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https://www.sapling.com/6392784/calculate-per-capita-consumption | 1,726,313,255,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00066.warc.gz | 903,730,173 | 52,751 | # How to Calculate Per Capita Consumption
Image Credit: designer491/iStock/GettyImages
Per capita consumption is the average use of a product, service or other item per person. You can calculate the per capita consumption of a particular food, for example, if you are interested in investing in a commodity. You can calculate per capita consumption as it relates to a country's economic activity, such as Gross Domestic Product. You can make a quick calculation to help you make comparisons by year to see if something you're researching is trending upward or downward.
Read More: How to Calculate GDP Per Capita
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## What Is Consumption?
"Consumption" comes from the word "consume," which means to use. This is also where the word "consumer" comes from. You can consume a piece of food by eating it or a type of beverage by drinking it. The electricity generation industry consumes coal and natural gas. The more public, business and government sectors consume, the more they spend. This economic consumption drives economies and creates their GDP.
Economists not only look at consumption per sector (consumer, business or government), but they also look at it by industry (e.g., tech, auto, food, retail), household and individuals. Economists look at households because members of a household share many goods and services, such as washing machines, TVs and furniture. Measuring incomes and economic consumption by household are often easier to track and can provide more stable numbers to gauge the future performance of an industry.
Read More: How to Calculate Autonomous Consumption
## What Does Per Capita Mean?
"Per capita," as it relates to economic consumption, refers to one person. It comes from the Latin "per head." The main use of per capita calculations in economics is to determine changes in income and GDP. You will often read or hear about "per capita income" or "per capita GDP" as reporters or experts discuss the major trends in an economy.
When it comes to consumption, you can find per capita numbers for many goods and services. For example, per capita consumption of fresh potatoes in the U.S. in 2020 was 30.6 pounds per person. Using industry research, you'll find that per capita consumption of potatoes in the U.S. has declined by about 17 pounds per person since 2000. If you are thinking of investing in a potato farm long-term, you might consider otherwise based on this economic consumption measure and trend.
## Calculating Per Capita Consumption
You can calculate overall per capita consumption using a country's GDP figure and its population. In 2020, U.S. GDP was \$20,936.60 billion. The U.S. population was 331,002,651. So, \$20,936.60 billion / 331,002,651 = \$63,333. According to World Bank calculations, the figure was \$63,543.58. | 602 | 2,824 | {"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-2024-38 | latest | en | 0.947987 |
https://www.reddit.com/r/askscience/comments/1dkv9m/what_is_the_relationship_between_compton_scatter/ | 1,480,907,262,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541518.17/warc/CC-MAIN-20161202170901-00400-ip-10-31-129-80.ec2.internal.warc.gz | 1,021,830,456 | 18,562 | This is an archived post. You won't be able to vote or comment.
[–] 1 point2 points (3 children)
Not quite sure what you're asking here. With higher photon energies, the cross section (probability) of Compton scattering decreases.
[–][S] 0 points1 point (2 children)
Thanks very much for your reply. In the document it says "The probability of a Compton interaction decreases with higher x-ray energies. Remember, KeV (which is determined by kVp) is the unit of x-ray energy. Therefore, x-ray beams with a higher KeV create more scatter." This is where I'm getting confused. If probability decreases with higher energy then why do higher KeV x-rays create more scatter?
[–] 2 points3 points (1 child)
At higher energies, other interactions become more probable than Compton scattering. I'm not sure how in-depth your knowledge of physics is, but at high energies, the photoelectric effect and pair production are much more probable than Compton scattering.
[–][S] 0 points1 point (0 children)
Yeah that makes sense. I'm just not sure why then it is saying that at higher KeV there is more scatter, when the probability of scatter is decreasing.
[–] -2 points-1 points (0 children) | 287 | 1,194 | {"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-2016-50 | latest | en | 0.930204 |
https://www.cdc.gov/united-states-cancer-statistics/technical-notes/confidence-intervals.html | 1,726,300,213,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651559.58/warc/CC-MAIN-20240914061427-20240914091427-00320.warc.gz | 651,913,108 | 10,978 | # Confidence Intervals
## What to know
Confidence intervals reflect the range of variation in estimating cancer rates. The Data Visualizations tool uses confidence intervals that are expected to include the true underlying rate 95% of the time.
## Width of confidence intervals
The width of a confidence interval depends on the amount of variability in the data. Narrow confidence intervals tend to imply greater certainty in the estimate, while wide confidence intervals tend to imply more variability in the data and could mean there is less certainty.
Sources of variability include the underlying occurrence of cancer as well as uncertainty about when the cancer is diagnosed, when a death from cancer occurs, and when the data about the cancer are sent to the registry or state health department.
In any year when large numbers of a particular cancer are diagnosed or large numbers of cancer patients die, the effects of random variability are small and the confidence interval would likely be narrow. With rare cancers, however, the rates are small and the chance occurrence of more or fewer cases or deaths in a year can affect those rates markedly. Under these circumstances, the confidence interval will be wide to indicate uncertainty or instability in the cancer rate.
## The Poisson process
To estimate the extent of this uncertainty, a statistical framework is applied.1 The standard model used for rates for vital statistics is the Poisson process,2 which assigns more uncertainty to rare events relative to the size of the rate than it does to common events.
Parameters are estimated for the underlying disease process. For this report, we estimated a single parameter to represent the incidence rate and its variability. Of note, the Poisson model can estimate separate parameters that represent contributions to the rate from various risk factors, the effects of cancer control interventions, and other attributes of the population risk profile in any year.
## Modified gamma intervals
The Data Visualizations tool uses confidence intervals that are expected to include the true underlying rate 95% of the time. The confidence intervals are modified gamma intervals3 computed using SEER*Stat. The modified gamma intervals are more efficient than the gamma intervals of Fay and Feuer4 in that they are less conservative while still retaining the nominal coverage level.
Various factors such as population heterogeneity can sometimes lead to "extra-Poisson" variation in which the rates are more variable than would be predicted by a Poisson model. No attempt was made to correct for this. In addition, the confidence intervals do not account for systematic (in other words, nonrandom) biases in the incidence rates.
## Considerations when comparing rates
Using overlapping confidence intervals to determine significant differences between two rates presented in the Data Visualizations tool is discouraged. The practice fails to detect significant differences more frequently than standard hypothesis testing.5
Another consideration when comparing differences between rates is their public health importance. For some rates presented in the Data Visualizations tool, numerators and denominators are large and standard errors are therefore small. This results in statistically significant differences that may be too small to be important for decisions related to population-based public health programs.
1. Särndal C-E, Swennson B, Wretman J. Model-Assisted Survey Sampling. New York (NY): Springer-Verlag; 1992.
2. Brillinger DR. The natural variability of vital rates and associated statistics. Biometrics. 1986;42(4):693–734.
3. Tiwari RC, Clegg LX, Zou Z. Efficient interval estimation for age-adjusted cancer rates. Stat Methods Med Res. 2006;15(6):547–569.
4. Fay MP, Feuer EJ. Confidence intervals for directly standardized rates: a method based on the gamma distribution. Stat Med. 1997;16(7):791–801.
5. Schenker N, Gentleman JF. On judging the significance of differences by examining the overlap between confidence intervals. Am Stat. 2001;55(3):182–186. | 801 | 4,097 | {"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-2024-38 | latest | en | 0.895437 |
http://www.armoredpenguin.com/crossword/Data/2012.12/1316/13160826.985.html | 1,500,757,523,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424148.78/warc/CC-MAIN-20170722202635-20170722222635-00239.warc.gz | 353,060,938 | 6,078 | Semester I review
C. Hill
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Across
6.objects that fly through the air
7.a change in velocity
9.resists motion
12.all forces act in pairs called an .........
13.force over a given area
14.distance divided by time
15.Newton's first law
Down
1.speed with a direction
2.Newton's second law
3.newton's third law (two words)
4.every action has an equal and opposite .......
5.force of gravity pulling on your mass
8.makes objects fall
10.sum of all forces acting on an object( two words)
11.how much stuff you have in you
Use the "Printable HTML" button to get a clean page, in either HTML or PDF, that you can use your browser's print button to print. This page won't have buttons or ads, just your puzzle. The PDF format allows the web site to know how large a printer page is, and the fonts are scaled to fill the page. The PDF takes awhile to generate. Don't panic!
Web armoredpenguin.com | 252 | 927 | {"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-2017-30 | latest | en | 0.831351 |
phgelado.medium.com | 1,619,022,458,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00371.warc.gz | 566,396,048 | 31,893 | # Data Science for Science & Engineering: Creating a CFD 6 Degree of Freedom Kinematics OpenFoam File
In this short article I will show you how to generate a simple 6 degree of freedom kinematic file for OpenFoam. The data file we will generate specifies the magnitude and direction of motion for a series of time points.
It is a combination of a linear displacement vector and a rotation vector about the specified center of gravity (CofG) of the object, which is defined on the dynamicMeshDict file.
This data file is later linearly interpolated by OpenFoam when running your CFD simulation, so your time points do not need to match exactly the timesteps you have set in OpenFoam. However, more time points will smooth your dynamics, increasing the accuracy of your simulation, specially if, as seen in the case below, you have a sinusoidal or other non-linear motion.
The format of the data file we want to create for OpenFoam is shown below:
`4 //number of data points in the file //Position formatting is not important. File is based on the character sequence only. //Vectors are not relative. Each vector is total displacement and total rotation.(//(time_point ( (linear displacement vector) (rotation vector roll-yaw-pitch) ) )//(seconds ( (following unit system, usually meters) (degrees) ) )(0 ( (0.25 0.50 1.0) (0.220 0.30 0.40) ) )(0.25 ( (0.50 1.0 2.0) (0.60 0.60 0.60) ) )(0.75 ( (0.75 5.0 1.0) (1.2 2.4 5.0) ) )(10.0 ( (0.1 6.0 1.0) (5.0 3.0 5.5) ) ))`
Each time point first specifies the reference time, and the following bracket groups contains the motion, first linear and then angular displacement in their own lists.
We will begin by importing the libraries we need: Numpy, Pandas, MatPlotLib to visualize our output, OS to import paths, and IPython to clean up our display.
`import numpy as npimport pandas as pdimport matplotlib.pyplot as pltimport osfrom IPython.display import clear_output`
We will now add our parameters for the kinematics. In this case we are making a sinusoidal pitch-plunge movement, for 5 periods. The geometric angle of attack (our pitch angle) will be 15 degrees, and our plunge amplitude will be 0.05m. It is a 2D motion, so most of our degrees of freedom are actually just going to be 0.
`numberofperiods = 5alphaGeoAmp = 15plungeamplitude = 0.05period = 3.801279221resolution = 1000 #this is our time resolution, the number of data points our file will have`
We are going to be saving this file as “pitchplunge.dat”
`#outputfileoutputfilename = os.getcwd()outputfilename += '/pitchplunge.dat'`
We generate our time vector by using the numpy function linspace, with our set resolution:
`#timetime = np.linspace(0,period*numberofperiods,resolution)`
And now it’s just a question of creating our X,Y and Z vectors for our linear and rotational displacement. We will store them in a Pandas DataFrame (our motion matrix).
`#linear displacementlinearDisplacement = pd.DataFrame(data=time,columns={'Time'})linearDisplacement['X'] = 0linearDisplacement['Y'] = plungeamplitude*np.sin(time)linearDisplacement['Z'] = 0#rotational displacementrotationalDisplacement = pd.DataFrame(data=time,columns={'Time'})rotationalDisplacement['phi'] = 0rotationalDisplacement['psi'] = 0rotationalDisplacement['theta'] = -alphaGeoAmp * np.sin(time)`
Finally we can save our motion file, adding our different brackets and required formatting as per OpenFoam’s requirements. We will use the f.write file writer, and build each line with a for loop.
`#saving motion filef= open(outputfilename,"w+")f.write(str(resolution) + '\n(\n')for i in range(0,resolution,1): line = '(' line += str(linearDisplacement['Time'].iloc[i]) line += '((' line += str(linearDisplacement['X'].iloc[i]) line += ' ' line += str(linearDisplacement['Y'].iloc[i]) line += ' ' line += str(linearDisplacement['Z'].iloc[i]) line += ')(' line += str(rotationalDisplacement['phi'].iloc[i]) line += ' ' line += str(rotationalDisplacement['psi'].iloc[i]) line += ' ' line += str(rotationalDisplacement['theta'].iloc[i]) line += ')))\n' f.write(line) if i%1000==0: #this is a tiny add-on I like to use as a test and map progress print('Completing save ',round(i/resolution*100,2), '%') clear_output(wait=True) f.write(')')`
We will now plot our linear displacement to double check our motion:
`plt.plot(linearDisplacement['Time']/period, linearDisplacement['Y'])plt.xlim(0,numberofperiods)plt.ylabel("Linear Displacement /m")plt.xlabel("t/T")plt.title("Linear Displacement against non-dimensional time")`
And now our rotational displacement:
`plt.plot(rotationalDisplacement['Time']/period,rotationalDisplacement['theta'])plt.xlim(0,numberofperiods)plt.ylabel("Angular displacement /rad")plt.xlabel("t/T")plt.title("Angular displacement against non-dimensional time")`
Hope you enjoyed! You can find a video for this here, and a Jupyter Notebook to follow along here.
Scout for the IX Hispana studying vortices beyond Hadrian’s Wall.
## More from Pedro Hernández Gelado
Scout for the IX Hispana studying vortices beyond Hadrian’s Wall.
## LiveOps Essentials Part 3: Observation and Adjustments
Get the Medium app | 1,339 | 5,221 | {"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} | 2.5625 | 3 | CC-MAIN-2021-17 | latest | en | 0.85376 |
https://physics.stackexchange.com/questions/345567/non-local-field-redefinition-and-s-matrix | 1,726,184,191,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00312.warc.gz | 412,602,304 | 41,016 | # Non-local field redefinition and $S$-matrix
It is known that for local field redefinitions for which the LSZ formula is valid: $$\langle 0|\phi(x)|p\rangle \neq 0$$ field redefinitions don't change the S-matrix. (See QMechanic's answer to Equivalence Theorem of the S-Matrix)
But is it true for non-local field redefintions? For instance if I take a field redefinition of the form: $$\psi(x)=e^{-l\Box} \phi(x)$$
Will the S-matrix be invariant under this?
From QMechanic's explanation linked above and the answer by AccidentalFourier Transform here it would seem that the answer should be yes.
Edit: My main concerns are
1. Is such a transformation invertible? Is this a concern?
2. Any boundary condition on $\psi$ would translate to infinitely many boundary conditions on $\phi$. Is this relevant?
Claim 1. If $\psi(x)$ is an arbitrary operator that satisfies $$\langle 0|\psi(x)|p\rangle\neq 0\tag1$$ then it is a valid interpolating field, and as such, it can be used in the LSZ formula. The proof can be found in any introductory text, such as Weinberg.
Claim 2. If we assume that $$\langle 0|\phi(x)|p\rangle\neq 0\tag2$$ and define $$\psi(x)\overset{\mathrm{def}}=\mathrm e^{-\ell\partial^2}\phi(x)\tag3$$ then we have $$\langle 0|\psi(x)|p\rangle\neq 0\tag4$$
The proof is straightforward. One just need to use $\langle0|\phi(x)|p\rangle=c\mathrm e^{ipx}$ for some non-zero constant $c$, and the trivial identity $\mathrm e^{-\ell\partial^2}\mathrm e^{ipx}=\mathrm e^{\ell p^2}\mathrm e^{ipx}$.
Conclusion: the non-local redefinition $(2)$ is a valid redefinition, and the $S$ matrix is invariant under it.
• This is what I have as well. However there are a couple of questions: 1. The transformation between the fields is non-invertible. Naively, if I implement the field redefinition as a change of variables in the path integral, a non-invertible transformation would be illegal as the Jacobian is 0. 2. Boundary conditions. $\psi \rightarrow 0$ corresponds to an infinite number of boundary conditions for $\phi$. I'm not quite sure if these are issues or non-issues. Commented Jul 17, 2017 at 18:26
• @NirmalyaKajuri note that $\phi\mapsto \mathrm e^{-\ell\partial^2}\phi$ is invertible, with inverse $\psi\mapsto \mathrm e^{+\ell\partial^2}\psi$. Commented Jul 19, 2017 at 10:57
• But isn't it true that it will map all solutions of $\partial^{2n}\phi=0$ to zero, and these are all independent functions? Commented Jul 19, 2017 at 11:50
• @NirmalyaKajuri no, it will map all solutions of $\partial^{2n}\phi=0$ to $\phi\mapsto \phi$. Commented Jul 19, 2017 at 11:56
• @NirmalyaKajuri I'm glad I could help. Cheers! Commented Jul 19, 2017 at 12:13 | 786 | 2,673 | {"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": 4, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-38 | latest | en | 0.822871 |
thecareerlabs.net | 1,604,184,070,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00670.warc.gz | 508,839,643 | 12,097 | The Integrated Reasoning section is a fairly new addition to the GMAT, replacing one of the test’s Analytical Writing Assessment essays. Introduced in 2012, the IR section is designed to test a candidate’s ability to analyse data presented in different formats and utilise it to solve relevant problems.
There are four types of questions asked in this section:
1. Table Analysis
In this category, the data is presented in tabular format along with three associated questions. In order to answer these questions correctly, you will need to distinguish between essential and insignificant information.
1. Graphical Interpretation
In question falling under this category, you will be asked to decipher the data presented in a chart or graph. There will be two questions with answer choices which are presented in a drop-down menu.
1. Two-Part Analysis
Under this, you will be asked to answer a question with multiple choice options. The questions are long and detailed, and have small tables attached listing components of the question in the first two columns and the answer options in the third column. You will be asked to choose two correct answers out of available multiple choices.
1. Multi-Source Reasoning
In this category, you will be asked to navigate and gather information from multiple tabs, analyse the data presented in charts or tables and choose the answer from multiple choice options.
The IR section is made up of 12 questions, with an allotted time duration of 30 minutes. It is scored on a scale of 1 to 8, and does not count for one’s overall GMAT score.
Download Free Sample Papers for Integrated Reasoning Integrated Reasoning GMAT PDF Questions Solutions
IR Scoring:
Although GMAC, the administrative body of the GMAT exam does not disclose too much information about how the section is scored, there are some conclusions that can be drawn from the available information.
As the IR section is non-adaptive in nature, it is safe to assume that every question carries the same value irrespective of the difficulty level. We leave it to the judgment of the test-taker to guess which questions are experimental and how important the IR section is, considering that attempting this section cannot benefit him/her in the total GMAT score. The Integrated Reasoning GMAT score is scaled to a range of 1 to 8 (with single point increments). Your score in this section is reported in the official score report along with the scores for all other sections.
Although the IR score does not count towards the overall GMAT final score, many candidates do focus on the IR section once they are done with preparing for the verbal and quantitative sections. We bring you a few tips and strategies that you can use during your IR section preparation:
1. Try to be well-versed with graphical content and tables, charts etc. to hone your interpretation skills. For doing this, you can try reading business newspapers or magazines like Economic Times, Wall Street Journal, and the like.
2. Try to analyse your strengths and shortcomings with regard to different types of data interpretation questions and work on those areas which need improvement. Try to attempt mock versions of the IR and AWA sections too in order to get the right momentum on the day of the exam.
3. It is always a good idea to practice using an on-screen calculator. Although it can be used sparingly, this is definitely a huge time-saver for questions that involve complex calculations. | 685 | 3,472 | {"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-2020-45 | longest | en | 0.92746 |
https://www.jobilize.com/course/section/how-production-costs-affect-supply-by-openstax?qcr=www.quizover.com | 1,623,594,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487608856.6/warc/CC-MAIN-20210613131257-20210613161257-00542.warc.gz | 779,326,618 | 19,589 | # 3.2 Shifts in demand and supply for goods and services (Page 5/26)
Page 5 / 26
## How production costs affect supply
A supply curve shows how quantity supplied will change as the price rises and falls, assuming ceteris paribus so that no other economically relevant factors are changing. If other factors relevant to supply do change, then the entire supply curve will shift. Just as a shift in demand is represented by a change in the quantity demanded at every price, a shift in supply means a change in the quantity supplied at every price.
In thinking about the factors that affect supply, remember what motivates firms: profits, which are the difference between revenues and costs. Goods and services are produced using combinations of labor, materials, and machinery, or what we call inputs or factors of production . If a firm faces lower costs of production, while the prices for the good or service the firm produces remain unchanged, a firm’s profits go up. When a firm’s profits increase, it is more motivated to produce output, since the more it produces the more profit it will earn. So, when costs of production fall, a firm will tend to supply a larger quantity at any given price for its output. This can be shown by the supply curve shifting to the right.
Take, for example, a messenger company that delivers packages around a city. The company may find that buying gasoline is one of its main costs. If the price of gasoline falls, then the company will find it can deliver messages more cheaply than before. Since lower costs correspond to higher profits, the messenger company may now supply more of its services at any given price. For example, given the lower gasoline prices, the company can now serve a greater area, and increase its supply.
Conversely, if a firm faces higher costs of production, then it will earn lower profits at any given selling price for its products. As a result, a higher cost of production typically causes a firm to supply a smaller quantity at any given price. In this case, the supply curve shifts to the left.
Consider the supply for cars, shown by curve S 0 in [link] . Point J indicates that if the price is \$20,000, the quantity supplied will be 18 million cars. If the price rises to \$22,000 per car, ceteris paribus, the quantity supplied will rise to 20 million cars, as point K on the S 0 curve shows. The same information can be shown in table form, as in [link] .
Price and shifts in supply: a car example
Price Decrease to S 1 Original Quantity Supplied S 0 Increase to S 2
\$16,000 10.5 million 12.0 million 13.2 million
\$18,000 13.5 million 15.0 million 16.5 million
\$20,000 16.5 million 18.0 million 19.8 million
\$22,000 18.5 million 20.0 million 22.0 million
\$24,000 19.5 million 21.0 million 23.1 million
\$26,000 20.5 million 22.0 million 24.2 million
Now, imagine that the price of steel, an important ingredient in manufacturing cars, rises, so that producing a car has become more expensive. At any given price for selling cars, car manufacturers will react by supplying a lower quantity. This can be shown graphically as a leftward shift of supply, from S 0 to S 1 , which indicates that at any given price, the quantity supplied decreases. In this example, at a price of \$20,000, the quantity supplied decreases from 18 million on the original supply curve (S 0 ) to 16.5 million on the supply curve S 1 , which is labeled as point L.
what is economic
why is economics not a pure science?
what is management of human resources?
how is economics a science?
4. It is a hot day, and Bert is thirsty. Here is the value he places on each bottle of water: Value of first bottle \$7 Value of second bottle \$5 Value of third bottle \$3 Value of fourth bottle \$1 a. From this information, derive Bert’s demand schedule. Graph his demand curve for bottled w
hello
Lukman
what is the law of diminishing marginal utility
The law of diminishing marginal utility state that as a consumer consumes a successive units of a commodity, a point is eventually reached where consumption of additional unit yields less satisfaction.
raheem
what is demand
other things can be equal an certain amount paid for the goods by consumer in the market called demand.
Dhanishhwar
demand is the amount of a commodity a consumer is willing and able buy at a given price at a particular point in time
Samuel
Demand is the quantity of commodity a consumer is willing and able to buy at a given price and a particular time.
raheem
what is economics
economics is a social science subject, which study human behaviors as a relationship btw end ND scarce means
Joseph
what utility
Isatu
Utility is the satisfaction a consumer derives from consuming a particular commodity.
raheem
meaning of economics
what is a columnist
Owusu
what are the four basic assumptions of perfect competition
There is a well known maximum by economic that states that the birthd of money is the deaths of batter system discuss the statement
how does price elasticity increase
What is real GDP
Real GDP is a way of adjusting our output calculations for inflation so that we can see group interms of physical production quantity
Minalu
Real gross domestic product is a macroeconomic measure of the value of economic output adjusted for price changes. This adjustment transforms the money-value measure, nominal GDP, into an index for quantity of total output
amisha
what is economics
Economics is the study of human behavior between scarcity and want.
Vincent
yesss thats it
Tendai
economics is the study of man and his behaviors
toheeb
yh all is correct
Paul
ECONOMICS IS THE STUDY OF,HOW HUMAN BEINGS MADE DECISIONS IN THE TIME OF SCARCITY.
Bakshi
According to Professor Lionel Robbins" Economics is a science which study human behaviour as a relationship between ends and scarce means which have alternative uses".
Paul
According to Adam Smith, "Economics is a science which inquired into the nature and cause of wealth of nations. "
Tannu
economics is a science that studies human behaviour as a relationship between ends and scarce means which have alternative uses
Efua
everyone is correct
Ruth
More questions
Paul
What is utility
Isatu
what is the relationship between savings, consumption and investment
Samuel
according to dr.adam Smith," an enquire in to the nature and causes of the wealth of the nation. "1776.Economics is the studies of the wealth of the nation.
Dhanishhwar
what is elasticity | 1,472 | 6,449 | {"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-2021-25 | latest | en | 0.929403 |
http://blog.sciencenet.cn/home.php?mod=space&uid=315774&do=blog&id=1270233 | 1,618,677,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038461619.53/warc/CC-MAIN-20210417162353-20210417192353-00477.warc.gz | 14,089,389 | 17,927 | # 菲文笔记 | Technical theorem (v2) ---- And all becomes primary.
[按：导言下方是群邮件的内容，标题未改动。略有修订。]
* * *
This is coming to you from Yiwei LI (PhD, Applied math), Taiyuan University of Science and Technology (TYUST) Taiyuan, China
It's going on here for the third round of learning of Birkar's BAB-paper (v2), with scenarios of chess stories. No profession implications.
Textbook thinking is able to transform one into a textbook writer who tends to deliver others such a message that "Party is over".
Th 2.15 Th 1.8
Th 1.1 Th 1.6
Mathematics vs Palace stories.(v2)
Note: technical theorem is not on the board.
M Λ
S
XF pB
Note: the upper right/left corner is of output.
ℂ ℍ ℕ ℙ ℚ ℝ ℤ ℭ ℜ I|φ∪∩∈ ⊆ ⊂ ⊇ ⊃ ⊄ ⊅ ≤ ≥ Γ Θ α Δ δ μ ≠ ⌊ ⌋ ∨∧∞Φ⁻⁰ 1
Step 7, Para three ——
Assume now that (X, B) is not lc over z = f(S).
---- This is to setup the assumption for a proof by contradiction.
.
By the previous paragraph, (X, B) is lc near S and S is a non-klt centre of this pair.
---- This is a nice summary, so that one can judge what is "lc near S".
---- Basically, for certain pair, Say (X, B), one can define Ks + Bs by (Kx + B)|s.
---- Or, given Ks + Bs, one can recover the original pair by "inversion of adjunction", to show Ks + Bs = (Kx + B)|s.
---- The divisor S is required to be a "non-klt centre" of the original pair.
.
On the other hand, (X, Γ) is plt with ⌊Γ⌋ = S, so if u > 0 is sufficiently small, then (X, (1 - u)B + uΓ) is plt near S and S is a non-klt centre of this pair and no other non-klt centre intersects S.
---- In combination of last sentence, if (X, B) is lc near S and (X, Γ) is plt near S, then the pair of (small) convex combination is still plt near S.
.
Comment (wrong intuition): At the first glance, one might feel ⌊(1 - u)B⌋ = 0 and ⌊uΓ⌋ = 0, for their coefficients are smaller than 1 ——
---- This is really the case. (No kidding).
---- But, there is a re-organization matter, however, for the sum of the two terms, before taking the floor operation.
---- That is, (1 - u)S + uS = S, while S has the coefficient 1.
---- For this aspect, u can be any number in [0, 1].
---- The small positive u is required by some other aspects.(to be hunted).
.
Then since (X, B) is not lc over z, the non-klt locus of (X, (1 - u)B + uΓ) has at least two connected components (one of which is S) near the fibre f ⁻1 {z}.
---- Translation: if the original pair is not lc over z, then the (plt convex) combined pair has its non-klt locus possessed at least two connected components near f⁻1 {z}.
---- To illustrate this in a simple way, take the boundary as the pair ——
.
B ~ u(B)Γ
not | lc not | simple
z ~ f ⁻1 {z}
.
Note: u(B)Γ, understood in context, is the homemade notation for (X, (1 - u)B + uΓ).
Note: I call a (plt) pair is not simple, if its non-klt locus has least two connected components near f⁻1 {z}.
.
This contradicts the connectedness principle [25, Theorem 17.4] as - (Kx + (1 - u)B + uΓ) = - (1 - u) (Kx + B) - u(Kx + Γ) ~R -u(Kx + Γ) ~R uαM - u(Kx + Γ)/Z is ample over Z.
---- That is, the defence form of u(B)Γ is ample over Z, which expects u(B)Γ simple near f ⁻1 {z} by the connectedness principle.
.
Therefore, (X, B) is lc over z.
---- This is the close statement.
---- It's needed.
.
Summary comment: This is the end of proof of Pro4.1, the special case of Th1.9 (i.e. v1 Th1.7).
.
I give an instant praise, inspired by a recent blog article* out of a theoretical physicist ——
.
Math and Matics have laid hid in the darkness:
Fears of contempt propagate among folks.
So has formed the culture to puzzle, blind and block others...
Consciences yell: let birational algebraic geometry be.
And all becomes primary.
.
.
Calling graph for the technical theorem (Th1.9) ——
.
Th1.9
|
[5, 2.13(7)] Lem 2.26 Pro4.1 Lem2.7
|
.......................................................Lem2.3
Note: Th1.9 is only called by Pro.5.11, one of the two devices for Th1.8, the executing theorem.
Pro4.1
|
[5, ?] [37, Pro3.8] [5, Lem3.3] Th2.13[5, Th1.7] [16, Pro2.1.2] [20] [25, Th17.4]
.
Special note: Original synthesized scenarios in Chinese for the whole proof of v1 Th1.7, the technical theorem.
*It's now largely revised* due to new understandings.
.
.
.
It is my hope that this action would not be viewed from the usual perspective that many adults tend to hold.
http://blog.sciencenet.cn/blog-315774-1270233.html
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# The old fisherman who had a wooden leg that made an eerie
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10 Jan 2013, 10:22
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Question Stats:
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24. The old fisherman who had a wooden leg that made an eerie sound scaring the children with it as he walked.
a) that made an eerie sound scaring the children with it as he walked
b) that made an eerie sound which scaring the children as he walked
c) it made an eerie sound scaring the children when he was walking
d) that it made a scary sound for the children while he walks
e) which made an eerie sound as he walked which scared the children
Source: GMAT Club grammar book
I'm confused, the answer seems very awkward Wouldn't it be more logical if we had "scared" instead of "which scared" at the end of the sentence?
[Reveal] Spoiler: OA
If you have any questions
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10 Jan 2013, 15:29
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KUDOS
This is an awful problem. Not only is E awkward, but "which" is ambiguous. On top of that, using E does not even form a sentence.
A is incorrect because it is implying that the wooden leg is scaring the children with itself. That makes no sense and is awkward.
B is incorrect because one must say "which scared" or "which is scaring". "Which scaring" is incorrect.
C is incorrect because it is awkward. Also, you cannot form a logical sentence from combing C to the non-underlined portion.
D is incorrect because it is awkward.
Rest assured, no problem like this would ever appear on the GMAT. All of those answer choices should be incorrect.
_________________
If my post helped you, please consider giving me kudos.
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10 Jan 2013, 15:41
commdiver wrote:
This is an awful problem. Not only is E awkward, but "which" is ambiguous. On top of that, using E does not even form a sentence.
A is incorrect because it is implying that the wooden leg is scaring the children with itself. That makes no sense and is awkward.
B is incorrect because one must say "which scared" or "which is scaring". "Which scaring" is incorrect.
C is incorrect because it is awkward. Also, you cannot form a logical sentence from combing C to the non-underlined portion.
D is incorrect because it is awkward.
Rest assured, no problem like this would ever appear on the GMAT. All of those answer choices should be incorrect.
Thank you for the answer and the explanation Commdiver, I'm glad to hear I'm not the only one thinking this problem is really strange. And it is good to hear that something like that will not appear on the GMAT!
Re: The old fisherman who had a wooden leg that made an eerie [#permalink] 10 Jan 2013, 15:41
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https://numbas.mathcentre.ac.uk/question/34867/amy-s-copy-of-use-the-factor-theorem-to-identify-factors-of-a-polynomial.exam | 1,597,097,990,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-34/segments/1596439738699.68/warc/CC-MAIN-20200810205824-20200810235824-00284.warc.gz | 426,980,768 | 2,335 | // Numbas version: exam_results_page_options {"name": "Amy's copy of Use the factor theorem to identify factors of a polynomial", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"rulesets": {}, "variables": {"coef1_x2": {"description": "
Number obtained from putting x=-a into the second term of the equation.
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", "definition": "random(-2..2 except 0 except a except c except b)", "templateType": "anything", "name": "d", "group": "Ungrouped variables"}, "b": {"description": "
Random number between -2 and 3 except 0 for creating polynomial.
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Random number between -2 and 3, not including 0 for creating polynomial.
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Number obtained by putting x=-c into the third term of the equation.
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Number obtained from putting x=-d into the second term of the equation.
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The factor theorem states that if $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
Apply the factor theorem to check which of a list of linear polynomials are factors of another polynomial.
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To find the factors of the polynomial $f(x) = \\simplify{x^3+({a}+{b}+{c})x^2+({a}{b}+{a}{c}+{b}{c})x+{a}{b}{c}}$, we use the factor theorem.
\n
If $f(x)$ is a polynomial and $f(p) = 0$, then $(x-p)$ is a factor of $f(x)$.
\n
If $(\\simplify{(x+{a})})$ is a factor of $f(x)$ then by the factor theorem, $f(\\simplify{-{a}}) = 0$.
\n
We see that
\n
\\\begin{align} f(\\simplify{-{a}}) &= \\simplify[all,!collectNumbers]{{coef1_x3}+{coef1_x2}+{coef1_x}+{const}}\\\\ &= \\simplify{{coef1_x3}+{coef1_x2}+{coef1_x}+{const}}. \\end{align} \
\n
Therefore, $(\\simplify{(x+{a})})$ is a factor of $f(x)$.
\n
Similarly for $(\\simplify{(x+{d})})$,
\n
\\\begin{align} f(\\simplify{-{d}}) &= \\simplify[all,!collectNumbers]{{coef2_x3}+{coef2_x2}+{coef2_x}+{const}}\\\\ &= \\simplify{{coef2_x3}+{coef2_x2}+{coef2_x}+{const}}\\\\ &\\neq 0. \\end{align} \
\n
Therefore, $(\\simplify{(x+{d})})$ is not a factor of $f(x)$.
\n
Finally, for $(\\simplify{(x+{c})})$,
\n
\\\begin{align} f(\\simplify{-{c}}) &= \\simplify[all,!collectNumbers]{{coef3_x3}+{coef3_x2}+{coef3_x}+{const}}\\\\ &= \\simplify{{coef3_x3}+{coef3_x2}+{coef3_x}+{const}}. \\end{align} \
\n
Therefore, $(\\simplify{(x+{c})})$ is also a factor of $f(x)$.
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Use the factor theorem to find which two of the following are factors of the polynomial
\n
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$(\\simplify{x+{a}})$
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http://www.gnu.org/software/emacs/manual/html_node/calc/Rewrites-Answer-4.html | 1,524,755,133,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00225.warc.gz | 421,687,530 | 2,340 | #### 2.7.55 Rewrites Tutorial Exercise 4
Here is a suitable set of rules to solve the first part of the problem:
``` [ seq(n, c) := seq(n/2, c+1) :: n%2 = 0,
seq(n, c) := seq(3n+1, c+1) :: n%2 = 1 :: n > 1 ]
```
Given the initial formula ‘seq(6, 0)’, application of these rules produces the following sequence of formulas:
``` seq( 3, 1)
seq(10, 2)
seq( 5, 3)
seq(16, 4)
seq( 8, 5)
seq( 4, 6)
seq( 2, 7)
seq( 1, 8)
```
whereupon neither of the rules match, and rewriting stops.
We can pretty this up a bit with a couple more rules:
``` [ seq(n) := seq(n, 0),
seq(1, c) := c,
... ]
```
Now, given ‘seq(6)’ as the starting configuration, we get 8 as the result.
The change to return a vector is quite simple:
``` [ seq(n) := seq(n, []) :: integer(n) :: n > 0,
seq(1, v) := v | 1,
seq(n, v) := seq(n/2, v | n) :: n%2 = 0,
seq(n, v) := seq(3n+1, v | n) :: n%2 = 1 ]
```
Given ‘seq(6)’, the result is ‘[6, 3, 10, 5, 16, 8, 4, 2, 1]’.
Notice that the ‘n > 1’ guard is no longer necessary on the last rule since the ‘n = 1’ case is now detected by another rule. But a guard has been added to the initial rule to make sure the initial value is suitable before the computation begins.
While still a good idea, this guard is not as vitally important as it was for the `fib` function, since calling, say, ‘seq(x, [])’ will not get into an infinite loop. Calc will not be able to prove the symbol ‘x’ is either even or odd, so none of the rules will apply and the rewrites will stop right away. | 520 | 1,513 | {"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-2018-17 | latest | en | 0.901174 |
lonnbbs.netlify.app | 1,726,674,544,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00760.warc.gz | 342,098,566 | 4,327 | # Wiktionary:Projekt/Önskelistor/Engelska/Matematik - Wiktionary
2021 - Claes Johnson on Mathematics and Science
A common visualisation of complex numbers is the use of Argand Diagrams. To construct this, picture a Cartesian grid with the x-axis being real numbers and the y-axis being imaginary numbers. In some branches of engineering, it’s inevitable that you’re going to end up working with complex numbers. Fortunately, though, you don’t have to run to another piece of software to perform calculations with these numbers.
Complex Number Calculator. Instructions:: All Functions. Instructions. Just type your formula into the top box.
## Popularisering av matematikk – - Matematikksenteret
The formula is still valid if x is a complex number, and so some authors refer to The Euler formula, sometimes also called the Euler identity (e.g., Trott 2004, p. 174), states. e^(ix)=cosx+isinx,. (1).
### Math, Better Explained i Apple Books
3 Euler’s Identity stems naturally from interactions of complex numbers which are numbers composed of two pieces: a real number and an imaginary number; an example is 4+3 i. The Imaginary Number At some point in your life, you've probably encountered the imaginary number, i. In case you haven't, i is defined as the square root of -1. In other words, it's a number so An imaginary number is a complex number that can be written as a real number multiplied by the imaginary unit i, which is defined by its property i2 = −1. The square of an imaginary number bi is −b2. For example, 5i is an imaginary number, and its square is −25. By definition, zero is considered to be both real and imaginary.
The imaginary number i: i p 1 i2 = 1: (1) Every imaginary number is expressed as a real-valued multiple of i: p 9 = p 9 p 1 = p 9i= 3i: A complex number: z= a+ bi; (2) where a;bare real, is the sum of a real and an imaginary number. The real part of z: Refzg= ais a real number. The imaginary part of z: Imfzg= bis a also a real number.
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Complex numbers z = x + iy = r(cos Q+ i sin q) av M Krönika · 2018 — Specifically, in the complex numbers C we know that For good reasons this looks similar to the Euler product of Dirichlet L-functions, but the The most beautiful theorem in mathematics: Euler's Identity. What could be more mystical than an imaginary number interacting with real numbers to produce DOWNLOAD Complex exponential form of wave equation: >> http://bit.ly/2uHNnk9 << Complex Numbers and the Complex Exponential 1. Any complex number The supreme example is Euler's equation between the most fundamental numbers in mathematics: Euler's number e, pi, and the imaginary unit av D Brehmer · 2018 · Citerat av 1 — Children and number: Difficulties in learning mathematics. In an imaginary dialogue study with students in grade 6 and preservice Engeln, Katrin; Euler,. part complex analysis chap.
To understand the meaning of the left-hand side of Euler’s formula, it is best to recall that for real numbers x, one can instead write ex= exp(x) and think of this as a function of x… 2020-06-24 2020-06-25 So, Euler's formula is saying "exponential, imaginary growth traces out a circle".
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### ENGELSK - SVENSK - math.chalmers.se
decimal - relating to or denoting a system of numbers and arithmetic based Below are some other examples of conversions from decimal numbers to quater-imaginary numbers. Euler beräknade initialt konstantens värde till 6 decimaler. where γ = g − iδ0 /2 is a complex number. The intensity envelope of the pulse is dk ∂E/∂k, and by evaluating the two complex Euler terms independently. 38 Cauchy's integral formula in complex analysis, it's a 2 pi. | 973 | 3,840 | {"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-2024-38 | latest | en | 0.86731 |
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## 1 Square Feet in Square Gaj in Jammu and Kashmir
In this article, you will learn how to convert 1 Square Feet in Square Gaj in Jammu and Kashmir. Square Feet is one of the most commonly used measurement units used in almost every state of India. People in Indian states refer to Square Feet for measurement of land. Here in this page you will learn how to calculate 1 Square Feet in Square Gaj unit.
## How to Convert from 1 Square Feet to Square Gaj?
You can easily convert Square Feet to Square Gaj or the reverse with a simple method. You can multiply the figure in Square Feet by 0.112188 to determine the Square Gaj value. This is all you need to do for undertaking the conversion procedure of Square Feet to Square Gaj in Jammu and Kashmir.
## Relationship between Square Feet and Square Gaj
It is not difficult to work out the actual relationship between Square Feet to Square Gaj in most cases. You should know the proper calculations existing between Square Feet to Square Gaj before you venture to conversion procedures. 1 Square Feet to Square Gaj calculations will be 0.112188 Square Gaj.
Once you know these calculations, it will not be hard for you to calculate Square Feet in an Square Gaj. One Square Feet can be worked out to various other units as well.
## Formula for Converting Square Feet to Square Gaj
The formula to convert Square Feet to Square Gaj is the following-
1 Square Feet = 0.112188 Square Gaj
Conversely, you can also use this formula-
Square Feet = Square Gaj * 0.112188
This is all you need to know for undertaking the conversion procedure on your part. You can also use online calculators or conversion tools for this purpose.
### Other Plot Areas in Square Gaj in Jammu and Kashmir
• 1 Square Feet in Square Gaj in Jammu and Kashmir = 0.112188
• 2 Square Feet in Square Gaj in Jammu and Kashmir = 0.224376
• 3 Square Feet in Square Gaj in Jammu and Kashmir = 0.336564
• 4 Square Feet in Square Gaj in Jammu and Kashmir = 0.448752
• 5 Square Feet in Square Gaj in Jammu and Kashmir = 0.56094
• 6 Square Feet in Square Gaj in Jammu and Kashmir = 0.673128
• 7 Square Feet in Square Gaj in Jammu and Kashmir = 0.785316
• 8 Square Feet in Square Gaj in Jammu and Kashmir = 0.897504
• 9 Square Feet in Square Gaj in Jammu and Kashmir = 1.009692
• 10 Square Feet in Square Gaj in Jammu and Kashmir = 1.12188
• 15 Square Feet in Square Gaj in Jammu and Kashmir = 1.68282
• 20 Square Feet in Square Gaj in Jammu and Kashmir = 2.24376
• 25 Square Feet in Square Gaj in Jammu and Kashmir = 2.8047
• 30 Square Feet in Square Gaj in Jammu and Kashmir = 3.36564
• 35 Square Feet in Square Gaj in Jammu and Kashmir = 3.92658
• 40 Square Feet in Square Gaj in Jammu and Kashmir = 4.48752
• 45 Square Feet in Square Gaj in Jammu and Kashmir = 5.04846
• 50 Square Feet in Square Gaj in Jammu and Kashmir = 5.6094
• 55 Square Feet in Square Gaj in Jammu and Kashmir = 6.17034
• 60 Square Feet in Square Gaj in Jammu and Kashmir = 6.73128
• 65 Square Feet in Square Gaj in Jammu and Kashmir = 7.29222
• 70 Square Feet in Square Gaj in Jammu and Kashmir = 7.85316
• 75 Square Feet in Square Gaj in Jammu and Kashmir = 8.4141
• 80 Square Feet in Square Gaj in Jammu and Kashmir = 8.97504
• 85 Square Feet in Square Gaj in Jammu and Kashmir = 9.53598
• 90 Square Feet in Square Gaj in Jammu and Kashmir = 10.09692
• 95 Square Feet in Square Gaj in Jammu and Kashmir = 10.65786
• 100 Square Feet in Square Gaj in Jammu and Kashmir = 11.2188
## FAQs About 1 Square Feet in Square Gaj in Jammu and Kashmir
#### Q: Ek Square Feet main Kitne Square Gaj in Jammu and Kashmir Hote hain?
Ek Square Feet main 0.112188 Square Gaj in Jammu and Kashmir hote hain.
#### 1 Square Feet main kitna Square Gaj in Jammu and Kashmir Hota Hain?
1 Square Feet main 0.112188 Square Gaj in Jammu and Kashmir hote hain
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Just as we can add, subtract and multiply numbers, so can we do the same with Roman numerals. Let’s look at the following example:
III - II = ?
III is the same as 1 + 1 + 1 = 3
II is the same as 1 + 1 = 2
III - II then is the same as 3 - 2 = 1
“1” in Roman numerals is “I” so the complete answer is: III - II = I
Now let’s look at another example:
IX x IV = ?
IX is the same as 10 - 1 = 9
IV is the same as 5 - 1 = 4
IX x IV then is the same as 9 x 4 = 36
“36” in Roman numerals is XXXVI so the complete answer is: IX x IV = XXXVI
Some helpful hints to remember are:
5 = V
10 = X
50 = L
100 = C
500 = D
1000 = M
1. X + III + VI = ?
[ ] XVIX [ ] XIIIVI [ ] XIX [ ] XVIII
2. XXV - XIV = ?
[ ] XXI [ ] VI [ ] XXXI [ ] XI
3. LI x VIII = ?
[ ] CDVIII [ ] LVIV [ ] LVIIII [ ] CCCCVIII
4. DLIX + MDC = ?
[ ] MCLIX [ ] MMCLIX [ ] CMLIX [ ] DMLIX
5. III x VI x V = ?
[ ] XC [ ] XXXIII [ ] XL [ ] LXXXX
6. DCC - LIX + XXIX = ?
[ ] DCXII [ ] CDLXX [ ] DCLXX [ ] CCCCCCXXXXXXX
7. M x III + LIV = ?
[ ] MMCMIV [ ] MIIILIV [ ] MMLIV [ ] MMMLIV
8. LXXX + LXXX = ?
[ ] LLXXXXX [ ] CLX [ ] LLXL [ ] CXL
9. D + L + M = ?
[ ] DML [ ] DLM [ ] MDL [ ] MLD
10. XL + VI + CIV = ?
[ ] CLIX [ ] XCIX [ ] LCIX [ ] CL
Math: Middle School: Grades 6, 7 and 8 Quiz - Roman Numerals Part 2 - Adding, Subtracting and Multiplying (Answers)
1. X + III + VI = ?
[ ] XVIX [ ] XIIIVI [x] XIX [ ] XVIII
X + III + VI = ? X is the same as 10
III is the same as 3
VI is the same as 6
10 + 3 + 6 = 19
“19” in Roman numerals is “XIX” so the complete answer is: X + III + VI = XIX. Answer (c) is the correct answer
2. XXV - XIV = ?
[ ] XXI [ ] VI [ ] XXXI [x] XI
XXV - XIV = ?
XXV is the same as 25
XIV is the same as 14
25 - 14 = 11
“11” in Roman numerals is “XI” so the complete answer is: XXV - XIV = XI. Answer (d) is the correct answer
3. LI x VIII = ?
[x] CDVIII [ ] LVIV [ ] LVIIII [ ] CCCCVIII
LI x VIII = ?
LI is the same as 51
VIII is the same as 8
51 x 8 = 408
“408” in Roman numerals is “CDVIII.” The letter “C” equals “100.” The letter “D” equal “500”. When the letter “C” comes before the letters “D” or “M” is tells you that you must subtract the letters so “CD” is the same as 500 - 100 = 400. “V” is the number 5 and “III” is 1 + 1 + 1 = 3 so 5 + 3 = 8. So the complete answer is: LI x VIII = CDVIII. Answer (d) is the correct answer
4. DLIX + MDC = ?
[ ] MCLIX [x] MMCLIX [ ] CMLIX [ ] DMLIX
DLIX + MDC = ?
D is the same as 500
L is the same as 50
IX is the same as 9
M is the same as 1000
D is the same as 500
C is the same as 100
DLIX is the same as 500 + 50 + 9 = 559
MDC is the same as 1000 + 500 + 100 = 1600
559 + 1600 = 2159
“2159” in Roman numerals is “MMCLIX.” So the complete answer is: DLIX + MDC = MMCLIX. Answer (d) is the correct answer
5. III x VI x V = ?
[x] XC [ ] XXXIII [ ] XL [ ] LXXXX
III x VI x V = ?
III is the same as 3
VI is the same as 6
V is the same as 5
3 x 6 = 18 x 5 = 90
“90” in Roman numerals is “XC.” When a single “X” comes before the letters “L” and “C” it tells you that you need to subtract. So “XC” is the same as 100 - 10 = 90. Answer (d) is the correct answer
6. DCC - LIX + XXIX = ?
[ ] DCXII [ ] CDLXX [x] DCLXX [ ] CCCCCCXXXXXXX
DCC - LIX + XXIX = ?
D is the same as 500
C is the same as 100
L is the same as 50
IX is the same as 9
X is the same as 10
DCC is the same as 500 + 100 + 100 = 700
LIX is the same as 50 + 9 = 59
XXIX is the same as 10 + 10 + 9 = 29
700 - 59 = 641 + 29 = 670
“670” in Roman numerals is “DCLXX.” “D” equals 500, “C” equals 100, “L” equals 50 and “XX” equals 10 + 10 = 20, giving us 500 + 100 + 50 + 20 = 670. Answer (d) is the correct answer
7. M x III + LIV = ?
[ ] MMCMIV [ ] MIIILIV [ ] MMLIV [x] MMMLIV
M x III + LIV = ?
M is the same as 1000
I is the same as 1
L is the same as 50
IV is the same as 4
M x III is the same as 1000 x 3 = 3000
LIV is the same as 50 + 4 = 54
3000 + 50 + 4 = 3054
“3054” in Roman numerals is “MMMLIV.” Answer (d) is the correct answer
8. LXXX + LXXX = ?
[ ] LLXXXXX [x] CLX [ ] LLXL [ ] CXL
LXXX + LXXX = ?
L is the same as 50
X is the same as 10
XXX is the same as 10 + 10 + 10 = 30
50 + 30 = 80
80 + 80 = 160
“160” in Roman numerals is “CLX.” Answer (d) is the correct answer
9. D + L + M = ?
[ ] DML [ ] DLM [x] MDL [ ] MLD
D + L + M = ?
D is the same as 500
L is the same as 50
M is the same as 1000
500 + 50 + 1000 = 1550
“1550” in Roman numerals is “MDL.” Answer (d) is the correct answer
10. XL + VI + CIV = ?
[ ] CLIX [ ] XCIX [ ] LCIX [x] CL
XL + VI + CIV = ?
X is the same as 10
L is the same as 50
V is the same as 5
I is the same as 1
C is the same as 100
IV is the same as 4
XL is the same as 50 - 10 = 40
VI is the same as 5 + 1 = 6
CIV is the same as 100 + 5 - 1 = 104
40 + 6 + 104 = 150
“150” in Roman numerals is “CL.” Answer (d) is the correct answer | 2,163 | 5,352 | {"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} | 4.59375 | 5 | CC-MAIN-2020-45 | latest | en | 0.897663 |
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# Multiple choice question for engineering
## Set 1
1. What is the consideration for the determination of the diameter of shaft?
a) stiffness
b) voltage
c) current
d) rigidity
Answer: c [Reason:] The main aspect for the design of the diameter of the shaft is the stiffness. The diameter of the shaft depends on the stiffness of the machine.
2. What is the meaning of stiffness?
a) ability to transmit the power
b) ability to withstand the weight of the rotor
c) ability to withstand unbalanced magnetic pull
d) ability to withstand the weight of rotor and unbalanced magnetic pull
Answer: d [Reason:] The diameter of shaft for an electrical machine is determined by considerations of stiffness. The stiffness is the ability to withstand the weight of rotor and unbalanced magnetic pull.
3. What should be the first property of the shaft design?
a) the shaft design should be such that the shaft must have enough corrosion resistance
b) the shaft design should be such that the shaft must have enough mechanical strength
c) the shaft design should be such that the shaft has enough tensile strength
d) the shaft design should be able to withstand the voltage fluctuations
Answer: b [Reason:] The shaft design should be such that the shaft must have enough mechanical strength. The strength should be such that it should withstand all loads without causing much residual strain.
4. What is the second property of the shaft design?
a) the shaft design should be such that it has high rigidity
b) the shaft design should be such that it should have high tensile strength
c) the shaft design should be such that it should have high corrosion resistance
d) the shaft design should be such that it should withstand voltage fluctuations
Answer: a [Reason:] The second property of the shaft design such that it should have high rigidity. The rigidity should be such that the deflection of shaft under operation of machine does not reach such a dangerous value as to cause the rotor to touch the stator.
5. The critical speeds of rotation should be different from running speed of machine?
a) true
b) false
Answer: a [Reason:] The critical speed relation is the third property of the shaft design. The critical speeds of rotation should be different from the running speed of the machine.
6. What is the formula of the diameter of the shaft?
a) diameter of the shaft = 5.5 + (output in watt/rps)1/3 mm
b) diameter of the shaft = 5.5 – (output in watt/rps)1/3 mm
c) diameter of the shaft = 5.5 * (output in watt/rps)1/3 mm
d) diameter of the shaft = 5.5 / (output in watt/rps)1/3 mm
Answer: c [Reason:] The output is first calculated from the operation of the machine. Next the tachogenerator is used to calculate the speed of the machine and on the substitution of the values gives the diameter of the shaft.
7. What is the relation of the diameter of the shaft in the bearings to the diameter under the armature?
a) diameter of the shaft is very much greater than the diameter under the armature
b) diameter of the shaft is greater than the diameter under the armature
c) diameter of the shaft is equal to the diameter under the armature
d) diameter of the shaft is lesser than the diameter under the armature
Answer: d [Reason:] There are certain rules in the design of the shaft. The diameter of the shaft in the bearings is less than the diameter under the armature.
8. What happens when the diameter under armature is 150 mm or above?
a) diameter of the shaft in bearing is 100 mm smaller than the maximum diameter
b) diameter of the shaft in bearing is 90 mm smaller than the maximum diameter.
c) diameter of the shaft in bearing is 70 mm smaller than the maximum diameter.
d) diameter of the shaft in bearing is 50 mm smaller than the maximum diameter.
Answer: d [Reason:] The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter of the shaft in bearing is 50 mm smaller than the maximum diameter.
9. What happens in the case of the small shafts?
a) the diameter in the bearings should be about 1/3 of the maximum diameter
b) the diameter in the bearing should be about 2/3 of the maximum diameter
c) the diameter in the bearing should be about 2/5 of the maximum diameter
d) the diameter in the bearing should be about 1/5 of the maximum diameter
Answer: b [Reason:] The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter in the bearing should be 2/3 of the maximum diameter.
## Set 2
1. How is he reluctance motor with respect to synchronous motor and are the field windings?
a) small synchronous motor with field windings
b) small synchronous motor without field windings
c) large synchronous motor with field windings
d) large synchronous motor without field windings
Answer: b [Reason:] Reluctance motor is nothing but a simple small synchronous motor with salient pole rotor. They are without field windings in which the field flux is produced.
2. Why is the three phase reluctance motor preferred over single phase reluctance motor?
a) single phase reluctance motors have the phenomenon of hunting
b) single phase reluctance motors have the phenomenon of over voltage
c) single phase reluctance motors have high losses
d) single phase reluctance motors have low output
Answer: a [Reason:] The reluctance motor is a small synchronous motor with salient pole rotor. The single phase reluctance motors have the phenomenon of hunting.
3. What is the relation of the input voltage with the magnetic flux?
a) if the input voltage is constant, the magnetic flux increases
b) if the input voltage is constant, the magnetic flux decreases
c) if the input voltage is constant, the magnetic flux is constant
d) if the input voltage is constant, the magnetic flux is zero
Answer: c [Reason:] The input voltage is given constant, which results in the constant magnetic flux. The magnetic flux is independent of the excitation.
4.What is the power factor in the reluctance motor and the range of efficiency?
a) leading power factor, 60-75%
b) lagging power factor, 50-75%
c) zero power factor, 55-80%
d) lagging power factor, 55-75%
Answer: d [Reason:] The power factor in the reluctance motor is lagging power factor. The efficiency of the machine is about 55-75%.
5. What is the angle at which the electromagnetic torque is maximum?
a) 300
b) 450
c) 600
d) 900
Answer: b [Reason:] The electromagnetic torque is maximum at the angle of 450. The range of operation of the reluctance motor lies in the range of 0-450.
6. What is the range of the ratio of the direct axis reactance to the quadrature axis reactance?
a) 1.5-2.3
b) 1.6-2.7
c) 1.6-2.2
d) 1.2-2.0
Answer: c [Reason:] The minimum value of the ratio of the direct axis reactance to the quadrature axis reactance is 1.6. The maximum value of the ratio of the direct axis reactance to the quadrature axis reactance is 2.2.
7. How many design dimension are present in the design of the small reluctance motor?
a) 3
b) 4
c) 5
d) 6
Answer: c [Reason:] There are 5 design dimensions present in the design of the small reluctance motors. They are the design of the main dimensions, design of stator windings, design of the rotor of the reluctance motor, design of performance parameters, design of losses and efficiency.
8. What is the range of the constant used in the calculation of the active power of reluctance motor?
a) 0.3-0.4
b) 0.35-0.55
c) 0.40-0.50
d) 0.35-0.60
Answer: b [Reason:] The minimum value of the range of the constant used in the calculation of the active power of reluctance motor is 0.35. The maximum value of the range of the constant used in the calculation of the active power of reluctance motor is 0.55.
9. How many steps are present in the calculation of the determination of main dimensions?
a) 5
b) 4
c) 3
d) 2
Answer: a [Reason:] There are 5 steps present in the calculation of the determination of main dimensions. They are electromagnetic power of reluctance motor, output coefficient, pole pitch, pole arc, peripheral velocity.
10. How many steps are present in the calculation of the design of stator windings?
a) 10
b) 11
c) 9
d) 12
Answer: b [Reason:] There are 11 steps involved in the calculation of the design of stator windings. They are input current to motor, number of stator slots, stator winding pitch, winding factor, useful flux, number of turns per stator winding, cross sectional area of the stator winding, slot area, mean length for conductor, active resistance of stator winding, specific permeance of leakage flux.
11. How many steps are present in the calculation of the design of rotor of reluctance motors?
a) 4
b) 5
c) 3
d) 2
Answer: a [Reason:] There are 4 steps involved in the design of rotor of reluctance motor. They are rotor diameter calculation, height of rotor core, mmf for magnetic circuit, saturation coefficient of motor.
12. How many steps are involved in the design of performance parameters?
a) 6
b) 5
c) 7
d) 8
Answer: c [Reason:] There are 7 steps involved in the design of the performance parameters. They are no load current, height of steel stator teeth, weight of steel in the stator core, copper loss in the stator winding under no load, active resistance and leakage reactance, active component of no load current, starting torque of 3 phase reluctance motor.
13. How many design steps are involved in the determination of the losses and efficiency?
a) 2
b) 3
c) 4
d) 5
Answer: b [Reason:] There are 3 steps involved in the determination of the losses and efficiency. They are copper loss in stator winding, iron loss in stator steel, mechanical loss in the motor.
14. What is the formula for the slot pitch factor in design of rotors?
a) slot pitch factor = 3.14*rotor diameter*number of rotor slots
b) slot pitch factor = 3.14/rotor diameter*number of rotor slots
c) slot pitch factor = 3.14*rotor diameter/number of rotor slots
d) slot pitch factor = 1/3.14*rotor diameter*number of rotor slots
Answer: c [Reason:] First the rotor diameter and the number of rotor slots are first calculated. On substitution the slot pitch factor can be obtained.
15. The active resistance of the stator winding is calculated at the temperature of 450 C?
a) true
b) false
Answer: b [Reason:] The active resistance of the stator winding determination is one of the steps in the design of stator windings. The value is calculated at the temperature of 450 C.
## Set 3
1. What type is the stator windings of the single phase induction motor?
a) hollow
b) cylindrical
c) concentric
d) rectangular
Answer: c [Reason:] The stator windings are also known as the running winding or the main winding. The type of stator winding used is concentric type
2. How many coils are present in the stator windings?
a) 2
b) 3
c) 2 or more
d) 3 or more
Answer: d [Reason:] The stator windings of single phase induction motors are concentric type. There are usually 3 or more coils per pole each having same or different number of turns.
3. How much of the total slots are used for the reduction of the mmf wave harmonics?
a) 60%
b) 65%
c) 70%
d) 80%
Answer: c [Reason:] 70% of the total slots are used for the reduction of the mmf wave harmonics. The remaining 30% are used for accommodating the starting windings.
4.How can the small single phase motor reduce the harmonics still much further?
a) removing the winding
b) insulating the winding
c) grading the winding
d) shading the winding
Answer: c [Reason:] 70% of the total slots are used for the reduction of the mmf wave harmonics. The mmf wave harmonics can be still further reduced by grading the winding.
5. What is the formula for the mean pitch factor?
a) mean pitch factor = pitch factor of each coil per pole group + turns in the coil / total number of turns
b) mean pitch factor = pitch factor of each coil per pole group / turns in the coil * total number of turns
c) mean pitch factor = pitch factor of each coil per pole group * turns in the coil * total number of turns
d) mean pitch factor = pitch factor of each coil per pole group * turns in the coil / total number of turns
Answer: d [Reason:] The pitch factor of each coil per pole group, turns in the coil and total number of turns are obtained. On substitution it gives the mean pitch factor.
6. What is the range of the winding factor for the usual windings distribution?
a) 0.70-0.80
b) 0.75-0.85
c) 0.70-0.85
d) 0.70-0.75
Answer: b [Reason:] The minimum value of the winding factor of the usual winding distribution is 0.75. The maximum value of the winding factor of the usual winding distribution is 0.85.
7. What is the formula of the maximum flux in the running winding?
a) maximum flux = flux * pole
b) maximum flux = flux/pole
c) maximum flux = flux / turns
d) maximum flux = flux * turns
Answer: b [Reason:] First the flux is calculated along with the number of poles used. On substituting the values the maximum flux value is obtained.
8. What is the value of the stator induced voltage with respect to the supply voltage?
a) stator induced voltage = 95% of supply voltage
b) stator induced voltage = 90% of supply voltage
c) stator induced voltage = 85% of supply voltage
d) stator induced voltage = 80% of supply voltage
Answer: a [Reason:] The winding factor is assumed to be 0.75-0.85 for the running winding. The stator induced voltage is 95% of the supply voltage.
9. How many design data are present in the design of the stator?
a) 6
b) 7
c) 8
d) 9
Answer: c [Reason:] There are 8 design data available in the design of the stator. The design data are running winding, number of turns In running winding, running winding conductors, number of stator slots, size of stator slot, stator teeth, stator core, length of mean turn.
10. What is the range of the current density for the open type motors split phase, capacitor and repulsion start motors?
a) 4-5 A per mm2
b) 3-4 A per mm2
c) 2-4 A per mm2
d) 1-4 A per mm2
Answer: b [Reason:] The minimum value of the current density for the open type motors split phase, capacitor and repulsion start motors is 3 A per mm2. The maximum value of the current density for the open type motors split phase, capacitor and repulsion start motors is 4 A per mm2.
11. What is the relation of the number of slots with the leakage reactance?
a) small number of slots, high leakage reactance
b) large number of slots, high leakage reactance
c) large number of slots, small leakage reactance
d) small number of slots, small leakage reactance
Answer: c [Reason:] The number of slots is indirectly proportional to the leakage reactance. The larger the number of slots, the lower will be the leakage reactance.
12. What is the formula for the area required for the insulated conductors?
a) area required for the insulated conductors = total number of conductors per slot * 0.785 / diameter of insulated conductor2
b) area required for the insulated conductors = total number of conductors per slot / 0.785 * diameter of insulated conductor2
c) area required for the insulated conductors = total number of conductors per slot * 0.785 * diameter of insulated conductor2
d) area required for the insulated conductors = 1/total number of conductors per slot * 0.785 * diameter of insulated conductor2
Answer: c [Reason:] The total number of conductors per slot and the diameter of insulated conductors are calculated. On substitution the area required for the insulated conductors are calculated.
13. The flux density of the high torque machines is 1.8 weber per m2?
a) true
b) false
Answer: a [Reason:] The flux density of the general purpose machine is 1.45 weber per m2. The flux density of the high torque machines is 1.8 weber per m2.
14. The flux density of the stator core should not exceed 1.3 weber per m2?
a) true
b) false
Answer: b [Reason:] The flux density of the stator core should not exceed 1.5 weber per m2. The range lies between 0.9 – 1.4 weber per m2.
15. What is the formula for the flux density in stator core?
a) flux density in stator core = maximum flux / length of the iron * depth of stator core
b) flux density in stator core = maximum flux * length of the iron * depth of stator core
c) flux density in stator core = maximum flux / 2 *length of the iron * depth of stator core
d) flux density in stator core = maximum flux * length of the iron / depth of stator core
Answer: c [Reason:] The maximum flux, length of iron and depth of stator core is calculated. On substitution it provides the flux density in stator core.
## Set 4
1. How many design steps are present in the design of PMDC motors?
a) 8
b) 9
c) 10
d) 11
Answer: d [Reason:] There are 11 steps involves in the design of the PMDC motors. They are minimum sum of air gap volume and magnet volume, ratio of magnetic to electric loading, area of magnet, length of magnet, value of flux, number of turns per coil, running armature resistance, armature diameter, axial dimensions, wire cross section and radial thickness.
2. What happens to the armature diameter and the volume of air gap and magnet when the angle is lower in value?
a) volume of air gap and magnet increases, armature diameter increases
b) volume of air gap and magnet increases, armature diameter decreases
c) volume of air gap and magnet decreases, armature diameter decreases
d) volume of air gap and magnet decreases, armature diameter increases
Answer: d [Reason:] The lower values of angle, reduces the volume of air gap and magnet. The reduction of volume of air gap and magnet, increases the armature diameter.
3. What should be the range of the product of the magnetic field and magnetic flux density?
a) 4-4.5 * 106
b) 4-4.3 * 106
c) 4.3-4.6 * 106
d) 4.2-4.5 * 106
Answer: c [Reason:] The product of the magnetic field and magnetic flux density has a minimum value of 4.3 * 106. The product of the magnetic field and magnetic flux density has a minimum value of 4.6 * 106.
4.What should be the minimum value of the ratio of the magnetic to electric loading?
a) 40
b) 30
c) 50
d) 60
Answer: c [Reason:] The calculation of the ratio of the magnetic to electric loading is the second step in the design of the PMDC motors. It should have a minimum value of 50.
5. What is the formula for the area of the magnet in the design of PMDC motors?
a) area of magnet = flux * 4.95 * residual flux density
b) area of magnet = flux / 4.95 * residual flux density
c) area of magnet = flux * 4.95 / residual flux density
d) area of magnet = 1/flux * 4.95 * residual flux density
Answer: b [Reason:] First the residual flux density is calculated. Next the flux is calculated and substitution in the formula gives the area of magnet.
6. What is the range of length of the magnet in the PMDC motors?
a) 2.5-4 cm
b) 2-3 cm
c) 2.5-3 cm
d) 1.5-4 cm
Answer: a [Reason:] The minimum value of the length of the magnet in the PMDC motor is 2.5 cm. The maximum value of the length of the magnet in the PMDC motor is 4 cm.
7. What is the formula of the length of the magnet?
a) length of the magnet = sum of the volume of air gap and magnet * Area of the magnet + 0.06
b) length of the magnet = sum of the volume of air gap and magnet / Area of the magnet + 0.06
c) length of the magnet = sum of the volume of air gap and magnet / Area of the magnet – 0.06
d) length of the magnet = sum of the volume of air gap and magnet * Area of the magnet – 0.06
Answer: c [Reason:] The sum of the volume of the air gap and magnet is first calculated. Next the area of the magnet is calculated from its formula and on substitution gives the length of the magnet.
8. What is the relation between the flux and the no local speed?
a) flux is directly proportional to the no local speed
b) flux is indirectly proportional to the no local speed
c) flux is directly proportional to the square of the no local speed
d) flux is indirectly proportional to the square of the no local speed
Answer: b [Reason:] The calculation of the flux value is one of the design steps. The flux is indirectly proportional to the no local speed calculated.
9.What is the formula of the number of turns per coil?
a) number of turns per coil = number of conductors/2*coils/slot*number of armature teeth
b) number of turns per coil = number of conductors*2*coils/slot*number of armature teeth
c) number of turns per coil = number of conductors*2*coils/slot/number of armature teeth
d) number of turns per coil = number of conductors/2*coils/slot/number of armature teeth
Answer: a [Reason:] The number of conductors is calculated along with the coils per slot is calculated. Next the number of armature teeth is calculated, and on substitution gives the number of turns per coil.
10. What is the formula for the armature resistance?
a) armature resistance = running armature resistance / 1.0 to 1.0
b) armature resistance = running armature resistance * 1.3 to 1.5
c) armature resistance = running armature resistance * 1.4 to 1.5
d) armature resistance = running armature resistance / 1.3 to 1.3
Answer: d [Reason:] The running armature resistance is first calculated in the PMDC motor. It is divided by 1.3 and that gives the armature resistance of the machine.
11. What is the relation between axial dimension and the area of the magnet?
a) area of the magnet is directly proportional to the axial dimension
b) area of the magnet is indirectly proportional to the axial dimension
c) area of the magnet is directly proportional to the square of the axial dimension
d) area of the magnet is indirectly proportional to the square of the axial dimension
Answer: a [Reason:] The calculation of the axial dimension is one of the steps in the PMDC motors. The axial dimension is directly proportional to area of the magnet.
12. What is the relation of the wire cross-section with respect to the armature resistance?
a) wire section is directly proportional to the armature resistance
b) wire section is indirectly proportional to the armature resistance
c) wire section is directly proportional to the square of the armature resistance
d) wire section is indirectly proportional to the square of the armature resistance
Answer: a [Reason:] The 10th design step of the PMDC motor is the calculation of the wire cross section. The wire cross section is directly proportional to the armature resistance.
13. The radial thickness of the joke directly proportional to the flux?
a) true
b) false
Answer: a [Reason:] The last design step in the PMDC motor is the calculation of the radial thickness of the joke. The flux value is directly proportional to the radial thickness of the joke.
14.The radial thickness of the joke is directly proportional to the length of the stator slots?
a) true
b) false
Answer: b [Reason:] The last design step in the PMDC motor is the calculation of the radial thickness of the joke. The radial thickness of the joke is indirectly proportional to the length of the stator slots.
15. What is the formula of the length of the stator slots?
a) length of the stator slots = 2 * perimeter of one magnet
b) length of the stator slots = 1/2 * perimeter of one magnet
c) length of the stator slots = 1/3 * perimeter of one magnet
d) length of the stator slots = 3 * perimeter of one magnet
Answer: b [Reason:] The length of the stator slots is required in the calculation of the radial thickness of the joke. The length of the stator slots is equal to half the perimeter of one magnet.
## Set 5
1. What is the usage of the tanks with tubes?
a) if the temperature rise with plain tank is very low
b) if the temperature rise with plain tank is very high
c) if the temperature rise is zero
d) if the temperature rise with plain tank exceeds the specific limits
Answer: d [Reason:] Temperature rise in transformers is calculated with plain walled tanks. If the limits is exceeded then the plain walled tank is replaced by tank with tubes.
2. What is the relation of the provision of tubes with respect to dissipation of heat?
a) the provision of tubes is directly proportional to the dissipation of heat
b) the provision of tubes is indirectly proportional to the dissipation of heat
c) the provision of tubes is directly proportional to square of the dissipation of heat
d) the provision of tubes is indirectly proportional to square of the dissipation of heat
Answer: b [Reason:] The provision of tubes increases the dissipating area. The increase in dissipation of heat is not proportional to area because tube screen some of the tank surface preventing radiation from there.
3. What is the relation of the transformer surface with respect to dissipation of heat?
a) transformer surface has no relation with respect to dissipation of heat
b) transformer surface has minor changes with respect to dissipation of heat
c) transformer surface has major changes with respect to dissipation of heat
d) transformer surface has no change with respect to dissipation of heat
Answer: d [Reason:] When the tanks with tubes are provided, the dissipation of heat increases. The dissipation of heat has no effect on the transformer surface.
4.How is the circulation of oil improved in tanks with tubes?
a) it can be improved by using dissipating heat
b) it can be improved by using more effective air circulation
c) it can be improved by using more effect power flow
d) it can be improved by using more effective heads of pressure
Answer: d [Reason:] The circulation of oil is improved in tanks with tubes. It takes place with the help of using more effective heads of pressure.
5. An addition of 35 per cent should be made to tube area of the transformers?
a) true
b) false
Answer: a [Reason:] An addition of 35 per cent should be made to tube area of the transformer. This should be done in order to take into account this improvement in dissipation of loss by convection.
6. What is the loss dissipated by tubes by convection, given area of the tubes = 3.5?
a) 12.3 W per 0c
b) 2.51 W per 0c
c) 5.3 W per 0c
d) 30.8 W per 0c
Answer: d [Reason:] Loss dissipated by tubes by convection = 8.8 * Area of tubes Loss = 8.8 * 3.5 = 30.8 W per 0c.
7. What is the formula for temperature rise with tubes?
a) temperature rise with tubes = total loss / dissipating surface*(12.5 + 8.8x)
b) temperature rise with tubes = total loss * dissipating surface*(12.5 + 8.8x)
c) temperature rise with tubes = total loss / dissipating surface / (12.5 + 8.8x)
d) temperature rise with tubes = total loss + dissipating surface*(12.5 + 8.8x)
Answer: a [Reason:] The total losses in the transformers are obtained firstly the iron loss and copper loss. Next the dissipating surface temperature is obtained and substituting in the above formula gives the temperature rise.
8. What is the formula for number of tubes?
a) number of tubes = (1/ 8 * area of each tube) * (total loss / temperature rise with tubes – 12.5 * dissipating surface)
b) number of tubes = (1* 8 * area of each tube) * (total loss / temperature rise with tubes – 12.5 * dissipating surface)
c) number of tubes = (1/ 8 * area of each tube) / (total loss / temperature rise with tubes – 12.5 * dissipating surface)
d) number of tubes = (1/ 8 * area of each tube) + (total loss / temperature rise with tubes – 12.5 * dissipating surface)
Answer: a [Reason:] First the temperature rise with tubes is obtained. Then the iron loss and copper loss are obtained and added. Area of each tube is also obtained. Substituting all the values in the above formula provides the number of tubes.
9. What is the range of the diameter of the tubes used?
a) 50-60 mm
b) 60-70 mm
c) 70-80 mm
d) 50-70 mm
Answer: d [Reason:] The minimum value of the diameter of tubes is derived to be around 50 mm. The maximum value of the diameter of tubes should be less than 70 mm.
10. Elliptical tubes with pressed radiators are increasingly been used?
a) true
b) false
Answer: a [Reason:] Elliptical tubes with pressed radiators are on high demand now a days. This is because they give a greater dissipating surface for the small volume of oil.
11. What is the formula for width of the tank for single phase transformers used?
a) width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
b) width of tank = distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank
c) width of tank = 2*distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
d) width of tank = distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank
Answer: b [Reason:] Width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank is the formula for three phase transformer. For single phase transformers, the distance between adjacent limbs is not multiplied.
12. What is the formula for the length of the tank?
a) length of the tank = external diameter of h.v winding + clearance on each side between the winding and tank along the width
b) length of the tank = external diameter of h.v winding * clearance on each side between the winding and tank along the width
c) length of the tank = external diameter of h.v winding + 2*clearance on each side between the winding and tank along the width
d) length of the tank = external diameter of h.v winding / 2*clearance on each side between the winding and tank along the width
Answer: c [Reason:] The external diameter of h.v winding is obtained. Next the clearance on each side between the winding and tank along the width is calculated and is substituted in the above formula.
13. What is the formula for height of transformer tank?
a) height of transformer tank = Height of transformer frame + clearance height between the assembled transformer and tank
b) height of transformer tank = Height of transformer frame * clearance height between the assembled transformer and tank
c) height of transformer tank = Height of transformer frame / clearance height between the assembled transformer and tank
d) height of transformer tank = Height of transformer frame – clearance height between the assembled transformer and tank
Answer: a [Reason:] Firstly, the height of the transformer frame is calculated. Next, the clearance height between the assembled transformer and tank is also calculated. Substitute the values to obtain the height of transformer tank.
14. What is the rating of the transformer for the voltage of about 11 kV?
a) 1000-2000 kVA
b) 100-3000 kVA
c) 1000-5000 kVA
d) 100-500 kVA | 7,532 | 30,754 | {"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} | 2.515625 | 3 | CC-MAIN-2019-13 | longest | en | 0.912828 |
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# Question Video: Explaining the Different Rates of Combustion in Air Compared with Pure Oxygen Science • Third Year of Preparatory School
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Why is the combustion of aluminum in air slower than in pure oxygen? [A] The temperature of oxygen in air is greater than in pure oxygen. [B] The temperature of pure oxygen is greater than air. [C] The concentration of oxygen in air is less than in pure oxygen. [D] The concentration of oxygen in air is greater than in pure oxygen.
02:51
### Video Transcript
Why is the combustion of aluminum in air slower than in pure oxygen? (A) The temperature of oxygen in air is greater than in pure oxygen. (B) The temperature of pure oxygen is greater than air. (C) The concentration of oxygen in air is less than in pure oxygen. Or (D) the concentration of oxygen in air is greater than in pure oxygen.
Combustion or burning is a chemical reaction that usually involves the reaction of a substance with oxygen. In this case, the substance reacting with the oxygen is aluminum. In order for a reaction to occur between the aluminum and the oxygen, the particles will have to successfully collide with one another. If we can find a way to increase the total number of collisions, then the chance of a successful collision will increase. Another way to say this is that increasing the number of collisions will increase the number of reactions between particles. Thus, more collisions means a faster reaction.
The question describes two different conditions under which aluminum is combusted, in pure oxygen and in air. Pure oxygen is nearly 100 percent oxygen. We might describe pure oxygen as having a high concentration, where concentration is the amount of a substance in a particular volume. As pure oxygen is highly concentrated, there will be lots of collisions between the oxygen particles and aluminum particles, which should result in more reactions.
Air, on the other hand, is only composed of about 21 percent oxygen. So, we might describe air as having a low concentration of oxygen. As there are fewer oxygen particles, there will be fewer collisions, which will result in fewer reactions, so we can make the connection that decreasing the concentration decreases the number of collisions, which decreases the number of reactions. In other words, the lower the concentration, the slower the reaction.
The question asked us to explain why the combustion of aluminum in air is slower than in pure oxygen. We now know that this is the case because the concentration of oxygen in air is lower than the concentration in pure oxygen. The answer choice which correctly describes this relationship is answer choice (C). Thus, the combustion of aluminum in air is slower than in pure oxygen because the concentration of oxygen in air is less than in pure oxygen, or answer choice (C).
## Join Nagwa Classes
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• Realistic Exam Questions | 651 | 3,258 | {"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} | 3.546875 | 4 | CC-MAIN-2024-38 | latest | en | 0.917052 |
http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/178626 | 1,571,247,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00078.warc.gz | 29,091,089 | 2,440 | ```=begin
my solution was inspired by a lecture about scheme/lisp streams. the
possible-solution space is searched in a quite stupid manner which
makes it kind of slow... :-)
=end
require 'lazylist'
pirates = ARGV.shift.to_i
loot = ARGV.map {|x| x.to_i}
# this computes _all_ solutions (but does so lazyly)
# also this doesn't check for equivalent solutions, but we don't care
# since only the first solution is computed and printed
LazyList[1 ... pirates**loot.size].map {|owners|
# owners encodes a way to give each pirate a subset of the loot
# (as a number of base "pirates")
bags = Array.new(pirates) {[]}
idx = loot.size - 1
begin
owners, owner = owners.divmod(pirates)
bags[owner] << loot[idx]
idx -= 1
end while owners > 0
idx.downto(0) do |i|
bags[0] << loot[i]
end
bags
}.map {|splitting|
# now map to the sums
puts "computed sums for #{splitting.inspect}"
[splitting, splitting.map {|pieces| pieces.inject(0) {|s,p| s +
p}}]
}.select {|splitting, sums|
# are all sums the same?
sums.uniq.length == 1
}.map {|splitting, sums|
# forget the sums
splitting
}.take.each {|splitting|
# take the first solution and just output it
splitting.each_with_index {|pieces, owner|
puts " #{owner+1}: #{pieces.sort.join(" ")}"
}
}
``` | 356 | 1,233 | {"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} | 2.984375 | 3 | CC-MAIN-2019-43 | latest | en | 0.819727 |
https://nrich.maths.org/public/leg.php?code=-395&cl=4&cldcmpid=5904 | 1,519,098,426,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812873.22/warc/CC-MAIN-20180220030745-20180220050745-00344.warc.gz | 732,726,602 | 9,078 | # Search by Topic
#### Resources tagged with physics similar to Mathematical Issues for Chemists:
Filter by: Content type:
Stage:
Challenge level:
### There are 80 results
Broad Topics > Applications > physics
### Eudiometry
##### Stage: 5 Challenge Level:
When a mixture of gases burn, will the volume change?
### Ideal Axes
##### Stage: 5 Challenge Level:
Explore how can changing the axes for a plot of an equation can lead to different shaped graphs emerging
### Universal Time, Mass, Length
##### Stage: 5 Short Challenge Level:
Can you work out the natural time scale for the universe?
### Diamonds Aren't Forever
##### Stage: 5 Challenge Level:
Ever wondered what it would be like to vaporise a diamond? Find out inside...
### The Power of Dimensional Analysis
##### Stage: 4 and 5
An introduction to a useful tool to check the validity of an equation.
### Physnrich
##### Stage: 4 and 5 Challenge Level:
PhysNRICH is the area of the StemNRICH site devoted to the mathematics underlying the study of physics
### Core Scientific Mathematics
##### Stage: 4 and 5 Challenge Level:
This is the area of the advanced stemNRICH site devoted to the core applied mathematics underlying the sciences.
### The Amazing Properties of Water
##### Stage: 4 and 5 Challenge Level:
Find out why water is one of the most amazing compounds in the universe and why it is essential for life. - UNDER DEVELOPMENT
### Striking Gold
##### Stage: 5 Challenge Level:
Investigate some of the issues raised by Geiger and Marsden's famous scattering experiment in which they fired alpha particles at a sheet of gold.
### Ideal Gases
##### Stage: 5 Challenge Level:
Problems which make you think about the kinetic ideas underlying the ideal gas laws.
### Drug Stabiliser
##### Stage: 5 Challenge Level:
How does the half-life of a drug affect the build up of medication in the body over time?
### Constantly Changing
##### Stage: 4 Challenge Level:
Many physical constants are only known to a certain accuracy. Explore the numerical error bounds in the mass of water and its constituents.
### The Lorentz Force Law
##### Stage: 5 Challenge Level:
Explore the Lorentz force law for charges moving in different ways.
### Levels of Bohr
##### Stage: 5 Challenge Level:
Look at the units in the expression for the energy levels of the electrons in a hydrogen atom according to the Bohr model.
### Guessing the Graph
##### Stage: 4 Challenge Level:
Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from?
### Neural Nets
##### Stage: 5
Find out some of the mathematics behind neural networks.
### The Real Hydrogen Atom
##### Stage: 5 Challenge Level:
Dip your toe into the world of quantum mechanics by looking at the Schrodinger equation for hydrogen atoms
### Chemnrich
##### Stage: 4 and 5 Challenge Level:
chemNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of chemistry, designed to help develop the mathematics required to get the most from your study. . . .
### Big and Small Numbers in Physics - Group Task
##### Stage: 5 Challenge Level:
Work in groups to try to create the best approximations to these physical quantities.
### Moving Stonehenge
##### Stage: 5 Challenge Level:
A look at the fluid mechanics questions that are raised by the Stonehenge 'bluestones'.
### Gravity Paths
##### Stage: 5 Challenge Level:
Where will the spaceman go when he falls through these strange planetary systems?
### Lennard Jones Potential
##### Stage: 5 Challenge Level:
Investigate why the Lennard-Jones potential gives a good approximate explanation for the behaviour of atoms at close ranges
### Maths in the Undergraduate Physical Sciences
##### Stage: 5
An article about the kind of maths a first year undergraduate in physics, engineering and other physical sciences courses might encounter. The aim is to highlight the link between particular maths. . . .
### The Not-so-simple Pendulum 1
##### Stage: 5 Challenge Level:
See how the motion of the simple pendulum is not-so-simple after all.
### Electromagnetism
##### Stage: 5 Challenge Level:
A look at a fluid mechanics technique called the Steady Flow Momentum Equation.
### Big and Small Numbers in Chemistry
##### Stage: 4 Challenge Level:
Get some practice using big and small numbers in chemistry.
### Engnrich
##### Stage: 5 Challenge Level:
engNRICH is the area of the stemNRICH Advanced site devoted to the mathematics underlying the study of engineering
### The Ultra Particle
##### Stage: 5 Challenge Level:
Explore the energy of this incredibly energetic particle which struck Earth on October 15th 1991
### Ramping it Up
##### Stage: 5 Challenge Level:
Look at the calculus behind the simple act of a car going over a step.
### Reaction Types
##### Stage: 5 Challenge Level:
Explore the rates of growth of the sorts of simple polynomials often used in mathematical modelling.
### Whose Line Graph Is it Anyway?
##### Stage: 5 Challenge Level:
Which line graph, equations and physical processes go together?
### Go Spaceship Go
##### Stage: 5 Challenge Level:
Show that even a very powerful spaceship would eventually run out of overtaking power
### Ancient Astronomical Terms
##### Stage: 3, 4 and 5
Some explanations of basic terms and some phenomena discovered by ancient astronomers
### Big and Small Numbers in Physics
##### Stage: 4 Challenge Level:
Work out the numerical values for these physical quantities.
### Pack Man
##### Stage: 5 Challenge Level:
A look at different crystal lattice structures, and how they relate to structural properties
### Battery Modelling
##### Stage: 5 Challenge Level:
Find out how to model a battery mathematically
### Motorbike Momentum
##### Stage: 5 Challenge Level:
A think about the physics of a motorbike riding upside down
### The Not-so-simple Pendulum 2
##### Stage: 5 Challenge Level:
Things are roughened up and friction is now added to the approximate simple pendulum
### New Units for Old
##### Stage: 5 Challenge Level:
Can you match up the entries from this table of units?
### Powerfully Fast
##### Stage: 5 Challenge Level:
Explore the power of aeroplanes, spaceships and horses.
### A Question of Scale
##### Stage: 4 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts?
### Mach Attack
##### Stage: 5 Challenge Level:
Have you got the Mach knack? Discover the mathematics behind exceeding the sound barrier.
### Pumping the Power
##### Stage: 5 Challenge Level:
What is an AC voltage? How much power does an AC power source supply?
### Approximately Certain
##### Stage: 4 and 5 Challenge Level:
Estimate these curious quantities sufficiently accurately that you can rank them in order of size
### Modelling Assumptions in Mechanics
##### Stage: 5
An article demonstrating mathematically how various physical modelling assumptions affect the solution to the seemingly simple problem of the projectile.
### Cobalt Decay
##### Stage: 5 Challenge Level:
Investigate the effects of the half-lifes of the isotopes of cobalt on the mass of a mystery lump of the element.
### Cannon Balls
##### Stage: 5 Short Challenge Level:
How high will a ball taking a million seconds to fall travel?
### Which Twin Is Older?
##### Stage: 5
A simplified account of special relativity and the twins paradox. | 1,633 | 7,585 | {"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-2018-09 | longest | en | 0.831529 |
http://aztekium.pl/procent.py?lang=en&title=calculating%20percentile | 1,571,627,016,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987751039.81/warc/CC-MAIN-20191021020335-20191021043835-00279.warc.gz | 22,803,554 | 4,124 | calculating percentile
5 percent of 1000
20% of 200
85 percent of 500
5 percent of 3000
55 percent of 1000
9% of 1000
60% of 5000
50% of 2000
50 percent of 1000
8 percent of 3000
2% of 100
4% of 1000
30 percent of 300
6% of 3000
20% of 200
85% of 1000
95 percent of 500
40% of 200
70 percent of 1000
45% of 1000
4% of 300
5 percent of 3000
15% of 100
40 percent of 5000
45% of 3000
5 percent of 300
2% of 5000
1 percent of 200
75% of 200
50 percent of 3000
http://Aztekium.pl/percent
# calculating percentile
A
Percentage of the number of:What is % of =
B
Calculating what percentage of one number is other number:Number is % of
C
Adding percentages to the number:Number + % =
D
Adding percentages to the number:Number + % =
E
Subtract percentages of the number:Number − % =
F
By what percentage has increased or decreased the number of:Number A Number B | 295 | 862 | {"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} | 3.421875 | 3 | CC-MAIN-2019-43 | latest | en | 0.847308 |
https://indigoengine.io/06-presentation/shapes/ | 1,720,921,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00744.warc.gz | 272,025,166 | 4,379 | # Shapes
Shapes allow you to draw simple graphics without needing images.
There are four `Shape` types that come shipped with Indigo at the time of writing, although we may add more in the future:
1. `Shape.Box` - Any rectangle
2. `Shape.Circle` - A circle centered around a point
3. `Shape.Line` - A straight line made of two points
4. `Shape.Polygon` - An arbitrary shape with up to 16 vertices.
They all share similar properties:
• They can all (aside from Line) have a `Fill` that is solid color or a linear or radial gradient.
• They can all have a `Stroke` applied to them with color and thickness.
• They can all be transformed in the usual ways
• They can all receive simple lighting effects (no normal mapping)
One limitation that shape strokes have, is that they are all solid and all use rounded corners. This is because of the way they are calculated. If there is interest we may look at adding more stroke effects like dashed lines or alternate capping types.
Below are examples of each to help get you started.
## Box
``````import indigo.*
Shape.Box(
Rectangle(Point(100, 100), Size(50, 50)),
Fill.Color(RGBA.White),
Stroke(4, RGBA.Blue)
)``````
## Circle
``````Shape.Circle(
center = Point(30, 30),
Point(20),
RGBA.Magenta,
Point.zero,
RGBA.Cyan
),
stroke = Stroke(3, RGBA.White)
)``````
## Line
``Shape.Line(Point(30, 80), Point(100, 20), Stroke(6, RGBA.Cyan))``
## Polygon
``````Shape.Polygon(
Fill.LinearGradient(Point(0), RGBA.Magenta, Point(45), RGBA.Cyan),
Stroke(4, RGBA.Black)
)(
Point(10, 10),
Point(20, 70),
Point(90, 90),
Point(70, 20)
)`````` | 437 | 1,589 | {"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-2024-30 | latest | en | 0.826237 |
http://www.reddit.com/r/cheatatmathhomework/comments/19z3gj/abstract_algebra_infinite_countable_sets/ | 1,438,719,284,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042991951.97/warc/CC-MAIN-20150728002311-00060-ip-10-236-191-2.ec2.internal.warc.gz | 657,474,533 | 15,255 | all 6 comments
[–] 2 points3 points (1 child)
sorry, this has been archived and can no longer be voted on
To prove that the set of arbitrary subsets of A is uncountable, represent a subset S (infinite or not) as an infinite string of 1s and 0s. The ith element is 1 if a_i is in the set, and 0 if it is not.
For example, S = {a_5, a_8, a_10} is represented as 0000100101000..., the empty subset is 000... and A itself is 111...
Suppose there is a claimed enumeration of all subsets {S_1, S_2, ..., S_n, ...}. You can use a diagonalization argument to construct a set S which is not on the list:
Set the first element of S to the opposite of the first element of S_1.
Set the second element of S to the opposite of the second element of S_2.
...
Set the ith element of S to the opposite of the ith element of S_i.
...
By construction, S is not equal to any of the S_i, so the list is incomplete.
[–] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
This is a good way to think about it and answers the first question since finite sets correspond to strings with infinitely many 0s which we can view a binary number. Which tells us that the finite sets can be injectively mapped into the natural numbers via the map sending the sequence (a_1,a_2,...) to [;\sum a_i 2i;]
[–] 1 point2 points (0 children)
sorry, this has been archived and can no longer be voted on
Since a is countable, consider an enumeration {a_1, a_2, ..., a_n, ...}
Consider an arbitrary finite subset S = {a_i_1, a_i_2, ... a_i_k}. Define a function f(S) = i_1 + i_2 + ... + i_k. For example, f({a_5, a_8, a_10}) = 5 + 8 + 10 = 23.
You can then create an enumeration of the finite sets:
• all subsets such that f(S) = 0 (there's only one, S = {})
• all subsets such that f(S) = 1 (there's only one, S = {a_1})
...
• all subsets such that f(S) = k
...
At each level, there are only finitely many possibilities, so the enumeration works.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Prove that the set of all finite subsets of A is countable
To prove that a set is countable, you need to prove that you can count its elements (which is another way of saying that there's a bijection with ℕ).
[Note: Here, I'm using the definition ℕ={1,2,3,...}, i.e. without 0. If you use the definition ℕ={0,1,2,3...}, the proof is pretty much the same.]
Since A is countable, there's a bijection between A and ℕ. Therefore, you can call one of its elements "1", the second "2", etc. In other words, if you prove the property for ℕ, you've proved it for A.
Now, let's count the subsets. The trick is, more often than not, to use finite "blocks". For example:
Take all the subsets of A that contain 1, and no number bigger than 1. There's only one: {1}.
Take all the subsets of A that contain 2, and no number bigger than 2. Sort them alphanumerically (i.e. like in a dictionary): {1,2} and {2}.
Take all the subsets of A that contain 3, and no number bigger than 3. Sort them alphanumerically (i.e. like in a dictionary): {1,2,3}, {2,3} and {2}.
Continue like that. In the end, you have:
#1: {1}
#2: {1,2}
#3: {2}
#4: {1,2,3}
#5: {2,3}
etc.
You've counted the finite subsets. To each number corresponds a subset; to each subset corresponds a number. In other words, there's a bijection between the set of finite subsets and ℕ.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
the set of all subsets of A is not countable.
By using infinite decimal expansions, prove that there's an injection from the interval (0,1) to P(A). Since (0,1) isn't countable, P(A) ain't, either.
If you need more help, my solution is here.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
If you don't want to consider infinite decimals, you can get the result easily with Cantor's theorem. | 1,136 | 3,951 | {"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} | 4.125 | 4 | CC-MAIN-2015-32 | latest | en | 0.918772 |
https://gpuzzles.com/mind-teasers/cross-the-river-mind-teaser/ | 1,493,165,290,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121000.17/warc/CC-MAIN-20170423031201-00544-ip-10-145-167-34.ec2.internal.warc.gz | 815,207,650 | 11,509 | • Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Cross The River Mind Teaser
Difficulty Popularity
A river should be crossed by a father, a mother and their two sons and two daughters.
There are some rules that should be followed while crossing the river. There can be only two people in the raft while crossing. The daughters cannot be with their father unless there is the presence of the mother. The sons cannot be with their mothers unless the father is present. Unless the guard is on the board, the criminals cannot be with any of the family members. Only the adults like the father, the guard, and the mother knows to use the raft.
How will they cross the river?
Discussion
Suggestions
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Logical Arrangement Bonfire Puzzle
Difficulty Popularity
30 classmates went for a bonfire. While sitting around the fire, they decided to play a game. To play it, they divided themselves into 5 teams in a way that there were 5 rows having 7 people each.
How can this arrangement be possible while sitting around the bonfire?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Trick Interesting Maths puzzle
Difficulty Popularity
There are three cars in a racing track. The track is made forming a perfect circle and is quite wide so that at one time, multiple cars can pass through it. The car which is leading is driving at 55 MPH and the slowest car is driving at 45 MPH. The car that is in middle of these two is driving in between the two speeds. For the time being you can say that the distance between the fastest car and the middle car is x miles and it is same between the middle car and the slowest car. Also, x is not equal to 0 or 1.
The car keeps running till the leading car catches up with the slowest car and then every car stops. Given the case, do you think that at any point, the distance between any two pairs will again become x miles?
• Views : 50k+
• Sol Viewed : 20k+
# Mind Teasers : Logic Coins Puzzle
Difficulty Popularity
Assuming i have an infinite supply of coins.
What is the fewest number of coins would be required in order to make sure each and every coin touched exactly three other coins.
• Views : 90k+
• Sol Viewed : 30k+
# Mind Teasers : Popular Interview Problem
Difficulty Popularity
I have two rectangular bars.
They have property such that when you light the fire from one end , it will take exactly 60 seconds to get completely burn.
However they do not burn at consistent speed (i.e it might be possible that 40 percent burn in 55 seconds and next 60 percent can burn in 10 seconds).
Problem is : How do you measure 45 seconds ?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Hardest Chess Problem
Difficulty Popularity
This is a classic chess puzzle and considered to be one of the hardest puzzles in history.
As shown in the picture below, White army is arranged in a classic chess board. You need to add black army i.e
1 king
1 queen
2 rooks
2 bishops
2 knights
8 pawns
in such a way that not a single piece of either color is under attack.
• Views : 80k+
• Sol Viewed : 20k+
# Mind Teasers : Smart Trick Puzzle
Difficulty Popularity
A local magician is performing his favorite trick to be under water for approx 10 minutes.
I went to the magician and told him that i can be under the water for 20 minutes.
He laughed and den responded "I will give you 1000 pounds , if you can do this"
I won the bet and get 1000 pounds.
what did i do ?
• Views : 40k+
• Sol Viewed : 10k+
# Mind Teasers : Logic Interview Riddle
Difficulty Popularity
prinka and tanu organized a small kitty party at home. They purchased a barrel of fruit juice.
Tanu told to prinka 'I bet this barrel of fruit juice is more than half full'.
'No, its less than half full' prinka replied.
I don't have any measuring instrument and without removing fruit juice from it , how can i determine which of us is right ?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Logical Cut Brain Teaser Riddle
Difficulty Popularity
Your friends are coming over to your birthday party. Your mum has bought this delicious giant doughnut for you guys. Unfortunately she had to go somewhere. Now you have to serve all of them including you.
There are a total of nine kids including you. None of you would mind a smaller piece. Can you cut this doughnut into nine pieces in just three straight cuts?
• Views : 60k+
• Sol Viewed : 20k+
# Mind Teasers : Sports Cricket Brain Teaser
Difficulty Popularity
Shihar Dhawan and Virat Kohli were batting on 94.
Seven runs were required to win by the team and only three balls were left.
At the end, both of them secured a century without being out.
How can this be feasible?
• Views : 70k+
• Sol Viewed : 20k+
# Mind Teasers : Popular Dice Probability Riddle
Difficulty Popularity
You roll two six sided dices together.
Can you calculate the probability that the first dice gives the number 2 while the second dice gives the number 5?
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An Indian Langoor, an African Monkey, an... | 1,340 | 5,609 | {"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} | 3.53125 | 4 | CC-MAIN-2017-17 | longest | en | 0.964636 |
https://www.traderji.com/community/threads/trading-tip-16-buyers-or-sellers.69341/ | 1,726,609,629,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00115.warc.gz | 947,207,452 | 14,986 | ##### Member
A question I often receive is, "How can there be more buyers or sellers at one price? Isn't there a buyer for every seller and a seller for every buyer?"
The answer is yes, but people are forgetting one important thing. There is a bid and an ask (or offer), and only one of them can be traded at a time.
A bid is an expression of willingness to buy at a price; an ask (or offer) is an expression to sell.
If the ES is trading at 1200.50, the bid is either 1200.25 or 1200.50. The answer depends on which way the market has just traded. Let's make it easy and simply say the ES is between 1200.25 & 1200.50, making the bid 1200.25. In order for the market to move from 1200.25 to 1200.50, someone must pay up to get filled.
You may not be in a hurry and attempt to wait to buy 1200.25, but that will usually only happen when the bid/ask drops to 1200.00 & 1200.25 and you are actually filled on the ask.
If you are trying to buy and really want to get filled, you must pay up at the offer or risk missing the trade. Conversely, if you really want to get filled on a sale, you must hit the bid, or reach down to get filled.
Sure, there is someone on the other side of the trade, but without you choosing to reach up and pay the offer the market stands still. Therefore when trades are executed at the offer it is said to be done by the buyers even though there are sellers at that price taking the other side.
Every buy will be filled on the offer and every sell will be filled on the bid, period.
Let's say we once more have a number of 1200.50 and we see that over time (sometimes just a few seconds) the fills were 100 x 1300. We can say that there were 1200 more buyers than sellers at 1200.50 because of how traders reacted to the bid/ask spread when it was at 1200.25 x 1200.50 and higher at 1200.50 x 1200.75 (called the spread.)
When the market was at the lower spread, 1300 buyers reached UP to pay the 1200.50 offer.
When the market was at the higher spread, 100 sellers reach DOWN to sell the 1200.50 bid.
When the spread traded around this price range there truly were more buyers than sellers at 1200.50.
Understanding bid and ask can open up other realms of technical analysis.
There are some traders who will look at the bid and ask order flows to try to get clues to potential movement in the market based on what buyers and sellers are doing. This is often referred to as reading order book flow or depth-of-market.
If you look at the number of orders for each bid and ask around the current market price you can see the probable number of transactions available at those levels. Reading this information is the key to certain kinds of volume based trading systems and other trading methods that follow the book order flow.
Larry Levin
Last edited:
#### kailashkumar
##### New Member
nice post but give me some example:thumb:
#### kailashkumar
##### New Member
nice but give me some explain | 719 | 2,935 | {"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 | 3 | CC-MAIN-2024-38 | latest | en | 0.970481 |
http://www.shmoop.com/points-vectors-functions/polar-theta-bounds-help.html | 1,472,449,064,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982952852.53/warc/CC-MAIN-20160823200912-00079-ip-10-153-172-175.ec2.internal.warc.gz | 686,530,976 | 11,234 | # At a Glance - Bounds on Theta
Polar function plots may leave your head spinning like you got off the Tilt-a-Whirl ride at the amusement park. But using them, it's possible to model that Tilt-a-Whirl ride, and they make some of the sweetest looking plots.
We need to place some bounds on the number of times the plot goes around the origin. These things are complicated; we will still need a calculator to help us plot. Think of it like we were placing a polar bear on a leash. You can try to tame it, but it may take you for a walk.
Sometimes we need to know which bounds on θ give a particular piece of a polar graph. The easiest way to find these bounds is to graph the function on the calculator and play with the bounds until we find the right piece of the graph.
In the example, we need to be careful to find the petal shown in the graph. Other choices for bounds might give us a single petal, but the wrong petal. If we take
we find one petal, but not the one asked for:
Depending on the problem, it might be important to find the specific petal mentioned.
It can also be important to find bounds on θ that trace out the graph exactly once.
### Sample Problem
The graph of r = cos θ for 0 ≤ θ ≤ 2π looks like this:
However, the graph of r = cos θ for 0 ≤ θ ≤ π looks the same.
If asked what values of θ are needed to describe the whole graph of r = cos θ, we would take the narrower bounds:
0 ≤ θ ≤ π.
When asked what bounds on θ give a particular portion of the graph, there are multiple correct answers. We know that r = cos(2θ) for looks like this:
If instead we take , we find the same piece of the graph:
How do we know if we've found correct bounds for θ? Put them in the calculator and draw the graph. As long as we find the correct portion of the graph, and the calculator traces it only once, then we've found correct bounds for θ.
#### Example 1
What bounds on θ produce the single petal of the graphr = cos(2θ)shown below?
#### Exercise 1
Find bounds on θ that trace out the specified piece of function only once.
#### Exercise 2
Find bounds on θ that trace out the specified piece of function only once.
#### Exercise 3
Find bounds on θ that trace out the specified piece of function only once. | 536 | 2,239 | {"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-2016-36 | latest | en | 0.927202 |
http://pennsylvania.car4car.top/adding-and-subtracting-fractions-worksheet-for-grade-4/ | 1,600,471,007,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400189264.5/warc/CC-MAIN-20200918221856-20200919011856-00490.warc.gz | 105,296,229 | 29,612 | In Free Printable Worksheets252 views
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http://scootle.edu.au/ec/search?topic=%22Geometry%22&browseBy=topic | 1,600,957,178,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00118.warc.gz | 116,733,643 | 14,259 | Close message Scootle will stop supporting resources that use the Adobe Flash plug-in from 18 Dec 2020. Learning paths that include these resources will have alerts to notify teachers and students that one or more of the resources will be unavailable. Click here for more info.
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### TIMES Module 20: Measurement and Geometry: parallelograms and rectangles - teacher guide
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### Delivering water
This is a mathematics unit of work about water: its world-wide availability and use; the time spent carrying it; the best shape for water tanks; and the area of land taken up around tanks and in paths and ditches to water sources. Intended for years 9 and 10 and written from a global education perspective, the resource ...
### Secondary mathematics: different representations
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### Measures: scaling down
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### Measures: scaling up solids
Compare the volumes of cubes and rectangular prisms before and after being scaled up (enlarged). Notice that 'similar solids' in the mathematical sense have the same shape but different volumes. Explore the relationship between side length and volume when scaling up solids. This learning object is the fifth in a series ...
### Measures: volumes
Compare the volumes of a range of rectangular prisms when scaling up (enlarging) side lengths by different ratios. Notice that the rectangular prisms are not similar in the mathematical sense, but it is possible to predict the effect on the volume produced by the scaling of the sides. Identify and describe the relationship ...
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### Exploring relationships of angles
Explore angles formed by a transversal line intersecting parallel lines. Look at illustrations showing pairs of angles: vertically opposite, corresponding and alternate angles. Name pairs of angles to score points and help a monkey drive to the supermarket to buy food.
### TIMES Module 22: Measurement and Geometry: introductory trigonometry - teacher guide
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### TIMES Module 9: Measurement and Geometry: introduction to plane geometry - teacher guide
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### TIMES Module 17: Measurement and Geometry: the circle - teacher guide
This is a 15-page guide for teachers. In the module the formulas for finding the circumference and area of a circle are introduced. The history and significance of the number pi is also included in this module. | 1,395 | 6,936 | {"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} | 3.265625 | 3 | CC-MAIN-2020-40 | latest | en | 0.870613 |
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Explore BrainMass
Simplify the Equation
Rewrite the rational expression on the left with the given new denominator.
28a/a*2 - 2a - 35 = ?/(a-7) (a + 5) (a+9)
What is the new numerator?
\$2.19 | 66 | 202 | {"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-2016-44 | latest | en | 0.703894 |
https://ask.libreoffice.org/t/if-row-contains-the-correct-month-then-add-amount-in-row-b-to-total-otherwise-dont/29381 | 1,675,183,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00298.warc.gz | 125,655,875 | 5,499 | # If row contains the correct month then add amount in row B to total. Otherwise don't
I know someone will have the answer to this, but I don’t know if I’ll make a hash of explaining it!.
Here goes …
I have a spreadsheet with my inventory of books that I sell, and I have the date sold in there as well as cost price, sale price and profit, etc - but I’m looking to create another sheet which will give me running totals on a month to month basis on my profit, preferably with a total of books sold too. So, if I have sheet 1 showing the date sold in column “B”, quantity in column “C” and the profit in column “D”, I’d like to have sheet 2 to give me totals, so for example, if I had 30 entries (rows) on sheet 1 showing I have sold 30 books in August, in sheet 2 I’d like it to show the total items sold (i.e. calculates totals in column “C” where the month “August” has been used in column “B”, and totals in column “D” where the month “August” has been used in column “B”. The same would be applied for the other months but once I have the formula for one month I’d be able to work out the rest.
Hope this all makes sense and look forward to the replies
All the best
Mark
Hopefully you will agree:
I changed the subject slightly. In common interpretation something like `"month"` is interpreted as meaning the literal string. However, you were talking of the month as a part of the date.
As I assume you enter the dates as usual, they will in addition be numeric. The month August should then also not be represented by the text “August” but by the number 8. This holds even if you formatted cells containg dates to show names of months.by `MMMM` format code.
Use SUMIFS() or SUMPRODUCT.
``````=SUMIFS(AmountsRange;MONTH(DateRange);ReferenceToCellContainingTheMonthNumber)
e.g. shown after entering for array-evaluation with Ctrl+Shift+Enter as:
{=SUMIFS(\$Sheet1.\$D\$2:\$D\$1001;MONTH(\$Sheet1.\$B\$2:\$B\$1001);\$F\$1)}
``````
See also this attached example.
(BTW: The variant using SUMPRODUCT() should even work if not explicitly entered for array-evaluation. It doesn’t, however. a wee little bug.)
Thank you for your answer, I did have a little trouble using your original formula and I’ve absolutely no doubt whatsoever it was me doing something wrong but strangely I managed to get the SUMPRODUCT working in the following way:
=SUMPRODUCT(MONTH(Books.C12:C1200)=8,Books.R12:R1200)
Where “Books” is the name of the sheet, “C” is the column for the dates, “8” is the month and R is the column for the amounts
Thanks again for your help, muchly appreciated
Mark | 642 | 2,583 | {"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-2023-06 | latest | en | 0.932289 |
https://metanumbers.com/3898 | 1,670,084,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00107.warc.gz | 437,278,879 | 7,246 | # 3898 (number)
3,898 (three thousand eight hundred ninety-eight) is an even four-digits composite number following 3897 and preceding 3899. In scientific notation, it is written as 3.898 × 103. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 1,948 positive integers (up to 3898) that are relatively prime to 3898.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 28
• Digital Root 1
## Name
Short name 3 thousand 898 three thousand eight hundred ninety-eight
## Notation
Scientific notation 3.898 × 103 3.898 × 103
## Prime Factorization of 3898
Prime Factorization 2 × 1949
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 3898 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,898 is 2 × 1949. Since it has a total of 2 prime factors, 3,898 is a composite number.
## Divisors of 3898
1, 2, 1949, 3898
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 5850 Sum of all the positive divisors of n s(n) 1952 Sum of the proper positive divisors of n A(n) 1462.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 62.434 Returns the nth root of the product of n divisors H(n) 2.6653 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,898 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 3,898) is 5,850, the average is 146,2.5.
## Other Arithmetic Functions (n = 3898)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1948 Total number of positive integers not greater than n that are coprime to n λ(n) 1948 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 543 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 1,948 positive integers (less than 3,898) that are coprime with 3,898. And there are approximately 543 prime numbers less than or equal to 3,898.
## Divisibility of 3898
m n mod m 2 3 4 5 6 7 8 9 0 1 2 3 4 6 2 1
The number 3,898 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion (3898)
Base System Value
2 Binary 111100111010
3 Ternary 12100101
4 Quaternary 330322
5 Quinary 111043
6 Senary 30014
8 Octal 7472
10 Decimal 3898
12 Duodecimal 230a
20 Vigesimal 9ei
36 Base36 30a
## Basic calculations (n = 3898)
### Multiplication
n×y
n×2 7796 11694 15592 19490
### Division
n÷y
n÷2 1949 1299.33 974.5 779.6
### Exponentiation
ny
n2 15194404 59227786792 230869912915216 899930920543511968
### Nth Root
y√n
2√n 62.434 15.7379 7.90152 5.22599
## 3898 as geometric shapes
### Circle
Diameter 7796 24491.9 4.77346e+07
### Sphere
Volume 2.48093e+11 1.90939e+08 24491.9
### Square
Length = n
Perimeter 15592 1.51944e+07 5512.6
### Cube
Length = n
Surface area 9.11664e+07 5.92278e+10 6751.53
### Equilateral Triangle
Length = n
Perimeter 11694 6.57937e+06 3375.77
### Triangular Pyramid
Length = n
Surface area 2.63175e+07 6.98006e+09 3182.7
## Cryptographic Hash Functions
md5 37db6bb5f1db992df92a919d20757eec 5383ce5f236412891d46ed5ed5c4992456b6a7b3 7f80e6caa06901e1173b174d213cd3341725b057108bda4013575b566ca5b211 dfbe2d73afdde84a1624be46d3a94004bf5cd88e9565dc8ed0b1a08eac575c069a10fae3fccc8e9fe922c7bd4338e1dea03e0c297064ef6d6a809d45462a3680 5bd01e9db1369cf42cc69f3b1f1d2d81fa3bade9 | 1,424 | 3,982 | {"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-2022-49 | latest | en | 0.82383 |
https://www.coursehero.com/file/6217304/Chapter5-B/ | 1,529,332,934,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860557.7/warc/CC-MAIN-20180618125242-20180618145242-00389.warc.gz | 792,520,475 | 42,649 | Chapter5_B
# Chapter5_B - Chapter 5 Section B Non-Numerical Solutions...
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Chapter 5 - Section B - Non-Numerical Solutions 5.1 Shown to the right is a PV diagram with two adi- abatic lines 1 2 and 2 3, assumed to inter- sect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce work. For the cycle U = 0, and therefore by the first law, Q + W = 0. Since W is negative, Q must be positive, indicating that heat is absorbed by the system. The net result is therefore a complete conversion of heat taken in by a cyclic process into work, in violation of Statement 1a of the second law (Pg. 160). The assumption of intersecting adiabatic lines is there- fore false. 5.5 The energy balance for the over-all process is written: Q = U t + E K + E P Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is negative, and by the preceding equation, so is Q . Thus heat is transferred to the surroundings. The total entropy change of the process is: S total = S t + S t surr Just as U t for the egg is zero, so is S t . Therefore, S total = S t surr = Q surr T σ = Q T σ Since Q is negative, S total is positive, and the process is irreversible. 5.6 By Eq. (5.8) the thermal efficiency of a Carnot engine is: η = 1 T C T H Differentiate: ∂η T C T H = − 1 T H and ∂η T H T C = T C T H 2 = T C T H 1 T H Since T C / T H is less unity, the efficiency changes more rapidly with T C than with T H . So in theory it is more effective to decrease T C . In practice, however, T C is fixed by the environment, and is not subject to control. The practical way to increase η is to increase T H . Of course, there are limits to this too. 5.11 For an ideal gas with constant heat capacities, and for the changes T 1 T 2 and P 1 P 2 , Eq. (5.14) can be rewritten as: S = C P ln T 2 T 1 R ln P 2 P 1 ( a ) If P 2 = P 1 , S P = C P ln T 2 T 1 If V 2 = V 1 , P 2 P 1 = T 2 T 1 Whence, S V = C P ln T 2 T 1 R ln T 2 T 1 = C V ln T 2 T 1 642
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Since C P > C V , this demonstrates that S P > S V . ( b ) If T 2 = T 1 , S T = − R ln P 2 P 1 If V 2 = V 1 , T 2 T 1 = P 2 P 1 Whence, S V = C P ln P 2 P 1 R ln P 2 P 1 = C V ln P 2 P 1 This demonstrates that the signs for S T and S V are opposite. 5.12 Start with the equation just preceding Eq. (5.14) on p. 170: dS R = C ig P R dT T d ln P = C ig P R dT T d P P For an ideal gas PV = RT , and ln P + ln V = ln R + ln T . Therefore, d P P + dV V = dT T or d P P = dT T dV V Whence, dS R = C ig P R dT T dT T + dV V = C ig P R 1 dT T + d ln V Because ( C ig P / R ) 1 = C ig V / R , this reduces to: dS R = C ig V R dT T + d ln V Integration yields: S R = T T 0 C ig V R dT T + ln V V 0 * * * * * * * * * * * * * * * * * * * * * * As an additional part of the problem, one could ask for the following proof, valid for constant heat capacities. Return to the original equation and substitute dT / T = d P / P + dV / V : dS R = C ig P R d P P + C ig P R dV V d P P = C ig V R d P P + C ig P R dV V Integration yields: S R = C ig V R ln P P 0 + C ig P R ln V V 0 5.13 As indicated in the problem statement the basic differential equations are: dW dQ H dQ C = 0 ( A ) dQ H dQ C = − T H
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Jill Tulane University ‘16, Course Hero Intern | 1,361 | 4,511 | {"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} | 3.625 | 4 | CC-MAIN-2018-26 | latest | en | 0.938063 |
http://idevgames.com/forums/printthread.php?tid=4949 | 1,516,411,743,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888341.28/warc/CC-MAIN-20180120004001-20180120024001-00411.warc.gz | 167,861,929 | 4,539 | Collision Response Question - Printable Version +- iDevGames Forums (http://www.idevgames.com/forums) +-- Forum: Development Zone (/forum-3.html) +--- Forum: Game Programming Fundamentals (/forum-7.html) +--- Thread: Collision Response Question (/thread-4949.html) Pages: 1 2 3 Collision Response Question - blobbo - Oct 2, 2005 03:05 PM Ok, so I need some help with the math of a collision response. The detection part is finished and works properly - but that's the easy part. The following is a diagram: Basically the ball has a direction vector of Vector A, and it's going to bounce off of the inside of the circle. I understand that it needs to reflect off the vector drawn between the large circle's origin and the point of collision (Vector B), thus becoming Vector C. I think Vector D gets involved somewhere in the calculations - that's why I've included it. What's the easiest way to do this? Seems a bit complicated, but I'm sure there's an easy solution to this one... Thanks in advance. Collision Response Question - OneSadCookie - Oct 2, 2005 03:14 PM http://mathworld.wolfram.com/Reflection.html Collision Response Question - unknown - Oct 2, 2005 03:21 PM maybe try and implement this with the new rational trigonometry, http://web.maths.unsw.edu.au/~norman/papers/Chapter1.pdf Just a thought, it might make things easier. Collision Response Question - willThimbleby - Oct 2, 2005 03:39 PM Or more succinctly: perpendicular = (A • B) × B parallel = A – perpendicular C = parallel × friction + perpendicular × –bounce where friction and bounce > 0 and < 1 and B is a unit vector Collision Response Question - blobbo - Oct 2, 2005 08:12 PM Ok, the Mathematica page looks great - but very confusing to someone who hasn't taken math courses in 6 years. Will: that explanation looks decent, but I have to wrap my mind around it for a bit. It's important to me that I *understand* this - makes debugging easier. Assuming there is no friction and the bounce is constant, would I use values of 1.0 for each? And when you suggest to multiply floats by vectors, do you mean simply to multiply the value by both elements of the vector? Excuse my math skills, or lack thereof. It's been a long time since high school math, and I find this all a bit confusing. Collision Response Question - willThimbleby - Oct 3, 2005 01:34 AM Yes when you multiply vectors by floats just multiply both elements of the vector. A bounce of 1 and the ball never stops bouncing. A bounce greater than 1 and it gets exponentially faster until something goes wrong. The maths above splits the balls velocity vector into two orthogonal components, one along the vector B perpendicular (pp) to the collision point, and one along the vector D parallel (p) to the collision point. It is easy to then combine these back to get the reflection vector. Collision Response Question - blobbo - Oct 3, 2005 07:02 AM Wow, thanks Will! Great diagram - what did you use to make it? I grow quickly tired of Sketch.app! I'll try to implement that tonight. Collision Response Question - Andrew - Oct 3, 2005 08:08 AM I was just reading on this page that the cross product can only be computed for 3D vectors. Since this is 2D space, what does (A • B) × B mean (how do you calculate it)? Collision Response Question - unknown - Oct 3, 2005 08:17 AM A dot B the dot product is a.x*b.x + a.y*b.y in 3D a.x*b.x + a.y*b.y + a.z*b.z an important property is that a.b/|ab| = cos (the angle between a and b) Collision Response Question - Andrew - Oct 3, 2005 08:37 AM unknown Wrote:A dot B the dot product is a.x*b.x + a.y*b.y in 3D a.x*b.x + a.y*b.y + a.z*b.z an important property is that a.b/|ab| = cos (the angle between a and b) But what about the cross product between the resulting scalar and B? Do I just multiply each of B's components by the scalar? For example, say the dot product turned out to be 3. Would I just stretch each of B's components by a factor of 3? Collision Response Question - unknown - Oct 3, 2005 09:45 AM Quote:But what about the cross product between the resulting scalar and B? You can only do the cross product in 3D. Quote:Would I just stretch each of B's components by a factor of 3? That would give you (A • B) x B. Collision Response Question - willThimbleby - Oct 3, 2005 10:48 AM Technically it is not a cross product. Cross-products are only between vectors. It is just a multiply. So yes just multiply each of B's components by the scalar. So (A • B) × B is just: B scaled to the magnitude of the dot product of A and B. Quote:Wow, thanks Will! Great diagram - what did you use to make it? I grow quickly tired of Sketch.app! thanks. I used Imprint, my new-as-yet-unreleased drawing program. http://will.thimbleby.net/draw If you are interested in beta-testing let me know. Collision Response Question - Andrew - Oct 3, 2005 10:31 PM my brother just told me it's probably better to write it like (A•B)B in order to avoid confusion Collision Response Question - unknown - Oct 4, 2005 05:06 AM Well the x means multiply, and you can also write it like that or * or •, the last one is probably quite confusing if your using it for dot product as well! Collision Response Question - blobbo - Oct 5, 2005 09:19 PM ARIGHT! It works! Thanks so much, Will! Code: PHP Code: `/* COLLISION RESPONSEGeneral mathematical idea:define vector B as vector from origin through point of collisionD = (A • B) × Bparallel = A – DC = parallel + D × –bounce */ // Define 1 temporary vector for calculations ...vector A = [ball direction];// Calculate the point where the ball is colliding with the play areavector pointOfCollision = (vector){ ([ball position].x + (A.x * [ball radius])), ([ball position].y + (A.y * [ball radius]))};// ... continue by defining 3 more temporary Vectors ...vector B = (vector){ (pointOfCollision.x - 250.0), (pointOfCollision.y - 250.0) };B = vectorNormalise(B);vector D = vectorMultiply(B, vectorDotProduct(A, B));vector parallel = vectorSubtract(A, D);// ... and finish by calculating the last "temporary" vector, which is the resulting vectorvector C = vectorAdd(parallel, vectorMultiply(D, -BOUNCE));C = vectorNormalise(C);// Apply the change to the ball object's direction vector[ball setDirection:C]; ` Binary (): http://www.owlnet.rice.edu/~ajm4979/BG.zip | 1,748 | 6,337 | {"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-2018-05 | latest | en | 0.902776 |
https://www.oreilly.com/library/view/signals-and-systems/9789332515147/xhtml/chapter005.xhtml | 1,571,248,822,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986669057.0/warc/CC-MAIN-20191016163146-20191016190646-00546.warc.gz | 1,045,733,172 | 8,364 | ## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
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## System Modelling
Since the cause and effect relationship between input and output of a system exists, we can define transfer function of the system mathematically which generally characterizes the input-output relationship of the system also. We can represent a system by its mathematical modelling. Therefore, it is possible to represent a system by mathematical equations. Block diagram and signal flow graph are useful to represent the system completely. Using the block diagram reduction technique or Mason’s gain formula in signal flow graph, we can get the overall transfer function of the system very easily. The aim of this chapter is to discuss the following:
• Transfer function of the system,
• block diagram representation,
• signal flow graph, ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required | 214 | 1,125 | {"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} | 2.5625 | 3 | CC-MAIN-2019-43 | latest | en | 0.897829 |
https://www.atozexams.com/test/aptitude/problems-on-ages.html | 1,563,388,214,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525374.43/warc/CC-MAIN-20190717181736-20190717203736-00087.warc.gz | 628,506,577 | 12,488 | Aptitude - Problems On Ages Test
Time Left : 00 : 30 : 00
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
A person's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present?
Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar's age at present?
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q's age?
Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?
John is 20 years younger than Tom. In two years Tom will be twice as old as John. How old are they now?
Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
5 years ago the average age of 4 members was 46 years. A new person joins them now and the average age of all the five members is 50 years. Then the age of new member is
Ayesha's father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years)
The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, son's age will be:
The age of father 10 years ago was thrice the age of his son. Ten years hence, father's age will be twice that of his son. The ratio of their present ages is:
A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:
At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun's age will be 26 years. What is the age of Deepak at present ?
Phil is Jack's father and Phil is 35 years old. Three years ago, Phil was four times as old as his son was then, How old is Jack now?
Note: | 750 | 2,785 | {"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} | 3.890625 | 4 | CC-MAIN-2019-30 | longest | en | 0.99269 |
http://oeis.org/A107639 | 1,597,040,672,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738609.73/warc/CC-MAIN-20200810042140-20200810072140-00021.warc.gz | 77,149,107 | 3,637 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A107639 Smallest prime a such that (a*b)^2 + a*b -1 is prime with b prime = 2^(2*n) - 2^n - 1, see A098845 for n. 0
5, 7, 31, 11, 5, 3, 809, 661, 139, 73, 521, 701, 131, 487, 89, 211, 733, 3011, 311, 8837, 2837, 8831, 1733, 401, 983, 17881, 1783, 19553, 13859 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS EXAMPLE 2^4-2^2-1=3=b(1) prime, (3*3)^2+3*3-1=89 prime so a(1)=3 2^8-2^4-1=239=b(2) prime, (7*239)^2+7*239-1=2800601 prime so a(2)=7 CROSSREFS Cf. A098845. Sequence in context: A081630 A307532 A135324 * A069688 A158323 A025119 Adjacent sequences: A107636 A107637 A107638 * A107640 A107641 A107642 KEYWORD hard,more,nonn AUTHOR Pierre CAMI, Jun 10 2005 STATUS approved
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Last modified August 10 02:24 EDT 2020. Contains 336365 sequences. (Running on oeis4.) | 454 | 1,224 | {"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} | 2.640625 | 3 | CC-MAIN-2020-34 | latest | en | 0.613667 |
https://www.transum.org/software/SW/Starter_of_the_day/starter_November27.asp | 1,642,356,629,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00413.warc.gz | 1,090,811,953 | 7,594 | ## A Maths Starter of The Day
Memorise:
• 7
• 16
• 25
• 34
• 43
• 16
• 20
• 24
• 28
• 32
• 11
• 18
• 25
• 32
• 39
There will be a quick quiz in a few minutes:
Test Your Memory
1. What was the largest number?
2. What number is to the left of the 43 in the top row?
3. What number is directly above the 18 in the bottom row?
4. Starting from the 18 in the bottom row. What number is two up and two to the right?
5. What is the difference between the numbers above and below the 16 in the middle row?
Answers
1. What was the largest number? 43
2. What number is to the left of the 43 in the top row? 34
3. What number is directly above the 18 in the bottom row? 20
4. Starting from the 18 in the bottom row. What number is two up and two to the right? 34
5. What is the difference between the numbers above and below the 16 in the middle row? 4
• 7
• 16
• 25
• 34
• 43
• 16
• 20
• 24
• 28
• 32
• 11
• 18
• 25
• 32
• 39
• Transum,
•
• Think of this starter as an exercise in remembering number patterns rather than trying to remember 15 numbers which is very difficult. Please let us know if you think this starter is too difficult!
• Mr J, Essex
•
• I always give the pupils a chance to memorise three rows by writing only 2 bits of information for each row. this forces them to write the start number and the difference between each number. Makes a nice intro to sequences and the term to term rule.
How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.
Click here to enter your comments.
Previous Day | This starter is for 27 November | Next Day
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Teacher, do your students have access to computers?Do they have iPads or Laptops in Lessons? Whether your students each have a TabletPC, a Surface or a Mac, this activity lends itself to eLearning (Engaged Learning).
Here a concise URL for a version of this page without the comments.
Transum.org/go/?Start=November27
Here is the URL which will take them to a related student activity.
Transum.org/go/?to=NumJig
For Students:
For All:
©1997-2022 WWW.TRANSUM.ORG | 852 | 3,457 | {"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} | 3.875 | 4 | CC-MAIN-2022-05 | latest | en | 0.934085 |
http://www.drumtom.com/q/x-3y-2-y-x-2-note-that-you-can-also-answer-no-solution-or-infinitely-many-solutions | 1,488,095,860,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171936.32/warc/CC-MAIN-20170219104611-00512-ip-10-171-10-108.ec2.internal.warc.gz | 387,473,913 | 8,749 | # X-3y=2 y=-x-2 Note that you can also answer "No solution" or "Infinitely many" solutions.?
• X-3y=2 y=-x-2 Note that you can also answer "No solution" or "Infinitely many" solutions.?
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solve systems of linear equations. by user. on 25 марта 2016. Category | 212 | 852 | {"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-2017-09 | longest | en | 0.912486 |
https://s7097.gridserver.com/46pjx/47bbd9-algebraic-topology-prerequisites | 1,656,386,119,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103347800.25/warc/CC-MAIN-20220628020322-20220628050322-00269.warc.gz | 567,502,431 | 7,692 | A certain deal of mathematical maturity is also needed; you should be comfortable in reading and writing rigorous proofs. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introductory video of the course on Introduction to Algebraic Topology. It only takes a minute to sign up. Knowledge is your reward. If you want to go into algebraic topology, you'll want to have a firm grasp of topol. The aim of the book is to introduce advanced undergraduate and graduate (master's) students to basic tools, concepts and results of algebraic topology. But my biggest advice is not worry about taking the course as quickly, if you don't feel safe. Can someone just forcefully take over a public company for its market price? Prerequisites are standard point set topology (as recalled in the first chapter), elementary algebraic notions (modules, tensor product), and some terminology from ca Math 215A: Algebraic Topology Xf(2s 1);p. Yg(1)) 1=2 s 1 g(2s 1) 1 s 1 = ( g(2s) 0 s 1=2 (p. X(f(2s 1);y. Use MathJax to format equations. What's in the Book? Ideas and tools from algebraic topology have become more and more important in computational and applied areas of mathematics. You should know the basics of point-set topology. The pre-requisites for an introductory algebraic topology course are a course in abstract algebra and general topology. Continue Reading. Prerequisites are standard point set topology (as recalled in the first chapter), elementary algebraic notions (modules, tensor product), and some terminology from category theory. Hatcher Algebraic Topology: I have all the prereqs, so why is this book unreadable for me? Leads To: MA4A5 Algebraic Geometry, MA5Q6 Graduate Algebra. In addition, PhD candidates must take Algebraic Topology (110.615) and Riemannian Geometry (110.645) by their second year. Preparing for “differential forms in algebraic topology”? There's no signup, and no start or end dates. Algebraic topology is a branch of mathematics that uses tools from abstract algebra to study topological spaces. Algebraic topology is the study of topological spaces using tools of an algebraic nature, such as homology groups, cohomology groups and homotopy groups. 168 views In addition to formal prerequisites, we will use a number of notions and concepts without much explanation. If these books are too brief books like the schaums one or. Courses More importantly, I wanted to know if the first chapter of the book Topology, Geometry and Gauge Fields by Naber or first 2 chapters of Lee's Topological Manifolds would be sufficient to provide me the necessary background for Hatcher. I was physicist. I was bitten by a kitten not even a month old, what should I do? The recommended prerequisites are commutative algebra at the level of Math 2510-2520, including familiarity with rings and modules, tensor product and localization, various ⦠You should also be familiar with abelian groups and at least be modestly familiar with abstract (non-abelian) groups up to quotient groups. I am doing a full masters level course of groups and rings, so I am pretty sure, I will have the algebra prerequisites. I think that all the point-set topology we will need (and a lot more) is reviewed in Bredon, Chapter I, Sections 1-13. This is an advanced undergraduate or beginning graduate course in algebraic topology. Have you read bout the hopf bundle?P.S: +1 for your answer. These days it is even showing up in applied mathematics, with topological data analysis becoming a larger field every year. What is the precise legal meaning of "electors" being "appointed"? In addition to formal prerequisites, we will use a number of notions and concepts without much explanation. These video lectures (syllabus here) follow Hatcher & I found the very little I've seen useful mainly for the motivation the guy gives. Why is it impossible to measure position and momentum at the same time with arbitrary precision? It depends. Does my concept for light speed travel pass the "handwave test"? There are no exams. Home The first-year algebra and analysis requirement can be satisfied by passing the corresponding written qualifying exam in September of the first year; these students must complete at least two courses each semester. Several sections include topics which have not appeared before in textbooks as well as simplified proofs for some important results. However, the more familiarity you have with algebra and topology, the easier this course will be. Prerequisites. Other than a little "mathematical maturity" there's not very many hard formal prerequisites for studying from Hatcher. My new job came with a pay raise that is being rescinded. Find materials for this course in the pages linked along the left. Algebraic topology Allen Hatcher. Topics covered include: singular homology; cell complexes and cellular homology; the Eilenberg-Steenrod axioms; cohomology; Along the way we will introduce the basics of homological algebra and category theory. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Our course will primarily use Chapters 0, 1, 2, and 3. Topics will include: simplicial, singular, and cellular homology; axiomatic descriptions of homology; cohomology, and cross and cup products; Universal coefficient and Künneth theorems; and Poincaré, Lefschetz, and Alexander duality. At the very least, a strong background from Math 120. I think that chapter 1 is good for you, is an intuitive approach for set-theory, since you are a physicist probably not like going too deeply into sets, but if you dont have time, skip it. Cables to serve a NEMA 10-30 socket for dryer Stack Exchange Inc ; user contributions licensed cc... My biggest advice is not worry about taking the course was conducted online of. In Chapters 2 and 3 privacy policy and cookie policy to measure position and momentum at the very,. Of differential topology, and cohomology operations find out more algebraic topology prerequisites to others. I did read part of the course on introduction to algebraic topology II » Syllabus, Lectures: sessions. The main prerequisite for this part of Nabers first chapter including the monopoles! Should I do, but not finding the time to do so much explanation strong background Math! Just wondering whether you read the section of Hopf Bundle, but not finding time... The fundamental group, and some basics of manifold topology ) to have a grasp... From Math 120 is to find out more or to teach others mathematical maturity is needed. More than 2,400 courses available, OCW is delivering on the promise open. A certain deal of mathematical maturity '' there 's no signup, and 3 and. References or personal experience read the section of Hopf Bundle, but not finding time... Universities one of over 2,200 courses on OCW professor skipped me on bonus. Hatcher algebraic topology from Munkres is actually used in the first two Chapters of Hatcher corresponds to chapter of... That you are intimately familiar with the subspace topology and the quotient topology or to teach others topology and... Front, you agree to our Creative Commons License and other terms of use: MA4A5 algebraic,... Lectures: 3 sessions / week, 1, 2, and.. Our Creative Commons License and other terms of use be written in a list containing both universities of... Materials is subject to our terms of service, privacy policy and policy! Exchange algebraic topology prerequisites ; user contributions licensed under cc by-sa will use a of... Which have not appeared before in textbooks as well as simplified proofs for some important results free download here Munkres... Company for its market price the lives of 3,100 Americans in a day... Advanced undergraduate or beginning graduate course in algebraic topology, and mathematical maturity.. At your own pace opinion ; back them up with references or personal experience prerequisites: the main for. Particular importance in homology Post your answer ”, you agree to our Creative Commons License other! Is it impossible to measure position and momentum at the Table of Contents and the quotient.... At least be modestly familiar with abelian groups and at least be modestly familiar point-set. Provided to cover details not discussed in lecture be modestly familiar with abelian groups and at be. The very least, a strong background from Math 120 v. Pennsylvania lawsuit supposed! To cite OCW as the source policy and cookie policy with a raise! Algebra ) courses available, OCW is delivering on the promise of open of! A pay raise that is being rescinded came with a Degree in mathematics of topology, you want... Be modestly familiar with the subspace topology and Geometry ” link to the download page the introductory chapter of?. The download page graduate-level textbook on algebraic topology, e.g no start or end dates: I all. This semester, we 'll cover Chapters 0-2 of the three or four basic graduate... ( and some basics of manifold topology with abstract ( non-abelian ) groups up to homotopy.... 3 sessions / week algebraic topology prerequisites 1, 2, and reuse ( just remember cite... Algebra ) signup, and some basics of manifold topology of algebraic topology ” topology homological... Lawsuit is supposed to reverse the election to download it in electronic form, follow this link the... electors '' being appointed '' of MIT courses, algebraic topology prerequisites the entire MIT curriculum Exchange a! Is a prerequisite at least be modestly familiar with the subspace topology and Geometry ” work, boss not. “ topology and the last half of the three or four basic graduate!, covering the entire MIT curriculum prerequisites, I agree formal requirements are some basic algebra, point-set topology algebraic! Day in American history, point-set topology, the more familiarity you have with algebra and general.... Advice is not worry about taking the course as quickly, if you want to familiar... Momentum at the Table of Contents and the last half of the COVID-19 pandemic topology ( 110.615 ) Riemannian! Our Creative Commons License and other terms of use this semester, we 'll Chapters. Few days, before I award you the bounty feel safe you read the introductory chapter of Naber job with. Cover the fundamental group, and cohomology operations ( 110.645 ) by their year... | 2,227 | 10,487 | {"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} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.933758 |
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# Tutor profile: Siv C.
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Value of x when 10/(1+10^-x)=2?
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When secx+tanx=3. Then value of tanx?
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Contact Siv | 197 | 659 | {"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} | 2.515625 | 3 | CC-MAIN-2020-16 | longest | en | 0.667922 |
https://www.aeon-toolkit.org/en/latest/api_reference/auto_generated/aeon.transformations.series.boxcox.BoxCoxTransformer.html | 1,701,271,781,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00823.warc.gz | 724,094,132 | 17,963 | # BoxCoxTransformer#
class BoxCoxTransformer(bounds=None, method='mle', sp=None)[source]#
Box-Cox power transform.
Box-Cox transformation is a power transformation that is used to make data more normally distributed and stabilize its variance based on the hyperparameter lambda. [1]
The BoxCoxTransformer solves for the lambda parameter used in the Box-Cox transformation given method, the optimization approach, and input data provided to fit. The use of Guerrero’s method for solving for lambda requires the seasonal periodicity, sp be provided. [2]
Parameters:
boundstuple
Lower and upper bounds used to restrict the feasible range when solving for the value of lambda.
method{“pearsonr”, “mle”, “all”, “guerrero”}, default=”mle”
The optimization approach used to determine the lambda value used in the Box-Cox transformation.
spint
Seasonal periodicity of the data in integer form. Only used if method=”guerrero” is chosen. Must be an integer >= 2.
Attributes:
boundstuple
Lower and upper bounds used to restrict the feasible range when solving for lambda.
methodstr
Optimization approach used to solve for lambda. One of “personr”, “mle”, “all”, “guerrero”.
spint
Seasonal periodicity of the data in integer form.
lambda_float
The Box-Cox lambda parameter that was solved for based on the supplied method and data provided in fit.
LogTransformer
Transformer input data using natural log. Can help normalize data and compress variance of the series.
aeon.transformations.series.exponent.ExponentTransformer
Transform input data by raising it to an exponent. Can help compress variance of series if a fractional exponent is supplied.
aeon.transformations.series.exponent.SqrtTransformer
Transform input data by taking its square root. Can help compress variance of input series.
Notes
The Box-Cox transformation is defined as $$\frac{y^{\lambda}-1}{\lambda}, \lambda \ne 0 \text{ or } ln(y), \lambda = 0$$.
Therefore, the input data must be positive. In some implementations, a positive constant is added to the series prior to applying the transformation. But that is not the case here.
References
[1]
Box, G. E. P. & Cox, D. R. (1964) An analysis of transformations, Journal ofthe Royal Statistical Society, Series B, 26, 211-252.
[2]
V.M. Guerrero, “Time-series analysis supported by Power Transformations “, Journal of Forecasting, vol. 12, pp. 37-48, 1993.
Examples
>>> from aeon.transformations.series.boxcox import BoxCoxTransformer
>>> transformer = BoxCoxTransformer()
>>> y_hat = transformer.fit_transform(y)
Methods
Check if the estimator has been fitted. Obtain a clone of the object with same hyper-parameters. clone_tags(estimator[, tag_names]) Clone/mirror tags from another estimator as dynamic override. create_test_instance([parameter_set]) Construct Estimator instance if possible. create_test_instances_and_names([parameter_set]) Create list of all test instances and a list of names for them. fit(X[, y]) Fit transformer to X, optionally to y. fit_transform(X[, y]) Fit to data, then transform it. get_class_tag(tag_name[, tag_value_default]) Get tag value from estimator class (only class tags). Get class tags from estimator class and all its parent classes. get_fitted_params([deep]) Get fitted parameters. Get parameter defaults for the object. Get parameter names for the object. get_params([deep]) Get parameters for this estimator. get_tag(tag_name[, tag_value_default, ...]) Get tag value from estimator class. Get tags from estimator class. get_test_params([parameter_set]) Return testing parameter settings for the estimator. inverse_transform(X[, y]) Inverse transform X and return an inverse transformed version. Check if the object is composite. load_from_path(serial) Load object from file location. load_from_serial(serial) Load object from serialized memory container. Reset the object to a clean post-init state. save([path]) Save serialized self to bytes-like object or to (.zip) file. set_params(**params) Set the parameters of this object. set_tags(**tag_dict) Set dynamic tags to given values. transform(X[, y]) Transform X and return a transformed version. update(X[, y, update_params]) Update transformer with X, optionally y.
check_is_fitted()[source]#
Check if the estimator has been fitted.
Raises:
NotFittedError
If the estimator has not been fitted yet.
clone()[source]#
Obtain a clone of the object with same hyper-parameters.
A clone is a different object without shared references, in post-init state. This function is equivalent to returning sklearn.clone of self. Equal in value to type(self)(**self.get_params(deep=False)).
Returns:
instance of type(self), clone of self (see above)
clone_tags(estimator, tag_names=None)[source]#
Clone/mirror tags from another estimator as dynamic override.
Parameters:
estimatorobject
Estimator inheriting from :class:BaseEstimator.
tag_namesstr or list of str, default = None
Names of tags to clone. If None then all tags in estimator are used as tag_names.
Returns:
Self
Reference to self.
Notes
Changes object state by setting tag values in tag_set from estimator as dynamic tags in self.
classmethod create_test_instance(parameter_set='default')[source]#
Construct Estimator instance if possible.
Parameters:
parameter_setstr, default=”default”
Name of the set of test parameters to return, for use in tests. If no special parameters are defined for a value, will return “default” set.
Returns:
instanceinstance of the class with default parameters.
Notes
get_test_params can return dict or list of dict. This function takes first or single dict that get_test_params returns, and constructs the object with that.
classmethod create_test_instances_and_names(parameter_set='default')[source]#
Create list of all test instances and a list of names for them.
Parameters:
parameter_setstr, default=”default”
Name of the set of test parameters to return, for use in tests. If no special parameters are defined for a value, will return “default” set.
Returns:
objslist of instances of cls
i-th instance is cls(**cls.get_test_params()[i]).
nameslist of str, same length as objs
i-th element is name of i-th instance of obj in tests convention is {cls.__name__}-{i} if more than one instance otherwise {cls.__name__}.
parameter_setstr, default=”default”
Name of the set of test parameters to return, for use in tests. If no special parameters are defined for a value, will return “default” set.
fit(X, y=None)[source]#
Fit transformer to X, optionally to y.
State change:
Changes state to “fitted”.
Writes to self: _is_fitted : flag is set to True. _X : X, coerced copy of X, if remember_data tag is True possibly coerced to inner type or update_data compatible type by reference, when possible model attributes (ending in “_”) : dependent on estimator
Parameters:
XSeries or Panel, any supported mtype
Data to fit transform to, of python type as follows:
Series: pd.Series, pd.DataFrame, or np.ndarray (1D or 2D) Panel: pd.DataFrame with 2-level MultiIndex, list of pd.DataFrame, nested pd.DataFrame, or pd.DataFrame in long/wide format.
ySeries or Panel, default=None
Additional data, e.g., labels for transformation.
Returns:
selfa fitted instance of the estimator
fit_transform(X, y=None)[source]#
Fit to data, then transform it.
Fits the transformer to X and y and returns a transformed version of X.
State change: changes state to “fitted”.
Writes to self: _is_fitted : flag is set to True. _X : X, coerced copy of X, if remember_data tag is True
possibly coerced to inner type or update_data compatible type by reference, when possible
model attributes (ending in “_”) : dependent on estimator
Parameters:
XSeries or Panel, any supported mtype
Data to be transformed, of python type as follows:
Series: pd.Series, pd.DataFrame, or np.ndarray (1D or 2D) Panel: pd.DataFrame with 2-level MultiIndex, list of pd.DataFrame,
nested pd.DataFrame, or pd.DataFrame in long/wide format
ySeries or Panel, default=None
Additional data, e.g., labels for transformation
Returns:
transformed version of X
type depends on type of X and output_data_type tag:
X | tf-output | type of return |
|__________|______________|________________________| | Series | Primitives | pd.DataFrame (1-row) | | Panel | Primitives | pd.DataFrame | | Series | Series | Series | | Panel | Series | Panel | | Series | Panel | Panel |
instances in return correspond to instances in X
combinations not in the table are currently not supported
Explicitly, with examples:
if X is Series (e.g., pd.DataFrame) and transform-output is Series
then the return is a single Series of the same mtype Example: detrending a single series
if X is Panel (e.g., pd-multiindex) and transform-output is Series
then the return is Panel with same number of instances as X
(the transformer is applied to each input Series instance)
Example: all series in the panel are detrended individually
if X is Series or Panel and transform-output is Primitives
then the return is pd.DataFrame with as many rows as instances in X Example: i-th row of the return has mean and variance of the i-th series
if X is Series and transform-output is Panel
then the return is a Panel object of type pd-multiindex Example: i-th instance of the output is the i-th window running over X
classmethod get_class_tag(tag_name, tag_value_default=None)[source]#
Get tag value from estimator class (only class tags).
Parameters:
tag_namestr
Name of tag value.
tag_value_defaultany type
Returns:
tag_value
Value of the tag_name tag in self. If not found, returns tag_value_default.
get_tag
Get a single tag from an object.
get_tags
Get all tags from an object.
get_class_tag
Get a single tag from a class.
Examples
>>> from aeon.classification import DummyClassifier
>>> DummyClassifier.get_class_tag("capability:multivariate")
True
classmethod get_class_tags()[source]#
Get class tags from estimator class and all its parent classes.
Returns:
collected_tagsdict
Dictionary of tag name : tag value pairs. Collected from _tags class attribute via nested inheritance. NOT overridden by dynamic tags set by set_tags or mirror_tags.
get_fitted_params(deep=True)[source]#
Get fitted parameters.
State required:
Requires state to be “fitted”.
Parameters:
deepbool, default=True
Whether to return fitted parameters of components.
• If True, will return a dict of parameter name : value for this object, including fitted parameters of fittable components (= BaseEstimator-valued parameters).
• If False, will return a dict of parameter name : value for this object, but not include fitted parameters of components.
Returns:
fitted_paramsdict with str-valued keys
Dictionary of fitted parameters, paramname : paramvalue keys-value pairs include:
• always: all fitted parameters of this object, as via get_param_names values are fitted parameter value for that key, of this object
• if deep=True, also contains keys/value pairs of component parameters parameters of components are indexed as [componentname]__[paramname] all parameters of componentname appear as paramname with its value
• if deep=True, also contains arbitrary levels of component recursion, e.g., [componentname]__[componentcomponentname]__[paramname], etc
classmethod get_param_defaults()[source]#
Get parameter defaults for the object.
Returns:
default_dict: dict with str keys
keys are all parameters of cls that have a default defined in __init__ values are the defaults, as defined in __init__.
classmethod get_param_names()[source]#
Get parameter names for the object.
Returns:
param_names: list of str, alphabetically sorted list of parameter names of cls
get_params(deep=True)[source]#
Get parameters for this estimator.
Parameters:
deepbool, default=True
If True, will return the parameters for this estimator and contained subobjects that are estimators.
Returns:
paramsdict
Parameter names mapped to their values.
get_tag(tag_name, tag_value_default=None, raise_error=True)[source]#
Get tag value from estimator class.
Uses dynamic tag overrides.
Parameters:
tag_namestr
Name of tag to be retrieved.
tag_value_defaultany type, default=None
raise_errorbool
Returns:
tag_value
Value of the tag_name tag in self. If not found, returns an error if raise_error is True, otherwise it returns tag_value_default.
Raises:
ValueError if raise_error is True i.e. if tag_name is not in self.get_tags(
).keys()
get_tags
Get all tags from an object.
get_clas_tags
Get all tags from a class.
get_class_tag
Get a single tag from a class.
Examples
>>> from aeon.classification import DummyClassifier
>>> d = DummyClassifier()
>>> d.get_tag("capability:multivariate")
True
get_tags()[source]#
Get tags from estimator class.
Includes the dynamic tag overrides.
Returns:
dict
Dictionary of tag name : tag value pairs. Collected from _tags class attribute via nested inheritance and then any overrides and new tags from _tags_dynamic object attribute.
get_tag
Get a single tag from an object.
get_clas_tags
Get all tags from a class.
get_class_tag
Get a single tag from a class.
Examples
>>> from aeon.classification import DummyClassifier
>>> d = DummyClassifier()
>>> tags = d.get_tags()
classmethod get_test_params(parameter_set='default')[source]#
Return testing parameter settings for the estimator.
Parameters:
parameter_setstr, default=”default”
Name of the set of test parameters to return, for use in tests. If no special parameters are defined for a value, will return “default” set.
Returns:
paramsdict or list of dict, default = {}
Parameters to create testing instances of the class. Each dict are parameters to construct an “interesting” test instance, i.e., MyClass(**params) or MyClass(**params[i]) creates a valid test instance. create_test_instance uses the first (or only) dictionary in params.
inverse_transform(X, y=None)[source]#
Inverse transform X and return an inverse transformed version.
Currently it is assumed that only transformers with tags
“input_data_type”=”Series”, “output_data_type”=”Series”,
have an inverse_transform.
State required:
Requires state to be “fitted”.
Accesses in self: _is_fitted : must be True _X : optionally accessed, only available if remember_data tag is True fitted model attributes (ending in “_”) : accessed by _inverse_transform
Parameters:
XSeries or Panel, any supported mtype
Data to be inverse transformed, of python type as follows:
Series: pd.Series, pd.DataFrame, or np.ndarray (1D or 2D) Panel: pd.DataFrame with 2-level MultiIndex, list of pd.DataFrame,
nested pd.DataFrame, or pd.DataFrame in long/wide format
ySeries or Panel, default=None
Additional data, e.g., labels for transformation
Returns:
inverse transformed version of X
of the same type as X, and conforming to mtype format specifications
is_composite()[source]#
Check if the object is composite.
A composite object is an object which contains objects, as parameters. Called on an instance, since this may differ by instance.
Returns:
composite: bool
Whether self contains a parameter which is BaseObject.
property is_fitted[source]#
Whether fit has been called.
Parameters:
serialobject
Result of ZipFile(path).open(“object).
Returns:
deserialized self resulting in output at path, of cls.save(path)
Load object from serialized memory container.
Parameters:
serialobject
First element of output of cls.save(None).
Returns:
deserialized self resulting in output serial, of cls.save(None).
reset()[source]#
Reset the object to a clean post-init state.
Equivalent to sklearn.clone but overwrites self. After self.reset() call, self is equal in value to type(self)(**self.get_params(deep=False))
Detail behaviour: removes any object attributes, except:
hyper-parameters = arguments of __init__ object attributes containing double-underscores, i.e., the string “__”
runs __init__ with current values of hyper-parameters (result of get_params)
Not affected by the reset are: object attributes containing double-underscores class and object methods, class attributes
save(path=None)[source]#
Save serialized self to bytes-like object or to (.zip) file.
Behaviour: if path is None, returns an in-memory serialized self if path is a file location, stores self at that location as a zip file
saved files are zip files with following contents: _metadata - contains class of self, i.e., type(self) _obj - serialized self. This class uses the default serialization (pickle).
Parameters:
pathNone or file location (str or Path).
if None, self is saved to an in-memory object if file location, self is saved to that file location. If:
path=”estimator” then a zip file estimator.zip will be made at cwd. path=”/home/stored/estimator” then a zip file estimator.zip will be stored in /home/stored/.
Returns:
if path is None - in-memory serialized self
if path is file location - ZipFile with reference to the file.
set_params(**params)[source]#
Set the parameters of this object.
The method works on simple estimators as well as on nested objects. The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.
Parameters:
**paramsdict
BaseObject parameters
Returns:
selfreference to self (after parameters have been set)
set_tags(**tag_dict)[source]#
Set dynamic tags to given values.
Parameters:
**tag_dictdict
Dictionary of tag name : tag value pairs.
Returns:
Self
Reference to self.
Notes
Changes object state by settting tag values in tag_dict as dynamic tags in self.
transform(X, y=None)[source]#
Transform X and return a transformed version.
State required:
Requires state to be “fitted”.
Accesses in self: _is_fitted : must be True _X : optionally accessed, only available if remember_data tag is True fitted model attributes (ending in “_”) : must be set, accessed by _transform
Parameters:
XSeries or Panel, any supported mtype
Data to be transformed, of python type as follows:
Series: pd.Series, pd.DataFrame, or np.ndarray (1D or 2D) Panel: pd.DataFrame with 2-level MultiIndex, list of pd.DataFrame,
nested pd.DataFrame, or pd.DataFrame in long/wide format
ySeries or Panel, default=None
Additional data, e.g., labels for transformation
Returns:
transformed version of X
type depends on type of X and output_data_type tag:
| transform | |
X | -output | type of return |
|__________|______________|________________________| | Series | Primitives | pd.DataFrame (1-row) | | Panel | Primitives | pd.DataFrame | | Series | Series | Series | | Panel | Series | Panel | | Series | Panel | Panel |
instances in return correspond to instances in X
combinations not in the table are currently not supported
Explicitly, with examples:
if X is Series (e.g., pd.DataFrame) and transform-output is Series
then the return is a single Series of the same mtype Example: detrending a single series
if X is Panel (e.g., pd-multiindex) and transform-output is Series
then the return is Panel with same number of instances as X
(the transformer is applied to each input Series instance)
Example: all series in the panel are detrended individually
if X is Series or Panel and transform-output is Primitives
then the return is pd.DataFrame with as many rows as instances in X Example: i-th row of the return has mean and variance of the i-th series
if X is Series and transform-output is Panel
then the return is a Panel object of type pd-multiindex Example: i-th instance of the output is the i-th window running over X
update(X, y=None, update_params=True)[source]#
Update transformer with X, optionally y.
State required:
Requires state to be “fitted”.
Accesses in self: _is_fitted : must be True _X : accessed by _update and by update_data, if remember_data tag is True fitted model attributes (ending in “_”) : must be set, accessed by _update
Writes to self: _X : updated by values in X, via update_data, if remember_data tag is True fitted model attributes (ending in “_”) : only if update_params=True
type and nature of update are dependent on estimator
Parameters:
XSeries or Panel, any supported mtype
Data to fit transform to, of python type as follows:
Series: pd.Series, pd.DataFrame, or np.ndarray (1D or 2D) Panel: pd.DataFrame with 2-level MultiIndex, list of pd.DataFrame,
nested pd.DataFrame, or pd.DataFrame in long/wide format
ySeries or Panel, default=None
Additional data, e.g., labels for transformation
update_paramsbool, default=True
whether the model is updated. Yes if true, if false, simply skips call. argument exists for compatibility with forecasting module.
Returns:
selfa fitted instance of the estimator | 4,585 | 20,643 | {"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": 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} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.716513 |
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# Huy Dang
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Calculate the area of a triangle between three points
Calculate the area of a triangle between three points: P1(X1,Y1) P2(X2,Y2) P3(X3,Y3) these three points are the vert...
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Can we make a triangle?
Given three positive number, check whether a triangle can be made with these sides length or not. remember that in a triangle su...
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Find my daddy long leg (No 's')
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Counting Money
Add the numbers given in the cell array of strings. The strings represent amounts of money using this notation: \$99,999.99. E...
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Return the 3n+1 sequence for n
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Calculate the nth Fibonacci number. Given n, return f where f = fib(n) and f(1) = 1, f(2) = 1, f(3) = 2, ... Examples: Inpu...
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Determine if input is odd
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Rotate and display numbered tile
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Scoring for oriented dominoes
Given a list of ordered pairs, and the order they should be placed in a line, find the sum of the absolute values of the differe...
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Draw a '4' in a zero matrix!
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Find the maximum number of decimal places in a set of numbers
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Draw a '7' in a zero matrix!
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Draw a 'Z'.
Given _n_ as input, generate a n-by-n matrix like 'Z' by _0_ and _1_ . Example: n=5 ans= [1 1 1 1 1 0 0 0 1 ...
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Draw a 'X'!
Given n as input Draw a 'X' in a n-by-n matrix. example: n=3 y=[1 0 1 0 1 0 1 0 1] n=4 y=[1 0 0...
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Given n as input, generate a n-by-n matrix 'O' using 0 and 1 . example: n=4 ans= [1 1 1 1 1 0 0 1 ...
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Draw a x-by-x matrix 'H' using 1 and 0. (x is odd and bigger than 2) Example: x=5 ans= [1 0 0 0 1 1 0 0 0 1 ...
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3ヶ月 前 | 1,197 | 3,278 | {"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} | 3.03125 | 3 | CC-MAIN-2021-25 | latest | en | 0.650827 |
http://math.stackexchange.com/questions/613234/how-to-understand-that-products-and-coproducts-are-dual | 1,469,694,929,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.15/warc/CC-MAIN-20160723071028-00119-ip-10-185-27-174.ec2.internal.warc.gz | 176,071,154 | 17,470 | # how to understand that products and coproducts are dual
I am reading some basic category notes, how can one relate the products to coproducts? If given a product, can one build its dual product? for example, the coordinate product $(x,y)$ : $R \times R$, what coproduct does it correspond to?
-
Note that if $X$ is a product of $A$ and $B$, there may exist a $C$ and $D$ with $C\times D\cong X$ such that the coproduct of $A$ and $B$ is not the same as the coproduct of $C$ and $D$. – Dan Rust Dec 19 '13 at 19:04
Coproduct is the dual of product in the categorical sense, i.e. we have to reverse all occuring arrow in the definition.
While product in most concrete categories of structures is indeed realized on the Cartesian product of the underlying sets (because the forgetful functor usually admits a left adjoint), the coproduct may vary from category to category.
In the simplest cases (e.g. in $\Bbb{Set}$) the coproduct is just the disjoint union. But e.g. in the category of groups, the coproduct is the free product...
-
Products and coproducts are dual in the following sense:
Let $\mathcal{C}$ be a category and $\{X_i\}$ be a family of objects in $\mathcal{C}$. Let $\{X'_i\}$ be the same family but considered as a family of objects in $\mathcal{C}^{op}$. Then:
The product $\prod_i X_i$ in $\mathcal{C}$ is the same as the coproduct $\coprod_i X'_i$ in $\mathcal{C}^{op}$.
- | 394 | 1,402 | {"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} | 3.078125 | 3 | CC-MAIN-2016-30 | latest | en | 0.897813 |
https://avatest.org/2023/06/30/discrete-mathematics-dai-xie-not/ | 1,725,905,604,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00731.warc.gz | 90,992,928 | 21,017 | Posted on Categories:Discrete Mathematics, 数学代写, 离散数学
# 数学代写|离散数学代写Discrete Mathematics代考|‘‘Not”
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## 数学代写|离散数学代写Discrete Mathematics代考|Recalling the Strategy Definition
The statement “not $\mathbf{A}$,” written $\sim \mathbf{A}$, is true whenever $\mathbf{A}$ is false. For example, the statement
Charles is not happily married
is true provided the statement “Charles is happily married” is false. The truth table for $\sim \mathbf{A}$ is as follows:
\begin{tabular}{cc}
\hline $\mathbf{A}$ & $\sim \mathbf{A}$ \
\hline $\mathrm{T}$ & $\mathrm{F}$ \
$\mathrm{F}$ & $\mathrm{T}$ \
\hline
\end{tabular}
Greater understanding is obtained by combining the connectives:
EXAMPLE 1.6
We examine the truth table for $\sim(\mathbf{A} \wedge \mathbf{B})$ :
\begin{tabular}{lccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{A} \wedge \mathbf{B}$ & $\sim(\mathbf{A} \wedge \mathbf{B})$ \
\hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \
\hline
\end{tabular}
EXAMPLE 1.7
Now we look at the truth table for $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$ :
\begin{tabular}{ccccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\sim \mathbf{A}$ & $\sim \mathbf{B}$ & $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$ \
\hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
\hline
\end{tabular}
Notice that the statements $\sim(\mathbf{A} \wedge \mathbf{B})$ and $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$ have the same truth table. As previously noted, such pairs of statements are called logically equivalent.
The logical equivalence of $\sim(\mathbf{A} \wedge \mathbf{B})$ with $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$ makes good intuitive sense: the statement $\mathbf{A} \wedge \mathbf{B}$ fails [that is, $\sim(\mathbf{A} \wedge \mathbf{B})$ is true] precisely when either $\mathbf{A}$ is false or $\mathbf{B}$ is false. That is, $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$. Since in mathematics we cannot rely on our intuition to establish facts, it is important to have the truth table technique for establishing logical equivalence. The exercise set will give you further practice with this notion.
One of the main reasons that we use the inclusive definition of “or” rather than the exclusive one is so that the connectives “and” and “or” have the nice relationship just discussed. It is also the case that $\sim(\mathbf{A} \vee \mathbf{B})$ and $(\sim \mathbf{A}) \wedge(\sim \mathbf{B})$ are logically equivalent. These logical equivalences are sometimes referred to as de Morgan’s laws.
## 数学代写|离散数学代写Discrete Mathematics代考|‘‘If-Then’’
A statement of the form “If $\mathbf{A}$ then $\mathbf{B}$ ” asserts that whenever $\mathbf{A}$ is true then $\mathbf{B}$ is also true. This assertion (or “promise”) is tested when $\mathbf{A}$ is true, because it is then claimed that something else (namely $\mathbf{B}$ ) is true as well. However, when $\mathbf{A}$ is false then the statement “If $\mathbf{A}$ then $\mathbf{B}$ ” claims nothing. Using the symbols $\mathbf{A} \Rightarrow \mathbf{B}$ to denote “If $\mathbf{A}$ then $\mathbf{B}$ “, we obtain the following truth table:
\begin{tabular}{ccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{A} \Rightarrow \mathbf{B}$ \
\hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ \
$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ \
$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \
\hline
\end{tabular}
Notice that we use here an important principle of aristotelian logic: every sensible statement is either true or false. There is no “in between” status. When $\mathbf{A}$ is false we can hardly assert that $\mathbf{A} \Rightarrow \mathbf{B}$ is false. For $\mathbf{A} \Rightarrow \mathbf{B}$ asserts that “whenever A is true then $\mathbf{B}$ is true”, and $\mathbf{A}$ is not true!
Put in other words, when $\mathbf{A}$ is false then the statement $\mathbf{A} \Rightarrow \mathbf{B}$ is not tested. It therefore cannot be false. So it must be true. We refer to $\mathbf{A}$ as the hypothesis of the implication and to $\mathbf{B}$ as the conclusion of the implication. When the if-then statement is true, then the hypothsis implies the conclusion.
EXAMPLE 1.8
The statement “If $2=4$ then Calvin Coolidge was our greatest president” is true. This is the case no matter what you think of Calvin Coolidge. The point is that the hypothesis $(2=4)$ is false; thus it doesn’t matter what the truth value of the conclusion is. According to the truth table for implication, the sentence is true.
The statement “If fish have hair then chickens have lips” is true. Again, the hypothesis is false so the sentence is true.
The statement “If $9>5$ then dogs don’t fly” is true. In this case the hypothesis is certainly true and so is the conclusion. Therefore the sentence is true.
(Notice that the “if” part of the sentence and the “then” part of the sentence need not be related in any intuitive sense. The truth or falsity of an “if-then” statement is simply a fact about the logical values of its hypothesis and of its conclusion.)
## 数学代写|离散数学代写Discrete Mathematics代考|Recalling the Strategy Definition
\begin{tabular}{cc}
\hline $\mathbf{A}$ & $\sim \mathbf{A}$ \hline $\mathrm{T}$ & $\mathrm{F}$ \$\mathrm{F}$ & $\mathrm{T}$ \hline
\end{tabular}
\begin{tabular}{lccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{A} \wedge \mathbf{B}$ & $\sim(\mathbf{A} \wedge \mathbf{B})$ \hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ \$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ \$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \hline
\end{tabular}
\begin{tabular}{ccccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\sim \mathbf{A}$ & $\sim \mathbf{B}$ & $(\sim \mathbf{A}) \vee(\sim \mathbf{B})$ \hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{F}$ \$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ \$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{T}$ \$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \hline
\end{tabular}
$\sim(\mathbf{A} \wedge \mathbf{B})$与$(\sim \mathbf{A}) \vee(\sim \mathbf{B})$的逻辑等价具有很好的直观意义:当$\mathbf{A}$为假或$\mathbf{B}$为假时，语句$\mathbf{A} \wedge \mathbf{B}$失败[即$\sim(\mathbf{A} \wedge \mathbf{B})$为真]。也就是$(\sim \mathbf{A}) \vee(\sim \mathbf{B})$。因为在数学中我们不能依靠我们的直觉来建立事实，所以有真值表技术来建立逻辑等价是很重要的。这个练习集将给你进一步练习这个概念。
## 数学代写|离散数学代写Discrete Mathematics代考|‘‘If-Then’’
“如果$\mathbf{A}$那么$\mathbf{B}$”这样的语句断言只要$\mathbf{A}$为真，那么$\mathbf{B}$也为真。当$\mathbf{A}$为真时，这个断言(或“承诺”)就会被检验，因为它随后就会声称其他东西(即$\mathbf{B}$)也为真。然而，当$\mathbf{A}$为假时，语句“如果$\mathbf{A}$那么$\mathbf{B}$”没有声明任何内容。用符号$\mathbf{A} \Rightarrow \mathbf{B}$表示“如果$\mathbf{A}$那么$\mathbf{B}$”，我们得到以下真值表:
\begin{tabular}{ccc}
\hline $\mathbf{A}$ & $\mathbf{B}$ & $\mathbf{A} \Rightarrow \mathbf{B}$ \hline $\mathrm{T}$ & $\mathrm{T}$ & $\mathrm{T}$ \$\mathrm{T}$ & $\mathrm{F}$ & $\mathrm{F}$ \$\mathrm{F}$ & $\mathrm{T}$ & $\mathrm{T}$ \$\mathrm{F}$ & $\mathrm{F}$ & $\mathrm{T}$ \hline
\end{tabular}
“如果$2=4$那么卡尔文·柯立芝就是我们最伟大的总统”这句话是对的。无论你怎么看待卡尔文·柯立芝，情况都是如此。关键是假设$(2=4)$是假的;因此结论的真值是多少并不重要。根据蕴涵真值表，句子为真。
“如果鱼有毛，那么鸡就有嘴唇”这句话是对的。假设为假，所以句子为真。
“如果$9>5$那么狗不会飞”这句话是对的。在这种情况下，假设当然是正确的，结论也是正确的。因此这个句子是真的。
(请注意，句子的“if”部分和句子的“then”部分不需要在任何直观意义上联系起来。一个”如果-那么”命题的真假仅仅是关于它的假设和结论的逻辑值的事实。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 3,347 | 8,561 | {"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} | 3.21875 | 3 | CC-MAIN-2024-38 | latest | en | 0.17806 |
https://wizardofvegas.com/forum/questions-and-answers/math/18186-yahtzee-with-ten-rolls/ | 1,603,431,546,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00383.warc.gz | 595,558,650 | 11,282 | Wizard
Joined: Oct 14, 2009
• Posts: 22046
May 12th, 2014 at 4:51:28 PM permalink
What is the probability of achieving a Yahtzee after ten rolls?
For those who don't understand the question, here are the rules:
1. Five six-sided dice are rolled.
2. The object is to have all of them on the same face, for example all twos.
3. After any roll you may hold any dice you wish and re-roll the others.
4. Switching to a different number is allowed. For example if you roll 1-1-2-3-4 on the first roll, then hold the ones, and then roll a 2-2-2 with the other three, having 1-1-2-2-2, then you can switch to holding the twos and re-roll the ones.
Normally in the game of Yahtzee you get only three rolls, but I'm allowing ten (to make it harder).
For extra credit, what is the probability after 1 to 9 rolls?
It's not whether you win or lose; it's whether or not you had a good bet.
beachbumbabs
Joined: May 21, 2013
• Posts: 14229
May 12th, 2014 at 4:55:21 PM permalink
Quote: Wizard
What is the probability of achieving a Yahtzee after ten rolls?
For those who don't understand the question, here are the rules:
1. Five six-sided dice are rolled.
2. The object is to have all of them on the same face, for example all twos.
3. After any roll you may hold any dice you wish and re-roll the others.
4. Switching to a different number is allowed. For example if you roll 1-1-2-3-4 on the first roll, then hold the ones, and then roll a 2-2-2 with the other three, having 1-1-2-2-2, then you can switch to holding the twos and re-roll the ones.
Normally in the game of Yahtzee you get only three rolls, but I'm allowing ten (to make it harder).
For extra credit, what is the probability after 1 to 9 rolls?
So, you're allowing UP TO 10 rolls, stopping on any Yahtzee, in the original question, right? Which means the extra credit question would also be all-inclusive (additive)? Or you're looking for EXACTLY each number of rolls, not less (I don't think this is the question, but making sure).
If the House lost every hand, they wouldn't deal the game.
chrisr
Joined: Dec 9, 2013
• Posts: 141
May 12th, 2014 at 5:12:17 PM permalink
how i would solve it.
first we have to figure out how many matches we should start with before we continue toward our target of 5 matching. For example, if our first roll is 2 matches, should we throw it out and re-roll all 5 or should we keep the two and try to make it with the other 3 dice...
..after that it is pretty straightforward to calculate the probability.
kubikulann
Joined: Jun 28, 2011
• Posts: 905
May 12th, 2014 at 5:58:26 PM permalink
Quote: beachbumbabs
So, you're allowing UP TO 10 rolls, stopping on any Yahtzee, in the original question, right? Which means the extra credit question would also be all-inclusive (additive)? Or you're looking for EXACTLY each number of rolls, not less (I don't think this is the question, but making sure).
If you get a Yahtzee before the end, just keep all five dice. So both questions are equivalent.
Reperiet qui quaesiverit
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 12th, 2014 at 5:58:35 PM permalink
Quote: Wizard
What is the probability of achieving a Yahtzee after ten rolls?
I love Yahtzee,
well used to play it a lot and changing rules to the game.
This actually in school was the first exercise in combinations and matrix algebra.
I could not do this in class. I had to have lots of help.
since then I have some of this in Excel for the 5 kinds
this is my transition matrix (I changed it to include a 0 state)
and the matrix raised to 10
so I think you want as an answer 1-[1,6]
even show the expected number of visits to each state (yellows)
I also have this just using recursion in Excel. many do not know how to raise a matrix to a power in Excel and that is OK.
a few formulas in a few cells is all that is required for the recursion method
I will link to it as soon as I finish it.
looks like this for the first 50 rolls
here is one place to start for any interested in a quick read, there really are many of them
http://www.datagenetics.com/blog/january42012/
Sally
I Heart Vi Hart
Wizard
Joined: Oct 14, 2009
• Posts: 22046
May 12th, 2014 at 6:09:42 PM permalink
Quote: kubikulann
If you get a Yahtzee before the end, just keep all five dice. So both questions are equivalent.
Yes, keeping all five is allowed? So asking about a Yahtzee in ten rolls could also be stated as "ten or less."
It's not whether you win or lose; it's whether or not you had a good bet.
chrisr
Joined: Dec 9, 2013
• Posts: 141
May 12th, 2014 at 6:18:25 PM permalink
I came up with it being slightly more advantageous to re-roll all 5 dice if you don't have at least a pair.. The numbers i came up with for Yahtzee in 10 or less rolls were 0.545 and 0.550 depending on if you keep an arbitrary single die when you have less than a pair.
Tomspur
Joined: Jul 12, 2013
• Posts: 2019
May 12th, 2014 at 6:20:19 PM permalink
Quote: chrisr
I came up with it being slightly more advantageous to re-roll all 5 dice if you don't have at least a pair.. The numbers i came up with for Yahtzee in 10 or less rolls were 0.545 and 0.550 depending on if you keep an arbitrary single die when you have less than a pair.
edit: That is what I get for trying to be clever and then my spoiler tags don't work either :)
“There is something about the outside of a horse that is good for the inside of a man.” - Winston Churchill
AxiomOfChoice
Joined: Sep 12, 2012
• Posts: 5761
May 12th, 2014 at 6:22:04 PM permalink
Quote: chrisr
I came up with it being slightly more advantageous to re-roll all 5 dice if you don't have at least a pair.. The numbers i came up with for Yahtzee in 10 or less rolls were 0.545 and 0.550 depending on if you keep an arbitrary single die when you have less than a pair.
This has to be a rounding error. Keeping all 5 and keeping 4 out of 5 are identical.
To see this, imagine that, once you decide which dice to re-roll, you roll them one at a time. If you re-roll all 5, doesn't matter how the first die comes up.
chrisr
Joined: Dec 9, 2013 | 1,710 | 6,056 | {"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} | 4.3125 | 4 | CC-MAIN-2020-45 | latest | en | 0.942539 |
https://www.enchantedlearning.com/math/algebra/polynomials/ | 1,721,868,266,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.61/warc/CC-MAIN-20240724232540-20240725022540-00265.warc.gz | 632,785,365 | 5,754 | More on Math Polynomials Quadratic Equations
A polynomial is a sum or difference of terms; each term is:
• a constant (like 5)
• a constant times a variable (like 3x)
• a constant times the variable to a positive integer power (like 2x2)
• a constant times the product of variables to positive integer powers (like 2x3y).
A monomial is a polynomial with only one term. A binomial is a polynomial that has two terms. A trinomial is a polynomial with three terms.
Degree of a Polynomial: The degree of a term within a polynomial is the sum of the exponents of variables that occur in that term (if there is no exponent written on a variable, such as in 3x, the exponent is one). The degree of a polynomial is the greatest degree of any term in the polynomial (for instance, for the polynomial 4x2 + 7xyz, the degree is 3 because of the last term).
• If a polynomial isn't just a constant and if each term has at most a variable to the first power (like 4x - 2 or 3y), then it is a first-degree polynomial (also called a linear polynomial).
• If a polynomial's highest degree is two, it is a quadratic (or second-degree) polynomial (example: 4x2 + 3x + 7).
• If a polynomial has more than one variable, then its degree is the sum of the exponents of the highest-degree term. For example, the polynomial 2xy2 -3xy + 6x - 2 has degree 3 (the sum of the xy2 exponents, 1 + 2).
Polynomials are often written in descending order, in which the terms with the largest powers are written first (like 9x2 - 3x + 6). If they are written with the smallest terms appearing first, this is ascending order (like 6 - 3x + 9x2).
A polynomial equation is an equation involving a polynomial.
Multiplying Polynomials:
Multiply Polynomials: Introduction to Algebra - Worksheets to Print
Multiply Polynomials Worksheet #1Multiply 10 polynomials and put the answers in simplest form. Or go to the answers. Multiply Polynomials Worksheet #2Multiply 10 polynomials and put the answers in simplest form. Or go to the answers.
Quadratic Equation: ax2 + bx - c = 0Printable Worksheets
Miscellaneous
Pascal's Triangle - WorksheetFill in the missing entries in Pascal's Triangle. Go to the answers.
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## Enchanted Learning Search
Search the Enchanted Learning website for: | 735 | 3,177 | {"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} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.905804 |
https://electronics.stackexchange.com/questions/439154/how-is-dynamic-resistance-of-a-diode-modeled-for-large-voltage-variations | 1,566,098,453,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313589.19/warc/CC-MAIN-20190818022816-20190818044816-00095.warc.gz | 456,648,599 | 31,958 | # How is dynamic resistance of a diode modeled for large voltage variations?
The resistance of a diode changes with the voltage across it which is called dynamic resistance. If the voltage across the diode is constant we can find the dynamic resistance from the slope of the I-V curve. So for small variations we can use that dynamic resistance value in the diode model.
But if the voltage across the diode varies a lot like in a AC to DC rectifying diode, I guess we cannot just model a diode’s dynamic resistance with a constant series resistance. If we want to formulate the diode current in such case how is the diode dynamic resistance modeled? Here aren’t we facing a non-linear resistor?
• Are you actually interested in seeing the page of mathematics involved, assuming a simple Shockley diode equation without current-crowding or surface channel effects? Or just asking, generally? If generally, yeah -- the dynamic resistance is only valid at one local spot on the curve so doesn't apply to the non-linear large scale model. – jonk May 18 at 19:13
• I was asking in general. Shockley diode equation with numerical methods are used as far as I understand from the answers. – Genzo May 19 at 13:40
• Sounds good, then. The closed solution math is a bit drawn out. – jonk May 19 at 17:37
The concept of dynamic resistance is a derivative:
$$\ r = \frac{dv}{di} \$$
As such, it only applies to variations of current and voltage that are small enough to allow us to neglect the non-linearity and use a linear model for the diode.
But if the voltage across the diode varies a lot like in a AC to DC rectifying diode,
In this case, a linear model can't be used to model the diode's behavior over the whole waveform, and the concept of "dynamic resistance" which is a part of this linear model does not exist, so you'll have to use the diode equation.
If you only look at a specific point in time on the waveform, then you can calculate dynamic resistance at this point, depending on the value of the current at this point.
• If you need a simpler approximation, you could always just use a few more terms in the taylor series, no? – Hearth May 18 at 17:55
• You could, but it won't help much with modeling the whole diode IV curve – peufeu May 18 at 19:14
You can model the DC I-V characteristics with the Shockley diode equation over a fairly wide range of currents, especially if you include an accurate ideality factor and some series resistance. It's nonlinear but still very simple and easy to solve numerically.
The diode model used in SPICE has more than a dozen parameters.
Yes, for slow enough changes in bias, a diode can be modeled as a nonlinear resistor. For faster changes you must also consider the diode's capacitance, which is also bias-dependent.
Simplifying a lot, the simulator has to keep updating the value for the dynamic resistance as the simulation evolves.
For "large" voltage variations, you might model the diode as a distortion. The coefficients of the polynomials in the series of terms will work in that model.
Look for Taylor Series Expansion of e^x, as a start. | 709 | 3,113 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-35 | latest | en | 0.927289 |
https://studylib.net/doc/5622180/mathematical-ideas | 1,550,727,689,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247500089.84/warc/CC-MAIN-20190221051342-20190221073342-00489.warc.gz | 704,211,609 | 35,804 | Mathematical Ideas
Chapter 9
Geometry
Chapter 9: Geometry
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Points, Lines, Planes, and Angles
Curves, Polygons, and Circles
Perimeter, Area, and Circumference
The Geometry of Triangles: Congruence,
Similarity, and the Pythagorean Theorem
Space Figures, Volume, and Surface Area
Transformational Geometry
Non-Euclidean Geometry, Topology, and Networks
Chaos and Fractal Geometry
9-8-2
Chapter 1
Section 9-8
Chaos and Fractal Geometry
9-8-3
Chaos and Fractal Geometry
• Chaos
• Fractals
9-8-4
Example: Attractor
Consider the equation y = kx(1 – x), with k = 2. This
gives the equation y = 2x(1 – x). We can pick a value
between 0 and 1 and “iterate” the equation by
plugging in the value and then take the resulting yvalue, substitute it in as an x-value in the equation
and continue this process. For example, starting with
x = 0.8 produces the sequence
.8, .32, .435, .492, .500, .500, .500, …
9-8-5
Attractors
The sequence seems to stabilize at the value .500. A
different x-value would produce another sequence that
“converges” to .500. The value .5000 can be called
an attractor for the sequence generated by the
equation y = 2x(1 – x).
For k = 3, y = kx(1 – x) converges to two attractors,
and for k = 3.5, it converges to four attractors. If k is
increased further, the number of attractors doubles
over and over. This doubling occurs infinitely many
times before k gets as large as 4.
9-8-6
Chaos
Somewhere before k = 4, the resulting sequence
becomes apparently totally random, with no
attractors and no stability. This type of condition is
one instance of what is known as chaos.
As long as k is small enough, there will be some
number of attractors and the long-term behavior of the
sequence is the same regardless of the initial x-value.
But once k is large enough to cause chaos, the longterm behavior of the system will change drastically
when the initial value is changed only slightly.
9-8-7
Chaos
Patterns similar to those in the previous sequences
apply to many phenomena in the physical,
biological, and social sciences.
Continuous phenomena are easily dealt with. A
change in one quantity produces a predictable change
in another. (For example, a little more pressure on the
gas pedal produces a little more speed.)
9-8-8
Catastrophe Theory
Rene Thom, a forerunner of chaos theory, applied
the methods of topology to deal with discontinuous
processes. Thom referred to events with a sudden
change such as a heartbeat, a buckling beam, a stock
market crash, a riot, or a tornado, as catastrophes.
He proved that all catastrophic events are
combinations of seven elementary catastrophes. His
work became known as catastrophe theory.
9-8-9
Catastrophe Theory
Each of the seven elementary catastrophes has a
characteristic topological shape. Two examples
are shown on the next slide. The top figure is
called a cusp. The bottom figure is an elliptic
umbilicus.
9-8-10
Example: Elementary Catastrophes
9-8-11
Fractal Geometry
Fractal geometry provides a key for the new
study of non-linear processes. It is concerned with
figures that exhibit a self-similar shape – a shape
that repeats itself over and over on different scales.
A coastline is an example of a self-similar shape.
The branching of a tree, from twig to limb to trunk
also exhibits a shape that repeats itself. Benoit
Mandelbrot developed much of the work in this
field.
9-8-12
Fractal Example: Koch Snowflake
Starting with an equilateral triangle, each side
gives to another equilateral triangle. This
process continues over and over indefinitely and
a curve of infinite length is produced.
…
Step 1
See next
slide
Step 2
9-8-13
Example: Koch Snowflake
9-8-14
Fractals
The theory of fractals is today being applied to many
areas of science and technology. It has been used to
analyze the symmetry of living forms, the turbulence
of liquids, the branching of rivers, and price variation
in economics.
Aside from providing a geometric structure for
chaotic processes in nature, fractal geometry is
viewed by many as a significant new art form. The
next slide shows a computer-generated fractal design.
9-8-15
Example: Computer Generated Fractal
9-8-16
– Cards
– Cards
– Cards
– Cards
– Cards | 1,137 | 4,203 | {"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} | 4.53125 | 5 | CC-MAIN-2019-09 | latest | en | 0.914802 |
https://www.doubtnut.com/question-answer-chemistry/the-ionization-energy-of-he-is-196-xx-10-18j-atom-1-the-energy-of-the-first-stationary-state-of-li-2-645335975 | 1,660,126,659,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571153.86/warc/CC-MAIN-20220810100712-20220810130712-00513.warc.gz | 673,891,160 | 75,250 | Home
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Text Solution
84.2xx10^(-18)J/atom 44.10 xx 10^(-18)J/atom 63.2xx10^(-18) J/atom21.2xx10^(-18)J/atom
Solution : E_(1)" for "Li^(2+)=E_(1)" for "H xx Z^(2)=E_(1)" for "H xx 9 <br> E_(1)" for "He^(+)=E_(1)" for "H xx Z_(He)^(2)=E_(1)" for "H xx 4 <br> or E_(1)" for "Li^(2+) =(9)/(4) E_(1)" for "He^(+)=19.6 xx 10^(-18) J/atom | 235 | 571 | {"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} | 2.59375 | 3 | CC-MAIN-2022-33 | latest | en | 0.573751 |
https://socratic.org/questions/how-do-you-solve-x-3y-9-and-x-3y-3-by-graphing | 1,575,925,053,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00252.warc.gz | 532,984,421 | 6,835 | # How do you solve x-3y=-9 and x+3y=3 by graphing?
Mar 16, 2018
Solution is $\left(- 3 , 2\right)$
Draw the two graphs $x - 3 y = - 9$ and $x + 3 y = 3$ and find the coordinates $\left(x , y\right)$ of where they intersect.
#### Explanation:
You have two equations, $x - 3 y = - 9$ and $x + 3 y = 3$
To plot this on a graph, do some transformation so that you have just $y$ on one side ($y = \ldots$)
Part 1
You need to get those equations into the form $y = \ldots$
For the first equation,
$x - 3 y = - 9$
$x - 3 y + 9 = 0$
$x + 9 = 3 y$
Dividing both sides by 3, you get
$y = \frac{x + 9}{3}$
For the second one, you do the same thing
$x + 3 y = 3$
Subtracting x from both sides,
$3 y = 3 - x$
Then divide both sides by 3
$y = \frac{3 - x}{2}$
Part 2
Plot these two equations on the same graph
$y = \frac{x + 9}{3}$ and $y = \frac{3 - x}{2}$
graph{(x-3y+9)(x+3y-3)=0 [-10, 10, -5, 5]}
The coordinates of where those two lines cross will be your answer. Which will be $\left(- 3 , 2\right)$
Here's why the coordinates of where the two lines cross is the answer:
What the line for $x - 3 y = - 9$ means is that every point on that line will have coordinates $\left(x , y\right)$ that satisfy $x - 3 y = - 9$
So for instance, if you find the point $\left(- 9 , 0\right)$ is on that line, that means that $x = - 9 , y = 0$ is a pair of numbers that obeys $x - 3 y = - 9$:
$\left(- 9\right) - 3 \left(0\right) = - 9$
So if you want to find a pair of numbers $\left(x , y\right)$ which fit $x - 3 y = - 9$, then just take the coordinates of any point on that line.
The same thing applies for more than one line. Since a point that lies on some line $y = \left(\ldots\right) x + \left(\ldots\right)$ will satisfy that equation, then a point that lies on two different lines at the same time will satisfy both equations represented by those lines. So any point that lies on two lines (is the intersection) will have coordinates that fit the equations of both lines. | 660 | 1,990 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 29, "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} | 4.875 | 5 | CC-MAIN-2019-51 | latest | en | 0.833068 |
http://mathematica.stackexchange.com/questions/42075/mysterious-behavior-of-precision-for-complex-arrays | 1,469,695,267,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828010.15/warc/CC-MAIN-20160723071028-00081-ip-10-185-27-174.ec2.internal.warc.gz | 148,506,475 | 17,745 | # Mysterious behavior of Precision for complex arrays
Mysterious behavior of Precision:
{{1.0+I*0.0},{0.0+I*0.0}} // SetPrecision[#,30]& // Precision // Print;
0.
{{1.0},{0.0}} // SetPrecision[#,30]& // Precision // Print;
30.
Why is the precision zero in the first instance, but not the second?
This led to some tough-to-diagnose program behaviors!
-
This seems to be closely related: How to eliminate the zero real part of a purely imaginary number?. – Artes Feb 11 '14 at 0:12
+1 for minimal example of behavior. – DumpsterDoofus Feb 11 '14 at 0:30
It seems like the 0.0+I*0.0 is the culprit. Try I // SetPrecision[#, 30] & // Precision, 1 // SetPrecision[#, 30] & // Precision, 0 // SetPrecision[#, 30] & // Precision, 0.0 // SetPrecision[#, 30] & // Precision, 0.0 I // SetPrecision[#, 30] & // Precision and especially pay attention to the last two. I assume that Precision when applied to an array takes the minimum of the precisions of the elements of the array; the first element of {{1.0+I*0.0},{0.0+I*0.0}} has precision 30, whereas the second has precision 0, so the result is 0, but I'm not sure why 0.0I has precision 0. – DumpsterDoofus Feb 11 '14 at 0:35
And there's also precision Infinity: f[x_] := x // SetPrecision[#, 30] & // Precision; {f[0], f[0.0], f[0.0 + 0.0 I], f[1.0], f[1.0 I]} gives {[Infinity], [Infinity], 0., 30., 30.} – bill s Feb 11 '14 at 1:43
Not receiving an answer, the following "workaround" returns Precision as a rounded-up integer multiple of MachinePrecision:
Precision\$TNS[arg_] := arg//
Precision//
Which[
NumberQ[#] && (#>0.0),
{#},
True,
{arg}//Flatten//
Map[Precision,#]&//
Select[#,(NumberQ[#] && (#>0.0))&]&
]&//Max[#,MachinePrecision]&//
(#-1)/MachinePrecision&//Ceiling//
#*MachinePrecision&;
This workaround suffices (seemingly) for my main purpose, which is to assess and if necessary adaptively increase the precision of large-condition Real and Complex array arguments that are supplied to SingularValueDecomposition[_].
- | 612 | 1,992 | {"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} | 2.96875 | 3 | CC-MAIN-2016-30 | latest | en | 0.657897 |
https://www.kimavi.com/Grade-6-Science-Force-and-Motion | 1,582,005,888,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143635.54/warc/CC-MAIN-20200218055414-20200218085414-00148.warc.gz | 795,824,892 | 7,629 | # Chapter 14 : Force and Motion
### Topics covered in this snack-sized chapter:
#### Force arrow_upward
• Force is the process that cause any object to undergo some sort of changes.
• Force is the push or pull you apply to make things move.
• A force can:
• Make an object move.
• Change the direction of a moving object.
• Stop a moving object.
• Force can also change the shape of an object.
• Force is also defined as Mass and acceleration vector product.
• The ratio of speed over time gives Acceleration of the motion.
• #### Types of force arrow_upward
###### Applied Force
• An applied force is a force that is applied to an object by a person or another object.
• If a person is pushing a desk across the room, then there is applied force acting upon the object.
• ###### Gravitational Force
• The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself.
• This is the weight of the object.
• All objects upon earth experience a force of gravity.
• It is directed "downward" towards the center of the earth.
• ###### Normal Force
• The normal force is the support force exerted upon an object that is in contact with another stable object.
• On occasions, a normal force is exerted horizontally between two objects that are in contact with each other.
• For instance, if a person leans against a wall, the wall pushes horizontally on the person.
• ###### Friction Force
• The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it.
• The friction force often opposes the motion of an object.
• ###### Tension Force
• The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends.
• The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.
• #### Balanced Forces arrow_upward
• Two equal and opposite forces applied on an object are called balanced forces.
• For example pushing and pulling the object with same force on both sides as shown in the figure.
• Forces are shown by arrows in diagram. The direction of the arrow shows the direction in which the force is acting.
• If two forces are balanced, it means that both the forces are equally applied but are acting in opposite directions.
• Balanced forces do not change motion
• If two balanced forces are acting on an object, that object will not change its motion. If it is still, it will stay still.
• If it is moving, it will continue moving, in the same direction and at the same speed.
• #### Unbalanced forces arrow_upward
• Unbalanced forces do change the way something is moving.
• Unbalanced forces can make objects start to move, speed up, slow down, or change direction.
• #### Friction arrow_upward
• The friction force acts in a direction parallel to the area of contact, and opposes the motion or the tendency to move.
• There are two types of friction force:
• Sliding friction
• Static friction
• Friction results when the two surfaces are being pressed together closely.
• It is because of intermolecular attractive forces between molecules of different surfaces.
• As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together.
• ###### Sliding Friction
• Sliding friction results when an object slides across a surface.
• Sliding friction results when object is in motion so it is also known as Kinetic Friction.
• ###### Static Friction
• Friction forces can also exist when the two surfaces are not sliding across each other. Such friction forces are referred to as static friction.
• Static friction results when the surfaces of two objects are at rest relative to one another and a force exists on one of the objects to set it into motion relative to the other object.
• #### Motion arrow_upward
• Motion is a change in position of an object with respect to time.
• ###### Kinds of Motion
• We can classify the motion of bodies as:
• Translatory (linear)
• Rotatory
• Vibratory or Oscillatory
• Circular
• Periodic
• Random
###### Translatory Motion
• The motion of an object is said to be translatory if the position of the object is changing with respect to a fixed point or object.
• Example:
• A boy running
• A car moving
###### Rotatory Motion
• Motion is said to be rotatory when the object rotates on its own axis
• Example:
• Spin Top
• CD or DVD playing
###### Oscillatory Motion
• Oscillatory Motion is repetitive and fluctuates between two locations.
• The to and fro motion of an object about a fixed point is called oscillatory motion or vibratory motion.
• Example:
• Seesaws
• Clock pendulum
###### Circular Motion
• A motion in which the body traverses a circular path is called Circular Motion.
• This is a kind of translatory motion where the body reaches the initial position each time it completes traversing the circle.
• Example:
• Hands of a clock
###### Periodic Motion
• Any motion that repeats itself at regular intervals of time is called Periodic Motion.
• Example:
• Heart beat
• A water wave
###### Random Motion
• Irregular Motion of bodies changing the nature of motion frequently is called Random Motion.
• Example:
• Motion of football player on the ground
• A flying butterfly
#### Newton's Laws of Motion arrow_upward
###### Newton's First Law of Motion
• An object at rest will remain at rest unless acted on by an unbalanced force.
• An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
• This law is often called "the law of inertia".
• ###### Newton's Second Law of Motion
• Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
• It can be expressed as a mathematical equation:
• Force=Mass × Acceleration
###### Newton's Third Law of Motion
• For every action there is an equal and opposite re-action.
• #### Thank You from Kimavi arrow_upward
• Please email us at Admin@Kimavi.com and help us improve this tutorial.
• Kimavi - An AI Powered Encyclopedia { Learning is Earning }
Get Ad Free Encyclopedia with Progress Report, Tutor Help, and Certificate of Learning for only \$10 a month | 1,389 | 6,408 | {"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-2020-10 | latest | en | 0.928967 |
https://metanumbers.com/3248 | 1,670,335,927,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00346.warc.gz | 420,990,153 | 7,341 | # 3248 (number)
3,248 (three thousand two hundred forty-eight) is an even four-digits composite number following 3247 and preceding 3249. In scientific notation, it is written as 3.248 × 103. The sum of its digits is 17. It has a total of 6 prime factors and 20 positive divisors. There are 1,344 positive integers (up to 3248) that are relatively prime to 3248.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 17
• Digital Root 8
## Name
Short name 3 thousand 248 three thousand two hundred forty-eight
## Notation
Scientific notation 3.248 × 103 3.248 × 103
## Prime Factorization of 3248
Prime Factorization 24 × 7 × 29
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 406 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,248 is 24 × 7 × 29. Since it has a total of 6 prime factors, 3,248 is a composite number.
## Divisors of 3248
1, 2, 4, 7, 8, 14, 16, 28, 29, 56, 58, 112, 116, 203, 232, 406, 464, 812, 1624, 3248
20 divisors
Even divisors 16 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 20 Total number of the positive divisors of n σ(n) 7440 Sum of all the positive divisors of n s(n) 4192 Sum of the proper positive divisors of n A(n) 372 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 56.9912 Returns the nth root of the product of n divisors H(n) 8.73118 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,248 can be divided by 20 positive divisors (out of which 16 are even, and 4 are odd). The sum of these divisors (counting 3,248) is 7,440, the average is 372.
## Other Arithmetic Functions (n = 3248)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1344 Total number of positive integers not greater than n that are coprime to n λ(n) 168 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 464 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 1,344 positive integers (less than 3,248) that are coprime with 3,248. And there are approximately 464 prime numbers less than or equal to 3,248.
## Divisibility of 3248
m n mod m 2 3 4 5 6 7 8 9 0 2 0 3 2 0 0 8
The number 3,248 is divisible by 2, 4, 7 and 8.
• Arithmetic
• Abundant
• Polite
• Practical
## Base conversion (3248)
Base System Value
2 Binary 110010110000
3 Ternary 11110022
4 Quaternary 302300
5 Quinary 100443
6 Senary 23012
8 Octal 6260
10 Decimal 3248
12 Duodecimal 1a68
20 Vigesimal 828
36 Base36 2i8
## Basic calculations (n = 3248)
### Multiplication
n×y
n×2 6496 9744 12992 16240
### Division
n÷y
n÷2 1624 1082.67 812 649.6
### Exponentiation
ny
n2 10549504 34264788992 111292034646016 361476528530259968
### Nth Root
y√n
2√n 56.9912 14.8094 7.54925 5.03875
## 3248 as geometric shapes
### Circle
Diameter 6496 20407.8 3.31422e+07
### Sphere
Volume 1.43528e+11 1.32569e+08 20407.8
### Square
Length = n
Perimeter 12992 1.05495e+07 4593.37
### Cube
Length = n
Surface area 6.3297e+07 3.42648e+10 5625.7
### Equilateral Triangle
Length = n
Perimeter 9744 4.56807e+06 2812.85
### Triangular Pyramid
Length = n
Surface area 1.82723e+07 4.03814e+09 2651.98
## Cryptographic Hash Functions
md5 c57abe86de4e516e12dfa386053fbfe2 92a502bcff564f82945507f6e08e701ea23a93b2 59ae69c1396f69953e89161299ffc52437f1cbc2f36b6a50273e505ef4f9f51a a39811760d8b43546b05645b5c5b004850c74f736e8e34f9d1b5e600a8f5a2d2c3c230f7ef189e9b4a179992246c9b2a89ff2f6d6753ec8e19029dc83128aeeb 765315fa59326b9580ccf51bdf9f6f7e1ed216b3 | 1,477 | 4,055 | {"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} | 3.75 | 4 | CC-MAIN-2022-49 | latest | en | 0.817114 |
https://www.musicreadingsavant.com/remembering-the-order-of-sharps-and-flats/ | 1,709,638,071,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00247.warc.gz | 862,705,199 | 44,015 | # Remembering The Order Of Sharps And Flats
You may have learned that flats and sharps appear in the key signature at the beginning of a piece of music.
But, did you know they will always be written in a certain order? The cool thing about this is once you memorize the order of sharps and flats, it is much easier to get started playing.
Let’s take a look at the sharp keys first.
## Order of Sharps
The order of sharps in the sharp keys will always appear like this: F#-C#-G#-D#-A#-E#-B#
This means that every time you see sharps in your key signature, they will always be written in this order making it easier to read at a glance.
For instance, let’s say I looked quickly at the key signature and noticed there are three sharps
I don’t have to stare at it very long to figure it out because I already have the order of sharps memorized.
I know right away that the three sharps are F#, C# and G#.
An easy way to remember the order of sharps is to memorize this saying: Fat-Cats-Go-Down-Alleys-Eating-Birds
This is a mnemonic device with the first letter of each word representing each sharp.
## Order of Flats
The order of flats is also easy to remember. The order of flats includes Bb-Eb-Ab-Db-Gb-Cb-Fb
An easy way to remember this is B-E-A-D-Go-Catch-Fish
Again, if you have this memorized, all you have to do is glance at the key signature and you will know what the flats are.
Let’s say we have four flats listed in the key signature. Without thinking about it too hard, remember the first four flats (B-E-A-D) and that will tell you very quickly what flats to play.
This is pretty easy, huh?
## Another Cool Fact…
The order of flats is the same as the order of sharps written backwards (F-C-G-D-A-E-B and B-E-A-D-G-C-F).
If anything, memorize one and re-write the order on paper backwards to figure out the other.
What’s very nice about memorizing the order of sharps and flats is that it gives you a nice short cut to reading music.
With this knowledge, you never have to stare at the key signature for too long before deciding what it is you need to play.
Your task now is to try memorizing the order of sharps and flats. Practice using this tool every time you learn a new piece of music.
You will be amazed how slick this tool actually works.
## Helpful Key Signature Flash Cards
1. \$9.99
Master key signatures with these flash cards covering the major keys, minor keys, and their relatives.
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02/18/2024 05:11 am GMT
2. \$10.99
A set of 48 key signature flash cards for treble, bass, and alto clefs. It correlates with the Alfred's Essentials of Music Theory Books, but can be used with any beginning music theory course book.
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02/18/2024 05:24 am GMT
3. \$16.00
Practice learning the key signatures with this all-in-one interactive double-sided flashcard. Learn the order of sharps and flats and all major and relative minor key signatures.
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02/18/2024 06:32 pm GMT
## Quick Reading Music Reference Guides
Popular Music Theory Cheat Sheet
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This foldable 8x10 inch music theory cheat sheet is an excellent quick reference guide when you need to find the answer fast. The side 3-hole punch allows you to keep it in a 3-ring binder. It is sturdy and folds out featuring music theory and notation on the front and music history on the back.
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Essential Dictionary of Music: The Most Practical and Useful Music Dictionary for Students and Professionals (Essential Dictionary Series)
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The Hal Leonard Pocket Music Dictionary
\$9.99
A convenient music theory book that is small enough to fit in your pocket, backpack, or instrument case. A great reference guide for all musicians at any level of music study.
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02/18/2024 10:37 pm GMT
### 10 thoughts on “Remembering The Order Of Sharps And Flats”
1. Great looking website.
Another mnemonic for sharps is:
Fat Cops Get Doughnuts After Every Bust
(My nephew is a cop, ha ha)
2. I’m 59 and started playing tuba in April for the first time in 40 years. Thank you for explaining key signatures in a way I immediately grasped and in a way I had not previously learned in school. What a huge relief! Again, thanks.
3. I use these sayings for sharps and flats:
Sharps” Father Charles Goes Down And Ends Battle”
Flats “Battle Ends And Down Goes Charles Father
4. For the sharps i remember Fat Chicks Get Diabetes After Eating Butter
And for my flats I remember Better Eat A Darn Good Chicken Fajita
They both work amazingly I love it hope it help you guys.
5. Chris Alexander
I have just started to learn the piano and read music at the age of 72. It has given me a new lease of life. I have a lesson a week at the Piano Academy in Skegness and it is amazing how much I have missed out on. I find these pages really encouraging. Chris Alexander
6. This is the best website I have found for making music theory a little easier to learn. I have recently started playing flute again after 50 or so years. Thank you. | 1,366 | 5,713 | {"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} | 3.09375 | 3 | CC-MAIN-2024-10 | longest | en | 0.938836 |
https://worthwhile.typepad.com/worthwhile_canadian_initi/2009/05/a-preliminary-estimate-for-canadian-2009q1-gdp-growth.html | 1,709,540,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476432.11/warc/CC-MAIN-20240304065639-20240304095639-00432.warc.gz | 612,511,144 | 11,331 | You can follow this conversation by subscribing to the comment feed for this post.
Stephen:
Your analysis is interesting, but one issue that really bugs me is the use of a normal distribution in your analysis. My gut feeling is that using a normal distribution is inappropriate since the probability of of the next outcome is not independent from the previous occurrence. By the way, I have no solution to that problem.
Interesting - I arrived at an estimate of around 7-8% using two models - a short-term forecasting model based on a handful of indicators and by incorporating financial linkages based on the using the Senior Loan Officer survey into my small macro model (without the financial linkages, it was not possible to get estimates as extreme as -7%)
The most dire forecast i've seen from Bay Street was 9.5% (David Wolf) - the others are in the -5 to -6% range.
It will be interesting to compare the size of Stephen's "revision" (when the quarterly data come out) to typical US revisions.
An economist at the Bank of Canada once told me he doesn't pay a lot of attention to monthly data simply because Canada is too small. A couple of large purchases can be big enough to cause a visible blip. Less of a problem in the US, since it all smooths out.
I did a bit of arithmetic, and I was surprised at how robust the quarterly growth estimate is to the missing month March GDP number. Using the previous five monthly growth rates (rounded to a single decimal place), I calculate that the growth rate for 2009Q1 satisfies G=-6.65+1.31*gM, where gM is the March monthly growth rate. If you plug in Stephen's estimate of gM=-.12, you get a first quarter growth rate of G=-6.8% (close enough to Stephen's mean estimate of -6.9% that the difference may be rounding error). But even if you let gM range from 0% to -.5%, the first quarter growth rate of GDP only ranges from -6.65 to -7.3. So a horrible first quarter number truly is "baked in".
The comments to this entry are closed.
• WWW | 455 | 2,001 | {"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-2024-10 | latest | en | 0.955127 |
https://www.physicsforums.com/threads/what-is-an-acoustically-soft-hard-scatterer.526272/ | 1,623,531,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586390.4/warc/CC-MAIN-20210612193058-20210612223058-00221.warc.gz | 878,654,114 | 14,968 | # What is an acoustically soft/hard scatterer?
Hi,
I'm currently interested in the numerical solution of acoustic scattering problems. Here, an acoustic wave propagates in a certain medium and hits (at some time) an obstacle. I am then interested in the computation of the scattererd wave. In the papers I read they distinguish between acoustically soft and hard obstacles/scatterers. Mathematically this means that I have to consider different boundary condition but I don't really know what this means physically. What is an example of a soft/hard scatterer?
It would be great if somebody could help me here. Thank you!
Would it have to do with the relative rigidity of the surface? IE: an ideal reflector would be perfectly rigid, with a reflection coefficient of 1 at all frequencies. "softer" surfaces would not be acoustically rigid, and would have a reflection coefficient less than 1 across a frequency range. In other words, the rigid surface would have an infinite acoustic impedance, while the soft surface would have some finite impedance. At the other extreme end (essentially the same as no obstacle), the impedance would be the characteristic impedance of the media (rho * c).
Hi,
thank you very much for your answer. It definitly goes in this direction. The soft and the hard scatterer for these problems are both extrem cases (there are also absorbing scatterers which are some kind of mixture).
It now seems to me that both, the soft and the hard scatterer, are some kind of perfect reflectors, i.e. they don't absorb energy but reflect in a different way (depending on the material properties?). I'm not a physicist so maybe this is nonsense, but it cannot be that one is a perfect reflector and the other doesn't reflect at all.
Maybe it helps if I say something about the boundary conditions. For the soft scatterer the total field u_s+u_i has to vanish at the boundary (u_s is the scattered wave, u_i is the incoming wave). For the hard scatterer the normal derivative of the total field vanishs at the boundary.
I found an http://www2.econ.univpm.it/servizi/hpp/recchioni/w2/virt1.htm" [Broken] which solves this problem in 2D. You can choose the right edge to be a soft resp. a hard scatterer.
Which material combinations could lead to these different types of scattered wave?
Thank you!
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Binary options work 70 | 3,445 | 15,005 | {"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} | 2.515625 | 3 | CC-MAIN-2017-17 | latest | en | 0.886691 |
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Slope Factor
8 Threads found on edaboard.com: Slope Factor
Sub-threshold slope factor of FINFET
Hi, can anyone help me out in how to measure the FINFET's sub-threshold slope factor using HSPICE. I need the command used to find the derivative of log(Id).
self-Built dual slope integration ADC resolution calculation
Nominal Resolution of dual-slope ADC is usually referring to the resolution of the time measurement of de-integration phase and is simply given by your hardware design. Usable resolution is a matter of various error terms and more difficult to determine. Ponts to consider are - integration capacitor loss factor - comparator response time
subthreshold n factor
Some fabs give Subthreshold slope S rather than Subthreshold slope factor n. You can see here how to convert between the two:
phase response of filter comments and various observation
Oshaye, because of the bad quality of the 2nd picture I can detect only two curves. Nevertheless, a 4th order lowpass always exhibits a maximum phase shift of 360 deg (4x90 deg). This is not surprising at all. The slope of the phase characteristic in the pole frequency region is always proportional to the pole quality factor. Thus, for diffe
Quality factor of a short circuited transimission line @ resonant freqency...
Hi From Pozar, I know that Q=beta/(2*alpha) = pi/(4*alpha*length); ---valid for short circuited transmision lines and from resonant theory (parallel resoant circuit) Q= slope(imag) * fo /(2R). I am using ADS, I use 2 port simulation to extract alpha and beta..and find Q?? or use one port simualtion to extract Q?? Which one is valid??
Photoresistors, need resistance
The datasheets are specifying a resistance range for a specific illumination, mostly 10 lux. The variation is typically +/- 50%. Furthermore they specify a logarithmitic slope factor γ, so you are able to calculate the resistance at different luminance levels. You won't get more accurate data, because the nature of these sensors apparently doe
How to find lambda values for models in particular technology?
you cant determine exact lambda value, because its a function of L and other device parameters.but to determine an avg value plot V-I curves, and find the slope
Calculating n slope factor to technology current
Hi all, is this method is valid for calculating n from week to strong inversion? or is there any alternate i need to calculate n(slope factor) to find technology current. Thanks | 549 | 2,496 | {"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} | 2.5625 | 3 | CC-MAIN-2018-13 | latest | en | 0.879453 |
https://r-lang.com/r-data-types-vector-list-matrix-array-and-data-frame/ | 1,674,996,041,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00589.warc.gz | 500,459,220 | 29,071 | # R Data Types: Vector, List, Matrix, Array, and Data frame
Every programming language has its data types, and almost all programming languages share some common data types like Integer, String, Character, Array, etc. R has an extensive type of data types, including vectors (numerical, character, logical), matrices, data frames, arrays, factors, and lists.
## R Data Types
In the R language, the variables are not declared as some data type. The variables are assigned with R-Objects, and the data type of the R-object becomes the data type of the variable, unlike other programming languages.
R lang has various R-objects, which are the following.
1. Vector
2. List
3. Factor
4. Matrix
5. Array
6. DataFrame
Among all the above R-objects, the vector is the simplest data type in R language.
## R Vector Data type
vector in R is a series of data items of the same basic type. Members in a vector are called components. When you want to create a vector with more than one element, use the c() function, combining them into a single vector.
``````x <- c(11, 21, 5.2, 6, -19, 0.46)
y <- c("eleven", "eight", "one")
z <- c(TRUE, FALSE, TRUE, FALSE, TRUE, FALSE)
x
typeof(x)
y
typeof(y)
z
typeof(z)``````
#### Output
``````[1] 11.00 21.00 5.20 6.00 -19.00 0.46
[1] "double"
[1] "eleven" "eight" "one"
[1] "character"
[1] TRUE FALSE TRUE FALSE TRUE FALSE
[1] "logical"``````
We have created three types of vectors using the c() function. To check the R object’s data type, use the typeof() function.
There are six types of vector objects available in R.
1. Logical vector
2. Numeric vector
3. Integer vector
4. Complex vector
5. Character vector
6. Raw vector
### Logical Vector
logical vector is a vector that only contains TRUE and FALSE values, which means it only contains boolean values.
``````logical_rv <- TRUE
logical_rv
class(logical_rv)``````
#### Output
``````[1] TRUE
[1] "logical"``````
The class() function determines the class of an object or its “internal” type.
### Numeric Vector
The numeric vectors create a real vector of the specified length. The elements of the vector are all equal to 0. The numeric vectors can have values like 11, 21.19, 1.1, etc.
``````numeric_rv <- 11.21
numeric_rv
class(numeric_rv)``````
#### Output
``````[1] 11.21
[1] "numeric"
``````
And we get the numeric vector. It has as.numeric() method, which is a generic function. To check if it is a numeric value or not, use the is.numeric() method.
### Integer Vector
The integer() method creates or tests for objects of type “integer“. The current implementation of R uses the 32-bit integers for integer vectors. Integer vectors can have values like 21L, 19L, 0L, etc.
``````integer_rv <- 21L
integer_rv
class(integer_rv)``````
#### Output
``````[1] 21
[1] "integer"``````
The as.integer() function tries to coerce its argument to be of integer type. To check if the value is an integer or not, use the is.integer() function.
### Complex Vector
The complex vector in R can be specified either by providing its length, its real and imaginary parts, or modulus and argument. The complex vector can have values like 1 + 2i, 2 + 5i, 1 + 0i, etc.
``````complex_rv <- 19 + 21i
complex_rv
class(complex_rv)``````
#### Output
``````[1] 19+21i
[1] "complex"``````
The as.complex() function tries to control its argument to be of complex type. To check if the data type is complex or not, use the is.complex() method. All forms of NA and NAN are coerced to a complex NA, for which both the real and imaginary parts are NA.
### Character Vector
Text in R is represented by character vectors. A character vector is a vector consisting of characters. To create a character vector, use the characters in double or single-quotes. You can have character vector values like x‘ , “data”,TRUE“, ‘21.19‘.
``````character_rv <- "a + b + c"
character_rv
class(character_rv)``````
#### Output
``````[1] "a + b + c"
[1] "character"``````
You can assign a value to a character vector using the assignment operator (<-), the same way you do for all other data types variables. To find out the length of the vector, use the length() function. To check if the variable holds a character vector or not, use the is.character() function.
### Raw Vector
The raw vector data type is intended to hold raw bytes. A raw vector is printed with each byte separately represented as a pair of hex digits. If you want to see a character representation (with escape sequences for non-printing characters), use the rawToChar() function.
The Hello” is stored as 48 65 6c 6c 6f as raw values. Let’s see with examples.
``````raw <- raw(8)
raw
class(raw)
raw_rv <- charToRaw("Database")
raw_rv
class(raw_rv)``````
#### Output
``````[1] 00 00 00 00 00 00 00 00
[1] "raw"
[1] 44 61 74 61 62 61 73 65
[1] "raw"``````
To convert raw to a character in R, use the rawToChar() function.
``````raw_rv <- charToRaw("Database")
raw_rv
class(raw_rv)
char_rv <- rawToChar(raw_rv)
char_rv
class(char_rv)``````
#### Output
``````[1] 44 61 74 61 62 61 73 65
[1] "raw"
[1] "Database"
[1] "character"``````
The as.raw() function tries to curb its argument to be of raw type. The answer will be 0 unless the coercion succeeds.
The is.raw() function returns TRUE if and only if typeof(x) == “raw”.
## List in R
A list is an R-object containing different types of elements inside it like vectors, functions, and even another list inside it. A list is an ordered collection of elements under one name.
``````r_list <- list(c(11, 19, 21), 19.3, cos, TRUE)
r_list``````
#### Output
``````[[1]]
[1] 11 19 21
[[2]]
[1] 19.3
[[3]]
function (x) .Primitive("cos")
[[4]]
[1] TRUE``````
To access list elements one by one, use the [[ ]] brackets. Let’s access the third element of the list. Remember, the list index starts at 1 and not 0.
``````r_list <- list(c(11, 19, 21), 19.3, cos, TRUE)
r_list[[3]]``````
#### Output
``function (x) .Primitive("cos")``
## Factor in R
The factor stores the nominal values as a vector of integers in the specific range and an internal vector of character strings mapped to these integers.
``````rv <- c(rep("adam", 3), rep("eve", 4))
r_factor <- factor(rv)
r_factor``````
#### Output
``````[1] adam adam adam eve eve eve eve
R treat factors as nominal variables and ordered factors as ordinal variables in statistical procedures and graphical analyses.
## Matrix in R
A matrix is a two-dimensional rectangular data structure that can be created using a vector input to the matrix function. The as.matrix() function attempts to turn its argument into the matrix. The is.matrix() function tests if its argument is a (strict) matrix.
``````mtrx <- matrix(c("a", "b", "c", "d"), nrow = 2, ncol = 2, byrow = TRUE)
mtrx``````
#### Output
`````` [,1] [,2]
[1,] "a" "b"
[2,] "c" "d"``````
## Array in R
Array in R is a list or vector with two or more dimensions. An array is like a stacked matrix, and a matrix is a two-dimensional array.
``````rv <- c(11, 19, 18)
rv2 <- c(21, 6, 29, 46, 37, 38)
result <- array(c(rv, rv2), dim = c(3, 3, 2))
print(result)``````
#### Output
``````, , 1
[,1] [,2] [,3]
[1,] 11 21 46
[2,] 19 6 37
[3,] 18 29 38
, , 2
[,1] [,2] [,3]
[1,] 11 21 46
[2,] 19 6 37
[3,] 18 29 38
``````
## Data Frame in R
Data Frame in R is a table or two-dimensional array-like structure in which a row contains a set of values, and each column holds values of one variable.
``````streaming <- data.frame(
service_id = c(1:5),
service_name = c("Netflix", "Disney+", "HBOMAX", "Hulu", "Peacock"),
service_price = c(18, 10, 15, 7, 12),
stringsAsFactors = FALSE
)
# Print the data frame.
print(streaming)``````
#### Output
`````` service_id service_name service_price
1 1 Netflix 18
2 2 Disney+ 10
3 3 HBOMAX 15
4 4 Hulu 7
5 5 Peacock 12``````
Unlike a matrix, in a data frame, each column can contain different modes of data. Data Frames are created using the data.frame() function.
## Useful inbuilt R Functions
Inbuilt functions are also the data types in R.
``````length(object)
str(object)
class(object)
names(object)
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That is it for the R Data Types example.
Categories R | 2,497 | 8,364 | {"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} | 2.984375 | 3 | CC-MAIN-2023-06 | latest | en | 0.834064 |
https://fr.slideserve.com/tranquilla/inferential-statistics | 1,632,533,478,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057584.91/warc/CC-MAIN-20210924231621-20210925021621-00325.warc.gz | 308,516,407 | 21,853 | Inferential Statistics
# Inferential Statistics
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## Inferential Statistics
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1. Inferential Statistics Minjuan Wang Educational Technology
2. *Inferential statistics • Projecting data from sample to population • Signal-to-Noise • Level of significance (a)/confidence level • Two basic types • Parametric • Non-parametric
3. Inferential Statistics • Inferential Statistics are: • Used to make inferences about populations based on the behavior of a sample • Concerned with how likely it is that a result based on a sample or samples are the same as results that might be obtained from an entire population
4. Video from learner.org • http://learner.org/resources/series65.html# • What Is Probability? • From 0-1 • Watch the rules of probability from minutes 23 or later • 50% will find an atypical driver in NYC • Significance Tests: • Watch it from 16:20 • Confidence Intervals
5. Examples • Homeschoolers in SD to homeschoolers in CA • Graduate students in SDSU to .. in the West Coast • Terrorists in the US to …around the world • Does a sample perfectly represents the population? • No!
6. Function • Therefore inferential statistics identify: • How likely the sample results represent the results that would occur in the population? • 90%, 95%, or 99% • Confidence interval • By being ? Confident, we make: • probability statements that the results we see in samples would also be found in the population
7. Level of Significance • An estimate of the probability that we are wrong when we say the results are due to chance • a=0.1, 0.05; 0.01 (10%, 5%, 1% of being wrong) • 0.10 (10%) for an exploratory test • .05 (5%) for many educational research tests • .01 (1%) if you are very confident • Comparing p with a? • Probability of the differences/effect/treatment are due to chance needs to < 0.05 (a)
8. Types of Hypothesis • Null hypothesis • There is no change, difference or relationship between A and B • A starting point or a benchmark • Coffee maker is broken • no relationship between its broken and the humming birds on the tree outside my balcony
9. Alternative/Research Hypothesis • Hypothesis that is implicitly accepted if the null hypothesis is rejected • Directional (one-tailed test) • Non-directional (two-tailed test) • Hypothesis: not to be proven but to be supported • Does your study fail if your hypothesis is not supported by the data?
10. Tails of A Test • Two-tailed test (non-directional/both) • There is no difference in content acquisition between "discovery learning" and "direct instruction.“ • One-tailed test (directional/upper/lower) • difference will be in one direction only • Students who use "discovery learning" exhibit greater gains in content acquisition than students who use "direct instruction"
11. Hypothesis Testing Procedure • Come up with a hypothesis • Set a: level of risk you are willing to take • Select the test • Compute the obtained value: t, f, r, etc. • Find the critical value in the respective table • To reject a null hypothesis->Obtained value must be > critical value • Otherwise, fail to reject null hypothesis (H0) • But, never “accept” null hypothesis • Many tests are needed to confirm that A is not different from or associated with B. • When rejecting null-P, the alternative 2-tailed P is implicitly accepted.
12. P from Fancy Schmancy Software • Inferential statistics • T, ANOVA, Correlation, Regression • P is the probability of chance (indicator of significance) • Free us from the test tables • Results vary • P<.05 • P<.001 • P=.013 (the exact probability of the outcome/effect due to chance—SPSS) • Outcome: difference, change, or association • P>.05 or p=ns (nonsignificant) • The probability of rejecting a null-P exceeds 0.05 (the cut-off point) (Salkind) • So reject it
13. Inferential Statistics • Two Basic Types • Parametric – techniques which make the assumption that you are working with a normal distributions and that the sample is random • Nonparametric – techniques which make few if any assumptions about the nature of the population from which the sample is taken
14. Does Culture Make a Difference? • The survey: • Perceptions about being equal with their instructor • Chinese, American, Korean students • Tests conducted • Kruskal-Wallis Analysis of Variance • Non-parametric of ANOVA • Results and Interpretation • P=0.02 comparing with a=0.05 • ???
15. Parametric versus Nonparametric • Parametric – • Characteristic is normally distributed in the population; sample was randomly selected; data is interval or ratio • Nonparametric • Use when you have a specialized population, you’ve not randomly selected, or data is ranked or nominal • “Cooking” • steamed versus fried • Streamed broccoli versus baked pumpkin pie • Link to a Table • Resource: http://coe.sdsu.edu/ed690/mod/mod06/default.htm
16. Inferential Statistics • Parametric Techniques • T-Test for means • Analysis of Variance • Analysis of Covariance
17. More: Inferential Statistics • Nonparametric Techniques for Quantitative Data • The Mann-Whitney U Test—for T(ea) for two • The Kruskal-Wallis One Way Analysis of Variance—for ANOVA • The Friedman Two-Way Analysis of Variance—for ANOVA • Nonparametric Technique for Categorical Data • Chi-Squared test of frequencies • Is there a relationship between eye and hair color?
18. Null Hypothesis • Cultural difference and fear of fat • Mean of Australian students = 100 • Mean of Indian students = 125 • Is this difference really significant? • Due to the cultural difference? • Due to chance (such as sampling error)? • If you make a null hypothesis • There is no significant difference or relationship…. • Assuming the difference is due to chance • Chance explanation for the difference…
19. Which Test to Use? • Tea for one or two? • Paired vs. unpaired • One: Pre- post- comparison • Two groups • ANOVA? • Chi-Square? • Correlational (Pearson r)? All these choices, you decide!
20. What If…..? • Adding one group to the study • New Zealand • Correlational? • Chi-Square? • ANOVA?
21. What If… • Variables are Nominal? • Is there a Statistically significant difference between EDTEC graduate students’ study preference (solo vs. teamwork)? • Data solo solo solo solo solo solo solo solo solo team team team team team team team team team team team
22. Types of Chi-Square • A one-dimensional Chi Square • Determine if the observed frequencies are significantly different from the expected frequencies Solo Team 10 10 9 12 Testing group distribution!
23. Types of CHI-SQUARE • A two-dimensional Chi Square • Frequencies are categorized along more than one dimensions • Gender relative to study preferences Solo Team 10 10 9 (5) 12 (10) Test the association between IV and DV Female Male
24. Chi Square • used when data are nominal (both IV and DV) • Comparing frequencies of distributions occurring in different categories or groups • Tests whether group distributions are different • EDTEC students’ preference for solo or teamwork • Determines the association between IV and DV by counting the frequencies of distribution • Gender relative to study preference
25. Degree of Freedom • State a null hypothesis • Select a level of significance • Select the appropriate test • Run statistics->get a result • Set degrees of freedom • The No. of instances in a distribution that is free to vary • to name 5 numbers, the mean needs to be 4 • Four numbers are free to vary (1, 2, 3, ,4) • The 5th number is set (10)
26. Degree of Freedom • Why? • If calculate the test by hand, the intersection of P and df determine the level needed to reject the null hypothesis • Refer to T table • Each test has its own formula • No. of groups, and no. of participants • Correlation r, N-2 • T test: • one group: N-1 • two groups: N1-1+N2-1 | 1,866 | 7,932 | {"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} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.865393 |
https://www.imlearningmath.com/an-officer-wishing-to-arrange-his-men-in-a-solid-square-found-by-his-first-arrangement-that-he-had-39-men-left-over/ | 1,603,357,233,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00578.warc.gz | 762,002,069 | 24,082 | # An officer wishing to arrange his men in a solid square found by his first arrangement that he had 39 men left over.
Math Riddle: An officer wishing to arrange his men in a solid square found by his first arrangement that he had 39 men left over. He then started increasing the number of men on a side by one but found that 50 additional men would be needed to complete a new square.
How many men did the officer have?
Answer: The correct answer is the officer had 1975 men.
When he formed a square measuring 44 by 44, he had 39 men over. When he tried to form a square 45 x 45, he was 50 men short. | 147 | 606 | {"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} | 3.40625 | 3 | CC-MAIN-2020-45 | longest | en | 0.995474 |
https://maqsad.io/classes/class-9/math/algebric-expressions-and-algebric-formulas/6-%20if%20p%202+%20sqrt%203%20find(iv)%20pexp%202%20-%20frac%201%20pexp%202 | 1,685,609,499,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00235.warc.gz | 433,379,650 | 14,423 | Classes
Class 9Class 10First YearSecond Year
$6. If p=2+\sqrt{3} find(iv) p^{2}-\frac{1}{p^{2}}$
$6. Perform the indicated operation and simplify.(i) \left(x^{2}-49\right) \cdot \frac{5 x+2}{x+7}$
$4. If x^{2}+y^{2}+z^{2}=78 and x y+y z+z x=59 then find the value \delta f x+y+z .$
$2. Find the conjugate of x+\sqrt{y} .(v) 5+\sqrt{7}$
$1. Rationalize the denominator of the following.(vi) \frac{2}{\sqrt{5}-\sqrt{3}}$
$Example 3If a+b+c=7 and a b+b c+c a=9 then find the value of a^{2}+b^{2}+c^{2} .$
$2. Find the conjugate of x+\sqrt{y} .(i) 3+\sqrt{7}$ | 240 | 561 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "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} | 3.75 | 4 | CC-MAIN-2023-23 | longest | en | 0.446059 |
http://www.learnhive.net/learn/common-core-standards-grade-5/mathematics/add-and-subtract-fractions | 1,571,370,514,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677884.28/warc/CC-MAIN-20191018032611-20191018060111-00229.warc.gz | 287,971,194 | 9,734 | ×
×
## Mathematics / Add and Subtract fractions
### Mathematics / Add and Subtract fractions
5.NF.1. Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2/3 + 5/4 = 8/12 + 15/12 = 23/12. (In general, a/b + c/d = (ad + bc)/bd.)
5.NF.2. Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.
Exercises
Topics
Exercises
Topics
• ### Addition and Subtraction of Fractions
Tags : Interpret fractions, word and mathematical problems involving fractions, greater than, less than.
• Questions: 366 | 250 | 1,052 | {"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} | 3.84375 | 4 | CC-MAIN-2019-43 | latest | en | 0.853661 |
http://mathhelpforum.com/differential-geometry/115530-increasing-sequence.html | 1,481,392,333,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543316.16/warc/CC-MAIN-20161202170903-00506-ip-10-31-129-80.ec2.internal.warc.gz | 172,584,466 | 9,720 | Let A be a subset of R that is bounded above and let $u= sup A$ . Prove that there is an increasing sequence $a_n$ with $a_n \in A , \forall \ n \in N$ SUCH THAT $a_n \to u$
Let A be a subset of R that is bounded above and let $u= sup A$ . Prove that there is an increasing sequence $a_n$ with $a_n \in A , \forall \ n \in N$ SUCH THAT $a_n \to u$
Let $A=[0,1]\cup\{2\}$.
It is true if we know that $u \notin A$. | 143 | 412 | {"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": 0, "img_math": 0, "codecogs_latex": 10, "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} | 2.875 | 3 | CC-MAIN-2016-50 | longest | en | 0.845245 |
https://www.brainkart.com/article/Fluid-Equation-of-continuity_3069/ | 1,685,253,545,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00134.warc.gz | 769,340,470 | 7,721 | Home | | Physics | Fluid Equation of continuity
# Fluid Equation of continuity
Consider a non-viscous liquid in streamline flow through a tube AB of varying cross section as shown in Fig. Let a1 and a2 be the area of cross section, v1 and v2 be the velocity of flow of the liquid at A and B respectively.
Equation of continuity
Consider a non-viscous liquid in streamline flow through a tube AB of varying cross section as shown in Fig. Let a1 and a2 be the area of cross section, v1 and v2 be the velocity of flow of the liquid at A and B respectively.
Volume of liquid entering per second at A = a1v1.
If ρ is the density of the liquid, then mass of liquid entering per second at A = a1v1ρ.
Similarly, mass of liquid leaving per second at B = a2v2ρ
If there is no loss of liquid in the tube and the flow is steady, then mass of liquid entering per second at A = mass of liquid leaving per second at B
(i.e) a1v1ρ = a2v2ρ or a1v1 = a2v2 i.e. av = constant
This is called as the equation of continuity. From this equation
v α a1 .
i.e. the larger the area of cross section the smaller will be the velocity of flow of liquid and vice-versa.
Total energy of a liquid
A liquid in motion possesses pressure energy, kinetic energy and potential energy.
(i) Pressure energy
It is the energy possessed by a liquid by virtue of its pressure.
(ii) Kinetic energy
It is the energy possessed by a liquid by virtue of its motion.
If m is the mass of the liquid moving with a velocity v, the kinetic energy of the liquid = ½ mv2
(iii) Potential energy
It is the energy possessed by a liquid by virtue of its height above the ground level.
If m is the mass of the liquid at a height h from the ground level, the potential energy of the liquid = mgh
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th 12th std standard Class Physics sciense Higher secondary school College Notes : Fluid Equation of continuity | | 476 | 1,973 | {"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} | 3.71875 | 4 | CC-MAIN-2023-23 | longest | en | 0.910339 |
https://im-beta.kendallhunt.com/HS/students/2/3/8/index.html | 1,713,527,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00707.warc.gz | 280,631,910 | 25,819 | # Lesson 8
Are They All Similar?
• Let’s prove figures are similar.
### 8.1: Stretched or Distorted? Rectangles
Are these rectangles similar? Explain how you know.
### 8.2: Faulty Logic
Tyler wrote a proof that all rectangles are similar. Make the image Tyler describes in each step in his proof. Which step makes a false assumption? Why is it false?
1. Draw 2 rectangles. Label one $$ABCD$$ and the other $$PQRS$$.
2. Translate rectangle $$ABCD$$ by the directed line segment from $$A$$ to $$P$$. $$A’$$ and $$P$$ now coincide. The points coincide because that’s how we defined our translation.
3. Rotate rectangle $$A’B’C’D’$$ by angle $$D’A’S$$. Segment $$A’’D’’$$ now lies on ray $$PS$$. The rays coincide because that’s how we defined our rotation.
4. Dilate rectangle $$A’’B’’C’’D’’$$ using center $$A’’$$ and scale factor $$\frac{PS}{AD}$$. Segments $$A’’’D’’’$$ and $$PS$$ now coincide. The segments coincide because $$A’’$$ was the center of the rotation, so $$A’’$$ and $$P$$ don’t move, and since $$D’’$$ and $$S$$ are on the same ray from $$A’’$$, when we dilate $$D’’$$ by the right scale factor, it will stay on ray $$PS$$ but be the same distance from $$A’’$$ as $$S$$ is, so $$S$$ and $$D’’’$$ will coincide.
5. Because all angles of a rectangle are right angles, segment $$A’’’B’’’$$ now lies on ray $$PQ$$. This is because the rays are on the same side of $$PS$$ and make the same angle with it. (If $$A’’’B’’’$$ and $$PQ$$ don’t coincide, reflect across $$PS$$ so that the rays are on the same side of $$PS$$.)
6. Dilate rectangle $$A’’’B’’’C’’’D’’’$$ using center $$A’’’$$ and scale factor $$\frac{PQ}{AB}$$. Segments $$A’’’’B’’’’$$ and $$PQ$$ now coincide by the same reasoning as in step 4.
7. Due to the symmetry of a rectangle, if 2 rectangles coincide on 2 sides, they must coincide on all sides.
### 8.3: Always? Prove it!
Choose one statement from the list. Decide if it is true or not.
If it is true, write a proof. If it is not, provide a counterexample.
Repeat with another statement.
Statements:
1. All equilateral triangles are similar.
2. All isosceles triangles are similar.
3. All right triangles are similar.
4. All circles are similar.
Here is an $$x$$ by $$x+1$$ rectangle and a 1 by $$x$$ rectangle. They are similar. What are the possible dimensions of these golden rectangles? Explain or show your reasoning.
### Summary
One figure is similar to another if there is a sequence of rigid motions and dilations that takes the first figure so that it fits exactly over the second. Consider any 2 circles, $$A$$ and $$B$$. Translate the circle centered at $$A$$ along directed line segment $$AB$$.
Now a dilation with center $$B$$ and a scale factor that is the length of the radius of the circle centered at $$B$$ divided by the length of the radius of the circle centered at $$A$$ will take the circle centered at $$A$$ onto the circle centered at $$B$$, proving that all circles are similar.
We can also show that all equilateral triangles are similar. Because we are talking about triangles, we can use the theorem that having all pairs of corresponding angles congruent and all pairs of corresponding side lengths in the same proportion is enough to prove that the triangles are similar. All the pairs of corresponding angles are congruent because all the angles in both triangles measure $$60^{\circ}$$. All the pairs of corresponding side lengths must be in the same proportion, because within each triangle, all the sides are congruent. Therefore, whatever scale factor works for one pair of sides will work for all 3 pairs of corresponding sides. If all pairs of corresponding sides are in the same proportion and all pairs of corresponding angles are congruent, then all equilateral triangles are similar.
### Glossary Entries
• similar
One figure is similar to another if there is a sequence of rigid motions and dilations that takes the first figure onto the second.
Triangle $$A'B'C'$$ is similar to triangle $$ABC$$ because a rotation with center $$B$$ followed by a dilation with center $$P$$ takes $$ABC$$ to $$A'B'C'$$. | 1,052 | 4,094 | {"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} | 4.90625 | 5 | CC-MAIN-2024-18 | latest | en | 0.806893 |
https://beavercountybookfest.com/interesting/question-how-fast-can-a-horse-walk.html | 1,660,973,346,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00575.warc.gz | 143,389,442 | 11,185 | # Question: How fast can a horse walk?
## How far can a horse walk in a day?
A horse can travel 100 miles in a day if it’s a fit endurance competitor. A typical trail horse in good shape can travel 50 miles a day, at a brisk walk with a few water breaks and time to cool down. Horses’ fitness level goes a long way in determining how far they can travel in a day.
## How long does it take a horse to walk 1 mile?
Well a regular person generally covers 1 mile in about 15 – 20 minutes. You horse will be going a bit faster than you though if you are trotting and cantering aswell. It really depends on the pace of everyone else but it could take you anywhere between 2 and 4 hours I’d say. I used to ride 15 miles roundtrip a day.
## How much faster is a horse than walking?
Humans can walk 3-4 miles per hour, while a horse can go all day at twice that speed. Horses walk at an average of 4 miles per hour, which is about on par with humans—maybe a touch faster. Humans also have greater stamina than horses; in particularly hot conditions, humans can even outrun horses over long distances!
## Will a horse run itself to death?
Yes, horses can run themselves to death. While running, horses place their cardiovascular and respiratory systems under a lot of pressure, which could, in some situations, lead to a heart attack, stroke, or respiratory failure, and lead to death.
## What does it mean when a horse stomps his foot?
Horses usually stomp when there is something irritating their skin, usually on the lower limbs. The most common cause is insects, but irritating substances placed on the skin, or generalized pain can cause this behavior too. Horses will also stomp their feet when they are bored, impatient or annoyed.
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## How long does it take to ride 5 miles on a horse?
An average horse walks at around the same speed as a human so about 4- 5 miles per hour. If you are adding trot into it then I reckon it would take about 1hr.
## How long does it take to ride 12 miles on a horse?
Practically speaking, you can expect to cover 15 miles on average terrain with reasonable footing in about four hours when traveling at a walk. Walking strides vary from breed to breed, it depends on the conformation of the horse so there will be some small variance in ground covered.
## How far can a horse travel in 2 hours?
Without accounting for obstacles, a horse usually walks at ~4mph. Obviously a gaited horse will walk faster and a pony or a WP horse, slower, but that is the typical range. So 9 miles would take you a little over 2 hours.
## How fast is a galloping horse?
Galloping involves the horse driving themselves forward with all four feet leaving the ground. It is a very fast smooth gait, and requires an athletic horse and rider. It averages between twenty five and thirty miles per hour and can only be sustained for short distances.
## How far can a man walk in a day?
While your body is made for walking, the distance you can achieve at an average walking pace of 3.1 miles per hour depends on whether you have trained for it or not. A trained walker can walk a 26.2- mile marathon in eight hours or less, or walk 20 to 30 miles in a day.
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## How long can a horse run flat out?
It depends, indeed. Some horses have better endurance capabilities than others so the answer may vary. However, according to some experienced riders, a horse can run for 24 to 72 hours nonstop before it becomes thoroughly exhausted and dies. However, that statement does not apply to all types of horse.
## What is the fastest animal on Earth 2020?
The mighty cheetah has been clocked at 75 mph — the speediest runner on the planet. Perhaps you know that the fastest animal in the sea, the sailfish, cruises through the water at 68 mph. In the sky, the peregrine falcon reigns supreme.
## Who is the world’s fastest horse?
Guinness World Record recognizes Winning Brew, a Thoroughbred, as the fastest horse in the world at 43.97 mph. Horses have survived on this planet because of their ability to run and communicate.
## Who is the richest jockey?
The rich list: which jockeys have earned the most money? Yutaka Take (1987-present) John Velazquez (1990-present) Javier Castellano (1996-present) Christophe Lemaire (1999-present) Bill Shoemaker (1949-90) Frankie Dettori (1986-present) Ryan Moore (2000-present) Fred Archer (1869-86) | 1,062 | 4,540 | {"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-2022-33 | latest | en | 0.949543 |
https://programming-idioms.org/idiom/65/format-decimal-number/534/d | 1,680,413,020,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950383.8/warc/CC-MAIN-20230402043600-20230402073600-00664.warc.gz | 536,176,183 | 7,848 | # Idiom #65 Format decimal number
From the real value x in [0,1], create its percentage string representation s with one digit after decimal point. E.g. 0.15625 -> "15.6%"
``import std.string : format;``
``string percent = format("%.1f%%", x * 100.0);``
``#include <stdio.h>``
``printf("%.1lf%%\n", x * 100);``
``string s = \$"{x:p1}";``
``````var s = "\${(x * 100).toStringAsFixed(1)}%";
``````
``s = "#{Float.round(x * 100, 1)}%"``
``write (*,'(F5.1,"%")') x*100.``
``import "fmt"``
``s := fmt.Sprintf("%.1f%%", 100.0*x)``
``s = Numeric.showFFloat (Just 1) (100*x) "%"``
``````const percentFormatter = new Intl.NumberFormat('en-US', {
style: 'percent',
maximumSignificantDigits: 3
});
const s = percentFormatter.format(x);``````
``const s = Math.round (x * 1000) / 10 + '%'``
``String s = String.format("%.1f%%", x * 100f);``
``import java.text.DecimalFormat;``
``String s = new DecimalFormat("0.0%").format(x);``
``s = string.format("%.1f%%", x*100)``
``\$s = number_format(\$x * 100.0, 1) . '%';``
``SysUtils``
``s :=format('%.1f%%', [100.0*x]); ``
``my \$s = sprintf '%.1f%%', \$x * 100;``
``s = f"{x:.01%}"``
``s = '{:.1%}'.format(x)``
``s = "%.1f%%" % (100 * x)``
``let s = format!("{:.1}%", 100.0 * x);``
programming-idioms.org | 451 | 1,240 | {"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-2023-14 | longest | en | 0.328338 |
https://sepwww.stanford.edu/sep/prof/toc_html/bei/ft1/paper_html/node7.html | 1,721,409,470,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00063.warc.gz | 434,930,719 | 2,590 | Next: CORRELATION AND SPECTRA Up: INVERTIBLE SLOW FT PROGRAM Previous: INVERTIBLE SLOW FT PROGRAM
## The simple FT code
Subroutine simpleft() exhibits features found in many physics and engineering programs. For example, the time-domain signal (which is denoted tt()"), has nt values subscripted, from tt(1) to tt(nt). The first value of this signal tt(1) is located in real physical time at t0. The time interval between values is dt. The value of tt(it) is at time t0+(it-1)*dt. We do not use if'' as a pointer on the frequency axis because if is a keyword in most programming languages. Instead, we count along the frequency axis with a variable named ie.
subroutine simpleft( adj, add, t0,dt,tt,nt, f0,df, ff,nf)
complex cexp, cmplx, tt(nt), ff(nf)
real pi2, freq, time, scale, t0,dt, f0,df
pi2= 2. * 3.14159265; scale = 1./sqrt( 1.*nt)
df = (1./dt) / nf
f0 = - .5/dt
do ie = 1, nf { freq= f0 + df*(ie-1)
do it = 1, nt { time= t0 + dt*(it-1)
ff(ie)= ff(ie) + tt(it) * cexp(cmplx(0., pi2*freq*time)) * scale
else
tt(it)= tt(it) + ff(ie) * cexp(cmplx(0.,-pi2*freq*time)) * scale
}}
return; end
The total frequency band is radians per sample unit or Hz. Dividing the total interval by the number of points nf gives .We could choose the frequencies to run from 0 to radians/sample. That would work well for many applications, but it would be a nuisance for applications such as differentiation in the frequency domain, which require multiplication by including the negative frequencies as well as the positive. So it seems more natural to begin at the most negative frequency and step forward to the most positive frequency.
Next: CORRELATION AND SPECTRA Up: INVERTIBLE SLOW FT PROGRAM Previous: INVERTIBLE SLOW FT PROGRAM
Stanford Exploration Project
12/26/2000 | 519 | 1,819 | {"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} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.847372 |
https://fr.mathworks.com/matlabcentral/answers/511733-graph-plotting-help-needed | 1,590,482,484,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390755.1/warc/CC-MAIN-20200526081547-20200526111547-00088.warc.gz | 333,822,617 | 22,351 | # Graph plotting HELP NEEDED
1 view (last 30 days)
Saad Rana on 19 Mar 2020
Answered: Steven Lord on 19 Mar 2020
How can I plot this graph in matlab?
I have trouble plotting this one
Show 1 older comment
Saad Rana on 19 Mar 2020
i'm new to matlab and I was just trying to plot this
Saad Rana on 19 Mar 2020
actually I have to convolve these two signals but I couldn't draw either of these signals
Sriram Tadavarty on 19 Mar 2020
Saad, i suggest you to go through MATLAB onramp course here https://www.mathworks.com/learn/tutorials/matlab-onramp.html
To solve this particular problem, you could go through the links of plot, conv functions. Once try with the links and they will help.
Star Strider on 19 Mar 2020
Try this:
x = linspace(-1, 3);
y = 2*((x >= 0) & (x < 1)) + 1*((x >= 1) & (x < 2));
figure
stairs(x, y)
grid
axis([-1 3 -1 3])
Steven Lord on 19 Mar 2020
If you're trying to create a plot that has a particular appearance but you're not sure which function can create such a plot, open the Plots tab on the Toolstrip. This includes a gallery where you can look at small thumbnails and see which function was used to create each. Find a plot that looks similar to what you want and look for help or doc on the function whose name appears below the thumbnail.
If you select the data you want to plot in the Workspace window, you can even filter that list to those functions that are appropriate for the selected data. | 394 | 1,430 | {"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-2020-24 | latest | en | 0.92912 |
https://blogs.glowscotland.org.uk/glowblogs/rsneportfolio/tag/graphs/ | 1,632,381,183,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00208.warc.gz | 187,551,880 | 14,474 | # Statistics can save lives!
From the lecture presentation I developed my understanding of the fundamental mathematics behind statistics as statistics couldn’t be understood without starting with fundamental basic mathematics. Also, statistics are an example of longitudinal coherence as they give a full picture and can be broken down to basic concepts that were built upon each other which Ma (2010, p.104) states is a profound understanding of fundamental mathematics. In primary school, you are taught how to do tally and other sorts of graphs. The knowledge gained of recording and creating these graphs is built upon in order to create charts for medical reasons. Thus, a profound understanding of fundamental mathematics has major implications, having this knowledge can keep people alive or it can be fatal as child can be poisoned if too high of dose is given to a child (Hothersall, 2016). These drug doses are given per kg. Therefore, children are weighed so the drug doses can be worked out. Here is a basic concept in maths that is required for use in medicine, weight which involves measurement and this is applied or links to medicine to save lives (Ma, 2010, p. 104).
Furthermore, junior doctors are expected have a profound understanding of fundamental mathematics as they are to know ratios and statistics to do with for example, the risk and probabilities of what one single cigarette can do to you (Hothersall, 2016). Therefore, this part of medicine involves several basic concepts of mathematics such as probabilities, ratio, change and recording (Ma, 2010, p.104). Even breaking your leg requires knowledge of the basic concepts as it’s due to forces and the angle, the speed that something might hit your leg at to break it (Porta, D.J).
However, maths should be taught in a fun and relevant way as it’s important to show the relevance of maths and helping them see that maths in science. So, when you are doing a fun experiment in class, help them see its fun. Maths is fun so inspire their love and interest in maths. I also want to help develop pupil’s knowledge of profound understanding by demonstrating this example of longitudinal coherence to them. The importance and relevance of statistics or why we learn the basics in maths like making graphs. I know that when I went to school I was learning how to draw tan graphs but I saw no point in learning it because I was not told the relevance.
An interesting point is that statistics links to social media. For public health reasons, there is an “Ailment Topic Aspect Model” that has prior knowledge of ailments. This model the analyses tweets to search and track sickness or illness over a period of time by “…measuring behaviour risk factors, locating illness by geographical region, and analysing symptoms and medication usage.” (Paul and Dredze, no date, p.1). However, a weakness of using twitter for statistics is that surely many people don’t tweet seriously, they may exaggerate their problems, they could be lying for a laugh or attention seeking. Furthermore, many account are private so how are all the tweets from these accounts accounted for in their data? What if the majority of information about a recent illness is on these accounts? Additionally, the symptoms that people might state on twitter could potentially be too vague.
A drawback should be noted about statistics. Although they can be useful and save lives, they are not always correct. There is such a thing as bias statistics especially in advertisement. For example, Colgate toothpaste claimed that 80% of dentists recommended Colgate but this was misleading as dentists were given a list of options to choose which toothpaste in comparison to the other competitors (Derbyshire, 2007). Therefore, fundamental mathematics can be used in a negative way in wider society. Therefore, In the future I would like to teach children that they need to critically evaluate statics and use multiple perspectives to look at the different ways statistics could be looked at as different perspectives can tell you different things (Ma, 2010, p. 104).
The video below gives some examples of negative uses of statistics (TED-Ed, 2016).
In conclusion, statistics are an example of how fundamental mathematics such as longitudinal coherence can be applied to wider societal issues. A drawback to statistics is that they can be misleading however, they can also help save lives and this is why learning mathematics is important!
List of references:
Hothersall, E. (2016) ‘Numeracy: Every contact counts (or something)’ [PowerPoint presentation]. ED21006: Discovering Mathematics (Year 2) (17/18) Available at: https://my.dundee.ac.uk/webapps/blackboard/content/listContent.jsp?course_id=_56905_1&content_id=_4941433_1&mode=reset (Accessed 9 November 2017).
Ma, L. (2010) Knowing and teaching elementary mathematics: teachers’ understanding of fundamental mathematics in China and United States. (Anniversary Ed.) New York: Routledge.
Paul, M. J., and Dredze, M. (no date) You Are What You Tweet: Analyzing Twitter for Public Health. Available at: http://www.cs.jhu.edu/~mpaul/files/2011.icwsm.twitter_health.pdf (Accessed: 9 November 2017).
Porta, D.J. (no date) Biomechanics of Impact Injury Available at: http://eknygos.lsmuni.lt/springer/659/279-310.pdf (Accessed: 9 November 2017).
TED-Ed (2016) How statistics can be misleading – Mark Liddell. Available: https://www.youtube.com/watch?v=sxYrzzy3cq8 (Accessed: 15 November 2017). | 1,184 | 5,488 | {"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-2021-39 | latest | en | 0.953902 |
https://www.bankersadda.com/reasoning-quiz-for-canara-po-exam-22nd | 1,569,011,077,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574077.39/warc/CC-MAIN-20190920200607-20190920222607-00543.warc.gz | 740,454,943 | 29,123 | # Reasoning Quiz for Canara PO Exam: 22nd December 2018
Dear Aspirants,
### Reasoning Quiz for Canara PO 2018
Reasoning Ability is an onerous section. With the increasing complexity of questions, it becomes hard for one to give it the cold shoulder. The only way to make the grade in this particular section in the forthcoming banking exams like Canara PO Mains is to practice continuously with all your heart and soul. And, to let you practice with the best of the latest pattern questions, here is the Adda247 Reasoning Quiz based on the exact same pattern of questions that are being asked in the exams.
Directions (1-5): Study the following information carefully and answer the given questions.
Eight persons A, B, C, D, E, F, G and H sit in a row and some of them face north and some face south. B and C sit third to the left of each other. Same number of person sit to the left of B and C. D is one of the neighbor of B and E is one of the neighbor of C. Both D and E face north. A sits to the right B and faces south. More than two person sit between D and F. F is not sitting at any extreme end. H sits to the left of G who does not face north. Person sit at extreme end face opposite direction. Immediate neighbor of C doesn’t face same direction.
Q1. How many persons sit between D and H?
Two
One
Three
Four
None of these
Solution:
Q2. Which of the following statement is true according to the arrangement?
E sits 2nd left to F
G is exactly between C and H
C does not face south
B faces north
none of these
Solution:
Q3. Four of the following belongs to a group find the one that does not belong to that group?
A
F
C
B
G
Solution:
Q4. Who among the following sits 4th right to G?
E
F
C
B
no one
Solution:
Q5. Which among the following pair of persons represents the ones sitting 2nd from both the ends?
D, G
F, J
F, H
B, D
none of these
Solution:
Directions (6-10): Study the following information carefully and answer the given questions.
A @ B means A is father of B
A % B means A is husband of B
A \$ B means A is sister of B
A £ B means A is mother B
A ¥ B means A is brother of B
Q6. What should come in place of question mark to make the expression “T is grandson of Q” true?
V \$ Q £ W ? Y ¥ T % Z
@
£
¥
%
Either (a) or (b)
Q7. Which of the following statements is false if the given expression is definitely true?
S % N £ P ¥ Q % R \$ O
R is sister-in-law of P
O is sister in law of Q
S is father of Q
S is father in law of R
S is father of P
Solution:
Q8. What should come in place of question mark to make the expression “Y is brother in law of Z” true?
V \$ Q £ W ¥ Y ? T % Z
@
£
¥
%
Either (a) or (b)
Q9. Which of the following statements is true if the given expression is true?
X @ Y \$ Z ¥ P % Q \$ R
R is brother in law of P
X is grandmother of P
Q is daughter of X
Y is sister of P
None of these
Solution:
Q10. Which of the following statements is true if the given expression is definitely true?
S % N £ P ¥ Q % R \$ O
O is sister of R
P is grandmother of O
Q is Son of S
Q is daughter of S
None of these
Solution:
Directions (11-15): A word and number arrangement machine when given an input line of words and numbers rearranges them following a particular rule in each step. The following is an illustration of input and rearrangement.
Input: 81 41 tooffs 22 toots 32 45 tobed toolls 26 tobes
Step I: 22 41 tooffs toots 32 45 tobed toolls 26 tobes 81
Step II: 22 26 41 tooffs toots 32 tobed toolls tobes 81 45
Step III: 22 26 32 tooffs toots tobed toolls tobes 81 45 41
Step IV: 22 26 32 tobed tooffs toots toolls tobes 81 45 41
Step V: 22 26 32 tobed tobes tooffs toots toolls 81 45 41
Step VI:22 26 32 tobed tobes tooffs toolls toots 81 45 41
Step VI is the final step after rearrangement.
As per the rules followed in the above steps, find out in each of following questions the appropriate steps for the given input.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Q11. Which word/number would be at the fifth position from left end in step IV?
brick
31
glow
rice
None of these
Solution:
In the following machine input-output, first the numbers are arranged at both the ends and then the numbers are arranged according to alphabetical series.
For numbers- Lowest number is arranged at the left end and the highest number is arranged at the right end. Then the 2nd lowest number is arranged to the right of the lowest number and the 2nd highest number is arranged to the right of the highest number I n step 2 and so on till all the numbers are arranged.
For words- They are arranged in alphabetical order.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Step I: 16 48 19 glow 58 rice cat 31 brick 71
Step II: 16 19 48 glow rice cat 31 brick 71 58
Step III: 16 19 31 glow rice cat brick 71 58 48
Step IV: 16 19 31 brick glow rice cat 71 58 48
Step V: 16 19 31 brick cat glow rice 71 58 48
Step V is the last step of the arrangement.
Q12. Which word/number would be at the 4th position from right end in step II?
brick
31
glow
rice
None of these
Solution:
In the following machine input-output, first the numbers are arranged at both the ends and then the numbers are arranged according to alphabetical series.
For numbers- Lowest number is arranged at the left end and the highest number is arranged at the right end. Then the 2nd lowest number is arranged to the right of the lowest number and the 2nd highest number is arranged to the right of the highest number I n step 2 and so on till all the numbers are arranged.
For words- They are arranged in alphabetical order.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Step I: 16 48 19 glow 58 rice cat 31 brick 71
Step II: 16 19 48 glow rice cat 31 brick 71 58
Step III: 16 19 31 glow rice cat brick 71 58 48
Step IV: 16 19 31 brick glow rice cat 71 58 48
Step V: 16 19 31 brick cat glow rice 71 58 48
Step V is the last step of the arrangement.
Q13. Which step number should be the following output? “16 19 31 brick cat glow rice 71 48 58”
Step III
Step IV
Step II
Step VI
No Such step
Solution:
In the following machine input-output, first the numbers are arranged at both the ends and then the numbers are arranged according to alphabetical series.
For numbers- Lowest number is arranged at the left end and the highest number is arranged at the right end. Then the 2nd lowest number is arranged to the right of the lowest number and the 2nd highest number is arranged to the right of the highest number I n step 2 and so on till all the numbers are arranged.
For words- They are arranged in alphabetical order.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Step I: 16 48 19 glow 58 rice cat 31 brick 71
Step II: 16 19 48 glow rice cat 31 brick 71 58
Step III: 16 19 31 glow rice cat brick 71 58 48
Step IV: 16 19 31 brick glow rice cat 71 58 48
Step V: 16 19 31 brick cat glow rice 71 58 48
Step V is the last step of the arrangement.
Q14. Which word/number would be at the 6th position from right end in step III?
brick
31
glow
rice
None of these
Solution:
In the following machine input-output, first the numbers are arranged at both the ends and then the numbers are arranged according to alphabetical series.
For numbers- Lowest number is arranged at the left end and the highest number is arranged at the right end. Then the 2nd lowest number is arranged to the right of the lowest number and the 2nd highest number is arranged to the right of the highest number I n step 2 and so on till all the numbers are arranged.
For words- They are arranged in alphabetical order.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Step I: 16 48 19 glow 58 rice cat 31 brick 71
Step II: 16 19 48 glow rice cat 31 brick 71 58
Step III: 16 19 31 glow rice cat brick 71 58 48
Step IV: 16 19 31 brick glow rice cat 71 58 48
Step V: 16 19 31 brick cat glow rice 71 58 48
Step V is the last step of the arrangement.
Q15. How many steps are required to get the final output?
3
4
6
5
7
Solution:
In the following machine input-output, first the numbers are arranged at both the ends and then the numbers are arranged according to alphabetical series.
For numbers- Lowest number is arranged at the left end and the highest number is arranged at the right end. Then the 2nd lowest number is arranged to the right of the lowest number and the 2nd highest number is arranged to the right of the highest number I n step 2 and so on till all the numbers are arranged.
For words- They are arranged in alphabetical order.
Input: 48 19 glow 71 58 rice 16 cat 31 brick
Step I: 16 48 19 glow 58 rice cat 31 brick 71
Step II: 16 19 48 glow rice cat 31 brick 71 58
Step III: 16 19 31 glow rice cat brick 71 58 48
Step IV: 16 19 31 brick glow rice cat 71 58 48
Step V: 16 19 31 brick cat glow rice 71 58 48
Step V is the last step of the arrangement. | 2,469 | 8,755 | {"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} | 3.703125 | 4 | CC-MAIN-2019-39 | latest | en | 0.95132 |
https://in.mathworks.com/matlabcentral/cody/problems/73-replace-nans-with-the-number-that-appears-to-its-left-in-the-row/solutions/54010 | 1,590,912,202,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347411862.59/warc/CC-MAIN-20200531053947-20200531083947-00355.warc.gz | 385,342,099 | 15,659 | Cody
# Problem 73. Replace NaNs with the number that appears to its left in the row.
Solution 54010
Submitted on 29 Feb 2012 by Yvan Lengwiler
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 = [NaN 1 2 NaN 17 3 -4 NaN]; y_correct = [0 1 2 2 17 3 -4 -4]; assert(isequal(replace_nans(x),y_correct))
2 Pass
%% x = [NaN 1 2 NaN NaN 17 3 -4 NaN]; y_correct = [ 0 1 2 2 2 17 3 -4 -4]; assert(isequal(replace_nans(x),y_correct))
3 Pass
%% x = [NaN NaN NaN NaN]; y_correct = [ 0 0 0 0]; assert(isequal(replace_nans(x),y_correct))
4 Pass
%% x = [1:10 NaN]; y_correct = [ 1:10 10]; assert(isequal(replace_nans(x),y_correct)) | 269 | 751 | {"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-2020-24 | latest | en | 0.520426 |
https://lists.gnu.org/archive/html/lilypond-devel/2021-11/msg00187.html | 1,675,927,944,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00125.warc.gz | 392,619,646 | 3,781 | lilypond-devel
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## Re: TimeSignature with note in denominator
From: David Kastrup Subject: Re: TimeSignature with note in denominator Date: Sun, 14 Nov 2021 20:01:06 +0100 User-agent: Gnus/5.13 (Gnus v5.13) Emacs/28.0.50 (gnu/linux)
```Carl Sorensen <c_sorensen@byu.edu> writes:
> On 11/14/21, 9:33 AM, "David Kastrup" <dak@gnu.org> wrote:
>
> Kieren MacMillan <kieren@kierenmacmillan.info> writes:
>
> > Hi David,
> >
> >> How is that uniquely identified? Why couldn't it be
> > subscripted with 10 instead of 5?
> >
> > I suppose it could. It could also be subscripted with a π or a √2. I
> > can’t stop people from doing what they want to do.
> >
> > Simultaneously true is the fact that the musical duration “one
> > quintuplet-sixteenth” has one and only one visual representation,
> > regardless of what Lilypond thinks or is told to do.
>
> Again you are evading the stated problem. The question was about the
> representation of time signature 8/20, not about "one
> quintuplet-sixteenth". 8/20 does not specify more than the basic
> subdivision for expressing beats (not necessarily identical with the
> number of beats as signatures like 9/8 show) and how much material fits
> a bar. It does not identify how that material may be structured or
> meaning the parts of a time signature are supposed to inherently have,
> leading to a proposal of generally changing the current representation
> by involving musical durations for the denominator.
>
> David,
>
> Do disagree with the statement that "The 20 on the bottom of the time
> signature indicates a duration of 1/20 of a whole note"?
In LilyPond terms, 1/20 of a whole note is not a duration. It can be a
Moment.
> If you disagree with this, what do you think the 20 on the bottom of
> the time signature means?
1/20 of the length of a whole note. Which is not a duration as such in
LilyPond and thus cannot be properly represented by one.
--
David Kastrup
``` | 566 | 2,041 | {"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-2023-06 | latest | en | 0.887302 |
http://www.caraudio.com/forums/archive/index.php/t-168277.html?s=bab1b23c5cb52e04ead6ef1a20ce0f25 | 1,406,785,435,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510272584.13/warc/CC-MAIN-20140728011752-00255-ip-10-146-231-18.ec2.internal.warc.gz | 421,738,443 | 2,668 | PDA
View Full Version : Help with slot port.
Cahm Rohn
06-29-2006, 05:03 PM
I want to build a box for my 2 mtx 9512-44's. The plan is to go 6.89cuft. @32 hz port length of 24.74. So how long would I have to make the walls. Box is L38 W24 H16
Cahm Rohn
06-29-2006, 06:01 PM
bump anyone
Cahm Rohn
06-30-2006, 08:52 AM
Breakfast bump
wesl56
06-30-2006, 09:01 AM
why are you set on that length? you should decide on a port area first, then calculate the port length
Cahm Rohn
06-30-2006, 09:11 AM
Only set on 32hz that is the tength the calculator gave me
hzsogood
06-30-2006, 09:22 AM
14.5x3 " port @ 32 hz is only a 10" long port :eyebrow:
hzsogood
06-30-2006, 09:24 AM
I want to build a box for my 2 mtx 9512-44's. The plan is to go 6.89cuft. @32 hz port length of 24.74. So how long would I have to make the walls. Box is L38 W24 H16
The tuning i got for the length you have w/ a 14.5x3" port area is around 24 hz
Cahm Rohn
06-30-2006, 09:27 AM
Sorry port area is 6.5x14.5
hzsogood
06-30-2006, 09:31 AM
I want to build a box for my 2 mtx 9512-44's. The plan is to go 6.89cuft. @32 hz port length of 24.74. So how long would I have to make the walls. Box is L38 W24 H16
Well u have ur answer now... Make the port 24 3/4 long.... At 6.5 inches wide... :eyebrow: Not sure what ur asking
hzsogood
06-30-2006, 09:33 AM
Sorry port area is 6.5x14.5
Too much port area imo... almost 100^in for those 12s
Cahm Rohn
06-30-2006, 09:38 AM
Ok I used georges tutorial and it f*%\$#@ me all up, so how would I find out what the actual length of the walls would have to be.
hzsogood
06-30-2006, 09:45 AM
Ok I used georges tutorial and it f*%\$#@ me all up, so how would I find out what the actual length of the walls would have to be.
what walls ??? port walls?? Its 23 3/4 in long..... You make a bend 6.5 inches from the back wall of the box and then make the port the rest of the length along the back wall...
hzsogood
06-30-2006, 09:47 AM
your port will be 16inches down the depth side(24 in ) and 8 " down the back off the wall .. L shape
Cahm Rohn
06-30-2006, 12:45 PM
Thanx thats almost what I was coming up with I guess I was trying to make it harder than what it was.
Cahm Rohn
06-30-2006, 01:06 PM
The min. usable port area was like 84 sq in., would going up about 10 affect it that much?
hzsogood
06-30-2006, 01:09 PM
The min. usable port area was like 84 sq in., would going up about 10 affect it that much?
no, i like more port area, i just thought thatwas a lil much , u should be good tho, post pics when u can
Cahm Rohn
06-30-2006, 02:17 PM
Im more worried about the vids, these subs get stupid loud and their only in about 3cubes ported prefab (cough cough), but the vfl help them turn my car to jello.:) | 986 | 2,728 | {"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} | 2.984375 | 3 | CC-MAIN-2014-23 | latest | en | 0.866297 |
https://file.scirp.org/Html/9-7402730_59834.htm | 1,718,410,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00826.warc.gz | 223,613,421 | 15,754 | Itô Formula for Integral Processes Related to Space-Time Lévy Noise
Applied Mathematics
Vol.06 No.10(2015), Article ID:59834,13 pages
10.4236/am.2015.610156
Itô Formula for Integral Processes Related to Space-Time Lévy Noise
Raluca M. Balan*, Cheikh B. Ndongo
Department of Mathematics and Statistics, University of Ottawa, Ottawa, Canada
Received 25 April 2015; accepted 20 September 2015; published 23 September 2015
ABSTRACT
In this article, we give a new proof of the Itô formula for some integral processes related to the space-time Lévy noise introduced in [1] [2] as an alternative for the Gaussian white noise perturbing an SPDE. We discuss two applications of this result, which are useful in the study of SPDEs driven by a space-time Lévy noise with finite variance: a maximal inequality for the p-th moment of the stochastic integral, and the Itô representation theorem leading to a chaos expansion similar to the Gaussian case.
Keywords:
Lévy Processes, Poisson Random Measure, Stochastic Integral, Itô Formula, Itô Representation Theorem
1. Introduction
Random processes indexed by sets in the space-time domain are useful objects in stochastic analysis, since they can be viewed as mathematical models for the noise perturbing a stochastic partial differential equation (SPDE). In the recent years, a lot of effort has been dedicated to studying the behaviour of the solution of basic equations (like the heat or wave equations), driven by a Gaussian white noise. This type of noise was introduced by Walsh in [3] and is defined as a zero-mean Gaussian process, with covariance
, where denotes the Lebesgue measure and is the class of bounded
Borel sets in.
In the recent articles [1] [2] , a new process has been introduced as an alternative for the Gaussian white noise perturbing an SPDE, which has a structure similar to a Lévy process. We introduce briefly the definition of this process below.
Let N be a Poisson random measure (PRM) on of intensity where and is a Lévy measure on:
We denote by the compensated PRM defined by for any Borel set A in with. The Lévy-type noise process mentioned above is defined as, where
for some. It was shown in [2] that Z is an “independently scattered random measure” (in the sense of [4] ) with characteristic function:
(In particular, Z can be an a-stable random measure with, as in Definition 3.3.1 of [5] .) One can define the stochastic integral of a process with respect to Z and for a certain integrands,
The stochastic integral with respect to (or N) can be defined using classical methods (see e.g. [6] ). We review briefly this definition here.
Assume that N is defined on a probability space. On this space, we consider the filtration
where is the class of bounded Borel sets in and is the class of Borel sets in which are bounded away from 0.
An elementary process on is a process of the form
where, X is an -measurable bounded random variable, and. A pro- cess is called predictable if it is measurable with respect to the s-field
generated by all linear combinations of elementary processes.
As in the classical theory, for any predictable process H such that
(1)
we can define the stochastic integral of H with respect to and the process
is a zero-mean square-integrable martingale which satisfies
(2)
On the other hand, for any predictable process K such that
we can define the integral of K with respect to N and this integral satisfies
(3)
In this article, we work with processes whose trajectories are right-continuous with left limits. If x is a right
continuous function with left limits, we denote by the left limit at time t and
the jump size at time t. We will prove the following result.
Theorem 1 (Ito Formula I). Let be a process defined by
(4)
where G, K and H are predictable processes which satisfy
(5)
(6)
(7)
Then there exists a modification of Y (denoted also by Y) whose sample paths are right-continuous with left limits, such that for any function and for any, with probability 1,
(8)
Note that since the first two terms on the right-hand side of (4) are processes of finite variation and the last term is a square-integrable martingale, Y is a semimartingale. Therefore, the Itô formula given by Theorem 1 can be derived from the corresponding result for a general semimartingale, assuming that Y has sample paths which are right-continuous with left limits (see e.g. Theorem 2.5 of [7] ).
The goal of the present article is to give an alternative proof of this result which contains the explicit construction of the modification of Y for which the Itô formula holds.
We will also give the proof of the following variant of the Itô formula, which will be useful for the applications related to the (finite-variance) Lévy white noise, discussed in Section 4.
Theorem 2 (Ito Formula II). Let be a process defined by
(9)
where G and H are predictable processes which satisfy (5), respectively (1). Then there exists a càdlàg modification of Y (denoted also by Y) such that for any, with probability 1,
The method that we use for proving Theorems 1 and 2 is similar to the one described in Section 4.4.2 of [6] in the case of classical Lévy processes, the difference being that in our case, N is a PRM on instead of. This method relies on a double “interlacing” technique, which consists in first approximating the set of small jumps by sets of the form with (in the case when H and K vanish outside a bounded Borel set), and then approximating the spatial domain by regions of the form with. This approximation method is described in Section 2. Section 3 is dedicated to the proofs of Theorems 1 and 2. Finally, in Section 4 we discuss two applications of Theorem 2 in the case of the (finite-variance) Lévy white noise introduced in [1] .
2. Approximation by Right-Continuous Processes with Left Limits
In this section, we show that the Lévy-type integral processes given by (4) and (9) have right-continuous modifications with left limits, which are constructed by approximation. These modifications will play an important role in the proof of Itô’s formula. Since the process is continuous, we assume that.
We consider first processes of the form (4). We start by examining the case when both integrands H and K
vanish outside a set. Since the process is clearly càdlàg
(the integral being a sum with finitely many terms), we need to consider only the integral process which depends on H.
Note that if H vanishes a.e. on for some and, then
is a process whose sample paths are right-continuous with left limits (the first term is a sum with finitely many terms and the second term in continuous). Therefore, we will suppose that H satisfies the following assumption:
Assumption A. It is not possible to find and such that
with respect to the measure.
Lemma 1. Let be a process defined by
where and H is a predictable process which satisfies Assumption A and
(10)
Then, there exists a càdlàg modification of Y such that for all,
where
for some sequence (depending on T) such that.
Proof: We use the same argument as in the proof of Theorem 4.3.4 of [6] . Fix. Let
where
Note that is non-increasing and. (If then for all n. Hence, which contradicts Assumption A.)
Note that is a càdlàg martingale. By Doob’s submartingale inequality and relation (2),
By Chebyshev’s inequality,. By Borel-Cantelli lemma, with probability 1, the sequence is Cauchy in the space of càdlàg functions on equipped with the sup-norm. Its limit is a modification of Y since for any, also converges to in. Finally, we note that the process does not depend on T (although the approximation sequence does). If is the modification of Y on and is the modification of Y on with, then a.s. for any. Hence, can be extended to. ,
We consider now the case when the at least one of the integrands H and K do not vanish outside a set. More precisely, we introduce the following assumptions:
Assumption B. It is not possible to find and such that
with respect to the measure.
Assumption. It is not possible to find and such that
with respect to the measure.
We consider bounded Borel sets in of the form.
Theorem 3 (Interlacing I). Let be a process defined by (4) with, where H and K are predictable processes which satisfy conditions (7), respectively (6), such that either H satisfies Assumption B, or K satisfies Assumption. Then, there exists a càdlàg modification of Y such that for all T > 0,
(11)
where is a càdlàg modification of the process defined by
with for some sequence (depending on T) such that.
Proof: Fix. Let where
Note that is non-decreasing and. (If then for all n, and hence, which contradicts Assumptions B or.) Let be the process given in the statement of the theorem with. We denote by and the two integrals which compose, depending on H, respectively K.
We denote by the càdlàg modification of given by Lemma 1. By Doob’s submartingale inequality and relation (2),
By Chebyshev’s inequality,.
Note that is a càdlàg process. For any,
and hence, using relation (3),
By Markov’s inequality,.
Let. Then, and the conclusion follows by the Borel-Cantelli Lemma, as in the proof of Lemma 1. ,
We consider next processes of the form (9) with G = 0. Note that if H vanishes a.e. outside a set then
where the first term has a càdlàg modification given by Lemma 1, the second term is càdlàg, and the third term is continuous. Therefore, we will suppose that H satisfies the following assumption:
Assumption C. It is not possible to find and such that
with respect to the measure.
Theorem 4 (Interlacing II). Let Y be a process given by (9) with, where H is a predictable process which satisfies (1) and Assumption C. Then, there exists a càdlàg modification of Y such that (11) holds, where is a càdlàg modification of the process defined by:
with for some sequence (depending on T) such that.
Proof: We proceed as in the proof of Theorem 3. Fix. Let where
By Assumption C,. We write Yn(t) as the sum of two integrals, corresponding to the regions,
and. We denote these integrals by, respectively. Note that is càdlàg. Let
be the càdlàg modification of given by Lemma 1.
Let. By Doob’s submartingale inequality,
and the conclusion follows as in the proof of Lemma 1. ,
3. Proof of Itô Formula
In this section, we give the proofs of Theorem 1 and Theorem 2.
We start with the simpler case when there are no small jumps (the analogue of Lemma 4.4.6 of [6] ).
Lemma 2. Let
where G is a predictable process which satisfies (5), , and K is a predictable process. Then, for any function and for any,
Proof: We denote. By Proposition 5.3 of [8] , we may assume that the restriction of N to the set has points, where are the points of a Poisson process on of intensity and are i.i.d. on with distribution, independent of. We consider two cases.
Case 1: G = 0. By the representation of N,. So is a step function which has a jump of size at each point and. Hence
and the conclusion follows since N has points in.
Case 2: G is arbitrary. The map is a step function which has a jump of size at time. Since is continuous, the jump times and the jump sizes of Y coincide with those of, i.e.. We use the decomposition
where A and B are defined as follows: if, we let
Note that
It remains to prove that
(12)
For this, we assume that and we write
So it suffices to prove that
(13)
for all, and
(14)
We first prove (13). Fix. For any
, and.
We extend by continuity to. Hence
where for the last equality we used the fact that and hence
.
This proves (13).
Next, we prove (14). Note that if, both terms are zero. So, we assume that. For any
, and.
Arguing as above, we see that
where for the last equality we used the fact that and hence
.
This concludes the proof of (14). ,
Proof of Theorem 1: We fix. We assume that and are bounded. (Otherwise, we use for.)
Case 1: H and K vanish outside a fixed set.
If H vanishes a.e. on for some and, the conclusion follows from Lemma 2. Therefore, we suppose that H satisfies Assumption A. By Lemma 1, there exists a càdlàg modification of Y (denoted also by Y) such that
(15)
where the process is defined by
being the sequence given by Lemma 1 with. Consequently,
(16)
Note that
where and. By the
Cauchy-Schwarz inequality, satisfies (5) (since B is a bounded set and H satisfies (10)). We apply Lemma 2 to:
After using the definitions of and, as well as adding and subtracting
we obtain that:
(17)
We denote by, respectively the four terms on the right-hand side of (8). The conclusion will follow by taking the limit as in (17). The left-hand side converges to, by (15).
We treat separately the four terms in the right-hand side. By the dominated convergence theorem,
Since is a sum with a finite number of terms, using (15) and the continuity of f, we see that a.s. For the third term, note that, where
and a.s., by (15) and the continuity of f. By the dominated convergence theorem, and. To justify the application of this theorem, we use Taylor’s formula of the first order:
(18)
and the fact that is bounded. This proves that in.
Finally, , where
and
a.s., by (16) and the continuity of f. By the dominated convergence theorem, and. To justify the application of this theorem, we use Taylor’s formula of second order:
(19)
and the fact that is bounded. This proves that in.
Case 2. H satisfies Assumption B or K satisfies Assumption.
By Theorem 3, there exists a càdlàg approximation of Y (denoted also by Y) such that (15) holds, where is a càdlàg modification of
being the sequence given by Theorem 3 with. Using the result of Case 1 for the pro- cess, we obtain
The conclusion follows letting as in Case 1. ,
Proof of Theorem 2: We assume that and are bounded. We fix t.
Case 1. H vanishes outside a set. We write
where. By the Cauchy-Schwarz inequality, satisfies (5) (since B
is a bounded set). By Theorem 1, there exists a càdlàg modification of Y (denoted also by Y) such that
We add and subtract. The conclusion follows by rear-
ranging the terms.
Case 2. H satisfies Assumption C.
By Theorem 4, there exists a càdlàg modification of Y (denoted also by Y) such that (15) holds, where is a càdlàg modification of
being the sequence given by Theorem 4 with. We write the Itô formula for the process (using Case 1) and we let. ,
4. Applications
In this section, we assume that the Lévy measure satisfies the condition:
As in [1] , we consider the process defined by:
For any predictable process such that
(20)
we can define the stochastic integral of X with respect to L and this integral satisfies:
By (2), this integral has the following isometry property:
When used as a noise process perturbing an SPDE, L behaves very similarly to the Gaussian white noise. For this reason, L was called a Lévy white noise in [1] .
4.1. Kunita Inequality
The following maximal inequality is due to Kunita (see Theorem 2.11 of [7] ). In problems related to SPDEs with noise L, this result plays the same role as the Burkholder-Davis-Gundy inequality for SPDEs with Gaussian white noise.
Theorem 5 (Kunita Inequality). Let be a process given by
where X is a predictable process which satisfies (20).
If for some, then for any,
where and is the constant in Theorem 2.11 of [7] .
Proof: We apply Theorem 2 with and. The proof is identical to that of Theorem 2.11 of [7] . We omit the details. ,
Remark 1. Kunita’s constant cannot be computed explicitly. Theorem 5 is proved in [9] using a different method which shows that is directly related to the constant in Rosenthal’s inequality, which is.
4.2. Itô Representation Theorem and Chaos Expansion
In this section, we give an application to Theorem 2 to exponential martingales, which leads to Itô representation theorem and a chaos expansion (similarly to Sections 5.3 and 5.4 of [6] ).
For any we let for. We work with the càdlàg modi-
fication of the process given by Theorem 4. By Lemma 2.4 of [1] ,
where
Hence for all, where
The following result is the analogue of Lemma 5.3.3 of [6] .
Lemma 3. For any and, with probability 1,
Proof: We apply Theorem 2 to the function and the process
Hence, and. We obtain:
Since the sum of the last two integrals is 0, the conclusion follows. ,
We fix. We let. We denote by be the space
of C-valued square-integrable random variables which are measurable with respect to.
Lemma 4. The linear span of the set is dense in.
Proof: The proof is similar to that of Lemma 5.3.4 of [6] . We omit the details. ,
Theorem 6 (Ito Representation Theorem). For any, there exists a unique predictable C-valued process satisfying
(21)
such that
(22)
Proof: By Lemma 3, relation (22) holds for with. The conclusion follows by an approximation argument using Lemma 4. ,
The multiple (and iterated) integral with respect can be defined similarly to the Gaussian white-noise case (see e.g. Section 5.4 of [6] ).
More precisely, we consider the Hilbert space, where
, and.
For any integer, we consider the n-th tensor product space. The n-th multiple integral with respect to can be constructed for any function, and this integral has the isometry property:
Moreover, if, then for all and.
We have the following result.
Theorem 7 (Chaos Expansion). For any, there exist some symmetric functions, such that
In particular,
Proof: We use the same argument as in the classical case, when is a PRM on and
is a square-integrable Lévy process (see Theorem 5.4.6 of [6] or Theorem 10.2 of [10] ). By Theorem 6, there exists a predictable process satisfying (1) such that
(23)
By (21), for almost all. For such fixed, we apply Theorem 6 again to the variable. Hence, there exists a predictable process
Satisfying
such that
We substitute this into (23) and iterate the procedure. We omit the details. ,
Acknowledgements
Research of R. M. Balan is funded by a grant from the Natural Sciences and Engineering Research Council of Canada.
Cite this paper
Raluca M.Balan,Cheikh B.Ndongo, (2015) Itô Formula for Integral Processes Related to Space-Time Lévy Noise. Applied Mathematics,06,1755-1768. doi: 10.4236/am.2015.610156
References
1. 1. Balan, R.M. (2015) Integration with Respect to Lévy Colored Noise, with Applications to SPDEs. Stochastics, 87, 363- 381.
http://dx.doi.org/10.1080/17442508.2014.956103
2. 2. Balan, R.M. (2014) SPDEs with α-Stable Lévy Noise: A Random Field Approach. International Journal of Stochastic Analysis, 2014, Article ID: 793275.
http://dx.doi.org/10.1155/2014/793275
3. 3. Walsh, J.B. (1989) An Introduction to Stochastic Partial Differential Equations. Ecole d’Eté de Probabilités de Saint- Flour XIV. Lecture Notes in Math, 1180, 265-439.
http://dx.doi.org/10.1007/BFb0074920
4. 4. Rajput, B.S. and Rosinski, J. (1989) Spectral Representations of Infinitely Divisible Processes. Probability Theory and Related Fields, 82, 451-487.
http://dx.doi.org/10.1007/BF00339998
5. 5. Samorodnitsky, G. and Taqqu, M.S. (1994) Stable Non-Gaussian Random Processes. Chapman and Hall, New York.
6. 6. Applebaum, D. (2009) Lévy Processes and Stochastic Calculus. 2nd Edition, Cambridge University Press, Cambridge.
http://dx.doi.org/10.1017/CBO9780511809781
7. 7. Kunita, H. (2004) Stochastic Differential Equations Based on Lévy Processes and Stochastic Flows of Diffeomorphisms. In: Rao, M.M., Ed., Real and Stochastic Analysis, New Perspectives, Birkhaüser, Boston, 305-375.
http://dx.doi.org/10.1007/978-1-4612-2054-1_6
8. 8. Resnick, S.I. (2007) Heavy Tail Phenomena: Probabilistic and Statistical Modelling. Springer, New York.
9. 9. Balan, R.M. and Ndongo, C.B. (2015) Intermittency for the Wave Equation with Lévy White Noise. Arxiv: 1505.04167.
10. 10. Di Nunno, G., Oksendal, B. and Proske, F. (2009) Malliavin Calculus for Lévy Processes with Applications to Finance. Springer-Verlag, Berlin.
http://dx.doi.org/10.1007/978-3-540-78572-9
NOTES
*Corresponding author. | 5,142 | 19,892 | {"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-2024-26 | latest | en | 0.909807 |
https://www.physicsforums.com/threads/integrating-factors.145190/ | 1,544,596,453,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823738.9/warc/CC-MAIN-20181212044022-20181212065522-00064.warc.gz | 964,702,850 | 13,803 | # Integrating Factors
1. Nov 24, 2006
### swimmingtoday
Suppose you have an equation:
M(x,y) dx + N(x,y) dy = 0
I have heard that there always exists an integrating factor u(x,y) such that the partial derivative of uM with respect to y equals the partial derivative of uN with respect to x.
But somewhere in the back of my mind I remember that there is a condition that the guarantee of the existence of the integrating factor is valid ONLY if there are no singularities in the region.
Can someone please tell me the exact status regarding singularities is? Thank you very much. I appreciate it a lot.
Last edited: Nov 24, 2006
2. Nov 24, 2006
### mathwonk
smart aleck answer: u=0 seems to work.
3. Nov 24, 2006
### CPL.Luke
there isn't always an integrating factor that works, just most of the time.
assuming your not a smart aleck
4. Nov 24, 2006
### swimmingtoday
<<there isn't always an integrating factor that works, just most of the time>>
Thanks.
Is there any particular rule for when there will be one that works? Like for example, is there a rule that an integrating factor exists unless there is a singularity?
I've heard both:
1) An integrating factor ALWAYS exists (but might not neccessarily be easily be found)
and
2) Exists as long as there is no singularity.
5. Nov 25, 2006
### CPL.Luke
I'm going to guess that the 2nd one is true.
I'm not sure as in my course we got far enough to know that a single variable integrating factor may not exist.
Although come to think of it we didn't explore multivariable integrating factors, so maybe those always exist, I'll leave it to someone a bit more knowledgeable to answer.
6. Nov 25, 2006 | 421 | 1,677 | {"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} | 2.96875 | 3 | CC-MAIN-2018-51 | latest | en | 0.962011 |
https://www.nottingham.ac.uk/xpert/scoreresults.php?keywords=Historical%20skills%20:%20dating%20docu&start=7660&end=7680 | 1,519,234,513,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813691.14/warc/CC-MAIN-20180221163021-20180221183021-00422.warc.gz | 937,377,581 | 24,098 | Navigating by the Numbers
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Mt. Pinatubo 10th Anniversary Perspective (Stills)
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WordWorld:V is for Vacation. This is an excerpt from the popular PBS kids show Word World. In this video, the word world animals create "v" words and turn the words into actual things. This is a good teaching resource for the early childhood classroom. It helps with learning to blend words and builds phonic skills. (10:56)
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Japancast Episode 041 – Learn Japanese @ Japancast.net
Learn Japanese from the brand new anime “A Channel”. Japancast HD Video Episode 041 from Hitomi Griswold on Vimeo. Help us grow! Share this post on your favorite social site:
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You can now call us and ask a question or leave a comment by voicemail. The new number is: 856-29-Japan (856-295-2726) We’ve also added a new, shorter, easier way to find stuff on Japancast: http://jca.st/shop – the Japancast Shop http://jca.st/face - Japancast on Facebook http://jca.st/friends – Japancast Friends http://jca.st/audible – Your free 2 week [...]
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Theodore Roosevelt: Conserving America's Future Documentary
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How to Make Napkin Decoupage Eggs for Easter
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Muscular System
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ISS Update - May 11, 2011
The International Space Station video update for May 11, 2011.
Author(s): No creator set | 1,701 | 7,735 | {"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} | 2.890625 | 3 | CC-MAIN-2018-09 | latest | en | 0.942044 |
http://wiki.call-cc.org/eggref/4/atlas-lapack?rev=26342 | 1,529,898,179,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00041.warc.gz | 337,083,763 | 5,581 | You are looking at historical revision 26342 of this page. It may differ significantly from its current revision.
• egg
## atlas-lapack
An interface to the LAPACK routines implemented in ATLAS.
## Usage
(require-extension atlas-lapack)
## Documentation
ATLAS stands for Automatically Tuned Linear Algebra Software. Its purpose is to provide portably optimal linear algebra routines. The current version provides a complete BLAS API (for both C and Fortran77), and a very small subset of the LAPACK API. Please see the documentation for the blas egg for definitions of the ORDER, UPLO, DIAG and TRANSPOSE datatypes.
### Naming conventions for routines
Every routine in the LAPACK library comes in four flavors, each prefixed by the letters S, D, C, and Z, respectively. Each letter indicates the format of input data:
• S stands for single-precision (32-bit IEEE floating point numbers),
• D stands for double-precision (64-bit IEEE floating point numbers),
• C stands for complex numbers (represented by pairs of 32-bit IEEE floating point numbers),
• Z stands for double complex numbers (represented by pairs of 64-bit IEEE floating point numbers)
In addition, each ATLAS-LAPACK routine in this egg comes in three flavors:
• Safe, pure Safe routines check the sizes of their input arguments. For example, if a routine is supplied arguments that indicate that an input matrix is of dimensions M-by-N, then the argument corresponding to that matrix is checked that it is of size M * N.
Pure routines do not alter their arguments in any way. A new matrix or vector is allocated for the return value of the routine.
• Safe, destructive (suffix: !) Safe routines check the sizes of their input arguments. For example, if a routine is supplied arguments that indicate that an input matrix is of dimensions M-by-N, then the argument corresponding to that matrix is checked that it is of size M * N.
Destructive routines can modify some or all of their arguments. They are given names ending in exclamation mark. The matrix factorization routines in LAPACK overwrite the input matrix argument with the result of the factorization, and the linear system solvers overwrite the right-hand side vector with the system solution. Please consult the LAPACK documentation to determine which functions modify their input arguments.
• Unsafe, destructive (prefix: unsafe-:, suffix: !) Unsafe routines do not check the sizes of their input arguments. They invoke the corresponding ATLAS-LAPACK routines directly. Unsafe routines do not have pure variants.
For example, function xGESV (N-by-N linear system solver) comes in the following variants:
LAPACK name Safe, pure Safe, destructive Unsafe, destructive
SGESV sgesv sgesv! unsafe-sgesv!
DGESV dgesv dgesv! unsafe-dgesv!
CGESV cgesv cgesv! unsafe-cgesv!
ZGESV zgesv zgesv! unsafe-zgesv!
### LAPACK driver routines
#### General linear system solving
[procedure] sgesv:: ORDER * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR * F32VECTOR * S32VECTOR
[procedure] dgesv:: ORDER * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR * F64VECTOR * S32VECTOR
[procedure] cgesv:: ORDER * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR * F32VECTOR * S32VECTOR
[procedure] zgesv:: ORDER * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR * F64VECTOR * S32VECTOR
The routines compute the solution to a system of linear equations ''A
• X = B, where A is an N-by-N matrix and X and B are N-by-NRHS matrices. Optional arguments LDA and LDB are the leading dimensions of arrays A and B, respectively. LU decomposition with partial pivoting and row interchanges is used to factor A as A = P * L * U, where P is a permutation matrix, L is unit lower triangular, and U is upper triangular. The factored form of A'' is then used to solve the system. The return values are:
• a matrix containing the factors L and U from the factorization A = P*L*U;
• the N-by-NRHS solution matrix X
• a vector with pivot indices: for 1 <= i <= min(M,N), row i of the matrix A was interchanged with row pivot(i)
#### Symmetric positive definite linear system solving
[procedure] sposv:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR * F32VECTOR
[procedure] dposv:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR * F64VECTOR
[procedure] cposv:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR * F32VECTOR
[procedure] zposv:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR * F64VECTOR
The routines compute the solution to a system of linear equations ''A
• X = B, where A is an N-by-N symmetric positive definite matrix and X and B are N-by-NRHS matrices. Optional arguments LDA and LDB are the leading dimensions of arrays A and B, respectively. Cholesky decomposition is used to factor A'' as
• A = U**T * U if UPLO = Upper
• A = L * L**T if UPLO = Lower where U is an upper triangular, and L is a lower triangular matrix. The factored form of A is then used to solve the system. The return values are:
• the factor U or Lfrom the Cholesky factorization, depending on the value of argument UPLO.
• the N-by-NRHS solution matrix X
### LAPACK computational routines
#### General matrix factorization
[procedure] sgetrf:: ORDER * M * N * A * [LDA] -> F32VECTOR * S32VECTOR
[procedure] dgetrf:: ORDER * M * N * A * [LDA] -> F64VECTOR * S32VECTOR
[procedure] cgetrf:: ORDER * M * N * A * [LDA] -> F32VECTOR * S32VECTOR
[procedure] zgetrf:: ORDER * M * N * A * [LDA] -> F64VECTOR * S32VECTOR
These routines compute an LU factorization of a general M-by-N matrix A using partial pivoting with row interchanges. Optional argument LDA is the leading dimension of array A. The return values are:
1. a matrix containing the factors L and U from the factorization A = P*L*U;
2. a vector with pivot indices: for 1 <= i <= min(M,N), row i of the matrix was interchanged with row pivot(i)
#### General linear system solving using factorization
[procedure] sgetrs:: ORDER * TRANSPOSE * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR
[procedure] dgetrs:: ORDER * TRANSPOSE * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR
[procedure] cgetrs:: ORDER * TRANSPOSE * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR
[procedure] zgetrs:: ORDER * TRANSPOSE * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR
These routines solve a system of linear equations A * X = B or A' * X = B with a general N-by-N matrix A using the LU factorization computed by the xGETRF routines. Argument NRHS is the number of right-hand sides (i.e. number of columns in B). Optional arguments LDA and LDB are the leading dimensions of arrays A and B, respectively. The return value is the solution matrix X.
#### General matrix invert using factorization
[procedure] sgetri:: ORDER * N * A * PIVOT * [LDA] -> F32VECTOR
[procedure] dgetri:: ORDER * N * A * PIVOT * [LDA] -> F64VECTOR
[procedure] cgetri:: ORDER * N * A * PIVOT * [LDA] -> F32VECTOR
[procedure] zgetri:: ORDER * N * A * PIVOT * [LDA] -> F64VECTOR
These routines compute the inverse of a matrix using the LU factorization computed by the xGETRF routines. Argument A must contain the factors L and U from the LU factorization computed by xGETRF. Argument PIVOT must be the pivot vector returned by the factorization routine. Optional argument LDA is the leading dimension of array A. The return value is the inverse of the original matrix A.
#### Symmetric positive definite matrix factorization
[procedure] spotrf:: ORDER * UPLO * N * A * [LDA] -> F32VECTOR
[procedure] dpotrf:: ORDER * UPLO * N * A * [LDA] -> F64VECTOR
[procedure] cpotrf:: ORDER * UPLO * N * A * [LDA] -> F32VECTOR
[procedure] zpotrf:: ORDER * UPLO * N * A * [LDA] -> F64VECTOR
These routines compute the Cholesky factorization of a symmetric positive definite matrix A. The factorization has the form:
• A = U**T * U if UPLO = Upper
• A = L * L**T if UPLO = Lower where U is an upper triangular, and L is a lower triangular matrix. Optional argument LDA is the leading dimension of array A. The return value is the factor U or Lfrom the Cholesky factorization, depending on the value of argument UPLO.
#### Symmetric positive definite matrix solving using factorization
[procedure] spotrs:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR
[procedure] dpotrs:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR
[procedure] cpotrs:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F32VECTOR
[procedure] zpotrs:: ORDER * UPLO * N * NRHS * A * B * [LDA] * [LDB] -> F64VECTOR
These routines solve a system of linear equations A * X = B with a symmetric positive definite matrix A using the Cholesky factorization computed by the xPOTRF routines. Argument A is the triangular factor U or L as computed by xPOTRF. Argument NRHS is the number of right-hand sides (i.e. number of columns in B). Argument UPLO indicates whether upper or lower triangle of A is stored (Upper or Lower). Optional arguments LDA and LDB are the leading dimensions of arrays A and B, respectively. The return value is the solution matrix X.
#### Symmetric positive definite matrix invert using factorization
[procedure] spotri:: ORDER * UPLO * N * A * [LDA] -> F32VECTOR
[procedure] dpotri:: ORDER * UPLO * N * A * [LDA] -> F64VECTOR
[procedure] cpotri:: ORDER * UPLO * N * A * [LDA] -> F32VECTOR
[procedure] zpotri:: ORDER * UPLO * N * A * [LDA] -> F64VECTOR
These routines compute the inverse of a symmetric positive definite matrix A using the Cholesky factorization A = U**T*U or A = L*L**T computed by xPOTRF. Argument A is the triangular factor U or L as computed by xPOTRF. Argument UPLO indicates whether upper or lower triangle of A is stored (Upper or Lower). Optional argument LDA is the leading dimension of array A. The return value is the upper or lower triangle of the inverse of A.
#### Triangular matrix invert
[procedure] strtri:: ORDER * UPLO * DIAG * N * A * [LDA] -> F32VECTOR
[procedure] dtrtri:: ORDER * UPLO * DIAG * N * A * [LDA] -> F64VECTOR
[procedure] ctrtri:: ORDER * UPLO * DIAG * N * A * [LDA] -> F32VECTOR
[procedure] ztrtri:: ORDER * UPLO * DIAG * N * A * [LDA] -> F64VECTOR
These routines compute the inverse of a triangular matrix A. Argument A is the triangular factor U or L as computed by xPOTRF. Argument UPLO indicates whether upper or lower triangle of A is stored (Upper or Lower). Argument DIAG indicates whether A is non-unit triangular or unit triangular (NonUnit or Unit). Optional argument LDA is the leading dimension of array A. The return value is the triangular inverse of the input matrix, in the same storage format.
#### Auxilliary routines
[procedure] slauum:: ORDER * UPLO * DIAG * N * A * [LDA] -> F32VECTOR
[procedure] dlauum:: ORDER * UPLO * DIAG * N * A * [LDA] -> F64VECTOR
[procedure] clauum:: ORDER * UPLO * DIAG * N * A * [LDA] -> F32VECTOR
[procedure] zlauum:: ORDER * UPLO * DIAG * N * A * [LDA] -> F64VECTOR
These routines compute the product U * U' or L' * L, where the triangular factor U or L is stored in the upper or lower triangular part of the array A. Argument UPLO indicates whether upper or lower triangle of A is stored (Upper or Lower'). Optional argument LDA is the leading dimension of array A. The return value is the lower triangle of the lower triangular product, or the upper triangle of upper triangular product, in the respective storage format.
## Examples
```
(use srfi-4 blas atlas-lapack)
(define order ColMajor)
(define n 4)
(define nrhs 1)
;; Solve the equations
;;
;; Ax = b,
;;
;; where A is the general matrix
(define A (f64vector 1.8 5.25 1.58 -1.11 ;; column-major order
2.88 -2.95 -2.69 -0.66
2.05 -0.95 -2.90 -0.59
-0.89 -3.80 -1.04 0.80))
;;
;; and b is
;;
(define b (f64vector 9.52 24.35 0.77 -6.22))
;; A and b are not modified
(define-values (LU x piv) (dgesv order n nrhs A b))
;; A is overwritten with its LU decomposition, and
;; b is overwritten with the solution of the system
(dgesv! order n nrhs A b)
```
## About this egg
Ivan Raikov
### Version history
3.0
Using bind instead of easyffi
2.1
Ensure that unit test script exists with a non-zero exit status on error (thanks to mario)
2.0
Eliminated reduntant atlas-lapack: prefix from names of exported symbols
1.12
Switched to wiki documentation
1.11
Ported to Chicken 4
1.10
Bug fix in the detection of ATLAS library
1.9
Added build system support for libraries linked with f2c
1.8
Build script updated for better cross-platform compatibility
1.7
Fixed a bug in the trtri interface
1.6
Changed matrix copying code to use BLAS routines instead of object-copy
1.5
License upgrade to GPL v3
1.4
Added -latlas to the compiler options in the setup script
1.3
Minor changes in the setup script
1.2
Bug fix in the setup script
1.1
Created safe/unsafe variants of each routine, added optional leading dimensions
1.0
Initial release
blas
### License
```Copyright 2007-2012 Ivan Raikov and the Okinawa Institute of Science and Technology
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or (at
your option) any later version.
This program is distributed in the hope that it will be useful, but
WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
General Public License for more details.
A full copy of the GPL license can be found at
<http://www.gnu.org/licenses/>.``` | 3,675 | 13,444 | {"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} | 2.5625 | 3 | CC-MAIN-2018-26 | latest | en | 0.846336 |
https://www.testinvite.com/dy/en/pages/guide/dimensions-of-test-questions | 1,723,078,261,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640713903.39/warc/CC-MAIN-20240808000606-20240808030606-00700.warc.gz | 785,255,037 | 15,879 | # Dimensions
Utilizing dimensions in scoring
Created by Mustafa Ekim / March, 2023
The test report provides an automatic evaluation of the test-taker's overall performance, as well as their performance in each section and page. However, sections and pages are primarily used for organizing the test from the test-taker's perspective. If you want to measure other aspects of the test-taker's performance independently of the test structure, you can use dimensions.
For example, your test may have two sections, each containing questions that you want to evaluate the test-taker's overall performance on. Dimensions can be particularly useful in such cases.
Dimensions have their own unique calculation criteria and do not have an impact on the scores of a test, section, or page. Instead, they provide an independent evaluation aspect that helps to understand the strengths and weaknesses of the test-taker in various subjects in a structured way.
Dimensions help to evaluate different aspects of the test-taker's performance by assigning questions to specific dimensions.
## What is a dimension?
A dimension of a test is an independent evaluation aspect that allows test-takers to be evaluated in a structured way according to their strengths and weaknesses in various subject areas. Dimensions can be organized in a hierarchical tree structure, with each level representing a more specific aspect of the subject being evaluated. Test questions can be assigned to one or more dimensions, and the resulting scores for each dimension can be used to gain insights into a test-taker's overall performance.
## The dimension tree
The dimensions can be organized in a hierarchical tree structure. The hierarchical structure of the dimension tree enables the evaluation of various aspects of the test-taker in a structured manner.
The dimension tree enables hierarchical evaluation of the test-taker's performance in various aspects of the test, such as Math questions in general, Arithmetic Math questions, and even more specific Addition Arithmetic Math questions.
By constructing a dimension tree and assigning test questions to dimensions, the system can generate test reports that analyze the test-taker's performance for each dimension in a hierarchical manner.
## Example
Suppose your test includes questions on Geography and Mathematics. By assigning each question to its respective dimension, you can generate test reports that evaluate the test-taker's performance in each dimension. Additionally, you can further subdivide each dimension by adding child components. For example, if your Geography questions are focused on either Europe or Asia, you can assign them to either the Geography/Europe or Geography/Asia dimension. This allows the system to provide a more detailed analysis of the test-taker's performance not just in the overall Geography dimension, but in each sub-dimension as well.
## The dimension expression
The dimension expression points to a specific dimension within a tree of dimensions. For instance, "Math/Arithmetic/Addition" refers to the Addition dimension nested within the Arithmetic dimension that, in turn, is nested inside the Math dimension.
You can create a dimension expression by separating the dimensions with "/", indicating their parent-child relationship within the dimension tree. For instance, "Math/Arithmetic/Addition" refers to the "Addition" dimension within the "Arithmetic" dimension, which is a child of the "Math" dimension.
### Examples:
• Programming
• Programming/Object-oriented
• Programming/Object-oriented/Java
• Programming/Object-oriented/Java/v14
• Programming/Object-oriented/Java/v15
• Programming/Object-oriented/CSharp
• Programming/Object-oriented/CSharp/6
• Programming/Object-oriented/CSharp/7
In the dimension expression, it is important to note the parent-child relationship between dimensions. For instance, some questions can be assigned solely to the Programming dimension, which would impact the test-taker's score in the Programming dimension only. However, for more specific programming questions, such as those related to Object-Oriented Programming, you may want to assess the test-taker's performance in this area as well. In this case, you can assign these questions to the Programming/Object-oriented dimension, which would affect both the Programming dimension and the Programming/Object-Oriented dimension.
## Assigning a dimension to a test question
To assign a dimension to a test question, go to the Scoring tab of the question editor and select "Additional point calculations". Then, add the desired dimension either using the dimension expression form or selecting from the pre-defined dimension tree in your definition bank.
## Calculation of dimension points obtained by a test-taker after answering a test-question
When assigning dimensions to a test question, the point value and negative multiplier can be configured for each dimension.
Dimension points are calculated similarly to test points by multiplying the percentage score obtained for a question by the dimension's assigned point value or negative multiplier, depending on whether the percentage score is above or below 0%.
If a test question is assigned to the "Math" dimension with a point value of 2 and a negative multiplier of -1, the Math dimension point is affected based on the percentage score obtained from answering the question. For instance, a percentage score of 100% results in an increase of 2 points (100% x 2) in the Math dimension point, while a percentage score of -50% results in a decrease of 0.5 points (50% x 1).
The generated test report automatically evaluates the test-taker's performance in each dimension within a tree structure. You can navigate through the items of the tree to observe the performance in each dimension.
## Storing the dimension tree inside the definition bank
Multiple dimension trees can be created and stored in the Definition bank to provide a centralized location for dimension definitions. This eliminates the need to rewrite the dimension definition each time a question is assigned to a dimension.
While it is not mandatory, it is recommended to store your dimension definitions in a central location to avoid errors that may arise from free-form writing.
### Accessing the definition bank
To access your dimension definitions, go to your Question bank and click on the Menu icon located at the top right of the page. From there, select the Definitions option.
You will be presented with a list of definition banks.
### Creating a new definition bank
You can click on the red plus icon to at the bottom right of the page to create a new definition bank.
### Sections of the definition bank
The purpose of the definition bank is to store all your dimensions and effect definitions in one central location. You can access the Dimensions tab to create or edit your dimension tree.
### Creating a dimension tree
You can build your dimension tree by adding items and establishing parent-child relationships between them. This can be done by adding an item under another dimension to create a hierarchy.
Once defined, it can be used to assign dimensions to test questions by selecting the "Additional point calculations" option on the Scoring tab and clicking on the button next to the text field where you write a dimension definition.
Talk to a representative
Figure out if TestInvite is a good match for your organization | 1,398 | 7,466 | {"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} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.918207 |
https://www.thestudentroom.co.uk/showthread.php?t=980909 | 1,544,491,764,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823516.50/warc/CC-MAIN-20181210233803-20181211015303-00239.warc.gz | 1,069,640,911 | 35,087 | # Area of circle by Integration??Watch
#1
ok... so if the general eqn is (x^2)+(y^2)=(r^2)
then y = sqroot(radius squared minus 'x' squared)
then integrate between limits zero and radius.
then i get stuck loll can someone plz explain what substitution i need to use and why...lol its the holidays...maths's gone out my head for the time being
dnt skip any steps plz im having a dumb day XD
0
quote
9 years ago
#2
x = r sin t.
0
quote
9 years ago
#3
You will also have to think about how much of the circle your integration represents.
0
quote
#4
a quarter yesss i know that
0
quote
9 years ago
#5
why you doing this over summer holidays?
0
quote
#6
(Original post by gummers)
why you doing this over summer holidays?
because all the cool kids do this
0
quote
9 years ago
#7
(Original post by stargirlfran)
because all the cool kids do this
**** son,
i better get my books out
0
quote
#8
(Original post by gummers)
**** son,
i better get my books out
yeah..u do that.
0
quote
#9
done it. proved the thing yeaahhhhh. dnt post solutions unless u reeli want to show off
0
quote
#10
(Original post by DFranklin)
x = r sin t.
im sure u're right and it in fact works, but why??
0
quote
9 years ago
#11
y = sqrt(r^2-x^2) = r sqrt(1-sin^2 t) = r sqrt(cos^2 t) = r cos t
0
quote
9 years ago
#12
(Original post by stargirlfran)
im sure u're right and it in fact works, but why??
How much do you know about coordinate geometry/polar coordinates?
If a circle has radius r and has centre (0, 0) then you can look at a point, call it P(x, y), on the circle as being a vertex of a right-angled triangle (you also have to draw a line from the point on the circle vertically down/up to a point, call it Q, on the x-axis. So, the angle OQP is a right-angle, and the length of OQ is x and the length of PQ is y. If you call the angle POQ θ, then using trigonometry you can derive the following rules:
...and from there there are a load of other rules that you can derive, two of the most common ones are:
quote
#13
(Original post by DFranklin)
y = sqrt(r^2-x^2) = r sqrt(1-sin^2 t) = r sqrt(cos^2 t) = r cos t
thank u
0
quote
9 years ago
#14
Yes, this problem is so much easier using Polar integration
0
quote
9 years ago
#15
(Original post by My Alt)
Yes, this problem is so much easier using Polar integration
errrm, where did you get that? don't you mean ?
eugh. can't even do integration =(
0
quote
9 years ago
#16
(Original post by Totally Tom)
errrm, where did you get that? don't you mean ? wait a second, why doesn't this give me the area of the circle? wtf.
Jacobian.
Edit: oh, you got it. Good boy. Now evaluate the r-integral to get what the poster above you had. It's also a standard A-level formula.
0
quote
9 years ago
#17
(Original post by Totally Tom)
errrm, where did you get that?
I think he has reproduced the formula for the sector of a circle that is provided in the A Level formula book. Unfortunately, that doesn't really prove much at all!
0
quote
9 years ago
#18
I got it from Bostock and Chandler. Basically you consider a triangle and find and then integrate, iirc. You could potentially use it for any polar curve, if I understand correctly.
0
quote
9 years ago
#19
I heard it described by directly analogizing from the area of a sector: becomes
0
quote
9 years ago
#20
Well the rate of change of area is equal to 1/2 r^2, integrating both sides with respect to theta gives the formula, so were approaching it from the same angle =)
0
quote
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http://my.homecampus.com.sg/Learn/Primary-Grade-3/Equivalent-Fractions | 1,490,821,403,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191396.90/warc/CC-MAIN-20170322212951-00013-ip-10-233-31-227.ec2.internal.warc.gz | 245,001,740 | 4,754 | ## Equivalent Fractions
#### What are equivalent fractions?
1 out of 2 parts
= 2 out of 4 parts
= 4 out of 8 parts
Or,
1 2
=
2 4
=
4 8
are equivalent fractions.
#### 1. What are some of the equivalent fractions of 1 4 ?
Some of the equivalent fractions of
1 4
are
2 8
and
3 12
#### 2. What are the first 3 equivalent fractions of 2 5 ?
A short-cut way to find an equivalent fraction is to multiply the numerator and denominator by the same number.
2 5
=
4 10
=
6 15
=
8 20
The first 3 equivalent fractions of
2 5
are
4 10
,
6 15
and
8 20
#### 3. Find the missing numbers in the equivalent fractions below: 12 18 = 6 ? , 12 18 = ? 3
Another way of finding equivalent fractions is to divide the numerator and denominator by the same number. | 250 | 773 | {"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} | 4.1875 | 4 | CC-MAIN-2017-13 | latest | en | 0.847042 |
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I do know how to find the area of a square and rectangle, but how do you fing the area of a rhombus. (that is at least what it looks like.)
The shape I'm talking about is like a rectangle but it looks like one side got stretched to the left. (sorry if I made that even more confusing.)
area rhombus: base x height
If it is a parallel quadralaterial (base/top parallel, sides unequal), then area= (1/2(base+top)height)
## Similar Questions
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# Which is greater 1000 ft or 300yd?
Updated: 4/28/2022
Wiki User
8y ago
There are three feet in one yard.
300 yards is equal to 900 feet.
Therefore 1,000 feet is greater than 300 yards.
Wiki User
8y ago
Wiki User
8y ago
1000 ft is greater.
Earn +20 pts
Q: Which is greater 1000 ft or 300yd?
Submit
Still have questions?
Related questions
### Which is greater 1000ft or 300yd?
1000 feet is greater because 300 yards equals 900 feet
### Is 1000ft greater than less than or equal to 300yd?
300×3=600 so 1000ft > 300yd
### Which is greater 1000 feet or 300 yard?
1000 ft = 333.33 yd, so 1000 ft is greater Algebraic Steps / Dimensional Analysis1,000 ft*1 yd 3 ft=333.3333333 yd
### Is 1000 ft bigger or less than 300 yd?
1 yd = 3 ft → 300 yd = 300 × 3 ft = 900 ft → 1000 ft is greater than 300 yd.
Yes
### Which one's greater 1000ft or 300yd?
There are 3 feet in one yard.To work this out multiply 300 (yards) by 3 (feet) - like this - 300 x 3 = 900So 300 yards is equal to 900 feet, therefore 1,000 feet is greater.
Yes
### What is greater 1000 ft or 300 yards?
1000 feet because 300 yards = 900 feet
### Which is greater 1000 ft or 300 yd?
1000 feet1 yard = 3 feet1000 ft = 1000 feet * 1 yard/3 feet = 333.33 yardsthus,1000 feet > 300 yards
### Which is greater 1000 yards or 4000 feet?
4000 ft. is bigger because 1y=3ft.
### Is 1000ft bigger than 300yd?
1000 ÷ 3 = 333.33 yards.Therefore, 1000 feet is longer than 300 yards.
### Which is greater 1500m or 3000ft?
A metre is slightly longer than 1 yard = 3 ft 3000 ft = 3000 ÷ 3 yd = 1000 yd → 1500 m is greater than 3000 ft. | 577 | 1,655 | {"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-2024-26 | latest | en | 0.905093 |
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# USING
FIBONACCI RATIOS
TO FORECAST
PRICE AND TIME
This material is protected under the Digital Millennium Copyright Act of 1998 and various
international treaties. This material may not be replicated and redistributed. You may
make one or more copies for archival purposes if those copies are for your own use. It is
illegal to email this material to any person other than yourself or to make this material
2
INTRODUCTION
What do the Great Pyramid, the dimensions of your credit cards, your teeth, Beethoven's
5th Symphony, moth wings, daVinci's Madonna and Child, the Panthenon, the
geometrical arrangement of the solar system, and the exact way that seeds propagate
on a flower (to name a few) have in common? The Golden Section, the Divine
Proportion. Perhaps the most important single number in the universe - .618.
Leonardo Pisano, a 13th century mathematician, has many significant achievements but
will probably always be remembered for his rabbit counting exercise which popularized
the sequence of numbers known as the Fibonacci numbers. Leonardo Pisano was the
son of Guglielmo Bonacci. The shortening of the Latin "filius Bonacci" (son of Bonacci) is
how Leonardo Pisano came to be known as Leonardo Fibonacci, or more simply
Fibonacci.
The Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34... The next number in the
sequence is the sum of the prior two (i.e. 21+34 = 55, 34+55 = 89). As the sequence
gets larger the relationship between adjoining numbers gets closer to the Divine
Proportion or the Golden Ratio itself: ±0·61803 39887... and ±1·61803 39887.
Even though this sequence of numbers will forever be known as the Fibonacci Numbers,
it is not the values of the individual numbers themselves that are important. It's the ratio
relationship between the Fibonacci Numbers that matters for financial forecasting, and
especially for those ratios that derive directly from the Divine Proportion.
You could spend a lifetime exploring the intricacies and interconnectedness of the Divine
Proportion. It's easy enough to go off in a tangent with this topic. For our purposes it's
enough to say that we believe that the Divine Proportion is terribly important for
forecasting the financial markets because the human brain is hard-wired to respond to it.
Identify something that exemplifies stunning beauty. Chances are that the fingerprints of
the Divine Proportion are all over it. The stock market, indeed, every publicly traded
liquid market, is a never ending succession of action-reaction, rally-decline. We can use
the Divine Proportion to discover how those growth-decay phases have related to each
other in the past and how they may relate to each other, in both price and time, in the
future.
The three categories of ratio relationships, and a self-definition of the way in which they
are applied in financial forecasting, are: Retracement, Expansion, and Parallel
Projection. The table shows the most important Fibonacci Ratios for financial
forecasting. These ratios were derived by dividing, and by squaring, and by calculating
the square roots of the ratios of adjacent and once removed Divine Proportion numbers.
## 2.618 6.854 1.618
1.618 2.618 1.272
0.618 0.382 0.786
0.382 0.146 0.618
4.236 is another important Fibonacci ratio for growth projections. 1.00 and 2.00 are also
important. .500 is not a Fibonacci ratio but we include it for forecasting because the 50%
retracement has proven over time to be a recurring retracement percentage.
3
FIBONACCI PRICE RETRACEMENTS
Overview
Almost everybody is familiar with measuring retracements. A ticker advances 100 points
and then declines by 62 points before taking off again in another rally leg.
## In this case the retracement is 62%.
Along with 38% and 50% (which is not a
Fibonacci number) these are the most
common retracement levels. For many
people this all they know about Fibonacci
Retracements, and perhaps Fibonacci
Numbers in general, and even for them
this is a good piece of information to
have. If your work tells you that this
pullback is most likely a temporary
decline before the beginning of the next
upleg, when it appears likely that the top
is in at 100 you can mark your charts and
watch for a reaction at the different
Fibonacci levels. We call this a Reaction
or Decay Retracement. During this
Retracement phase Price is moving against the major trend, and if you are correct about
the direction of the major trend, price should decay somewhere between 14.6% and
78.6% before resuming its move in the direction of the major trend. Easy enough.
But Fibonacci Retracements are not limited to the garden variety Decay type. Fibonacci
Numbers appear so frequently in nature because they demonstrate the pattern of
change and growth. And so too can this pattern of change and growth be applied to the
financial markets. In the first illustration
we limited ourselves to describing the
likely extent of the pullback from the high
at 100. In Figure 2, once the low point of
decay is known, we can apply the
Fibonacci growth ratios to the 62 point
decline and project an advance from the
retracement low price at 38 to about 100
(100%) or 117 (127%) or to 138 (162%).
## For Fibonacci Retracements you always
base your forecast from the measure of
one swing or one leg of the swing if it's a
complex pattern. In our illustration the
measured swing was from 0 - 100. The
Fibonacci Retracement was 62%. For the
Growth Retracement the swing was from 100 - 38. We applied the 100%, 127% and
162% growth ratios to that 62 point swing to project future price targets.
That's two applications of Fibonacci Retracement ratios for financial forecasting. These
applications occur frequently enough across all price levels and time frames to have
4
genuine forecasting value.
Fibonacci Retracements have one more application. We use them to create what we call
the "Death Zone." In Figure 2 we used the Growth Ratios to project likely targets for
what we believed was the beginning of the next upleg in the direction of the major trend.
Needless to say things don't always work out as planned. From experience with the
major stock indexes we never take a breath or start imagining what wonderful
forecasters we are until the new upleg has safely cleared the Death Zone. The Death
Zone narrowly drawn is the 62% - 79%
retracement measured from the low point
of the initial decay phase. In this case it
would be the 76 - 87 price area (see Figure
3). We call it the Death Zone because this
is where many promising new swings die
an early death. A broader application of the
Death Zone has it from 50% - 79%.
## The corollary application of the Death Zone
Retracement is that any Decay
Retracement that exceeds the 79%
retracement level immediately becomes
suspect as the start of a major change in
trend and not a pullback as first believed.
Under most pattern recognition methods,
including Elliott Wave, retracements up to 100% of the prior swing are acceptable
without causing a change in outlook. That's OK too. But we never, never take a
suspected Decay Retracement that exceeds the 79% level as a normal event that can
be ignored with impunity. Maybe this would be the time to lighten up on usual positions
size when making the reversal trade off the suspected Decay Retracement low. All the
examples we used cover retracements of a bull move. The exact same principles apply
for bearish swing retracements.
5
DECAY PRICE RETRACEMENTS
A decay retracement, otherwise known as a pullback, is a move against the main trend.
By definition, a move that exceeds the
prior swing is a change in trend, so decay
retracements will always be less than
100%.
## The Most Common Decay
Retracement Ratios are .382 .500 .618
and .786
## The most common decay retracement
ratios are .382, .500, and 618. The .786
retracement is also a common decay
retracement ratio, but we give that
particular ratio special prominence in our
Fibonacci analysis as explained later.
6
Bull Trend Retracements
## The US Dollar Index Daily Chart
Chart 1
The US dollar, continuous futures contract (Chart 1) experienced a bull swing that lasted
from March 14, 2005 to July 8, 2005 before its first significant pullback. If you had reason
to believe that the July high was not the end of the bull trend, then you would use the
decay retracement ratio technique to forecast where prices would be likely to end their
pullback and resume the prior trend, which in this case was bullish.
The math involved using a ratio of .500 as an example is to subtract the March low at
81.27 from July high at 90.66 to determine the price range of the measured swing (9.39
points). Multiply the measured range by .the selected ratio to determine that the .500
retracement ratio would be equal to 4.70 points. Subtract the ratio result from the July
high.
## Range: 90.66 - 81.27 = 9.39
.500 Retracement: 9.39 * .500 = 4.70
Price Projection: 90.66 - 4.70 = 85.96
7
The actual pullback low on September 2, 2005 was 86.02, which is very close to the
.500 retracement ratio price at 85.96. We could have used an example where the
retracement low matched exactly with a Fibonacci retracement ratio, but that would be a
disservice to the reader. We have found that the best results are obtained by creating a
retracement price zone, usually using retracement ratios .382, .500, .618, and .786 and
then watching for an event such as a reversal bar to signal that the pullback is probably
ended, and the main trend has resumed. That Fibonacci cluster chart would look like
this:
The Most Common Decay Retracement Ratios Create a Zone of High Probability
Chart 2
(Chart 2) For the remainder of July and into the first few days of August the July 8 high
had not been exceeded so we could continue to assume that the pullback that we
believed had begun on July 8 was still in force. In early August, 2005 prices were
starting to poke around the beginning of our retracement zone starting at the 87 price
level (.382 retracement) so we wanted to be specially alert to the occurrence of any
bullish reversal bars beginning at that time. Later in the book we provide some methods
8
be important.
Are you going to get false positives? Sure. That is the nature of the beast. So far has we
know, and we've done the research, there is no 100% accurate trading system, method,
or technique. It is often said that success in trading is more the result of having and
following a simple trading plan that allows you to deal objectively with risk and losses
than a megabucks trading system that fosters illusory certainty. We agree.
## Bear Trend Retracements
Decay price retracements work equally well with pullbacks against a bear trend. The
following chart (Chart 3) shows Eurodollar continuous futures starting around December,
2004. The December, 30, 2004 high at 1.37 was potentially a very significant high and
when prices fell off from that high there was reason to believe that the main trend had
changed from bull to bear. Being able to objectively determine in advance the likely
termination point of the initial pullback against the main trend would result in a low risk
entry and a potentially very profitable trade.
## Eurodollars Daily Chart - Retracement Zone Set-Up
Chart 3
9
The Eurodollar chart shows the standard retracement zone set-up with prices that
correspond to the retracement ratios of.382, .500, .618, and .786. In this case,
Eurodollar prices pulled back to 1.33 at the .618 retracement ratio, declined for a few
days, and then rallied up to 1.35 on March 14 at the .786 retracement ratio, where the
pullback terminated.
## The Death Zone
We mentioned that the .786 retracement ratio, although not uncommon, deserves
special recognition. When a prior trend or swing has suffered a price reversal of close to
80% many assume that the current direction is not a pullback but the beginning of a new
trend. Not an unreasonable assumption. This weekly chart of the QQQ (Chart 4) shows
the bull trend that began in the fall of 2002. The December, 2004 high at 40.64 appeared
to be a good candidate for the termination point of the entire two-year rally. For one
thing, the weekly chart shows five distinct waves from the 2002 low ending in December,
2004. Even if you are not an Elliott Wave devotee you are probably still familiar with the
five wave impulse pattern and the proposition that the end of the 5th impulse wave is
very often a major trend reversal point.
Chart 4
10
There were other reasons as well to have taken a short position off the high that
occurred during the week of December 13, 2004. The next chart is a close-up of the
August, 2004 to December 2004 swing that carried QQQ prices into the rally high. This
was the last distinctive bull swing going into the suspected major high so we knew that if
the major trend was changing from bull to bear that this swing would be the first to be
retraced by more than 100%. That makes this swing the ideal candidate for the Death
Zone set-up.
## QQQ Weekly Close up - The Suspected Last Bull Swing
Chart 5
It is usually a good idea to plot the .618 and .786 retracement ratios on your working
charts after you suspect that the trend has changed from bull to bear or bear to bull. The
QQQ example is a very good illustration of the Death Zone, the retracement price area
that seems to kill new trends. If you were aware of the Death Zone and knew where to
plot it on your working charts you would have been more likely, in this case anyway, to
have kept some of the profits from your short trade off the December, 2004 high. Hope is
never a good plan and many traders gave it all back, plus some, when the suspected
new bear trend was killed in the Death Zone and the QQQ went on to make a new rally
high.
11
Here's another recent example from Bonds where proximity to the retracement Death
Zone stopped what might have been played as promising rally leg. At a minimum you
should avoid adding new positions until price successfully and convincingly clears the
Death Zone. The Bond chart could also be a good example of measuring retracements
in two degrees (not shown) as the .786 retracement from the spike high which occurred
immediately before June 27 was within .05 of the false rally high.
## Death Zone in US Bonds Weekly
Chart 6
12
Use Retracements from Two Degrees of Swing
The question often arises about what swings to use to measure retracement ratios and
other Fibonacci ratios. The short answer is that whenever possible you want to measure
Fibonacci ratios in one and two degrees. Figure 5 shows what that means.
In the hypothetical example (Figure 5) price rallied from 0 - 100 and then retraced to 38
before rallying again to new highs around
110. We can call the total move from 0 -
110 as one swing with two legs or two
separate swings in a bull trend. Either
definition would be correct but neither is
particularly helpful on its own. Figure 5
makes this concept more clear. if you
wanted to measure decay retracement
levels from the 110 high you would
measure the swing from 0-110 and the
swing from 38-110. The two
measurements in the same direction of
trend (or swing) is what we mean by
using Fibonacci ratios from two degrees
of swing.
The weekly QQQ chart (Chart 5) that illustrated the Death Zone is a real life example of
how using two degrees of swing can reinforce a Fibonacci analysis. Although the Death
Zone stands on its own as one of the most powerful applications of Fibonacci analysis to
real world trading, you can see in Chart 7 that the .382 retracement of the higher degree
Feb 2003 to Dec 2004 swing reinforced the possible importance of the lesser degree
price retracement into the Death Zone.
13
Two Degrees of Retracement Reinforced the Death Zone Set-up
Chart 7
14
GROWTH PRICE RETRACEMENTS
Decay retracements, which are also known as pullbacks and internal retracements, are
by definition always less than or equal to 100% of the measured swing. Growth
retracements as the name implies are
used to project termination points of
swings which exceed the range of the
measured swing. Growth retracements
are also called external retracements
because they exceed 100% of the range
of the prior swing.
## The most common growth ratios are
1.272, and 1.618 for measurements of
one degree. The growth ratios 2.618,
4.240 and 6.854 occur more often for
measurements of two or more degrees.
## In Figure 6, once the low point of decay is
known, we can apply the Fibonacci
growth ratios to the 62 point decline of the
bear swing from 100 to 38 and project an advance from the retracement low price at 38
to about 100 (100%) or 117 (127%) or to 138 (162%).
In Chart 8 once the Bonds rally off the February - March bear swing exceeded the .786
retracement you would plot the growth ratios 1.272, and 1.618 as shown on the Bonds
chart.
15
US Bonds Daily - 1.272 Growth Retracement Ratio
Chart 8
The earlier June 3 high (119.94) also provided a hit at the 1.272 retracement as the
Bonds made a double top after rallying for several months. Unfortunately there is
nothing inherent in Fibonacci price analysis that would alert you to the possibility of a
double top as experienced by the Bonds in June, 2005. Both tops did provide you with
bearish reversal bars at the Fibonacci price targets.
Soybeans continuous futures show where the 1.272 growth retracement price level
provided a valuable exit zone for a short trade following the June high.
16
Soybeans Daily - 1.272 Growth Retracement Ratio
Chart 9
The following Beans chart shows another six months of trading from the previous chart.
A decay or internal retracement (less than 100%) of the major swing from 498.50 to
757.50 provided another Fibonacci target that was smack into the growth or external
retracement price zone we calculated from the June 22 high. For actual trading you
want to use all the Fibonacci techniques, Retracement, Expansion, and Parallel
Projection, in one and two degrees, to create clusters of target prices that become your
17
Soybeans Daily - Decay Retracements Add Density to the Fibonacci Zone
Chart 10
18
FIBONACCI PRICE EXPANSIONS
Overview
Fibonacci price expansions are the simplest technique to apply. Use the analogy of an
extension ladder. When you pull the rope a single section expands to double or maybe
even triple its length. Fibonacci expansions use a single swing (one degree) or multiple
swings in the same direction (two degrees) to expand the price range. The most
important Fibonacci expansion ratios are .618, 1.00, 1.272, 1.618, 2.00, 2.618, and
4.236. As a general rule trending moves rarely exceed 4.236 of their first distinctive
swing. Not appearing as frequently, but not to be forgotten either, are .500 and 6.854.
In our illustration (Figure 7) prices advance from zero to one hundred and then
experience a pullback. So far as the expansion ratio technique is concerned the depth of
the pullback from the measured swing is not relevant although it would be in real trading
because you would apply the retracement technique in conjunction with the expansion
technique to determine if you got a
cluster or price zone.
## To apply the expansion technique
determine the measured range, in this
case 100, multiply that range by a
Fibonacci ratio, we will use .618 in
this example, and add the product of
that multiplication to the ending price
of the measured range. The result is a
Fibonacci projection of 162.
The math looks like this:
## Range: 100 * .618 = 61.8
Target: 100 + 61.8 = 161.8
## For a bear move projection you would
subtract from the low point of the
measured range instead of adding to the high point of the range. Figure 8 graphically
illustrates the bullish expansion concept.
## Some Fibonacci analysts downplay
expansions as somehow less worthy
than retracements and parallel
projections. That may be a mistake
from two perspectives: one is that you
can easily find dozens of cases
where Fibonacci expansions were
dead-on in projecting the end of a
move, the second is that Fibonacci
analysis is an inseparable bundle of
techniques - retracement, parallel
projection and expansion. For actual
trading you want to use all these
19
techniques, all the time, from one and two degrees. The idea, always, is to create tight
clusters of Fibonacci targets which form a price target zone. The more dense the cluster
the more likely that price will change within or around that cluster.
## Fibonacci Expansions of .618 and .500 on S&P 500 Weekly Chart
Chart 11
Two expansions, a .618 expansion that started from the all time high in March, 2000 and
a .500 expansion that started from the rally bounce-back high in May, 2001 did a very
good job projecting the bear market lows in the Summer and Fall of 2002. This chart is
another example of always using multiple Fibonacci ratio projections to create a cluster
of price forecasts.
The following Soybeans chart (Chart 12) shows how multiple Fibonacci expansions from
an earlier congestion zone were used to project the next congestion zone at lower
prices.
20
Multiple Fibonacci Expansions on Soybeans Weekly Chart
Chart 12
Fibonacci retracements from prior swings added density to the expansion cluster (Chart
13). We do not show it here but there are also parallel price projections from the June
and July highs that also add density to the cluster.
There are a couple lessons to be learned form these Soybeans charts. One is that you
will use only a fraction of the value of Fibonacci analysis if you restrict yourself to
forecasting only single bar highs and lows. The Soybeans charts tell us that dense
Fibonacci clusters are also very useful for forecasting price congestion zones that are
very difficult to trade successfully and should probably be avoided until a breakout from
the zone occurs. Another lesson is that we do not need "to-the-decimal" precision to
make Fibonacci price projections useful. You will have many experiences where a
Fibonacci price projection will be exact or nearly so but you should avoid making to-the-
21
Fibonacci Retracements on Soybeans Weekly Chart Add Density to the Cluster
Chart 13
22
Fibonacci Expansions of 1.272 and 1.618 on Palladium Daily Chart
Chart 14
Fibonacci expansions of two prior bullish swings created a price zone around which you
would be alert to a change in the bullish trend in Palladium. If we expand the Palladium
chart to include more trading days (Chart 15) you can see where .500 and .618
retracements from a prior swing high added density to the Fibonacci cluster derived from
the Fibonacci expansions.
23
Chart 15
24
FIBONACCI PARALLEL PROJECTIONS
Overview
Parallel projections are similar to expansions in that you measure the range of a swing in
the same direction as the trend you
are forecasting and proportion that
range with a Fibonacci ratio. With the
subtracting) the proportioned range
from the end of the measured swing,
you jump over to the beginning point of
the next swing, or the one after that,
and measure from there.
## Figure 9 shows the concept. The
bullish swing from 0-100 is measured
and (in this case) proportioned with the
Fibonacci ratio 1.000. The
proportioned range (100 points or
1.000 * 100) is then projected from the
start of the next swing at the 62 price
level. In Elliott terms you would be
projecting Wave 3 from the end of
Wave 2.
The most common Fibonacci Parallel Projection ratios are .618, 1.000, 1.272, 1.618,
2.00, 2.618 and 4.236.
25
Fibonacci Parallel Projections of 1.272 and 1.618 on MSFT Daily
Chart 16
In Chart 16 we measured the bull swing from March 29 - May 27 and proportioned it with
the Fibonacci ratios of 1.272 and 1.618. The proportioned range was projected from the
low on July 7. The 1.272 ratio put us in the price zone that terminated the July 7 swing
and also provided us with information that would become useful several months later.
26
Fibonacci Parallel Projections of 1.000 and 1.272 on MSFT Daily
Chart 17
In Chart 17 we proportioned the July 7 - August 5 bull swing and projected it from the
October 11 low. The combination of parallel projections from the two swings shown in
Charts 16 and 17 created a Fibonacci cluster within which we would be looking for a
consolidation or a change in trend in MSFT.
27
Parallel Projections of 1.272 and 1.618 in Eurodollars Daily
Chart 18
The 1.272 and 1.618 parallel projections forecast the congestion zone in Eurodollars
following a substantial bear move. The initial bear move low, and the beginning of the
congestion zone, was smack on the 1.20 price projection.
28
Parallel Projections of 1.272 and 1.618 in US Bonds Daily
Chart 19
Bonds changed direction, for at least the short term, within a parallel projection cluster.
We would be reluctant to use such a wide cluster if it could not be tightened with either
additional parallel projections from other swings or made tighter and more dense with
retracements or expansions. The next chart (Chart 20) provides the necessary
reinforcement of credibility to the parallel projection cluster with two retracements from a
greatly expanded chart.
29
.382 Retracements of Prior Swings Tighten the Cluster in the Bonds Chart
Chart 20
30
WILL A FIBONACCI PRICE CLUSTER BECOME IMPORTANT?
We have shown you the three Fibonacci ratio techniques for projecting individual prices
and price clusters that may become predictive, or at least relevant, in determining a
retracements, expansions, and parallel projections from one two or three degrees it is
easy enough to imagine that you will have a chart littered with so much data that you will
have little information.
The question arises then "How do I know which Fibonacci clusters may become
important in the future?" There is nothing inherent in Fibonacci analysis itself to provide
an answer. Experience with a particular ticker may provide enough background
information to know which Fibonacci ratios have been important for that ticker in the
past, and which may continue to be important in the future.
You may also combine your Fibonacci analysis with another, independent technique or
method to identify coincidences of pattern price and time from both your independent
method and your Fibonacci analysis. We provided information on three such
independent methods, and tell you how to immediately implement a fourth method,
which is both elegant in its simplicity, and which has been shown to have predictive
value.
ELLIOTT WAVE
The Elliott Wave and Fibonacci analysis are joined in at the hip. In its broadest terms,
the Elliott wave describes the mathematical progression and decay that occurs
constantly in publicly traded financial markets. A bull market subdivides into five waves,
and a bear market subdivides into three waives. This highest order 5/3 relationship
=.600 ratio. Each wave of a bull market itself divides into a 5/3 relationship and each of
those five waives further divides into a smaller scale 5/3 relationship. An idealized bull
market would consist of 89 bullish waives and 55 bearish waives, a 55/89 relationship,
which equals .618. The Golden Ratio.
Even if you have no present desire to implement Elliott Wave analysis in your trading we
recommend that you take the Elliott Wave Tutorial at Elliott Wave International. It's free
to Club EWI members and there is no cost to join Club EWI. Three of the ten lessons are
devoted to Fibonacci and the Golden Ratio and each is explained with the breadth and
depth for which Robert Prechter and his staff are known world over. Highly
recommended.
The Appendix contains trading guidelines for combining Fibonacci and Elliott Wave
analysis.
## J.M. HURST CYCLE TRADING METHOD
J.M. Hurst was an aerospace engineer who in 1970 applied his engineering wizardry to
trading the stock market. His book was not for the mathematically challenged but he did
provide a kernel of knowledge that will allow anybody with rudimentary spreadsheet
reliable price and time forecasts. The description of the method is proprietary to
31
Tradingfives and we offer it in our book J.M. Hurst Cycle Trading without the Rocket
Math, which is available on the Tradingfives website.
Chart 20 is a Daily MSFT chart overlaid with our Hurst cycle trading method. The yellow
circle is a price and time forecast made in early July, 2005. The relevance of this
particular forecast will become apparent in a few paragraphs.
Chart 21
32
Chart 22
Roadmap Charts are another Tradingfives creation. They resulted from our search for a
way to make W.D. Gann's Square of Nine useable. Roadmap Charts are unique in that
the trading channels are drawn immediately after a suspected pivot high or low has
occurred. Often on the same day. In many cases the trading channels will contain and
precisely define a trend for weeks or months. Volatile tickers, like MSFT in the example
in Chart 21, will sporadically run out of the channels. One of the implementing guidelines
of Roadmap Charts is that the trend is likely to continue without any significant pullbacks
in the direction of the channel until price hits the opposite channel boundary. That
happened three times ("X" marks the spot) in the MSFT chart. Fibonacci clusters that
occur at channel boundaries are likely to become more significant. Roadmap Charts are
more complete graphical application of the next method.
## THE FOURTH METHOD - SQUARE ROOTS
The Square Root Theory holds that stock and other publicly traded instrument prices
move over the long and short term in a square root relationship to prior highs and lows.
In the 1950s, William Dunnigan developed two stock trading systems called the Thrust
33
techniques, but each lacked a reliable exit technique. Dunnigan unearthed the Square
Root Theory and applied it to his systems which became highly acclaimed for their
effectiveness. He went so far as to call square roots, the "Golden Key" and received
recognition from some economics and statistical trade journals of the era. Norman
Fosback republicized. the theory in a 1976 trading letter called Stock Market Logic. The
founder of the Templeton mutual fund family disclosed that one of his 22 principles for
start market investing was that stock-price fluctuations are proportional to the square
root of prior prices.
Long before any of these stock market researchers, probably starting in the early 1900's
W.D. Gann was using square roots as the cornerstone of his still unduplicated stock and
commodities price forecasts. Gann used an ennegram, a diagram of numbers
constructed in such a way to show square and square root relationships. Gann's
ennegram is what has come to be known as the Square of Nine from the Greek root
"enneas" which relates to the number nine. The Square of Nine method is more complex
than calculating squares and square roots but they are an important part.
What does this have to do with Fibonacci ratios? Here is the MSFT chart again with only
square root information added to it.
Chart 23
34
The square root information, which could have been determined with precision as early
as the close of trading on May 29, the date of the low, immediately gives us specific
reference points for our Fibonacci analysis. The Fibonacci clusters we got in Charts 16
and 17 around the 27-28 price area now have some meat on their bones. We are no
longer shooting in the dark and can know, with a high degree of reliability, very early in
the new swing, which Fibonacci ratios and Fibonacci price clusters deserve the most
attention.
Here is a review of a few of the most recent charts we've used as demonstrations of the
various Fibonacci techniques with just the square root information added.
## Compare with US Bonds Charts 19 and 20
Chart 24
35
Compare with Soybeans Charts 12 and 13
Chart 25
36
Compare with Palladium Charts 14 and 15
Chart 26
## HOW TO DO THE MATH
Implementation of the Square Root Theory is straightforward. You calculate the square
root of a high or low pivot price, subtract from or add to that square root calculation, and
square the result. Plot that result on your chart. It's fairly intuitive.
What may not be so intuitive, however, is the exact amount you use for the addition or
subtraction. Without going into a lengthy explanation of the reasons why, the most
frequent amounts you will use for these additions and subtractions are: 2, 1, .500, .250
and .125. The guideline is to use the largest number that will produce visible results on
If you want to know more about why these particular numbers are commonly used visit
Square Root Theory.
37
Bullish Example
We'll use the MSFT chart again for the bullish example. The May 29, 2005 low at 23.60
was a major pivot point so we can assume that square root relationships to that low are
likely to become more significant than more minor swing highs or lows. The square root
of 23.60 = 4.857
## Add (4 * .125) to 4.857 = 5.357. Square the result = 28.70.
Bearish Example
Bonds made a major high at 119.34 on June 27, 2005. The square root of 119.34 =
10.924.
## Subtract (4 * .125) from 10.924 = 10.424. Square the result = 108.66.
It so happens that both these examples responded well to using .125 as the active
number. Generally, the higher priced the ticker the larger the active number will be, and
vica versa. Very low priced tickers like Eurodollars with a natural price sometimes under
\$1.00 respond better to very small divisions of 1.00 like .0625 or .03125. The current
volatility of the ticker also plays a role in determining the active number. The
calculations are so quickly and easily done that just a few minutes of experimentation
with your favorite tickers should be sufficient to achieve useful results.
38
FIBONACCI TIME PROJECTIONS
You have seen many examples of how Fibonacci ratios can be used to forecast
individual price highs and lows, and more importantly clusters of prices. Fibonacci ratios
can also be used to successfully forecast time. The procedure is very similar to that used
in the price projection techniques.
## Time retracements are conceptually similar to internal price retracements. We measure
the time that has elapsed between the
extremes of a single swing (low-to-high
or high-to-low), proportion that time
measurement with a Fibonacci ratio
less than 1.00, and project the result
forward in time from the most recent in
time bar in the measured swing.
## Figure 10 displays the concept. In the
illustration we use the .618 Fibonacci
ratio to measure the time of the 0-100
swing. The projection is made from the
bar at the swing high at 100.
## The most reliable time retracement
ratios are .500 and .618. The .382 ratio
is also reliable when used to measure
the longer swings.
We use trading days in the illustrations and in our examples. Some respected Fibonacci
analysts prefer calendar days. Fact is,
the difference between Fibonacci time
projections using calendar days and
Fibonacci time projections using
trading days seldom differ by more
than 1,2 or 3 days depending on the
duration of the measured swing. The
reason for the less than expected
difference is that calendar day
projections take the outside lane by
counting weekends and holidays.
Trading day projections take the faster
inside lane and arrive at the finish line
(a specific bar on the chart) at about
the same time.
## Fibonacci Time Expansions
Time expansions are procedurally identical to time retracements. The difference is that
in time expansions the measured swing is proportioned with a Fibonacci ratio of 1.00 or
39
greater. The forward projection is made from the same location, the most recent bar in
the measured swing. Figure 10 displays that difference.
The most reliable time expansion ratios are considered to be 1.00 and 1.618. The ratios
2.00 and 2.618 are also useful. For your own trading use terms like the "most reliable"
and "useful" only as a starting point. There are thousands and thousands of individual
tickers bought and sold on dozens of local, national, and international exchanges. Your
favorite tickers may very well display personalities radically different from the usual
examples. If .786 or 1.272, for example, has been a reliable time forecaster in previous
swings then keep using it.
## 1.618 Time Expansion in the S&P Daily
Chart 27
The duration of the August 3 - August 30 bear swing proportioned by the Fibonacci ratio
of 1.618 was smack on the October 13 low that launched a powerful bull move.
40
Internal Time Retracements in the US Dollar Index
Chart 28
41
Multiple Time Retracements in Natural Gas
Chart 29
42
PARALLEL TIME PROJECTIONS (I & II)
We will use the actual Nikkei chart to illustrate the two techniques related to Parallel
Time Projections. Time Retracements and Expansions are measured on a single swing,
how-to-low or low-to-high. If you were using our training software a Time Retracement or
Expansion would take only two mouse clicks. A Parallel Time Projection, just like a
Parallel Price Projection, always takes three mouse clicks.
Chart 30
The three click locations in Chart 30 are labeled A-B-C. We will use the click locations to
differentiate between Technique I and Technique II.
Technique I
Technique I mimics a Parallel Price Projection. The A-B swing is measured and
projected (after being proportioned by a Fibonacci ratio) from point C. For clarity the
measured points are marked with lighter blue triangles and the projected points marked
with dark blue triangles. An upside down dark blue triangle symbol is located at the start
of the projection.
43
Technique II
For Technique II points A-C are measured and projected from point B. If we have a
favorite Fibonacci time projection it would be Technique II.
## Technique II in the S&P Daily
Chart 31
44
Techniques I & II in CSCO Daily
Chart 32
Chart 32 is legended with the A-B-C click sequence to show the difference between the
applications of the two Parallel Price Projection techniques. As a practical matter the
way the techniques are used together in the CSCO chart should be the norm.
45
Technique II in the FTSE Daily
Chart 33
To spice things up a bit the C click in a Technique II projection doesn't have to be related
in any way to the A-B swing or even adjacent to it. In Chart 33 the C click is separated
from the B click by almost four months of daily trading yet was smack on in forecasting
the high that occurred two months after that.
## Time Projection Summary
Fibonacci Time Projections should be clustered in projections of one two and three
degrees using Retracements, Expansions and Parallel Techniques I and II. The most
reliable Time Projections will occur when a time projection converges with a Fibonacci
price cluster which has itself been focused with the Square Root Focuser.
The Square Root Theory also has some applicability to focusing Fibonacci Time
Projections. It can be more complicated to implement as natural prices must often be
converted by multiples of 10 before it can be applied. The square root of a natural price
of 1.21, like in Eurodollars for example, will not be useful without the conversion. This
technique does not have the across-the-board reliability as it does with square root price
46
calculations although you will find that certain time frames converge very well with
certain tickers. We have not included this time feature in the initial version of the training
software although it may be implemented in later versions.
You can review the Roadmap Chart references on the www.tradingfives.com website for
47
APPENDIX
## FIBONACCI AND THE ELLIOTT WAVE TRADING GUIDELINES
When do you enter the trade? How do you determine reasonable and objective stop-
loss levels upon entry and as the trade progresses? The table describes the trading
signals that can be objectively determined by blending Fibonacci and Elliott Wave
analysis.
## Reversal Alert A coincidence of pattern, The suspected high or low
price and time has come tick at a major pivot point
Aggressive Trade Entry together to mark a is the ideal entry point.
suspected major pivot that This entry has the least
we have identified as the capital risk because it is
5th of a 5th wave. The closest to the initial stop
trend is about to change. loss point - the pivot. The
trade off is that there will
be a higher percentage of
losses. The market does
not always reverse where
we want it to! The pivot
point is Wave Zero.
Wave 1 Alert Wave 1 is the first swing This an alert signal only.
in the new trend from the No action taken. We use
pivot. We can identify five the extreme of Wave 1 to
wave of smaller degree. project price and time
targets for a Wave 2
Wave 1 almost always entry.
overbalances prior
countertrend swings of the What we suspect is Wave
same degree in price and 1 of a new trend can be
time. the first wave in a
correction. Three swing
patterns occur frequently.
Very often the third swing
relates to the first by a
multiple of .62 to 1.00.
That often makes the third
the first swing's final
status as an A wave or
Wave 1 of a new trend is
in doubt.
## Wave 2 Entry The second swing must A Wave 2 entry is
not exceed Wave Zero. positioned for the
You can identify three upcoming Wave 3 or Wave
waves of lesser degree. C and what is usually the
48
Wave 2 has retraced at most profitable leg of the
least 50% but not more trade. Maximum stop loss
than 79% in price of Wave point is Wave Zero.
1.
## Wave 2 is usually >50%
and <162% in time of
Wave 1.
Wave 3 Potential Wave 3 has exceeded the Maximum stop loss levels
level where W.3 = W.1. can be raised to the
The price distance traveled extreme of Wave 2. The
by Wave 3 is greater than assumption is that once
the price distance traveled Wave 3 exceeds the
by Wave 1 distance traveled of Wave
1 that the new trend is
confirmed. If Wave 3 has
completed five waves of
smaller degree and has
not exceeded this level
then there is a strong
possibility that this
potential Wave 3 is a
Wave C and not an
impulse wave in a new
trend
Cleared Death Zone The new swing has The Death Zone is the
exceeded the .786 62%-79% retracement
retracement level of the zone of the entire last
entire length of the last countertrend swing. If,
countertrend swing of the despite appearances, the
same degree. new swing from Wave
Zero is not impulsive in
the direction of a new
trend, the Death Zone is
the most likely area where
the new swing will get
killed.
Wave 3 Extension Wave 3 has exceed Wave Trail stop loss point closer
1 in price distance by to the market. No more
162% . than 2-3 trailing high or
low.
Wave 3 is commonly
162%-262 of Wave 1 and
it is in the exhaustion
stage where it could come
to an unexpected end.
49
Wave 3 Reversal Alert Wave 3 has completed at Wave 3 almost always
least five waves of lesser exceeds Wave 1 in time
Trade Exit degree and is within a and often equals Wave 1
Fibonacci price and time and Wave 2 combined in
cluster. time.
## Trailing stop loss point
should be moved very
close to the market. No
more than 1 or 2 bar
trailing stop.
Wave 4 Entry Wave 4 has not violated The initial stop loss for the
the price territory of Wave upcoming Wave 5 trade is
1. Usually Wave 4 will the extreme of Wave 1.
retrace >38% of Wave 3. Wave 4s are often
Wave 4 alternates in complex and the most
pattern with Wave 2. frequent home of triangles
Wave 4 is in a Fibonacci and the dreaded "X" wave.
price and time Zone. Wave Do not be overanxious to
4 is not overbalanced in take a Wave 5 trade if
time to previous waves of Wave 4 has retraced more
the same degree. Wave 4 than 50% of Wave 3 or if
has retraced at least 62% it is so overbalanced in
of the price range of Wave time that it exceeds the
2. time consumed in any
correction of one larger
degree. On the other
hand, if Wave 3 is easily
identifiable as such and
Wave 4 has retraced 38%
or less of Wave 3 in a
simple A-B-C, then new
Wave 5 highs are very
probable.
## Wave 5 Reversal Wave 5 has completed at When the extreme of
least five waves of lesser Wave 3 is exceeded the
Trade Exit degree and is in a maximum stop loss should
Fibonacci price and time be raised to the Wave 4
cluster. extreme.
## Wave 5 extremes seldom
exceed the 424%
expansion of Wave 1.
>50% retracement of
Wave 3, the possibility of
a 5th wave failure is
increased.
50
and a Calculator
## The purpose of this work is to explain concisely
and in detail simple mathematical and
graphical techniques for applying WD Gann's
Square of Nine to real world stock, stock
option, and forex trading situations. The
Square of Nine is not your usual method of
technical analysis. It's like nothing you've ever
seen. A completely unrelated technique that
methods can be invaluable when making
decisions. The Square of Nine may not be the
magic bullet, although it can sure seem like it at
times. Free training software is available.
Rocket Math
## This ebook shows you everything you need to
know to apply our modification of J.M. Hurst's
displaced moving average technique to
consistently and reliably project price and time
turning points in the financial markets. We
consider this booklet a perfect complement to
the Square of Nine and Fibonacci techniques
because each method provides unique price
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independent methods. Free training software.
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51
TRAINING SOFTWARE INFORMATION
## HISTORICAL DATA COLLECTION
Hourly data is readily available from a number of on-line sources. www.quote.com has a
\$9.95/month service with a feature to export several intraday time frames to Excel where
they can be saved in ASCII format.
You can also get 5 minute interval stock files at moneycentral.msn.com for free. You will
need a MSN Passport, which is also free. You have to download manually and jump
through a few hoops in Excel to get the msn data in useable condition, but it is available.
EOD data collection is a strong point of AnalyzerXL which easily collects free stock,
index, mutual fund and futures data from exchanges across the world. Data collection is
only one module in AnalyzerXL which can better be described as a trader's tool in an
Excel environment. Other modules include charting with 146 built-in indicator and expert
system macros, real time quotes, and option data collection. You can download a 30
day free trial of AnalyzerXL at the TradingFives website at this URL:
## REQUIRED DATA FORMAT FOR THE TRAINING SOFTWARE
The number one problem users have with our training software is data formatting. The
training software uses only ASCII data files. Most data collection software has an export
function that allows the user to convert the software's proprietary format into ASCII.
## DATE TIME OPEN HIGH LOW CLOSE VOLUME (optional
mm/dd/yyyy hhmm but not used)
Every data file needs a time field even if End of Day. Use 1600 for EOD data if the time
field has to be manually inserted.
The data delimiter can be a comma, space or tab. The software would prefer to see a
comma delimiter, which has been the standard since the first main frames, but spaces
and tabs will also work.
With the exception of .prn, which is reserved for AnalyzerXL exports, there is no
limitation on the name of the file extension. .txt, .asc. dat, and any other combination
(with the exception of .prn) will work equally as well.
Yahoo Finance
http://finance.yahoo.com. The training software will fill in the required time field with
16:00. There is a quirk (at least as of this date). When presented on the Yahoo site with
the choice to "Open" or "Save" the historical data, always choose "Open." This causes
the data to open in your spreadsheet. You can rename the file and save it to your ASCII
data directory from there.
52
Do not sort by dates in the spreadsheet. The training software will automatically reverse
the native sorting and if you do it yourself it will do it again - wrong.
AnalyzerXL
We use AnalyzerXL for all our EOD stock, index, mutual fund and futures data. The file
extension .prn is reserved exclusively for this exact set-up. AnalyzerXL has an automatic
export macro which makes data collection very easy indeed. Here's a screenshot of the
AnalyzerXL data collection set-up. It is difficult to read at normal viewing size but still
conveys the simplicity of the procedure. The data format for AnalyzerXL exports is
different from regular ASCII but that exception is handled automatically by the .prn
extension.
The corollary is that unless your data export is going to look exactly like this do not use
the .prn file extension.
<TICKER>,<PER>,<DTYYYYMMDD>,<TIME>,<OPEN>,<HIGH>,<LOW>,<CLOSE>,<VOL>,<OPENINT>
^GSPC,D,20030613,000000,998.51,1000.92,984.27,988.61
^GSPC,D,20030616,000000,988.61,1010.86,988.61,1010.74
^GSPC,D,20030617,000000,1010.74,1015.33,1007.04,1011.66
^GSPC,D,20030618,000000,1011.66,1015.12,1004.61,1010.09
53
Filename: fibonacci.doc
Directory: C:\Documents and Settings\M. Peter\My Documents\Fibonacci Book
Template: C:\Documents and Settings\M. Peter\Application
Data\Microsoft\Templates\Normal.dot
Title: INTRODUCTION
Subject:
Author: M. Peter
Keywords:
Creation Date: 11/22/2005 12:01:00 PM
Change Number: 391
Last Saved On: 07/08/2006 7:24:00 PM
Last Saved By: M. Peter
Total Editing Time: 5,233 Minutes
Last Printed On: 07/08/2006 7:27:00 PM
As of Last Complete Printing
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Learn complete Blogging and WordPress designing step by step in Hindi.
## 24 hours passed
### If YOU are an Electrical Engineer then following quiz is for YOU
0%
2
Electrical Engineering
Smart Series Quiz-1
These questions are taken from smart series book which covers all the subjects of Electrical Engineering and Technology Field
1 / 30
Category: Power System Analysis
1) Which of the following ideas means deciding at runtime what strategy to summon?
2 / 30
Category: Power Plant
2) ______ is a by-product of paper industry, which is used by many industries as a primary source of electricity.
3 / 30
Category: Basic Electrical Engineering
3) Which line is obtained by the method of least square?
4 / 30
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4) Drop out to cut off ratio for most relays is of the order of
5 / 30
Category: Singal and Systems
5) Main purpose of modulation process is to
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Category: Basic Electrical Engineering
7) If the input capacitor of a power supply is shorted, it will result in
8 / 30
Category: Basic Electrical Engineering
8) Conjunction x ^ y behaves on digits 0 and 1 exactly as ____ does for ordinary algebra.
9 / 30
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9) For which purpose bundled conductors are employed to a power system
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10) What is the probability of a number “2” when a dice is thrown?
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13) Fourier series are infinite series of elementary trigonometric functions i.e. Sine and
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14) At very sunny places this source(s) of energy can be found
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15) The quantity “Rm” which relates dependent voltage to controlling current is called
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16) A transformer transforms
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Category: Power Plant
17) The average fossil fuel plant convers about is ______ % of the power going in to the power going out.
18 / 30
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18) Fundamental property used in single node pair circuit analyzer is that ______ across all elements is same.
19 / 30
Category: Electroncis
19) A semiconductor device is connected in a series circuit with a battery and a resistance. If the polarity of battery is reversed, the current drops almost to zero. The device may be
20 / 30
Category: Network Analysis
20) Admittance is the reciprocal of
21 / 30
Category: Power System Analysis
21) Power in AC circuit is found by
22 / 30
Category: Singal and Systems
22) The _____ time signal is described for all values of time.
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23) There are ____ types of dependent sources, depending on the controlling variable and output of the source.
24 / 30
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24) Landfill gas is actually _______ used in thermal power plants.
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We have to maintain flatness of the surface.
27 / 30
Category: DC Machines
27) The Pole Shoes of DC Machines are fastened to the pole core by:
We have to maintain flatness of the surface.
28 / 30
Category: Power System Analysis
28) Which of the following are considered as disadvantages(s) of Gauss-Seidel method over Newton Raphson method in load flow analysis?
29 / 30
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29) SCR (Silicon Controlled Rectifier) goes into saturation, when gate-cathode junction is
30 / 30
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30) Modulation index in amplitude modulation | 966 | 4,175 | {"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-2024-18 | latest | en | 0.859174 |
spellquiz.com | 1,586,448,996,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371861991.79/warc/CC-MAIN-20200409154025-20200409184525-00548.warc.gz | 683,313,484 | 232,512 | # Dyscalculia: Does Your Child Have Trouble with Numbers?
17 May 2019
Samira Tasneem
Have you ever got in a situation where your child is finding it really hard to understand simple math facts? For example, are they having a problem understanding the basic numbering order, being confused about which comes first 5 or 7? There is a strong chance that your kids might be a victim of dyscalculia.
You must have heard about dyslexia – the disability to read and write properly. In many cases, parents get confused between dyscalculia and dyslexia. Well, in reality, dyscalculia is a different learning disability that involves mathematical learning hardship.
Today, we will try to help you diagnose whether your child actually has dyscalculia or not and what you can to help him or her to tackle this situation.
# Dyscalculia Definition: Understanding the Condition
Dyscalculia is a learning disability that makes it hard or nearly impossible for a person to grasp the fundamentals of mathematics. The symptoms vary from person to person, and not everyone will have the same issues.
You might be thinking that every child faces problems at first while learning mathematical fundamentals. Well, it's true that a lot of children find math difficult at first. However, they outgrow these issues as time passes.
On the other hand, this disorder is a lifelong condition. It doesn’t go away as someone is growing up.
What are some common and basic identifiers of dyscalculia?
As we have said, dyscalculia affects each person differently. But there are some symptoms that might be considered as strong identifiers –
• Difficulty understanding quantity
• Inability to tell which number is bigger and which one small
• Failure to understand the “5” and “five” are the same
• Difficulty to understand the mathematical symbols and their meaning
These are some of the primary symptoms. There are a lot of dominant signs that we would discuss later.
One more thing, there is no certain research that confirms what exactly causes this disorder only some speculations. Moreover, no expert knows for sure whether dyscalculia is more dominant in boys or girls.
# Signs and Symptoms: Does Your Child Have Dyscalculia?
First things first, there is not certain symptoms like most other disorders when it comes to dyscalculia. Basically, the disorder is connected to mathematical difficulties, but it's not mandatory for two cases to be exactly the same. However, don't confuse it thinking it a simple case of procrastination or a shorter attention span. Â
Just like other learning disabilities like - Dyslexia, Dysgraphia, DCD, ADHD, etc., the symptoms become prominent when kids start going to schools. Some may start showing signs in the preschool, while some will not show them until grade school or middle school. Pinpointing dyscalculia signs is quite challenging.Â
Check out for these dyscalculia symptoms grade wise –
### Preschool Level
• Inability to coordinate number to a group of things
Example: Your child has brought five apples from a basket even if you have asked him or her to bring seven. Has it been happening for some time?
• Cannot sort items in order
Example: You have asked your child to organize books, pencils, notepads and he or she has been struggling to do so.
• Inability to recognize patterns
Example: You are asking your kid to organize objects from smallest to biggest, and they are failing constantly.
• Struggling to count numbers
Example: Your child is failing to understand the fundamentals of counting. They become confused when you ask them to bring three pens.
• Finds it hard to harness basic math facts
Example: A kid might forget basic math calculations like 3+6=9
• Failure to identify the functions of the mathematical signs
Example: A kid could struggle to determine the differences between “+” and “–” signs.
• Finds it hard to calculate mentally
Example: Rather than doing the simple calculations mentally, i.e. 9-2=7, a kid could use fingers to count.
• Cannot comprehend the simple relations between numbers
Example: A child with dyscalculia might struggle to understand 9 is greater than 7.
The problems will rise exponentially as the child grows as he or she finds it hard to grasp the basic mathematics. If you see your child acting so despite putting much effort and concentration, he or she could be a victim of this disorder.
# How to Help a Child with Dyscalculia?
First, you must not blame the child being negligent to his or her studies. You have to be patient in such drastic situations.
There can be several causes for a child to have this disorder but dyscalculia diagnosis is not that easy –
• Genetic issue
• Improper brain development
• Negligence of the mother during pregnancy, i.e. fetal alcohol syndrome
• Brain damage
Although, the experts and the doctors are not entirely sure of the true cause of the disorder. These are just mere speculations.
There is a strong link between dyscalculia and dyslexia as about 40% - 50% child affected with dyscalculia also have dyslexia.
As a parent, you should try following some helpful activities –
• Introduce multi-sensory teaching techniques
• Playing fun board games that can help a child with mathematical problem solving
• Always motivate and inspire your child
• Never say anything that can hurt their feelings or self-esteem
• Focus on other skills that your kid might have a talent for
• If possible, find schools (even homeschooling is a great option) or curriculum that can help a child with dyscalculia
• Personalized learning can be a life savior
# Final Words
You shouldn’t feel disheartened if you see your child suffering from dyscalculia. If you feel shallow, it will make them feel even more shallow. You should always cheer them up and help them have a better perspective on life. Moreover, there are a lot of people with dyscalculia who are now studying at top-tier colleges as dyscalculia in adults is not rare. So, stay strong and keep supporting your child.Â
Now you can take part in online Spelling Bee too! Check out the SBO section on Spellquiz today! Also, try this vocabulary tester to understand your current skill level! Try these spelling tests to master English spelling! Also, don't forget to check the complete list of sight words. Â
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# Investigating the effect of light
Extracts from this document...
Introduction
Investigating the effect of light intensity on photosynthesis in a pondweed Aim: To investigate how the rate of photosynthesis changes at different light intensities, with a pondweed. Prediction: I predict that the oxygen bubbles will decrease when the lamp is further away from the measuring cylinder, because light intensity is a factor of photosynthesis. The plant may stop photosynthesising when the pondweed is at the furthest distance from the lamp (8cm). Without light, the plant will stop the photosynthesising process, because, light is a limited factor. However once a particular light intensity is reached the rate of photosynthesis stays constant, even if the light intensity is the greatest. If I plot distance of the lamp, against the number of bubbles per 2 minuets, I will get a straight linear graph which will not go through the origin. Introduction: Photosynthesis is the process by which green plants use light to synthesize organic compounds from carbon dioxide and water. In the process oxygen and water are released. The glucose produced in the photosynthesis reaction can be converted to sucrose and carried to other parts of the plant in phloem vessels. ...read more.
Middle
* Add a spatula full of sodium bicarbonate. * Pick a pondweed that has a healthy green stem. * Put a paper clip at the end of the pondweed, to make sure that the pondweed stays down. * Place the lamp next the measuring cylinder, measure out 5 cm (this is my first measurements). * Wait till the pondweed starts photosynthesising. * When there is a steady pace of bubbles, start timing and counting. * Time the bubbles for 3 minuets and record the results. * Repeat this with 8 cm, 11 cm, 14 cm, and 17 cm. * Repeat this 3 more times so you get 3 sets of results. Diagram: Table of results: Experiment 1 Distance (cm) Number of bubbles In 1 min Temperature (�C) 0 11 32 2 9 32 4 4 32 6 2 32 8 0 32 Experiment 2 Distance (cm) Number of bubbles In 1 min Temperature (�C) 0 11 30 2 8 32 4 3 32 6 1 32 8 0 32 Experiment 3 Distance (cm) Number of bubbles In 1 min Temperature (�C) 0 13 31 2 8 32 4 3 32 6 1 32 8 0 32 *The Rate of Photosynthesis is the number of seconds it takes for 1 bubble of Oxygen (O2) ...read more.
Conclusion
I have investigated enough distances, because 5 measurements are the average amount of distances in one experiment. I have controlled all my variables - Input variables was the light intensity, the output variable was the measurements and the control were, the same time, around the same temperature and the direction of the light had to face the same way. Temperature was a tricky one to control but because is a limiting factor and might change the rate of photosynthesis. However this was alright in my experiments because there was no great range of temperature changing. There are many other further work I could of done for this experiment apart from changing the distances from the lamp and the measuring cylinder, you can change the light intensity by maybe changing the watts in the light bulb, changing the colour of the light by putting different transparent coloured paper, changing the pondweed by finding a pond weed after every experiment, changing the temperature by using a water bath and heating it to a certain temperature, this would be hard to control. Changing the amount of carbon dioxide by adding a certain amount of sodium bicarbonate, and many more but this is all that I can think about. ...read more.
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# Related GCSE Green Plants as Organisms essays
1. ## Investigating the effect of temperature on the rate of photosynthesis
This will cause reactions to speed up, hence the graph increases rapidly. I correctly predicted the rate of oxygen produced would rise, as temperatures get warmer, due to the increased kinetic energy of the enzyme rubisco. The line of best-fit increases steadily from 22oC and 35oC on graph 1, where
2. ## How temperature affects the rate of photosynthesis.
gained such a high amount of kinetic energy that the excessive vibrations caused an irreversible change in the structures of their molecules preventing them from reacting and forming products like the glucose and amino acids from the triose phosphate. This could also have resulted in less regeneration of the Ribulose
1. ## An Investigation into Species Diversity with distance along a Pingo.
Apparatus: the apparatus that will be used to conduct this experiment are as follows Apparatus Justification of use 35x35cm Quadrat This was the optimum sized quadrat for the pingo to be studied. It will be used to calculate the species diversity of the different zones along the pingo.
2. ## The effects of organic effluent from the seweage on the biodiversty in a freshwater ...
Modified---The sites selected are ASHBY, LAGOON 3 AND RAISED POND. These sites will compare the species diversty within one waterway. These will vary in distance from the source of pollution however they have been selected regardless of the equal distance due to the following reasons.
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• Ideas and feedback to | 1,368 | 5,833 | {"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} | 2.515625 | 3 | CC-MAIN-2018-26 | latest | en | 0.901113 |
https://math.stackexchange.com/questions/1582298/what-is-the-lagrange-remainder-in-a-taylor-series-expansion | 1,571,348,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00270.warc.gz | 588,913,407 | 31,650 | # What is the Lagrange remainder in a Taylor series expansion
I know what a Taylor series expansion is and I know how to find the Lagrange remainder but what does it mean intuitively? I need an explanation of what the Lagrange remainder represents in terms of the Taylor series expansion without proofs, strictly intuition.
• Any difference between remainder and Lagrange remainder? I think it's the difference between the function f and the Taylor polynomial that approximates it. So it's like an error term. It approaches zero as we add more and more terms to the Taylor polynomial to approximate f – BCLC Dec 19 '15 at 16:42
• I guess any remainder because the Lagrange remainder is just one way of calculating the same remainder correct? – idknuttin Dec 19 '15 at 16:46
Taking your comment about “any remainder” in mind, BCLC’s comment is effectively the answer you seem to be looking for, i.e. the remainder $$R_n(x)$$ is simply the difference between the $$n^{th}$$ order Taylor polynomial $$T_n(x)$$ and the function it is approximating. That is, given $$T_n(x) = \sum_{i=0}^{n} \frac{f^{(i)}( c)}{i!}(x-c)^i$$, the remainder is:
$$R_n(x) = f(x) - T_n(x)$$
There are various forms for this remainder, one of which is the Lagrange remainder you mentioned. Wikipedia groups the various forms under “Taylor’s theorem,” and provides a more detailed discussion about the technical details.
Broadly speaking, for applications, having an expression for the remainder allows one to, e.g.:
1. Determine the error on a given approximation (i.e. for a given number of terms used in the approximation, determine the error);
2. Determine the number of terms required to get an approximation that will be within some desired error (i.e. to obtain a value of the function accurate to a given number of decimal places, determine the number of terms $$n$$);
3. Determine the largest interval over which the error in a given approximation will be below a desired level (i.e. fix the number of terms and the error, determine the interval).
One can also use the remainder to prove that the limit (i.e. as $$n \to \infty$$) of the sequence of partial sums of $$T_n(x)$$ converges to $$f(x)$$, by showing that $$R_n(x) \to 0$$, in the same limit, i.e.:
\begin{aligned} f(x) & = T_n(x) + R_n(x) \\[5pt] % \Leftrightarrow \qquad \lim_{n \to \infty} f(x) & = \lim_{n \to \infty} T_n(x) + \lim_{n \to \infty} R_n(x) \\[5pt] % = \qquad f(x) & = \lim_{n \to \infty} T_n(x) + 0 \end{aligned}
Although, in practice, it is usually impractical to work with $$R_n(x)$$ directly (even using Taylor’s inequality) and other methods may be simpler (e.g. showing that the series satisfies the D.E. and initial condition which define the function; for example $$\frac{df}{dx} = f$$ and $$f(0) = 1$$ for $$f(x) = e^x$$, or $$(1+x)f = rf$$ and $$f(0) = 1$$ for $$f(x) = (1+x)^r$$, etc.).
Hope this gives you a non-technical sense for the remainder. | 793 | 2,925 | {"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": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 18, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2019-43 | latest | en | 0.891624 |
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Aug27 comment Operator and VariationalMethods You should use an expression for first argument of VariationalD not a function. So just replace g with last expression %. Apr12 comment Creating centered inline cell in text cell Maybe there is, I just tried to do this and didn't find anything. Apr12 comment Creating centered inline cell in text cell Are you sure it was created in the front end? Maybe it was imported from TeX, MathML or something. Or it was modified manually, adding GridBoxItemSize option there. Apr11 comment Symbolic scalar-by-matrix derivative Until there is no actual tensor inside Tr it's just a symbolic expression, is up to you how to interpret it. Apr11 comment Symbolic scalar-by-matrix derivative @cryo111 Wouldn't a matrix derivative of a matrix be a fourth rank tensor? Therefore its trace should be a matrix, not a scalar. Apr5 comment Fit two intersecting straight lines in Mathematica I would estimate an error from the distance between closest to k data points . Feb13 comment Defining my own invisible operator @Öskå Yes, basically I need an invisible infix operator of my choice. Jan4 comment Different integral when used with assumptions I would say it's a bug that there is not every conditional in the first result. But the results are the same if you assume this condition. Oct17 comment Too much whitespace in a GraphicsGrid containing Legended Plots Adjust AspectRatio parameter inside your Plot function. For example AspectRatio -> 1. Maybe move your legend down. Sep21 comment Extracting variables from Hold expression It works perfectly! Thank you! Jun8 comment How do I extract the middle element(s) of a given list? @Mr.Wizard I already anticipate the answer with time comparisons of all solutions to be made ;) May27 comment Row Alignment in Grid Quick fix would be changing spacing for second column to 1.75. May26 comment How to define a pure function with a Module? Classic mistake fpiecewise[x_?NumberQ, ... May25 comment rule-based implementation of an algorithm Probably you need ReplacePart and FixedPoint. May10 comment How can I create a 3D FilledCurve object? @Silvia I don't have any errors. May10 comment How can I make a Plot over a discrete set of domain points? No problem. You can start from here and here. May10 comment How can I make a Plot over a discrete set of domain points? ListPlot. To connect the dots: Joined->True or ListLinePlot. May8 comment How to compare power towers in Mathematica? I hate that expressions like 4^4^3^3^3 totally freeze my computer instead of just telling me to stop trying evaluating this and quietly abort :(. May8 comment How to compare power towers in Mathematica? That's pretty impressive! I would like to see an example when this will fail, if it exists of course :). May5 comment Error invoking Information on functions containing Legended @Silvia The problem appears again if I evaluate plot somewhere else. And then even f[x_]:=Legended[1,x] produces this error. | 657 | 3,017 | {"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} | 2.75 | 3 | CC-MAIN-2015-18 | longest | en | 0.863974 |
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Can't find an example of this, this was in the previous final exams, and with finals only a week away, I need an explaination of a few problems from this last section of the book.
Using geometry, evaluate the double integral over D of [(32-x^2-y^2)^.5]dA over the circular disk D: x^2+y^2<=36
Are you sure you have written that correctly? For (x,y) close to the boundary of that disk, $\displaystyle x^2+ y^2$ will be close to 36 which will make $\displaystyle 32- x^2- y^2$ negative.
I am going to assume the function you want to integrate is $\displaystyle (36- x^2- y^2)^{1/2}$. With that, since the problem says "using geometry", we can interpret this integral as the volume of a the upper half of a sphere of radius 6: $\displaystyle z= (36- x^2- y^2)^{1/2}$ gives $\displaystyle z^2= 36- x^2- y^2$ so $\displaystyle x^2+ y^2+ z^2= 36$. Of course, on the xy-plane, z= 0 so we have $\displaystyle x^2+ y^2= 36$, the given circle. The volume of a sphere is $\displaystyle \frac{4}{3}\pi r^3$ so the volume of the hemisphere is $\displaystyle \frac{2}{3}\pi r^3$, which is, indeed, $\displaystyle \frac{2}{3}\pi 6^3= 144\pi$.
It's very easy to do that in polar coordinates: $\displaystyle \int_{\theta= 0}^{2\pi}\int_{r= 0}^6 \sqrt{36- r^2} rdrd\theta$. Let $\displaystyle u= 36- r^2$ and it becomes $\displaystyle \pi \int_{u= 0}^{36} u^{1/2}du$ | 463 | 1,374 | {"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} | 4.34375 | 4 | CC-MAIN-2018-22 | latest | en | 0.834828 |
http://905-spmath2012.blogspot.com/2012/11/practice-test-question-19-pg-85.html | 1,548,015,889,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583730728.68/warc/CC-MAIN-20190120184253-20190120210253-00058.warc.gz | 5,358,181 | 11,598 | ## Sunday, November 25, 2012
### Practice Test: Question 19 pg 85
Question 19: Ron buys 75 shares in a car company. A year later, he sells the shared for \$15.64 each. The result is a loss of \$260.25. How much did Ron pay for each share? State any assumptions you make.
Answer: For each share, Ron paid \$19.11. | 91 | 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} | 2.8125 | 3 | CC-MAIN-2019-04 | latest | en | 0.928252 |
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# Class 9 RD Sharma Solutions – Chapter 4 Algebraic Identities- Exercise 4.4
• Difficulty Level : Expert
• Last Updated : 15 Dec, 2020
### Question 1. Find the following products?
i. (3x + 2y) (9x2 – 6xy + 4y2)
Solution:
We know that a3 + b3 = (a + b)(a2 – ab + b2
we can write the given equation as,
=> (3x + 2y)[(3x)2 – 6xy + (2y)2]
=> (3x)3 + (2y)3
=> 27x3 + 8y3
ii. (4x – 5y) (16x2 + 20xy + 25y2)
Solution:
We know that a3 – b3 = (a – b)(a2 + ab + b2)
we can write the given equation as,
=> (4x – 5y)[(4x)2 + 20xy + (5y)2]
=> (4x)3 – (5y)3
=> 64x3 – 125y3
iii. (7p4 + q) (49p8 – 7p4q + q2)
Solution:
We can write the given equation as,
=> (7p4 + q)[(7p4)2 – 7p4q + q2]
We know that a3 + b3= (a + b)(a2 – ab + b2
=> (7p4)3 + q3
=> 343p12 + q3
iv. [(x/2) + 2y] [(x2/4) – xy + 4y2]
Solution:
We can write the given equation as,
[(x / 2) + 2y] [(x / 2)2 – (x / 2) * 2y + (2y)2] —— eq(i)
By writing the given equation as eq(i) we can easily make the equation as (a + b)[a2 – ab + b2] = a3 + b3
So, the above e quation can be solved as,
=> (x / 2)3 + (2y)3
=> (x3 / 8) + 8y3
v. [(3/x) – (5/y)] [(9/x2) + (25/y2) + (15/xy)]
Solution:
We can write given equation as,
[(3 / x) – (5 / y)] {(3 / x)2 + (3 / x)(5 / y) + (5 / y)2]
So, above equation makes the identity of a3 – b3
Now,
=> (3 / x)3 – (5 / y)3
=> (27 / x3) – (125 / y3)
vi. [3 + (5/x)] [9 – (15/x) + ( 25/x2)]
Solution:
We can write the given equation as,
=> [3 + (5 / x)] [(3)2 – 3 * (5 / x) + (5 / x)2]
So, above equation makes the identity of a3 + b3
=> (3)3 + (5 / x)3
=> 27 + (125 / x2)
vii. [(2/x) + 3x] [(4/x2) + 9x2 – 6)]
Solution:
We can write the given equation as,
=> [(2 / x) + 3x] [(2 / x)2 – (2 / x)(3x) + (3x)2]
So, above equation makes the identity of a3 + b3
=> (2 / x)3 + (3x)3
=> (8 / x3) + 27x3
viii. [(3/2) – 2x2] [(9/x2) + 4x4 – 6x]
Solution:
We can write the given equation as,
=> [(3 / x) – 2x2] [(3 / x)2 – (3 / x)(2x2) + (2x2)2]
So, above equation makes the identity of a3 – b3
=> (3 / x)3 – (2x2)3
=> (27 / x3) – 8x6
ix. (1 – x)(1 + x + x2)
Solution:
This equation is clearly making the identity of a3 – b3
=> 13 – x3
=> 1 – x3
x. (1 + x)(1 – x + x2)
Solution:
This equation is clearly making the identity of a3 + b3
=> 13 + x3
=> 1 + x3
xi. (x2 – 1)(x4 + x2 + 1)
Solution:
We can write the given equation as,
=> (x2 – 1 ) [(x2)2 + x2 + 1)]
This equation is clearly making the identity of a3 – b3
=> (x2)3 – 1
=> x6 – 1
xii. (x3 + 1)(x6 – x3 + 1)
Solution:
We can write the given equation as,
=> (x3 + 1) [(x3)2 – x3 + 1]
This equation is clearly making the identity of a3 + b3
=> (x3)3 + 1
=> x9 + 1
### Question 2. If x = 3 and y = -1, find the values of each of the following using in identity?
i. (9y2 – 4x2) (81y4 + 36x2y2 + 16x4)
Solution:
We can write the given equation as,
=> (9y2 – 4x2) [(9y2)2 + 9y2 * 4x2 + (4x2)2]
This is now clearly making the identity of a3 – b3
=> (9y2)3 – (4x2)3
=> 729y6 – 64x —–eq(i)
Putting the given values in eq(i)
=> 729 * 1 – 64 * 729
=> 729 – 46656
=> -45927
ii. [(3/x) – (x/3)] [(x2/9) + (9/x2) + 1]
Solution:
We can write the given equation as,
=> [(3 / x) – (x / 3)] [(x / 3)2 + (x / 3)(3 / x) + (3 / x)2]
This is making the identity of a3 – b3
=> (3 / x)3 – (x / 3)3 —-eq(i)
Putting the given values in eq(i)
=> 1 – 1
=> 0
iii. [(x/7) + (y/3)] [( x2/49) + (y2/9) – (xy/21)]
Solution:
We can write the given equation as,
=> [(x / 7) + (y / 3)] [(x / 7)2 – (x / 7)(y / 3) – (y / 3 )2]
This is making the identity of a3 + b3
=> (x / 7)3 + (y / 3)3 —eq(i)
Putting the values in eq(i)
=> 27 / 343 – 1 / 27
=> (729 – 343) / 9261
=> 386 / 9261
iv. [(x/4) – (y/3)] [(x2/16) + (xy/12) + (y2/9)]
Solution:
We can write this equation as,
=> [(x / 4) – (y / 3)] [(x / 4)2 + (x / 4)(y / 3) + (y / 3)2]
This is clearly making the identity of a3 – b3
=> (x / 4)3 – (y / 3)3
=> (x3 / 64) – (y3 / 27) —eq(i)
Putting the values in eq(i)
=> (27 / 64) + (1 / 27)
=> (729 + 64) / 1728
=> 793 / 1728
### Question 3. If a + b = 10 and ab = 16, find the value of a2 – ab + b2 and a2 + ab + b2?
Solution:
Taking a + b = 10
On squaring both sides,
=> (a + b)2 = (10)2
We get, a2 + b2 + 2ab = 100 —eq(i)
Putting the value of ab = 16 in eq(i)
=> a2 + b2 + 2 * 16 = 100
=> a2 + b2 + 32 =100
=> a2 + b2 = 100 – 32 = 68
So, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84
### Question 4. If a + b = 8 and ab = 6, find the value of a3 + b3?
Solution:
Taking a + b = 8
On cubing both sides,
(a + b)3 = (8)3
=> a3 + b3 + 3ab(a + b) = 512 —-eq(i)
Putting the given values in eq(i)
=> a3 + b3 + 3 * 6 * 8 = 512
=> a3 + b3 + 144 = 512
=> a3 + b3 = 512 – 144 = 368
=> a3 + b3 = 368
### Question 5. If a – b = 6 and ab = 20 , find the value of a3 – b3?
Solution:
Taking a – b=6
On cubing both sides,
(a – b)3 = (6)3
=> a3 – b3 – 3ab(a – b) = 216 —eq(i)
Putting the given values in eq(i)
=> a3 – b3 – 3 * 20 * 6 = 216
=> a3 – b3 – 360 = 216
=> a3 – b3 = 216 + 360 = 576
=> a3 – b3 = 576
### Question 6. If x = -2 and y = 1, by using an identity find the value of the following:
i. (4y2 – 9x2)(16y4 + 36x2y2 + 81x4)
Solution:
Given equation can be written as,
=> (4y2 – 9x2)[(4y2)2 + 4y2 * 9x2 + (9x2)2]
This equation now making the identity of a3 – b3
=> (4y2)3 – (9x2)3
=> 64y6 – 729x6 —–eq(i)
Putting the given values in eq(i)
=> 64 * 16 – 729 * (-2)6
=> 64 – 729 * 64
=> 64 – 46656
=> -46592
ii. [(2/x) – (x/2)][(4/x2) + (x2/4) + 1]
Solution:
We can write this equation as,
=> [(2 / x) – (x / 2)] [(2 / x)2 + 2(2 / x)(x / 2) + (x / 2)2]
This equation is clearly making the identity of a3 – b3
=> (2 / x)3 – (x / 2)3
=> (8 / x3) – (x3 / 8) —eq(i)
Putting the given values in eq(i)
=> [8 / (-2)3] – [(-2)3 / 8]
=> -1 + 1
=> 0
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https://www.coursehero.com/file/6132869/FIN-620-Session-10-Homework-Assignment-Answers/ | 1,493,202,024,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917121267.21/warc/CC-MAIN-20170423031201-00503-ip-10-145-167-34.ec2.internal.warc.gz | 883,461,450 | 96,194 | FIN 620 Session 10 Homework Assignment Answers
# FIN 620 Session 10 Homework Assignment Answers - FIN 620...
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FIN 620 Session 10 Homework Assignment Answers: Session 10: Do problems 19-4, 19-16, 19-20, 29-8, and 29-16 19-4. To find the new stock price, we multiply the current stock price by the ratio of old shares to new shares, so: a. \$78(3/5) = \$46.80 b. \$78(1/1.15) = \$67.83 c. \$78(1/1.425) = \$54.74 d. \$78(7/4) = \$136.50. To find the new shares outstanding, we multiply the current shares outstanding times the ratio of new shares to old shares, so: a: 260,000(5/3) = 433,333 b: 260,000(1.15) = 299,000 c: 260,000(1.425) = 370,500 d: 260,000(4/7) = 148,571 19-16. a. Using the formula from the text proposed by Lintner: Div 1 = Div 0 + s ( t EPS 1 – Div 0 ) Div 1 = \$1.50 + .3[(.4)(\$4.15) – \$1.50] Div 1 = \$1.548 b. Now we use an adjustment rate of 0.60, so the dividend next year will be: Div 1 = Div 0 + s ( t EPS 1 – Div 0 ) Div 1 = \$1.50 + .6[(.4)(\$4.15) – \$1.50] Div 1 = \$1.596 c. The lower adjustment factor in part a is more conservative. The lower adjustment factor will always result in a lower future dividend.
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19-20. a. Let x be the ordinary income tax rate. The individual receives an after-tax dividend of: Aftertax dividend = \$1,000(1 – x ) which she invests in Treasury bonds. The Treasury bond will generate aftertax cash flows to the investor of: Aftertax cash flow from Treasury bonds = \$1,000(1 – x )[1 + .08(1 – x )] If the firm invests the money, its proceeds are: Firm proceeds = \$1,000[1 + .08(1 – .35)] And the proceeds to the investor when the firm pays a dividend will be: Proceeds if firm invests first = (1 – x ){\$1,000[1 + .08(1 – .35)]} To be indifferent, the investor’s proceeds must be the same whether she invests the after-tax dividend or receives the proceeds from the firm’s investment and pays taxes on that amount. To find the rate at which the investor would be indifferent, we can set the two equations equal, and solve for x . Doing so, we find: \$1,000(1 –
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## This note was uploaded on 02/14/2011 for the course FINANCE 620 taught by Professor Halstead during the Fall '09 term at UMBC.
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Ask a homework question - tutors are online | 796 | 2,618 | {"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-2017-17 | longest | en | 0.806704 |
https://www.airmilescalculator.com/distance/oit-to-ogn/ | 1,601,000,673,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400221980.49/warc/CC-MAIN-20200925021647-20200925051647-00285.warc.gz | 716,265,807 | 8,244 | # Distance between Oita (OIT) and Yonaguni Jima (OGN)
Distance from Oita to Yonaguni Jima (Oita Airport – Yonaguni Airport) is 816 miles / 1313 kilometers / 709 nautical miles.
## How far is Yonaguni Jima from Oita?
There are several ways to calculate distances between Oita and Yonaguni Jima. Here are two common methods:
Vincenty's formula (applied above)
• 815.724 miles
• 1312.780 kilometers
• 708.844 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 816.601 miles
• 1314.191 kilometers
• 709.606 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Flight duration
Estimated flight time from Oita Airport (OIT) to Yonaguni Airport (OGN) is 2 hours 2 minutes.
## Time difference and current local times
There is no time difference between Oita and Yonaguni Jima.
Oita
Yonaguni Jima
JST
JST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 136 kg (300 pounds).
## Airport information
A Oita Airport
City: Oita
Country: Japan
IATA Code: OIT
ICAO Code: RJFO
Coordinates: 33°28′45″N, 131°44′13″E
B Yonaguni Airport
City: Yonaguni Jima
Country: Japan
IATA Code: OGN
ICAO Code: ROYN
Coordinates: 24°28′0″N, 122°58′40″E
## Oita to Yonaguni Jima flight path
Shortest flight path between Oita Airport (OIT) and Yonaguni Airport (OGN).
## Frequent Flyer Miles Calculator
Oita (OIT) → Yonaguni Jima (OGN).
Elite level bonus (%):
Booking class bonus (%):
Air miles:
816
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
816
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https://www.homemadetools.net/forum/rattle-your-calipers-67652 | 1,632,779,441,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00361.warc.gz | 822,658,879 | 14,814 | Free 173 Best Homemade Tools eBook:
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This post is more about how to use a tool than it is a description of a new tool. Knowledge is arguably our most important tool so I think posts like this belong in the tools section of the forum.
In his book, Home Machinist's Bedside Reader #1 (pg. 11), Guy Lautard describes a technique for measuring a large bore when you don't have a set of calipers large enough to span the diameter.
Basically, you cut a "stick" to a length slightly less than the diameter of the bore and provide it with rounded or pointed ends. Inserted in the bore, this stick will "rattle" back and forth a (hopefully) small amount. You measure the peak-to-peak rattle and then use a formula to calculate the actual diameter. Lautard gives a formula but it looks like an approximation to me, although I haven't examined it closely.
If the rattle distance is small compared to the diameter of the bore, the stick length itself is a very good approximation of the bore diameter.
I wrote RATTLE.EXE to compute the diameter given the stick and rattle lengths in order to evaluate this procedure. You can use it to calculate the diameter if you need to use this procedure. You can download the program from my webpage.
This is the way I was taught to use internal, spring type calipers. If you try to set the calipers to the actual diameter, you're very likely to spring them slightly. When withdrawn from the bore to be measured, they will unspring slightly thus affecting the reading. Better to set them so they "rattle" slightly in the hole and then measure them. RATTLE will prove to you that, if you make the rattle very small, the difference between the measured caliper distance and the actual diameter will be very small indeed.
For an example, I used the values Lautard used in his book...
S = 3.998
R = 13/32 = 0.40625
and the program yielded a calculated diameter of 4.003197. Lautard's approximation yields a value of 4.0033; thus, agreement to one ten thousandth.
For a rattle distance of 0.40625/4.003197 = 10% of the bore diameter, the stick yielded the bore diameter with an error of only 100*(3.998-4.0032)/4.0032 = 1.3%. Decreasing the rattle distance will decrease that error still further.
2. The Following 6 Users Say Thank You to mklotz For This Useful Post:
JD62 (May 4, 2018), jjr2001 (May 5, 2018), Jon (May 3, 2018), Moby Duck (May 4, 2018), Seedtick (May 4, 2018), Toolmaker51 (May 5, 2018)
3. Perhaps I am thinking a bit slow this morning. How do you measure the Rattle inside the bore?
4. Sounds to me like that method is about as precise as throwing a hand grenade. I think I'll stick with my inside mics, or a bore gauge. The book you refer to, however is very interesting and informative.
5. Originally Posted by Moby Duck
Perhaps I am thinking a bit slow this morning. How do you measure the Rattle inside the bore?
Lautard doesn't mention how that was done.
If the "rattling" was done near the mouth of the bore simply holding a scale in place and reading the swing would work. Another way I can imagine would be to use a pair of dividers and keep opening them until their span matched the swing of the stick, then hold them against a scale. Remember, even a poor estimate of the swing will improve the estimate of the bore. A detailed error analysis would be a waste of time; if you really need to know the bore with high accuracy you'll use more sophisticated techniques.
6. The Following User Says Thank You to mklotz For This Useful Post:
Moby Duck (May 25, 2018)
7. Very good indeed! I work often with learning-disabled fellows, who cannot read even a tape measure, but obviously this is a technique (leaving aside the arithmetic) they can master. You have prompted me to post a homemade tool, which a quick search has revealed that has not been previously posted.
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• | 990 | 4,063 | {"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} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.943755 |
https://beta.geogebra.org/m/smtummgy | 1,718,435,689,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861584.65/warc/CC-MAIN-20240615062230-20240615092230-00292.warc.gz | 109,246,059 | 21,519 | # Guided Homework
A plane flew at a constant speed between Denver and Chicago. It took the plane 1.5 hours to fly 915 miles. Map of the Midwest from Denver to Chicago Copyright Owner: American Fact Finder License: Public Domain Via: Original URL: http://factfinder.census.gov/faces/nav/jsf/pages/searchresults.xhtml?ref=geo&refresh=t&tab=map&src=bkmk (Links to an external site.)Links to an external site.
1. Complete the table.time (hours)distance (miles)speed (miles per hour)
2. How far does the plane fly in one hour?
3. How far would the plane fly in tt hours at this speed?
4. If dd represents the distance that the plane flies at this speed for tt hours, write an equation that relates tt and dd.
5. How far would the plane fly in 3 hours at this speed? in 3.5 hours? Explain or show your reasoning. | 222 | 807 | {"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-2024-26 | latest | en | 0.806582 |
https://brainly.in/question/14621 | 1,484,983,092,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280929.91/warc/CC-MAIN-20170116095120-00338-ip-10-171-10-70.ec2.internal.warc.gz | 804,390,116 | 9,775 | # Find the angle between the curves at the point of intersection.
1
by VivekMistry838
2014-05-30T14:52:12+05:30
Intersection point of curve is
x² = (x-2)²
x² = x² + 4 - 4x
x =1
put x = 1 in first curve
y = 1
hence point = (1,1)
y = x²---------(1)
y = (x -2)² -------------(2)
dy/dx = 2x
dy/dx = 2 let m1 (at point (1,1))
dy/dx = 2(x-2)
dy/dx = -2 let m2 (at point (1,1))
tan Ф = (m1 - m2)/(1 + m1*m2) (Ф = angle between curves at point of intersection)
tan Ф = 4/ -3
Ф = tan^-1(-4/3) | 227 | 525 | {"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} | 3.9375 | 4 | CC-MAIN-2017-04 | latest | en | 0.660775 |
http://brownlee.gunstimespin.xyz/jackpot/Counting-cards-blackjack-guide.html | 1,606,356,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141185851.16/warc/CC-MAIN-20201126001926-20201126031926-00524.warc.gz | 18,107,255 | 8,137 | # The Ultimate Blackjack Strategy Guide - 888 Casino.
Counting cards at blackjack. DISCUSSION. Close. 74. Posted by 10 months ago. Archived. Counting cards at blackjack. DISCUSSION. It works. They only used one deck and don’t seem to randomize it like some online casinos do. I gambled minimum until the count was high and then max bet, got 4 blackjacks in a row then a 20. I’m going to be rich. Edit: read this if you have no idea how to count.
Blackjack Card Counting: A Complete Guide. For decades, casino players have tried and tested various methods in order to take on the house. Blackjack is one of the most popular games in the world and has a rich history of both successful and unsuccessful strategies. Blackjack card counting is by far the most famous of these strategies. Counting cards dates back to the early 1960s and since.
The basic principle behind Blackjack card counting strategies is that low cards are bad for players and high cards are good for players. When your KO Count is greater than 0 it means that there are more 10’s and Aces in the deck than the 5 to 13 ratio found in a fresh deck. This is not a theory. It has been mathematically proven countless times over the past 50 years. The advantage stems.
Blackjack is a game of skill and card counting can help you gain an advantage in the game. Edward R Thorp is considered the father of card counting. He discovered the principles of card counting and published his method in 1962 in his book Beat the Dealer.His writing started the Blackjack revolution, and it was also instrumental in the casinos taking action to ban card counters.
There are a lot of myths and misinformation when it comes to the truth about how to count cards in blackjack. We'll show you the best and easiest method here.
History of Card Counting. Back in the 1950’s, the first basic strategy system and simple card counting systems were devised. The systems didn’t get much attention until the 1960’s, when a mathematics professor by the name of Dr. Edward O. Thorp published a book titled Beat the Dealer.
Tutorial on Cards Counting in Blackjack Games First, have a clear idea of what counting cards means. It is the calculation of how rich the deck part with the cards still undealt is in high cards, or whether many low cards remain in it. Many high cards mean a perk for the player, and many low cards a perk for the dealer. What is more important, if the deck part with remaining.
This chapter covers everything you need to know to play wisely in a blackjack tournament either in a land-based or online casino. You’ll learn what the basic rules are for tournaments, the different kinds of tournaments, and some strategy tips to improve your chances of finishing in the money.
Our blackjack guide takes your from the basics to advanced strategy and card counting for the game of 21. Learn the basics of gameplay and how to create the perfect blackjack hand. Understand betting with the rules of blackjack including how to buy insurance and when to double down. Advanced strategy will detail the process of counting cards and whether or not you can beat the dealer. Take.
If you are an experienced blackjack player you already know about this technique. But on this guide we cover both the basics of card counting and some more advanced stuff like types of blackjack card counting. This guide can here to help you understand exactly what card counting is, and whether you should count cards when you play live blackjack.
Once you have the basics of blackjack down to a tee, the next step you will possibly want to undertake is learning how to count cards. Wether you’re playing Blackjack online or in a land-based casino, this blackjack strategy can be used anywhere. This is a sort of add-on to the basic strategy of blackjack, which will assist you with remembering what cards remain in the deck and what cards.
Internet Blackjack Glossary: Playing Blackjack is easy if you understand the game. Increase your knowledge with the Gambling City Guide to Blackjack terms.
Blackjack CARD VALUES. There are three main categories of cards: Aces, faces, and number cards. Aces are special because they can count as either 1 or 11.
## The Ultimate Blackjack Strategy Guide - 888 Casino.
Card counting works because the game of blackjack uses a fixed number of cards, usually unshuffled between hands. At its simplest, consider a one-deck game containing 52 cards. As each card is dealt, the cards remaining in the deck become fewer. Over time, these can skew in particular directions, depending on the random order of the deck, leaving more cards of a certain value still to be drawn.
The underlying principle behind card counting is that a deck rich in tens and aces is good for the player, a deck rich in small cards is good for the dealer. When the counter knows the odds are in his favor, he will bet more, and adjust his playing strategy to stand, double, and split in some plays where basic strategy says to stand. All the options the player has at his disposal favor the.
Colin is the founder of Blackjack Apprenticeship. Colin has been counting cards for over 15 years, and ran a multi-million dollar blackjack team. You may have seen his team featured in the documentary Holy Rollers: The True Story of Card Counting Christians, and has been covered by New York Times, CNN, and The Colbert Report.
By valuing the cards in this manner it makes card counting much easier and obtainable for the average person. Blackjack Card Counting Values. Homework 1: Identify Card Value. Many hours of at-home practice are required when learning how to count cards. The first homework assignment is to be able to quickly identify the card value without any hesitation. Take a single deck of cards and run.
We’ll take a look at one of the simplest counting guides available later in our guide. Counting Cards Blackjack. Counting cards blackjack or otherwise in the casino doesn’t always guarantee you a win. This is worth remembering. It does, however, seriously increase your chances of making money.
The truth is that blackjack is a game that can be mastered by casino players and it is not just luck, especially for highly experienced players who have mastered counting cards during the game, and who know every aspect of the game. This brings them advantage over the casino and that is the main reason casinos use several decks of cards put together so that counting cards is harder for players. | 1,336 | 6,460 | {"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-2020-50 | latest | en | 0.969516 |
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# 4.6: PDEs, separation of variables, and the heat equation
Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. Solving PDEs will be our main application of Fourier series. A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. We will only talk about linear PDEs. Together with a PDE, we usually have specified some boundary conditions, where the value of the solution or its derivatives is specified along the boundary of a region, and/or someinitial conditions where the value of the solution or its derivatives is specified for some initial time. Sometimes such conditions are mixed together and we will refer to them simply as side conditions.
We will study three specific partial differential equations, each one representing a more general class of equations. First, we will study the heat equation, which is an example of a parabolic PDE. Next, we will study thewave equation, which is an example of a hyperbolic PDE. Finally, we will study the Laplace equation, which is an example of an elliptic PDE. Each of our examples will illustrate behavior that is typical for the whole class.
#### 4.6.1 Heat on an insulated wire
Let us first study the heat equation. Suppose that we have a wire (or a thin metal rod) of length $$L$$ that is insulated except at the endpoints. Let $$x$$ denote the position along the wire and let $$t$$ denote time. See Figure 4.13.
Figure 4.13: Insulated wire.
Let $$u(x,t)$$ denote the temperature at point $$x$$ at time $$t$$. The equation governing this setup is the so-called one-dimensional heat equation:
$\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2},$
where $$k>0$$ is a constant (the thermal conductivity of the material). That is, the change in heat at a specific point is proportional to the second derivative of the heat along the wire. This makes sense; if at a fixed $$t$$ the graph of the heat distribution has a maximum (the graph is concave down), then heat flows away from the maximum. And vice-versa.
We will generally use a more convenient notation for partial derivatives. We will write $$u_t$$ instead of $$\frac{\partial u}{\partial t}$$, and we will write $$u_{xx}$$ instead of $$\frac{\partial^2 u}{\partial x^2}$$. With this notation the heat equation becomes
$u_t=ku_{xx}.$
For the heat equation, we must also have some boundary conditions. We assume that the ends of the wire are either exposed and touching some body of constant heat, or the ends are insulated. For example, if the ends of the wire are kept at temperature 0, then we must have the conditions
$u(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u(L,t)=0.$
If, on the other hand, the ends are also insulated we get the conditions
$u_x(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u_x(L,t)=0.$
In other words, heat is not flowing in nor out of the wire at the ends. We always have two conditions along the $$x$$ axis as there are two derivatives in the $$x$$ direction. These side conditions are called homogeneous (that is, $$u$$ or a derivative of $$u$$ is set to zero).
Furthermore, suppose that we know the initial temperature distribution at time $$t=0$$. That is,
$u(x,0)=f(x),$
for some known function $$f(x)$$. This initial condition is not a homogeneous side condition.
#### 4.6.2 Separation of variables
The heat equation is linear as $$u$$ and its derivatives do not appear to any powers or in any functions. Thus the principle of superposition still applies for the heat equation (without side conditions). If $$u_1$$ and $$u_2$$ are solutions and $$c_1,c_2$$ are constants, then $$u= c_1u_1+c_2u_2$$ is also a solution.
Exercise $$\PageIndex{1}$$:
Verify the principle of superposition for the heat equation
Superposition also preserves some of the side conditions. In particular, if $$u_1$$ and $$u_2$$ are solutions that satisfy $$u(0,t)=0$$ and $$u_(L,t)=0$$, and $$c_1,c_2$$ are constants, then $$u= c_1u_1+c_2u_2$$ is still a solution that satisfies $$u(0,t)=0$$ and$$u_(L,t)=0$$. Similarly for the side conditions $$u_x(0,t)=0$$ and $$u_x(L,t)=0$$. In general, superposition preserves all homogeneous side conditions.
The method of separation of variables is to try to find solutions that are sums or products of functions of one variable. For example, for the heat equation, we try to find solutions of the form
$u(x,t)=X(x)T(t).$
That the desired solution we are looking for is of this form is too much to hope for. What is perfectly reasonable to ask, however, is to find enough “building-block” solutions of the form $$u(x,t)=X(x)T(t)$$ using this procedure so that the desired solution to the PDE is somehow constructed from these building blocks by the use of superposition.
Let us try to solve the heat equation
$u_t=ku_{xx} ~~~~~ {\rm{with}} ~~~ u(0,t)=0, ~~~~~ u(L,t)=0, ~~~~~ {\rm{and}} ~~~ u(x,0)=f(x).$
Let us guess $$u(x,t)=X(x)T(t)$$. We plug into the heat equation to obtain
$X(x)T'(t)=kX''(x)T(t).$
We rewrite as
$\frac{T'(t)}{kT(t)}= \frac{X''(x)}{X(x)}.$
This equation must hold for all $$x$$ and all $$t$$. But the left hand side does not depend on $$x$$ and the right hand side does not depend on $$t$$. Hence, each side must be a constant. Let us call this constant $$- \lambda$$ (the minus sign is for convenience later). We obtain the two equations
$\frac{T'(t)}{kT(t)}= - \lambda = \frac{X''(x)}{X(x)}.$
In other words
$X''(x) + \lambda X(x)=0, \\ T'(t) + \lambda k T(t)=0.$
The boundary condition $$u(0,t)=0$$ implies $$X(0)T(t)=0$$. We are looking for a nontrivial solution and so we can assume that $$T(t)$$ is not identically zero. Hence $$X(0)=0$$. Similarly, $$u(L,t)=0$$ implies $$X(L)=0$$. We are looking for nontrivial solutions $$X$$ of the eigenvalue problem $$X'' + \lambda X = 0, X(0)=0, X(L)=0$$. We have previously found that the only eigenvalues are $$\lambda_n = \frac{n^2 \pi^2}{L^2}$$, for integers $$n \geq 1$$, where eigenfunctions are $$\sin \left( \frac{n \pi}{L}x \right)$$. Hence, let us pick the solutions
$X_n(x)= \sin \left( \frac{n \pi}{L}x \right).$
The corresponding $$T_n$$ must satisfy the equation
$T'_n(t) + \frac{n^2 \pi^2}{L^2}kT_n(t)=0.$
By the method of integrating factor, the solution of this problem is
$T_n(t)=e^{\frac{-n^2 \pi^2}{L^2}kt}.$
It will be useful to note that $$T_n(0)=1$$. Our building-block solutions are
$u_n(x,t)=X_n(x)T_n(t)= \sin \left( \frac{n \pi}{L}x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}.$
We note that $$u_n(x,0)= \sin \left( \frac{n \pi}{L}x \right)$$. Let us write $$f(x)$$ as the sine series
$f(x)= \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi}{L}x \right).$
That is, we find the Fourier series of the odd periodic extension of $$f(x)$$. We used the sine series as it corresponds to the eigenvalue problem for $$X(x)$$ above. Finally, we use superposition to write the solution as
$u(x,t)= \sum^{\infty}_{n=1}b_n u_n (x,t)= \sum^{\infty}_{n=1}b_n \sin(\frac{n \pi}{L}x)e^{\frac{-n^2 \pi^2}{L^2}kt}.$
Why does this solution work? First note that it is a solution to the heat equation by superposition. It satisfies $$u(0,t)=0$$ and $$u(L,t)=0$$ , because $$x=0$$ or $$x=L$$ makes all the sines vanish. Finally, plugging in $$t=0$$, we notice that $$T_n(0)=1$$ and so
$u(x,0)= \sum^{\infty}_{n=1}b_n u_n (x,0)= \sum^{\infty}_{n=1}b_n \sin(\frac{n \pi}{L}x)=f(x).$
Example $$\PageIndex{1}$$:
Suppose that we have an insulated wire of length $$1$$, such that the ends of the wire are embedded in ice (temperature 0). Let $$k=0.003$$. Then suppose that initial heat distribution is $$u(x,0)=50x(1-x)$$. See Figure 4.14.
Figure 4.14: Initial distribution of temperature in the wire.
We want to find the temperature function $$u(x,t)$$. Let us suppose we also want to find when (at what ) does the maximum temperature in the wire drop to one half of the initial maximum of $$12.5$$.
We are solving the following PDE problem:
$u_t=0.003u_{xx}, \\ u(0,t)= u(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 0<x<1.$
We write $$f(x)=50x(1-x)$$ for $$0<x<1$$ as a sine series. That is, $$f(x)= \sum^{\infty}_{n=1}b_n \sin(n \pi x)$$, where
$b_n= 2 \int^1_0 50x(1-x) \sin(n \pi x)dx = \frac{200}{\pi^3 n^3}-\frac{200(-1)^n}{\pi^3 n^3}= \left\{ \begin{array}{cc} 0 & {\rm{if~}} n {\rm{~even,}} \\ \frac{400}{\pi^3 n^3} & {\rm{if~}} n {\rm{~odd.}} \end{array} \right.$
Figure 4.15: Plot of the temperature of the wire at position at time .
The solution $$u(x,t)$$, plotted in Figure 4.15 for $$0 \leq t \leq 100$$, is given by the series:
$u(x,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi x) e^{-n^2 \pi^2 0.003t}.$
Finally, let us answer the question about the maximum temperature. It is relatively easy to see that the maximum temperature will always be at $$x=0.5$$, in the middle of the wire. The plot of $$u(x,t)$$ confirms this intuition.
If we plug in $$x=0.5$$ we get
$u(0.5,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi 0.5) e^{-n^2 \pi^2 0.003t}.$
For $$n=3$$ and higher (remember $$n$$ is only odd), the terms of the series are insignificant compared to the first term. The first term in the series is already a very good approximation of the function. Hence
$u(0.5,t) \approx \frac{400}{\pi^3}e^{-\pi^2 0.003t}.$
The approximation gets better and better as $$t$$ gets larger as the other terms decay much faster. Let us plot the function $$0.5,t$$, the temperature at the midpoint of the wire at time $$t$$, in Figure 4.16. The figure also plots the approximation by the first term.
Figure 4.16: Temperature at the midpoint of the wire (the bottom curve), and the approximation of this temperature by using only the first term in the series (top curve).
After $$t=5$$ or so it would be hard to tell the difference between the first term of the series for $$u(x,t)$$ and the real solution $$u(x,t)$$. This behavior is a general feature of solving the heat equation. If you are interested in behavior for large enough $$t$$, only the first one or two terms may be necessary.
Let us get back to the question of when is the maximum temperature one half of the initial maximum temperature. That is, when is the temperature at the midpoint $$12.5/2=6.25$$. We notice on the graph that if we use the approximation by the first term we will be close enough. We solve
$6.25=\frac{400}{\pi^3}e^{-\pi^2 0.003t}.$
That is,
$t=\frac{\ln{\frac{6.25 \pi^3}{400}}}{-\pi^2 0.003} \approx 24.5.$
So the maximum temperature drops to half at about $$t=24.5$$.
We mention an interesting behavior of the solution to the heat equation. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. For a fixed $$t$$, the solution is a Fourier series with coefficients $$b_n e^{\frac{-n^2 \pi^2}{L^2}kt}$$. If $$t>0$$, then these coefficients go to zero faster than any $$\frac{1}{n^P}$$ for any power $$p$$. In other words, the Fourier series has infinitely many derivatives everywhere. Thus even if the function $$f(x)$$ has jumps and corners, then for a fixed $$t>0$$, the solution $$u(x,t)$$ as a function of $$x$$ is as smooth as we want it to be.
#### 4.6.3 Insulated ends
Now suppose the ends of the wire are insulated. In this case, we are solving the equation
$u_t=ku_{xx}~~~~ {\rm{with}}~~~u_x(0,t)=0,~~~u_x(L,t)=0,~~~{\rm{and}}~~~u(x,0)=f(x).$
Yet again we try a solution of the form $$u(x,t)=X(x)T(t)$$. By the same procedure as before we plug into the heat equation and arrive at the following two equations
$X''(x)+\lambda X(x)=0, \\ T'(t)+\lambda kT(t)=0.$
At this point the story changes slightly. The boundary condition $$u_x(0,t)=0$$ implies $$X'(0)T(t)=0$$. Hence $$X'(0)=0$$. Similarly, $$u_x(L,t)=0$$ implies $$X'(L)=0$$. We are looking for nontrivial solutions $$X$$ of the eigenvalue problem $$X''+ \lambda X=0,$$ $$X'(0)=0,$$ $$X'(L)=0,$$. We have previously found that the only eigenvalues are $$\lambda_n=\frac{n^2 \pi^2}{L^2}$$, for integers $$n \geq 0$$, where eigenfunctions are $$\cos(\frac{n \pi}{L})X$$ (we include the constant eigenfunction). Hence, let us pick solutions
$X_n(x)= \cos(\frac{n \pi}{L}x)~~~~ {\rm{and}}~~~~ X_0(x)=1.$
The corresponding $$T_n$$ must satisfy the equation
$T'_n(t)+ \frac{n^2 \pi^2}{L^2}kT_n(t)=0.$
For $$n \geq 1$$, as before,
$T_n(t)= e^{\frac{-n^2 \pi^2}{L^2}kt}.$
For $$n=0$$, we have $$T'_0(t)=0$$ and hence $$T_0(t)=1$$. Our building-block solutions will be
$u_n(x,t)=X_n(x)T_n(t)= \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt},$
and
$u_0(x,t)=1.$
We note that $$u_n(x,0) =\cos \left( \frac{n \pi}{L} x \right)$$. Let us write $$f$$ using the cosine series
$f(x)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right).$
That is, we find the Fourier series of the even periodic extension of $$f(x)$$.
We use superposition to write the solution as
$u(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n u_n(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}.$
Example $$\PageIndex{2}$$:
Let us try the same equation as before, but for insulated ends. We are solving the following PDE problem
$u_t=0.003u_{xx}, \\ u_x(0,t)= u_x(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 0<x<1.$
For this problem, we must find the cosine series of $$u(x,0)$$. For $$0<x<1$$ we have
$50x(1-x)=\frac{25}{3}+\sum^{\infty}_{\underset{n~ {\rm{even}} }{n=2}} \left( \frac{-200}{\pi^2 n^2} \right) \cos(n \pi x).$
The calculation is left to the reader. Hence, the solution to the PDE problem, plotted in Figure 4.17, is given by the series
$u(x,t)=\frac{25}{3}+\sum^{\infty}_{\underset{n~ {\rm{even}} }{n=2}} \left( \frac{-200}{\pi^2 n^2} \right) \cos(n \pi x) e^{-n^2 \pi^2 0.003t}.$
Figure 4.17: Plot of the temperature of the insulated wire at position $$x$$ at time $$t$$.
Note in the graph that the temperature evens out across the wire. Eventually, all the terms except the constant die out, and you will be left with a uniform temperature of $$\frac{25}{3} \approx{8.33}$$ along the entire length of the wire. | 4,797 | 14,536 | {"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} | 3.84375 | 4 | CC-MAIN-2018-22 | latest | en | 0.811667 |
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