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https://www.worksheetanswers.com/adding-subtracting-scientific-notation-worksheet/ | 1,701,450,717,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00369.warc.gz | 1,216,393,315 | 15,965 | # Adding Subtracting Scientific Notation Worksheet
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Adding Subtracting Scientific Notation Worksheet. Select the document you wish to signal and click on Upload. Scientific Notation Recognize and use scientific notation. Now on this case we can’t subtract these from one another as a outcome of our exponents aren’t the identical. Use a useful resource on which you’ll base your lesson on base 10 and scientific notation.
Here are 910 actions that help college students practice operations with scientific notation. You most likely won’t have time to do all of them, but you’ll find a way to decide and choose what works greatest in your college students. Adding/Subtracting when the Exponents are DIFFERENT • When including or subtracting numbers in scientific notation, the exponents have to be the identical. • If they’re totally different, you must transfer the decimal so that they will have the same exponent.
• You could select issues with multiplication, division, or merchandise to a power.
• The powers for very large numbers are expressed using constructive exponents and very small numbers using unfavorable exponents.
• Students should specific each whole numbers and decimals in scientific notation.
• This maze consists of eleven numbers that college students should convert from commonplace notation to scientific notation.
• The numbers are shortened and multiplied in the energy of 10.
And due to its cross-platform nature, signNow works nicely on any gadget, desktop or mobile, regardless of the OS. Scientific Notation worksheets right here teach you the finest way to precise very giant numbers and really small numbers. The numbers are shortened and multiplied in the energy of 10. The powers for very giant numbers are expressed using optimistic exponents and really small numbers using unfavorable exponents. In these worksheets, study to specific standard numbers in scientific kind and scientific numbers in standard form. Also scientific notation Addition, Subtraction, Multiplication, and Division worksheets have been ready for superior learning.
Practice this pdf worksheet to become familiar with writing very giant numbers and small numbers in scientific notation. Each number is given in the usual type, and children specific the numbers in scientific notation. Give this hands-on activity a strive for a fun approach to apply writing small and huge numbers using scientific notation.
### Including And Subtracting Scientific Notation Worksheet
This is an including and subtracting scientific notation RIDDLE worksheet. Positive and adverse exponents.Decimal places as a lot as the thousandths.Worksheets are copyright material and are intended for use in the classroom solely. Purchased worksheets could NOT be posted on the web, together with but not limited to instructor net pages. These exponents worksheets introduce problems with powers of ten. The worksheets begin out introducing easy powers of ten phrases, together with ones that should be memorized.
Explains tips on how to add and subtract numbers written in scientific notation, whether or not they’ve the identical exponent. For this particular topic I used a couple of problems from my task playing cards for operations with scientific notation. I did this at the beginning of sophistication and we did 3-4 problems. It was an attractive way to get college students serious about what we had done the day before.
### Studying Resource Type
Addition and subtraction with scientific notation worksheet computers. You can take them in all places and even use them whereas on the go as long as you have a steady connection to the internet. Therefore, the signNow internet utility is a must-have for finishing and signing including and subtracting scientific notation worksheet with reply key pdf on the go. In a matter of seconds, receive an digital document with a legally-binding signature.
When you add two or more exponents, the powers ought to be made equal first. Its is all the time higher to transform the smaller energy to the very best energy. Generally mathematicians, scientists and engineers take care of very massive and really small numbers on every day foundation. They additionally do exponent addition, subtraction, multiplication, and division. Each worksheet has ten problems expressing decimals in both commonplace and scientific notation. In this set of pdf worksheets, categorical every number in normal notation.
If omitted, the information is returned for the last modified cell on the sheet. The tutorial exhibits tips on how to use the CELL perform in Excel to retrieve numerous information about a cell corresponding to cell tackle, contents, formatting, location, and more. Performing mathematical computations with scientific notation. ‘Subtracting Scientific Notation Calculator’ is an online software that helps to calculate the difference between the given two scientific notations.
Go to the Chrome Web Store and add the signNow extension to your browser. Click on the hyperlink to the document you need to design and select Open in signNow. Select the document you wish to signal and click on on Upload. Here is a list of the most typical buyer questions. If you can’t discover an answer to your question, please don’t hesitate to achieve out to us. Now it’s potential to print, save, or share the document.
## Scientific Notation Pdf Pdf,doc ,Images
Coloring activities have turn out to be one of my students’ favorites. A couple of years in the past once I first started seeing these activities I truthfully thought they appeared like a waste of time. So, when I started making a few of my own I made the image smaller than a number of the others I’ve seen and students don’t spend that much time coloring, perhaps 4-5 minutes. For this subject there are two coloring pages that I use. We do that exercise two days in a row, completing one worksheet each day. Are you looking for ways to follow operations with scientific notation?
Add one to the exponent, should you transfer the decimal to the left by a single place, and subtract one from it, when you transfer the decimal to the right by one place. If you’re seeking to take this concept further and produce it into the actual world you might wish to do this activityby Robert Kaplinsky. Students will actually get to see how scientific notation with operations are utilized in real life. This exercise may take a complete class period for faculty kids to work by way of. I haven’t used it yet, but I’m excited to make use of it after state testing after I have time to get previous ability building and can show students concepts in action. When adding or subtracting numbers in scientific notation, the exponents should be the same.
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https://www.physicsforums.com/threads/a-strange-wave-function-of-the-hydrogen-atom.994391/ | 1,716,578,810,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058736.10/warc/CC-MAIN-20240524183358-20240524213358-00348.warc.gz | 819,398,072 | 17,967 | # A strange wave function of the Hydrogen atom
• omegax241
In summary, the conversation discusses the possible values of the measurement of L_z, the probability of a measurement of L^2 giving the value 6 \hbar^2, and the minimum value of an energy measurement for an electron in a H atom. The only possible measurement for L_z is \hbar, but there is uncertainty about the reasoning behind this. To solve the problem, the wavefunction can be rewritten as a linear combination of stationary states, but the square on the r part and the exponentiation of the angular part pose difficulties. To find the coefficient of a particular eigenfunction, one does not need to completely decompose the wavefunction. However, it is unclear how to find a definite number with the n dependence.
omegax241
I am trying to solve the following exercise.
In a H atom the electron is in the state described by the wave function in spherical coordinates:
$$\psi (r, \theta, \phi) = e^{i \phi}e^{-(r/a)^2(1- \mu\ cos^2\ \theta)}$$
With $a$ and $\mu$ positive real parameters. Tell what are the possible values of the measurement of $L_z$, what is the probability that a measurement of $L^2$ gives the value $6 \hbar^2$, and then what is the minimum value of an energy measurement.
First I've observed that $m=1$, so the only possible measurement for $L_z$ is $\hbar$. (I'm pretty confident with this reasoning, but not so much.)For the other questions my idea was to rewrite the wavefunction as a linear combination of stationary states of the form $\phi_{n,l,1}$, but the problem is the square on the $r$ part, and the fact that the angular part $\theta$ is exponentiated.
In fact I've even started to ask myself is if necessary at all to find this decomposition to solve the problem, but nothing comes to mind.
Thank you for listening, every bit of help is appreciated.
omegax241 said:
I am trying to solve the following exercise.
In a H atom the electron is in the state described by the wave function in spherical coordinates:
$$\psi (r, \theta, \phi) = e^{i \phi}e^{-(r/a)^2(1- \mu\ cos^2\ \theta)}$$
With $a$ and $\mu$ positive real parameters. Tell what are the possible values of the measurement of $L_z$, what is the probability that a measurement of $L^2$ gives the value $6 \hbar^2$, and then what is the minimum value of an energy measurement.
First I've observed that $m=1$, so the only possible measurement for $L_z$ is $\hbar$. (I'm pretty confident with this reasoning, but not so much.)For the other questions my idea was to rewrite the wavefunction as a linear combination of stationary states of the form $\phi_{n,l,1}$, but the problem is the square on the $r$ part, and the fact that the angular part $\theta$ is exponentiated.
In fact I've even started to ask myself is if necessary at all to find this decomposition to solve the problem, but nothing comes to mind.
Thank you for listening, every bit of help is appreciated.
To find the coefficient of a particular eigenfunction you do not need to completely decompose the wavefunction. Do you know how to do that?
omegax241 said:
Thank you for listening, every bit of help is appreciated.
I've had a look at this. I must admit I don't see how to tackle it. That wave function looks horrible. Sorry, I've no ideas either.
PeroK said:
To find the coefficient of a particular eigenfunction you do not need to completely decompose the wavefunction. Do you know how to do that?
Yes, I should evaluate the product:
$$\langle \phi_{n, l ,m} | \psi \rangle$$
But suppose I want to tackle the second question, if a value of $6 \hbar$ is found for $L^2$ this means that $l = 2$, and then $m = \pm 1 ; \pm 2 ; 0$. So I should evaluate the coefficents
$$\langle \phi_{n, 2, \pm 1} | \psi \rangle$$
$$\langle \phi_{n, 2, \pm 2} | \psi \rangle$$
$$\langle \phi_{n, 2, 0} | \psi \rangle$$
But how can I find a definite number with this n dependence ?
vanhees71 and PeroK
omegax241 said:
Yes, I should evaluate the product:
$$\langle \phi_{n, l ,m} | \psi \rangle$$
But suppose I want to tackle the second question, if a value of $6 \hbar$ is found for $L^2$ this means that $l = 2$, and then $m = \pm 1 ; \pm 2 ; 0$. So I should evaluate the coefficents
$$\langle \phi_{n, 2, \pm 1} | \psi \rangle$$
$$\langle \phi_{n, 2, \pm 2} | \psi \rangle$$
$$\langle \phi_{n, 2, 0} | \psi \rangle$$
But how can I find a definite number with this n dependence ?
You know that ##m = 1##. It's the variable ##n## that's the problem.
omegax241
Doesn't parity tell you that an eigenstate of l=2 must be even? The wave function provided is odd.
PhDeezNutz and vanhees71
## 1. What is a wave function?
A wave function is a mathematical representation of the quantum state of a particle, which describes the probability of finding the particle in a particular location or state.
## 2. Why is the wave function of the Hydrogen atom considered strange?
The wave function of the Hydrogen atom is considered strange because it does not have a definite shape or size like classical objects. Instead, it exists in a superposition of multiple states, making it difficult to visualize or understand in traditional terms.
## 3. How is the wave function of the Hydrogen atom used in quantum mechanics?
The wave function of the Hydrogen atom is used in quantum mechanics to calculate the probability of finding an electron in a particular energy level or orbital. It also helps to predict the behavior of the atom and its interactions with other particles.
## 4. Can the wave function of the Hydrogen atom be observed or measured?
No, the wave function of the Hydrogen atom cannot be directly observed or measured. It is a mathematical concept that represents the quantum state of the atom, and its values can only be obtained through calculations and experiments.
## 5. How does the wave function of the Hydrogen atom relate to the uncertainty principle?
The wave function of the Hydrogen atom is directly related to the uncertainty principle, which states that it is impossible to know both the exact position and momentum of a particle simultaneously. The wave function represents the probability of finding the particle in a particular location, and the uncertainty principle explains why the position of an electron cannot be precisely determined.
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3K | 1,645 | 6,444 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.908354 |
https://www.sparkcodehub.com/numpy-nanmin-explained | 1,701,674,106,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00512.warc.gz | 1,092,810,617 | 16,883 | # Mastering NumPy nanmin: Delving Into Minimum Value Computation with NaNs
## Introduction
Python’s NumPy library stands as a pillar in the field of data manipulation, providing robust functions to work with arrays. Among its arsenal of tools is ` np.nanmin ` , a function that specializes in calculating the minimum value in an array while intelligently ignoring NaNs (Not a Number). This function is indispensable for analyses that require clean and accurate metrics despite the presence of incomplete or corrupt data. Let’s unpack the workings of ` np.nanmin ` and how it can be leveraged in various scenarios.
## What is ` np.nanmin ` ?
` np.nanmin ` serves as a guardian against the disruptive influence of NaNs when computing the minimum of array values. It ensures that the presence of NaNs does not distort the statistical calculations which often form the bedrock of data analysis projects.
### Syntax of ` np.nanmin `
``numpy.nanmin(a, axis=None, out=None, keepdims=np._NoValue, *, where=np._NoValue) ``
Here, ` a ` is the input array, ` axis ` specifies the axis to reduce, ` out ` is an alternative output array to place the result, and ` keepdims ` dictates whether the output should maintain the dimensionality of the original array.
## Utilizing ` np.nanmin ` in Data Analysis
### Simple Array Example
Consider an array replete with both real numbers and NaNs:
``````import numpy as np
# Array with NaN values
data = np.array([5, 1, np.nan, 3, np.nan])
# Determining the minimum
min_val = np.nanmin(data)
print(f"The minimum value, discarding NaNs, is {min_val}") ``````
### Multi-dimensional Array Analysis
` np.nanmin ` extends its functionality to n-dimensional arrays:
``````# 2D array example
data_2d = np.array([[np.nan, 4, 2], [8, np.nan, 1], [7, 6, np.nan]])
# Minimum along columns
min_val_col = np.nanmin(data_2d, axis=0)
print(f"Column-wise minimums: {min_val_col}")
# Minimum along rows
min_val_row = np.nanmin(data_2d, axis=1)
print(f"Row-wise minimums: {min_val_row}") ``````
### The Role of ` keepdims `
Maintaining the original shape of data can be critical for aligned computations, and ` keepdims ` accomplishes this:
``````# Preserve the array dimensions
min_val_keepdims = np.nanmin(data_2d, axis=1, keepdims=True)
print(min_val_keepdims) ``````
## Conclusion
The ` np.nanmin ` function is a testament to NumPy's commitment to providing comprehensive solutions for data analysis. Through its capacity to omit NaNs from its operations, it allows analysts and scientists to derive meaningful insights from data that might otherwise be considered unusable. As datasets grow increasingly complex and riddled with gaps, the ability to perform such clean statistical operations becomes ever more crucial. ` np.nanmin ` is, therefore, not just a function but a facilitator of more accurate and reliable data analysis. | 681 | 2,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2023-50 | longest | en | 0.749034 |
https://www.vedantu.com/question-answer/find-all-the-zeros-of-x3-+-6x2-+-11x-+-6if-x-+-class-9-maths-cbse-5f5da6488f2fe2491853a495 | 1,726,531,617,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651714.51/warc/CC-MAIN-20240916212424-20240917002424-00189.warc.gz | 979,196,419 | 26,788 | Courses
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# Find all the zeros of ${x^3} + 6{x^2} + 11x + 6\;$if $(x + 1)\;$is a factor.
Last updated date: 16th Sep 2024
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Hint: We start with taking $(x + 1)\;$ common from the equation ${x^3} + 6{x^2} + 11x + 6\;$ as this is a factor of the cubic equation. Then we will try to factorize the result we get by the splitting of terms. In this way, we reach the roots of the equation which we need to find.
Complete step by step answer:
Our given equation is, ${x^3} + 6{x^2} + 11x + 6\;$
Since $(x + 1)\;$is a factor of the given cubic equation, it will completely divide the given equation.
Now, we will factorize the given equation
$\left( {{x^3} + 6{x^2} + 11x + 6} \right)$
On expansion of terms we get,
${x^3} + {x^2} + 5{x^2} + 5x + 6x + 6$
On taking factors common we get,
${x^2}(x + 1) + 5x(x + 1) + 6(x + 1)$
$\Rightarrow \left( {{x^2} + 5x + 6} \right)\left( {x + 1} \right)$
On expansion we get,
$({x^2} + 3x + 2x + 6)(x + 1)$
On taking factors common we get,
$[x(x + 3) + 2(x + 3)](x + 1)$
$= \left( {x + 2} \right)\left( {x + 3} \right)\left( {x + 1} \right)$
∴ now, the zeros of $\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\;$would be,
For, $(x + 1) = 0$ we get, $x = - 1$
As,
$x + 1 = 0 \Rightarrow x = - 1$
For, $(x + 2) = 0$ we get, $x = - 2$
As,
$x + 2 = 0 \Rightarrow x = - 2$
For, $(x + 3) = 0$ we get, $x = - 3$
As,
$x + 3 = 0 \Rightarrow x = - 3$
Therefore, the zeros of ${x^3} + 6{x^2} + 11x + 6\;$if $(x + 1)\;$ is a factor are, $- 1, - 2, - 3$
Note: A cubic equation is an equation which can be represented in the form ${\text{a}}{{\text{x}}^{\text{3}}}{\text{ + b}}{{\text{x}}^{\text{2}}}{\text{ + cx + d}}$, where ${\text{a}} \ne {\text{0}}$
As a cubic polynomial has three roots (not necessarily distinct) by the fundamental theorem of algebra, at least one root must be real. | 789 | 1,954 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.716354 |
https://garyherstein.com/2021/07/22/the-infinite/2/ | 1,679,350,555,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943562.70/warc/CC-MAIN-20230320211022-20230321001022-00314.warc.gz | 315,441,146 | 26,956 | Tags
How do you make sense of a claim that one infinity is different from, much less bigger than another? Well the trick is in two parts. First, it is taken for granted that when you talk about “one infinity,” you’re speaking about a collection of “objects” that cannot be counted by any finite number. (That’s a significant constraint, but we’ll come back to that.) Thus, we all know that the integers (0, 1, 2, 3, … [note the ellipsis]) have no largest number; they go on “forever,” they are infinite. Say then you have another collection, and you wish to test if it is infinite. The test is, can you correlate each element of this new collection with that of the integers, on a one for one basis, that is without any repeats or repetitions? If you can do that with the new collection, then it too is infinite. But what happens if there are more elements in this new collection than there are integers. Well, that is a really BIG infinite. And that’s the real numbers.
Funny fact, one of many, the fractions, the x/y numbers? Seems like there’s a lot of them, right? I mean, between 0 and 1 there are infinitely many fractions. And between any two fractions, you can always find infinitely many more. So that’s a BIG collection, right?
Yeah, not so much. Turns out there is a trick where you can correlate every fraction, one to one, with a unique integer. Those two collections are actually the same size. With infinities, appearances are almost always deceiving. Our brains aren’t born to think about such things. With training, discipline, practice, and whatever else keeps you going, one can educate both thought and intuition to follow these relational patterns. But it’s like hitting an overhead top-spin serve in tennis; you can’t even picture it until you’ve spent years with a mentor learning how.
So the collection of real numbers – you know, π, e, √2, crazies that can’t be written down in any finite list – is bigger than the integers and the fractions (which are the same size as one another). Once mathematicians, following geniuses like Georg Cantor, discovered these relationships, it became clear that there were infinities even bigger than these. (Also, hearkening back to my previous two posts, you might begin to sense how and why the solvable problems – which can be placed in a one-to-one correlation with the integers (or fractions) are so catastrophically overwhelmed by the unsolvable problems, which are one-to-one with the reals.)
This hierarchy of infinities made set theorists – the mathematicians who specialize in studying such things – very happy. But in the early ‘60’s Paul Cohen, who worked in an area of logic called “model theory,” demonstrated using his array of tools that one could prove there were infinitely many infinities in between the integers and the real numbers. For set theorists, this made no sense whatsoever, and for many years this led to many noses being thumbed at one another across this mathematical divide. (Set theorists tend to be very “ontological” in their view of things: their formulae and proofs are not merely logically coherent, they represent the REAL as such. Model theorists are content with logical coherence.) So the infinite, which was already bloody crazy, got crazier still.
But amidst all of these crazies, several of which are wildly incompatible with one another, there remains one commonality: they are all composed of points – “atoms,” if you will – that are the minimal units that make it possible to count, and thus determine, how big the collection is. But what if you’re dealing with a kind of “gunk” that has no minimal, nor any maximal, “unit”? A kind of “atomless gunk”?
Lest you accuse me of being silly, that is the technical name for a topic in “mereology,” the study of the logical relations of part-and-whole. What if there is no least part, no “atom” or basic element to the collection one is dealing with? To avoid prejudicing matters with biased terminology, this was called “atomless gunk,” and the name stuck (rather the way gunk always does.) This is a different kind of infinity, because it is no longer a matter of enumerating the size of the collection, but rather an operational infinity in which no amount of division or accretion can ever achieve a least or greatest element.ii Like the Koch snowflake, as one looks at smaller and smaller bits, one never encounters a smallest bit. Compare this to numbers where, say, the numbers “7” or “π.” These are absolutely atomic bits, “points” on the number line, in their respective infinite wholes. Which is to say, you can find a number smaller (or larger) than 7, but the point on the number line that corresponds to, or is named, “7” cannot be divided into smaller pieces. Similarly with other points on the integer, rational, or real number lines.
Now, with this little bit (so to speak) under our belts, what can be say about yet other things that are called infinite? For example, when someone says “God is infinite,” what (if anything) might that mean? Well, here we are dealing with the generic sloppiness of language, because until mathematicians and logicians got their prickly mitts on the topic, “infinite” was just the Latin version of “ἄπειρον,” and dogged by just as much vagueness.
The first stab at this might be to say that God is unbounded ( ἄπειρον), but this might not work. For consider the surface of a sphere: it is unbounded, and yet it is finite. It is not enough to say that God is of “unlimited extent,” for that is a spatial/temporal metaphor (which is already a mistake), and our friend the Koch snowflake (for example) is of unlimited extent, and yet it is strictly bounded. One can try to go the route of power – God is infinite in knowledge and ability – but that simply opens one to the problem of evil (why doesn’t God do something, when he is the author of it all?) and that is a rabbit hole that leads to no solutions.
I submit that this is the wrong direction to go with such a question; that is, size, power, BIG, is just wrong headed when one turns to theological issues. The Greeks thought the infinite, the infinite, the ἄπειρον, was evil because it exceeded the possibility of human interpretation. But perhaps that is not evil at all, any more than understanding how the real numbers exceed the rationals. There is a line of thought that says one’s reach ought to exceed one’s grasp. So even if the idea of God is beyond any of our concepts, we can still stretch ourselves by approaching – even if we never achieved – a concept of something we can never fully understand or represent.iii
– – – – – – – – – –
i Simply proving that √2 was not a fraction was a magnificent achievement. According to legend, the genius who produced this demonstration was rewarded for his effort by having a stone tied around his ankles and then tossed off a ship into the Mediterranean sea. The Greeks considered this kind of infinity a sign of profound evil.
ii Whitehead’s theory of extension – a central part of his entire philosophy of process – is built around a mereology of atomless gunk, a point that does not appear to be widely recognized in the scholarship.
iii This is very much of a piece with Plato’s dialog on love, the Symposium. The image is of a spiral that always approaches, but never reaches, the center point. Iris Murdoch, in her brilliant Metaphysics as a Guide to Morals, picked up this idea. She went well beyond just talking about it, and exemplified it as well in the very structure of her discussion. | 1,661 | 7,517 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-14 | latest | en | 0.95315 |
https://www.idtech.com/courses/tutoring-statistics | 1,695,504,322,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00769.warc.gz | 907,955,161 | 58,504 | # Statistics
Pricing
Starting at \$60/lesson
Ages
13 - 19 (Must turn 13 by 12-31-22)
Skill Level
Beginner–Intermediate
Tools
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Program
Online Private Lessons | lessons, 1 hour per lesson
Path
Math
Prerequisite
None
## Statistics
If you’re sitting in a room with 23 other people, the chance of two of them sharing a birthday is about 50%! Statistics and probability help you understand why that’s the case. Here, you’ll analyze how data is represented and distributed, whether or not the data fits common models, and even if the data fits a trend or a conclusion. Using data, you’ll calculate the probability of simple and compound events, evaluate the validity of published studies, and most importantly, learn how numbers and statistics are used to tell stories.
## In this course, you will:
• Learn how data is represented and distributed
• Use scatter plots, histograms, and other plots to represent data
• Evaluate the validity of published statistics | 221 | 977 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.860821 |
https://math.answers.com/other-math/What_is_3.8_of_950_g | 1,718,551,889,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861665.97/warc/CC-MAIN-20240616141113-20240616171113-00453.warc.gz | 341,484,703 | 47,580 | 0
# What is 3.8 of 950 g?
Updated: 4/28/2022
Alexis0927
Lvl 1
10y ago
It is 3.8*950 = 3610 g
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10y ago
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Q: What is 3.8 of 950 g?
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Related questions
### What is 38 times 25?
The product is 950
### Why is 38 times 25 a factor of 950?
Since 38 x 25 = 950, 38 and 25 are factors of 950 (38,25) is a factor pair of 950. 38 x 25 is a factor string of 950.
### What is the LCM of 50 and 38?
LCM(50, 38) = 950.
### What is the least common multiple of 25 and 38?
25 = 5x5 and 38 = 2x19 There is no number except 1 which divides both 25 and 38. So, L.C.M. is equal to the product of 25 and 38. L.C.M. = 25 x 38 = 950.
### How many kg is equal to 950 g?
950 g = 0.95 kgTo convert from g to kg, divide by 1000.
950
950
### What is 25 feet by 38 feet?
25*38=950 square feet
950
### What is the LCM of 25 and 38?
The least common multiple of the numbers 25 and 38 is 950.
47.5g.
### What are the whole number factors of 950?
1, 2, 5, 10, 19, 25, 38, 50, 95, 190, 475, 950 | 399 | 1,034 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-26 | latest | en | 0.907159 |
https://se.mathworks.com/help/pde/ug/pde.conductionresults.interpolatecurrentdensity.html | 1,726,382,307,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00101.warc.gz | 454,851,282 | 20,103 | # interpolateCurrentDensity
Interpolate current density in DC conduction result at arbitrary spatial locations
Since R2022b
## Syntax
``Jintrp = interpolateCurrentDensity(results,xq,yq)``
``Jintrp = interpolateCurrentDensity(results,xq,yq,zq)``
``Jintrp = interpolateCurrentDensity(results,querypoints)``
## Description
````Jintrp = interpolateCurrentDensity(results,xq,yq)` returns the interpolated current density values at the 2-D points specified in `xq` and `yq`.```
example
````Jintrp = interpolateCurrentDensity(results,xq,yq,zq)` uses 3-D points specified in `xq`, `yq`, and `zq`.```
example
````Jintrp = interpolateCurrentDensity(results,querypoints)` returns the interpolated current density values at the points specified in `querypoints`.```
example
## Examples
collapse all
Create an `femodel` object for DC conduction analysis and include a 2-D geometry of a plate with a hole into the model.
```model = femodel(AnalysisType="dcConduction", ... Geometry="PlateHolePlanar.stl");```
Plot the geometry.
`pdegplot(model.Geometry,EdgeLabels="on");`
Specify the conductivity of the material.
```model.MaterialProperties = ... materialProperties(ElectricalConductivity=6e4);```
Apply the voltage boundary conditions on the top and bottom edges of the plate.
```model.EdgeBC(3) = edgeBC(Voltage=100); model.EdgeBC(2) = edgeBC(Voltage=200);```
Specify the surface current density on the edge representing the hole.
`model.EdgeLoad(5) = edgeLoad(SurfaceCurrentDensity=200000);`
Generate a mesh.
`model = generateMesh(model);`
Solve the problem.
`R = solve(model);`
Plot the electric potential and current density.
```figure pdeplot(R.Mesh,XYData=R.ElectricPotential,ColorMap="jet", ... FlowData=[R.CurrentDensity.Jx R.CurrentDensity.Jy]) axis equal```
Interpolate the resulting current density to a grid covering the central portion of the geometry.
```[X,Y] = meshgrid(2:0.25:8,8:0.25:12); Jintrp = interpolateCurrentDensity(R,X,Y)```
```Jintrp = FEStruct with properties: Jx: [425x1 double] Jy: [425x1 double] ```
Reshape `Jintrp.Jx` and `Jintrp.Jy`, and plot the resulting current density.
```JintrpX = reshape(Jintrp.Jx,size(X)); JintrpY = reshape(Jintrp.Jy,size(Y)); quiver(X,Y,JintrpX,JintrpY,Color="red") axis equal```
Alternatively, you can specify the grid by using a matrix of query points.
```querypoints = [X(:),Y(:)]'; Jintrp = interpolateCurrentDensity(R,querypoints);```
Create an `femodel` object for DC conduction analysis and include a geometry representing a 10-by-10-by-1 solid plate into the model.
```model = femodel(AnalysisType="dcConduction", ... Geometry="Plate10x10x1.stl");```
Plot the geometry.
`pdegplot(model.Geometry,FaceLabels="on",FaceAlpha=0.3)`
Specify the conductivity of the material.
```model.MaterialProperties = ... materialProperties(ElectricalConductivity=6e4);```
Apply the voltage boundary conditions on the two faces of the plate.
`model.FaceBC([1 3]) = faceBC(Voltage=0);`
Specify the surface current density on the top of the plate.
`model.FaceLoad(5) = faceLoad(SurfaceCurrentDensity=100);`
Generate a mesh.
`model = generateMesh(model);`
Solve the problem.
`R = solve(model);`
Plot the electric potential.
```figure pdeplot3D(R.Mesh,ColorMapData=R.ElectricPotential)```
Plot the current density.
```figure pdeplot3D(R.Mesh,FlowData=[R.CurrentDensity.Jx, ... R.CurrentDensity.Jy, ... R.CurrentDensity.Jz])```
Interpolate the resulting current density to a coarser grid.
```[X,Y,Z] = meshgrid(0:10,0:10,0:0.5:1); Jintrp = interpolateCurrentDensity(R,X,Y,Z)```
```Jintrp = FEStruct with properties: Jx: [363x1 double] Jy: [363x1 double] Jz: [363x1 double] ```
Reshape `Jintrp.Jx`, `Jintrp.Jy`, and `Jintrp.Jz`.
```JintrpX = reshape(Jintrp.Jx,size(X)); JintrpY = reshape(Jintrp.Jy,size(Y)); JintrpZ = reshape(Jintrp.Jz,size(Z));```
Plot the resulting current density.
```figure quiver3(X,Y,Z,JintrpX,JintrpY,JintrpZ,Color="red")```
## Input Arguments
collapse all
Solution of a DC conduction problem, specified as a `ConductionResults` object. Create `results` using the `solve` function.
x-coordinate query points, specified as a real array. `interpolateCurrentDensity` evaluates the current density at the 2-D coordinate points `[xq(i) yq(i)]` or at the 3-D coordinate points `[xq(i) yq(i) zq(i)]` for every `i`. Because of this, `xq`, `yq`, and (if present) `zq` must have the same number of entries.
`interpolateCurrentDensity` converts the query points to column vectors `xq(:)`, `yq(:)`, and (if present) `zq(:)`. It returns current density values as a column vector of the same size. To ensure that the dimensions of the returned solution are consistent with the dimensions of the original query points, use `reshape`. For example, use `Jintrp = reshape(Jintrp,size(xq))`.
Example: `xq = [0.5 0.5 0.75 0.75]`
Data Types: `double`
y-coordinate query points, specified as a real array. `interpolateCurrentDensity` evaluates the current density at the 2-D coordinate points `[xq(i) yq(i)]` or at the 3-D coordinate points `[xq(i) yq(i) zq(i)]` for every `i`. Because of this, `xq`, `yq`, and (if present) `zq` must have the same number of entries.
`interpolateCurrentDensity` converts the query points to column vectors `xq(:)`, `yq(:)`, and (if present) `zq(:)`. It returns current density values as a column vector of the same size. To ensure that the dimensions of the returned solution are consistent with the dimensions of the original query points, use `reshape`. For example, use `Jintrp = reshape(Jintrp,size(yq))`.
Example: `yq = [1 2 0 0.5]`
Data Types: `double`
z-coordinate query points, specified as a real array. `interpolateCurrentDensity` evaluates the current density at the 3-D coordinate points `[xq(i) yq(i) zq(i)]`. Therefore, `xq`, `yq`, and `zq` must have the same number of entries.
`interpolateCurrentDensity` converts the query points to column vectors `xq(:)`, `yq(:)`, and `zq(:)`. It returns current density values as a column vector of the same size. To ensure that the dimensions of the returned solution are consistent with the dimensions of the original query points, use `reshape`. For example, use `Jintrp = reshape(Jintrp,size(zq))`.
Example: `zq = [1 1 0 1.5]`
Data Types: `double`
Query points, specified as a real matrix with either two rows for 2-D geometry or three rows for 3-D geometry. `interpolateCurrentDensity` evaluates the current density at the coordinate points `querypoints(:,i)` for every `i`, so each column of `querypoints` contains exactly one 2-D or 3-D query point.
Example: For a 2-D geometry, ```querypoints = [0.5 0.5 0.75 0.75; 1 2 0 0.5]```
Data Types: `double`
## Output Arguments
collapse all
Current density at query points, returned as an `FEStruct` object with the properties representing the spatial components of the current density at the query points. For query points that are outside the geometry, `Jintrp.Jx(i)`, `Jintrp.Jy(i)`, and `Jintrp.Jz(i)` are `NaN`. Properties of an `FEStruct` object are read-only.
## Version History
Introduced in R2022b | 1,973 | 7,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2024-38 | latest | en | 0.550592 |
https://www.cheenta.com/largest-area-amc-8-2003-problem-22/?mode=grid | 1,596,636,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00147.warc.gz | 631,898,729 | 34,698 | Select Page
Try this beautiful problem from Geometry based Largest area.
## Largest area – AMC-8, 2003 – Problem 22
The following figures are composed of squares and circles. Which figure has a shaded region with largest area?
• $A$
• $B$
• $C$
### Key Concepts
Geometry
Circle
Square
But try the problem first…
Answer:$C$
Source
AMC-8 (2003) Problem 22
Pre College Mathematics
## Try with Hints
First hint
To find out the largest area at first we have to find out the radius of the circles . all the circles are inscribed ito the squares .now there is a relation between the radius and the side length of the squares….
Can you now finish the problem ……….
Second Hint
area of circle =$\pi r^2$
can you finish the problem……..
Final Step
In A:
Total area of the square =$2^2=4$
Now the radius of the inscribed be 1(as the diameter of circle = side length of the side =2)
Area of the inscribed circle is $\pi (1)^2=\pi$
Therefore the shaded area =$4- \pi$
In B:
Total area of the square =$2^2=4$
There are 4 circle and radius of one circle be $\frac{1}{2}$
Total area pf 4 circles be $4 \times \pi \times (\frac{1}{2})^2=\pi$
Therefore the shaded area =$4- \pi$
In C:
Total area of the square =$2^2=4$
Now the length of the diameter = length of the diagonal of the square=2
Therefore radius of the circle=$\pi$ and lengthe of the side of the square=$\sqrt 2$
Thertefore area of the shaded region=Area of the square-Area of the circle=$\pi (1)^2-(\sqrt 2)^2$=$\pi – 2$ | 419 | 1,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-34 | latest | en | 0.838477 |
https://www.jiskha.com/display.cgi?id=1338156746 | 1,501,239,872,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448198.69/warc/CC-MAIN-20170728103510-20170728123510-00485.warc.gz | 797,800,011 | 4,186 | # math
posted by .
A specfic model of a car with an original value of \$32500 depreciates at a rate of \$7000 per year, as represented in the equation below: V(y) = 32500 - 7000 t
• math -
and the question is...?
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Post a New Question | 611 | 2,196 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-30 | longest | en | 0.911168 |
https://justhindi.in/hot-chip-challenge-scoville/ | 1,721,426,237,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00259.warc.gz | 295,672,806 | 17,358 | Question
1. # HOT CHIP CHALLENGE SCOVILLE: How many scoville units is in the one chip challenge?
Recently, a global challenge called the HOT CHIP CHALLENGE was launched. The goal of the challenge is to consume as many spicy foods as possible in one hour. For some people, this may sound like an easy task – after all, who can resist a little heat? But for others, the thought of trying spicy food all at once may be quite daunting. For those who are considering participating in the challenge, we wanted to provide you with some information about what scoville units are in a chip and how many there are in one cup. As it turns out, there are a whopping 1 million scoville units in a single chip! So if you’re feeling up for it, be sure to give the HOT CHIP CHALLENGE a try!
## What is the One Chip Challenge?
The One Chip Challenge is a gaming competition in which contestants must eat a single chip that is hot enough to make them blister. The chip is placed in the mouth and then removed after 30 seconds. The chip with the most blisters after the 30 seconds is considered the winner.
The current record for eating the hottest chip is 54 degrees Celsius (129 degrees Fahrenheit). This was achieved by AMD sponsored athlete Will Macpherson in July of 2017.
## How many scoville units are in a chip?
Hot Chip challenge scoville units:
There are 125 scoville units in a chip.
## Results of the One Chip Challenge
The One Chip Challenge is a competition where participants have to taste a small number of hot peppers and guess their scoville unit rating.
The winner of the challenge was determined to be Brandon Shipp, who was able to taste and rate 43 different peppers on a scale from 1 to 10,000. This makes the one chip challenge the hottest food challenge ever held! The average scoville unit rating for these 43 peppers was 2,711, making the one chip challenge hotter than even some chili peppers!
## What do the results mean?
The Hot Chip Challenge is a taste test that uses spicy chili peppers as its main ingredient. The challenge consists of eating a chip that contains a certain amount of scoville units. So far, the record holder for most scoville units eaten in one sitting is Tony Kim with 256.8 scoville units! This means that the chip has over 25,000 times the heat of a jalapeño pepper. How high can you get?
The Hot Chip Challenge was created in 2007 by two food entrepreneurs, Adam Smolen and Rick Harrison. Their goal was to create an event that would challenge people’s senses and bring together people from around the world. Participants are given three minutes to eat as many chips as possible and then rate them on a scale of 1-10 according to how hot they feel. The winner is the person who eats the most chips in three minutes!
So far, there have been over 180 different participants from all over the world who have taken part in the challenge. It is currently being held in London every year and it is estimated that it generates £2 million for charity each time it is held! What do the results mean? Well, first and foremost, it’s a great way to get people excited about spicy food! Second of all, it provides insights into how people rank spicy foods according to their own personal heat tolerance. And finally, it can help us develop new flavors and ingredients for our food products!
## Conclusion
The HOT CHIP CHALLENGE is a spicy food challenge where participants have to eat as many hot chips in a minute as possible. The chip with the most Scoville units (a measure of heat) is declared the winner. As you can imagine, this challenge can be pretty spicy! In fact, one chip has been known to have over 1 million Scoville units! So, how do you handle a challenge like this? Read our article to find out!
2. The Hot Chip Challenge Scoville is a new trend among foodies and spice fanatics alike. It’s a challenge that tests one’s endurance for the heat of spicy food by seeing how many scoville units they can handle with just one chip. The goal of the challenge is to find out exactly how much heat each participant can withstand, and it all comes down to understanding exactly how many scoville units are contained in the chip.
The Hot Chip Challenge Scoville contains anywhere from 1,500 to 3 million scoville units, depending on the brand and type of pepper used in manufacturing the chip. Some brands even go beyond this range with some chips containing upwards of 6 million scoville units!
3. Do you know what the Hot Chip Challenge is? If you don’t, then you’re missing out on a delicious and fiery snack experience. The Hot Chip Challenge involves eating a single chip that has been doused in a spicy sauce known as the Scoville sauce. Depending on the sauce, the chip can range from mild to extremely hot.
So, how many scoville units (a measure of the spiciness of a sauce) are in the Hot Chip Challenge? Well, it depends on the sauce and the person eating it, but generally, the Hot Chip Challenge falls somewhere between 2,500 and 4,500 scoville units. That means the chip is pretty hot, but not necessarily unbearably so.
For those who want to take the Hot Chip Challenge to the next level, they can opt to use a chip that is doused in a sauce with a scoville rating higher than 4,500. This is definitely not for the faint of heart, as the chip can be incredibly spicy and may even cause a burning sensation in the mouth.
No matter which chip you choose for your Hot Chip Challenge, you can be sure that you’ll be in for an unforgettable snacking experience. 🌶️🔥 So, if you’re looking for a fun and fiery challenge, the Hot Chip Challenge is definitely for you! 🤩
4. The One Chip Challenge, created by the company Paqui, features a single tortilla chip made with the Carolina Reaper pepper, which is known to be one of the hottest peppers in the world. The Scoville scale measures the heat or spiciness of chili peppers and their products. According to Paqui, the One Chip Challenge chip has an estimated Scoville rating of around 1.5 million units.
To put this into perspective, jalapeno peppers typically have a Scoville rating ranging from 2,500 to 8,000 units. So you can imagine just how intense and fiery the One Chip Challenge can be! It’s definitely not for the faint-hearted or those who are sensitive to spicy foods. | 1,408 | 6,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.957928 |
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Published at Wednesday, July 29th 2020, 09:25:24 AM. 3rd Grade Math Worksheets. By Nancey Garnier.
For the kids who love stickers they are not left behind, with plenty of birthday printables to choose from stickers are abundant. For those who love nature, they will get butterfly, rainbow, flowers and plenty of others just to decorate their bedroom or anything they desire. It will really look like a natural landscape once they are done. Books can be decorated with clouds, smiling faces and lots of toys to choose from, if not yet filled up. A room can be transformed into a cozy personal space with love stickers that surely will fill the room with a warm fuzzy look and feel.
Homeschool worksheets are a vital part of the student has homeschool experience. They allow the child to test his or her knowledge, and they offer them a practical application for their learning. Worksheets also, when used properly, provide both the students and parent / tutor immediate feedback as to the child has progress. This means they can be used to point out areas where the student needs further reinforcement. Homeschool worksheets fortunately will not over-tax your budget. There are many places where you can get them at extremely low costs. In fact, several websites offer printable worksheets for free.
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Coloring pages can provide enrichment by providing pictures of numbers, letters, animals, and words, so that your child will expand their knowledge in a number of areas. Your child will receive the greatest educational benefit when the coloring pages bring all of these elements together, such as when there is a picture of an animal with it has name written on the page, or when numbers are shown with the name of the number written out, or even a two-sided coloring page with one side showing the letter, number or word, and the other side of the page depicting an animal whose shape resembles the letter or number on the previous page. Take for example a coloring page that has a number one, along with a giraffe that is reminiscent of a number one in it has shape. Your child will not only have fun coloring the number and animal, but their lesson will also include writing the number one and being introduced to a giraffe.
Coloring pages can provide enrichment by providing pictures of numbers, letters, animals, and words, so that your child will expand their knowledge in a number of areas. Your child will receive the greatest educational benefit when the coloring pages bring all of these elements together, such as when there is a picture of an animal with it has name written on the page, or when numbers are shown with the name of the number written out, or even a two-sided coloring page with one side showing the letter, number or word, and the other side of the page depicting an animal whose shape resembles the letter or number on the previous page. Take for example a coloring page that has a number one, along with a giraffe that is reminiscent of a number one in it has shape. Your child will not only have fun coloring the number and animal, but their lesson will also include writing the number one and being introduced to a giraffe.
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# Chapter 4 Fundamental Data Types - PowerPoint PPT Presentation
Chapter 4 Fundamental Data Types. Chapter Goals. To understand integer and floating-point numbers To recognize the limitations of the numeric types To become aware of causes for overflow and roundoff errors To understand the proper use of constants . Continued…. Chapter Goals.
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## Chapter 4 Fundamental Data Types
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1. Chapter 4Fundamental Data Types
2. Chapter Goals • To understand integer and floating-point numbers • To recognize the limitations of the numeric types • To become aware of causes for overflow and roundoff errors • To understand the proper use of constants Continued…
3. Chapter Goals • To write arithmetic expressions in Java • To use the String type to define and manipulate character strings • To learn how to read program input and produce formatted output
4. Assignment • Review Exercises pp. 171-173 • #’s R4.1-R4.12, R4.16 • Due Tuesday, October 16, 2012 • Programming Exercises pp. 174 – 179 • P4.1 – P4.6, • Due Thursday, October 18, 2012 • P4.11 – P4.13, P4.15 – 4.18 • Due Wednesday, October 24, 2012 • String Lab Due TBA • First Quarter Test – Wednesday October 23, 2012 • Chapter 4 Test – October 29 & 30, 2012
5. Number Types • int: integers, no fractional part • double: floating-point numbers (double precision) 1, -4, 0 0.5, -3.11111, 4.3E24, 1E-14
6. Number Types • A numeric computation overflows if the result falls outside the range for the number type • Java: 8 primitive types, including four integer types and two floating point types int n = 1000000;System.out.println(n * n); // prints -727379968
7. Primitive Types Continued…
8. Primitive Types
9. Number Types: Floating-point Types • Rounding errors occur when an exact conversion between numbers is not possible • Java: Illegal to assign a floating-point expression to an integer variable double f = 4.35;System.out.println(100 * f); // prints 434.99999999999994 double balance = 13.75; int dollars = balance; // Error Continued…
10. Number Types: Floating-point Types • Casts: used to convert a value to a different type • Cast discards fractional part. • Math.roundconverts a floating-point number to nearest integer of type long. int dollars = (int) balance; // OK long rounded = Math.round(balance); // if balance is 13.75, then // rounded is set to 14
11. Syntax 4.1: Cast (typeName) expression Example: (int) (balance * 100) Purpose: To convert an expression to a different type
12. Self Check 1. Which are the most commonly used number types in Java? Ans: intand double • When does the cast (long)xyield a different result from the call Math.round(x)? Ans: When the fractional part of x is ≥ 0.5 • How do you round the double value x to the nearest intvalue, assuming that you know that it is less than 2 · 109? Ans: By using a cast: (int) Math.round(x)
13. Constants: final • A finalvariable is a constant • Once its value has been set, it cannot be changed • Named constants make programs easier to read and maintain • Convention: use all-uppercase names for constants final double QUARTER_VALUE = 0.25;final double DIME_VALUE = 0.1;final double NICKEL_VALUE = 0.05;final double PENNY_VALUE = 0.01;payment = dollars + quarters * QUARTER_VALUE + dimes * DIME_VALUE + nickels * NICKEL_VALUE + pennies * PENNY_VALUE;
14. Constants: static final • If constant values are needed in several methods, declare them together with the instance fields of a class and tag them as static and final • Give static final constants public access to enable other classes to use them public class Math{ . . . public static final double E = 2.7182818284590452354; public static final double PI = 3.14159265358979323846;} double circumference = Math.PI * diameter;
15. Syntax 4.2: Constant Definition In a method: final typeName variableName = expression ; In a class: accessSpecifier static final typeName variableName = expression; Example: final double NICKEL_VALUE = 0.05; public static final double LITERS_PER_GALLON = 3.785; Purpose: To define a constant in a method or a class
16. File CashRegister.java(page 143 – textbook) 01: /** 02: A cash register totals up sales and computes change due. 03: */ 04:public classCashRegister 05: { 06: /** 07: Constructs a cash register with no money in it. 08: */ 09:public CashRegister() 10: { 11: purchase = 0; 12: payment = 0; 13: } 14: Continued…
17. File CashRegister.java 15: /** 16: Records the purchase price of an item. 17: @param amount the price of the purchased item 18: */ 19:public void recordPurchase(double amount) 20: { 21: purchase = purchase + amount; 22: } 23: 24: /** 25: Enters the payment received from the customer. 26: @param dollars the number of dollars in the payment 27: @param quarters the number of quarters in the payment 28: @param dimes the number of dimes in the payment 29: @param nickels the number of nickels in the payment 30: @param pennies the number of pennies in the payment 31: */ Continued… Continued…
18. File CashRegister.java 32:public void enterPayment(int dollars, int quarters, 33:int dimes, int nickels, int pennies) 34: { 35: payment = dollars + quarters * QUARTER_VALUE + dimes * DIME_VALUE 36: + nickels * NICKEL_VALUE + pennies * PENNY_VALUE; 37: } 38: 39: /** 40: Computes the change due and resets the machine for the next customer. 41: @return the change due to the customer 42: */ Continued… Continued…
19. File CashRegister.java 43:public double giveChange() 44: { 45:double change = payment - purchase; 46: purchase = 0; 47: payment = 0; 48:return change; 49: } 50: 51: public static final double QUARTER_VALUE = 0.25; 52: public static final double DIME_VALUE = 0.1; 53: public static final double NICKEL_VALUE = 0.05; 54: public static final double PENNY_VALUE = 0.01;55: 56:private double purchase; 57:private double payment; 58: } Continued…
20. File CashRegisterTester.java 01: /** 02: This class tests the CashRegister class. 03: */ 04:public class CashRegisterTester 05: { 06:public static void main(String[] args) 07: { 08: CashRegister register = new CashRegister(); 09: 10: register.recordPurchase(0.75); 11: register.recordPurchase(1.50); 12: register.enterPayment(2, 0, 5, 0, 0); 13: System.out.print("Change="); 14: System.out.println(register.giveChange()); 15: Continued…
21. File CashRegisterTester.java 16: register.recordPurchase(2.25); 17: register.recordPurchase(19.25); 18: register.enterPayment(23, 2, 0, 0, 0); 19: System.out.print("Change="); 20: System.out.println(register.giveChange()); 21: } 22: } Output Change=0.25 Change=2.0
22. Self Check • What is the difference between the following two statements?and Ans:The first definition is used inside a method, the second inside a class final double CM_PER_INCH = 2.54; public static final double CM_PER_INCH = 2.54;
23. Self Check continued… • What is wrong with the following statement? Ans: (1) You should use a named constant, not the "magic number" 3.14(2) 3.14 is not an accurate representation of double circumference = 3.14 * diameter;
24. Assignment, Increment, and Decrement • Assignment is not the same as mathematical equality:items = items + 1; • items++is the same as items =items + 1 • items-- subtracts 1 from items
25. Assignment, Increment, and Decrement Cont… • The increment and decrement operators are arithmetic and operate on one operand • The increment operator (++) adds one to its operand • The decrement operator (--) subtracts one from its operand • The statement items--; is functionally equivalent to: items = items - 1;
26. Assignment, Increment and Decrement Figure 1:Incrementing a Variable
27. More Assignment Operators • Often we perform an operation on a variable, and then store the result back into that variable • Java provides assignment operators to simplify that process • For example, the statement num += count; is equivalent to: num = num + count;
28. Operator += -= *= /= %= Example x += y x -= y x *= y x /= y x %= y Equivalent To x = x + y x = x - y x = x * y x = x / y x = x % y Assignment Operators • There are many assignment operators, including the following:
29. Self Check • What is the meaning of the following statement? Ans: The statement adds the amount value to the balance variablethen assigns the sum back to balance again. balance += amount;
30. Self Check continued… 7. What is the value of n after the following sequence of statements? n--;n++;n--; Ans: One less than it was before
31. What is the Value? • intnumberOfCoins = 10; • intmoreCoins = 12; • numberOfCoins = numberOfCoins + 1; • numberOfCoins ++; • numberOfCoins+= moreCoins; • numberOfCoins --;
32. Arithmetic Operations • / is the division operator • If both arguments are integers, the result is an integer. The remainder is discarded 7.0 / 4 yields 1.757 / 4 yields 1 • Get the remainder with % (pronounced "modulo") 7 % 4 is 3
33. Arithmetic Operations final int PENNIES_PER_NICKEL = 5;final int PENNIES_PER_DIME = 10;final int PENNIES_PER_QUARTER = 25;final int PENNIES_PER_DOLLAR = 100;// Compute total value in penniesint total = dollars * PENNIES_PER_DOLLAR + quarters * PENNIES_PER_QUARTER+ nickels * PENNIES_PER_NICKEL + dimes * PENNIES_PER_DIME + pennies;// Use integer division to convert to dollars, centsint dollars = total / PENNIES_PER_DOLLAR;int cents = total % PENNIES_PER_DOLLAR;
34. What is the Result and Type?
35. The Math class • Math class: contains methods like sqrt and pow • To compute xn, you write Math.pow(x, n) • However, to compute x2 it is significantly more efficient simply to compute x * x • To take the square root of a number, use the Math.sqrt; for example, Math.sqrt(x) Continued…
36. The Math class • In Java, can be represented as (-b + Math.sqrt(b * b - 4 * a * c)) / (2 * a)
37. Mathematical Methods in Java
38. Analyzing an Expression Figure 3:Analyzing an Expression
39. Self Check • What is the value of 1729 / 100? Of 1729 % 100? Ans: 17 and 29 • Why doesn't the following statement compute the average of s1, s2, and s3? Ans: Only s3 is divided by 3. To get the correct result, use parentheses. Moreover, if s1, s2, and s3 are integers, you must divide by 3.0 to avoid integer division: double average = s1 + s2 + s3 / 3; // Error
40. Self Check continued… • What is the value of in mathematical notation? Ans: Math.sqrt(Math.pow(x, 2) + Math.pow(y, 2))
41. Calling Static Methods • A static method does not operate on an object • Static methods are defined inside classes • Naming convention: Classes start with an uppercase letter; objects start with a lowercase letter double x = 4;double root = x.sqrt(); // Error MathSystem.out
42. Syntax 4.3: Static Method Call ClassName. methodName(parameters) Example: Math.sqrt(4) Purpose: To invoke a static method (a method that does not operate on an object) and supply its parameters
43. Self Check • Why can't you call x.pow(y) to compute xy? Ans: x is a number, not an object, and you cannot invoke methods on numbers • Is the call System.out.println(4) a static method call? Ans: No–the println method is called on the object System.out
44. Strings • A string is a sequence of characters • Strings are objects of the String class • String constants: • String variables: • String length: • Empty string: "Hello, World!" String message = "Hello, World!"; int n = message.length(); ""
45. Concatenation • Use the + operator: • If one of the arguments of the + operator is a string, the other is converted to a string String name = "Dave";String message = "Hello, " + name; // message is "Hello, Dave" String a = "Agent";int n = 7;String bond = a + n; // bond is Agent7
46. Concatenation in Print Statements • Useful to reduce the number of System.out.print instructions versus System.out.print("The total is ");System.out.println(total); System.out.println("The total is " + total);
47. Converting between Strings and Numbers • Convert to number: • Convert to string: int n = Integer.parseInt(str);double x = Double.parseDouble(x); String str = "" + n;str = Integer.toString(n);
48. Substrings String greeting = "Hello, World!";String sub = greeting.substring(0, 5); // sub is "Hello" • Supply start and “past the end” position • First position is at 0 Continued… Figure 3:String Positions
49. Substrings • Substring length is “past the end” - start Figure 4:Extracting a Substring
50. Self Check • Assuming the String variable s holds the value "Agent", what is the effect of the assignment s = s +s.length()? Ans: s is set to the string Agent5 | 3,447 | 13,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2019-35 | latest | en | 0.819451 |
https://studylib.net/doc/10407954/problem-1.-the-following-matrix-a-is-diagonalizable.-find... | 1,680,200,673,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00665.warc.gz | 617,440,486 | 12,245 | # Problem 1. The following matrix A is diagonalizable. Find e method.
```Problem 1.
The following matrix A is diagonalizable.
method.
−3 −8
A=
−4 −8
−4 −10
Problem 2.
The following matrix is nilpotent.
0
0
N =
0
0
Find eAt by the diagonalization
10
11
13
Find etN .
1 0 0
0 1 0
.
0 0 1
0 0 0
Problem 3.
The following matrix J is in Jordan Canonical Form. Find the matrices S
and N so that J = S + N , S is diagonalizable (diagonal, in this case), N is
nilpotent, and SN = N S. Compute eJt = eSt eN t by computing etS , etN and
multiplying.
1 1 0 0 0 0
0 1 1 0 0 0
0 0 1 0 0 0
J =
0 0 0 2 1 0
0 0 0 0 2 0
0
0
0
1
0
0
3
Problem 4.
Consider the matrix
568 −347 −31
1257 −1079 −215
791 −485 −61
1767 −1504 −312
−681
419
45
−1517
1299
262
A=
.
−620
393
37
−1393
1209
229
−686
432
46
−1542
1331
259
129 −60 −6
261 −204 −59
A. Find a basis of each of the generalized eigenspaces of A.
B. Find the matrices S and N so that A = S + N , S is diagonalizable, N
is nilpotent, and SN = N S. Verify that your answers for S and N really
satisfy these conditions. (It is not necessary to go all the way to the Jordan
Canonical Form of A.)
C. Use S and N to compute eAt = etS etN .
2
Problem 5.
In each part, use Leonard’s Algorithm to find eAt .
A.
5 −1
A=
12
−2
3 −1
B.
0
0
2
1 1 −2
A=
−3 5 −6
0 1 −1
C.
1 1 −2
−3
5
−6
A=
0 1 −1
D.
−79 −4 −36
614
7
A=
262
4
71 −2
76
−558
137 −240
43 −61
325
Problem 6. Recall that we have shown
et(A+B) = etA etB
provided that A and B commute. To see what happens if A and B do not
commute, consider the function
Ψ(t) = et(A+B) e−tB e−tA ,
so if A and B commute, Ψ(t) = I for all t. To see how much Ψ differs from
the identity in the general case, find the first two nonzero terms in the Taylor
expansion of Ψ(t), centered at t = 0.
Problem 7.
Consider the following mechanical system.
3
We have two weights of masses m1 and m2 connected by three springs to a
stationary frame. The spring constants of the three springs are k1 , k2 and k3 , as
indicated in the diagram. The weights are constrained to move in a horizontal
line. The deviation of the two weights from their equilibrium positions are y1
and y2 respectively, measured positive to the right.
A. Assuming there is no friction, write down the system of linear equations that
governs the motion of the weights, using Hooke’s law for the force produced
by the springs. By very careful with your signs! Make this system into a first
order system by introducing new dependent variables z1 and z2 and adding
the equations y10 = z1 and y20 = z2 (see Section 3.1 in the book). Write the
resulting system in matrix form dy/dt = Ay.
B. Assume the values
m1 = 1,
m2 = 2,
k1 = 1,
k2 = 2,
k3 = 2.
Find eAt , where A is the coefficient matrix of the system, using one of the
methods developed in class.
Find the solution for the initial conditions
y1 (0) = 1,
y2 (0) = −1,
y10 (0) = 0,
y20 (0) = 0.
Provide a graph of y1 and y2 as functions of time.
C. Assume now that the weights are subject to frictional forces proportional to
their velocities, i.e., the force on the first weight due to friction is −b1 y10 and
the force on the second weight due to friction is −b2 y20 , where b1 and b2 are
positive constants. Find the new system of equations dy/dt = By.
D. Assume the values
m1 = 1, m2 = 1
k1 = 1, k2 = 1, k3 = 1
b1 = 1/2, b2 = 1/2.
Find eBt , using one of the methods developed in class. Solve the system
subject to the initial conditions
y1 (0) = 1,
y2 (0) = 0,
y10 (0) = 0 y20 (0) = 0.
Provide a graph showing y1 and y2 as functions of time.
4
Problem 8.
Do Problem 20 on page 190 of the book. See Section 1.7 in the book for how
to model electric circuits.
5
EXAM
Practice Exam # 2
Math 3351, Spring 2003
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• ## Why the p values from my linear regression model don't agree with the ones from ANOVA?
Hi everyone, here I'm confused: I think ANOVA and Linear Regression Model are supposed to be the same thing, but...
Last edited by Yuan Chang; 08 Dec 2018, 22:08.
• ## Two line frequency graph help
Hello everyone.
I've been trying to figure this out on my own but I haven't been very successful.
I am working on trying to make a...
Working...
X | 796 | 3,500 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-13 | longest | en | 0.886632 |
http://www.physicsforums.com/showthread.php?s=707fcee61ea9d263643324d8b6075e94&p=3820574 | 1,408,651,639,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500821289.49/warc/CC-MAIN-20140820021341-00195-ip-10-180-136-8.ec2.internal.warc.gz | 541,467,572 | 6,433 | # How to evaluate this integral to get pi^2/6:
by hb1547
Tags: evaluate, integral, pi2 or 6
P: 35 $\int_0^\infty \frac{u}{e^u - 1}$ I know that this integral is $\frac{\pi^2}{6}$, just from having seen it before, but I'm not really sure if I can evaluate it directly to show that. I know that: $\zeta(x) = \frac{1}{\Gamma(x)} \int_0^\infty \frac{u^{x-1}}{e^u -1} du$ Does the value $\frac{\pi^2}{6}$ come from using other methods of showing the result for $\zeta(2)$ and solving the equation, or is that integral another way of evaluating $\zeta(2)$?
Quote by hb1547 $\int_0^\infty \frac{u}{e^u - 1}$ I know that this integral is $\frac{\pi^2}{6}$, just from having seen it before, but I'm not really sure if I can evaluate it directly to show that. I know that: $\zeta(x) = \frac{1}{\Gamma(x)} \int_0^\infty \frac{u^{x-1}}{e^u -1} du$ Does the value $\frac{\pi^2}{6}$ come from using other methods of showing the result for $\zeta(2)$ and solving the equation, or is that integral another way of evaluating $\zeta(2)$? | 341 | 1,022 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2014-35 | latest | en | 0.893434 |
https://medium.com/quotidians-from-rajendran-dandapani/how-exams-should-be-quotidian-368-3fd0a543c0c?source=post_internal_links---------5---------------------------- | 1,657,137,754,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104676086.90/warc/CC-MAIN-20220706182237-20220706212237-00289.warc.gz | 442,250,975 | 41,699 | # How Exams Should Be! — Quotidian — 368
(Transcript of video originally posted 08 Jan 2022)
Some of you must have received this Whatsapp forward titled “Why I hate mathematics”.
This is what the student opines:-
While in class, this is what they want me to do. Five plus Five is what. Okay, that can be done.
And then, they give me homework. They want me to evaluate this complex expression. This is doable too…
That is alright too.. While the classwork and the homework are like this, how fair is it to burden me with an exam like this..! “With two sheep flying, one yellow and the other headed right, how much does a kilo of cement cost, given that the cow is 10 years old?”
Whaaaat? What is the problem with our exams, and can they be structured in some other way? This is a topic that we may not be able to do justice to, even spanning ten full Quotidians, but I am going to just touch a little about it today.
(Intro Music)
Namaste! This game, is called Zaxxon. My first game. The first time I saw a computer, and went “Aha”, “Wow”, “A Game? Playable on a computer?”.. I lost badly too, pretty soon! No lives. Just an infinite scroller. Getting harder and harder. You just have to manoeuver your plane through all those obstacles. That’s all. One chance. All depends on how well you do.
Games evolved after that. Some of you will remember Dangerous Dave. This is Level One of Dangerous Dave. You just have to go along straight as an arrow, you climb, you lower yourself, and open the door, and exit! And, if you want, some diamonds to pick up along the way. No trouble. Easy Peesy. How is Level Two? How is Level Five? How is Level Ten?
A lot of monsters. A lot of timing. There is a gun involved. There is a jetpack that you will have to use judiciously. In a graded level of difficulty, the Tenth is harder than the First. Is it resembling school yet? Are you able to draw some connections between how education progresses exam after exam in school and how Dangerous Dave progresses level after level? The gameplay involved in this pastime, is there something that we can take, we can borrow, from this world. Do you see those three lives up above and to the right? Two of them… are gone. One life..! Can we give our students a second chance if they lose the first time?
Games evolved even more. Some of you will remember Wolfenstein 3D. That was probably the first pseudo-3D game. In that game, yes, there were levels. But, within a level, there were some spaces that were pretty hard to cross. Too many enemies that could pick you off too easily. So, at strategically placed hidden spots, (and I still don’t get why a prison should have medpacks hidden inside it!) you would get medical first aid kits. If you touched them, you would get healthier. How do we fashion suck medpacks in our school curriculum? Can we let people who have injured themselves get stronger without having to restart the level?
And, while on the topic of “restarting a level”, I am reminded of even more evolved games. Here is one more called “Dark Souls”. In this game, there would be a huge level. The hero would be exploring a huge cave. In a large castle, the hero would be defeating all enemies and proceeding fast. Here and there, there would be bonfires. If he goes to the bonfire and warms his hands there, it is called a Save Point. You don’t have to go through that whole exploration again, if you have reached that Save Point. The more Save Points there are, the easier it is for the players. For students, the more Save Points there can be, the easier will education be. Think about that.
Games evolved even more. There is a game called God of War. In this epic game called God of War, this hero is fighting the nearly undefeatable monster blocking his path. There is no Save Point there. There is no Med Pack. He will fight, and if he is injured, he adopts one important trick. It is called Life Regen. He will just run away from the enemy. From that battlefield area, he just has to move away from the action a bit, and just given the time, and space, our hero will regenerate his power. He fights. He gets injured. He runs away from the battlefield. To survive. To regen. To recoup. And comes back. What lesson can we take from here and incorporate into our education system, into our examination system?
This person’s name is Carleton Washburne. In 1919, he came up with something called the Winnetka Plan. An educationist. He was the Principal of a school. And, he came up with an idea. But, even a full hundred years afterwards, his principles, amazing principles, haven’t been implemented anywhere. There are facilities in place for it. The technology is readily available. People have even proved that it is a viable plan. But, somehow the Winnetka Plan never got implemented. What is this Winnetka Plan?
He said, students can either be asked to finish something by some time. Whatever mastery you get, finish it by this time. What.. You have scored 60 percent in Maths? Okay, 40 is the cutoff. You’ve got 60. So, please go to the next level. NOT FAIR. Why? This person might have skipped probability or permutations and combinations or statistics, and the next year, statistics is going to be even more important. So, … basing it on a fixed time period, and telling the student to achieve whatever mastery they can in that allotted period and keep running to the next one, .. a sort of conveyor belt mechanism of education and examination is wrong — says Washburne. He says, instead, given them flexi-time, but freeze on the mastery. “You have to achieve hundred percent mastery on this subject. But, take as long as you want.” If he had proposed this in 1919, what a foresightful genius he should have been?
But today, this is possible. With the smartphone in your hand, with the ubiquitous internet, with most classrooms already online because of Covid, “take your time, achieve mastery!”
There is another twist to this thing. The teacher gets to study not just the subject, the teacher gets to study the student, and customise the program per student. To suit every student, in a way they will understand it right, the program can be customised. That was the Winnetka Plan. And that is my dream. I wish educational institutions adopt or probably at least adapt this model.
One of them has actually already done it. Some of you may already know him. We have seen him already. Sal Khan. Salman Khan. Not the Bollywood actor Salman Khan, but the founder of Khan Academy — Salman Khan. He says knowledge itself is like a tree. Everything connected to everything else. Each cascading to further, deeper, broader, more fundamental concepts. If you progress based on fixed time, you will not have enough fundamentals for you to build a proper structure, so give yourself as much time as you want. That was the beginning of the Flipped Classroom Approach. Give them videos. Give them lectures. Let them watch it as many times as they want. Let them watch it in 2X speed. Let them watch it again and again if they want. Let them come back and in the classroom the teacher is not a sage delivering lecture after lecture, the teacher is a facilitator. Students sit around, question, debate, expand, what else can we do with the subject, give the opportunity for such discussions. And, his tests, his examination methodology follows two interesting approaches.
One, some of you may already know, is the Adaptive Testing approach. If you do very well, the next level gets harder. Progressively harder. Until you hit a blockade, and the next level is lowered to suit you. Your current level of difficulty, for a problem you have been given, is dependent on the previous four or five questions that you have answered. Many international evaluations already use the Adaptive Mode.
He also introduced one more concept called Streaking. He said, “I am not going to say you did something very well because you got something right, or you did something very badly because you didn’t get one thing right. No. I am going to look for streaks. If you do ten problems, consecutively, ten graded adaptive problems, correctly, then, you have mastered (Winnetka), we can move to the next level”.
One of my favorites.. Perhaps even the all time favourite App is called Elevate. It is calling itself a Brain Trainer App. But, basically, it is a vocabulary-building and mathematics — simple maths, algebra, probably some number sense training App. In this App, they have a system called the EPQ. Elevate Proficiency Quotient. That EPQ is across all these aspects. And, at various levels. Whatever level you are in, to move to the next level is harder progressively. If you are in the Beginner Mode, Beginner to Intermediate may be an easy move. But, if you are in the Advanced Mode, to move from Advanced to Expert will be harder. If you make one mistake, they will forgive you. If you consecutively make two or three mistakes, they won’t punish you, they will help you, by lowering your level. Do some research on this. Really wonderful, beautiful, graded, .. it would almost seem as if there is some complex Machine Learning going on inside the App all the time.
What should we look for? If we are going to conduct an evaluation, .. should we expect them to remember everything we have taught them… that’s not what we should look for. Instead, can we look for these things.. Can we look for what they have learned? Can we look for whether they are able to think critically? See how their responses are? Can we see whether they can come up with new responses that they haven’t been trained for? That creating.. can we look for that? Is there an emotional quotient also involved that we can evaluate? Also, how well can they work with others? It is not about competition in the real world.. Can they work with others. And finally, can they lead others? Can they inspire?
How are we going to track all these things in an exam? If we can find methodologies that will evaluate these, only then probably, exams will become relevant. And… if we don’t evaluate for these, we will be stuck with situations like this one. Just a year ago, IIT Delhi came out with an advertisement like this. They are looking for a Dog Handler, and they want a B.Tech or equivalent undergraduate degree.. Seriously?! A B.Tech undergraduate degree for a dog handler? If things shouldn’t deteriorate to this, please, please, examine your exams! Thank you!
And, yes, we will meet tomorrow! Bye Bye!
--
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## More from Quotidians From Rajendran Dandapani
Welcome to Quotidians — a humble attempt to bring a smile to your face… as I connect the commonplace everyday nuggets into meaningfully connected insights.
## Rajendran Dandapani
Business Solutions Evangelist at Zoho Corp. President at The Zoho Schools Of Learning. | 2,424 | 10,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2022-27 | latest | en | 0.965604 |
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=RJI | 1,524,194,495,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937113.3/warc/CC-MAIN-20180420022906-20180420042906-00477.warc.gz | 426,728,156 | 71,646 | Back to list of Stocks See Also: Seasonal Analysis of RJIGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
Fourier Analysis of RJI (ELEMENTS Rogers Intl Cmdty TR ETN)
RJI (ELEMENTS Rogers Intl Cmdty TR ETN) appears to have interesting cyclic behaviour every 37 weeks (.2066*sine), 46 weeks (.1665*sine), and 55 weeks (.1582*cosine).
RJI (ELEMENTS Rogers Intl Cmdty TR ETN) has an average price of 7.52 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Fourier Analysis
Using data from 10/19/2007 to 4/16/2018 for RJI (ELEMENTS Rogers Intl Cmdty TR ETN), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
07.52073 0
1-.72565 1.4169 (1*2π)/549549 weeks
21.04005 .70561 (2*2π)/549275 weeks
3.94931 .5567 (3*2π)/549183 weeks
4.06834 .92013 (4*2π)/549137 weeks
5.02385 .78858 (5*2π)/549110 weeks
6.05516 .65452 (6*2π)/54992 weeks
7-.39885 .66613 (7*2π)/54978 weeks
8-.33691 .11453 (8*2π)/54969 weeks
9-.25178 -.10855 (9*2π)/54961 weeks
10-.15823 .0802 (10*2π)/54955 weeks
11-.04093 -.06541 (11*2π)/54950 weeks
12.10109 -.16655 (12*2π)/54946 weeks
13.06441 .06753 (13*2π)/54942 weeks
14.11004 .06586 (14*2π)/54939 weeks
15.0875 .2066 (15*2π)/54937 weeks
16.01688 .14488 (16*2π)/54934 weeks
17.03881 .13342 (17*2π)/54932 weeks
18.00745 .11758 (18*2π)/54931 weeks
19-.01907 .02965 (19*2π)/54929 weeks
20-.06797 .14895 (20*2π)/54927 weeks
21.0173 .01422 (21*2π)/54926 weeks
22.01988 .09795 (22*2π)/54925 weeks
23-.0758 -.01291 (23*2π)/54924 weeks
24.03294 .04574 (24*2π)/54923 weeks
25.03983 .03544 (25*2π)/54922 weeks
26.06383 .06441 (26*2π)/54921 weeks
27-.0126 .07626 (27*2π)/54920 weeks
28.0179 .05022 (28*2π)/54920 weeks
29-.00758 .05866 (29*2π)/54919 weeks
30.04301 .0197 (30*2π)/54918 weeks
31.02147 .06971 (31*2π)/54918 weeks
32-.00641 .04917 (32*2π)/54917 weeks
33.06138 .0446 (33*2π)/54917 weeks
34.05951 .06229 (34*2π)/54916 weeks
35-.00051 .07592 (35*2π)/54916 weeks
36.00463 .08839 (36*2π)/54915 weeks
37-.02231 .06036 (37*2π)/54915 weeks
38-.02071 .04954 (38*2π)/54914 weeks
39-.00224 .05181 (39*2π)/54914 weeks
40-.02537 .05743 (40*2π)/54914 weeks
41-.0056 -.00945 (41*2π)/54913 weeks
42.01636 .02341 (42*2π)/54913 weeks
43.01555 .02555 (43*2π)/54913 weeks
44.00951 .01313 (44*2π)/54912 weeks
45.0177 .0273 (45*2π)/54912 weeks
46.05434 .01799 (46*2π)/54912 weeks
47.01195 .04691 (47*2π)/54912 weeks
48-.01742 .0419 (48*2π)/54911 weeks
49.02705 .03229 (49*2π)/54911 weeks
50-.00676 .01916 (50*2π)/54911 weeks
51.00549 .00933 (51*2π)/54911 weeks
52.01146 .04816 (52*2π)/54911 weeks
53.01124 .03023 (53*2π)/54910 weeks
54-.02031 .02429 (54*2π)/54910 weeks
55-.00916 .0071 (55*2π)/54910 weeks
56.01005 -.01887 (56*2π)/54910 weeks
57.03579 -.01509 (57*2π)/54910 weeks
58.05319 .03634 (58*2π)/5499 weeks
59.022 .03498 (59*2π)/5499 weeks
60.04798 .06836 (60*2π)/5499 weeks
61-.01503 .04146 (61*2π)/5499 weeks
62.01478 .04173 (62*2π)/5499 weeks
63.02081 .05615 (63*2π)/5499 weeks
64-.0106 .03043 (64*2π)/5499 weeks
65-.02304 .04155 (65*2π)/5498 weeks
66-.00413 .02983 (66*2π)/5498 weeks
67.00662 .03358 (67*2π)/5498 weeks
68.00935 .05109 (68*2π)/5498 weeks
69-.00761 .01418 (69*2π)/5498 weeks
70.0257 .01666 (70*2π)/5498 weeks
71.00321 .01963 (71*2π)/5498 weeks
72-.00165 .02417 (72*2π)/5498 weeks
73-.00371 .02616 (73*2π)/5498 weeks
74-.02371 .02636 (74*2π)/5497 weeks
75.01654 -.02122 (75*2π)/5497 weeks
76.03135 .03262 (76*2π)/5497 weeks
77-.0244 .01976 (77*2π)/5497 weeks
78-.00448 .0042 (78*2π)/5497 weeks
79.0255 .02168 (79*2π)/5497 weeks
80-.00232 .01178 (80*2π)/5497 weeks
81.01254 .00952 (81*2π)/5497 weeks
82.00716 .03169 (82*2π)/5497 weeks
83-.01111 -.00659 (83*2π)/5497 weeks
84.02418 .00175 (84*2π)/5497 weeks
85-.00972 .0259 (85*2π)/5496 weeks
86-.00398 -.01466 (86*2π)/5496 weeks
87.01957 .00534 (87*2π)/5496 weeks
88.00722 .02423 (88*2π)/5496 weeks
89.03389 -.00077 (89*2π)/5496 weeks
90.03899 .03075 (90*2π)/5496 weeks
91.00769 .03601 (91*2π)/5496 weeks
92-.01322 .0202 (92*2π)/5496 weeks
93.00503 .01193 (93*2π)/5496 weeks
94-.00455 .01692 (94*2π)/5496 weeks
95.01293 .01707 (95*2π)/5496 weeks
96-.01254 .01612 (96*2π)/5496 weeks
97-.01056 .00952 (97*2π)/5496 weeks
98-.0093 .00483 (98*2π)/5496 weeks
99.02447 .004 (99*2π)/5496 weeks
100.01728 .00296 (100*2π)/5495 weeks
101.01395 .01196 (101*2π)/5495 weeks
102.00115 .00209 (102*2π)/5495 weeks
103.00797 .02037 (103*2π)/5495 weeks
104.01207 .01154 (104*2π)/5495 weeks
105.01125 .00868 (105*2π)/5495 weeks
106.00164 .00706 (106*2π)/5495 weeks
107.00468 .00784 (107*2π)/5495 weeks
108-.001 .00422 (108*2π)/5495 weeks
109.01125 .00087 (109*2π)/5495 weeks
110.02282 .00386 (110*2π)/5495 weeks
111.01304 .02148 (111*2π)/5495 weeks
112.01935 .01967 (112*2π)/5495 weeks
113.0206 .01037 (113*2π)/5495 weeks
114.01881 .03202 (114*2π)/5495 weeks
115.00067 .01989 (115*2π)/5495 weeks
116.00008 .02678 (116*2π)/5495 weeks
117-.0082 .0197 (117*2π)/5495 weeks
118-.00542 .01814 (118*2π)/5495 weeks
119-.00416 -.00332 (119*2π)/5495 weeks
120.00245 .00017 (120*2π)/5495 weeks
121.01761 -.00128 (121*2π)/5495 weeks
122.0124 .01358 (122*2π)/5495 weeks
123.01317 .01868 (123*2π)/5494 weeks
124.00897 -.00008 (124*2π)/5494 weeks
125.01037 .0204 (125*2π)/5494 weeks
126.00437 .01642 (126*2π)/5494 weeks
127.00294 .00924 (127*2π)/5494 weeks
128.00827 .00362 (128*2π)/5494 weeks
129-.00134 .00466 (129*2π)/5494 weeks
130.00522 -.007 (130*2π)/5494 weeks
131.02388 .00582 (131*2π)/5494 weeks
132.02037 .0003 (132*2π)/5494 weeks
133.0067 .00569 (133*2π)/5494 weeks
134.01026 -.00389 (134*2π)/5494 weeks
135.03021 .01747 (135*2π)/5494 weeks
136.01015 .018 (136*2π)/5494 weeks
137.01664 .00186 (137*2π)/5494 weeks
138.0113 .03397 (138*2π)/5494 weeks
139-.0202 .01762 (139*2π)/5494 weeks
140-.00392 .00524 (140*2π)/5494 weeks
141-.00214 -.00236 (141*2π)/5494 weeks
142.00167 -.00851 (142*2π)/5494 weeks
143.00867 .00381 (143*2π)/5494 weeks
144.00659 .00071 (144*2π)/5494 weeks
145.01698 -.00529 (145*2π)/5494 weeks
146.0148 .00313 (146*2π)/5494 weeks
147.01376 .00534 (147*2π)/5494 weeks
148.02026 .01405 (148*2π)/5494 weeks
149.00823 .02193 (149*2π)/5494 weeks
150.01346 .01974 (150*2π)/5494 weeks
151-.00919 .02172 (151*2π)/5494 weeks
152-.0126 .0085 (152*2π)/5494 weeks
153.00411 .01062 (153*2π)/5494 weeks
154.00094 .00152 (154*2π)/5494 weeks
155.00407 -.00007 (155*2π)/5494 weeks
156.01004 .00983 (156*2π)/5494 weeks
157-.00124 .00435 (157*2π)/5493 weeks
158.01598 -.01003 (158*2π)/5493 weeks
159.00983 .01509 (159*2π)/5493 weeks
160-.00266 .00701 (160*2π)/5493 weeks
161.0109 -.00686 (161*2π)/5493 weeks
162.01888 .0106 (162*2π)/5493 weeks
163.01184 .0159 (163*2π)/5493 weeks
164.00369 .00452 (164*2π)/5493 weeks
165.00967 .01116 (165*2π)/5493 weeks
166-.00042 .01541 (166*2π)/5493 weeks
167.00154 .00915 (167*2π)/5493 weeks
168.00358 .00689 (168*2π)/5493 weeks
169.00175 -.00269 (169*2π)/5493 weeks
170.01453 .00925 (170*2π)/5493 weeks
171-.00017 .00261 (171*2π)/5493 weeks
172.0032 .00777 (172*2π)/5493 weeks
173.00635 .01044 (173*2π)/5493 weeks
174.00399 .00173 (174*2π)/5493 weeks
175.00956 .00062 (175*2π)/5493 weeks
176.0049 .00321 (176*2π)/5493 weeks
177.00336 .01252 (177*2π)/5493 weeks
178-.0001 -.0026 (178*2π)/5493 weeks
179.01291 .003 (179*2π)/5493 weeks
180.0012 .00109 (180*2π)/5493 weeks
181.01383 -.00527 (181*2π)/5493 weeks
182.0132 .00404 (182*2π)/5493 weeks
183.00716 -.00396 (183*2π)/5493 weeks
184.01046 -.00405 (184*2π)/5493 weeks
185.00817 -.0025 (185*2π)/5493 weeks
186.00724 .00649 (186*2π)/5493 weeks
187.00309 -.00562 (187*2π)/5493 weeks
188.01655 -.00625 (188*2π)/5493 weeks
189.00783 .01586 (189*2π)/5493 weeks
190.01625 -.00109 (190*2π)/5493 weeks
191.0218 .00466 (191*2π)/5493 weeks
192.00356 .01501 (192*2π)/5493 weeks
193.011 .01067 (193*2π)/5493 weeks
194.0013 .01426 (194*2π)/5493 weeks
195-.0119 .00677 (195*2π)/5493 weeks
196-.00845 -.00825 (196*2π)/5493 weeks
197.00616 -.01385 (197*2π)/5493 weeks
198.0067 -.00194 (198*2π)/5493 weeks
199.01737 -.00447 (199*2π)/5493 weeks
200.01346 .00985 (200*2π)/5493 weeks
201.01377 .00487 (201*2π)/5493 weeks
202.00977 .00824 (202*2π)/5493 weeks
203.00516 -.00294 (203*2π)/5493 weeks
204.01223 .00997 (204*2π)/5493 weeks
205-.00353 .00471 (205*2π)/5493 weeks
206.00769 .00033 (206*2π)/5493 weeks
207.00518 -.00104 (207*2π)/5493 weeks
208.00886 -.0118 (208*2π)/5493 weeks
209.02382 .00071 (209*2π)/5493 weeks
210.01706 .01272 (210*2π)/5493 weeks
211.01221 .01276 (211*2π)/5493 weeks
212.01419 .00442 (212*2π)/5493 weeks
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5471.04005 -.70561 (547*2π)/5491 weeks | 9,568 | 21,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-17 | latest | en | 0.782707 |
https://forum.allaboutcircuits.com/threads/thermal-cycler-circuit-interpretation.75036/ | 1,548,019,838,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583739170.35/warc/CC-MAIN-20190120204649-20190120230649-00416.warc.gz | 513,111,621 | 17,021 | # thermal cycler circuit interpretation
Discussion in 'The Projects Forum' started by justtrying, Sep 28, 2012.
1. ### justtrying Thread Starter Active Member
Mar 9, 2011
330
836
I am trying to brush up on my circuit reading skills. I am at a loss with four mosfets. I also have never seen "peltier", did some research on it, it is temperature controller which makes sense given the application.
I really can only go so far as if motor+ goes above 5V, U1 should go high and "close" 3904 (T1). If we assume that the other motor results in T2 being open, would current then flow through M1 because gate is now tied to ground via short? Then I think T4 and M4 would be active too and other transistors will be off. I really am quite lost...
I will try to go through this again but will appreciate any help.
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2. ### wayneh Expert
Sep 9, 2010
15,233
5,573
What are you trying to do?
That circuit appears to be an "H-bridge" motor controller (which allows for reversing the motor) adapted to a TEC (or peltier). It can send current thru the TEC in either direction, for heating or cooling, based on the control signal it receives.
3. ### justtrying Thread Starter Active Member
Mar 9, 2011
330
836
I am trying to understand figure out the current flow for different conditions. I've worked with H-bridge circuit once, but transistors are not my strong point. This circuit has p and n-channel mosfets and I want to be able to trace out current given different conditions (say as above) .
I am guessing that the current should flow, depending on conditions, through the peltier, meaning that pairs of mosfets that are on at the time time are M1/M4 and M2/M3? Were my assumptions above wrong?
This is to brush up on some weak points and to prepare for something that I have coming up...
4. ### wayneh Expert
Sep 9, 2010
15,233
5,573
T1 and T4 are on together (base voltage higher than emitter), as are T2 and T3 (on with base voltage below collector).
When T3 or T4 are ON, the gates of M3 and M4 go high and they conduct (in either direction).
T1 and T2 ON pulls the gates of M1 and M2 low, which causes them to conduct.
justtrying likes this. | 567 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-04 | latest | en | 0.953131 |
http://www.stata.com/statalist/archive/2005-02/msg00264.html | 1,495,675,435,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607960.64/warc/CC-MAIN-20170525010046-20170525030046-00474.warc.gz | 670,094,157 | 3,028 | # st: RE: Hannan-Quinn Criterion for nonstationary series
From "Scott Merryman" To Subject st: RE: Hannan-Quinn Criterion for nonstationary series Date Tue, 8 Feb 2005 18:03:39 -0600
```Sean,
I would think you would be able to use -varsoc- for each cross-section to
find the optimal lag length.
Below, is an example where -varsoc- is used to determine the optimal lag
length for each county and then these values are fed into -levinlin-.
Hope this helps,
Scott
use http://fmwww.bc.edu/ec-p/data/hayashi/sheston91.dta,clear
qui {
keep if country < 13
preserve
levels country, local(levels)
foreach l of local levels {
varsoc rgdp if country == `l'
matrix A = r(stats)
if `l' > 1 {
matrix C = C \ A
}
else {
matrix C = A
}
}
svmat C, name(col)
keep if HQ < .
egen id = fill(1 1 1 1 1 2 2 2 2 2)
egen minh = min(H), by(id)
gen optimal_lag = lag if minh == HQ
sort id opt
forv i = 1(5)60 {
local lag = opt[`i']
local opt_lag "`opt_lag' `" "' `lag'"
}
restore
}
levinlin rgdp, lag(`opt_lag')
> -----Original Message-----
> From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-
> statalist@hsphsun2.harvard.edu] On Behalf Of Sean MC allister
> Sent: Tuesday, February 08, 2005 4:35 AM
> To: statalist@hsphsun2.harvard.edu
> Subject: st: Hannan-Quinn Criterion for nonstationary series
>
> Hi
> I'm performing a convergence test on panel data based on the unit root
> test
> of Levin and Lin. I face the problem of selecting, for each country, the
> lag's length which eliminates residual autocorrelation. I wanted to use
> the
> Hannan-Quinn's criterion but the only command I found to get it is
> "varsoc"
> which is used for VAR. I worry because my sample is very small (T=11 and
> N=15). Is there a command that give this criterion for short nonstationary
> time series?
> Yours sincerely,
> Sean
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 610 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2017-22 | latest | en | 0.725794 |
https://au.mathworks.com/matlabcentral/cody/problems/1035-generate-a-vector-like-1-2-2-3-3-3-4-4-4-4/solutions/1573960 | 1,606,324,956,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141183514.25/warc/CC-MAIN-20201125154647-20201125184647-00516.warc.gz | 194,602,161 | 17,728 | Cody
# Problem 1035. Generate a vector like 1,2,2,3,3,3,4,4,4,4
Solution 1573960
Submitted on 3 Jul 2018 by noam gridish
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 2; y_correct = [1 2 2]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2
2 Pass
x = 5; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5
3 Pass
x = 10; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10
4 Pass
x = 12; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 66 11 11 11 11 11 11 11 11 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 78 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12
5 Pass
x = 9; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9
6 Pass
x = 7; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7
7 Pass
x = 15; y_correct = [1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ... 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 ... 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ... 11 11 11 11 11 11 11 11 11 11 11 ... 12 12 12 12 12 12 12 12 12 12 12 12 ... 13 13 13 13 13 13 13 13 13 13 13 13 13 ... 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ... 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15]; assert(isequal(your_fcn_name(x),y_correct))
ans = 1 ans = 1 2 2 ans = 1 2 2 3 3 3 ans = 1 2 2 3 3 3 4 4 4 4 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 ans = 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 36 8 8 8 8 8 8 8 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 45 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 55 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 66 11 11 11 11 11 11 11 11 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 78 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 91 13 13 13 13 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 105 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 ans = Columns 1 through 29 1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 Columns 30 through 58 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 10 10 10 10 10 10 10 10 10 10 11 11 11 Columns 59 through 87 11 11 11 11 11 11 11 11 12 12 12 12 12 12 12 12 12 12 12 12 13 13 13 13 13 13 13 13 13 Columns 88 through 116 13 13 13 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 15 15 15 15 15 15 15 15 15 15 15 Columns 117 through 120 15 15 15 15
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# BBTopic5_08 - MARGINALIST GROWTH THEORY I: SOLOW - SWAN...
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MAIN POINTS : Growth without technical progress in a marginalist framework: full-employment in the steady state Diminishing returns and the steady state The steady state growth rate independent of the saving propensity Distribution and growth in a marginalist model Harrod’s unemployment problem seen as a lack of flexibility in relative factor prices Technical progress in the Solow-Swan model: exogenous growth in steady state output per worker MARGINALIST GROWTH THEORY I: SOLOW - SWAN
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Assume constant returns to scale production technology using capital and labour: Growth without technical progress Assume α + β =1 => = 1 - . In per worker terms (dividing through by L) Ignore technical progress => in equilibrium output and the capital stock grow at the same rate, g e Ignoring physical depreciation, % K = I / K (with depreciation at the rate δ => % K = I / K - δ ) ( 29 L AK L K F Y = = , ( 29 - - - = = 1 1 L AK L L AK L Y L AK L AK = = - - 1 1 k A y . = ( 29 k f y = where k = K / L , or ........ (4.1) ........ (4.2) ........ (4.3)
In this model, g e = g n (i.e. % working age pop.) i.e. in equilibrium, output and the capital stock grow at the same rate as the labour force Writing for v v s g K I Y Y g n e = = = = y k Y L L K Y K v= = = . n g k y s = . = K Y Y S . ( 29 n g k k f s = . ( 29 s k g k f n . = => ........ (4.4) ........ (4.5) ........ (4.6) In turn, from equation (4.6) ( 29 n g k k f s = . n g K I K Y s = = . n g L K L Y s = . => => ...(4.7) Along the equilibrium path, S , I , Y and K all grow at the same rate - g n
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An example: assume g L = g n (i.e. labour employment grows at the same rate as the labour force – why this is, we will take up shortly) Suppose at a point in time K = 1000,
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## Conditional Reasoning Online with Mastermind
As mathematical patterns become more complex, students' conditional reasoning skills need to be nurtured so that students continue to critique, construct, and persevere in making sense of these complexities. This article describes a mathematical task designed around the online version of the game Mastermind to safely foster conditional reasoning.
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## Escape Room Math: Luna's Lines
An escape room can be a great way for students to apply and practice mathematics they have learned. This article describes the development and implementation of a mathematical escape room with important principles to incorporate in escape rooms to help students persevere in problem solving.
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## GPS: Equivalent Expressions
Equality is one of the main concepts in K–12 mathematics. Students should develop the understanding that equality is a relationship between two mathematical expressions. In this month's GPS, we share tasks asking students one main question: how do they know whether or not two mathematical expressions are equivalent?
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## Investigating Using Lesson Study
Our middle school mathematics department used lesson study to investigate how to introduce fractions division to our sixth-grade students. We highlight our learnings during the Study and Plan phases, describe our observations during the lesson, and provide tips for educators interested in using lesson study to study their own content.
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## Mathematizing Multilevel Marketing
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## PD For Your Earbuds: Making a Math Moments That Matter Podcast
Wondering how to create a classroom culture where students don't want to stop exploring mathematics when the bell rings? We were too and that's why we teammed up to uncover how we can Make Math Moments That Matter for every student in the math classroom with a weekly podcast.
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## Discourse Actions to Promote Student Access
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## Giving Birth to Inferential Reasoning
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## GPS: Spring Frogs
April 2020's GPS department provides tasks for each grade band that invite students to reason with age-appropriate number theoretic concepts. | 651 | 3,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-31 | latest | en | 0.914779 |
https://community.nxp.com/t5/Kinetis-Microcontrollers/How-to-use-the-Kinets-LTC-ECC-HW-to-accelerate-Curve25519/ta-p/1103375 | 1,701,803,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00460.warc.gz | 215,995,047 | 39,466 | # How to use the Kinets LTC ECC HW to accelerate Curve25519.
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# How to use the Kinets LTC ECC HW to accelerate Curve25519.
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## How to use the Kinets LTC ECC HW to accelerate Curve25519.
Curve22519 is a Montgomery elliptic-curve. Such as Apple HomeKit, most of network and IoT software use it
in Diffie-Hellman algorithm for key exchanging.
On the Security Kinets MCU chip,if we use just the software algorithm (base on mbedTLS), Curve25519 will spend
180ms for calculation of the shared security.
It is faster than other 256bit elliptic-curve with software algorithm, Because of the shared security
calculation will take more than 1200ms with a Weierstrass’s BP256R1curve when use software algorithm.
With LTC ECC HW acceleration, it take only 16ms to calculate the shared security on 256bit elliptic-curve.
Whatever you do, the speed of hardware acceleration always faster than the software algorithm.
Now that we should also want to use the LTC to accelerate the Curve22519. The LTC, however, only
supported Weierstrass form curve, but Curve22519 is a Montgomery curve…
Although, we can't use LTC in Curve22519 directly, we can use it by mapping it to a Weierstrass form
to use it. As below, we gave parameters of these curves, transform formulas, example code and test result
to show how and why to do it.
1. Curve parameter:
Cuvre22519 in Montgomery form:
Y^2 = X^3 + A*X^2 + X
Fp = 0x7fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffed
A= 486662
Gx = 9
Gy = 0x20ae19a1b8a086b4e01edd2c7748d14c923d4d7e6d7c61b229e9c5a27eced3d9
Order of G point = 0x1000000000000000000000000000000014def9dea2f79cd65812631a5cf5d3ed
Cuvre22519 in Weierstrass form :
Y^2 = X^3 + a*X + b
Fp = 0x7fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffed
a = 0x2aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa984914a144L
b = 0x7b425ed097b425ed097b425ed097b425ed097b425ed097b4260b5e9c7710c864L
Gy = 0x20ae19a1b8a086b4e01edd2c7748d14c923d4d7e6d7c61b229e9c5a27eced3d9
Order of G point = 0x1000000000000000000000000000000014def9dea2f79cd65812631a5cf5d3ed
2. Calculation formula:
x_w – x-coordinate value in Weierstrass form
y_w – y-coordinate value in Weierstrass form
x_m - x-coordinate
value in Montgomory form
y_m - we don’t care y-coordinate value in Weierstrass mode
a_m – a coefficient
of Montgomery equation ( Y^2 = X^3 + a_m * X^2 + X)
a_w – a coefficient
of Weierstrass equation ( Y^2 = X^3 + a*X + b )
b_w – a coefficient
of Weierstrass equation ( Y^2 = X^3 + a*X + b )
a) x_w = (x_m + a_m/3) % p
b) y_w ^2 = x_w ^ 3 + a_w*x_w + b_w
c) x_m = (x_w - a_m/3) % p
You could reference these document as below:
https://en.wikipedia.org/wiki/Curve25519
https://en.wikipedia.org/wiki/Montgomery_curve
3. example code:
// public and private at Montgomery end
#define M255_Qx "0x3BA5048381744348D84E754B9944ABE080B37F7D4158DCE60CD79F66B98AB89E"
// public and private at Weierstrass end
#define WTS255_d "0x09CC5CCF43C656C1309EE5A3491D5A8361607CEEB0C9B2B31A575E0FEF2B8835"
#define WTS255_Qx "0x3F4BDE110EE7AF71EF428D1018D188E35BAFB019F34F84E6465C5194B363DC2D"
#define WTS255_Qy "0x7540577CE6F920354E2A9D38CE88847D7447E66FA4D188AC75CB63C17210B718"
#define WTS255_Qx_TO_M255_Qx "0x14A13366643D04C74497E2656E26DE38B105056F48A4DA3B9BB1A6EA08B6B7DC"
int ecdh_wts_curve_end( )
{
unsigned int ticks;
int ret = 0;
size_t blen = 0, blen_peer = 0;
ecdh_context ecdh;
ecdh_context ecdh_peer; // to_wts255
ecdh_context ecdh_peer_m255;
mpi R;
mpi_init(&R);
ecdh_init( &ecdh);
ecdh_init( &ecdh_peer);
ecdh_init( &ecdh_peer_m255);
MPI_CHK(ecp_use_known_dp( &ecdh.grp, ECP_DP_WTS25519 ));
MPI_CHK(ecp_use_known_dp( &ecdh_peer.grp, ECP_DP_WTS25519 ));
MPI_CHK(ecp_use_known_dp( &ecdh_peer_m255.grp, ECP_DP_M255 ));
blen = set_hash_buff(/*TEST_ECP_GRP_ID*/ECP_DP_WTS25519, &secret_buf, ecp_name);
if(blen == 0) {
ret = -1;
goto cleanup;
}
mpi_lset(&ecdh.Q.Z, 1);
mpi_init(&ecdh_peer_m255.Q.Y);
mpi_lset(&ecdh_peer_m255.Q.Z, 1);
// map M255 point to WTS255 point
my_timer_start();
mpi_mod_mpi(&ecdh_peer.Q.X, &ecdh_peer.Q.X, &ecdh_peer_m255.grp.P);
mpi_lset(&R, 3);
mpi_exp_mod (&ecdh_peer_m255.Q.Y , &ecdh_peer.Q.X, &R, &ecdh_peer_m255.grp.P, NULL);
mpi_mul_mpi(&R, &ecdh_peer.grp.A, &ecdh_peer.Q.X);
mpi_mod_mpi(&R, &R, &ecdh_peer.grp.P);
mpi_mod_mpi(&ecdh_peer_m255.Q.Y, &ecdh_peer_m255.Q.Y, &ecdh_peer.grp.P);
mpi_mod_sqrt(&ecdh_peer.Q.Y, &ecdh_peer_m255.Q.Y, &ecdh_peer_m255.grp.P);
// z = 1
mpi_lset(&ecdh_peer.Q.Z, 1);
MPI_CHK(ecp_copy(&ecdh.Qp, &ecdh_peer.Q));
MPI_CHK(ecdh_calc_secret_wts2mont( &ecdh, &blen, secret_buf, blen, myrand, NULL));
mpi_sub_mpi(&ecdh_peer_m255.Q.X, &ecdh.Q.X, &R);
mpi_mod_mpi(&ecdh_peer_m255.Q.X, &ecdh_peer_m255.Q.X, &ecdh_peer_m255.grp.P);
ticks = my_timer_stop();
// print out message
mpi_printf_string( &ecdh.z, 16);
polarssl_printf("%s ecdh peer to peer: %lu ticks, %d ms (%d) \n", ecp_name , ticks, ticks / (CLOCK_SYS_GetPitFreq(0) / 1000),CLOCK_SYS_GetPitFreq(0) );
cleanup:
if( ret !=0 )
polarssl_printf( "%s test Unexpected error, return code = %08X\n", ecp_name, ret );
mpi_free(&R);
ecdh_free( &ecdh);
ecdh_free( &ecdh_peer);
ecdh_free( &ecdh_peer_m255);
return( 0 );
}
int ecdh_mont_curve_end( )
{
int verbose = 1;
unsigned int ticks;
int ret = 0;
size_t blen = 0, blen_peer = 0;
ecdh_context ecdh;
ecp_point Q_peer; // peer public point
ecdh_init( &ecdh);
ecp_point_init( &Q_peer);
MPI_CHK(ecp_use_known_dp( &ecdh.grp, ECP_DP_M255 ));
blen_peer = set_hash_buff(ECP_DP_M255, &secret_buf_peer, ecp_name);
if(blen_peer == 0) {
ret = -1;
goto cleanup;
}
mpi_init(&ecdh.Q.Y); // don't care Y, only init it
mpi_lset(&ecdh.Q.Z, 1);
mpi_init(&Q_peer.Y);
mpi_lset(&Q_peer.Z, 1);
MPI_CHK(ecp_copy(&ecdh.Qp, &Q_peer));
my_timer_start();
MPI_CHK(ecdh_calc_secret( &ecdh, &blen_peer, secret_buf_peer, blen_peer, myrand, NULL));
ticks = my_timer_stop();
polarssl_printf("%s ecdh peer to peer: %lu ticks, %d ms (%d) \n", ecp_name , ticks, ticks / (CLOCK_SYS_GetPitFreq(0) / 1000),CLOCK_SYS_GetPitFreq(0) );
mpi_printf_string( &ecdh.z, 16);
cleanup:
if( ret !=0 && verbose != 0 )
polarssl_printf( "%s test Unexpected error, return code = %08X\n", ecp_name, ret );
ecdh_free( &ecdh);
ecp_point_free( &Q_peer);
if( verbose != 0 )
return( 0 );
}
4. Test result:
1. Test result of curv25519 in Weierstrass form with LTC:
2. Test result of curve25519 in Montgomery form with software algorithm:
We could see that the shared security both in Weierstrass form with LTC and Montgomery form are “0x1454BDCD6A94D6336AA5A76F3CB40BBE12B65A2CDC9DA6B478948906638896D1”.
But the calculation speed with LTC was ten times faster than other one.
Labels (1)
• ### Kinetis K Series MCUs
Version history
Last update:
09-10-2020 01:48 AM
Updated by: | 3,075 | 7,497 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-50 | latest | en | 0.790753 |
https://samacheerkalvi.guru/samacheer-kalvi-9th-maths-chapter-2-ex-2-6/ | 1,709,586,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00085.warc.gz | 512,295,409 | 24,708 | # Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.6
## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.6
Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5$$\sqrt { 3 }$$ + 18$$\sqrt { 3 }$$ – 2$$\sqrt{3}$$
(ii) 4$$\sqrt [ 3 ]{ 5 }$$ + 2$$\sqrt [ 2 ]{ 5 }$$ – 3$$\sqrt [ 3 ]{ 5 }$$
(iii) 3$$\sqrt { 75 }$$ + 5$$\sqrt { 48 }$$ – $$\sqrt { 243 }$$
(iv) 5$$\sqrt [ 3 ]{ 40 }$$ + 2$$\sqrt [ 3 ]{ 625 }$$ – 3$$\sqrt [ 3 ]{ 320 }$$
Solution:
(i) 5$$\sqrt { 3 }$$ + 18$$\sqrt { 3 }$$ – 2$$\sqrt { 3 }$$ = (5 + 18 – 2)$$\sqrt { 3 }$$ = 21$$\sqrt { 3 }$$
(ii) 4$$\sqrt [ 3 ]{ 5 }$$ + 2$$\sqrt [ 2 ]{ 5 }$$ – 3$$\sqrt [ 3 ]{ 5 }$$ = (4 + 2 – 3)$$\sqrt [ 3 ]{ 5 }$$ = 3$$\sqrt [ 3 ]{ 5 }$$
(iii) 3$$\sqrt { 75 }$$ + 5$$\sqrt { 48 }$$ – $$\sqrt { 243 }$$
(iv) 5$$\sqrt [ 3 ]{ 40 }$$ + 2$$\sqrt [ 3 ]{ 625 }$$ – 3$$\sqrt [ 3 ]{ 320 }$$
Question 2.
Simplify the following using multiplication and division properties of surds :
(i) $$\sqrt{3} \times \sqrt{5} \times \sqrt{2}$$
(ii) $$\sqrt { 35 }$$ ÷ $$\sqrt { 7 }$$
(iii) $$\sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125}$$
(iv) (7$$\sqrt { a }$$ – 5$$\sqrt { b }$$)(7$$\sqrt { a }$$ + 5$$\sqrt { b }$$)
(v) $$[\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}] \div \sqrt{\frac{16}{81}}$$
Solution:
Question 3.
If $$\sqrt { 7 }$$ =1.414, $$\sqrt { 3 }$$ = 1.732, $$\sqrt { 5 }$$ = 2.236, $$\sqrt { 10 }$$ = 3.162, then find the values of the following correct to 3 places of decimals.
(i) $$\sqrt{40}-\sqrt{20}$$
(ii) $$\sqrt{300}+\sqrt{90}-\sqrt{8}$$
Solution:
Question 4.
Arrange surds in descending order :
(i) $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$
(ii) $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$
Solution:
(i) $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$
5$$\frac { 1 }{ 3 }$$
∴ The order of the surds $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$ are 3, 9, 6.
4$$\frac { 1 }{ 9 }$$
3$$\frac { 1 }{ 6 }$$ l.c.m of 3, 9, 6 is 18
∴ The descending order of $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$ is $$(15625)^{\frac{1}{18}}>(27)^ {\frac{1}{18}} >16^{\frac{1}{18}} \text { i.e. } \sqrt[3]{5}>\sqrt[6]{3}>\sqrt[9]{4}$$
(ii) $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$
The order of the surds $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$ are 6, 12, 4
l.c.m of 6, 12, 4 is 12
Question 5.
Can you get a pure surd when you find
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
Question 6.
Can you get a rational number when you compute
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes (5 – $$\sqrt{3}$$) + (5 + $$\sqrt{3}$$) = 10, a rational number
(ii) Yes $$(5+\sqrt[3]{7})-(-6+\sqrt[3]{7})=11$$, a rational number
(iii) Yes (5 + $$\sqrt{3}$$ ) (5 – $$\sqrt{3}$$ ) = 25 – 3 = 22, a rational number
(iv) Yes $$\frac{5 \sqrt{3}}{\sqrt{3}}=5$$ ,a rational number | 1,386 | 3,097 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-10 | longest | en | 0.302111 |
http://mathhelpforum.com/calculus/133263-solved-local-min-max.html | 1,521,313,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645280.4/warc/CC-MAIN-20180317174935-20180317194935-00315.warc.gz | 192,278,120 | 11,380 | # Thread: [SOLVED] local min and max
1. ## [SOLVED] local min and max
let f(x)= (x-1)^(4/5) x find the local minimum and maximum
2. Originally Posted by cole123
let f(x)= (x-1)^(4/5) x find the local minimum and maximum
$\displaystyle f(x) = (x - 1)^{\frac{4}{5}}x$
To find local maxima and minima, you need to find the derivative, set it equal to 0 and solve for $\displaystyle x$.
So $\displaystyle f'(x) = (x - 1)^{\frac{4}{5}}\frac{d}{dx}(x) + x\frac{d}{dx}[(x - 1)^{\frac{4}{5}}$
$\displaystyle = (x - 1)^{\frac{4}{5}} + \frac{4}{5}x(x - 1)^{-\frac{1}{5}}$.
Can you go from here?
3. I am on the next step :
I have (4/5)x + (x-1)= 0
I know there is both a min and a max just dont know what to do from here
4. Originally Posted by cole123
I am on the next step :
I have (4/5)x + (x-1)= 0
I know there is both a min and a max just dont know what to do from here
No.
It's $\displaystyle (x - 1)^{\frac{4}{5}} + \frac{4}{5}x(x - 1)^{-\frac{1}{5}} = 0$.
5. I guess I dont then!!
6. Can you help me?
7. Prove It, cole123 was heading in the right direction. After you multiply $\displaystyle (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0$ by $\displaystyle (x-1)^{1}{5}$, you are left with $\displaystyle (x-1)+ \frac{4}{5}x= 0$.
cole123, solve that: $\displaystyle x+ \frac{4}{5}x= 1$.
Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.
8. Originally Posted by cole123
I am on the next step :
I have (4/5)x + (x-1)= 0
I know there is both a min and a max just dont know what to do from here
I did that and someone said it was wrong
9. so is x = 5/9 my only answer? Ive only answered max/min for parabolas. and I believe there could be another answer?!
10. Originally Posted by HallsofIvy
Prove It, cole123 was heading in the right direction. After you multiply $\displaystyle (x-1)^{\frac{4}{5}}+ \frac{4}{5}x(x-1)^{-\frac{1}{5}}= 0$ by $\displaystyle (x-1)^{1}{5}$, you are left with $\displaystyle (x-1)+ \frac{4}{5}x= 0$.
cole123, solve that: $\displaystyle x+ \frac{4}{5}x= 1$.
Remember that the local max and min must be at a point where the derivative is 0, or where there is NO derivative.
OK point taken.
Except when you multiply both sides by $\displaystyle (x - 1)^{\frac{1}{5}}$ the equation actually becomes
$\displaystyle (x - 1)^4 + \frac{4}{5}x = 0$
$\displaystyle x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0$.
Now simplify and solve for $\displaystyle x$.
11. Well, whoever told you that was wrong!
I suspect they graphed that on a calculator and saw no graph for x< 1.
Many perfectly good calculators (like my TI 89) will not handle odd roots of negative numbers correctly because they do them using a logarithm which is not defined for $\displaystyle x\le 0$.
If you graph y1= (x-1)^(4/5)*x on a TI calculator, for example, you will see a graph that begins at (1, 0) and rises to the right.
If you also graph y2= (1- x)^(4/5)*x, you will see the rest of the graph which has maximum at x= 5/9.
12. Originally Posted by Prove It
OK point taken.
Except when you multiply both sides by $\displaystyle (x - 1)^{\frac{1}{5}}$ the equation actually becomes
$\displaystyle (x - 1)^4 + \frac{4}{5}x = 0$
$\displaystyle x^4 - 4x^3 + 6x^2 - 4x + 1 + \frac{4}{5}x = 0$.
Now simplify and solve for $\displaystyle x$.
No! $\displaystyle (x- 1)^{\frac{4}{5}}(x-1)^{\frac{1}{5}}= (x- 1)^1$, not $\displaystyle (x- 1)^4$. | 1,216 | 3,438 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-13 | latest | en | 0.887419 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/462/2/p/b/ | 1,632,688,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00221.warc.gz | 872,079,598 | 63,471 | # Properties
Label 462.2.p.b Level $462$ Weight $2$ Character orbit 462.p Analytic conductor $3.689$ Analytic rank $0$ Dimension $16$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$462 = 2 \cdot 3 \cdot 7 \cdot 11$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 462.p (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$3.68908857338$$ Analytic rank: $$0$$ Dimension: $$16$$ Relative dimension: $$8$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\mathbb{Q}[x]/(x^{16} - \cdots)$$ Defining polynomial: $$x^{16} - 8 x^{15} + 74 x^{14} - 378 x^{13} + 1878 x^{12} - 6718 x^{11} + 22086 x^{10} - 56904 x^{9} + 130215 x^{8} - 239606 x^{7} + 378750 x^{6} - 477124 x^{5} + 493030 x^{4} - 386266 x^{3} + 223844 x^{2} - 82874 x + 13417$$ Coefficient ring: $$\Z[a_1, \ldots, a_{7}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{15}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q -\beta_{12} q^{2} + \beta_{11} q^{3} + ( 1 + \beta_{13} ) q^{4} + ( 1 + \beta_{9} + \beta_{13} ) q^{5} + q^{6} + ( 1 + \beta_{1} + \beta_{8} + \beta_{11} + \beta_{13} - \beta_{15} ) q^{7} + ( -\beta_{11} - \beta_{12} ) q^{8} -\beta_{13} q^{9} +O(q^{10})$$ $$q -\beta_{12} q^{2} + \beta_{11} q^{3} + ( 1 + \beta_{13} ) q^{4} + ( 1 + \beta_{9} + \beta_{13} ) q^{5} + q^{6} + ( 1 + \beta_{1} + \beta_{8} + \beta_{11} + \beta_{13} - \beta_{15} ) q^{7} + ( -\beta_{11} - \beta_{12} ) q^{8} -\beta_{13} q^{9} + ( \beta_{7} - \beta_{11} - \beta_{12} ) q^{10} + ( -1 - \beta_{1} - \beta_{2} - \beta_{5} + \beta_{10} + \beta_{11} + \beta_{12} ) q^{11} -\beta_{12} q^{12} + ( -1 - \beta_{5} - \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} + \beta_{10} - \beta_{13} + \beta_{14} + \beta_{15} ) q^{13} + ( 1 - \beta_{3} - \beta_{4} + \beta_{5} - \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{12} - \beta_{14} ) q^{14} + ( -\beta_{10} - \beta_{12} ) q^{15} + \beta_{13} q^{16} + ( -1 + \beta_{3} + \beta_{6} - \beta_{7} + \beta_{8} - \beta_{9} - \beta_{13} + \beta_{14} - \beta_{15} ) q^{17} + \beta_{11} q^{18} + ( -1 - \beta_{1} - \beta_{4} - 2 \beta_{8} + \beta_{9} - 2 \beta_{11} - \beta_{12} - 2 \beta_{13} ) q^{19} + ( -\beta_{8} + \beta_{9} + \beta_{13} ) q^{20} + ( -\beta_{3} - \beta_{4} + \beta_{5} - \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{12} - \beta_{13} ) q^{21} + ( -\beta_{1} - \beta_{3} - \beta_{4} + \beta_{5} + \beta_{6} - \beta_{7} - \beta_{8} - \beta_{10} - \beta_{13} ) q^{22} + ( 2 \beta_{1} + \beta_{3} + 3 \beta_{4} - 2 \beta_{5} - \beta_{6} + 2 \beta_{7} + \beta_{8} - \beta_{9} + 2 \beta_{10} + \beta_{12} + 2 \beta_{13} + \beta_{15} ) q^{23} + ( 1 + \beta_{13} ) q^{24} + ( \beta_{1} + \beta_{2} + \beta_{3} + \beta_{5} + 2 \beta_{6} - \beta_{7} + 2 \beta_{9} - \beta_{12} + \beta_{13} - \beta_{15} ) q^{25} + ( -\beta_{1} - \beta_{2} + \beta_{3} - \beta_{5} - \beta_{6} - \beta_{9} + \beta_{10} + \beta_{12} + \beta_{14} ) q^{26} + ( \beta_{11} + \beta_{12} ) q^{27} + ( 1 + \beta_{1} + \beta_{2} + \beta_{8} - \beta_{12} + \beta_{13} - \beta_{15} ) q^{28} + ( 1 + \beta_{1} + \beta_{2} + \beta_{3} + 2 \beta_{4} - 2 \beta_{5} - 2 \beta_{6} + \beta_{7} + 3 \beta_{8} - 3 \beta_{9} + \beta_{12} - \beta_{13} + \beta_{14} - \beta_{15} ) q^{29} + ( 1 + \beta_{9} + \beta_{13} ) q^{30} + ( 1 - \beta_{3} - \beta_{4} + \beta_{5} + \beta_{6} - 2 \beta_{7} + \beta_{9} - 2 \beta_{10} + \beta_{11} - \beta_{12} - \beta_{14} - \beta_{15} ) q^{31} -\beta_{11} q^{32} + ( -1 - \beta_{1} - \beta_{4} - \beta_{5} - \beta_{8} - \beta_{11} - \beta_{13} + \beta_{14} ) q^{33} + ( -\beta_{2} - \beta_{3} - \beta_{7} + 2 \beta_{11} + \beta_{12} - \beta_{14} + \beta_{15} ) q^{34} + ( 2 + 4 \beta_{1} + \beta_{2} + 2 \beta_{4} + 3 \beta_{8} + \beta_{9} - \beta_{12} + 2 \beta_{13} - \beta_{14} - \beta_{15} ) q^{35} + q^{36} + ( 1 + \beta_{1} + 2 \beta_{2} + 2 \beta_{3} + 3 \beta_{4} - 2 \beta_{5} - 2 \beta_{6} + 2 \beta_{7} + 2 \beta_{8} - 3 \beta_{9} - 4 \beta_{11} - \beta_{12} - 2 \beta_{13} + 2 \beta_{14} ) q^{37} + ( -1 - \beta_{5} + \beta_{7} + 2 \beta_{10} + 2 \beta_{11} + \beta_{12} + \beta_{13} ) q^{38} + ( -1 - \beta_{1} - \beta_{2} + \beta_{3} + \beta_{7} - \beta_{8} + 2 \beta_{10} + \beta_{12} - \beta_{13} + \beta_{15} ) q^{39} + ( \beta_{7} + \beta_{10} - \beta_{11} ) q^{40} + ( -3 + \beta_{1} - \beta_{2} + \beta_{3} + 2 \beta_{6} - \beta_{7} + \beta_{8} + \beta_{9} + \beta_{11} + \beta_{13} - \beta_{14} - \beta_{15} ) q^{41} + ( 1 + \beta_{1} + \beta_{8} + \beta_{11} + \beta_{13} - \beta_{15} ) q^{42} + ( 1 + \beta_{1} + 2 \beta_{4} - \beta_{8} + \beta_{9} - \beta_{11} - \beta_{12} + 3 \beta_{13} ) q^{43} + ( \beta_{1} + \beta_{4} + \beta_{6} - \beta_{7} + \beta_{8} + \beta_{11} - \beta_{15} ) q^{44} + ( 1 + \beta_{8} ) q^{45} + ( -1 - 2 \beta_{1} + \beta_{3} + \beta_{5} - \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} - 3 \beta_{11} - \beta_{13} + \beta_{14} + \beta_{15} ) q^{46} + ( 3 \beta_{1} + \beta_{2} + 2 \beta_{3} + \beta_{4} + \beta_{5} + 3 \beta_{8} + 2 \beta_{9} - \beta_{10} + 2 \beta_{13} - \beta_{14} - 3 \beta_{15} ) q^{47} + ( -\beta_{11} - \beta_{12} ) q^{48} + ( 1 - \beta_{1} - 2 \beta_{3} - 2 \beta_{4} - \beta_{5} - 2 \beta_{6} + \beta_{7} - \beta_{10} + 2 \beta_{11} + \beta_{15} ) q^{49} + ( 1 + \beta_{1} + 2 \beta_{4} - \beta_{5} - \beta_{6} + 3 \beta_{7} + \beta_{8} - \beta_{9} + 2 \beta_{10} + \beta_{12} + \beta_{13} - \beta_{14} + \beta_{15} ) q^{50} + ( -\beta_{1} - \beta_{3} - \beta_{6} - \beta_{8} - \beta_{9} + \beta_{12} - \beta_{13} + \beta_{15} ) q^{51} + ( -1 - \beta_{1} - \beta_{2} - \beta_{3} - 2 \beta_{4} - \beta_{7} - \beta_{8} - \beta_{12} - \beta_{13} + \beta_{15} ) q^{52} + ( -2 + \beta_{3} + \beta_{4} + \beta_{5} + 3 \beta_{6} - 3 \beta_{7} + \beta_{8} - \beta_{9} + 2 \beta_{11} + 2 \beta_{12} - 2 \beta_{13} + \beta_{14} - \beta_{15} ) q^{53} -\beta_{13} q^{54} + ( -2 \beta_{2} + \beta_{3} + \beta_{4} - 2 \beta_{5} + \beta_{7} + 3 \beta_{8} - 2 \beta_{9} + 2 \beta_{10} + \beta_{11} + 4 \beta_{12} + \beta_{13} + 2 \beta_{14} ) q^{55} + ( 1 + \beta_{13} - \beta_{14} ) q^{56} + ( 1 - \beta_{5} - \beta_{6} + 2 \beta_{7} + \beta_{10} + \beta_{11} + 2 \beta_{12} + 2 \beta_{13} ) q^{57} + ( -1 - 2 \beta_{1} - \beta_{2} - 2 \beta_{4} + \beta_{5} + \beta_{6} - 2 \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} + 2 \beta_{11} - \beta_{13} - \beta_{14} + \beta_{15} ) q^{58} + ( -4 - 3 \beta_{1} - 3 \beta_{2} + 2 \beta_{3} - \beta_{4} - 2 \beta_{5} + \beta_{6} + \beta_{7} - 2 \beta_{8} + \beta_{9} + 4 \beta_{10} - 2 \beta_{11} + 2 \beta_{12} - \beta_{14} + 2 \beta_{15} ) q^{59} + ( \beta_{7} - \beta_{11} - \beta_{12} ) q^{60} + ( -\beta_{3} - \beta_{4} - 3 \beta_{7} - \beta_{8} + \beta_{9} - 3 \beta_{10} + 3 \beta_{11} + 3 \beta_{13} - \beta_{15} ) q^{61} + ( 2 + \beta_{1} + \beta_{2} - \beta_{3} + \beta_{5} + \beta_{6} - \beta_{7} + \beta_{8} + \beta_{9} - \beta_{10} + \beta_{11} - \beta_{12} + \beta_{13} - \beta_{14} - \beta_{15} ) q^{62} + ( -\beta_{2} + \beta_{11} + \beta_{12} ) q^{63} - q^{64} + ( -8 - 2 \beta_{1} - \beta_{2} - \beta_{3} - \beta_{4} - \beta_{5} - 2 \beta_{8} - \beta_{9} + \beta_{10} + 2 \beta_{12} - 5 \beta_{13} + \beta_{14} + 2 \beta_{15} ) q^{65} + ( -1 - \beta_{1} - \beta_{2} - \beta_{5} + \beta_{10} + \beta_{11} + \beta_{12} ) q^{66} + ( -4 + \beta_{3} + \beta_{5} + 3 \beta_{6} - 2 \beta_{7} + \beta_{8} - \beta_{9} - 2 \beta_{11} - 5 \beta_{12} - 4 \beta_{13} + \beta_{14} - \beta_{15} ) q^{67} + ( 1 + \beta_{2} + \beta_{5} - \beta_{7} + \beta_{8} - \beta_{9} - 2 \beta_{10} - \beta_{11} - \beta_{12} - \beta_{13} + \beta_{14} - \beta_{15} ) q^{68} + ( \beta_{2} + \beta_{3} + 2 \beta_{5} + 2 \beta_{6} - 2 \beta_{11} - 2 \beta_{12} ) q^{69} + ( 1 + \beta_{2} + 2 \beta_{5} - \beta_{6} + \beta_{7} - 2 \beta_{10} - \beta_{11} - \beta_{12} + \beta_{13} - \beta_{14} ) q^{70} + ( -2 + 2 \beta_{2} - 2 \beta_{3} - 2 \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} - 3 \beta_{11} + 2 \beta_{12} - \beta_{13} + \beta_{14} + \beta_{15} ) q^{71} -\beta_{12} q^{72} + ( 6 + 2 \beta_{1} + 2 \beta_{2} + 2 \beta_{5} + 2 \beta_{6} + 2 \beta_{9} + 2 \beta_{11} + 2 \beta_{12} + 6 \beta_{13} - 2 \beta_{14} ) q^{73} + ( -3 - 2 \beta_{1} - 2 \beta_{2} + 2 \beta_{3} - \beta_{5} + \beta_{7} + 2 \beta_{10} + 2 \beta_{11} + \beta_{12} + \beta_{13} + 2 \beta_{15} ) q^{74} + ( 2 + \beta_{2} - \beta_{3} + \beta_{4} + \beta_{5} - \beta_{6} + \beta_{7} - \beta_{9} - 2 \beta_{10} - \beta_{11} - \beta_{12} - \beta_{14} ) q^{75} + ( 1 - \beta_{1} - \beta_{8} - \beta_{9} - \beta_{11} + \beta_{12} - \beta_{13} ) q^{76} + ( -3 + \beta_{1} + \beta_{2} - 2 \beta_{3} + \beta_{4} - \beta_{5} - 2 \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} - \beta_{10} - \beta_{11} + 4 \beta_{12} - 3 \beta_{13} ) q^{77} + ( -1 - \beta_{5} - \beta_{6} + \beta_{7} - \beta_{8} - \beta_{9} + \beta_{10} - \beta_{13} + \beta_{14} + \beta_{15} ) q^{78} + ( 8 + 2 \beta_{2} - 2 \beta_{3} + 2 \beta_{4} + 2 \beta_{5} - 2 \beta_{6} + \beta_{7} - 2 \beta_{9} - 3 \beta_{10} - \beta_{11} + 2 \beta_{13} - 2 \beta_{14} ) q^{79} + ( -1 - \beta_{8} ) q^{80} + ( -1 - \beta_{13} ) q^{81} + ( 1 + \beta_{2} - 2 \beta_{3} + \beta_{5} - \beta_{6} - \beta_{8} + \beta_{9} - \beta_{10} + 2 \beta_{12} - \beta_{14} + \beta_{15} ) q^{82} + ( 1 - 2 \beta_{1} + \beta_{5} - \beta_{6} - \beta_{10} - 2 \beta_{11} + \beta_{12} ) q^{83} + ( 1 - \beta_{3} - \beta_{4} + \beta_{5} - \beta_{7} - \beta_{8} + \beta_{9} - \beta_{10} - \beta_{12} - \beta_{14} ) q^{84} + ( -4 - \beta_{1} - \beta_{2} - \beta_{3} - 2 \beta_{4} + \beta_{5} + \beta_{6} - 5 \beta_{7} - 2 \beta_{8} + 2 \beta_{9} - 3 \beta_{10} + 3 \beta_{11} + \beta_{12} - 6 \beta_{13} + \beta_{14} - \beta_{15} ) q^{85} + ( 2 \beta_{5} + \beta_{6} + \beta_{7} + \beta_{10} - 3 \beta_{11} - \beta_{12} + \beta_{13} ) q^{86} + ( -1 - \beta_{3} - 2 \beta_{4} + \beta_{5} + \beta_{6} - 2 \beta_{7} - \beta_{8} + \beta_{9} + 2 \beta_{11} + \beta_{12} - \beta_{13} - \beta_{14} + \beta_{15} ) q^{87} + ( 1 - \beta_{3} + 2 \beta_{5} + \beta_{6} - \beta_{7} - \beta_{10} + \beta_{11} - \beta_{14} ) q^{88} + ( -2 - 2 \beta_{3} - 2 \beta_{4} - 4 \beta_{6} - 2 \beta_{8} - 2 \beta_{13} + 2 \beta_{15} ) q^{89} + ( -\beta_{10} - \beta_{12} ) q^{90} + ( -6 - 4 \beta_{1} - 3 \beta_{2} + \beta_{3} - 5 \beta_{4} - \beta_{5} - 4 \beta_{8} + \beta_{9} + 3 \beta_{10} - 4 \beta_{11} - 4 \beta_{12} - 3 \beta_{13} + \beta_{14} + 2 \beta_{15} ) q^{91} + ( -3 + \beta_{1} - \beta_{2} + \beta_{3} - 2 \beta_{5} + \beta_{7} + 2 \beta_{10} + \beta_{12} + \beta_{14} + \beta_{15} ) q^{92} + ( 1 - \beta_{3} - \beta_{4} - \beta_{7} + \beta_{8} - \beta_{10} + \beta_{11} - \beta_{15} ) q^{93} + ( \beta_{1} + \beta_{2} - 2 \beta_{3} - 2 \beta_{4} + \beta_{5} - \beta_{6} + 2 \beta_{7} - 2 \beta_{8} + 3 \beta_{9} + \beta_{11} + \beta_{12} - 3 \beta_{14} + 2 \beta_{15} ) q^{94} + ( -4 - 4 \beta_{1} + 2 \beta_{3} - 2 \beta_{6} + 2 \beta_{7} - 2 \beta_{8} - 2 \beta_{9} - 6 \beta_{11} + 2 \beta_{14} + 2 \beta_{15} ) q^{95} + \beta_{13} q^{96} + ( -5 - \beta_{1} - 2 \beta_{2} - 2 \beta_{3} - 2 \beta_{4} + 2 \beta_{5} + 2 \beta_{6} + \beta_{7} - 3 \beta_{8} + 3 \beta_{9} + \beta_{10} - \beta_{11} - \beta_{12} - 7 \beta_{13} - \beta_{14} + \beta_{15} ) q^{97} + ( 2 - \beta_{1} + \beta_{3} - \beta_{4} - \beta_{5} - \beta_{7} + \beta_{9} - 2 \beta_{10} - \beta_{11} - \beta_{12} + \beta_{14} - 2 \beta_{15} ) q^{98} + ( -1 - 2 \beta_{1} - \beta_{2} - \beta_{4} - \beta_{5} - \beta_{6} + \beta_{7} - \beta_{8} + \beta_{10} + \beta_{12} + \beta_{15} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$16q + 8q^{4} + 12q^{5} + 16q^{6} + 6q^{7} + 8q^{9} + O(q^{10})$$ $$16q + 8q^{4} + 12q^{5} + 16q^{6} + 6q^{7} + 8q^{9} - 2q^{10} - 4q^{11} + 8q^{14} - 4q^{15} - 8q^{16} + 10q^{19} + 4q^{21} + 2q^{22} - 4q^{23} + 8q^{24} + 10q^{25} + 12q^{26} + 12q^{30} + 6q^{31} + 4q^{33} + 8q^{35} + 16q^{36} + 14q^{37} - 12q^{38} + 12q^{39} + 2q^{40} - 32q^{41} + 6q^{42} + 4q^{44} + 12q^{45} - 18q^{46} - 24q^{47} - 6q^{49} - 6q^{51} + 8q^{54} + 14q^{55} + 4q^{56} - 2q^{60} - 28q^{61} + 8q^{62} + 6q^{63} - 16q^{64} - 72q^{65} - 4q^{66} - 16q^{67} - 30q^{70} - 56q^{71} + 44q^{73} - 24q^{74} - 12q^{75} + 20q^{76} - 52q^{77} + 30q^{79} - 12q^{80} - 8q^{81} - 12q^{82} - 8q^{83} + 8q^{84} - 12q^{86} - 2q^{88} - 36q^{89} - 4q^{90} - 8q^{91} - 8q^{92} + 4q^{93} - 14q^{94} - 72q^{95} - 8q^{96} + 40q^{98} - 8q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{16} - 8 x^{15} + 74 x^{14} - 378 x^{13} + 1878 x^{12} - 6718 x^{11} + 22086 x^{10} - 56904 x^{9} + 130215 x^{8} - 239606 x^{7} + 378750 x^{6} - 477124 x^{5} + 493030 x^{4} - 386266 x^{3} + 223844 x^{2} - 82874 x + 13417$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$($$$$-3325577 \nu^{14} + 23279039 \nu^{13} - 218220364 \nu^{12} + 1006694677 \nu^{11} - 4890679917 \nu^{10} + 15780182142 \nu^{9} - 48699356733 \nu^{8} + 110238860505 \nu^{7} - 222125567883 \nu^{6} + 336742486204 \nu^{5} - 415275809861 \nu^{4} + 371398223899 \nu^{3} - 230411582664 \nu^{2} + 86434816533 \nu - 13304183911$$$$)/ 3261539424$$ $$\beta_{2}$$ $$=$$ $$($$$$14496748 \nu^{15} + 165765669 \nu^{14} - 1387066295 \nu^{13} + 16433893584 \nu^{12} - 89388557825 \nu^{11} + 460987458741 \nu^{10} - 1704964101954 \nu^{9} + 5511712419345 \nu^{8} - 14215412394141 \nu^{7} + 30601359218347 \nu^{6} - 53771673709840 \nu^{5} + 74664957471809 \nu^{4} - 81879734308403 \nu^{3} + 64273622949996 \nu^{2} - 34535277031465 \nu + 8734585660679$$$$)/ 316369324128$$ $$\beta_{3}$$ $$=$$ $$($$$$14496748 \nu^{15} - 383216889 \nu^{14} + 2455811611 \nu^{13} - 20082727968 \nu^{12} + 79753758709 \nu^{11} - 375114507753 \nu^{10} + 1016663085714 \nu^{9} - 3103062983781 \nu^{8} + 5617482562641 \nu^{7} - 11089350694487 \nu^{6} + 11620451848424 \nu^{5} - 12919822583029 \nu^{4} + 2845093331383 \nu^{3} + 825486570276 \nu^{2} - 5168168247283 \nu + 1933997835005$$$$)/ 316369324128$$ $$\beta_{4}$$ $$=$$ $$($$$$-28661011 \nu^{15} + 376248067 \nu^{14} - 3019246022 \nu^{13} + 19609879751 \nu^{12} - 91540961349 \nu^{11} + 380021763240 \nu^{10} - 1215316946595 \nu^{9} + 3449204606697 \nu^{8} - 7754522847207 \nu^{7} + 14949580551542 \nu^{6} - 22774230237823 \nu^{5} + 27760712142815 \nu^{4} - 25755258117138 \nu^{3} + 16706236742943 \nu^{2} - 7914233731523 \nu + 1766457326490$$$$)/ 316369324128$$ $$\beta_{5}$$ $$=$$ $$($$$$-30920939 \nu^{15} - 47258327 \nu^{14} - 211956648 \nu^{13} - 7396232221 \nu^{12} + 30361391909 \nu^{11} - 217943855310 \nu^{10} + 703406190993 \nu^{9} - 2539649789049 \nu^{8} + 5922057053211 \nu^{7} - 13228223786684 \nu^{6} + 21168894176673 \nu^{5} - 29136194430747 \nu^{4} + 28877960180876 \nu^{3} - 21487115641149 \nu^{2} + 10220884783775 \nu - 2047012171496$$$$)/ 158184662064$$ $$\beta_{6}$$ $$=$$ $$($$$$-30920939 \nu^{15} + 511072412 \nu^{14} - 4120271821 \nu^{13} + 28521203647 \nu^{12} - 134335126050 \nu^{11} + 572900637369 \nu^{10} - 1834246369401 \nu^{9} + 5281654047894 \nu^{8} - 11789222038392 \nu^{7} + 23033657253157 \nu^{6} - 34641417899267 \nu^{5} + 43240535558476 \nu^{4} - 39267317510445 \nu^{3} + 26479017902163 \nu^{2} - 10659357632440 \nu + 1740262265133$$$$)/ 158184662064$$ $$\beta_{7}$$ $$=$$ $$($$$$7915220 \nu^{15} - 40953259 \nu^{14} + 414028833 \nu^{13} - 1460644352 \nu^{12} + 7486016479 \nu^{11} - 18277094811 \nu^{10} + 59268665334 \nu^{9} - 96106377663 \nu^{8} + 208572400371 \nu^{7} - 200115377413 \nu^{6} + 311565411864 \nu^{5} - 169504852815 \nu^{4} + 313514371381 \nu^{3} - 241017203316 \nu^{2} + 288697703863 \nu - 132462757465$$$$)/ 28760847648$$ $$\beta_{8}$$ $$=$$ $$($$$$-7979 \nu^{15} + 68912 \nu^{14} - 628546 \nu^{13} + 3357412 \nu^{12} - 16625487 \nu^{11} + 61197354 \nu^{10} - 201214665 \nu^{9} + 526067616 \nu^{8} - 1200045765 \nu^{7} + 2196214414 \nu^{6} - 3408471857 \nu^{5} + 4104840184 \nu^{4} - 3987532800 \nu^{3} + 2721817314 \nu^{2} - 1298790385 \nu + 285645792$$$$)/18627492$$ $$\beta_{9}$$ $$=$$ $$($$$$-7979 \nu^{15} + 68912 \nu^{14} - 628546 \nu^{13} + 3357412 \nu^{12} - 16625487 \nu^{11} + 61197354 \nu^{10} - 201214665 \nu^{9} + 526067616 \nu^{8} - 1200045765 \nu^{7} + 2196214414 \nu^{6} - 3408471857 \nu^{5} + 4104840184 \nu^{4} - 3987532800 \nu^{3} + 2721817314 \nu^{2} - 1317417877 \nu + 304273284$$$$)/18627492$$ $$\beta_{10}$$ $$=$$ $$($$$$-6671 \nu^{15} + 41254 \nu^{14} - 402799 \nu^{13} + 1707247 \nu^{12} - 8612564 \nu^{11} + 26205779 \nu^{10} - 85020497 \nu^{9} + 186738624 \nu^{8} - 410011674 \nu^{7} + 634496219 \nu^{6} - 934157663 \nu^{5} + 967933718 \nu^{4} - 916480415 \nu^{3} + 580147099 \nu^{2} - 304374922 \nu + 82787651$$$$)/7787936$$ $$\beta_{11}$$ $$=$$ $$($$$$24636003 \nu^{15} - 180367241 \nu^{14} + 1683650240 \nu^{13} - 8042787727 \nu^{12} + 39684270259 \nu^{11} - 133306414374 \nu^{10} + 429081442695 \nu^{9} - 1032074337351 \nu^{8} + 2272009170957 \nu^{7} - 3839497364232 \nu^{6} + 5691504149431 \nu^{5} - 6374489557805 \nu^{4} + 5941303854100 \nu^{3} - 3888260852079 \nu^{2} + 1867377806769 \nu - 463004009716$$$$)/ 28760847648$$ $$\beta_{12}$$ $$=$$ $$($$$$24636003 \nu^{15} - 189172804 \nu^{14} + 1745289181 \nu^{13} - 8640627827 \nu^{12} + 42470004626 \nu^{11} - 147354845901 \nu^{10} + 475256763393 \nu^{9} - 1181342381406 \nu^{8} + 2620154110008 \nu^{7} - 4588380736521 \nu^{6} + 6885584559083 \nu^{5} - 8007122573884 \nu^{4} + 7544240352965 \nu^{3} - 5042857783047 \nu^{2} + 2373229705776 \nu - 503813289929$$$$)/ 28760847648$$ $$\beta_{13}$$ $$=$$ $$($$$$15958 \nu^{15} - 119685 \nu^{14} + 1130119 \nu^{13} - 5530551 \nu^{12} + 27795985 \nu^{11} - 96035346 \nu^{10} + 317610396 \nu^{9} - 791678301 \nu^{8} + 1812360285 \nu^{7} - 3211318028 \nu^{6} + 5028988886 \nu^{5} - 5993427763 \nu^{4} + 5979185161 \nu^{3} - 4172380467 \nu^{2} + 2121539399 \nu - 518381770$$$$)/18627492$$ $$\beta_{14}$$ $$=$$ $$($$$$-731076296 \nu^{15} + 5564679969 \nu^{14} - 51759660791 \nu^{13} + 255254450652 \nu^{12} - 1262781933077 \nu^{11} + 4374603032517 \nu^{10} - 14191982654406 \nu^{9} + 35285931287397 \nu^{8} - 78739325671977 \nu^{7} + 138259928421331 \nu^{6} - 209114172126772 \nu^{5} + 245045998417421 \nu^{4} - 234039065550755 \nu^{3} + 159802254505512 \nu^{2} - 77971936362277 \nu + 18336335470607$$$$)/ 316369324128$$ $$\beta_{15}$$ $$=$$ $$($$$$791321153 \nu^{15} - 5699211694 \nu^{14} + 53761281981 \nu^{13} - 254911441289 \nu^{12} + 1268932240096 \nu^{11} - 4251240695721 \nu^{10} + 13809836508279 \nu^{9} - 33245971930596 \nu^{8} + 74020750115370 \nu^{7} - 125650206707929 \nu^{6} + 189347503067457 \nu^{5} - 214060374517662 \nu^{4} + 205152860375869 \nu^{3} - 136001913845709 \nu^{2} + 68927049496618 \nu - 16525376998129$$$$)/ 316369324128$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$-\beta_{9} + \beta_{8} + 1$$ $$\nu^{2}$$ $$=$$ $$\beta_{15} + \beta_{14} - \beta_{13} - 2 \beta_{11} - \beta_{10} - 2 \beta_{9} + \beta_{7} - 2 \beta_{6} - 2 \beta_{1} - 4$$ $$\nu^{3}$$ $$=$$ $$2 \beta_{15} + \beta_{14} - 5 \beta_{13} - \beta_{12} - 2 \beta_{11} - 3 \beta_{10} + 9 \beta_{9} - 12 \beta_{8} - 2 \beta_{7} - 3 \beta_{6} - 4 \beta_{4} - \beta_{3} - \beta_{2} - 5 \beta_{1} - 14$$ $$\nu^{4}$$ $$=$$ $$-8 \beta_{15} - 10 \beta_{14} + 2 \beta_{13} + 5 \beta_{12} + 17 \beta_{11} + 10 \beta_{10} + 31 \beta_{9} - 13 \beta_{8} - 16 \beta_{7} + 20 \beta_{6} - 2 \beta_{5} - 8 \beta_{4} - 2 \beta_{3} - 2 \beta_{2} + 15 \beta_{1} + 19$$ $$\nu^{5}$$ $$=$$ $$-28 \beta_{15} - 22 \beta_{14} + 64 \beta_{13} + 24 \beta_{12} + 23 \beta_{11} + 58 \beta_{10} - 69 \beta_{9} + 119 \beta_{8} + 24 \beta_{7} + 52 \beta_{6} - 8 \beta_{5} + 43 \beta_{4} + 12 \beta_{3} + 12 \beta_{2} + 74 \beta_{1} + 162$$ $$\nu^{6}$$ $$=$$ $$52 \beta_{15} + 75 \beta_{14} + 61 \beta_{13} - 74 \beta_{12} - 170 \beta_{11} - 65 \beta_{10} - 411 \beta_{9} + 264 \beta_{8} + 228 \beta_{7} - 176 \beta_{6} + 31 \beta_{5} + 149 \beta_{4} + 29 \beta_{3} + 53 \beta_{2} - 76 \beta_{1} - 5$$ $$\nu^{7}$$ $$=$$ $$325 \beta_{15} + 298 \beta_{14} - 657 \beta_{13} - 450 \beta_{12} - 323 \beta_{11} - 852 \beta_{10} + 373 \beta_{9} - 1066 \beta_{8} - 167 \beta_{7} - 719 \beta_{6} + 219 \beta_{5} - 344 \beta_{4} - 139 \beta_{3} - 55 \beta_{2} - 886 \beta_{1} - 1701$$ $$\nu^{8}$$ $$=$$ $$-228 \beta_{15} - 448 \beta_{14} - 1368 \beta_{13} + 585 \beta_{12} + 1711 \beta_{11} - 150 \beta_{10} + 5023 \beta_{9} - 3987 \beta_{8} - 3036 \beta_{7} + 1380 \beta_{6} - 128 \beta_{5} - 2090 \beta_{4} - 422 \beta_{3} - 746 \beta_{2} - 67 \beta_{1} - 1786$$ $$\nu^{9}$$ $$=$$ $$-3506 \beta_{15} - 3490 \beta_{14} + 5934 \beta_{13} + 6780 \beta_{12} + 4769 \beta_{11} + 10592 \beta_{10} + 824 \beta_{9} + 8212 \beta_{8} - 336 \beta_{7} + 9182 \beta_{6} - 3490 \beta_{5} + 1801 \beta_{4} + 1418 \beta_{3} - 544 \beta_{2} + 9812 \beta_{1} + 16429$$ $$\nu^{10}$$ $$=$$ $$-775 \beta_{15} + 1126 \beta_{14} + 20300 \beta_{13} - 160 \beta_{12} - 15824 \beta_{11} + 13530 \beta_{10} - 56633 \beta_{9} + 52830 \beta_{8} + 37407 \beta_{7} - 8116 \beta_{6} - 3201 \beta_{5} + 25763 \beta_{4} + 5773 \beta_{3} + 7951 \beta_{2} + 10306 \beta_{1} + 36293$$ $$\nu^{11}$$ $$=$$ $$35787 \beta_{15} + 37693 \beta_{14} - 47572 \beta_{13} - 87239 \beta_{12} - 66603 \beta_{11} - 115639 \beta_{10} - 67910 \beta_{9} - 44026 \beta_{8} + 36821 \beta_{7} - 110918 \beta_{6} + 43269 \beta_{5} + 5326 \beta_{4} - 12756 \beta_{3} + 18132 \beta_{2} - 101639 \beta_{1} - 143099$$ $$\nu^{12}$$ $$=$$ $$39984 \beta_{15} + 26340 \beta_{14} - 258878 \beta_{13} - 94424 \beta_{12} + 124230 \beta_{11} - 288350 \beta_{10} + 581676 \beta_{9} - 643364 \beta_{8} - 424150 \beta_{7} + 706 \beta_{6} + 98482 \beta_{5} - 289332 \beta_{4} - 73746 \beta_{3} - 66104 \beta_{2} - 215666 \beta_{1} - 542199$$ $$\nu^{13}$$ $$=$$ $$-342678 \beta_{15} - 375858 \beta_{14} + 321008 \beta_{13} + 984284 \beta_{12} + 862898 \beta_{11} + 1100808 \beta_{10} + 1390953 \beta_{9} - 90875 \beta_{8} - 832560 \beta_{7} + 1270202 \beta_{6} - 441664 \beta_{5} - 348692 \beta_{4} + 98730 \beta_{3} - 303334 \beta_{2} + 963960 \beta_{1} + 1048389$$ $$\nu^{14}$$ $$=$$ $$-729089 \beta_{15} - 672801 \beta_{14} + 3058731 \beta_{13} + 2237884 \beta_{12} - 659178 \beta_{11} + 4637781 \beta_{10} - 5273480 \beta_{9} + 7301078 \beta_{8} + 4374105 \beta_{7} + 1058408 \beta_{6} - 1805654 \beta_{5} + 2971940 \beta_{4} + 895656 \beta_{3} + 343122 \beta_{2} + 3421446 \beta_{1} + 7051750$$ $$\nu^{15}$$ $$=$$ $$3012048 \beta_{15} + 3370541 \beta_{14} - 1368219 \beta_{13} - 9668935 \beta_{12} - 10414656 \beta_{11} - 8491987 \beta_{10} - 21648505 \beta_{9} + 7773878 \beta_{8} + 14106296 \beta_{7} - 13734847 \beta_{6} + 3483128 \beta_{5} + 6940034 \beta_{4} - 565701 \beta_{3} + 3983641 \beta_{2} - 7907405 \beta_{1} - 4830300$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/462\mathbb{Z}\right)^\times$$.
$$n$$ $$155$$ $$199$$ $$211$$ $$\chi(n)$$ $$1$$ $$-\beta_{13}$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
241.1
0.5 − 3.32851i 0.5 − 0.0286340i 0.5 + 0.921602i 0.5 + 3.43554i 0.5 − 3.19339i 0.5 − 1.56688i 0.5 + 1.35798i 0.5 + 2.40229i 0.5 + 3.32851i 0.5 + 0.0286340i 0.5 − 0.921602i 0.5 − 3.43554i 0.5 + 3.19339i 0.5 + 1.56688i 0.5 − 1.35798i 0.5 − 2.40229i
−0.866025 + 0.500000i −0.866025 0.500000i 0.500000 0.866025i −2.13257 + 1.23124i 1.00000 0.941950 2.47239i 1.00000i 0.500000 + 0.866025i 1.23124 2.13257i
241.2 −0.866025 + 0.500000i −0.866025 0.500000i 0.500000 0.866025i 0.725202 0.418696i 1.00000 −2.44037 + 1.02205i 1.00000i 0.500000 + 0.866025i −0.418696 + 0.725202i
241.3 −0.866025 + 0.500000i −0.866025 0.500000i 0.500000 0.866025i 1.54813 0.893814i 1.00000 −0.165362 2.64058i 1.00000i 0.500000 + 0.866025i −0.893814 + 1.54813i
241.4 −0.866025 + 0.500000i −0.866025 0.500000i 0.500000 0.866025i 3.72526 2.15078i 1.00000 1.43173 + 2.22489i 1.00000i 0.500000 + 0.866025i −2.15078 + 3.72526i
241.5 0.866025 0.500000i 0.866025 + 0.500000i 0.500000 0.866025i −2.01555 + 1.16368i 1.00000 1.31629 + 2.29508i 1.00000i 0.500000 + 0.866025i −1.16368 + 2.01555i
241.6 0.866025 0.500000i 0.866025 + 0.500000i 0.500000 0.866025i −0.606961 + 0.350429i 1.00000 1.82993 1.91085i 1.00000i 0.500000 + 0.866025i −0.350429 + 0.606961i
241.7 0.866025 0.500000i 0.866025 + 0.500000i 0.500000 0.866025i 1.92604 1.11200i 1.00000 −2.45660 0.982398i 1.00000i 0.500000 + 0.866025i 1.11200 1.92604i
241.8 0.866025 0.500000i 0.866025 + 0.500000i 0.500000 0.866025i 2.83045 1.63416i 1.00000 2.54243 + 0.732142i 1.00000i 0.500000 + 0.866025i 1.63416 2.83045i
439.1 −0.866025 0.500000i −0.866025 + 0.500000i 0.500000 + 0.866025i −2.13257 1.23124i 1.00000 0.941950 + 2.47239i 1.00000i 0.500000 0.866025i 1.23124 + 2.13257i
439.2 −0.866025 0.500000i −0.866025 + 0.500000i 0.500000 + 0.866025i 0.725202 + 0.418696i 1.00000 −2.44037 1.02205i 1.00000i 0.500000 0.866025i −0.418696 0.725202i
439.3 −0.866025 0.500000i −0.866025 + 0.500000i 0.500000 + 0.866025i 1.54813 + 0.893814i 1.00000 −0.165362 + 2.64058i 1.00000i 0.500000 0.866025i −0.893814 1.54813i
439.4 −0.866025 0.500000i −0.866025 + 0.500000i 0.500000 + 0.866025i 3.72526 + 2.15078i 1.00000 1.43173 2.22489i 1.00000i 0.500000 0.866025i −2.15078 3.72526i
439.5 0.866025 + 0.500000i 0.866025 0.500000i 0.500000 + 0.866025i −2.01555 1.16368i 1.00000 1.31629 2.29508i 1.00000i 0.500000 0.866025i −1.16368 2.01555i
439.6 0.866025 + 0.500000i 0.866025 0.500000i 0.500000 + 0.866025i −0.606961 0.350429i 1.00000 1.82993 + 1.91085i 1.00000i 0.500000 0.866025i −0.350429 0.606961i
439.7 0.866025 + 0.500000i 0.866025 0.500000i 0.500000 + 0.866025i 1.92604 + 1.11200i 1.00000 −2.45660 + 0.982398i 1.00000i 0.500000 0.866025i 1.11200 + 1.92604i
439.8 0.866025 + 0.500000i 0.866025 0.500000i 0.500000 + 0.866025i 2.83045 + 1.63416i 1.00000 2.54243 0.732142i 1.00000i 0.500000 0.866025i 1.63416 + 2.83045i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 439.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
77.i even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 462.2.p.b yes 16
3.b odd 2 1 1386.2.bk.b 16
7.c even 3 1 3234.2.e.b 16
7.d odd 6 1 462.2.p.a 16
7.d odd 6 1 3234.2.e.a 16
11.b odd 2 1 462.2.p.a 16
21.g even 6 1 1386.2.bk.a 16
33.d even 2 1 1386.2.bk.a 16
77.h odd 6 1 3234.2.e.a 16
77.i even 6 1 inner 462.2.p.b yes 16
77.i even 6 1 3234.2.e.b 16
231.k odd 6 1 1386.2.bk.b 16
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
462.2.p.a 16 7.d odd 6 1
462.2.p.a 16 11.b odd 2 1
462.2.p.b yes 16 1.a even 1 1 trivial
462.2.p.b yes 16 77.i even 6 1 inner
1386.2.bk.a 16 21.g even 6 1
1386.2.bk.a 16 33.d even 2 1
1386.2.bk.b 16 3.b odd 2 1
1386.2.bk.b 16 231.k odd 6 1
3234.2.e.a 16 7.d odd 6 1
3234.2.e.a 16 77.h odd 6 1
3234.2.e.b 16 7.c even 3 1
3234.2.e.b 16 77.i even 6 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{13}^{8} - 64 T_{13}^{6} + 8 T_{13}^{5} + 836 T_{13}^{4} - 1168 T_{13}^{3} - 592 T_{13}^{2} + 1216 T_{13} - 128$$ acting on $$S_{2}^{\mathrm{new}}(462, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( 1 - T^{2} + T^{4} )^{4}$$
$3$ $$( 1 - T^{2} + T^{4} )^{4}$$
$5$ $$35344 - 29328 T - 66900 T^{2} + 62244 T^{3} + 118381 T^{4} - 154872 T^{5} + 25139 T^{6} + 39396 T^{7} - 11772 T^{8} - 8472 T^{9} + 4381 T^{10} + 312 T^{11} - 484 T^{12} + 12 T^{13} + 47 T^{14} - 12 T^{15} + T^{16}$$
$7$ $$5764801 - 4941258 T + 2470629 T^{2} - 168070 T^{3} - 328937 T^{4} + 187964 T^{5} - 13916 T^{6} - 28112 T^{7} + 16958 T^{8} - 4016 T^{9} - 284 T^{10} + 548 T^{11} - 137 T^{12} - 10 T^{13} + 21 T^{14} - 6 T^{15} + T^{16}$$
$11$ $$214358881 + 77948684 T - 15944049 T^{2} - 3221020 T^{3} + 6163861 T^{4} + 1277760 T^{5} - 338074 T^{6} + 29304 T^{7} + 71702 T^{8} + 2664 T^{9} - 2794 T^{10} + 960 T^{11} + 421 T^{12} - 20 T^{13} - 9 T^{14} + 4 T^{15} + T^{16}$$
$13$ $$( -128 + 1216 T - 592 T^{2} - 1168 T^{3} + 836 T^{4} + 8 T^{5} - 64 T^{6} + T^{8} )^{2}$$
$17$ $$576 + 6624 T + 69528 T^{2} + 159780 T^{3} + 538417 T^{4} - 880544 T^{5} + 2791005 T^{6} - 1314372 T^{7} + 661234 T^{8} - 134304 T^{9} + 43709 T^{10} - 5340 T^{11} + 2082 T^{12} - 136 T^{13} + 53 T^{14} + T^{16}$$
$19$ $$2768896 - 3940352 T + 12875776 T^{2} + 2782208 T^{3} + 23727360 T^{4} - 8339456 T^{5} + 6396800 T^{6} - 1621056 T^{7} + 860080 T^{8} - 205120 T^{9} + 72936 T^{10} - 13172 T^{11} + 3489 T^{12} - 530 T^{13} + 107 T^{14} - 10 T^{15} + T^{16}$$
$23$ $$24039882304 + 37979937888 T + 51979552888 T^{2} + 18938176388 T^{3} + 8371183769 T^{4} + 1395574872 T^{5} + 569006503 T^{6} + 73933044 T^{7} + 25326352 T^{8} + 2171968 T^{9} + 697889 T^{10} + 47696 T^{11} + 13392 T^{12} + 532 T^{13} + 143 T^{14} + 4 T^{15} + T^{16}$$
$29$ $$12810617856 + 11931674112 T^{2} + 3667420672 T^{4} + 491528720 T^{6} + 33087905 T^{8} + 1203204 T^{10} + 23910 T^{12} + 244 T^{14} + T^{16}$$
$31$ $$262144 - 1376256 T + 2318336 T^{2} + 473088 T^{3} - 2094848 T^{4} - 308736 T^{5} + 1998848 T^{6} - 866688 T^{7} - 81488 T^{8} + 108960 T^{9} + 3296 T^{10} - 11520 T^{11} + 1441 T^{12} + 294 T^{13} - 37 T^{14} - 6 T^{15} + T^{16}$$
$37$ $$125622042624 - 34637930496 T + 40451561472 T^{2} - 12484740096 T^{3} + 9620770048 T^{4} - 2631917056 T^{5} + 994476160 T^{6} - 189887424 T^{7} + 54018160 T^{8} - 8801536 T^{9} + 1843896 T^{10} - 199348 T^{11} + 27961 T^{12} - 2062 T^{13} + 263 T^{14} - 14 T^{15} + T^{16}$$
$41$ $$( 256 + 8064 T + 2848 T^{2} - 6824 T^{3} - 4815 T^{4} - 952 T^{5} + 2 T^{6} + 16 T^{7} + T^{8} )^{2}$$
$43$ $$23084548096 + 25298960384 T^{2} + 8477803520 T^{4} + 1124077312 T^{6} + 71086992 T^{8} + 2303384 T^{10} + 38761 T^{12} + 318 T^{14} + T^{16}$$
$47$ $$1344737098384 + 3457746300816 T + 3090667648540 T^{2} + 326590505388 T^{3} - 363709353443 T^{4} - 50476519572 T^{5} + 44065903899 T^{6} + 5594306520 T^{7} - 527953996 T^{8} - 90988980 T^{9} + 5374933 T^{10} + 1115364 T^{11} - 2484 T^{12} - 5976 T^{13} - 57 T^{14} + 24 T^{15} + T^{16}$$
$53$ $$26488213504 - 61179127808 T + 111058636544 T^{2} - 72713118848 T^{3} + 40171071120 T^{4} - 9183945376 T^{5} + 2778511712 T^{6} - 171246848 T^{7} + 163718908 T^{8} - 2083056 T^{9} + 3016384 T^{10} - 36036 T^{11} + 40865 T^{12} - 232 T^{13} + 235 T^{14} + T^{16}$$
$59$ $$109280491776 + 294630488064 T + 190208538240 T^{2} - 201062028288 T^{3} + 24074195440 T^{4} + 15710742144 T^{5} - 2352978552 T^{6} - 1019215632 T^{7} + 245570017 T^{8} + 8528784 T^{9} - 4201594 T^{10} - 70032 T^{11} + 52915 T^{12} - 266 T^{14} + T^{16}$$
$61$ $$3489501184 - 3714447360 T + 11848985344 T^{2} + 9559013504 T^{3} + 16635270224 T^{4} + 2680338048 T^{5} + 1536277900 T^{6} + 219443376 T^{7} + 100720417 T^{8} + 12716608 T^{9} + 3059300 T^{10} + 399224 T^{11} + 69975 T^{12} + 6928 T^{13} + 620 T^{14} + 28 T^{15} + T^{16}$$
$67$ $$2403352576 - 7497142272 T + 23480118784 T^{2} - 4344754048 T^{3} + 7183184528 T^{4} + 129355824 T^{5} + 1933517692 T^{6} + 56306316 T^{7} + 86888329 T^{8} + 6732460 T^{9} + 3037910 T^{10} + 195332 T^{11} + 38595 T^{12} + 2344 T^{13} + 350 T^{14} + 16 T^{15} + T^{16}$$
$71$ $$( 4193232 + 2297712 T - 45068 T^{2} - 251544 T^{3} - 58124 T^{4} - 4360 T^{5} + 97 T^{6} + 28 T^{7} + T^{8} )^{2}$$
$73$ $$119793516544 + 942044872704 T + 7545799049216 T^{2} - 961323204608 T^{3} + 623445557248 T^{4} - 102344933376 T^{5} + 38692855808 T^{6} - 6019252224 T^{7} + 1260078592 T^{8} - 189543936 T^{9} + 30397952 T^{10} - 3545344 T^{11} + 349904 T^{12} - 24272 T^{13} + 1316 T^{14} - 44 T^{15} + T^{16}$$
$79$ $$1130708969104 + 4406943704592 T + 6164867690476 T^{2} + 1712977494492 T^{3} - 246923168675 T^{4} - 138506030502 T^{5} + 15304968387 T^{6} + 4732344210 T^{7} - 620713636 T^{8} - 102519378 T^{9} + 25951621 T^{10} - 1615398 T^{11} - 32868 T^{12} + 6390 T^{13} + 87 T^{14} - 30 T^{15} + T^{16}$$
$83$ $$( 5604 - 12516 T - 263 T^{2} + 8396 T^{3} + 1031 T^{4} - 780 T^{5} - 141 T^{6} + 4 T^{7} + T^{8} )^{2}$$
$89$ $$96546188427264 + 151241105866752 T + 105379288645632 T^{2} + 41364533084160 T^{3} + 9579425505280 T^{4} + 1178899292160 T^{5} + 26915948544 T^{6} - 10521747456 T^{7} - 350878976 T^{8} + 253334784 T^{9} + 37971200 T^{10} + 1587840 T^{11} - 71984 T^{12} - 6336 T^{13} + 256 T^{14} + 36 T^{15} + T^{16}$$
$97$ $$68597371921 + 325271379408 T^{2} + 446119112140 T^{4} + 163854186512 T^{6} + 4910736438 T^{8} + 58793584 T^{10} + 340460 T^{12} + 944 T^{14} + T^{16}$$ | 16,686 | 32,634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-39 | latest | en | 0.403446 |
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# Kiran Ramsaroop
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Human Resources Analyst interview questions shared by candidates
## Top Interview Questions
Sort: Relevance|Popular|Date
HR Analyst was asked...May 27, 2018
### Tell me more about yourself?
I can spot site anywhere supervisor
Pekerja keras, disiplin, tidak mudah putus asa, mudah beradaptasi, mau belajat dengan hal yang baru, jujur, teliti, dan suka olahraga Less
Good attitude with company employee store keeper and bosses also costomar diling for the company Less
### ¿If you were an object, which one would you want to be and which one you wouldn't be? Same question with an animal.
I would be a piano, but I never want to be a toilet
Because it carry you anywhere, has the control of all its parts but obey o the owner Less
A Car..
HR Analyst was asked...November 23, 2020
### you buy a racket and a tennis ball for 1,20Euro. and the racket cost 1/3 more than the ball. what is the cost of the racket...
Sorry, 33,3% more*
Lets say cost of ball = x Then cost of racket = x + x/3 x + (x +x/3) = 1.2 7x/3 = 1.2 x = 0.53 (approx) = cost of =ball Cost of racket = 0.5 + 0.5/3 = 0.67 (approx) Less
3+3+1=7 120/7=17…. 17*3=51{ball} so around 69{racket}
### I was asked for my current company's EBITDA. Considering they are a private company, I felt ashamed I was even asked, considering I have no reason to provide it.
I felt that if I said I could not provide it, I would have been dismissed right off the bat. I made up numbers that were ridiculous and unreasonable instead. Less
I agree the asked me a lot of questions that were completely out of scope
### My experience in the specific area that the job was responsible for.
How long was the entire process?
From the multiple choice > to panel interview > receiving feedback?
No
No
Yes
### How do you handle multiple tasks and responsibilities?
Take care one at a time in a timely fashion
Who do you consider to be your clients?
### Q: If at 5pm on Friday you received a request from stakeholder for report etc how would you deal with this.
Clarify requirements, timelines, ensure specifics and requirements are clear and understood by both sides. Act as appropriate to timescale given to you by stakeholder. If appropriate complete the task before leaving. Less
Clarify with them what it was that was needed and Dependent on if the request could be dealt with straight away with any information that i had to hand . it would be dealt with before leaving if it was something that i did not have to hand . i would investigate what would be needed for me deal with this request straight away . Less
### Give an example of when you had a conflict with a manager. What was the scenario and how was it resolved?
To resolve any dispute between the manager and any of his subordinates needs someone who has a key role in the process, which is not an easy task. As it is his responsibility to reach agreement on the basic rules of discussions aimed at showing facts and reducing aggressive behavior. It must also monitor ways of expressing negative views during discussions and encourage the parties to identify problems, causes and motives for the purpose of reaching a mutually satisfactory solution. It is also the duty of this person not to be prejudiced by any party to the dispute. He should also adopt a personal consultation method so that he / she listens attentively and observes that one helps the parties to understand and identify the problem by asking questions about the subject of the dispute, identifying the views and allowing the parties to express them, , Helps the parties develop plans to implement and provide advice and assistance in case of request. It should also be noted that the dispute itself is associated with progress and change. What we should regret is our failure to use the dispute constructively. There is no doubt that a practical solution to problems and a constructive response to them will help resolve differences in a fair manner and open channels of discussion and cooperative behavior. (Differences can contribute to the general issue if they are resolved by integration rather than through control or settlement) Less
"M. Akram"- You did not answer the question; you had given an extended complicated definition of the fundamentals of management. My conflict with a manager was when I had identified a Hydrogen gas leak but was told by my manager that it was a nitrogen line. I personally verified it again to be a Hydrogen leak and had explained in detail how I am identifying it to my manager; I was then told that I was correct and it was indeed a hydrogen leak and we had it repaired over the weekend. Less | 1,046 | 4,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.960247 |
https://fuxuemingzhu.cn/leetcode/392.html | 1,726,806,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652130.6/warc/CC-MAIN-20240920022257-20240920052257-00408.warc.gz | 238,479,337 | 9,998 | # # 【LeetCode】392. Is Subsequence 解题报告（Python）
## # 题目描述：
Given a string s and a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).
``````Example 1:
s = "abc", t = "ahbgdc"
Return true.
Example 2:
s = "axc", t = "ahbgdc"
Return false.
``````
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?
## # 解题方法
``````class Solution(object):
def isSubsequence(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
queue = collections.deque(s)
for c in t:
if not queue: return True
if c == queue[0]:
queue.popleft()
return not queue
``````
``````class Solution(object):
def isSubsequence(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
si, ti = 0, 0
while si < len(s) and ti < len(t):
if t[ti] == s[si]:
si += 1
ti += 1
return si == len(s)
``````
## # 日期
2018 年 3 月 15 日 --雾霾消散，春光明媚 | 435 | 1,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.758714 |
http://blog.piston.rs/2017/02/06/the-mathematics-of-infinite-things-space/ | 1,534,749,843,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215858.81/warc/CC-MAIN-20180820062343-20180820082343-00440.warc.gz | 57,826,148 | 4,582 | I am currently doing some research on procedurally generated 3D by inflating structures.
This blog post is about how the mathematics work.
### Visualizing Infinite-Things-Space
Vertex data in a 3D structure is often stored in an array.
Ignoring the position and other attributes of vertices, each vertex can be represented as a natural number which refers to the index in the array of vertices.
To connect two vertices, I use a math formula that takes an index and outputs a new one.
The nice thing about this method is that you can compactly represent an infinite number of structures.
When the function returns the same index, I visualize it as a loop for each natural number from 0 to infinity.
By adding by one, `f(x) = x + 1`, the loops are disconnect from themselves and attached to each other.
To create circles with 4 vertices, use the formula `f(x) = x - x % 4 + (x+1) % 4`.
Mathematical operations on the natural numbers are equivalent to operations on these loops.
For any mathematical expression of a single natural number, there is an equivalent shape in Infinite-Things-Space.
In group theory, these functions are called “generators”.
### The Super Function
The Super function is a higher order function that takes a generator and produces a new generator. It preserves the structure of the input generator, but now “scaled up” with a factor of `n`.
``````super(n, f: N -> N) = \(x: N) = f(x / n) * n + x % n
``````
The Super function is the key to combine multiple generators.
``````f_0(x) = x - x % 4 + (x + 1) % 4
f_1(x) = super(4, \(x) = x + 1)
``````
You can use the Super function with any generator, for example circles to create a donut topology (torus):
``````f_0(x) = x - x % n + (x + 1) % n
f_1 = super(n, \(x) = x - x % m + (x + 1) % m)
``````
When `n` and `m` in the example above goes to infinity, you can start at 0,
add infinity to take an infinitesimal step in the rotation direction of the torus.
Take an Aleph-1 step (a higher infinity), and you jump from the 0-point at the first torus
to the 0-point (modulus Aleph-1) to the second torus.
### Inflating 3D structures
To inflate a 3D structure created by generators, I use an algorithm that takes a list of generators
and walks along all edges for a fixed range, generating a list of spring constraints.
If a generator leads back to the same vertex, it gets ignored.
The spring constraints are passed to a simulator that inserts random positions for vertices.
An extra spring constraint drives inflation by being connected temporarily between two random vertices `A` and `B` with a target distance `10 * (A - B).length().sqrt()`.
Inflation causes an internal pressure outward on the whole shape, so it unfolds into its “natural form”.
The square root operation is to avoid too much acceleration between distant points,
which might lead to stretching the shape like spaghetti.
Here is a shape inflated by connecting cubes with some internal support to edges of a pentagon,
which in turn is connected by a super hundredgon (a circle shape with 100 vertices).
As you see in the picture above, the shapes do not become perfect.
This is because the springs are not possible to satisfy unless the differential topology is complete.
Because the inner circle is the same size as the outer one, the shape above twists and stretches the pentagons.
Finding a complete differential topology could be done by “tuning” the constraints while simulating,
or perhaps relaxing the target distances based on a function of the tension.
Alternatively, some new constraints could be added to keep edges at a fixed angle relative to each other.
The twisting effect can be used on purpose, e.g. on a cylinder with sinus modified distances:
Have you seen how the sweets with double twisted paper get wrinkles? This seems to be the same effect! | 904 | 3,829 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-34 | latest | en | 0.867518 |
https://brainly.com/question/170278 | 1,485,272,877,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00443-ip-10-171-10-70.ec2.internal.warc.gz | 794,217,870 | 9,342 | If 2 people share a cost of a meal in the ratio of 2:3 and 1 person pays £52.20. What is the total cost of the meal ?
2
by alishabarker66
and which of these persons pays 52.20?
the second person
Does it say that in the question, that the 2nd person pays 52.20?
2014-10-31T11:25:01-04:00
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
52.20/3 = 17.4
17.4 x 5 = 87
total cost= £ 87
• ZackFry
• Moderators United Runner Up
2014-10-31T11:28:26-04:00 | 206 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2017-04 | latest | en | 0.927036 |
http://intermath.coe.uga.edu/tweb/CPTM1/mbburns/GSPlesson2forbeginners/GSPlesson2forbeginners.htm | 1,516,772,013,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084893397.98/warc/CC-MAIN-20180124050449-20180124070449-00727.warc.gz | 187,361,155 | 4,374 | ## Georgia Learning Connections
*TITLE: Geometry Sketchpad for Beginners – Lesson 2
*Annotation:
This lesson plan is a technology lesson and is designed for students who are learning to use Geometry Sketchpad and have some understanding of Microsoft Word. Students will be paired up and given laptop computers to continue to learn some of the features of Geometry Sketchpad. After reviewing familiar features of Geometry Sketchpad, students will experiment with additional features of GSP. Students will then be given an assessment evaluation form to follow and complete on Microsoft Word to demonstrate their knowledge and understanding of additional features of Geometry Sketchpad.
## *Primary Learning Outcome:
Students will be able to increase (1) their knowledge of geometric concepts, and (2) their technology understanding of running software programs and editing, saving, and exiting them.
Students will work together to complete the primary learning objectives.
*Assessed QCC:
Math QCC 8.16 – Applies formulas (e.g., area, perimeter, circumference).
Math QCC 8.27 – Measures and draws angles and classifies angles by their measures (e.g., acute, obtuse, right, straight, complementary, supplementary).
*Total Duration (Estimated Time): one 50-minute class period
Procedures:
## Step One:Write the words AREA, PERIMETER, CIRCUMFERENCE, ANGLE, ACUTE, OBTUSE, RIGHT, STRAIGHT, COMPLEMENTARY, and SUPPLEMENTARY on the chalkboard. State the desired objectives of this lesson (see Math QCC 8.16 and QCC 8.27). Ask students to tell what they think the terms mean. List their ideas on the board.
Step Two: Randomly divide the class into paired teams. Distribute laptop computers to each team. Tell the students to open the software program GSP (Geometry Sketchpad) and begin to explore the features of the program concentrating on the words listed on the chalkboard.
Step Three: After students have had a sufficient time of exploration, ask for questions the students might have concerning the features of the program. Demonstrate, through the use of a TV-ator, any features the students had difficulty manipulating or demonstrating.
Step Four: Distribute an assessment evaluation form to each team. Discuss the required elements of the assessment form. Answer any questions concerning the assignment.
Step Five: Teams complete the assignment following the guidelines of the assessment evaluation form and the instructions to operating GSP and Microsoft Word.
Lesson Materials to be Attached:
# Assessment evaluation form
*Assessment:
Students will print a hard copy of their work that will then be evaluated according to the assessment evaluation form.
Extension:
This lesson is for students who are beginning to learn the features of Geometry Sketchpad. However, to make this a more challenging assignment, students could complete the work with no assistance or guided instruction by the teacher.
Remediation:
For lower level students, the teacher could “walk-through step by step” the assessment evaluation form guiding the students from GSP to Microsoft Word to printing a hard copy.
Accommodation:
The teacher could follow the guidelines under Remediation, but extend the lesson over a period of days.
Modification:
The teacher could simplify the assessment evaluation form and choose fewer concepts to evaluate at a time. | 680 | 3,399 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-05 | latest | en | 0.882298 |
http://www.emathematics.net/polinomios.php?a=&ejercicio=foil | 1,369,217,833,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701577515/warc/CC-MAIN-20130516105257-00057-ip-10-60-113-184.ec2.internal.warc.gz | 449,435,239 | 13,654 | User:
Documento sin título
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FOIL
Multiplication of binomials, the FOIL method.
The FOIL method is a way to multiply binomials. "FOIL" is an acronym to remember a set of rules to perform this multiplication. To FOIL you multiply together all of the following:
• F: First terms
• O: Outer terms
• I: Inner terms
• L: Last terms
and write them as a polynomial.
Multiply (3x+1)(x + 4)
(3x+1)(x + 4)=3x2+12x+x+4=3x2+13x+4
$$$4x^2-3x$$\cdot$$-3x$$=$
Solution: x3 x2 x | 592 | 2,789 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2013-20 | longest | en | 0.77314 |
https://www.physicsforums.com/threads/solving-polarization-formula-problem-get-help-now.72554/ | 1,723,709,784,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00892.warc.gz | 725,037,441 | 18,695 | # Solving Polarization Formula Problem: Get Help Now
• sghaussi
In summary, the conversation discusses a problem involving three stacked polarizing filters with varying angles of polarizing axis. The incident intensity of unpolarized light is given as 76.5 W/cm^2 after passing through the stack. The question is posed on what the intensity would be after removing the second filter. The solution involves using Malus' Law to calculate the intensity for two cases and comparing them. The conversation also clarifies some confusion regarding the positioning of the first filter and the use of 76.5 W/cm^2 as the maximum intensity.
sghaussi
Hello! I'm having difficulty solving this problem, I was wondering if you can help me solve it.
Three polarizing filters are stacked, with the polarizing axis of the second and third filters at angles of 28.4 and 57.0, respectively, to that of the first. If unpolarized light is incident on the stack, the light has an intensity of 76.5 W/cm^2 after it passes through the stack.
If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?
I know I need to use the formula: I = I_max cos^2 phi
Do I solve for I_max first given the intensity 76.5 W/cm^2 then change the angle to fit the second situation? And what do they mean by "light is incident" on the stack? Does that mean I am looking for I or I_max?
I've read the section on Polarization 5 times so far and I still don't understand this problem. I hope you can help me. Thank you. =)
Last edited by a moderator:
Here's one way to approach it. Call the intensity of the light after passing through the first filter $I_1$. Now use your formula (which is Malus' Law) to calculate the intensity of the light for two cases:
(a) after passing through filter 2, then through filter 3
(b) after passing directly through filter 3
Compare (a) with (b), realizing that the intensity you calculate for (a) is given as 76.5 W/cm^2.
Does this mean that filter one is positioned at zero degrees? Also, am I supposed to be using 76.5 W/cm^2 as my I_max or is that just the intensity given after the light has passed through all three filters? Do I just use the given intensity for find I_max? Thanks for your quick response - it's greatly appreciated!
sghaussi said:
Does this mean that filter one is positioned at zero degrees?
Yes.
Also, am I supposed to be using 76.5 W/cm^2 as my I_max or is that just the intensity given after the light has passed through all three filters?
As the problem stated: "...the light has an intensity of 76.5 W/cm^2 after it passes through the stack"
Do I just use the given intensity for find I_max?
Do what I suggested and see what happens.
## 1. What is a polarization formula problem?
A polarization formula problem refers to a mathematical equation used to calculate the degree of polarization of light passing through a polarizing filter. It involves understanding the properties of polarized light and using mathematical formulas to determine its intensity and direction.
## 2. Why is solving polarization formula problems important?
Solving polarization formula problems is important because it helps us understand the behavior of polarized light, which is essential in many scientific fields such as optics, physics, and astronomy. It also allows us to make accurate predictions and calculations in various applications, including polarized sunglasses, LCD screens, and polarimeters.
## 3. What are some common methods for solving polarization formula problems?
There are several methods for solving polarization formula problems, including Malus' law, Brewster's law, and the Jones calculus. Malus' law states that the intensity of polarized light passing through a polarizer is proportional to the square of the cosine of the angle between the polarizer and the incident light. Brewster's law relates the angle of incidence and the refractive index of a material to the angle at which light is polarized. The Jones calculus uses matrices to calculate the intensity and direction of polarized light passing through multiple polarizers.
## 4. Can I solve polarization formula problems without advanced mathematical knowledge?
Yes, it is possible to solve polarization formula problems without advanced mathematical knowledge. Basic algebra, trigonometry, and vector operations are usually sufficient for most problems. However, a solid understanding of polarized light and the properties of polarizing filters is necessary.
## 5. Where can I get help with solving polarization formula problems?
You can get help with solving polarization formula problems from various sources, such as textbooks, online resources, and scientific forums. You can also seek assistance from a tutor or your professor if you are a student. Additionally, some scientific calculators have built-in functions for solving polarization formula problems.
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2K | 1,215 | 5,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-33 | latest | en | 0.950859 |
https://www.mathmusic.org/solving-linear-inequalities.html | 1,544,551,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823674.34/warc/CC-MAIN-20181211172919-20181211194419-00584.warc.gz | 950,541,162 | 10,096 | Algebra Tutorials! Tuesday 11th of December
Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Number of inequalities to solve: 23456789
Ineq. #1:
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# Solving Linear Inequalities
These examples illustrate the properties that we use for solving inequalities.
## Properties of Inequality
If the same number is added to both sides of an inequality, then the solution set to the inequality is unchanged.
Multiplication Property of Inequality
If both sides of an inequality are multiplied by the same positive number, then the solution set to the inequality is unchanged.
If both sides of an inequality are multiplied by the same negative number and the inequality symbol is reversed, then the solution set to the inequality is unchanged.
Because subtraction is defined in terms of addition, the addition property of inequality also allows us to subtract the same number from both sides. Because division is defined in terms of multiplication, the multiplication property of inequality also allows us to divide both sides by the same nonzero number as long as we reverse the inequality symbol when dividing by a negative number.
Equivalent inequalities are inequalities with the same solution set. We find the solution to a linear inequality by using the properties to convert it into an equivalent inequality with an obvious solution set, just as we do when solving equations.
Example 1
Solving inequalities
Solve each inequality. State and graph the solution set.
a) 2x - 7 < -1
b) 5 - 3x < 11
Solution
a) We proceed exactly as we do when solving equations:
2x - 7 < -1 Original inequality 2x < 6 Add 7 to each side x < 3 Divide each side by 2.
The solution set is written in set notation as {x | x < 3} and in interval notation as (-∞, 3). The graph is shown below:
b) We divide by a negative number to solve this inequality.
5 - 3x < 11 Original equation -3x < 6 Subtract 5 from each side x > -2 Divide each side by -3 and reverse the inequality symbol
The solution set is written in set notation as {x | x > -2} and in interval notation as (-2, ∞). The graph is shown below:
Example 2
Solving inequalities
Solve State and graph the solution set.
≥ -4 Original inequality ≤ -5(-4) Multiply each side by -5 and reverse the inequality symbol. 8 + 3x ≤ 20 Simplify 3x ≤ 12 Subtract 8 from each side. x ≤ 4 Divide each side by 3.
The solution set is (-∞, 4], and its graph is shown below: | 717 | 2,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-51 | longest | en | 0.910544 |
http://axishost.net/error-analysis/error-analysis-physics-book.php | 1,537,532,737,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157070.28/warc/CC-MAIN-20180921112207-20180921132607-00077.warc.gz | 25,715,335 | 5,396 | Home > Error Analysis > Error Analysis Physics Book
# Error Analysis Physics Book
## Contents
For the error estimates we keep only the first terms: DR = R(x+Dx) - R(x) = (dR/dx)x Dx for Dx ``small'', where (dR/dx)x is the derivative of function R with Was this review helpful to you? But, as already mentioned, this means you are assuming the result you are attempting to measure. Skip to main content University Libraries WMU Research Guides By Course PHYS 3100: Intro to Modern Physics Lab Find Background Info - Course Reserves Search this Guide Search PHYS 3100: Intro my review here
Yes No Sending feedback... Examples are the age distribution in a population, and many others. Sign up for a free 30min tutor trial with Chegg Tutors Dismiss Notice Dismiss Notice Join Physics Forums Today! A further problem with this accuracy is that while most good manufacturers (including Philips) tend to be quite conservative and give trustworthy specifications, there are some manufacturers who have the specifications navigate to these guys
## Error Analysis Physics Lab Report
The Chi-Squared Test for a Distribution I'm not sure what parts of Chapters 8 - Chapter 11 I will need. Note that all three rules assume that the error, say x, is small compared to the value of x. In this case the precision of the result is given: the experimenter claims the precision of the result is within 0.03 m/s.
Fulfillment by Amazon (FBA) is a service we offer sellers that lets them store their products in Amazon's fulfillment centers, and we directly pack, ship, and provide customer service for these Suppose we are to determine the diameter of a small cylinder using a micrometer. Note: This assumes of course that you have not been sloppy in your measurement but made a careful attempt to line up one end of the object with the zero of How To Calculate Error Analysis In Physics However, they were never able to exactly repeat their results.
In[17]:= Out[17]= The function CombineWithError combines these steps with default significant figure adjustment. Error Analysis In Physics Experiments Further Reading Introductory: J.R. In[19]:= Out[19]= In this example, the TimesWithError function will be somewhat faster. But this is not a 'bug', it's a 'feature'!.
Nonetheless, in this case it is probably reasonable to accept the manufacturer's claimed accuracy and take the measured voltage to be 6.5 ± 0.3 V. How To Do Error Analysis In Physics In general, there are two different types of experimental data taken in a laboratory and the question of rejecting measurements is handled in slightly different ways for each. In[7]:= Out[7]= In the above, the values of p and v have been multiplied and the errors have ben combined using Rule 1. Wolfram Data Framework Semantic framework for real-world data.
## Error Analysis In Physics Experiments
Wird geladen... https://phys.columbia.edu/~tutorial/ Sprache: Deutsch Herkunft der Inhalte: Deutschland Eingeschränkter Modus: Aus Verlauf Hilfe Wird geladen... Error Analysis Physics Lab Report There is a mathematical procedure to do this, called "linear regression" or "least-squares fit". Error Analysis Physics Example There was an error retrieving your Wish Lists.
Share Facebook Twitter Pinterest Hardcover from \$20.70 Paperback from \$2.98 Other Sellers from \$2.98 More Buying Choices 3 New from \$71.04 36 Used from \$2.98 39used&newfrom\$2.98 See All Buying Options This this page Wird verarbeitet... It also varies with the height above the surface, and gravity meters capable of measuring the variation from the floor to a tabletop are readily available. In[39]:= In[40]:= Out[40]= This makes PlusMinus different than Datum. Error Analysis In Physics Pdf
Another advantage of these constructs is that the rules built into EDA know how to combine data with constants. Do you have a recommendation for a reference or slightly more advanced book to follow Taylor? Rule 3: Raising to a Power If then or equivalently EDA includes functions to combine data using the above rules. get redirected here Now, for Bevington [3rd edition] I'm looking at covering: Chapter 1 - Uncertainties in Measurements Chapter 2 - Probability Distributions Chapter 3 - Error Analysis Chapter 4 - Estimates of Means
Support FAQ Wolfram Community Contact Support Premium Support Premier Service Technical Services All Support & Learning » Company About Company Background Wolfram Blog News Events Contact Us Work with Us Careers Error Propagation Physics In[12]:= Out[12]= To form a power, say, we might be tempted to just do The reason why this is wrong is that we are assuming that the errors in the two The standard deviation has been associated with the error in each individual measurement.
## The friendliest, high quality science and math community on the planet!
Applying the rule for division we get the following. Please try the request again. An example is the calibration of a thermocouple, in which the output voltage is measured when the thermocouple is at a number of different temperatures. 2. Percent Error Physics By declaring lists of {value, error} pairs to be of type Data, propagation of errors is handled automatically.
Lectures and textbooks often contain phrases like: A particle falling under the influence of gravity is subject to a constant acceleration of 9.8 m/. The task of ever-increasing observational and analytical precision is both an art and a science in and of itself, and one of the tasks of any scientist is to discover where Comment 2 people found this helpful. useful reference Such a procedure is usually justified only if a large number of measurements were performed with the Philips meter.
We repeat the measurement 10 times along various points on the cylinder and get the following results, in centimeters. Elwin.Martin, Mar 16, 2012 (Want to reply to this thread? Please try again Report abuse 5.0 out of 5 starsExcellent! How about 1.6519 cm?
For example, assume you are supposed to measure the length of an object (or the weight of an object). In[1]:= In[2]:= Out[2]= In[3]:= Out[3]= In[4]:= Out[4]= For simple combinations of data with random errors, the correct procedure can be summarized in three rules. or its affiliates v Forums Search Forums Recent Posts Unanswered Threads Videos Search Media New Media Members Notable Members Current Visitors Recent Activity New Profile Posts Insights Search Log in or Each data point consists of {value, error} pairs.
Instructors often put books and other reading materials in Course Reserves so everyone in a class can access them. Taylor (Author), John Taylor (Author) 4.5 out of 5 stars 4 customer reviews ISBN-13: 978-0935702101 ISBN-10: 0935702105 Why is ISBN important? Wenn du bei YouTube angemeldet bist, kannst du dieses Video zu einer Playlist hinzufĂĽgen. Log in or Sign up here!) Show Ignored Content Know someone interested in this topic?
Uncertainty due to Instrumental Precision Not all errors are statistical in nature. In[29]:= Out[29]= In[30]:= Out[30]= In[31]:= Out[31]= The Data and Datum constructs provide "automatic" error propagation for multiplication, division, addition, subtraction, and raising to a power. Thus, we can use the standard deviation estimate to characterize the error in each measurement. In this case the meaning of "most", however, is vague and depends on the optimism/conservatism of the experimenter who assigned the error.
Chapter 4 deals with error propagation in calculations. You can change this preference below. Please try again Report abuse See all 4 customer reviews (newest first) Write a customer review Customer Images Search Customer Reviews Search What Other Items Do Customers Buy After Viewing This Read more Published 10 months ago by JeepGen 5.0 out of 5 starsAn Introduction to Error Analysis: The Study of Ordered for a college class.
However, fortunately it almost always turns out that one will be larger than the other, so the smaller of the two can be ignored. | 1,723 | 7,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-39 | latest | en | 0.91299 |
https://blog.ssis.edu.vn/104687/category/math/ | 1,540,278,870,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516117.80/warc/CC-MAIN-20181023065305-20181023090805-00270.warc.gz | 640,716,030 | 7,641 | ## Multiplication Unit
In math, we are learning about multiplication, division and the strategies we can use to figure out the product. I learned that there are many different ways to solve a multiplication problem. Some different strategies are The Lattice Method, The Traditional Algorithm, The Area Model and The Partial Products Method.
29×84=___?___
____Partial Products: 20×80 = 1,600
________________ 20×4 = 800
________________ 9×80 = 720
________________ 9×4 =+___36__
_________________ _2,556_
_________________ +1 | 148 | 663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-43 | latest | en | 0.498191 |
https://www.physicsforums.com/threads/translating-quantified-statements.698149/ | 1,503,576,098,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886133449.19/warc/CC-MAIN-20170824101532-20170824121532-00439.warc.gz | 942,156,753 | 16,604 | # Translating Quantified statements.
1. Jun 21, 2013
### Bashyboy
1. The problem statement, all variables and given/known data
Let L(x, y) be the statement “x loves y,” where the do main for both x and y consists of all people in the world. Use quantifiers to express each of these statements.
g) There is exactly one person whom everybody loves.
2. Relevant equations
3. The attempt at a solution
I couldn't determine the answer myself, so I looked to the answer key for aid. According to the answer key, $\exists x[(\forall y L(y,x) \wedge \forall z ( \forall w L(w,z)) \implies z = x))]$ is the proper translation. However, I the introduction of the variable w is extraneous, that the answer to be simplified roughly to $\exists x \forall y [(L(y,x) \wedge \forall z L(y,z)) \implies z = x)]$ Would you care to share your opinion?
2. Jun 21, 2013
### haruspex
I disagree. First, let's make the first form a bit clearer (not sure what the standard binding order is, and a parenthesis is unmatched):
$\exists x[(\forall y L(y,x)) \wedge (\forall z ( \forall w L(w,z) \implies z = x))]$ .
In your version, the implication at the end has the wrong relationship to L(y,x). Worse, what do you think this translates to: $\forall z L(y,z)$?
3. Jun 22, 2013
### verty
Some authors define the quantifier $\exists!$ to mean there exists uniquely, there exists one and only one. It translates this way:
$\exists! x P(x)$ <==> $\exists x (P(x) \wedge \forall y (P(y) → y = x))$ where y is not a free variable of P.
You should verify that the answer given is just a version of $\exists! x \forall y L(y,x)$. | 446 | 1,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2017-34 | longest | en | 0.848525 |
https://documen.tv/question/question-what-are-the-solutions-to-3-2-12-6-0-hint-use-the-quadratic-formula-21518339-59/ | 1,716,151,737,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00426.warc.gz | 178,092,338 | 16,426 | ## Question: What are the solutions to 3x^2 + 12x + 6 = 0? Hint:: use the quadratic formula
Question
Question:
What are the solutions to 3x^2 + 12x + 6 = 0?
Hint:: | 59 | 166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.836567 |
https://www.physicsforums.com/threads/help-with-finding-parallel-vectors.380313/ | 1,508,651,083,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00332.warc.gz | 951,577,767 | 15,856 | # Help with finding parallel vectors
1. Feb 21, 2010
1. The problem statement, all variables and given/known data
The question states "let u= 2i+mj-10k and v=i-3j+nk, find the value of n and m such that u,v are parallel", the second part states asks the same but "u,v are perpendicular"
2. Relevant equations
3. The attempt at a solution
I attempted to use a dot product solution I guess, because vectors u+v should equal 0 when perpendicular. I'm lost on what to actually do. Thanks in advance.
2. Feb 21, 2010
### RoyalCat
Remember the definition of the dot product:
$$\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z$$
As you correctly stated, for two perpendicular vectors, the dot product is 0.
Let two vectors, $$\vec w, \vec q$$ be parallel. That means that they're both in the same direction, and therefore, they only differ by some scalar factor R (For instance, $$\vec q$$ could be 2 times longer than $$\vec w$$ (R=2) or it could be the same length, but anti-parallel (A negative R value of -1 would achieve that goal) or 2 times longer, but anti-parallel (R=-2)).
So in general, we can write: $$\vec q=R\vec w$$
Note that we've written a vector equation. That's actually 3 scalar equations in one. Simply solve for your three variables, $$R, m, n$$ and you're done.
What's important is that you understand how we've identified parallel\anti-parallel vectors and perpendicular ones. Are these two points clear to you?
3. Feb 21, 2010
### ehild
If two vectors
u=x1 i +y1 j + z1 k and v=x2 i +y2 j + z2 k
are parallel, then one of them is a multiple of the other
u=a*v (a is a scalar).
That means the same for all components:
x1=a*x2
y2=a*y2
z1=a*z2.
The two vectors are perpendicular if their dot product is 0 which means
x1x2 + y1y2+ z1z2 = 0.
ehild
4. Feb 21, 2010
I somewhat understand what you are saying, so are you saying I need to isolate the unknowns? I still don't fully understand this.
5. Feb 21, 2010
### Juan Pablo
Yes, solve the linear system of equations.
6. Feb 21, 2010
Ok, trying this with the perpendicular qustion it'd be "(2i+mj-10k) . (i-3j+nk)=0". I would have 2 unknowns, how would I solve this. I still don't know how to go about the parallel question. | 647 | 2,218 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2017-43 | longest | en | 0.922697 |
https://blackdiamondgames.blogspot.com/2013/10/game-store-on-budget-boom.html | 1,619,178,219,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00500.warc.gz | 239,423,408 | 19,293 | ## Friday, October 4, 2013
### Game Store on a Budget (The Boom)
Lets assume we're building an average game store in Averageville, USA. We've got a decent population base, perhaps a college or military base, and we're hoping for modest, lets say... average ... gross sales of around \$200,000 a year. My average salary will be about \$30,000 plus profits, which with \$200K/year will be \$12,000. A \$42,000 salary is actually the national average. Good for you!
How do you get to your average salary? That requires that \$200K gross sales, which requires inventory. We're going to assume again that you're average, and average inventory turns in the game trade, when things are normal, is three. We'll also assume average cost of goods of around 55%. The opposite of this number, your gross margin, is 100% minus 55%, or 45%. These numbers get confusing, just remember they're mirror images of each other.
So this isn't so terribly hard. We now have a simple math problem. \$200,000 divided by three turns is \$66,666. 55% of that number is \$36,666. That's your inventory number. You'll probably spend an equal amount or more getting started, but it's possible to get going for a lot less. That inventory number is key though. It's your economic engine. You won't hit your numbers without it, unless you boost your turns, which is what's happening right now.
So how does this look? We'll assume your first year you won't open your doors with \$36,666 in inventory, as you really don't know what customers want. You'll add inventory over that first year as you narrow it down, and because of that, your turn rates will be low.
Year Inventory @ Retail Turns Gross Sales 1 66,666.66 2 \$133,333.32 2 66,666.66 2.5 \$166,666.65 3 66,666.66 3 \$199,999.98
So before you can make your average salary, you'll need to have some money put away to cover the losses you'll incur during the first couple of years, what we call startup losses. How much? Good question, since sales projections are just that, projections. I would be lying if I threw a number out.
So why are stores popping up like crazy right now? It's the turns! Magic is huge at the moment, and turns on CCGs are nothing like your regular game sales. CCG's which account for 50% or more of a lot of store's income right now, turn in the 8-16 range. Now lets look at our calculations, assuming 50% of our sales are CCGs:
Year Reg Inventory Turns Gross Sales CCG Inventory Turns Gross Sales Total Sales 1 33,333.00 2 \$66,666.00 33,333.00 4 \$133,332.00 \$199,998.00 2 33,333.00 2.5 \$83,332.50 33,333.00 6 \$199,998.00 \$283,330.50 3 33,333.00 3 \$99,999.00 33,333.00 8 \$266,664.00 \$366,663.00
Look at that! Your average \$200,000/year store is now at \$366,663/year. Your average salary is now well over \$55,000/year. I would like to say that's what motivates new game store owners, the math, but it's really more abstract, as in "Magic! Shiny!"
So is there a down side? Well, if you built your store to be a \$200,000 store, no. You've just gotten a kick start for your small business. However, if you built a \$366,663 store, with \$366,663 expenses, expecting the CCG boom to continue, assuming it's your baseline, you will, without a doubt, have serious trouble down the road. Lots of stores are going to fail.
So what do you do if the majority of sales are Magic and you're worried about a fall? First would be not to overextend yourself beyond what can be easily scaled back when sales begin to flag. That primarily means your harder expenses, like rent, but also wages. By all means hire the people you need, but know there will come a very uncomfortable day. Also delay buying that fancy car and if it was possible, I would say delay buying a house (I don't think game store owners can get houses anymore).
I can't say don't take advantage of what's happening, because that's what business is about. I can say build muscles in other areas or work out atrophied muscles if you had a store before the boom. Build communities in other game departments, diversify inventory and work to sell it, even if it seems kind of futile with all that Magic money around. The reason is it takes a lot of time to build muscle after it's atrophied, and your store will be dead if it's only muscle is the Magic muscle. If you have no idea what to do, just bank the cash. Cash solves a lot of problems, although no amount of cash can save a failed business model.
Personally, we've paid off all our debt, upgraded our infrastructure, and, I'll admit, I've had a year of fun. The future will be about expansion in one form or another, with the assumption that things will return to normal soon. | 1,179 | 4,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-17 | longest | en | 0.965886 |
http://yourfinancebook.com/minimum-balance-saving-bank-account/ | 1,480,702,216,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540409.8/warc/CC-MAIN-20161202170900-00109-ip-10-31-129-80.ec2.internal.warc.gz | 805,718,354 | 12,375 | # Understanding Minimum Balance Requirements for saving accounts
Till date RBI has not stipulated any minimum balance requirements that need to be maintained by a customer in his saving or current account.
Banks are prescribing these requirements based on the cost involved in maintaining accounts. If you do not maintain your minimum balance then bank will levy specific charges for it. In private banks these charges are higher compare to nationalized banks.
Minimum balance in a saving account is calculated either monthly (MAB) or quarterly (QAB). If your bank’s requirement is to maintain MAB then bank will take all day’s closing balance and divide it by the number of days in a month to get your average minimum balance that you are required to maintain in that month.
If its higher or equal to the minimum requirements then you are safe. If it is lower than the minimum requirement then as per bank’s policy, you will be charged penalty for not maintaining such balance with them.
Similarly QAB is calculated by taking all day’s closing balance and divided by number of days in a quarter.
Formula for MAB = Total of all day’s closing balance / Number of days in a month
Formula for QAB = Total of all day’s closing balance / Number of days in a quarter
Minimum balance is actually an average that you need to maintain with the bank. If your MAB is Rs. 5, 000 then you need to maintain an average balance of Rs. 5, 000 throughout the month. To do that, you can even deposit Rs. 1, 50,000 for a day and maintain zero amount for the rest of the period in that month as average balance will be Rs.5, 000 i.e. 1, 50,000/30.
Similarly QAB is also calculated. If your QAB is Rs. 5, 000 then you can deposit Rs. 4, 50,000 for a day and maintain zero amount for rest of the period in that quarter.
Compare to private banks, nationalised banks like SBI, PNB, BOB, UBI and BOI have lower minimum balance requirements. Even non maintenance of minimum balance, penalty is very low.
Recently most of these private banks moved from QAB requirements to MAB by which you have to maintain higher balance in these banks. However, RBI has made it very clear that banks cannot change its minimum balance requirements without intimating the account holder. In some cases, individuals while opening their account or after opening the account are not aware about such requirements. When they do not maintain such balance in their account, bank charge them penalty. If you have any concern on minimum balance then you can report it to RBI.
## One thought on “Understanding Minimum Balance Requirements for saving accounts”
1. Tulika Majumdar
Had gone through a terrible experience with Icici bank. I had a savings bank account in icici bank indirapuram branch, Ghaziabad. While opening the bank account in the year 2013, bank didn’t give any information regarding Mab charges. From last 3 months bank was deducting RS 510 from my account without any intimation. I came to know only when I checked my bank statement 2 days back. Immediately went to bank and inquired and in return the only solution I got from them was to close the account. They didn’t even bother about customers grievances. These private banks r very pathetic. | 697 | 3,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-50 | longest | en | 0.937977 |
https://www.wallstreetprep.com/knowledge/fixed-asset-turnover-ratio/ | 1,719,118,210,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00774.warc.gz | 914,452,977 | 32,048 | Welcome to Wall Street Prep! Use code at checkout for 15% off.
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# Fixed Asset Turnover Ratio
Step-by-Step Guide to Understanding Fixed Asset Turnover Ratio
Last Updated February 8, 2024
## How to Calculate Fixed Asset Turnover Ratio
The fixed asset turnover ratio tracks how efficiently a company’s assets are being used (and producing sales), similar to the total asset turnover ratio.
However, the distinction is that the fixed asset turnover ratio formula includes solely long-term fixed assets, i.e. property, plant & equipment (PP&E), rather than all current and non-current assets.
Common examples of fixed assets that provide long-term economic benefits (>1 year) include the following:
• Land and Property
• Buildings and Offices
• Machinery
• Equipment
• Vehicles
Therefore, the fixed asset turnover ratio determines if a company’s purchases of fixed assets – i.e. capital expenditures (Capex) – are being spent effectively or not.
## Fixed Asset Turnover Ratio Formula
The formula to calculate the fixed asset turnover ratio compares a company’s net revenue to the average balance of fixed assets.
Fixed Asset Turnover Ratio = Net Revenue ÷ Average Fixed Assets
Where:
• Net Revenue = Gross Revenue – Returns – Discounts – Allowances
• Average Fixed Assets = (Beginning + Ending Fixed Assets) ÷ 2
## What is a Good Fixed Asset Turnover Ratio?
The fixed asset turnover ratio answers, “How much revenue is generated per dollar of fixed asset owned?”
If a company’s fixed asset turnover is 2.0x, it is implied that each dollar of fixed assets owned results in \$2.00 of revenue.
In general, the higher the fixed asset turnover ratio, the better, as the company is implied to be generating more revenue per dollar of long-term assets owned.
• High Fixed Asset Turnover → The company is implied to be purchasing long-term assets efficiently.
• Low Fixed Asset Turnover → The company is NOT receiving sufficient value (i.e. revenue) in return from its long-term assets.
Given how costly fixed asset purchases can be – on the initial date of purchase, as well as the associated maintenance (or replacement) expenses – Capex decisions must be carefully made.
Otherwise, operating inefficiencies can be created that have significant implications (i.e. long-lasting consequences) and have the potential to erode a company’s profit margins.
## How to Interpret Fixed Asset Turnover by Industry?
After calculating the fixed asset turnover ratio, the efficiency metric can be compared across historical periods to assess trends.
Comparisons to the ratios of industry peers can gauge how a company fares against its competitors regarding its spending on long-term assets (i.e. whether it is more efficient or lagging behind peers).
But to be useful, the ratio must be compared to industry comparables, or companies with similar characteristics as the target company, such as similar business models, target end markets, and risks.
The average ratio varies substantially across different industries. For instance, comparisons between capital-intensive (“asset-heavy”) industries cannot be made with “asset-lite” industries, since their business models and reliance on long-term assets are too different.
In particular, Capex spending patterns in recent periods must also be understood when making comparisons, as one-time periodic purchases could be misleading and skew the ratio.
## Fixed Asset Turnover Ratio Calculator
We’ll now move to a modeling exercise, which you can access by filling out the form below.
Submitting...
## 1. Operating Assumptions
Suppose an industrials company generated \$120 million in net revenue in the past year, with \$40 million in PP&E.
After that year, the company’s revenue grows by 10%, with the growth rate then stepping down by 2% per year.
The company’s PP&E, the only fixed asset on its balance sheet, falls by 8% after Year 0 – with the growth rate then stepping up by 1% each year in each subsequent period (i.e. to negative 4% by the end of the projection period).
• Year 0 → Year 1: 10% Revenue Growth; (8%) PP&E Growth
• Year 1 → Year 2: 8% Revenue Growth; (7%) PP&E Growth
• Year 2 → Year 3: 6% Revenue Growth; (6%) PP&E Growth
• Year 3 → Year 4: 4% Revenue Growth; (5%) PP&E Growth
• Year 4 → Year 5: 2% Revenue Growth; (4%) PP&E Growth
From Year 0 to the end of Year 5, the company’s net revenue expanded from \$120 million to \$160 million, while its PP&E declined from \$40 million to \$29 million.
In our hypothetical scenario, we can assume that the company’s revenue model is shifting from being predominantly comprised of one-time expensive purchases to recurring component purchases and services related to maintenance.
Unlike the initial equipment sale, the revenue from recurring component purchases and services provided to existing customers requires less spending on long-term assets.
For example, inventory purchases or hiring technical staff to service customers are cheaper than major Capex.
## 2. Fixed Asset Turnover Calculation Example
We can now calculate the fixed asset turnover ratio by dividing the net revenue for the year by the average fixed asset balance, which is equal to the sum of the current and prior period balance divided by two.
• Fixed Asset Turnover = Net Revenue ÷ Average (Current, Prior Period Fixed Asset Balance)
The calculated fixed turnover ratios from Year 1 to Year 5 are as follows.
• Fixed Asset Turnover, Year 1 = 3.4x
• Fixed Asset Turnover, Year 2 = 4.0x
• Fixed Asset Turnover, Year 3 = 4.6x
• Fixed Asset Turnover, Year 4 = 5.0x
• Fixed Asset Turnover, Year 5 = 5.4x
Despite the reduction in Capex, the company’s revenue is growing – higher revenue is generated on lower levels of Capex purchases.
Since the company’s revenue growth remains robust across the 5-year forecast period, while its Capex spending declined in the same period, the fixed asset turnover ratio trends upward.
Step-by-Step Online Course
### Everything You Need To Master Financial Modeling
Enroll in The Premium Package: Learn Financial Statement Modeling, DCF, M&A, LBO and Comps. The same training program used at top investment banks. | 1,390 | 6,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.8643 |
http://www.expertsmind.com/library/calculating-cost-of-preferred-stock-51297897.aspx | 1,618,563,979,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00518.warc.gz | 128,586,552 | 12,152 | ### Calculating cost of preferred stock
Assignment Help Finance Basics
##### Reference no: EM131297897
1. Calculating Cost of Equity. Stock in CDB Industries has a beta of .90. The market risk premium is 7 percent, and T-bills are currently yielding 3.5 percent. CDB's most recent dividend was \$1.80 per share, and dividends are expected to grow at a 5 percent annual rate indefinitely. If the stock sells for \$47 per share, what is your best estimate of CDB's cost of equity?
2. Calculating Cost of Preferred Stock. Sixth Fourth Bank has an issue of preferred stock with a \$6.25 stated dividend that just sold for \$108 per share. What is the bank's cost of preferred stock?
3. Calculating Cost of Debt. ICU Window, Inc., is trying to determine its cost of debt. The firm has a debt issue outstanding with seven years to maturity that is quoted at 108 percent of face value. The issue makes semiannual payments and has an embedded cost of 6.1 percent annually. What is ICU's pretax cost of debt? If the tax rate is 38 percent, what is the aftertax cost of debt?
4. Finding the WACC. Given the following information for Janicek Power Co., find the WACC. Assume the company's tax rate is 35 percent.
Debt: 8,500 7.2 percent coupon bonds outstanding, \$1,000 par value, 25 years to maturity, selling for 118 percent of par; the bonds make semiannual payments.
Common stock: 225,000 shares outstanding, selling for \$87 per share; beta is 1.15.
Preferred stock: 15,000 shares of 4.8 percent preferred stock outstanding, currently selling for \$98 per share.
Market: 7 percent market risk premium and 3.1 percent risk-free rate.
5. EBIT and Leverage. Kaelea, Inc., has no debt outstanding and a total market value of \$125,000. Earnings before interest and taxes, EBIT, are projected to be \$10,400 if economic conditions are normal. If there is strong expansion in the economy, then EBIT will be 20 percent higher. If there is a recession, then EBIT will be 35 percent lower. Kaelea is considering a \$42,000 debt issue with a 6 percent interest rate. The proceeds will be used to repurchase shares of stock. There are currently 6,250 shares outstanding. Ignore taxes for this problem.
a. Calculate earnings per share, EPS, under each of the three economic scenarios before any debt is issued. Also, calculate the percentage changes in EPS when the economy expands or enters a recession.
b. Repeat part (a) assuming that Kaelea goes through with recapitalization. What do you observe?
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Where do the components of the changes in retained earnings appear in the statement of cash flows? Assume the indirect method is used to prepare the statement of cash flows. | 971 | 4,315 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-17 | latest | en | 0.920554 |
https://www.streetauthority.com/18567/the-market-has-a-liquidity-problem-for-now/ | 1,675,374,975,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00215.warc.gz | 1,029,170,498 | 19,426 | # The Market Has A Liquidity Problem (For Now…)
Trading was unusually light last week and likely be light this week. In fact, there’s not enough liquidity for me to recommend a trade in one of my premium services, which is unusual. In fact, this is the first time I haven’t had enough confidence in the market’s liquidity to recommend a trade.
Let me explain why this matters…
### Liquidity Is Good For Markets
Liquidity describes how easily a trade can be completed. For example, when we sell a put, we need someone to buy that contract in order to complete the trade.
This doesn’t mean we need someone who expects the stock to decline. Although buying puts is one way to benefit from an expected decline, there are other reasons traders buy the puts that we sell.
The most common reason there is a market for our trades is high-frequency trading (HFT), which now accounts for more than half of all trades.
Many individual investors believe HFT firms take advantage of them. In my opinion, HFT firms are among the most positive developments for individual investors since commissions were deregulated in 1975.
When we place an order to sell, our broker sends the order to the market. The market is no longer a group of men standing around the floor of the exchange like it was 50 years ago. Now, the exchange is a collection of computers.
The order displays on a computer, and HFT firms compete to fill the order. The firms execute trades with high-speed computers and algorithms that allow the firms to profit from trades priced in fractions of a penny.
Many of the algorithms use a formula to define the relationship between the three products an investor could buy – the stock, a put, and a call. The correct value of a stock in relation to its options is defined by the put-call parity equation. The simple version of the formula says that the value of a put plus a share of the stock is equal to the value of a call and the present value of the option strike price.
P + S = C + PV(A)
where…
P = price of put with strike price A
S = price of stock
C = price of call with strike price A
PV(A) = net present value of the option’s strike price
Basically, it says the value of a put and the stock is equal to a call and some cash.
When we enter an order to sell a put, the firm needs to balance that equation to make a small profit. They do this thousands of times a day and it can be quite profitable. Most importantly, from my perspective, they allow us to execute trade.
It means we can be sure we are getting fair prices for the options we trade. If we were not getting fair prices, arbitrage traders would jump into the market and push the mispriced option back in line with the put-call parity equation.
Arbitrage traders look for assets that are equivalent to each other but are traded in different markets. If one market is mispricing the asset, the arbitrage trader can make a risk-free profit by buying in the cheaper market and selling in the more expensive market. For Wall Street firms, this is easier than it sounds.
Years ago, the classic example of arbitrage focused on New York and Chicago. Futures on the S&P 500 were traded in Chicago while the individual stocks in the index traded in New York.
If futures prices deviated from the value of the 500 stocks, arbitrage traders would buy the market that was underpriced and sell the overpriced market. They would then reverse the trade with stocks and futures priced at the correct value. The result was a riskless and guaranteed profit.
This trade required access to real-time market data, large amounts of trading capital to buy and sell 500 stocks at a time, and high-speed computers to spot the trading opportunities. Only the largest firms could compete at first, but eventually other firms found a way to take part in arbitrage trading. Those other firms are the HFT firms that allow us to trade.
### Action To Take
To be honest, the type of trading we typically do would not have been so consistently profitable even in the early 2000s. That’s because our trades would’ve depended on the ability of the men — and a few women by that time — on the exchange floor to balance our orders so they could be filled.
But there are times when the liquidity just isn’t there. Generally, these periods are restricted to a few hours or a few days at a time. Last week, with the holiday-shortened schedule, volume is low and the puts I had signals in are not trading. I don’t believe there is enough liquidity that we could get into the trade at a fair price, and overpaying to open a trade is one of the quickest ways to lose money.
If you’re trading on a regular basis, hopefully this can be an illustrative lesson for you. Sometimes you just need to back away and let the market come to you. Either way, I’m certain this is temporary and expect it to be resolved by next week. By then, traders will be back from holiday breaks and go back to providing liquidity for end-of-year portfolio rebalancing.
P.S. Regardless of what happens in the market in 2021, I have a plan to ensure that we get paid either way…
I like to think of it as an “insurance” plan — because it allows us to get paid instantly, rather than sit around and wait. My subscribers and I have been trading this way for years, allowing us to pocket hundreds (or even thousands) of dollars with very little effort.
I think every investor who’s looking for income owes it to themselves to learn more about how this works.
That’s why I just released a brand new report that gives you all the details. You can access it right here. | 1,195 | 5,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-06 | latest | en | 0.954821 |
https://help.altair.com/hwsolvers/os/topics/solvers/os/global_search_option_r.htm | 1,716,151,999,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057922.86/warc/CC-MAIN-20240519193629-20240519223629-00518.warc.gz | 265,025,523 | 14,601 | # Global Search Option
A common discussion that arises when an optimization problem is solved is whether or not the obtained optimum is a local or global optimum.
Local approximation based methods (gradient-based optimizations) are more susceptible to finding a local optimum, while global approximation methods (response surface methods) and exploratory techniques (genetic algorithms) are less susceptible than local approximation based methods to finding a local optimum. In other words, these techniques improve the chances of finding a more global optimum. However, no algorithm can guarantee that the optimum found is in fact the global optimum. An optimum can be guaranteed to be the global optimum only if the optimization problem is convex. For a convex optimization problem, the objective function and feasible domain need to be convex. Unfortunately, in reality, most engineering problems being solved cannot be shown to be convex. Therefore, for practical problems, a global optimum remains elusive. Different algorithm types simply alter the chances of finding a more global optimum, not guarantee it. With that consideration, it is important to keep in mind that algorithms which improve the chances come at a computational cost, and most often this can be significant.
Figure 1 illustrates the concept of a convex problem as discussed above. A convex optimization problem has just one minimum (or maximum). This minimum (Point A in the image) is the global minimum.
In the case of non-convex problems solved using gradient-based techniques, the inherent behavior is that the optimized result obtained is dependent on the initial design starting point. This makes these types of algorithms all the more susceptible to finding local optimum. Recently implemented in OptiStruct version 11.0, is a new global search algorithm - an extension to the gradient-based optimization approach. The approach is called Multiple Starting Points Optimization. This global search algorithm performs an extensive search of the design space for multiple starting points to improve the chances of finding a more global optimum. Being dependent on the initial design starting point, n different design starting points could potentially result in n different optimum solutions. It is also highly likely that different design starting points could result in the same optimum solution. However, this does not mean that the optimum solution found is the global optimum. This concept is illustrated in Figure 2.
Consider the non-convex function, f(x), bounded by -a ≤ x ≤ b. Optimizing a design from design starting point A will result in the optimum solution, P. Similarly, optimizing a design from starting point B will result in the same optimum solution, P. On the other hand, optimizing a design from initial design starting point C will result in the optimum solution, Q. From this, it can be seen that through the multiple starting points approach, a global optimum cannot be guaranteed (as with any other algorithm), but at the same time, the chances of finding a more global optimum are improved.
The Global Search Option (GSO) in OptiStruct is supported for Size and Shape Optimization.
## Input
Identifying a global search optimization study is done through the DGLOBAL entry in the I/O Options section of the input deck, and the parameters required to setup and run a GSO are defined on the DGLOBAL Bulk Data Entry.
Parallelization
The Global Search Optimization can be run in parallel with the help of the Domain Decomposition Method (DDM). In this process, the starting points (NPOINT) are assigned in parallel to the DDM MPI processes (first level of parallelization). If enough number of cores are available on the machine, you can use PARAM,DDMNGRPS,<ngrps> to divide the number of MPI processes (-np) into groups. The starting points are again assigned sequentially, but now they are assigned to the MPI process groups (instead of individual MPI processes). Each MPI process group will now solve a starting point via DDM geometric partitioning (second level of parallelization).
For additional information, refer to DDM Level 2 – Parallelization of Geometric Partitions in Domain Decomposition Method (DDM).
## Output
If multiple starting points from defined via DGLOBAL entry converge to multiple unique designs, then the output files corresponding to each unique design are retained in your working directory. If all starting points converge to a single design, then only one set of output files are retained, corresponding to this common design. | 867 | 4,567 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-22 | longest | en | 0.896804 |
http://mca.nowgray.com/2017/04/decidability-of-language-containing.html | 1,502,927,506,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102757.45/warc/CC-MAIN-20170816231829-20170817011829-00179.warc.gz | 273,992,765 | 33,220 | Cheap and Secure Web Hosting Provider : See Now
Decidability of language containing strings of length equal to that of some substring of 1s in $\pi$?
, ,
Problem Detail:
I read that the following language is decidable.
$\{w | w \in \{0, 1\}^* \text{ and } 1^{|w|} \text{ is a substring in binary expansion of } \pi \}$
The proof has been given by considering the two possible cases
• Suppose it is true that there exists a minimal integer $N$ such that the binary expansion of $\pi$ only contains substrings of 1 having length at most $N$. That is, it contains substring $1^N$ but not $1^{N + 1}$. Define a corresponding turing machine $M_N$ such that it accepts the input string if it is of length $N$, otherwise rejects it. This machine decides the given language.
• Suppose the binary expansion of $\pi$ contains substrings of $1$s of all lengths. Then the Turing machine that accepts all strings decides the given language.
We are not able to construct a single Turing machine that decides the language given, but one of these machines thus constructed will decide it but we do not know which. I don't quite understand how this language is decidable?
Answered By : Andrej Bauer
You are confusing knowledge and truth. Something may be true without us knowing precisely why. An object may exist without us being able to construct it.
If you belive in classical logic, then it is true that the maximum length of consequtive 1's in the expansion of $\pi$ is either finite or infinite. In each case, there exists a decision procedure for your language. Therefore a decision procedure exists.
But we do not know which of the cases happens, and so we not not know a particular decision procedure, even though one exists. If you don't like this line of reasoning, then you should ask for a constructive proof of decidability of your language.
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Question Source : http://cs.stackexchange.com/questions/66947
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Let us know your responses and feedback | 463 | 2,025 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-34 | latest | en | 0.8966 |
https://www.physicsforums.com/threads/trigonometric-function-with-specific-properties.314058/ | 1,695,456,346,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506480.35/warc/CC-MAIN-20230923062631-20230923092631-00687.warc.gz | 1,071,109,248 | 15,931 | # Trigonometric function with specific properties
Is there any trigonometric function f(theta) where theta is an angle such that
f(Pi-theta)=-f(theta) and such that its square (i.e f(theta)^2) is maximum at Pi/2?
Thanks
try theta = PI/2
then f(PI-(PI/2)) = f(PI/2)
so:
if f(PI-theta)=-f(theta) for all theta, then f(PI/2) must equal zero.
and you also said that:
f(PI/2)^2 is its maximum. but f(PI/2)^2 is 0.
so f(theta) = zero, for all theta. A boring and useless function.
edit: Presuming f(theta) cannot be infinite
Last edited:
(have a pi: π and a theta: θ )
Is there any trigonometric function f(theta) where theta is an angle such that
f(Pi-theta)=-f(theta) and such that its square (i.e f(theta)^2) is maximum at Pi/2?
if f(PI-theta)=-f(theta) for all theta, then f(PI/2) must equal zero.
It depends exacty what the question envisages …
if f is allowed to be infinite at π/2, then you could have f(π/2)- = -f(π/2)+ = ±∞
if f is allowed to be infinite at π/2, then you could have f(π/2)- = -f(π/2)+ = ±∞
Ahh, I missed that. How about the tan() function then?
Thanks a lot guys..f cannot be infinite as it is a wave function in a physics system I am researching on..I just wanted to make sure that there aint any solutions except the f(theta)=0 as tiny-tim pointed out..Thanks again...
sin(Pi-theta) = sin(PI)cos(theta) - sin(theta)cos(pi) = -sin(theta)
And [sin(x)]^2 has local maxima at PI/2 + nPI.
Am I missing something? | 454 | 1,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-40 | latest | en | 0.894638 |
http://blog.al.com/breaking/2012/05/columnist_smarter_than_a_fifth.html | 1,542,629,907,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745761.75/warc/CC-MAIN-20181119105556-20181119131556-00106.warc.gz | 39,467,449 | 20,400 | 403 Forbidden
The question is, if I may borrow from the backwards format of a TV quiz show on which I'd at least stand half a chance, "Are you smarter than a fifth grader?"
I pretty much knew I wasn't all along. And for much of my adult life I've been reminded I'm not smarter than the teenagers under my roof, too.
Specifically, I know I'm not smarter than a Horizon Elementary School fifth grader named Isaiah Jackson.
If it's any consolation, nobody else was either.
The Madison Optimist Club, as part of its scholarship awards banquet Thursday evening, held a take-off of the TV show, "Are You Smarter Than A Fifth Grader?" that stars the comedian Jeff Foxworthy.
As if regular public humiliation in this column isn't enough, the club's Jim Gilchrist invited me to play, along with Dr. Dee Fowler, superintendent of Madison City Schools; Ray White, president of the Madison City Board of Education; and Tommy Overcash, president of the0 Madison City Council.
Our opponents were students of Ms. Elizabeth Bero at Horizon. Along with Isaiah were Aiden Shaw, Hannalyn Wilks and Keara Lacy.
I pretty much had them pegged as those obnoxious kids we were all in school with who, when the bell rang, would remind the teacher she'd forgotten to assign homework. But these kids seemed far too nice.
Even with a "lifeline" we were hopelessly outclassed. Kelsey Kokes, a Bob Jones student and winner of one of the Optimist Club scholarships, was allowed to help on one of the five questions.
I'm not smarter than a 12th grader, either.
My questions were:
1. What did Paul Revere really say on his famous ride?
2. Who was Lincoln's opponent in the 1860 presidential election?
3. What is on the underside of a fern's fronds?
4. What mathematical operations are indicated by the phrase, "Please Excuse My Dear Aunt Sally?"
5. What is the number 6 on Moh's mineral hardness scale?
I'll give you a few paragraphs to come up with the answers.
Isaiah nailed them all.
Then, to add to the ignominy, as he finished scribbling down his answers, he would punch a big red button. It had a robotic voice that boasted, "That was easy."
Isaiah is the oldest of three children of Michael and Rhonda Jackson. When he was four, Rhonda said, "he started teaching himself to read."
Not only is he smarter than a newspaper columnist, he's also a better quote than a football coach or politician.
The students' preparation included a study of many of the questions that might be answered. Isaiah described his process wonderfully:
"We had to get memorized in our heads. We had to do these over and over and over until they stuck in our head. It was like dried glue-on paper."
And, "It was pretty memorize-able. My brain is like a sponge."
My brain is like a stale, one-pound block of Swiss cheese.
I had to don a dunce cap after my five academic airballs, having failed to know:
1. "The regulars are coming out!"
2. John Breckenridge.
3. Sori. ("I thought that was Tom Cruise's kid," I pleaded.)
4. Parenthesis, exponent, multiplication, division, addition and subtraction.
5. Orthoclase.
Isaiah had no small amount of delight in placing the cap back on my head to pose for a couple of pictures when the event ended.
He then offered me some faint consolation.
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https://coolgyan.org/ncert-solutions/ncert-solutions-class-6-maths-chapter-5-understanding-elementary-shapes/ | 1,679,784,578,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945376.29/warc/CC-MAIN-20230325222822-20230326012822-00106.warc.gz | 226,104,928 | 33,777 | # NCERT Solutions For Class 6 Maths Chapter 5 : Understanding Elementary Shapes
NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes are available here so that students can learn better and more effectively. These materials in PDF format can be downloaded by the students to practise offline. CoolGyan’S offers online learning materials such as notes, question papers, exemplar problems, etc. for almost all the classes which students can use as worksheets to prepare well for exams. NCERT Solutions for Class 6 Chapter 5 Understanding Elementary Shapes are provided here to support the students in their exam preparation. Practising these solved problems of NCERT Class 6 Chapter 5 will help the students in understanding the techniques to solve the different types of questions.
## NCERT Solutions for Class 6 Chapter 5: Understanding Elementary Shapes Download PDF
All the learning materials available in CoolGyan’S are designed by our subject experts with respect to CBSE syllabus. The NCERT Solutions for class 6 Maths contain solved questions which have been described in the best possible methods. Students can use these methods to solve problems in textbooks, sample papers and previous year question papers.
### NCERT Solutions Class 6 Maths Chapter 5 – Understanding Elementary Shapes
Exercise 5.1 Solutions
Exercise 5.2 Solutions
Exercise 5.3 Solutions
Exercise 5.4 Solutions
Exercise 5.5 Solutions
Exercise 5.6 Solutions
Exercise 5.7 Solutions
Exercise 5.8 Solutions
Exercise 5.9 Solutions
## Access NCERT Solutions for Class 6 Chapter 5: Understanding Elementary Shapes
Exercise 5.1 page no: 88
1. What is the disadvantage in comparing line segments by mere observation?
Solutions:
By mere observation we can’t compare the line segments with slight difference in their length. We can’t say which line segment is of greater length. Hence, the chances of errors due to improper viewing are more.
2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?
Solutions:
While using a ruler, chances of error occur due to thickness of the ruler and angular viewing. Hence, using divider accurate measurement is possible.
3. Draw any line segment, say . Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?
Solutions:
Since given that point C lie in between A and B. Hence, all points are lying on same line segment
. Therefore for every situation in which point C is lying in between A and B we may say that
AB = AC + CB
For example:
AB is a line segment of length 7 cm and C is a point between A and B such that AC = 3 cm and CB = 4 cm.
Hence, AC + CB = 7 cm
Since, AB = 7 cm
∴ AB = AC + CB is verified.
4. If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?
Solutions:
Given AB = 5 cm
BC = 3 cm
AC = 8 cm
Now, it is clear that AC = AB + BC
Hence, point B lies between A and C.
5. Verify, whether D is the mid point of .
Solutions:
Since, it is clear from the figure that AD = DG = 3 units. Hence, D is the midpoint of
6. If B is the mid point of and C is the mid point of , where A, B, C, D lie on a straight line, say why AB = CD?
Solutions:
Given
B is the midpoint of AC. Hence, AB = BC (1)
C is the midpoint of BD. Hence, BC = CD (2)
From (1) and (2)
AB = CD is verified
7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.
Solutions:
Case 1. In triangle ABC
AB= 2.5 cm
BC = 4.8 cm and
AC = 5.2 cm
AB + BC = 2.5 cm + 4.8 cm
= 7.3 cm
As 7.3 > 5.2
∴ AB + BC > AC
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 2. In triangle PQR
PQ = 2 cm
QR = 2.5 cm
PR = 3.5 cm
PQ + QR = 2 cm + 2.5 cm
= 4.5 cm
As 4.5 > 3.5
∴ PQ + QR > PR
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 3. In triangle XYZ
XY = 5 cm
YZ = 3 cm
ZX = 6.8 cm
XY + YZ = 5 cm + 3 cm
= 8 cm
As 8 > 6.8
∴ XY + YZ > ZX
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 4. In triangle MNS
MN = 2.7 cm
NS = 4 cm
MS = 4.7 cm
MN + NS = 2.7 cm + 4 cm
6.7 cm
As 6.7 > 4.7
∴ MN + NS > MS
Hence, the sum of any two sides of a triangle is greater than the third side.
Case 5. In triangle KLM
KL = 3.5 cm
LM = 3.5 cm
KM = 3.5 cm
KL + LM = 3.5 cm + 3.5 cm
= 7 cm
As 7 cm > 3.5 cm
∴ KL + LM > KM
Hence, the sum of any two sides of a triangle is greater than the third side.
Therefore we conclude that the sum of any two sides of a triangle is always greater than the third side.
Exercise 5.2 page no: 91
1. What fraction of a clockwise revolution does the hour hand of a clock turn through, when it goes from
(a) 3 to 9
(b) 4 to 7
(c) 7 to 10
(d) 12 to 9
(e) 1 to 10
(f) 6 to 3
Solutions:
We know that in one complete clockwise revolution, hour hand will rotate by 3600
(a) When hour hand goes from 3 to 9 clockwise, it will rotate by 2 right angles or 1800
∴ Fraction = 1800 / 3600
= 1 / 2
(b) When hour hand goes from 4 to 7 clockwise, it will rotate by 1 right angle or 900
∴ Fraction = 900 / 3600
= 1 / 4
(c) When hour hand goes from 7 to 10 clockwise, it will rotate by 1 right angle or 900
∴ Fraction = 900 / 3600
= 1 / 4
(d) When hour hand goes from 12 to 9 clockwise, it will rotate by 3 right angles or 2700
∴ Fraction = 2700 / 3600
= 3 / 4
(e) When hour hand of a clock goes from 1 to 10 clockwise, it will rotate by 3 right angles or 2700
∴ Fraction = 2700 / 3600
= 3 / 4
(f) When hour hand goes from 6 to 3 clockwise, it will rotate by 3 right angles or 2700
∴ Fraction = 2700 / 3600
= 3 / 4
2. Where will the hand of a clock stop if it
(a) starts at 12 and makes 1 / 2 of a revolution, clockwise?
(b) starts at 2 and makes 1 / 2 of a revolution, clockwise?
(c) starts at 5 and makes 1 / 4 of a revolution, clockwise?
(d) starts at 5 and makes 3 / 4 of a revolution, clockwise?
Solutions:
We know that one complete clockwise revolution, hour hand will rotate by 3600
(a) When hour hand of a clock starts at 12 and makes 1 / 2 revolution clockwise, it will rotate by 1800.
Hence, the hour hand of a clock will stop at 6.
(b) When hour hand of a clock starts at 2 and makes 1 / 2 revolution clockwise, it will rotate by 1800
Hence, the hour hand of a clock will stop at 8.
(c) When hour hand of a clock starts at 5 and makes 1 / 4 revolution clockwise, it will rotate by 900
Hence, hour hand of a clock will stop at 8.
(d) When hour hand of a clock starts at 5 and makes 3 / 4 revolution clockwise, it will rotate by 2700
Hence, hour hand of a clock will stop at 2
3. Which direction will you face if you start facing
(a) east and make 1 / 2 of a revolution clockwise?
(b) east and make 1 ½ of a revolution clockwise?
(c) west and make 3 / 4 of a revolution anti – clockwise?
(d) south and make one full revolution?
(should we specify clockwise or anti – clockwise for this last question? Why not?)
Solutions:
Revolving one complete round in clockwise or in anti – clockwise direction we will revolve by 3600 and two adjacent directions are at 900 or 1 / 4 of a complete revolution away from each other.
(a) If we start facing towards East and make 1 / 2 of a revolution clockwise, we will face towards West direction.
(b) If we start facing towards East and make 1 ½ of a revolution clockwise, we will face towards West direction
(c) If we start facing towards West and make 3 / 4 of a revolution anti – clockwise, we will face towards North direction
(d) If we start facing South and make one full revolution, again we will face the South direction.
In case of revolving 1 complete revolution, either clockwise or anti-clockwise we will be back at the original position.
4. What part of a revolution have you turned through if you stand facing
(a) east and turn clockwise to face north?
(b) south and turn clockwise to face east
(c) west and turn clockwise to face east?
Solutions:
By revolving one complete revolution either in clockwise or in anti-clockwise direction, we will revolve by 3600 and two adjacent directions are at 900 or 1 / 4 of a complete revolution away from each other
(a) If we start facing towards East and turn clockwise to face North, we have to make 3 / 4 of a revolution
(b) If we start facing towards South and turn clockwise to face East, we have to make 3 / 4 of a revolution
(c) If we start facing towards West and turn clockwise to face East, we have to make 1 / 2 of a revolution
5. Find the number of right angles turned through by the hour hand of a clock when it goes from
(a) 3 to 6
(b) 2 to 8
(c) 5 to 11
(d) 10 to 1
(e) 12 to 9
(f) 12 to 6
Solutions:
The hour hand of a clock revolves by 3600 or it covers 4 right angles in one complete revolution
(a) If hour hand of a clock goes from 3 to 6, it revolves by 900 or 1 right angle
(b) If hour hand of a clock goes from 2 to 8, it revolves by 1800 or 2 right angles
(c) If hour hand of a clock goes from 5 to 11, it revolves by 1800 or 2 right angles
(d) If hour hand of a clock goes from 10 to 1, it revolves by 900 or 1 right angle
(e) If hour hand of a clock goes from 12 to 9, it revolves by 2700 or 3 right angles
(f) If hour hand of a clock goes from 12 to 6, it revolves by 1800 or 2 right angles
6. How many right angles do you make if you start facing
(a) south and turn clockwise to west?
(b) north and turn anti – clockwise to east?
(c) west and turn to west?
(d) south and turn to north?
Solutions:
By revolving one complete round in either clockwise or anti-clockwise direction, we will revolve by 3600 and two adjacent directions are at 900 away from each other.
(a) If we start facing towards South and turn clockwise to West, we have to make one right angle
(b) If we start facing towards North and turn anti-clockwise to East, we have to make 3 right angles
(c) If we start facing towards West and turn to West, we have to make one complete round or 4 right angles
(d) If we start facing towards South and turn to North, we have to make 2 right angles
7. Where will the hour hand of a clock stop if it starts
(a) from 6 and turns through 1 right angle?
(b) from 8 and turns through 2 right angles?
(c) from 10 and turns through 3 right angles?
(d) from 7 and turns through 2 straight angles?
Solutions:
We know that in 1 complete revolution in either clockwise or anticlockwise direction, hour hand of a clock will rotate by 3600 or 4 right angles
(a) If hour hand of a clock starts from 6 and turns through 1 right angle, it will stop at 9
(b) If hour hand of a clock starts from 8 and turns through 2 right angles, it will stop at 2
(c) If hour hand of a clock starts from 10 and turns through 3 right angles, it will stop at 7
(d) If hour hand of a clock starts from 7 and turns through 2 straight angles, it will stop at 7
Exercise 5.3 page NO: 94
1. Match the following:
(i) Straight angle (a) Less than one-fourth of a revolution
(ii) Right angle (b) More than half a revolution
(iii) Acute angle (c) Half of a revolution
(iv) Obtuse angle (d) One-fourth of a revolution
(v) Reflex angle (e) Between 1 / 4 and 1 / 2 of a revolution
(f) One complete revolution
Solutions:
(i) Straight angle = 1800 or half of a revolution
(ii) Right angle = 900 or one-fourth of a revolution
(iii) Acute angle = less than 900 or less than one-fourth of a revolution
(iv) Obtuse angle = more than 900 but less than 1800 or between 1 / 4 and 1 / 2 of a revolution
(v) Reflex angle = more than 1800 but less than 3600 or more than half a revolution
2. Classify each one of the following angles as right, straight, acute, obtuse or reflex:
Solutions:
(i) The given angle is acute angle it measures less than 900
(ii) The given angle is obtuse angle as it measures more than 900 but less than 1800
(iii) The given angle is right angle as it measures 900
(iv) The given angle is reflex angle as it measures more than 1800 but less than 3600
(v) The given angle is straight angle as it measures 1800
(vi) The given angle is acute angle as it measures less than 900
Exercise 5.4 page no: 97
1. What is the measure of
(i) a right angle?
(ii) a straight angle
Solutions:
(i) The measure of a right angle is 900
(ii) The measure of a straight angle is 1800
2. Say True or False:
(a) The measure of an acute angle < 900
(b) The measure of an obtuse angle < 900
(c) The measure of a reflex angle > 1800
(d) The measure of one complete revolution = 3600
(e) If m ∠A = 530 and m ∠B = 350, then m ∠A > m ∠B.
Solutions:
(a) True, the measure of an acute angle is less than 900
(b) False, the measure of an obtuse angle is more than 900 but less than 1800
(c) True, the measure of a reflex angle is more than 1800
(d) True, the measure of one complete revolution is 3600
(e) True, ∠A is greater than ∠B
3. Write down the measures of
(a) some acute angles
(b) some obtuse angles
(give at least two examples of each)
Solutions:
(a) The measures of an acute angle are 500, 650
(b) The measures of obtuse angle are 1100, 1750
4. Measures the angles given below using the protractor and write down the measure.
Solutions:
(a) The measure of an angle is 450
(b) The measure of an angle is 1200
(c) The measure of an angle is 900
(d) The measures of an angles are 600, 900 and 1300
5. Which angle has a large measure? First estimate and then measure.
Measure of Angle A =
Measure of Angle B =
Solutions:
The measure of angle A is 400
The measure of angle B is 680
∠B has a large measure than ∠A
6. From these two angles which has larger measure? Estimate and then confirm by measuring them.
Solutions:
The measures of these angles are 450 and 550. Hence, angle shown in second figure is greater.
7. Fill in the blanks with acute, obtuse, right or straight:
(a) An angle whose measure is less than that of a right angle is _____
(b) An angle whose measure is greater than that of a right angle is ____
(c) An angle whose measure is the sum of the measures of two right angles is _______
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is _____
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ______
Solutions:
(a) An angle whose measure is less than that of a right angle is acute angle
(b) An angle whose measure is greater than that of a right angle is obtuse angle (but less than 1800)
(c) An angle whose measure is the sum of the measures of two right angles is straight angle
(d) When the sum of the measures of two angles is that of a right angle, then each one of them is acute angle
(e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be obtuse angle.
8. Find the measure of the angle shown in each figure. (First estimate with your eyes and then find the actual measure with a protractor).
Solutions:
The measures of the angles shown in above figure are 400, 1300, 650 and 1350
9. Find the angle measure between the hands of the clock in each figure:
Solutions:
The angle measure between the hands of the clock are 900, 300 and 1800
10. Investigate
In the given figure, the angle measure 300. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
Solutions:
The measure of an angle will not change by viewing through a magnifying glass
11. Measure and classify each angle:
Angle Measure Type ∠AOB ∠AOC ∠BOC ∠DOC ∠DOA ∠DOB
Solutions:
Angle Measure Type ∠AOB 400 Acute ∠AOC 1250 Obtuse ∠BOC 850 Acute ∠DOC 950 Obtuse ∠DOA 1400 Obtuse ∠DOB 1800 Straight
Exercise 5.5 page no: 100
1. Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.
Solutions:
(a) The adjacent edges of a table top are perpendicular to each other.
(b) The lines of a railway track are parallel to each other.
(c) The line segments forming the letter ‘L’ are perpendicular to each other
(d) The sides of letter V are inclined forming an acute angle.
Therefore (a) and (c) are models for perpendicular lines.
2. Let be the perpendicular to the line segment . Let and intersect in the point A. What is the measure of ∠PAY?
Solutions:
From the figure it is clear that the measure of ∠PAY is 900
3. There are two set squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Solutions:
The measure of angles in one set square are 300, 600 and 900
The other set square has a measure of angles 450, 450 and 900
Yes, the angle of measure 900 is common in between them
4. Study the diagram. The line l is perpendicular to line m
(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG
(ii) CD = GH
(iii) BC < EH.
Solutions:
(a) Yes, since, CE = 2 units and EG = 2 units respectively
(b) Yes. Since, CE = EG as both are of 2 units. Hence PE bisect CG
(c) and are the line segments for which PE is the perpendicular bisector
(d) (i) True. Since AC = 2 units and FG = 1 unit
∴ AC > FG
(ii) True because both are of 1 unit
(iii) True. Since, BC = 1 unit and EH = 3 units
∴ BC < EH
Exercise 5.6 page no: 103
1. Name the types of following triangles:
(a) Triangle with lengths of sides 7 cm, 8 cm and 9 cm.
(b) ∆ABC with AB = 8.7 cm, AC = 7 cm and BC = 6 cm.
(c) ∆PQR such that PQ = QR = PR = 5 cm.
(d) ∆DEF with ∠D = 90°
(e) ∆XYZ with ∠Y = 90° and XY = YZ.
(f) ∆LMN with ∠L = 30°, ∠M = 70° and ∠N = 80°.
Solutions:
(a) Scalene triangle
(b) Scalene triangle
(c) Equilateral triangle
(d) Right angled triangle
(e) Right angled isosceles triangle
(f) Acute angled triangle
2. Match the following:
Measures of Triangle Type of Triangle
(i) 3 sides of equal length (a) Scalene
(ii) 2 sides of equal length (b) Isosceles right angled
(iii) All sides are of different length (c) Obtuse angled
(iv) 3 acute angles (d) Right angled
(v) 1 right angle (e) Equilateral
(vi) 1 obtuse angle (f) Acute angled
(vii) 1 right angle with two sides of equal length (g) Isosceles
Solutions:
(i) Equilateral triangle
(ii) Isosceles triangle
(iii) Scalene triangle
(iv) Acute angled triangle
(v) Right angled triangle
(vi) Obtuse angled triangle
(vii) Isosceles right angled triangle
3. Name each of the following triangles in two different ways: (you may judge the nature of the angle by observation)
Solutions:
(i) Acute angled and isosceles triangle
(ii) Right angled and scalene triangle
(iii) Obtuse angled and isosceles triangle
(iv) Right angled and isosceles triangle
(v) Equilateral and acute angled triangle
(vi) Obtuse angled and scalene triangle
4. Try to construct triangles using match sticks. Some are shown here. Can you make a triangle with
(a) 3 matchsticks?
(b) 4 matchsticks?
(c) 5 matchsticks?
(d) 6 matchsticks?
(Remember you have to use all the available matchsticks in each case)
Name the type of triangle in each case. If you cannot make a triangle, think of reasons for it
Solutions:
(a) By using three match sticks we may make a triangle as shown below
The above triangle is an equilateral triangle
(b) By using 4 match sticks we cannot make a triangle, since we know that sum of the lengths of any two sides of a triangle is always greater than the third side.
(c) By using 5 match sticks we may make a triangle as shown below
The above triangle is an isosceles triangle
(d) By using 6 match sticks we may make a triangle as shown below
The above triangle is an equilateral triangle
Exercise 5.7 page no: 106
1. Say True or False:
(a) Each angle of a rectangle is a right angle.
(b) The opposite sides of a rectangle are equal in length.
(c) The diagonals of a square are perpendicular to one another.
(d) All the sides of a rhombus are of equal length.
(e) All the sides of a parallelogram are of equal length.
(f) The opposite sides of a trapezium are parallel.
Solutions:
(a) True, each angle of a rectangle is a right angle
(b) True, the opposite sides of a rectangle are equal in length.
(c) True, the diagonals of a square are perpendicular to one another
(d) True, all the sides of a rhombus are of equal length
(e) False, all the sides of a parallelogram are not equal
(f) False, the opposite sides of a trapezium are not parallel
2. Give reasons for the following:
(a) A square can be thought of as a special rectangle.
(b) A rectangle can be thought of as a special parallelogram.
(c) A square can be thought of as a special rhombus.
(d) Squares, rectangles, parallelograms are all quadrilaterals.
(e) Square is also a parallelogram.
Solutions:
(a) A rectangle in which all the interior angles are of same measure i.e 900 and only opposite sides of the rectangle are of same length whereas in square all the interior angles are of 900 and all the sides of the square are of same length. Hence, a rectangle with all sides equal becomes a square. Therefore square is a special rectangle.
(b) In a parallelogram opposite sides are parallel and equal. In a rectangle opposite sides are parallel and equal. The interior angles of the rectangle are of same measure i.e 900. Hence, a parallelogram with each angle as right angle becomes a square. Therefore a rectangle is a special parallelogram
(c) All sides of a rhombus and square are equal but in case of square all interior angles are of 900. A rhombus with each angle as right angle becomes a square. Therefore a square is a special rhombus
(d) Since, all are closed figures with 4 line segments. Hence all are quadrilaterals
(e) Opposite sides of a parallelogram are equal and parallel whereas in a square opposite sides are parallel and all 4 sides are of same length. Therefore a square is a special parallelogram.
3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral?
Solutions:
Square is a regular quadrilateral because all the interior angles are of 900 and all sides are of same length.
Exercise 5.8 page no: 108
1. Examine whether the following are polygons. If any one among them is not, say why?
Solutions:
(i) It is not a closed figure. Hence, it is not a polygon.
(ii) It is a polygon made of six sides
(iii) No it is not a polygon because it is not made of line segments.
(iv) It is not a polygon as it is not made of line segments.
2. Name each polygon.
Make two more examples of each of these.
(a) It is a closed figure and is made of four line segments. Hence, the given figure is a quadrilateral. Two more examples are
(b) The given figure is a triangle as it is a closed figure with 3 line segments. Two more examples are
(c) The given figure is a pentagon as this closed figure made of 5 line segments. Two more examples are
(d) The given figure is an octagon as it is a closed figure made of 8 line segments. Two more examples are
3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn.
Solutions:
We can draw an isosceles triangle by joining three of vertices of a hexagon as shown in below figure
4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon.
Solution:
The below figure is a regular octagon in which a rectangle is drawn by joining four of the vertices of the octagon.
5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals.
Solutions:
From the figure we may find AC, AD, BD, BE and CE are the diagonals
Exercise 5.9 page no: 111
1. Match the following:
Give two new examples of each shape.
Solutions:
(a) An ice cream cone and birthday cap are examples of cone
(b) Cricket ball and tennis ball are examples of sphere
(c) A road roller and lawn roller are examples of cylinder
(d) A book and a brick are examples of cuboid
(e) A diamond and Egypt pyramids are examples of pyramid
2. What shape is
(b) A brick?
(c) A match box?
Solutions:
(a) The shape of an instrument box is cuboid
(b) The shape of a brick is cuboid
(c) The shape of a match box is cuboid
(d) The shape of a road roller is cylinder
(e) The shape of a sweet laddu is sphere
## Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 5
### Define angles and its types covered under NCERT Solutions for Class 6 Maths Chapter 5?
An angle is made up of any two rays which have the same endpoint or starting point. It can be better understood by the movement of clock-hands. When a clock hand moves, it forms an angle. To measure the angle, a protractor is used. It should be noted that a straight angle is 180 degrees while a right angle is 90 degrees.
Based on the degree, an angle can be classified into 3 main types:
1. Acute angle: When an angle measure is less than a right angle, it is called an acute angle.
2. Obtuse angle: When an angle measures more than a right angle but less than a straight angle, it is called an obtuse angle.
3. Reflex angle: When an angle measures more than a straight angle, it is called a reflex angle.
### Explain solid shapes covered under NCERT Solutions for Class 6 Maths.
A solid shape or three-dimensional shape (3D shape) can be defined as the objects which can be measured in three directions i.e. length, breadth, and height. Examples of 3d shapes are cylinder, cube, cuboid, sphere, etc.
### Does CoolGyan’S give the most reliable answers in Chapter 5 of NCERT Solutions for Class 6 Maths?
The most accurate and reliable NCERT Solutions for Class 6 Maths Chapter 5 are available on CoolGyan’S. Students can easily download the solutions which are present in PDF format and access to them during their exam preparation. The solutions are framed and compiled by a set of expert faculty who possess numerous years of experience in the respective subjects. The most detailed solutions of the exercise wise problems are curated with the aim of helping students to ace the exam without fear. | 7,123 | 27,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-14 | latest | en | 0.901794 |
https://www.jcchouinard.com/pca-plot-3d-pca-graph-python-2/ | 1,702,108,922,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100873.6/warc/CC-MAIN-20231209071722-20231209101722-00716.warc.gz | 894,489,024 | 30,703 | # How to Plot a 3D PCA Scatterplot (with Python Example)
As part of the series of tutorials on PCA with Python, we will learn how to plot a 3D PCA graph (scatter plot) on the Iris Dataset with Python, Scikit-learn and Matplotlib.
## What is 3D PCA Scatter plot?
A 3D PCA (Principal Component Analysis) scatter plot is a PCA visualization that shows the distribution of data points in a 3D space after reducing a dataset to 3 PCA features.
## How to Plot a 3D PCA Graph in Python?
To plot a 3D PCA scatter plot in Python, reduce the number of features to 3 principal components. After, use `matplotlib` to generate a three dimensional scatterplot from the data.
Here are the detailed steps to plot a 3D PCA scatter plot in Python:
1. Load the required Python Libraries
3. Set up a 3D plotting environment
4. Assign PCA Features to their own Axes of the Scatter Plot
5. Plot the 3D PCA Graph using scatter3D
6. Interpret the 3D PCA Scatterplot
```import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn import datasets
from sklearn.preprocessing import StandardScaler
from sklearn.decomposition import PCA
plt.style.use('default')
```
To start, let’s load the Iris dataset in Python.
```# load features and targets separately
X = iris.data
y = iris.target
```
From this data, we will learn various ways to plot the 3D PCA graph with Python.
### 3. Scale and Reduce the Number of Features Using PCA
Next, do some preprocessing and use PCA to reduce the dataset to 3 features. Scale the data before applying PCA, and select the `n_component` to be equal to 3. To learn what this means, follow our tutorial on PCA with Python.
```# Data Scaling
x_scaled = StandardScaler().fit_transform(X)
# Reduce from 4 to 3 features with PCA
pca = PCA(n_components=3)
# Fit and transform data
pca_features = pca.fit_transform(x_scaled)
```
### 4. Set up a 3D Plotting Environment in Matplotlib
Sett up a 3D plotting environment in `matplotlib` using `plt.axes(projection='3d')`.
``ax = plt.axes(projection='3d')``
Let’s see an example by plotting our selected features into a 3D graph.
```# Prepare 3D graph
fig = plt.figure()
ax = plt.axes(projection='3d')
```
### 5. Assign PCA Features to their own Axes of the Scatter Plot
Before we can plot the data, we need to set-up the data for the x, y and z axes of the 3D scatter plot. Each feature will be on its own axis.
```# Plot scaled features
xdata = pca_features[:,0]
ydata = pca_features[:,1]
zdata = pca_features[:,2]
```
### 6. Plot the 3D PCA Graph using scatter3D
To plot the 3D PCA graph in Python, use `ax.scatter3D` with the x, y and z data as its argument, mapping each PCA feature to its own axes in the scatter plot.
```# Plot 3D plot
ax.scatter3D(xdata, ydata, zdata, c=zdata, cmap='viridis')
# Plot title of graph
plt.title(f'3D Scatter of Iris')
# Plot x, y, z even ticks
ticks = np.linspace(-3, 3, num=5)
ax.set_xticks(ticks)
ax.set_yticks(ticks)
ax.set_zticks(ticks)
# Plot x, y, z labels
ax.set_xlabel('sepal_length', rotation=150)
ax.set_ylabel('sepal_width')
ax.set_zlabel('petal_length', rotation=60)
plt.show()
```
### 6. Interpret the 3D PCA Scatterplot
When plotting a 3D graph, it is clearer that there is less variance in `Petal length` of Iris flowers than in `Sepal length` or `Sepal width`, almost making a flat 2D pane inside the 3D graph. That shows that the intrinsic dimension of the data is essentially 2 dimensions instead of 4.
Reducing these 3 features to 2 would not only make the model faster but the visualizations more informative without losing too much information.
## Next Steps
After plotting a 3D PCA Scatterplot, it is interesting to learn how to plot a 3D PCA Biplot.
## Full Code
```import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from sklearn import datasets
from sklearn.preprocessing import StandardScaler
from sklearn.decomposition import PCA
plt.style.use('default')
# load features and targets separately
X = iris.data
y = iris.target
# Data Scaling
x_scaled = StandardScaler().fit_transform(X)
# Dimention Reduction
pca = PCA(n_components=3)
pca_features = pca.fit_transform(x_scaled)
# Prepare 3D graph
fig = plt.figure()
ax = plt.axes(projection='3d')
# Plot scaled features
xdata = pca_features[:,0]
ydata = pca_features[:,1]
zdata = pca_features[:,2]
# Plot 3D plot
ax.scatter3D(xdata, ydata, zdata, c=zdata, cmap='viridis')
# Plot title of graph
plt.title(f'3D Scatter of Iris')
# Plot x, y, z even ticks
ticks = np.linspace(-3, 3, num=5)
ax.set_xticks(ticks)
ax.set_yticks(ticks)
ax.set_zticks(ticks)
# Plot x, y, z labels
ax.set_xlabel('sepal_length', rotation=150)
ax.set_ylabel('sepal_width')
ax.set_zlabel('petal_length', rotation=60)
plt.show()
```
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https://courseworkhero.co.uk/questions-2/ | 1,623,731,521,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487616657.20/warc/CC-MAIN-20210615022806-20210615052806-00268.warc.gz | 188,746,784 | 7,898 | # Questions
2. Suppose equilibrium price in the market is \$30, and the marginal revenue is \$20. What is the price elasticity of demand
3. The David Company’s demand curve for the company’s product is P = 2,000 – 20Q, where P = price and Q = the number sold per month.
. a .Derive the marginal revenue curve for the firm.
b.. At what output is the demand for the firm’s product price elastic?
c. If the firm wants to maximize its dollar sales volume, what price should it charge?
4. Rebel Sole is a rapidly expanding shoe company. The following is the demand estimate for its popular shoes. The estimate was done using 40 observations.
Q = 10 – 10 P + 4 A + 0.42I + 0.25Py
(3) (1.8) (0.7) (0.1) (0.1)
F = 93, s = 6, R2 = 93%
Q is quantity sold (in thousands), P is shoe price, A is advertising expenditure (in thousands), the numbers in parentheses are standard errors, I is disposable income per capita (thousands of dollars), and Py is the price of related goods.
a. Evaluate the model based on F, R2.
b. Test the significance of Py.
c. If P = \$5, A = \$30,000, I = 50,000, and Py = \$6, calculate advertising elasticity.
d. Given the information in c. above, calculate the 95% confidence interval for Q.
Calculator
Total price:\$26
Our features | 355 | 1,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.908696 |
https://universitymcqs.com/graphs-hasse-diagrams/ | 1,723,571,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641082193.83/warc/CC-MAIN-20240813172835-20240813202835-00812.warc.gz | 448,329,180 | 42,944 | Graphs – Hasse Diagrams
1. Hasse diagrams are first made by ______
a) A.R. Hasse
b) Helmut Hasse
c) Dennis Hasse
d) T.P. Hasse
2. If a partial order is drawn as a Hasse diagram in which no two edges cross, its covering graph is called ______
a) upward planar
b) downward planar
c) lattice
d) biconnected components
3. If the partial order of a set has at most one minimal element, then to test whether it has a non-crossing Hasse diagram its time complexity __________
a) NP-complete
b) O(n2)
c) O(n+2)
d) O(n3)
.
4. Which of the following relation is a partial order as well as an equivalence relation?
a) equal to(=)
b) less than(<)
c) greater than(>)
d) not equal to(!=)
5. The relation ≤ is a partial order if it is ___________
a) reflexive, antisymmetric and transitive
b) reflexive, symmetric
c) asymmetric, transitive
d) irreflexive and transitive
6. In which of the following relations every pair of elements is comparable?
a) ≤
b) !=
c) >=
d) ==
7. In a poset (S, ), if there is no element nS with m<n, then which of the following is true?
a) an element n exists for which m=n
b) An element m is maximal in the poset
c) A set with the same subset of the poset
d) An element m is minimal in the poset
.
8. In a poset P({v, x, y, z}, ) which of the following is the greatest element?
a) {v, x, y, z}
b) 1
c) ∅
d) {vx, xy, yz}
.
9. Suppose P1 is a partially ordered class and a cut of P1 is pair (D, T) of nonempty subclasses of P1 satisfies which of the following properties?
a) D∩T=Ø
b) D∪T=P1
c) xyz∈T
d) z∈T and zx∈D
.10. Let G be the graph defined as the Hasse diagram for the ⊆ relation on the set S{1, 2,…, 18}. How many edges are there in G?
a) 43722
b) 2359296
c) 6487535
d) 131963 | 564 | 1,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-33 | latest | en | 0.862532 |
https://www.geotechcontrol.de/lt-m-sand-rate/23281.html | 1,603,692,984,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890586.57/warc/CC-MAIN-20201026061044-20201026091044-00269.warc.gz | 745,986,169 | 6,234 | # Lt M Sand Rate
See more of civil work on facebookog inrreate new accountee more of civil work on facebookog inriver sand 823 kg m 3 total water 185 kg m 3 fresh concrete density 2398 kgm 3 m20 1 2 3 water 1000 ltm3 05ement 1440 kgm3 06gallon 4 litres 07ink 8" 200mm.
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• ### Civil Work CONCRETE GRADE M5 148 M10
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Waste-water treatment systems use loading rates as a measure for determining whether the system will have a tendency to clogecommended loading rates exist for different types of material such as sands, soil and domestic sewageydraulic and organic loading rates can both be used as units of measurementeparate.
• ### Calculate Cement Sand Aggregate For Concrete
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• ### CHOPS Production Rate Increase Mechanisms
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• ### What Does IT Stand For
Looking for the definition of it find out what is the full meaning of it on abbreviationsm italy is one option -- get in to view more the webs largest and most authoritative acronyms and. | 1,613 | 6,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-45 | latest | en | 0.889378 |
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# Time Travel - PowerPoint PPT Presentation
Time Travel. By B. Justin Scholfield. What is the nature of Time?. Time is used to relate events We define two actions as simultaneous when we see the two actions at the same time. Special & General Relativity. Special Relativity. This is the boost matrix, or Lorentz tensor.
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### Time Travel
By B. Justin Scholfield
• Time is used to relate events
• We define two actions as simultaneous when we see the two actions at the same time
This is the boost matrix, or Lorentz tensor.
It describes the translation of position and time to another frame moving at speed v along one axis
Its called special relativity because it can’t handle acceleration that well.
It can be used for doing relativistic mechanics as long as gravity is not involved
Fun Fact: It took Einstein only 8 weeks to come up with special relativity.
This is the Einstein Field Equations
It is described as:
“16 coupled hyperbolic-elliptic nonlinear partial differential equations “
Meaning its very, very hard to solve.
In expanded form it is three pages long, just to define the problem
The Field Equations are used to predict how a distribution of mass will effect space
All the time machines I will mention will be based on solutions to this equation
Fun Fact: it took Einstein 8 years to create this
Minkowski diagrams and Light cones
Minkowski diagrams are used to
show a particle traveling thru space
time, they have one time axis and
one space axis.
The particles in a diagram form a world line which is the line that a particle takes thru space time
Light Cones are used to show how a particle can move. The edge of the cone can’t be broken because that would mean that the particle has some how gone faster than light
The following forms of time travel are all future ward and all of them have been experimentally verified
The equation shown is the formula for time dilation at high speeds
For example at .999999c the dilation factor is about 707
The difficulty with a practical application is that to reach such high speeds requires a good deal of acceleration
General Relativity shows that time dilation is created by acceleration. The acceleration can be anything from gravity to linear acceleration to angular acceleration
• Fun Facts:
• There is a 7% difference in the time measured on earth and on a neutron star
• On a rotating disk with a 66,000 g centripetal acceleration an atomic clock was used to measure the time dilation for an experiment
That rotation leads to time dilation implies that a possible time machine could be a high powdered dryer
One twin leaves the earth and travels into space on a space ship (a fast one) and the other stays behind. The space twin is younger than the earth twin when he returns. Why can’t it be the other way around?
Answer: one has a curved world line, the other doesn't.
Gödel's Idea was to make a Universe that would enable time travel. He dose this by using frame dragging and a rotating Universe
Frame Dragging happens when a rotating mass drags space with it, it there by causes light cones to tip.
Fun Fact: Yesterday (10/21/2004) the first experimental verification of frame dragging was published
The Tippler cylinder is related to the Gödel universe. It uses an infinitely long rotating cylinder to achieve the same effect.
Worm Holes are solutions to the field equations that allow space to be non simply connected
Once a Worm Hole is made one end of it can be placed in a high gravity field or accelerated to a high speed then slowed down again. This provides a time machine for as long as the Worm Hole exists.
The a difficulty arises in doing this because worm holes tend to be unstable and my be destroyed if moved or placed into a high gravity field.
A worm hole requires an lot of tension to be kept open, for a 3 foot diameter worm hole it is on the order of 10^36 tones. Something that could hold at this level of tension would be classified as can rightly be called exotic matter.
No one is sure how to make a worm hole. There are many suggested methods of worm hole creation but most require a black hole or an ability to inflate one form the quantum foam.
A Worm Hole that passes thru time could also make a unbounded energy resonance that would destroy the worm hole.
The diagram is a picture of a billiard ball going thru a worm hole time machine. This is a machine version of the grandfather paradox.
The billiard ball has an infinite number of possible paths, all self consistent. The solution to what the path of the ball will be is as a group of probabilities and can’t be solved until we have a better grasp of quantum gravity.
All The time machines That I have mentioned are allowed by general relativity. There are no known physics that don’t allow this.
Sources
1.Time Machines: Time Travel and Physics, Metaphysics, and Science-fiction. Nahin, Paul J. 1999, Springer-Verlag New York, Inc
2. How to Build aTime Machine. Davies, Paul. 2001, Viking Penguin.
3. Black Holes & Time Warps, Throne, Kip S. 1994, H.W. Norton & Company.
4. Factors, Tensors, and the Basic Equations of Fluid Mechanics. Aris, Rutherford. 1962 Dover Books
5. The Einstein Equations. Author unknown. http://archive.ncsa.uiuc.edu/Cyberia/NumRel/EinsteinEquations.html#intro. Accessed 10/21/2004
6. Spinning Earth Twists Space. News report. http://www.nature.com/news/2004/041018/full/041018-11.html. Accessed 10/21/2004 | 1,376 | 6,090 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-34 | longest | en | 0.908606 |
http://www.traditionaloven.com/building/masonry/cement/convert-gram-g-of-cement-to-dram-dr-of-cement.html | 1,529,557,451,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00191.warc.gz | 531,609,290 | 11,887 | Portland Cement 1 gram mass to drams converter
# Portland cement conversion
## Amount: 1 gram (g) of mass Equals: 0.56 drams (dr) in mass
Converting gram to drams value in the Portland cement units scale.
TOGGLE : from drams into grams in the other way around.
## Portland cement from gram to dram Conversion Results:
### Enter a New gram Amount of Portland cement to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other Portland cement measuring units - complete list.
Conversion calculator for webmasters.
## General Portland cement
General or common purpose Portland cement type (not any other weaker/cheaper cement replacement-version). It's the primary masonry binder hence bonding agent for mortars and concretes consisting of building sand, stones or other gravel aggregate, mixed with water.
By standard practice, when freshly poured, Portland cement has unit volume mass of 94 lbs/cu-ft - 1506 kg/m3 (but it becomes denser as the storage time is prolonged, when it gets compressed or vibrated; in such situations its weight per volume can increase to as high as 104 lbs/cu-ft). This calculator is based on the fresh form Portland cement w/ the standard mass properties of 94 pounds to 1 cubic foot.
Convert Portland cement measuring units between gram (g) and drams (dr) but in the other reverse direction from drams into grams.
conversion result for Portland cement: From Symbol Result To Symbol 1 gram g = 0.56 drams dr
# Converter type: Portland cement measurements
This online Portland cement from g into dr converter is a handy tool not just for certified or experienced professionals.
First unit: gram (g) is used for measuring mass.
Second: dram (dr) is unit of mass.
## Portland cement per 0.56 dr is equivalent to 1 what?
The drams amount 0.56 dr converts into 1 g, one gram. It is the EQUAL Portland cement mass value of 1 gram but in the drams mass unit alternative.
How to convert 2 grams (g) of Portland cement into drams (dr)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.564383391193 * 2 (or divide it by / 0.5)
QUESTION:
1 g of Portland cement = ? dr
1 g = 0.56 dr of Portland cement
## Other applications for Portland cement units calculator ...
With the above mentioned two-units calculating service it provides, this Portland cement converter proved to be useful also as an online tool for:
1. practicing grams and drams of Portland cement ( g vs. dr ) measuring values exchange.
2. Portland cement amounts conversion factors - between numerous unit pairs.
3. working with - how heavy is Portland cement - values and properties.
International unit symbols for these two Portland cement measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for gram is:
g
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for dram is:
dr
### One gram of Portland cement converted to dram equals to 0.56 dr
How many drams of Portland cement are in 1 gram? The answer is: The change of 1 g ( gram ) unit of Portland cement measure equals = to 0.56 dr ( dram ) as the equivalent measure for the same Portland cement type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in g - grams for portland cement amount, the rule is that the gram number gets converted into dr - drams or any other Portland cement unit absolutely exactly.
Conversion for how many drams ( dr ) of Portland cement are contained in a gram ( 1 g ). Or, how much in drams of Portland cement is in 1 gram? To link to this Portland cement gram to drams online converter simply cut and paste the following.
The link to this tool will appear as: Portland cement from gram (g) to drams (dr) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 973 | 4,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | latest | en | 0.839628 |
https://budgeting.thenest.com/calculate-total-percent-investment-return-over-multiple-year-period-27435.html | 1,723,698,365,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641151918.94/warc/CC-MAIN-20240815044119-20240815074119-00823.warc.gz | 118,479,316 | 23,126 | # How to Calculate the Total Percent Investment Return Over a Multiple Year Period
The best way to measure how well your investments are doing is to compare their gains or losses over a period of time as a percentage of your initial investment. Using percentages allows you to accurately compare investments of different sizes. For example, while \$5,000 might sound like a great return on an investment, the investment was significantly better if you only invested \$10,000 than if you only made a \$5,000 return on a \$1 million investment.
#### TL;DR (Too Long; Didn't Read)
In order to determine the total percent investment return you have accrued, you will need to divided your gains or losses over the multi-year period by your initial investment. This figure can then be converted into a percentage value to provide the desired results.
## Total Return Percentage
To figure the total percentage investment return for the entire time you’ve held an investment, you need to know what you paid for the investment and how much you received from the investment through selling it. When you calculate your total return, include any dividends you’ve been paid over the time you owned the investment because that adds to your total return. First, subtract what you paid for the investment from your total return to find your gain or loss. Second, divide your gain or loss by your initial investment. Third, multiply the result by 100 so you can convert it to a percentage.
For example, say that you invested \$2,400 in a stock ten years ago and in total, you earn \$5,000 from selling the stock. First, subtract \$2,400 from \$5,000 to find you earned \$2,600. Second, divide \$2,600 by \$5,000 to get 0.52. Third, multiply 0.52 by 100 to find you earned a total return of 52 percent.
## Annualized Percentage Return
While finding your overall return is useful, it doesn’t help you compare the rates of return for investments for different periods of time. For example, if one investment grew by 18 percent over a four-year period, you don’t know whether that’s better or worse than a 40 percent return over eight years. To make an accurate comparison, you must calculate the average annual report.
To do that, divide the final value by the initial investment. Then, raise the result to the power of 1 divided by the number of years you invested the money. You’ll need to use the exponent key on your calculator. Next, subtract 1. Last, multiply by 100.
## Exploring an Example
For example, if you invested \$2,400 and ten years later your investment is worth \$5,000, divide \$5,000 by \$2,400 to get 2.08333. Then, raise 2.08333 to the 1/5 power to get 1.158. Next, subtract 1 to get 0.158. Last, multiply 0.158 by 100 to find the average percentage investment return over the five-year period equals 15.8 percent. | 637 | 2,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.947528 |
https://www.teacherspayteachers.com/Product/St-Patricks-Day-Math-3rd-Grade-Bump-Games-Bundle-3503204 | 1,537,837,641,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160853.60/warc/CC-MAIN-20180925004528-20180925024928-00546.warc.gz | 892,087,792 | 21,450 | # St. Patrick's Day Math 3rd Grade+ Bump Games Bundle
Subject
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Common Core Standards
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Product Description
St. Patrick's Day Math Bump Games for 3rd Grade+
LOW PREP & 100% black and white - all sheets are practical, engaging, and Common Core based
==========
Click Here for all of our March themed 3rd Grade material
==========
Fluency + Strategy = Bump
Bump (or "Roll and Cover”) is a fast paced math game combining tricky subjects with a competition aspect. It is played with a partner and can be used as part of a math center also. It is also tons of fun when hyper-themed to the current month and holidays.
If you feel games are a crucial part of the learning process, my themed Bump games are perfect for your class. I priced them affordably so you can purchase multiple months as needed. This set is themed for St. Patrick's Day and includes:
* Rounding by 100s
* Rounding by 10s
* Multiplying by 20
* Multiplying by 30
* Multiplying by 40
* Multiplying by 50
* Multiplying by 60
* Multiplying by 70
* Multiplying by 80
* Division by 2
* Division by 3
* Division by 4
* Division by 5
* Division by 6
* Division by 7
* Division by 8
* Division by 9
* Division by 10
* Division by 11
* Division by 12
* Division by 13
* Telling Time
* Fractions
Meets Common Core Standards:
- 3.MD.A.1 Tell and write time to the nearest minute
- 3.NBT.A.1 Use place value understanding to round whole numbers
- 3.NBT.A.3 Multiply one digit whole numbers by multiples of 10
- 3.OA.C.7 Fluently multiply and divide within 100
- Fractions
An engaging way to teach some tricky 3rd grade math topics. Also works as a review for 4th Grade or for advanced 2nd graders.
BONUS: Each topic contains 2 different versions for variety.
Warning! Your kiddos will want to “Bump” every week after playing these.
******************************
How To Use
******************************
Sometimes your incoming students are lagging in fluency on some key skills. You need quick and ready games to help your kids practice these skills - not bore them.
By combining Fluency with Strategy, Bump lets students who might lose interest in other games become focused and engaged. It’s competitive enough to interest them without becoming “un-fun” for other students.
And as more districts get away from timed math work, it’s up to you to find more open-ended engaging ways to reinforce key concepts. Roll And Cover ensures kiddos are constantly practicing their facts.
You can use these every month if you would like. This listing is made for March, but each month is themed to stay engaging with holidays/events going on. Whenever possible I have made the answer sets different so your students are always figuring something new out.
Bump / Roll and Cover Directions: Print out sets, lamination is optional. Works with 1 or 2 normal 6 sided dice. Students will need some kind of marker to mark their positions. Each game sheet also includes instructions specific to that version.
***********************************************************
Chalk and Challenge™
Be the best teacher your students have ever seen!
***********************************************************
Graphics by A Hughes Design, https://www.teacherspayteachers.com/Store/Mai-Huynh
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Tags: march3rd
Total Pages
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Sign Up | 1,001 | 4,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-39 | latest | en | 0.936308 |
https://www.exammain.com/ce6008-groundwater-engineering-impotent-questions-question-bank-syllabus-model-and-previous-question-papers-download-pdf/ | 1,696,055,634,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00773.warc.gz | 852,916,455 | 13,801 | ## CE6008 Groundwater Engineering
Subject Informations :
University : Anna University
Department : Civil Engineering
Semester : 07 th sem / ELECTIVE II
Year : 04 th year
Regulation : R2013
Subject Code : CE6008
Subject Name : Groundwater Engineering
Are you Searching about Anna University Exams Important Questions ? Exammain.com is the right place to get all semester Anna University Important Questions Download PDF.
CE6008 Groundwater Engineering | Impotent Questions
CE6008 GROUNDWATER ENGINEERING – REGULATION – 2013
OBJECTIVES :
• To introduce the student to the principles of Groundwater governing Equations and
Characteristics of different aquifers,
• To understand the techniques of development and management of groundwater.
UNIT I HYDROGEOLOGICAL PARAMETERS
Introduction – Water bearing Properties of Rock – Type of aquifers – Aquifer properties –
permeability, specific yield, transmissivity and storage coefficient – Methods of Estimation– Ground water table fluctuation and its interpretations – Groundwater development and
Potential in India – GEC norms.
UNIT II WELL HYDRAULICS
Objectives of Groundwater hydraulics – Darcy’s Law – Groundwater equation – steady state flow
– Dupuit Forchheimer assumption – Unsteady state flow – Theis method – Jacob method -Slug
tests – Image well theory – Partial penetrations of wells.
UNIT III GROUNDWATER MANAGEMENT
Need for Management Model – Database for groundwater management –groundwater balance
study – Introduction to Mathematical model – Conjunctive use – Collector well and Infiltration
gallery.
UNIT IV GROUNDWATER QUALITY
Ground water chemistry – Origin, movement and quality – Water quality standards – Health and
aesthetic aspects of water quality – Saline intrusion – Environmental concern and Regulatory
requirements
UNIT V GROUNDWATER CONSERVATION
Artificial recharge techniques – Remediation of Saline intrusion– Ground water management
studies – Protection zone delineation, Contamination source inventory, remediation schemes
– Ground water Pollution and legislation.
TOTAL : 45 PERIODS
OUTCOMES:
• Students will be able to understand aquifer properties and its dynamics after the
completion of the course. It gives an exposure towards well design and practical problems
of groundwater aquifers.
• Students will be able to understand the importance of artificial recharge and groundwater
quality concepts.
TEXTBOOKS:
1. Raghunath H.M., “Ground Water Hydrology”, New Age International (P) Limited, New
Delhi, 2010.
2. Todd D.K., “Ground Water Hydrology”, John Wiley and Sons, New York, 2000.
REFERENCES:
1. Fitts R Charles, “Groundwater Science”. Elsevier, Academic Press, 2002.
2. Ramakrishnan, S, Ground Water, K.J. Graph arts, Chennai, 1998.
Also check :
Check Anna University – Exam Result | Internal Mark >>
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CE6008 Groundwater Engineering Question Bank | 941 | 4,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-40 | latest | en | 0.657515 |
https://www.aqua-calc.com/calculate/weight-to-volume/substance/tar | 1,607,048,468,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141733120.84/warc/CC-MAIN-20201204010410-20201204040410-00548.warc.gz | 526,129,399 | 6,737 | # Volume of Tar
## tar: convert weight to volume
### Volume of 100 grams of Tar
centimeter³ 86.73 milliliter 86.73 foot³ 0 oil barrel 0 Imperial gallon 0.02 US cup 0.37 inch³ 5.29 US fluid ounce 2.93 liter 0.09 US gallon 0.02 meter³ 8.67 × 10-5 US pint 0.18 metric cup 0.35 US quart 0.09 metric tablespoon 5.78 US tablespoon 5.87 metric teaspoon 17.35 US teaspoon 17.6
### The entered weight of Tar in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
• 1 153 kilograms [kg] of Tar fit into 1 cubic meter
• 71.97944 pounds [lbs] of Tar fit into 1 cubic foot
• Tar weighs 1.153 gram per cubic centimeter or 1 153 kilogram per cubic meter, i.e. density of tar is equal to 1 153 kg/m³. In Imperial or US customary measurement system, the density is equal to 71.98 pound per cubic foot [lb/ft³], or 0.6665 ounce per cubic inch [oz/inch³] .
• Bookmarks: [ weight to volume | volume to weight | price | density ]
#### Foods, Nutrients and Calories
TURKEY PEPPERONI, TURKEY, UPC: 037600188081 contain(s) 233 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Florida Crushed Coral weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Plutonium, alpha form [Pu] weighs 19 860 kg/m³ (1 239.8193 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Rapeseed oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
The microgram per square micron surface density measurement unit is used to measure area in square microns in order to estimate weight or mass in micrograms
The online number conversions include conversions between binary, octal, decimal and hexadecimal numbers.
g/m³ to short tn/l conversion table, g/m³ to short tn/l unit converter or convert between all units of density measurement.
#### Calculators
Calculate the volume and area of the surface of a quarter cylinder | 669 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-50 | latest | en | 0.588737 |
http://soft-matter.seas.harvard.edu/index.php?title=Interaction_Forces_between_Colloidal_Particles_in_Liquid:_Theory_and_Experiment&oldid=13080 | 1,611,323,588,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529331.99/warc/CC-MAIN-20210122113332-20210122143332-00010.warc.gz | 91,471,205 | 7,732 | Interaction Forces between Colloidal Particles in Liquid: Theory and Experiment
Overview
• [1] Yuncheng Liang, Nidal Hilal, Paul Langston, and Victor Starov, Advances in Colloid and Interface Science 134-135, 151-166 (2007).
• Keywords: van der Waals Forces, Electric Double Layer Forces, Solvation Forces, Hydrophobic Forces, Steric Forces, Atomic Force Microscopy, Surface Forces Apparatus
Summary
Liang et al. write a review article summarizing the major findings and contributions of 158 publications. Most of the major contributions to this field were made since 1940. This review article includes information on the theories of colloidal interaction forces, experiments testing the theories, and the historical development of the field.
Timeline of events compiled from article:
1936-1937 de Boer and Hamaker publish papers on theoretical aspects of "dispersion forces acting between colloidal objects (p. 152)."
1941 Derjaguin and Landau publish first big paper leading to DLVO theory
1948 Verwey and Overbeek publish second big paper contributing to DLVO theory
1956 Lifshitz publishes "The Theory of Molecular Attractive Forces between Solids"
1964 Derjaguin et. al. make experimental discovery of one of the first non-DLVO forces
1978 First accurate macroscopic surface measurements in aqueous solution carried out by Israelachivili and Adams (using surface forces apparatus)
1991 Atomic force microscopy adapted for measuring interactions of a colloidal particle and a solid surface
1) van der Waals Forces
Dispersion forces between colloidal spheres yielding an attractive interaction energy of the form (equation 1 from [1]):
$V_A(D)=-\frac{A_H}{6}\left(\frac{2a^2}{D^2+4aD}+\frac{2a^2}{(D+2a)^2}+ln\left(1-\frac{4a^2}{(D+2a)^2}\right)\right)$
$A_H$= Hamaker Constant, $a$= Sphere Radius, $D=$ Interparticle Distance
2) Electric Double Layer Forces
Repulsive forces between colloidal spheres with like charge with an interaction energy of the form (for $\kappa a>5$ and $h<<a$ (equation 5 from [1]):
$V_R=\frac{128\pi a_1 a_2 n_\infty kT}{(a_1+a_2)\kappa^2} \gamma_1 \gamma_2 e^{(-\kappa h)}$
$a_1, a_2$= Particle Radii, $n_\infty$= Bulk Density of Ions, $h=$ Interparticle Distance, $\kappa$= Reciprocal Debye-Huckel Length, $\gamma_1, \gamma_2$= Measure of Surface Potentials (see article equation 6 for details)
3) Solvation Forces
When surfaces are only a few nanometers apart, the molecular, discrete nature of a solvent becomes important. DLVO theory based on continuum theories cannot describe short-range interparticle forces. Solvation forces arise when only a few layers of solvent molecueles remain in between surfaces. The solvent molecules become ordered into layers and create a force which oscillates between attractive and repulsive as the separation distance varies. If the solvent is water, these forces are called hydration forces.
4) Hydrophobic Forces
Hydrophobic forces are attractive forces between two hydrophobic surfaces in water. Water molecules trapped between the two surfaces are not free to orient themselves as they would naturally and are expelled from the gap between the surfaces. The range of this force is accepted to extend out to separation distances greater than 10nm. “Unfortunately, so far no generally accepted theory has been developed for these forces... (p.157)."
5) Steric Forces
Steric forces arise when polymer-coated surfaces come close enough for the polymers to overlap. The overlapping of the polymers leads to an entropy-driven repulsive force. The magnitude of the force depends on many parameters such as polymer packing density on the surfaces and polymer solubility in the solvent of the system. “There is no simple, comprehensive theory available as steric forces are complicated and difficult to describe (p. 157)."
The second half of this review article discusses experimental evidence for and against the forces listed above. There are still open questions in this field and room for further experiments to distinguish between proposed mechanisms.
Soft Matter Details
Surface Properties:
The topic of this review article falls within the major field of colloidal science within soft matter. The forces between colloidal particles are due to surface properties and solvent properties which are important in both applied physics and chemistry. Understanding interparticle forces helps us understand the stability of colloids (when the particles will stay dispersed compared to when they will aggregate and sediment). The small distances between particles which are interesting in colloidal science draw out interesting questions about when the continuum treatment of a solvent breaks down.
Experimental Methods:
The main experimental techniques focused on in this article are atomic force microscopy and using the surface forces apparatus. Both tools measure very small forces between two surfaces only nanometers apart.
History:
How much does knowing the historical development of science help us in doing current research? This review article presents theories in the context of when they were discovered. Particularly in the context of such new work (since 1940), I think it could be very useful to understand the chronology of the work in the field. Not only does knowing the history help you understand the motivations behind developments, but it also gives a good frame of reference when reading other works from within the same time period. | 1,199 | 5,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.794582 |
http://mathhelpforum.com/algebra/106417-simple-algebra-question.html | 1,524,487,115,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00065.warc.gz | 207,477,596 | 9,682 | 1. simple algebra question
Hi
Shame on me for asking this but
$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2$
Do you multiply everything out before subtracting ?
I got the answer n+1 but it's wrong.
2. Hello Jones
Originally Posted by Jones
Hi
Shame on me for asking this but
$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2$
Do you multiply everything out before subtracting ?
I got the answer n+1 but it's wrong.
No, not if you want the easiest method!
Note that each term has a common factor $\displaystyle \frac{n+1}{2}$, so take it out (not forgetting that it has to be squared):
$\displaystyle \bigg(\frac{(n+1)(n+2)}{2}\bigg)^2 - \bigg(\frac{n(n+1)}{2}\bigg)^2= \Big(\frac{n+1}{2}\Big)^2\Big((n+2)^2-n^2\Big)$
Even now, there's no need to multiply out. Use 'difference of two squares' in the second bracket. | 305 | 895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-17 | latest | en | 0.810083 |
http://math.stackexchange.com/questions/499321/why-is-the-determinant-zero-iff-the-column-vectors-are-linearly-dependent | 1,464,331,447,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276543.81/warc/CC-MAIN-20160524002116-00140-ip-10-185-217-139.ec2.internal.warc.gz | 198,478,744 | 17,882 | # Why is the determinant zero iff the column vectors are linearly dependent?
The determinant of a square matrix is zero if and only if the column vectors are linearly dependent.
I see a lot of references to this all over the web, but I can't find an actual explanation for this anywhere.
-
What definition of determinant are you using? – KReiser Sep 20 '13 at 5:38
Do you know it is zero if you can find linearly dependent "rows" inside the matrix? – Babak S. Sep 20 '13 at 6:13
Hint: proofwiki.org/wiki/… – dls Sep 20 '13 at 6:24
@dls: The answer will be covered by the points in the link but the OP asked about columns, so he/she does not know that the ranks of spaces which rows and columns create are the same. – Babak S. Sep 20 '13 at 6:30
Well, $\det$ is multilinear and anti-symmetric. Done. – Michael Hoppe Sep 20 '13 at 9:02
The reason why is because if you have a matrix whose column vectors are linearly independent then when you look at reduced row echelon form there will end up being a zero row which means that parameter for the system can be any value you like. Meaning the system has infinitely many solutions. Also recall in reduced row echelon form the diagonal elements will be 1's excluding the row of zeros. Finally, the determinant of a upper triangular matrix is the product of the diagonal elements, therefore the determinant will be zero. It would look something like $$A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c\\ 0& 0 & 0 \end{pmatrix}.$$ $Det(A) = 1\times1\times0 = 0$. | 397 | 1,502 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2016-22 | latest | en | 0.882435 |
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# Number 271737008656
two hundred seventy one billion seven hundred thirty seven million eight thousand six hundred fifty six
### Properties of the number 271737008656
Factorization 2 * 2 * 2 * 2 * 19 * 19 * 19 * 19 * 19 * 19 * 19 * 19 Divisors 1, 2, 4, 8, 16, 19, 38, 76, 152, 304, 361, 722, 1444, 2888, 5776, 6859, 13718, 27436, 54872, 109744, 130321, 260642, 521284, 1042568, 2085136, 2476099, 4952198, 9904396, 19808792, 39617584, 47045881, 94091762, 188183524, 376367048, 752734096, 893871739, 1787743478, 3575486956, 7150973912, 14301947824, 16983563041, 33967126082, 67934252164, 135868504328, 271737008656 Count of divisors 45 Sum of divisors 555739923951 Previous integer 271737008655 Next integer 271737008657 Is prime? NO Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? square(521284) Binary 11111101000100110010011011001000010000 Octal 3750462331020 Duodecimal 447b8211414 Hexadecimal 3f44c9b210 Square 7.3841001873311E+22 Square root 521284 Natural logarithm 26.328100555571 Decimal logarithm 11.434148790279 Sine -0.62548907597869 Cosine -0.78023292408827 Tangent 0.80166967666687
Number 271737008656 is pronounced two hundred seventy one billion seven hundred thirty seven million eight thousand six hundred fifty six. Number 271737008656 is a composite number. Factors of 271737008656 are 2 * 2 * 2 * 2 * 19 * 19 * 19 * 19 * 19 * 19 * 19 * 19. Number 271737008656 has 45 divisors: 1, 2, 4, 8, 16, 19, 38, 76, 152, 304, 361, 722, 1444, 2888, 5776, 6859, 13718, 27436, 54872, 109744, 130321, 260642, 521284, 1042568, 2085136, 2476099, 4952198, 9904396, 19808792, 39617584, 47045881, 94091762, 188183524, 376367048, 752734096, 893871739, 1787743478, 3575486956, 7150973912, 14301947824, 16983563041, 33967126082, 67934252164, 135868504328, 271737008656. Sum of the divisors is 555739923951. Number 271737008656 is not a Fibonacci number. It is not a Bell number. Number 271737008656 is not a Catalan number. Number 271737008656 is not a regular number (Hamming number). It is a not factorial of any number. Number 271737008656 is an abundant number and therefore is not a perfect number. Number 271737008656 is a square number with n=521284. Binary numeral for number 271737008656 is 11111101000100110010011011001000010000. Octal numeral is 3750462331020. Duodecimal value is 447b8211414. Hexadecimal representation is 3f44c9b210. Square of the number 271737008656 is 7.3841001873311E+22. Square root of the number 271737008656 is 521284. Natural logarithm of 271737008656 is 26.328100555571 Decimal logarithm of the number 271737008656 is 11.434148790279 Sine of 271737008656 is -0.62548907597869. Cosine of the number 271737008656 is -0.78023292408827. Tangent of the number 271737008656 is 0.80166967666687
### Number properties
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Examples: 3628800, 9876543211, 12586269025 | 1,096 | 2,973 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-25 | latest | en | 0.506042 |
https://bz.vita-aidelos.com/comments/ | 1,659,977,508,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570868.47/warc/CC-MAIN-20220808152744-20220808182744-00191.warc.gz | 160,950,743 | 16,413 | In the cafeteria where I usually go to breakfast for coffee with milk and two toasts cost € 2. However, if I have a toast and two coffees with milk, the price is € 2.20. Today I had a toast and a coffee with milk. How much did it cost me in total? Solution If we call T toast and C to coffee with milk we have to: C + 2T = € 2 2C + T = € 2.20 If we add both equations we have to: 3C + 3T = € 4.20 If we divide by 3 both sides of the equation we have that: C + T = € 1.4 which is the price of a coffee with milk and a toast.
In 1999 the fourth centenary of the birth of Velázquez was celebrated. The year he died is a multiple number of 5, whose tens figure is not a prime number. If all the figures in that number add up to 13, how many years did the great painter live? Solution He lived 61 years. The year he died has to start at 16 and end at 0 or 5.
Alberto is thirty-five years old, Beatriz twelve and Cecilia fifteen. When will the sum of the ages of Betriz and Cecilia be equal to Alberto's age? Solution In 8 years Alberto will be 43, Beatriz will be 20, and Cecilia 23. (20 + 23 = 43)
The hare and the turtle depart from the goal to walk a track in a circular stadium. While the hare goes around and a quarter the turtle only travels a third of a turn. How many laps should each one take to match the goal again? Solution When the turtle reaches the finish line for the first time, the hare is in the third quarter of the stadium, after three laps.
After four days of crossing a cargo ship sank and only part of the crew could be saved in the lifeboat. There they found provisions and water for thirteen days, one liter per person and day. On the third day after the wreck they lost part of the water from a sea stroke and at the end of the fifth day one of the men died.
When I had already gone down 7 of the steps of the staircase that overlooks the street I ran into the neighbor's son who climbed the steps two at a time so that when I still had 4 steps left to get down the boy had already climbed All the stairs. What agility! I thought. For every step that I go under it goes up two.
Agustín, Benito, Carlos, Diego, Esteban and Federico are six collectors of paintings of which two of them are brothers. One day they went to an exhibition together and bought as follows: Agustin bought 1 painting, Benito bought 2, Carlos 3, Diego 4, Esteban 5 and Federico 6 paintings. The two brothers paid the same amount of money for each of the paintings they bought.
Using the numbers from 1 to 9, without changing their order or putting them together and using the basic arithmetic operations: addition, subtraction, multiplication and division, are you able to get 100 as a result? For example in this attempt: 1 - 2 + 3 + 4 + (5 × 6) + (7 × 8) + 9, we get as a result 101 Solution There are many solutions.
Monica's cat is lighter than Sandra's, but more lazy and fat than Josemi's, which is even more lazy than Victoria's, which is thinner than Monica's, which is fatter than Sandra's, which is darker than Victoria's, although Josemi's is less lazy and lighter than Sandra's.
Among this harmless pile of puzzle pieces hides one of the fiercest animals on earth. Can you find it Solution If you move away from the image you will see more easily the head of a tiger.
Ames's room was invented by the American ophthalmologist Adelbert Ames, Jr. It is constructed in such a way that when viewed from the front it appears to be a common room of a cubic shape, with a back wall and two lateral sides parallel to each other and perpendicular to the horizontal plane of the floor and ceiling.
Watch this video that shows four roads that converge in the center. Apparently a ball placed on one of the roads should fall out of the circuit, but sometimes things are not what they seem.
When Pepito and his wife sat down to have tea with their sister-in-law and daughter-in-law, each one ate a different number of cakes (no one was left without eating) and in total they devoured eleven. Pepito's wife ate two and sister-in-law four. How many cakes did Pepito eat? Pepito solution ate five cakes. The only way to add the accounts is that Pepito's sister-in-law and daughter-in-law are the same person.
Four rugby players enter an elevator that can carry a maximum of 380 kilos. Paul weighs the most, if each of the others weighs as much as he does, the alarm would stop the elevator. Carlos is the lightest, the elevator could go up to five like him. Ramón weighs 14 kilos less than Pablo and only six less than Jesus weighing 17 kilos more than Carlos.
I have an invoice that tells me that eight astrodeniums, a bartofón, three cartunes and three dosefríos were bought for my company a month and a half ago and we were charged a total of 350 euros. Just one month ago, we have another invoice from the same supplier worth 250 euros for buying five astrodennes, two bartofones, two cartunes and a dosefrio.
In a certain city, in the face of a brutal plague of mice, cats that each hunted the same number of rodents were used. It is known that 1,111,111 were the hunted mice and less than 1,111 the cats involved. Although there was no physical evidence on the exact number of mice that each cat hunted, this data is recorded in the official records.
In this drawing we can see a man who fell asleep with a cup in his hand, and who is leaning on a round table full of books, glasses, writing devices, grapes and some drawing devices ... However, as with All visual anamorphic optical illusions, the hidden motif is not visible to the naked eye the original drawing.
Three friends bought a box full of wine bottles from a street vendor. At the time of making the distribution they saw that the box contained 21 equal bottles of which only 7 were fully filled. There were 7 others that were exactly in the middle and another 7 empty. Despite the scam, they decided to distribute the wine and bottles among all so that each one took the same amount of wine and bottles.
Ana, Carmen and Teresa are three friends who live in neighboring homes. Carmen is the one who lives in the middle house. One of them likes the theater a lot, the other goes by bicycle and the last swim, but we don't know to whom each of the hobbies corresponds. However, a neighbor of the town tells us the following: The one who likes to go by bicycle from time to time takes Teresa's dog for a walk.
The square of the age of Mary is the fourth part of the square of the age of John that is half of 20. What is the age of Mary? Solution Maria is 5 years old. The square of the age of Mary: it is the fourth part of the square of the age of John that is half of 20
We have five rectangles whose sides have the following lengths: 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 although not necessarily in this order. With these five rectangles, we can form a larger square. What is the length of said square? Solution The image shows the square formed by the different rectangles. | 1,666 | 6,929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.973713 |
https://vifleem.info/relationship-between-and/relationship-between-tangential-and-centripetal-acceleration.php | 1,568,808,887,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00417.warc.gz | 716,834,104 | 9,639 | # Relationship between tangential and centripetal acceleration
### Tangential Acceleration vs. Centripetal Acceleration - Physics Help Forum
Hello, I'm having a very hard time understanding the difference between tangential and centripetal acceleration (even though I've been warned. In rotational motion, tangential acceleration is a measure of how quickly a tangential velocity changes. It always acts perpendicular to the centripetal acceleration. Tangential Acceleration vs Centripetal Acceleration. Acceleration is the rate of change of velocity, and when expressed using calculus, it is the.
You might say, 'no, I'm not, I'm sitting still'. Assuming you are on Earth, you are accelerating radially.
## Tangential Acceleration Formula
A radial acceleration is the type of accelerating that is always present when something moves in a circular path and it is directed towards the center of the circular path. For an object experiencing curve-linear motion there can be two types of acceleration: Tangential Acceleration If an object is moving in a circular path and either speeding up or slowing down, it is accelerating tangentially. Tangential acceleration is like linear acceleration, but is slightly different from straight-line linear acceleration.
An object is linearly accelerating if it is traveling in a straight-line path. For example, an airplane accelerating down the runway during takeoff. Contrast that with a car accelerating around a curve in the road. The car is accelerating tangentially to the curve of its path. Using the car's motion we will investigate tangential acceleration a bit more.
Red angular velocity magnitudes are increasing with time. The link between the angular acceleration and the tangential acceleration is the radius of the curve the object is traveling along. The larger the radius, the larger the tangential acceleration. The radius of its path is 20 meters. We can calculate the angular acceleration using a rotational kinematics equation.
Now we can calculate the tangential acceleration of the car using the angular acceleration and the radius of the curved path.
## Tangential & Radial Acceleration in Curve-Linear Motion
If somebody is spinning the ball over their head then imagine you are above looking down from above onto the rotation. Now as it hits That little arrow is the velocity vector of the ball and it is pointing tangentially to the path the ball is orbiting in. Now let time move again and then stop it again at the 3: Now look down at the arrow you had placed on the ball it should now be pointing completely south, or downwards only.
Maybe the easiest way to imagine the tangent line the ball would like to travel in if let go is to look at it differently. If you draw a circle on a piece of paper such that the center of the circle is exactly at the origin of an x,y coordinate system.
Now you have the x line passing though the 3: If somebody asks you to show them the line tangent to the circle at the 90 degree position, 90 degrees from the x axis means we are talking about a point on the circle that would be at the OK, the y axis passes through the point where we want to find the line tangent to. When you want a tangent line to a circle the easiest way is to draw a line from the origin out in a radial direction to that point, like the y axis passing through our Here's a "cheesy" example of a tangent line.
I will type the letter o here in a second followed by the symbolthis will show a line tangent to a point on a circle at the 3: Here it is Othat little " " just to the right of the circle is a line that is tangent to the 3: That is why I has said that simply drawing a line from the origin to a point on the circle followed by another line that crosses it at a right angle will give you the line tangent to that point on the circle.
Anyway, the point is, an orbiting object if not held in orbit by some inward acting force centripetal force will fly away from the orbit in a direction that is tangent to the point where the inward force was removed. So because velocity is a vector that is constantly changing its direction that means the velocity itself is changing all the time. A change in velocity over some period of time is an acceleration.
As the object in orbit changes position from the 3: So tangential acceleration is, again, for uniform orbiting speed which need not be the case a measure of the rate of change in the tangential velocity vector over some period of time. If it is orbiting at the same rate of speed, its tangential velocity will be changing at a constant rate, and changing tangential velocity is tangential acceleration, so the tangential acceleration is going to be a constant value, since acceleration is just a measure of the rate of change in velocity, if the velocity is changing at a constant rate, then it is accelerating at a constant value.
### Tangential Acceleration Formula
So centripetal acceleration is due to the force at the center of the orbiting object keeping it in orbit and not flying away tangentially from where it was released. This acceleration acts towards the origin because that is where the force that is keeping the orbiting object in orbit is coming from.
Tangential acceleration is the rate of change in the tangential velocity over some period of time. Typically the orbiting object orbits at the same speed, but its position is always changing, causing the tangential velocity to be continually changing. | 1,114 | 5,470 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-39 | latest | en | 0.915149 |
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# PLZZZZZZZZZZZ analyze my awa s... :?: :|
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17 Jul 2010, 02:25
ARGUMENT
"Since a competing lower-priced newspaper, The Bugle, was started 5 years ago, The Mercury's circulation has declined by 10,000 readers. The best way to get more people to read The Mercury is to reduce its price below that of The Bugle, at least until circulation increases to former levels. The increased circulation of The Mercury will attract more businesses to buy advertising space in the paper."
Though it might seem that the argument sounds promising enough in the way The Mercury would attract more number of readers than the present number, by reducung its price than The Bugle's and the way this strategy would work in attracting more businesses to buy advertising space in the paper, it would have been more convincing if the argument had provided more information to substantiate itself.
My first reason not to agree with the argument would be considering the possibility where there might be different reasons for a reader to choose a particular newspaper than just the price. It could be the quality of the paper that allured a reader more, or even the availability of more number of interesting topics or sections like a more colorfull sports or entertainment column in The Bugle when compared to the one in The Mercury.
Second, The Bugle could have taken advantage of the situation, where it attracted The Mercury's readers, and managed to attract more readers through cheaper schemes where in the readers would get a high %of discount in taking a long-term prescription of the newspaper. In such a case, even if The Mercury would have reduced its prices, these readers would not or will not be able to prefer The Mercury to The Bugle.
Another factor which was not considered in the argument, regarding the way more businesses would be attracted towards The Mercury to buy advertising space, is the possibilty that these businesses could have made long-term deals with The Bugle already to cast their advertisements in The Bugle.
In conclusion, I would like to state that if the argument included the above factors as well then it would have sounded more consistent and convincing. Trying to follow a business plan without considering all the factors involved would usually put the business at risk.
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September 24, 2003, 06:36 Air humidity modelling #1 Chaz Guest Posts: n/a Sponsored Links Dear.... How to model air humidity in room in segregrate solver? Thank very much T_T Help me pls.
September 25, 2003, 04:39 Re: Air humidity modelling #2 Chetan Kadakia Guest Posts: n/a You can use the VOF model, but I am uncertain how accurate it will be. You will have define two phases. The primary phase is air, and the secondary phase is water vapor. But I'm not sure how you can dipserse the water vapor in the air.
September 25, 2003, 07:16 Re: Air humidity modelling #3 thomas Guest Posts: n/a Hey, I do not know what is the importance of using a seggregated solver for you. According to me you should create a Scalar (UDS) corresponding to the air humidity fraction throught the domain. Like that this scalar will be transport through out your domain. In that case the mixture model is one of the solution possible. Now maybe Fluent already takes into account air humidity transport in a multiphase model, you should check. Hope this help thomas
September 25, 2003, 10:03 Re: Air humidity modelling #4 Chaz Guest Posts: n/a The important of using seggregrated solver because I familiar with this solver than others. Thank you very much for your (2) answers I'll tryyyyyy
September 25, 2003, 14:51 Re: Air humidity modelling #5 emre Guest Posts: n/a Hi, Air humidity can be processed by solving the species transport equations, you dont need to go to multiphase models. If youre looking for predicted mean vote or predicted percent dissatisfied, you should explicitly write them. Regards emre
October 16, 2003, 04:10 Re: Air humidity modelling #6 raluca Guest Posts: n/a Hi, for a humidity model I tried with 2 speces: air and water vapour. You define allways your properties as fonctions of the mass fraction and I assure you that it works. I validate this kind of model by experimental data. good luck!
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Posted by Jay J -Moderator on October 9th, 2001 08:22 AM
In reply to pocket doors by Ron J. on October 9th, 2001 07:51 AM [Go to top of thread]
Hi Ron,
1) The wall is either built w/a header in it (when the 'pocket' is in a load-bearing wall), or 2) a cripple is built when the wall is NOT a load bearing wall. In BOTH cases, neither the pocket nor its framing is 'carrying' any load. In #1, the header is carrying the load and in #2 there is no load; just framing ALL around the door (except on the bottom since the bottom plate needs to be removed. If it's not removed, the pocket will sit up 1-1/2" off of the floor.)
3) If you turned studs sideways, and you did the math, you'd realize that 2 studs turned sideways totals up to 3". A 2" x 4" is 1-1/2" x 3-1/2", thus, 2 studs turned sideways = 3". This means that the door can only be 1/2" thick. And you can be SURE that is w/o any tolerance on the sides. In other words, it's a TIGHT fit.
Now, you're (probably) going to say (or thinking) that you only need 1 stud turned sideways. No, not good. This means that you're supporting what's above the stud is only being supported on one SIDE (or 'end') of the header. Well, a header is usually 2, 2 x 'somethings', up there. What about the 2nd 2 by ? ??? DEFINITELY not a good situation to be in! Even if the header was an I-beam, a 2 by would still need to be perpendicular to the beam to be of any use.
RE: Metal studs - The pocket door 'kit' doesn't use metal studs. I am being more 'particular' here but it's simply the pocket framing that's metal. Again, they're not called studs. First, you 'frame' the wall to 'house' the pocket kit. Then you install the pocket kit (or frame) into the stud framing. The pocket kit is ATTACHED to the 2 x 4 framing. I guess to me, studs implies more vs. pocket framing.
Please, don't think that the pocket is what's 'holding up' your wall (in either a load or non-load bearing situation.) It's not. It's the wall's framing (including the header if applicable) that's holding up the wall. The pocket is 'just there'. It's just housing a sliding door, and nothing else. I hope this alleviates any concerns and clarifies the entire construction situation. (I've said a lot so it's possible I did nothing but confuse you all ...)
My best to ya and hope this helps.
Jay J -Moderator
PS: Visit the Pocket Door Links I provided in my initial How To follow-up to Melissa.
PPS: God Bless America! | 671 | 2,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2017-04 | latest | en | 0.946898 |
https://community.powerbi.com/t5/DAX-Commands-and-Tips/For-every-row-return-the-Last-Year-and-Quarter-ASP/m-p/2951810 | 1,680,116,820,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949025.18/warc/CC-MAIN-20230329182643-20230329212643-00219.warc.gz | 234,078,654 | 98,208 | cancel
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Frequent Visitor
## For every row return the Last Year and Quarter ASP
Hi guys,
I have a fact table like the one below:
Date Revenue Quantity
31/01/2021 1000 500
28/02/2021 2000 500
....
31/12/2021 3000 100
31/01/2022 4000 200
28/02/2022 5000 100
...
The fact table is connected to a data table: my goal is calculate the ASP (Revenue / Quantity) and set the ASP from the 31/12/2021 (30.00) for every row in the 2022 having a result like the following
Date ASP
31/12/2021 30.00
31/01/2022 30.00
28/02/2022 30.00
Can someone help me to figure out a simple measure to be used with the same logic in a report?
E
1 ACCEPTED SOLUTION
Frequent Visitor
I solved my issue with the following formula
ASP_Baseline =
calculate ( [Revenue] / [Quantity], DATESBETWEEN( 'DateTable'[Date], DATE( 2022, 04, 01), DATE( 2022, 06, 30) ) )
The one above is the last quarter in a fiscal year of course.
Thanks for the hint @A_Ben_Chaoued
2 REPLIES 2
Frequent Visitor
I solved my issue with the following formula
ASP_Baseline =
calculate ( [Revenue] / [Quantity], DATESBETWEEN( 'DateTable'[Date], DATE( 2022, 04, 01), DATE( 2022, 06, 30) ) )
The one above is the last quarter in a fiscal year of course.
Thanks for the hint @A_Ben_Chaoued
New Member
ASP_21 = calculate ( Revenue / Quantity, 'Table'[DATE] =31/12/2021) return
IF ('Table'[DATE] <=31/12/2021,Revenue / Quantity, ASP_21) | 491 | 1,493 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-14 | latest | en | 0.637742 |
http://semesters.in/frames-of-reference/ | 1,544,476,619,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823442.17/warc/CC-MAIN-20181210191406-20181210212906-00569.warc.gz | 249,606,913 | 25,435 | # Frames of Reference
Before considering the difference between inertial and non-inertial frames of reference it is firstly important to consider what is a frame of reference and why is it important to understanding motion. A frame of reference is most easily defined as being that which other things are measured from. That means, for example, that we only know how fast we are going because we are able to compare it to those things around us. Consider the following, if a person is standing on the surface of the Earth next to a large tree, how fast are they moving? Most people would respond that they are not moving, that is because they have a velocity of zero relative to the tree and surrounds. However, they are actually moving relative to other frames of reference. For example they could be argued to be moving at nearly 460 m/s relative to the core of the Earth (this is the speed of the Earth’s rotation), or they could be said to be travelling at over 170,000 km/h relative to the Sun (that is the speed of the Earth’s orbit). So the original question is missing something about what frame of reference the question is considering.
The experience of relativity is often felt when you are on a train and your sole vision is of another train. You sometimes feel that you are moving, despite being stationary relative to the station, as the other train begins to leave. In such a situation there are two important components to considering what is happening. Firstly, who and where is the observer (O) and what frame or reference (R) are they referring to?
### Feedback is important to us.
#### Shiva
I love to write technical articles.
error: Content is protected !! | 345 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-51 | longest | en | 0.973426 |
http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/247139 | 1,539,856,343,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00264.warc.gz | 44,956,921 | 3,601 | ```Here is my solution. Again I've used Test::Unit to test my solution,
as well as providing a fairly nice command-line operation. I too did
the extra credit (I actually coded it that from the beginning, seemed
easy enough.)
I borrowed some code for the operator permutations, but after writing
my own code to create the numerical groups for the equations I
wondered if it might be possible to use one basic algorithm for both.
But I don't have time to fix this up at the moment. I may submit
another solution later this week.
# Based on:
# Author: Endy Tjahjono
class String
def perm
return [self] if self.length < 2
ret = []
0.upto(self.length - 1) do |n|
rest = self.split(//u) # for UTF-8 encoded strings
picked = rest.delete_at(n)
rest.join.perm.each { |x| ret << picked + x }
end
ret
end
end
# All the rest is Ryan Leavengood code :)
class EquationSolver
attr_reader :num_seq, :operators, :result
def initialize(num_seq = "123456789", operators = "--+", result_wanted = 100)
@num_seq, @operators, @result_wanted = num_seq, operators, result_wanted
end
SEP = '*' * 24
def solve
equations = 0
ops = op_combinations(@operators).map{|a| a.split('')}
generate_groups(@num_seq).each do |group|
ops.each do |op|
eq = group.zip(op).join(' ')
result = eval(eq)
puts SEP if result == @result_wanted
puts "#{eq}= #{result}"
puts SEP if result == @result_wanted
equations += 1
end
end
puts "#{equations} possible equations tested"
end
def op_combinations(operators)
operators.perm.uniq
end
# Returns an array of numeric strings representing how the given
# number can be split into the given number of groups
def num_split(number, num_groups)
return [number.to_s] if num_groups == 1
return ["1" * num_groups] if number == num_groups
result = []
((number + 1)-num_groups).times do |i|
cur_num = i + 1
num_split(number - cur_num, num_groups - 1).each do |group|
result << "#{cur_num}#{group}"
end
end
result
end
def generate_groups(num_seq, num_groups = @operators.length+1)
num_split(num_seq.length, num_groups).map do |split_on|
# Turn the result from num_split into a regular expression,
# with each number becoming grouped dots
reg_exp = split_on.split('').map{|n| "(#{'.' * n.to_i})"}.join
num_seq.scan(/#{reg_exp}/).first
end
end
end
require 'test/unit'
class SolverTester < Test::Unit::TestCase
def setup
@es = EquationSolver.new
end
def test_string_perm
assert_equal(["1"],
"1".perm)
assert_equal(["12","21"],
"12".perm)
assert_equal(["123", "132", "213", "231", "312", "321"],
"123".perm)
end
def test_op_combinations
assert_equal(["1"],
@es.op_combinations("1"))
assert_equal(["12","21"],
@es.op_combinations("12"))
assert_equal(["123", "132", "213", "231", "312", "321"],
@es.op_combinations("123"))
assert_equal(["223", "232", "322"],
@es.op_combinations("223"))
assert_equal(["--+", "-+-", "+--"],
@es.op_combinations("--+"))
end
def test_num_split
assert_equal(["11"],
@es.num_split(2,2))
assert_equal(["111"],
@es.num_split(3,3))
assert_equal(["12", "21"],
@es.num_split(3,2))
assert_equal(["13", "22", "31"],
@es.num_split(4,2))
assert_equal(["112", "121", "211"],
@es.num_split(4,3))
end
def test_generate_groups
assert_equal([["1", "2", "34"], ["1", "23", "4"], ["12", "3", "4"]],
@es.generate_groups("1234", 3))
end
end
if \$0 == __FILE__
# By default do not run the test cases
Test::Unit.run = true
if ARGV.length > 0
if ARGV[0] == 'ut'
# Run the test cases
Test::Unit.run = false
else
begin
if ARGV.length != 3
raise
else
if ARGV[0] =~ /^\d*\$/ and
ARGV[1] =~ /^[\+\-\*\/]*\$/ and
ARGV[2] =~ /^-?\d*\$/
EquationSolver.new(ARGV[0], ARGV[1], ARGV[2].to_i).solve
else
raise
end
end
rescue
puts "Usage: #\$0 <number sequence> <string of operators>
<result wanted>"
puts "\tOr just the single parameter 'ut' to run the test cases."
exit(1)
end
end
else
# Solve the default case
EquationSolver.new.solve
end
end
__END__
Regards,
Ryan
``` | 1,161 | 3,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-43 | latest | en | 0.66805 |
http://www.math.columbia.edu/~dejong/wordpress/?p=697 | 1,716,591,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058751.45/warc/CC-MAIN-20240524214158-20240525004158-00847.warc.gz | 39,092,994 | 8,442 | # More flattening
This is a continuation of previous post on flattening stratifications. The experts reading this blog could probably tell that I hadn’t really understood what is going on at all. I still haven’t mastered the subject but I think I know a little bit more now.
Let f : X —> S be a morphism of schemes. Consider the functor F : (Sch)^{opp} —> (Sets) which to a scheme T associates the set of morphisms T —> S such that the base change X_T is flat over T. Clearly the map F —> S is a monomorphism. We propose to introduce the following
Definition: We say the flattening stratification of f exists if F is an algebraic space.
What I added to the stacks project last Friday is the following: Assume S is the spectrum of a Noetherian complete local ring and f is of finite type. Then there exists a biggest closed subscheme Z of S such that X_Z —> Z is flat at all the points of the closed fibre. Moreover, Z satisfies a universal property which is formulated in terms of local morphisms of local schemes and flatness at points of the special fibre. If in addition X —> S is closed, then it follows that X_Z —> Z is flat as the set of points where X_Z —> Z is flat is an open set.
Assume S Noetherian and f of finite type and proper. In terms of Artin’s axioms for F the result in the previous paragraph takes care of the existence of a formal versal deformation. I think there is a straightforward little argument which takes care of openness of versality (but I did not write this out completely). Since f is of finite presentation, it follows that F is of finite presentation by the usual arguments on limits and flatness. Relative representability is OK too. Hence, if S is excellent then F is an algebraic space by Artin’s theorem. But of course we can descend X —> S to a situation of finite type over Z and hence we get the result in general (with same hypotheses). In fact, using limit arguments we may be able to prove the same thing when S is arbitrary and f proper and of finite presentation.
Still, my answer to Jason’s question here was a bit premature. Some of the above may work exactly as stated in the generality of Jason’s question. But I was trying to prove flattening stratifications exist without using Artin’s theorem. In particular, it should be possible to avoid using general N\’eron desingularization.
The reason I started looking at flattening stratifications was to construct Quot and Hilbert schemes/spaces/stacks. And the reason to discuss those was that Artin’s trick uses Hilbert spaces. However, it only uses the Hilbert space parametrizing closed subschemes of length n on a space. Of course I could take the easy way out and just use one of the explicit constructions of Hilb^n. But once I started looking at the problem of constructing flattening stratifications (which is related to descent of flat modules) I just couldn’t stop myself.
## 3 thoughts on “More flattening”
1. I think you can use Raynaud-Gruson, Theorem 4.1.2 to avoid using general N\’eron desingularization. This theorem basically says that the result of the 4th paragraph above works over a henselian local ring. But I think there is a much easier way in the proper finite presentation case.
2. Pingback: Raynaud-Gruson « Stacks Project Blog | 749 | 3,265 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-22 | latest | en | 0.932166 |
https://www.wyzant.com/resources/blogs/357281/taylor_expansion_and_approximating_roots_of_polynomials_over_the_rational_field | 1,606,273,713,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141180636.17/warc/CC-MAIN-20201125012933-20201125042933-00303.warc.gz | 934,229,788 | 12,512 | # Taylor Expansion and approximating roots of polynomials over the rational field...
I answered a nice question on WyzAnt about approximating roots to a quadratic equation that had real but irrational roots.
Calculus enables us to find rational approximations to these irrational roots. I will re-present the solution to the problem, and continue to find a general expression to do this for an arbitrary polynomial.
"Problem: approximate roots of -9x^2+8x+5.
There is a nice approach using calculus to estimate/approximate a function without a square root and calculator.
We can use the concept of moments to get an approximation to a function. For this example, we have a quadratic function in (x) with coefficients, a=-9, b=8, and c=5, as indicated in a previous solution.
Thus f(x)=ax^2+bx+c.
First we need to find a general idea of where a root (an x where f(x)=0) is located. To do this we can check the value of the function for some easy numbers, x=0 and x=1 are always good choices:
f(1)=a+b+c=4
f(0)=c=5
We notice the function is increasing from 0 to 1 (the value decreased as x decreased). So if we try x>1 maybe we will find a 0...next evaluate f(2)
f(2)=a*4+b*2+c=-36+16+5 < 0
Awesome, since this is a continuous function we know there must be a root between 1 and 2...lets call this root x=1+\eps, for some little \eps that is between 0 and 1.
We are now ready to use a linear approximation and a first derivative to approximate where this root is!
Let (x_0,y_0)=(1,4), and we set \delta x=\epsilon.
We need to recall the derivative f'(x) denotes the slope of the function at a point, so we will approximate the derivative at the root using f'(1)...Since f'(x)=2ax+b...f'(1)=2a+b=-10...
So near to x=1, say like the root we seek at x=1+\epsilon, the slope of a line approximating this function should be about -10...thus when y=0 at x=1+\epsilon a line approximating the curve should be:
y-y_0=m(x-x_0)...0-4=-10(\epsilon)...so \epsilon=4/10=2/5...and the root should be close by to x=1.4.
Now evaluate f(1.4)= -9*1.4^2+8*1.4+5 = -9*1.96+11.2+5=-17.64+16.2 = -1.44...
So we overshot by a little, our \epsilon was too large, it made our y value decrease from 4 to something past 0, a negative number...thus the true \epsilon should of been smaller. We have a choice now, with given possible values above we can quickly see which choice has an x greater than 1 and less than 1.4 {1.31,-.42}.
Of course evaluating all the choices initially works too...but this allows us to apply calculus to get an approximation, in case no choices were available and a linear approximation would be acceptable.
We can use this new point (x_1,y_1)=(1.4,-1.44)=(7/5,-36/25) to find a better approximation.
Use x=7/5-\delta, for some small \delta between 0 and 1, and repeat the procedure:
y-y_1=m(x-x_1), where we now let m be the deviated at x_1=7/5...
Thus 0-(-36/25)=f'(7/5)*(-\delta)...and \delta=18/215...so x=1+\epsilon-\delta=1+2/5-18/215=(215+86-18)/215=283/215...doing long divison to a few decimal places yields 1.316.
This shows how we can by hand...using no machinery, calculators, excel sheets...produce a better approximation over the field of rational numbers than the choices given offers!"
We see that in general for a function y=f(x) that we are looking for a root of...if we find an x_0 where f(x_0)>0 and f(x_0+1)<0 (or vice versa)...we know there exists an \epsilon in (0,1), with x=x_0+\epsilon and f(x)=0.
We can find this by writing down the linear approximation:
0-f(x_0)=f'(x_0)(\epsilon), and solve for \epsilon=-f(x_0)/f'(x_0)...this is the first moment, an approximation of first order:
root ~ x_0-f(x_0)/f'(x_0)
We can now evaluate f at x_0+\epsilon=x_1 and find a new base point to approximate near....x_1=x_0-f(x_0)/f'(x_0)...
We first need to check if f(x_1) is positive or negative and then proceed by adding or subtracting appropriately...but we end up with a telescoping sum of epsilons and deltas that get better and better approximations for our root...
WLOG assume that f(x_1)<0, meaning our epsilon was too large and we should subtract off a \delta in (0,1), and let
x = x_1 - \delta = x_0 + \epsilon - \delta.
We then have a linear approximation of this x by looking at the line her x_1=x_0+\epsilon:
0-f(x_1)=f'(x_1)(-\delta)...so \delta = f(x_1)/f'(x_1)...and a second order approximation for the irrational root becomes:
root ~ x_0+\epsilon-\delta = x_0-f(x_0)/f'(x_0)-f(x_1)/f'(x_1)
Note how this is a sort of generating function for moments...substitution our formula for x_1 in terms of x_0, x_1=x_0-f(x_0)/f'(x_0), yields:
root ~ x_0+\epsilon-\delta = x_0 - f(x_0)/f'(x_0) - f(x_0-f(x_0)/f'(x_0))/f'(x_0-f(x_0)/f'(x_0))
Note the nested derivative in the second order term here.
Untangling this using what we know about derivatives leads to a general taylor expansion for a function near a point x=a when the function is infinitely differentiable at x=a.
\$90p/h
Casey W.
Mathematics (and Science) Instruction by a Mathematician!
100+ hours
if (isMyPost) { } | 1,497 | 5,040 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-50 | latest | en | 0.891428 |
https://mathsuggestion.com/class-seven-math-suggestion-2017/ | 1,674,934,646,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00530.warc.gz | 430,233,619 | 11,592 | # Class Seven Math Suggestion 2017
Class Seven Math Suggestion 2017 for the student of Class Six from www.mathsuggesiton.com. Because The Class Seven is second step of High School Level, after they have passed in PSC or PECE or Ebtadayee. And the next step is second center exam of a student life in Bangladesh Curriculum Education System. That’s JSC Examination. Before the JSC examination, a student must be mentally prepared. So class Seven really carries importance. We will give them Class Seven Math Suggestion 2017 totally free. Because, we want to serve the students whom will not capable continue their study various reason such as Weakness in Math.
They have total 11thnumber of chapter in their Bangladesh Curriculum Mathematics Text Book. Again, 11thChapter is divided by three division. There are the Arithmetic, the Algebra and the Geometry. That division is very familiar education system of High School Level for Bangladesh Curriculum Education Board.
We was already arranged Class Seven Math Suggestion 2017 for their (Class Seven’s Students) upcoming exam. We’ve got it in Four parts. Then we will discuss the whole questions by chapter-based creative questions. Along with this, we created creative questions by looking at all the boards. So that one question can be raised if they do a question. We think that if students can understand the figures, they will have a good creation in creative mathematics question.
Completed mathematics are divided into 4th categories. There are –
1. Arithmetic: Rational and irrational numbers, equality and profit loss, measurement. A creative question will be examined from each section. But must be one question examined from last chapter.
2. Algebra: Algebraic Multiplication and division, Algebraic sources and applications, Algebraic Fraction, Simple Equation. A creative question will be examined from each section. (Note: but one question must be included from Simple Equation)
3. Geometry: Parallel Straight Line, Triangle, consensus and similarity fraction. Two question must be included this section.
4. Information and Data: A question will be examined from this section. | 429 | 2,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-06 | longest | en | 0.949582 |
https://www.physicsforums.com/threads/what-is-the-acceleration-of-the-car.435977/ | 1,725,815,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00127.warc.gz | 917,655,323 | 18,527 | # What is the acceleration of the car?
• tica86
In summary, the conversation discusses the acceleration of a car sliding down an icy driveway and the calculation of its acceleration based on the forces acting on it. The answer to the question is determined to be 1.7 m/s^2. The conversation also briefly touches on finding the net force required to accelerate a bike and rider from rest to a final velocity in a given time frame.
tica86
A car is slides down an icy driveway that slopes at an angle of 10 degrees. What is the acceleration of the car?
If it is can someone explain how they figure it out, thanks!
Forces acting on the car:
Weight = mg
Normal = mg cos 10 (deg)
Horizontal = mg sin 10
macar = mg sin 10
cancel out m,
acar = g sin 10 = 1.7 m s-2
So yes, 1.7.
Ambidext said:
Forces acting on the car:
Weight = mg
Normal = mg cos 10 (deg)
Horizontal = mg sin 10
macar = mg sin 10
cancel out m,
acar = g sin 10 = 1.7 m s-2
So yes, 1.7.
Ok, thanks. I have another question:
Find the net force (in Newtons) required to accelerate a 110 kg bike and rider from rest to a final velocity of 22.7 m/s in 10 seconds?
since f=ma 110(22.7)=2497 do I divide that by 10 seconds??
not exactly...
first solve for average acceleration using equations of motion...
The correct answer for the acceleration of the car is 1.70 m/s^2. To calculate this, we can use the formula for acceleration, which is a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time taken for the change in velocity.
In this scenario, we can assume that the car starts from rest (vi = 0) and reaches a final velocity as it slides down the icy driveway. The final velocity can be calculated using the equation vf = vicosθ, where vi is the initial velocity and θ is the angle of the slope (10 degrees in this case).
Thus, vf = (0)m/s x cos10 = 0 m/s. This means that the car's velocity remains constant throughout its motion down the slope.
Next, we need to calculate the time taken for the car to reach the bottom of the slope. This can be done using the equation t = d/vf, where d is the distance traveled by the car and vf is the final velocity.
As we do not have the distance traveled, we can use the trigonometric relationship tanθ = opposite/adjacent to calculate the distance. In this case, the opposite side is the height of the driveway and the adjacent side is the length of the driveway. So, tan10 = h/L, where h is the height of the driveway and L is the length.
Rearranging this equation, we get h = Ltan10. We can assume that the driveway is 10 meters long and the height is 1.7 meters (as tan10 = 0.176 and 0.176 x 10 = 1.76). So, the distance traveled by the car is 1.7 meters.
Now, we can plug in the values in the equation t = d/vf. So, t = 1.7m/0m/s = undefined. This means that the time taken for the car to reach the bottom of the slope is 0 seconds.
Finally, we can use the formula for acceleration a = (vf - vi)/t to calculate the acceleration. As we have already calculated that vf = 0 m/s and vi = 0 m/s, the equation becomes a = (0 - 0)/0, which is undefined.
However, in physics, we can use the concept of limits to calculate the acceleration in this scenario.
## 1. What is acceleration?
Acceleration is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction.
## 2. How is acceleration calculated?
Acceleration is calculated by dividing the change in velocity by the change in time. This can be represented by the formula a = (v2 - v1) / (t2 - t1), where a is acceleration, v1 and v2 are the initial and final velocities, and t1 and t2 are the initial and final times.
## 3. What units is acceleration measured in?
Acceleration is measured in units of distance per time squared, such as meters per second squared (m/s^2) or feet per second squared (ft/s^2).
## 4. How does acceleration affect the motion of a car?
Acceleration affects the motion of a car by changing its speed and/or direction. When a car accelerates, it will either speed up, slow down, or change direction, depending on the direction of the acceleration and the initial velocity of the car.
## 5. What factors can affect the acceleration of a car?
The acceleration of a car can be affected by various factors, such as the engine power, the weight of the car, the friction between the car and the road, and external forces like wind or incline of the road. Additionally, the acceleration can also be affected by the driver's actions, such as pressing the gas or brake pedal.
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2K | 1,249 | 4,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.906046 |
https://www.wyzant.com/resources/answers/8959/what_does_n_mean | 1,524,595,901,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125947033.92/warc/CC-MAIN-20180424174351-20180424194351-00001.warc.gz | 926,173,604 | 18,179 | 0
# What does n! mean
In the algebra problem: for any positive integers n,(n+1)!/n! -n=0 what does the ! mean?
### 4 Answers by Expert Tutors
Miles A. | An enthusiastic game designer with experience teaching childrenAn enthusiastic game designer with exper...
5.0 5.0 (2 lesson ratings) (2)
0
The ! in n! means n-factorial. A factorial is the product of all positive integers less than n.
Matthew S. | Statistics, Algebra, Math, Computer Programming TutorStatistics, Algebra, Math, Computer Prog...
4.9 4.9 (22 lesson ratings) (22)
0
n! refers to a factorial, a product of n numbers, each one less than the preceding value. You can write a factorial n! by starting with the number n, multiplying it by one less than the previous number, and repeat until you reach 1, at which time you can stop. So 5! can be written as follows: 5! = 5 * 4 * 3 * 2 * 1 = 120. Similarly, 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720. Note all the repetition here... it turns out 6! = 6 * 5! That's important because you can re-write (n+1)! as (n+1) * n! So when you divide (n+1)! by n! as you have in the first term, the n! cancels out in both numerator and denominator leaving only (n+1). So as Robert points out, subtracting n from the first term would leave 1, not 0. So it appears the second term should be (n+1) to be valid.
Edgardo O. | Math, ballroom dancing and ki ho' alu tutoring available. Math, ballroom dancing and ki ho' alu tu...
0
The ! is the factorial sign. Example 4! = 4 x 3 x 2 x 1
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
4.6 4.6 (13 lesson ratings) (13)
0
(n+1)!/n! -n ≠ 0
Do you mean (n+1)!/n! - (n+1) = 0?
Since (n+1)!/n! = (n+1)n!/n! = n+1, (n+1)!/n! - (n+1) = 0 is an identity for any positive integers n. | 607 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2018-17 | latest | en | 0.866624 |
https://cstheory.stackexchange.com/questions/21409/what-relations-are-there-between-a-problem-hardness-and-the-hardness-of-verifyin | 1,621,382,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00467.warc.gz | 207,508,244 | 38,461 | # What relations are there between a problem hardness and the hardness of verifying a witness?
Suppose you are given a Dominating Set instance, $<G,k>$.
Now suppose I give you a set of vertices $D$ of size $k$. Deciding whether $D$ is a dominating set of $G$ requires linear time in the size of $G$, and doesn't seem to be possible only by looking at the vertices of $D$.
In contrast, if we have a $k-path$ instance $<G,k>$ (asking whether a simple path of length $k$ exists in $G$), and I give you a tuple $P$ of $k$ vertices, you can verify that $P$ is a k-path by reading merely $k$ bits of the adjacency table.
In general, assume you have some graph problem that includes a parameter$L\subset \mathcal{G}\times \mathbb{N}$, whose witness is a set of $k$ vertices/edges ($k$ is the parameter of the problem).
Now we can define the set of problems which are "easy" to verify as:
$EasyVer=\{L\subset \mathcal{G}\times \mathbb{N}|$ a witness $w$ of $L$'s instance can be verified in $poly(k)$ time$\}$, i.e. independent of $|V|,|E|$.
It seems, for example, that $EasyVer \not\subset FPT$ and $FPT \not\subset EasyVer$, as
1. $VC\in FPT, VC \notin EasyVer$.
2. $Clique\in EasyVer$, while Clique is $W[1]-hard$.
• Does my definition even make sense?
• Are there known complexity classes with similar meaning?
• Any other complexity relations to known classes?
• Does it makes more sense to define $EasyVer$ in terms of the number of bits a verification algorithm needs to read from the adjacency matrix?
• Does generalizing the definition to $Ver_{f(n,k)}=\{L\subset \mathcal{G}\times \mathbb{N}|$ a witness $w$ of $L$'s instance can be verified in time $O(f(|G|,k))$ $\}$ makes sense?
• This is a neat observation and a nice question. Perhaps explicitly state that poly(k) in the definition of EasyVer should be independent of n and m. Also maybe observe that parameterized hardness does not imply anything about EasyVer. Clique in W[1] is in EasyVer, but DomSet in W[2] is not in EasyVer. Also, don't state that Clique is not in FPT since it is dependent on the open problem of whether FPT = W[i] for any i>0. – JimN Mar 6 '14 at 20:28
• If you're willing to relax "independent of n" to "only weakly interacting with the input" there's an active body of work on this topic. – Suresh Venkat Mar 7 '14 at 6:09
• @SureshVenkat, can you please elaborate? – R B Mar 7 '14 at 8:49
• Just some quick thoughts: (1) This seems related to the PCP theorem, but without any randomness. (2) I think that asking about witness size is very close to asking about the nondeterministic time complexity of the algorithm, i.e. you want those problems solvable in sublinear nondeterministic time and space (in fact time/space parameterized by k) in RAM or some such model. This maybe relates to your final question (#5), depending on if your model of computation requires you to scroll all the way through the adjacency list to get to some particular entry. – usul Mar 7 '14 at 22:05
• There are two ways people have studied to reduce interaction with the input: (1) only allow sublinear number of probes of the input (2) only allow streaming access to the input (with sublinear working space). In these cases, there are results characterizing what you can compute in a single round or even in multiple prover/verifier rounds. In all cases the verifier is randomized though. – Suresh Venkat Mar 8 '14 at 6:10 | 919 | 3,396 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2021-21 | latest | en | 0.888958 |
https://www.mathdoubts.com/factorize-x-power-4-plus-64/ | 1,721,102,792,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00342.warc.gz | 782,590,966 | 11,123 | Factorize $x^4+64$
The real number $64$ is added to a variable $x$ raised to the power of $4$ for representing an unknown in a polynomial. In this problem, the given algebraic expression has to be factored.
$x^4+64$
Express the expression in sum of squares
The variable $x$ raised to the power $4$ and the real number $64$ can be written in square form. So, they can be written as square of $x$ squared and $8$ squared.
$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$
Simplify expression as square of a Binomial
The algebraic expression is converted as the sum of squares. The sum of squares is actually part of the expansion of square of either sum or difference of two terms. According to them, the polynomial in square form should have a term additionally and it is $2 \times 8 \times x^2$. So, add it to the expression and subtract the same term from their sum for mathematical acceptance.
$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$ $+$ $2 \times x^2 \times 8$ $-$ $2 \times x^2 \times 8$
$=\,\,\,$ $\big(x^2\big)^2$ $+$ $8^2$ $+$ $2 \times 8 \times x^2$ $-$ $2 \times 8 \times x^2$
According to the square of sum of two terms formula, the first three terms can be simplified as the square of sum of $x$ square and $8$.
$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $16 \times x^2$
Look at the second term in the expression, the second factor is in square form. So, try to express the real number $16$ in square form for expressing their product in square form.
$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $4^2 \times x^2$
According to the power of a product rule, the product of squares can be simplified as a square of product of their bases.
$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $(4 \times x)^2$
$=\,\,\,$ $\big(x^2+8\big)^2$ $-$ $(4x)^2$
Factorise by the difference of squares method
The given algebraic expression $x^4+64$ is successfully converted as $\big(x^2+8\big)^2$ $-$ $(4x)^2$. The simplified expression represents difference of the terms and it can be factored as per the factorization by the difference of two squares.
$=\,\,\,$ $\big(x^2+8+4x\big)\big(x^2+8-4x\big)$
$=\,\,\,$ $\big(x^2+4x+8\big)\big(x^2-4x+8\big)$
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Get the latest math updates from the Math Doubts by subscribing us. | 734 | 2,441 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-30 | latest | en | 0.875805 |
https://www.univerkov.com/in-a-rectangular-trapezoid-the-bases-are-05-and-17-cm-and-the-large-side-side-is-13-cm-find-the-area-of-the-trapezoid/ | 1,723,354,169,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00016.warc.gz | 804,570,199 | 6,326 | # In a rectangular trapezoid, the bases are 05 and 17 cm, and the large side side is 13 cm, find the area of the trapezoid.
Let’s draw the height of the CH trapezoid. The formed quadrangle ABSN is a rectangle, since AB and CH are perpendicular to the bases of the trapezoid.
Then AH = BC = 5 m.
The length of the segment DH = AD – AH = 17 – 5 = 12 cm.
From the right-angled triangle СDН, we determine the length of the leg СН.
CH ^ 2 = CD ^ 2 – DH ^ 2 = 169 – 144 = 25.
CH = 5 cm.
Determine the area of the trapezoid.
Savsd = (ВС + АD) * СН / 2 = (5 + 17) * 5/2 = 55 cm2.
Answer: The area of the trapezoid is 55 cm2.
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# Is 73 An Odd Number?
• Yes, number 73 is an odd number.
• Seventy-three is an odd number, because when it is divided by 2 it will leave a comma spot.
## Is 73 Odd Or Even?
• Number 73 is an odd number.
## How To Calculate Odd Numbers
• All numbers that cannot be divided exactly by 2 are odd numbers.
• Odd numbers can't be divided evenly into groups of two. The number five can be divided into two groups of two and one group of one. Odd numbers always end with a digit of 1, 3, 5, 7, or 9.
## Mathematical Information About Numbers 7 3
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
• About Number 3. Three is the first odd prime number and the second smallest right after number two. At the same time it is the first Mersenne prime (2 ^ 2-1), the first Fermat prime (2 ^ {2 ^ 0} +1), the second Sophie Germain prime and the second Mersenne prime exponent. It is the fourth number of the Fibonacci sequence and the second one that is unique. The triangle is the simplest geometric figure in the plane. With the calculation of its sizes deals trigonometry. Rule of three: If the sum of the digits of a number is a multiple of three, the underlying number is divisible by three.
## What Is An Odd Number?
• An odd number is an integer which is not a multiple of two. If it is divided by two the result is a fraction. One is the first odd positive number. The next four bigger odd numbers are three, five, seven, and nine. An integer that is not an odd number is an even number.
• An even number is defined as a whole number that is a multiple of two. If an even number is divided by two, the result is another whole number. On the other hand, an odd number, when divided by two, will result in a fraction. Since odd and even numbers are defined only in reference to the set of integers (..., -3, -2, -1, 0, 1, 2, 3, ...), all negative numbers can also be either odd or even.
© Mathspage.com | Privacy | Contact | info [at] Mathspage [dot] com | 671 | 2,735 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-27 | latest | en | 0.910996 |
fiftysix.scot | 1,619,087,931,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00126.warc.gz | 352,564,563 | 4,884 | # Time
The current time is:
## : : .
(F : 07 : A . 71 9E) reload
• (optional) javascript to keep the clock updated
• time conversion form
# Definition
The unit of this time system is days.
It has “seconds”, “minutes”, “days”, “months” and “years”.
Each aspect is described in detail below.
## Naming
#TODO
• 1/65536 Day (second): ?
• 1/256 Day (minute): ?
• 1 Day (day): ?
• 8 Days (week): ?
• 16 Days (month): ?
• 1 Year (year): ?
## Days
The day is exactly 86400 SI seconds.
The day is split into 256 minutes. In base-16 this reads similarly to a percentage.
The minute is split into 256 seconds.
The first second + hour of the day is _ (00_00) and the last second is _ (FF_FF)
The day begins at (nearly) the same time as a UTC day (difference in leap seconds causes some offset)
## Months
There are a total of 23 months per year.
The first 22 months have exactly 16 days.
The last month continues to the end of the year, and has the task of dealing with leap days and leap seconds. It has either 13 or 14 days.
Months are identified primarily by numbers.
The month may be split into “weeks” which must align with months.
Weeks may be 2*8 days, 4*4 days or even 7+7+2 days
## Years
A year begins in timezone 0 at the beginning of the day of the solstice. (In other timezones the solstice will sometimes be the last day of the year).
Year 0 began at 2005-12-22 00:00:00 (UTC-10:30? (timezones not completed yet))
This number was chosen randomly, as I can’t think of a logical beginning.
## Time “zones”
Time zones are defined as lines instead of zones. Countries choose their nearest line.
The earth is split into (16) “times”, each is (16) minutes apart.
The first time is +. (+0), which is equal to… (UTC-10.30?). The solstice is always on the first day of the year.
The last time zone is +. (+0.F), which is equal to… (UTC+12?). The solstice is always on the last day of the year.
The “International Date Line” might stay the same?
Countries can chose their own timezone, so they can recreate the current mess and avoid splits through the middle of countries.
Some countries may also want to chose a time zone outside the range 0-1, as they already do in the current system (-12 to +14).
#TODO: Diagram
## Leap days
leap years are not chosen using a basic calculation, but are simply a result of the time between solstices, meaning they will always ensure the solstice is the first day (at time line+0). They are still usually 4 years apart.
This results in leap years being delayed by one year roughly every 32 years.
This image shows the current leap years relative to the solstice. The solstice goes over about 2.5 days of time, whereas in this system, the solstice would never go over more than one day of time.
(source: https://en.wikipedia.org/wiki/File:Gregoriancalendarleap_solstice.svg)
## Leap seconds
Leap seconds occur at the end of the year. They occur to stay as close to either UTC or “actual” time (#TODO) as possible, but are not copied directly from UTC due to the difference in length of the seconds.
They can be calculated with the following calculation, where `LS` = number of original leap seconds at beginning of (new) year, `S` = number of new leap seconds this whole year, `SI` = length of original second, `SS` = length of new second.
`S = round(SI/SS*LS)` or `S = round(512/675*LS)`
Since there “cannot” be a second with 5 digits, the last second of the year will be repeated twice. If the second needs to be referenced precisely, it will require an imaginary date, the 14th or 15th day of the month.
## Format
I have chosen this format to indicate months, days, hours and seconds, while also showing it as a single number (if the dashes/spaces are ignored).
Dashes or spaces can be used as a separator. The year is it’s own number, and is therefore separated more prominently by a colon.
`Y:MM-D.HH-SS` / `Y: MM D. HH SS`
Example: :-.- / : . (E:14-F.FB-E7 / E: 14 F. FB E7)
It could also be shortened to `Y:MMD.HHSS` (Eg `E:14E.DAE1`) when readability is less important.
Since all measurements below months match the base, this means that the month, day, hour and second can be written as a single number which represents the number of days since the start of the year, but in which the month, day hour and second can still be easily read. | 1,269 | 4,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-17 | latest | en | 0.915672 |
http://roguebasin.roguelikedevelopment.org/index.php?title=Diffusion-limited_aggregation&direction=prev&oldid=22305 | 1,585,594,247,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370497301.29/warc/CC-MAIN-20200330181842-20200330211842-00098.warc.gz | 155,036,539 | 8,945 | # Diffusion-limited aggregation
Many roguelikes use rectangular rooms connected by straight corridors for their maps, which after a while might get boring (or might not). Relatively few alternatives have been documented (apart from cellular automata, an excellent article), so here's another radically different method.
Diffusion-limited aggregation is a process which grows tree-like structures from a central seed "." . Particles "&" are placed at random on the grid, and go on a random walk (they repeatedly move one step in a random direction) until they hit the seed or a "frozen" particle, at which point they "freeze" in place and become a floor grid:
```##### ##### ##### ##&## ##### ##&## ##### #####
&#### #&### ##### ##### ##&## ##### ##&## ##.##
##.## ##.## #&.## #..## #..## #..## #..## #..##
##### ##### ##### ##### ##### ##### ##### ####&
##### ##### ##### ##### ##### ##### ##### #####
```
etc.
Varying parameters and details of the algorithm can produce different-looking maps; for instance, here's one generated by random walkers who move both diagonally and orthogonally:
```##########################################
#######.#########.########################
#####.#.########.#############.#.#########
###.##.#.########.#############..#########
####.#....######.################.####.###
#####.##....##...##############.#..#.....#
######..##..##..################..##....##
#######.###...###############.#........###
######.#####....########..##....#..##...##
############.#.########...##.#...###..##.#
##########......#######..##.#..##..##.####
##########....#.#####.#.####...#..####.###
########...##..##.###....#.......#########
#######...#........###....#...#.##########
########..#.##.....####...#..##.##########
########...#######.#.....#....############
########..######.##.....##...#############
#######...#.####...#..#####..####.########
#######....###.#..#..#.#.##...##.#.#######
########.######.#####.#..#...#.......#####
#################.##...#.............#####
#################.....##...###.#..#.######
################....#..##.....#.#....#####
############.###...#...####...#.....######
#########.#.##.##..##...###...##.#########
#########.#.....#.###...###..#############
##.#####..#...#..#.#......#.##############
##..#.##......##.#..#.#.......############
####....#..###..#.####..#.....############
###..###.###..#.#####....#...#############
#######.####...#.####..##...##############
####..#......#....####.#..#....###..######
####.#.#..#..#.########...##....##..######
#######........#########.#.#.#....########
########.#...#.########.#.#..#.#.#########
######..#.###.##########.###.#..##########
##.#...#.####..###########################
#........###..############################
#.######.#################################
##########################################
```
And here's one produced by walkers who can only move orthogonally:
```#############################################
############################.################
#########################.##.################
########################..#..################
##############.#########.....################
##############..#######.......###############
#############....#######.......##############
###############..####........################
##############.#..##.....#.##################
#############........##..####################
###################..#...####################
###############......##..###.################
################......##..#..#...############
######.#.########.#...##.##..#...############
###..#....###.##.....##..##.###..##########.#
#...........#..##..#.#..........###.#.#.##..#
####...###.....###.........#.#....#.#......##
####.##........#.#...........#.#.........####
####.####..............##.##.....####...#####
########....#......##...#........############
######......#.#..##...#.##..##.##############
#######.....######.......#..#.###############
########...#####.#.............##############
#########.#####..#...#..##......##..#########
################........#..#######.##########
###############....##........#.##....#.######
##############...####.....#............######
##########......####...##.#.#....#...########
###########....####...#####.#..#.#....#######
###########...#######...##.....##..##.#######
##################....#.####...##...#########
##################..###..##.....#..##########
###################.###..###.#...#.##########
##################..###..###......###########
#######################.########.############
##################......#####################
#################....#..#####################
######################...####################
#####################...#####################
####################..#######################
#####################.#######################
#############################################
```
Obviously, the orthogonal-only version produces much "neater" maps, and wider tunnels. You could also try freezing particles whenever they pass next to the dug-out section, rather than just when they try to walk into it (as was used in the above examples). This would probably speed execution up somewhat. A few other things are immediately noticeable (they may or may not be good things, depending on what you want from your gameplay):
• The algorithm is guaranteed to produce a connected map.
• Loops are quite infrequent, large loops are generally not present, and dead ends are common. If you want more loops, they can be added by placing a tunnelling random walker or forcing placement of extra corridors (according to the level's flavour).
• There are lots of places to corner a player or monster. (Or to hide in!)
• There are no well-defined rooms, but a vague impression of corridors
• There are lots of obstacles to ranged combat
It's not visible in the above because I trimmed it, but there may be a lot of unused space on the perimeter of the level. You can fix this by tailoring the number of particles to your level size, or by running regular checks that the cave hasn't yet reached the edge.
However, building a whole dungeon by putting every floor tile on its own random walk can take a lot of time, especially at the beginning when a particle won't stop until it hits the precise spot in the centre of the map. This can be reduced by:
• making the program more eager to freeze particles,
• beginning with a larger "seed", for instance a room,
• weighting the random walk so that they drift towards the centre (this may change the look of your levels significantly!),
• resizing the grid as the level grows (this could be difficult),
• using blocks of tiles instead of single tiles as particles.
This last option can have great effect on the appearance of the level, depending on the set of blocks used and the weights assigned to them; for instance, if all the blocks are relatively large rectangles, and long rows and columns, this method could be used to make traditional roguelike dungeons (and collision-checking would reduce to checking the edges of the block). Placing Angband-style vaults would also be easy in this model: simply make sure they have a wall surrounding them, and change the way particles are initially placed so that they don't interfere with the vault's innards. Block aggregation will look very different depending on whether particles are allowed to spawn in a position where they would immediately freeze.
Other ideas for block aggregation:
```############################################################
#######################################..#.......###########
#####...###############################..#.#####.###########
#####.#.#########################........#.#####.###########
#####.#.###############.......###........#.#####.###########
#####.#.###############.#####.#####.##...#.#####.###########
#####.#.###############.#####.#####.##...#.#####.###########
#####.#.###############.#####.#####.##...#.#####.###########
#####.#.......#########.......#####......#.......###########
#####....####.################..######........##############
########.####.################..######.###....##############
########.####.#######........#..######.###.##.##############
########.####.#######.######.#..######.###.##.##############
########.####.#######.######.#..######.###....##############
########.####.#######.######.#..######...####.##############
########......#######....###.#..######........##############
##########.....#.............#..##.................#########
##########.....#.........#####...........##.######.#########
############......###.....####..............######.#########
############........#.###.######...##..#.##.######.#########
############.####.#.#.###.#..............##........#########
############......#.#.#...........###.....##################
############.######.#.#....#####..###.....##################
####.....###.######.#.....######..####.#..#####...##########
####.##.......#####.######........####.#..#####.#.##########
####.##........####.######.####...####.......##.#.##########
####.##....................####...##.........##.#.##########
####...........#####.##.........####.###........#.##########
######...##.......................##.###...###..#.##########
######.#.##.##.##..##.#...#.#..##.##.###...###....##########
######.#.##........##.#...#.#..................#############
######.#.#####.##.........#.#...#........###################
######...####.........#.#.#........#.#...###################
######..........................#........###################
######...###........##.##.....#..........###################
######.......#.....##.........#..........###################
##############.....##..#####....#........###########.....###
##############.....##........##.......#..#########....##.###
############.......###.###...##.##.##.#........###.#..##.###
############....#..###........#.##.##.....##.#.###.#..##.###
##########..............##.#..#....#####.....#.###.#.....###
##########.#...#...####.......####.................##.######
##########.#...#.#########.#.#####..##.##.###.###..##.######
##########.........#######.#.#####........###.###..##.######
########..##....##.#######...#####.###.######.........######
########..##.......###############.###.#####################
########..###.#.##################.###.#####################
########..###.#.######....########.....#####################
########..###.#........##.##################################
####...#..###.#........##.##################################
#.......#####...######.##.##################################
#.##.#..##############....##################################
#.##.#..####################################################
#.##.#..####################################################
#.##.#..####################################################
#.......####################################################
############################################################
```
• Corridors only
```############################################################
###############################################.############
######################################.########.############
###################################....########.############
######################################.########.############
###########...########################........#.############
###########......###################....####....############
###########.#####......###############.######.##############
###########.######.##########################.##############
###########.######.################.....##....##############
###########.#.####.#################.#.....##.##....###.....
###########.#.####.############.#.##.########.####.##.#.####
###########.#.####.############.#.##.....####.####.#..#.####
##########....#.#.#############.#.##..#######.####.#....####
#############.#.#.###.#########.#.##..#######.###....#..####
#############.#.#.###.##......#....#..#######.###..#.#...###
#############.#.#.###.########.........##########..###.#.###
#############.......#.##.......#...##........####..###....##
#############.######......#####.......######.####.####....##
#############.######...########...##########.####.####....##
#############.######...########...#.....###...............##
#############.##........####.##...###.#####.#####..#####..##
########.......#.....#.#####.##.#####.#####.#####..#.####.##
###############....#.......#.##.##.##.#####.####.....#######
####################.##.###..........####....#######.#######
####################.##.###.....#..#.###.#.......###.#######
#######################.#####........###.###########.#######
#######################......####....###.###########.#######
#######################.####.........###.###########.#######
############################.#####...###############.#######
############################.#####....##############.#######
############################.#####.#..##############.#######
############################.#####.#..##############.#######
############################.###......####.#################
#############################...##.#...###.#################
###########################........##...##.#################
##################################.####......###############
##################################.#########################
################################....########################
############################################################
```
```############################################################
########################################.###################
#######################################.####################
###################.##################.#####################
########.###########.################.######################
#########.#########.#..##.###.######.#######################
##########.#########...#.###.####...########################
######.####.########..#.###.####...#########################
#######.###..######.##.###.####...##########################
########.###..#######.#.#.#.##...###########################
#########.###..######.##.###..##############################
########.#####..######..###.#.##############################
#########.#####..######.##.###.########.####################
#########..#####..####.#..####..######.########.############
########.##.#####..##.##.####.#...###.##.#####.#.###########
#######.####.#####.###..######.#...#..###.###.###.##########
######..###.###.#.####..#######.###...##.#.#.#####.#########
###.#.##.#.##.##..###..#.#####.#.#...#.##.#.#######.########
##.#.####.####....#..##.#.###.#.#...#.#..#.#.#######.#######
#.######...###..##.####..###.###.....#..####..#######.######
#######.##..#..#..####.##.#.###.#.#.....#####.##############
######.####.......###.####.###.#.#.#..########..############
#.###.#####..#...#.#.##.#.###.#.#.##.##########..###########
##.#.#####.##.......##.#..##.#.##########.#######.##########
###.#####.###..#....#.###..##.#######.##.#########.#########
########.###.#####...###.##...######.##.###########.########
###########.#######...#.###..#.####.##.######.#####.########
##########.#######.##..##.###.#.##..#.######.##.##.#########
#########.#####.#.###.##.#####.#...#.#.####.####..######.###
################.###.#...######...###.#.#..#####..#####.####
##############..#.#.#...####.##..#.###.#.#..###.##.##..#####
##############..##.#...####.#...#.#.###.#.#..#.####..#.#.###
#############..#.##...####.##..#.###.##..###...####...#.####
############...###...###..##.##.#####.##.####...##.#...#####
############.##.#.#.#.#..##.##########.##.######..##########
###########.####...###..####.####.#####.##..####..##########
##########.######..#.#.####..#####.#####.#.#####.#.#########
#########.######......####.##.#####.#####.#####.##..########
##########.####......#.##.####.#####.###.#####.##.##########
#########..###.#......#..############.#.#####.##.###########
########..#####.#..#.#...#############.#.###.###############
######...#####.#..###..################.#.##################
##.###..##.####..#####.###.##############..#################
###.#.##..####..######..#..#############.##.################
#.##.###...#.#.######.##..#############.####.###############
##..#.#.#...#.######.###.#############.#####################
##..##.#.###.#.####.###.####################################
####..#.#####.#.##.#########################################
#####..#.#####.#..##.#######################################
###############.#.#.########################################
##################.#########################################
#################.##########################################
############################################################
```
## Code
Pseudocode for block aggregation:
• Fill a level with wall
• Dig a seed tile or block, probably in the centre
• Do until the level is considered sufficiently full (predefined number of blocks, edge proximity or number of tiles dug):
• Choose a block, either from a given set or using a function
• Choose a location from which to begin the block's walk
• If necessary, limit this choice to locations that don't interfere with already-dug sections
• Do until the block neighbours or collides with the dug-out section:
• Move the block one step in a random direction
• Dig out the tiles covered by the block | 3,784 | 17,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-16 | latest | en | 0.502504 |
http://web.mit.edu/6.02/www/s2009/handouts/labs/lab10_random_graph.py | 1,679,717,023,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-14/segments/1679296945315.31/warc/CC-MAIN-20230325033306-20230325063306-00188.warc.gz | 50,177,930 | 1,423 | import random,sys,math """Random graph generator """ class RandomGraph: def __init__(self,numnodes=12): self.numnodes = numnodes if self.numnodes > 26: print "Maximum number of nodes = 26" self.numnodes = 26 elif self.numnodes < 12: print "Minimum number of nodes = 12" self.numnodes = 12 self.names = ['S', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'R', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'] self.maxRows = math.ceil(math.sqrt(self.numnodes)) self.maxCols = math.ceil(math.sqrt(self.numnodes)) def getCoord(self, i): x= i % self.maxCols y = math.floor(i/self.maxCols) return (x,y) def getIndex(self, x, y): if x<0 or y < 0 or x>=self.maxCols or y>=self.maxRows: return -1 ind = y*self.maxCols + x if ind < self.numnodes: return ind else: return -1 def getAllNgbrs(self, i): (x,y) = self.getCoord(i) ngbrs = [] ngbrsX = [x-1, x, x+1] ngbrsY = [y-1, y, y+1] for nx in ngbrsX: for ny in ngbrsY: if not (nx==x and ny == y): ind = self.getIndex(nx, ny) if ind>=0: ngbrs.append(ind) return ngbrs def checkLinkExists(self, links, a, b): for (c,d) in links: if a==c and b==d: return True if a==d and b==c: return True return False def genGraph(self): NODES = [] LINKS = [] for i in range(self.numnodes): (x,y) = self.getCoord(i) name = self.names[i] NODES.append((name,x,y)) for i in range(self.numnodes): ngbrs = self.getAllNgbrs(i) outdeg = int(random.random()*len(ngbrs)) + 1 sampleNgbrs = random.sample(ngbrs, outdeg) for n1 in sampleNgbrs: n = int(n1) if not self.checkLinkExists(LINKS, self.names[i], self.names[n]): LINKS.append((self.names[i], self.names[n])) return (NODES, LINKS) | 566 | 1,622 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-14 | latest | en | 0.295782 |
https://socratic.org/questions/how-do-you-solve-and-graph-abs-4b-2-3-12 | 1,585,498,813,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00231.warc.gz | 698,039,416 | 6,425 | # How do you solve and graph abs((4b-2)/3)<12?
Apr 19, 2017
See the entire solution process below:
#### Explanation:
The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.
$- 12 < \frac{4 b - 2}{3} < 12$
First, multiply each segment of the system of inequalities by $\textcolor{red}{3}$ to eliminate the fraction while keeping the system balanced:
$\textcolor{red}{3} \cdot - 12 < \textcolor{red}{3} \cdot \frac{4 b - 2}{3} < \textcolor{red}{3} \cdot 12$
$- 36 < \cancel{\textcolor{red}{3}} \cdot \frac{4 b - 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} < 36$
$- 36 < 4 b - 2 < 36$
Next, add $\textcolor{red}{2}$ to each segment to isolate the $b$ term while keeping the system balanced:
$- 36 + \textcolor{red}{2} < 4 b - 2 + \textcolor{red}{2} < 36 + \textcolor{red}{2}$
$- 34 < 4 b - 0 < 38$
$- 34 < 4 b < 38$
Now, divide each segment by $\textcolor{red}{4}$ to solve for $b$ while keeping the system balanced:
$- \frac{34}{\textcolor{red}{4}} < \frac{4 b}{\textcolor{red}{4}} < \frac{38}{\textcolor{red}{4}}$
$\frac{2 \cdot - 17}{\textcolor{red}{2 \cdot 2}} < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} b}{\cancel{\textcolor{red}{4}}} < \frac{2 \cdot 19}{\textcolor{red}{2 \cdot 2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot - 17}{\textcolor{red}{\cancel{2} \cdot 2}} < b < \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \cdot 19}{\textcolor{red}{\cancel{2} \cdot 2}}$
$- \frac{17}{2} < b < \frac{19}{2}$
Or
$b > - \frac{17}{2}$ and $b < \frac{19}{2}$
Or
$\left(- \frac{17}{2} , \frac{19}{2}\right)$ | 643 | 1,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 19, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2020-16 | longest | en | 0.526791 |
https://socratic.org/questions/two-corners-of-an-isosceles-triangle-are-at-2-3-and-1-4-if-the-triangle-s-area-i#324781 | 1,716,975,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059221.97/warc/CC-MAIN-20240529072748-20240529102748-00854.warc.gz | 448,397,488 | 6,420 | # Two corners of an isosceles triangle are at (2 ,3 ) and (1 ,4 ). If the triangle's area is 64 , what are the lengths of the triangle's sides?
Oct 21, 2016
The 3 sides are $90.5 , 90.5 , \mathmr{and} \sqrt{2}$
#### Explanation:
Let b = the length of the base from $\left(2 , 3\right)$ to $\left(1 , 4\right)$
$b = \sqrt{{\left(1 - 2\right)}^{2} + {\left(4 - 3\right)}^{2}}$
$b = \sqrt{2}$
This cannot be one of the equal sides, because the maximum area of such a triangle would occur, when it is equilateral, and specifically:
$A = \frac{\sqrt{3}}{2}$
This conflicts with our given area, $64 u n i t {s}^{2}$
We can use the Area to find the height of the triangle:
$A r e a = \left(\frac{1}{2}\right) b h$
$64 = \frac{1}{2} \sqrt{2} h$
$h = 64 \sqrt{2}$
The height forms a right triangle and bisects the base, therefore, we can use the Pythagorean theorem to find the hypotenuse:
${c}^{2} = {\left(\frac{\sqrt{2}}{2}\right)}^{2} + {\left(64 \sqrt{2}\right)}^{2}$
${c}^{2} = 8192.25$
$c \approx 90.5$ | 366 | 1,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2024-22 | latest | en | 0.740341 |
https://de.mathworks.com/help/images/derive-statistics-from-glcm-and-plot-correlation.html | 1,685,365,663,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00710.warc.gz | 229,241,637 | 18,390 | Derive Statistics from GLCM and Plot Correlation
This example shows how to create a set of Gray-Level Co-Occurrence Matrices (GLCMs) and derive statistics from them. The example also illustrates how the statistics returned by `graycoprops` have a direct relationship to the original input image.
Read an image into the workspace and display it. The example converts the truecolor image to a grayscale image and then, for this example, rotates it 90 degrees.
```circuitBoard = rot90(im2gray(imread("board.tif"))); imshow(circuitBoard)```
Define offsets of varying direction and distance. Because the image contains objects of a variety of shapes and sizes that are arranged in horizontal and vertical directions, the example specifies a set of horizontal offsets that only vary in distance.
`offsets0 = [zeros(40,1) (1:40)'];`
Create the GLCMs. Call the `graycomatrix` function specifying the offsets.
`glcms = graycomatrix(circuitBoard,"Offset",offsets0);`
Derive statistics from the GLCMs using the `graycoprops` function. The example calculates the contrast and correlation.
`stats = graycoprops(glcms,["Contrast" "Correlation"]);`
Plot correlation as a function of offset.
```figure, plot([stats.Correlation]); title("Texture Correlation as a function of offset"); xlabel("Horizontal Offset") ylabel("Correlation")```
The plot contains peaks at offsets 7, 15, 23, and 30. If you examine the input image closely, you can see that certain vertical elements in the image have a periodic pattern that repeats every seven pixels. | 350 | 1,539 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-23 | longest | en | 0.800864 |
https://chemistry.coach/general-chemistry-2/chemical-reactivity/exercise/59 | 1,721,242,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00559.warc.gz | 143,772,061 | 19,532 | # Balancing chemical equations - 1 | Fundamentals of Chemical Reactions
The unbalanced chemical equation of the combustion of octane is:
C8H18 (l) + O2 (g) → CO2 (g) + H2O (l)
Balance the equation.
What mass of O2 will be needed to burn 50.8 g of octane?
How many moles of water are produced?
2 C8H18 (l) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (l)
According to the stoechiometry of the balanced equation:
$\frac{{\mathrm{n}}_{\mathrm{C}8\mathrm{H}18}}{2}$ = $\frac{{\mathrm{n}}_{\mathrm{O}2}}{25}$ ⇒ $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ = $\frac{25}{2}$ x $\frac{{\mathrm{m}}_{\mathrm{C}8\mathrm{H}18}}{{\mathrm{M}}_{\mathrm{C}8\mathrm{H}18}}$
mO2 = 178 g
$\frac{{\mathrm{n}}_{\mathrm{C}8\mathrm{H}18}}{2}$ = $\frac{{\mathrm{n}}_{\mathrm{H}2\mathrm{O}}}{18}$ ⇒ nH2O $\frac{18}{2}$ x $\frac{{\mathrm{m}}_{\mathrm{C}8\mathrm{H}18}}{{\mathrm{M}}_{\mathrm{C}8\mathrm{H}18}}$
nH2O = 4.01 mol | 387 | 920 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-30 | latest | en | 0.356174 |
http://www.itdaan.com/blog/2016/08/19/a9543c5ad21fed3caa99994a3bfe29e2.html | 1,555,751,865,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529472.24/warc/CC-MAIN-20190420080927-20190420102927-00375.warc.gz | 250,554,904 | 6,689 | ### codeforces CF703D Mishka and Interesting sum 树状数组
1 0 0 0
1 2 0 0
0 2 1 0
0 0 1 2
``````#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define N 1000005
using namespace std;
int n,m,c[N],a[N],sum[N],last[N];
struct data{int l,r,ans,id;}q[N];
struct hehe{int a,id;}b[N];
bool cmp(data a,data b)
{
return a.r<b.r;
}
int lowbit(int x)
{
return x&(-x);
}
void ins(int x,int y)
{
if (!x) return;
while (x<=n)
{
c[x]^=y;
x+=lowbit(x);
}
}
int find(int x)
{
if (!x) return 0;
int s=0;
while (x)
{
s^=c[x];
x-=lowbit(x);
}
return s;
}
bool cmp1(data a,data b)
{
return a.id<b.id;
}
bool cmp2(hehe a,hehe b)
{
return (a.a<b.a||a.a==b.a&&a.id<b.id);
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b[i].a=a[i];b[i].id=i;
sum[i]=sum[i-1]^a[i];
}
sort(b+1,b+n+1,cmp2);
for (int i=2;i<=n;i++)
if (b[i].a==b[i-1].a) last[b[i].id]=b[i-1].id;
scanf("%d",&m);
for (int i=1;i<=m;i++)
{
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
sort(q+1,q+m+1,cmp);
int point=1;
for (int i=1;i<=n;i++)
{
if (last[i]) ins(last[i],a[i]);
ins(i,a[i]);
while (q[point].r==i)
{
q[point].ans=find(q[point].r)^find(q[point].l-1)^sum[i]^sum[q[point].l-1];
point++;
}
if (point>m) break;
}
sort(q+1,q+m+1,cmp1);
for (int i=1;i<=m;i++)
printf("%d\n",q[i].ans);
return 0;
}`````` | 558 | 1,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2019-18 | latest | en | 0.149474 |
http://news.stanford.edu/news/2011/november/devlin-ranked-voting-110711.html | 1,436,144,340,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375097757.36/warc/CC-MAIN-20150627031817-00297-ip-10-179-60-89.ec2.internal.warc.gz | 176,027,687 | 8,065 | # Potential for odd outcomes in San Francisco mayoral election with ranked-choice voting system, says Stanford mathematician
Ranked choice voting – in which voters rank candidates in order of preference – is an increasingly popular alternative to voting for only one candidate in each race. But whether ranked voting does a better job of producing a winner that most voters are happy with is a matter of debate. What would a mathematician say?
L.A. Cicero
In the San Francisco mayoral race this year, voters select their top three choices out of a field of 16 candidates and rank them in order of preference.
"Instant runoff" voting – which San Franciscans will use this week to choose their new mayor, county sheriff and district attorney – requires voters to rank their three top choices in each race, instead of simply voting for their first choice, with a later run-off if needed.
Advocates say instant runoff does a better job than a traditional election of producing a winner who truly represents most of the voters' preference; opponents say the opposite. It all boils down to what you think of the math, which raises an obvious question: What does a mathematician think?
We posed the question to Stanford mathematician (and NPR Math Guy) Keith Devlin.
"There is no perfect voting method," said Devlin. "A famous 1950 result by the Nobel Prize–winning mathematician Kenneth Arrow, who spent much of his career at Stanford, tells us that there is no voting system that meets all basic democratic requirements.
"One advantage of the system we use in our national elections, the 'first past the post' method, is that it is easy to understand. Everyone votes for one candidate and the one with the most votes wins. But ease of understanding is about the only thing in its favor, and all the experts agree it is just about the worst possible system.
"For one thing, in a field with at least three candidates, split votes between two top candidates can result in a third person being elected who is disliked by a large majority of the electorate. Another problem is more social than numerical, in that it forces candidates into opposing camps, slugging it out in a verbal prizefight.
"That may have a lot to do with the current dysfunction in Washington," Devlin suggested, "where no one is willing to compromise or make deals."
The method being used in San Francisco is a variant of the "instant runoff" process, also known as "ranked choice" voting. Here, voters select their top three choices out of the field of candidates and rank them in order of preference. In the mayoral race this year, that means ranking three out of 16 contenders.
Courtesy of Keith Devlin
'With ranked-choice voting, you can get a winner who is the first choice of only a relatively small minority of the voters,' said Stanford mathematician Keith Devlin.
All the first choice votes are counted and then an iterative tallying process begins.
First, the contender with the lowest number of first-choice votes is dropped from the competition. Each voter who had ranked that candidate as his No. 1 choice then has his vote given to whichever candidate he selected as his second choice. The votes are re-tallied and, as before, the contender with the lowest vote total is eliminated.
This process continues for as many rounds as needed until one candidate has over 50 percent of the votes tallied in a round, at which point he or she is declared the winner.
### Costs vs. benefits
Advocates point out that this method can save costs. With the one-vote approach previously used in San Francisco, if no candidate received over 50 percent of the votes in the election, a special runoff election had to be held, entailing the expense of printing new ballots and paying poll workers. With the instant runoff approach, a second election is not needed.
Opponents of the method point out that voters whose choices are repeatedly eliminated effectively get to vote several times, and moreover the process gives equal value to a person's third-place ranking of a candidate and someone else's top-choice vote.
But there are other problems, Devlin pointed out. "For example, with ranked choice voting, you can get a winner who is the first choice of only a relatively small minority of the voters.
"Undesirable outcomes such as this can arise," he explained, "because the candidates are eliminated and their votes reassigned one after another, and the order in which that happens can make a huge difference.
"A shift of a large block of votes in an early round can eliminate a candidate who would have gone on to win had she survived until a later round and then picked up more votes to boost her tally."
### In math, if it can happen, it will
Devlin's example is not a just theoretical possibility. In the 2010 race for supervisor in San Francisco's District 10, the eventual winner received just 11.8 percent of the first-place votes, ultimately edging out the candidate who had the most first-place votes, according to reports in the San Francisco Chronicle.
There is also Jean Quan's win over Don Perata last year for mayor of Oakland. Quan had only 24.4 percent of first-choice votes; Perata had 33.7 percent, the Chronicle reported.
The known vagaries of the voting method have resulted in some candidates trying new approaches to getting votes. For example, mayoral candidate Michela Alioto-Pier sent out a mailing urging voters to "please consider at least making Michela your number 2 choice for Mayor."
A mailing from the San Francisco Republican Party suggested two mayoral "candidates to avoid."
Devlin said he thinks that as candidates become more aware of the election math and the possibilities it opens up, more of these tactics are likely to be seen, adding, "Particularly in the era of social media, where it is possible for large numbers of voters to coordinate their actions."
Is this a misuse of election math? Devlin does not think so. "Arrow's theorem tells us that the only mathematically ideal system is a dictatorship, and no American wants that.
### It isn't rocket science
"So voting is not like physics or engineering, where we have to do what the math tells us. Rather, it is one of those cases where we can make the math work for us – to use it to achieve our own ends as a society. The voters will make the selection, but the math we choose can shape the kind of government we get. Do we want politics to be about partisanship and fighting, where half the electorate will always end up as losers and we just keep seesawing between the two, or do we encourage cooperation and compromise, where no one gets everything but everyone gets something?" Devlin asked.
Devlin said that ranked choice voting now in use in San Francisco almost certainly encourages coalition building and reduces negative campaigning. "The question is, do you think that is a good thing? I have my opinion, but there I am being a citizen, not a mathematician."
Keith Devlin, Human-Sciences and Technologies Advanced Research Institute: (650) 329-9742; devlin@stanford.edu
Louis Bergeron, Stanford News Service: (650) 725-1944, louisb3@stanford.edu | 1,488 | 7,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2015-27 | longest | en | 0.9699 |
https://www.elevenplusexams.co.uk/forum/11plus/viewtopic.php?f=4&t=33184 | 1,516,518,162,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890314.60/warc/CC-MAIN-20180121060717-20180121080717-00284.warc.gz | 956,416,634 | 9,593 | It is currently Sun Jan 21, 2018 7:02 am
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Post subject: Need Answer Explanation for Letts Non VerbalPosted: Fri Aug 23, 2013 12:40 pm
Joined: Thu Jun 14, 2012 8:51 pm
Posts: 26
Paper C Section 5 Question 6
Paper D Section 3 Question 1
Paper D Section 3 Question 12
Paper D Section 4 Question 9
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Post subject: Re: Need Answer Explanation for Letts Non VerbalPosted: Sat Aug 24, 2013 5:21 pm
Joined: Sat Jan 12, 2013 10:02 pm
Posts: 88
kavitag wrote:
Paper C Section 5 Question 6
Paper D Section 3 Question 1
Paper D Section 3 Question 12
Paper D Section 4 Question 9
I'll try my best to explain the answers to these questions as clearly as possible.
*Test C - S5, Q6: A is the odd one out because it is the only shape that has a shaded square.
*Test D - S3, Q1: Moving from left to right, a line is added in a clockwise direction in each box to eventually form a triangle in the third box. As you can see in the 5th box there is only one line, so instead of adding lines after the third box, we are now removing lines. Therefore moving from the 3rd to the 4th box a line will be removed and since we are still going in a clockwise direction, it is the bottom part of the triangle that is removed. The answer is therefore C.
*Test D - S3, Q12: Moving from left to right, an arrow is added in a clockwise direction in each box. Therefore the 2nd box could be A or C. Now looking at the middle shapes from the 3rd to the 5th box, they are rotating 45 degrees anti-clockwise. Therefore moving from the 1st to the 2nd box, the middle shapes will look like C.
*Test D - S4, Q9: B is the odd one out because it is the only clock half past an hour. All the other clocks are on the hour.
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Post subject: Re: Need Answer Explanation for Letts Non VerbalPosted: Mon Aug 26, 2013 12:15 pm
Joined: Thu Jun 14, 2012 8:51 pm
Posts: 26
Thanks a lot Rainbow petals.
I got all answers except Section3 Question 12,
If arrow is adding from left to right clockwise thn that way option D is also possible.
Thanks for ur time and effort.
Top
Post subject: Re: Need Answer Explanation for Letts Non VerbalPosted: Mon Aug 26, 2013 1:43 pm
Joined: Sat Jan 12, 2013 10:02 pm
Posts: 88
kavitag wrote:
Thanks a lot Rainbow petals.
I got all answers except Section3 Question 12,
If arrow is adding from left to right clockwise thn that way option D is also possible.
Thanks for ur time and effort.
Hi
You're welcome, no worries. If you look at the smaller arrows in boxes 4 and 5 it is clearer to spot the order in which the arrows are added, therefore we must follow the same rule for box 1 to 2. We started with an arrow on the left-handside of box 1, then the next arrow is at the top, so in box 2 the next arrow must be on the right-handside. Does this make sense?
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Post subject: Re: Need Answer Explanation for Letts Non VerbalPosted: Tue Aug 27, 2013 12:00 pm
Joined: Thu Jun 14, 2012 8:51 pm
Posts: 26
exactly, if we add the arrow to right hand side then the answer will be d, caz in answer C the arrow is added down.
I hope i m not troubling u.
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Post subject: Re: Need Answer Explanation for Letts Non VerbalPosted: Tue Aug 27, 2013 12:16 pm
Joined: Sat Jan 12, 2013 10:02 pm
Posts: 88
kavitag wrote:
exactly, if we add the arrow to right hand side then the answer will be d, caz in answer C the arrow is added down.
I hope i m not troubling u.
Don't worry, it's kind of difficult to explain NVR online where you can't illustrate answers. If we add the arrow to the right handside the answer will be C, we want the arrow to point downwards. I will PM you in a moment to show you what I mean.
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Jump to: Select a forum ------------------ FORUM RULES Forum Rules and FAQs 11 PLUS SUBJECTS VERBAL REASONING MATHS ENGLISH NON-VERBAL REASONING CEM 11 Plus GENERAL GENERAL 11 PLUS TOPICS 11 PLUS APPEALS 11 PLUS TUTORS INDEPENDENT SCHOOLS 11 PLUS CDs/SOFTWARE 11 PLUS TIPS PRIMARY SEN and the 11 PLUS EVERYTHING ELSE .... 11 PLUS REGIONS Berkshire Bexley and Bromley Birmingham, Walsall, Wolverhampton and Wrekin Buckinghamshire Devon Dorset Essex Essex - Redbridge Gloucestershire Hertfordshire (South West) Hertfordshire (Other and North London) Kent Lancashire & Cumbria Lincolnshire Medway Northern Ireland Surrey (Sutton, Kingston and Wandsworth) Trafford Warwickshire Wiltshire Wirral Yorkshire BEYOND 11 PLUS Beyond 11 Plus - General GCSEs 6th Form University | 1,448 | 5,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-05 | latest | en | 0.928409 |
https://www.coursehero.com/file/9072990/Lecture-2/ | 1,544,603,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00304.warc.gz | 864,513,890 | 121,375 | # Lecture 2 - 2 Multistep Methods Up to now all methods we...
• biscuit1993
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2 Multistep Methods Up to now, all methods we studied were single step methods, i.e., the value y n +1 was found using information only from the previous time level t n . Now we will consider so-called multistep methods , i.e., more of the history of the solution will affect the value y n +1 . 2.1 Adams Methods Consider the first-order ODE y ( t ) = f ( t, y ( t )) . If we integrate from t n +1 to t n +2 we have t n +2 t n +1 y ( τ ) = t n +2 t n +1 f ( τ, y ( τ )) or y ( t n +2 ) - y ( t n +1 ) = t n +2 t n +1 f ( τ, y ( τ )) dτ. (31) As we saw earlier for Euler’s method and for the trapezoidal rule, different numer- ical integration rules lead to different ODE solvers. In particular, the left-endpoint rule yields Euler’s method, while the trapezoidal rule for integration gives rise to the trapezoidal rule for IVPs. Incidentally, the right-endpoint rule provides us with the backward Euler method. We now use a different quadrature formula for the integral in (31). Example Instead of viewing the slope f as a constant on the interval [ t n , t n +1 ] we now represent f by its linear interpolating polynomial at the points τ = t n and τ = t n +1 given in Lagrange form, i.e., p ( τ ) = τ - t n +1 t n - t n +1 f ( t n , y ( t n )) + τ - t n t n +1 - t n f ( t n +1 , y ( t n +1 )) = t n +1 - τ h f ( t n , y ( t n )) + τ - t n h f ( t n +1 , y ( t n +1 )) , where we have used the stepsize t n +1 - t n = h . FIGURE Therefore, the integral becomes t n +2 t n +1 f ( τ, y ( τ )) t n +2 t n +1 p ( τ ) = t n +2 t n +1 t n +1 - τ h f ( t n , y ( t n )) + τ - t n h f ( t n +1 , y ( t n +1 )) = f ( t n , y ( t n )) - 1 2 ( t n +1 - τ ) 2 h + f ( t n +1 , y ( t n +1 )) ( τ - t n ) 2 2 h t n +2 t n +1 = 3 h 2 f ( t n +1 , y ( t n +1 )) - h 2 f ( t n , y ( t n )) . 42
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Thus (31) motivates the numerical method y n +2 = y n +1 + h 2 [3 f ( t n +1 , y n +1 ) - f ( t n , y n )] . (32) Since formula (32) involves two previously computed solution values, this method is known as a two-step method . More precisely, is is known as the second-order Adams- Bashforth method (or AB method) dating back to 1883. Remark 1. We will establish later that this method is indeed of second order ac- curacy. 2. Note that the method (32) requires two initial conditions. Since the IVP will give us only one initial condition, in the Matlab demo script ABDemo.m we take the second starting value from the exact solution. This is, of course, not realistic, and in practice one often precedes the Adams-Bashforth method by one step of, e.g., a second-order Runge-Kutta method (see later). However, even a single Euler step (which is also of order O ( h 2 )) can also be used to start up (and maintain the accuracy of) the second-order AB method. This approach can also be used in ABDemo.m by uncommenting the corresponding line. Example The Matlab script ABDemo.m compares the convergence of Euler’s method (the one-step AB method) with the two-step AB method (32) for the IVP y ( t ) = - y 2 ( t ) , y (0) = 1 on the interval [0 , 10] with different stepsizes N = 50 , 100 , 200 and 400. The exact solution of this problem is y ( t ) = 1 t + 1 .
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dtrtrs.f
Go to the documentation of this file.
1*> \brief \b DTRTRS
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8*> \htmlonly
10*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.tgz?format=tgz&filename=/lapack/lapack_routine/dtrtrs.f">
11*> [TGZ]</a>
12*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.zip?format=zip&filename=/lapack/lapack_routine/dtrtrs.f">
13*> [ZIP]</a>
14*> <a href="http://www.netlib.org/cgi-bin/netlibfiles.txt?format=txt&filename=/lapack/lapack_routine/dtrtrs.f">
15*> [TXT]</a>
16*> \endhtmlonly
17*
18* Definition:
19* ===========
20*
21* SUBROUTINE DTRTRS( UPLO, TRANS, DIAG, N, NRHS, A, LDA, B, LDB,
22* INFO )
23*
24* .. Scalar Arguments ..
25* CHARACTER DIAG, TRANS, UPLO
26* INTEGER INFO, LDA, LDB, N, NRHS
27* ..
28* .. Array Arguments ..
29* DOUBLE PRECISION A( LDA, * ), B( LDB, * )
30* ..
31*
32*
33*> \par Purpose:
34* =============
35*>
36*> \verbatim
37*>
38*> DTRTRS solves a triangular system of the form
39*>
40*> A * X = B or A**T * X = B,
41*>
42*> where A is a triangular matrix of order N, and B is an N-by-NRHS
43*> matrix. A check is made to verify that A is nonsingular.
44*> \endverbatim
45*
46* Arguments:
47* ==========
48*
49*> \param[in] UPLO
50*> \verbatim
51*> UPLO is CHARACTER*1
52*> = 'U': A is upper triangular;
53*> = 'L': A is lower triangular.
54*> \endverbatim
55*>
56*> \param[in] TRANS
57*> \verbatim
58*> TRANS is CHARACTER*1
59*> Specifies the form of the system of equations:
60*> = 'N': A * X = B (No transpose)
61*> = 'T': A**T * X = B (Transpose)
62*> = 'C': A**H * X = B (Conjugate transpose = Transpose)
63*> \endverbatim
64*>
65*> \param[in] DIAG
66*> \verbatim
67*> DIAG is CHARACTER*1
68*> = 'N': A is non-unit triangular;
69*> = 'U': A is unit triangular.
70*> \endverbatim
71*>
72*> \param[in] N
73*> \verbatim
74*> N is INTEGER
75*> The order of the matrix A. N >= 0.
76*> \endverbatim
77*>
78*> \param[in] NRHS
79*> \verbatim
80*> NRHS is INTEGER
81*> The number of right hand sides, i.e., the number of columns
82*> of the matrix B. NRHS >= 0.
83*> \endverbatim
84*>
85*> \param[in] A
86*> \verbatim
87*> A is DOUBLE PRECISION array, dimension (LDA,N)
88*> The triangular matrix A. If UPLO = 'U', the leading N-by-N
89*> upper triangular part of the array A contains the upper
90*> triangular matrix, and the strictly lower triangular part of
91*> A is not referenced. If UPLO = 'L', the leading N-by-N lower
92*> triangular part of the array A contains the lower triangular
93*> matrix, and the strictly upper triangular part of A is not
94*> referenced. If DIAG = 'U', the diagonal elements of A are
95*> also not referenced and are assumed to be 1.
96*> \endverbatim
97*>
98*> \param[in] LDA
99*> \verbatim
100*> LDA is INTEGER
101*> The leading dimension of the array A. LDA >= max(1,N).
102*> \endverbatim
103*>
104*> \param[in,out] B
105*> \verbatim
106*> B is DOUBLE PRECISION array, dimension (LDB,NRHS)
107*> On entry, the right hand side matrix B.
108*> On exit, if INFO = 0, the solution matrix X.
109*> \endverbatim
110*>
111*> \param[in] LDB
112*> \verbatim
113*> LDB is INTEGER
114*> The leading dimension of the array B. LDB >= max(1,N).
115*> \endverbatim
116*>
117*> \param[out] INFO
118*> \verbatim
119*> INFO is INTEGER
120*> = 0: successful exit
121*> < 0: if INFO = -i, the i-th argument had an illegal value
122*> > 0: if INFO = i, the i-th diagonal element of A is zero,
123*> indicating that the matrix is singular and the solutions
124*> X have not been computed.
125*> \endverbatim
126*
127* Authors:
128* ========
129*
130*> \author Univ. of Tennessee
131*> \author Univ. of California Berkeley
132*> \author Univ. of Colorado Denver
133*> \author NAG Ltd.
134*
135*> \ingroup trtrs
136*
137* =====================================================================
138 SUBROUTINE dtrtrs( UPLO, TRANS, DIAG, N, NRHS, A, LDA, B, LDB,
139 \$ INFO )
140*
141* -- LAPACK computational routine --
142* -- LAPACK is a software package provided by Univ. of Tennessee, --
143* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
144*
145* .. Scalar Arguments ..
146 CHARACTER DIAG, TRANS, UPLO
147 INTEGER INFO, LDA, LDB, N, NRHS
148* ..
149* .. Array Arguments ..
150 DOUBLE PRECISION A( LDA, * ), B( LDB, * )
151* ..
152*
153* =====================================================================
154*
155* .. Parameters ..
156 DOUBLE PRECISION ZERO, ONE
157 parameter( zero = 0.0d+0, one = 1.0d+0 )
158* ..
159* .. Local Scalars ..
160 LOGICAL NOUNIT
161* ..
162* .. External Functions ..
163 LOGICAL LSAME
164 EXTERNAL lsame
165* ..
166* .. External Subroutines ..
167 EXTERNAL dtrsm, xerbla
168* ..
169* .. Intrinsic Functions ..
170 INTRINSIC max
171* ..
172* .. Executable Statements ..
173*
174* Test the input parameters.
175*
176 info = 0
177 nounit = lsame( diag, 'N' )
178 IF( .NOT.lsame( uplo, 'U' ) .AND. .NOT.lsame( uplo, 'L' ) ) THEN
179 info = -1
180 ELSE IF( .NOT.lsame( trans, 'N' ) .AND. .NOT.
181 \$ lsame( trans, 'T' ) .AND. .NOT.lsame( trans, 'C' ) ) THEN
182 info = -2
183 ELSE IF( .NOT.nounit .AND. .NOT.lsame( diag, 'U' ) ) THEN
184 info = -3
185 ELSE IF( n.LT.0 ) THEN
186 info = -4
187 ELSE IF( nrhs.LT.0 ) THEN
188 info = -5
189 ELSE IF( lda.LT.max( 1, n ) ) THEN
190 info = -7
191 ELSE IF( ldb.LT.max( 1, n ) ) THEN
192 info = -9
193 END IF
194 IF( info.NE.0 ) THEN
195 CALL xerbla( 'DTRTRS', -info )
196 RETURN
197 END IF
198*
199* Quick return if possible
200*
201 IF( n.EQ.0 )
202 \$ RETURN
203*
204* Check for singularity.
205*
206 IF( nounit ) THEN
207 DO 10 info = 1, n
208 IF( a( info, info ).EQ.zero )
209 \$ RETURN
210 10 CONTINUE
211 END IF
212 info = 0
213*
214* Solve A * x = b or A**T * x = b.
215*
216 CALL dtrsm( 'Left', uplo, trans, diag, n, nrhs, one, a, lda, b,
217 \$ ldb )
218*
219 RETURN
220*
221* End of DTRTRS
222*
223 END
subroutine xerbla(srname, info)
Definition cblat2.f:3285
subroutine dtrsm(side, uplo, transa, diag, m, n, alpha, a, lda, b, ldb)
DTRSM
Definition dtrsm.f:181
subroutine dtrtrs(uplo, trans, diag, n, nrhs, a, lda, b, ldb, info)
DTRTRS
Definition dtrtrs.f:140 | 2,054 | 6,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.432787 |
https://web2.0calc.com/questions/trapezoid_72045 | 1,721,241,600,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00730.warc.gz | 547,378,518 | 5,777 | +0
# Trapezoid
0
14
1
+355
Given an isosceles trapezoid with bases of 8 and 18 and an area of 120 square units, what is the number of units in the length of one of the non-parallel sides?
Mar 18, 2024
#1
+945
0
Given an isosceles trapezoid with bases of 8 and 18 and an area of 120 square units, what is the number of units in the length of one of the non-parallel sides?
Area of a trapezoid is the height times half the sum of the top base plus the bottom base.
A = (h) • (btop + bbottom) / 2
120 = (h) • 26 / 2
from this we get h = 120 / 13 = 9.2308
The height dropped from the top left corner intersects the bottom base at 5 from the left.
The non-parallel side of the trapezoid is the hypotenuse of the right triangle thus formed.
Use Pythagoras Theorem c2 = (9.2308)2 + (5)2 = 110.207
side = sqrt(110.207) = 10.5
When I was doing this on the calculator, I didn't round the numbers.
.
Mar 19, 2024 | 320 | 967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.791401 |
https://byjus.com/questions/the-mass-of-an-empty-density-bottle-is-30g-it-is-75g-when-filled-completely-with-water-and-65g-when-filled-completely-with-a-liquid-find-a-volume-of-density-bottle-b-density-of-liquid/ | 1,624,535,534,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00253.warc.gz | 157,302,695 | 37,477 | # The mass of an empty density bottle is 30g, it is 75g when filled completely with water and 65g when filled completely with a liquid. Find : a) volume of density bottle b) density of liquid and c) relative density of liquid
Mass of empty density bottle, M1 = 30 grams
MAss of bottle and water, M2 = 75 grams
Mass of liquid and liquid x, M3 = 65 grams
Mass of water = M2 – M1 = 45 grams
a) Volume of density bottle = Mass of water = 45 grams
b) Density of liquid x, D = mass of liquid/mass of water = 35/45
Mass of iquid = M3 – M1 = 65 – 30 = 35 grams
D = Mass of liquid/mass of water = 35/45 = 0.77 grams/cm3
c) Mass of water in the density bottle = 75 – 30 = 45 grams
Therefore, the volume of water in density bottle = 45 cc
Mass of the liquid whose volume is equal to the density bottle = 65 – 30 = 35 grams
Therefore, relative density of the liquid = mass of 45 cc of liquid/mass of 45 cc of water = 35/45 = 7/9 = 0.77 | 288 | 935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-25 | latest | en | 0.879431 |
https://www.sawaal.com/quantitative-aptitude-arithmetic-ability-questions-and-answers/in-these-questions-a-question-is-given-followed-by-information-in-three-statements-you-have-to-consi_18992 | 1,653,216,039,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00774.warc.gz | 1,140,633,666 | 14,769 | 1
Q:
# In these questions, a question is given followed by information in three statements. You have to consider the information in all the three statements and decide the information in which of the statement(s) is not necessarily required to answer the question and therefore can be dispensed with. Indicate your answer accordingly How many students from institute 'A' got placement?I. Number of students studying in institutes A and B are in the ratio of 3:4 respectively.II. Number of students who got placement from institute B is 120% of the number of students who got placement from institute AIII. 80% of the students studying in institute B got placement.
A) None of the statements can be dispensed with B) Only I C) Only II D) Any one of the three
Explanation:
From I No. of students studying in A and B are 3x and 4x respectively.
From II No. of students studying in B who got placement = (4x × 80) ÷ 100 = 16x/5
Hence Question cannot be answered even with the information in all the three statements
Hence option 5 is correct
Q:
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
The given Bar Graph presents the Imports and Exports of an item (in tonnes) manufactured by a company for the five financial years, 2013-2014 to 2017-2018.
A) 2016-2017 B) 2014-2015 C) 2015-2016 D) 2017-2018
Explanation:
0 68
Q:
The ratio of the areas of two triangles ABC and PQR is 4: 5 and the ratio of their heights is 5 : 3. The ratio of the bases of triangle ABC to that of triangle PQR is:
A) 25 : 12 B) 12 : 25 C) 11 : 15 D) 15 : 11
Explanation:
0 61
Q:
The ratio between the speeds of two trains is 5 : 7. If the first train covers 300 km in 3 hours, then the speed (in km/h) of the first train is:
A) 120 B) 140 C) 150 D) 100
Explanation:
0 32
Q:
Select the Venn diagram that best illustrates the relationship between the following classes.
Metal, Iron, Mercury
A) B) C) D)
Explanation:
0 79
Q:
The average age of fifteen persons is 32 years. If two more persons are added, then the average is increased by 3 years. The new persons have age difference of 7 years. The age (in years) of the younger among the new persons is:
A) 50 B) 58 C) 54 D) 61
Explanation:
0 113
Q:
Two numbers are in the ratio 3 : 4. On increasing each of them by 30, the ratio becomes 9 : 10. The sum of the numbers is:
A) 35 B) 30 C) 32 D) 25
Explanation:
0 70
Q:
The given Bar Graph presents the number of students of two schools for six years.
What is the ratio of students taken for all years together from School B to that from school A ?
A) 370 : 429 B) 415 : 401 C) 401 : 415 D) 429 : 370
Explanation:
0 490
Q:
The given Bar Graph presents the number of students of two schools for six years.
In which year, the percentage increase in students in school B is the highest in comparison to its previous year?
A) 2010 B) 2013 C) 2009 D) 2011 | 838 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-21 | latest | en | 0.88974 |
https://mcpt.ca/problem/triples | 1,585,662,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00264.warc.gz | 581,582,576 | 6,260 | ## Triples
View as PDF
Points: 7
Time limit: 1.0s
Memory limit: 64M
Problem type
Given an array of integers, , compute the number of unique ways to satisfy the equation , where and . To be unique, each counted equation after being filled in must never have been counted before.
#### Input Specification
The first line will contain a single integer , .
The next line will contain space separated integers, describing the array.
#### Output Specification
The output will contain a single line, the number of unique triples that can be formed by the array.
#### Sample Input 1
4
1 5 3 2
#### Sample Output 1
2
#### Explanation for Sample 1
The unique equations are and
#### Sample Input 2
5
1 1 1 2 3
#### Sample Output 2
2
#### Explanation for Sample 2
The unique equations are and | 199 | 800 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-16 | longest | en | 0.788381 |
https://forums.bzflag.org/viewtopic.php?f=103&t=18594&p=166863 | 1,603,684,633,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890273.42/warc/CC-MAIN-20201026031408-20201026061408-00436.warc.gz | 319,601,873 | 7,179 | ## Statistics class: summary, quiz and match prediction
Discussion for GU League Players
An SQUERRILz
Private First Class
Posts: 91
Joined: Wed Apr 25, 2007 2:08 am
### Statistics class: summary, quiz and match prediction
I did a study and it was good.
Summary of matches at bzbattleground over the last four months:
1.8 official matches per day
46% of matches are 20 minutes
102 unique players have offi'd
Red win rate is 45%, purple is 50% [Slime says purple bullets harder to see]
Quiz: Here are some extra statistics graphs, you have to guess what they are of. Think statistics, things that would be interesting to know and are easy to calculate.
I made a website http://gu.zapto.org/ to predict the outcome of GU offi and FMs. This is a glimpse of the rating system I created, SQelo (read more background at http://forums.bzflag.org/viewtopic.php? ... 75#p166289). It has scored 67% on over a hundred real matches. If you don't think 67% is very good then I challenge you to take a match prediction quiz. (Consider that draws can't really be predicted, technically you must predict the same thing for the same matchup, many re-matches have different outcomes such that even swaying by 4 caps is common [hint graph], lots of new players who are allergic to photos...)
sn0w_m0nkey
Private First Class
Posts: 293
Joined: Tue Dec 24, 2002 4:26 pm
### Re: Statistics class: summary, quiz and match prediction
That is some very cool work AS!
oi!
Snake12534
Private First Class
Posts: 216
Joined: Thu Oct 04, 2012 9:41 pm
Location: Austin, Texas
### Re: Statistics class: summary, quiz and match prediction
Very good work there AnSqu, I will conduct a test by matching 10 teams using the SQElo and see how accurate it is.
I will report my conclusion here, I might also trying fitting a 3vs2 in there, just to see the accuracy.
retired | 476 | 1,845 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-45 | latest | en | 0.938461 |
https://it.mathworks.com/matlabcentral/cody/problems/1107-find-max/solutions/2985622 | 1,610,872,867,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703511903.11/warc/CC-MAIN-20210117081748-20210117111748-00731.warc.gz | 404,149,538 | 16,615 | Cody
Problem 1107. Find max
Solution 2985622
Submitted on 24 Sep 2020 by Farhan Gony
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = magic(5); y_correct = 25; assert(isequal(your_fcn_name(x),y_correct))
2 Pass
x = [2 4 9 0 7 19;3 4 1 2 0 6]; y_correct = 19; assert(isequal(your_fcn_name(x),y_correct))
3 Pass
x = [2 4 9 0 7 19;3 4 1 2 0 6]'; y_correct = 19; assert(isequal(your_fcn_name(x),y_correct))
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