -->2$ as part of the integration variable:
$$\int <!--\; \int \; -->\frac{1}{-<!--\; -\; -->0.5x}dx=-<!--\; -\; -->2\ln <!--\; \; -->(-<!--\; -\; -->0.5x)$$
This seems to obtain two different answers. Which method is correct? Or is there something that I'm missing?

Modeling following constraints in MILP
I want to know how I should formulate the following constraints in my MIP problem?
$$x=x_1{z}_1+\cdots <!--\; \cdots \; -->+x_n{z}_n\text{ and }{y}_1\le <!--\; \le \; -->y\le <!--\; \le \; -->y_n\text{ and }{z}_1+\cdots <!--\; \cdots \; -->+z_n=1$$
OR
$$y=y_1{w}_1+\cdots <!--\; \cdots \; -->+y_n{w}_n\text{ and }{x}_1\le <!--\; \le \; -->x\le <!--\; \le \; -->x_n\text{ and }{w}_1+\cdots <!--\; \cdots \; -->+w_n=1$$
x and y are continuous variables. $z_1,\dots <!--\; \dots \; -->,z_n$ and $w_1,\dots <!--\; \dots \; -->,w_n$ are binary decision variables. $x_1,\dots <!--\; \dots \; -->,x_n$ and $y_1,\dots <!--\; \dots \; -->,y_n$ are parameters.

Proof that a polynomial has a minimum in R
I have to prove to following statement and I am having a really hard time here.
There it is:
Prove that the following polynomial has a minimum in R
$$p(x)=x^4+a_3{x}^3+a_2{x}^2+a_{1}x+a_0$$
I tried to make the following proof, but I am stuck:
The polynomial can be shown like this:
$$x^4(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4})$$
Let
$$g(x)=(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4})$$
So we can now write the polynomial like
$$p(x)=x^{4}g(x)$$
Since $x,a\in <!--\; \in \; -->\mathbb{R}$ we can say that $\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}=\underset{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}=1$ (from limit arithmetic).
We know that it is always the case that $x^4>0$, so the sign of
$$p(x)=x^{4}g(x)$$
is determined by g(x). For certain values, $g(x)<0$ so we can say that there is some $a,b\in <!--\; \in \; -->\mathbb{R}$ such that $g(a)<0$ and $g(b)>0$.
p(x) is continouos everywhere, certainly in [a,b], so from the Extreme Value Theorem, the function has a minimum in [a,b].
There are two problems in my proof:
I can't find any value in which g(x) is negative, only if there is some a which is negative. What if a is always positive?
If the first problem is solved, I managed to prove the statement for some [a,b]. Is it just enough? It seems to me that it isn't.

re-arrange equation $L=2^{10(v-<!--\; -\; -->1)}{v}^2$
Is it possible to re-arrange this equation to make v the subject?
$$L=v^2.2^{10(v-<!--\; -\; -->1)}$$
If so, what is the answer?
If it helps (which by excluding zero it should)...
$$0<v<1$$
I have tried pages of scribbling and got nowhere. In desperation I have tried solving each product separately (easy enough), and then tried to get the right overall result from some combination of $$\sqrt{L}$$ and $(log2(L)+10)/10$ but that hasn't got me anywhere.
I am out of my depth. Please help.

Finding out the logarithmic function for the situation below
The situation reads as follows:
There are 3000 barbs in a pond and every year 20% of the barbs die and then 1000 new barbs come to the pond. A logarithmic function needs to be plotted to graph this change in population.
I worked through a part of the above situation and arrived at the function:
$$y=3000({0.8}^x)+1000({0.8}^{x-<!--\; -\; -->1})+{0.8}^{x-<!--\; -\; -->2}+\cdots <!--\; \cdots \; -->+{0.8}^1+{0.8}^0$$
How do I convert this above equation into a logarithmic function?

Solving a logarithmic equation with variables on each side
Okay, so while doing a problem for my calculus class I was required to graph two functions in order to see where they intersect, as according to my teacher there is no way to solve it analytically.
This really bothers me and there must be a way to solve it. I did research online to try to solve it, but I had no idea where to even start.
Here is the equation.
$$y=\ln <!--\; \; -->(5-<!--\; -\; -->y)$$

Find the derivative of the following function: ${\displaystyle y(x)=(\ln <!--\; \; -->(2x){)}^{5x},x>\frac{1}{2}}$

Integration problem: $\int <!--\; \int \; -->\ln <!--\; \; -->(\sin <!--\; \; -->(\sqrt{x})+\cos <!--\; \; -->(\sqrt{x}))dx$
I really don't know how to start solving this problem; any tips or solutions will be greatly appreciated.
Thank you in advance.

The cost of a bottle of orange juice is equal to the cost of two bottles of milk. 3 bottles of orange juice and 5 bottles of milk equals 66 dollar
How much is the cost of a bottle of milk

Is there any expansion for $\log <!--\; \; -->(1+x)$ when $x><!--\; >\; -->1$?
Is there any expansion for $\log <!--\; \; -->(1+x)$ when $x><!--\; >\; -->1$?

Showing an inequality with ln
I have to show that the following inequation is true:
$\frac{\ln <!--\; \; -->(x)+\ln <!--\; \; -->(y)}{2}\le <!--\; \le \; -->\ln <!--\; \; -->(\frac{x+y}{2})$
I transformed it into
$\frac{\ln <!--\; \; -->(x\cdot <!--\; \cdot \; -->y)}{2}\le <!--\; \le \; -->\ln <!--\; \; -->(x+y)-<!--\; -\; -->\ln <!--\; \; -->(2)$
because I thought that I better can show the inequation here, but I don't know how to proceed.
How can I proceed or am I completely wrong?

Is there 'Algebraic number' which cannot display with Arithmetic operation and root
Let $a+bi$ be an algebraic number. Then there is polynomial which coefficients are rational number and one of root is $a+bi$.
I think..
$$x=a+bi$$
we can subtract $c_1$ (which is rational number) from both sides.
$$x-<!--\; -\; -->c_1=a-<!--\; -\; -->c_1+bi$$
and we can power both side.
$$(x-<!--\; -\; -->c_1{)}^n=(a-<!--\; -\; -->c_1+bi{)}^n$$
and repeat we can get polynomial which coefficients are rational number and one of root is $a+bi$.
Therefore, I think there is no 'Algebraic number' which cannot be displayed with arithmetic operations and roots.

Exponents with the same power
I've wanted to practice solving simple operations on exponents, so I've made a couple of equations to which I know the answers.
$5^x-<!--\; -\; -->4^x=9$
I feel really stupid, because I can't solve this one (obviously the answer is 2). I can't separate the x's after using logarithms.
Any hints? Thanks.

Finding the leading exponent of a binary number
Let's say that the binary representation of a number $k$ is $2^{X_n}+2^{X_{n-<!--\; -\; -->1}}+\cdots <!--\; \cdots \; -->+2^{X_0}$ with each term in this polynomial having a $1$ or $0$ multiplied to it.
Now, given only $k$, and no other information, is there a way that I can find the number $X_n$(which is the highest degree of the binary representation of $k$)?

An investment counselor calls with a hot stock tip. He believes that if the economy remains​ strong, the investment will result in a profit of ​$ 40 comma 40,000. If the economy grows at a moderate​pace, the investment will result in a profit of ​$ 10 comma 000 10,000. ​However, if the economy goes into​ recession, the investment will result in a loss of $40,000. You contact an economist who believes there is a 30​% probability the economy will remain​ strong, a 60​% probability the economy will grow at a moderate​ pace, and a 10​% probability the economy will slip into recession.

How to solve the following equation? ${\log }_3<!--\; \; -->({\log }_x<!--\; \; -->({\log }_4<!--\; \; -->16))=-<!--\; -\; -->1$
I am trying to solve this equation for $x$. This is what I have so far:
$${\log }_3<!--\; \; -->({\log }_x<!--\; \; -->2)=-<!--\; -\; -->1.$$
Okay, now I have this:
$$\log <!--\; \; -->2=(1/3)\log <!--\; \; -->x$$
How do I isolate x from here?