Algebra Word Problem Solutions

Convert 12 kg to pounds used $1kg\approx <!--\; \approx \; -->2.2lb$

Write the equation of the ellipse formed by foci (6,2) and (10,2) and having a minor axis of length 10.

Solve this quadratic equation by completing the square:
$-2t^2+3t+2=0$

"Let $\mathrm{\alpha <!--\; \alpha \; -->}$ be an algebraic integer and p(x) be its minimal (monic) polynomial
$p(x)=\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}{a}_i{x}^i$
Such that $p(\mathrm{\alpha <!--\; \alpha \; -->})=0$ and $a_i\in <!--\; \in \; -->\mathbf{\text{Z}}$ (Z is the set of integers) and $a_n=1$.
"The extension of a ring A by the element a is the set $A[\mathrm{\alpha <!--\; \alpha \; -->}]$ of all complex numbers of the form
$\underset{j=0}{\overset{n-<!--\; -\; -->1}{\sum <!--\; \sum \; -->}}{c}_j{\mathrm{\alpha <!--\; \alpha \; -->}}^j$
such that $c_j\in <!--\; \in \; -->A$, with all the operations inherited from A.
"The degree of the extension is the degree of the polynomial."
I completely understand the 'algebraic integer' and the 'minimal polynomial' and the concept of set/ring extension, at least I think. My issue is mostly with the middle sentence; when it says 'all the complex numbers of form', but isn't there only one minimal polynomial so there's only one element in the set $A[\mathrm{\alpha <!--\; \alpha \; -->}]$? Is that single element basically just $p(\mathrm{\alpha <!--\; \alpha \; -->})-<!--\; -\; -->x^n$? Or are the $c_j$ related to the $a_i$ at all? The same n is mentioned twice. If not then why introduce p(x) at all? And what is ring A? Is that an initially empty set? And what does 'all the operations inherited from A' mean? Honestly, I can't find an explanation online for the $A[\mathrm{\alpha <!--\; \alpha \; -->}]$ sort of notation or any of my other questions online?

Prove that f(x) is differentiable $U(x)=x^2+f(x)$ is differentiable for all x
Here's my proof please let me know if it's OK:
$x^2$ is differentiable for all x because it's a polynomial. Thus if f(x) is differentiable for all x then U(x) is differentiable for all x. Let U(x) be differentiable for all x. Thus according to addition arithmetic $U(x)-<!--\; -\; -->x^2$ is differentiable for all x.
But $U(x)-<!--\; -\; -->x^2=x^2+f(x)-<!--\; -\; -->x^2=f(x)$. Thus the addition that differentiable itself equals f(x) therefore f(x) is differentiable as well.
I have to admit that there was a very similar exercise in my class and from which I used the "addition arithmetic" however I'm not sure what it would refer to. Does this refer to the fact that according to derivatives arithmetic the result of addition of 2 function which are differentiable themselves is differentiable? But we didn't actually derive these functions.

One of the factors of $8\times <!--\; \times \; -->3–800x$ is ____

Limit of logarithmic function using l'Hospital
How can I find the following limit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+\mathrm{\alpha <!--\; \alpha \; -->}x)}{\ln <!--\; \; -->(\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}))}$
where $\mathrm{\alpha <!--\; \alpha \; -->},\mathrm{\beta <!--\; \beta \; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$
My first guess was to use l'Hospital:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+\mathrm{\alpha <!--\; \alpha \; -->}x)}{\ln <!--\; \; -->(\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}))}=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})\mathrm{\alpha <!--\; \alpha \; -->}}{(1+\mathrm{\alpha <!--\; \alpha \; -->}x){\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}\mathrm{\beta <!--\; \beta \; -->}}$
But what can I do now? Is my approach correct or is there a simpler method?
Edit:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})\mathrm{\alpha <!--\; \alpha \; -->}}{(1+\mathrm{\alpha <!--\; \alpha \; -->}x){\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}\mathrm{\beta <!--\; \beta \; -->}}=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})\mathrm{\alpha <!--\; \alpha \; -->}}{(1+\mathrm{\alpha <!--\; \alpha \; -->}x)\mathrm{\beta <!--\; \beta \; -->}}\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(1+{\text{e}}^{-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}x})$
Applying L'Hospital a second time on the first fraction,
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\ln <!--\; \; -->(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})\mathrm{\alpha <!--\; \alpha \; -->}}{(1+\mathrm{\alpha <!--\; \alpha \; -->}x)\mathrm{\beta <!--\; \beta \; -->}}\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(1+{\text{e}}^{-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}x})=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{\beta <!--\; \beta \; -->}{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}}{(1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x})\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{\beta <!--\; \beta \; -->}}\cdot <!--\; \cdot \; -->1=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}}{1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}}\cdot <!--\; \cdot \; -->1$
Now let's apply L'Hospital one final time:
$\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}}{1+{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}}\cdot <!--\; \cdot \; -->1=\underset{x\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}\mathrm{\beta <!--\; \beta \; -->}}{{\text{e}}^{\mathrm{\beta <!--\; \beta \; -->}x}\mathrm{\beta <!--\; \beta \; -->}}\cdot <!--\; \cdot \; -->1=1$
Is this correct?

What is the explicit formula for the arithmetic sequence?

Divide polynomial by x + 2 using synthetic division.
$2x^4+5x^3-<!--\; -\; -->16x^2-<!--\; -\; -->45x-<!--\; -\; -->18$

Is the amount of irrationals numbers in between consecutive integers always the same? is this amount infinite?

Find the 8th term of the sequence if consider the sequence -2, -4, -8, -16,...

Spliced probability distribution modeling
I have a simulation model for insurance claims that works as follows:
1. Assume random values come from two distributions defined as
$\begin{array}{rcl}{f}_1(x)& ,& 0<x\le <!--\; \le \; -->c\\ f_2(x)& ,& x>c\end{array}$
A random value will fall in $f_1(x)$ with probability p and $f_2(x)$ with probability $1-<!--\; -\; -->p$.
So the model (in Excel) uses rand() to determine which distribution to sample from. $f_2(x)$ is shifted at c so the minimum value from this distribution is c while $f_1(x)$ is set to be capped at c. A determined number of values is generated in the model and then summed to give an aggregate value.
I'm trying express the limited expected value of this distribution and am wondering is the distribution simply defined as:
$f_X(x)=\{\begin{array}{ll}p\cdot <!--\; \cdot \; -->f_1(x),& 0<x\le <!--\; \le \; -->c\\ (1-<!--\; -\; -->p)\cdot <!--\; \cdot \; -->f_2(x),& x>c\end{array}$
or am I missing something? This distribution will not be continuous at c unless I force it to through solving for p but assume that's not a requirement.

Modeling Exponential Equation with 2 given points?
So let's say that the population of rabbits increase according to the law exponential growth.
If a certain population of rabbits has 100 rabbits after the second day and 300 after the fourth day, can we figure out how many rabbits that there were to begin with, or do we not have enough information?

Use the polynomial division algorithm to write the rational function in the form Q(x) + R(x)/D(x), where the degree of R is less than the degree of D. $x^2+x+1/x^2+x-<!--\; -\; -->2$

What is the number if twelve less than 4 times a number is 20

Find a possible value for the missing term of the geometric sequence 1250,__,50,...

Find the slope of any line perpendicular to the line passing through (−3,19) and (−14,12)

How to solve logarithm word problem given the exponential equation?
The question is
The world population in 2000 was approximately 6.08 billion. The annual rate of increase was about 1.26%. The function that models this is
$y=6.08(10{)}^{.0052t}$
where y represents the population in billions and t is the time in years after 2000. Write and solve a logarithmic equation to determine what year the world population will exceed 7.5 billion.
So is the question asking me to convert the given equation into logarithmic equation than
$7.05=6.08(10{)}^{.0052t}$
( 7.5 billion is y right?)
$7.05/6.08={10}^{.0052t}$
$1.15={10}^{.0052t}$
$log(1.15)=.0052t$
How would i even solve after that? is the way I've done so far good? how would i proceed after this than?

Consider $x$ a rational number. Let $\mathrm{ϵ<!--\; ϵ\; -->}\ge <!--\; \ge \; -->0$ be the minimal value such that x+ϵ is irrational, and let also $\mathrm{\gamma <!--\; \gamma \; -->}>0$ be the minimal value such that $x+\mathrm{\gamma <!--\; \gamma \; -->}$ is rational. How to compute $\mathrm{\gamma <!--\; \gamma \; -->}/\mathrm{ϵ<!--\; ϵ\; -->}$?

Write the point slope form of the equation given (7,-6) and m=1/2