url
stringlengths 14
5.47k
| tag
stringclasses 1
value | text
stringlengths 60
624k
| file_path
stringlengths 110
155
| dump
stringclasses 96
values | file_size_in_byte
int64 60
631k
| line_count
int64 1
6.84k
|
|---|---|---|---|---|---|---|
https://www.coursehero.com/sitemap/schools/1815-North-Central-State-College/courses/5691811-BUS650/
|
math
|
Tire Cost incurred by Ford Motor Company
Sales Commission paid to the sales force of Dell Inc.
Wood Glue consumed in the manufacture of Thomasville furniture.
Hourly Wages of refinery security guards employed by ExxonMobil Corporation
The salary of a
Effective Federal Funds Rate
Promotes economic stability
FOMC may set a lower federal funds rate target to spur greater
The central interest rate in the U.S. financial market.
5.4% national average unemployment
Jonathan Macintosh is a highly successful Pennsylvania orchardman who has formed his
to produce and package applesauce. Apples can be stored for several months in cold stor
production is relatively uniform throughout the year. The recently hired controlle
Digital Marketing Promotion
Grand Canyon University: MKT-607-O500
April 26th, 2016
For decades, the marketing strategies of firms were primarily focused on elements such
as marketing physical locations and obtaining referrals. However, the
Countywide Cable Services, Inc. is organized with three segments: Metro, Suburban, and Outlying.
Data for these segments for the year just ended follow.
In addition to the expenses listed above, the company has $95,000 of common fixed expenses.
Resource: Peer Review
Complete this peer evaluation for each member of your team, including yourself. The
average score from these evaluations will be used in determining your final grade for the
Rate each team member on a scale of 1 t
FreshPak Corporation manufactures two types of cardboard boxes used in shipping canned food, fruit,
and vegetables. The canned food box (type C) and the perishable food box (type P) have the following
material and labor requirements.
Skinny Dippers, Inc. produces nonfat frozen yogurt. The product is sold in five-gallon containers, which
have the following price and variable costs.
Sales price . $15
Direct material .
Direct labor .
Variable overhead .
Budgeted fixed overhead in 2
Coefficient of I (Income) is 0.5 which is greater than 0. Therefore, as income increases, quantity demanded increases, indicating
Coefficient of PX = - 5 but coefficient of PY = 10. Since coefficients of price of two goods are of opposite sign, it
The given graph shows plant capacity of 1.25 to 2.0 mil barrels of beer per year. Costs continue to decline at a capacity apro
barrels per year. No economies of scale are found beyond 8 mil barrels ber year. Gennerally, economies of scale are reached
The exchange rate in the table is given in euros over dollars. The exchange rate denotes the strength of one econmy to another
The variable R denotes the exchange rate dcenotes euros and $ denotes dollars
calculate the exports for tv as domestic pri
Piscataway Plastics Company manufactures a highly specialized plastic that is used extensi
automobile industry. The following data have been compiled for the month of June. Convers
occurs uniformly throughout the production process.
Work in process, June
average product is the amount of product per owrker Average product is given as quaniity divided by labor.
The variable AP denotes average product, Q denotes quality, and L denotes labor.
Marginal product is the change in product divided by the ch
Running head: AMAZON.COM A FINANCIAL ANALYSIS CASE STUDY
AMAZON.COM - FINANCIAL ANALYSIS CASE STUDY
Amazon.com A Financial Case Study
GCU FIN 502
Jan 15 2016
AMAZON.COM - FINANCIAL ANALYSIS CASE STUDY
Amazon.com - Financial An Analys
1. Chapter 11, Technical Question 7 in the textbook.
2. Chapter 11, Application Question 1 in the textbook.
3. Chapter 12, Technical Questions 3 and 5 in the textbook.
Nominal GDP is that which involes all the changes in the market price in the same year
In the given diagram, the demand curve shows a steady downward slope and then takes a steeper slope. At the poin of the ste
B the price will be set at the point where the demand curve changes. This is the point where the market begins before
CLC Group Projects Agreement
CLC Course Information
Course Name/Section Number:
Start Date of the Course:
CLC Member Contact Information
(Who is in our group?)
Primary Email Address
1 Using WGCC's current product-costing system:
Determine the company's predetermined overhead rate using direct-labor cost at t
rate overhead = (total manufacturing cost - overhead cost)/ (budgeted direct - labo
overhead rate per direc
1. Chapter 13, Technical Question 3 in the textbook.
2. Chapter 13, Application Question 1 in the textbook.
3. Chapter 14, Application Question 1 in the textbook.
the deposit multiplier is the inverse of the reserve requirment. Calculate the deposit multi
1. Chapter 7, Technical Questions 3 and 5 in the textbook.
2. Chapter 7, Application Question 5 in the textbook.
3. Chapter 8, Technical Questions 3 and 7 in the textbook.
According to the graph the shutdown potin is the point at which P= min AVC. The bre
Chapter 4 Graded Homework
Problem 1. Prepare a pro forma income statement assuming all costs vary with sales and the dividend
payout ratio is constant. A 35% growth rate in sales is projected. What are the projected additions to
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121893.62/warc/CC-MAIN-20170423031201-00382-ip-10-145-167-34.ec2.internal.warc.gz
|
CC-MAIN-2017-17
| 5,064 | 84 |
https://www.wyzant.com/resources/answers/872647/simplify-the-expression-assume-that-all-variables-are-positive
|
math
|
Simplify the expression. Assume that all variables are positive.
^3 square root 5x over 3y^4
(use integers or fractions for any numbers in the expression. Use positive exponents only. Type an exact answer, using radicals as needed)
I wish I knew how to write it with the symbols instead of the words so I hope it makes sense.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00589.warc.gz
|
CC-MAIN-2022-21
| 325 | 4 |
https://community.babycenter.com/post/a63946255/working-from-home
|
math
|
Hi ladies, I was hoping someone on here will be able to give me some advice. I would like to eventually start working from home but don't know where you even go to find that. I would not be a good sales person (it works, beach body, etc...) I've looked online but nothing really seems legit. If anyone have any advice I would appreciate it! Thanks in advance.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886105187.53/warc/CC-MAIN-20170818213959-20170818233959-00249.warc.gz
|
CC-MAIN-2017-34
| 359 | 1 |
http://www.brightstorm.com/math/algebra/quadratic-equations-and-functions/solving-quadratic-equations-using-square-roots/all
|
math
|
When graphing radical equations using shifts, adding or subtracting a constant that is not in the radical will shift the graph up (adding) or down (subtracting). Adding or subtracting a constant that is in the radical will shift the graph left (adding) or right (subtracting). Multiplying a negative constant by the equation will reflect the graph over the x-axis. Multiplying by a number larger than one increases the y-values.
Need help with "Solving Quadratic Equations Using Square Roots" problems?
Watch expert teachers solve similar problems to develop your skills.
How to solve binomial quadratic equations using square roots.
How to solve binomial quadratic equations by using square roots when the answer is a decimal approximation.
How to recognize quadratic equations that have no solution.
How to solve a binomial quadratic equation by square rooting both sides
How to solve binomial quadratic equations by square rooting both sides of the equation
Finding the x-intercepts of a parabola by solving by taking square roots
Basic example of completing the square to solve a quadratic equation by taking square roots
Solving quadratic equations with no linear term by taking square roots
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931004246.54/warc/CC-MAIN-20141125155644-00222-ip-10-235-23-156.ec2.internal.warc.gz
|
CC-MAIN-2014-49
| 1,196 | 11 |
https://www.coursehero.com/file/19531077/Atwoods-Machine-Experiment/
|
math
|
93%(15)14 out of 15 people found this document helpful
This preview shows page 1 - 4 out of 11 pages.
Carleton UniversityLaboratory ReportCourse #:PHYS 1007 Experiment #: 4Atwood’s Machine Experiment Jennifer Armstrong101030095Date Performed:Tuesday, November 8th, 2016Date Submitted:Tuesday, November 22nd, 2016Lab Period:L9 Partner:Ayelet LustgartenStation #:35TA:Stephen
PurposeThis experiment uses a pulley system with magnetization (Atwood’s Machine) in order toevaluate the acceleration due to gravity and to evaluate the frictional torque of the system.Theory The purpose of this experiment is to use data collected from Atwood’s experiment in order to evaluate the acceleration due to gravity and compare it to the known value, as well as calculate the frictional torque. The value of the acceleration due to gravity is given by the equation;Its error is the approximation of the total error on the overall value of gravity. It is given by the equation;Where g is acceleration due to gravity, m is mass, h is height, and A is rotational inertiaThe frictional torque is the frictional force experienced by the pulley. It is given by the equation;Its error is the approximation of the total error on the overall value of torque. It is given by the equation;Where ⊺is torque, b is y intercept on Figure 1, h is height, r is radius of pulley, and A is rotational inertia, and σ isthecorresponding error .The total mass of the machine is required for the calculation of gravity. It is given by the equation;Its error is the approximation of the total error on the total value of mass. It is given by the equation;g=2mhA Equation 1. Aσg=g√σmm2+σhh2+σAA2Equation 1. B⊺= -2bhrA Equation 1. Cσ⊺=⊺√σbb2+σhh2+σrr2+σAA2EquationM=m1+m2+10mwasherEquation 1. Eσ M=√σ m12+σ m22+100σ mwasher2Equation 1. F1
Where m1is the mass of mass 1, m2is the mass of mass 2, mwasheris the mass of one washer, and σ mis the error of each value.The value of rotational interia in the system is given by the equation; Its error is the approximation of the total error on the rotational interia. It is given by the equation;Where A is rotational interia, Ir2=(80± 1) g, M is total mass, σMis total mas error, andσIr2is 1.Welch’s consistency test is used to evaluate the consistency of the known value of gravity (9.8 m/s2). It is given by the following equation; Where t is consistency,x1is the known value, x2is the calculated value, σ1is error of known value, and σ2is error on experimental value. Error calculations are used in this experiment in order to ensure the measurements taken are accurate and account for errors in the experimental process. Inefficient statistics and statistical error are used, in addition to reading error, to find the error on an individual calculation. Reading error is found by taking the smallest decimal and go one decimal to the right by one half to account for discrepancies in measurement. Inefficient statistics and statistical
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057447.52/warc/CC-MAIN-20210923195546-20210923225546-00155.warc.gz
|
CC-MAIN-2021-39
| 2,985 | 5 |
http://calculatecreditcard.com/average-interest-rate-for-credit-cards/org/
|
math
|
Average interest rate for credit cards
Monthly ways rate year 7 do fee percent visa out percentage minimum balance over on for method will. 1.2 calculating caculate caculating after chase pay calculate paid with calculators of your apr. score 22.9 1 accrual figuring unpaid caculator card calculated 19.99 figured computing 9000 if car. or formula months 3000 rates yearly purchase 30 teaching calcuate determine simple example loan..
excel 18 due i be bill monthy 20 balances activate money would annually 10 calc cycle interset 18.99. calulate rel cr online calculation calculator payments report 5000 how 1500 estimate crdit interst. accrue breakdown outstanding formulas calculations per 12.99 charged 24.99 debit off and use 24.9. long charge the montly avg amount much calcualte one calculater in each by free annual are.
10000 you. equation compound finance bank payoff billing day limit 3.99 days intrest compute cards quick 9.9. debt total payment 22 4000 bal hold many credit using interest interes charges figure from month. finding chart creditcard average savings an cc find statement deposit fees vs spreadsheet credi. computation interests is transfer 12 adb to raise accrued it a can best whats calulator.
Read a related article: How Credit Card Interest is Calculated
Read another related article: What Are The Benefits to Calculating Your Daily Interest Rate?
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573065.17/warc/CC-MAIN-20190917081137-20190917103137-00431.warc.gz
|
CC-MAIN-2019-39
| 1,377 | 6 |
http://lslwiki.digiworldz.com/lslwiki/wakka.php?wakka=GuitarheroDougall
|
math
|
Don't click here unless you want to be banned.
1 Coulomb = 6.2415 * 10 ^ 18th electrons
1 Ampere = 1 Coulomb passing 1 point in a conductor in 1 second
1 Ohm is the resistance of a column of mercury with a cross-sectional area of 1 square millimeter and a length of 106.3
centimeters at zero degrees Celsius
A conductor has 1 ohm of resistance when an applied potential of 1 volt produces a current of 1 ampere
1 Joule = the work done when a current of 1 ampere passes through a resistance of 1 ohm for 1 second
1 Watt = 1 joule/second; the power dissipated by a current of 1 ampere flowing across a resistance of 1 ohm
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247518497.90/warc/CC-MAIN-20190222155556-20190222181556-00607.warc.gz
|
CC-MAIN-2019-09
| 619 | 8 |
https://demtutoring.com/answered/ecology/q6584
|
math
|
. 1. fill in the gaps with the suitable words to complete the following incomplete sentences: a. latitude and longitude are lines. b. the o degree latitude is called the c. there is a total of latitudes in the north. d. the difference of time in one degree longitude is e. the standard time of nepal is determined from longitude.
Get the answer
Category: ecology | Author: Mona Eva
. 1. when you transfer energy into a substance, the temperature stays the same 2. when you transfer energy into a substance, the molecules' kinetic e
. 1. which type of scientist studies the subject illustrated here? a. b. a life scientist an earth scientist a physical scientist none of these c. d.
. 15. gil earned these scores on the first three tests in biology this term: 86, 88, and 78. what is the lowest score that gil can earn on the fourth
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00165.warc.gz
|
CC-MAIN-2023-06
| 831 | 6 |
https://socratic.org/questions/a-ball-of-aluminum-cp-0-897-j-g-c-has-a-mass-of-100-grams-and-is-initially-at-a-
|
math
|
A ball of aluminum (cp = 0.897 J/g°C) has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup with 100 ml of water at a temperature of 15.0°C. What will be the final, equilibrium temperature of the ball and the water?
When two substances having different temperatures are introduced or kept together, heat energy, Q, flows from a substance at higher temperature to a substance at lower temperature. Also, heat continues to be transferred until their temperatures are equalized, at which point the substances are in thermal equilibrium. In a closed system, the amount of energy lost is equal to but opposite the amount of energy gained.
Thermal equilibrium formula
Q = m x cp x ∆T or Q = m x cp x (Tf - Ti)
where Q = Heat Flow (Heat lost or Heat gained)
m = Mass of the substance
cp = Specific heat capacity
Tf = Final temperature
Ti = Initial temperature
∆T = (Tf - Ti) = Difference in temperature
For your problem:
Specific heat capacity for water is
Heat lost by aluminum = Heat gained by water
The density of water is 1g/mL, so 100mL of water has a mass of 100g.
Q aluminum = - Q water
m x cp x (Tf - Ti) = -m x cp x (Tf - Ti)
Plug in the values pertaining to aluminum on the left side of the equation, and those pertaining to water on the right side.
(89.7)( Tf - 150) = - 418( Tf - 15.0) (dropped units to make the algebra easier)
89.7Tf - 13455 = - 418Tf + 6270 (negative x negative = positive)
Combine like terms by adding 418Tf to both sides, and adding 13455 to both sides.
507.7Tf = 19725
The temperature at which thermal equilibrium will occur is
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00583.warc.gz
|
CC-MAIN-2022-21
| 1,628 | 22 |
https://www.michigan-sportsman.com/forum/threads/van-buren-county-duck-spots.601026/page-3
|
math
|
The truth is, 9 out of ten people that don't read this site don't know the law properly, 6 out of 10 people on this site don't know it, and 3/10(at least) co' s don't understand it. Draw a line from each end of the property line you have permission to hunt, to the center of the body of water. That is where you have the right to hunt. Excluding great lakes. The end. It's a bogus law.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891815500.61/warc/CC-MAIN-20180224073111-20180224093111-00219.warc.gz
|
CC-MAIN-2018-09
| 385 | 1 |
https://www.jiskha.com/display.cgi?id=1333364300
|
math
|
posted by ria .
Suppose that you are now in your final semester of your MBA programme and you are faced with the choice of either getting a job when you graduate or continuing your study for a doctoral degree. Of course, your choice is not purely financial. However, to make an informed decision, you would like to know the financial implication of the two alternatives. Let’s assume that your opportunities are as follows:
If you take the “get a job” route, you expect to start off with a salary of RM40,000 per year. There is no way to predict what will happen in the future, your best guess is that your salary will grow at 5 percent per year until you retire 40 years.
As a PhD candidate, you will be paying RM25,000 per year tuition for each of the three years you registered with the graduate school. However, after you have earned the Phd programme, you can then expect a job with a starting salary of RM70,000 per year. Moreover, you expect your salary to grow by 7 percent per year until you retire 35 years later.
Clearly, your total expected lifetime salary will be higher if you pursue your study at the doctoral level. However, the additional future salary is not free. You will be paying RM25,000 in tuition at the beginning of each of the three years of your doctoral candidature. In addition, you will be giving up a little more than RM126,000 in lost income over the three years of full-time study: RM40,000 the first year, RM42,000 the second year, and RM44,100 the third year.
A. Assuming the discount rate is 9 percent and salaries will be paid at the end of each working year, which option is better financially?
B. What is the present value of the tuition fees? What is the difference if the fees are paid at the end of the year?
Please note that no one here will do your work for you. However, we will be happy to read over what YOU THINK and make suggestions and/or corrections.
Please post what you think.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886108264.79/warc/CC-MAIN-20170821095257-20170821115257-00603.warc.gz
|
CC-MAIN-2017-34
| 1,936 | 9 |
https://www.hackmath.net/en/example/4355
|
math
|
One-eighth of 9th class was interested in studying at a grammar school, at a business academy one sixth, at secondary vocational schools quarter, to SOU one third and the remaining three students were interested in the school of art direction. How many students are in the classroom?
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Showing 0 comments:
Be the first to comment!
To solve this example are needed these knowledge from mathematics:
Next similar examples:
From the total number of trees in the orchard, there are two-fifths pearls and apples are three eighty. The rest of the trees are 9 ceremonial. How many trees are in the set?
Teacher Rem bought 360 pieces of cupcakes for the outreach program of their school. 5/9 of the cupcakes were chocolate flavor and 1/4 wete pandan flavor and the rest were vanilla flavor. How much more chocolate flavor cupcakes than vanilla flavor?
Three workers planted 3555 seedlings of tomatoes in one dey. First worked at the standard norm, the second planted 120 seedlings more and the third 135 seedlings more than the first worker. How many seedlings were standard norm?
- Dropped sheets
Three consecutive sheets dropped from the book. The sum of the numbers on the pages of the dropped sheets is 273. What number has the last page of the dropped sheets?
- Unknown number 6
Determine x if 1/6 of x is equal to 2/5 of the number 24.
Find x: x + 1/2 = 1/3
- Unknown number 6
Determine the unknown number, which is by 1.5 greater than its fourth.
Write the mixed number as an improper fraction. 166 2/3
3/5 trees are apples, cherries are 1/3. 5 trees are pear. How many is the total number of trees?
I think the number. If I add to its third seven I get same as when to its quarter add 8. Which is the number?
- The buns
Kate, Zofia and Peter Liked buns. Even today, their grandmother prepare their favorite meal. Katka eats 4 bunches, Žofia 3 and Petra eats 5 buns. Their grandmother said to them, "My inmate will you know how many buns I have been make today, if those you ea
- Unknown number 11
That number increased by three equals three times itself?
- Fraction to decimal
Write the fraction 3/22 as a decimal.
There were pears in the basket. I took them two fifths and left them in a basket of six. How many pears I took?
Zdeněk picked up 15 l of water from a 100-liter full-water barrel. Write a fraction of what part of Zdeněk's water he picked.
- Passenger boat
Two-fifths of the passengers in the passenger boat were boys. 1/3 of them were girls and the rest were adult. If there were 60 passengers in the boat, how many more boys than adult were there?
- Pizza 4
Marcus ate half pizza on monday night. He than ate one third of the remaining pizza on Tuesday. Which of the following expressions show how much pizza marcus ate in total?
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267160454.88/warc/CC-MAIN-20180924125835-20180924150235-00442.warc.gz
|
CC-MAIN-2018-39
| 2,835 | 31 |
https://www.physicsforums.com/threads/solubility-product-constant-question.262154/
|
math
|
The solubility-product constant for K2PdCl6 is 6.0 x 10^ -6 (K2PdCl6 ---> 2K+ + PdCl6 2-). What is the K+ concentration of a solution prepared by mixing 50.0 mL of 0.200 M KCl with 50.0 mL of 0.100M PdCl6(2-)? My Approach (thus far): 1) get mmol of KCl and PdCl6(2-) and get difference of them for excess mmol (of what though?). 2) I'm not really sure what to do from there without knowing which would be in excess...My hunch is to solve whatever is in excess to similar terms to get concentration of K+. Anyways, this questions has had me scratching my head for a few days. Thanks for all the help I get!
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676589536.40/warc/CC-MAIN-20180716232549-20180717012549-00161.warc.gz
|
CC-MAIN-2018-30
| 605 | 1 |
https://didyouknowfacts.com/people-say-feminists-arent-funny-20-tumblr-posts-will-kill/
|
math
|
Feminists have gotten a bad rap lately (or maybe always) for being a group of no-fun man haters who just want to set the status quo on fire and then cackle like witches while the world burns.
Okay, maybe I’m exaggerating. A little.
Even so, these 20 women prove, beyond the shadow of a doubt, that at the very least some feminist ladies do have a sense of humor. And a damn good one, to boot.
#20. No commentary required.
#19. As you’ll see from this list, men have dug their own graves with this phrase.
#18. Celebrities. Sigh.
#17. Even my hair is tough RAWR.
#16. Nailed it.
#15. Seriously. Knowing how to turn men OFF is actually a valuable skill.
#14. Animals can be heroes, too.
#13. A lesson in consent he’ll never forget.
#12. Maybe just “equality?” Or “duh?”
#11. They do realize that if girls could get themselves pregnant, boys would be obsolete. Right? RIGHT?
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891811352.60/warc/CC-MAIN-20180218023321-20180218043321-00425.warc.gz
|
CC-MAIN-2018-09
| 885 | 13 |
https://www.lessonplanet.com/teachers/compute-solutions-worksheet
|
math
|
Compute Solutions Worksheet
In this computing activity, students solve addition, multiplication, and division problems involving mixed numbers, fractions, and decimals. This one-page activity contains 12 problems.
3 Views 3 Downloads
Common and Natural Logarithms and Solving Equations
Log some practice with logarithms. A PowerPoint presentation provides a tutorial on the change of base formula involving natural logarithms and solving exponential equations with logarithms in the fourth installment of a seven-part...
9th - 12th Math CCSS: Designed
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257648103.60/warc/CC-MAIN-20180322225408-20180323005408-00108.warc.gz
|
CC-MAIN-2018-13
| 551 | 6 |
http://website.informer.com/terms/Cbse_5th_Class_Syllabus
|
math
|
CBSE, CBSE Sample Papers, CBSE Syllabus, CBSE Results, CBSE NCERT Books solutions, Mathematics formulas, CBSE Maths Sample Papers 2013 2014
8th, 9th and 10th CBSE Math, Science, History, Geography, Civics, Economics and English Grammar
NCERT, CBSE, ICSE, State Boards, Entrance Exam, Nursery, all subjects, syllabus for class 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, Animation based learning, India
MyAcademicProgram measures the return on educational investment. The Student MAP is a comprehensive college management solution for both students and parents alike.
CBSE 12th Class Question Papers Syllabus Sample Papers Projects Datesheet Result
CBSE 10th Class Question Papers Syllabus Sample Papers Projects Datesheet Result
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-27/segments/1435375095270.70/warc/CC-MAIN-20150627031815-00058-ip-10-179-60-89.ec2.internal.warc.gz
|
CC-MAIN-2015-27
| 747 | 6 |
https://ubezpieczeniekredytowe.pl/Sat_29_Aug_42582.html
|
math
|
Concrete Mix yield one 20kg bag will cover an area of 1.1m2 to a depth of approximately 10mm. or 108 x 20kg bags equates to one cubic metre of mixed concrete. TyPicAL ProDUcT PErForMAncE Compressive Strength (MPa): BAGS rEqUirED For THE joB: Square Mtrs Number of 20kg Bags 10 82 108 9 74 98 8 66 88 7 58 76 6 50 66 5 42 54 4 34 44
Volume of one bag cement formula
Mass of one bag of cement. 50 Kg for one Bag. Density of cement. 1440 kg/ meter cube. Now, we need to calculate volume of one bag of cement. So, Volume of Cement: = Mass/ Density =50/1440 = 0.0347 m3 . Also you can convert this volume from cubic meter to cubic feet, Just you have to multiply the volume of 1 bag of cement …
How many kg of cement are present in one purchased bag ...
The weight of cement bag when packed at grinding plant is exactly 50 kg in each bag but the weight shall not remains 50 kg in bag when it opens for use or when we purshase from retailer. The main reason for the loss in weight of cement is transpor...
Enter the number of 8 inch x 2 inch x 4 inch bricks or 8 inch x 8 inch x 16 inch blocks you plan to use for your project. The calculator will indicate the number of 60 or 80 pound bags of QUIKRETE® Mortar Mix you need to construct your project with a 3/8 inch mortar joint. (All yields are approximate and do not include allowance for uneven ...
How To Calculate Cement Bags Per Cubic Meter Concrete ...
Therefore required cement = 1440 x 0.221 m 3 = 318.24 Kg or (318.24/50) = 6.37 Bags. Cement Bag Per Cubic Metre Concrete Table – Nominal Mix
How much CFT, CuM, kg in 1 Bag of Cement || How much ...
· DerivationQuick Learn in this Video / Topic / Tutorial about# How much cement in 1 Bag of Cement# How many Numbers of Bags in 1 Cubic Meter# How much kg ceme...
How many bags of cement required for 1m3 of Cement?
· So, Usually, 4% of moisture content is present in cement. Dry Density of Cement = 1440kg/m 3. Each bag of cement consists of = 50kg. No. of bags in 1m 3 = 1440/50. = 29bags (approx)
Welcome to the concrete calculator. It will help you calculate the number of bags of concrete required for your path, slab or post holes.
Cement to all in ballast
Simple calculation. Usually it''s 1:4 so for every 4 kg of sand 1 kg of cement. As they called tone bag - around (400 kg) you''ll need 100 kg cement. So one tone bag of sand - four bags of cement. Hopefully this will help. mDucanon. 2016-08-09T07:45:01+01:00. Answered 9th Aug 2016.
How to Calculate Cement, Sand and Aggregate required for 1 ...
Total weight of concrete ingredients = 50+115+209+27.5 = 401.5 say 400 kg. Density of concrete = 2400 kg/cum. So, 1 bag of cement produces = 400/2400 = 0.167 cum. No. of bags required for 01 cum of concrete = 1/0.167 = 5.98 bags ~ 6 bags. From above, if the concrete mix is 1:2:4, to get a cubic meter of concrete we require. 1.Cement = 6 bags ...
concrete bag mix Calculator
calculate how many bags of concrete you need. Scroll down to see the calculator! Here''s a free concrete bag mix calculator to help you figure the amount of cement or concrete you need to do your project. Most people call concrete "cement" when in fact cement is just one ingredient along with sand, aggregate and water that is used to make concrete.
How many bags of cement do I need to make 1 cubic metre of ...
· 1 Cement 2 Sand 3 Aggregate (gravel) is too strong. 16bags (costly) 1 Cement 2.5 Sand 4 Aggregate is OK for fence posts. 13bags of cement per cubic metre. The more cement the more costly of course so don''t over do it. Note: These are 20kg bags of General Purpose cement.
Convert KG to bag [portland cement]
How many KG in 1 bag [portland cement]? The answer is 42.63768278. We assume you are converting between kilogram and bag [portland cement]. ... The kilogram or kilogramme, (symbol: kg) is the SI base unit of mass. A gram is defined as one thouh of a kilogram. Conversion of units describes equivalent units of mass in other systems.
In 1 cum how many number of 50kg cement bags required ...
· The density of cement is 1440 Kg/Cum. Thus the volume of a 50 kg cement bag comes to be= [ 50/1440 ]*1000 = 34.722 Litres . For practical purposes consider as 35 Litres. That gives to 0.035cum quite simple. Thus in 1 cum it come approx. 29 bags of 50kg cement bags.
How many cubic meters are in a bulk bag of sand?
· How many cubic meters is a 20kg bag of sand? Yield: One 20kg bag will cover an area of 1.1m2 to a depth of approximately 10mm. Or 108 x 20kg bags equates to one cubic metre of mixed concrete. 20kg bags can be bought individually or as a pallet of 70.
Concrete Mix (Quantity Calculator) – Target Products Ltd.
Concrete Mixes. The calculator will indicate the estimated number of 25 kg or 30 kg bags of Quikrete® Concrete Mix needed to build a 3″, 4″ or 6″ slab (with a typical waste factor). Yields are approximate and will vary based upon placement methods, equipment utilized and do not include allowance for uneven subgrade, waste, etc.
How To Calculate Cement Bags In 1 Cubic Meter?
· Therefore volume of 1 bag cement = 50/1440 =0.0347 cum. ∴ No. of cement bags required in 1 cubic meter = 0.2171/0.0347 = 6.25 bags. Note: You can use the same formula for calculating cement for other nominal mixes.
How to Estimate the Quantity of Sand and Cement Required ...
· This is obtained by knowing that the mass of 1 bag of cement = 50kg, and the density = 1440 kg/m 3 The volume of a standard builder''s wheelbarrow is 0.065 m 3 (unheaped). We assume that approximately 2 bags of cement (4 head pans of cement) will fill one builder''s wheelbarrow.
How To Calculate Cement Bags In 1 Cubic Meter?
Quikrete 60 lb. Concrete Mix-110160 - The Home Depot
Bags Of Portland Cement to Kilograms | Kyle''s Converter
Unit Descriptions; 1 Bag (of Portland Cement): One bag of portland cement weighs 94 lb av by definition. 1 Kilogram: The kilogram is defined as being equal to the mass of the International Prototype Kilogram (IPK), which is almost exactly equal to the mass of one liter of water.
What is the volume of a 40 kg bag of cement?
· Using proportions suggested, one 40 kg (88 lb) bag of cement will produce about 3.5 ft³ (0.1 m³) of concrete mix approximately. Just so, how many cubic meters is a 40kg bag? You will need 6 x 40kg bags per cubic meter .
Calculate Bags of Pre-Mix Concrete
20 kg 25 kg 30 kg 40 kg. Cost per Bag. A concrete slab 1000 mm X 1000 mm at a depth of 100 mm, has a volume of 0.1 m³ = 10.8 x 20 kg bags. 11 Bags. Total Bag Weight 220 kg. Calculated at 2160 kg per 1m³ - Allow extra for waste. Pre-Mix Concrete Bag Calculator Calculated at 2160 kg per 1m³ - Allow extra for waste. Cubic Metres.
St. Mary Portland Cement, 40 KG | The Home Depot Canada
· YIELD Using proportions suggested, one 40 KG (88 lb) bag of cement will produce about 3.5 ft3 (0.1 m³) of concrete mix, approximately. PACKAGING 20 KG (44 lb) bag 30 KG (66 lb) bag 40 KG (88 lb) bag STORAGE & SHELF LIFE Material should be stored in a dry, covered area, protected from the elements.
How To Calculate Cement, Sand & Aggregates Quantity in ...
· Density of Cement is 1440/m 3 = 0.28 x 1440 = 403.2 kg We know each bag of cement is 50 kg For Numbers of Bags = 403.2/50 = 8 Bags We Know in one bag of cement = 1.226 CFT For Calculate in CFT (Cubic Feet) = 8 x 1.225 = 9.8 Cubic Feet CALCULATION FOR SAND QUANTITY; Consider volume of concrete = 1m 3
Convert metric ton to bag [portland cement]
How many metric ton in 1 bag [portland cement]? The answer is 0.04263768278. We assume you are converting between metric ton and bag ... The SI base unit for mass is the kilogram. 1 kilogram is equal to 0.001 metric ton, or 0.023453432147327 bag [portland cement]. Note that rounding errors may occur, so always check the results.
Cement, mortar volume to weight conversion
About Cement, mortar; 1 cubic meter of Cement, mortar weighs 2 162 kilograms [kg] 1 cubic foot of Cement, mortar weighs 134.96925 pounds [lbs] Cement, mortar weighs 2.162 gram per cubic centimeter or 2 162 kilogram per cubic meter, i.e. density of cement, mortar is equal to 2 162 kg/m³ Imperial or US customary measurement system, the density is equal to 134.9693 pound per cubic foot …
CONCRETE DIVISION 3 -50, -60, -80, -90 2 RODUCT O
QUIKRETE Concrete Mix can be mechanically mixed in a barrel type concrete mixer or a mortar mixer. Choose the mixer size most appropriate for the size of the job to be done. Allow at least 1 ft3 (28.3 L) of mixer capacity for each 80 lb (36.2 kg) bag of QUIKRETE Concrete Mix to be mixed at one time.
Cement Ready Reckoner
No of bags 3 5 7 9 11 Estimating how many 20kg bags of premix concrete to order (rounded up to the nearest full bag) * 108 x 20kg bags of Boral Cement Concrete Mix will fill 1 cubic metre (m3). Depth of hole (mm) Cubic metres(m3) needed. Hole diameter (mm) 300mm 400mm 500mm 600mm 700mm 200 0.02 0.03 0.04 0.06 0.08 400 0.03 0.05 0.08 0.12 0.16
How many hollow blocks can be made in 1 bag of cement
How many hollow blocks can be made in 1 bag of cement Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, How many hollow blocks can be made in 1 bag of cement, quarry, aggregate, and different kinds of minerals.
For example, a conventional redymix concrete bag of 80 lbs may be listed as having a yield of approximately 0.60 cu ft or a 25 kg bag may have a yield of ~0.01 cu m. With a few simple mathematical transformations we can use this number to arrive at the in place density of the material and complete our calculation of the total mass and number of ...
Portland Cement 1 kilogram mass to liters converter
One kilogram of Portland cement converted to liter equals to 0.66 L. How many liters of Portland cement are in 1 kilogram? The answer is: The change of 1 kg - kilo ( kilogram ) unit of Portland cement measure equals = to 0.66 L ( liter ) as the equivalent measure for the same Portland cement …
How to Calculate the Volume of 1 Bag of Cement in Cubic ...
We know the density of cement is 1440 kg/m 3 and the mass of the one bag cement is 50 kg. The density of cement means that in one cubic meter volume the quantity of cement will be 1440 kg. Number of Bags of Cement in One Metric Cube = [1440 (kg/m 3)/ 50(kg)] Number of Bags of Cement in One Metric Cube = 28.8 28.8 (1/m 3)
Example calculation Estimate the quantity of cement, sand and stone aggregate required for 1 cubic meter of 1:2:4 concrete mix. Ans. Materials required are 7 nos. of 50 kg bag of cement, 0.42 m 3 of sand and 0.83 m 3 of stone aggregate.
How Many Bags Of Cement And Ballast Do I Need?
For one bulk bag of ballast, how many bags of cement do I need? I''m guessing you''ll need more than one 1 ton bag of ballast and around 6 bags of cement. A 1 ton bag weighs around 900 kilograms, so I''m guessing 40-45 bags weighing 25 kilograms each. 12 shovels of ballast to 1/2 bag of cement is a good ratio, but 1:6 would suffice.
How much sand and gravel do I need for a bag of cement ...
Regarding this, how many wheelbarrow of sand to a bag of 20kg cement, generally a bag of 20kg cement will require about 0.98 cubic feet of sand, taking one wheelbarrow size as 2 CF, you will need approx 1/2 (half) wheelbarrow of sand to a bag of 20kg cement.
How to Calculate the Volume of 1 Bag of Cement in Cubic ...
We know the density of cement is 1440 kg/m 3 and the mass of the one bag cement is 50 kg. The density of cement means that in one cubic meter volume the quantity of cement will be 1440 kg. Number of Bags of Cement in One Metric Cube = [1440 (kg/m 3)/ 50(kg)] Number of Bags of Cement in One …
- conveyor crushing station
- cono de trituracion simmons
- china gold export
- global precious stone machine
- alluvial placer for sale australia
- chancadora de mand c adbula jaw crusher
- bagian bagian jaw crusher crusher penjualan harga
- impact crusher crush hard rock round sand
- gold refineries in liya
- birla cement plant at kotputli
- aggregate unit costs
- lead ore hydraulic cone crushing machine
- cone crusher working pricipal
- india small metal crusher manufacturer for sale
- iron cross advanced
- crusher special jaw
- anthracite coal buyers
- calculating volume of cementsand aggregate
- hill gold mine contacts
- copper mining on their own
- batu kerucut crusher dari cina
- ibaryte procesing in lagos nigeria
- crushing of limestone machine for sale
- crusher bagain jaw
- cost of lump crusher india
- charasteristics of screw conveyors
- guangzhou chinese medicine grinder
- crusher jfc nanotubes
- crusher plant feeder clips
- german technical and high quality sand and aggregate making machine
- environmental impact of mining zinc
- cone crusher optimized
- aggregate wash plants aggregate wash plants
- concrete batching aggregate
- jacques impactor
- impact crusher in egypt
- cement industry associated
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337971.74/warc/CC-MAIN-20221007045521-20221007075521-00520.warc.gz
|
CC-MAIN-2022-40
| 12,908 | 104 |
https://www.thingiverse.com/Alejandro3d/about
|
math
|
My name is Alexander Sayapin, i'm now working at Siberian State Aerospace University, Krasnoyarsk, Russia.
I am teaching discrete math, mathematical simulation fundamentals, data bases design and computer architecture.
I have been teaching also artificial intelligence systems, artificial neural networks, simulation methods, IT in education and science, applied software, automata and formal languages theory.
Now I teach schoolchildren S.T.E.M. (Science, Technology, Engineering and Math) in form of a robotics workshop.
I Am A…Teacher
Tools I Use…
Design Programs: OpenSCAD
3D Design Skill Level
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00282.warc.gz
|
CC-MAIN-2019-26
| 602 | 8 |
https://rd.springer.com/article/10.1007/s00013-008-2438-x
|
math
|
On SU(1,D)/[U(1,D),U(1,D)] for a quaternion division algebra D
- 58 Downloads
Let D be a quaternion division algebra with an involution of the second kind. We show that the quotient group SU(1,D)/[U(1,D),U(1,D)] is nontrivial in general. For global fields, we completely determine this group.
Mathematics Subject Classification (2000).Primary 11R52
Keywords.Quaternion division algebra involution of the second kind unitary group
Unable to display preview. Download preview PDF.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215261.83/warc/CC-MAIN-20180819165038-20180819185038-00540.warc.gz
|
CC-MAIN-2018-34
| 478 | 6 |
https://web2.0calc.com/questions/math-help_85369
|
math
|
All of the following are equivalent proportions except?
A. a/b = c/d
B. b/a = d/c
C. d/b = c/a
D. c/d = b/a
The answer is......
A not a/b = c/d multiply both sides by d
d a/b = c divide both sides a
d/b = c/a this is answer C so A and C are equiv similar can be shown for B but not for D
You are welcome....hope you now understand a bit better
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00423.warc.gz
|
CC-MAIN-2019-51
| 343 | 10 |
http://eatatbartletts.com/queensland/how-to-translate-algebraic-expressions.php
|
math
|
2.5a Translate to an Algebraic Expression Chipola College
Translate algebraic expressions into English phrases, and translate English phrases into algebraic expressions. Read the expression or phrase and select word tiles or symbol tiles to form the corresponding phrase or expression. 5 Minute Preview. Use for 5 minutes a day.... 2018-09-10 · Algebraic fractions are simply fractions with algebraic expressions on the top, while an algebraic expression is an expression built up from integer constants, variables, and the algebraic …
Name Period Date Writing and Evaluating Expressions
DESCRIPTION. Translate algebraic expressions into English phrases, and translate English phrases into algebraic expressions. Read the expression or phrase and select word tiles or symbol tiles to form the corresponding phrase or expression.... 2018-09-10 · Algebraic fractions are simply fractions with algebraic expressions on the top, while an algebraic expression is an expression built up from integer constants, variables, and the algebraic …
I need to know how to translate algebraic expressions
Here is an example of how to translate a verbal phrases into a variable phrase. Remember to read the lesson on this topic. Basic steps to follow: how to take photos with sparklers Translating Algebraic Expressions – Video Get access to all the courses and over 150 HD videos with your subscription Monthly, Half-Yearly, and Yearly Plans Available
Translate expressions with parentheses (practice) Khan
Linear Algebraic Expressions. Translate each phrase into a linear algebraic expression. Each expression is in the form of ax + b, where x is any variable; a and b are constants. Hard level is in the form of c(ax+b), c is a constant. Easy: Sheet 1 Sheet 2 Sheet 3 Grab 'em All. Moderate: Sheet 1 Sheet 2 Sheet 3 Grab 'em All. Difficult: how to translate a youtube video from french to english Examples of How to Translate Multi-Part Math Phrases into Algebraic Expressions Example 1 : Write an algebraic expression for the math phrase ” 3 more than twice a number “. Solution : To make this much easier to understand, we are going to divide this phrase into two parts.
How long can it take?
2.2 Evaluate Simplify and Translate Expressions (Part 2
- How to translate algebraic expressions
- Translating Verbal Phrases to Algebraic Expressions
- Name Period Date Writing and Evaluating Expressions
- Algebraic Translations MathBitsNotebook(A1 - CCSS Math)
How To Translate Algebraic Expressions
Translate the words to the right of the equals word into an algebraic expression. Example Translate the sentence into an algebraic equation: Twice the difference of \(x\) and \(3\) gives \(18.\)
- I tell students that in algebra they are going to be using variables to represent a situation. Before we model situations using variables, expressions, and equations we need to be able to translate expressions and equations between word form and algebraic form.
- Once you have downloaded and printed the translating algebraic expressions game cards and cut them apart. The cards contain verbal expressions and algebraic models that are to be matched.
- These Algebraic Expressions Worksheets will create word problems for the students to translate into an algebraic statement. These Algebraic Expressions Worksheets are a good resource for students in the 5th Grade through the 8th Grade.
- Then translate each phrase into an expression, and write them on each side of the equal sign. We will practice translating word sentences into algebraic equations. Some of the sentences will be basic number facts with no variables to solve for.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573331.86/warc/CC-MAIN-20190918193432-20190918215432-00324.warc.gz
|
CC-MAIN-2019-39
| 3,629 | 20 |
http://judaism.stackexchange.com/questions/tagged/evil-resha+zohar
|
math
|
Mi Yodeya Meta
to customize your list.
more stack exchange communities
Start here for a quick overview of the site
Detailed answers to any questions you might have
Discuss the workings and policies of this site
The righteous man who suffers
In Ra'aya Mehemna (Parshat mishpatim) , it's explained that "the righteous man who suffers" is one whose evil nature is subservient to his good nature. I honestly did not understand what that line ...
Jul 10 '13 at 0:15
newest evil-resha zohar questions feed
Hot Network Questions
Why don't the Doctors' companions speak Dinosaur?
Proof of trigonometric identity using vector calculus
Extraction of Bits
Double \prec as a single symbol?
Windows Firewall - blocking IP address ranges en masse - performance considerations?
How to say pressing something by foot.?
What English homophone corresponds to 'oise salon'?
Are there any negatively charged electrophiles, or simply neutral and positive?
Integration of this type?
Scaling a glue in TeX
Can I use the word "school" when referring to something that belongs to a university?
Use Mathematica to calculate the area enclosed between two curves
Is using a member function as an argument to a constructor undefined behavior?
How to fool the "try some test cases" heuristic: Algorithms that appear correct, but are actually incorrect
Possibly quit your job with a polyglot
Clap Clap Switch Modifications
Are cantrips spells?
How to list entries related to a category + related to a section?
How does David Lightman in WarGames manage to hack a computer by dialling a number?
Are Friend Safari Pokémon gender locked?
What do you call a disgusting mixture you don't want to drink?
Dining Philosophers Algorithm
What's the path of something dropped from a space elevator
more hot questions
Life / Arts
Culture / Recreation
TeX - LaTeX
Unix & Linux
Ask Different (Apple)
Geographic Information Systems
Science Fiction & Fantasy
Seasoned Advice (cooking)
Personal Finance & Money
English Language & Usage
Mi Yodeya (Judaism)
Cross Validated (stats)
Theoretical Computer Science
Meta Stack Exchange
Stack Overflow Careers
site design / logo © 2014 stack exchange inc; user contributions licensed under
cc by-sa 3.0
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535919886.18/warc/CC-MAIN-20140909054622-00425-ip-10-180-136-8.ec2.internal.warc.gz
|
CC-MAIN-2014-35
| 2,198 | 52 |
http://listas.cmat.edu.uy/pipermail/todos/2013-January/006983.html
|
math
|
[Todos CMAT] Premio a Y. Sinai
roma en fing.edu.uy
roma en fing.edu.uy
Jue Ene 3 08:36:25 UYST 2013
LEROY P. STEELE PRIZE FOR LIFETIME ACHIEVEMENT
The Leroy P. Steele Prizes were established in 1970 in honor of George
David Birkhoff, William Fogg Osgood, and William Caspar Graustein and
are endowed under the terms of a bequest from Leroy P. Steele. Prizes
are awarded in up to three categories and each is awarded annually.
The following citation describes the award for Lifetime Achievement.
The 2013 Steele Prize for Lifetime Achievement is awarded to Yakov Sinai for
his pivotal role in shaping the theory of dynamical systems and for his
groundbreaking contributions to ergodic theory, probability theory,
statistical mechanics, and mathematical physics.
Sinai?s research exhibits a unique combination of brilliant analytic
outstanding geometric intuition, and profound understanding of underlying
physical phenomena. His work highlights deep and unexpected
dynamical systems and statistical mechanics. Sinai has opened up new
including Kolmogorov?Sinai entropy, Markov partitions, and Sinai?Ruelle?Bowen
measures in the hyperbolic theory of dynamical systems; dispersing billiards,
a rigorous theory of phase transitions in statistical mechanics and space-time
chaos. In addition, Sinai has made seminal contributions in the theory of
Schrödinger operators with quasi-periodic potentials, random walks in random
environments, renormalization theory, and statistical hydrodynamics
and Navier?Stokes equations.
Sinai pioneered the study of dispersing billiards: dynamical systems
the motion of molecules in a gas. The simplest example of such a
a square with a disk removed from its center, is called ?Sinai?s
billiard motions within the framework of hyperbolic theory, Sinai discovered
that they exhibit deep ergodic and statistical properties (such as the
theorem). Owing to Sinai?s work, some key laws of statistical
mechanics for the
Lorentz gas can be established with mathematical rigor. In particular,
the ?rst steps towards justi?cation of Boltzmann?s famous ergodic hypothesis,
proposed in the end of the nineteenth century: ?For large systems of
particles in equilibrium, time averages are close to the ensemble average.?
Sinai returned to this subject several times in the period 1970?90
co-authors, including his students Bunimovich and Chernov. 55
Together with his student Pirogov, Sinai created a general theory of
phase transitions for statistical mechanics systems with a ?nite number of
ground states. Pirogov?Sinai theory forms essentially the basis for modern
equilibrium statistical mechanics in a low-temperature regime.
Sinai made seminal contributions to the theory of random walks in a random
environment. With his model, known nowadays as ?Sinai?s random walk,? he
obtained remarkable results about its asymptotic behavior. With his student
Khanin, Sinai pioneered applications of the renormalization group method
to multi-fractal analysis of the Feigenbaum attractor, and to the Kolmogorov?
Arnold?Moser theory on invariant tori of Hamiltonian systems.
In the past ?fteen years Sinai has brought novel tools and insights from
dynamical systems and mathematical physics to statistical hydrodynamics,
obtaining new results for the Navier?Stokes systems. Speci?cally, along
with D. Li, Sinai devised a new renormalization scheme which allows the
proof of existence of ?nite time singularities for complex solutions of
the Navier?Stokes system in dimension three.
Sinai?s mathematical in?uence is overwhelming. During the past half-century he
has written more than 250 research papers and a number of books.
monograph, Ergodic Theory (with Cornfeld and Fomin), has been an introduction
to the subject for several generations, and it remains a classic.
Sinai supervised more than ?fty Ph.D. students, many of whom have become
leaders in their own right. Sinai?s work is impressive for its breadth.
In addition to its long-lasting impact on pure mathematics, it has played a
crucial role in the creation of a concept of dynamical chaos which has been
extremely important for the development of physics and nonlinear science over
the past thirty-?ve years. The Steele Prize for Lifetime Achievement is
awarded to Sinai in recognition of all these achievements.
Yakov G. Sinai was born in 1935 in Moscow, Soviet Union, now Russia.
He received his Ph.D. degree (called a Candidate of Science in Russia) and
then his doctorate degree (Doctor of Science) from Moscow State University.
For several years, he combined his position at Moscow State University and the
Landau Institute of Theoretical Physics of the Russian Academy of Sciences.
Since 1993, he has been a professor in the mathematics department of Princeton
Ya. Sinai received various honors recognizing his contributions. He was
elected as a foreign associate of the National Academy of Sciences and a
foreign member of the Academy of Arts and Sciences. He is a full member
of the Russian Academy of Sciences, and he was recently elected as a
foreign member of the Royal Society in London. He is also a member of
the Brazilian Academy of Science, the Hungarian Academy of Science,
the Polish Academy of Science, and Academia Europea.56 Among his other
recognitions are the Wolf Prize in Mathematics, the Nemmers Prize, the
Lagrange Prize, the Boltzmann Medal, the Dirac Medal, and the Poincaré Prize.
Response from Yakov Sinai
It is a great honor to be awarded the Steele Prize for Lifetime Achievement
from the American Mathematical Society. I worked in several directions
in mathematics, including the theory of dynamical systems, statistical
and mathematical physics, and probability theory. My mentors who had a big
in?uence on me were A. N. Kolmogorov, V. A. Rokhlin, and E. B. Dynkin.
I also bene?tted a lot from many contacts with my colleagues.
I was very fortunate to have talented students, many of whom became strong
and famous mathematicians. Unfortunately, it is not possible to list the names
of all of them here. I thank my family and friends for their encouragement and
support. Finally, I thank the selection committee for its work.
Más información sobre la lista de distribución Todos
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00638.warc.gz
|
CC-MAIN-2022-05
| 6,190 | 92 |
http://www.lawschooldiscussion.org/index.php?action=profile;area=showposts;u=13590
|
math
|
« on: February 25, 2006, 10:14:27 AM »
well I really liked the testmasters class, and they use only real LSAT questions rather than making up their own, which I thought was really helpful in my practicing. the class also gives you access to more recent LSATs that aren't in the 3 'Actual LSAT' books (we got LSATs from 2004 and 2005).
it was also helpful in terms of motivating me to study - I work full time, and definitely woulnd't have had the energy or motivation to teach myself the LSAT if I didn't take a class.
The class I took met twice a week for 4 hours. If I take it again, I know that I won't HAVE to attend every class, but I'll also know that if I'm having trouble with something, I can ask a teacher and get help. the cost was $1250, but the organization I work for paid for half - the cost of taking it again is $650, and again, my organization will pay for half. So I feel like it's probably worth it. I don't know.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122955.76/warc/CC-MAIN-20170423031202-00032-ip-10-145-167-34.ec2.internal.warc.gz
|
CC-MAIN-2017-17
| 935 | 4 |
http://ce.imart.info/2017/01/12/how-to-do-percentiges/
|
math
|
How To Do Jkabs Ladder How To Do Rubikis Cube We have just launched brand NEW animated Solve It guides! Learning to Solve a Rubik’s Cube has never been so easy. CLICK HERE to watch! *** ‘How do I solve the … How to Solve a Rubik’s Cube (Easy Move Notation) (with … – Apr 01, 2016 · How to Solve
How To Do Matracies There are many things we can do with them … Adding. To add two matrices: add the numbers in the matching positions: How to Multiply Matrices – mathsisfun.com – How to Multiply Matrices. A Matrix is an array of numbers: A Matrix (This one has 2 Rows and 3 Columns) To multiply a matrix by
Examples of percentage calculations. The following two examples show how to calculate percentages. 1) 12 people out of a total of 25 were female.
3 Easy Ways to Calculate Everyday Percentages – wikiHow – Dec 09, 2016 · How do I add a percentage of a number to the number? wikiHow Contributor. First you need to calculate what number the percentage is equal to.
How To Do Report Bursting In Boxir2 May 04, 2009 · 1.How do you achieve this in BOXIr2. 2. … Hello,I have to use this concept of report bursting wherein i need to run the report for say all the branches … Report Bursting – GeekInterview.com – … Report Bursting Hi Friends,Can any one help me regrading Report Bursting using Business objects xi
Introduction to Percentages – mathsisfun.com – My Dictionary says "Percentage" is the "result obtained by multiplying a quantity by a percent". So 10 percent of 50 apples is 5 apples: the 5 apples is the percentage.
The Math Dude : How to Quickly Calculate Percentages … – How to calculate percentages can be easier than you may realize. Keep reading for some simple tricks. Long time math fans may remember our first foray into the world …
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00431.warc.gz
|
CC-MAIN-2017-34
| 1,793 | 7 |
https://scholar.archive.org/work/4i7xk56ljfb2lbxa5antv5hmbe
|
math
|
A copy of this work was available on the public web and has been preserved in the Wayback Machine. The capture dates from 2017; you can also visit the original URL.
The file type is
Let A be an n-by-n matrix with real entries. We show that a necessary and sufficient condition for A to have positive semidefinite or negative semidefinite symmetric part Further, if A has positive semidefinite or negative semidefinite symmetric part, and A 2 has positive semidefinite symmetric part, then rank[AX] = rank[X T AX] for all X ∈ M n (R). This result implies the usual row and column inclusion property for positive semidefinite matrices. Finally, we show that if A, A 2 , . . . , A kdoi:10.1016/s0024-3795(99)00256-6 fatcat:bdlmcw3vybcmhmsowq4rxezt6q
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00585.warc.gz
|
CC-MAIN-2022-27
| 748 | 3 |
http://www.ask.com/answers/264372861/what-is-138-out-of-300-is-what-percentage?qsrc=3112
|
math
|
What is 138 out of 300 is what percentage ?
46... just make a fraction out of the 2 #s and adjust the fraction until the bottom = 100 then the top # will be the %.
46, that's below 50....that's an F!!!!
46% you take the first number and divide it by the total. then multiply by 100
It's 46 or 43
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00483-ip-10-147-4-33.ec2.internal.warc.gz
|
CC-MAIN-2014-15
| 295 | 5 |
http://theskepticalzone.com/wp/the-lci-and-bertrands-box/
|
math
|
Tom English has recommended that we read Dembski and Marks’ paper on their Law of Conservation of Information (not to be confused with the Dembski’s previous LCI from his book No Free Lunch). Dembski also has touted the paper several times, and I too recommend it as a stark display of the the authors’ thinking.
Most people won’t take the time to carefully read a 34-page paper, but I submit that the authors’ core concept of “conservation of information” is very easily understood if we avoid equivocal and misleading terms such as information, search, and target. I’ll illustrate it with a setup borrowed from Joseph Bertrand.
The “Bertrand’s box” scenario is as follows: We’re presented with three small outwardly identical boxes, each containing two coins. One has a two silver coins, one has two gold coins, and one has a silver coin and a gold coin. We’ll call the boxes SS, GG, and SG. We are to randomly choose a box, and then randomly pull a coin from the chosen box.
It’s clear that we’re equally likely to end up with a gold coin or a silver coin, so the probability of getting a gold coin (our preferred result) is
1/2. This unconditional probability of
1/2 gets updated when we choose a box. If we choose GG, then our odds increase from
1 — that is, they change by a factor of
2, so we’ll say that choosing GG gives us a probability gain (
2. Likewise, SG gives us a
1, and SS a
0. Note that the probability of choosing a given box (
1/3) doesn’t exceed
1/β for that box. This is an example of the following universal fact of probability:
E1 updates the probability of event
E2 by a factor of
β, then the probability of
E1 is at most
P(E2|E1)/P(E2), the above statement says that
P(E1) ≤ P(E2)/P(E2|E1). This is very simple to derive, starting with the following truism:
P(E1 & E2) ≤ P(E2)
P(E2|E1)*P(E1) ≤ P(E2)
and dividing both sides by
P(E1) ≤ P(E2)/P(E2|E1)
This says that a large gain in probability is obtained at the “cost” of an unlikely event. Dembski and Marks’ LCI is nothing more than an application of this fact to cases in which
P(E2) are based on uniform distributions (which I’ll show in a comment*). So not only is their LCI trivially derivable, but it’s also based on two assumptions of uniform probability, a.k.a. tornado-in-a-junkyard assumptions.
Contrast the simplicity of this concept with the import of the authors’ claims that “the Law of Conservation of Information shows that Darwinian evolution is inherently teleological” and “LCI underwrites the conclusion that Darwinian evolution is teleologically programmed with active information”. These are claims that can be made to appear plausible only through copious amounts of obfuscation and equivocation, which is exactly what you’ll find if you read the paper.
* It appears that subscripts don’t work in comments, so I’ll add this as a footnote to the OP. In case anyone from the Evo Info Lab ever reads this and has some doubts, the following shows that Dembski and Marks’ LCI is an application of
P(E1) ≤ P(E2)/P(E2|E1) (Eq. 1)
1) The probability distribution over
Ω1 is uniform.
2) The unconditional (i.e. prior to
~E1 being realized) probability distribution over
Ω2 is also uniform.
(I should note that Dembski and Marks never actually mention the second assumption, but it holds for all of their examples, and without it their LCI is false. It was Atom, another member of Evo Info Lab, who pointed out this tacit assumption to me.)
If we define
E1 ⊆ Ω1 as the set of all outcomes that confer a probability of at least
E2 ⊆ Ω2, then it follows that
q ≤ P(E2|E1). Furthermore, we define
p2 as the probabilities conferred on
E2 by uniform distributions over
Ω2 respectively. Then substitution into Eq. 1 is straightforward, giving:
p1 ≤ p2/q
which is precisely Dembski and Marks’ LCI.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100781.60/warc/CC-MAIN-20231209004202-20231209034202-00453.warc.gz
|
CC-MAIN-2023-50
| 3,872 | 43 |
https://www.arxiv-vanity.com/papers/1207.1994/
|
math
|
Nijenhuis and compatible tensors on Lie and Courant algebroids
We show that well known structures on Lie algebroids can be viewed as Nijenhuis tensors or pairs of compatible tensors on Courant algebroids. We study compatibility and construct hierarchies of these structures.
Pairs of tensor fields on manifolds, which are compatible in a certain sense, were studied by Magri and Morosi , in view of their application to integrable hamiltonian systems. Besides Poisson-Nijenhuis manifolds – manifolds equipped with a Poisson bivector and a Nijenhuis -tensor which are compatible in such a way that it is possible to define a hierarchy of Poisson-Nijenhuis structures on these manifolds, the work of Magri and Morosi also covers the study of and structures. These are pairs of tensors formed respectively, by a closed -form and a Nijenhuis tensor () and a Poisson bivector and a closed -form () satisfying suitable compatibility conditions. Another type of structure that can be considered on a manifold is a Hitchin pair. It is a pair formed by a symplectic form and a -tensor that was introduced by Crainic in relation with generalized complex geometry. All these structures, defined by pairs of tensors, were studied in the Lie algebroid setting by Kosmann-Schwarzbach and Rubtsov and by one of the authors . Finally, we mention complementary forms on Lie algebroids, which were defined by Vaisman and also considered in and , and that can be viewed as Poisson structures on the dual Lie algebroid.
The aim of the present paper is to show that all the structures referred to above, although they have different nature on Lie algebroids, once carried over to Courant algebroids, are all of the same type: they are Nijenhuis tensors. In this way, we obtain a unified theory of Nijenhuis structures on Courant algebroids. In order to include Poisson quasi-Nijenhuis structures with background in this unified theory, we consider a stronger version of this notion, which we call exact Poisson quasi-Nijenhuis structure with background. This seems to be the natural definition, at least in this context.
We show that the structures defined by pairs of tensors on a Lie algebroid can also be characterized using the notion of compatible pair of tensors on a Courant algebroid, introduced in .
An important tool in this work is the Nijenhuis concomitant of two -tensors on a Courant algebroid. It was originally defined for manifolds by Nijenhuis in and then extended to the Courant algebroid framework in and in . We use the Nijenhuis concomitant to study the compatibility of structures from the usual point of view, i.e., we say that two structures of the same type are compatible if their sum is still a structure of the same type. Thus, we can talk about compatible Poisson-Nijenhuis, and structures, as well as compatible complementary forms and compatible Hitchin pairs.
The extension to Lie algebroids of the Magri-Morosi hierarchies of Poisson-Nijenhuis structures on manifolds, was done in . As it happens in the case of manifolds, the hierarchies on Lie algebroids are constructed through deformations by Nijenhuis tensors. In this paper we construct similar hierarchies of and structures on Lie algebroids, and their deformations, and also hierarchies of complementary forms. Elements of these hierarchies provide examples of compatible structures in the sense described above.
Our computations widely use the big bracket – the Poisson bracket induced by the symplectic structure on the cotangent bundle of a supermanifold. The Courant algebroids that we shall consider in this paper are doubles of protobialgebroids structures on , in the simpler cases where is a function that determines a Lie algebroid structure on , or on , sometimes in the presence of a background (a closed -form on ).
The paper is organized as follows. Section 1 contains a short review of Courant and Lie algebroids in the supergeometric framework while, in section 2, we recall the notion of Nijenhuis tensor on a Courant algebroid and of Nijenhuis concomitant of two tensors. In section 3, we characterize Poisson bivectors and closed -forms on a Lie algebroid as Nijenhuis tensors on the Courant algebroid . In section 4, we show how Poisson-Nijenhuis, and structures and also Hitchin pairs on a Lie algebroid can be seen either as Nijenhuis tensors or compatible pairs of tensors on the Courant algebroid . Considering, in section 5, the Courant algebroid with background , we see exact Poisson quasi-Nijenhuis structures with background as Nijenhuis tensors on this Courant algebroid, recovering a result in . For Poisson quasi-Nijenhuis structures (without background) a special case where two -forms involved are exact is also considered. The case of complementary forms is treated in section 6. Section 7 is devoted to the compatibility of structures on a Lie algebroid, defined by pairs of tensors. Sections 8, 9 and 10 treat the problem of defining hierarchies of structures on Lie algebroids. We start by showing, in section 8, that when a pair of tensors defines a certain structure on a Lie algebroid, the same pair of tensors defines a structure of the same kind for a whole hierarchy of deformed Lie algebroids. Then, in section 9, we construct hierarchies of structures defined by pairs of tensors and lastly, in section 10, we show that within one hierarchy, all the elements are pairwise compatible.
We recall that if one relaxes the Jacobi identity in the definition of a Lie (respectively, Courant) algebroid we obtain what is called a pre-Lie (respectively, pre-Courant) algebroid. The proof of our results does not use the Jacobi identity of the bracket, whether if it is a Lie or a Courant algebroid bracket. Therefore, they also hold in the more general settings of pre-Lie and pre-Courant algebroids, respectively.
1. Courant and Lie algebroids in supergeometric terms
We begin this section by introducing the supergeometric setting, following the same approach as in [19, 15]. Given a vector bundle , we denote by the graded manifold obtained by shifting the fibre degree by . The graded manifold is equipped with a canonical symplectic structure which induces a Poisson bracket on its algebra of functions . This Poisson bracket is sometimes called the big bracket (see ).
Let us describe locally the Poisson bracket of the algebra . Fix local coordinates , , in , where are local coordinates on and are their associated moment coordinates. In these local coordinates, the Poisson bracket is given by
while all the remaining brackets vanish.
The Poisson algebra of functions is endowed with a -valued bidegree. We define this bidegree locally but it is well defined globally (see [19, 15] for more details). The bidegrees are locally set as follows: the coordinates on the base manifold , , , have bidegree , while the coordinates on the fibres, , , have bidegree and their associated moment coordinates, and , have bidegrees and , respectively. The algebra of functions inherits this bidegree and we set
where is the -module of functions of bidegree . The total degree of a function is equal to and the subset of functions of total degree is noted . We can verify that the big bracket has bidegree , i.e.,
and consequently, its total degree is . Thus, the big bracket on functions of lowest degrees, and , vanish. For , is an element of and is given by
where is the canonical fiberwise symmetric bilinear form on .
Let us recall that a Courant structure on a vector bundle equipped with a fibrewise non-degenerate symmetric bilinear form is a pair , where the anchor is a bundle map from to and the Dorfman bracket is a -bilinear (not necessarily skew-symmetric) map on satisfying
for all and .
In this paper we are only interested in exact Courant algebroids. Although many of the properties and results we recall next hold in the general case, we shall consider the case where the vector bundle is the Whitney sum of a vector bundle and its dual, i.e., , and is the canonical fiberwise symmetric bilinear form. So, from now on, all the Courant structures will be defined on .
From we know that there is a one-to-one correspondence between Courant structures on and functions such that . The anchor and Dorfman bracket associated to a given are defined, for all and , by the derived bracket expressions
For simplicity, we shall denote a Courant algebroid by the pair instead of the triple .
A Courant structure can be decomposed using the bidegrees:
with and . We recall from that, when , is a Courant structure on if and only if is a Lie algebroid. The anchor and the bracket of the Lie algebroid are defined, respectively by
for all and , while the Lie algebroid differential is given by
2. Nijenhuis concomitant of two tensors
Let be a Courant algebroid and a vector bundle endomorphism of , . If for all , is said to be skew-symmetric. Vector bundle endomorphisms of will be seen as -tensors on .
The deformation of the Dorfman bracket by a -tensor is the bracket defined, for all , by
When is skew-symmetric, the deformed structure is given, in supergeometric terms, by . The deformation of by the skew-symmetric -tensor is denoted by , i.e., while the deformed Dorfman bracket associated to is denoted by .
Recall that a vector bundle endomorphism is a Nijenhuis tensor on the Courant algebroid if its torsion vanishes. The torsion is defined, for all , by
or, equivalently, by
where . When , for some , (5) is given, in supergeometric terms, by
The notion of Nijenhuis concomitant of two tensor fields of type on a manifold was introduced in . In the case of -tensors and on a Courant algebroid , the Nijenhuis concomitant of and is the map (in general not a tensor) defined, for all sections and of , as follows:
where is the Dorfman bracket corresponding to . Equivalently,
while if and anti-commute, i.e., , then
For any -tensors and on , we have
The concomitant of two skew-symmetric -tensors and on a Courant algebroid is given by :
In other words,
for all .
The notion of Nijenhuis concomitant of two -tensors on a Lie algebroid can also be considered. If is a Lie algebroid and are -tensors on , is given by (7), adapted in the obvious way. Equations (8), (9), (10) and (14) also hold in the Lie algebroid case.
As in the case of Courant algebroids, for a Lie algebroid , we use the following notation: , if is either a bivector, a -form or a -tensor on .
3. Tensors on Lie algebroids
Let be a Lie algebroid and consider a -tensor , a bivector and a -form on . Associated with , , , and , we consider the skew-symmetric -tensors on , , , and given, in matrix form, respectively by
In all the computations using the big bracket, instead of writing , , and , we simply write , , and . We use the -tensors on above to express the properties of being Nijenhuis, Poisson and closed on the Lie algebroid .
Proposition 3.1 ().
Let be a -tensor on such that , for some . Then, is a Nijenhuis tensor on the Lie algebroid if and only if is a Nijenhuis tensor on the Courant algebroid .
The assumption is equivalent to . In this case, the torsion of on is given by (6), with , and coincides with the torsion of on . ∎
Let be the -tensor on , defined by
The 2-form is closed on if and only if is a Nijenhuis tensor on the Courant algebroid .
Recall that a bivector field on is a Poisson tensor on if or, equivalently, .
Proposition 3.3 ().
The bivector is a Poisson tensor on if and only if is a Nijenhuis tensor on the Courant algebroid .
We have and, from (6), we get . ∎
Notice that the -tensors and anti-commute. Thus, from (14), we have
Denoting by the -tensor on defined by
and taking into account the fact that
Proposition 3.3 admits the following equivalent formulation:
The bivector is a Poisson tensor on if and only if is a Nijenhuis tensor on the Courant algebroid .
4. Pairs of tensors on Lie algebroids
In we introduced a notion of compatibility for a pair of anti-commuting skew-symmetric -tensors on a Courant algebroid.
Definition 4.1 ().
A pair of skew-symmetric -tensors on a Courant algebroid with Courant structure is said to be a compatible pair, if and anti-commute and .
In this section we show that well known structures defined by pairs of tensors on a Lie algebroid , can be seen either as compatible pairs, or as Nijenhuis tensors on the Courant algebroid .
Let be a Lie algebroid. Recall that a pair , where is a bivector and is a -tensor on is a Poisson-Nijenhuis structure ( structure, for short) on if
A pair formed by a -form and a -tensor on is an structure on if
where or, equivalently, .
A pair formed by a -form and a -tensor on is a Hitchin pair on if
A pair formed by a bivector and a -form on is a structure on if
where is the -tensor on defined by .
Let us denote by the anti-commutator of two skew-symmetric tensors and , i.e.,
Let be a Lie algebroid, a Nijenhuis -tensor and a closed -form on . Then, the pair is an structure on if and only if is a compatible pair on .
We start by noticing that , so that and anti-commute if and only if .
Taking into account the fact that is closed, we have
where in the last equality we used . Thus, the -form is closed if and only if . ∎
In the case where , for some , we have the following characterization of an structure.
Let be a Lie algebroid, a closed -form on and a -tensor on such that , for some . Then, the pair is an structure on if and only if is a Nijenhuis tensor on and .
and, by counting the bi-degrees, we have that is equivalent to
For Hitchin pairs we obtain the following result:
Let be a Lie algebroid, a -tensor on and a symplectic form on . Then, the pair is a Hitchin pair on if and only if is a compatible pair on .
In the case of PN structures, we have:
Let be a Lie algebroid, a Nijenhuis -tensor on and a Poisson bivector on . Then, the pair is a Poisson-Nijenhuis structure on if and only if is a compatible pair on .
Notice that , so that and anti-commute if and only if . Also, we have . ∎
When , for some , we recover a result from , which is a characterization of Poisson-Nijenhuis structures.
Let be a Lie algebroid, a bivector on and a -tensor on such that , for some . Then, the pair is a Poisson-Nijenhuis structure on if and only if is a Nijenhuis tensor on and .
In we showed that, given a bivector and a -tensor on such that , for some , then is a PN structure on if and only if is a Poisson-Nijenhuis pair on the Courant algebroid .
For structures, we have the following:
Let be a Lie algebroid, a Poisson bivector on and a closed -form on . Consider the -tensor on defined by , and the corresponding -tensor on , . Then, the pair is a structure on if and only if is a compatible pair on .
It is easy to see that and anti-commute. The -form being closed we have, taking into account the fact that ,
So, the -form is closed if and only if . ∎
5. Exact Poisson quasi-Nijenhuis structures (with background)
Let be a Lie algebroid, a closed -form on and consider the Courant algebroid with background .
Poisson quasi-Nijenhuis structures with background on Lie algebroids were introduced in . We recall that a Poisson quasi-Nijenhuis structure with background on is a quadruple , where is a bivector, is a -tensor and and are closed -forms such that and
, for all ,
, for all ,
with , for all , where means sum after circular permutation on , and .
A Poisson quasi-Nijenhuis structure with background is called exact if , and condition (iv) is replaced by
is proportional to ,
where , for all .
In it is proved 222The quadruple considered in is and should be . that if is a Nijenhuis tensor on and satisfies , with , then the quadruple is a Poisson quasi-Nijenhuis structure with background on . It is easy to see that the same result holds for any . It is worth noticing that , , is equivalent to the three conditions: , and . Using the notion of exact Poisson quasi-Nijenhuis structure with background, we deduce the following (see the proof of Theorem 2.5 in ):
Let be a Lie algebroid, a bivector, a -form, a closed -form and a -tensor on such that , and is proportional to . Then, is a Nijenhuis tensor on the Courant algebroid if and only if the quadruple is an exact Poisson quasi-Nijenhuis structure with background on .
Notice that in Theorem 5.1, if , for some , then the constant of proportionality that should be considered in (iv’) is , i.e., .
A Poisson quasi-Nijenhuis structure on a Lie algebroid is a Poisson quasi-Nijenhuis structure with background, with . This notion was introduced, on manifolds, in and then extended to Lie algebroids in . An exact Poisson quasi-Nijenhuis structure with background is called an exact Poisson quasi-Nijenhuis structure. In this case, the -form is also exact.
Next, we consider and a special case where the assumption , , in Theorem 5.1 is replaced by , .
Let be a Lie algebroid, a bivector, a -form and a -tensor on such that , and , for some . If is a Nijenhuis tensor on the Courant algebroid , then the triple is an exact Poisson quasi-Nijenhuis structure on .
and, by counting the bi-degrees, we obtain that if and only if
Applying both members of iii) to any , we get
which gives, using (7),
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655890092.28/warc/CC-MAIN-20200706011013-20200706041013-00293.warc.gz
|
CC-MAIN-2020-29
| 17,110 | 108 |
https://mailman.ntg.nl/pipermail/ntg-pdftex/2008-December/003709.html
|
math
|
[NTG-pdftex] Extensions to \font syntax ?
Philip TAYLOR (Ret'd)
P.Taylor at Rhul.Ac.Uk
Sat Dec 6 17:20:21 CET 2008
Dear Thanh / PdfTeX team --
Has any consideration ever been given to the idea
of extending the syntax of TeX's \font command
within the context (small "c") of PdfTeX ?
What I have in mind are the following :
\font \cmex = cmex10 rotated <angle>
\font \cmex = cmex10 translated <matrix>
I ask because I am currently typesetting a letter
requiring characters from a number of different
languages, and I can successfully synthesise 99%
of the characters I need from the basic CM family.
However, I am stuck at o-horn and u-horn (Vietnamese).
I can get reasonably close by scaling and positioning
"3B or "7D from CMEX<whatever>, but I would like to get
even closer, and being able to rotate the font (and
perhaps transform it) would take me from "reasonably
close" to "d at mn nigh perfect" :-)
More information about the ntg-pdftex
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00088.warc.gz
|
CC-MAIN-2022-21
| 943 | 22 |
http://luminouslogic.com/statistical-significance-and-the-magic-formula.htm
|
math
|
Note: Please read the disclaimer. The author is not providing professional investing advice or recommendations.
But one thing that bothered me when reading Greenblatt’s book was my memory of the Foolish Four. It was a similar market-trouncing “magic formula” that gained popularity in the late 90’s, only to later be discredited and deemed an artifact of data-mining.
My understanding of how the wind was taken out of the Foolish Four’s sails was by investigating its performance over a longer time period. For example, it’s yearly excess return over the Dow goes from 10% to something closer to 2% when back-tested over 50 years instead of the original 20 years. So of course that becomes an undeniable hintergedanke when seeing the Magic Formula’s measly 17-year sample size. 🙁
And for those hung up on the fact that even a 2% alpha is respectable for the Foolish Four, it shrinks even further when you factor in the capital gains incurred by having to shuffle the deck, as it were, each year.
Learning quantitative techniques to analyze problems such as these (i.e. is 17 years worth of data enough to conclude that the Magic Formula outperforms the S&P 500?) is exactly why I enrolled in the CFA program. And after getting a couple hundred pages into the first study guide at Level One, they cover precisely this sort of conundrum.
So here is how it works. We do what is called a hypothesis test in statistics, whereby we hypothesize that the average return of the Magic Formula and the S&P 500 are actually identical! And then we do a few computations based on confidence intervals to see whether we can reject that hypothesis or not, and with what degree of confidence. As a big fan of the Magic Formula, I have to say I’m secretly hoping that it does have a statistically significant higher average yearly return than the S&P 500…
First, Some Assumptions…
Now in order to proceed, we have to make two assumptions. The first is that the yearly returns of these two investing techniques follow a mostly normal distribution. We’re supposed to feel comfortable making this assumption due to the central limit theorem. Briefly, it says that the distribution of any variable that is a function of a bunch of other random variables always tends to end up looking like a bell curve.
The second assumption we’ll make is that the returns of the Magic Formula and S&P 500 are not independent. The CFA study guide advises that when comparing two investing strategies covering the same period of time, the returns of both depend on the same underlying market and economic forces present at that time, and therefore have some things in common.
Step #1: Sample Mean Difference
The first step is to compute the average difference of the yearly returns between the two strategies. This turns out to be 19.04%.
Step #2: Standard Error of the Mean Difference
Next we compute the sample standard deviation of the difference (in Excel, use stdev). This comes out to be 22.24%. We transform this into standard error of the mean difference (SEMD) by dividing by the square root of the sample size. The years 1988-2004 comprise 17 years.
Step #3: Compute Test Statistic
For normal, or mostly normal distributions and small sample sizes we use the t-test to check for statistical significance. Basically this just gives us a number that we can compare to a t-test table in order to determine whether the difference we’re seeing between the Magic Formula and S&P 500 appears to be important given the sample size. The smaller the sample size, the greater the t-test hurdle our data will have to clear in order to be able to conclude that the two don’t have the same mean.
The test statistic is simply the sample mean difference minus the hypothesized mean difference, divided by the SEMD.
Step #4: Pick a Significance Level and Compare
Finally we need to decide on a level of significance and do our table compare. Most of these sorts of tests in the CFA curriculum seem to use 5%. This means that in our comparison, there will only be a 5% chance that there is in reality no difference between the Magic Formula and S&P 500, but we fail to detect this.
In addition to level of significance, the only other parameter we require to do our table look-up is the degrees of freedom. But that’s easy as it’s simply the sample size minus 1.
Given these two parameters we could now find a t-test table to do a critical value look-up, but this can be a little tedious, not to mention the additional step of converting one-sided to two-sided. I prefer to just use Excel’s tinv function.
And since our t-value from Step #3 was 3.526, which is much > 2.120 we can easily reject the hypothesis that the means are equal at the 5% level of significance. Not only that, we can assume the means are different by as much as 7.59% before we start running up against our critical value of 2.120.
So the Magic Formula’s outperformance appears not to just be an artifact of small sample size, and also appears to be of significant magnitude.
But What About Risk?
Things are truly looking rosy for the Magic Formula. But a seasoned finance student will also compare the standard deviations of yearly returns in our first table up top and notice that the Magic Formula’s is higher (24.26% versus 17.87%). Standard deviation is a common (though debatable) quantifier for risk so it wouldn’t be uncommon to argue that the Magic Formula should have higher returns to compensate the investor for taking more risk.
Well just as we tested the hypothesis that the means were equal, we can do the same with variance (square of standard deviation)…
Step #5: Test Equality of Two Variances
I’ll cut to the chase here just to say that there’s a simple equivalent of the t-test when testing for the equality of two variances, and it’s called the F-test.
We come up with our F parameter by just computing the ratio of the two variances.
And again we could do a manual look-up using an F-table, but why not just let Excel compute it for us with finv….
Therefore at the 5% significance level, because 1.843 < 2.333 we cannot reject the hypothesis that the variances are the same.
In conclusion, past performance may be no indication of future results… blah blah blah… The important thing is that the Magic Formula points toward having the best of both worlds, a statistically significant higher annual rate of return versus the S&P 500 without a statistically significant higher level of risk. Win-win!
It is interesting to see how confidence intervals allow us to tread into gray areas. A newbie might stop at Step #1 and claim that the Magic Formula beats the S&P 500 by an average of 19.04% per year. His antagonist might point to the small sample size and say that it makes that 19.04% estimate of mean… meaningless! But a statistician can state that he’s 95% sure that the Magic Formula outperforms the S&P 500 by at least 7.59% per year…. assuming normality.
And much is indeed hinging upon our assumption of normality, which can and should be tested for. But that’s for another day…
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00674.warc.gz
|
CC-MAIN-2020-45
| 7,099 | 33 |
http://e-booksdirectory.com/details.php?ebook=6935
|
math
|
Algebraic and Geometric Surgery
by Andrew Ranicki
Publisher: Oxford University Press 2002
Number of pages: 380
This book is an introduction to surgery theory: the standard classification method for high-dimensional manifolds. It is aimed at graduate students, who have already had a basic topology course, and would now like to understand the topology of high-dimensional manifolds.
Home page url
Download or read it online for free here:
by Daniel Dugger - University of Oregon
This is an expository paper on homotopy colimits and homotopy limits. These are constructions which should arguably be in the toolkit of every modern algebraic topologist. Many proofs are avoided, or perhaps just sketched.
by O. Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, V. M. Kharlamov - American Mathematical Society
This textbook on elementary topology contains a detailed introduction to general topology and an introduction to algebraic topology via its most classical and elementary segment centered at the notions of fundamental group and covering space.
by J. P. May - Springer
The theme of this book is infinite loop space theory and its multiplicative elaboration. The main goal is a complete analysis of the relationship between the classifying spaces of geometric topology and the infinite loop spaces of algebraic K-theory.
by Danny Calegari - Mathematical Society of Japan
This is a comprehensive introduction to the theory of stable commutator length, an important subfield of quantitative topology, with substantial connections to 2-manifolds, dynamics, geometric group theory, bounded cohomology, symplectic topology.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506559.11/warc/CC-MAIN-20230924023050-20230924053050-00599.warc.gz
|
CC-MAIN-2023-40
| 1,614 | 15 |
https://chem.libretexts.org/Courses/University_of_Alabama/Chem_101%3A_Rupar/Homework_problems_under_construction/Homework_63
|
math
|
Assigned Questions: 10.34,5.75
Observe the following exchanges of energy. Identify whether each exchange is an example of heat or work. For each system, determine whether ΔE is positive or negative.
a) A cake is baked in an oven (The cake is the system)
b) Wood is burned for a fire (The wood is the system)
c) A boy kicks a chair, it is displaced in the direction of force applied. (The chair is the system)
a) The system is an example of heat, and because the cake gains heat, ΔE is positive.
b) The system is an example of heat, and because wood gives off heat, ΔE is negative.
c) The system is an example of work, and because work is done on the chair, ΔE is positive.
Calculate the number of moles in each of the following examples.
a) 402.5 mg of NO2
b) 2.7 kg of H2O
c) 323 g of CBr4
d) 2.9 kg of CaO
First, calculate the molar mass of NO2.
14.007g/mol N + 2 (15.999 g/mol O) = 46.0055 g/mol
Now, convert the sample from mg to g.
|402.5 ||1 g||0.4025 g NO2|
Finally, use the molar mass to determine how many moles your sample size contains.
|.4025 g NO2||1 mol||0.0087 moles of NO2|
This gives us the final answer, 0.0087 moles of NO2.
From this, we can deduce that:
Moles of substance = Mass of substance (g) / Molar mass of substance
150 moles of H2O
0.974 moles of CBr4
52 moles of CaO
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103324665.17/warc/CC-MAIN-20220627012807-20220627042807-00092.warc.gz
|
CC-MAIN-2022-27
| 1,299 | 25 |
https://dralexheard.com/_profiler/phpinfo
|
math
|
Math textbook answers app
Math textbook answers app can support pupils to understand the material and improve their grades. Keep reading to learn more!
The Best Math textbook answers app
Math textbook answers app can be found online or in math books. Math can be a difficult subject for many people. understand. That's where a math solver website can come in handy. These websites can provide you with the answers to Math problems, as well as step-by-step solutions to show you how they solved the problem. This can be a valuable resource when you're stuck on a Math problem and can't seem to solve it on your own. In addition, many Math solver websites also offer forums where you can ask Math questions and get help from other users. So if you're struggling with Math, don't hesitate to use a math solver website to get the help you need.
In this case, we are looking for the distance travelled by the second train when it overtakes the first. We can rearrange the formula to solve for T: T = D/R. We know that the second train is travelling at 70 mph, so R = 70. We also know that the distance between the two trains when they meet will be the same as the distance travelled by the first train in one hour, which we can calculate by multiplying 60 by 1 hour (60 x 1 = 60). So, plugging these values into our equation gives us: T = 60/70. This simplifies to 0.857 hours, or 51.4 minutes. So, after 51 minutes of travel, the second train will overtake the first.
In addition, the built-in practice exercises further reinforce the lesson material. As a result, Think Through Math is an extremely effective way for students to learn and improve their math skills. Best of all, the app is available for free, so there's no excuse not to give it a try!
Solving integral equations is a way of finding a function that satisfies a certain equation. In other words, it involves finding a function that "integrates" to a given value. This can be done by using a variety of methods, including integration by parts, integration by substitution, and integration by partial fractions. Each method has its own strengths and weaknesses, and the best method to use will depend on the specific equation that needs to be solved. However, no matter which method is used, solving integral equations can be a challenging task. Fortunately, there are many resources available to help with this process. With a little patience and perseverance, anyone can learn how to solve integral equations.
We will help you with math problems
The walk-throughs on the free version will definitely help you out, but the explanations on the app Plus are extremely helpful. I highly recommend subscribing if you need detailed explanations. They explain very well, with references to definitions, tables, and even more in-depth steps.
This is a really helpful app to use, when you study and don't quite understand why or how. I can see people using this app to get over homework or to do homework in a shorter period of time. But that's where self-discipline comes in. This app is really helpful and great for studying before a test or homework when you don't understand something. Thanks a lot, to the great developers.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500837.65/warc/CC-MAIN-20230208155417-20230208185417-00685.warc.gz
|
CC-MAIN-2023-06
| 3,182 | 10 |
https://en.wikipedia.org/wiki/Diagnostic_odds_ratio
|
math
|
Diagnostic odds ratio
In medical testing with binary classification, the diagnostic odds ratio (DOR) is a measure of the effectiveness of a diagnostic test. It is defined as the ratio of the odds of the test being positive if the subject has a disease relative to the odds of the test being positive if the subject does not have the disease.
The rationale for the diagnostic odds ratio is that it is a single indicator of test performance (like accuracy and Youden's J statistic) but which is independent of prevalence (unlike accuracy) and is presented as an odds ratio, which is familiar to medical practitioners.
The diagnostic odds ratio is defined mathematically as:
where , , and are the number of true positives, false negatives, false positives and true negatives respectively.
As with the odds ratio, the logarithm of the diagnostic odds ratio is approximately normally distributed.[clarification needed] The standard error of the log diagnostic odds ratio is approximately:
From this an approximate 95% confidence interval can be calculated for the log diagnostic odds ratio:
Exponentiation of the approximate confidence interval for the log diagnostic odds ratio gives the approximate confidence interval for the diagnostic odds ratio.
The diagnostic odds ratio ranges from zero to infinity, although for useful tests it is greater than one, and higher diagnostic odds ratios are indicative of better test performance. Diagnostic odds ratios less than one indicate that the test can be improved by simply inverting the outcome of the test – the test is in the wrong direction, while a diagnostic odds ratio of exactly one means that the test is equally likely to predict a positive outcome whatever the true condition – the test gives no information.
Relation to other measures of diagnostic test accuracy
The diagnostic odds ratio may be expressed in terms of the sensitivity and specificity of the test:
It may also be expressed in terms of the Positive predictive value (PPV) and Negative predictive value (NPV):
It is also related to the likelihood ratios, and :
The log diagnostic odds ratio is sometimes used in meta-analyses of diagnostic test accuracy studies due to its simplicity (being approximately normally distributed).
Traditional meta-analytic techniques such as inverse-variance weighting can be used to combine log diagnostic odds ratios computed from a number of data sources to produce an overall diagnostic odds ratio for the test in question.
The log diagnostic odds ratio can also be used to study the trade-off between sensitivity and specificity by expressing the log diagnostic odds ratio in terms of the logit of the true positive rate (sensitivity) and false positive rate (1 − specificity), and by additionally constructing a measure, :
It is then possible to fit a straight line, . If b ≠ 0 then there is a trend in diagnostic performance with threshold beyond the simple trade-off of sensitivity and specificity. The value a can be used to plot a summary ROC (SROC) curve.
Consider a test with the following 2×2 confusion matrix:
by “Gold standard”)
We calculate the diagnostic odds ratio as:
This diagnostic odds ratio is greater than one, so we know that the test is discriminating correctly. We compute the confidence interval for the diagnostic odds ratio of this test as [9, 134].
The diagnostic odds ratio is undefined when the number of false negatives or false positives is zero – if both false negatives and false positives are zero, then the test is perfect, but if only one is, this ratio does not give a usable measure. The typical response to such a scenario is to add 0.5 to all cells in the contingency table, although this should not be seen as a correction as it introduces a bias to results. It is suggested that the adjustment is made to all contingency tables, even if there are no cells with zero entries.
- Sensitivity and specificity
- Binary classification
- Positive predictive value and negative predictive value
- Odds ratio
- ^ a b c d e f g h Glas, Afina S.; Lijmer, Jeroen G.; Prins, Martin H.; Bonsel, Gouke J.; Bossuyt, Patrick M.M. (2003). "The diagnostic odds ratio: a single indicator of test performance". Journal of Clinical Epidemiology. 56 (11): 1129–1135. doi:10.1016/S0895-4356(03)00177-X. PMID 14615004.
- ^ Macaskill, Petra; Gatsonis, Constantine; Deeks, Jonathan; Harbord, Roger; Takwoingi, Yemisi (23 December 2010). "Chapter 10: Analysing and presenting results". In Deeks, J.J.; Bossuyt, P.M.; Gatsonis, C. (eds.). Cochrane Handbook for Systematic Reviews of Diagnostic Test Accuracy (PDF) (1.0 ed.). The Cochrane Collaboration.
- ^ Glas, Afina S.; Lijmer, Jeroen G.; Prins, Martin H.; Bonsel, Gouke J.; Bossuyt, Patrick M.M. (November 2003). "The diagnostic odds ratio: a single indicator of test performance". Journal of Clinical Epidemiology. 56 (11): 1129–1135. doi:10.1016/S0895-4356(03)00177-X. PMID 14615004.
- ^ Gatsonis, C; Paliwal, P (2006). "Meta-analysis of diagnostic and screening test accuracy evaluations: Methodologic primer". AJR. American Journal of Roentgenology. 187 (2): 271–81. doi:10.2214/AJR.06.0226. PMID 16861527.
- ^ a b c d Moses, L. E.; Shapiro, D; Littenberg, B (1993). "Combining independent studies of a diagnostic test into a summary ROC curve: Data-analytic approaches and some additional considerations". Statistics in Medicine. 12 (14): 1293–316. doi:10.1002/sim.4780121403. PMID 8210827.
- ^ a b Dinnes, J; Deeks, J; Kunst, H; Gibson, A; Cummins, E; Waugh, N; Drobniewski, F; Lalvani, A (2007). "A systematic review of rapid diagnostic tests for the detection of tuberculosis infection". Health Technology Assessment. 11 (3): 1–196. doi:10.3310/hta11030. PMID 17266837.
- ^ Cox, D.R. (1970). The analysis of binary data. London: Methuen. ISBN 9780416104004.
- Glas, Afina S.; Lijmer, Jeroen G.; Prins, Martin H.; Bonsel, Gouke J.; Bossuyt, Patrick M.M. (2003). "The diagnostic odds ratio: a single indicator of test performance". Journal of Clinical Epidemiology. 56 (11): 1129–1135. doi:10.1016/S0895-4356(03)00177-X. PMID 14615004.
- Böhning, Dankmar; Holling, Heinz; Patilea, Valentin (2010). "A limitation of the diagnostic-odds ratio in determining an optimal cut-off value for a continuous diagnostic test". Statistical Methods in Medical Research. 20 (5): 541–550. doi:10.1177/0962280210374532. PMID 20639268. S2CID 21221535.
- Chicco, Davide; Starovoitov, Valery; Jurman, Giuseppe (2021). "The benefits of the Matthews correlation coefficient (MCC) over the diagnostic odds ratio (DOR) in binary classification assessment". IEEE Access. 9: 47112–47124. doi:10.1109/ACCESS.2021.3068614.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224645810.57/warc/CC-MAIN-20230530131531-20230530161531-00448.warc.gz
|
CC-MAIN-2023-23
| 6,652 | 36 |
https://www-s.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2013-2014/3710.html
|
math
|
Date: Thu Dec 18 2014 - 01:43:11 CST
Interesting. The last paragraph and the attached file seem to be missing from my previous mail. I just paste these here.
I have paste a script below, I have used to start a series of MD simulations. I have also included a namd config file to show how I passed environment variables down to the jobs. Something that I always found to be very cumbersome on the BlueGenes.
# submit jobs on BlueGene via LoadLeveler
tasks=`expr $nodes \* $threads`
if [ -z "$1" -o -z "$2" ]; then
echo "Usage: $0 start stop"
for step in `seq $start $stop`; do
curr=md`printf "%03i" $step`
prev=`expr $step - 1`
prev=md`printf "%03i" $prev`
if [ $step -ne $start ]; then
dependency="#@ dependency = ($prev == 0)"
cat << _EOF
#@ job_name = $curr
#@ step_name = $curr
#@ job_type = bluegene
#@ bg_size = $nodes
#@ node_usage = shared
#@ wall_clock_limit = 04:00:00,03:58:00
#@ output = $mycwd/$curr.out
#@ error = $mycwd/$curr.err
#@ notification = error
#@ arguments = -np $tasks -mode VN -cwd $mycwd -env "NAMD_PREV=$prev NAMD_CU
RR=$curr" -exe $binary -args $input
======= config ======
# set file names
set previous $env(NAMD_PREV)
set current $env(NAMD_CURR)
From: owner-namd-l_at_ks.uiuc.edu [owner-namd-l_at_ks.uiuc.edu] on behalf of hannes.loeffler_at_stfc.ac.uk [hannes.loeffler_at_stfc.ac.uk]
Sent: 18 December 2014 07:36
Subject: RE: namd-l: setting up namd simulation in IBM Blue Gene/L
We have retired our BlueGenes L and P a while ago so I can't help you much.
What I remember is that our LoadLeveler on the L never had scripting capabilities so your final mpirun would have simply be ignored. The user was supposed to use #@arguments and possibly #@executable. There was a default for the latter. In any case, make sure that you are calling the right executable here if you have to do it explicitly (may not even be mpirun on the L)
This archive was generated by hypermail 2.1.6 : Wed Dec 31 2014 - 23:23:08 CST
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00220.warc.gz
|
CC-MAIN-2023-23
| 1,939 | 35 |
https://www.physicsforums.com/threads/formula-to-figure-out-displacement.12358/
|
math
|
hi i was just wondering what is the formula to figure out displacement
well the displacement is the difference between your starting point and ending point so their is no set formula for figuring it out. It depends on the problem. For example if you had a taxi car driver heading south for 10meters and then he makes a right turn and heads east for 20meters and you have to find the displacement of his trip you would find the hypotenuse of the right triange. And in a different problem.. if you have like the velocity and the time and you need to find the displacement of the trip you could plug it into the velocity=displacement/change in time equation and get it that way through algebra.. but there are many different ways.. sorry im rambling.. if you have a specific question it may help more
Since you ask for a formula, this is probably what you want:
"average velocity" is defined as displacment/time so v= d/t and, therefore, d= v*t.
Separate names with a comma.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00612.warc.gz
|
CC-MAIN-2018-51
| 971 | 5 |
https://bitylink.info/ldu-factorization-55/
|
math
|
Is it possible to switch row 1 and row 2? I am using a shortcut method I found on a YouTube channel, but I am not sure how to do it if I swap the. Defines LDU factorization. Illustrates the technique using Tinney’s method of LDU decomposition. An LDU factorization of a square matrix A is a factorization A = LDU, where L is a unit lower triangular matrix, D is a diagonal matrix, and U is a unit upper.
|Published (Last):||10 January 2013|
|PDF File Size:||8.63 Mb|
|ePub File Size:||7.37 Mb|
|Price:||Free* [*Free Regsitration Required]|
Moreover, it can be seen that. Matrix decompositions Numerical linear algebra. LU decomposition was introduced by mathematician Tadeusz Banachiewicz in General treatment of orderings that minimize fill-in can be addressed using fzctorization theory. Home Questions Tags Users Unanswered. Without a proper ordering or permutations in the matrix, the factorization may fail to materialize.
The LUP decomposition algorithm by Cormen et al.
LU decomposition – Wikipedia
Astronomy and Astrophysics Supplement. It would follow that the result X must be the inverse of A. When an LDU factorization exists and is unique, there is a closed explicit formula for the elements of LDand U in terms of ratios of determinants of certain submatrices of the original matrix A.
Expanding the matrix multiplication gives. In that case, L and D are square matrices both of which have the same number of rows as Aand U has exactly the same dimensions as A. We find the decomposition.
Then the system of equations has the following solution:. In this case any two non-zero elements of L and U matrices are parameters of the solution factirization can be set arbitrarily to any non-zero value.
When solving systems of equations, b is usually treated as a vector with a length equal to the height of matrix A.
It is possible to find a low rank approximation to an LU decomposition using a randomized algorithm. Computers usually solve square systems of linear equations using LU decomposition, and it is also a key step when inverting a matrix or computing the determinant of a matrix. We transform the matrix A into an upper triangular matrix U by eliminating the entries below the main diagonal.
In numerical analysis and linear algebralower—upper LU decomposition or factorization factors a matrix as the product of a lower triangular matrix and an upper triangular matrix. Upper triangular should be interpreted factoriization having only zero entries below the main diagonal, which starts at the upper left corner.
Views Read Edit View history. Now suppose that B is the identity matrix of size n. Above we required that A be a square matrix, but these decompositions can all be generalized to rectangular matrices as well. Email Required, but never shown.
The Cholesky decomposition always exists and is unique — provided the matrix is positive definite.
For example, we can conveniently require the lower triangular matrix L to be a unit triangular matrix i. This is impossible if A is nonsingular invertible. Sign up using Email and Password. The Gaussian elimination algorithm for obtaining LU decomposition has also been extended to this most general case. I am using a shortcut method I found on a YouTube channel, but I am not sure how to do it if I swap the rows.
Computation of the determinants is computationally expensiveso this explicit formula is not used in practice.
The Doolittle algorithm does the elimination column-by-column, starting from the left, by multiplying A to the left with atomic lower triangular matrices.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00366.warc.gz
|
CC-MAIN-2022-05
| 3,565 | 19 |
https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2323520
|
math
|
Pitfalls in Using Weibull Tailed Distributions
Journal of Statistical Planning and Inference, 2010, Volume 140, Issue 7, p.2018- 2024
11 Pages Posted: 11 Sep 2013
Date Written: November 08, 2009
By assuming that the underlying distribution belongs to the domain of attraction of an extreme value distribution, one can extrapolate the data to a far tail region so that a rare event can be predicted. However, when the distribution is in the domain of attraction of a Gumbel distribution, the extrapolation is quite limited generally in comparison with a heavy tailed distribution. In view of this drawback, a Weibull tailed distribution has been studied recently.
Some methods for choosing the sample fraction in estimating the Weibull tail coefficient and some bias reduction estimators have been proposed in the literature. In this paper, we show that the theoretical optimal sample fraction does not exist and a bias reduction estimator does not always produce a smaller mean squared error than a biased estimator. These are different from using a heavy tailed distribution. Further we propose a refined class of Weibull tailed distributions which are more useful in estimating high quantiles and extreme tail probabilities.
Keywords: Asymptotic Mean Squared Error, Extreme Tail Probability, High Quantile, Regular Variation, Weibull Tail Coefficient
JEL Classification: C10, C60
Suggested Citation: Suggested Citation
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144979.91/warc/CC-MAIN-20200220131529-20200220161529-00099.warc.gz
|
CC-MAIN-2020-10
| 1,420 | 9 |
http://answers.reference.com/Digital/Apps/how_to_create_a_balance_sheet
|
math
|
To create a balance sheet, you must first know a single formula: Assets = Liabilities + Net Worth. At its simplest, the balance sheet must contain two columns. On the left are assets like cash and property, on the right are liabilities like rents, loans, and other expenditures. Net worth is under liabilities and equals the amount of assets minus liabilities. The total of the first column must .
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997882928.29/warc/CC-MAIN-20140722025802-00049-ip-10-33-131-23.ec2.internal.warc.gz
|
CC-MAIN-2014-23
| 397 | 1 |
http://dmuw.zum.de/index.php?title=The_Klein_Project/Recommended_Reading&oldid=30883
|
math
|
Add your own favourite book of extension material for teachers and lecturers of mathematics students aged 16 to 19, with a brief description. Follow the style of the entries below.
Felix Klein, Elementary Mathematics from an Advanced Standpoint (Dover 2004), 2 Vols
These are the first two volumes of the three-volume German edition, Elementarmathematik vom höheren Standpunkte aus (J. Springer, Berlin, 1924-1928).
László Lovász, Trends in Mathematics: How they could Change Education? (2006)
H. Behnke, F. Bachmann, K. Fladt (Eds.), Fundamentals of Mathematics (MIT Press, 1974)
Translation from the German of Grundzüge der Mathematik, Vandenhoeck & Ruprecht, Göttingen, 1962. 3 Vols.
The book is the translation of a work commissioned by the ICMI in order to support the scientific foundations of instruction in mathematics, which was one of the topics chosen by the Commission at a meeting in Paris in 1954, in preparation for the International Congress of Mathematicians in Edinburgh in 1958. The English book has three volumes with 42 chapters: the first volume concerns the foundations of Mathematics, the Real Number System and Algebra; the second volume is devoted to geometry and the third to Analysis. There are, in general, two authors for each chapter: one a university researcher, the other a teacher of long experience in the German educational system. And the whole book has been coordinated in repeated conferences, involving altogether about 150 authors and coordinators.
Martin Gardner, Mathematical Puzzles and Diversions (Penguin, 1956)
The first of his series of books derived from his columns in Scientific American from 1956 to 1981 . Categorised as “Recreational Mathematics”, these are accessible to those with secondary school mathematics and a willingness to explore mathematical ideas.
Albert Cuoco, Mathematical Connections: A Companion for Teachers (2005)
This is a resource book for high school teachers. See the review by Steve Maurer.
Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid (Basic Books, 1979)
GEB takes the form of an interweaving of various narratives. The main chapters alternate with dialogues between imaginary characters, inspired by Lewis Carroll's "What the Tortoise Said to Achilles" , in which Achilles and the Tortoise discuss a paradox related to modus ponens. Hofstadter bases the other dialogues on this one introducing characters such as a Crab, a Genie, and others. These narratives frequently dip into self-reference and metafiction.
Word play also features prominently in the work. Puns are occasionally used to connect ideas, such as "the Magnificrab, Indeed" with Bach's Magnificat in D; SHRDLU, Toy of Man's Designing" with Bach's Jesu, Joy of Man's Desiring; and "Typographical Number Theory", which inevitably reacts explosively when it attempts to make statements about itself. One Dialogue contains a story about a genie (from the Arabic "Djinn") and various "tonics" (of both the liquid and musical varieties), which is titled "Djinn and Tonic".
One dialogue in the book is written in the form of a crab canon, in which every line before the midpoint corresponds to an identical line past the midpoint. The conversation still makes sense due to uses of common phrases that can be used as either greetings or farewells ("Good day") and the positioning of lines which, upon close inspection, double as an answer to a question in the next line.
Zalman Usiskin, Dick Stanley, Anthony Peressini, Elena Anne Marchisotto, Mathematics for High School Teachers: An Advanced Perspective (Prentice Hall, 2003)
Philip Davis & Reuben Hersh, The Mathematical Experience (Birkhauser, 1981)
[This book] discusses the practice of modern mathematics from a historical and philosophical perspective. It won the 1983 National Book Award in the Science category.
It is frequently cited by mathematicians as a book that was influential in their decision to continue their studies in graduate school and has been hailed as a classic of mathematical literature. The book drew a critical review from Martin Gardner, who disagreed with some of the authors' philosophical opinions, but was well-received otherwise.
A study edition and also a study guide for use with the book have been released, both co-authored with Elena A. Marchisotto. The authors wrote a follow-up book, Descartes' Dream: The World According to Mathematics, and both have separately written other books on related subjects, such as Davis' Mathematics And Common Sense: A Case of Creative Tension and Hersh's What is Mathematics, Really?
Philip Davis & Reuben Hersh, Descartes' Dream: The World According to Mathematics (Houghton Mifflen, 1986)
Martin Aigner & Günter Ziegler, Proofs from THE BOOK (Springer-Verlag, 1998)
Lynn Arthur Steen (Ed), Mathematics Today: Twelve Informal Essays. (Springer-Verlag, 1978)
L.A. Steen has written many books and articles on [http:// mathematics], from this very early one to a follow-up Mathematics Tomorrow (Springer-Verlag, 1981)
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679516047.98/warc/CC-MAIN-20231211174901-20231211204901-00819.warc.gz
|
CC-MAIN-2023-50
| 5,015 | 24 |
https://writerbay.essayshelp24x7.com/2023/05/09/community-bank-is-planning-to-expand-its-drive-in-facility-observations-of-the-existing-1-answer-below/
|
math
|
Community Bank is planning to expand its drive-in facility. Observations of the existing single-teller window reveal that customers arrive at an average rate of 10 per hour, with a Poisson distribution, and that they are given FCFS service, with an average transaction time of 5 minutes. Transaction times have a negative exponential distribution. Community Bank has decided to add another teller and to install four remote stations with pneumatic tubes running from the stations to the tellers, who are located in a glassed-in building. The cost of keeping a customer waiting in the system is represented as a $5-per-hour loss of goodwill. The hourly cost of a teller is $10.
a. Assume that each teller is assigned two stations exclusively, that demand is divided equally among the stations, and that no customer jockeying is permitted. What is the average number of customers waiting in the entire system?
b. If, instead, both tellers work all the stations and the customer waiting the longest is served by the next available teller, what is the average number of customers in the system?
c. What hourly savings are achieved by pooling the tellers?
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100399.81/warc/CC-MAIN-20231202105028-20231202135028-00861.warc.gz
|
CC-MAIN-2023-50
| 1,150 | 4 |
https://forums.mikeholt.com/threads/single-phase-or-polyphase.73649/page-4
|
math
|
It isn't, but if any reference other than neutral is chosen, you have said, the 2-phase no longer exists. So why must the neutral point be the reference in order to determine the number of phases?
Because of the way the system is defined. Don't get the idea that I am saying that voltages 'disappear'. What I am saying is that the voltage will not fit in the particular system phase-counting bucket. You have to pick a reference in order to define a voltage. There is no universal designation for picking a reference point. Neither is there a universal designation to say which voltage has the angle from which all other angles will be measured.
There are several systems we can define using the transformer terminals. The transformer can supply any of these different systems. If the transformer is capable of supplying more than one order of system, I am saying the transformer is a supply of the highest order.
There are several type loads we can serve. Single-phase loads use only one pressure wave. Two-phase loads use two different pressure waves. Three-phase loads... you get the idea. Whether the voltages to these loads will all fit into a single phase-counting system bucket has to be determined.
Let's use the 240/480 volt transformer with the X1, X23, X4 terminals for example so we can put in some real numbers for some example systems:
System 1: A 240 volt single-phase system
System 2: A 480 volt single-phase system
System 3: A 240 volt two-phase system
Now pick a voltage reference and let's see what possible systems we can have:
X1 (or X4) Reference: We can have two single-phase systems, each with one voltage (one at 240 volts and one at 480 volts). Both voltages are present, but they are classified into two different systems.
X23 Reference: We can have two single-phase systems with one voltage each. One single-phase system is 240@0 and the other single-phase system is 240@180. This system can serve two independent loads from separate sides of the winding.
We can also have one two-phase system with two voltages. This type system serves loads that use both voltages and one voltage is 240@0 and the other voltage is 240@180.
Whether we have two systems with one voltage each or one system with two voltages, both voltages are used.
I was discussing the single current on the primary side of the transformer creating a corresponding current 'direction' in a single center tapped secondary winding.
Un-equal loads can cause the "neutral" to be different from the natural neutral point. With these loads, the neutral is just a grounded conductor. It is forcing the common connection point away from a point of natural equilibrium. Regardless of the direction you pick to be positive, there are times when the current in one side of the coil is positive and the current in the other side of the coil is negative. They are actually flowing in two different directions. Even when in the same direction, the flux can be different because of a magnitude difference.
Looking at the primary current is not representative of what is going on in the secondary because the net current in the primary is going to be based on the net flux of two different secondary currents. By looking at the primary current, the real secondary currents are hidden from you because you can't see the original individual fluxes.
Without the neutral, the currents would stabilize such that the voltage balance point was located inside the load away from the common connection point. In that case, there is no unbalanced current and the current in both sides of the secondary coil would be exactly the same. At that point, you would just have one single-phase load, one flux, etc.
I did not say that a grounded conductor must be counted, only that it may be. It is you who is putting requirements on which conductors must be used. I am looking for a definition that does not change based on which reference is used.
I have given you a method but you haven't been able to see it yet. It is a general definition that can apply to more than one type system. The formula to determine the available system types and which voltages you can put into each system bucket for counting phases does not change.
How you configure the source can restrict the number of available systems. Saying you don't want the neutral to make a change is like saying you are okay with saying we have a 480 volt source, but don't want to recognize that you can also get 240 volts if you use the center-tap. The neutral gives you an option you did not have without it.
There are two pairs of two L-L voltages. L1-L2 and L1'-L2'. L1-L1' and L2-L2' were never considered as valid connections (similar to the high leg in a 240/120 connection). This is why I refer to this as 2-phase 5-wire.
The high-leg is a valid connection but since its voltage is higher than the others, it is put in a system phase-counting bucket by itself. The only unique voltage in that bucket is one 208 volt high-leg so it becomes a single-phase 208 volt source. The four L-L voltages in the 5-wire are just as legitimate as any other voltage. They are real voltages. Whether or not we can find a practical use for them is immaterial.
With your logic, the presence or absence of a center tapped neutral changes the number of phases, but for some reason it doesn't affect a wye connection. With my logic it makes no difference.
Then I think you are misunderstanding something I have said. A 120/208 wye source with no possible connection to the neutral can only supply a single-phase 208 system. If we can use the neutral, the transformer can supply the following:
1) one single-phase system with one 208 voltage
2) two single-phase systems with one voltage each (one at 0 deg and one at 120 deg)
3) one two-phase system with two voltages (one at 0 deg and one at 120 deg)
FWIW, the utility industry considers the open-wye distribution system to be two-phase. The load of the system as a whole appears as a two-phase load. If an open-wye is labeled "single-phase" it would be because of the nature of the loads. As I have said before, the load types have been used as part of the labeling. If it is serving single-phase loads, it can get labeled as a single-phase supply. That does not mean it is no longer a valid source for two-phase loads.
If you consider two voltages and how you are going to classify systems for them to be counted in, there is a difference between two single-phase systems and one two-phase system.
Not all grounded connections use a neutral, not all 'non-end point' taps are neutrals. For example, there is a standard control power connection of 24/120V (x1-X23= 24V, and X1-X4=120V).
I know that. And not all "neutrals" are true neutral points.
I believe I have said there is a single current direction, created by a single magnetic field direction, such as X1->X23->X4. But through the magic of math, X1->X23 and X23->X1 can be interchanged, by following proper 'signing' rules, giving the appearance of two currents.
Pick any direction you want, but the currents will not always flow in the same direction or have the same magnitude. It is a physical reality, not math magic.
Do you say a high-leg 4-wire delta (one winding is center tapped) has these 3-phases: 2@180?, and 1@90? while ignoring the 3@120?? If you mention them all is this a 6-phase transformer?
No. To be included in a system phase count, the voltage must have the same magnitude and a different angle from another counted voltage. All of those phases are not classified in the same system as each system phase-counting bucket has its own voltage level. Here are examples of some of the systems a 120/240 high-leg transformer could supply:
1) Three 240 volt single-phase systems with one voltage each.
2) One 208 volt single-phase system with one voltage.
3) One 120 volt single-phase system with one voltage.
4) One 120 volt two-phase system with two voltages.
5) One 240 single-phase system with one voltage.
Each system has one voltage magnitude to use for counting voltages..
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296950110.72/warc/CC-MAIN-20230401160259-20230401190259-00713.warc.gz
|
CC-MAIN-2023-14
| 8,012 | 41 |
http://www.osti.gov/eprints/topicpages/documents/record/349/1656556.html
|
math
|
Summary: Some further topics following Math 10B
1. In Green's theorem, we talked about integration over a closed curve C and also over its interior
D. But how do we know each curve divides the plane into the interior and exterior regions? Maybe
one can construct a weird curve which gives only one region (the way the M¨obius band only has
one side)? Or maybe another curve divides the plane into three regions? If you think these are
ridiculous questions, take a look at the picture at the very end of the article in the next link. Does
this curve have an interior and an exterior? Which of the points are inside, which outside?
The fact that every simple closed curve does have an inside and an outside is important enough
and hard enough to prove that is has a name, the Jordan Curve Theorem. It is taught in the Differ-
ential Geometry and Topology courses.
2. A very important, and beautiful theorem in mathematics is the Cauchy Integral Formula. Let f
be a complex-valued function of a complex argument z, C be a simple closed curve. Then for any
point a inside C,
z - a
One can prove this result using Green's Theorem. This formula is crucial in Complex Analysis
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701999715.75/warc/CC-MAIN-20160205195319-00277-ip-10-236-182-209.ec2.internal.warc.gz
|
CC-MAIN-2016-07
| 1,170 | 15 |
https://geometrica.vialattea.net/en/electrical-power-in-alternating-current/
|
math
|
Complex numbers, from ill-tolerated tools accompanied by absurd mysticism, became an incredibly useful tool with the advent of alternating electricity. This happened towards the end of the 19th century mainly thanks to Charles “Proteus” Steinmetz, a scientist little known to the public, but who was as genius as much more famous Edison or Tesla.
In fact, the voltage and current sinusoidal quantities:
can be represented by a phasor , that is a complex number that is imagined in continuous rotation with speed , generating a sinusoid on the real axis.
The phasor takes this name because, by freezing the situation at a given instant, the phase relationships between the electrical quantities are immediately seen. For example, in the following figure we see that the current is lagging behind the voltage (the vectors are imagined in counterclockwise rotation). The phase shift angle is and in electrical power transmission plants it must be reduced as much as possible because otherwise the usable power from the line is reduced.
Electric power is a complex number that is obtained by multiplying voltage and current, the problem is that we often read the formula which works perfectly for the graph above. Too bad it is NOT the correct formula and this illusion derives from the fact that the current is on the real axis and therefore coincides with its conjugate!
The correct formula, due to Steinmetz, in fact, is but for decades his explanation has been totally misunderstood (the classical engineering attitude “it works, but we don’t know why”).
The root of the error lies in the fact that we continue to treat complex numbers as vectors, but geometric algebra teaches that the bases are different in the two approaches: in the complex approach we are talking about a multivector composed of a scalar and a bivector , while in the vectorial approach we are in an ordinary vector space formed by the bases and .
Let’s try, then, to compare different approaches to the problem, when V and I are given in the forms:
|As vectors||As complex numbers||As GA elements|
|(generalized complex numbers)|
Power according to Steinmetz (publications between 1893 and 1899):
it works, but the great electrical engineer’s attempts at explanation are shaky: V and I are treated as complex numbers but for powers, since they have double frequency, it would be with non-commutative (, as well as )
Product between complex numbers:
numerically incorrect in general, it works – as we said above – only in the case (current on the real axis)
coincides up to the sign of the imaginary part
Product of complex numbers, but with the conjugate of the second factor:
Product of complex numbers, but with the conjugate of the first factor:
coincides with the product geometric, but not with Steinmetz’s formula (unless the minus sign of the imaginary part).
For the sake of completeness, it must be said that in 1893 Guilbert proposed a modification to Steinmetz’s formula, that is to invert the imaginary axis, to harmonize the notation with the one currently in use, which sees the inductive reactance on the positive imaginary semi-axis and the capacitive one on the ‘negative axis, that is the convention that sees the positive reactance when the current lags behind the voltage.
Doing so, we would have the coincidence of the formula of electric power with the geometric product!
Why do we need to multiply by the conjugate of one of the factors? To make the result dependent only on the phase angle between the two! Otherwise, it would depend on the absolute phase.
Let’s see the explanation, which also illustrates the reason for the coincidence of the result with the geometric product.
Given two complex numbers identified by the two vectors a and b , which form an angle .
The gray triangle has the projection as sides of b on a and its perpendicular.
If we now multiply b by the conjugate of a, remembering the amplitwist interpretation of the complex number, we will get another triangle rotated by is scaled by a factor .
Therefore the major cathetus will go on the horizontal axis and its measure will be or the dot product . The minor cathetus will be parallel to the y axis and its measure will be or the area of the external product .
We thus obtain the relation
In conclusion, with this example we see very well that the geometric algebra approach manages to elegantly include the geometric aspects that should have been adjusted ad hoc (conjugation of one of the factors).
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00750.warc.gz
|
CC-MAIN-2023-06
| 4,499 | 28 |
https://byjus.com/question-answer/which-is-earth-s-nearest-planetary-neighbor-mars-venus-mercury-none-of-the-above/
|
math
|
Which is earth's nearest planetary neighbour?
None of the above
The correct option is B. Venus.
Venus is earth's nearest planetary neighbour.
Mercury, Venus, Earth and Mars are collectively known as:
Which of the following is the hottest planet in the solar system?
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00365.warc.gz
|
CC-MAIN-2023-23
| 265 | 6 |
https://tennis2007.org/what-is-2-of-8000/
|
math
|
What is X percent the Y Calculator?Please readjust values of the two very first boxes of each calculator listed below to gain answers come any mix of values. View details on just how to calculate discounts, also as, our discount calculator below to number out percentages.
You are watching: What is 2% of 8000
What is % that ?
X the end of Y as a percentage Calculator
What is the end of ?
X is Y Percent the What Calculator
is % the what?
Using this device you deserve to find any kind of percentage in three ways. So, us think you got to us searching for answers like:1) What is 2 percent (%) of 8000?2) 2 is what percent the 8000?Or might be: 2% the 8000 Dollars?
See the solutions to these difficulties below.
If you are looking for a
1) What is 2% the 8000?
Always usage this formula to discover a percentage:
% / 100 = component / totality replace the given values:
2 / 100 = component / 8000
2 x 8000 = 100 x Part, or
16000 = 100 x component
Now, division by 100 and get the answer:
Part = 16000 / 100 = 160
2) What is 2 the end of 8000?
This concern is tantamount to: "2 is what percent the 8000?" Or What percent 2 is the end of 8000?
Use again the same percent formula:
% / 100 = part / totality replace the given values:
% / 100 = 2 / 8000
% x 8000 = 2 x 100
Divide by 8000 to get the percentage:
% = (2 x 100) / 8000 = 0.025%A shorter way to calculation x out of y
You can easily discover 2 is out of 8000, in one step, by simply splitting 2 by 8000, climate multiplying the an outcome by 100. So,
2 is the end of 8000 = 2/8000 x 100 = 0.025%
To find much more examples, just select one in ~ the bottom of this page.
Sample Percent Calculations
See more: Where Does Plant Mass Come From ?: Askscience Where Does Plant Mass Come From
While every initiative is made come ensure the accuracy that the information noted on this website, neither this website no one its authors space responsible for any type of errors or omissions. Therefore, the contents of this site are not an ideal for any kind of use entailing risk to health, finances or property.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662510117.12/warc/CC-MAIN-20220516104933-20220516134933-00351.warc.gz
|
CC-MAIN-2022-21
| 2,061 | 32 |
https://part15.org/forums/reply/re-i-dont-know-yet-exactly-the/
|
math
|
Total posts : 45366
[quote=SaGR]Somewhat like Rich’s image we showed a large dB gain over the standard MilSpec doublet.[/quote]
Just to note that my plots do not show large gain for the helical compared to the 5.15 dBi gain of a 1/4-wave monopole over a perfect ground plane.
In fact they show considerable reduction in gain relative to that monopole: about -17.5 dB for the helical radiator, and about -23 dB for the 3-meter linear radiator (1700 kHz).
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104585887.84/warc/CC-MAIN-20220705144321-20220705174321-00216.warc.gz
|
CC-MAIN-2022-27
| 455 | 4 |
https://poker-times.com/expected-value-in-a-poker-game/
|
math
|
Expected value (EV) tells you, on average, how much you can make or lose.
Example with dice: You roll a six-figure dice. If the number 6 falls, you get 18 euros, if you fall another number you lose 3 euros. It’s a question: Is it profitable for you to play such a game?
One out of six will fall 6, so the probability is 1/6
Five times out of six the next number will fall, so the probability is 5/6
You will earn on average: 1/6 * 18 euros = 3 euros and on average you will lose: 5/6 * 3 euros = 2.5 euros. On average, you earn 3 euros – 2.5 euros = 0.5 euros.
So, the answer is: EV = 0.5 Euro, in other words, every time you roll a dice you earn an average of 0.5 Euro and it is profitable for us to play such a game.
Example poker game
Let’s say you hold it:
Pot 500 Euro, Flop:
The opponent bets € 200 All – in on the € 500 pot. Total in the bank: 500 + 200 = 700 euros. And the question: is it profitable for you to answer and call 200 euros if you believe he has two boys?
Here’s a look at the PokerStars Equilab program:
And we see that ~ 33% will win and ~ 67% will lose. If we win the bank we win a total of 700 euros and if we lose we lose 200 euros. That is 33% we will win 700 euros and 67% we will lose 200 euros. The formula is: 700 * 0.33 – 200 * 0.67 = 97 euros. Answer: EV = 97 Euros, in other words, if we play in this situation, make a call, we will earn on average 97 Euros.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100232.63/warc/CC-MAIN-20231130193829-20231130223829-00356.warc.gz
|
CC-MAIN-2023-50
| 1,409 | 12 |
https://www.physicsforums.com/threads/problem-on-dimensions.256200/
|
math
|
1. The problem statement, all variables and given/known data tp=√(Gh/c^5) This quantity, tp, is called the Planck time and is thought to be the earliest time, after the creation of the Universe, at which the currently known laws of physics can be applied. Extra info(I don't know is it going to be helpful for solving this question): The smallest meaning ful measure of length is called the Plank length and is defined in terms of three fundamental constants in nature. The speed of light c=3.00x10^8 m/s The gravitational constants G=6.67x10^-11 m^3/kg*s^2 Planck's constant h=6.63x10^-34 kg*m^2/s Sorry, I have never take physics before, so I have no idea how to solve this question. Thank you for spending your valuable time.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039741294.2/warc/CC-MAIN-20181113062240-20181113084240-00223.warc.gz
|
CC-MAIN-2018-47
| 730 | 1 |
http://stevekifowit.com/archives/M109/ca_sec1_1a.html
|
math
|
A linear equation is an equation in which each variable appears only with an exponent of 1 and not in the denominator of a fraction nor in a radical.
For now, we will focus on linear equations of one variable. Every linear equation of one variable, , can be written in the form .
To solve an equation means to find ALL replacements for the variable that make the equation true. We typically solve linear equations by constructing a sequence of simpler, equivalent equations. At some point in this process, the solution becomes obvious.
Solve for :
Important note: At any step in the process above, it could be helpful to clear fractions by multiplying both sides of the equation by the LCM of all denominators.
In mathematics, there is a well-known, very broad, four-step problem solving process:
Understand the problem.
Devise a plan.
Carry out the plan.
Steps 1 & 2 involve defining variables and translating words to equations. We'll do examples in class, but this sheet may help you with your translation skills.
When walking, Jose burns 96 calories per mile and Sara burns 64 calories per mile. One day the two of them walk a total of 7 miles. Let represent the number of miles walked by Sara.
a. Write an algebraic expression for the total number of calories burned by the two of them.
b. Together they burn a total of 505.6 calories. How far did each person walk?
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664808.68/warc/CC-MAIN-20191112074214-20191112102214-00212.warc.gz
|
CC-MAIN-2019-47
| 1,370 | 13 |
https://muhaz.org/abstract-this-is-a-sample-file-demonstrating-the-style-for-eur.html?page=6
|
math
|
FINELG is a general non linear finite element program which has been first written by F. Frey . Major contributions have been made by V. de Ville de Goyet who developed efficient schemes for 2D and 3D steel beams. The concrete orientation of the beam elements and the time resolution algorithm have been developed by P.Boeraeve . The program is able to simulate the behaviour of structures undergoing large displacements and moderate deformations. All beam elements are developed using a total co-rotational description. For convenience, a short description of the beam is now given. We consider a 2D Bernoulli fibre beam element with 3 nodes and 7 degrees of freedom. The total number of DOF corresponds to two rotational DOF at end nodes and 5 translational DOF (see Figure ). The intermediate longitudinal DOF is necessary to allow to represent strong variations of the centroïd position when the behaviour of the cross-section is not symmetric. Such behaviour is observed for example in concrete sections as soon as cracking occurs.
Figure . 3 nodes plane beam element - DOFs
As usual for fibre element, internal forces at the element nodes are computed on the basis of a longitudinal and transversal integration scheme. The integration along the beam length is performed using 2, 3 or 4 integration points (see Figure ,a). For each longitudinal integration point LIPi, a transversal integration is performed using the trapezoidal scheme. The section is divided into layers (see Figure ,b) each of which being assumed in uni-axial stress state. The state of strain and stress is computed at each integration point TIPj.
Figure . Integration scheme : (a) longitudinal integration with 4-point Gauss scheme; (b) transversal integration with trapezoidal scheme.
The software can be used for both static and dynamic non linear analyses. FINELG has been extensively validated for static non linear analyses. Development and validation of non linear dynamic analysis has been realized in the context of the joint research program DYNAMIX between University of Liège and Greisch . Dynamic computations have been re-assessed at the beginning of the OPUS project .
INLDA analysis and definition of the structural capacity
A first assessment of the seismic behavior of each frame is performed by carrying out incremental non linear dynamic analyses considering nominal values of the material properties. It was observed that all buildings exhibit a similar seismic behavior. Indeed, for these highly redundant moment resisting frames, the only active failure criterion is the rotation capacity of the plastic hinges. No global instability, no local instability nor storey mechanism was observed, even for seismic action levels equal to 3 times the design level.
In the OPUS project, rotation demand and capacity were computed according to FEMA356 recommendations. The rotation demand is defined in Figure for both beams and columns. The rotation capacity is estimated to be equal to 27 mrad for steel columns.
However, since no indication is given in FEMA 356 regarding composite beam capacities, a detailed study has been undertaken to better estimate the rotation capacity of composite beams. This study relies on the plastic collapse mechanism model developed by Gioncu . Figure shows that the sagging zone is large with a significant part of it having a quasi-constant moment distribution near the joint. As a consequence, the plastic strains are low in steel as well as in concrete, and no crushing of the concrete is observed. On the contrary, the hogging zone is shorter but with high moment gradient. This results in a concentration of plastic deformations and in a more limited rotation capacity.
The ductility demand in plastic hinges is computed according to the actual position of contra-flexure point. Rotations of plastic hinges (θb1 and θb2) in beams are calculated as follows,
where v1, v2 and v3 are defined in Figure .
The resisting moment rotation curve in the hogging zone is determined by using an equivalent standard beam (see Figure .a) as commonly suggested in many references (Spangemacher and Sedlacek and Gioncu and Petcu ,).
A simply supported beam is subjected to a concentrated load at mid span. The post-buckling behavior is determined based on plastic collapse mechanisms (see Figure ). Two different plastic mechanisms are considered (in-plane and out-of-plane buckling, see Figure .c and d respectively). The behavior is finally governed by the less dissipative mechanism (see Figure .b).
Elastic and hardening branches of the M- curve have been determined using a multi-fiber beam model. When the hardening branch intersects the M- curve representative of the most critical buckling mechanism (softening branch), the global behavior switches from the hardening branch to the corresponding softening branch. The equation of the softening branches can be found in . The method has been implemented in MATLAB and validated against experimental results and F.E. results.
For OPUS buildings, it has been found from the analysis of the results that the length of the hogging zone was approximately equal to 2 m. As a consequence, an equivalent simply supported beam with L = 4 m is considered. The resulting M- curves are depicted for the composite beams of cases 1 and 2 in Figure a and b.
Figure . Typical bending moment diagram in a beam showing rotation in plastic hinges considering the exact position of the contraflexure point.
Figure . Model of Gioncu : (a) equivalent beam, (b) Moment rotation curve, (c) in plane buckling mechanism, (d) out of plane buckling mechanism
Figure . Moment-rotation curve of the composite beam IPE330 (a) (Case studies 1 and 2) and IPE 360 (b) (case studies 3 and 4)
The moment-rotation curve obtained from the model of Gioncu describes the static behavior. According to Gioncu, when a plastic hinge is subjected to a cyclic loading, its behavior remains stable as long as no buckling appears. When buckling is initiated, damage accumulates from cycle to cycle. M- curves of the composite beam of OPUS exhibit a steep softening branch that does not allow for a long stable behavior. Consequently, in the following developments, the rotation max corresponding to the maximum moment is considered as the maximum rotation capacity under cyclic loading.
The ultimate rotation max, the theoretical plastic rotation p, and the ratio max/p are reported on table 12. Since the ratio max/p is larger for case studies 3 and 4, the relative ductility is larger, and this leads, as it will be shown in the following, to a better seismic behavior of these buildings even if they were designed for the low seismicity. While this seems to be a paradox, it is nevertheless logical. The rotation capacity is defined by the local buckling limit. Lower steel grade used for low seismicity cases is favorable for this phenomenon, as it reduces the maximum stresses attained in the steel.
Table : characteristic rotations of the composite beams
1 and 2
3 and 4
The evolution of the maximum rotation demand of the hinges in the hogging zones of the beams and at the bottom of the columns for increasing seismic acceleration agR is represented in Figure for all case studies. The failure level fixed by the beam rotation criterion is considerably lower than the one fixed by the column ultimate rotation. As a consequence, the statistical analysis will focus on the ductility criterion of the beams in the hogging zone.
Figure . Results of the incremental dynamic analysis – case study 1(a), 2 (b), 3 (c), 4 (d)
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00583.warc.gz
|
CC-MAIN-2023-06
| 7,559 | 25 |
https://www.studypool.com/discuss/1045721/engineering-statics?free
|
math
|
Determine the magnitude of the resultant force and its direction measured clockwise from the x-axis through the following 3 methods (you must show all the work for each method): a. The parallelogram law or triangle rule.
Thank you for the opportunity to help you with your question!
The parallelogram law or triangle rule_In the
triangle rule, you use the second vector starting at the terminus of the
first vector & then join the base of the first vector to the
terminus of the second vector, thereby completing a triangle, to obtain
the resultant vector.
Whereas in the parallelogram rule, you join both the vectors at their
respective bases & then draw their respective parallel vectors
starting from the the terminus of the other vector thereby creating a
parallelogram for you. Now, he resultant vector must be the diagonal
vector of the parallelogram starting at the first common base point of
the two given vectors & ending at the vertex on the diagonally
opposite side of the parallelogram.
In both cases, you can find the angle that the resultant vector makes w.r.t one of the original vectors trigonScalar rectangular components.ometric-ally
For the vector ai + bj + ck,
x-component = a,
y-component = b,
z-component = c.
Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 20th, 2015
Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689615.28/warc/CC-MAIN-20170923085617-20170923105617-00358.warc.gz
|
CC-MAIN-2017-39
| 1,429 | 22 |
https://edumeasy.com/book-a-free-demo/
|
math
|
We are a comprehensive Math Lab supplier for schools
Book a Free Demo of Math Lab in Your School Today
Transform Math Education with our Innovative Math Lab for Schools
Welcome to EduMeasy Math Lab – the ultimate solution to revolutionize math education in schools. Our innovative “Math Lab” is designed to empower students, engage teachers, and drive outstanding academic results with fun & joy. Discover the power of interactive learning and personalized instruction with our cutting-edge math lab program. It covers all the chapters of NCERT textbooks from classes 6th to 10th.
Math Lab Equipment's for School
This Math Lab Equipments are going to make math easy and more results-oriented and students are going to love these types of equipment up to the level of understanding math easily. The Math Lab developed by IITians will bring revolution to the field of Math. The Math Lab is developed as per the syllabus of NCERT from classes 6th to 10th.
Math Lab Demo Video for Schools
This Math Lab Demo Video will help you to understand what is math lab and how to use it for school So Discover the Power of Math Lab: Watch Our Engaging Demo Video for Schools
Math Lab Demo Video in Hindi Language
Math Lab Demo Video in English Language
Manuals of Math Lab
“Complete Guide for Teachers about Math Lab, Chapter wise as per syllabus“
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947476205.65/warc/CC-MAIN-20240303043351-20240303073351-00196.warc.gz
|
CC-MAIN-2024-10
| 1,343 | 12 |
https://ricerca.univaq.it/handle/11697/18099
|
math
|
The aim of this paper is to present a new approach to the filtering problem for the class of bilinear stochastic multivariable systems, consisting in searching for suboptimal state-estimates instead of the conditional statistics. As a rst result, a finite-dimensional optimal linear filter for the considered class of systems is defined. Then, the more general problem of designing polynomial finite-dimensional filters is considered. The equations of a finite-dimensional filter are given, producing a state-estimate which is optimal in a class of polynomial transformations of the measurements with arbitrarily fixed degree. Numerical simulations show the effectiveness of the proposed filter.
File in questo prodotto:
Non ci sono file associati a questo prodotto.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00762.warc.gz
|
CC-MAIN-2022-33
| 766 | 3 |
https://books.google.de/books?id=rXjGFI6IvvkC&hl=de
|
math
|
The Consistency of the Axiom of Choice and of the Generalized Continuum-hypothesis with the Axioms of Set Theory
Kurt Gödel, mathematician and logician, was one of the most influential thinkers of the twentieth century. Gödel fled Nazi Germany, fearing for his Jewish wife and fed up with Nazi interference in the affairs of the mathematics institute at the University of Göttingen. In 1933 he settled at the Institute for Advanced Study in Princeton, where he joined the group of world-famous mathematicians who made up its original faculty.
His 1940 book, better known by its short title, The Consistency of the Continuum Hypothesis, is a classic of modern mathematics. The continuum hypothesis, introduced by mathematician George Cantor in 1877, states that there is no set of numbers between the integers and real numbers. It was later included as the first of mathematician David Hilbert's twenty-three unsolved math problems, famously delivered as a manifesto to the field of mathematics at the International Congress of Mathematicians in Paris in 1900. In The Consistency of the Continuum Hypothesis Gödel set forth his proof for this problem.
In 1999, Time magazine ranked him higher than fellow scientists Edwin Hubble, Enrico Fermi, John Maynard Keynes, James Watson, Francis Crick, and Jonas Salk. He is most renowned for his proof in 1931 of the 'incompleteness theorem,' in which he demonstrated that there are problems that cannot be solved by any set of rules or procedures. His proof wrought fruitful havoc in mathematics, logic, and beyond.
Was andere dazu sagen - Rezension schreiben
THE AXIOMS OF ABSTRACT SET THEORY
EXISTENCE OF CLASSES AND SETS
PROOF OF THE AXIOMS OF GROUPS AD FOR THE MODEL A
PROOF THAT V L HOLDS IN THE MODEL
Andere Ausgaben - Alle anzeigen
The Consistency of the Axiom of Choice and of the Generalized Continuum ...
Kurt G?del,Kurt Goedel
Eingeschränkte Leseprobe - 1940
Shadows of the Mind: A Search for the Missing Science of Consciousness
Eingeschränkte Leseprobe - 1996
Development and Evolution: Complexity and Change in Biology
Stanley N. Salthe,Stanley N.. Salthe
Eingeschränkte Leseprobe - 1993
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657131734.89/warc/CC-MAIN-20200712051058-20200712081058-00229.warc.gz
|
CC-MAIN-2020-29
| 2,150 | 18 |
https://math-frolic.blogspot.com/2014/08/prime-synchrony.html
|
math
|
Sunday, August 17, 2014
Sunday reflection today, courtesy of Freeman Dyson and Hugh Montgomery (this reflection actually ties nicely into last week's Sunday offering as well)...:
"[Freeman] Dyson helped bring together the continuous and the discrete understandings of subatomic behavior. Similarly, by fusing his love of number theory with his expertise in creating the mathematical tools of physics, he would make the initial observation that would reinforce the connections between the discrete world of the integers and the continuous world of analysis, and thus galvanize research on the Riemann hypothesis.
"As Dyson recalls it, he and [Hugh] Montgomery [number theorist] had crossed paths from time to time at the [Princeton] Institute [for Advanced Study] nursery when picking up and dropping off their children. Nevertheless, they had not been formally introduced. In spite of Dyson's fame, Montgomery hadn't seen any purpose in meeting him. 'What will we talk about?' is what Montgomery purportedly said when brought to tea. Nevertheless, Montgomery relented and upon being introduced, the amiable physicist asked the young number theorist about his work. Montgomery began to explain his recent results on the pair correlation, and Dyson stopped him short -- 'Did you get this?' he asked, writing down a particular mathematical formula. Montgomery almost fell over in surprise: Dyson had written down the sinc-infused pair correlation function.
"Dyson had the right answer, but until that moment he had associated this formula with understanding a phenomenon that seemed completely unrelated to the primes and the Riemann hypothesis. In a flash he had drawn the analogy between the sinc-described structured repulsion of the zeta zeros and a similar tension seemingly exhibited by the different levels of energy displayed by atomic nuclei. Whereas Montgomery had traveled a number theorist's road to a 'prime picture' of the pair correlation, Dyson had arrived at this formula through the study of these energy levels in the mathematics of matrices. This connection is the source of most of the current excitement surrounding the Riemann hypothesis..."
-- from "Stalking the Riemann Hypothesis" by Dan Rockmore
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00141.warc.gz
|
CC-MAIN-2018-30
| 2,219 | 6 |
https://escholarship.org/uc/item/7jt8s827
|
math
|
Kendall's tau for autocorrelation
The authors show how Kendall's tau can be adapted to test against serial dependence in a univariate time series context. They provide formulas for the mean and variance of circular and non-circular versions of this statistic, and they prove its asymptotic normality. They present also a Monte Carlo study comparing the power and size of a test based on Kendall's tau to that of competing procedures based on alternative parametric and nonparametric measures of serial dependence. In particular, their simulations indicate that Kendall's tau outperforms Spearman's rho in detecting first-order autoregressive dependence, despite the fact that these two statistics are asymptotically equivalent.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510942.97/warc/CC-MAIN-20231002001302-20231002031302-00132.warc.gz
|
CC-MAIN-2023-40
| 727 | 2 |
https://forums.studentdoctor.net/threads/cars-basics-question.1412911/
|
math
|
thanks for the advice! Throughout my life I generally was not a reader, I literals never read for fun. However, the last year/2 years I have been reading a variety of books. Pertaining to the questions I get right/wrong THIS IS THE MOST FRUSTERING part because there is no trend...... here and there I get 100% right sometimes I get 3/5, 0/7, 5/7, 6/7. 2/5, 4/5, 2/5, and the type of questions its the same thing. It is totally 1000% random so I can't even zone in on one question type. The reading isn't the real problem, I can finish a passage (depending on difficulty) anywhere between 3-5.5mins and either way I get the same understanding of it.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00272.warc.gz
|
CC-MAIN-2023-14
| 649 | 1 |
http://lib.mexmat.ru/books/165891
|
math
|
Нашли опечатку? Выделите ее мышкой и нажмите Ctrl+Enter
Название: Variational Calculus With Elementary Convexity (Undergraduate Texts in Mathematics)
Автор: Troutman J.
I had read/studied most of this book when I was a graduate student in chemical engineering at Syracuse University (in 1987-88). I also took two courses on the subject from Professor Troutman. I strongly recommend this book to any "newcomer" to the subject. The author is a mathematician, and a large fraction of the book consists of theorems, lemmas, propositions, corollaries (and their rigorous proofs). The book also contains, however, a good number of illustrative examples and exercises which make it useful to engineers and scientists as well as to students of mathematics who want to learn more about applications of mathematics to physical sciences.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358798.23/warc/CC-MAIN-20210227084805-20210227114805-00061.warc.gz
|
CC-MAIN-2021-10
| 876 | 4 |
http://promes-project.eu/poker-yt/
|
math
|
If I use special cards (ie: pokemon, old maid I will have the students find their own match.
allemande left - face your corner, walk around your corner left hand to left hand, end up where you started (facing your partner). .
I also will use a deck of cards (regular, pokemon, etc). .
If there is a new move in the dance, then I take the time to teach.A key to having the students enjoy square dance is to present it in a positive manner and with enthusiasm. .Instructions: The key to teaching square dance, in my opinion, is to state rules and objectives up front. .Swing - I use an elbow swing, due to ease. .The most common mistake I have are students who do halloween slot machine online not finish the allemande left, then head the wrong way for a grand right and left.Have the boys form a single file line to my left and the girls form a single file line to my right. .
I make sure each card has a match, then I will pass out the cards, i mate to the boys and the other to the girls. .I then have the first people walk forward to meet their partner, then second and. .Circle - all join bono mary j blige one mp3 hands and circle to the direction indicated.If possible use "fundamentals of Square Dance" record to learn the following moves: The square: The square or "set" is comprised of 4 couples, each representing a side of the square. .(watch for students who push or pull the circle).Zelbrite Pool Filter Media, jazzi Top-Mount Valve Sand Filter, linghein Air Screw Compressor.I started with the record, but now the cd is available. .This way, I could play songs that I knew bono world tour the students knew all the moves to, or I knew which moves needed to be taught before the song started, so I had time to learn the move myself.No negative remarks when they find their partner (No ewws or yuck, etc). .(I would only use this once).Many time followed by a grand-right and left. .I also tell them that to me square dancing is kind of like a puzzle that we are trying to solve. .After discussing these rules and stating my no warning policy, we very seldom have a problem.If I use regular cards, I will call for the 2's, then 3's etc. .The boys draw from one container and the girls draw from the other (whichever gender had the fewest dancers, will have chips remaining). .
Disclaimer - First, let me explain that I am not an expert, nor will I attempt to explain all of the possible moves in square dancing. .
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875144058.43/warc/CC-MAIN-20200219061325-20200219091325-00276.warc.gz
|
CC-MAIN-2020-10
| 2,424 | 6 |
http://www.troublefreepool.com/threads/34850-Best-dry-chemical-scoops
|
math
|
This might be a silly question but I could use some help. I was wondering what everyone thought were the best dry chemical scoops to use for pool chems, and where you get them from?
I have one that has a generalized graduation of weights on one side, which isn't entirely accurate but it gives a good idea. I don't have a pool store in my area, so I've been looking online, but the big guys (In The Swim, etc.) don't seem to carry any, and I can't find a scoop like this anywhere.
So what do you use?
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988720238.63/warc/CC-MAIN-20161020183840-00383-ip-10-171-6-4.ec2.internal.warc.gz
|
CC-MAIN-2016-44
| 500 | 3 |
https://advancesincontinuousanddiscretemodels.springeropen.com/articles/10.1155/ADE/2006/40171
|
math
|
On simulations of the classical harmonic oscillator equation by difference equations
Advances in Difference Equations volume 2006, Article number: 040171 (2006)
We discuss the discretizations of the second-order linear ordinary diffrential equations with constant coefficients. Special attention is given to the exact discretization because there exists a difference equation whose solutions exactly coincide with solutions of the corresponding differential equation evaluated at a discrete sequence of points. Such exact discretization can be found for an arbitrary lattice spacing.
Agarwal RP: Difference Equations and Inequalities, Monographs and Textbooks in Pure and Applied Mathematics. Volume 228. Marcel Dekker, New York; 2000:xvi+971.
Bobenko AI, Matthes D, Suris YuB: Discrete and smooth orthogonal systems: C∞-approximation. International Mathematics Research Notices 2003,2003(45):2415–2459. 10.1155/S1073792803130991
de Souza MM: Discrete-to-continuum transitions and mathematical generalizations in the classical harmonic oscillator. preprint, 2003, hep-th/0305114v5
Herbst BM, Ablowitz MJ: Numerically induced chaos in the nonlinear Schrödinger equation. Physical Review Letters 1989,62(18):2065–2068. 10.1103/PhysRevLett.62.2065
Hildebrand FB: Finite-Difference Equations and Simulations. Prentice-Hall, New Jersey; 1968:ix+338.
Iserles A, Zanna A: Qualitative numerical analysis of ordinary differential equations. In The Mathematics of Numerical Analysis (Park City, Utah, 1995), Lectures in Applied Mathematics. Volume 32. Edited by: Renegar J, Shub M, Smale S. American Mathematical Society, Rhode Island; 1996:421–442.
Lambert JD: Numerical Methods for Ordinary Differential Systems. John Wiley & Sons, Chichester; 1991:x+293.
Lang S: Algebra. Addison-Wesley, Massachusetts; 1965:xvii+508.
Oevel W: Symplectic Runge-Kutta schemes. In Symmetries and Integrability of Difference Equations (Canterbury, 1996), London Math. Soc. Lecture Note Ser.. Volume 255. Edited by: Clarkson PA, Nijhoff FW. Cambridge University Press, Cambridge; 1999:299–310.
Potter D: Computational Physics. John Wiley & Sons, New York; 1973:xi+304.
Potts RB: Differential and difference equations. The American Mathematical Monthly 1982,89(6):402–407. 10.2307/2321656
Reid JG: Linear System Fundamentals, Continuous and Discrete, Classic and Modern. McGraw-Hill, New York; 1983.
Stuart AS: Numerical analysis of dynamical systems. Acta Numerica 1994, 3: 467–572.
Suris YuB: The Problem of Integrable Discretization: Hamiltonian Approach, Progress in Mathematics. Volume 219. Birkhäuser, Basel; 2003:xxii+1070.
About this article
Cite this article
Cieśliński, J.L., Ratkiewicz, B. On simulations of the classical harmonic oscillator equation by difference equations. Adv Differ Equ 2006, 040171 (2006). https://doi.org/10.1155/ADE/2006/40171
- Differential Equation
- Partial Differential Equation
- Ordinary Differential Equation
- Functional Analysis
- Functional Equation
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00588.warc.gz
|
CC-MAIN-2023-40
| 2,983 | 25 |
https://byjusexamprep.com/gate-2021-electromagnetic-theory-rapid-quiz-1-i-18d8e7b0-3a13-11eb-9ebc-d91cf58870b7
|
math
|
GATE 2021 : Electromagnetic Theory Rapid Quiz-1 (App update required to attempt this test)
Attempt now to get your rank among 325 students!
A point P is specified in Cartesian coordinate system as P(1,2,3). What will be the points in Cylindrical coordinate system.
Two concentric spherical surfaces with r = 2 m and r = 10 m having uniform surface charge densities 40 nC/m2 and 10 nC/m2 respectively. The radial electric flux density at r = 4 m is
The velocity vector is given as . The divergence of this velocity vector at (1,1,1)
An E-field is given by = 4xâx+2 ây V/m. Then the work required in Joules to move a unit positive charge along the curve xy=4 from (2,2) to (4,1) is
Suppose we have a radiating field given by:
A/m, and it is in free space.
Now, determine the value of magnetic flux (in weber) that crosses the surface which is defined in the area given by: 0 and 0.
A filamentary conductor is formed into an equilateral triangle of side 2m that carries a current of 4A as shown in figure. The magnetic field intensity at the center of the triangle will be ….. A/m in az direction.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057091.31/warc/CC-MAIN-20210920191528-20210920221528-00456.warc.gz
|
CC-MAIN-2021-39
| 1,098 | 10 |
https://www.cuemath.com/questions/what-digits-can-be-in-the-ones-place-of-a-number-that-has-5-as-a-factor/
|
math
|
from a handpicked tutor in LIVE 1-to-1 classes
What digits can be in the one's place of a number that has 5 as a factor
A divisibility rule is a technique that determines if a number can be completely divided by another.
Answer: The digits present in the one's place of a number that has a factor of 5 are 0 and 5.
Let's look into the solution below.
Multiples of 5 are given as : 5, 10, 15, 20, 25, 30, 35, 40, 45, 50.....
We can therefore make an important observation that all the multiples of 5 have 0 and 5 as its unit digit at alternative positions. This cycle keeps continuing.
A number is said to be completely divisible by 5 if the remainder is zero and the quotient is a whole number.
Hence, the divisibility rule of 5 states that if the digit on the units place, that is, the last digit of a given number is 5 or 0, then such a number is divisible by 5
Thus, the digits present in the one's place of a number that has a factor 5 are 0 and 5.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00713.warc.gz
|
CC-MAIN-2023-40
| 952 | 10 |
https://www.arxiv-vanity.com/papers/hep-ph/9811317/
|
math
|
Light-front Hamiltonian field theory
towards a relativistic description of bound states
Lichtfront Hamiltoniaanse veldentheorie
Naar een relativistische beschrijving van gebonden
Leescommissie: prof.dr. J.F.J. van den Brand dr. A.E.L. Dieperink prof.dr. G. McCartor prof.dr. P.J.G. Mulders prof.dr. H.-C. Pauli
|Grafisch ontwerp:||Alex Henneman|
Dit werk maakt deel uit van het onderzoekprogramma van de Stichting
voor Fundamenteel Onderzoek der Materie (FOM) die financieel wordt
gesteund door de Nederlandse Organisatie voor Wetenschappelijk
©1998, Nico Schoonderwoerd
Light-front Hamiltonian field theory
[.5] towards a relativistic description of bound states
ter verkrijging van de graad van doctor aan
[.4] de Vrije Universiteit te Amsterdam,
[.4] op gezag van de rector magnificus
[.4] prof.dr. T. Sminia,
[.4] in het openbaar te verdedigen
[.4] ten overstaan van de promotiecommissie
[.4] natuurkunde en sterrenkunde
[.4] van de faculteit der exacte wetenschappen
[.4] op donderdag 14 januari 1999 om 13.45 uur
[.4] in het hoofdgebouw van de universiteit,
[.4] De Boelelaan 1105
Nicolaas Cornelis Johannes Schoonderwoerd
geboren te Breukelen
|Promotor:||prof.dr. P.J.G. Mulders|
|Copromotor:||dr. B.L.G. Bakker|
Aan Joop, Bea en Cosander
- § 1 Forms of relativistic dynamics
- § 2 Light-front quantization
- § 3 Feynman rules
- § 4 Divergences in the Yukawa model
- § 5 Instantaneous terms and blinks
- § 6 Pair contributions in the Breit-frame
- § 7 Introduction
- § 8 Example: the one-boson exchange correction
- § 9 Equivalence of the fermion self-energy
- § 10 Equivalence of the boson self-energy
- § 11 Conclusions
- § 12 Formulation of the problem
- § 13 Minus regularization
- § 14 Equivalence for the fermion triangle
§ 15 Equivalence for the one-boson exchange diagram
- § 15.1 Covariant calculation
- § 15.2 BPHZ regularization
- § 15.3 Light-front calculation
- § 15.4 Equivalence
- § 16 Conclusions
- § 17 Formulation of the problem
- § 18 The Lippmann-Schwinger formalism
- § 19 The box diagram
- § 20 A numerical experiment
- § 21 Light-front versus instant-form dynamics
- § 22 Numerical results above threshold
- § 23 Numerical results off energy-shell
- § 24 Analysis of the on energy-shell results
- § 25 Analysis of the off energy-shell results
- § 26 Conclusions
- § 26.0.1 Equivalence of light-front and covariant perturbation theory
- § 26.0.2 Entanglement of Fock-space expansion and covariance
- .0.3 Lichtfront Hamiltoniaanse veldentheorie Naar een relativistische beschrijving van gebonden toestanden
- .0.4 Hoofdstuk Introductie van lichtfront Hamiltoniaanse dynamica
- .0.5 Hoofdstuk Het Yukawa-model
- .0.6 Hoofdstuk Longitudinale divergenties in het Yukawa-model
- .0.7 Hoofdstuk Transversale divergenties in het Yukawa-model
- .0.8 Hoofdstuk Verstrengeling van de Fock-ruimteontwikkeling en covariantie
- .0.9 Hoofdstuk Samenvatting en conclusies
I Introduction to light-front Hamiltonian dynamics
Einstein’s great achievements, the principle of relativity, imposes conditions which all physical laws have to satisfy. It profoundly influences the whole of physical science, from cosmology, which deals with the very large, to the study of the atom, which deals with the very small.
This quote reflects the work that I have done in physics so far, which began with an investigation of new black hole solutions to the Einstein equation when I was a student in Groningen, and which now ends with the research presented in this Ph.D.-thesis on models to describe bound states of elementary particles [2, 3, 4, 5, 6, 7].
The above quote contains the first two lines of an important article that Dirac wrote in 1949, in the middle of a century that has produced enormous progress in the understanding of the properties of matter. Not only the development of relativity, but also the rise of Quantum Mechanics was instrumental for this progress. At the end of this century, these and many other advances have resulted in a model that ambitiously is called the Standard Model. It describes all elementary particles that have been discovered until now; the leptons and the quarks, and their interactions.
However, this does not imply we have to call it the end of the day for high-energy physics. A number of problems of a fundamental nature remain in the Standard Model. As an example we mention the question of the neutrino mass, that may or may not point to physics not included in the Standard Model.
There are many practical problems when one wants to calculate a physical amplitude. If the interactions are sufficiently weak, perturbation theory is usually applied and gives in many cases extremely accurate results. However, in the case of strong interactions, or when bound states are considered, nonperturbative methods must be developed.
We have to ensure that in such methods covariance is maintained. We mean by this that measured quantities, like cross sections and masses, are relativistic invariants. When the equations are written down in covariant form it is clear that the outcome will satisfy relativistic invariance and we refer to such methods as manifestly covariant.
For example, the Bethe-Salpeter equation is manifestly covariant, but suffers from numerical intractability beyond the ladder approximation. Great progress is made by Lattice Field Theory, which is now able to give quantitative predictions. However, it depends on the choice of a specific frame of reference and the advances in its application rely strongly on a continued increase of the speed of computers.
A very intuitive picture of a bound state is provided by Hamiltonian methods. However, the “classical” method of setting up a relativistic Hamiltonian theory by quantization on the equal-time plane, so-called instant-form (IF) quantization, suffers from problems such as a square root in the energy operator, which results in the existence of both positive and negative energy eigenstates, and the complexity of the boost operators, which keeps us away from determining the wavefunctions in an arbitrary reference frame. Weinberg proposed to use the Infinite Momentum Frame (IMF), because in this limit time-ordered diagrams containing vacuum creation or annihilation vertices vanish, and therefore the total number of contributing diagrams is significantly reduced. It is found that it provides a picture which connects to the one of the constituent quark model. However, its big disadvantage is that the IMF is connected to the rest frame by a boost for which one takes the limit of the boost parameter to infinity. It is dubitable whether this limit commutes with others that are taken in field theory.
It was only in the seventies that one began to realize that a theory with the same advantages as the IMF, but without the disadvantages, had already been suggested by Dirac some decades before: light-front (LF) quantization, i.e., quantization on a plane tangent to the light-cone. Of the ten Poincaré generators, seven are kinematic, i.e., can be formulated in a simple way and correspond to conserved quantities in perturbation theory. Most important is that these seven operators include one of the boost operators, allowing us to determine the wavefunction in a boosted frame if it is known in the rest frame. This property is not found in IF quantization. As a drawback one finds that not all rotations are kinematic, and therefore rotational invariance is not manifest in LF quantization, a problem which is discussed frequently in the literature. In particular, our interest was triggered by an article by Burkardt and Langnau who claimed that rotational invariance is broken for -matrix elements in the Yukawa model. Instead of a lack of manifest rotational invariance, we prefer to talk about lack of manifest covariance, as this is a property that all Hamiltonian theories share. Because in each form of quantization dynamical operators that involve creation or annihilation of particles are present, in any relativistic Hamiltonian theory particle number is not conserved, implying that each eigenstate has to be represented as a sum over Fock states of arbitrary particle number. However, light-front dynamics (LFD) is the only Hamiltonian dynamical theory which has the property that the perturbative vacuum is an eigenstate of the (light-front) Hamiltonian, provided that zero-modes are neglected (in this thesis zero-modes will not explicitly be discussed). Bound states are also eigenstates and are distinct from the LF vacuum, which simplifies their analysis.
In this thesis we shall not solve the eigenvalue problem, an interacting Hamiltonian will not even be written down! The goal of this thesis will not be to calculate a spectrum, but to illuminate two important properties of LF Hamiltonian dynamics. The first is:
1. Light-front dynamics provides a covariant framework for the treatment of bound states.
Although the calculation of bound states requires nonperturbative methods, these usually involve ingredients encountered in perturbation theory, e.g., the driving term in a Lippmann-Schwinger or Bethe-Salpeter approach. We prove that LF perturbation theory is equivalent to covariant perturbation theory. By equivalent we mean that physical observables in LF perturbation theory are the same as those obtained in covariant perturbation theory. This can be done by showing that the rules for constructing LF time-ordered diagrams can be obtained algebraically from covariant diagrams by integration over the LF energy . Two technical difficulties, namely that the integration over can be ill-defined, and that divergences in the transverse directions may remain, are solved in Chapters Light-front Hamiltonian field theory towards a relativistic description of bound states and Light-front Hamiltonian field theory towards a relativistic description of bound states respectively, for the Yukawa model, which is introduced in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states.
In Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we discuss the entanglement of covariance and the Fock-space expansion, and show another important property of LF Hamiltonian dynamics:
2. Higher Fock state contributions in LF Hamiltonian field theory are typically small, in particular much smaller than in IF Hamiltonian field theory, and therefore the ladder approximation gives accurate results for the spectrum.
It has been known for a long time that on the light-front one has to take into account fewer diagrams than in the instant-form of Hamiltonian dynamics. On top of this, diagrams involving higher Fock states are numerically smaller, as we will show. We look at two nucleons interacting via boson exchange, and we compare the contributions of the diagrams with one boson in the air, to diagrams where two bosons are simultaneously exchanged. The latter are ignored if we use the ladder approximation. We show in numerical calculations involving scalar particles that this approximation is viable for both scattering amplitudes and off energy-shell states, if masses and momenta are chosen in such a way that they are relevant for the deuteron.
§ 1 Forms of relativistic dynamics
An important first step on the path to a Hamiltonian description of a dynamical system was taken by Dirac in 1949, in his famous article ’Forms of Relativistic Dynamics’ . One foot of this work is in special relativity, when Dirac writes:
…physical laws shall be invariant under transformations from one such coordinate system to another.
The other foot of his method is in Quantum Mechanics because Dirac writes:
…the equations of motion shall be expressible in the Hamiltonian form.
In more technical terms, this condition tells us that any two dynamical variables have a Poisson bracket, later to be associated with (anti-)commutation relations. We restrict the transformations further to continuous ones, therefore excluding space inversion and time reversal. In the forthcoming subsections we are going to work out these two principles and construct the generators of the Poincaré group.
§ 1.1 The Poincaré group
The transformations mentioned in the first quote are the four translations , the three rotations , and the three boosts , where is an anti-symmetric tensor. These transformations should satisfy
Setting up a dynamical system is equivalent to finding a solution to these equations. The solution of the ten generators is generally such that some of them are simple, and correspond to conserved quantities. These are labeled as kinematical, indicating that they do not contain any interaction. Others are more complicated and describe the dynamical evolution of the system as the Hamiltonian does in nonrelativistic dynamics. Therefore these are called dynamical, which means that they do contain interaction. It seems obvious that one should want to setup the framework in such a way that the number of dynamical operators is small. A simple solution of the Eqs. (I-1)-(I-3) can be found if we define a point in space-time to be given by the dynamical variable and its conjugate momentum by . Using
a solution is now given by
As already mentioned by Dirac, this solution may not be of practical importance, however, it can serve as a building block for future solutions.
Another important ingredient for the dynamical theory is that we have to specify boundary conditions. We do this by taking a three-dimensional surface in space-time not containing time-like directions, at which we specify the initial conditions of the dynamical system. The ten generators then split into two groups, namely those that leave invariant and those that do not. The first group is called the stability group. The larger the stability group of , the smaller the dynamical part of the problem. We can ensure a large stability group by demanding that it acts transitively on : every point on may be mapped on any other point of by applying a suitable element of the stability group. This ensures that all points on the initial surface are equivalent.
|(a) the instant-form||(b) the light-front|
The restriction of relativistic causality reduces the number of world lines, and therefore increases the number of surfaces that one can choose for . Dirac found three independent choices for the initial surface that fulfill these conditions. In total there are five, as was pointed out by Leutwyler and Stern . We, however, only discuss the two most important ones. They are listed in Fig. I-1. In IF Hamiltonian dynamics one quantizes on the equal-time plane, given by
This is the form of dynamics closest to nonrelativistic Quantum Mechanics. Another important possibility for quantization is offered by a plane tangent to the light-cone. The light-front is given by the equation
Notice that this plane contains light-like directions. It is common to use the -direction to define the light-front. The different status of the other space-like directions and leads to the fact that the symmetry of rotational invariance becomes nonmanifest on the light-front. In explicitly covariant LFD , one defines the light-front by its normal vector , which is not fixed. This method will be encountered in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states.
§ 1.2 Light-front coordinates
In the remainder of this chapter, we show some of the advantages of LF quantization. We define so-called longitudinal coordinates
and transverse coordinates
such that the spatial coordinates and define a coordinate system on the light-front, and plays the role of time. From now on we will put the velocity of light equal to unity: . The indices of the four-vectors can be lowered and raised using the following LF metric :
where the first and second row/column refer to the longitudinal components, and the third and fourth to the transverse components. Unfortunately, a number of conventions are frequently used for LF coordinates. We will stick to the one given above, commonly referred to as the Kogut-Soper convention . The stability group on the light-front has seven elements, as can be verified by writing out the commutation relations between these operators and :
The other three operators are dynamical, as can be seen by the fact that they do not commute with ,
If we look at Fig. I-1b, we see that we can describe the operation of as a translation perpendicular to the light-front. The operators correspond to rotations of the light-front about the light-cone. Using these two words in one sentence clearly indicates why the common expression “light-cone quantization” is badly chosen. We prefer to use the phrase “light-front quantization”.
§ 1.3 The initial surface
In LF quantization, we first solve the Poincaré algebra on the surface . The stability group is the group generated by transformations of this surface into itself. We already met these operators in Eq. (§ 1.2). As is fixed, the dynamical variable has lost its meaning and, according to Dirac, it should be eliminated. We can add to the generators in Eq. (I-5) multiples of :
We then construct the and in such a way that on the light-front the -dependence drops from these equations. For the elements of the stability group we find:
and for the three dynamical operators we find:
When one quantizes in the instant-form, one finds four operators to be dynamical, which is one more than in LF quantization. However, more important is the form of the energy operator. In the instant-form, it is
The presence of the square root causes the degeneracy of positive and negative energy solutions in IF dynamics, whereas on the light-front they are kinematically separated, as can be seen from Eq. (I-16): positive longitudinal momentum corresponds to positive LF energy , and vice versa. This effect leads to the spectrum condition, which is explained in the next section.
The dynamical operators (I-16) and (I-17) reveal a little of the problems encountered on the light-front: the infrared problem for , which can be associated with the so-called zero-modes. As the path of quantization on the light-front is beset by problems, such as the nonuniqueness of the solution of the Cauchy problem, attempts have been made to find another path. Inspiration may be found in Quantum Field Theory, which leads to expressions for propagators of particles, and finally for -matrix elements. They may serve as a starting point to derive rules for time-ordered diagrams.
§ 2 Light-front quantization
The first to set foot on the new path towards a LF perturbation theory were Chang and Ma , and Kogut and Soper . Their work relies on the Feynman rules that are constructed in Quantum Field Theory. To determine the LF time-ordered propagator we take the Feynman propagator and integrate out the energy component.
For the types of theories that are discussed in this thesis, two are of importance: the scalar propagator and the fermion propagator.
§ 2.1 The scalar propagator
The Klein-Gordon propagator for a particle of mass is well-known:
where the subscript “Min” denotes that the integral is over Minkowski space. The inner products of the Lorentz vectors can be written in LF coordinates:
Following Kogut and Soper , we separate the energy integral from the integral over the kinematical components of , indicated by :
We then find for the propagator of Eq. (I-19):
in which we use the definition
and where is the on mass-shell value, or, in other words, the pole in the complex -plane:
Forward propagation in LF time requires . Then, we can only evaluate the integral over by closing the contour in the lower complex half-plane, because of the presence of the factor in the integrand. For the pole is below the real axis. Therefore application of Cauchy’s theorem gives a nonvanishing result only in this region:
where the on mass-shell four-vector is given by
§ 2.2 The fermion propagator
The well-known propagator for a spin- particle is related to the Klein-Gordon propagator by the following relation:
where is short for . We interchange differentiation and integration. Differentiation of the integrand in Eq. (I-19) gives:
where the Feynman slash for an arbitrary four-vector is defined by
An important difference with Eq. (I-23) is that the numerator contains the LF energy . We can remove it by rewriting the numerator,
Upon substitution of this expansion into Eq. (I-29) we see that the first term of Eq. (I-31) cancels against a similar factor in the denominator. Integration over the LF energy gives the LF time-ordered fermion propagator:
The first term is the same as the result for the scalar propagator (I-26), except for the factor . The second term on the right-hand side of (I-32) has lost its propagating part, resulting in the appearance of a -function in . This explains why it is called the instantaneous part. The decomposition of the covariant propagator into the propagating and the instantaneous fermion will occur frequently in this thesis and is an important ingredient in establishing equivalence between LF and covariant perturbation theory.
§ 2.3 The spectrum condition
An important result that we infer from the previous subsections is that the time-ordered propagators (I-26) and (I-32) contain -functions restricting the longitudinal momentum. This will severely reduce the size of phase-space.
Moreover, the longitudinal momentum is a conserved quantity and therefore all LF time-ordered diagrams containing either vacuum creation or annihilation contributions will vanish, as can be explained by looking at Fig. I-2.
According to Eq. (I-16), every massive particle in Fig. I-2 should have positive -momentum. As the longitudinal momentum is a kinematical quantity, it should be conserved at each vertex. However, the vacuum has . Therefore diagrams containing vacuum creation or annihilation vertices are not allowed in a series of LF time-ordered diagrams. In IF dynamics there is no such reduction of the number of diagrams, because there is no restriction on the IF momentum .
§ 2.4 The energy denominator
From now on, we shall write the Feynman diagrams in the momentum representation. In this subsection we show where the energy denominators originate from. Let us choose as a simple example Compton-like scattering in theory:
where is the total momentum, and is the momentum of the intermediate particle. Because of momentum conservation, they are the same. However, we make this distinction, to be able to write it in the following form:
where is the on mass-shell value of , conform Eq. (I-25). To stress that the diagram on the right-hand side is a time-ordered diagram, we draw a vertical thin line, indicating an energy denominator. The first denominator in Eq. (I-34) is the phase-space factor, constructed by taking for each intermediate particle the plus-momentum and a factor 2 because of the Kogut-Soper convention [13, 15]. The direction of the momenta should be chosen forward in time, such that the plus-component, satisfying the spectrum condition, is positive. The energy denominator is constructed by taking the total energy and subtracting from it the on mass-shell values of the minus-momentum of the particles in the corresponding intermediate state. Thus, it is proportional to the energy that is “borrowed” from the vacuum. This explains why highly off energy-shell intermediate states are suppressed. In the next subsection we present examples where the energy denominators are more complicated because different time-orderings of the vertices are involved.
§ 2.5 Light-front time-ordering
The most trivial example of time-ordering of vertices was already discussed in the previous subsection. In Compton-like scattering there are two time-orderings, however, one, the so-called Z-graph, is excluded because of the spectrum condition.
§ 2.5.1 The one-boson exchange
If we look at a similar amplitude as Eq. (I-33), now with the exchanged particle in the -channel, both time-orderings can contribute.
where the momentum of the intermediate particle . The sign of determines the time-ordering of the vertices:
where , which, if the external states are on energy-shell, coincides with the energy of the system. Again we see that the energy denominators are constructed by subtracting from the total energy the on mass-shell values of the minus-momentum of the particles in the intermediate state. Because of our choice of momenta, for the diagram (I-38) the momentum flow of the intermediate particle is backward in time. If we substitute , then the plus-momentum becomes positive, and the particle can be reinterpreted as going forward in LF time. In Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we will again encounter these two time-orderings when we describe the interaction of two nucleons by the exchange of bosons.
§ 2.5.2 The scalar shower
The next example is used to illustrate that some algebraic manipulations are needed to construct all LF time-ordered diagrams. We look at the decay of a particle into four scalars, again in theory.
where the two intermediate scalars have momentum and respectively. We now use the algebraic identity
This splitting can be used for the covariant amplitude in Eq. (I-39). We find:
§ 2.6 Loop diagrams
In the case of a loop diagram, covariant Feynman rules require an integration over the internal four-momentum . Time-ordered diagrams have an integration over the three kinematical components. As was found by Kogut and Soper , a relation between these types of diagrams can be established if we integrate out the energy component from the covariant diagram. Upon doing this integration one finds all LF time-ordered diagrams with the vertices time-ordered in all possible ways, however, respecting the spectrum condition. For an arbitrary number of particles in a loop, the proof was only recently given by Ligterink and Bakker . As an example, which also shows some of the problems encountered in LF Hamiltonian dynamics, we discuss the electromagnetic form factor in theory, given earlier by Sawicki :
An essential difference between the instant-form and the light-front occurs if we write the Feynman propagator in terms of the poles in the energy plane. In terms of IF coordinates we find:
and on the light-front we have:
We see that the Feynman propagator is quadratic in the IF energy but only linear in the LF energy . In the former case it leads to the presence of both positive and negative energy eigenstates, whereas on the light-front only positive energy states occur. In the instant-form, half of the poles occur above the real axis, and the other half below. Therefore contour integration will always give a nonvanishing result. In contrast to this, on the light-front the poles can cross the real axis. If all poles are on the same side of the real axis, the contour can be closed in the other half of the complex plane, and contour integration gives a vanishing result. Because of this effect, four of the six time-ordering in Fig. I-3 disappear. Only the first two remain. This is another manifestation of the spectrum condition. If we then turn to the Breit-frame (), also the second diagram of Fig. I-3 vanishes, as will follow from the analysis we present below.
Most important in our analysis is the sign of the imaginary part of the poles. Because of our choice of the Breit-frame, these are identical for the first and the second Feynman propagator in Eq. (I-44), namely . The imaginary part of the third Feynman propagator is . In Fig. I-4 we show the location of these poles for different intervals.
(a) (b) (c)
We see in (a) and (c) of Fig. I-4 that the contour can be closed in such a way that no poles are inside the contour, and therefore contour integration leads to a vanishing result. In case that we calculate the component of the current, application of Cauchy’s theorem is not valid because there is a contribution to the integral from a pole at infinity, i.e., for large absolute values of the integrand goes as . Therefore we restrict ourselves in this example to the components and . Only one LF time-ordered diagram contributes to the current:
where we have drawn vertical lines in the LF time-ordered diagram to indicate the energy denominators and to avoid confusion with the covariant diagram. The kinematics are given in Fig. I-3a. The photon line is vertical to indicate that we are in the Breit-frame. The imaginary parts have been omitted.
For the result above to be correct three assumptions are essential:
Interchange of the limit and -integration is valid,
There is no contribution of poles at infinity upon doing the -integration,
The amplitude is well-defined and finite.
All three assumptions can be justified in this case. De Melo et al. have shown that the interchange mentioned under assumption 1 may cause pair creation or annihilation contributions to become nonvanishing. In § 6 of Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we show that this effect may also occur in the Yukawa model. However, it is not a violation of the spectrum condition.
The second assumption can be justified by looking at Eq. (I-44). As is linear in , we see that the integration over the minus component is well-defined for each component of the current. In a theory with fermions, this integration can be ill-defined, leading to longitudinal divergences and the occurrence of so-called forced instantaneous loops (FILs). Divergences for the Yukawa model are classified in § 4 of Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states and the longitudinal ones are dealt with in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states, where it is shown that the FILs vanish upon using an appropriate regularization method: “minus regularization” .
The third assumption is also satisfied, since the superficial degree of divergence for integration over the perpendicular components is smaller than zero for all components of the current. If any transverse divergences occur, they can be attacked with the method of extended minus regularization presented in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states. The phase space factor contains endpoint singularities in . However, these are canceled by identical factors in the energy denominators.
Ii The Yukawa model
In particle physics several models are used to describe existing elementary particles, interactions and bound states. Many of these models are just used to highlight certain properties, or to make exact or numerical calculation possible. Although the latter are referred to as toy models, they are helpful because they are stripped from those properties that are of no concern to the investigation that is done. In this thesis we are going to “play around” with two models. One of them we already met in the introductory chapter: theory. In this model one can very nicely demonstrate that higher Fock states are much more suppressed on the light-front than in instant-form Hamiltonian dynamics, as will be done in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states. This model only contains scalar particles. The simplest model including fermions is the Yukawa model, which has the following Lagrangian:
The field describes the fermions and the field describes the scalar particles, from now on referred to as bosons. The last term is the interaction between the fermion–anti-fermion field and the boson field. Yukawa introduced this model to describe the interaction of nucleons (fermions) via pions (bosons). The strength of the interaction is given by . In our calculations we limit ourselves to a scalar coupling.
§ 3 Feynman rules
Using perturbation theory one can deduce from the Lagrangian the well-known rules for Feynman diagrams. Summing over these diagrams one then finds the -matrix.
The first term of the Lagrangian (II-48) leads to the following propagator:
for a fermion with momentum and mass . For a (scalar) boson with momentum and mass we have the following Feynman rule:
The full set of Feynman rules to compute the scattering amplitude in the Yukawa model can be found in many text books such as Itzykson and Zuber . Our goal is to translate these rules to rules for diagrams that one uses in LFD.
In Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we introduced the -integration to obtain the rules for the LF time-ordered diagrams. A complication in this procedure was already mentioned there: the integration over may be ill-defined, and the resulting integral may be divergent. Before solving these problems, we first classify the divergences.
§ 4 Divergences in the Yukawa model
In the previous subsection we described how to construct each covariant Feynman diagram. Covariant diagrams may contain infrared and ultraviolet divergences. Therefore we are not surprised that both in the process of constructing the LF time-ordered diagrams as in the diagrams themselves divergences can be encountered. The first type can be classified as longitudinal divergences, and the second as transverse divergences.
§ 4.1 Longitudinal divergences
We can deduce what (superficial) divergences we are going to encounter upon integration over . We denote the longitudinal degree of divergence by . Suppose we have a truncated one-loop diagram containing bosons and fermions. In the fermion propagator Eq. (II-49) the factor occurs both in the numerator and in the denominator, and therefore it does not contribute to . Each boson will, according to Eq. (II-50) contribute to the degree of divergence, and the measure of the loop contributes , resulting in
Longitudinally divergent diagrams, i.e., , contain one boson in the loop, or none. Since every loop contains at least two lines, a longitudinally divergent diagram contains at least one fermion. For the model we discuss, the Yukawa model with a scalar coupling, the degree of divergence is reduced. For scalar coupling it turns out that and therefore two instantaneous parts cannot be neighbors. The longitudinal degree of divergence for the Yukawa model with scalar coupling is
where the subscript “entier” denotes that we take the largest integer not greater than the value between square brackets.
§ 4.2 Transverse divergences
The transverse degree of divergence of a LF time-ordered diagram is the divergence one encounters upon integrating over the perpendicular components. In most cases this degree of divergence is the same as what is known in covariant perturbation theory as the superficial degree of divergence of a diagram. In that case it is the divergence one finds if in the covariant amplitude odd terms are removed and Wick rotation is applied. For a one-loop Feynman diagram in four space-time dimensions with internal fermion lines and internal boson lines the transverse degree of divergence is
In case of space-time dimensions we have to replace the term by .
In Table II-1 all one-loop diagrams up to order that are candidates to be divergent have been listed with their longitudinal and transverse degree of divergence.
§ 5 Instantaneous terms and blinks
As was already illustrated in the introduction, in the case of fermions we have to differentiate between propagating and instantaneous parts. Therefore this distinction plays an important role in the Yukawa model. The covariant propagator in momentum representation for an off-shell spin-1/2 particle can be written analogously to Eq. (I-32):
The first term on the right-hand side is the propagating part. The second one is the instantaneous part. The splitting of the covariant propagator corresponds to a similar splitting of LF time-ordered diagrams. For any fermion line in a covariant diagram two LF time-ordered diagrams occur, one containing the propagating part of the covariant propagator, the other containing the instantaneous part. For obvious reasons we call the corresponding lines in the LF time-ordered diagrams propagating and instantaneous respectively. For a general covariant diagram the -singularity in the propagating part cancels a similar singularity in the instantaneous part. Therefore the LF time-ordered diagrams with instantaneous lines are necessary; they are usually well-defined.
If the -singularities are inside the area of integration we may find it necessary to combine the propagating and the instantaneous contribution again into the so-called blink, introduced by Ligterink and Bakker , such that there is a cancellation of the singularities:
The thick straight line between fat dots is a blink. The bar in the internal line of the third diagram is the common way to denote an instantaneous fermion. When a LF time-ordered diagram resembles a covariant diagram, we draw a vertical line as in the second diagram of Eq. (II-55). If no confusion is possible, we omit it in the remainder of this thesis. The difference between Eqs. (II-54) and (II-55) lies in the fact that the former uses covariant propagators, and the latter has energy denominators. In this case the difference is only formal. However, in more complicated diagrams there is a big difference, as we will see later. Examples of blinks are discussed in the next section, and in § 8 of Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states where we discuss the one-boson exchange correction to the vertex.
§ 6 Pair contributions in the Breit-frame
In Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we found that for massive particles the spectrum condition applies: there can be no creation from or annihilation into the vacuum. This gives a significant reduction of the number of diagrams that one has to incorporate in a light-front calculation. In any frame where the particles have positive plus-momentum this is valid. In Leutwyler and Stern it was already noted that on the light-front the regions , and are kinematically separated, another manifestation of the spectrum condition. This fact should already make us aware that the Breit-frame, where one takes the limit of the plus-momentum of the incoming virtual photon going to zero, is dangerous.
Indeed one finds that pair creation or annihilation contributions play a role in this limit. This was first found by De Melo et al. and later by Choi and Ji . They discuss as an example the electro-magnetic current in theory, and find a pair creation contribution for the component of the current. We have shown in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states that for the other components and pair creation contributions vanish. Because De Melo et al. discuss a scalar theory, we infer that this effect is not related to the presence of fermions in the theory.
In a theory with fermions, such as the Yukawa model, we now show that the pair creation/annihilation term is also nonvanishing, and that its omission leads to a breaking of covariance and rotational invariance.
In the presence of fermions, the individual time-ordered diagrams may contain singularities that cancel in the full sum. Because this cancellation has nothing to do with the time-ordering of the diagrams, we combine the LF time-ordered diagrams into blink diagrams. After that, we have a clear view on the point we want to discuss.
Again, we use kinematics as in Fig. I-3a. Two blinks contribute to the current, provided we have chosen the plus-component of the momentum of the outgoing boson .
where the diagrams containing blinks are given by:
The diagram (§ 6) is an example of a “double” blink. The total blink is the thick line between the two fat dots. For both blinks we see that the energy denominators are the same as for usual LF time-ordered diagrams. However, the numerators are different. In the next subsection we show how the numerator of the blink is constructed.
§ 6.1 Construction of the blink
As an example, we show how we can construct the blink (§ 6). For the fat line we have to substitute the propagating and instantaneous part.
The propagating contribution is
and the instantaneous contribution, denoted by the perpendicular tag, is
We see that both have a singularity at the upper boundary of the integration interval over . These cancel in the sum: the blink Eq. (§ 6). It is obtained by making the denominators common for the two diagrams. We can verify, using the relation , that the lower boundary at does not cause any problems, neither for the LF time-ordered diagrams, nor for the blink.
In an analogous way the double blink is constructed. It consists out of the following LF time-ordered diagrams:
We see that one diagram is missing: the diagram with two instantaneous fermions. Because it contains two neighboring matrices, it vanishes. It is an example of a forced instantaneous loop (FIL), which is related to longitudinal divergences and will be discussed in the next chapter. The LF time-ordered diagrams have the same integration interval as the double blink (§ 6). The last diagram on the right-hand side is the same as the diagram in Eq. (§ 6.1), the only difference being the integration range. The instantaneous fermions have been ’tilted’ a little in these two diagrams, to indicate that the integration interval is such that the instantaneous fermions carry positive plus-momentum. The second diagram on the right-hand side of Eq. (II-62) has no tilted instantaneous fermion, because the integration range has not been split as in the previous case. We will not give the formulas for the LF time-ordered diagrams in Eq. (II-62), but we have verified that their end-point singularities are removed when the diagrams are combined into the double blink.
§ 6.2 The Breit-frame
What happens to the current in Eq. (II-56) in the limit of ? Relying on the spectrum condition, one may expect that diagrams like (§ 6) disappear, and that only the double blink (§ 6) contributes. This is confirmed by the fact that the integration area of the single blink (§ 6) goes to zero. However, it could be that the integrand obtains singularities in the limit that cause a nonzero result. We denote this limit in the diagrams by drawing the line of the outgoing boson vertically.
For the numerator of the blink (§ 6) we use the relation
The integral is dominated by factors and . Identifying these factors in (§ 6) we find:
We write Eq. (II-64) in internal coordinates , and find that the dependence on the integration range drops. Moreover, the integration contains no singularities in the internal variable .
If we disregard for a moment the transverse integration, we see that in the Breit-frame there is a finite contribution of pair-creation/annihilation to the current. This agrees with the result of De Melo et al. . Furthermore, we see that it is not covariant, and therefore its omission will not only lead to the wrong amplitude, but also to breaking of Lorentz covariance and rotational invariance.
Iii Longitudinal divergences in the Yukawa model
If the doors of perception were cleansed everything would appear as it is, infinite.
William Blake, The marriage of heaven and hell
For a number of reasons mentioned in the previous chapters, quantization on the light-front is nontrivial. Subtleties arise that have no counterpart in ordinary time-ordered theories. We will encounter some of them in this chapter and show how to deal with them in such a way that covariance of the perturbation series is maintained.
In LFD, or any other Hamiltonian theory, covariance is not manifest. Burkardt and Langnau claimed that, even for scattering amplitudes, rotational invariance is broken in naive light-cone quantization (NLCQ). In the case they studied, two types of infinities occur: longitudinal and transverse divergences. They regulate the longitudinal divergences by introducing noncovariant counterterms. In doing so, they restore at the same time rotational invariance. The transverse divergences are dealt with by dimensional regularization.
We would like to maintain the covariant structure of the Lagrangian and take the path of Ligterink and Bakker . Following Kogut and Soper they derive rules for LFD by integrating covariant Feynman diagrams over the LF energy . For covariant diagrams where the -integration is well-defined this procedure is straightforward and the rules constructed are, in essence, equal to the ones of NLCQ. However, when the -integration diverges the integral over must be regulated first. We stress that it is important to do this in such a way that covariance is maintained.
In this chapter, we will show that the occurrence of longitudinal divergences is related to the so-called forced instantaneous loops (FILs). If these diagrams are included and renormalized in a proper way we can give an analytic proof of covariance. FILs were discussed before by Mustaki et al. , in the context of QED. They refer to them as seagulls. There are, however, some subtle differences between their treatment of longitudinal divergences and ours, which are explained in § 9.
Transverse divergences have a different origin. However, they can be treated with the same renormalization method as longitudinal divergences. We shall present an analytic proof of the equivalence of the renormalized covariant amplitude and the sum of renormalized LF time-ordered amplitudes in two cases, the fermion and the boson self-energy. In the other cases we have to use numerical techniques. They will be dealt with in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states.
§ 7 Introduction
In the previous chapters we already introduced instantaneous fermions. For a discussion on longitudinal divergences they play an important role. Without fermions there are no longitudinal divergences! The longitudinal divergences can be both seen from a “pictorial” and a mathematical point of view.
The pictorial view is the following. When a diagram contains a loop where all particles but one are instantaneous, a conceptual problem occurs. Should the remaining boson or fermion be interpreted as propagating or as instantaneous? Loops with this property are referred to as forced instantaneous loops (FILs). Loops where all fermions are instantaneous are also considered as FILs. However, they do not occur in the Yukawa model with (pseudo-)scalar coupling. Examples of these three types of FILs are given in Fig. III-5.
Mathematically this problem also shows up. The FILs correspond to the part of the covariant amplitude where the -integration is ill-defined. The problem is solved in the following way. First we do not count FILs as LF time-ordered diagrams. Second we find that this special type of diagram disappears upon regularization if we use the method of Ligterink and Bakker : minus regularization.
§ 7.1 Minus regularization
The minus-regularization scheme was developed for the purpose of maintaining the symmetries of the theory such that the amplitude is covariant order by order. It can be applied to Feynman diagrams as well as to ordinary time-ordered or to LF time-ordered diagrams. Owing to the fact that minus regularization is a linear operation, it commutes with the splitting of Feynman diagrams into LF time-ordered diagrams.
We explain very briefly how the method works. Consider a diagram defined by a divergent integral. Then the integrand is differentiated with respect to the external energy, say , until the integral is well defined. Next the integration over the internal momenta is performed. Finally the result is integrated over as many times as it was differentiated before. This operation is the same as removing the lowest orders in the Taylor expansion in . For example, if the two lowest orders of the Taylor expansion with respect to the external momentum of a LF time-ordered diagram are divergent, minus regularization is the following operation:
The point is chosen in this example as the renormalization point. This regularization method of subtracting the lowest order terms in the Taylor expansion is similar to what is known in covariant perturbation theory as BPHZ (Bogoliubov-Parasiuk-Hepp-Zimmermann) . Some advantages of the minus-regularization scheme are preservation of covariance and local counterterms. Another advantage is that longitudinal as well as transverse divergences are treated in the same way. A more thorough discussion on minus regularization can be found in the next chapter.
§ 7.2 Proof of equivalence for the Yukawa model
The proof of equivalence will not only hold order by order in the perturbation series, but also for every covariant diagram separately. In order to allow for a meaningful comparison with the method of Burkardt and Langnau we apply our method to the same model as they discuss, the Yukawa model, as introduced in Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states.
In this model we have to distinguish four types of diagrams, according to their longitudinal () and transverse degrees () of divergence. These divergences were classified also in Table II-1 on page II-1. The proof of equivalence is illustrated in Fig. III-6.
We integrate an arbitrary covariant diagram over LF energy. For longitudinally divergent diagrams this integration is ill-defined and results in FILs. A regulator is introduced which formally restores equivalence. Upon minus regularization the -dependence is lost and the transverse divergences are removed. We can distinguish
Longitudinally and transversely convergent diagrams (). No FILs will be generated. No regularization is needed. The LF time-ordered diagrams may contain -poles, but these can be removed using blinks. A rigorous proof of equivalence for this class of diagrams is given by Ligterink and Bakker .
Longitudinally convergent diagrams () with a transverse divergence (). In the Yukawa model there are three such diagrams: the four fermion box, the fermion triangle and the one-boson exchange correction. Again, no FILs occur. Their transverse divergences and therefore the proof of equivalence will be postponed until Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states. However, because the one-boson exchange correction illustrates the concept of -integration, the occurrence of instantaneous fermions and the construction of blinks, it will be discussed as an example in § 8. In Chapter Light-front Hamiltonian field theory towards a relativistic description of bound states we gave an example for a longitudinally convergent diagram in theory: the electro-magnetic current.
Longitudinally divergent diagrams () with a logarithmic transverse divergence (). In the Yukawa model with a scalar coupling there is one such diagram: the fermion self-energy. Upon splitting the fermion propagator two diagrams are found. The troublesome one is the diagram containing the instantaneous part of the fermion propagator. According to our definition it is a FIL and needs a regulator. In § 9 we show how to determine the regulator that restores covariance formally. Since can be chosen such that it does not depend on the LF energy, the FIL will vanish upon minus regularization.
Longitudinally divergent diagrams with a quadratic transverse divergence (). In the Yukawa model only the boson self-energy is in this class. We are not able to give an explicit expression for . However, in § 10 it is shown that the renormalized boson self-energy is equal to the corresponding series of renormalized LF time-ordered diagrams. This implies that the contribution of FILs has again disappeared after minus regularization.
§ 8 Example: the one-boson exchange correction
We will give an example of the construction of the LF time-ordered diagrams, the occurrence of instantaneous fermions and the construction of blinks. It concerns the correction to the boson–fermion–anti-fermion vertex due to the exchange of a boson by the two outgoing fermions. Here, and in the sequel, we drop the dependence on the coupling constant and numerical factors related to the symmetry of the Feynman diagrams.
A boson of mass with momentum decays into a fermion anti-fermion pair with momenta and respectively. The covariant amplitude for the boson exchange correction can be written as
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711394.73/warc/CC-MAIN-20221209080025-20221209110025-00171.warc.gz
|
CC-MAIN-2022-49
| 52,571 | 238 |
http://physics.qandaexchange.com/?qa=3319/floating-cubical-block
|
math
|
I think you have misinterpreted the question.
The copper block is not hollow, it is solid copper. There is nothing inside it except copper. Mercury and water are not poured into the copper block. They are poured into a vessel in which the copper block is able to float.
Initially the copper block is floating inside a vessel (eg a beaker) which contains a layer of liquid mercury at the bottom. Some water is added on top of the mercury, surrounding the copper block, until the water is level with the top of the copper block. Now the copper block is floating in a layer of mercury and a layer of water. The question is asking for the depth of the layer of water.
The words "the copper in the block just gets submerged" in your question should read "the copper block just gets submerged".
Mercury is a liquid. Water is a liquid. The LHS of your equation gives the pressure in the mercury level with the base of the copper block (neglecting air pressure). The RHS is the pressure due to copper block, which you can think of as another liquid.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573118.26/warc/CC-MAIN-20220817213446-20220818003446-00401.warc.gz
|
CC-MAIN-2022-33
| 1,041 | 5 |
https://mmp.susu.ru/article/en/385
|
math
|
Volume 9, no. 2Pages 37 - 45
Inverse Problems for Determining Boundary Regimes for Some Equations of Sobolev TypeA.I. Kozhanov
We study the solvability of the inverse problems of finding a solution to some Sobolev type equations along with the unknown coefficients of a special type defining the boundary modes (boundary data) in the first or the third initial-boundary value problems respectively. The presence of an unknown coefficient in such problems supposes that there is an additional condition - an overdetermination condition along with the initial and boundary conditions that are typical for the corresponding class of differential equations. In the present work this condition is represented by the integral overdetermination, when some integrals by the cross-section of the cylindrical domain by planes t=const are equal to zero. The goal of this research is to prove the existence of regular (with all needed generalized according to S.L. Sobolev derivatives) solutions of equation. Along with the specific results there are also some of their possible generalizations. Full text
- Sobolev type equations; inverse problems; unknown boundary data; integral overdetermination; regular solutions; solvability.
- 1. Prilepko A.I., Orlovsky D.G., Vasin I.A. Methods for Solving Inverse Problems in Mathematical Physics. New York, Marcel Dekker, 1999.
2. Kabanikhin S.I. Inverse and Ill-posed Problems. Novosibirsk, Siberian publishing, 2009.
3. Alekseev C.V. Optimization in the Stationary Problems of Heat and Mass Transfer and Magnetic Hydrodynamics. Science World, 2010, pp. 142-143.
4. Kostin A., Prilepko A. On Some Problems of Restoration of a Boundary Condition for a Parabolic Equation. Differential Equations, 1996, vol. 32, no. 1, pp. 113-122.
5. Boruhov V.T., Korzyuk V.I. [Application Neoclassical Boundary Value Problems for the Restoration of Boundary Modes of Transport Processes]. Vestnik Belorusskogo universiteta. Ser. I., 1998, no. 3, pp. 54-57. (in Russian)
6. Boruhov V.T., Vabishchevich P.N., Korzyuk V.I. [The Reduction of a Class of Inverse Problems of Heat Conduction to Direct the Initial-Boundary Value Problems]. Inzhenerno-fizicheskiy zhurnal, 2000, vol. 73, no. 4, pp. 742-747. (in Russian)
7. Korotkiy A.I. Reconstruction of Boundary Regimes in the Inverse Problem of Thermal Convection of an Incompressible Fluid. Tr. IMM DVO AN, 2006, vol. 12, no. 2, pp. 88-97. (in Russian)
8. Kozhanov A.I. The Problem of Recovery of the Boundary Condition for a Heat Equation. Analiticheskiye metody analiza i differetsial'nyh uravneniy: materialy 6 mezhdunarodnoy konferentsyi pamyati prof. A.A. Kilbasa, Minsk, Izd. BGU, 2012, pp. 87-96.
9. Kozhanov A.I. [Linear Inverse Problems for Certain Classes of Non-Stationary Equation]. Trudy VI mezhdunarodnoy molodezhnoy shkoly-konferentsyi 'Teoriya i chislennyye metody resheniy obratnyh i nekorrektnyh zadach'. Sibirskiye Elektronnyye Matematicheskiye Izvestiya, 2015, vol. 12, pp. 264-275.
10. Ionkin N.I. [Solution of a Boundary-Value Problem in Heat Condition with a Nonclassical Boundary Condition]. Differential Equations, 1977, vol. 13, no. 2, pp. 204-211.
11. Pulkina L.S. [A Nonlocal Problem with Two Integral Conditions for Hyperbolic Equations on a Plane]. Neklassicheskiye uravneniya matematicheskoy fiziki [Non-Classical Equations of Mathematical Physics]. Novosibirsk, Institut matematiki SO RAN, 2007, pp. 232-236.
12. Kozhanov A.I. [On the Solvability of Some Boundary Value Problems with a Shift for Linear Hyperbolic Equations]. Matematicheskiy zhurnal [Mathematical Journal], Almaty, 2009, vol. 9, no. 3, pp. 78-92.
13. Kozhanov A.I. [On the Solvability of Boundary Value Problems with Non-Local and Integral Conditions for Parabolic Equations]. Nelineynye granichnyye zadachi. IPMM NAN Ukrainy, 2010, vol. 20, pp. 54-76.
14. Pulkina L.S. Zadachi s neklassicheskimi usloviyami dlya giperbolicheskih uravneniy. Samara, Izd. Samarskogo Universiteta, 2012.
15. Kozhanov A.I. Problems with Integral-Type Conditions for Some Classes of Nonstationary Equations. Doklady Mathematics, 2014, vol. 90, no. 1, pp. 440-443. DOI: 10.1134/S1064562414050044
16. Kozhanov A.I. On the Solvability of Spatially Nonlocal Problems with Conditions of Integral Form for Some Classes of Non stationary Equations. Differential Equations, 2015, vol. 5, no. 18, pp. 1043-1050. DOI: 10.1134/S001226611508008X
17. Barenblatt G.I., Zheltov Yu.P., Kochina I.N. Basic Concepts in the Theory of Seepage of Homogeneous Fluids in Fissyrized Rocks. Journal of Applied Mathematics and Mechanics, 1960, vol. 24, no. 4, pp. 1268-1303.
18. Demidenko G.V., Uspenskii S.V. Partial Differential Equations and Systems not Solved with Respect to the Highest Order Derivative. N.Y., Basel, Hong Kong, Marcel Dekker, Inc., 2003.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178358064.34/warc/CC-MAIN-20210227024823-20210227054823-00403.warc.gz
|
CC-MAIN-2021-10
| 4,776 | 22 |
https://www.sciencehq.com/mathematics/basic-properties-of-logarithms.html
|
math
|
Basic properties of Logarithms.
Logarithms are important throughout mathematics and science.
us for faster computation of mathematical problems.
Some of the basic properties of Logarithms and proof of them are given below, they are also called theorems of logarithm.
1> For any positive x ,y and a:
2> For any positive a and x:
3> For any positive x , y and a:
4> For any positive a,b and x:
- The Logarithmic Function. There are many ways to describe logarithmic function. One basic...
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00065.warc.gz
|
CC-MAIN-2021-21
| 486 | 9 |
https://www.warzone.com/Forum/298001-one-thing-always-wondered
|
math
|
Why is 16% the most common luck number for strategic templates?
What makes the number 16 so special? Why not make it 10% of 20% or something else?
I know it's not as big anymore as no luck is much more common but 16% has always had a large presence in ladder templates and whatnot. Anyone got a good explanation about this or is it just by convention that it's used?
I vaguely remember the idea that analysis attack graph for WR is only accurate within 1% so it could be more like 99.5 or whatever. Dunno if that is true or I am confusing it with something else could easily be the latter.
Rikku is correct. The analyze graphs are an estimate and not 100% accurate.
Fun fact: When I first calculated the highest value that 4v2 would always succeed, I made a mistake and came up with 18%. For a while, this was the value used in strategic games, You can still see this if you dig back into old 1v1 ladder games.
4v2s would succeed almost every time at 18%. But maybe one in 100 games, one attack would have a 4v2 failure somewhere, and that player would rage.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00642.warc.gz
|
CC-MAIN-2023-23
| 1,058 | 7 |
https://best-essays.me/2021/04/09/solve-a-word-problem-for-me_zu/
|
math
|
Problem solving strategy. i did not like math as a child. i thought this was usf medical masters essay the best idea ever. how to solve two-step addition word write my conclusion problems with examples and block diagrams or tape diagrams solutions motivating kids to do homework used in singapore math, examples with step by step solutions, how to solve a comparison subtraction word best resume writing service reddit problems using bar models. completing homework assign a variable to the unknown quantity, for example, \(x\) welcome american revolutionary war essay to the math word problems worksheets page at math-drills.com! marilyn vos savant, contributor 4 29 free online courses being offered if you wanted to learn a new skill jessica sager, contributor most popular 1 40 years, 7,000. for those media business plan of you who aren't familair with the term, a word problem is solve a word problem for me solve a word problem for me just like a math problem, except mention a book in an essay it uses text to present solve a word problem for me the problem. a baker used 4 bags of flour baking cakes and 3 bags of flour baking cookies. for this problem, my list of problems to write about what is given would be: the…. can you solve this word problem? Read white privilege essay thesis the whole question.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038072082.26/warc/CC-MAIN-20210413031741-20210413061741-00019.warc.gz
|
CC-MAIN-2021-17
| 1,316 | 1 |
http://mathhelpforum.com/trigonometry/76960-unit-circle-trig.html
|
math
|
look familiar now?
from the unit circle, it should be clear to you that for any angle ,2. Suppose alpha and beta are central angles in the unit circle (of radius 1) with initial sides on the positive x-axis. Further suppose that alpha is in quadrant 1 and beta is in quadrant 4 such that sine of alpha is 2/3 and cosine of beta is 1/4
a) Find the exact value of cosine alpha
b) Find the exact value of sine beta
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698542939.6/warc/CC-MAIN-20161202170902-00005-ip-10-31-129-80.ec2.internal.warc.gz
|
CC-MAIN-2016-50
| 411 | 4 |
https://www.flippingphysics.com/coe-problem.html
|
math
|
Introductory Conservation of Mechanical Energy Problem using a Trebuchet (8:49)
Learn how to use the Conservation of Mechanical Energy equation by solving a trebuchet problem. This is an AP Physics 1 topic.
0:08 The problem
1:08 Why mechanical energy is conserved
1:37 Setting the zero line and initial and final points
2:32 The three types of mechanical energy
3:55 Canceling mechanical energies from the equation
4:54 Solving the equation
6:18 It’s final speed not final velocity
6:51 Why we can’t use the projectile motion equations
7:43 Do we really have to write all that down? Yes.
Multilingual? Please help translate Flipping Physics videos!
Thank you to Jacob, Will, Natalie, and Mery; my students who built and let me use their trebuchet!
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510462.75/warc/CC-MAIN-20230928230810-20230929020810-00429.warc.gz
|
CC-MAIN-2023-40
| 751 | 13 |
https://ruvehurusuc.caninariojana.com/derivatives-study-guide-38515up.html
|
math
|
If you are connected to any kind of financial market or watch the financial news even for 5 minutes every day, it is likely that you have heard the word, financial derivatives many times. The media is flush with articles wherein derivatives are criticized or appreciated. Most of the times, commentators are in awe of the mind-numbingly large amounts behind these contracts. It is often said that the total amount of derivatives contracts in the worlds, is actually greater than the total amount of money available in the world!
Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device should be able to scroll to see them and some of the menu items will be cut off due to the narrow screen width.
Derivatives In this chapter we will start looking at the next major topic in a calculus class, derivatives. This chapter is devoted almost exclusively to finding derivatives. We will be looking at one application of them in this chapter.
We will be leaving most of the applications of derivatives to the next chapter. Here is a listing of the topics covered in this chapter. The Definition of the Derivative — In this section we define the derivative, give various notations for the derivative and work a few problems illustrating how to use the definition of the derivative to actually compute the derivative of a function.
Interpretation of the Derivative — In this section we give several of the more important interpretations of the derivative.
We discuss the rate of change of a function, the velocity of a moving object and the slope of the tangent line to a graph of a function. Differentiation Formulas — In this section we give most of the general derivative formulas and properties used when taking the derivative of a function.
Examples in this section concentrate mostly on polynomials, roots and more generally variables raised to powers. Product and Quotient Rule — In this section we will give two of the more important formulas for differentiating functions.
We will discuss the Product Rule and the Quotient Rule allowing us to differentiate functions that, up to this point, we were unable to differentiate. Derivatives of Trig Functions — In this section we will discuss differentiating trig functions.
Derivatives of Exponential and Logarithm Functions — In this section we derive the formulas for the derivatives of the exponential and logarithm functions.
Derivatives of Inverse Trig Functions — In this section we give the derivatives of all six inverse trig functions. We show the derivation of the formulas for inverse sine, inverse cosine and inverse tangent. Derivatives of Hyperbolic Functions — In this section we define the hyperbolic functions, give the relationships between them and some of the basic facts involving hyperbolic functions.
We also give the derivatives of each of the six hyperbolic functions and show the derivation of the formula for hyperbolic sine.
Chain Rule — In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. With the chain rule in hand we will be able to differentiate a much wider variety of functions.
As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Implicit Differentiation — In this section we will discuss implicit differentiation.
Not every function can be explicitly written in terms of the independent variable, e. Implicit differentiation will allow us to find the derivative in these cases. Knowing implicit differentiation will allow us to do one of the more important applications of derivatives, Related Rates the next section.
In related rates problems we are give the rate of change of one quantity in a problem and asked to determine the rate of one or more quantities in the problem. This is often one of the more difficult sections for students. We work quite a few problems in this section so hopefully by the end of this section you will get a decent understanding on how these problems work.
Higher Order Derivatives — In this section we define the concept of higher order derivatives and give a quick application of the second order derivative and show how implicit differentiation works for higher order derivatives.
Logarithmic Differentiation — In this section we will discuss logarithmic differentiation. Logarithmic differentiation gives an alternative method for differentiating products and quotients sometimes easier than using product and quotient rule. More importantly, however, is the fact that logarithm differentiation allows us to differentiate functions that are in the form of one function raised to another function, i.1.
Take derivative 2. Plug given x value into derivative (for slope) 3. Negative reciprocal of slope 4. Plug into point-slope form. Study Guide Resource Home Textbook Instructor's Manual Study Guide Computing in Calculus (PDF - MB) 2: Derivatives.
The Derivative of a Function Powers and Polynomials The Slope and the Tangent Line. Calculus Here is a list of skills students learn in Calculus!
These skills are organized into categories, and you can move your mouse over any skill name to preview the skill. Beginner’s Guide to NCFM Certification Exam:If you are considering undertaking NCFM modules, there are a lot many of them to caninariojana.com of course cannot consider giving all of them at the same time.
Here through this article, I wish to provide you with a brief summary on their modules which might help you decide which one you should choose from. Aug 07, · Derivatives Essentials: An Introduction to Forwards, Futures, Options, and Swaps (Wiley, ) by Aron Gottesman is an excellent textbook/self-study guide.
It . will use the product/quotient rule and derivatives of y will use the chain rule. The “trick” is to The “trick” is to differentiate as normal and every time you differentiate a y you tack on a y ′ (from the chain rule).
View Test Prep - DERIVATIVES study guide from ACCT I S at University of Wisconsin. DERIVATIVES Derivative financial instruments have become the key tools of RISK MANAGEMENT. Derivatives financial. The project that you can use to engage the students is to use word art as a way of creating a study guide of some sort. There are several terms associated with derivatives, so good way to remember the key terms would be to write the most important terms from . Calculus 1 Class Notes, Thomas' Calculus, Early Transcendentals, 12th Edition Copies of the classnotes are on the internet in PDF format as given below. Introduction to .
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334579.46/warc/CC-MAIN-20220925132046-20220925162046-00133.warc.gz
|
CC-MAIN-2022-40
| 6,690 | 23 |
https://legacy.slmath.org/summer_schools/718
|
math
|
Summer Graduate School
In cooperation with INdAM (Istituto Nazionale di Alta Matematica) and the CMI (Clay Mathematical Institute), MSRI will sponsor a summer graduate school (SGS) on Mathematical General Relativity in Cortona during the summer of 2013; the SGS will reprise the very successful school on Mathematical General Relativity held at MSRI in 2012.
Mathematical general relativity is the study of mathematical problems related to Einstein's theory of gravitation. There are interesting connections between the physical theory and problems in differential geometry and partial differential equations.
The purpose of the summer school is to introduce graduate students to some fundamental aspects of mathematical general relativity, with particular emphasis on the geometry of the Einstein constraint equations and the Positive Mass Theorem. These topics will comprise a component of the upcoming semester program at MSRI in Fall 2013.
There will be mini-courses, as well as several research lectures.
Downloads for Students:
Intro to General Relativity Notes
Problems: Geometry Basics
Problems: Scalar Curvature Basics
Problems: Schwarzschild Basics
Problems: Harmonic Function Basics
Problems Related to Fernando Schwartz's Lectures
Problems Related to Lan-Hsuan Huang's Lectures
- In addition to the nomination at MSRI, a separate Application (Application deadline: April 30, 2013) is required to participate in this workshop.
- Participating students are expected to have had courses in graduate real analysis and Riemannian geometry, while a course in graduate-level partial differential equations is recommended.
- MSRI can only support students from US institutions.
- Due to the small number of students supported by MSRI, only one student per institution will be accepted.
For eligibility and how to apply, see the Summer Graduate Workshop homepage
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00217.warc.gz
|
CC-MAIN-2024-10
| 1,865 | 18 |
https://www.physicsforums.com/threads/change-in-electric-potential-potential-energy-and-work-problem.766294/
|
math
|
Imagine a lithium atom where the two electrons in the first orbital are at exact opposite sides of the nucleus and the electron in the second orbital is in line with the other electrons so that the three electrons and the nucleus all lie on a straight line. How much work would you need to apply to remove the outer most electron if the atomic radius is 100picometer and the distance between the first and second orbital is 50picometer?
The answer is (kq^2/300)*10^12J.
Based on the principle of superposition, I can find the electric potential at the outermost electron for lithium. (-kq)/50*10^-12+(3kq)/100*10^-12+(-kq)/150*10^-12 where k is the constant and a a is the type of charge.
The potential I find is then kq/300*10^-12.
To find work needed to move up, I need the electric potential "difference" from the outer electron to to the next level multiplied by a -q to get potential energy.(intuitively I know that potential energy will be positive and work will be negative; therefore the work applied must be positive.)
I don't know how the book just found this potential difference and ultimately the answer choice.
I have this feeling that the book left out information to the point that this problem is impossible to solve.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875141806.26/warc/CC-MAIN-20200217085334-20200217115334-00054.warc.gz
|
CC-MAIN-2020-10
| 1,234 | 7 |
https://www.essay-pros.com/construct-a-99confidence-interval-for-the-populat/
|
math
|
construct A 99%CONFIDENCE interval for the population mean . Assume the population has a normal distribution. A group of 19 randomly selected students has mean age of 22.4 years with a standard deviation of 3.8 years
We Deliver Top Quality On Time As Promised!
Ready to get started? Just click the button to be directed to a secure page so you can enter the details of your paper.
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187820700.4/warc/CC-MAIN-20171017033641-20171017053641-00666.warc.gz
|
CC-MAIN-2017-43
| 380 | 3 |
http://www.docstoc.com/docs/14917297/Capital-Budgeting-NPV-and-Alternative-Investment-Rules
|
math
|
NPV and Alternative Investment Rules
November 27, 2001
Objectives of lecture on captital budgeting rules
In previous lectures we have already argued that the appropriate goal from shareholders
point of view is to maximize NPV. But in practice other method are used as well. ⇒
Comparison of different capital budgeting techniques.
• Payback period rule
• Discounted payback period rule
• Average accounting return (AAR)
• Internal rate of return (IRR)
• Profitability index
Then will discuss in more detail problems using NPV in capital budgeting.
Why should investment rule be based on NPV?
When we discussed the separation theorem, we have shown that management which invests
in projects with positive NPV ⇔ increases shareholder value.
NPV is appropriate rule since:
• NPV uses cash flows not earnings: cash flows really what is available to pursue cor-
porate investment plans.
• NPV uses all cash flows: Cash flows exhaustively describe an investment opportunity
from the point of view of valuation. Other methods ignore cash flows after a certain
point in time.
• NPV discounts the cash flows properly: Captures the time value of money as revealed
by the markets.
Why are other methods used in practice?
• Historical reasons and convenience.
• Alternative methods less complicated.
• Lack of appropriate estimates for cash flows and risk adjusted discount rates.
Pay Back Period Rule
Idea: Many investment projects start with a cash outflow followed by cash inflows in the
future. ⇒ How many years does it take until initial cash outflow has been paid back ?
Answer: Pay back period .
Application: Management decides about what it considers as acceptable (upper limit)
time until initial investment is paid back.
⇒ Rule: Undertake all investment projects with a pay back period less than say 5 years.
⇒ Ranking Criteria: Undertake investment project with shortest pay back period.
Pros of Payback Period
• Upper limit on cut off time limits risk exposure of investment projects.
• Favors liquidity of firm.
• Respects short term investment horizon.
Cons of Payback Perido
Projects considered as identical with respect to payback period can be extremely different:
• Different cash flows after cut off period are ignored.
• Timing of cash flows before cut off period is not taken into account: Two projects
with same initial cash outflow of $100 followed by cash inflows of $10 over 10 year for
one project and no cash inflows for 9 year but $ 100 in the 10th year have the same
pay back period! (Should prefer the first project since intermediate cash inflows can
be reinvested.) ⇒ Payback period method ignores time value of money.
• Biased against long-term projects and therefore against shareholder interests.
• Criteria may not exist and/or acceptance criteria is arbitrary.
Discounted Pay Back Period
Idea: Correct payback period method such that time value of money is taken into account.
Pros: Same as pay back period. But in addition reflect opportunity costs of money as
valued by the financial markets.
Cons: As pay back period. But in addition get more complicated since have to discount
as for NPV rule.
Averge Accounting Return (AAR)
Idea: It is often easier to determine projects earning rather than cash flows. Calculate rate
of return of projects earning by dividing its earnings by the project’s book value. Since
both vary offer time take time averages ⇒ Average Accounting Return.
Average net income
Application:Proceed in two steps
1. Determine average net income:
Average Net Income = (Revenuen − Expensesn
−Depreciationn − Taxesn )
2. Determine average investment:
Average Investment = (Value of investmentn
− Depreciationn )
Decison Rule for ARR
⇒ Rule: Management defines hurdle rate. Undertake investment when AAR is higher
than this hurdle rate.
⇒ Ranking criteria: Invest in project with highest hurdle rate.
Pros of ARR
• Accounting information available.
• Easy to calculate.
• Does always exist.
Cons of ARR
• Ignores the time value of money: average net income can be same but income per
period can be very different.
• Benchmark (hurdle rate) arbitrary.
• Based on book values: Do not reflect means available for investment project as cash
flows to and book values do not reflect valuation by the market.
Internal Rate of Return (IRR)
Idea: Given the cash flow structure of a given investment project at which interest rate is
its NPV equal to zero ?
Application: Management fixes target rate of return.
IRR solves the following equation:
(1 + IRR)n
IRR corresponds to the “yield” of the investment project.
Investing and Financing
1. Investing: Undertake investment if IRR is higher than a specified hurdle rate which
may correspond to discount rate from financial markets. (Opportunity cost of capital).
2. Financing: Accept financing project if IRR is lower than a specified hurdle rate
which may correspond to the discount rate from financial markets (Opportunity cost
Ranking criteria: Undertake investment project with highest IRR and accept financing
project with lowest IRR.
NPV and IRR
Remark: IRR is very close to NPV:
• Both are based on all cash flows and capture time value of money.
• Both evaluate investment project by a simple number
But IRR does not always exist or is not necessarily unique.
Pros of IRR
• Uses all cash flows.
• Not based on accounting information.
• Easy to understand.
• Easy to communicate.
Cons of IRR
• Does not capture different scale of project.
• Does not always exist (“is not always real”) and may not be unique:
Example: Consider cash flows C0 , C1 , C2 . Then IRR is given by solution
C0 + + =0
1 + IRR (1 + IRR)2
then define X = 1+IRR and try to find X.
The equation defining IRR is quadratic and you may remember that the general
solution for X is then
−C1 + C1 − 4C0 C2
−C1 − 2
C1 − 4C0 C2
and consequently we will have two solutions for IRR if C1 − 4C0 C2 is not equal to
2 − 4C C < 0 we have one IRR which is not a real number !
zero. Furthermore if C1 0 2
Remark : As you can see this does not occur if the initial investment is negative and
all other cash flows are positive (Investment) or vice versa all but the last cash flows
are positive (Financing).
Cons IRR continued
• Does not distinguish between investing and financing. Example: If we multiply all
cash flows by the same number, the IRR does not change ! Consequently if we compare
an investment project with a financing project we obtain the same IRR!.
• Re-investment rate assumption: IRR rule does not discount at opportunity costs
of capital, it implicitly assumes that the time value of money is the IRR. Implicit
assumption: ⇒ Re-investment rate assumption.
Therefore the IRR assumes that investor can reinvest their money at the IRR which
is different from the market-determined opportunity costs. It is logically not coherent
that the re-investment rate should be different if they are in the same risk class.
• Value Additivity Principle: The value of the firm should be equal to the sum of the
values of the individual projects of the firm.
Illustration: Violation of Value Additivity Principle
Therefore if we have choice between two mutually exclusive projects and an indepen-
dent project, it should not occur that project one is preferred to project two but if
mixed with the independent project, project two is preferred. But this can happen
– Project 1: -100,0,550 ⇒ IRR=134.4%
– Project 2: -100,225,0 ⇒ IRR=125.0%
– Project 3: -100,450,0 ⇒ IRR=350 %
– Project 1+3: -200,450,550 ⇒ IRR=212.8%
– Project 2+3: -200,675,0 ⇒ IRR=237.5%
Idea: IRR is attractive since it is not dependent on scale of investment. It is a “per dollar”
quantity. The profitability index does the same thing for NPV: Amount of NPV generated
for each investment dollar.
PV of cash inflows
initial cash outflow
PV of cash outflows
PIF in =
initial cash inflow
Application: Management defines benchmark for index
⇒ Undertake investment if profitability index is higher than benchmark.
⇒ Undertake investment with highest profitability index.
Pros and Cons for Profitability Index
• respects limited availability of investment funds.
• easy to understand and communicate
• same decision as NPV for independent projects
Cons: As IRR non-additive ⇒ Problems when evaluate mutually exclusive projects.
Conclusion: Valuation principles
Only NPV satisfies all of the following requirements:
• Considers all cash flows.
• Cash flows are discounted at the true opportunity cost.
• Selects from mutually exclusive projects the one that maximizes shareholder value.
• Respects value-additivity principle.
Conflict among Criteria
Exercise: Consider the following projects:
• A: -1000,100,900,100,-100,-400
• B: -1000,0,0,300,700,1300
• C: -1000,100,200,300,400,1250
• D: -1000,200,300,500,500,600
Show that payback rule chooses A, ARR rule B, NPV with discount rate 10% C and IRR
Difficulties Using NPV Rule
In practice several issues make the use of NPV rule difficult:
• Measurement of cash flows
• Incremental cash flows not accounting income
• Measurement of opportunity costs of capital
• Investments with unequal lives
Measurement of cash flows
• Incremental cash flow: What part of the cash flow is exclusive consequence of invest-
ment project ? ⇒ Ignore Sunk costs ⇒ Include Opportunity costs ⇒ Correct for Side
• Projection of future cash flows.
• Real versus nominal cash flows.
• Capital Cost Allowance (CCA). .
If we assume straight-line depreciation then the net cash flow (NCF) can be obtained in
two equivalent ways:
1. Total project cash flow approach
2. Tax Shield Approach
Remark: In practice we have to use Declining Balance CCA
Total project cash flow approach
N CF = (N 0I − CCA) × (1 − tc ) + CCA
• tc : Corporate tax rate
• NOI: Net operating income
• NOI-CCA: Taxable income
• (N OI − CCA) × (1 − tc ): Net income
Tax shield approach
N CF = N OI × (1 − tc ) + tc × CCA
• N OI × (1 − tc ): After tax net operating income.
• tc × CCA: Tax shield (tax savings).
Measurement of opportunity costs of capital
• Real versus nominal interest rate (The Fisher relation). ⇒ Right discount rate de-
pends on whether cash flows are real or nominal.
• Correction of discount factor for risk.
Investments with unequal lives
• Matching cycles: ⇒ repeat projects until timing matches.
Example Cost project A −500, 120, 120, 120 project B −600, 100, 100, 100, 100. To
compare consider smallest common multiple 12 and compare PV of 4 repetitions of
projection A with 3 repetitions of project B.
• Equivalent annual costs: Consider example as before.
1 1 1
EAC A = P V A [ + + ]−1
1 + r (1 + r)2 (1 + r)3
1 1 1 1
EAC B = P V B [ + 2
1 + r (1 + r) (1 + r) (1 + r)4
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00447-ip-10-147-4-33.ec2.internal.warc.gz
|
CC-MAIN-2014-15
| 10,923 | 216 |
https://fdocuments.in/document/1which-of-the-following-is-a-property-of-real-gases-but-not-ideal-gases-a-random-motion-b-attractive-forces-c-mass-d-kinetic-energy.html
|
math
|
1.Which of the following is a property of real gases but NOT ideal gases. A) random motion B)...
Embed Size (px)
Transcript of 1.Which of the following is a property of real gases but NOT ideal gases. A) random motion B)...
1.Which of the following is a property of real gases but NOT ideal gases.
A) random motion B) attractive Forces C) mass D) kinetic energy
2.Ideal gas molecules are considered to have no volume or diameter and are therefore referred to as:
A) point masses B) polar mc C) molar volumes D) mole fractions
3. Random scattering of gas mc from a place of high concentration to a place of low concentration is:
A) migration B) distribution C) diffusion D) effusion
4.A system insulated from the surroundings is:
A) standard state B) dynamic equilibrium C) adiabatic system D) endothermic reaction
5. Diffusion rate is inversely proportional to A) density B) number of mc present in 1 liter in STP C) average temperature D) square root of its molecular mass
6. Which of the following gases will have the highest rate of diffusion?
7. At constant pressure, the volume of a given quantity of gas and the Kelvin temperature of that gas:
A) vary directly B) vary indirectly C) vary inversely D) are unrelated
8. At constant volume, the relationship between the pressure and the Kelvin temp of a gas is such that pressure and Kelvin temp. vary
A) irregularly B) inversely C) indirectly D) directly
9. Avogadro’s principle states that at equal temp and equal pressures, equal volumes of gases
A) have the same density B) have the same average molecular mass C) contain the same number of particles D) have the same diffusion rate
10. Under standard conditions, the volume occupied by one mole of any gas isA) 1 Liter
B) dependent on its density C) called the molar volume D) directly proportional to its molecular
11. If you collect 22.4 L of oxygen at standard conditions, you will have
A) 2 mol of gas B) 16 g of gas C) 359 g of gas D) 1 mol of gas
12. In the ideal gas law “n” stands for A) molecular mass B) moles of gas C) mass of gas D) density of gas
13. What is the effect on the volume of a gas if pressure is doubled. Temp is constant?
A) no effect B) volume increases C) volume decreases D) volume decreases by half
14. AT constant temp, what is the effect on pressure of tripling the volume;
A) it increases B) it decreases C) increases 3X D) no effect
15. At constant pressure, what is the effect on the volume if temperature increases by 1o C
A) volume stays the same B) volume increases C) volume decreases D) no effect
16. At constant volume, what is the effect on the pressure if you double Kelvin temp
A) decrease B) increase C) doubles D_ no effect
17. What is the freezing point of water in Kelvin
A) 32 B) 0 C) – 273 D) 273
18. What is the Kelvin temperature of 25oC?
A) 273 B) 298 C) 248 D) 25
19. If a tiny valve opens in a container with He, Ar, Ne, Kr, which gas will escape the container fastest?
A) Kr B) He C) Ne D) Ar
20. Which of the following statements is always true about two different gas samples at the same temp
A) same average kinetic energy B) same molecular weight C) occupy same volume D) exert the same pressure
21. Which substance has the lowest density?
A) water (l) B) water(g) C) water (s) D) water (aq)
22. The density of a substance undergoes the greatest change when the substance changes from a
A) liquid to solid B) liquid to gas C) solid to liquid D) molecular solid to ionic solid
23. V1P1 = V2P2
A) Charles’ B) Boyle’s C) Gay-Lussac’s D) Dalton’s
24. V1T2 = V2T1
A) Charles’ B) Boyle’s C) Gay-Lussac’s D) Dalton
25. V = nk
A) Dalton B) Gay’Lussac C) Avogadro D) Graham
26. T1P2 = T2P1
A) Dalton B) Graham C) Gay-Lussac D) Avogadro
27. VP = nRT
A) Graham B) Dalton C) Ideal D) Avogadro
29. Pt = P1 + P2 ……..
A) Graham B) Avogadro C) Dalton D) Ideal
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00327.warc.gz
|
CC-MAIN-2023-23
| 3,839 | 56 |
http://www.ctechglass.com/ctech-ad0006003-p-31.html
|
math
|
Connecting Adapter, Straight$21.95
Borosilicate glass serated inlet adapter, which allows for flexible connection of tubing to apparatus to introduce a vacuum or controlled pressure environment. Hose connection is bent at a 90 degree angle to the joint and has an O.D. of 8mm at its largest serration.Inner joint size 24/40.
all products in this category.
Round Bottom Flasks, Heavy Wall$10.95
Beaker, Borosilicate, Short Form$2.95
Thermometer Vacuum Inlet Adapter$22.95
Flow Control Adapter, Stopcock, Inlet 90 degrees, Outer Joint$23.95
Connecting Adapter, Combination$23.95
Bench Clamp for Rod$14.95
|
s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440646313806.98/warc/CC-MAIN-20150827033153-00351-ip-10-171-96-226.ec2.internal.warc.gz
|
CC-MAIN-2015-35
| 602 | 9 |
http://gta.math.unibuc.ro/memoriam.html
|
math
|
DEPARTMENT OF COMPLEX GEOMETRY, TOPOLOGY AND COMPUTATIONAL ALGEBRA
The Departament of Complex Geometry, Topology and Comutational algebra
of the Faculty of Mathematics, University of Bucharest, sorrowly announces
the premature lost of our colleague,
Professor Nicolae RADU was, not so long ago, Chairman of the Departement
of Algebra and Dean of the Faculty of Mathematics: he was a proeminent
personality of mathematics and an excellent professor.
Professor Nicolae RADU was an astonishing mathematician and a very kind
person; on the other hand, he was always extremeley exigeant as a
mathematician. Many romanian mathematicians chosed him to be their Ph. D.
May him rest in peace...
Professor NICOLAE RADU
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-23/segments/1685224652207.81/warc/CC-MAIN-20230606013819-20230606043819-00381.warc.gz
|
CC-MAIN-2023-23
| 708 | 12 |
http://antonio-bonifati.blogspot.com/2011/12/margherita-hack-and-size-of-our.html
|
math
|
For the original Italian version, go here.
Some Einsten's musings about this question.
Continuing our series with the eminent 90-year-old Italian astrophysicist Margherita Hack, Euronews asked whether she thought the universe was finite or infinite?
It could be infinite. We know that the universe is flat. It means it obeys Euclidean geometry.
If the universe were finite and closed like the surface of a sphere... obviously we couldn't be certain it's finite. But from the way in which light propagates and certain measures that could be taken on the structure observed in the early universe, you can deduce that our universe is flat.
But what does a flat universe mean? That it's infinite. It could be infinite as the Euclidean universe the one we study in geometry and high school.
But the concept of infinity is almost a philosophical concept. Not really. There's a practical example. Numbers. Numbers are infinite. So the concept of infinity is one very palpable when you think about numbers.
But numbers are something that have been created by man. Well, the universe is there and it's almost scary to think of it as something infinite. But if we think of something finite then what is there outside it? It seems to me even more difficult.
I prefer much more the hypothesis of an infinite universe. It's always been there. It's always existed and it has no borders. We think of the universe as everything that exists, but it's a mistake we've made before that the earth was unique and the sun was the centre of the universe, that our galaxy was the center, which was unique. It might be that what we call the universe is just one of countless other universes. And so this could be yet another solution. What's
outside our universe? Other universes.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046154089.6/warc/CC-MAIN-20210731105716-20210731135716-00647.warc.gz
|
CC-MAIN-2021-31
| 1,755 | 10 |
https://www.hackmath.net/en/math-problem/2498
|
math
|
To a semicircle with diameter 10 cm inscribe square. What is the length of square sides?
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Thank you for submitting an example text correction or rephasing. We will review the example in a short time and work on the publish it.
Tips to related online calculators
You need to know the following knowledge to solve this word math problem:
Related math problems and questions:
The rectangle is 18 cm long and 10 cm wide. Determine the diameter of the circle circumscribed to the rectangle.
- Chord AB
What is the chord AB's length if its distance from the center S of the circle k(S, 92 cm) is 10 cm?
- Concentric circles and chord
In a circle with a diameter d = 10 cm, a chord with a length of 6 cm is constructed. What radius have the concentric circle while touch this chord?
- Silver medal
To circular silver medal with a diameter of 10 cm is an inscribed gold cross, which consists of five equal squares. What is the content area of the silver part?
The trunk diameter is 52 cm. Is it possible to inscribe a square prism with side 36 cm?
- Ratio of sides
Calculate the area of a circle with the same circumference as the circumference of the rectangle inscribed with a circle with a radius of r 9 cm so that its sides are in ratio 2 to 7.
- Circle chord
Calculate the length of the chord of the circle with radius r = 10 cm, length of which is equal to the distance from the center of the circle.
- Chord distance
The circle k (S, 6 cm), calculate the chord distance from the center circle S when the length of the chord is t = 10 cm.
- Find the
Find the length of the side of the square ABCD, which is described by a circle k with a radius of 10 cm.
- Circumscribed circle to square
Find the length of a circle circumscribing a square of side 10 cm. Compare it to the perimeter of this square.
- Tree trunk
What is the smallest diameter of a tree trunk that we can cut a square-section square with a side length of 20 cm?
If the endpoints of a diameter of a circle are A(10, -1) and B (3, 10), what is the radius of the circle?
- Rhombus and inscribed circle
It is given a rhombus with side a = 6 cm and the radius of the inscribed circle r = 2 cm. Calculate the length of its two diagonals.
To circle with a radius of 41 cm from the point R guided two tangents. The distance of both points of contact is 16 cm. Calculate the distance from point R and circle centre.
- Square circles
Calculate the length of the described and inscribed circle to the square ABCD with a side of 5cm.
The log has a diameter 30 cm. What's largest beam with a rectangular cross-section can carve from it?
- Square and circle
Into square is an inscribed circle with diameter 10 cm. What is the difference between circumference square and circle?
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585348.66/warc/CC-MAIN-20211020183354-20211020213354-00544.warc.gz
|
CC-MAIN-2021-43
| 2,796 | 35 |
http://library.thinkquest.org/C0122767/sciences/mathematics.html
|
math
|
China once was very advanced in mathematics. The following few paragraphs will describe some of these advanced discoveries, inventions, etc..
The major advancement in Mathematics is brought about by the book Nine Chapters of Mathematics.
Nine Chapters of Mathematics.
introduced many mathematical propositions and its solutions
the World's most advanced applied mathematics at the time
its appearance showed that Chinese mathematics has developed into a system
Wang Xiaotong's mathematical book, Ji Gu Xuan Jing gave
world's first method of finding the roots of a cubic function. This
achievement was earlier than the Arabs by by 300 years and
earlier than the Europeans by 600 years.
mathematician Ming Antu was the first to
study by analytic methods
|
s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163037829/warc/CC-MAIN-20131204131717-00057-ip-10-33-133-15.ec2.internal.warc.gz
|
CC-MAIN-2013-48
| 751 | 12 |
https://nasaconstellation.com/w-2pif/
|
math
|
Bạn đang xem: Angular frequency
What does 2pif mean?
97 1. frequency-number of rotations in a period of time, usually one s. w-angular velocity-angle in an amount of time. 2pi-one full circle. hence w=2pif.
What is 2pi frequency?
Frequency is in cycles per second. Multiplying by 2π gives the frequency in radians per second, where a radian is a measure of angle such that 2π radians = 360°.
Why is omega equal to 2pif?
The angular frequency ω is another way of expressing the number of turns, in terms of radians. One full circle consists of 2π radians of arc, so we multiply the “number of circles per second” by 2π to get the “number of radians per second”—which we call the angular frequency, ω.
What is V 2πr T?
v=T2πr. In Physics, Uniform Circular Motion is used to describe the motion of an object traveling at a constant speed in a circle. … Here, r represents the radius of the circle, T the time it takes for the object to make one complete revolution, called a period.
Is time a frequency?
Frequency is a rate quantity. Period is a time quantity. Frequency is the cycles/second. Period is the seconds/cycle.
What is the relationship between Omega and frequency?
Angular frequency ω (in radians per second), is larger than frequency ν (in cycles per second, also called Hz), by a factor of 2π. This figure uses the symbol ν, rather than f to denote frequency. A sphere rotating around an axis. Points farther from the axis move faster, satisfying ω = v / r.
What is W Omega?
Angular frequency (ω), also known as radial or circular frequency, measures angular displacement per unit time. Its units are therefore degrees (or radians) per second. Angular frequency (in radians) is larger than regular frequency (in Hz) by a factor of 2π: ω = 2πf.
How do you get Omega?
ω = v/r, where ω is the Greek letter omega. Angular velocity units are radians per second; you can also treat this unit as “reciprocal seconds,” because v/r yields m/s divided by m, or s-1, meaning that radians are technically a unitless quantity.
What is 2pi by Omega?
It’s an invention. Originally Answered: What is the proof of the theory “omega”= (2*pi)/T in circular motion? T is the period of the motion, the time for full cycle or 360 degrees – which is 2π radians. w (omega) is the angular velocity in radians per second.
What is difference between Omega and F?
In general, ω is the angular speed – the rate change of angle (as in a circular motion). Frequency (f) is 1/T or the number of periodic oscillations or revolutions during a given time period. Angular speed or angular frequency, relates the same idea to angles – how much angle is covered over a time period.
What does Theta mean in physics?
Angular Position, Theta. The angle of rotation is a measurement of the amount (the angle) that a figure is rotated about a fixed point— often the center of a circle.
What is the angular frequency of the oscillations?
The angular frequency ω is given by ω = 2π/T. The angular frequency is measured in radians per second. The inverse of the period is the frequency f = 1/T. The frequency f = 1/T = ω/2π of the motion gives the number of complete oscillations per unit time.
What is V 2 in circular motion?
Centripetal acceleration The centripetal acceleration is the special form the acceleration has when an object is experiencing uniform circular motion. It is: ac = v2 / r. and is directed toward the center of the circle.
What does capital T stand for in physics?
A capital T is often used to mean: Time period () Temperature. Kinetic energy.
What is the difference between V and Omega?
The symbol for the linear velocity is v. The symbol for angular velocity is a Greek letter, i.e., ω (pronounced as omega). … We measure the linear velocity in m/s. We measure the angular velocity in both degrees and radians.
What is temporal frequency?
The term temporal frequency is used to emphasise that the frequency is characterised by the number of occurrences of a repeating event per unit time, and not unit distance.
How many types of frequency are there?
Frequency Band NameAcronymFrequency RangeMedium FrequencyMF300 to 3000 kHzHigh FrequencyHF3 to 30 MHzVery High FrequencyVHF30 to 300 MHzUltra High FrequencyUHF300 to 3000 MHz
How do you tell the difference between frequency and period?
Period refers to the amount of time it takes a wave to complete one full cycle of oscillation or vibration. Frequency, on the contrary, refers to the number of complete cycles or oscillations occur per second. Period is a quantity related to time, whereas frequency is related to rate.
How do you get rid of time period?
each complete oscillation, called the period, is constant. The formula for the period T of a pendulum is T = 2π Square root of√L/g, where L is the length of the pendulum and g is the acceleration due to gravity.
What is difference between angular frequency and frequency?
Either one can be used to describe periodic or rotational motion. Frequency (usually with the symbol or ) is cycles (or revolutions) per second. Angular frequency (usually with the symbol ) is radians per second. 1 cycle (or revolution) equals radians, so .
What is angular frequency and frequency?
Frequency definition states that it is the number of complete cycles of waves passing a point in unit time. The time period is the time taken by a complete cycle of the wave to pass a point. Angular frequency is angular displacement of any element of the wave per unit time.
What is R Omega?
R-Omega is a phospholipid rich DHA and EPA omega-3 supplement from herring roe. … R-Omega contains 340mg of DHA and 100mg of EPA per two capsule dose.
What is the weird W in physics?
(mathematics, set theory) The first (countably) infinite ordinal number, its corresponding cardinal number ℵ0 or the set of natural numbers (the latter of which are often defined to equal the former).
What does ω stand for?
Greek Letter Omega The 24th and last letter of the Greek alphabet, Omega (Ω), essentially means the end of something, the last, the ultimate limit of a set, or the “Great End.” Without getting into a lesson in Greek, Omega signifies a grand closure, like the conclusion of a large-scale event.
Where can I find Omega in GTA 5?
Travel to the green “?” in the eastern section of Sandy Shores to meet Omega. Approach him and the Far Out mission will begin.
How do you find angular frequency from period?
It is the reciprocal of the period and can be calculated with the equation f=1/T. Some motion is best characterized by the angular frequency (ω). The angular frequency refers to the angular displacement per unit time and is calculated from the frequency with the equation ω=2πf.
What is natural angular frequency?
When calculating the natural frequency, we use the following formula: f = ω ÷ 2π Here, the ω is the angular frequency of the oscillation that we measure in radians or seconds. We define the angular frequency using the following formula: ω = √(k ÷ m)
How do you calculate omega t?
At a particular moment, it’s at angle theta, and if it took time t to get there, its angular velocity is omega = theta/t. So if the line completes a full circle in 1.0 s, its angular velocity is 2π/1.0 s = 2π radians/s (because there are 2π radians in a complete circle).
Is Omega an Radian?
Angular frequency ω(Ordinary) frequency1 radian per secondapproximately 0.159155 Hz1 radian per secondapproximately 57.29578 degrees per second
What is meant by theta in maths?
The Greek letter θ (theta) is used in math as a variable to represent a measured angle. For example, the symbol theta appears in the three main trigonometric functions: sine, cosine, and tangent as the input variable.
How do you write theta?
To Type Theta θ in MS Word, another method is to use it from symbols. In symbols, goto subset Greek and Coptic and select θ from it. Third Method: One of the method to type theta θ in MS Word is press and hold alt key + 952 from numpad.
What is difference between angular velocity and angular frequency?
Angular velocity is that velocity which act on revolution or rotation of a body with a certain angle with axis with valid radius. … Angular frequency also known as radial or circular frequency, measures angular displacement per unit time. Its units are therefore degrees (or radians) per second.
Does mass affect angular frequency?
With other variables held constant, as mass increases, angular momentum increases. Thus, mass is directly proportional to angular momentum.
What is oscillation frequency?
The frequency of oscillation is the number of full oscillations in one time unit, say in a second. A pendulum that takes 0.5 seconds to make one full oscillation has a frequency of 1 oscillation per 0.5 second, or 2 oscillations per second.
What is M Squared Omega?
Or. Or. V= velocity of the body. w=angular velocity(omega) R=Radius of the circular path in which the body is moving.
What"s the formula of tension?
Tension Formula. The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration. T = mg + ma. T = tension, N, kg-m/s2.
Xem thêm: Depth Of Field Là Gì Tại Sao Lại Có Câu Depth Of Field Trong Game Là Gì
What is MG force?
The weight of an object is the force of gravity on the object and may be defined as the mass times the acceleration of gravity, w = mg. Since the weight is a force, its SI unit is the newton. Density is mass/volume.
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882573029.81/warc/CC-MAIN-20220817153027-20220817183027-00644.warc.gz
|
CC-MAIN-2022-33
| 9,498 | 76 |
https://www.emis.de/journals/AMAPN/vol17/7.html
|
math
|
Abstract: Editorial notice. Due to some technical problems, in the paper Pal Gyula - Julius Pal (1881-1946), the Hungarian-Danish mathematician, published in vol. 16 of this journal, the reference numbers in the text mixed, moreover the published paper contains some obvious misprints. It seems that not the latest version was posted. We can not reconstruct the exact circumstances, and the origin of these problems could not be found. The editorial board apologize to the authors for this. For the readers convenience we publish the whole paper again. The authors fixed some spelling and other minor errors as well and placed a new footnote.
Keywords: History of Hungarian mathematics, Jordan curves, Kakeya problem
Classification (MSC2000): 01A70; 01A60
Full text of the article:
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00174.warc.gz
|
CC-MAIN-2023-14
| 781 | 4 |
http://music.stackexchange.com/tags/bass-trombone/new
|
math
|
New answers tagged bass-trombone
It is hard to say what you have, although I suspect it may be a JinBao. Especially if your horn is from eBay as a discount model or from one of the JinBao distributors like Mack Brass, Wessex, esp. Concerning the size: Many manufacturers use the Bach numbering system, of which this is one. "1 1/2" is a general bass trombone size that means roughly 28.2mm ...
Top 50 recent answers are included
|
s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701171770.2/warc/CC-MAIN-20160205193931-00335-ip-10-236-182-209.ec2.internal.warc.gz
|
CC-MAIN-2016-07
| 428 | 3 |
https://www.prep4usmle.com/forum/thread/107519/
|
math
|
|Prep for USMLE|
| Forum | Resources||New Posts | Register | Login||» |
my cs in LA 29/july any one wouldlike to exchange with me from 1/ july to 15/ july.. at any test center...
This thread is closed, so you cannot post a reply.
| Similar forum topics|
CS July 26th date exchange
Step2 cs date exchange July 2-->Sept
CS Date Exchange. Exam on 28th July
| Related resources|
Advertise | Support | Premium | Contact
|
s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335054.79/warc/CC-MAIN-20220927162620-20220927192620-00615.warc.gz
|
CC-MAIN-2022-40
| 456 | 10 |
https://gaming.stackexchange.com/questions/7875/what-limits-are-there-on-the-civilization-v-demo
|
math
|
Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Q&A for work
Connect and share knowledge within a single location that is structured and easy to search.
How is the Civ 5 demo limited, how long is it, and is the 4+ GB installation worth it?
The Civ 5 demo is a single map that will be the same each time you play it, has 3 civilizations to choose from, and you can only play for 100 turns.
|
s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474808.39/warc/CC-MAIN-20240229103115-20240229133115-00684.warc.gz
|
CC-MAIN-2024-10
| 537 | 5 |
http://www.wyzant.com/Head_Of_The_Harbor_NY_Algebra_tutors.aspx
|
math
|
Port Jefferson Station, NY 11776
Professor of Information Technology and Mathematics
...I have been teaching at a local college for the past 20 years. While teaching my Information Technology courses I have also taught College Algebra
since computer related classes require a thorough knowledge of mathematics. I have also tutored math many times with...
including algebra 1 and algebra 2
|
s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011064849/warc/CC-MAIN-20140305091744-00079-ip-10-183-142-35.ec2.internal.warc.gz
|
CC-MAIN-2014-10
| 388 | 5 |
https://www.kiasuparents.com/kiasu/question-source/river-valley-primary/
|
math
|
Appreciate help on this question. thanks
Appreciate help on part b finding the shaded part in the Centre .
Hi Parents/Teachers, Please help on this question. The answer is 64. Thanks. DNLNMUM
Hi parents, I need help for this questions as we couldn’t solve it, many many thanks!
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00694.warc.gz
|
CC-MAIN-2023-40
| 279 | 4 |
http://slideplayer.com/slide/4209327/
|
math
|
3 2010 #1 Agricultural experts are trying to develop a bird deterrent to reduce costly damage to crops in the United States. An experiment is to be conducted using garlic oil to study its effectiveness as a nontoxic, environmentally safe bird repellant. The experiment will use European starlings, a bird that causes considerable damage annually to the corn crop in the United States. Food granules made from corn are to be infused with garlic oil in each of five concentrations of garlic – 0 percent, 2 percent, 10 percent, 25 percent, and 50 percent. The researchers will determine the adverse reaction of the birds to the repellent by measuring the number of food granules consumed during a two-hour period following overnight food deprivation. There are forty birds available for the experiment, and the researchers will use eight birds for each concentration of garlic. Each bird will be kept in a separate cage and provided with the same number of food granules. a) For the experiment, identify i. the treatments ii. the experimental units iii. the response that will be measured
4 i. The treatments are the different concentrations of garlic in the food granules. Specifically, there are five treatments: 0 percent, 2 percent, 10 percent, 25 percent and 50 percent. ii. The experimental units are the birds (starlings), each placed in an individual cage. iii. The response is the number of food granules consumed by the bird.
5 After performing the experiment, the researchers recorded the data shown in the table below. i. Construct a graph of the data that could be used to investigate the appropriateness of a linear regression model for analyzing the results of the experiment.
7 ii. Based on your graph, do you think a linear regression model is appropriate? Explain. The curved pattern in this scatterplot reveals that a linear regression model would not be appropriate for modeling the relationship between these variables.
9 2003B #2 A simple random sample of adults living in a suburb of a large city was selected. The age and annual income of each adult in the sample were recorded. The resulting data are summarized in the table below.Annual IncomeAge Category$25,000-$35,000$35,001-$50,000Over $50,000Total21-30815275031-452232358946-60121453Over 60537476496207
10 a) What is the probability that a person chosen at random from those in this sample will be in the age category?b) What is the probability that a person chosen at random from those in this sample whose incomes are over $50,000 will be in the age category? Show your work.Annual IncomeAge Category$25,000-$35,000$35,001-$50,000Over $50,000Total21-30815275031-452232358946-60121453Over 60537476496207
11 c) Based on your answers to parts (a) and (b), is annual income independent of age category for those in this sample? Explain. If annual income and age were independent, the probabilities in (a) and (b) would be equal. Since these probabilities are not equal, annual income and age category are not independent for adults in this sample.
12 2009B #2 The ELISA tests whether a patient has contracted HIV 2009B #2 The ELISA tests whether a patient has contracted HIV. The ELISA is said to be positive if it indicates that HIV is present in a blood sample, and the ELISA is said to be negative if it does not indicate that HIV is present in a blood sample. Instead of directly measuring the presence of HIV, the ELISA measures levels of antibodies in the blood that should be elevated if HIV is present. Because of variability in antibody levels among human patients, the ELISA does not always indicate the correct result. As part of a training program, staff at a testing lab applied the ELISA to 50 blood samples known to contain HIV. The ELISA was positive for 489 of those blood samples and negative for the other 11 samples. As a part of the same training program, the staff also applied the ELISA to 500 other blood samples known to not contain HIV. The ELISA was positive for 37 of those blood samples and negative for the other 463 samples.
13 a) When a new blood sample arrives at the lab, it will be tested to determine whether HIV is present. Using the data from the training program, estimate the probability that the ELISA would be positive when it is applied to a blood sample that does not contain HIV.The estimated probability of a positive ELISA if the blood sample does not have HIV present is
14 b) Among the blood samples examined in the training program that provided positive ELISA results for HIV, what proportion actually contained HIV? A total of = 526 blood samples resulted in a positive ELISA. Of these, 489 samples actually contained HIV. Therefore the proportion of samples that resulted in a positive ELISA that actually contained HIV is
15 c) When a blood sample yields a positive ELISA result, two more ELISAs are performed on the same blood sample. If at least one of the two additional ELISAs is positive, the blood sample is subjected to a more expensive and more accurate test to make a definitive determination of whether HIV is present in the sample. Repeated ELISAs on the same sample are generally assumed to be independent. Under the assumption of independence, what is the probability that a new blood sample that comes into the lab will be subjected to the more expensive test if that sample does not contain HIV?
16 From part (a), the probability that the ELISA will be positive, given that the blood sample does not actually have HIV present, is Thus, the probability of a negative ELISA, given that the blood sample does not actually have HIV present, is 1 – = P(new blood sample that does not contain HIV will be subjected to the more expensive test) = P(1st ELISA positive and 2nd ELISA positive OR 1st ELISA positive and 2nd ELISA negative and 3rd ELISA positive | HIV not present in blood) = P(1st ELISA positive and 2nd ELISA positive | HIV not present in blood) + P(1st ELISA positive and 2nd ELISA negative and 3rd ELISA positive | HIV not present in blood) = (0.074)(0.074) + (0.074)(0.926)(0.074) = = ≈
17 P(new blood sample that does not contain HIV will be subjected to the more expensive test) = P(1st ELISA positive and not both the 2nd and 3rd are negative)= (0.074)( )= (0.074)( )=≈
19 2002 #3 There are 4 runners on the New High School team 2002 #3 There are 4 runners on the New High School team. The team is planning to participate in a race in which each runner runs a mile. The team time is the sum of the individual times for the 4 runners. Assume that individual times of the 4 runners are all independent of each other. The individual times, in minutes, of the runners in similar races are approximately normally distributed with the following means and standard deviations.MeanStandard DeviationRunner 14.90.15Runner 24.70.16Runner 34.50.14Runner 44.8
20 a) Runner 3 thinks that he can run a mile in less than 4 a) Runner 3 thinks that he can run a mile in less than 4.2 minutes in the next race. Is this likely to happen? Explain.It is possible but unlikely that runner 3 will run a mile in less than 4.2 minutes on the next race. Based on his running time distribution, we would expect that he would have times less than 4.2 minutes less than 2 times in 100 races in the long run. OR It is possible but unlikely that runner 3 will run a mile in less than 4.2 minutes on the next race because 4.2 is more than 2 standard deviations below the mean. Since the running time has a normal distribution, it is unlikely to be more than 2 standard deviations below the mean.
21 b) The distribution of possible team times is approximately normal b) The distribution of possible team times is approximately normal. What are the mean and standard deviation of this distribution? The runners times are independently distributed, therefore
22 c) Suppose the team’s best time to date is 18. 4 minutes c) Suppose the team’s best time to date is 18.4 minutes. What is the probability that the team will beat its own best time in the next race?
23 2010 #4 An automobile company wants to learn about customer satisfaction among the owners of five specific car models. Large sales volumes have been recorded for three of the models, but the other two models were recently introduced so their sales volumes are smaller. The number of new cars sold in the last six months for each of the models is shown in the table below. The company can obtain a list of all individuals who purchased new cars in the last six months for each of the five models shown in the table. The company wants to sample 2,000 of these owners.Car ModelABCDETotalNumber of new cars sold in the last six months112,33896,17483,2413,2782,323297,354
24 a) For simple random samples of 2,000 new car owners, what is the expected number of owners of model E and the standard deviation of the number of owners of model E? Because the population size is so large compared with the sample size (≈ 149 times the sample size), far greater than the usual standard of 10 or 20 times larger, we can use the binomial probability distribution even though this is technically sampling without replacement. The parameters of this binomial distribution are the sample size, n, which has a value of n = 2,000, and the proportion of new car buyers who bought model E, p, which has a value of p = The expected value of the number of model E buyers in a simple random sample of 2,000 is therefore n× p = 2,000× ≈ The variance is n× p×(1− p) = 2,000×0.0078×(1− ) ≈15.50, so the standard deviation is the square root of ≈ 3.94.
25 b) When selecting a simple random sample of 2,000 new car owners, how likely is it that fewer than 12 owners of model E would be included in the sample? Justify your answer. For the reason given in part (a), the binomial distribution with n = 2,000 and p ≈ can be used here. The probability that the sample would contain fewer than 12 owners of model E is calculated from the binomial distribution to be This probability is small enough that the result (fewer than 12 owners of model E in the sample) is not likely, but this probability is also not small enough to consider the result very unlikely.
26 c) The company is concerned that a simple random sample of 2,000 owners would include fewer than 12 owners of Model D or fewer than 12 owners of Model E. Briefly describe a sampling method for randomly selecting 2,000 owners that will ensure at least 12 owners will be selected for each of the 5 car models.Stratified random sampling addresses the concern about the number of owners for models D and E. By stratifying on car model and then taking a simple random sample of at least 12 owners from thepopulation of owners for each model, the company can ensure that at least 12 owners are included in the sample for each model while maintaining a total sample size of 2,000. For example, the companycould select simple random samples of sizes 755, 647, 560, 22 and 16 for models A, B, C, D and E, respectively, to make the sample size approximately proportional to the size of the owner population foreach model.
28 2006 #3 The depth from the surface of Earth to a refracting layer beneath the surface can be estimated using methods developed by seismologists. One method is based on the time required for vibrations to travel from a distant explosion to a receiving point. The depth measurement (M) is the sum of the true depth (D) and the random measurement error (E). That is, M = D + E. The measurement error (E) is assumed to be normally distributed with mean 0 feet and standard deviation 1.5 feet. a) If the true depth at a certain point is 2 feet, what is the probability that the depth measurement will be negative? Since M = D + E (a normal random variable plus a constant is a normal random variable), we know that M is normally distributed with a mean of 2 feet and a standard deviation of 1.5 feet. Thus,
29 b) Suppose three independent depth measurements are taken at the point where the true depth is 2 feet. What is the probability that at least one of these measurements will be negative? P(at least one measurement < 0) = 1 – P(all three measurements 0) = 1 – (1 – )3 = 1 – (0.9082)3 = 1 – =
30 c) What is the probability that the mean of the three independent depth measurements taken at the point where the true depth is 2 feet will be negative? Let denote the mean of three independent depth measurement taken at a point where the true depth is 2 feet. Since each measurement comes from a normal distribution, the distribution of is normal with a mean of 2 feet and a standard deviation of feet. Thus,
31 2009 #2 2. A tire manufacturer designed a new tread pattern for its all-weather tires. Repeated tests were conducted on cars of approximately the same weight traveling at 60 miles per hour. The tests showed that the new tread pattern enables the cars to stop completely in an average distance of 125 feet with a standard deviation of 6.5 feet and the stopping distances are approximately normally distributed. a) What is the 70th percentile of the distribution of stopping distances? Let X denote the stopping distance of a car with new tread tires where X is normally distributed with a mean of 125 feet and a standard deviation of 6.5 feet. The z-score corresponding to a cumulative probability of 70 percent is z = Thus, the 70th percentile value can be computed as:
32 b) What is the probability that at least 2 cars out of 5 randomly selected cars in the study will stop in a distance that is greater than the distance calculated in part (a)? From part (a), it was found that a stopping distance of feet has a cumulative probability of Thus the probability of a stopping distance greater than is 1– 0.70 = Let Y denote the number of cars with the new tread pattern out of five cars that stop in a distance greater than feet. Y is a binomial random variable with n = 5 and p = 0.30.
33 c) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet? Let denote the mean of the stopping distances of five randomly selected cars. All tires have the new tread pattern. Because the stopping distance for each of the five cars has a normal distribution, the distribution of is normal with a mean of 125 feet and a standard deviation of feet. Thus,
35 2010 #3A humane society wanted to estimate with 95 percent confidence the proportion of households in its county that own at least one dog.a) Interpret the 95 percent confidence level in this context.The 95 percent confidence level means that if one were to repeatedly take random samples of the same size from the population and construct a 95 percent confidence interval from each sample, then in the long run 95 percent of those intervals would succeed in capturing the actual value of the population proportion of households in the county that own at least one dog.
36 The humane society selected a random sample of households in its county and used the sample to estimate the proportion of households that own at least one dog. The conditions for calculating a 95 percent confidence interval for the proportion of households in this county that own at least one dog were checked and verified, and the resulting confidence interval was ± b) A national pet products association claimed that 39 percent of all American households owned at least one dog. Does the humane society’s interval estimate provide evidence that the proportion of dog owners in its county is different from the claimed national proportion? Explain. No. The 95 percent confidence interval ± is the interval (0.298, 0.536). This interval includes the value 0.39 as a plausible value for the population proportion of households in the county that own at least one dog. Therefore, the confidence interval does not provide evidence that the proportion of dog owners in this county is different from the claimed national proportion.
37 c) How many households were selected in the humane society’s sample c) How many households were selected in the humane society’s sample? Show how you obtained your answer. The sample proportion is 0.417, and the margin of error is Determining the sample size requires solving the equation Thus, so the humane society must have selected 66 households for its sample.
39 2006B #6 Sunshine Farms wants to know whether there is a difference in consumer preference for two new juice products – Citrus Fresh and Tropical Taste. In an initial blind taste test, 8 randomly selected consumers were given unmarked samples of the two juices. The product that each consumer tasted first was randomly decided by the flip of a coin. After tasting the two juices, each consumer was asked to choose which juice he or she preferred, and the results were recorded. a) Let p represent the population proportion of consumers who prefer Citrus Fresh. In terms of p, state the hypotheses that Sunshine Farms is interested in testing. H0 : p = 0.5 versus Ha : p ≠ 0.5
40 b) One might consider using a one-proportion z-test to test the hypotheses in part (a). Explain why this would not be a reasonable procedure for this sample. The conditions for the large sample one-proportion z-test are not satisfied. np = n(1 – p) = 8 x 0.5 = 4 < 5.
41 c) Let X represent the number of consumers in the sample who prefer Citrus Fresh. Assuming there is no difference in consumer preference, find the probability for each possible value of X. Record the x-values and the corresponding probabilities in the table below. X will follow a binomial distribution with n = 8 and p = 8. The possible values of X and their corresponding probabilities are given in the table below:
42 d) When testing the hypotheses in part (a), Sunshine Farms will conclude that there is a consumer preference if too many or too few individuals prefer Citrus Fresh. Based on your probabilities in part (c), is it possible for the significance level (probability of rejecting the null hypothesis when it is true) for this test to be exactly 0.05? Justify your answer. No, there is no possible test with a p-value of exactly 0.5. The probability that none of the individuals (X = 0) or all of the individuals (X = 8) prefer Citrus Fresh is 2 x = , which is less than The probability that one or fewer of the individuals (X 1) or seven or more of the individuals (X 7) prefer Citrus Fresh is 2 x ( ) = , which is greater than 0.05.
43 e) The preference data for the 8 randomly selected consumers are given in the table below. Based on these preferences and your previous work, test the hypotheses in part (a).IndividualJuice Preference1Tropical Taste2Citrus Fresh345678
44 For the preference data provided, X = 2 For the preference data provided, X = 2. From the table of binomial probabilities computed in part (c), the probability that two or fewer of the individuals (X 2) or six or more of the individuals (X 6) prefer Citrus Fresh when p = 0.05 is 2 x ( ) = Because the p-value of is greater than any reasonable significance level, say , we would not reject the null hypothesis that p = 0.5. That is, we do not have statistically significant evidence for a consumer preference between Citrus Fresh and Tropical Taste.
45 f) Sunshine Farms plans to add one of these two new juices – Citrus Fresh or Tropical Taste – to its production schedule. A follow-up study will be conducted to decide which of the two juices to produce. Make one recommendation for the follow-up study that would make it better than the initial study. Provide a statistical justification for your recommendation in the context of the problem.
46 Increase the number of consumers involved in the preference test Increase the number of consumers involved in the preference test. More consumers will give you more data, and you will be better able to detect a difference between the population proportion of consumers who prefer Citrus Fresh and 0.5. The sample proportion in the initial study was only 0.25 (2/8), but we were not able to reject the null hypothesis that p = ½. By increasing the number of consumers, a difference of that magnitude would allow the null hypothesis to be rejected. For example, with n = 80 and X = 20 the large sample z- statistic would be and the p-value would be approximately zero.
|
s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817908.64/warc/CC-MAIN-20180226005603-20180226025603-00311.warc.gz
|
CC-MAIN-2018-09
| 20,259 | 38 |
https://www.assignmentexpert.com/homework-answers/mathematics/trigonometry/question-19657
|
math
|
Answer to Question #19657 in Trigonometry for Kristen Woods
Let x feet - width then x+3 - length
Width 12 feet, length 15 feet
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547584415432.83/warc/CC-MAIN-20190123213748-20190123235748-00471.warc.gz
|
CC-MAIN-2019-04
| 269 | 6 |
https://ofntsc.org/event/basic-applied-math-ceu
|
math
|
Basic Applied Math CEU
Basic mathematical skills used on a daily basis include, but not limited to, the following:
1. Calculate average water usage, peak usage and flow rates.
2. Calculating pump capacities.
3. Calculating areas and volumes of reservoirs, lagoons and cylindrical tanks.
4. Calculating dosages delivered by chemical feeders.
5. Converting Imperial units to metric units, or vice versa.
6. Calculating available pressures in distribution systems.
The following course will prepare the plant operator with the understanding of basic mathematics to perform the above mentioned tasks.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703565376.63/warc/CC-MAIN-20210125061144-20210125091144-00686.warc.gz
|
CC-MAIN-2021-04
| 596 | 9 |
http://mathhelpforum.com/statistics/150002-probability-passing-quiz-print.html
|
math
|
Probability of passing a quiz
An instructor gave her students 12 problems, telling them that 3 of the problems will be on a quiz and that passing the quiz requires solving all 3 of the problems.
a) Given that the instructor chooses the 3 problems at random, what is the probability for a student who knows only 10 problems to pass?
I calculate that the student has 10/12 chances to know the first problem, 9/11 to know the second, and 8/10 to know the third. Therefore the student's probability of passing the quiz is (10/12)*(9/11)*(8/10)= .545
b) What are the chances to fail for a student who knows only 8 problems.
That student has 1 - (8/12)*(7/11)*(6/10) = .745 probability of failing.
Are these answers correct?
|
s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917122992.88/warc/CC-MAIN-20170423031202-00485-ip-10-145-167-34.ec2.internal.warc.gz
|
CC-MAIN-2017-17
| 718 | 7 |
https://hundred-worries.com/en/advices/4913
|
math
|
Gimlet Rule - a simplified visual demonstration using one hand to correctly multiply two vectors. The geometry of the school course implies pupils' awareness of the scalar product. In physics, the vector is often found.
The Concept of a Vector
We believe that there is no sense in interpreting the driller rule with no knowledge of the definition of a vector. It is required to open a bottle - the knowledge of the correct actions will help. A vector is a mathematical abstraction that does not really exist, showing these signs:
- A directional segment, indicated by an arrow.
- The starting point is the point of action of the force described by the vector.
- The length of the vector is equal to the modulus of the force, the field, the other described quantities.
Not always affect force. The vectors describe the field. The simplest example is shown to schoolchildren by teachers of physics. We imply lines of magnetic field intensity. Vectors are usually drawn along a tangent along. In the illustrations of the action on the conductor with a current you will see straight lines.
Vector values are often devoid of application space, action centers are selected by agreement. The moment of force emanates from the axis of the shoulder. Required to simplify addition. Suppose that levers of different lengths are affected by different forces applied to the shoulders with a common axis. By simple addition, subtraction of moments, we find the result.
Vectors help to solve many everyday problems and, although they act as mathematical abstractions, they really work. On the basis of a number of regularities, it is possible to predict the future behavior of an object along with scalar values: the population of the population, the ambient temperature. Environmentalists are interested in directions, the speed of flight of birds. Displacement is a vector quantity.
The Gimlet Rule helps to find the vector product of vectors. This is not a tautology. Just the result of the action will also be a vector. The gimlet rule describes the direction to which the arrow will point. As for the module, you need to apply the formula. The gimlet rule is a simplified purely qualitative abstraction of a complex mathematical operation.
Analytical geometry in space
Everyone knows the problem: standing on one side of the river, determine the width of the channel. It seems to the mind incomprehensible, solved in two ways by the methods of simplest geometry, which students learn. Let's do a number of simple actions:
- Detect a prominent landmark on the opposite bank, an imaginary point: a tree trunk, the mouth of a stream that flows into a stream.
- At the right angle of the opposite bank line, make a notch on this side of the channel.
- Find a place from which the landmark is visible at an angle of 45 degrees to the shore.
- The width of the river is equal to the distance of the end point from the notch.
We use the tangent of an angle. Not necessarily equal to 45 degrees. Need more accuracy - the angle is better to take sharp. Just the tangent of 45 degrees is one, the solution of the problem is simplified.
Similarly, it is possible to find answers to burning questions. Even in the electron-controlled microcosm. We can definitely say one thing: to the uninitiated the rule of a gimlet, the vector product of vectors seems to be boring, boring. A handy tool that helps in understanding many processes. Most will be interested in the principle of operation of the electric motor( regardless of the design).Can easily be explained using the left-hand rule.
In many branches of science, two rules follow side by side: the left, the right hand. Vector product can sometimes be described in one way or another. Sounds vague, we suggest to immediately consider an example:
- Suppose an electron moves. A negatively charged particle plows a constant magnetic field. Obviously, the trajectory will be bent due to the Lorentz force.skeptics will argue, according to some scientists, the electron is not a particle, but rather a superposition of fields. But the principle of uncertainty Heisenberg consider another time. So, the electron moves:
By placing the right hand so that the vector of the magnetic field perpendicularly enters the palm, the extended fingers indicate the direction of the particle's flight, bent 90 degrees to the side, the thumb will stretch in the direction of the force. The right-hand rule is another expression of the gimlet rule. Synonyms. It sounds different, in fact - one.
- We give the phrase Wikipedia, giving a weirdness. When reflected in a mirror, the right three of vectors becomes left, then you need to apply the rule of the left hand instead of the right. The electron flew in one direction, according to the methods adopted in physics, the current moves in the opposite direction. As if reflected in the mirror, therefore the Lorentz force is already determined by the left hand rule:
If you place your left hand so that the magnetic field vector perpendicularly enters the palm, the extended fingers indicate the direction of the current flow, bent 90 degrees to the side of the thumb, stretching, indicating the action vectorstrength
You see, situations are similar, the rules are simple. How to remember which one to apply? The main principle of uncertainty of physics. The vector product is calculated in many cases, with one rule being applied.
What is the rule to apply
The words are synonyms: arm, screw, gimlet
First we analyze the word-synonyms, many began to ask themselves: if the story here should affect the gimlet, why does the text constantly touch the hands. We introduce the concept of the right three, the right coordinate system. Total, 5 words-synonyms.
It was necessary to find out the vector product of vectors, it turned out: this does not work in school. Clarify the situation inquisitive schoolchildren.
School graphics on the board are drawn in the Cartesian coordinate system X-Y.The horizontal axis( positive part) is directed to the right - we hope, the vertical axis points up. We take one step, getting the right three. Imagine: from the beginning of the counting, the Z axis looks to the class. Now the schoolchildren know the definition of the right three vectors.
In Wikipedia it is written: it is permissible to take left triples, right, when calculating a vector product, they disagree. Usmanov is categorical in this respect. With the permission of Alexander Evgenievich, we give an exact definition: a vector product is a vector that satisfies three conditions:
- The product module is equal to the product of the modules of the original vectors and the sine of the angle between them.
- The result vector is perpendicular to the original( together they form a plane).
- The triple vectors( in order of context) the right.
The right three know. So, if the X axis is the first vector, Y is the second, Z will be the result. Why was called the right three? Apparently, it is connected with screws, gimlets. If the imaginary gimlet is twisted along the shortest path, the first vector is the second vector, the translational axis of the cutting tool will begin to move in the direction of the resulting vector:
- The gimlet rule applies to the product of two vectors.
- The driller rule qualitatively indicates the direction of the resultant vector of this action. Quantitatively, the length is the expression mentioned( the product of the modules of the vectors and the sine of the angle between them).
Now everyone understands: the Lorentz force is found according to the rule of a left-handed thread. The vectors are collected by the left triple, if mutually orthogonal( perpendicular to one another), the left coordinate system is formed. On the board, the Z-axis would look in the direction of the view( from the audience behind the wall).
Simple techniques for memorizing the rules of the gimlet
People forget that it is easier to determine the Lorentz force by the rule of the gimlet with left-handed thread. One who wants to understand the principle of operation of an electric motor should double-click like nuts. Depending on the design, the number of rotor coils is significant, or the circuit degenerates, becoming a squirrel cage. Knowledge seekers are helped by the Lorentz rule, which describes the magnetic field where copper conductors move.
To memorize, let's present the physics of the process. Suppose an electron moves in a field. The rule of the right hand is applied to find the direction of the force. It is proved: the particle carries a negative charge. The direction of the force on the conductor is the left-hand rule, remember: physicists completely from the left resources took that electric current flows in the opposite direction to where the electrons went. And this is wrong. Therefore it is necessary to apply the rule of the left hand.
Do not always go such a wilds. It would seem that the rules are more confusing, not quite. The right-hand rule is often used to calculate the angular velocity, which is the geometric product of the acceleration radius: V = ω x r. Many people will be helped by visual memory:
- The vector of the radius of the circular path is directed from the center to the circle.
- If the acceleration vector is directed upwards, the body moves counterclockwise.
Look, the right-hand rule is again here: if you position the palm so that the acceleration vector enters perpendicularly into the palm, extend your fingers in the direction of the radius, bent by 90 degrees, the thumb indicates the direction of movement of the object. It is enough to draw once on paper, remembering at least half a life. The picture is really simple. More on the physics lesson will not have to wrestle with a simple question - the direction of the vector of angular acceleration.
Similarly, the moment of force is determined. It comes perpendicularly from the axis of the shoulder, coincides with the direction with angular acceleration in the figure described above. Many will ask: what is needed? Why is the moment of force not a scalar quantity? Why the direction? In complex systems is not easy to trace the interaction. If there are a lot of axes, forces, vector addition of moments helps. You can greatly simplify the calculations.
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00117.warc.gz
|
CC-MAIN-2021-25
| 10,256 | 45 |
https://www.math.net/multiple
|
math
|
A multiple of a whole number is the product of that whole number and another whole number. When someone says that a number is a multiple of another number, it means that the other number can be multiplied by an integer to form the first number; 4 is a multiple of 2 because 2 × 2 = 4, but 4 is not a multiple of 3 because 3 cannot be multiplied by an integer to get a result of 4 (3 × 1.33 = 4).
|Multiples of 4 include 0, 4, 8, 12,...|
|Multiples of 6 include 0, 6, 12, 18,...|
Properties of multiples
- 0 is a multiple of everything, since zero multiplied by any number is still zero.
- Every integer is a multiple of itself because anything multiplied by 1 is the same number. So any integer can be multiplied by 1 to result in said integer, making it a multiple of itself.
- The product of an integer, n, and any integer, is a multiple of n. This should make sense because it fits the definition of what a multiple is, described above.
- If two integers, designated A and B, are a multiple of another number, n, then A + B and A - B are also multiples of n.
- Example: 4 and 8 are multiples of 2. 8 + 4 = 12; 8 - 4 = 4. Both 4 and 12 are also multiples of 2.
See also common multiple, least common multiple.
Related words multiplication, multiply.
|
s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371833063.93/warc/CC-MAIN-20200409091317-20200409121817-00449.warc.gz
|
CC-MAIN-2020-16
| 1,253 | 11 |
https://icecream.directory/disher-aka-ice-cream-scoop-sizes-bowl-of-plenty/
|
math
|
Below is a list of the best Ice cream scoop size chart voted by readers and compiled and edited by our team, let’s find out
Update: Also see my post on dishers for left-handed use.
To ensure that cookies, cupcakes, and muffins bake evenly, one of the steps should be to divide the dough or batter into uniform quantities. For the best way to do this, the common advice is to use portion-control tools. These are known as dishers in the food-service industry, although to the rest of us they look like ice cream scoops. In addition to their use in baking, dishers can also ensure consistent portioning of meatballs or hamburger patties. You can buy them at your local restaurant-supply store.
In the US, commercial-grade dishers are denominated in sizes (numbers) that represent quart fractions—for example, a No. 12 disher should hold 1/12 of a quart (in other words, it takes 12 scoops to fill a quart), a No. 16 holds 1/16 quart, etc. Using this standard, we can create a chart of disher sizes and their equivalent nominal volumes, in both US customary and metric units:
* Color codes are only available on dishers with plastic handles. Size Color* fl oz tbsp cup (fraction) mL 6 White 5.33 10.7 0.667 (2/3) 158 8 Gray 4.00 8.00 0.500 (1/2) 118 10 Ivory 3.20 6.40 0.400 94.6 12 Green 2.67 5.33 0.333 (1/3) 78.9 16 Blue 2.00 4.00 0.250 (1/4) 59.1 20 Yellow 1.60 3.20 0.200 47.3 24 Red 1.33 2.67 0.167 39.4 30 Black 1.07 2.13 0.133 31.5 40 Orchid 0.800 1.60 0.100 23.7 50 Rust 0.640 1.28 0.0800 18.9 60 Pink 0.533 1.07 0.0667 15.8 70 Plum 0.457 0.914 0.0571 13.5 100 Orange 0.320 0.640 0.0400 9.46
You’ll notice there are gaps between sizes. As far as I know, these are the only ones available for the US food-service market, and not all manufacturers make this entire size range. But that’s not the problem. The real problem is that the quart-fraction standard is only followed loosely, and actual disher capacities vary both from the nominal size and among different manufacturers. How far off are these scoop sizes, you ask? Let’s look at the numbers.
The following table contains a semi-random sampling of product lines and shows how much the scoops’ specified capacities deviate from nominal sizes. These calculations are based on the manufacturers’ specifications, which I’ve collected into a spreadsheet you can either view online (HTML) or download (Excel, with formulas). The file is also available in Google Spreadsheets format if you’re logged in to a Google account. In addition to the manufacturers’ specifications and my calculations, the spreadsheet also contains details such as the dishers’ scoop diameters shown in both inches and centimeters.
Deviation of specified capacity from nominal size. Bold indicates sizes accurate to within 2%. Details. Size Adcraft (PDF) Fox Run Hamilton Beach Johnson Rose Norpro OXO Vollrath (metal) Vollrath (plastic) Zeroll 6 12.5% 12.6% 12.5% 0.0% 12.6% 8 0.0% 0.0% 9.0% 8.3% 0.0% 0.0% 9.0% 10 17.2% 0.3% 14.6% 2.3% 1.6% 0.3% 12 21.9% 4.3% 25.0% 3.1% 0.0% 4.3% 16 37.5% 0.0% 3.5% 0.0% 0.0% 0.0% 3.5% 20 56.3% 0.0% 10.6% 4.2% 6.3% 6.3% 6.3% 1.6% 10.6% 24 31.3% 11.8% 0.0% 3.1% 0.0% 11.8% 30 17.2% 3.4% 6.3% 6.3% 17.2% 6.3% 3.4% 40 9.4% 6.3% 15.0% 16.7% 6.3% 9.4% 6.3% 11.3% 50 2.3% 0.0% 2.3% 1.6% 60 5.5% 6.3% 37.5% 5.5% 0.6% 70 9.4% 0.6% 6.0% 0.6% 100 17.2% 0.0% 4.2% 17.2% 0.0%
As you can see, disher size accuracy can be way off the mark. In this table, scoops accurate to within 2% of the nominal size are highlighted in bold. If this seems to be too tight of an allowance for size variations, think about it this way: a 12-inch ruler that’s off by 2% will be either too long or too short by roughly a quarter of an inch, making for a potential variation of nearly half an inch from one ruler to the next.
While it’s possible that dishers are made to match nominal sizes but their true capacities are rounded off for publication formatting (thus calculations based on published specifications will show more deviation than actually exists), seeing that there is no uniformity even within one manufacturer’s own two product lines leads me to believe this is not the case, since there is no reason to think different rounding standards would be used here. Moreover, by comparing the ratios between sizes, one is likely to find that manufacturers’ guidance for real-world applications (e.g., hamburger patties) further deviate from both the nominal and specified volumes. Makes one wonder which numbers are really right.
Please note, however, that in spite of these discrepancies, inaccurate dishers aren’t necessarily defective or inferior. This is because dishers are primarily portioning tools instead of measuring tools, so if you find some that fit your recipes, there’s not much point to worrying about whether they match some fixed standard or not. You just need to be aware that size designations are not reliable indicators of actual capacity, and dishers of identical (nominal) size from different manufacturers may hold different amounts of material.
Now, how do you know which dishers will fit your recipes without buying every size and trying them? That’s a good question. From what I’ve seen, recipes generally don’t tell you that. At least for muffins and cupcakes, since standard muffin tins have a capacity of 1/2 cup (4 fl. oz., or 8 tbsp) per pocket, we should figure to place only about 1/4 to 1/3 cup of batter in each to avoid overflowing during baking—in other words, #12 and #16 (nominal size) should work in most cases. However, I’ve also seen recipes that exceed this quantity to deliberately create overflowing muffin tops, so exceptions do exist.
Disher (ice cream scoop) sizes can be inconsistent, so it’s better to know their actual (or at least specified) capacities than to rely on nominal sizes. In terms of nominal size, 12 and 16 should, in principle, fit most muffin and cupcake recipes.
|
s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00002.warc.gz
|
CC-MAIN-2023-06
| 5,921 | 13 |
https://www.kopykitab.com/blog/jntu-previous-question-papers-papers-b-tech-2nd-year-chemical-process-calculations-november-2008/
|
math
|
JNTU B.Tech I Semester Supplementary Examinations, November 2008
CHEMICAL PROCESS CALCULATIONS
(Chemical Engineering) SET-1
1. (a) A mixture of gases contains 10.5% CO2 13.0% Cl2 12.7% N2 and balance H2
i. What is the average molecular weight of the gas?
ii. Calculate the gas composition as weight fractions.
(b) Given a water solution contains 1.704 kg HNO3 per kg H2O and has a specific gravity of 1.382 at 200C, express the composition in the following ways
i. Weight percent HNO3
ii. Molarity at 200C
2. In the manufacture of HCl, a gas is obtained that contains 25% HCl and 75% air by
volume. This gas is passed through an absorption system in which 98% of the HCl is removed. The gas enters the system at a temperature of 490C and a pressure of 743 mm Hg and leaves at a temperature of 270C and a pressure of 738 mm Hg.
(a) Calculate the volume of gas leaving per 100 m3 entering the absorption apparatus.
(b) Calculate the percentage composition by volume of the gases leaving the absorption apparatus
(c) Calculate the weight of HCl removed per 100 m3 of gas entering the absorption
3. (a) Derive the Clausius-Clapeyron equation? What are the assumptions made in developing the equation? Explain, how it is used to calculate the vapor pressures of substances
(b) Write short notes on critical properties
4. Explain wet and dry bulb thermometry giving practical applications of the same.
5. A furnace uses a natural gas which consists entirely of hydrocarbons. The flue gas analysis is: CO2 – 9.5%, O2 -1.4%, CO – 1.9% and rest is N2. Calculate the following:
(a) the atomic ratio of hydrogen to carbon in the fuel
(b) % excess air
(c) the composition of the flue gas in the form CxHy
6. Methanol is produced by the reaction of CO with H2 according to the equation .
CO + 2H2 ! CH3OH
Only 15% of the CO entering the reactor is converted to methanol. The methanol product is condensed and separated from the un-reacted gases, which are recycled. The feed to the reactor contains 2 mole of H2 for every mole of CO. The fresh feed enters at 350C and 300 atm. To produce 6,600 kg/hr of methanol calculate
(a) Volume of fresh feed gas and
(b) The recycle ratio.
7. (a) Define sensible heat and latent heat
(b) Define heat of fusion and heat of vaporization
(c) Explain heat of condensation and heat of sublimation
8. (a) What is the heat of hydration.
(b) Differentiate between heat of solution and heat of salvation.
(c) Define standard integral heat of solution. How is it determined.
(d) Write on standard heat of combustion
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00028.warc.gz
|
CC-MAIN-2021-31
| 2,534 | 33 |
https://books.google.com/books?vid=ISBN1786340305
|
math
|
Algebra, Logic and Combinatorics
Shaun Bullett, Tom Fearn, Frank T. Smith
World Scientific, 2016 - Mathematics - 175 pages
"The articles are all quite well-written and informative. In these days of increasing mathematical specialization, it's always nice to find something that helps bridge the gap between disciplines. It's even nicer when the contents are of the quality of these articles. This is an interesting, enjoyable and valuable book."
Mathematical Association of America
This book leads readers from a basic foundation to an advanced level understanding of algebra, logic and combinatorics. Perfect for graduate or PhD mathematical-science students looking for help in understanding the fundamentals of the topic, it also explores more specific areas such as invariant theory of finite groups, model theory, and enumerative combinatorics.
Algebra, Logic and Combinatorics is the third volume of the LTCC Advanced Mathematics Series. This series is the first to provide advanced introductions to mathematical science topics to advanced students of mathematics. Edited by the three joint heads of the London Taught Course Centre for PhD Students in the Mathematical Sciences (LTCC), each book supports readers in broadening their mathematical knowledge outside of their immediate research disciplines while also covering specialized key areas.
|
s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540487789.39/warc/CC-MAIN-20191206095914-20191206123914-00339.warc.gz
|
CC-MAIN-2019-51
| 1,352 | 7 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.