Expert Help with High School Geometry

Geometry Proof: Convex Quadrilateral
A quadrilateral ABCD is formed from four distinct points (called the vertices), no three of which are collinear, and from the segments AB, CB, CD, and DA (called the sides), which have no intersections except at those endpoints labeled by the same letter. The notation for this quadrilateral is not unique- e.g., quadrilateral $ABCD=$ quadrilateral CBAD.
Two vertices that are endpoints of a side are called adjacent; otherwise the two vertices are called opposite. The remaining pair of segments AC and BD formed from the four points are called diagonals of the quadrilateral; they may or may not intersect at some fifth point. If X, Y, Z are the vertices of quadrilateral ABCD such that Y is adjacent to both X and Z, then angle XYZ is called an angle of the quadrilateral; if W is the fourth vertex, then angle XWZ and angle XYZ are called opposite angles.
The quadrilaterals of main interest are the convex ones. By definition, they are the quadrilaterals such that each pair of opposite sides, e.g., AB and CD, has the property that CD is contained in one of the half-planes bounded by the line through A and B, and AB is contained in one of the half-planes bounded by the line through C and D.
a) Using Pasch's theorem, prove that if one pair of opposite sides has this property, then so does the other pair of opposite sides.
b) Prove, using the crossbar theorem, that the following are equivalent:
1. The quadrilateral is convex.
2. Each vertex of the quadrilateral lies in the interior of the opposite angle.
3. The diagonals of the quadrilateral meet.

Let $m=\frac{1}{2}$ and $n=\frac{1}{2}$ be the probabilities of success and failure respectively and let p(k) be a probability mass function such that $p(k)=m\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}{n}^i+n\underset{i=1}{\overset{k}{\sum <!--\; \sum \; -->}}{m}^i.$
Is this a valid probability mass function?
Say we carry out this test infinitely many times, does $p(k)=1$?
We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).
Say we are summing from 0 to $\mathrm{\infty <!--\; \infty \; -->}$, that is let
$S_0=\underset{i=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{n}^i.$
Then this is simply evaluated as $S_0=1+n+n^2+n^3+\cdots <!--\; \cdots \; -->=\frac{1}{1-<!--\; -\; -->n}$
which is perfectly valid as $n=\frac{1}{2}$. If, then, I start this sequence at $i=1$, is it valid to say that
$S_1=\underset{i=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{n}^i=\frac{1}{1-<!--\; -\; -->n}-<!--\; -\; -->1=\frac{1-<!--\; -\; -->(1-<!--\; -\; -->n)}{1-<!--\; -\; -->n}=\frac{n}{1-<!--\; -\; -->n}$
and as such that, as $k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, we have something like
$\frac{mn}{1-<!--\; -\; -->n}+\frac{mn}{1-<!--\; -\; -->m}=1.$
Have I made any ridiculous logical jumps or is this valid?

Two circles intersect at C and D, and their common tangents intersect at T. CP and CQ are the tangents at C to the two circles; prove that CT bisects $\mathrm{\angle <!--\; \angle \; -->}PCQ$.

Probability problem involving hyper-geometric distribution
n balls are chosen randomly and without replacement from an urn containing N white balls and M black balls. Give the probability mass function of the random variable X which counts the number of white balls chosen. Show that the expectation of X is $\frac{Nn}{(M+N)}$.
Hint: Do not use the hypergeometric distribution. Instead, write $X=X_1+X_2+...+X_N$ where $X_i$ equals 1 if the ith white ball was chosen.

Finding the total surface area of cuboid model using volume
You are to design a cuboid model with a square base that has a volume of $12m^3$. In order to have a minimum total surface area for the cuboid, what are the values for the height of the cuboid and the length of a side in the square base?
So far, this is what I have done:
$\text{Volume = }L.B.H=B^2.H$
$12=B^2.H$
$H=\frac{12}{B^2}$
$\text{S.A. = }2.B^2+4.B.H$
$=2.B^2+4.B.(\frac{12}{B^2})$
$=2.B^2+\frac{48}{B}$

On a coordinate grid what is the distance from C(5, 8) to D(5, 1)?

How is this angle measured in the triangle?
Given the image below:

Let the angle at vertex A be called $\mathrm{\alpha <!--\; \alpha \; -->}$. Polya says by looking at the diagram and the isoceles triangles $\mathrm{△<!--\; △\; -->}ACE$ and $\mathrm{△<!--\; △\; -->}ABD$ we can deduce that $\mathrm{\angle <!--\; \angle \; -->}DAE=\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{2}+90°$.
Where will I get $90°$ from?

Does the intersection of two concave polygons without holes result in a set of several polygons without holes?
I did not find a proof anywhere, so here is my assumption :
If I have 2 concave and simple polygons without holes, then the intersection yields a set of different polygons without any holes.
If it can result to a polygon with a hole inside, could you provide an example?
Edit: the polygons are simple. edges of one polygone don't intersect, except consecutive edges, which intersect in their common vertex. The polygons in input do not have holes. I consider the intersection of the interiors of the polygons. The resulting set is all the points that are included both in interior of polygon P1 and the interior of the polygon P2.

Evaluating the Volume of a Cupola-Shaped Set by Integration
Let $f(x,y)=1-<!--\; -\; -->\frac{x^2}{4}-<!--\; -\; -->y^2$ and $\mathrm{\Omega <!--\; \Omega \; -->}=\{(x,y)\in <!--\; \in \; -->{\mathbb{R}}^2:<!--\; :\; -->f(x,y)\ge <!--\; \ge \; -->0\}.$
Compute the volume of the set $A=\{(x,y,z)\in <!--\; \in \; -->{\mathbb{R}}^3:<!--\; :\; -->(x,y)\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->},0\le <!--\; \le \; -->z\le <!--\; \le \; -->f(x,y)\}.$
My idea is to slice the set along the z-axis, obtaining a set $E_z$ - in fact, an ellipse - and computing the volume as ${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_{E_z}dxdydz$.
However, I am stuck finding a way to describe $E_z$. What is the best strategy to do that?

Finding Volume of object in 3D space rotating on x-axis
Find the volume of the solid, generated by revolving the region bounded by $y=\sqrt{\sin <!--\; \; -->(4x)},y=0$ and $0\le <!--\; \le \; -->x\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$ about the x-axis.
To solve for volume, I have the following integral
$V={\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}\mathrm{\pi <!--\; \pi \; -->}r(x{)}^{2}dx=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}\sin <!--\; \; -->(4x)dx=\mathrm{\pi <!--\; \pi \; -->}{[-<!--\; -\; -->\frac{\cos <!--\; \; -->(4x)}{4}]}_0^{\mathrm{\pi <!--\; \pi \; -->}/4}=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}.$
That is not one of the solution options. What did I do wrong?

Least Squares Circumcenter of Polygons
It is well known that the circumcenter of a polygon exists if and only if the polygon is cyclic.
I would like to extend the definition of an circumcenter for noncyclic polygons. Namely, let us define the least squares circumcenter as the point A $(x_0,y_0)$ such that the point A minimizes the sum of the squares of the residuals.
Let us consider the case for a noncyclic quadrilateral with vertices P $(x_1,y_1)$, Q$(x_2,y_2)$, R$(x_3,y_3)$, and S $(x_4,y_4)$. Let us also define the origin O(0,0).
How would we solve for the point A in this case? I was thinking of using matrices and solving $A^{\mathsf{T}}A\stackrel{^<!--\; ^\; -->}{x}=A^{\mathsf{T}}b$, although any methods are welcome.

Distance between (6,0,4) and (-1,4,-2)?

Find the volume of the solid bounded by
$x^2+y^2–2y=0,z=x^2+y^2,z=0$. I have to calculate volume using triple integrals but I struggle with finding intervals. I calculated $x_1=\sqrt{2y-<!--\; -\; -->y^2}$ and $x_2=-<!--\; -\; -->\sqrt{2y-<!--\; -\; -->y^2}$. I think z will have interval from 0 to 2y, but I don't know what to do next.

Finding the volume of an object with 3 parameters
I know how to find the volume of a sphere/ball around x-axis using:
$V=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_a^b{f}^2(x)dx$
Lets say if:
$x^2+y^2=r^2$
We do: $y=\sqrt{r^2-<!--\; -\; -->x^2}$
So now: $V=\mathrm{\pi <!--\; \pi \; -->}{\int <!--\; \int \; -->}_{-<!--\; -\; -->r}^r(\sqrt{r^2-<!--\; -\; -->x^2}{)}^{2}dx$
But the problem starts here:
Im given an equation of ellipsoid, it has 3 parameters:
$x^2+4y^2+4z^2\le <!--\; \le \; -->4$
What do i do with the z parameter? How do i build y now? how does it fit to the equation by integrals of V?
I would like an explanation and not just a solution - because its homework.

Choose the letter of the word or phrase that best completes each statement: The ratio of the lengths of two corresponding sides of two similar polygons is the __________ . Choose from: a. scale factor, b. AA Similarity Postulate, c. SSS Similarity Theorem, d. SAS Similarity Theorem, e. Symmetric Property of Similarity, f. midsegment, g. dilation, h. enlargement, i. reduction, j. Transitive Property of Similarity.

Calculate the length of an arc of a circle with a radius of 4 cm if its degree measure is $${120}^{\circ <!--\; \circ \; -->}$$. What is the area of the circular sector corresponding to the given arc?

Find the area of a circle whose radius is 6 cm

The perimeter of a square circumscribed about a circle is 16 in. find the perimeter of a regular pentagon inscribed in the same circle

There are colored pencils in a box, 9 pencils of each color. It is known that there are a natural number of tens and a natural number of dozens of pencils in the box, while there are less than 300 pencils in the box. How many pencils are in the box?