A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude 25ft and base edge of 10ft. how much will the surface rise when 6,779 liters of water is added?

enmobladatn
2022-07-25
Answered

A liquid vessel which is initially empty is in the form of an inverted regular hexagonal pyramid of altitude 25ft and base edge of 10ft. how much will the surface rise when 6,779 liters of water is added?

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asked 2022-07-18

Newton polygons

This question is primarily to clear up some confusion I have about Newton polygons.

Consider the polynomial ${x}^{4}+5{x}^{2}+25\in {\mathbb{Q}}_{5}[x]$. I have to decide if this polynomial is irreducible over ${\mathbb{Q}}_{5}$

So, I compute its Newton polygon. On doing this I find that the vertices of the polygon are (0,2), (2,1) and (4,0). The segments joining (0,2) and (2,1), and (2,1) and (4,0) both have slope $-\frac{1}{2}$, and both segments have length 2 when we take their projections onto the horizontal axis.

Am I correct in concluding that the polynomial ${x}^{4}+5{x}^{2}+25$ factors into two quadratic polynomials over ${\mathbb{Q}}_{5}$, and so is not irreducible?

I am deducing this on the basis of the following definition of a pure polynomial given in Gouvea's P-adic Numbers, An Introduction (and the fact that irreducible polynomials are pure):

A polynomial is pure if its Newton polygon has one slope.

What I interpret this definition to mean is that a polynomial $f(x)={a}_{n}{x}^{n}+...+{a}_{0}$ $\in {\mathbb{Q}}_{p}[x]$ (with ${a}_{n}{a}_{0}\ne 0$) is pure, iff the only vertices on its Newton polygon are $(0,{v}_{p}({a}_{0}))$ and $(n,{v}_{p}({a}_{n}))$. Am I right about this, or does the polynomial ${x}^{4}+5{x}^{2}+25$ also qualify as a pure polynomial?

This question is primarily to clear up some confusion I have about Newton polygons.

Consider the polynomial ${x}^{4}+5{x}^{2}+25\in {\mathbb{Q}}_{5}[x]$. I have to decide if this polynomial is irreducible over ${\mathbb{Q}}_{5}$

So, I compute its Newton polygon. On doing this I find that the vertices of the polygon are (0,2), (2,1) and (4,0). The segments joining (0,2) and (2,1), and (2,1) and (4,0) both have slope $-\frac{1}{2}$, and both segments have length 2 when we take their projections onto the horizontal axis.

Am I correct in concluding that the polynomial ${x}^{4}+5{x}^{2}+25$ factors into two quadratic polynomials over ${\mathbb{Q}}_{5}$, and so is not irreducible?

I am deducing this on the basis of the following definition of a pure polynomial given in Gouvea's P-adic Numbers, An Introduction (and the fact that irreducible polynomials are pure):

A polynomial is pure if its Newton polygon has one slope.

What I interpret this definition to mean is that a polynomial $f(x)={a}_{n}{x}^{n}+...+{a}_{0}$ $\in {\mathbb{Q}}_{p}[x]$ (with ${a}_{n}{a}_{0}\ne 0$) is pure, iff the only vertices on its Newton polygon are $(0,{v}_{p}({a}_{0}))$ and $(n,{v}_{p}({a}_{n}))$. Am I right about this, or does the polynomial ${x}^{4}+5{x}^{2}+25$ also qualify as a pure polynomial?

asked 2022-09-25

Regular polygons meeting at a point

N regular polygons with E edges each meet at a point with no intervening space. Show that $N=\frac{2E}{E-2}\phantom{\rule{2em}{0ex}}E=\frac{2N}{N-2}$.

I did this part considering that the internal angles must add up to $2\pi $, i.e. $(1-{\displaystyle \frac{2}{E}})\pi \times N=2\pi $. I am unable to explain the following:

Using the result given above, show that the only possibilities are $E=3,4,6$

N regular polygons with E edges each meet at a point with no intervening space. Show that $N=\frac{2E}{E-2}\phantom{\rule{2em}{0ex}}E=\frac{2N}{N-2}$.

I did this part considering that the internal angles must add up to $2\pi $, i.e. $(1-{\displaystyle \frac{2}{E}})\pi \times N=2\pi $. I am unable to explain the following:

Using the result given above, show that the only possibilities are $E=3,4,6$

asked 2022-09-24

If I have a circle that has ten dots on it and I have to make as many polygons using those dots, the answer should be 10. Connecting each dot starting with triangle, to square, all the way to the max of a 10-sided polygon should give me only 10 polygons that I can make. Am I missing something here?

asked 2022-08-11

Scan line algorithm for intersecting polygons

Given two sets of polygons ${P}_{1}=\{{s}_{1},...,{s}_{m}\}$ and ${P}_{2}=\{{s}_{m}+1,...,{s}_{n}\}$ with total number of n segments, the previous and next segment on it's polygon can be determined in O(1). Describe a scan-line algorithm that computes all points of ${P}_{1}\cap {P}_{2}$ in O(n).

Given two sets of polygons ${P}_{1}=\{{s}_{1},...,{s}_{m}\}$ and ${P}_{2}=\{{s}_{m}+1,...,{s}_{n}\}$ with total number of n segments, the previous and next segment on it's polygon can be determined in O(1). Describe a scan-line algorithm that computes all points of ${P}_{1}\cap {P}_{2}$ in O(n).

asked 2022-08-11

Constructible polygons

I know certain polygons can be constructed while others cannot. Here is Gauss' Theorem on Constructions:

$\mathrm{cos}(2\pi /n)$ is constructible iff $n={2}^{r}{p}_{1}{p}_{2}\xb7\xb7\xb7{p}_{k}$, where each ${p}_{i}$ is a Fermat prime.

Can this is be used to determine the constructibility of a regular ${p}^{2}$ polygon? If so how and what would be the $\mathrm{cos}(2\pi /n)$ here?

I know certain polygons can be constructed while others cannot. Here is Gauss' Theorem on Constructions:

$\mathrm{cos}(2\pi /n)$ is constructible iff $n={2}^{r}{p}_{1}{p}_{2}\xb7\xb7\xb7{p}_{k}$, where each ${p}_{i}$ is a Fermat prime.

Can this is be used to determine the constructibility of a regular ${p}^{2}$ polygon? If so how and what would be the $\mathrm{cos}(2\pi /n)$ here?

asked 2022-07-25

Find the lateral surface area of a regular hexagonal pyramid whose edge measures 20 cm and the radius of a circle inscribed in the base is 9 squareroot of 3 cm.

asked 2022-08-11

Hyperbolic Angle Measure of Polygons

Is there a way to determine the angle measure of a regular polygon in hyperbolic space? I know that this depends on the length of the sides. A an example, I know that for an equilateral triangle with side length a and angle A, then $\mathrm{sec}A=1+\frac{2{e}^{a}}{{e}^{2a}+1}$.

Is there a similar formula for higher regular polygons?

Is there a way to determine the angle measure of a regular polygon in hyperbolic space? I know that this depends on the length of the sides. A an example, I know that for an equilateral triangle with side length a and angle A, then $\mathrm{sec}A=1+\frac{2{e}^{a}}{{e}^{2a}+1}$.

Is there a similar formula for higher regular polygons?