Algebra Word Problem Solutions

Doing some computations and plottings I've found out that the function
${}_2{F}_1(2s-<!--\; -\; -->1,s-<!--\; -\; -->\frac{1}{2};s;-<!--\; -\; -->1)$
behaves for large real s like $4^{-<!--\; -\; -->s}$. More precisely: It seems that
${}_2{F}_1(2s-<!--\; -\; -->1,s-<!--\; -\; -->\frac{1}{2};s;-<!--\; -\; -->1)4^s$
grows, but very very slowly. No exponential growth or decay at all. If you modify the 4 only slightly this of course changes and you get exponential groth or decay. So my question is to explain this phenomenon. Is
$\underset{s\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{}_2{F}_1(2s-<!--\; -\; -->1,s-<!--\; -\; -->\frac{1}{2};s;-<!--\; -\; -->1)4^s=\mathrm{\infty <!--\; \infty \; -->}?$
Is there an easy to detemine for every x>0 the respective b(x)>0 such that you have
${}_2{F}_1(2s-<!--\; -\; -->1,s-<!--\; -\; -->\frac{1}{2};s;-<!--\; -\; -->x)\sim <!--\; \sim \; -->b^{-<!--\; -\; -->s}?$
What is
$\underset{s\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{}_2{F}_1(2s-<!--\; -\; -->1,s-<!--\; -\; -->\frac{1}{2};s;-<!--\; -\; -->x)b^s$
then?

Basic calculus question about epidemic modeling
I am a medical student attempting to build mathematical models of disease, but am struggling with the calculus. In particular, I am very confused by this equation shown in attached photo, and feel worse because it is called "elementary". How is the second equation the integral of the first?
Feel free to ignore the first sentence, before "we consider."
The assumption (i) requires a fuller mathematical explanation, since the assumption of a recovery rate proportional to the number of infectives has no clearepidemiological meaning. We consider the "cohort" of members who were all infected at one time and let u(s) denote the number of these who are still infectivea time units after having been infected. If a fraction a of these leave the infective class in unit time then
$u^{\prime }=\mathrm{\alpha <!--\; \alpha \; -->}u,$
and the solution of this elementary differential equation is
$u(s)=u(0)e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}s}$

Proof of an $\frac{a}{b}=\frac{c}{d}⟺\frac{a+b}{a-b}=\frac{c+d}{c-d}$
but going from $\frac{a}{b}=\frac{c}{d}$
I already proved it backwards
$\frac{a+b}{a-b}=\frac{c+d}{c-d}$ I multiply each side by (a-b)(c-d)
$ac-<!--\; -\; -->ad+bc-<!--\; -\; -->bd=ac-<!--\; -\; -->bc+ad-<!--\; -\; -->bd$ so $2bc=2ad$ which by dividing both sides by $2bd$ gives us :
$\frac{a}{b}=\frac{c}{d}$
any ideas how to proceed ?
Edit : this equivalence only works when $a\ne <!--\; \ne \; -->b$ and $c\ne <!--\; \ne \; -->d$

Radicals and rational exponents. simplifying $4^{5/3}$

Some doctors recommend that men drink 3 liters of water every day . There are approximately 29.6 milliliters in 1 fluid ounce. Which measurement is closest to the number of fluid ounces in 3 liters ?
A. $89\text{ fl oz}$
B. $101\text{ fl oz}$
C. $10\text{ fl oz}$
D. $33\text{ fl oz}$

Potential such that the time-independent Schroedinger equation has an explicit solution
Consider the time-independent Schroedinger equation
${\mathrm{\varphi <!--\; \varphi \; -->}}^{″}(x)+V(x)\mathrm{\varphi <!--\; \varphi \; -->}(x)=\mathrm{\lambda <!--\; \lambda \; -->}\mathrm{\varphi <!--\; \varphi \; -->}(x)$
For testing my numerics, I would like to now:
For which choices of V are there explicit expressions for the solutions $(\mathrm{\varphi <!--\; \varphi \; -->},\mathrm{\lambda <!--\; \lambda \; -->})$?
For example, if $V(x)=x^2$, then the solutions are products of Hermite polynomials and $exp(-<!--\; -\; -->x^2)$ (up to correct scales) and arithmetic progressions (the same works in higher dimensions with the analogue potential).

$33\frac{1}{3}$ percent of 600

Real world use of even and odd functions
What is the real world use of knowing whether a function is odd or even? Any practical examples?
For example, quadratic equations, differential equations and calculus in General is used for, among other things, determining motion of bodies. Just knowing that a function is symmetrical or not, help how? In terms of solving practical problems with the knowledge.

Modeling drunkeness over time
So you have the function of drunkenness,D(t), the amount of alcohol in your blood, for $t>0$, over time,t in hours:
$D(t)=T-<!--\; -\; -->ct$
c being the arbitrary amount of alcohol your liver will process in 1 hour. T being the total amount of alcohol you have drunk at time t.
So T does depend on t, but it still a totally arbitrary amount.
As a piece wise function based on $T=c_{n}t$. $c_n=nc$ with n being some non negative number, representing the number of drinks of strength c had in 1 hours time:
$T=\begin{array}{cc}\{& \begin{array}{cc}t=1& n_{1}c\\ t=2& n_{2}c\\ ...\\ t=n& n_{n}c\end{array}\end{array}$
So to find T for a certain time, t,
$T(t=n)=n_{1}c+n_{2}c...+...n_{n}c=(n_1+n_2+n_n)c$
The number of drinks had each hour is totally arbitrary. Now what I want to do is to compare this between different values of c. Like comparing the same amount of drinking between a light beer (call it 1c) and a craft beer (2c). Is there a way to generalize this.
For example:
$T=\begin{array}{cc}\{& \begin{array}{cc}t=1& 2c\\ t=2& 1c\\ t=3& 3c\end{array}\end{array}$
$T(t=3)=6c$
So
$D(t=3)=3c$
but if $c=2c$
$T(t=3)=12c$
and
$D(t=3)=9c$
So drunkenness is not a linear relationship related to the strength of alcohol you are drinking.

How to know that irrational numbers never repeat?

Write the equation in point slope form given (0,9) and (2,0)

Find the common ratio for the following geometric sequence 4, 12, 36, 108...

Bob is 8 years younger than charlie. Doug is 2 years older than charlie. The sum of Charlie and Dougs ages is 4 times bobs age. How do you find the age of all boys. math problem?

For what class of functions is steepest descent guaranteed to converge in finite number of iterations?
Is there any classification of ${\mathbb{R}}^n\to <!--\; \to \; -->\mathbb{R}$ functions with respect to finite convergence of steepest descent (and similar first-order methods, e.g., nonlinear conjugate gradients)?
By finite convergence I mean that algorithm reaches a point $x_k$ with $||\mathrm{\nabla <!--\; \nabla \; -->}f(x_k)||=0$ after a finite number of iterations, given that no round-off errors occur (and maybe assuming that line search is exact).
It is known that algorithms are globally convergent, i.e. $li{m}_{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}||\mathrm{\nabla <!--\; \nabla \; -->}f(x_k)||=0$, but there is no guarantee that it happens for finite k.
For example, conjugate gradient method is guaranteed to converge in n steps for positive definite quadratic functions. Are there any similar results for steepest descent?
What about (strongly) convex general functions? I was only able to find results of the type
$k=O\left(\frac{||x_0-<!--\; -\; -->x^{\ast <!--\; \ast \; -->}|{|}^2}{\mathrm{ϵ<!--\; ϵ\; -->}}\right)$
to achieve $f(x_k)-<!--\; -\; -->f(x^{\ast <!--\; \ast \; -->})\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}$, where f is smooth convex and x∗ is a stationary point.
As far as I understand, it does not imply finite convergence for $\mathrm{ϵ<!--\; ϵ\; -->}=0$. Is it known that finite convergence does not happen even for strongly convex functions?

How many inches are equal to 10 centimeters? Remember, 2.54 cm = 1 in. Round your answer to the nearest hundredth. Do not include units in your answer.

Are there many more irrational numbers than rational?

Find answers of this system of equations in real numbers
$\{\begin{array}{c}x+\frac{2}{y}=3\\ y+\frac{2}{z}=3\\ z+\frac{2}{x}=3\end{array}$

If $a,b,m$ and $n$ are positive integers such that $\sqrt[m]{a}$ and $\sqrt[n]{b}$ are irrational numbers, how can we prove that the sum $\sqrt[m]{a}+\sqrt[n]{b}$ is also irrational?

Find three consecutive odd integers such that when five times the third minus the second is added to the first, the result is 23

A geometric inequality in a triangle
If $x,y,z>0,$ could it be:
$\frac{z(2z+x)}{z+x}{b}^2+\frac{y(2y+x)}{y+x}{c}^2>\frac{yz}{y+z}{a}^2$
for a triangle $ABC,$ with $a=BC,$$b=AC,$$c=AB?$
If $b^2+c^2=a^2$, then the inequality holds, since
$\frac{z(2z+x)}{z+x}-<!--\; -\; -->\frac{yz}{y+z}=\frac{z^2(x+y+2z)}{(z+x)(y+z)},\frac{y(2y+x)}{y+x}-<!--\; -\; -->\frac{yz}{y+z}=\frac{y^2(x+2y+z)}{(z+x)(y+z)}$
If $b^2+c^2>a^2$, then the inequality holds, since
$\frac{z(2z+x)}{z+x}{b}^2+\frac{y(2y+x)}{y+x}{c}^2>\frac{yz}{y+z}{b}^2+\frac{yz}{y+z}{c}^2>\frac{yz}{y+z}{a}^2$
I can’t handle the case $b^2+c^2<a^2$ and I need some help