Expert Help with High School Geometry

Geometric Brownian motion probability question
Let S(t), $t\ge <!--\; \ge \; -->0$ be a geometric Brownian motion with drift parameter $\mathrm{\mu <!--\; \mu \; -->}=0.3$ and volatility parameter $\mathrm{\sigma <!--\; \sigma \; -->}=0.3$. Find
$(a)P(S(1)>S(0))$
$(b)P(S(2))>S(0))$

Finding the Vertices of an Ellipse Given Its Foci and a Point on the Ellipse
The question is as follows:
The focal points of an ellipse are (12,0) and (-12,0), and the point (12,7) is on the ellipse. Find the points where this curve intersects the coordinate axes.
I know that the center of the ellipse would be (0,0) because that is the midpoint of the foci. However, I am not sure as to how this information will help me in finding the intersections on the coordinate axes (or the vertices). Any help will be greatly appreciated.

Find the distance from the point P(0,0,1) to the surface z=x^2+2y^2

Finding the volume of the solid generated by revolving the given curve.
The objective is to find the volume of the solid generated by revolving the curve $y={\displaystyle \frac{a^3}{a^2+x^2}}$ about its asymptote.
Observing the given function yields that $y\ne <!--\; \ne \; -->0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about $x-<!--\; -\; -->axis$ is gives as $V=2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}(f(x{)}^2-<!--\; -\; -->0)dx$.
Which gives: $2\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{\displaystyle \frac{a^6}{(a^2+x^2{)}^2}}dx$
Now, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form ${\displaystyle \frac{1}{x^4+2x^2+1}}$ for $a=1$, which is transformed into ${\displaystyle \frac{\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}}=2[{\displaystyle \frac{1+\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}}}-<!--\; -\; -->{\displaystyle \frac{(1-<!--\; -\; -->\frac{1}{x^2})}{x^2+2+\frac{1}{x^2}}}]$ which can be integrated, but this integral is tending to infinity.

Probability of two successes, possible geometric distribution problem
Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula $P(Y=k)=(k-<!--\; -\; -->1)(p^2)(1-<!--\; -\; -->p{)}^{k-<!--\; -\; -->2}$.
Okay I know the outcome space is $Y:=2,3,4,...$, the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra $(k-<!--\; -\; -->1)\cdot <!--\; \cdot \; -->p$ in front.

Minkowski sum of convex sets in the plane which are not polygons
Can the Minkowski sum of two convex sets in the plane which are not polygons be a polygon? Explicitly my convex set is of the form $C=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:x,y\ge <!--\; \ge \; -->0\text{ , }\sqrt{y}+\sqrt{x}\ge <!--\; \ge \; -->1\text{ and }y\le <!--\; \le \; -->1-<!--\; -\; -->x\}$
I am interested in knowing if there is a convex set C' such that the Minkowski sum
$C+C^{\prime }=P$
where $P=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:x,y\ge <!--\; \ge \; -->0\text{ and }1/2\le <!--\; \le \; -->x+y\le <!--\; \le \; -->1\}.$

Finding volume by rotating cos x and sin x about $y=-<!--\; -\; -->1$, $x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$
I'm working on a problem that is asking for the volume of $y=\cos <!--\; \; -->x$ when rotated about the line $y=1$, with a restricted domain of $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. The range is [0, 2].
There's a problem I'm working on that is asking for the volume of $y=\cos <!--\; \; -->x$ and $y=\sin <!--\; \; -->x$ when rotated about line $y=-<!--\; -\; -->1$ with a domain of $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{4})$
I don't even know where to begin. First off; which method would be most effective in this case? A shell, washer or disc?

Finding volume of a revolution
I want to find the volume of the revolution that occurs when the region bounded by $y=x^2$ and $y=1$ is revolved around the line $y=2$. The problem is that it is not solid and I cannot understand how the disk method could be used. How can one attack the problem?

Distance between (3,13,10) and (3,-17,-1)?

How can you tell when you need to use the binomial coefficient for probabilities?
For example, the probability of being dealt a hand with 3 spades and 2 hearts from a standard deck of cards. I would think the geometric probability is:
${\displaystyle \frac{13\ast <!--\; \ast \; -->12\ast <!--\; \ast \; -->11\ast <!--\; \ast \; -->13\ast <!--\; \ast \; -->12}{52\ast <!--\; \ast \; -->51\ast <!--\; \ast \; -->50\ast <!--\; \ast \; -->49\ast <!--\; \ast \; -->48}}$ because I usually don't use the binomial coefficient with geometric probabilities, but the answer is incorrect unless I multiply by the binomial coefficient (which is a factor of 10 here).

Distance between (9,2,0) and (8,6,2)?

Find the smallest nonnegative integer that sasfied congruence $$y\equiv <!--\; \equiv \; -->41(mod4)$$

Write a proof for each theorem. Transitive Property of Angle Congruence

Comparing between regular polygons
In this experiment, I am adding the inradius (let's call it A) and circumradius (let's call it B) of different polygons with equal sides each equal 1 (starting with a square and adding one side each time). The result is $A+B=C$ when side of polygon $=1$.
When comparing the C of one polygon with the C of a polygon with one side more, the difference seems to go smaller, as if approaching a version of $\mathrm{\pi <!--\; \pi \; -->}$ number with 0. before (possibly such as 0.314159265359...).
Can anyone confirm it or elaborate on it?
I can not go over a polygon with 1000 sides in my computation power, and would like to know what to expect while going towards a polygon with infinity sides.
Here are some examples:
4 sided polygon: $0.5+0.707106781=1.207106781$
5 sided polygon: $0.68819096+0.850650808=1.5388417680000002$ (Difference of 0.33173498700000015 from previous result)
6 sided polygon: $0.866025404+1=1.866025404$ (Difference of 0.3271836359999998 from previous result)
7 sided polygon: $1.0382607+1.15238244=2.1906431399999997$ (Difference of 0.32461773599999977 from previous result)
8 sided polygon: $1.20710678+1.30656296=2.51366974$ (Difference of 0.3230266000000004 from previous result)
9 sided polygon: $1.37373871+1.4619022=2.83564091$ (Difference of 0.32197116999999986 from previous result)
10 sided polygon: $1.53884177+1.61803399=3.15687576$ (Difference of 0.3212348500000002 from previous result)
11 sided polygon: $1.70284362+1.77473277=3.47757639$ (Difference of 0.3207006299999997 from previous result)
12 sided polygon: $1.8660254+1.93185165=3.79787705$ (Difference of 0.32030066 from previous result)
13 sided polygon: $2.02857974+2.08929073=4.11787047$ (Difference of 0.31999341999999986 from previous result)
14 sided polygon: $2.19064313+2.2469796=4.43762273$ (Difference of 0.31975226000000045 from previous result)
15 sided polygon: $2.35231505+2.40486717=4.757182220000001$ (Difference of 0.3195594899999996 from previous result)
...
999 sided polygon: $158.995264+158.99605=317.991314$
1000 sided polygon: $159.154419+159.155205=318.309624$ (Difference of 0.31830999999999676 from previous result)

Suppose we roll a fair dice repeatedly. Find the probability that we get our 4th prime number on the 25th roll. I think it should be 4 successes and 21 failures but I feel it incorrect.

Geometric distribution with unequal probabilities for trials
I am researching an engineering problem in which I want to model the probability distribution of the number X of independent trials needed to get one success. If the probability of success at each trial is a constant p, the solution clearly reduces to the simple geometric distribution. However, in this particular case, the probability varies with each trial in a way that I can predict.
I feel sure there must already be a body of work on this type of problem that I can refer to, but I have been unable to find it. I have come across the Poisson Binomial distribution which models the probability distribution of the number of successes in a sequence of X independent yes/no experiments with different success probabilities. So I naturally searched for the Poisson Geometric or Geometric Poisson distribution, but this in fact refers to something quite different (a variation on the Poisson rather than Geometric distribution).
Can anyone help with the name of the distribution I am looking for?

Pentagon with two right angles (aka Van Aubel's Theorem)
Given that ABCDE is a convex pentagon such that $AB=BC$, $CD=DE$, M is middle point of side EA and the angles $\stackrel{^<!--\; ^\; -->}{ABC}=\stackrel{^<!--\; ^\; -->}{CDE}=90°$, find the measure of the angle $\stackrel{^<!--\; ^\; -->}{BMD}$.

Finding volume inside a paraboloid and inside a sphere
I want to find the volume inside the paraboloid $z=x^2+y^2$ and inside the sphere $x^2+y^2+z^2=12$.
So, I wrote $x=r\cos <!--\; \; -->t,y=r\sin <!--\; \; -->t$, find out $0\le <!--\; \le \; -->r\le <!--\; \le \; -->\sqrt{3}$ and $0\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$. We also have $z=r^2$ the paraboloid and $z=\sqrt{12-<!--\; -\; -->r^2}$. Now, why is the desired volume is given as:
${\displaystyle {\int <!--\; \int \; -->}_0^{2\mathrm{\pi <!--\; \pi \; -->}}{\int <!--\; \int \; -->}_0^{\sqrt{3}}(\sqrt{12-<!--\; -\; -->r^2}-<!--\; -\; -->r^2)rdrdt?}$
Why do we subtract two z values in this way and not use just one of them?