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https://zacksinvest.com/88d63pub5/nP38872xQ/ | 1,611,763,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00051.warc.gz | 1,005,966,713 | 9,824 | # Halloween Math Worksheets For Preschool And Kindergarten Students Teachersmag Preschool Algebra
Jessamyn Louise December 25, 2020 Worksheet
Rather than using worksheets, a better method is to use individual size white boards and have the child writing entire facts many times. Having a child writing 9 x 7 = 7 x 9 = 63 while saying ”nine times seven is the same as seven times nine and is equal to sixty-three” is many times more successful than a worksheet with 9 x 7 = ___ and the student just thinks the answer once and then writes that answer on the duplicate problems.
There are some new materials being developed now based on what we are learning about how the brain learns. These brain-friendly materials should be an improvement over what has existed. I recently bought a book by Marcia L. Tate titled ”Mathematics Worksheets Don’t Grow Dendrites.” I highly recommend her book. She gives a great deal of information on alternative activities that are better for your child’s brain development and for learning.
As a parent, I’m very aware of what my own children are learning in school. For the most part, I’ve been happy with their progress, but as they rise in grade level, I’m starting to see more emphasis on a loose understanding of the concepts and less emphasis on skills–particularly skills with arithmetic of fractions. The main problem with what I see with my students and my own children is that kids are taught ”concepts” and are not taught skills–unless they’re lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic.
Learning math requires repetition that is used to memorize concepts and solutions. Studying with math worksheets can provide them that opportunity; Math worksheets can enhance their math skills by providing them with constant practice. Working with this tool and answering questions on the worksheets increases their ability to focus on the areas they are weaker in. Math worksheets provide your kids’ the opportunity to analytical use problem solving skills developed through the practice tests that these math worksheets simulate.
### Functional Relationship Math Worksheets
Dec 25, 2020
#### Free Fun 3rd Grade Math Worksheets
Dec 25, 2020
##### Get The Message Math Worksheet Dd-25
Dec 25, 2020
###### Math Equations Worksheets Grade 7
Dec 25, 2020
Numerous research studies have found that when students are actively engaged with the content, they have a much better chance of understanding and remembering what they have learned. Unfortunately, math worksheets tend to bore most students, especially those who need the most help in math. Engagement entails much more than rote repetition of a procedure. Math worksheets tend to present very similar problem types over and over, leading to mundane practice of disassociated skills. For students who understand the material and successfully complete an assignment, another worksheet becomes meaningless. On the other hand, for the students who don’t understand the material, an alternative method of instruction is what’s needed. Another worksheet simply adds to the student’s frustration, or worse, contributes to a belief that ”I’ll never understand math.” A cute image or a ”fill-in-the-blanks” riddle does nothing to increase engagement or learning (and let’s face it, those riddles are not funny!). Instead, teachers need to increase engagement by providing students with exercises in which they discover patterns and relationships, solve problems, or think creatively about math relationships.
Some students are unable to access tools that many of us take for granted when they try to complete worksheets. They may be unable to grasp pencils, control their movements within the limited spaces provided on the sheet, or be able to simply stabilize their paper while writing. Other students, including those for whom English is not their primary language or who struggle with reading, have difficulty reading the directions, words, and math terminology on the worksheets. Still other students require different visual representations or methods of engagement in order acquire an understanding the content. Most math worksheets do not provide information in multiple formats so they are inaccessible to students with a wide variety of learning styles and abilities. Well-designed technology can provide these students with access to excellent content. For example, these fractions tools and supplemental curriculum allow students with physical disabilities to access fractions content using a variety of assistive technology devices. Instructions, prompts and feedback can be read aloud, while visual models, cues combined with sounds support a wide range of learning styles and abilities.
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Dec 25, 2020 | 1,093 | 5,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2021-04 | latest | en | 0.957111 |
https://www.cbc.ca/radio/spark/380-phantom-traffic-jams-catfishing-scams-and-smart-speakers-1.4482967/sick-of-traffic-jams-stop-tailgating-1.4482984 | 1,660,339,611,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00783.warc.gz | 658,142,159 | 33,945 | Spark
# Sick of traffic jams? Stop tailgating
The sudden backups on highways that appear even when there's no accident ahead have annoyed drivers for ages. Now, MIT researchers have found a simple solution to these "phantom traffic jams."
Almost everyone has sat through a traffic jam. It's awful in so many ways — the constant stopping, cars as far as the eye can see and worst of all, the lack of explanation for what caused it once you drive up to the front.
Turns out there's a term to describe these traffic bottlenecks that appear even when there's no accident ahead: phantom traffic jams.
These unexplained backups happen when a car suddenly decides to brake, causing the car behind it to slow down. This causes a ripple effect down the lane and builds up until it creates a full-blown traffic jam.
If everyone stopped tailgating, that would already dramatically improve the situation.- Berthold Horn, professor of computer science and engineering at Massachusetts Institute of Technology
Berthold Horn is a professor of computer science and engineering at the Massachusetts Institute of Technology. He drives along U.S. Interstate 93 to work every day, and one day decided he was fed up with the congestion. He got to work researching phantom traffic jams and came up with a simple solution to fix it by simulating the scenario using math models.
What he discovered was that if every car on the road keeps an equal distance between the car immediately ahead and behind, it'll cut driving time by almost half.
"There will still be disturbances travelling down the line of traffic [when a car suddenly brakes]," Berthold explains, "but they get attenuated down the line, and eventually die away."
In traffic situations today, the opposite happens: "The disturbances get amplified until you reach a full stop."
Phantom traffic jams were first studied in the 1930s and since then more than 1,000 papers have been published to explain the phenomenon. However, little has been done to solve it.
Past solutions tried to limit the number of cars on the road at a given time by using traffic lights and speed limit signs.
"Keeping the density down does reduce the incidence of traffic jams," Berthold says. "But the problem is you're not making full use of the road."
Berthold believes his solution is more effective and it's easy to implement. There's no need to build new roads or develop new technologies. A simple update to the current adaptive cruise control system available in certain brands of cars will do the trick.
Adaptive cruise control works by using sensors at the front of the car to detect and keep a safe distance from the car ahead. To implement Berthold's solution, "all you need to do is to add additional sensors to the rear end of the car and then replace the current adaptive cruise control algorithm with our bilateral control algorithm."
Berthold says Toyota funded his research and might be implementing this new two-way model of adaptive cruise control in the future.
However, to see significant changes in traffic times, there needs to be wide adoption and that won't happen instantaneously, Berthold says.
"It's going to require some impetus such as a mandate."
So, until a new kind of cruise control system becomes standard, what drivers can do now is stop tailgating the vehicle ahead if there's a large gap behind their own vehicle, Berthold says. Constantly checking the rearview mirror is not a good idea since it distracts drivers from what's ahead and increases the chance of an accident.
"But if everyone stopped tailgating, that would already dramatically improve the situation."
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now | 964 | 4,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-33 | latest | en | 0.953971 |
https://oeis.org/A071760 | 1,702,261,805,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103464.86/warc/CC-MAIN-20231211013452-20231211043452-00363.warc.gz | 471,618,058 | 4,330 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A071760 Denominators of record absolute values of partial sums of Sum (mu(n)/n). 1
2, 6, 30, 105, 2310, 58644190679703485491635, 5364750833138837555449767529261714317873456270532298668855 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS a(1) = 2 because the first minimum of sum(mu(n)/n), n>1, is 1-1/2 = 1/(2). LINKS Table of n, a(n) for n=1..7. PROG (PARI) t = 0; t1 = 1; v = []; for( n = 1, 200, t = t + moebius( n) / n; if( ( t / t1)^2 < 1, t1 = t; v = concat( v, denominator( t)), )); v CROSSREFS Cf. A071759. Sequence in context: A034501 A203461 A071758 * A304579 A036752 A065563 Adjacent sequences: A071757 A071758 A071759 * A071761 A071762 A071763 KEYWORD nonn,frac AUTHOR Donald S. McDonald, May 18 2002 STATUS approved
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Last modified December 10 19:44 EST 2023. Contains 367717 sequences. (Running on oeis4.) | 484 | 1,533 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-50 | latest | en | 0.686773 |
https://aovgame.org/faq/quick-answer-how-to-play-rubber-bridge.html | 1,656,366,681,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00190.warc.gz | 150,594,014 | 23,339 | # Quick Answer: How To Play Rubber Bridge?
• In a standard game of Rubber Bridge, four players take part, in two fixed partnerships. Tradition is to refer to the players according to their position at the table, based on the main compass points. So, in clockwise order, one player will be North, one will be East, one will be South and one will be West.
Rubber bridge is played with a standard deck of 52 cards. From high to low, the cards are ranked A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3, and 2. Suits are ranked Spades (♠), Hearts (♥), Diamonds (♦), Clubs (♣). Four players play in two partnerships, with partners sitting opposite each other.
## What are the rules of rubber bridge?
Rubber bridge is a form of contract bridge played by two competing pairs using a particular method of scoring. A rubber is completed when one pair becomes first to win two games, each game presenting a score of 100 or more contract points; a new game ensues until one pair has won two games to conclude the rubber.
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## How is the game of bridge played?
The essential features of all bridge games, as of whist, are that four persons play, two against two as partners; a standard 52-card deck of playing cards is dealt out one at a time, clockwise around the table, so that each player holds 13 cards; and the object of play is to win tricks, each trick consisting of one
## How do you score points in rubber bridge?
For a completed rubber in two games, a rubber bonus of 700 points. For a completed rubber in three games, a rubber bonus of 500 points. a part-score it scores 50 points. At the end of the rubber all the points above and below the line are added up.
## Why is rubber bridge called rubber?
The word’s origins can be traced back to the 1600s, when the English used the term “rubber” as a reference to an object used for rubbing, or cleaning. The term found its way into the popular 16th century game of lawn bowling, which is similar to bocce ball.
## Why is rubber bridge so called?
The original form of contract bridge; so called because bonuses are awarded for scoring sufficient points to win games and thereby a rubber which is the best of 3 games.
## Is bridge difficult to learn?
It takes only rudimentary knowledge to begin playing and enjoying bridge, but be forewarned: this is not an easy game to learn, and it’s even more difficult (most say impossible) to master. No matter how many years you play, you’ll always find new challenges, and the learning process will never end.
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## What is the best way to learn bridge?
The best way to master the card-playing part of bridge is to learn any four-person card game where each person plays a single card in turn and the best card wins. The most popular trick-taking game is Spades (a great game in itself).
## What is the object of bridge?
The object in bridge play is to win tricks for your side. A trick consists of four cards, one from each player in turn, clockwise around the table. Hence, there are 13 tricks to be won on each deal. The first card played to each trick is called the lead.
## Can bridge be played with 2 players?
Two players use a 52 card pack. Suits and cards in each suit rank as in Bridge. After 13 tricks each player still has 13 cards, and if they have very good memories they will know each other’s hands. The players now bid as in Contract Bridge (doubles and redoubles are allowed), until one player passes.
## What is bridge Chess?
Bridge is largely about communication, and every message a player sends—by bidding or playing a significant card—is broadcast to the player’s partner and his opponents. The difficulty of weighing truth and lies is one reason that computers don’t win at bridge, whereas at the highest level of chess they do very well.
## Is playing bridge good for your brain?
Bridge stimulates the brain. Bridge is one of the best ways to practice the “use it or lose it” advice for maintaining mental sharpness in older age. Research has shown that regular bridge playing improves reasoning skills and long- and short-term memory.
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## How many points is a game of bridge?
When to bid “Game.” You need at least 26 points between you and your partner to bid “Game.”
## What is the difference between rubber bridge and Chicago bridge?
Learn about this topic in these articles: Rubber bridge is the simplest form for four players and is frequently played in casual games among friends. Chicago, or four-deal bridge, is most often used for small card parties in which several tables are used.
## What is the difference between rubber bridge and duplicate bridge?
Duplicate bridge stands in contrast to rubber bridge where each hand is freshly dealt and where scores may be more affected by chance in the short run. In duplicate bridge, a player normally plays with the same partner throughout an event. The two are known as a “pair”. | 1,131 | 5,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | latest | en | 0.970897 |
https://quant.stackexchange.com/questions/4235/running-a-simple-alpha-estimation-test-for-statistical-significance-of-a-signal | 1,563,533,813,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526210.32/warc/CC-MAIN-20190719095313-20190719121313-00078.warc.gz | 513,356,059 | 36,109 | # Running a simple alpha estimation test for statistical significance of a signal
I'm looking for some direction on testing whether a simple entry signal has statistical significance.
Let's say this is my simple entry signal:
Buy when some indicator has a positive slope and is above zero on 3 time frames (5M,15M,60M)
Sell when the indicator has a negative slope and is below zero on 3 time frames.
How do I go about statistically testing this signal to see if there's any alpha to be extracted?
I was thinking something along these lines:
1. plot the P&L distribution for the signal should it exit 1,2,3, N bars after the entry.
2. plot the P&L distribution if the entry point had been random and the exit had been again 1,2,3, N bars after the entry.
I'm not really sure where to go from there.
How do I go about creating a "random" entry signal? Perhaps taking the time difference between the first and last quote and creating a pseudo random time to enter. The number of entries would have to be equal to the number of non-random signals created from the indicator. Also, I say pseudo random because it probably wouldn't be desirable if all these "random" signals happened to cluster in one short time interval.
Then, what characteristics of these distributions should I be looking for to establish if the indicator signal has any statistical significance in terms of potential alpha that can be extracted.
Any direction or reference would be appreciated. Being that there are so many different types of statistical tests that can be done, it's hard to find anything particularly relevant.
• They way you ask your question makes me think that you are looking for somekind of bootstrap test. A related paper on that would be : Pairs Trading: Performance of a Relative-Value Arbitrage Rule Evan Gatev. Of course this will not actually tell you if you have found alpha. For this I think you should be looking at parametrical test. – Zarbouzou Oct 2 '12 at 9:31
• Thanks for your reply. Looking into parametrical tests for trading signals I found "Trading with a 't' Signal Extraction Using Non-Parametric Newey-West Style t-statistics" (Cameron Rookley). It seems this may be a good example of what I'm looking for but I'll have to brush up on some statistics concepts before I can understand this paper fully. If anyone can propose a simple, practical parametric (or otherwise) test in the meanwhile that I can use to evaluate statistical significance of my particular signal, I would be much obliged. – Vazgen Oct 2 '12 at 21:32
I think the simplest way to achieve what you're looking for is through regression coefficient hypothesis testing.
1. Perform linear regression on returns (y-axis) vs. dates (x-axis) over the desired time frames (do it once for 5 months, once for dataset w/15 months worth of data, and once for 60 months worth of data).
2. As a result of regression, you will get coefficients for $y = mx + b$. To check if the slope is significantly different from 0.0, you would perform a t-test on m: $$t = \frac{m - 0.0}{SE(m)} = \frac{m}{SE(m)},$$ where $SE(m)$ is standard error of $m$. Depends on how you're doing the regression, but it may be reported by the software.
3. Using $n-2$ degrees of freedom (where $n$ is the number of points included in regression, look up $t$ values for the desired confidence level (typ., $\alpha=0.05$ for 95% confidence interval, CI). Let this value be $t_{critical}$.
4. $m$ is significantly different from 0.0 if $t > t_{critical}$.
Slope's sign can be determined by adjusting the null hypothesis ($m > 0$, $m < 0$) & picking different $t_{critical}$ values.
...unless I misunderstood your question, that's one of the ways I see achieving what you're after.
• Thank you for taking the time to help. It seems the test you proposed could be what I need. If I understand correctly, m is the regression coefficient and if it varies significantly from 0.0 then there is a correlation between the trade returns and time. But how does that relationship imply my signal is statistically significant? Perhaps I'm misunderstanding the interpretation of step 4. – Vazgen Oct 4 '12 at 5:15
• I ultimately want to be able regress across a number of user defined metrics that are recorded right before the signal (bar range right before the signal, distance to previous high, value of a different indicator right before signal, etc) to see if there are relationships there that can yield new information about entries with a statistically high chance of turning profitable. – Vazgen Oct 4 '12 at 5:16 | 1,044 | 4,565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-30 | longest | en | 0.945686 |
https://gurujiclasses.com/daily-math-practice-test-4number-system/ | 1,695,768,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00068.warc.gz | 326,020,296 | 16,519 | 1. Find the greatest number of four digits which must be added to 5,231 so that the final number becomes exactly divisible
by 12, 15, 27, 32 and 40.
(a) 7,729 (b) 7,829 (c) 7,929 (d) 9,729
2. Find the greatest number of six digits, which on being divided by 6, 7, 8, 9 and 10 leaves 4, 5, 6, 7 and 8 as remainder,
respectively.
(a) 9,97,920 (b) 9,97,918 (c) 9,98,918 (d) 9,99,918
3. A heap of stones can be made up into groups of 21. When made up into groups of 16, 20, 25 and 45, there are 3 stones
left in each case. How many least stones can there be in the heap?
(a) 2,403 (b) 3,603 (c) 4,803 (d) 7,203
4. What is the least number of pieces of equal length that can be cut out of two lengths, 10 m, 857 mm and 15 m 87 mm?
(a) 174 (b) 172 (c) 164 (d) 184.
5. The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.
(a) 66 (b) 99 (c) 1,119 (d) 909
6. What is the least multiple of 17, which leaves a remainder of 1 when divided by each of the first twelve integers excepting
unity?
(a) 1,38,599 (b) 1,38,601 (c) 27,719 (d) 27,720
7. The LCM and GCM of two numbers are 1,530 and 51, respectively. Find how many such pairs are possible:
(a) 4 (b) 3 (c) 2 (d) 1
8. Find the greatest four-digit number which, when divided by 12, 18, 21 and 28 leaves a remainder 3 in each case.
(a) 9,830 (b) 9,831 (c) 9,835 (d) 9,836
9. The LCM of 54, 90 and a third number is 1,890 and their GCM is 18. What is the third number?
(a) 36 (b) 126 (c) 108 (d) 180
10. Find the least number which on being divided by 5, 6, 8, 9, 12 leaves in each case a remainder 1, but when divided by
13 leaves no remainder.
(a) 3,595 (b) 3,600 (c) 3,601 (d) 3,602
11. Three men start together to travel the same way around a circular track of 11 km circumference. Their speeds are 4, 5
and 8 km/h, respectively. When will they meet at the starting point?
(a) 11 h (b) 12 h (c) 22 h (d) 220 h
12. When 5 or 6 dozens of oranges were packed in each box, three dozens were remaining. Therefore, bigger boxes were
taken to pack 8 or 9 dozens of oranges. However, still three dozens of oranges remained. What was the least number of
dozens of oranges to be packed?
(a) 216 (b) 243 (c) 363 (d) 435
13. Four bells toll at intervals of 6, 8, 12 and 18 minutes, respectively. If they start tolling together at 12 a.m., after what
interval will they toll together again and how many times will they toll together in 6 hours?
(a) 6 times (b) 5 times (c) 4 times (d) Data inadequate
14. Which of the following is a pair of co-primes?
(a) (14, 35) (b) (18, 25) (c) (32, 62) (d) (31, 93)
15. The GCM of two numbers is 38 and their LCM is 98,154. If one of the numbers is 1,558, the other number is
(a) 3,450 (b) 2,395 (c) 2,394 (d) 1,260
July 6, 2023
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## Fast Track Practice-MATH
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https://www.puzzlegamemaster.com/pythagorea-60-janes-triangle-level-25-1-25-16-solutions-answers/ | 1,566,660,307,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00342.warc.gz | 945,987,248 | 8,676 | # Pythagorea 60 degree Level 25.1 25.2 25.3 25.4 25.5 25.6 25.7 25.8 25.9 25.10 25.11 25.12 25.13 25.14 25.15 25.16
Pythagorea 60 janes triangle is an android/iOS app developed by Horis International Limited. This game is mostly focused on geometric puzzles and construction. The workspace is divided into triangular grids lines aligned 60° to each other, that’s why this game names as pythagorea 60°. You should know all the basic Math operations. All lines and shapes are drawn on a grid whose cells are equilateral triangles. Most of the game levels can be answered using natural intuition and bye some basic laws of geometry.
If you are here for levels other than ‘janes triangle’ Go to directory of all other levels at : http://www.puzzlegamemaster.com/pythagorea-60-complete/
Pythagorea 60 Jane’s Triangle 25.1 Solution:
Pythagorea 60 Jane’s Triangle 25.2 Solution:
Pythagorea 60 Jane’s Triangle 25.3 Solution:
Pythagorea 60 Jane’s Triangle 25.4 Solution:
Pythagorea 60 Jane’s Triangle 25.5 Solution:
Pythagorea 60 Jane’s Triangle 25.6 Solution:
Pythagorea 60 Jane’s Triangle 25.7 Solution:
Pythagorea 60 Jane’s Triangle 25.8 Solution:
Pythagorea 60 Jane’s Triangle 25.9 Solution:
Pythagorea 60 Jane’s Triangle 25.10 Solution:
Pythagorea 60 Jane’s Triangle 25.11 Solution:
Pythagorea 60 Jane’s Triangle 25.12 Solution:
Pythagorea 60 Jane’s Triangle 25.13 Solution:
Pythagorea 60 Jane’s Triangle 25.14 Solution:
Pythagorea 60 Jane’s Triangle 25.15 Solution:
Pythagorea 60 Jane’s Triangle 25.16 Solution: | 446 | 1,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-35 | longest | en | 0.643164 |
https://m.hanspub.org/journal/paper/15038 | 1,638,425,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00427.warc.gz | 422,891,372 | 10,364 | 基于改进层次模糊评判的智能变电站二次系统状态评估
# 基于改进层次模糊评判的智能变电站二次系统状态评估Status Assessment of Secondary Equipment for Smart Substation Based on Improved Hierarchical Fuzzy Evaluation
Abstract: For smart substation secondary equipments’ failure mechanism are complex, uncertain and diffi-cult to be effectively evaluated, a comprehensive evaluation method based on improved analytic hierarchy process and fuzzy synthetic evaluation is proposed in this paper. On the basis of run maintenance procedures and test specification and taking operating experience, work environment and maintenance records into consideration, this paper establishes the quantitative and qualitative state evaluation index system of secondary equipment and gives the corresponding weight according to the relative importance of each indicator. Through building the scoring model of state indicator by the semi-trapezoidal distribution function and introducing the relative deterioration degree to characterize the membership function of secondary equipment status transformation degree and to establish a hierarchical fuzzy evaluation matrix. Thus, a condition assessment model of smart substation secondary system based on improved analytic hierarchy process and fuzzy synthetic evaluation is established. Example analysis shows that this method is effective and feasible and provides a new way for the health assessment of smart substation secondary system.
[1] 潘华君, 许晓峰, 许东, 吴焕 (2014) 基于模糊评判法的智能变电站二次系统状态评估. 沈阳工程学院学报(自然科学版), 2, 143-147.
[2] 王德文, 王艳, 邸剑 (2011) 智能变电站状态监测系统的设计方案. 电力系统自动化, 18, 51-56.
[3] 张晓华, 刘跃新, 刘永欣, 孙嘉, 邱俊宏 (2011) 智能变电站二次设备的状态监测技术研究. 电气技术, 4, 41-44.
[4] 吴杰余, 张哲, 尹项根, 胡文平 (2002) 电气二次设备状态检修研究. 继电器, 2, 22-24.
[5] 刘力丰, 高中德, 刘百震, 曹利安 (1997) 电力系统继电保护设计专家系统的模糊知识处理. 电力系统自动化, 6, 34-37.
[6] 李瑞生, 李燕斌, 周逢权 (2010) 智能变电站功能架构及设计原则. 电力系统保护与控制, 21, 24-27.
[7] 倪益民, 杨宇, 樊陈, 郭艳霞, 窦仁晖, 黄国方 (2014) 智能变电站二次设备集成方案讨论. 电力系统自动化, 3, 194-199.
[8] 廖瑞金, 王谦, 骆思佳, 廖玉祥, 孙才新 (2008) 基于模糊综合评判的电力变压器运行状态评估模型. 电力系统自动化, 3, 70-75.
[9] 卢绪祥, 李录平, 张晓玲, 黄竹青 (2006) 基于相对劣化度模型的大型汽轮机状态综合评价. 动力工程, 4, 507-510.
[10] Jiang, H. and Ruan, J.H. (2009) Fuzzy evaluation on network security based on the new algorithm of membership degree transformation-M(1, 2, 3). Journal of Networks, 4, 324-331.
[11] Song, Z.Y., Zhu, H.Q., Jia, G.W. and He, C.N. (2014) Comprehensive evaluation on self-ignition risks of coal stockpiles using fuzzy AHP approaches. Journal of Loss Prevention in the Process Industries, 32, 78-94.
[12] Zhang, J.C. (2012) The theory of membership degree of conclusion in several n-valued logic systems. American Journal of Operations Research, 2, 147-152.
[13] 段若晨, 王丰华, 顾承昱, 满玉岩, 傅正财, 刘亚东 (2014) 采用改进层次分析法综合评估 500 kV 输电线路防雷改造效果. 高电压技术, 1, 131-137.
[14] Zhang, Y., Zhang, H., Gao, X., et al. (2007) Improved AHP method and its application in lake environmental comprehensive quality evaluation—A case study of Xuanwu Lake, Nanjing, China. Chinese Journal of Oceanology and Limnology, 25, 427-433.
[15] (2010) Q/GDW441-2010智能变电站继电保护技术规范.
[16] (2010) Q/GDW428-2010智能变电站智能终端技术规范.
[17] (2010) Q/GDW427-2010智能变电站测控单元技术规范.
[18] (2010) Q/GDW429-2010智能变电站网络交换机技术规范.
[19] 袁宇波, 高磊, 卜强生, 宋亮亮, 等 (2013) 智能变电站集成测试技术与应用. 中国电力出版社, 北京.
[20] 何建军, 徐瑞林, 陈涛 (2012) 智能变电站系统测试技术. 中国电力出版社, 北京.
Top | 1,243 | 3,241 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.58726 |
https://www.scribd.com/document/41991676/After-ANOVA | 1,569,135,530,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00246.warc.gz | 1,023,846,960 | 72,782 | You are on page 1of 13
# AFTER ANOVA ... WHAT THEN ?
Learning Objective
## At the end of this lesson, the participants should be able to:
• distinguish between planned and unplanned mean comparisons;
• gain skills in partitioning treatment sums of squares; and
• perform trend analysis for quantitative treatment levels.
The result of the Analysis of Variance verifies the existence of some differences among
treatments tested but does not specify which treatment means significantly differ from
one another. To obtain this information, pairwise mean comparison is usually performed
after the ANOVA. However, there are statistical techniques that can extract more
information on the treatment means other than the pairwise mean comparison.
Some of the statistical techniques useful for further analysis after ANOVA are:
• Pairwise comparison
- Planned
- Unplanned
## • Partitioning of sums of squares (PSS)
- Group comparison
- Factorial comparison
- Orthogonal Polynomials
## After Anova ... What Then? 181
Pairwise Comparison
Pair comparison is the simplest and most commonly used comparison in agricultural
research. There are two types: planned and unplanned pairwise comparison.
## Planned pairwise comparison - in which the specific pair of treatments to be compared
is identified before the start of the experiment. A common example is comparison of the
control treatment with each of the other treatments.
Example 1:
## In an experiment where the objective is to determine the effectiveness of a new herbicide
in controlling weeds relative to three (3) existing herbicides where the treatments are as
follows:
## Treatments: H1 = non-weeded treatment
H2 = Butachlor 1.0
H3 = 2,4-D
H4 = Thiobencarb 1.0
H5 = New Herbicide
H1 vs H5
H2 vs H5
H3 vs H5
H4 vs H5
## 182 Pairwise Comparison
Example 2:
As another example, if the objective is to determine if any of the following three neem
seed kernel water extract (NSKWE) treatments are effective in controlling the incidence
of major insect pests:
Treatments:
application)
T1 vs T4
T2 vs T4
T3 vs T4
## After Anova ... What Then? 183
Unplanned pairwise comparison - in which no specific pairs of treatments are
identified for comparison before the start of the experiment. Instead, every possible pair
of treatment means is compared to identify those that are significantly different
Example 3:
## In an experiment where the objective is to evaluate extracts of 4 promising plants for
insecticidal properties against primary pests of rice after harvest, all possible pairs of
treatments are to be compared:
## Neem Oil vs Margosan Oil
Neem Oil vs Acorus Oil
Neem Oil vs Turmeric Oil
Margosan Oil vs Acorus Oil
Margosan Oil vs Turmeric Oil
Acorus Oil vs Turmeric Oil
This situation of unstructured qualitative treatments is, however, extremely rare and
unplanned pairwise comparisons are hardly ever appropriate.
The two most commonly used test procedures for pair comparisons are the least
significant difference (LSD) test which is suited for planned pair comparisons, and the
Duncan’s multiple range test (DMRT) which is suited for unplanned pair comparisons.
LSDα is the value against which the difference between any two means is compared. If
the difference exceeds this value, then it is significant at a probability α. The LSD is
computed as LSDα = tα sd. In the LSD test, a single LSD value, at a prescribed level of
significance α serves as the critical value between significant and non-significant
difference between any pair of treatment means. This test is not suitable for comparing
all possible pairs of means when number of treatments is large (say, greater than 5)
because in experiments where no real difference exists among the treatments, the
probability that at least one pair will have a difference that exceeds the LSD value,
increases with the number of treatments being tested.
## 184 Unplanned Pairwise Comparison
For example, it can be shown that the numerical difference between the largest and the
smallest treatment means is expected to exceed the LSD.05 even when there is no
difference between treatments with the following frequency:
## Number of treatments Percent of the time
Range exceeds LSD(.05)
5 29%
10 63%
15 83%
Hence, the LSD test must be used only when the F-test for treatment effect is significant
and the number of treatments tested is small, say less than six.
Duncan’s Multiple Range Test (DMRT) is useful in experiments that require the
comparison of all possible pairs of treatment means. The procedure for applying the
DMRT is similar to that for the LSD test but DMRT involves the computation of critical
values that allow for the rank differences between the two treatment means being
compared at each stage.
Multiple comparison procedures like LSD and DMRT are best suited for experiments
with unstructured qualitative treatments. Usually, the objective for this type of
experiment is to select the best treatment among a set of qualitative treatments. Example
of this is variety trials. However, multiple comparison is never appropriate when 1) the
treatments are graded levels of a quantitative variable, 2) for factorial combination of two
or more factors at two or more levels, and 3) for quantitative treatments where previously
formulated linear combinations of treatments are of particular interest. Indiscriminant use
of multiple comparison tests may lead to loss of information and reduced efficiency when
more appropriate procedures are available.
## After Anova ... What Then? 185
Partitioning of Sum of Squares
Main effects or interaction sums of squares in the Analysis of Variance (ANOVA) can
often be partitioned into two or more components based on the structure of the
treatments. Each component is designed to test a specific hypothesis about the
treatments. Such procedure is called the partitioning of sum of squares or PSS.
If the decision was made before the experiment is performed, i.e., without studying the
data, then tests based on PSS are perfectly valid no matter what the status of the overall
F-test for the partitioned effect. These tests are independent provided that the contrasts
are orthogonal. When factor levels are amounts of quantitative input, the form of the
response to that factor can be tested through PSS using orthogonal polynomials.
For example, in a variety trial with three varieties A, B, and C, variety A may be the
standard variety and B and C new varieties. Immediately, the researcher would like to
know if the new varieties are different from the standard one, and whether the new ones
differ between themselves. These are special questions that have been explicitly or
implicitly asked by the researcher prior to doing the experiment.
Such questions are expressed as contrasts between treatment means as follows: Is the
value of (2 x mean of A - mean of B - mean of C) significantly different from zero?
These comparisons are termed contrasts because the coefficients of the means: 2, -1, -1,
always sum to zero. In general given a factor with k levels and it is the intention to test a
contrast among these levels with coefficients C1, C2, ..., Ck, then ΣCi = 0 and proceed as
follows:
The SS for the contrast is computed as: CSS = (Σ CiTi)2 / (r Σ Ci2) where Ti are the
treatment totals and r is the number of replications of each treatment. A contrast SS has 1
dF, and test is made about the null hypothesis that the contrast is zero using F =
CSS/EMS with 1 dF and error dF.
## 186 Partitioning of Sum of Squares
Example 4:
In the RCB analysis, it is of interest to compare the standard variety A with the new
varieties B and C.
Variety A B C
Totals 16.0 18.8 18.0
Contrast Coefficients for standard against new 2 -1 -1
varieties
Contrast Coefficients for variety B against C 0 1 -1
## F = 0.960/0.089 = 10.7 with 1 and 6 dF which is significant at 0.01
The other obvious contrast among these varieties is to compare the two new varieties, B
and C. The coefficients for this contrast would be: 0, +1 , -1 so that CSS = (0*16.0 +
1*18.8 - 1*18.0)2/(4*2) = 0.080 which leads to an F computed less than one and so
cannot be significant.
Notice, in this case the two contrast SSs add up to the variety SS, 0.960 + 0.080 = 1.040.
In effect the variety SS has been partitioned into two separate SSs for testing specific
hypotheses.
In general any SS with k dF can be partitioned into k contrast SSs provided the contrasts
are orthogonal. If the contrasts are orthogonal then the sum of products of the
coefficients of any two of the contrasts will always be zero, as in this case: (2*0) + (-1*1)
+ (-1*-1) = 0. When the contrasts are orthogonal the resulting tests are statistically
independent and so the test level (α) is preserved over the set of tests.
## After Anova ... What Then? 187
Frequently the contrast tests are included in the ANOVA table and for the example above,
the table would look as follows:
SV dF SS MS F
Blocks 3 0.587 0.196
Varieties 2 1.040 0.520 5.84*
A vs B&C 1 0.960 0.960 10.70**
B vs C 1 0.080 0.080 0.89
Error 6 0.533 0.089
Total 11 2.160
A special case of these comparisons is always suggested when the factor has a set of
quantitative levels such as the amount of fertilizer, or the temperature of a mixture. It is
almost always desired to consider whether the response to the levels of the factor follows
a straight line or some other specific curve, perhaps quadratic or cubic.
The contrast coefficients for these tests are called orthogonal polynomials and provided
the levels of the factor are equally spaced, the coefficients can be found in common
statistical tables. They are different for each number of different levels. The most
common ones are given below. They are used exactly the same way as the examples
above.
No. of Trend SS
Levels
3 Linear -1 0 +1 2
3 Quadratic +1 -2 +1 6
4 Linear -3 -1 +1 +3 20
4 Quadratic +1 -1 -1 +1 4
4 Cubic -1 +3 -3 +1 20
5 Linear -2 -1 0 +1 +2 10
5 Quadratic +2 -1 -2 -1 +2 14
5 Cubic -1 +2 0 -2 +1 10
## 188 Partitioning of Sum of Squares
There are three (3) ways of Partitioning Sums of Squares
1. Group Comparison
## • Between-Group Comparison, in which treatments are classified into s meaningful
groups, each group consisting of one or more treatments, and the aggregated mean
of each group is compared to that of the others.
## • Within-Group Comparison, which is primarily designed to compare treatments
belonging to a subset of all the treatments tested.
## 2. Factorial Comparison is applicable only to factorial treatments in which specific sets
of treatment means are compared to investigate the main effects of the factors tested
and, in particular, the nature of their interaction.
## 3. Orthogonal Polynomials are partitioning designed to examine the functional
relationship between factor levels and treatment means. As a result, it is suited for
treatments that are quantitative, such as rate of herbicide application, rate of fertilizer
application, and planting distance or spacing.
Example 5:
For group comparison, consider a varietal trial involving 5 varieties and 4 replications
layed out in RCB.
Varieties : V1
Japonica group
V2
V3
V4 Indica group
V5
## After ANOVA ... What Then? 189
With these, the ANOVA table will be of the following form:
SV dF
Block 3
Variety 4
(V1, V2) vs (V3, V4, V5) (1)
V1 vs V2 (1)
V3 vs V4 vs V5 (2)
Error 12
Total 19
Example 6:
## For factorial comparison, consider an experiment where in the objective is to determine
the effect of prilled urea (PU) sulfur coated urea (SCU) and urea super granules (USG) at
three rates of application on rice yield.
1 (control) 0 -
2 29 PU
3 29 SCU
4 29 USG
5 58 PU
6 58 SCU
7 58 USG
8 87 PU
9 87 SCU
10 87 USG
## 190 Partitioning of Sum of Squares
And further assume that this experiment is layed-out in RCB with 4 replications, then the
detailed analysis of variance table should have the following form:
SV dF MS F
Rep 3 0.7064 5.00**
Treatment 9 1.1638 8.23*
Ctrl vs Trtd (1) 4.0541 28.68**
Among Trtd (8) 0.8025 5.68**
Rate (R) 2 1.0653 7.54**
Form (F) 2 0.2028 1.43ns
RxF 4 0.9710 6.87**
Error 27 0.1413
Example 7:
For orthogonal polynomials, consider the following yield trial involving 4 N-rates and
3 replications layed out in RCB. The objective is to determine the optimum N-rate that
will maximize rice yield.
Treatments: N-rates
1. 0 kg N/ha
2. 60 kg N/ha
3. 90 kg N/ha
4. 120 kg N/ha
## The following is an outline of the ANOVA table for this trial:
SV dF MS F
Block 2 0.104381 <1
N-rate 3 3.055244 21.07**
Linear (1) 8.974925 61.09**
Cubic (1) 0.172552 1.19ns
Error 6 0.145011
c.v. = 5.7%
## After Anova ... What Then? 191
Results show that N-rate is highly significant which indicate that much of the variation in
yield is affected by the differential rates in N application. The partitioning of the N-rate
SS verifies a linear effect of N-rate on yield within the range of 0 kg N/ha to 120 N kg/ha.
References :
Cochran, W.G., and Cox, G.M. (1957). Experimental Designs, Second Edition. John
Wiley, New York, New York.
Gomez, K.A., and Gomez, A.A. (1984). Statistical Procedures for Agricultural Research.
John Wiley, New York, New York.
Petersen, R.G. (1977), Use and Misuse of Multiple Comparison Procedures, Agronomy
Journal, Vol. 69, March-April, 205-208.
Searle, S.R. (1987). Linear Models for Unbalanced Data. John Wiley, New York, New
York.
Snedecor, G.W., and Cochran, W.G. (1989). Statistical Methods, Eighth Edition. Iowa
State University, Ames, Iowa.
192 Exercise
After ANOVA ... What Then? 193 | 3,479 | 13,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-39 | latest | en | 0.894318 |
http://hackage.haskell.org/package/vector-fftw-0.1.3.1/docs/Numeric-FFT-Vector-Unitary.html | 1,477,702,301,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476990033880.51/warc/CC-MAIN-20161020190033-00449-ip-10-142-188-19.ec2.internal.warc.gz | 111,295,921 | 3,234 | vector-fftw-0.1.3.1: A binding to the fftw library for one-dimensional vectors.
Numeric.FFT.Vector.Unitary
Contents
Description
This module provides normalized versions of the transforms in `fftw`.
All of the transforms are normalized so that
• Each transform is unitary, i.e., preserves the inner product and the sum-of-squares norm of its input.
• Each backwards transform is the inverse of the corresponding forwards transform.
(Both conditions only hold approximately, due to floating point precision.)
Synopsis
# Creating and executing `Plan`s
run :: (Vector v a, Vector v b, Storable a, Storable b) => Transform a b -> v a -> v bSource
Create and run a `Plan` for the given transform.
plan :: (Storable a, Storable b) => Transform a b -> Int -> Plan a bSource
Create a `Plan` of a specific size. This function is equivalent to `planOfType Estimate`.
execute :: (Vector v a, Vector v b, Storable a, Storable b) => Plan a b -> v a -> v bSource
Run a plan on the given `Vector`.
If `planInputSize p /= length v`, then calling `execute p v` will throw an exception.
# Complex-to-complex transforms
A discrete Fourier transform. The output and input sizes are the same (`n`).
`y_k = (1/sqrt n) sum_(j=0)^(n-1) x_j e^(-2pi i j k/n)`
An inverse discrete Fourier transform. The output and input sizes are the same (`n`).
`y_k = (1/sqrt n) sum_(j=0)^(n-1) x_j e^(2pi i j k/n)`
# Real-to-complex transforms
A forward discrete Fourier transform with real data. If the input size is `n`, the output size will be `n `div` 2 + 1`.
A normalized backward discrete Fourier transform which is the left inverse of `dftR2C`. (Specifically, `run dftC2R . run dftR2C == id`.)
This `Transform` behaves differently than the others:
• Calling `plan dftC2R n` creates a `Plan` whose output size is `n`, and whose input size is `n `div` 2 + 1`.
• If `length v == n`, then `length (run dftC2R v) == 2*(n-1)`.
# Discrete cosine transforms
Some normalized real-even (DCT). The input and output sizes are the same (`n`).
A type-2 discrete cosine transform. Its inverse is `dct3`.
`y_k = w(k) sum_(j=0)^(n-1) x_j cos(pi(j+1/2)k/n);` where `w(0)=1/sqrt n`, and `w(k)=sqrt(2/n)` for `k>0`.
A type-3 discrete cosine transform which is the inverse of `dct2`.
`y_k = (-1)^k w(n-1) x_(n-1) + 2 sum_(j=0)^(n-2) w(j) x_j sin(pi(j+1)(k+1/2)/n);` where `w(0)=1/sqrt(n)`, and `w(k)=1/sqrt(2n)` for `k>0`.
A type-4 discrete cosine transform. It is its own inverse.
`y_k = (1/sqrt n) sum_(j=0)^(n-1) x_j cos(pi(j+1/2)(k+1/2)/n)` | 790 | 2,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-44 | longest | en | 0.703493 |
https://premiumessaymillers.com/he-supermarket-displays-the-unit-price-for-the-12-ounce-jar-in-terms-of-cost-per-2/ | 1,638,088,077,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00087.warc.gz | 524,329,251 | 9,143 | He Supermarket Displays The Unit Price For The 12 Ounce Jar In Terms Of Cost Per
He supermarket displays the unit price for the 12-ounce jar in terms of cost per ounce, but displays the unit price for the 18-ounce jar in terms of cost per quart. Assuming 32 ounces in a quart, what are the unit prices, to the nearest cent, given by the supermarket?
Posted in Uncategorized | 87 | 375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-49 | latest | en | 0.842026 |
https://acmewriters.com/mathematics-matlab/ | 1,679,421,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00485.warc.gz | 129,568,964 | 12,222 | # Mathematics Matlab
• To solve the questions below you need to use ONLY Matlab or mathematica
Project :The demand function for a monopolist’s product is and the average cost per unit for producing units is , where = price, and = quantity demand.
Don't use plagiarized sources. Get Your Custom Essay on
Mathematics Matlab
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Questions
1. Develop the total revenue, total cost (if not given), and profit functions. Explain these functions in few sentences.
2. Develop a 2D plot for demand versus the price. What do you notice?
3. Develop a 2D plot for total cost versus the quantity; average cost versus the quantity; marginal cost versus the quantity; and marginal revenue versus quantity on the same graph. What do you notice?
4. Compute the demand elasticity when price equal to 4. Interpret your result.
5. Find the quantities that maximize the total revenue and profit. Interpret your result.
6. Find the quantity that minimize the total cost function. Interpret your results.
7. Discuss the results of 6 and 7.
8. What is the price that maximizes the profit? Interpret your result.
9. What is the maximum profit? Explain clearly your result.
10. Provide a simple conclusion that would explain what you have learned during the group project. Each of the member of the selected team must relate his/her own experience in the conclusion.
275 words
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Type of paper | 436 | 2,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-14 | latest | en | 0.856253 |
https://www.namastekadapa.com/2016/05/31/bioinformatics-and-computational-biology-bio-computing-computational-biology/ | 1,713,575,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817463.60/warc/CC-MAIN-20240419234422-20240420024422-00587.warc.gz | 825,109,633 | 51,433 | Current Affairs
Bioinformatics and Computational Biology – Bio Computing – Computational Biology
Bioinformatics and Computational Biology
Bioinformatics and computational biology each maintain close interactions with life sciences to realize their full potential. Bioinformatics applies principles of information sciences and technologies to make the vast, diverse, and complex life sciences data more understandable and useful. Computational biology uses mathematical and computational approaches to address the oretical and experimental questions in biology. Although bioinformatics and computational biology are distinct, there is also significant overlap and activity at their interface.
Bio Computing
Bio computing is often used as a catch-all term covering all this area at the intersection of and Computation although many other terms are used to name the same area. We distinguish can in to sub-fields:
• • Bioinformatics this includes management of biological databases, data mining and data modeling, as well as for data visualization
• • Computational Biology this includes efforts to solve biological problems with computational tools (such as modeling, algorithms, heuristics)
• • DNA computing and nano-engineering this includes models and experiments to use DNA (and other) to perform computations
• • Computations in living organisms this is concerned with constructing computational components in living cells, as well as with studying computational processes taking place daily in living organisms
Computational Biology
Computational Biology is application of core technology of computer science (eg. algorithms. artificial intelligence, databases etc) to problems arising from biology. Computational biology is particularly exciting today because the problems are large enough to motivate the efficient algorithms and moreover the demand of biology on computational science is increasing.
The most pressing tasks in bioinformatics involve the analysis of sequence information. Computational Biology is the name given to this process, and it involves the following:
• • Finding the genes in the DNA sequences of various organisms
• • Developing methods to predict the structure and/or function of newly discovered proteins and structural RNA sequences
• • Clustering protein sequences into families of related sequences and the development of protein models.
• • Aligning similar proteins and generating phylogenetic trees to examine evolutionary relationships. | 415 | 2,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-18 | longest | en | 0.904107 |
https://transum.org/Software/sw/Starter_of_the_day/Starter_June5.asp | 1,638,727,964,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00326.warc.gz | 612,389,526 | 13,305 | Kayla bought 5 pineapples and 3 peppers which altogether cost £32.
On another occasion she bought 8 pineapples and 9 peppers which cost £68.
Assume the cost of pineapples and peppers stays the same. How much does one pineapple cost? How much does one pepper cost?
Hint: The answers are whole numbers.
## A Mathematics Lesson Starter Of The Day
• Transum,
•
• This is one of those starters that is different every time you load the page so you can use it many times with the same class. At a guess most pupils seeing this starter will not be familiar with simultaneous equations so a trial and improvement (involving lots of calculations) may be used. It is a good problem to develop a systematic way of working so that it is possible to home in on the correct answer.
• Mrs Walker, St.James' School
•
• My year 6 class found this fairly easy once they had a discussion with their neighbours. Satisfactory.
• Louisa, Greentrees School
•
• I did this with my Year 4 class and they found it very easy! They got the answer right! Slightly harder next time Transum. To see if it would be a harder one the next time we reloaded the page but it wasn't.
[Transum: Scroll down the page Louisa and you will find buttons which enable you to select different levels of difficulty.]
How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.
Previous Day | This starter is for 5 June | Next Day
This activity is suitable for students of mathematics all around the world. Use the button below to change the currency symbol used to make it more relevant to your students. You may wish to choose an unfamiliar currency to extend your students' experience.
:::
:::
Generate another problem:
Level 1:
Eg. 3x + 2y = 18
5x + 2y = 26
(same coefficient of second variable)
Level 2:
Eg. 3x + 2y = 18
5x + 4y = 32
(Simple multiple coefficient of second variable)
Level 3:
Random coefficients
Note to teacher: Doing this activity once with a class helps students develop strategies. It is only when they do this activity a second time that they will have the opportunity to practise those strategies. That is when the learning is consolidated. Click the button above to regenerate another version of this starter from random numbers.
Christmas Present Ideas
It is often very difficult choosing Christmas presents for family and friends but so here are some seasonal, mathematics-related gifts chosen and recommended by Transum Mathematics.
## Equate board game
Here's a great board game that will give any family with school-aged kids hours of worthwhile fun. Christmas is a time for board games but this one will still be useful at any time of year. Games can be adapted to suit many levels of Mathematical ability.
For Maths tutors working with just one or small groups of pupils this game has proved to be an excellent activity for a tutorial. Deciding on the best moves can spark pertinent discussions about mathematical concepts.
Equate looks a bit like Scrabble--for aspiring mathematicians, that is. Designed by a real mathematician, it works like this: You put down tiles on a board and make points by correctly completing simple equations. Your nine tiles include both numbers and mathematical symbols; you can add on to previous plays both vertically and horizontally. more...
## How Not To Be Wrong
The maths we learn in school can seem like an abstract set of rules, laid down by the ancients and not to be questioned. In fact, Jordan Ellenberg shows us, maths touches on everything we do, and a little mathematical knowledge reveals the hidden structures that lie beneath the world's messy and chaotic surface. In How Not to be Wrong, Ellenberg explores the mathematician's method of analyzing life, from the everyday to the cosmic, showing us which numbers to defend, which ones to ignore, and when to change the equation entirely. Along the way, he explains calculus in a single page, describes Gödel's theorem using only one-syllable words, and reveals how early you actually need to get to the airport.
What more could the inquisitive adult want for Christmas? This book makes a cosy, interesting read in front of the fire on those cold winter evenings. more...
## Graphic Display Calculator
This handheld device and companion software are designed to generate opportunities for classroom exploration and to promote greater understanding of core concepts in the mathematics and science classroom. TI-Nspire technology has been developed through sound classroom research which shows that "linked multiple representation are crucial in development of conceptual understanding and it is feasible only through use of a technology such as TI-Nspire, which provides simultaneous, dynamically linked representations of graphs, equations, data, and verbal explanations, such that a change in one representation is immediately reflected in the others.
For the young people in your life it is a great investment. Bought as a Christmas present but useful for many years to come as the young person turns into an A-level candidate then works their way through university. more...
The analytics show that more and more people are accessing Transum Mathematics via an iPad as it is so portable and responsive. The iPad has so many other uses in addition to solving Transum's puzzles and challenges and it would make an excellent gift for anyone.
The redesigned Retina display is as stunning to look at as it is to touch. It all comes with iOS, the world's most advanced mobile operating system. iPad Pro. Everything you want modern computing to be. more...
## Aristotle's Number Puzzle
It’s a bit of a tradition to give puzzles as Christmas Gifts to nieces and nephews. This puzzle is ideal for the keen puzzle solver who would like a challenge that will continue over the festive period (at least!).
This number puzzle involves nineteen numbers arranged into a hexagon. The goal of the puzzle is to rearrange the numbers so each of the fifteen rows add up to 38. It comes in a wooden style with an antique, aged look.
Keep the Maths in Christmaths with this reasonably priced stocking filler. more...
## The Story Of Maths [DVD]
The films in this ambitious series offer clear, accessible explanations of important mathematical ideas but are also packed with engaging anecdotes, fascinating biographical details, and pivotal episodes in the lives of the great mathematicians. Engaging, enlightening and entertaining, the series gives viewers new and often surprising insights into the central importance of mathematics, establishing this discipline to be one of humanity s greatest cultural achievements. This DVD contains all four programmes from the BBC series.
Marcus du Sautoy's wonderful programmes make a perfect Christmas gift more...
## Christmas Maths
This book provides a wealth of fun activities with a Christmas theme. Each photocopiable worksheet is matched to the Numeracy Strategy and compatible with the Scottish 5-14 Guidelines. This series is designed for busy teachers in the late Autumn term who are desperate for materials that are relevant and interesting and that can be completed with minimun supervision.
All the activities are suitable for use by class teachers, supply teachers, SEN teachers and classroom assistants and cover topics such as 'How many partridges did the true love give all together?' and 'Filling a sleigh with presents by rolling a dice!'. Children will have lots of fun working through the Christmas Maths themes but also gain valuable skills along the way.
A great source of ideas and another reasonably priced stocking filler. more...
## A Compendium Of Mathematical Methods
How many different methods do you know to solve simultaneous equations? To multiply decimals? To find the nth term of a sequence?
A Compendium of Mathematical Methods brings together over one hundred different approaches from classrooms all over the world, giving curious mathematicians the opportunity to explore fascinating methods that they've never before encountered.
If you teach mathematics to any age group in any country, you are guaranteed to learn lots of new things from this delightful book. It will deepen your subject knowledge and enhance your teaching, whatever your existing level of expertise. It will inspire you to explore new approaches with your pupils and provide valuable guidance on explanations and misconceptions. more...
I had been tutoring the wonderful Betsy for five years. When the day came for our last ever session together before the end of her Year 13, I received this beautiful book as a gift of appreciation.
This a very readable book by Ben Orlin. I'm really enjoying the humour in the writing and the drawings are great.
Ben Orlin answers maths' three big questions: Why do I need to learn this? When am I ever going to use it? Why is it so hard? The answers come in various forms-cartoons, drawings, jokes, and the stories and insights of an empathetic teacher who believes that mathematics should belong to everyone. more...
Click the images above to see all the details of these gift ideas and to buy them online.
## Maths T-Shirts
Your access to the majority of the Transum resources continues to be free but you can help support the continued growth of the website by doing your Amazon shopping using the links on this page. Below is an Amazon search box and some items I have chosen and recommend to get you started. As an Amazon Associate I earn a small amount from qualifying purchases which helps pay for the upkeep of this website.
Teacher, do your students have access to computers?Do they have iPads or Laptops in Lessons? Whether your students each have a TabletPC, a Surface or a Mac, this activity lends itself to eLearning (Engaged Learning).
Transum.org/go/?Start=June5
Here is the URL which will take them to a related student activity.
Transum.org/go/?to=simsol
## Curriculum Reference
See the National Curriculum page for links to related online activities and resources.
For Students:
For All: | 2,066 | 10,216 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2021-49 | longest | en | 0.940765 |
https://sciblogs.co.nz/physics-stop/2011/09/13/the-problem-with-having-odd-shaped-balls/ | 1,516,643,846,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891530.91/warc/CC-MAIN-20180122173425-20180122193425-00612.warc.gz | 782,845,204 | 19,053 | # The problem with having odd-shaped balls…
By Marcus Wilson 13/09/2011
…by which, of course, I mean rugby balls.
To be precise, a rugby ball is a prolate ellipsoid – that is, something that is like a 3d version of an ellipse, but having a cross-section along its long axis of a circle. (A flying-saucer would be an oblate ellipsoid)
Rugby balls behave awkwardly. In one sense that’s obvious – their awkward shape will lead to awkward movement, on the ground at least. But what about when in the air? Why is it hard to get a rugby ball to spin nicely and stay in one orientation (like the professionals can do?) Or, put another way, why is it so easy to make a rugby ball tumble all over the place in flight, making it more difficult for the opposing full-back to catch?
Once in the air, the rugby ball undergoes torque-free movement. (We’ll neglect things like the Magnus effect). That means there’s no external forces trying to create extra spin on the ball. The behaviour of an object in this situation can be described by Euler’s equations (after the Swiss mathematician Leonhard Euler). I won’t write them in a blog, but they are not that difficult to interpret if you know about rotational inertia and angular velocities.
These equations can be solved fairly simply in some cases. It turns out, that if you have a rugby ball that is spinning on its long axis (like you try to do when you pass to one of your teammates) the rotation is stable. That means, once you have established it rotating in this manner, it will remain doing so. If the ball is perturbed slightly off this spin (perhaps by a piece of flying mud) it will return to this direction.
However, the same is not true if you spin the ball about a short axis (so it tumbles end on end). In this case, the motion is critically stable – meaning that any perturbation is not corrected. The ball can easily lose this manner of motion, and start moving in another way. If you don’t start it exactly in this way of movement, it’s unlikely to stay there. Find yourself a rugby ball and try it.
Perhaps more interesting is the case when you have an object with three axes of different lengths (e.g. a paperback book). You’ll find that if you try to spin the book about its long axis, or its short axis, these are stable, but about the middle axis the rotation is unstable. Try it. Take a book (a cuboid with three different length axes), hold it with its cover the right way up and facing you, and grip its bottom two corners. Now flip it in the air – try to get it to spin 360 degrees and catch it again, by its bottom two corners. See what happens. It’s very difficult to do – the book will tumble around, because rotation about this middle axis is unstable. Any small perturbation from it will be magnified as the book spins. However, you try the same experiment spinning about one of the other axes of the book, and you’ll find the motion stable.
Incidentally, if you take an object with three identical axes (e.g. a soccer ball), its rotations are critically stable in all directions, which is one reason why a soccer ball in flight can spin (and bend) in all kinds of whacky ways. | 715 | 3,163 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-05 | longest | en | 0.936598 |
https://maker.pro/forums/threads/i-am-pretty-sure-my-answer-is-right.276179/ | 1,716,289,571,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058442.20/warc/CC-MAIN-20240521091208-20240521121208-00729.warc.gz | 324,505,653 | 64,442 | # I am pretty sure my answer is right!!
#### Integrator741
Jun 16, 2013
125
Good evening,
I am practising my ability to manipulate units and stuff, and I came across this question that doesn't sound that hard but I just can't get the right answer.
"The electricity meter on a plant initially reads 6032 kWh. If the plant now runs for 73 minutes at a steady power consumption of 320 kW determine a) Energy consumed in MJ b) The final reading of the meter"
Answers:. a) 1401.6 MJ; b) 6421.3 kWh
My working out is:
320 kWh/60min = 5.3 kW per min.
5.3 kW per min * 73 min = 386.9 kWh
so the final reading will be 6032 + 386.9 = 6419 kWh
now to convert 386.9 to MJ:
1kWh = 3.6 MJ
386.9/3.6 = 107.472 MJ;
I am close with a) but b)... no idea.
Thanks
#### davenn
Moderator
Sep 5, 2009
14,273
now to convert 386.9 to MJ:
1kWh = 3.6 MJ
386.9/3.6 = 107.472 MJ;
I am close with a) but b)... no idea.
why did you divide by 3.6 ? it should be multiply
1kWh = 3.6 ...... ( 3.6MJ for each kWh)
2kWh = 7.2
10kWh = 36
etc
386.9 x 3.6 = ..... ?
you now work that out
cheers
Dave
#### dorke
Jun 20, 2015
2,342
Very simple:
Do not round the results!
a) 320*(73/60)*3.6= ??
b) 320*(73/60)+6032 = ??
Last edited by a moderator:
Moderator
Sep 5, 2009
14,273
Dorke
#### Integrator741
Jun 16, 2013
125
Very simple:
Do not round the results!
a) 320*(73/60)*3.6=1401.6
b) 320*(73/60)+6032 =6421.3
Thanks, I didn't think that it would make such a significant difference. Again thanks. Maybe you'd be able to help me with another question, about some copper wire coiled around a tube.
"Copper wire (density 8900) of cross-section 1.7mm^2 is coiled onto a long cylindrical former. The resulting coil has 820 turns, and diameter 0.13m. Find the mass of wire in KG."
What I did:
First I need to know the length of the catble - length of one turn is (pi)*Diameter of the cylinder. (pi)*0.13m = 0.408m;
Total length of the cable is: 0.408m * 820 turns = 334.9m;
then I need to calculate the volume of the cable in order to calculate its mass.
To find volume of the cylinder - I need diameter.
I know that area=(pi)*r^2 so 0.13=(pi)*r^2; (solve for r)
r=(square root)0.13/(pi) = 0.2m - so diameter is 0.4m;
Volume of the cylinder is: ((pi)*d^2*l)/4 = I got 42.08m^3, then to find mass I multiply my answer by the given density and I get: 374512kg and the actual answer is 5.0669 kg - what the hell?
#### davenn
Moderator
Sep 5, 2009
14,273
I like Dorke's way, at least it gives the expected answer
his gives 389.333, where as yours gives 386.9 and of course that made the answer
wrong for you and me as I used your number
#### dorke
Jun 20, 2015
2,342
Thanks, I didn't think that it would make such a significant difference. Again thanks. Maybe you'd be able to help me with another question, about some copper wire coiled around a tube.
"Copper wire (density 8900) of cross-section 1.7mm^2 is coiled onto a long cylindrical former. The resulting coil has 820 turns, and diameter 0.13m. Find the mass of wire in KG."
What I did:
First I need to know the length of the catble - length of one turn is (pi)*Diameter of the cylinder. (pi)*0.13m = 0.408m;
Total length of the cable is: 0.408m * 820 turns = 334.9m;
then I need to calculate the volume of the cable in order to calculate its mass.
To find volume of the cylinder - I need diameter.
I know that area=(pi)*r^2 so 0.13=(pi)*r^2; (solve for r)
r=(square root)0.13/(pi) = 0.2m - so diameter is 0.4m;
Volume of the cylinder is: ((pi)*d^2*l)/4 = I got 42.08m^3, then to find mass I multiply my answer by the given density and I get: 374512kg and the actual answer is 5.0669 kg - what the hell?
I think they are talking about a coil raped around a cylinder.
That means the cylinder is hollow:
there is only copper wire on it's "outer- case" .
So you have a copper wire with a cross-section of
1.7mm^2 wrapped around the cylinder of diameter 0.13m
1.You did o.k with calculating the wire length,
to be more accurate you need to add the diameter of the wire to the diameter of the cylinder!)-again don't round numbers.
2.You need to convert the cross-section to meters and multiply by the length to find the wires volume in meters^3
3.Last multiply by the density (in meters^3) to get the result.
#### Integrator741
Jun 16, 2013
125
I think they are talking about a coil raped around a cylinder.
That means the cylinder is hollow:
there is only copper wire on it's "outer- case" .
So you have a copper wire with a cross-section of
1.7mm^2 wrapped around the cylinder of diameter 0.13m
1.You did o.k with calculating the wire length,
to be more accurate you need to add the diameter of the wire to the diameter of the cylinder!)-again don't round numbers.
2.You need to convert the cross-section to meters and multiply by the length to find the wires volume in meters^3
3.Last multiply by the density (in meters^3) to get the result.View attachment 23010
Thank you - I got it now.
One question though, I see that I get my answer in grams, why is that?
#### dorke
Jun 20, 2015
2,342
"I see that I get my answer in grams, why is that?
#### Integrator741
Jun 16, 2013
125
Cable length = turns*circumference = 820 * ((pi)*0.13) = 334.89
Volume = (1.7*10^-3) * 334.89 = 0.56932 m^3
Volume*density = 0.56932 * 8900 = 5066.9
#### davenn
Moderator
Sep 5, 2009
14,273
Volume*density = 0.56932 * 8900 = 5066.9
One question though, I see that I get my answer in grams, why is that?
because the 8900 is 8900 grams / m3 ( 8.9 g / cm3 )
so you ended up with 5066.9 grams = 5.066kg
also again you are still rounding off ... stop it !!
you should be using 8.96 g / cm3
#### dorke
Jun 20, 2015
2,342
Dave,
You both got it wrong...
The density of copper is indeed 8900Kg/m^3 not gr.
A 1x1x1 meter cube of copper weighs about 9tons not 9Kg...
The problem is in transferring the wire diameter from mm^2 to m^2
the factor should be 1x 10^-6 and not 1x10^-3...
#### davenn
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Sep 5, 2009
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DOH ... that will teach me for doing maths at 6 am !!!
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1K | 1,958 | 6,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-22 | latest | en | 0.853867 |
https://ebrary.net/1016/economics/heteroskedasticity_diagnostics | 1,695,889,705,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510368.33/warc/CC-MAIN-20230928063033-20230928093033-00504.warc.gz | 235,664,201 | 9,576 | # Heteroskedasticity and diagnostics
The classical assumption required for the OLS estimator to be efficient states that the variance of the error term has to be constant and the same for all observations. This is referred to as a homoskedastic error term. When that assumption is violated and the variance is different for different observations we refer to this as heteroskedasticity. This chapter will discuss the consequences of violating the homoskedasticity assumption, how to detect any deviations from the assumption and how to solve the problem when present.
## Consequences of using OLS
The classical assumptions made on the error terms are that they are uncorrelated, with mean zero and constant varianceoU2 . In technical terms this means that
Assumption (9.1) is in use to make the OLS estimators unbiased and consistent. Assumptions (9.2) and (9.3) are important for the OLS estimator to be efficient. Hence, if (9.2) is ignored and the residual variance is heteroskedastic we can no longer claim that our estimator is the best estimator among linear unbiased estimators. This means that it is possible to find another linear unbiased estimator that is more efficient.
Heteroskedasticity implies that
The OLS estimators of the population parameters are still unbiased and consistent. The usual standard errors of the estimated parameters are biased and inconsistent.
It is important to understand that the violation of (9.2) makes the standard errors of the OLS estimators and the covariances among them biased and inconsistent. Therefore tests of hypothesis are no longer valid, since the standard errors are wrong. To see this, consider the variance of the estimator for the slope coefficient of the simple regression model:
The expression given by (9.4) represents the correct variance that should be used. Unfortunately it involves the unknown population variance of the error term which is different for different observations.
Since the error term is heteroskedastic, each observation will have a different error variance. The expression will therefore deviate from the variance estimated under homoskedasticity, that is:
As can be seen in (9.5) the variance of the OLS estimator is different from the expected value of the sample variance of the estimator that works under the assumption of a constant variance.
An important use of the regression equation is that of making predictions and forecasts of the future. Since the OLS estimators are unbiased and consistent, so will the forecasts. However, since the estimators are inefficient, the uncertainty of the forecasts will increase, and the confidence interval of the forecast will be biased and inconsistent.
## Detecting heteroskedasticity
Since we know that heteroskedasticity invalidate test results it is very important to investigate whether our empirical model is homoskedastic. Fortunately there are a number of test, graphical as well as statistical that one can apply in order to receive an answer to the question. Below the most commonly used test will be discussed.
### Graphical methods
A natural starting point in detecting possible deviations from homoskedasticity is to plot the data. Since we are interested in the behavior of the error term and its variation, two obvious scatter plots are given in Figure 9.1 that comes from a simple linear regression model. In Figure 9.1a the dependent variable is plotted against its explanatory variable X. Here we can see a clear pattern of heteroskedastidity which is driven by the explanatory variable. That is, the larger the value of x, the larger is the variance of the error term.
As an alternative to Figure 9.1a we can also plot the estimated residuals directly against X, as is done in Figure 9.1b. In the simple regression case with just one explanatory variable these two graphs are always very similar. However, when using a multiple regression models the picture might be different, since the residual is a linear combination of all variables included in the model. Since it is the partial effect of x on the residual that is of primary interest, it is advised that the residual should be plotted against all involved variables separately. If it is possible to find a systematic pattern that give indications of differences of the variances over the observations, one should be concerned. Graphical methods are useful, but sometimes it is difficult to say if heteroskedasticy is present and found harmful. It is therefore necessary to use statistical test. The graphical method is therefore merely a first step in the analysis that can give a good picture of the nature of the heteroscedasticity we might have, which will be helpful later on when correcting for it.
Figure 9.1 Scatter plots to detect heteroskedasticity
Example 9.1
We are interested in the rate of the return to education and estimate the coefficients of the following human capital model:
We use a sample of 1483 individuals with information on hourly wages in logarithmic form (lnY), years of schooling (ED), and years of work experience (year). Both explanatory variables are also squared to control for any non linear relation between the dependent variable and the two explanatory variables. Using OLS we received the following results with f-values within parenthesis:
In Y should be interpreted as the predicted value of lnY. We observe that all coefficients are significantly different from zero. The squared terms have very small coefficients, even though their f-values are sufficiently large to make them significant. Observe that the coefficient for the square of year is different from zero. Since the value is very small and we only report the first three decimals it appears to be zero. Its f-value shows that the standard error is even smaller.
We suspect that our residual might be heteroskedastic and we would like to investigate this by looking at a graph between the residual and the two explanatory variables. Sometimes, to enlarge the possible differences in variance among the residuals it is useful to square the estimated residual. If we do that we receive the graphs given in Figure 9.2.
Figure 9.2 Graphical analysis of error variance
In Figure 9.2a we see the squared error term against the number of years of schooling, and a pattern can be identified. The error variance seems to be larger for lower years of schooling than for more years of schooling. This is of course just an indication that we need to investigate further using formal statistical test.
In Figure 9.1b the picture is less obvious. If ignoring the top outlier that makes it look likes the variance is higher around 35 year of work experience, it is difficult to say if there is some heteroskedasticity to worry about. Since it is an unclear case, we still need to investigate the issue further, holding in mind that hypothesis testing is meaningless if the standard errors of the parameter estimates are wrong. Below we will go through the basic steps of the three most popular tests discussed in the textbook literature. | 1,411 | 7,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-40 | latest | en | 0.907172 |
https://www.coursehero.com/file/p4ci5hgd/You-can-easily-change-this-to-hours-by-multiplying-R-by-60-minutes-to-get-hours/ | 1,553,418,442,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203409.36/warc/CC-MAIN-20190324083551-20190324105551-00283.warc.gz | 743,951,295 | 100,505 | Little's Law Problems(Group 1).docx
# You can easily change this to hours by multiplying r
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was in terms of minutes. You can easily change this to hours by multiplying R by 60 minutes to get hours (2 x 60 = 120 per hour. 5/120 = .0417 hours) or to days, or to months to find the rate that best fits your needs, but REMEMBER: INVENTORY (I) REMAINS CONSTANT. Note, that the Little’s Law of T=I/R is nothing more than a unit conversion, converting numbers into time. It turned 5 units of inventory into 2.5 minutes of inventory. Suppose we have 100 units of item A, and 1000 units of item B. What item do we have more of? By the count dimension, item B has a higher inventory. Now suppose we use 4 units of item A per day (Ra= 4/ day ), and 200 units of item B per day (Rb=200/ day ). By the time dimension we have 25 days of inventory for item A (100/4 = T), and 5 days of inventory for item B (1000/200 = T). The inventory of item A, from the time perspective, is larger than that of item B.
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Now, suppose there are two lines. Suppose R is still 2 customers per minute and still on average there are 5 customers in the first line to pay for their order and get their non-exotic coffee. In addition, suppose that 4 out of the five customers place their order and then go on to wait in the exotic order (latte, cappuccino, etc.) line, i.e. 40% of the customers place exotic orders. What is the flow of time for a person who orders latte, cappuccino, etc.? Answer: What is the throughput time (T) of customers in the 1 st line? Each customer spends 2.5 minutes in the first line (determined by you from the first problem). What is the throughput rate (R) of customers in the exotic order line? Throughput rate of the second line is 40% of 2 customers per minute, or .8 customers per minute [R = 0.4(2) = 0.8]. What is the inventory (I) of the exotic order line? Inventory of the second line is 4. You can now determine on average how long an exotic order customer waits using the Little’s Law. Units Value Inventory Time 40% 60% 40% Exotic order customers All customers Ordinary order customers
R x T = I .8 x T = 4 T(second line)= 5 T(simple order) = 2.5 minutes. T(exotic order) = 2.5 + 5 = 7.5 minutes. What is the flow time of all customers? This is the average flow time of all customers that enter your coffee shop, both simple order and exotic order. Answer: Procedure 1 - 60% simple order: T = 2.5 40% exotic order: T=2.5+5= 7.5 An average customer is 60% a simple order person (2.5 min.) and 40% an exotic order person (7.5 min). T = 0.6(2.5) + 0.4(7.5) = 4.5 minutes Procedure 2 - Everyone goes through the first process and spends 5 minutes. 60% spend no additional time and leave. 40% spend 5 additional minutes. 0.6(0) ordinary order customers + 0.4(5) exotic order customers = 2 minutes 2.5 (every customer) + 2 (exotic orders) = 4.5 minutes. Procedure 3 – (Simplest solution) Throughput rate of the system is 2 per minute. There are 9 people in the system (5 at the register and 4 in the second line). R x T= I 2 x T = 9 T = 4.5 minutes Throughput rate of the coffee shop (system) is 2 per minute or 120 per hour or 720 per day (assuming 6 hours per day). But inventory in the system is ALWAYS 9.
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Throughput time of the coffee shop (system) is 4.5 minutes per customer (inventory).
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Jill Tulane University ‘16, Course Hero Intern | 1,185 | 4,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2019-13 | latest | en | 0.928765 |
https://testbook.com/question-answer/directionsin-the-question-below-are-given-t--5a60564cee09c50fa1459777 | 1,638,858,641,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363336.93/warc/CC-MAIN-20211207045002-20211207075002-00467.warc.gz | 598,985,652 | 31,706 | # Directions: In the question below are given three statements followed by four conclusions I, II, III and IV. You have to take the given statements to be true even if they seem to be at variance from commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.Statements:All tables are chairs.No chair is a bottle.Some bottles are blue.Conclusions:I. Some tables being blue is a possibility.II. All chairs are blue.III. At least some blue are tables.IV. All bottles are tables.
1. II and III follow
2. I follows
3. III and IV follow
4. II and IV follow
5. None follow
Option 2 : I follows
## Detailed Solution
From the given statements we draw the following least possible Venn diagram.
Conclusions:
I. Some tables being blue is a possibility → possibility is true.
II. All chairs are blue → it’s possible but not definite, hence false.
III. At least some blue are tables → it’s possible but not definite, hence false.
IV. All bottles are tables → it’s possible but not definite, hence false.
Thus, only conclusion I follow.
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100 Questions 100 Marks 60 Mins | 279 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2021-49 | latest | en | 0.876078 |
https://www.jiskha.com/display.cgi?id=1282515698 | 1,511,388,981,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806676.57/warc/CC-MAIN-20171122213945-20171122233945-00526.warc.gz | 833,351,808 | 3,892 | # phy
posted by .
vector B has a magnitude of 186. By is 0. What are the two possibilities for Bx?
• phy -
Bx= 186, or -186
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http://peeterjoot.com/archives/math2007/index.html | 1,568,655,017,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572879.28/warc/CC-MAIN-20190916155946-20190916181946-00059.warc.gz | 154,165,261 | 4,951 | This is an archival listing of my notes from 2007 and earlier.
October 22, 2007 Radial components of vector derivitives.
This and the other October docs below are all based on very ad-hoc notes I dumped into the Wikipedia Geometric algebra page , which had very little actual content in it. As I tried to learn GA from the much too advanced treatments I found on the web, I put my notes in the wiki as I puzzled things out myself. Eventually I bought some books and if I had done so earlier, I wouldn’t have needed to figure out a lot of this on my own. It was however time well spent as I learned things more thoroughly trying to figure it out from the basic properties. I wanted to add details to the wiki pages that didn’t really fit with a wiki page, or were hard to do with wiki markup compared to latex, so eventually I stopped trying to make my notes in wiki and just typed them up directly here. I never did go back and see if there was anything extra that I added to these that would be worth re-adding to the wiki pages. What I have here is all based on my wiki page additions (and I was the only contributor at the time), and excludes the minimal bits that were there previously.
October 22, 2007 Rotational dynamics. Angular velocity.
Angular velocity as radial component of velocity. Kepler’s law from Salas/Hille reworked with GA. Acceleration, and circular motion special case expressed in GA form, noting how close that ends up being to the scalar equivalent, but vector product and inversion encodes the direction too.
October 13, 2007 Maxwell’s equations expressed with Geometric Algebra.
Taking the hint that Maxwell’s equations could be expressed naturally using GA. I didn’t get it as far as the GAFP book since I didn’t see how to also formulate the spacetime gradient in the STA basis. This is probably a treatment that can be understood without too much GA prep.
October 13, 2007 My wikipedia Geometric Algebra notes.
Basic identities, and comparisions to non GA formulations. Wedge vs. Cross. Norm using GA square compared equivalent dot product method. Lagrange identity in cross and wedge product form. Determinant bivector and cross product expansions. Cross and wedge product formula for planes. Projection and rejection. Parallelogram area, and parallopiped volume. Vector angle. Vector inversion. Symetric and antisymetric product representations of dot and wedge products. Reversion. Complex numbers. Rotation in arbitrary plane. Cross product as wedge dual.
October 16, 2007 Cramer’s rule using wedge products.
An example showing how Cramer’s rule is just a special case of linear equation solution using the wedge product. Hestenes uses examples like this in his book. Without using GA, the grassman algebra book also has a worked example of this which is nice. I find the GA approach is easier to understand than the Grassman book approach. The grassman book avoids the dot product to use the metric free regressive product approach. I find the explicit use of the metric and associated dot product in GA more naturally when trying to bridge from traditional vector identities to a wedge product treatment.
October 13, 2007 Torque expressed with geometric algebra.
This contains scalar and bivector formulations of torque based on rate of change of work with respect to angle.
October 16, 2007 Derivatives of a unit vector.
This is a derivation of the unit (radially expressed) vector time derivative in terms of the GA wedge product.
May 10, 2007 The cross product in three and more dimensions
Revisiting my cross product treatment, but after discovering differential forms and the wedge product (Harley Flanders. “Differential forms and Applications to the Physical Sciences”). This was curtosy of Tor who was smart enough to search for it.
I had trouble with relating the wedge product on differential quantities to vectors. I could see this was related to my old attempt to generalize the vector dot product. I went back to my old notes and tried to lead from those ideas to the wedge product in a natural way.
This doc contains a few interesting bits.
A coordinate description of vector rejection (difference from projection) in terms of determinants is developed. This provides a way to calculate the normal component of a vector with respect to a spanning set.
This is used to calculate the area and volume of an N dimensional hyper-parallelogram, or hyper-parallopiped.
Also included is a calculation of a general normal to a set of vectors based on Null space calculation. Of particular interest is the description of the case when such a normal cannot necessarily be unambiguously defined.
The ideas here are then used to try to intuitively introduce the wedge product and a dot product on pairs of wedge products, without also introducing differentials. The wedge product and the normal and rejection concepts are all related by the determinant (in fact the determinant really ought to be viewed as generated by the wedge product).
I end with an poor attempt to introduce differential forms for geometric area and volume elements. When attempting to find a treatment of differnential forms that I could understand I ended up discovering Geometric Algebra. That subject has exactly what I was looking for as the “answer” for all this generalized vector algebra. Due to the time required to study that algebraic toolbox I have not yet gotten back to the differential forms that started me down that path. Soon I hope.
March 25, 2000 Various formulations of Maxwell’s equations.
Integral and differential forms of Maxwell’s equations in differential, and integral form, as well as explicit normal and tangential form (as in my Dad’s old 1960s Encyclopedia Britanica). Derivation of the wave equation in free space and in presence of matter using the normal triple cross product vector relation identity method. Followup with a four vector complex number treatment that puts the equations in a much more symmetrical form, that highlights the complex number factor that is naturally associated with the magnetic field.
I later followed up on this idea of complex representation in more detail in an email to Tor (at which point I got excited about it and wondered why this wasn’t in all the E&M books since it was so simple and natural and appropriate seeming). I should dig up and write that up since it would be a nice bridge to the GA ideas.
January 17, 2000. Triangulating distance to a star from orbital angle measurements. Scan a picture and add to this old note from Feynman's lectures so I can toss the old beat up paper sitting in the back of the book.
Circa ~1999 Old cross product generalization musings.
The three dimensionality (and even non-2 dimensionality) of the cross product always bugged me. I never saw any hint (or if I did I didn’t recognize it) about existing generalizations or degeneralizations of the cross product, even with four years of Engineering classes. So, one day I felt like “mathing” as my wife and kids call now it, and played with generalizing it.
Starting point was looking at torque in a plane (aka: Feynman Lectures, Vol I), and considering the work done by an incremental rotation, then doing the same for three dimensions. You end up with the cross product as an operator, and it takes the form of a completely antisymmetric matrix. I now know that this naturally expresses the antisymmetry of the wedge product (ie: cross product as the dual of a bivector).
In these notes I attempted to take this cross product operator form and extract some of the structure. In particular I factored this antisymmetric matrix into the product of diagonal and permutation product matrixes, and tried to use that as a way to define a generalized cross product. I was able to create a 4D cross product generalization in matrix form that had at least some of the 3D cross product properties (and properties of the 2D de-generalization of the cross product). However, the core problem here is that one doesn’t know how to define the normal that is intrinsic to the cross product, so the whole approach is kind of busted.
After starting with generalizing the cross product using incremental rotation as a starting point, I went back to the other starting point in the cross product definition, where the cross product is defined in terms of Euclidean normal vector properties (where one such orientation of the general normal between two vectors is picked).
There is a final attempt to generalize this further to 5D, but I don’t think I did a particularily good job at it. The most notable issue is that the 4D version wasn’t really well defined, and worse, I invented a 4D cross product generalization that I didn’t know any applications for.
Perhaps even worse than inventing a cross product generalization for was that I spent a bunch of time working out stuff that others had already figured out much more completely than I, and I didn’t go looking to see what others had done first to ensure I wasn’t (badly) reinventing the wheel. My friend Tor Aamodt (Prof at UBC), says this is one of the biggest reasons that grad students have thesis advisors. If I had stayed in school and done this in an academic context I would have had the good fortune to have somebody help point out that I was wasting time.
The final bits in this (long) set of notes was as far as I went with these ideas for quite a while and ended up dropping math as a hobby again for a number of years. | 1,977 | 9,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.979452 |
https://www.instructables.com/How-to-Draw---Basic-Linear-Perspective/ | 1,721,212,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00648.warc.gz | 739,043,094 | 36,162 | ## Introduction: How to Draw - Basic Linear Perspective
In this first installment of my ongoing series of "How to Draw" Instructables, I will show you how to create real-looking three dimensional shapes.
Linear Perspective is the most basic form of perspective in which all objects with faces parallel to the horizon, appear to converge in the distance at a single point on the horizon (the vanishing point).
To learn what on Earth this possibly means, grab yourself:
and dare follow me to the next step.
* Newsprint is fine to start. As you get better, you will want to invest in some quality drawing paper.
(Also note that some of the links on this page contain Amazon affiliate links. This does not change the price of any of the items for sale. However, I earn a small commission if you click on any of those links and buy anything. I reinvest this money into materials and tools for future projects. If you would like an alternate suggestion for a supplier of any of the parts, please let me know.)
## Step 1: Vanishing Horizons.
If you were to stand on a plane and look out into the distance, the imaginary line that demarcates between the Earth and the sky is considered the horizon. For argument's sake, the horizon is a straight line (even though in actual space it is slightly curved).
Now, if you were to stare straight ahead at the horizon, the point on the horizon directly in front of you would be considered the vanishing point. It's called the vanishing point since all objects seem to vanish towards it as they go back into the distance.
## Step 2: A Single Point. a Single Perspective.
One-point perspective is marked by the fact that all objects seem to converge towards one solitary point on the horizon. In order for all objects to converge at a single point, their closest face has to appear to be parallel to horizon.
In other words, if there was a cube between you and the horizon, the face of the cube closest to you would have two horizontal lines parallel to the horizon. In fact, everything viewed in this perspective must have horizontal lines parallel to the horizon.
If horizontal lines are no longer parallel, you have just gained a whole new perspective (but lets not worry about that for now).
Turn your paper sideways (landscape) and measure 9" up on each side and make a mark. Connect both marks with a line. You should have just successfully bisected your paper.
Next, you want to find the midpoint on the sheet of paper.
The easiest way to find the midpoint is to connect each opposite corner on the sheet of paper. Where the two lines converge is the center point (the center point of any parallelogram - rectangle, square - can be found this way).
Your center point should fall right in the center of the line you have just drawn. Enlarge this point and label it V.P. for vanishing point.
Erase the diagonal lines, but leave the horizontal line since that is now your horizon.
You should be now left with a horizon with a vanishing point centered on it.
The vanishing point must always be located on the horizon!
## Step 4: L7 Square
Measure 3" from the left edge and make a mark above and below the horizon. Repeat this step now with a measurement of 5".
Connect each set of points with a vertical line so that you have two vertical parallel lines.
Now measure 1" up from the horizon and make a mark on both lines and then measure 1" down from the horizon line and do the same.
Connect both new sets of dots to form a square.
## Step 5: The Thirrrrrd Diiiimension!
Right now you should have a two-dimensional box. To make this book look three-dimensional, it has to be appear to have depth.
And of course, anything with depth must appear to travel back towards the vanishing point.
To give your box depth trace the top and bottom right-edges back towards the vanishing point (where they should converge).
You now have a mighty long box. You're probably going to want a box that looks a little more reasonable.
On one the lines you just drew, pick a point that is about half-way between the original square and the vanishing point. Make a small mark and measure how far it is from the left edge of the paper.
Once you know the distance, move your ruler vertically up or down from this point and make another dot.
Connect these two dots and extend the vertical line to fully intersects diagonal lines that you have just drawn.
You should be left with something that resembles a three-dimensional box. Erase all lines until you are left with just a box sitting on the horizon (with a vanishing point).
## Step 6: In Good Company
Now would be a good time to draw four more rectangles on your sheet of paper using the method we used to make the first one in "Step 4".
One rectangle should be above the horizon line (to the right or left of the vanashing point). One rectangle should be below the horizon line (to the right or left of the vanashing point). The third rectangle should be above or below the horizon and positioned so that it is also above or below the vanishing point. The last rectangle should be on the horizon line, but on the opposite side of the vanishing point as the first box. Don't allow any of the rectangles to intersect.
## Step 7: A Perspective on Boxing
Place your ruler on the vanishing point and connect lines to all of the corners of the rectangles you just drew that don't require drawing a line over the face of the rectangle (see secondary image). If done correctly, you should have just drawn 10 diagonal lines.
Starting with the rectangle above both the horizon and vanishing point, draw a horizontal line connecting the diagonal lines. This should complete the box. Erase all unnecessary lines.
Next find the rectangle that is also above the horizon (but not above the vanishing point). Draw a horizontal line between the diagonals projecting from the bottom two corners and the vanishing point. At the point where the right diagonal line meets the newly created horizontal line, draw a vertical that connects to the diagonal that has been drawn from the top corner of the square. You should now have something that looks like a box. Erase all excess lines until you have a clean box.
Now, with the rectangle below the horizon, you are going to do similarly. Create a horizontal connecting the diagonals coming off the top corners and then drop a vertical from this intersection to the line drawn from the bottom corner.
The rectangle on the right should be completed in a similar manner to the one shown in "Step 5"
You should now have 4 new three-dimensional boxes.
## Step 8: Demystifying the Horizon
Draw a 6th rectangle that contains the vanishing point within it (see below).
From the image below we can now assert a couple of things.
1) All boxes at least have one visible face. This is the closest rectangular face to you, the viewer, and it is parallel to to the horizon line.
2) Any box sitting on the horizon line and is located to the left or right of the vanishing point will have 1 additional visible face that fades into the distance. This additional face is a side-ace. (visible faces in total: 2 - a front face and a side face)
3) Any box sitting above the horizon line and is located to the left or right of the vanishing point will have 2 additional visible faces that fade into the distance. These additional faces are a side-face and a bottom-face. (visible faces in total: 3 - a front face, a side face and a bottom face)
4) Any box sitting below the horizon line and is located to the left or right of the vanishing point will have 2 additional visible faces that fade into the distance. These additional faces are a side-face and a top-face. (visible faces in total: 3 - a front face, a side face and a top face)
5) Any box sitting in front of the vanishing point will only have one visible face. You will be unable to see the top, bottom, left or right-faces since all vanishing lines are hidden from view. (visible faces in total: 1 - a front face)
6) Any box sitting directly above the vanishing point (and horizon) will have 1 additional visible faces that fade into the distance. This additional face is a bottom-face. (visible faces in total: 2 - a front face and a bottom face)
7) Any box sitting directly below the vanishing point (and horizon) will have 1 additional visible faces that fade into the distance. This additional face is a top-face. (visible faces in total: 2 - a front face and a top face)
In strict one-point perspective, no solid object can be drawn and/or viewed outside the parameters just listed.
Using these guidelines, you can accurately (re)present any forward facing three-dimensional object that you may find in life.
## Step 9: Moving On
However, before I let you loose into the world to draw advanced shapes and the like, you may want to take a gander at "How to draw - Advanced Linear Perspective"
Did you find this useful, fun, or entertaining? | 1,921 | 8,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.959011 |
http://math.stackexchange.com/questions/353010/is-this-pointwise-convergence-sequence-also-uniform-convergence | 1,469,294,583,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823133.4/warc/CC-MAIN-20160723071023-00290-ip-10-185-27-174.ec2.internal.warc.gz | 162,073,013 | 16,411 | # Is this pointwise convergence sequence also uniform convergence?
$f_{n}$ and $f$ are continuous functions and $f_{n}\rightarrow f$ pointwise. Which of the following are correct?
1. $\int _{0}^{x}F_{n}\left( t\right) dt\rightarrow\int _{0}^{x}F\left( t\right) dt$
2. $\int _{0}^{x}f_{n}\left( t\right) dt\rightarrow\int _{0}^{x}f\left( t\right) dt$
3. $F_{n}'\left( x\right) \rightarrow f\left( x\right)$
where $F_{n}\left( x\right) =\int f_{n}\left( x\right) dx$ and $F\left( x\right) =\int f\left( x\right) dx$
I think the key to this problem is to deduce whether $f_{n}\rightarrow f$ uniformly or not. I got stucked in how to prove that.
Any counter-examples for 1 and 2？
-
You need to precise the $F_n$'s by setting $F_n(x):=\int_{x_0}^xf_n$. The formula $\int f_n$ is only defined up to a constant. – 1015 Apr 6 '13 at 15:25
Your $t(x)$ should be $f(x)$, and then $3$ is trivially true. – 1015 Apr 6 '13 at 15:28
$f_n\not{\rightarrow}f$. uniformly. – user45099 Apr 6 '13 at 15:28
@user57 The assumption is: $f_n\rightarrow f$ pointwise. And $3$ does not mention uniform convergence. – 1015 Apr 6 '13 at 15:29
No, sorry, I was addressing OP's last sentence. – user45099 Apr 6 '13 at 15:30 | 437 | 1,201 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-30 | latest | en | 0.761205 |
https://www.thestudentroom.co.uk/showthread.php?page=17&t=4308402 | 1,501,124,839,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426951.85/warc/CC-MAIN-20170727022134-20170727042134-00127.warc.gz | 859,468,769 | 46,062 | You are Here: Home >< Maths
# Maths C3 - Trigonometry... Help?? Watch
1. for part (e) where do I go from?...
2. (Original post by Philip-flop)
for part (e) where do I go from?...
Have you tried anything from here? Do some manipulation of the equation and see what happens.
Post our working if you get stuck.
3. (Original post by notnek)
Have you tried anything from here? Do some manipulation of the equation and see what happens.
Post our working if you get stuck.
I still couldn't manage to work it out so I moved on to the other questions
4. (Original post by Philip-flop)
I still couldn't manage to work it out so I moved on to the other questions
Multiply both sides by
Then you reaarange and eventually you have something in the form
Where a, b and c are constants,
Now you factorise or use quadratic fotmula
5. (Original post by asinghj)
Multiply both sides by
Then you reaarange and eventually you have something in the form
Where a, b and c are constants,
Now you factorise or use quadratic fotmula
OMG thank you so much!! Managed to work it out now
6. Where have I gone wrong for part (b)?...
Attachment 585308585310
Attached Images
7. (Original post by Philip-flop)
Where have I gone wrong for part (b)?...
Attachment 585308585310
Recheck how you got from 3rd line to 4th line
8. (Original post by asinghj)
Recheck how you got from 3rd line to 4th line
Thank you so much!! I was stressing about this question so much to the point where I couldn't even think straight! I got there in the end
9. Can someone help me? I'm stuck on part (h) and (i)...
For part (h) I've done....
For part (i) I've done...
<<<or should I re-arrange this line to give...???
<<< I know the double angle formula... ... but this can't be used here, right?
Don't know where to go from there for either of them though
Edit: I've managed to work out part (i) now by rearranging the second line from my workings above
10. (Original post by Philip-flop)
Can someone help me? I'm stuck on part (h) and (i)...
For h, multiply top and bottom by and use identities for double angles. Also it will be useful to rewrite as
11. (Original post by Philip-flop)
Can someone help me? I'm stuck on part (h) and (i)...
For part (h) I've done....
For part (i) I've done...
<<<or should I re-arrange this line to give...???
<<< I know the double angle formula... ... but this can't be used here, right?
Don't know where to go from there for either of them though
Edit: I've managed to work out part (i) now by rearranging the second line from my workings above
For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with
12. (Original post by RDKGames)
For h, multiply top and bottom by and use identities for double angles. Also it will be useful to rewrite as
Ok so this is what I've done for part (h) now...
But then where from here?
(Original post by notnek)
For h), the quickest way is to use the tan/sec identity on the denominator and then compare what you've got with
Ok, following your way for part (h) I've done...
But then I'm having trouble comparing it to the double angle formula...
I can only vaguely see why it becomes...
Is it because from the identity you would have to times by -1 and divide by 2 to get the identity into the same form as ???
13. (Original post by Philip-flop)
Ok so this is what I've done for part (h) now...
But then where from here?
Ok, following your way for part (h) I've done...
But then I'm having trouble comparing it to the double angle formula...
Multiply this by 2:
Then multiply it by -1, which is the same as multiplying the denominator by -1:
Does that help?
A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.
14. (Original post by Philip-flop)
Is it because from the identity you would have to times by -1 and divide by 2 to get the identity into the same form as ???
That's correct - you got it before my reply
15. (Original post by notnek)
Multiply this by 2:
Then multiply it by -1, which is the same as multiplying the denominator by -1:
Does that help?
A lot of this exercise is about comparing expressions with the double angle identities. This question is one of the trickier ones.
Thank you so much! I had a feeling I might have to compare it to that double-angle formula, but it took me a while to realise it at first!
16. (Original post by Philip-flop)
Ok so this is what I've done for part (h) now...
But then where from here?
If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.
17. (Original post by RDKGames)
If you turn tan into sine over cosine then you are left with sin(x)cos(x). As I mentioned above, manipulating this slightly will enable you to use double angle formulae.
Omg yes!! Thank you!! I've managed to get there now I seized up at first, but then re-arranged tan like you said and it was pretty simple from there. Really appreciate the help
18. Ok, so I spent way too much time trying to solve this question last night without any luck....
I've tried comparing to the double angle formula for
I can see that needs to be divided by 4 in order to get
so this means...
Divide by 4 gives...
.... right??
And then I have to sub into the equation I've just found above??
19. (Original post by Philip-flop)
Ok, so I spent way too much time trying to solve this question last night without any luck....
I've tried comparing to the double angle formula for
I can see that needs to be divided by 4 in order to get
so this means...
Divide by 4 gives...
.... right??
And then I have to sub into the equation I've just found above??
You are not dividing tan by 4, you are dividing the angle. So
20. (Original post by RDKGames)
You are not dividing tan by 4, you are dividing the angle. So
Thanks for making realise that! I'm still a little stuck on this question though
If how do I find ?
Sorry if I'm sounding pretty silly, but I've never come across a question like this
Updated: April 23, 2017
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Top tips from students who have already aced their exams | 1,653 | 6,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-30 | latest | en | 0.946956 |
https://www.dataunitconverter.com/kilobyte-to-mebibit/24 | 1,701,556,762,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00638.warc.gz | 819,714,982 | 15,037 | # Kilobytes to Mebibits - 24 kB to Mibit Conversion
Input Kilobyte (kB) - and press Enter.
You are converting Kilobyte, a decimal unit, to Mebibit, a binary unit.
verified_user RESULT =
24 kB = 0.18310546875 Mibit
( Equal to 1.8310546875E-1 Mibit )
content_copy
Calculated as → 24 x (8x1000) ÷ 10242... - View details
## Kilobyte (kB) to Mebibit (Mibit) Conversion Formula and Steps
kB to Mibit Calculator Tool allows you to easily convert from Kilobyte (kB) to Mebibit (Mibit). This converter uses the below formula and steps to perform the conversion.
The formula of converting the Kilobyte (kB) to Mebibit (Mibit) is represented as follows :
diamond Mibit = kB x (8x1000) ÷ 10242
Source Data Unit Target Data Unit
Kilobyte
Equal to 1000 bytes
(Decimal Unit)
Mebibit
Equal to 1024^2 bits
(Binary Unit)
Now let us apply the above formula and see how to manually convert Kilobyte (kB) to Mebibit (Mibit). We can further simplify the formula to ease the calculation.
FORMULA
Mebibits = Kilobytes x (8x1000) ÷ 10242
STEP 1
Mebibits = Kilobytes x (8x1000) ÷ (1024x1024)
STEP 2
Mebibits = Kilobytes x 8000 ÷ 1048576
STEP 3
Mebibits = Kilobytes x 0.00762939453125
If we apply the above Formula and steps, conversion from 24 Kilobyte (kB) to Mebibit (Mibit) will be processed as below.
1. = 24 x (8x1000) ÷ 10242
2. = 24 x (8x1000) ÷ (1024x1024)
3. = 24 x 8000 ÷ 1048576
4. = 24 x 0.00762939453125
5. = 0.18310546875
6. i.e. 24 kB is equal to 0.18310546875 Mibit.
Note : Result rounded off to 40 decimal positions.
You can use above formula and steps to convert Kilobytes to Mebibits using any of the programming language such as Java, Python or Powershell.
### Unit Definitions
#### Kilobyte
A Kilobyte (kB) is a decimal unit of digital information that is equal to 1000 bytes (or 8,000 bits) and commonly used to express the size of a file or the amount of memory used by a program. It is also used to express data transfer speeds and in the context of data storage and memory, the binary-based unit of kibibyte (KiB) is used instead.
arrow_downward
#### Mebibit
A Mebibit (Mib or Mibit) is a binary unit of digital information that is equal to 1,048,576 bits and is defined by the International Electro technical Commission(IEC). The prefix 'mebi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'megabit' (Mb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Kilobyte (kB) to Mebibit (Mibit)
Apply the formula as shown below to convert from 24 Kilobyte (kB) to Mebibit (Mibit).
A B C
1 Kilobyte (kB) Mebibit (Mibit)
2 24 =A2 * 0.00762939453125
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Kilobyte (kB) to Mebibit (Mibit) Conversion
You can use below code to convert any value in Kilobyte (kB) to Kilobyte (kB) in Python.
kilobytes = int(input("Enter Kilobytes: "))
mebibits = kilobytes * (8*1000) / (1024*1024)
print("{} Kilobytes = {} Mebibits".format(kilobytes,mebibits))
The first line of code will prompt the user to enter the Kilobyte (kB) as an input. The value of Mebibit (Mibit) is calculated on the next line, and the code in third line will display the result.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,055 | 3,476 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2023-50 | latest | en | 0.751252 |
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# Probability - Confidence Interval for population proportion
Question :
From a population of 1600 documents a random sample of 100 documents is drawn and found 10 defects in the sample. Find the representative number of defects in the total population of 1600 documents at 80% confidence level, at 90% confidence level and at 95% confidence level.
Please see the question file attached.
#### Solution Preview
Therefore, we can be 95% confident that the number of defects in the total population of 1600 documents would be between 69 ...
#### Solution Summary
S.E of p = 0.029056
95% confidence limits of defects .....
Representative number of defects in the total population = (68.8 , 251 )
90% confidence limits ............
80% confidence limits .............
95% confident that the number of defects ..........
\$2.19 | 188 | 854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2016-44 | longest | en | 0.870544 |
https://www.daylight.com/meetings/mug04/Delany/user_sim.html | 1,642,327,098,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299852.23/warc/CC-MAIN-20220116093137-20220116123137-00704.warc.gz | 812,809,055 | 2,189 | Improvements to Daylight Clustering
DAYLIGHT Chemical Information Systems, Inc. Mission Viejo, CA USA
User-defined Similarity Measures
Daylight currently supports three different similarity measures: Tanimoto, Tversky, and Euclidean distance. There are numerous other measures described and studied in the literature (Holliday).
At EMUG01 John spoke about a general facility for computing similarity values, which has been implemented in the clustering package. Nearneighbors, spherex, and kmodes all accept the user-specified similarity measure. Also, Daycart has a new function, "user_similarity()", which accepts a user-provided similarity measure and performs the generic similarity search.
OBJECT B 0 1 d b b+ d a c a + c = A a + d b + c = B n
a is the count of bits on in object A but not in object B.
b is the count of bits on in object B but not in object A.
c is the count of the bits on in both object A and object B.
d is the count of the bits off in both object A and object B.
Currently, there are a number of symbolic named expressions which are built into the expression evaluation:
Symbolic name Expression
TANIMOTO c/(a+b+c)
EUCLID sqrt((c+d)/(a+b+c+d))
DICE (2.0*c)/(a+c+b+c)
COSINE c/sqrt((a+c)*(b+c))
KULCZYNSKI 0.5*((c/(a+c))+(c/(b+c)))
JACCARD c/(a+b+c)
RUSSELL/RAO c/(a+b+c+d)
MATCHING (c+d)/(a+b+c+d)
HAMMAN ((c+d)-(a+b))/(a+b+c+d)
ROGERS/TANIMOTO (c+d)/((a+b)+(a+b+c+d))
FORBES (c*(a+b+c+d))/((a+c)*(b+c))
SIMPSON c/min((a+c),(b+c))
PEARSON (c*d-a*b)/sqrt((a+c)*(b+c)*(a+d)*(b+d))
YULE (c*d-a*b)/(c*d+a*b)
MANHATTAN (a+b)/(a+b+c+d)
For programs, there are two new options: -expr and -comparison. If not specified, the default for -expr is TANIMOTO. The option -comparison should be either "DISTANCE" or "SIMILARITY". If unspecified, it will choose based on two limit cases: a = b = 1, c = d = 0, vs a = b = 0, c = d = 1.
```\$ nearneighbors -expr FORBES wdi.fp
nearneighbors: expression (FORBES) identified as a SIMILARITY | 589 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.783297 |
https://edurev.in/course/quiz/attempt/-1_Test-Nyquist-Stability-Criterion/362f287c-785b-4771-b4db-aefb75dc5b0d | 1,624,052,580,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487641593.43/warc/CC-MAIN-20210618200114-20210618230114-00043.warc.gz | 217,343,065 | 41,858 | Courses
# Test: Nyquist Stability Criterion
## 10 Questions MCQ Test | Test: Nyquist Stability Criterion
Description
This mock test of Test: Nyquist Stability Criterion for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 10 Multiple Choice Questions for Electrical Engineering (EE) Test: Nyquist Stability Criterion (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Nyquist Stability Criterion quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Test: Nyquist Stability Criterion exercise for a better result in the exam. You can find other Test: Nyquist Stability Criterion extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1
### Which principle specifies the relationship between enclosure of poles & zeros by s-plane contour and the encirclement of origin by q(s) plane contour?
Solution:
Explanation: Argument principle specifies the relationship between enclosure of poles & zeros by s-plane contour and the encirclement of origin by q(s) plane contour.
QUESTION: 2
### If a Nyquist plot of G (jω) H (jω) for a closed loop system passes through (-2, j0) point in GH plane, what would be the value of gain margin of the system in dB?
Solution:
Explanation: Gain Margin is calculated by taking inverse of the gain where the Nyquist plot cuts the real axis.
QUESTION: 3
### For Nyquist contour, the size of radius is _______
Solution:
Explanation: For Nyquist contour, the size of radius is ∞.
QUESTION: 4
Consider a feedback system with gain margin of about 30. At what point does Nyquist plot crosses negative real axis?
Solution:
Explanation: Gain Margin is always inverse of the point which cuts the Nyquist on the real axis.
QUESTION: 5
According to Nyquist stability criterion, where should be the position of all zeros of q(s) corresponding to s-plane?
Solution:
Explanation: According to Nyquist stability criterion zeroes must lie on the left half on the s plane.
QUESTION: 6
If the system is represented by G(s) H(s) = k (s+7) / s (s +3) (s + 2), what would be its magnitude at ω = ∞?
Solution:
Explanation: On calculating the magnitude of the system and putting the value of frequency one gets the magnitude as 0.
QUESTION: 7
Consider the system represented by the equation given below. What would be the total phase value at ω = 0?
200/[s3 (s + 3) (s + 6) (s + 10)].
Solution:
Explanation: The phase can be calculated by the basic formula for calculating phase angle
QUESTION: 8
Due to an addition of pole at origin, the polar plot gets shifted by ___ at ω = 0 ?
Solution:
Explanation: Addition of pole causes instability to the system.
QUESTION: 9
In polar plots, if a pole is added at the origin, what would be the value of the magnitude at Ω = 0?
Solution:
Explanation: Addition of pole causes instability to the system.
QUESTION: 10
In polar plots, what does each and every point represent w.r.t magnitude and angle?
Solution: | 705 | 3,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2021-25 | latest | en | 0.862785 |
https://sites.google.com/a/jeffcoschools.us/6th-grade-top-dogs/math-questions-1 | 1,545,045,944,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828501.85/warc/CC-MAIN-20181217091227-20181217113227-00151.warc.gz | 720,931,048 | 8,141 | ### Math Videos - Unit 5
Unit 5, Lesson 4 * I noticed that I made (at least) one mistake in this video. If you notice a mistake, write down what my mistake was and at exactly what time in the video it happened and you will get a ticket! **Please be sure to complete the questions at the end of the video in your journal. You will be sharing your equation and diagram in class tomorrow. Unit 5, Lesson 9 In this lesson, we are reviewing and practicing simplifying expressions, analyzing expressions, using properties, and recognizing equivalent expressions. So, you have two options for this lesson: 1) Watch the Lesson 9 Show Me using this link: Unit 5, Lesson 9 video 2) If you feel like you have mastered the concepts listed above, you can do Destination Math for Lesson 8 instead. (if you don't remember how to get there, follow the instructions on the "Math" page. ***You must choose one of these options and write examples in your journal for the choice you made. (I will be checking journals taking a grade on your notes.) Unit 5, Lesson 17, Multiplication and Division equations * Will need HW p. 131 to complete
Subpages (1):
Comments | 335 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-51 | latest | en | 0.523497 |
http://kgargar.tk/dimension-math-formula-sheet.html | 1,560,972,810,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999040.56/warc/CC-MAIN-20190619184037-20190619210037-00459.warc.gz | 89,636,953 | 4,217 | # Dimension math formula sheet
Sheet formula
## Dimension math formula sheet
” Learning objectiveswere explicitly communicated to the students and, even though difficult for. Dimensional sheet Analysis It is a useful technique. Basic Conversion Cheat Sheet • Three basic units of measurement length mass ( weight) volume o The basic unit of length is: METER o The basic unit of volume is: LITER o dimension The basic unit of mass dimension ( weight) is: GRAM • The following are some of the sheet prefixes for the metric system. We would like to show you a description here but the site won’ t allow us. The main concern dimension of every student about this subject is the Geometry Formula.
Weight per cubic inch ( Density) sheet of steel dimension is. Dimension: Number of elements in a basis. You can choose formulas from different pages. Geometry Formulas. Find Missing Dimensions of Rectangles 35. Steel Weight Formulas Based on theoretical nominal weights and considered approximate; used for estimating only. Free math lessons math homework help from basic math to algebra, geometry beyond. Choose from sheet 500 different sets of study sheet math flashcards on Quizlet.
Welcome to the Math Salamanders' Geometry Formula Sheet area. Move a formula to another location. They are used to calculate the length, perimeter. The only danger is that you may end up thinking that chemistry is simply a math problem - which it dimension definitely is not. They are based on. Here you will find a range of sheet different Geometry formulas for 2d and 3d shapes. Formula Reference Sheet ; Instruction & Assessment. T- TESS Observation Evidence Sheet 7th Grade Math ( Pi and Circumference of Circles) Domain: Instruction. It contains a list of basic math formulas commonly used dimension when doing basic math computation.
Assess sheet Close Another Dimension Have students explain in writing how to find the missing dimension of a rectangle that has a length of 7 inches an area of 42 square inches. Here you will find our support page about different Geometry formulas including triangles, quadrilaterals , polygons, circles as well as 3d shapes. Eventually, formulas are used to provide mathematical solution for real world problems. Unlike copying a formula the cell dimension references in the formula dimension don’ t change, when dimension you move a formula to dimension another location in the same , another worksheet regardless of what type of cell reference you used. Math Formulas: Planes in three dimensions. For Exercises 1- 8, see Answer Appendix p. Tips about the usage of the formulas can math be found below. can use the perimeter formula of a rectangle to find sheet the perimeter of a square.
Math formula shows how things work out with the help of some equations like the equation for force or acceleration. Recursive Formulas Used with permission from sheet Dave' s Math Tables Formula Derivations - ( High School + ) Derivations of area more for 2 , volume , perimeter 3 dimensional figures. A comprehensive list of the most commonly used basic math formulas. Students , parents, teachers everyone can find solutions to their math problems instantly. After you have selected all the formulas which you would like to include in cheat sheet, click the sheet " Generate PDF" button. If you are looking for a formula to solve your basic math problems, your formula is likely here. Dimension math formula sheet. Learn study sheet math with free interactive flashcards. length of one side Perimeter of a rectangle: l + w + l + w l: length w: width Perimeter of a triangle: a + b + c a b, c: lengths of dimension the 3. For real eigenvalues, use the. Formula Reference Sheet Formulas for Area ( A) and Circumference ( C) Triangle A 1 2 bh 1 2 sheet base height Trapezoid A 1 2 ( b 1 + 2) h 1 2 sum of bases height Parallelogram A bh base height. Point- Slope Formula: Coordinate Geometry Formulas: Let ( x 1 y 1) , ( x 2 y 2) be two points in the plane. Dimension math formula sheet. Dimension Evidence math Rating Achieving Expectations The teacher models performance expectations by sharing her “ story of a sheet square.
Select the cell that has dimension the formula you want to move. Math 54 Cheat Sheet Vector spaces Subspace: If u v are in W, then u + v are in W . If you are looking for a formula to solve your basic math problems, your formula is likely here - Homepage; Basic Mathematics blog;.
## Math formula
Draw a prism that has a trapezoid as its base. Use your formula to fi nd the volume of the prism. Use what you learned about the volumes of prisms to complete Exercises 4 – 6 on page 302. Work with a partner. A ream of paper has 500 sheets.
dimension math formula sheet
Does a single sheet of paper have a volume? This is the standard formula sheet given to you in any Additional Mathematics ( A Maths) exams. | 1,041 | 4,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-26 | latest | en | 0.849871 |
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Re: [xml-dev] Infinity
• From: "Norman Gray" <norman@astro.gla.ac.uk>
• To: "Peter Hunsberger" <peter.hunsberger@gmail.com>
• Date: Sun, 04 Mar 2018 21:27:13 +0000
Peter, hello.
On 3 Mar 2018, at 22:05, Peter Hunsberger wrote:
On Sat, Mar 3, 2018 at 7:33 AM Norman Gray <norman@astro.gla.ac.uk>
It will be, but since there are as many elements in that set as there
are positive integers (they can be put into a one-to-one
correspondence), it is no bigger or smaller an infinity than the number
of integers. In contrast, the number of real numbers is a 'larger
infinity' than the number of integers. If you wish to further explore
this rabbit hole, see <https://en.wikipedia.org/wiki/Aleph_number> and
work outwards...
and reals are both of cardinality Aleph naught. The easiest way to
conceptualize this equivalence is to think of them both as being mappable
to a set of points on a line.
I'm fairly sure the set of real numbers has a larger cardinality than the integers (I say this with some diffidence, though, since I've never covered this formally, so I'm basing this on a mixture of incidental reading and Wikipedia).
(By the way, I take it that we are both taking 'real number' to mean the mathematical reals rather than floating point numbers -- Liam touches on this).
The Wikipedia page I quoted [1] mentions that \aleph_1 is the cardinality of the ordinal numbers, and explicitly states that 'The cardinality of the set of real numbers [...] is 2^{\aleph_0}' (and goes on to imply that this is indeed larger than \aleph_0 given certain hypotheses).
Also, Cantor's diagonal argument [2] explicitly shows (if I recall and understand it correctly) that there is no one-to-one correspondence between the integers and the reals. That is, although the integers can indeed be mapped to a set of a points on a real line, they can be mapped only to a _subset_ of those points, and in any such mapping there will be points on the real line which do not correspond to an integer.
There's a one-to-one correspondence from integers to rationals, and to the set of algebraic numbers (the set of solutions to polynomials), so both of those sets are of cardinality \aleph_0. The latter set of course excludes the transcendental numbers, but I don't _think_ the main point depends directly on the existence or not of transcendental numbers.
There are a number of subtleties here which I would be reluctant to speak confidently about, but I think the main statement ('more reals than integers') stands.
Best wishes,
Norman
[1] https://en.wikipedia.org/wiki/Aleph_number
[2] https://en.wikipedia.org/wiki/Cantor's_diagonal_argument
--
Norman Gray : https://nxg.me.uk
SUPA School of Physics and Astronomy, University of Glasgow, UK
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Copyright 1993-2007 XML.org. This site is hosted by OASIS | 810 | 3,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2019-13 | latest | en | 0.899392 |
https://historicsweetsballroom.com/does-solid-sodium-chloride-conduct-electricity/ | 1,653,510,101,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00134.warc.gz | 356,281,806 | 5,729 | The physical properties of sodium chloride
This page defines the relationship in between the setup of the ions in a typical ionic solid like sodium chloride and its physical properties - melting point, boil point, brittleness, solubility and also electrical behavior. It also explains why cesium chloride has actually a various structure from sodium chloride also though sodium and also cesium room both in team 1 that the routine Table.
You are watching: Does solid sodium chloride conduct electricity
## The structure of a common ionic hard - sodium chloride
Sodium chloride is taken together a common ionic compound. Compounds favor this covers a giant (endlessly repeating) lattice of ions. So salt chloride (and any other ionic compound) is explained as having a giant ionic structure.
You have to be clean that gigantic in this context does not simply mean very large. It means that girlfriend can"t state precisely how many ions over there are. There can be billions of sodium ions and also chloride ions packed together, or trillions, or whatever - it merely depends how big the crystal is. That is various from, say, a water molecule which always contains specifically 2 hydrogen atoms and one oxygen atom - never an ext and never ever less. A little representative little bit of a salt chloride lattice looks choose this:
If friend look in ~ the chart carefully, you will see that the sodium ions and chloride ions alternating with each other in every of the three dimensions. This diagram is easy enough to draw with a computer, but extremely challenging to attract convincingly by hand. Us normally attract an "exploded" variation which looks favor this:
api/deki/files/15274/draw1.GIF?revision=1" />
Now attract an identical square behind this one and also offset a bit. You can have to practice a bit to get the placement of the two squares right. If you obtain it wrong, the ions acquire all tangled up with each other in your final diagram.
Turn this right into a perfect cube by joining the squares together:
Now the tricky bit! Subdivide this big cube into 8 tiny cubes by joining the mid suggest of each edge come the mid point of the edge opposite it. To complete the procedure you will likewise have to join the mid allude of each face (easily found once you"ve join the edges) to the mid point of opposing face.
Now all you have to do is placed the ion in. Use different colors or different sizes because that the two various ions, and don"t forget a key. It does not issue whether you end up v a sodium ion or a chloride ion in the facility of the cube - all the matters is the they alternate in all 3 dimensions.
You should be able to draw a perfectly sufficient free-hand map out of this in under 2 minutes - much less than one minute if you"re not as well fussy!
## The different structure the cesium chloride
We"ll look first at the arrangement of the ions and then talk about why the structures of salt chloride and cesium chloride are different afterwards. Imagine a great of chloride ions as shown below. The individual chloride ions aren"t touching each other. That"s really crucial - if they to be touching, there would be repulsion.
Now let"s ar a an in similar way arranged great of cesium ion on height of these.
If you now think about a cesium ion sandwiched between the two layers of chloride ions, that is touching four chloride ions in the bottom layer, and another 4 in the optimal one. Each cesium ion is touched by eight chloride ions. We say the it is 8-coordinated. If we added another layer of cesium ions, friend could an in similar way work out that every chloride ion was emotional eight cesium ions. The chloride ion are also 8-coordinated. Overall, then, cesium chloride is 8:8-coordinated.
See more: How Many Cups In 25 Ounces To Cups, Ounces (Oz) To Cups Conversion
The last diagram in this succession takes a slightly tilted view of the framework so the you can see how the layers build up. This diagrams space quite challenging to draw without it looking together if ions of the very same charge are emotional each other. Lock aren"t!
By reversing the colour (green chloride ion in the center, and also orange cesium ions bordering it), friend would have actually an specifically equivalent diagram for the setup of cesium ions about each chloride ion. | 919 | 4,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | longest | en | 0.921978 |
https://klikyballs.com/electromagnetism/what-is-the-meaning-of-electromagnetic-theory.html | 1,618,645,238,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00539.warc.gz | 439,673,691 | 10,656 | # What is the meaning of electromagnetic theory?
Contents
: a theory in physics: light consists of electromagnetic oscillations perpendicular to the direction of travel of the wave motion.
## What is the electromagnetic theory?
Electromagnetic Theory covers the basic principles of electromagnetism: experimental basis, electrostatics, magnetic fields of steady currents, motional e.m.f. and electromagnetic induction, Maxwell’s equations, propagation and radiation of electromagnetic waves, electric and magnetic properties of matter, and …
## Why electromagnetic theory is needed?
EM theory is an essential basis for understanding the devices, methods, and systems used for electrical energy. Both electric and magnetic fields are defined in terms of the forces they produce. … All engineering study related to electrical energy and power relies on key concepts from EM theory.
## What are the main points of electromagnetic wave theory?
The main points of this theory are:
1)The energy is emitted from any source continuously in the form of radiations and is called the radiant energy. 2)The radiations consist of electric and magnetic fields oscillating perpendicular to each other and both perpendicular to the direction of propagation of the radiation.
## What are the applications of electromagnetic field?
Nowadays electromagnetic fields play a key role in advanced medical equipments such as hyperthermia treatments for cancer, implants and magnetic resonance imaging (MRI). RF range frequencies are mostly used in medical applications.
## Why is it called electromagnetic?
Scientists call them all electromagnetic radiation. The waves of energy are called electromagnetic (EM) because they have oscillating electric and magnetic fields. … If it has low frequency, it has less energy and could be a TV or radio wave. All EM energy waves travel at the speed of light.
## How does electromagnetism affect our daily life?
Exposure to electric, magnetic and electromagnetic fields (EMF), if they are strong enough, can lead to short term health effects. Exposure to low frequency fields that are strong enough can lead to dizziness, seeing light flashes and feeling tingling or pain through stimulation of nerves.
## Who proposed electromagnetic theory?
George Green was the first person to create a mathematical theory of electricity and magnetism and his theory formed the foundation for the work of other scientists such as James Clerk Maxwell, William Thomson, and others.
## Is an electromagnetic wave?
Definition: Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields. … They are also perpendicular to the direction of the EM wave.
## How do we use electromagnetic energy?
Up to the end of the microwave spectrum, most all modern conveniences that use electromagnetic energy in one way or another are in the lower frequency region, including millimeter waves, cell phones, WiFi, microwave ovens, space and terrestrial communications, radar for airports and military uses, AM and FM radio, …
IT IS INTERESTING: Your question: What do you mean by electromagnet with example?
## What do all electromagnetic waves have in common?
All electromagnetic waves have two wavefronts, which are an oscillating electric field and an oscillating magnetic field.
## How safe are electromagnetic waves?
There is no doubt that short-term exposure to very high levels of electromagnetic fields can be harmful to health. … Despite extensive research, to date there is no evidence to conclude that exposure to low level electromagnetic fields is harmful to human health.
## What are the two applications of electromagnet?
Applications of electromagnets
• Motors and generators.
• Transformers.
• Pickups.
• Relays.
• Electric bells and buzzers. | 735 | 3,928 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-17 | latest | en | 0.907301 |
https://projecteuler.chat/viewtopic.php?f=3&t=3814&p=41449 | 1,590,522,023,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00544.warc.gz | 525,040,368 | 7,431 | ## Flipping objects in a bounded region
Shape and space, angle and circle properties, ...
Bubbler
Posts: 28
Joined: Sat Feb 23, 2013 6:12 am
### Flipping objects in a bounded region
(Sorry, this problem is kind of computational geometry, not math-oriented...)
A convex polygonal region R with k vertices is given on a 2D plane, and there are n convex polygons Pi, with k vertices each, which are pairwise disjoint and completely inside R. These polygons do not sum to R; in other words, there are some gaps inside R that are not part of any of Pi.
Now you want to flip each Pi - place the mirror image Pi' of Pi so that the position of its center of mass remains constant. Each Pi' must still lie inside R and be pairwise disjoint. Rotation of Pi' around the center of mass is freely allowed.
Is it possible to computationally determine whether each polygon can be successfully flipped or not?
jaap
Posts: 550
Joined: Tue Mar 25, 2008 3:57 pm
Contact:
### Re: Flipping objects in a bounded region
I don't doubt that it can be done, but it will be very difficult. You will have to reduce it to a combinatorial problem.
That means that the n degrees of freedom, which are the mirror angles for the n pieces, each have to be restricted to a finite set of values in such a way that a solution is present whenever the original problem has a solution.
However even if this reduction can be done, it may be the case that this class of combinatorial problem is difficult. It would not surprise me if it is a PSPACE-complete problem, even if all the pieces are only allowed to be mirrored vertically or horizontally.
I have written a page about PSPACE completeness for a different kind of puzzle here: http://www.jaapsch.net/puzzles/pspace.htm
It feels to me that there will probably be a way to define wires, AND and OR widgets, etcetera using only horizontal/vertical mirroring of pieces. If that is the case, then it means that if you were to actually find a program that can solve it quickly (in polynomial time) then you will win a million dollars. | 477 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-24 | latest | en | 0.931677 |
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Dynamics and Control 3, 107-125 (1993) 9 1993 Kluwer Academic Publishers, Boston. Manufactured in The Netherlands.
### Multirigid-Body Systems
Department of Mathematics, Bilkent University, 06533 Bilkent, Ankara, Turkey Received April 17, 1991; revised December 3, 1991 and February 4, 1992 Editor: W, Stadler
Abstract. The mathematical model of a rigid body in three dimensional motion developed in [1] is used to for- mulate the equations of motion for the systems o f rigid bodies connected to form a special type of open kinematic chain. In this interconnection pattern of rigid bodies, each rigid body is considered as a 3-port component, and for the sake of generality, initially no constraints are imposed on the joints used to interconnect the rigid bodies. The system is considered as an (m + 1)-port component and the corresponding terminal equations are obtained in closed form. As an appliction of these equations, a three-link plane manipulator is considered.
1. Introduction
In an earlier article [1], a mathematical model of a rigid body in three-dimensional motion as a multi-terminal component is derived. In this derivation, an approach widely considered and applied in the analysis of electrical networks is used. In this article the same approach is applied to the formulation of the equations of motion for a system of interconnected rigid bodies. To be more specific, here the application is restricted to a system of special type of interconnected rigid bodies possessing an open kinematic chain. Such a system actually appears in various kinds of manipulators. The system itself is considered as an (m + 1)- port mechanical component and modeled in such a way that the corresponding terminal equations are obtained in closed form.
The simplest form of the terminal equations of a rigid body corresponding to a star-like tree (a terminal graph) with k branches is given by equation (42) of [1]. In what follows, these equations are reproduced for k = 2. With this restriction we actually assume that each rigid body (6lj) in the system is to be connected to, at most, two adjacent rigid bodies ((Rj_0 and ((Rj+I) at its designated ports (O
### Aj) and (0 Bj).
Therefore, each rigid body ((Rj) (see Figure la) will be considered as a
### three-port
component which is represented by the schematic diagram shown in Figure lb, and its terminal equations corresponding to the terminal graph in Figure lc will be in the following form:
Y~ 012,12
### L xtj J
012,1 u~ (1)
(2)
where all the submatrices
### i mJ, l
Kaj = RGaj I , Kbj = RGb j I , UGj = 0
### [ ~ 1 7 6
Waj = e G j ~ + QG 1,Pay = , Q~j =
### o %/G s
(2)
contain 6 rows, x . , y . , UG. are column matrices, while the 6 x 6 matrices are patitione PJ Pi y into 3 • 3 submatrices. If the rigid body is assumed to be ideal, in equation (1) the 6 x 6 operator matrix WG,(d/dt) and the column matrix u~ are replaced by appropriate zero matrices yielding the terminal equations of the ideal rigid body ((R0j) as
O12,12 ]
- -
### t
Y~ ] -Kaj --Kb~ [ 06,6
(3)
This observation permits us to represent the three-port rigid body as the interconnection of an ideal rigid body ((R0j), and an active one-port mechanical component with the ter- minal equations of the form
= WGj ' (4)
Y6 i xa s + UG i
and as indicated in the schematic diagram in Figure ld, where the active one-port compo- nent is further decomposed into the parallel interconnection of two one-ports, (Woj) and (ua), the latter being a through driver (source) corresponding to the force and torque due to dae gravity. Stated differently, the kinetic property of rigid body ((Rj) can be represented by the addition of an active one-port component, having the terminal equations in (4), to the (OGj) port of an ideal rigid body ((Roy). One may also be interested in knowing whether the active component can be moved to any one of the other ports (OBj) or (OAj) of the ideal rigid body ((Roj), so that the new interconnection configuration still represents the same rigid body ((Ry). This is indeed possible, however, with the distinction that the coef- ficientmatrices WG: and u@ in equation (4) are now replaced by different ones as indicated in the schematic diagrams of Figures le and lf. This property follows from the fact that in equation (1) both of the submatrices Ka. and Kb. are nonsingular and the terminal equa-
l/ 3 . . .
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NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 109
### - 4
012,12 (K~I) r --gbbj 1 WBj
O12,1 +
(5) and
+
### 1
O12,1 _ K ~ I K~ -1 ~j UAj (6) where
### i I 01 [, 01 [, 01
KAj = KaJ Kaj = Rrbj I R6a j I Rj I " (7)
The expressions of the other submatrices in equations (5) and (6) are given by equations (49) and (50), in the appendix, respectively.
The forms of the terminal equations of a three-port rigid body given in equations (1), (5), and (6) are all equivalent. In a systematic formulation of the equations of a system of rigid bodies, it may be necessary to select a particular and perhaps dafferent form of the terminal equations for each rigid body in the system. As indicated in [1], the ideal rigid body (fft0j) with terminal equations given in (3) corresponds to a three-port ideal transformer in network theory. Therefore, a topological necessary condition which is well established in network theory [2-7] for the unique solvability of the networks containing ideal transformers can also be applied for the systems of rigid bodies. Considering the three possible forms of the terminal equations of a rigid body in equations (1), (5), and (6), we then state that in the system graph drawn for a system of rigid bodies there must exist a formulation tree T such that only two of the edges of the terminal graph for each rigid body (shown in Figure lc) should be contained in T while the third edge be contained in the co-tree T' of T.
There is another topological necessary condition on the classifications of the edges cor- responding to the drivers is [2, 8]: For the existence of a unique solution of the system equations, all the edges in the system graph corresponding to the across drivers must be included in every formulation tree Twhile all the edges corresponding to the through drivers must be in the co-tree T' of T.
(4)
( o ) ( b ) (c)
9 ~ (WGj) ,
_,___L .~_ o
### B~ A~
( UAj ) ( d ) Ce) ( f )
### Figure 1.
(a) Three-port rigid body (d{j) with the joint terminals
and the terminal
### Gj
corresponding to the center of mass of
### ((~-j).
(b) Pictorial representation of (~j).
(c) The terminal graph of the three-port (~j).
(d) Three-port rigid body (61j) is represented as an interconnection of ideal rigid body
### ((Roj)
with the terminal equations in (3) and an active one-port component with the terminal equations in (4), (e) Active one-port component at port (Gj0) is moved to the port
### (BjO).
(f) Active one-port component at port
### (GjO)
is moved to the port
### (AjO).
2. The systems of rigid bodies
In this section the system of rigid bodies ((Rj) ( j = 1, 2 . . . m) (see Figure 2a) will be of interest where each rigid body is regarded as a three-port component with ports
### 0), (Bj, O) and (Gj, 0).
Such systems are useful in robotic studies [9]. In this system, the terminal
### Aj
o f ((Rj) is assumed to b e connected to the terminal
### Bj+I
o f ((Rj+I) ( j = 0, 1 . . . . , m - 1) to form an open kinematic chain. In the following, this system will be modeled as an (m + 1)-port component, one of the ports being
### (Am,
O) and the remain- ing ports corresponding to the interconnected terminal pairs
### (Aj, Bj+ 0
of the rigid bodies ((Rj) and ((Rj+I), respectively; i.e., in the model, the terminal graph will of the form shown in Figure 2b. Note, the terminal graph is disconnected (separated). At the beginning, we shall save the mass centers of the rigid bodies as additional terminals; therefore, we first consider the enlarged terminal graph given in Figure 2c. In order to obtain the terminal equations in the desired closed form corresponding to the terminal graph in Figure 2c,
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NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 111
this system must be excited by (2m + 1) drivers (1"), (2*) . . . . , (m*), (m + 1)* (1") ', (2*) ', 9 (m*)' at the ports which are dictated by the edges of this terminal graph. The schematic diagram of the system augmented with the (2m + 1) drivers is given in Figure 2d and the corresponding system graph will be as in Figure 2e. Note, in an actual system no connec- tions are made to the ports (Gj, O) and, hence, even if we assume that the drivers (j*)'
are connected to these ports, the through variables (y*)' (i.e., forces ( f T ) ' and torques (M 7 ) ' ) of the drivers are identically equal to zero. However, as can be seen from the system graph (see Figure 2e), the across variables of the drivers are equal to the aross variables xGi at the ports (Gj, O). Although these across variables are to be eliminated in the final expression of the terminal equations, their use as auxiliary variables will result in some simplifications in the expressions we will be dealing with. From the system graph (see Figure 2e), the following observations can be made:
I
### o!'t k
o ... b? ?" o ~ 3 ~o
(c j-~ f i . ' f
i m;1
(3")
### (e)
Figure 2. (a) The system of m rigid bodies.
Co) The terminal graph associated with the terminal equations in (20). (c) The terminal graph associated with the terminal equations in (21).
(d) A pictorial diagram of the system of rigid bodies in Figure l(a) augmented with the through (force- moment) drivers (1"), (2"), ..., (mr*) and the across (velocity) drivers (1") ', (2*) ', .... (m*) ', and (m + 1)*.
(6)
In the system, for each rigid body, let the form of the terminal equations: (i) in equation (1) be used.
(ii) in equation (5) be used. (iii) in equation (6) be used.
Then the topological conditions for the existence of a solution, as discussed at the end of the Introduction, implies for each case that
(i) the edges (a i, o), (bi, o) q = 1, 2, . . . , m) are included in a formulation tree T while the edges (gi, o) (i = 1, 2 . . . m) are in the co-tree T'. Therefore the edges (1"), (2"), . . . , (m*) are necessarily in T ' while the edges (1") ', (2*) ', . . . , (m*) ', and (m + 1)* are included in T. These two groups of edges must correspond to through and across drivers, respectively.
(ii) The edges (a i, o), (gi, o) (i = 1, 2, . . . , m ) are included in T while the edges (bi,
o) (i = 1, 2, . . . , m) are in T'. In this case the edges (1"), (2"), . . . , (m*) must corre- spond to across and the edges (1") ', (2*) ', . . . . (m*)' and (m + 1)* must correspond to through drivers.
(iii) In this case the edges (bi, o), (gi, o) (i = 1, 2 . . . . , m) will be in T and the edges
(ai, o) (i = 1, 2, . . . , m) will be in T '. Hence the edges (2"), (3*) . . . (m + 1)* correspond to across while the edges (1"), (1")', (2*) ', . . . , (m*)' correspond to through drivers.
The desired terminal equations can be obtained by considering any one of the three cases considered above. In application, however, at the joints the displacement variables and at the port (Am, O) the mechanical load (the force fm+l and torque Mm+ 1) variables are specified. Therefore, the proper choice of drivers is that of case (ii). However, the operator matrix WB in equation (5) has rather complex expression compared to Wa in equation
J . . . ~ .
(1). For this reason, the following formulation wall consider the type of excitation discussed in case (i); i.e., in the system graph, the formulation tree will necessarily be as shown by the thick lines in Figure 2e.
Note, to keep the analysis valid under fairly general interconnection conditions, the types of joints (element pairs) to be used for connecting the rigid bodies in the system are not specified in advance. The special cases where the joints are spherical, revolute and prismatic (sliding pair) are considered after the establishment of the general system model. Although the closed form equations are useful for theoretical investigations, an algorithmic approach to the solution of these equations is preferable [10-14]. This is the main reason for keep- ing the across variables xc. associated with the ports (Gj, 0 ) .
From the system graph ~see Figure 2e), the fundamental circuit equations are
X * I =
### t
Xj* = Xbj - - X a ( j _ l ) ( j = 2, 3, . . . , m) x(j.), = X Gj ( j = 1, 2 . . . m) X ( m + l ) * ~
### (8)
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NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 113
and the fundamental cut-set equations are,
Yj* +Ybj = 0 ( j = 1 , 2 , . . . , m ) - ]
### Yaj
= yb(;+u. ( j = 1, 2, . . . , m -- I)
### )
YGj + Y~j*)' = 0 ( j = 1 , 2 . . . m) . Ya m + Y(m+l)* = 0 YC/*)' -- 0 ( j = 1, 2 . . . m) (9)
F r o m the first six rows of equation (1) we get r.-T . K - 1 . Z
Xaj = l~ajl, by ) Xbj = (Kbj 1 Kaj)Txbj =
### Kra~Xbj
(10) and the circuit equation (8) and the relations in equation (10) thus yield
Xl* X2* Xm* X(m+l)* I -K~ t T --K~l(m_l ) I
### x 2[
Xbm A = [KllXB'K2
### (11)
Note that the 6m • 6m submatrix K 1 is nonsingular, while K2 is a 6 • 6m submatrix. By considering the second row block in equation (1), we write
x s = Xb Xb 2 xbm
### x~ 1
x ~ XGm = KdXo (12) where the coefficient matrix Kd is nonsingular. Substituting equation (12) into equation (11) we obtain
### 6m Ex* 1 I ll
(6) x(m+ 0* = K2 KdxG = K ' x 6 (13)
where the submatrix K = KIKd is nonsingular. On the other hand, f r o m the last six rows of equation (1) we have
(8)
If this set of equations is put into one matrix equation form and considering the fundamen- tal cut-set equations in equation (9) we obtain
XG 1
Kb~ L Y,~* UG 1
b I - K a 1
- K a ( m _ l )
Y(m+
UG 1 UG 2
### (15)
or in a more compact form
= - - ua. (16)
### [KT (K')r]
Y~m+l)*
Since the terminal variables and the variables corresponding to the drivers are related by
x(m+
Xl X2 Xra x(m+
Xm*
Yl
Y(m+
Ytrt*
### Y(m+ 1)
(17) from equations (13) and (16) we obtain the intermediate form of the terminal equations as
(6) XCm+l) K '
(6)
Y(m+l)
- - U G
### (18)
Note, as stated above, the 6m • 6m submatrix K = K1K d is nonsingular and the explicit
form of its transpose appears in (15). Therefore, if we let S = K -l, then the expressions of the 6 • 6 submatrices Spq ( p ,
### q = 1, 2, ..., m)
of S are given as follows,
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NETWORK M O D E L APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 115
Spq = O p < q ) S p q = I p = q Spq = (Kffp)(K:4(p_I~)(KA(,_2))...( - 1 T T T KTA~ p > q
(19)
In equation (18) the auxiliary variables x G can be eliminated. Indeed, from the first rela- tion in equation (18) we have xo = K - i x = Sx and Xm+l = K 'Sx, hence the second rela- tion becomes Wc, Sx = Kry + (K 3ry(m+l ) - u G. These relations may be written together in the following single matrix form
(6)
X(m + 1) K 'S 0
### I x ]+IS luo
Y(m+ 1)
These are the terminal equations corresponding to the terminal graph in Figure 2b of the rigid body system considered as (m + 1)-port component. However, the terminal equa- tions of the rigid body system, augmented by the ports (OGj) ( j = 1, 2 . . . m) corre- sponding to the terminal graph in Figure 2c, can also be given. Indeed, assuming that the yGjs are not zero, and since xo = Sx, we have
### ,6m, lyl
(6) XCm+a ) = 0 0
### lxl li t
Y(m+l) +
(6m) x6 0 0 Ya
uo. (21)
Taking Yo =- 0, this equation reduces to equation (20).
2.1. Forward dynamics
The equations (18) are called forward dynamics equations of the rigid body system. In- deed, if all the through variables (y, Ym+l and u6) are specified, then the motion of the system is determined; i.e., from these equations, once the across variables (velocities) x6 associated with the mass centers are determined, the joint velocities are obtained. In fact, if we let W = SrWG S = Sr(Pc~tt + O6)S = p d + Q dt where
### d )
Q = ST"(Qc, S + Pc--~tt S)
then the first 6m rows of equation (20) can be put into the form
(22)
(23)
e d x + (Qx - (K'S)ry,n+l-y) + SrUG = 0
(10)
which represents the equations of motion of the system in closed form. However, to obtain the final form of these equations, position and velocity dependent coefficient matrices together with the joint velocities x in equation (24) must be expressed in terms of generalized coordinates which may be chosen for this system and represented by a column matrix q of order 6m, i.e., 3m components of q correspond to the joint displacements ~, and the other 3m components correspond to the joint angular displacements/9. More specifically, from equation (2) and the expression for W in equation (22), it follows that the entries in P, (K 'S) T and S are functions of q while the entries in Q are functions of both q and
x = d/dt q; i.e., the relations in equation (24) yields 6m second order differential equa-
tions in 6m generalized coordinates. Note the square coefficient matrix P in equation (24), the inertia matrix of the system, is symmetric and positive definite. This property is evi- dent from equation (23) and the expressions of PGj and W G in equations (2) and (16), respectively where PGj is symmetric and positive definite.
2.2. Explicit forms of the coefficient matrices in equation (24)
The explicit forms of P and Q in equation (24) are rather cumbersome as will be seen from the following expressions. Considering equations (19) and (23) we have
### PG2
\$21 S22 . (25)
r Smt S, n2 . S,,~
S~n PG~ ""
Therefore a typical 6 • 6 block entry Ppq of P is
### ~ ol = max(p, q).
(26)
Note that since j -> p or q, equations (7) and (19) together result in
### E
I ~ k ~ q R ~ + R c , oj
= "K-IxTKT K~
--- (27)
### \ bj
J A(J -1) A(J - 2 ) ' " 0 I "
Substituting the expressions of PGj appearing in equation (2) together with equation (27) into equation (26) yields the following explicit expression for Ppq:
### 1
= m "--1R DT ~:~.~-1 DT " Ppq ~=,~ mj(,YJk=p - 1
### Rk
+ Rrc, b) JGj + mj(Z/k=p k + l~'Gbjlk k=q'"k + RGbj)
(28)
'-1
RTbj and rj-1
### R T
The submatrices ~=p k + ~=q k -[- RGbj in equation (28) can be modified
somewhat. Indeed, considering Figure 3, we define
= "-1 R
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NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 117 A~ B1 A2
### "Gi_; ~, Ai_~
Gj "-FGb} gj 1 -rG,:i o
Vectors
### rj, rGj and ~j
associated with the system in Figure la.
where the 3 x 3 skew symmetric matrix ~k represents the position vector (k of B k relative to Ak-1 while Rp ~] corresponds to the position vector of Gj relative toBp. T h e term Rq G. iS similarly defined9 In the case of spherical joints [15] ~k - 0 and using equation (29) with Ek - 0 in equation (28), we have
### 1
m r ; a = m a x ( p , q).
epq = Z?=et mjRe,Gj J~j + ,,.jRp,GRq,6j
### (30)
Note that in equations (28) and (30) m denotes the number of rigid bodies in the system, while mj is the body mass. Carrying out similar operations, explicit expression for the coefficient matrix Q in equation (23) can be given. However, in order to keep its com- plicated appearance relatively low, derivatives of matrices in equation (23) are not replac- ed by the relations d/dt R = fiR - Rf~ given in A. 12 of the appendix of [1]. Note, the second term Rfl on the right may be omitted due to its multiplication by o~ in the final expressions; i.e.,
### E ~
~(..0 "~- (.0 3 0 --W 1 O~ 2 ~ 0 9 - - ~ 2 0)1 0 W 3
"-t r d "-1 T ;
### 0 ~G/Gj -[- mj(~k=pg k + eGbj)--~(~k=qek + gGbj)
o~ = max(p, q) (31) In the case of spherical joints, we have
d T
d r
### ~G/Gj + mjep,Gj-~ttgq,G j
; ot = max(p, q). (32)
(12)
On the other hand, when the rows of the last column matrix u =
### Sruc
in equation (24) are partitioned into m column matrices of order 6, the p-th one is
(33)
### Up
or in the case of spherical joints,
### g.
(34) To complete the explicit expressions of various submatrices appearing in equation (20), consider the last block of rows:
X(m+l ) - - - - (K'S) = Kam[Sml Sm2 . . . Smm ] x . r (35)
Utilizing the expression in equation (27) with j and q replaced by m and j, respectively, equation (35) takes on the form
X(m+l) =
~7=lSmjX j =
### wjVj
(36)
Note, in the case of spherical joints, since the joint velocities
### vj
are zero, the first column blocks of the coefficient matrices appearing in equations (28), (30), (31), (32), and (36) may be omitted. In the case of sliding joints, the second column blocks of the same coeffi- cient matrices may be omitted. In the case of spherical joints, the position vector of
relative to
### Bj
is
Rj, m = ~tff=jR k
and the relation in equation (36) becomes
X(m+I) = [ Vm+l ? = I R ~ I m " ' " O)m+ 1 " 9 9 (37) Rmm
(.0 m
### (38)
Furthermore, if we let 60j = njwj where
### nj
indicates the unit vector in the direction of wj, then equation (38) may be put into a more compact form
X ( m + l ) =
### 39
V RTrnnl r 601 1_ nl 9 . . n m
r m
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NETWORK M O D E L APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 119
### 2.3. Inverse dynamics
The terminal equations (20) can be interpreted in various directions. First, if the linear and angular velocities at the joints are specified (i.e., x is given), then the velocities
### Vm+l,
r at the load port are automatically obtained from equation (20) irrespective of the ex- istence of mechanical load and also the forces due to gravity (Forward Kinematics). However, the determination of forces and torques (moments) at the joints (Inverse Dynamics) re- quires additional information about the specification of load force
### fm+l
Mm+l.
Second, if the motion of the load terminal
### A m
is prescribed (e.g.,
### Xm+ 1
is given), in general, a unique determination of joint velocities (Inverse Kinematics) is not possible since (K 'S) is a 6 • 6m rectangular matrix. In this case (redundant manipulator), the concept of a generalized inverse may be used to obtain a solution in the sense of least squares which minimizes
[15].
### 2.4. Recursive computations
The closed form of the terminal equations (20), corresponding to a system of m rigid bodies interconnected in a special way to form an open kinematic chain, may not be preferable from the computational point of view. In this case one may consider the equivalent set, equations (18), containing the auxiliary variables x 6 for a recursive computation. This is indeed possible, since the 1-st, j-th and the last row blocks of the first equation in (18) yield the forward recursive kinematic equations:
XG1 = (K~I)Txl
=1 T T
XGj = (Kb'j ) (Kao-" 1) ) xG'(j-1) -['- (Kbj)Txj ( J = 2, 3, . . . , m)
X(m+l ) = KamXGm
### (40)
and the inverse recursive equations are obtained from the second equation in (18) as
y j :
### K~I(WGdXGj -F
Kay0-+1 ) + utj) j = (1, 2 . . . m - 1)
### J
Ym = K;m~(WG~XGm - Ka.y(m+l) + u % )
(41)
w h e r e -Y(m+l) represents the load through variables. The recursive relations can be displayed as a block diagram (see Figure 4a). Furthermore, the blocks Woj =
### Po~(d/dt)
+ Qc. can be replaced by the detailed blocks in Figure 4b. Due to the presence of the derivaJtive operator blocks
computations of
### (d/dO xqi
are also required. These ex- pressions can be derived from equation (40) where the derivatives of the matrices
### (K~l)r(Ka(.
x)) T will appear which may again be replaced by simpler forms by the use of the relatioh-obtained in A. 12 of the appendix of [1]; i.e.,
### (d/dOR
= f i r - Rfl, where the second term may be omitted due to the multiplication of this expression by w.
(14)
XG,
### X •
G~m'~ [~[ m-, m Gt
### ~
m IG m XG~ IGj (a) (b)
Figure 4. (a) Block diagram corresponding to forward recursive kinematic and inverse recursive dynamics equations in (40) and (41), respectively.
(b) Replacement of WGj block by an equivalent block diagram.
3. Example
As an application of the closed form equations (20), the three-link plane manipulator with revolute joints is considered (see Figure 5a). All the rotation axes at the joints are along the z - axis, normal to the paper surface [17]. It is assumed that the links are homogeneous and prismatic in shape so that the centers of gravity occur at the mid points of the links. The lengths of the links are la, 12 and 13, respectively. At the firsst three joints the ter- minal velocities vl, v2, v3 are zero and the angular velocities r 1, 6o2, 6o3 have only z - components, 601,602, c03, respectively. We are also interested in only the z components of the torques M1, M2, M3 at the joints, indicated by M1, ME, M3, respectively. The manipulator is subject to the loads, f4 and M4, at the terminal A 3 (the z component o f ~ is zero while M4 has only a z - component, M4). Under these conditions, the manipulator as a four-port (scalar six-port) component with the schematic diagram and the terminal graph given in Figures 5b and 5c will have the following terminal equations which can be deduced directly from (20)
M1 M2 M3 V4x Vay 604 ')/11 "Y21 ")'31 ~ 1 1 ~ 1 2 ~13 ')'12 "Y13 I --~11 --~12 --~13 ")'22 "I"23 [ --~21 - ~ 2 2 - ~ 2 3 7 3 2 "Y33 ] --~31 --~32 - ~ 3 3 62t 6 3 1 1 0 0 0 ~22 ~32 I 0 0 0 ~23 ~33 [ 0 0 0 60 3 ______ -[- - - - - o _ 0 _1 (42)
(15)
NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 121 M4 y 0 3 / / / _ A1 ,B (m~) [3 3 M3 { ~3 ',, b2 B 2 ~ : = p A 3 2 M2( 11,4 4 G)2t~ I I (04 a~ ~ - I f~ ~ B,o5- I v~ b,o u31 0
## t!i
o 4 ( a ) ( b ) ( c )
Figure 5. (a) A three-link plane manipulator.
(b) Pictorial diagram of the manipulator as the four-port mechanical component.
(c) The terminal graph of the manipulator (port 4 represents both the translational and rotational ter- minal variables).
where the scalar entries 7pq, ~ q and ~p are the r e d u c e d forms of the 3 x 3 submatrices o f W = SrWc, S, (K'S) and S'uG in equation (20), respectively. To obtain these reduced forms, in the case o f spherical joints, we should consider only the last rows and columns o f the submatrices o f equation (2):
### 1
J% = * * , Rp,aj = 0 0 -(rp,@x 9 * J: -(re,@y (rp,@x 0
### I~
Rq,Gj = 0 0 ( )x ~Gj 0 = -- (rq'GJ)r --(rq'G: )x rq~j , = ~OjO 0 ' g 0 g (43) where we define d q/pq = Cpq ~ Jr" dpq
and where the coefficients are obtained from equations (30) and (32) as
3
C?q = ~j=~ {Jj + mj[(rp,aj)x(rq,@x + (rp,@y(rq,@y]} (or = m a x ( p , q))
and
### (44)
(16)
where the dot indicates the time derivative. The entries ~pq in equation (42) will be ob- tained from equation (36). Since
=
1 0
### 0
0 1
the first two and the last rows and columns of equation (35) yield I(311 ~21 ~31 7
r2y -I-
--(F2y +
### r3y ) --F3y
612 6Z2 b32 ~ = ~
### 613 (~23 r
1 1 1
(46) Finally, from the last rows of equations (33) and (34) we obtain
(47)
### m3(r3,G3)x
More explicit expressions for the entries
### Cpq,
dpq, ~pq and/3p can be obtained by using the parameters indicated in Figure 5a. Note that for the links, we have Jy =
### (1/12)mjl~ (j =
1, 2, 3). Let C i = cosOi, S i = sinOi, Cij = cos(Oi + Oj), Sij = sin(O i -]- Oj), Cij k -- cos(• i
### + Oj + Ok) and Sijk = sin(Oi + Oj + Ok),
then all the entries calculated from equations (44)-(47) are given in Table 1.
For a two-link manipulator with no loads, the terminal equations (42) take the following form
E M1
### 48,
where the entries
~-- Cpq
### (d/dt) + dpq,/3p
can be obtained from Table 1 with m3 replaced by zero. These coefficients are given in Table 2.
4. Conclusions
Although the procedure presented in this article is only applied to obtain the terminal equa- tions (20) for a system of m rigid bodies connected in a special form, it is general. It can be applied to other rigid body systems having an arbitrary interconnection pattern which may exhibit an open or closed kinematic chain. Indeed, in the general case, the only change that occurs is in writing the fundamental circuit and fundamental cut-set equations obtained directly from the system graph representing the interconnection pattern of the rigid bodies.
(17)
N E T W O R K M O D E L A P P R O A C H T O T H E A N A L Y S I S O F M U L T I R I G I D - B O D Y S Y S T E M S 123 Table 1. Cll = ml(l/31~) + m2(l ~ + lll2C 2 + ~I~) + m3(l~ + 122 + 1/312 + 21112C 2 + 1213C 3 + 1113C23 ) c12 = c2~ = m 2 ( ~ t ~ + 8 9 z) + m3(l~ + 1At~ + 1213C 3 + l~t2C 2 + V2t113C23 c13 = c31 = m 3 ( ~ l ~ + 1/21213C3 + 1/21113C23) c22 = m2(%12) + m3(1Al ~ + l~ + 1213C3) c23 = c32 = m3(1/al 2 + 1/21213C3) c33 = m3(1/3l z) dtt = --rrb2(V2ltl2Szw2) -- m31l12SzoJ2 + 89 2 + w 3) dl2 = -m21AlllzS2(% + ~ - m3Vz1213S3~ - m31112Sz(~ + ~~ + V21l13S23(~ + ~2 + ~ 613 = -m3(1/2121383 + 1/21113823)(o~ 1 + 602 + 603)
d21 = m2(1/21112S20Ol) + m3(lllzS 2 q- 1/21113S23)o01 - 1/21213S3oo 3
d22 = -m3(1/21213S3oo3) d23 = -m3(1/ZlE13S3)(~ + ~ + ~3) d31 = m31/21113S23oo I + 1/21213S3(001 + W 2) d32 = -m3(1/21213C3)(oo I -b 6o2) , d33 = 0 till = -(11S1 + 12S12 q- 13S123), t~21 = -(12S12 + 13S123), ~31 = -/3S123 t~12 = llC1 + 12C12 q- 13C123 , t~22 = 12C12 "}- 13C123, ~32 = 13C123 ~13 = (~23 = (~33 = 1 a n d (1/g)fl I = m1(1/2llC1) + m2(llC 1 -b 1,~12C12 ) -b m3(11C 1 + 12C12 + 1/213C123) (1/g)/~ 2 : 1qb2(1/212C12) + m3(12C12 q- 1/212C123) (1/g)fl 3 = m3(1/213C123) Table 2. Cll c12 6"22 d l l d12 d21 d = 31 &
= ml(~Al~) + m2(l~ + lal2C 2 + 1Al22)
= c21 = m2(%l 2 + 1/21112C2) = m2(V3122) = -~(VzlllaS2~o2) = -m2[ 89 -1- w2) ] = m2(V21112S2Wl) = 0 = mlg(V211C1) + m2g(llC 1 + 1/212C12) = m2g(Vzl2C12 ) I f l 1 = l 2 = l t h e n t h e s e e q u a t i o n s r e d u c e to t h o s e g i v e n i n [17] o n p a g e s 102, 124 a n d 142.
(18)
Appendix
The explicit expressions of the submatrices appearing in the terminal equations (5) and (6) are given, respectively, in the following:
and with and and with
Gbj
aj
mj I mj R %
, QA~ =
0 P a J =
= K _ 1
### mj R~a i g
(49)
(19)
NETWORK MODEL APPROACH TO THE ANALYSIS OF MULTIRIGID-BODY SYSTEMS 125
References
1. Y. Tokad, "A Network Model for Rigid Body Motion," Dynamics and Control, vol. 2, no. 1, pp. 59-82, 1992. 2. H.E. Koenig, Y. Tokad, and H.K. Kesavan, Analysis of Discrete Physical Systems, McGraw-Hilh New
York, 1967.
3. Y. Tokad, "On the Existence of Mathematical Models of Multi-terminal Linear Lumped Time Invariant Networks," Proc. of the (1981) European Conference on Circuit Theory and Design (ECCTD'81), pp. 886-891, The Hague, The Netherlands, 1981.
4. Y. Tokad, "On the Topological Conditions for Linear Lumped Time Invariant Networks Containing Multi- terminal Components," Bulletin of the Technical University oflstanbul, vol. 40, no. 2, pp. 479-496, 1987. 5. K. Abdullah and Y. Tokad, "On the Existence of Mathematical Models for Multiterrninal RCF Networks,"
IEEE Trans. Circuit Theory, vol. CT-19, no. 5 pp. 419-424, 1972.
6. P.R. Bryant and J. Tow, "The A-Matrix of Linear Passive Reciprocal Networks," J. Franklin Inst., vol. 293, no. 6 pp. 401-419, 1972.
7. A. Recski, Matroid Theory and Its Applications, Springer-Verlag: Budapest, 1989.
8. S. Seshu and M.B. Reed, Linear Graphs and Electrical Networks, Addison-Wesley Publishing Co., Inc., MA, 1961.
9. M. Vukobratovic and V. Potkonjak, Dynamics of Manipulation Robots (Scientific Fundamentals of Robotics; 1), Springer-Verlag: Berlin, Heidelberg, 1982.
10. J.Luh, M. Walker, and R. Paul, "On-line Computational Scheme for Mechanical Manipulators," J. Dynamic Systems, Measurements, and Control, vol. 102, no. 6, pp. 69-76, 1980.
11. J.M. Hollerbach, " A Recursive Lagrangian Form~ation of Manipulator Dynamics and Comparative Study of Dynamic Formulation Complexity," 1EEE Trans. Syst., Man, Cybern., vol. SMC-10, no. 11, pp. 730-736, 1980.
12. M.W. Walker and D.E. Orin, "Efficient Dynamic Computer Simulation of Robotic Mechanism," J. Dynamic Systems, Measurement, and Control, vol. 104, pp. no. 9, 205-211, 1982.
13. L.T. Wang and B. Ravanl, "Recursive Computations of Kinematic and Dynamic Equations for Mechanical Manipulators," 1EEE Journal of Robotics and Automation, vol. RA-1, no. 3, pp. 124-131, 1985. 14. R. Nigam and C.S.G. Lee, " A Microprocessor-Based Controller for the Control of Mechanical Manipulators,"
IEEE Journal of Robotics and Automation, vol. RA-1, no. 4, pp. 173-182, 1985.
15. J.M. Hollerbach and K.C. Suh, "Redundancy Resolution of Manipulators Through Torque Optimization," 1EEE Journal of Robotics and Automation, vol. RA-3, no. 4, pp. 308-316, 1987.
16. R.P. Paul, Robot Manipulators: Mathematics, Programming, and Control, The MIT Press: Cambridge, Massachusetts, 1981.
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### Site Tools
howtos:toolcoding:scaling_normalizing_to_a_control_total
Suppose you have a variable `estimate[a,b,t]` which you want scale to match `targetControl_tot[t]`. Two equivalent statements are:
`target[a,b,t] = norm (estimate[a,b,t]; dim1=a, dim2=b) * targetControl_tot[t]`
and
`target[a,b,t] = targetControl_tot[t] / estimate_tot[t] * estimate[a,b,t]`
where
```estimate_tot[t] = sum (estimate[a,b,t]; dim1=a, dim2=b)
targetControl_tot[t] = sum (target[a,b,t]; dim1=a, dim2=b)```
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#### AR.Math.Content.HSN.RN.A: Extend the properties of exponents to rational exponents
AR.Math.Content.HSN.RN.A.1: Explain how extending the properties of integer exponents to rational exponents provides an alternative notation for radicals.
#### AR.Math.Content.HSN.RN.B: Use properties of rational and irrational numbers
AR.Math.Content.HSN.RN.B.4: Simplify radical expressions. Perform operations (add, subtract, multiply, and divide) with radical expressions. Rationalize denominators and/or numerators.
### AR.Math.Content.HSN.CN: The Complex Number System
#### AR.Math.Content.HSN.CN.A: Perform arithmetic operations with complex numbers
AR.Math.Content.HSN.CN.A.1: Know there is a complex number i such that i² = -1, and every complex number has the form a + bi with a and b real.
AR.Math.Content.HSN.CN.A.2: Use the relation i² = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers.
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#### AR.Math.Content.HSN.VM.A: Represent and model with vector quantities
AR.Math.Content.HSN.VM.A.1: Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v).
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https://number.academy/230126 | 1,660,304,571,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571692.3/warc/CC-MAIN-20220812105810-20220812135810-00318.warc.gz | 405,854,399 | 12,335 | # Number 230126
Number 230,126 spell 🔊, write in words: two hundred and thirty thousand, one hundred and twenty-six . Ordinal number 230126th is said 🔊 and write: two hundred and thirty thousand, one hundred and twenty-sixth. Color #230126. The meaning of number 230126 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 230126. What is 230126 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 230126.
## What is 230,126 in other units
The decimal (Arabic) number 230126 converted to a Roman number is (C)(C)(X)(X)(X)CXXVI. Roman and decimal number conversions.
#### Weight conversion
230126 kilograms (kg) = 507335.8 pounds (lbs)
230126 pounds (lbs) = 104384.5 kilograms (kg)
#### Length conversion
230126 kilometers (km) equals to 142994 miles (mi).
230126 miles (mi) equals to 370353 kilometers (km).
230126 meters (m) equals to 754998 feet (ft).
230126 feet (ft) equals 70144 meters (m).
230126 centimeters (cm) equals to 90600.8 inches (in).
230126 inches (in) equals to 584520.0 centimeters (cm).
#### Temperature conversion
230126° Fahrenheit (°F) equals to 127830° Celsius (°C)
230126° Celsius (°C) equals to 414258.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
230126 seconds equals to 2 days, 15 hours, 55 minutes, 26 seconds
230126 minutes equals to 5 months, 2 weeks, 5 days, 19 hours, 26 minutes
### Codes and images of the number 230126
Number 230126 morse code: ..--- ...-- ----- .---- ..--- -....
Sign language for number 230126:
Number 230126 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 230126
### Multiplications
#### Multiplication table of 230126
230126 multiplied by two equals 460252 (230126 x 2 = 460252).
230126 multiplied by three equals 690378 (230126 x 3 = 690378).
230126 multiplied by four equals 920504 (230126 x 4 = 920504).
230126 multiplied by five equals 1150630 (230126 x 5 = 1150630).
230126 multiplied by six equals 1380756 (230126 x 6 = 1380756).
230126 multiplied by seven equals 1610882 (230126 x 7 = 1610882).
230126 multiplied by eight equals 1841008 (230126 x 8 = 1841008).
230126 multiplied by nine equals 2071134 (230126 x 9 = 2071134).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 230126
Half of 230126 is 115063 (230126 / 2 = 115063).
One third of 230126 is 76708,6667 (230126 / 3 = 76708,6667 = 76708 2/3).
One quarter of 230126 is 57531,5 (230126 / 4 = 57531,5 = 57531 1/2).
One fifth of 230126 is 46025,2 (230126 / 5 = 46025,2 = 46025 1/5).
One sixth of 230126 is 38354,3333 (230126 / 6 = 38354,3333 = 38354 1/3).
One seventh of 230126 is 32875,1429 (230126 / 7 = 32875,1429 = 32875 1/7).
One eighth of 230126 is 28765,75 (230126 / 8 = 28765,75 = 28765 3/4).
One ninth of 230126 is 25569,5556 (230126 / 9 = 25569,5556 = 25569 5/9).
show fractions by 6, 7, 8, 9 ...
230126
### Advanced math operations
#### Is Prime?
The number 230126 is not a prime number. The closest prime numbers are 230123, 230137.
#### Factorization and factors (dividers)
The prime factors of 230126 are 2 * 13 * 53 * 167
The factors of 230126 are
1 , 2 , 13 , 26 , 53 , 106 , 167 , 334 , 689 , 1378 , 2171 , 4342 , 8851 , 17702 , 115063 , 230126
Total factors 16.
Sum of factors 381024 (150898).
#### Powers
The second power of 2301262 is 52.957.975.876.
The third power of 2301263 is 12.187.007.156.440.376.
#### Roots
The square root √230126 is 479,714498.
The cube root of 3230126 is 61,280443.
#### Logarithms
The natural logarithm of No. ln 230126 = loge 230126 = 12,346382.
The logarithm to base 10 of No. log10 230126 = 5,361966.
The Napierian logarithm of No. log1/e 230126 = -12,346382.
### Trigonometric functions
The cosine of 230126 is -0,365588.
The sine of 230126 is -0,930777.
The tangent of 230126 is 2,545972.
### Properties of the number 230126
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 230126 in Computer Science
Code typeCode value
PIN 230126 It's recommendable to use 230126 as a password or PIN.
230126 Number of bytes224.7KB
CSS Color
#230126 hexadecimal to red, green and blue (RGB) (35, 1, 38)
Unix timeUnix time 230126 is equal to Saturday Jan. 3, 1970, 3:55:26 p.m. GMT
IPv4, IPv6Number 230126 internet address in dotted format v4 0.3.130.238, v6 ::3:82ee
230126 Decimal = 111000001011101110 Binary
230126 Decimal = 102200200012 Ternary
230126 Decimal = 701356 Octal
230126 Decimal = 382EE Hexadecimal (0x382ee hex)
230126 BASE64MjMwMTI2
230126 MD52c02cf87ef776f560018398dd484582b
230126 SHA13ab75bf290633fdd56e99f8ec0e9e4e102655c1d
230126 SHA22425830fba35228f94fb03e73c9bd79d63bc02203c7870af002c0e2970
230126 SHA256ba0c7c5c50ca9ae0df14fc48dfc9894990fbf925f219301747bdeb9502542382
230126 SHA38404fb4b520fbff8f61e9610c4bdd7dec93f017d42d6ab5a311af3cda271a0aa5edddae108db25fc42655f35165f376cab
More SHA codes related to the number 230126 ...
If you know something interesting about the 230126 number that you did not find on this page, do not hesitate to write us here.
## Numerology 230126
### Character frequency in number 230126
Character (importance) frequency for numerology.
Character: Frequency: 2 2 3 1 0 1 1 1 6 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 230126, the numbers 2+3+0+1+2+6 = 1+4 = 5 are added and the meaning of the number 5 is sought.
## Interesting facts about the number 230126
### Asteroids
• (230126) 2001 MQ23 is asteroid number 230126. It was discovered by NEAT, Near Earth Asteroid Tracking from Haleakala Observatory on 6/27/2001.
## № 230,126 in other languages
How to say or write the number two hundred and thirty thousand, one hundred and twenty-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 230.126) doscientos treinta mil ciento veintiséis German: 🔊 (Anzahl 230.126) zweihundertdreißigtausendeinhundertsechsundzwanzig French: 🔊 (nombre 230 126) deux cent trente mille cent vingt-six Portuguese: 🔊 (número 230 126) duzentos e trinta mil, cento e vinte e seis Chinese: 🔊 (数 230 126) 二十三万零一百二十六 Arabian: 🔊 (عدد 230,126) مئتانثلاثون ألفاً و مائة و ستة و عشرون Czech: 🔊 (číslo 230 126) dvěstě třicet tisíc sto dvacet šest Korean: 🔊 (번호 230,126) 이십삼만 백이십육 Danish: 🔊 (nummer 230 126) tohundrede og tredivetusindethundrede og seksogtyve Dutch: 🔊 (nummer 230 126) tweehonderddertigduizendhonderdzesentwintig Japanese: 🔊 (数 230,126) 二十三万百二十六 Indonesian: 🔊 (jumlah 230.126) dua ratus tiga puluh ribu seratus dua puluh enam Italian: 🔊 (numero 230 126) duecentotrentamilacentoventisei Norwegian: 🔊 (nummer 230 126) to hundre og tretti tusen, en hundre og tjue-seks Polish: 🔊 (liczba 230 126) dwieście trzydzieści tysięcy sto dwadzieścia sześć Russian: 🔊 (номер 230 126) двести тридцать тысяч сто двадцать шесть Turkish: 🔊 (numara 230,126) ikiyüzotuzbinyüzyirmialtı Thai: 🔊 (จำนวน 230 126) สองแสนสามหมื่นหนึ่งร้อยยี่สิบหก Ukrainian: 🔊 (номер 230 126) двiстi тридцять тисяч сто двадцять шiсть Vietnamese: 🔊 (con số 230.126) hai trăm ba mươi nghìn một trăm hai mươi sáu Other languages ...
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If you know something interesting about the number 230126 or any natural number (positive integer) please write us here or on facebook. | 2,619 | 7,697 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-33 | latest | en | 0.654113 |
https://www.chegg.com/homework-help/sentence-skills-with-readings-4th-edition-chapter-2.8-solutions-9780073533261 | 1,529,474,979,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00290.warc.gz | 786,007,984 | 14,146 | # Sentence Skills with Readings (4th Edition) View more editions Solutions for Chapter 2.8
• 1697 step-by-step solutions
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Chapter: Problem:
Change the verbs where needed in the following selection so that they are consistently in the past tense. Cross out each incorrect verb and write the correct form above it, as shown in the example. You will need to make ten corrections.
Making a foul shot that won a basketball game was a special moment for me. For most of the year, I sat on the bench. The coach put me on the team after the tryouts and then about me. Then my chance appears near the end of the Rosemont High School game. The score was tied 65 to 65. Because of injuries and foul-outs, most of the substitutes, except me, were in the game. Then our last first-stringer, Larry Toner, got an elbow in the eye and leaves the game. The coach looked at me and said, “Get in there, Watson.” The clock showed ten seconds to go. Rosemont had the ball when, suddenly, one of their players misses a pass. People scramble for the ball; then our center, Kevin, grabbed it and starts down the court. He looked around and saw me about twenty feet from the basket. I caught his pass, and before I could decide whether to shoot or pass, a Rosemont player fouls me. The referee’s whistle blew, and I had two free throws with two seconds left in the game. My stomach churns as I stepped to the foul line. I almost couldn’t hold the ball because my hands were so damp with sweat. I shot and missed, and the Rosemont crowd sighs with relief. My next shot would mean a win for us or overtime. I looked at the hoop, shot, and waited for what seemed like forever. The ball circles the rim and dropped in, and then the buzzer sounded. Everyone on the team slapped me on the back and the coach smacks my rear end, saying, “All right, Watson!” I’ll always remember that moment.
Sample Solution
Chapter: Problem:
• Step 1 of 1
appeared, left, missed, scrambled, started, fouled, churned, sighed, circled, smacked
Corresponding Textbook
Sentence Skills with Readings | 4th Edition
9780073533261ISBN-13: 0073533262ISBN: John LanganAuthors:
Alternate ISBN: 9780077422707 | 532 | 2,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-26 | latest | en | 0.963736 |
https://1828.mshaffer.com/d/word/quantity/simple/ | 1,621,298,512,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991650.73/warc/CC-MAIN-20210518002309-20210518032309-00020.warc.gz | 89,576,788 | 2,214 | # quantity
QUAN'TITY, n. [L. quantitas, from quantus, how much, or as much as.]
1. That property of any thing which may be increased or diminished.
This definition is defective, and as applicable to many other properties as to quantity. A definition strictly philosophical cannot be given. In common usage, quantity is a mass or collection of matter of indeterminate dimensions, but consisting of particles which cannot be distinguished, or which are not customarily distinguished, or which are considered in the aggregate. Thus we say, a quantity of earth, a quantity of water, a quantity of air, of light, of heat, of iron, of wood, of timber, of corn, of paper. But we do not say, a quantity of men, or of horses, or of houses; for as these are considered as separate individuals or beings, we call an assemblage of them, a number of multitude.
2. An indefinite extent of space.
3. A portion or part.
If I were sawed into quantities. [Not in use.]
4. a large portion; as a medicine taken in quantities, that is, in large quantities.
5. In mathematics, any thing which can be multiplied, divided or measured.
Thus mathematics is called the science of quantity. In algebra, quantities are known and unknown. Known quantities are usually represented by the first letters of the alphabet, as a, b, c, and unknown quantities are expressed by the last letters, x, y, z, &c. Letters thus used to represent quantities are themselves called quantities. A simple quantity is expressed by one term, as + a, or - abc; a compound is expressed by more terms than one, connected by the signs, + plus, or -minus, as a + b, or a - b + c. quantities which have the sign + prefixed, are called positive or affirmative; those which have the sign - prefixed are called negative.
6. In grammar, the measure of a sullable; that which determines the time in which it is pronounced.
7. In logic, a category, universal, or predicament; a general conception.
8. In music, the relative duration of a note or syllable.
Quantity of matter, in a body, is the measure arising from the joint consideration of its magnitude and density.
Quantity of motion, in a body, is the measure arising from the joint consideration of its quantity of matter and its velocity. | 505 | 2,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-21 | latest | en | 0.96118 |
https://www.shaalaa.com/question-bank-solutions/hardness-water-15000-litres-hard-water-was-passed-through-zeolite-softener-exhausted-zeolite-required-120-litres-naci-having-stream-30g-l-nacl-calculate-hardness-water_58073 | 1,560,641,826,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997501.61/warc/CC-MAIN-20190615222657-20190616004657-00308.warc.gz | 898,637,463 | 12,740 | BE Production Engineering Semester 1 (FE First Year)University of Mumbai
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# 15000 Litres of Hard Water Was Passed Through a Zeolite Softener. the Exhausted Zeolite Required 120 Litres of Naci Having Stream of 30g/L of Nacl Calculate the Hardness of Water. - BE Production Engineering Semester 1 (FE First Year) - Applied Chemistry 1
#### Question
15000 litres of hard water was passed through a zeolite softener. The exhausted zeolite required 120 litres of NaCI having stream of 30g/l of NaCL Calculate the hardness of water.
#### Solution
1 litre of NgCl solution = 30 gm of NaCl.
∴ 120 litre of NaCl = 120x30x103 mg of NaCl.
∴ 3600 x103 mg of NaCl. = 3600 x103 x 50/58.5
= 30.7692 xx 10^5 "mg of " CaC0_3 equivalent.
15000 litres of water = 30.7692 xx10^5 " mg of" CaC0_3 equivalent.
∴ 1 litres of water = (30.7692 xx10^5)/15000 = 205.1 ppm
∴ Hardness of water samQ.Ie = 205. 1 ppm
Is there an error in this question or solution?
#### APPEARS IN
Solution 15000 Litres of Hard Water Was Passed Through a Zeolite Softener. the Exhausted Zeolite Required 120 Litres of Naci Having Stream of 30g/L of Nacl Calculate the Hardness of Water. Concept: Hardness of Water.
S | 386 | 1,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-26 | longest | en | 0.630162 |
https://purrr.tidyverse.org/reference/index.html | 1,709,459,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00558.warc.gz | 467,514,326 | 6,126 | ## Map family
The map(.x, .f) functions transforms each element of the vector .x with the function .f, returning a vector defined by the suffix (_lgl, _chr() etc). walk() is a variant for functions called primarily for their side-effects; it returns .x invisibly.
As well as functions, .f, can take numbers and characters (used as a shorthand for [[), and formulas (used as a succint function definition). See details in as_mapper()
map() map_lgl() map_int() map_dbl() map_chr() map_vec() walk()
Apply a function to each element of a vector
as_mapper()
Convert an object into a mapper function
## Map variants
A rich set of variants builds on the basic idea of map(): modify() modifies “in place”, returning a vector the same type as .x; map2() iterates over two vectors in parallel; pmap() (parallel map) iterates over a list of vectors; imap() (index map) is a shortcut for the common pattern map2(x, names(x)).
map_if() map_at()
Apply a function to each element of a vector conditionally
map_depth() modify_depth()
Map/modify elements at given depth
map2() map2_lgl() map2_int() map2_dbl() map2_chr() map2_vec() walk2()
Map over two inputs
pmap() pmap_lgl() pmap_int() pmap_dbl() pmap_chr() pmap_vec() pwalk()
Map over multiple input simultaneously (in "parallel")
modify() modify_if() modify_at() modify2() imodify()
Modify elements selectively
modify_tree()
Recursively modify a list
imap() imap_lgl() imap_chr() imap_int() imap_dbl() iwalk()
Apply a function to each element of a vector, and its index
lmap() lmap_if() lmap_at()
Apply a function to list-elements of a list
## Predicate functionals
A predicate function is a function that either returns TRUE or FALSE. The predicate functionals take a vector and a predicate function and do something useful.
detect() detect_index()
Find the value or position of the first match
every() some() none()
Do every, some, or none of the elements of a list satisfy a predicate?
has_element()
Does a list contain an object?
head_while() tail_while()
Find head/tail that all satisfies a predicate.
keep() discard() compact()
Keep/discard elements based on their values
keep_at() discard_at()
Keep/discard elements based on their name/position
## Plucking
Getting or setting a single element.
pluck() pluck<-() pluck_exists()
Safely get or set an element deep within a nested data structure
chuck()
Get an element deep within a nested data structure, failing if it doesn't exist
pluck_depth()
Compute the depth of a vector
modify_in() assign_in()
Modify a pluck location
attr_getter()
Create an attribute getter function
## Other vector transforms
A grab bag of useful tools for manipulating vectors.
accumulate() accumulate2()
Accumulate intermediate results of a vector reduction
list_c() list_cbind() list_rbind()
Combine list elements into a single data structure
list_flatten()
Flatten a list
list_assign() list_modify() list_merge()
Modify a list
list_simplify()
Simplify a list to an atomic or S3 vector
list_transpose()
Transpose a list
reduce() reduce2()
Reduce a list to a single value by iteratively applying a binary function
## Superseded
Superseded functions have been replaced by superior solutions, but due to their widespread use will not go away. However, they will not get any new features and will only receive critical bug fixes.
flatten() flatten_lgl() flatten_int() flatten_dbl() flatten_chr() flatten_dfr() flatten_dfc()
Flatten a list of lists into a simple vector
map_dfr() map_dfc() imap_dfr() imap_dfc() map2_dfr() map2_dfc() pmap_dfr() pmap_dfc()
Functions that return data frames
as_vector() simplify() simplify_all()
Coerce a list to a vector
transpose()
Transpose a list.
Adverbs modify the action of a function; they take a function as input and return a function with modified action as output.
auto_browse()
Wrap a function so it will automatically browse() on error
compose()
Compose multiple functions together to create a new function
insistently()
Transform a function to wait then retry after an error
negate()
Negate a predicate function so it selects what it previously rejected
partial()
Partially apply a function, filling in some arguments
possibly()
Wrap a function to return a value instead of an error
quietly()
Wrap a function to capture side-effects
safely()
Wrap a function to capture errors
slowly()
Wrap a function to wait between executions
## Misc
array_branch() array_tree()
Coerce array to list
rate_delay() rate_backoff() is_rate()
Create delaying rate settings
progress_bars
Progress bars in purrr | 1,071 | 4,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-10 | latest | en | 0.619643 |
https://hubpages.com/family/Easter-Themed-Minute-to-Win-it-Games | 1,566,082,551,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313501.0/warc/CC-MAIN-20190817222907-20190818004907-00214.warc.gz | 502,200,028 | 28,964 | # Easter Themed Minute to Win it Games
Updated on March 11, 2016
Want to plan an easter themed evening of fun? Minute to Win it games can be easily adapted to the theme of Easter and provides fun for all ages. Using resources you probably have at home already or bought at the dollar store the games can be put together cheaply and quickly. Most of the games suggested below are suitable for young and old, and the games can be run as player against player or team against team. A timer or stopwatch is needed for every game.
## Egg Toss
Equipment Needed
Four hoops
16 plastic eggs, taped closed to prevent opening.
Marker to show standing position
Egg toss can be played in several ways. Using some hoops (these ones are from a twister game) and plastic eggs, players need to get all the eggs in the hoops in a minute. Other ways to play to make the game more challenging could be throwing the correct color egg into the corresponding hoop. If playing team against team you can play for the highest number of eggs in each hoop thrown in a minute to find the wining team.
## Jellybean Sort
Equpiment Needed
Tweezers or chop sticks
Jellybeans in 5 different colors
sorting cups
In this game a set of jellybeans for a each color are placed in one container. The player using tweezers or chopsticks must move each jellybean to sort it into a container of the correct color. The winner is the person who sorts all the jelly beans into the correct color containers in one minute.
## Stacking Peeps
Equipment Needed
Peeps
Flat surface
How many peeps can you stack in a minute? Stack the Peeps one on top of the other to make a tower. If the stack falls over then start again. The winner if the player with the most stacked Peeps in a minute.
## Stacking Cups
Equipment Needed
10 cups of one color
1 cup of a different color
Using a series of cups in one color and one cup in another color, the cups are stacked with the different colored cup at the bottom of the stack of cups. As shown below the player has to move the blue colored cup to the top of the stack, but the cups can only be moved from top to bottom. So the top cup moves to the bottom of the pile, the next cup on top is moved to the bottom of the pile and so on until the green cup reaches the top. The objective is to get the cup from the bottom of the stack to the top in one minute.
## Peeps in the Nest
Equipment Needed
Pre determined number of Peeps
Marker for batting position
To play Peeps in the nest the player stands by the marker and when the stopwatch starts they bat Peeps into the basket. If all the Peeps land in the basket before the minute is up the player wins. It is helpful to have a few people retrieving any Peeps that do no make it to the basket to keep restocking the player.
## Egg Hunt
Equipment Needed
Styrofoam eggs
Tub
Grass or shredded paper
Bowl for the found eggs
The player picks as many eggs as they can find hidden in the grass. The objective is to find a pre specified number of eggs before the minute is up. This can be made harder or easier by adapting the number of eggs used and the size of the container. It is a good idea to play this game on a table or on a large cloth to catch any escaping shredded paper.
## Egg Race
Equipment Needed
Sraws - reusable ones are better if possible
Plastic eggs
Marker
Hoops
The objective of this game is to move three eggs to the hoop by blowing them with a straw. To win the game the player needs to get the three eggs touching the hoop in one minute.
## Flying Peeps
Equipment Needed
Peeps
Tubes
Eggs
The objective of the game is to knock down a series of eggs on tubes using Peeps as the 'ball'. The player has to knock down a pre determined number of eggs by throwing peeps at the eggs in a minute.
## Peeps on the Move
EquIpment Needed
Paper Peeps
Straws
Hoops
Marker
The objective of the game is to move the paper Peeps from the marker to the hoop using by sucking up on a straw. the winner is the person who moves all the Peeps to the hoop using only the straw in one minute.
All of these games can be easily adapted for any time of year or event and are lots of fun! The games are easily resources from items you may have around the home. | 969 | 4,238 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-35 | latest | en | 0.942831 |
https://www.teacherspayteachers.com/Product/Common-Core-First-Grade-Commutative-Associative-Properties-1OA3-3789492 | 1,532,135,197,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592001.81/warc/CC-MAIN-20180720232914-20180721012914-00124.warc.gz | 990,756,040 | 16,071 | # Common Core First Grade Commutative, Associative Properties 1.OA.3
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This file contains pages to practice the Commutative and Associative properties CC1.OA.B.3. These pages include 6 printable activity sheets to use as guided practice and independent work.
CCSS.Math.Content.1.OA.3 Apply properties of operations as strategies to add and subtract.2 Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Total Pages
9 pages
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Teaching Duration
Lifelong tool
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 255 | 1,008 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-30 | latest | en | 0.85323 |
http://www.docstoc.com/docs/74923477/Grade-4-Mental-Math-Strategies | 1,432,476,005,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928019.31/warc/CC-MAIN-20150521113208-00303-ip-10-180-206-219.ec2.internal.warc.gz | 435,751,862 | 39,940 | # Grade 4 Mental Math Strategies by gjjur4356
VIEWS: 278 PAGES: 3
• pg 1
``` Grade 4
Mental Math Strategies
Skip Counting: Children should review counting by 2’s, 5’s, 10’s, 25’s and 100’s.
to 10 000 and their corresponding subtractions (limited to 3 and 4-digit
numerals) by
• using personal strategies for adding and subtracting
• estimating sums and differences
• solving problems involving addition and subtraction
Some strategies to consider: using benchmarks, rounding, front-end addition and
subtraction (left-to-right calculations), and clustering of compatible numbers.
benchmarks: 207 – 126 would give an answer between 75 (200 - 125), and 85
(210 - 125)
rounding: 439 + 52 is approximately 440 + 50.
front-end: 138 + 245 = 370 (200 + 100 is 300, 30 + 40 is 70 for an estimate
of 370). Some students may include the ones in their estimate making their
front-end subtraction: 476 – 348 = 130 (400 – 300 is 100, 70 – 40 is 30, 6
and 8 are about the same so I’ll ignore them; my estimate is 130.
clustering: cluster the 29, 35, and 42 together to make 100.
compatibles: 225 + 68 + 75 = (225 + 75) + 68 = 368
Lessons 10, 11, 13, 15 to 18, and 21
SCO:N5: Describe and apply mental mathematics strategies, such as:
skip counting from a known fact
using doubling or halving
using doubling or halving and adding or subtracting one more group
using patterns in the 9s facts
using repeated doubling to determine basic multiplication facts to 9 × 9 and
related division facts.
doubling, e.g., for 4 3, think 2 3 = 6, so 4 3 = 6 + 6
doubling and adding one more group, e.g., for 3 7, think 2 7 = 14, and
14 + 7 = 21
use ten facts when multiplying by 9, e.g., for 9 6, think 10 6 = 60, and
60 – 6 = 54; for 7 9,
think 7 10 = 70, and 70 – 7 = 63
halving, e.g., if 4 6 is equal to 24, then 2 × 6 is equal to 12
relating division to multiplication, e.g., for 64 ÷ 8, think 8 = 64.
multiplying single-digit numbers by 10 and 100
SCO:N11: Demonstrate an understanding of addition and subtraction of
decimals (limited to hundredths) by:
• using compatible numbers
• estimating sums and differences
• using mental math strategies to solve problems.
Use estimation strategies including:
Compatible numbers: e.g., 0.72 + 0.23 are close to 0.75 and 0.25 which are
compatible numbers so the sum of the decimal numbers must be close to 1.
Front-end addition: e.g., 32.3 + 24.5 + 14.1; a student might think “30 + 20 +
10 is 60 and the ones and tenths clustered together make about another 10
for a total of 70.”
Front-end subtraction: e.g., 4.76 – 3.48; a student might think “4 ones – 3
ones is 1 and 7 tenths – 4 tenths is 3 tenths for a difference of
approximately 1 and 3 tenths.”
Rounding: e.g., 4.39 + 5.2 is approximately 4+ 5 for an estimate of 9.
Use mental computation strategies including:
Compatible numbers: e.g., 3.55 + 6.45 or \$3 and \$6 would be \$9 while 55
cents and 45 cents would make another dollar, for a sum of \$10 or 10.
Front-end strategy: e.g., 7.69 – 2. 45 A student might think “7 ones subtract
2 ones is 5 ones, 6 tenths subtract 4 tenths is 2 tenths and 9 hundredths
subtract 5 hundredths is 4 hundredths, so the difference would be 5.24”
Compensate: e.g., \$4.99 + 1.98 + 0.99 could be calculated by finding the sum
of 5 + 2 + 1 and then subtracting 0.04 or 4 cents. The sum would be \$7.96.
Counting on/ counting back: \$2 – 1.48; a student might think, “2 more pennies
would make \$1.50 and 50 cents more makes \$2 so the difference (change) is
52 cents
Renaming: think of 3.2 + 0.9 as 32 tenths + 9 tenths.
```
To top | 1,243 | 3,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2015-22 | longest | en | 0.859501 |
https://en.sorumatik.co/t/how-to-determine-when-easter-is-each-year/4891 | 1,723,298,079,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00144.warc.gz | 187,770,916 | 7,364 | # How to determine when easter is each year
how to determine when easter is each year
@LectureNotes is absolutely correct, determining the date of Easter each year can be a bit tricky as it is not a fixed date and varies annually. However, there is a formula used to calculate the date of Easter, known as the Computus. The Computus is based on a combination of astronomical and ecclesiastical calculations.
The date of Easter is determined as follows:
1. Find the vernal equinox: The date of the vernal equinox is the day when the sun crosses the celestial equator and day and night are of equal length. In the Gregorian calendar, this is usually around March 20th.
2. Find the full moon: The next step is to find the first full moon after the vernal equinox. However, it is not the astronomical full moon, but the ecclesiastical full moon, calculated using a slightly modified lunar cycle known as the Golden Number and the Epact. The Golden Number represents the year in a 19-year cycle, and the Epact represents the age of the moon on January 1st.
3. Determine the Sunday after the full moon: Easter is celebrated on the first Sunday after the ecclesiastical full moon. If the full moon falls on a Sunday, Easter is celebrated on the following Sunday.
4. Adjust for the paschal limit: The date of Easter should not be before March 22nd or later than April 25th. If the calculated date falls outside of this range, adjustments are made to bring it within the acceptable period.
While this may seem complex, it is primarily performed by religious authorities and through computer algorithms nowadays. Many online resources and calendars provide the date of Easter each year, making it easily accessible for everyone.
It’s important to note that different Christian churches may use slightly different calculations, resulting in different dates for Easter. The Western Christian tradition follows the Gregorian calendar, while the Eastern Orthodox Church uses the Julian calendar.
In conclusion, determining the date of Easter each year involves a combination of astronomical and ecclesiastical calculations, resulting in a variable date. It is best to consult reliable sources or calendars to find the exact date of Easter for a particular year. | 465 | 2,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-33 | latest | en | 0.942339 |
https://www.physicsforums.com/threads/differential-geoemtry-tangent-lines-parallel-proof.468936/ | 1,590,751,367,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00208.warc.gz | 853,196,996 | 15,950 | # Differential geoemtry tangent lines parallel proof
## Homework Statement
Prove that a(s) is a straight line if and only if its tangent lines are all parallel.
## Homework Equations
Frenet serret theorem
## The Attempt at a Solution
I'm confused on the direction "if the tangent lines are parallel then a(s) is a straight line".
Assume all the tangent lines of a(s) are parallel. So the tangent vector T is the same for all points xo on the curve a(s) and the values of T(s) of any two points on the curve are parallel. Thus T(s) is constant, and T'(s)=0 which implies that the curvature is zero, and thus a(s) must be a straight line.
Last edited:
Related Calculus and Beyond Homework Help News on Phys.org
lanedance
Homework Helper
so you know the direction fo the tanget vector at all times, eg.
T(t) = a.g(t)
where a is a constant vector and g is a scalar function. Think about integrating this to get the original curve and/or the effect on other parameters, curvature etc.
Is the way I did it incorrect?
Dick | 246 | 1,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-24 | longest | en | 0.875559 |
https://www.electriciansforums.net/threads/question-about-induction-motor.2927/ | 1,582,902,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147234.52/warc/CC-MAIN-20200228135132-20200228165132-00138.warc.gz | 696,651,777 | 19,354 | Discuss question about induction motor in the Industrial Electrician Talk area at ElectriciansForums.net
S
#### subhpoto
Welcome to ElectriciansForums.net - The American Electrical Advice Forum
How Current vary by increasing/ decreasing the number of poles of an induction motor?. in other word what is relation ship b/w number of poles and motor current.
This official sponsor may provide discounts for members
M
#### montybaber
I know that to calculate the synchronous rpm of a motor is
rpm = 120 x f/Np
to calculate the horsepower
HP = rpm x T (torque)/5252
and to calculate the current
I = HP x 746/V x Eff x PF
so by adding some values i have roughly worked out the following
with 415v(delta) efficiency of 90% and pf of 88% and torque of 4 ft/ib
(I1) 2 poles= 5.17A
4 poles= 2.593A
6 poles= 1.729A
8 poles= 1.296A
so as you can see that the current is inversely proportional to the number of poles, remember though you are also reducing the rpm with increase in poles
in this case the current after number of poles added (I2) =
(I2) = (I1) / (No of poles/2)
but then again what do i know...it could all just be bollox lol
Last edited by a moderator:
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Reply to question about induction motor in the Industrial Electrician Talk area at ElectriciansForums.net
This official sponsor may provide discounts for members | 347 | 1,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-10 | longest | en | 0.925051 |
https://www.thepunterspage.com/kickform/rayo-vallecano-vs-real-betis-balompie/ | 1,709,032,476,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474674.35/warc/CC-MAIN-20240227085429-20240227115429-00382.warc.gz | 1,032,073,105 | 22,274 | Your tip. Scientifically founded.
Calculated 100,000 times!
# Vallecano vs Real Betis Predictions, Odds and Betting Tips
## Vallecano vs Real Betis match prediction on the 17.03.2024.
17.03.2024 18:30
Not yet known
On the 17.03.2024 the match between Vallecano and Real Betis will take place at the "Estadio de Vallecas". We analysed the match with the help of our Football-Formula for you beforehand and calculated the probabilities for a win, a draw or a loss. Our prediction shows you furthermore the three most likely results. For you to predict or bet more successful, we also offer you statistics of the last matches, the bookmakers’ odds, a Form – Check of both teams as well as current news on either team. With the help of the scientifically sound KickForm prediction you can better bet on the match of Vallecano against Real Betis.
Prognosis / Trend
Probability for win, draw and loss:
34% Home
28% Draw
38% Away
The trend prognosis states the probability for a win, a draw or a loss from the home team’s perspective. At a glance it becomes visible whether there is a frontrunner or whether the teams face each other on equal footing.
Result / Top 3
The three most likely results according to our formula.:
1:1 13%
0:1 11%
1:0 11%
The reut prediction shows the three most likely end results. Additionally the probability of occurrence of this exact outcome is featured for each of the three results.
Over / Under
How many goals are scored in the game?
68%
32%
Over 1.5
Under 1.5
42%
58%
Over 2.5
Under 2.5
21%
79%
Over 3.5
Under 3.5
over/under display shows the probability that more or less than 1.5, 2.5 or 3.5 goals will be shot in regular playing time.
Both teams to score
The probability that both teams will score a goal.
48%
Yes
52%
No
The "Both teams to score"-Value indicates the probability that both teams will score a goal in regular time.
Current Table / 29. Match Day
Team P.
The last 5 matches
Rayo Vallecano
• L
• D
• L
• L
• L
Real Betis
• W
• D
• W
• D
• W
Clear graphic illustration of the last 5 matches of the respective team. Wins (W), draws (D) and losses (L) are displayed.
Last 5 matches Vallecano
• Girona 3 : 0 Vallecano
La Liga 2023/2024
26.02.2024
• Vallecano 1 : 1 Real Madrid
La Liga 2023/2024
18.02.2024
• Mallorca 2 : 1 Vallecano
La Liga 2023/2024
11.02.2024
• Vallecano 1 : 2 Sevilla
La Liga 2023/2024
05.02.2024
• Atletico 2 : 1 Vallecano
La Liga 2023/2024
31.01.2024
Last 5 matches Real Betis
• Real Betis 3 : 1 Bilbao
La Liga 2023/2024
25.02.2024
• Real Betis 0 : 0 Alaves
La Liga 2023/2024
18.02.2024
• Cadiz 0 : 2 Real Betis
La Liga 2023/2024
09.02.2024
• Real Betis 1 : 1 FC Getafe
La Liga 2023/2024
04.02.2024
• Mallorca 0 : 1 Real Betis
La Liga 2023/2024
27.01.2024
Last lineup RAY
Girona vs. Vallecano
• 1
Dimitrievs ...
• 5
Hernández
• 24
Lejeune
• 20
Balliu
• 12
Espino
• 17
López
• 18
García
• 23
Valentín
• 15
Crespo
• 22
Tomás
• 19
Frutos
Last lineup BET
4-2-3-1Real Betis vs. Bilbao
• 13
Silva
• 6
Pezzella
• 23
Sabaly
• 24
Ruibal
• 28
• 18
Fornals
• 4
Johnny
• 27
Altimira
• 12
José
• 9
Ávila
• 8
Fekir
News Vallecano
Form-Check Vallecano
In the last 5 matches there were 0 win(s), 1 draw(s) and 4 loss(es).
Most recently they lost with 0:3 against Girona in the away match, before that Vallecano played against Real Madrid . The match resulted in a draw.
Form-Check Real Betis
In the last 5 matches there were 3 win(s), 2 draw(s) and 0 loss(es).
Most recently they won with 3:1 against Bilbao in front of their home crowd, before that Real Betis played against Alaves . The match resulted in a draw.
News Real Betis
Spanish La Liga 23/24
20 Clubs - Round 26/38
# Player PositionPos Team Starts (subs)Starts (subs) GoalsGoals PensPens ShotsShots Shots on TargetSOT AssistsAssists
1
Willian Jose
Willian JoséReal Betis
FWD Real Betis Real Betis 16.0616 (6)7030173
2
Alvaro Garcia
Álvaro GarcíaRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 22.0122 (1)6025121
3
Isco
IscoReal Betis
MID Real Betis Real Betis 22.0022 (0)6137223
4
Ayoze Perez
Ayoze PérezReal Betis
FWD Real Betis Real Betis 19.0019 (0)4034141
5
Isi Palazon
Isi PalazónRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 24.0124 (1)3237241
6
Aitor Ruibal
Aitor RuibalReal Betis
FWD Real Betis Real Betis 8.068 (6)20721
7
Bebe
BebéRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 0.140 (14)21840
8
Assane Diao Diaoune
Assane Diao DiaouneReal Betis
FWD Real Betis Real Betis 13.0413 (4)201330
9
Marc Roca
Marc RocaReal Betis
MID Real Betis Real Betis 21.0121 (1)201472
10
Sergio Camello
Sergio CamelloRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 13.0913 (9)201581
11
Pablo Fornals
Pablo FornalsReal Betis
MID Real Betis Real Betis 3.013 (1)10210
12
Sergi Altimira Clavell
Sergi Altimira ClavellReal Betis
MID Real Betis Real Betis 6.066 (6)10210
13
Abdul Mumin
Abdul MuminRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 10.0110 (1)10210
14
Johnny
JohnnyReal Betis
MID Real Betis Real Betis 6.006 (0)10311
15
Chimy Avila
Chimy ÁvilaReal Betis
FWD Real Betis Real Betis 2.012 (1)10310
16
FWD Rayo Vallecano Rayo Vallecano 0.140 (14)11530
17
Guido Rodriguez
Guido RodríguezReal Betis
MID Real Betis Real Betis 13.0113 (1)10740
18
Randy Nteka
Randy NtekaRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 3.083 (8)11741
19
Kike Perez
Kike PérezRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 8.128 (12)10950
20
Oscar Valentin
Óscar ValentínRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 23.0123 (1)10960
21
Jorge De Frutos
Jorge De FrutosRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 5.195 (19)101160
22
German Pezzella
Germán PezzellaReal Betis
DEF Real Betis Real Betis 25.0025 (0)101230
23
Pathe Ciss
Pathé CissRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 11.0911 (9)101560
24
Unai Lopez
Unai LópezRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 19.0419 (4)101632
25
Abde Ezzalzouli
Abde EzzalzouliReal Betis
FWD Real Betis Real Betis 5.125 (12)101641
26
Raul de Tomas
Raúl de TomásRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 12.0712 (7)112192
27
Jose Angel Pozo
José Ángel PozoRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 0.010 (1)00000
28
Miguel Crespo
Miguel CrespoRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 1.031 (3)00000
29
Bellerin
BellerínReal Betis
DEF Real Betis Real Betis 17.0117 (1)00000
30
Salvi
SalviRayo Vallecano
FWD Rayo Vallecano Rayo Vallecano 0.010 (1)00000
31
Stole Dimitrievski
Stole DimitrievskiRayo Vallecano
GK Rayo Vallecano Rayo Vallecano 26.0026 (0)00000
32
Claudio Bravo
Claudio BravoReal Betis
GK Real Betis Real Betis 7.007 (0)00000
33
Diego Mendez Molero
Diego Méndez MoleroRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 0.010 (1)00000
34
Paul Akouokou
Paul AkouokouReal Betis
MID Real Betis Real Betis 0.020 (2)00000
35
Rui Silva
Rui SilvaReal Betis
GK Real Betis Real Betis 17.0017 (0)00000
36
Sokratis
SokratisReal Betis
DEF Real Betis Real Betis 7.017 (1)00110
37
Abner
AbnerReal Betis
DEF Real Betis Real Betis 13.0413 (4)00100
38
Cedric Bakambu
Cédric BakambuReal Betis
FWD Real Betis Real Betis 0.010 (1)00100
39
Bartra
BartraReal Betis
DEF Real Betis Real Betis 3.003 (0)00100
40
Youssouf Sabaly
Youssouf SabalyReal Betis
DEF Real Betis Real Betis 4.004 (0)00101
41
Andrei Ratiu
Andrei RatiuRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 4.024 (2)00110
42
MID Real Betis Real Betis 4.084 (8)00200
43
Luiz Felipe
Luiz FelipeReal Betis
DEF Real Betis Real Betis 4.004 (0)00200
44
Juan Cruz
Juan CruzReal Betis
FWD Real Betis Real Betis 0.050 (5)00200
45
Aridane
AridaneRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 16.0016 (0)00300
46
Pep Chavarria
Pep ChavarríaRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 5.075 (7)00401
47
William Carvalho
William CarvalhoReal Betis
MID Real Betis Real Betis 2.112 (11)00422
48
Borja Iglesias
Borja IglesiasReal Betis
FWD Real Betis Real Betis 4.074 (7)00501
49
Ivan Balliu
Ivan BalliuRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 22.0122 (1)00630
50
Oscar Trejo
Oscar TrejoRayo Vallecano
MID Rayo Vallecano Rayo Vallecano 12.1012 (10)00620
51
Juan Miranda
Juan MirandaReal Betis
DEF Real Betis Real Betis 12.0212 (2)00731
52
DEF Real Betis Real Betis 13.0213 (2)00730
53
Rodri
RodriReal Betis
MID Real Betis Real Betis 7.127 (12)00840
54
Alfonso Espino
Alfonso EspinoRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 24.0024 (0)001141
55
Nabil Fekir
Nabil FekirReal Betis
MID Real Betis Real Betis 5.045 (4)001142
56
Luiz Henrique Andre Rosa da Silva
Luiz Henrique André Rosa da SilvaReal Betis
FWD Real Betis Real Betis 6.086 (8)001243
57
Florian Lejeune
Florian LejeuneRayo Vallecano
DEF Rayo Vallecano Rayo Vallecano 26.0026 (0)0031101 | 3,217 | 8,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.871063 |
https://www.amathsteacherwrites.co.uk/radians/ | 1,659,986,527,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00797.warc.gz | 574,125,556 | 13,359 | A picture paints a thousand words
Most people and pupils* are pretty comfortable with using degrees to measure angles. Whilst 360o might seem an arbitrary number to split a complete rotation into, it makes sense to most.
But in A-Level maths (and beyond) the humble degree doesn’t always cut it, and students are introduced to Radians, often shortened to ‘Rads’, as a means of measuring angles.
The definition of a radian goes something like this:
The radian is the standard unit of angular measure, used in many areas of mathematics. An angle’s measurement in radians is numerically equal to the length of a corresponding arc of a unit circle, so one radian is just under 57.3 degrees (when the arc length is equal to the radius).
Which is all well and good, but is a bit dry and less than illuminating.
Which is why I was so delighted to stumble across the animation at the top of this post which quickly, visually and with the minmum of fuss explains what a radian is – in this case, a picture really is worth a thousand words.
The picture isn’t mine – it was created by LucasVB, you can read his blog post about radians here and you can see the gallery of the animations he has created for Wikipedia here. A talented bloke.
I wanted to post the pic in my blog for two reasons:
1. (Rather selfishly) I wanted a copy of the animation that I would always be able to find
2. Its such a great explanation of Radians that I figure it deserves the widest possible audience
So, if you find yourself teaching radians, or your son or daughter asks “What’s a rad” – show them the image above and all will become clear.
*Not a very good sentence as, of course, pupils are a subset of people.
1. Posted 08/09/2013 at 2:31 am | Permalink
I’m thinking of showing the animation first and then letting students write their own definition of a radian. Then compare and contrast with the textbook definition. Here’s another website the advertises 7 animations that will make you understand trigonometry. I found it very helpful as well. Thanks for your post.
• Posted 08/09/2013 at 9:39 am | Permalink
I think that’s a great idea: showing the animation first and then asking students to come up with a definition for a radian – crucially, it’ll will get them to think which will help them to understand and retain that understanding, even if they don’t come up with the exact definition of a radian in the first place.
I’m a great believer in not just spoon feeding pupils – let them explore, investigate, make mistakes; its the best way to learn.
Thanks for the link to the website with trigonometry animations – well worth a visit.
Let us know how you get on,
Thanks again,
A Maths Teacher | 606 | 2,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-33 | latest | en | 0.955313 |
http://blog.csdn.net/zhaohh/article/details/238190 | 1,519,019,323,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812405.3/warc/CC-MAIN-20180219052241-20180219072241-00351.warc.gz | 47,688,712 | 11,402 | # 常用的一些判断
dim c,words,word,i,wnum
function IsValiddomin(word)
IsValiddomin = true
words = Split(word, ".")
wnum=UBound(words)
if words(0)="www" then
IsValiddomin = IsValidword(words(1))
IsValiddomin = IsValidword2(words(2))
if words(wnum)="cn" then
if wnum<>3 then
IsValiddomin = false
exit function
end if
else
if wnum<>2 then
IsValiddomin = false
exit function
end if
end if
else
IsValiddomin = IsValidword(words(0))
IsValiddomin = IsValidword2(words(1))
if words(wnum)="cn" then
if wnum<>2 then
IsValiddomin = false
exit function
end if
else
if wnum<>1 then
IsValiddomin = false
exit function
end if
end if
end if
end function
function IsValidword2(word)
IsValidword2 = true
IsValidword2 = IsValidword(word)
if word<>"net" and word<>"com" and word<>"cc" and word<>"org" and word<>"info" and word<>"gov" then ' 自己添加
IsValidword2 = false
exit function
end if
end function
function IsValidword(word)
IsValidword = true
if Len(word) <= 0 then
IsValidword = false
exit function
end if
for i = 1 to Len(word)
c = Lcase(Mid(word, i, 1))
if InStr("abcdefghijklmnopqrstuvwxyz-", c) <= 0 and not IsNumeric(c) then
IsValidword = false
exit function
end if
next
end function
if IsValiddomin("wrclub.net.cn") then
response.write "right"
else
response.write "wrong"
end if
function nothaveChinese(para)
dim str
nothaveChinese=true
str=cstr(para)
for i = 1 to Len(para)
c=asc(mid(str,i,1))
if c<0 then
nothaveChinese=false
exit function
end if
next
end function
function isChinese(para)
on error resume next
dim str
dim i
if isNUll(para) then
isChinese=false
exit function
end if
str=cstr(para)
if trim(str)="" then
isChinese=false
exit function
end if
for i=1 to len(str)
c=asc(mid(str,i,1))
if c>=0 then
isChinese=false
exit function
end if
next
isChinese=true
if err.number<>0 then err.clear
end function
function IsValidEmail(email)
dim names, name, i, c
'Check for valid syntax in an email address.
IsValidEmail = true
names = Split(email, "@")
if UBound(names) <> 1 then
IsValidEmail = false
exit function
end if
for each name in names
if Len(name) <= 0 then
IsValidEmail = false
exit function
end if
for i = 1 to Len(name)
c = Lcase(Mid(name, i, 1))
if InStr("abcdefghijklmnopqrstuvwxyz_-.", c) <= 0 and not IsNumeric(c) then
IsValidEmail = false
exit function
end if
next
if Left(name, 1) = "." or Right(name, 1) = "." then
IsValidEmail = false
exit function
end if
next
if InStr(names(1), ".") <= 0 then
IsValidEmail = false
exit function
end if
i = Len(names(1)) - InStrRev(names(1), ".")
if i <> 2 and i <> 3 then
IsValidEmail = false
exit function
end if
if InStr(email, "..") > 0 then
IsValidEmail = false
end if
end function
function IsValidTel(para)
on error resume next
dim str
dim l,i
if isNUll(para) then
IsValidTel=false
exit function
end if
str=cstr(para)
if len(trim(str))<7 then
IsValidTel=false
exit function
end if
l=len(str)
for i=1 to l
if not (mid(str,i,1)>="0" and mid(str,i,1)<="9" or mid(str,i,1)="-") then
IsValidTel=false
exit function
end if
next
IsValidTel=true
if err.number<>0 then err.clear
end function
• 本文已收录于以下专栏：
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举报原因： 您举报文章：常用的一些判断 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 1,534 | 4,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-09 | latest | en | 0.199889 |
https://measuringstuff.com/11-things-that-are-10-meters-long/ | 1,716,835,294,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00177.warc.gz | 324,935,551 | 47,429 | 11 Things That Are 10 Meters Long
In this article, you will see 11 things that are 10 meters long.
Meters are often used when measuring longer items like baseball bats and rooms in a house.
1 meter is equal to 39 inches long, therefore 10 meters equals 390 inches long.
1 meter equals 3 feet 3 inches.
10 meters equals 32.5 feet.
Check out the following items that are 10 meters long.
Half of a tennis court
A regulation-sized tennis court is 23.77 meters long. Although 10 meters is not exactly half of a tennis court, it will give you a good idea of how long 10 meters is.
If you were to measure 10 meters from the backline of the court, you would be around 3 feet short of the net.
5 Full-sized sofas
Although a full-sized sofa is not a standard size and can vary in length, it’s common for a sofa to measure between 72 – 96 inches or 2 – 2.5 meters long.
Picturing 5 sofas of this size lined up together will give you an example of something that is 10 meters long.
Half of a tractor-trailer with cab
Most trailers being pulled by a semi-truck are 53 feet or 16 meters long.
If you include the cab of the truck, the length will be closer to 65 feet or 20 meters long.
If you are looking to find something that is 10 meters long, picture half of a tractor-trailer with the cab will be very close to that length depending on the type of cab.
10 Walking steps
Using adult walking steps is a great way to determine the length of something that is larger in size using meters.
Most adult steps will measure around 3 feet each which is right around 1 meter long.
10 adult walking steps will equal very close to 10 meters in length.
10 Adult walking steps = 10 meters
5 Queen size mattresses
The queen-sized bed is considered to be the most common bed size for many people.
The standard size for a queen-size mattress is 60 inches wide x 80 inches long.
This is equal to 5 feet wide x 6.67 feet long or 1.5 meters wide x 2 meters long.
Therefore if you can imagine 5 queen-sized mattresses lined up together in a row, they would equal around 10 meters long.
Telephone pole
Depending on where you are located, telephone poles can vary in height. In residential areas, they can range from 30 – 60 feet or 9 – 18 meters tall and along highways, they can reach 120 feet or 36 meters tall.
CHECK OUT How Long Is 40 Inches Compared To An Object?
Considering the size of a telephone pole in a residential area, 1 pole would equal close to 10 meters long.
3 Alligators
Many full-grown alligators will grow to about 10 – 13 feet or 3 – 4 meters in length and although it’s very rare to see an alligator over 12 feet long, it’s possible.
If you are looking to find something that is 10 meters long, consider the length of 3 full-grown alligators which will equal close to 32.5 feet or 10 meters.
11 Baseball bats
The length of a baseball bat can vary between 24 – 36 inches in length which is less than 1 meter. It’s common for a Major League baseball player to use a 34-inch or 0.86-meter long bat.
If you can picture 11 of these baseball bats placed together in a row, they would be very close to measuring 10 meters long.
6 Hockey sticks
Another piece of sports equipment that is helpful for measuring in meters is the hockey stick.
The length of a hockey stick can vary from person to person as not all players are the same height. But on average, a hockey stick will be around 5 feet or 1.5 meters long.
Placing 6 hockey sticks together lengthwise will equal close to 10 meters long and give you a good indication of something that is 10 meters in length.
4 Pieces of 2×4 lumber
A common length for a 2×4 board is 8 feet or 2.5 meters. Most hardware stores and lumber yards will have 2×4’s available for many different uses.
If you have 4 of these boards available, place them in a row and they will equal 32 feet long which is just about 10 meters long.
5 Garage doors
Many houses will have a garage and therefore have a garage door. It’s common for single and double car garages to have doors that are 7 feet tall.
Using these doors is a great item to use for measuring other items.
As most garage doors are 7 feet tall, if you imagine 5 of these doors placed together in a row, they would equal around 35 feet long which is just over 10 meters in length. | 1,033 | 4,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.950274 |
https://wikimili.com/en/Infinite_compositions_of_analytic_functions | 1,631,887,900,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00611.warc.gz | 637,276,052 | 37,735 | # Infinite compositions of analytic functions
Last updated
In mathematics, infinite compositions of analytic functions (ICAF) offer alternative formulations of analytic continued fractions, series, products and other infinite expansions, and the theory evolving from such compositions may shed light on the convergence/divergence of these expansions. Some functions can actually be expanded directly as infinite compositions. In addition, it is possible to use ICAF to evaluate solutions of fixed point equations involving infinite expansions. Complex dynamics offers another venue for iteration of systems of functions rather than a single function. For infinite compositions of a single function see Iterated function. For compositions of a finite number of functions, useful in fractal theory, see Iterated function system.
## Contents
Although the title of this article specifies analytic functions, there are results for more general functions of a complex variable as well.
## Notation
There are several notations describing infinite compositions, including the following:
Forward compositions:${\displaystyle F_{k,n}(z)=f_{k}\circ f_{k+1}\circ \dots \circ f_{n-1}\circ f_{n}(z).}$
Backward compositions:${\displaystyle G_{k,n}(z)=f_{n}\circ f_{n-1}\circ \dots \circ f_{k+1}\circ f_{k}(z).}$
In each case convergence is interpreted as the existence of the following limits:
${\displaystyle \lim _{n\to \infty }F_{1,n}(z),\qquad \lim _{n\to \infty }G_{1,n}(z).}$
For convenience, set Fn(z) = F1,n(z) and Gn(z) = G1,n(z).
One may also write ${\displaystyle F_{n}(z)={\underset {k=1}{\overset {n}{\mathop {R} }}}\,f_{k}(z)=f_{1}\circ f_{2}\circ \cdots \circ f_{n}(z)}$ and ${\displaystyle G_{n}(z)={\underset {k=1}{\overset {n}{\mathop {L} }}}\,g_{k}(z)=g_{n}\circ g_{n-1}\circ \cdots \circ g_{1}(z)}$
## Contraction theorem
Many results can be considered extensions of the following result:
Contraction Theorem for Analytic Functions [1] Let f be analytic in a simply-connected region S and continuous on the closure S of S. Suppose f(S) is a bounded set contained in S. Then for all z in S there exists an attractive fixed point α of f in S such that:
${\displaystyle F_{n}(z)=(f\circ f\circ \cdots \circ f)(z)\to \alpha .}$
## Infinite compositions of contractive functions
Let {fn} be a sequence of functions analytic on a simply-connected domain S. Suppose there exists a compact set Ω ⊂ S such that for each n, fn(S) ⊂ Ω.
Forward (inner or right) Compositions Theorem {Fn} converges uniformly on compact subsets of S to a constant function F(z) = λ. [2]
Backward (outer or left) Compositions Theorem {Gn} converges uniformly on compact subsets of S to γ ∈ Ω if and only if the sequence of fixed points {γn} of the {fn} converges to γ. [3]
Additional theory resulting from investigations based on these two theorems, particularly Forward Compositions Theorem, include location analysis for the limits obtained here . For a different approach to Backward Compositions Theorem, see .
Regarding Backward Compositions Theorem, the example f2n(z) = 1/2 and f2n−1(z) = −1/2 for S = {z : |z| < 1} demonstrates the inadequacy of simply requiring contraction into a compact subset, like Forward Compositions Theorem.
For functions not necessarily analytic the Lipschitz condition suffices:
Theorem [4] Suppose ${\displaystyle S}$ is a simply connected compact subset of ${\displaystyle \mathbb {C} }$ and let ${\displaystyle t_{n}:S\to S}$ be a family of functions that satisfies
${\displaystyle \forall n,\forall z_{1},z_{2}\in S,\exists \rho$ :\quad \left|t_{n}(z_{1})-t_{n}(z_{2})\right|\leq \rho |z_{1}-z_{2}|,\quad \rho <1.}
Define:
{\displaystyle {\begin{aligned}G_{n}(z)&=\left(t_{n}\circ t_{n-1}\circ \cdots \circ t_{1}\right)(z)\\F_{n}(z)&=\left(t_{1}\circ t_{2}\circ \cdots \circ t_{n}\right)(z)\end{aligned}}}
Then ${\displaystyle F_{n}(z)\to \beta \in S}$ uniformly on ${\displaystyle S.}$ If ${\displaystyle \alpha _{n}}$ is the unique fixed point of ${\displaystyle t_{n}}$ then ${\displaystyle G_{n}(z)\to \alpha }$ uniformly on ${\displaystyle S}$ if and only if ${\displaystyle |\alpha _{n}-\alpha |=\varepsilon _{n}\to 0}$.
## Infinite compositions of other functions
### Non-contractive complex functions
Results [5] involving entire functions include the following, as examples. Set
{\displaystyle {\begin{aligned}f_{n}(z)&=a_{n}z+c_{n,2}z^{2}+c_{n,3}z^{3}+\cdots \\\rho _{n}&=\sup _{r}\left\{\left|c_{n,r}\right|^{\frac {1}{r-1}}\right\}\end{aligned}}}
Then the following results hold:
Theorem E1 [6] If an ≡ 1,
${\displaystyle \sum _{n=1}^{\infty }\rho _{n}<\infty }$
then FnF is entire.
Theorem E2 [5] Set εn = |an−1| suppose there exists non-negative δn, M1, M2, R such that the following holds:
{\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }\varepsilon _{n}&<\infty ,\\\sum _{n=1}^{\infty }\delta _{n}&<\infty ,\\\prod _{n=1}^{\infty }(1+\delta _{n})&
Then Gn(z) → G(z) is analytic for |z| < R. Convergence is uniform on compact subsets of {z : |z| < R}.
Theorem GF3 [4] Suppose ${\displaystyle f_{n}(z)=z+\rho _{n}\varphi (z)}$ where there exist ${\displaystyle R,M>0}$ such that ${\displaystyle |z| implies ${\displaystyle |\varphi (z)| Furthermore, suppose ${\textstyle \rho _{k}\geq 0,\sum _{k=1}^{\infty }\rho _{k}<\infty }$ and ${\textstyle R>M\sum _{k=1}^{\infty }\rho _{k}.}$ Then for ${\textstyle R*
${\displaystyle G_{n}(z)\equiv \left(f_{n}\circ f_{n-1}\circ \cdots \circ f_{1}\right)(z)\to G(z)\qquad {\text{ for }}\{z:|z|
Theorem GF4 [4] Suppose ${\displaystyle f_{n}(z)=z+\rho _{n}\varphi (z)}$ where there exist ${\displaystyle R,M>0}$ such that ${\displaystyle |z| and ${\displaystyle |\zeta | implies ${\displaystyle |\varphi (z)| and ${\displaystyle |\varphi (z)-\varphi (\zeta )|\leq r|z-\zeta |.}$ Furthermore, suppose ${\textstyle \rho _{k}\geq 0,\sum _{k=1}^{\infty }\rho _{k}<\infty }$ and ${\textstyle R>M\sum _{k=1}^{\infty }\rho _{k}.}$ Then for ${\textstyle R*
${\displaystyle F_{n}(z)\equiv \left(f_{1}\circ f_{2}\circ \cdots \circ f_{n}\right)(z)\to F(z)\qquad {\text{ for }}\{z:|z|
Theorem GF5 [5] Let ${\displaystyle f_{n}(z)=z+zg_{n}(z),}$ analytic for |z| < R0, with |gn(z)|Cβn,
${\displaystyle \sum _{n=1}^{\infty }\beta _{n}<\infty .}$
Choose 0 < r < R0 and define
${\displaystyle R=R(r)={\frac {R_{0}-r}{\prod _{n=1}^{\infty }(1+C\beta _{n})}}.}$
Then FnF uniformly for
Example GF1: ${\displaystyle F_{40}(x+iy)={\underset {k=1}{\overset {40}{\mathop {R} }}}\left({\frac {x+iy}{1+{\tfrac {1}{4^{k}}}(x\cos(y)+iy\sin(x))}}\right),\qquad [-20,20]}$
Example GF2: ${\displaystyle G_{40}(x+iy)={\underset {k=1}{\overset {40}{\mathop {L} }}}\,\left({\frac {x+iy}{1+{\tfrac {1}{2^{k}}}(x\cos(y)+iy\sin(x))}}\right),\qquad [-20,20]}$
### Linear fractional transformations
Results [5] for compositions of linear fractional (Möbius) transformations include the following, as examples:
Theorem LFT1 On the set of convergence of a sequence {Fn} of non-singular LFTs, the limit function is either:
1. a non-singular LFT,
2. a function taking on two distinct values, or
3. a constant.
In (a), the sequence converges everywhere in the extended plane. In (b), the sequence converges either everywhere, and to the same value everywhere except at one point, or it converges at only two points. Case (c) can occur with every possible set of convergence. [7]
Theorem LFT2 [8] If {Fn} converges to an LFT, then fn converge to the identity function f(z) = z.
Theorem LFT3 [9] If fnf and all functions are hyperbolic or loxodromic Möbius transformations, then Fn(z) → λ, a constant, for all ${\textstyle z\neq \beta =\lim _{n\to \infty }\beta _{n}}$, where {βn} are the repulsive fixed points of the {fn}.
Theorem LFT4 [10] If fnf where f is parabolic with fixed point γ. Let the fixed-points of the {fn} be {γn} and {βn}. If
${\displaystyle \sum _{n=1}^{\infty }\left|\gamma _{n}-\beta _{n}\right|<\infty \quad {\text{and}}\quad \sum _{n=1}^{\infty }n\left|\beta _{n+1}-\beta _{n}\right|<\infty }$
then Fn(z) → λ, a constant in the extended complex plane, for all z.
## Examples and applications
### Continued fractions
The value of the infinite continued fraction
${\displaystyle {\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+\cdots }}}}}$
may be expressed as the limit of the sequence {Fn(0)} where
${\displaystyle f_{n}(z)={\frac {a_{n}}{b_{n}+z}}.}$
As a simple example, a well-known result (Worpitsky Circle* [11] ) follows from an application of Theorem (A):
Consider the continued fraction
${\displaystyle {\cfrac {a_{1}\zeta }{1+{\cfrac {a_{2}\zeta }{1+\cdots }}}}}$
with
${\displaystyle f_{n}(z)={\frac {a_{n}\zeta }{1+z}}.}$
Stipulate that |ζ| < 1 and |z| < R < 1. Then for 0 < r < 1,
${\displaystyle |a_{n}|, analytic for |z| < 1. Set R = 1/2.
Example.${\displaystyle F(z)={\frac {(i-1)z}{1+i+z{\text{ }}+}}{\text{ }}{\frac {(2-i)z}{1+2i+z{\text{ }}+}}{\text{ }}{\frac {(3-i)z}{1+3i+z{\text{ }}+}}\cdots ,}$${\displaystyle [-15,15]}$
Example. [5] A fixed-point continued fraction form (a single variable).
${\displaystyle f_{k,n}(z)={\frac {\alpha _{k,n}\beta _{k,n}}{\alpha _{k,n}+\beta _{k,n}-z}},\alpha _{k,n}=\alpha _{k,n}(z),\beta _{k,n}=\beta _{k,n}(z),F_{n}(z)=\left(f_{1,n}\circ \cdots \circ f_{n,n}\right)(z)}$
${\displaystyle \alpha _{k,n}=x\cos(ty)+iy\sin(tx),\beta _{k,n}=\cos(ty)+i\sin(tx),t=k/n}$
### Direct functional expansion
Examples illustrating the conversion of a function directly into a composition follow:
Example 1. [6] [12] Suppose ${\displaystyle \phi }$ is an entire function satisfying the following conditions:
${\displaystyle {\begin{cases}\phi (tz)=t\left(\phi (z)+\phi (z)^{2}\right)&|t|>1\\\phi (0)=0\\\phi '(0)=1\end{cases}}}$
Then
${\displaystyle f_{n}(z)=z+{\frac {z^{2}}{t^{n}}}\Longrightarrow F_{n}(z)\to \phi (z)}$.
Example 2. [6]
${\displaystyle f_{n}(z)=z+{\frac {z^{2}}{2^{n}}}\Longrightarrow F_{n}(z)\to {\frac {1}{2}}\left(e^{2z}-1\right)}$
Example 3. [5]
${\displaystyle f_{n}(z)={\frac {z}{1-{\tfrac {z^{2}}{4^{n}}}}}\Longrightarrow F_{n}(z)\to \tan(z)}$
Example 4. [5]
${\displaystyle g_{n}(z)={\frac {2\cdot 4^{n}}{z}}\left({\sqrt {1+{\frac {z^{2}}{4^{n}}}}}-1\right)\Longrightarrow G_{n}(z)\to \arctan(z)}$
### Calculation of fixed-points
Theorem (B) can be applied to determine the fixed-points of functions defined by infinite expansions or certain integrals. The following examples illustrate the process:
Example FP1. [3] For |ζ| ≤ 1 let
${\displaystyle G(\zeta )={\frac {\tfrac {e^{\zeta }}{4}}{3+\zeta +{\cfrac {\tfrac {e^{\zeta }}{8}}{3+\zeta +{\cfrac {\tfrac {e^{\zeta }}{12}}{3+\zeta +\cdots }}}}}}}$
To find α = G(α), first we define:
{\displaystyle {\begin{aligned}t_{n}(z)&={\cfrac {\tfrac {e^{\zeta }}{4n}}{3+\zeta +z}}\\f_{n}(\zeta )&=t_{1}\circ t_{2}\circ \cdots \circ t_{n}(0)\end{aligned}}}
Then calculate ${\displaystyle G_{n}(\zeta )=f_{n}\circ \cdots \circ f_{1}(\zeta )}$ with ζ = 1, which gives: α = 0.087118118... to ten decimal places after ten iterations.
Theorem FP2 [5] Let φ(ζ, t) be analytic in S = {z : |z| < R} for all t in [0, 1] and continuous in t. Set
${\displaystyle f_{n}(\zeta )={\frac {1}{n}}\sum _{k=1}^{n}\varphi \left(\zeta ,{\tfrac {k}{n}}\right).}$
If |φ(ζ, t)|r < R for ζS and t ∈ [0, 1], then
${\displaystyle \zeta =\int _{0}^{1}\varphi (\zeta ,t)\,dt}$
has a unique solution, α in S, with ${\displaystyle \lim _{n\to \infty }G_{n}(\zeta )=\alpha .}$
### Evolution functions
Consider a time interval, normalized to I = [0, 1]. ICAFs can be constructed to describe continuous motion of a point, z, over the interval, but in such a way that at each "instant" the motion is virtually zero (see Zeno's Arrow): For the interval divided into n equal subintervals, 1 ≤ kn set ${\displaystyle g_{k,n}(z)=z+\varphi _{k,n}(z)}$ analytic or simply continuous – in a domain S, such that
${\displaystyle \lim _{n\to \infty }\varphi _{k,n}(z)=0}$ for all k and all z in S,
and ${\displaystyle g_{k,n}(z)\in S}$.
#### Principal example [5]
{\displaystyle {\begin{aligned}g_{k,n}(z)&=z+{\frac {1}{n}}\phi \left(z,{\tfrac {k}{n}}\right)\\G_{k,n}(z)&=\left(g_{k,n}\circ g_{k-1,n}\circ \cdots \circ g_{1,n}\right)(z)\\G_{n}(z)&=G_{n,n}(z)\end{aligned}}}
implies
${\displaystyle \lambda _{n}(z)\doteq G_{n}(z)-z={\frac {1}{n}}\sum _{k=1}^{n}\phi \left(G_{k-1,n}(z){\tfrac {k}{n}}\right)\doteq {\frac {1}{n}}\sum _{k=1}^{n}\psi \left(z,{\tfrac {k}{n}}\right)\sim \int _{0}^{1}\psi (z,t)\,dt,}$
where the integral is well-defined if ${\displaystyle {\tfrac {dz}{dt}}=\phi (z,t)}$ has a closed-form solution z(t). Then
${\displaystyle \lambda _{n}(z_{0})\approx \int _{0}^{1}\phi (z(t),t)\,dt.}$
Otherwise, the integrand is poorly defined although the value of the integral is easily computed. In this case one might call the integral a "virtual" integral.
Example.${\displaystyle \phi (z,t)={\frac {2t-\cos y}{1-\sin x\cos y}}+i{\frac {1-2t\sin x}{1-\sin x\cos y}},\int _{0}^{1}\psi (z,t)\,dt}$
Example. [13] Let:
${\displaystyle g_{n}(z)=z+{\frac {c_{n}}{n}}\phi (z),\quad {\text{with}}\quad f(z)=z+\phi (z).}$
Next, set ${\displaystyle T_{1,n}(z)=g_{n}(z),T_{k,n}(z)=g_{n}(T_{k-1,n}(z)),}$ and Tn(z) = Tn,n(z). Let
${\displaystyle T(z)=\lim _{n\to \infty }T_{n}(z)}$
when that limit exists. The sequence {Tn(z)} defines contours γ = γ(cn, z) that follow the flow of the vector field f(z). If there exists an attractive fixed point α, meaning |f(z) − α| ≤ ρ|z − α| for 0 ≤ ρ < 1, then Tn(z) → T(z) ≡ α along γ = γ(cn, z), provided (for example) ${\displaystyle c_{n}={\sqrt {n}}}$. If cnc > 0, then Tn(z) → T(z), a point on the contour γ = γ(c, z). It is easily seen that
${\displaystyle \oint _{\gamma }\phi (\zeta )\,d\zeta =\lim _{n\to \infty }{\frac {c}{n}}\sum _{k=1}^{n}\phi ^{2}\left(T_{k-1,n}(z)\right)}$
and
${\displaystyle L(\gamma (z))=\lim _{n\to \infty }{\frac {c}{n}}\sum _{k=1}^{n}\left|\phi \left(T_{k-1,n}(z)\right)\right|,}$
when these limits exist.
These concepts are marginally related to active contour theory in image processing, and are simple generalizations of the Euler method
### Self-replicating expansions
#### Series
The series defined recursively by fn(z) = z + gn(z) have the property that the nth term is predicated on the sum of the first n 1 terms. In order to employ theorem (GF3) it is necessary to show boundedness in the following sense: If each fn is defined for |z| < M then |Gn(z)| < M must follow before |fn(z) z| = |gn(z)| ≤ n is defined for iterative purposes. This is because ${\displaystyle g_{n}(G_{n-1}(z))}$ occurs throughout the expansion. The restriction
${\displaystyle |z|0}$
serves this purpose. Then Gn(z) → G(z) uniformly on the restricted domain.
Example (S1). Set
${\displaystyle f_{n}(z)=z+{\frac {1}{\rho n^{2}}}{\sqrt {z}},\qquad \rho >{\sqrt {\frac {\pi }{6}}}}$
and M = ρ2. Then R = ρ2 − (π/6) > 0. Then, if ${\displaystyle S=\left\{z:|z|0\right\}}$, z in S implies |Gn(z)| < M and theorem (GF3) applies, so that
{\displaystyle {\begin{aligned}G_{n}(z)&=z+g_{1}(z)+g_{2}(G_{1}(z))+g_{3}(G_{2}(z))+\cdots +g_{n}(G_{n-1}(z))\\&=z+{\frac {1}{\rho \cdot 1^{2}}}{\sqrt {z}}+{\frac {1}{\rho \cdot 2^{2}}}{\sqrt {G_{1}(z)}}+{\frac {1}{\rho \cdot 3^{2}}}{\sqrt {G_{2}(z)}}+\cdots +{\frac {1}{\rho \cdot n^{2}}}{\sqrt {G_{n-1}(z)}}\end{aligned}}}
converges absolutely, hence is convergent.
Example (S2): ${\displaystyle f_{n}(z)=z+{\frac {1}{n^{2}}}\cdot \varphi (z),\varphi (z)=2\cos(x/y)+i2\sin(x/y),>G_{n}(z)=f_{n}\circ f_{n-1}\circ \cdots \circ f_{1}(z),\qquad [-10,10],n=50}$
#### Products
The product defined recursively by
${\displaystyle f_{n}(z)=z(1+g_{n}(z)),\qquad |z|\leqslant M,}$
has the appearance
${\displaystyle G_{n}(z)=z\prod _{k=1}^{n}\left(1+g_{k}\left(G_{k-1}(z)\right)\right).}$
In order to apply Theorem GF3 it is required that:
${\displaystyle \left|zg_{n}(z)\right|\leq C\beta _{n},\qquad \sum _{k=1}^{\infty }\beta _{k}<\infty .}$
Once again, a boundedness condition must support
${\displaystyle \left|G_{n-1}(z)g_{n}(G_{n-1}(z))\right|\leq C\beta _{n}.}$
If one knows n in advance, the following will suffice:
${\displaystyle |z|\leqslant R={\frac {M}{P}}\qquad {\text{where}}\quad P=\prod _{n=1}^{\infty }\left(1+C\beta _{n}\right).}$
Then Gn(z) → G(z) uniformly on the restricted domain.
Example (P1). Suppose ${\displaystyle f_{n}(z)=z(1+g_{n}(z))}$ with ${\displaystyle g_{n}(z)={\tfrac {z^{2}}{n^{3}}},}$ observing after a few preliminary computations, that |z| ≤ 1/4 implies |Gn(z)| < 0.27. Then
${\displaystyle \left|G_{n}(z){\frac {G_{n}(z)^{2}}{n^{3}}}\right|<(0.02){\frac {1}{n^{3}}}=C\beta _{n}}$
and
${\displaystyle G_{n}(z)=z\prod _{k=1}^{n-1}\left(1+{\frac {G_{k}(z)^{2}}{n^{3}}}\right)}$
converges uniformly.
Example (P2).
${\displaystyle g_{k,n}(z)=z\left(1+{\frac {1}{n}}\varphi \left(z,{\tfrac {k}{n}}\right)\right),}$
${\displaystyle G_{n,n}(z)=\left(g_{n,n}\circ g_{n-1,n}\circ \cdots \circ g_{1,n}\right)(z)=z\prod _{k=1}^{n}(1+P_{k,n}(z)),}$
${\displaystyle P_{k,n}(z)={\frac {1}{n}}\varphi \left(G_{k-1,n}(z),{\tfrac {k}{n}}\right),}$
${\displaystyle \prod _{k=1}^{n-1}\left(1+P_{k,n}(z)\right)=1+P_{1,n}(z)+P_{2,n}(z)+\cdots +P_{k-1,n}(z)+R_{n}(z)\sim \int _{0}^{1}\pi (z,t)\,dt+1+R_{n}(z),}$
${\displaystyle \varphi (z)=x\cos(y)+iy\sin(x),\int _{0}^{1}(z\pi (z,t)-1)\,dt,\qquad [-15,15]:}$
#### Continued fractions
Example (CF1): A self-generating continued fraction. [5]
{\displaystyle {\begin{aligned}F_{n}(z)&={\frac {\rho (z)}{\delta _{1}+}}{\frac {\rho (F_{1}(z))}{\delta _{2}+}}{\frac {\rho (F_{2}(z))}{\delta _{3}+}}\cdots {\frac {\rho (F_{n-1}(z))}{\delta _{n}}},\\\rho (z)&={\frac {\cos(y)}{\cos(y)+\sin(x)}}+i{\frac {\sin(x)}{\cos(y)+\sin(x)}},\qquad [0
Example (CF2): Best described as a self-generating reverse Euler continued fraction. [5]
${\displaystyle G_{n}(z)={\frac {\rho (G_{n-1}(z))}{1+\rho (G_{n-1}(z))-}}\ {\frac {\rho (G_{n-2}(z))}{1+\rho (G_{n-2}(z))-}}\cdots {\frac {\rho (G_{1}(z))}{1+\rho (G_{1}(z))-}}\ {\frac {\rho (z)}{1+\rho (z)-z}},}$
${\displaystyle \rho (z)=\rho (x+iy)=x\cos(y)+iy\sin(x),\qquad [-15,15],n=30}$
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## References
1. P. Henrici, Applied and Computational Complex Analysis, Vol. 1 (Wiley, 1974)
2. L. Lorentzen, Compositions of contractions, J. Comp & Appl Math. 32 (1990)
3. J. Gill, The use of the sequence ${\displaystyle F_{n}(z)=f_{n}\circ \dots \circ f_{1}(z)}$ in computing the fixed points of continued fractions, products, and series, Appl. Numer. Math. 8 (1991)
4. J. Gill, A Primer on the Elementary Theory of Infinite Compositions of Complex Functions, Comm. Anal. Th. Cont. Frac., Vol XXIII (2017) and researchgate.net
5. J. Gill, John Gill Mathematics Notes, researchgate.net
6. S.Kojima, Convergence of infinite compositions of entire functions, arXiv:1009.2833v1
7. G. Piranian & W. Thron,Convergence properties of sequences of Linear fractional transformations, Mich. Math. J.,Vol. 4 (1957)
8. J. DePree & W. Thron,On sequences of Mobius transformations, Math. Z., Vol. 80 (1962)
9. A. Magnus & M. Mandell, On convergence of sequences of linear fractional transformations,Math. Z. 115 (1970)
10. J. Gill, Infinite compositions of Mobius transformations, Trans. Amer. Math. Soc., Vol176 (1973)
11. L. Lorentzen, H. Waadeland, Continued Fractions with Applications, North Holland (1992)
12. N. Steinmetz, Rational Iteration, Walter de Gruyter, Berlin (1993)
13. J. Gill, Informal Notes: Zeno contours, parametric forms, & integrals, Comm. Anal. Th. Cont. Frac., Vol XX (2014) | 8,367 | 26,417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 105, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-39 | latest | en | 0.820441 |
https://www.physicsforums.com/threads/basis-proof.642014/#post-4105147 | 1,639,044,250,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00161.warc.gz | 1,008,360,563 | 14,998 | Basis Proof
I attached the problem and its solution.
I was looking at this solution and got a little confused. Why did they say that "Assume that S = {v1, v2, · · · , vn} is a basis for V and c is a nonzero scalar. Let S1 = {cv1, cv2, · · · , cvn}. Since S is a basis for V , V has dimension n. Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S is independent."
That seems redundant. The problem statement already stated that S is a basis, which means that the vectors in S are linearly independent.
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Dick
Homework Helper
I attached the problem and its solution.
I was looking at this solution and got a little confused. Why did they say that "Assume that S = {v1, v2, · · · , vn} is a basis for V and c is a nonzero scalar. Let S1 = {cv1, cv2, · · · , cvn}. Since S is a basis for V , V has dimension n. Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S is independent."
That seems redundant. The problem statement already stated that S is a basis, which means that the vectors in S are linearly independent.
It's clearly a minor typo. They meant to say, "Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S1 is independent."
It's clearly a minor typo. They meant to say, "Since S1 has n vectors, it suffices (by Part 1 of Theorem 4.12) to prove that S1 is independent."
Oh okay. So by saying that, wouldn't that have proved S1 is a basis? The next part (writing out all the linear combinations) just proved that it's linearly independent again. Do you need to do that?
Dick | 464 | 1,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-49 | latest | en | 0.976106 |
https://socratic.org/questions/what-is-the-final-concentration-of-50-0-ml-of-a-2-00m-solution-diluted-to-500-0- | 1,576,279,476,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540569332.20/warc/CC-MAIN-20191213230200-20191214014200-00179.warc.gz | 501,278,238 | 6,180 | # What is the final concentration of 50.0 mL of a 2.00M solution, diluted to 500.0 mL?
$\text{Concentration"="Moles of stuff"/"Volume of solution}$ $=$ $0.20 \cdot m o l \cdot {L}^{-} 1$
$\frac{50.0 \times {10}^{- 3} \cancel{L} \times 2.00 \cdot m o l \cdot \cancel{{L}^{-} 1}}{500.0 \times {10}^{-} 3 \cdot L}$ $=$ ??mol*L^-1
Note that the answer is reasonable. If you read the question you have diluted the starting solution from $50.0 \cdot m L$ to $500.0 \cdot m L$, a tenfold dilution, so the final solution shoould be $10 \times$ as dilute. | 196 | 547 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-51 | longest | en | 0.714385 |
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# What is the least common multiple of 26 and 55?
Updated: 9/16/2023
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14y ago
The LCM of 26 and 55 is 1430.
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The Least Common Multiple of 42 and 55 is 2,310. | 425 | 1,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2024-10 | latest | en | 0.927615 |
https://ift6266h17.wordpress.com/2017/03/26/q2-basics-of-neural-networks/ | 1,501,209,232,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436321.71/warc/CC-MAIN-20170728022449-20170728042449-00434.warc.gz | 657,014,749 | 36,239 | # Q2 – Basics of Neural Networks
Contributed by Philippe Lacaille.
Under a given training dataset $\mathit{D}=\{(x^{(1)}, y^{(1)}), (x^{(2)}, y^{(2)}), \dots, (x^{(n)}, y^{(n)}) \}$, where $x^{(i)} \in \mathbb{R}^d$ and $y^{(i)} \in \{ 0, 1 \}$. Let’s further consider a
standard feedforward network with a hidden layer $h$. For the rest of the exercise, we denote $h^a$ as the pre-activated layer and $h^s$ as the post-activated layer.
1. Assuming the network’s goal is to do binary classification (with the
detailed structure above), what would be an appropriate:
1. Activation function for the output layer $f(x)$? What does the output represent under this activation function
2. Loss function, represented by $L(f(x, \theta), y)$, that will be used in training? (You can write it as a function of $f(x, \theta)$).
2. Now let’s assume the following about the network, while still using the same dataset:
• Network goal is now to do multi-class classification $y^{(i)} \in \{ 1, 2, 3 \}$
• The set of parameters $\theta = \{ W^{(1)}, b^{(1)}, W^{(2)}, b^{(2)} \}$
• $x^{(i)}$ has 10 features and there are 15 units at the hidden layer $h$.
1. What would be an appropriate activation function for the output layer $f(x)$ under this structure? What does the output vector represent under this activation function?
2. What are the number of dimensions for each of the parameters in $\theta$?
3. Using gradient descent, what is the update function for the parameters $\theta$ (no regularization)? You can write it as a function of $L(f(x, \theta), y)$.
1. if using batch gradient descent?
2. if using mini-batch (size $m$) gradient descent?
3. if using stochastic gradient descent?
3. Under a classification problem, why isn’t the network trained to directly minimize the classification error?
## 5 thoughts on “Q2 – Basics of Neural Networks”
1.
a) For binary classification, we should use a sigmoid activation, so that the output is between 0 and 1. It can be interpreted as a posterior probability in the binary case $p(y=1|x)$
b) BCE (binary cross entropy) would be an appropriate loss.
$L(f(x,\theta),y) = -y * log(f(x,\theta)) - (1-y) * log(1 - f(x,\theta))$
2.
a) For multiclass classification, we should use a softmax activation. It can be interpreted as the posterior probability of choosing one class $p(y=c|x)$
b)
— W(1) is of dimension (10, 15)
— b(1) is of dimension (15)
— W(2) is of dimension (15, 3)
— b(2) is of dimension (3)
c) $\theta = \theta - \alpha \Delta$
For batch gradient descent, $\Delta = \frac{1}{n} \sum_{i=1}^n \nabla_{\theta} L(f(x^{(i)}, \theta), y^{(i)})$
For mini-batch gradient descent, $\Delta = \frac{1}{m} \sum_{i=1}^m \nabla_{\theta} L(f(x^{(i)}, \theta), y^{(i)})$
For stochastic gradient descent, $\Delta = \nabla_{\theta} L(f(x^{(i)}, \theta), y^{(i)})$
3. Minimizing classification error would be ideal, but the function is non-smooth, so we need to use a surrogate loss instead.
Liked by 1 person
• Stéphanie Larocque says:
3. Furthermore, there is no gradient with classification error.
Liked by 2 people
2. In my answer of item 3 I just said that the classification error is not a differentiable function.
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• Stéphanie Larocque says:
Classification error is a sum of 0/1 loss for each data point so it’s differentiable at every point except at the points where the 0 jumps to 1. So I think this is not the main problem (like for relu, I think that if it’s not differentiable for only a countable number of points, that’s not “really” a problem since we most likely won’t fall on those points). I think the problem is more because the gradient is 0 everywhere (except where 0 jumps to 1) so we can’t learn. Do you think it’s correct?
Liked by 3 people
• I think so! I should have mentioned this aspect in my answer. Thanks.
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# Math Question - The Penny Challenge
barney15c | 05:00 Mon 23rd Jan 2023 | Quizzes & Puzzles
Money saving Expert has a penny challenge where you save 1p on the first day and increment it by 1 p a day for a full year so 1p,2p 3p up to £3.65 for the last day, by that time you would save £665.95 .
Say you had 7 people who decided to club together and each person was to pay in once a week in rotation what would be the fairest way of doing it - obviously if you kept the same sequence the last person would be putting in more money than the 1st over the course of time - i was thinking that once the 7th person paid they also pay the 1st payment of the next week that way it would even out as the 1st person would pay in the most the nextweek, then you would reverse it again the next week, it would mean person 4 will always retain that position, would that work or is there a better way.
I was thinking of doing this for 7 friends going on a holiday and having a bit of a pot to start the holiday with for going out.
1 to 5 of 5
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Question Author
Sorry £667.95
Interesting idea, and definitely more affordable to have multiple people involved. I'd personally just go for bunging in a set amount each week. I know that I'd be able to manage for the first few months, however after the 6-month point it would start to get a lot more difficult.
I'd be interested to know how it goes.
Seems a faff for around £90 each. You could just put £13 each in the pot once a month.
// £90 each. You could just put £13 each in the pot once a month //
Problem - Over the year each person will have contributed £156 which is 73.3% more.
Bigger pot of £1092, little hardship, no problem
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# Physics Homework
0
321
7
I have a physic homework on concave and convex lens and how the light refracts when it goes through it. And I'm stuck on the part where I have to find the height of the object. The only information on the question is that it has the magnification factor of 2.3x and that the object is 3m away from the lens. Thank you so much if you can tell me how to solve this. I need to draw a diagram on it so I need it asap!!
physics
Guest Jan 4, 2016
### Best Answer
#7
+8826
+5
Now it is correct!
Omi67 Jan 4, 2016
Sort:
### 7+0 Answers
#1
0
dont do it
Guest Jan 4, 2016
#2
+8826
+5
Hallo guest, I hope is is ok so?
Omi67 Jan 4, 2016
#3
+8826
+5
Look here
file:///C:/Users/Christel/AppData/Local/Temp/phet-geometric-optics/geometric-optics_de.html
Omi67 Jan 4, 2016
#4
+8826
+5
This is the right side, sorry
https://phet.colorado.edu/de/simulation/legacy/geometric-optics
Omi67 Jan 4, 2016
#5
+91045
0
That looks great Omi 67
Melody Jan 4, 2016
edited by Melody Jan 4, 2016
#6
+26329
+5
You have found the distance of the image Omi, but the question asks for the height of the object. There isn't enough information in the question to determine this (other than it will be 2.3 times smaller than the image!).
Alan Jan 4, 2016
#7
+8826
+5
Best Answer
Now it is correct!
Omi67 Jan 4, 2016
### 5 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 520 | 1,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-47 | latest | en | 0.851769 |
http://www.carrawaydavie.com/how-many-liters-in-a-ml/ | 1,558,255,478,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232254731.5/warc/CC-MAIN-20190519081519-20190519103519-00452.warc.gz | 236,023,844 | 8,726 | ## What is a gallon how many liters in a ml?
Gallon is a royal as well as United States normal measurement system volume device. how many liters in a ml? There is one kind of gallon in the imperial system and also 2 kinds (fluid and completely dry) in the US customary dimension system. 1 US liquid gallon is specified as 231 cubic inches, 1 US dry gallon is 268.8 cubic inches and 1 royal gallon is 277.4 cubic inches. The abbreviation is gall. how many liters in a ml?
Gallons Conversion:
1 Gallon = 4 Quarts
1 Gallon = 8 Pints
1 Gallon = 16 Cups
1 Gallon = 256 Tablespoons
1 Gallon = 768 Teaspoons
how many liters in a ml
## What is a liquid ounce (fl oz)how many liters in a ml?
A liquid ounce is an imperial as well as US customary measurement system volume unit. 1 US fluid ounce equals to 29.5735 mL and 1 royal (UK) fluid ounce equals to 28.4131 mL. The acronym is fl oz. So, how many liters in a ml?
There are 8 fluid ounces in a United States mug and also 10 imperial liquid ounces in an imperial cup.
The list for your referral:
1 United States Fluid Gallon = 128 US Fluid Ounces
1 US Dry Gallon = 148.946 United States Fluid Ounces
1 Imperial Gallon (UK) = 160 Imperial Fluid Ounces (UK).
how many liters in a ml
## Liquid Ounce how many liters in a ml
Liquid Ounce is utilized for volume, Ounce for mass, and also they are various.
As an example, 1 liquid ounce of honey has a mass of about 1,5 ounces!
But also for water, 1 liquid ounce has a mass of about 1 ounce.
how many liters in a ml
If you imply an ounce of fluid says fluid ounce (fl oz).
In Summary:.
1 gallon = 4 quarts = 8 pints = 16 mugs = 128 fluid ounces.
how many liters in a ml | 460 | 1,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-22 | latest | en | 0.909273 |
https://www.in2013dollars.com/UK-inflation-rate-in-1773 | 1,585,843,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00390.warc.gz | 963,995,673 | 20,720 | # U.K. inflation rate in 1773: 0.00%
## UK Inflation Calculator
£
### Inflation in 1773 and Its Effect on Pound Value
Purchasing power decreased by 0.00% in 1773 compared to 1772. On average, you would have to spend 0.00% more money in 1773 than in 1772 for the same item.
In other words, £1 in 1772 is equivalent in purchasing power to about £1 in 1773.
The 1772 inflation rate was 10.45%. The inflation rate in 1773 was 0.00%. The 1773 inflation rate is lower compared to the average inflation rate of 2.06% per year between 1773 and 2020.
Inflation rate is calculated by change in the composite price index (CPI). The CPI in 1773 was 7.40. It was 7.40 in the previous year, 1772. The difference in CPI between the years is used by the Office for National Statistics to officially determine inflation.
Average inflation rate 0.00% Converted amount (£1 base) £1 Price difference (£1 base) £0.00 CPI in 1772 7.400 CPI in 1773 7.400 Inflation in 1772 10.45% Inflation in 1773 0.00%
GBP Inflation since 1750
Annual Rate, the Office for National Statistics CPI
### How to Calculate Inflation Rate for £1, 1772 to 1773
This inflation calculator uses the following inflation rate formula:
CPI in 1773CPI in 1772
×
1772 GBP value
=
1773 GBP value
Then plug in historical CPI values. The U.K. CPI was 7.4 in the year 1772 and 7.4 in 1773:
7.47.4
×
£1
=
£1
£1 in 1772 has the same "purchasing power" or "buying power" as £1 in 1773.
To get the total inflation rate for the 1 years between 1772 and 1773, we use the following formula:
CPI in 1773 - CPI in 1772CPI in 1772
×
100
=
Cumulative inflation rate (1 years)
Plugging in the values to this equation, we get:
7.4 - 7.47.4
×
100
=
0%
### Data Source & Citation
Raw data for these calculations comes from the composite price index published by the UK Office for National Statistics (ONS). A composite index is created by combining price data from several different published sources, both official and unofficial. The Consumer Price Index, normally used to compute inflation, has only been tracked since 1988. All inflation calculations after 1988 use the Office for National Statistics' Consumer Price Index, except for 2017, which is based on The Bank of England's forecast.
You may use the following MLA citation for this page: “Inflation Rate in 1773 | UK Inflation Calculator.” Official Inflation Data, Alioth Finance, 27 Mar. 2020, https://www.officialdata.org/UK-inflation-rate-in-1773. | 698 | 2,461 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-16 | longest | en | 0.91553 |
http://iridl.ldeo.columbia.edu/SOURCES/.LEVITUS94/.MONTHLY/.temp/%5BZ+T+%5Ddetrend-bfl/X+integral/%5BY+Z+T+%5Ddetrend-bfl/X+differences/%5BX+T+%5Ddetrend-bfl/%5BY+%5Drmsover/X+partial/%5BT+%5Dminover/Z+partial/Z+integral/doc+unitsnote.html | 1,582,749,879,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146485.15/warc/CC-MAIN-20200226181001-20200226211001-00068.warc.gz | 75,143,864 | 2,801 | Convert units of \$integral dZ\$ \$partialdiff sub Z\$ min \$partialdiff sub X\$ root mean sq detrended-bfl detrended-bfl \$integral dX\$ detrended-bfl [ LEVITUS94 MONTHLY temp ] from degree_Celsius to
# Units in the Data Library
This Data Library makes some effort to interpret and manipulate the units given for each variable. The code is based on the udunits package, though we had to make a number of changes in order to arrive at something suitable for our needs. Some of this documentation is taken from the udunits documentation.
A unit is the amount by which a physical quantity is measured. For example:
Physical QuantityPossible Unit
timeweeks
distancecentimeters
powerwatts
For the most part, units are purely multiplicitive, i.e. if we multiply a velocity in m/s times the time in s, we end up with a distance in m. This technique of dimensional analysis is a powerful constraint in checking scientific calculations.
Some units (also called scales) also have an implicit origin or base value. For example, zero on the Celsius scale corresponds to 273.15 on the Kelvin scale. So if we want to convert a Celsius temperature to a Kelvin temperature, we have to add 273.15. On the other hand, if we have a Celsius temperature anomaly (i.e. deviation from its normal value), it is already a Kelvin temperature anomaly, and requires no conversion. If we remove the mean, for example, from a temperature, it then loses its origin and becomes a 'anomaly' unit.
Unfortunately, we use the word Celsius or Kelvin to refer both to the scale and the anomaly unit. To help diminish the impact of this ambiguity, we have chosen the convention to refer to temperature units as Celsius_scale or Kelvin_scale, while the temperature anomaly equivalent is degree_Celsius or degree_Kelvin. (The plain names are considered to be scales.)
Note that scales (i.e. units with origins) do not really work properly in dimensional analysis, a reflection of the fact that in most cases a reference value should be removed before such a quantity is manipulated.
To create a units grammer that is readily manipulated by machine, we follow the convention that units are separated by spaces, / denotes division, and m2 corresponds to meters squared while m2 s-1 would be meters squared per second. The origin mentioned earlier is denoted by above, i.e. Celsius is defined as degree_Kelvin above 273.15. @, from, and since are all synonyms for above. | 531 | 2,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.851631 |
http://mathhelpforum.com/math-software/61836-ode45-matlab.html | 1,480,734,232,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540804.14/warc/CC-MAIN-20161202170900-00459-ip-10-31-129-80.ec2.internal.warc.gz | 169,411,116 | 12,584 | 1. ## ode45 Matlab
Hello everyone!
I need to solve a IVP, for example
$\dot{x} = x + 2\dot{y} + y$
$\dot{y} = y - 2\dot{x} + 5x$
x(0) = 0.995
y(0) = 0
(I actually don't know if there is a solution...)
But how to solve it by using ode45? $\dot{x}$ depends on $\dot{y}$ and y ...
By the way,
I know how to solve
$\dot{v}(t) = 10 - 5/2*v^2(t), t\in [0, 10]$
v0 = v(0) = 0
for example:
[ t , v ] = ode45 (@( t , v ) 10-5v^2/2 , [0, 10], v0 ) ;
----
How to solve the IVP with 2 equations?
Best regards,
Rapha
2. Originally Posted by Rapha
Hello everyone!
I need to solve a IVP, for example
$\dot{x} = x + 2\dot{y} + y$
$\dot{y} = y - 2\dot{x} + 5x$
x(0) = 0.995
y(0) = 0
(I actually don't know if there is a solution...)
But how to solve it by using ode45? $\dot{x}$ depends on $\dot{y}$ and y ...
By the way,
I know how to solve
$\dot{v}(t) = 10 - 5/2*v^2(t), t\in [0, 10]$
v0 = v(0) = 0
for example:
[ t , v ] = ode45 (@( t , v ) 10-5v^2/2 , [0, 10], v0 ) ;
----
How to solve the IVP with 2 equations?
Best regards,
Rapha
Eliminate the $\dot{y}$ term in the first equation by substituting from the second, and $\dot{x}$ for the second by substituting from the first:
$\dot{x}=\frac{11x+2y}{5}$
$\dot{y}=\frac{-2x-y}{5}$
So now writing this as a first order vector ODE:
$
\dot{\bold{x}}=\left[\begin{array}{cc}
11/5 & 2/5\\
-2/5 & -1/5
\end{array}\right] {\bold{x}}
$
where:
$
{\bold{x}}=
\left[\begin{array}{c}
x\\
y
\end{array}\right]
$
So now can you get ode45 to work?
CB
3. Hello.
So now can you get ode45 to work?
that was very helpful, thank you very much (and yes, i could get ode45 to work).
But actually i kinda screwed it(sorry, i did not exactly realize what my problem was and asked for something different), because my problem is :
$\ddot{x} = x + 2 \dot{y} - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$
$\ddot{y} = y - 2 \dot{x} - 0.9 \frac{y}{D_1} - 0.1 \frac{y}{D_2}$
with $D_1 = \sqrt{(((x+0.1)^2+y^2)^3)}$
$D_2 = \sqrt{(((x-0.9)^2+y^2)^3)}$
i tried to find the IVP 1. order:
$x_1 = x$
$\dot{x_1} = \dot{x} = x_2$
$y_1 = y$
$y_2 = \dot{y_1} = \dot{y}$
=>
$\dot{x_2} = x_1 + 2 y_2 - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$
$\dot{y_2} = y_1 - 2 x_2 - 0.9 \frac{y_1}{D_1} - 0.1 \frac{y_1}{D_2}$
I did not realize that there are 4 functions: $y_2(t), y_1(t), x_1(t), x_2(t)$
I still want to solve this by using ode45.
Regards
Rapha
4. Originally Posted by Rapha
Hello.
that was very helpful, thank you very much (and yes, i could get ode45 to work).
But actually i kinda screwed it(sorry, i did not exactly realize what my problem was and asked for something different), because my problem is :
$\ddot{x} = x + 2 \dot{y} - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$
$\ddot{y} = y - 2 \dot{x} - 0.9 \frac{y}{D_1} - 0.1 \frac{y}{D_2}$
with $D_1 = \sqrt{(((x+0.1)^2+y^2)^3)}$
$D_2 = \sqrt{(((x-0.9)^2+y^2)^3)}$
i tried to find the IVP 1. order:
$x_1 = x$
$\dot{x_1} = \dot{x} = x_2$
$y_1 = y$
$y_2 = \dot{y_1} = \dot{y}$
=>
$\dot{x_2} = x_1 + 2 y_2 - 0.9 \frac{x+0.1}{D_1} - 0.1 \frac{x-0.9}{D_2}$
$\dot{y_2} = y_1 - 2 x_2 - 0.9 \frac{y_1}{D_1} - 0.1 \frac{y_1}{D_2}$
I did not realize that there are 4 functions: $y_2(t), y_1(t), x_1(t), x_2(t)$
I still want to solve this by using ode45.
Regards
Rapha
Now we have as state vector:
${\bold{x}}=\left[ \begin{array}{c}x\\y\\ \dot{x}\\ \dot{y} \end{array} \right]$
and derivative:
$
\dot{\bold{x}}=\left[ \begin{array}{c}
\bold{x}_3\\
\bold{x}_4\\
\bold{x}_1-2\bold{x}_4-0.9(\bold{x}_1+0.1)/D_1-0.1(\bold{x}_1-0.9)/D_2 \\
\bold{x}_2-2\bold{x}_3-0.9\bold{x}_2/D_1-0.1\bold{x}_2/D_2
\end{array} \right]
$
where:
$D_1 = \sqrt{(((\bold{x}_1+0.1)^2+\bold{x}_2^2)^3)}$
$D_2 = \sqrt{(((\bold{x}_1-0.9)^2+\bold{x}_2^2)^3)}$
You will need to write a Matlab function to evaluate the derivative and pass that to ode45 (the derivative is now too complicated to be easily passed as an anonymous function).
(You cound simplify the derivative by collecting terms)
CB
5. Hello CaptainBlack!
Originally Posted by CaptainBlack
You will need to write a Matlab function to evaluate the derivative and pass that to ode45 (the derivative is now too complicated to be easily passed as an anonymous function).
(You cound simplify the derivative by collecting terms)
I tried it but it still doesn't work.
Code:
function dx = ivp(t, x)
dx = \dot{x} * x;
clearly represented, I wrote $\dot{x}$ for
$
\dot{\bold{x}}=\left[ \begin{array}{c}
\bold{x}_3\\
\bold{x}_4\\
\bold{x}_1-2\bold{x}_4-0.9(\bold{x}_1+0.1)/D_1-0.1(\bold{x}_1-0.9)/D_2 \\
\bold{x}_2-2\bold{x}_3-0.9\bold{x}_2/D_1-0.1\bold{x}_2/D_2
\end{array} \right]
$
then:
[t, x] = ode45(@ivp, [0, 10], [??,??,??,??] )
where $t \in [0, 10]$ and ?? = [x_3(0); x_4(0);, ...(0); ...(0)]
this is stupid, but I don't know the initial values :-(
6. Originally Posted by Rapha
Hello CaptainBlack!
I tried it but it still doesn't work.
Code:
function dx = ivp(t, x)
dx = \dot{x} * x;
clearly represented, I wrote $\dot{x}$ for
$
\dot{\bold{x}}=\left[ \begin{array}{c}
\bold{x}_3\\
\bold{x}_4\\
\bold{x}_1-2\bold{x}_4-0.9(\bold{x}_1+0.1)/D_1-0.1(\bold{x}_1-0.9)/D_2 \\
\bold{x}_2-2\bold{x}_3-0.9\bold{x}_2/D_1-0.1\bold{x}_2/D_2
\end{array} \right]
$
then:
[t, x] = ode45(@ivp, [0, 10], [??,??,??,??] )
where $t \in [0, 10]$ and ?? = [x_3(0); x_4(0);, ...(0); ...(0)]
this is stupid, but I don't know the initial values :-(
Somewere in the staement of the problem there should be initial values.
But you can try any initial values to test the code [0;0;1;1] should do.
the following seems to work on FreeMat:
Code:
function dx=deriv(t,x)
d1=sqrt(((x(1)+0.1)^2+x(2)^2)^3);
d2=sqrt(((x(1)-0.9)^2+x(2)^2)^3);
dx=zeros(4,1);
dx(1)=x(3);
dx(2)=x(4);
dx(3)=x(1)-2*x(4)-0.9*(x(1)+0.1)/d1-0.1*(x(1)-0.9)/d2;
dx(4)=x(2)-2*x(3)-0.9*x(2)/d1-0.1*x(2)/d2;
with calling statement:
Code:
[t,x]=ode45(@deriv ,[0,10],[0;0;1;1]);
(note the absurdty that we have state and derivative as colum vectors, but the output is an array of row vectors, one for each time point, but it does not work if we have row vectors for the state and derivative!!)
CB
7. Originally Posted by CaptainBlack
Somewere in the staement of the problem there should be initial values.
Oooops
I forgot to post them :-(
Originally Posted by CaptainBlack
the following seems to work on FreeMat:
Code:
function dx=deriv(t,x)
d1=sqrt(((x(1)+0.1)^2+x(2)^2)^3);
d2=sqrt(((x(1)-0.9)^2+x(2)^2)^3);
dx=zeros(4,1);
dx(1)=x(3);
dx(2)=x(4);
dx(3)=x(1)-2*x(4)-0.9*(x(1)+0.1)/d1-0.1*(x(1)-0.9)/d2;
dx(4)=x(2)-2*x(3)-0.9*x(2)/d1-0.1*x(2)/d2;
with calling statement:
Alright, thank you. You are great!
Rapha | 2,774 | 6,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 48, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2016-50 | longest | en | 0.826979 |
https://stonespounds.com/365-5-stones-in-stones-and-pounds | 1,643,362,986,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00697.warc.gz | 608,226,629 | 4,909 | # 365.5 stones in stones and pounds
## Result
365.5 stones equals 365 stones and 7 pounds
You can also convert 365.5 stones to pounds.
## Converter
Three hundred sixty-five point five stones is equal to three hundred sixty-five stones and seven pounds (365.5st = 365st 7lb). | 74 | 279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2022-05 | latest | en | 0.839581 |
http://slideplayer.com/slide/5740155/ | 1,715,992,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057216.39/warc/CC-MAIN-20240517233122-20240518023122-00761.warc.gz | 25,706,073 | 17,942 | # Graph linear inequalities with one variable EXAMPLE 2 Graph ( a ) y < – 3 and ( b ) x < 2 in a coordinate plane. Test the point (0,0). Because (0,0) is.
## Presentation on theme: "Graph linear inequalities with one variable EXAMPLE 2 Graph ( a ) y < – 3 and ( b ) x < 2 in a coordinate plane. Test the point (0,0). Because (0,0) is."— Presentation transcript:
Graph linear inequalities with one variable EXAMPLE 2 Graph ( a ) y < – 3 and ( b ) x < 2 in a coordinate plane. Test the point (0,0). Because (0,0) is not a solution of the inequality, shade the half- plane that does not contain (0,0). a. Graph the boundary line y = – 3. Use a solid line because the inequality symbol is <.
Graph linear inequalities with one variable EXAMPLE 2 b. Graph the boundary line x = 2.Use a dashed line because the inequality symbol is <. Test the point (0,0). Because (0,0) is a solution of the inequality, shade the half- plane that does not contains (0,0).
Graph linear inequalities with two variables EXAMPLE 3 Graph (a) y > – 2x and (b) 5x – 2y ≤ – 4 in a coordinate plane. a. Graph the boundary line y = – 2x. Use a dashed line because the inequality symbol is >. Test the point (1,1). Because (1,1) is a solution of the inequality, shade the half- plane that contains (1,1).
Graph linear inequalities with two variables EXAMPLE 3 Test the point (0,0). Because (0,0) is not a solution of the inequality, shade the half-plane that does not contain (0,0). b. Graph the boundary line 5x –2y = –4.Use a solid line because the inequality symbol is <.
GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 5. y > –1 SOLUTION Test the point (0,0). Because (0,0) is a solution of the inequality, shade the half- plane that contains (0,0). Graph the boundary line y = – 1.Use a solid line because the inequality symbol is >.
GUIDED PRACTICE for Examples 2 and 3
GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 6. x > –4 SOLUTION Graph the boundary line x = – 4.Use a dashed line because the inequality symbol is >. Test the point (0,0). Because (0,0) is a solution of the inequality, shade the half- plane that contains (0,0).
GUIDED PRACTICE for Examples 2 and 3
GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 7. y > –3x SOLUTION Graph the boundary line y = – 3x.Use a dashed line because the inequality symbol is >. Test the point (1,1). Because (1,1) is a solution of the inequality, shade the half- plane that contains (1,1).
GUIDED PRACTICE for Examples 2 and 3
GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 8. y < 2x +3 SOLUTION Test the point (0,0). Because (0,0) is a solution of the inequality, shade the half-plane that contains (0,0). Graph the boundary line y = 2x + 3.Use a solid line because the inequality symbol is <.
GUIDED PRACTICE for Examples 2 and 3
GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 9. x + 3y < 9 SOLUTION Test the point (0,0). Because (0,0) is a solution of the inequality, shade the half-plane that contains (0,0). Graph the boundary line x + 2y = 9.Use a solid line because the inequality symbol is <.
GUIDED PRACTICE for Examples 2 and 3
Graph the boundary line 2x – 6y = 9.Use a solid line because the inequality symbol is >. GUIDED PRACTICE for Examples 2 and 3 Graph the inequality in a coordinate plane. 10. 2x – 6y > 9 SOLUTION Test the point (0,0). Because (0,0) is not a solution of the inequality, shade the half- plane that does not contains (0,0).
GUIDED PRACTICE for Examples 2 and 3
Download ppt "Graph linear inequalities with one variable EXAMPLE 2 Graph ( a ) y < – 3 and ( b ) x < 2 in a coordinate plane. Test the point (0,0). Because (0,0) is."
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# PPSC Lecturers Commerce Preparation Materials at the one place 1Nov2020
Respected Members,
Here is some data for the preparation. Hope it will help you to be prepare for the exam of Punjab Public Service Commission ...
Agr ap k pass b koch data hay to plz share kr dijiyee ga,, ta kay baKi Logon ka b faida ho saky or tiyari achi ho saky..
PPSC special for Lecturer Commerce
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something more
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ammm,,, or koch yeah
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ammm,,, or koch yeah
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hmmm ook
okkkkkkk
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esko 6 mints agy kr k tiyari krain :P ,,
Price Elasticity of Demand=P€d = Percentage change in Quantity Demanded/ Percentage change in Price
Price Elasticity of supply=P€s = Percentage change in Quantity Supplied/ Percentage change in Price
Income Elasticity of Demand: Price Elasticity of Supply: Y€d = Percentage change in Quantity demanded/ Percentage change in Income
Cross-Price Elasticity of Demand: Pb€da = Percentage change in Demand for good a/ Percentage change in Price of good b
The quadratic demand function: Qd = 60 – 15P + P2
The formula of elasticity = (dQ / dP) (P/Q)
MARGINAL RATE OF SUBSTITUTION= MRS= dY = MUX
dX MUY
Profit=TR-TC
Average physical product =APP = TPPF/QF
Marginal physical product= ΔTPPF/ΔQF or MPPx=MPPL
PX PL
MARGINAL RATE OF TECHNICAL SUBSTITUTION= MRTS = Δ K/ Δ L
Total Cost (TC)= FC + VC
Average variable cost (AVC) = TVC/Q
Average fixed cost (AFC)= TFC/Q
MARGINAL COST (MC)= ΔTC/ΔQ
Total revenue=TR = P x Q.
Average revenue= AR = TR/Q
Marginal revenue= ΔTR/ΔQ
Slope of AR = dAR / dQ
Slope of MR = dMR / dQ
Net present value = PV – Purchase cost
Value added or GDP: Value of transaction
GDP = Sum of the value added by each of the firms
Growth rate of per capita income = Growth rate of total output - Growth rate of population
GDP Deflator = Nominal GDP / Real GDP
NNP = GNP – Depreciation allowance
GNP = GDP + Net factor incomes from abroad
NDP = GDP – Depreciation allowance
GDP at factor cost = GDP at market price – Indirect taxes
Per capita income= national income/ Total population of that country
NNP = GNP – Depreciation allowance
Real GDP year a = Nominal GDP year a X (Price Index base year / Price Index year a)
Growth rate in nominal GDP = Nominal GDP in year2- Nominal GDP in year1
AD = C + I + G + (X-M)
AD = AS, C + S + T = C + I + G + X – M
Y = AD = AS
Income = Expenditure = Output
Withdrawal = Injection= S + T + M = I + G + X
Marginal Propensity to Consume (MPC)= 1 – MPS or ΔC / ΔYd
Average propensity to consume (APC)= 1 – APS or C / Yd
Marginal propensity to save= MPS = 1 – MPC or MPS = ΔS / ΔYd
Average propensity to save= APS = 1 – APC or S / Yd
Real exchange rate= C = RER = PF x NER
PD
MULTIPLIER=k=1/(1-MPC)
uedu ky notes hen kya??????????
+++Shamp y RANA Girl++ .. nahi mera subject wo nahi too,, oskay collect nahi kiyee,,
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https://www.airmilescalculator.com/distance/wju-to-cjj/ | 1,611,673,344,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704800238.80/warc/CC-MAIN-20210126135838-20210126165838-00696.warc.gz | 649,774,817 | 11,493 | # Distance between Wonju (WJU) and Cheongju (CJJ)
Flight distance from Wonju to Cheongju (Wonju Airport – Cheongju International Airport) is 56 miles / 90 kilometers / 49 nautical miles. Estimated flight time is 36 minutes.
Driving distance from Wonju (WJU) to Cheongju (CJJ) is 81 miles / 130 kilometers and travel time by car is about 1 hour 35 minutes.
## Map of flight path and driving directions from Wonju to Cheongju.
Shortest flight path between Wonju Airport (WJU) and Cheongju International Airport (CJJ).
## How far is Cheongju from Wonju?
There are several ways to calculate distances between Wonju and Cheongju. Here are two common methods:
Vincenty's formula (applied above)
• 55.895 miles
• 89.954 kilometers
• 48.571 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 55.954 miles
• 90.050 kilometers
• 48.623 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Wonju Airport
City: Wonju
Country: South Korea
IATA Code: WJU
ICAO Code: RKNW
Coordinates: 37°26′17″N, 127°57′35″E
B Cheongju International Airport
City: Cheongju
Country: South Korea
IATA Code: CJJ
ICAO Code: RKTU
Coordinates: 36°42′59″N, 127°29′56″E
## Time difference and current local times
There is no time difference between Wonju and Cheongju.
KST
KST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 33 kg (73 pounds).
## Frequent Flyer Miles Calculator
Wonju (WJU) → Cheongju (CJJ).
Distance:
56
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
56
Round trip? | 495 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.830022 |
https://www.edplace.com/worksheet_info/maths/keystage2/year5/topic/269/12652/can-you-calculate-the-missing-measurements | 1,718,717,416,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861752.43/warc/CC-MAIN-20240618105506-20240618135506-00623.warc.gz | 651,695,559 | 12,368 | # Work Out the Missing Measurements in Composite Shapes
In this worksheet, students will work out the missing measurements in composite rectangles.
Key stage: KS 2
Curriculum topic: Measurement
Curriculum subtopic: Measure Shapes in Metric
Difficulty level:
#### Worksheet Overview
In this activity, we're going to be using information provided to work out missing measurements.
Things we need to know:
The perimeter of a shape is the total distance around the outside.
A composite/compound rectangle is any shape that is made up of two or more rectangles (including squares).
We can usually use the information provided to work out that which is not provided.
Opposite lengths of a rectangle are equal.
Here is a composite rectangle where one of the measurements is missing.
We know that opposite sides of a rectangle are equal, so we can see that the 4 cm + ? must equal 10 cm.
10 cm - 4 cm = 6 cm
The missing measure is 6 cm.
Top Tips:
Check the unit of measure and remember to include it in your answer.
If asked to work out the perimeter, check the number of sides each shape has and make sure you have included them all.
Some shapes might have two measures missing!
Now it's your turn to have a go.
Good luck!
### What is EdPlace?
We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them.
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### Drone safety lesson: How close, how far?
This lesson provides an authentic context to develop skills of estimation and measuring length. It provides an opportunity for students to connect decimal representations to the metric system and convert from centimetres to metres, and metres to kilometres. It also provides a context to investigate and become familiar with ...
### Rates and ratios: Year 8 – planning tool
This planning resource for Year 8 is for the topic of Rates and ratios. Students recognise and use rates to solve problems including, constant rates, rate of pay, cost per kilogram, recipes, simple interest and average rates. Students will also be required to make comparisons between two related quantities of different ...
### Modelling climate changes
There is a saying: 'climate is what you expect and weather is what you get'. |Understanding climate change is very difficult for most people, especially when the weather we experience is different from the information we are given by scientists about the climate changing. The difference is that weather reflects short-term ...
### Catalyst: Measuring our coastline
How long is the Australian coastline? See Dr Derek Muller and Simon Pampena discussing the perimeter of the Australian coastline. Find out how the accuracy of that measurement depends on the length of the 'measuring stick' used. They discuss how a coastline is much like a fractal such as 'Koch's Snowflake'!
### Catalyst: Small scale measurements
What units of measurements do we use to describe incredibly small things like blood cells and atoms? Watch as you are taken on a journey to explain the different units of measurement that we use to describe the very small.
### Volume and its units of measurement
This is a website designed for both teachers and students that refers to the study of volume and capacity and their units. It contains material on finding the volume of rectangular prisms, the units of volume and capacity, and understanding the connection between volume and capacity. There are pages for both teachers and ...
### Sites2See – measurement for primary
Selected links to a range of interactive and print resources for Measurement topics in K-6 Mathematics.
### Numeracy wrap: The long and the short
In this resource students measure objects of different length in centimetres and millimetres, order lengths from shortest to longest, convert between millimetres, centimetres, metres and kilometres. | 506 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-38 | latest | en | 0.932028 |
http://mymathforum.com/algebra/43034-factor.html | 1,563,721,536,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527048.80/warc/CC-MAIN-20190721144008-20190721170008-00504.warc.gz | 103,814,910 | 9,350 | My Math Forum factor
Algebra Pre-Algebra and Basic Algebra Math Forum
April 19th, 2014, 09:48 PM #1 Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8 factor factor x^12-x^4-x^3+1
April 20th, 2014, 10:46 AM #2 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Divide by x-1
April 21st, 2014, 05:42 PM #3 Newbie Joined: Apr 2014 From: South Florida Posts: 8 Thanks: 1 $\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
April 21st, 2014, 08:08 PM #4
Math Team
Joined: Oct 2011
Posts: 14,597
Thanks: 1038
Quote:
Originally Posted by scomora $\displaystyle x(x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5-x^2-x-1)$
Shouldn't that be:
$\displaystyle (x-1)(x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4-x^2-x-1)$
?
April 22nd, 2014, 07:26 PM #5 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 > factor x^12 - x^4 - x^3 + 1 Re-write: (x^12 - x^4) - (x^3 - 1) Using geometric series: u = (x^12 - x^4) / (x - 1) = x^4 + x^5 .... + x^11 v = (x^3 - x^0) / (x - 1) = x^0 + x^1 + x^2 So factoring results: (x - 1)(u - v) = (x - 1)(x^11 + x^10 ... + x^4 - x^2 - x^1 - x^0) Is that a valid "way"? Never seen it before...was just playing around... Thanks from agentredlum and Olinguito
April 23rd, 2014, 04:11 AM #6 Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 I like it and i've never seen it before either Denis.
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Contact - Home - Forums - Cryptocurrency Forum - Top | 728 | 1,836 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-30 | longest | en | 0.8228 |
https://forex-station.com/viewtopic.php?p=1295523663 | 1,721,225,790,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514771.72/warc/CC-MAIN-20240717120911-20240717150911-00855.warc.gz | 218,411,314 | 16,455 | ## Moving Average indicators for MT4
### Re: Moving Average indicators for MT4
3121
VWAP Candlesticks Oscillator
This is a vwap candles indicator using Mladen's average of the candles Keltner channel idea.
### Re: Moving Average indicators for MT4
3122
mrtools wrote: Fri Sep 15, 2023 1:42 am
If you're interested in using VWAP - albeit in different form - there is an interesting book I came across (just out):
Maximum Trading Gains with Anchored VWAP: The Perfect Combination of Price, Time, and Volume by Brian Shannon
### Re: Moving Average indicators for MT4
3123
Regularized Relative Moving Average
This is regularized relative moving average, on the chart picture is the regularized ema (in blue and orange) for comparison.
What is a Regularized Moving Average?
Chris Satchwell's "Regularized" Moving Average is a faster responding MA which adds just enough smoothing to it's calculation to avoid the introduction of lag by using "Lambda".
Mathematically, it's seen as a superior, fitting replacement over the standard Moving Average and Exponential Moving Average.
What makes it different to any other Moving Average?
Traditionally, analysts will reduce the length (period) of an indicator's settings in order to combat lag, but by introducing "Regularization" to the code equips it with a mathematical factor called Lamda (also written as λ-calculus) that provides control over the moving average's untimely "twitch" by adjusting it's Lambda period.
Note: Adjust the Lambda setting on smaller intraday timeframes by increasing it to 6.0 or higher. When doing so, be sure to take note when backtesting see how false signals are filtered out.
### Re: Moving Average indicators for MT4
3124
mrtools wrote: Tue Sep 19, 2023 6:08 am This is regularized relative moving average, on the chart picture is the regularized ema (in blue and orange) for comparison.
Heyhey man I love the write up thank u mrtoolz for our new mathematical indi's!!! I am using them & in simple terms I'm calling these "better than smoothing filter" indicator...... it uses Lambda calculus
Lambda setting of 8.0 on M5 chart. A bit laggy but much smoother than SMA
See a GIF with Forex-station.com on it? I probably made it
The best divergence indicator in the world.
Real news exists: Infowars.com
### Re: Moving Average indicators for MT4
3125
mrtools wrote: Tue Sep 19, 2023 6:08 am
Thanks, MrTools,
looks like a really interesting development.
Keep them coming...
### Re: Moving Average indicators for MT4
3126
josi wrote: Tue Sep 19, 2023 7:39 pm Thanks, MrTools,
looks like a really interesting development.
Keep them coming...
Moving Average Ribbon based on Regularized Relative Moving Average
In case of experimenting I made this ribbon based on regularized relative MA (Thanks for idea from Mrtools).
Also have choice use ema or fema in menu.
Arrows/alerts are based on cross or its slopes.
### Re: Moving Average indicators for MT4
3127
kvak wrote: Thu Mar 16, 2023 8:47 pm My version of 3 ma cross.
I'm also attaching the onchart version, it's somewhere in this topic..not updated versions
Could you add the psar as a filter......it would be much appreciated.
### Re: Moving Average indicators for MT4
3128
256robertm wrote: Fri Sep 22, 2023 12:38 am Could you add the psar as a filter......it would be much appreciated.
Double 2x MA Cross filtered with PSAR (Parabolic Stop & Reverse)
Hello, somethink like this?
This is 2 x MA cross filtered with PSAR.
If you want histogram type, I am able to do...
Other versions
### Re: Moving Average indicators for MT4
3129
kvak wrote: Sat Sep 23, 2023 9:29 am Hello, somethink like this?
This is 2 x MA cross filtered with PSAR.
If you want histogram type, I am able to do...
nice kvak
i have something interesting to open positions with stoch & rsi setting..
"There is NO GOD higher than TRUTH" - Mahatma Gandhi
### Re: Moving Average indicators for MT4
3130
kvak wrote: Sat Sep 23, 2023 9:29 am Hello, somethink like this?
This is 2 x MA cross filtered with PSAR.
If you want histogram type, I am able to do...
Respected kvak,
You are Superb.
Thanks for all your mods, experiments, shares. | 1,031 | 4,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.879381 |
http://wikisori.org/index.php?title=Binomial_cube&oldid=1842 | 1,561,640,135,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001138.93/warc/CC-MAIN-20190627115818-20190627141818-00334.warc.gz | 181,074,975 | 6,454 | Binomial cube
Binomial cube
<videoflash>D9Bvi00xlns&NR</videoflash>
3-6
Materials
A box containing eight wooden blocks:
• One red cube
• One blue cube
• Three red and black rectangular prisms
• Three blue and black rectangular prisms
All the blocks fit into a natural wood box. Each side of the cube has the same dimensions and pattern, and represents the square of (a + b). The faces of the small blocks are color coded: a2 is always red, b2 is always blue, and "ab" is always black.
a2 + 2ab + b2 = (a + b)2
Preparation
Algebraic expression: (a + b)3 = (a + b)(a + b)(a + b) = a3+3a2b+3ab2+b3
This is an individual exercise (Note: work cycle to be observed)
The child takes the cube apart beginning with a3 and lays out the pieces as shown, according to the formula. The child reconstructs the cube, matching red faces, black faces, and blue faces, beginning with a3.
Presentation
1. The Directress shows the child how to carry the box to the table. She sits besides the child, open up the box and lays out the blocks in the following pattern that will make it easy for rebuilding the cube in two layers.
2. The Directress then shows the child how to rebuild the cube. The colors printed on the lid act as a guide.
Control Of Error
The color codes.
The cube can not be built if incorrectly assembled.
Points Of Interest
At this sensorial stage, the child is not given the plan.
Purpose
• Provide the child with tangible experience of the way in which the cube can be divided and sub-divided.
• Lay the foundations for the later study of algebra.
• The child is at the stage of the absorbent mind. The child is not asked to understand the formula, but is using the cube in a mathematical way. The child will build up a predisposition to enjoy and understand mathematics later. | 448 | 1,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2019-26 | longest | en | 0.896646 |
https://www.physicsforums.com/threads/what-kind-of-math-is-this.472685/ | 1,590,936,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413406.70/warc/CC-MAIN-20200531120339-20200531150339-00263.warc.gz | 842,216,456 | 15,940 | # What kind of math is this?
## Main Question or Discussion Point
Given the acceleration of a particle as a function of its position (x) along a straight line, how can one create a function to represent its acceleration as a function of time (t)?
I haven't covered differential equations yet and was wondering if this is an example of a kind of problem that subject deals with.
For example,
Let's say the particle moves with A(x)=M/X^2
taking d(A)/d(x) can give the change in Acceleration at any distance x, but what about time?
In calc 1 we did some 'related rates' problems that were similar to this, but they always had a special relationship that let you solve it (right triangles etc).
So: how do you solve this case in particular, what subject matter deals with this, and where can I get more info on other questions like it?
Thanks for the help! :)
chiro
Typically what happens is that you parametrize the system.
I'll give an example of a something going in a circle.
Say we have a system x^2 + y^2 = 1.
We can use the substitution
x = cos(theta)
y = sin(theta)
Now using trigonometric identities we can verify this works since
x^2 + y^2 = cos^2(theta) + sin^2(theta) = 1 as required.
You basically do the same sort of thing and put things in terms of a parameter (namely time), but its not always easy to do so (as the circle example)
HallsofIvy
Homework Helper
Given the acceleration of a particle as a function of its position (x) along a straight line, how can one create a function to represent its acceleration as a function of time (t)?
I haven't covered differential equations yet and was wondering if this is an example of a kind of problem that subject deals with.
For example,
Let's say the particle moves with A(x)=M/X^2
taking d(A)/d(x) can give the change in Acceleration at any distance x, but what about time?
A is the velocity at position x? Then the "acceleration" is the derivative with respect to t, not x, whether at a given x or t. Using the chain rule, the acceleration would be given by
$$\frac{dA}{dt}= \frac{dA}{dx}\frac{dx}{dt}$$
Now, since A is velocity, you are saying that $dx/dt= A$
so that equation would be the same as
$$\frac{dA}{dt}= A\frac{dA}{dx}$$
In particular, if A(x)= M/x^2, as you give, then
$$\frac{dA}{dt}= \left(\frac{M}{x^2}\right\left(\frac{-3M}{x^3}\)= -\frac{3M}{x^5}$$
In calc 1 we did some 'related rates' problems that were similar to this, but they always had a special relationship that let you solve it (right triangles etc).
So: how do you solve this case in particular, what subject matter deals with this, and where can I get more info on other questions like it?
Thanks for the help! :)
That's just Calculus- in particular the application of the chain rule. | 710 | 2,749 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-24 | longest | en | 0.949911 |
http://openstudy.com/updates/50b017d5e4b0e906b4a5e441 | 1,449,020,234,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398525032.0/warc/CC-MAIN-20151124205525-00017-ip-10-71-132-137.ec2.internal.warc.gz | 162,418,111 | 9,575 | ## runfreeandlive 3 years ago what is the mid-point of the line formed by points (0,10) and (5,5)?
1. jasonxx
(0+5)/2 and (10+5)/2 2.5,7.5
2. jasonxx
$\frac{ X _{1}+X _{2} }{ 2 },\frac{ Y _{1}+Y _{2}}{ 2 }$ | 100 | 210 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2015-48 | longest | en | 0.690768 |
https://iitbrain.com/2232-a-small-body-of-mass-m030kgm030kgnbspstarts-sliding-down-from-the-top-of-a-smooth.html | 1,606,581,604,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195687.51/warc/CC-MAIN-20201128155305-20201128185305-00431.warc.gz | 335,216,113 | 6,510 | ## Question
A small body of mass m=0.30kg$m=0.30kg$ starts sliding down from the top of a smooth sphere of radius R=1.00m$R=1.00m$. The sphere rotates with a constant angular velocity ω=6.0rad/s$\omega =6.0rad/s$ about a vertical axis passing through its centre. Find the centrifugal force of inertia and the Coriolis force at the moment when the body breaks off the surface of the sphere in the reference frame fixed to the sphere. | 118 | 433 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2020-50 | latest | en | 0.720394 |
https://www.doubtnut.com/qa-hindi/645690359 | 1,660,579,924,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00171.warc.gz | 665,854,912 | 99,855 | Home
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अभिक्रिया 2N_(2) O_(5) to 4NO_...
Updated On: 27-06-2022
Get Answer to any question, just click a photo and upload the photo and get the answer completely free,
k = 2k , k = k k = 2k , k = k//2k = 2k , k = 2k k = k , k = k
Solution : दर = -1/2 (d[N_(2)O_(5)])/(dt) =1/4 (d[NO_(2)])/(dt) =(d[O_(2)])/(dt) <br> = 1/2 k[N_(2)O_(5)] = 1/4 k [N_(2)O_(5)] = k [N_(2)O_(5)] <br> = k/2 = k/4 = k <br> :. K = 2k , k = k/2 | 249 | 497 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2022-33 | longest | en | 0.590616 |
http://www.reference.com/browse/preclosure | 1,432,384,210,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927592.52/warc/CC-MAIN-20150521113207-00296-ip-10-180-206-219.ec2.internal.warc.gz | 669,389,382 | 16,237 | Related Searches
Definitions
# Preclosure operator
In topology, a preclosure operator, or Čech closure operator is a map between subsets of a set, similar to a topological closure operator, except that it is not required to be idempotent. That is, a preclosure operator obeys only three of the four Kuratowski closure axioms.
## Definition
A preclosure operator on a set $X$ is a map $\left[quad\right]_p$
$\left[quad\right]_p:mathcal\left\{P\right\}\left(X\right) to mathcal\left\{P\right\}\left(X\right)$
where $mathcal\left\{P\right\}\left(X\right)$ is the power set of $X$.
The preclosure operator has to satisfy the following properties:
1. $\left[varnothing\right]_p = varnothing !$ (Preservation of nullary unions);
2. $A subseteq \left[A\right]_p$ (Extensivity);
3. $\left[A cup B\right]_p = \left[A\right]_p cup \left[B\right]_p$ (Preservation of binary unions).
The last axiom implies the following:
4. $A subseteq B$ implies $\left[A\right]_p subseteq \left[B\right]_p$.
## Topology
A set $A$ is closed (with respect to the preclosure) if $\left[A\right]_p=A$. A set $Usubset X$ is open (with respect to the preclosure) if $A=Xsetminus U$ is closed. The collection of all open sets generated by the preclosure operator is a topology.
The closure operator cl on this topological space satisfies $\left[A\right]_psubseteq operatorname\left\{cl\right\}\left(A\right)$ for all $Asubset X$.
## Examples
### Premetrics
Given $d$ a prametric on $X$, then
$\left[A\right]_p=\left\{xin X : d\left(x,A\right)=0\right\}$
is a preclosure on $X$.
### Sequential spaces
The sequential closure operator $\left[quad\right]_mbox\left\{seq\right\}$ is a preclosure operator. Given a topology $mathcal\left\{T\right\}$ with respect to which the sequential closure operator is defined, the topological space $\left(X,mathcal\left\{T\right\}\right)$ is a sequential space if and only if the topology $mathcal\left\{T\right\}_mbox\left\{seq\right\}$ generated by $\left[quad\right]_mbox\left\{seq\right\}$ is equal to $mathcal\left\{T\right\}$, that is, if $mathcal\left\{T\right\}_mbox\left\{seq\right\}=mathcal\left\{T\right\}$. | 665 | 2,139 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2015-22 | latest | en | 0.714397 |
https://goteachthis.com/index.php/product/tables-bookmark-scaffold-6x/ | 1,669,651,823,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00521.warc.gz | 313,187,870 | 24,706 | # Tables Bookmark Scaffold – 6x
Home » Shop » Tables Bookmark Scaffold – 6x
# Tables Bookmark Scaffold – 6x
6 Times Table Bookmark – a free printable resource to help students focus on multiplication facts anywhere anytime.
\$0.00
# Free Tables Bookmark Scaffold
## 6 Times Table
6x Tables Bookmark
#### Possible Ways to Use the Math Fact Scaffold:
– My favorite way of using this resource is to have a Number Talk / Interview with each student and ask them which table they feel they need to focus on. We then print out the bookmark appropriate to their needs.
– Next, discuss with the student the facts the students have committed to memory and which ones they need to work on. I like to use the term, ‘See it. Say it’. If a student can look at an algorithm and say the answer immediately they have achieved automacy or ‘See it. Say It’. I have the students put a small tick next to the number facts they have committed to memory as a way to help them break the task into smaller, achievable tasks.
NB The way the scaffold is designed, students can look at the side without the answers and use mental strategies to work out a solution to the algorithms. They can then turn the 6x tables bookmark over to confirm if their answer is correct.
#### Another Way to Use the Tables Bookmark:
Some people like to make a collection of these bookmarks on a sort of ‘key ring’. By using a curtain ring to bind all the word lists together.
#### Yet Another Way to Use the Tables Bookmark:
As a scaffold – when playing Multiplication Games have the bookmark handy so students can check each others answers.
#### How to Make Your Tables Bookmark:
1. Print out the tables bookmark. I prefer 200 weight paper as it stands up to the rigors of classroom life well 😉
2. Cut out the bookmarks and fold them over.
3. Laminate the tables bookmark.
4. Use a hole punch to put a hole through the top of the bookmark.
5. Plait some wool or colored string and attach it to the top of the bookmark. | 434 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | longest | en | 0.903813 |
https://www.kramerphotography.nl/JiCJJY | 1,611,765,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704828358.86/warc/CC-MAIN-20210127152334-20210127182334-00238.warc.gz | 816,862,555 | 8,746 | # Water Required For 1m3 Concrete
## Calculate Cement Sand Aggregate and Water for Concrete
WATER REQUIREMENT FOR DIFFERENT CONCRETE GRADES OF NOMINAL MIX CONCRETE AS PER IS456 CODE. For design mix concrete water content is calculated from Wate/Cement ratio which depends on various factors like the weight of cement, workability etc. But for nominal mix concrete IS456 codebook suggests the following quantities per bag of cement.
## Concrete mix design - planete-tp : All about public works
* and lastly, if required, the amount of additives (a few Kilograms for 1m3 of concrete). One cubic metre of concrete weighs 2.5 Tonnes. Typically, 1m3 of concrete is made up of 350Kg of cement, 700Kg of sand, 1,200Kg of chippings and 150 Litres of water. The mix design process can either be conducted on the basis of charts or experimentally ...
## How many sand, cement, stone and water is in 1m3 of concrete?
How can such a poorly expressed question be answered? How about working with the person who submitted the question to help him/her express it coherently. I was an editor once, and, particularly if English was not the person's native language, I wo...
## Mix Design For M35 Grade Of Concrete
The mix design for M35 Grade Of Concrete for pile foundations provided here is for reference purpose only. Actual site conditions vary and thus this should be adjusted as per the location and other factors.
## Testing of Concrete Blocks
Concrete can be converted into precast masonry units such as Hollow and Solid normal and light weight concrete blocks of suitable size to be used for load and non-load bearing units for wallings. Use of such concrete blocks are more appropriate in region where soil bricks are costly, poor in strength and are not available.
## How many bags of cement to make 1m3 of concrete?
How many bags of cement to make 1m3 of concrete. This is one of the basic answers that every civil engineer should know still, a lot of students don’t know how many bags of cement to make 1m3 of concrete.
## Practical Method for Mix Design of Cement-based Grout
water-cement ratio is required if more sand is used in the mix to maintain the grout flowability. For instance the 1 : 2 mix must have a water-cement ratio of 0.65 to pass the flow cone test, meanwhile the 1 : 1.5 and 1 : 1 mixes only need water-cement ratios of 0.52 and 0.50 respectively. 2.2. Gradation of Sand
## How Much Water Required To Make 1M3 Of Structural Concrete
How Much Water Required To Make 1M3 Of Structural Concrete. We are a large-scale manufacturer specializing in producing various mining machines including different types of sand and gravel equipment, milling equipment, mineral processing equipment and building materials equipment.
## To Study Properties of Concrete by Replacing Fresh Water with Treated Waste Water
2009, 186 liters water is required for 1m3 of concrete. On an average 150 liters water is required for 1m3 of concrete. The construction of 10,000 sq.mt. multi-storied structure can require about 10 million liters water for production, curing and site ...
## Concrete calculator mix-on-site
Concrete calculator mix-on-site . This calculator estimates the required amount of cement, sand and aggregate (stone) for mixing on site for a given area of concrete.
## Making Concrete - A Layman's Guide to Clean Water
* The amount of water used to mix the ingredients is the most important factor in determining the final strength of the concrete. Use the least amount of water that will still give a workable mix. Too much water weakens the concrete and leads to cracking. The 1:0.5 cement:water ratio noted above is merely a guide for estimating when making ...
## How many sand, cement, stone and water is in 1m3 of concrete
How can such a poorly expressed question be answered? How about working with the person who submitted the question to help him/her express it coherently. I was an editor once, and, particularly if English was not the person's native language, I wo...
## how we calculate of Sand, cement and aggregate of M20 ratio
For 1m3 of M20 (1:1.5:3) ... coarse agrregate and water, It required to do mix design & for this initially it is requird to find out the properties of all ingradients used for preparing concrete ...
## Mixing Concrete
Mixing Concrete . The three main options for mixing concrete are: Mixing by hand - this is probably only suitable where less than about ¼m³ of concrete is required for a job providing that a reasonably fit person is available. Using a cement mixer - this is generally suitable where between about ¼m³ and 2m³ of concrete is required. It will ...
## What does 1m3 of water weigh
What does 1m3 of water weigh? Unanswered Questions Why does Greg Gutfeld wear a ring on the ... How much quantaty of sand and metal is required for 1m3 concrete m25 grade? One cubic metre of ...
## How to Calculate Cement, Sand and Aggregate required for 1
The proportions of cement, sand, coarse aggregate and water plays an important role in determining the fresh and hardended properties of concrete. So care should be taken while calculating the amount of Cement, Sand and Aggregate required for 1 Cubic meter of Concrete.
## concrete, plaster & mortar mixes for builders
The amount of water added to a mix must be enough to make the mix workable and plastic. 2. Common cement complying with SANS 50197 may be used for concrete, plaster or mortar.
## Batching by volume - DIY
Water should be added to obtain a workable but non-segregating mix, the total free water being roughly 170 litres per cubic metre of concrete. Mix A is suitable for general-purpose projects, such as paths, patios and general slabs. Mix B is suitable for concrete
## FAQs
Cement is the binding agent in concrete, normally the most active component, and usually the most costly. Its selection and proper use are important in obtaining the balance of properties required for a particular concrete mixture and in minimising the cost of that ...
## Ready Mix Concrete Volume Calculator
Water - this is needed to chemically react with the cement and to make the concrete workable. Aggregate - The fines are sand and gravel or crushed stone is the coarse aggregate (10, 20 & 40mm) in most mixes. Admixtures in Concrete Accelerators - added to the concrete to reduce overall setting times and gain early strength. Calcium chloride is ...
## How to Calculate Quantities of Cement, Sand and Aggregate
In the step 3 of “How To Calculate Quantities Of Cement, Sand And Aggregate For Nominal Concrete Mix (1:2:4)” you have calculated that: 01 cum of concrete will require Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags Sand required = 115/0.167 = 688 Kgs or 14.98 cft Aggregate required = 209/0.167 = …
## How to calculate Cost of Concrete work
Compaction of concrete 4. Reinforcement. For reinforced concrete work, the cost of steel and labour charges for the cutting and tieing of rebars is also calculated and added to the concrete works. Reinforcement in Concerte 5. FormWork or Shuttering. Concreting work in beam, columns or slab, the formwork required to hold the concrete is also added.
## How to Calculate Quantities of Cement, Sand and Aggregate for
In the step 3 of “How To Calculate Quantities Of Cement, Sand And Aggregate For Nominal Concrete Mix (1:2:4)” you have calculated that: 01 cum of concrete will require Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags Sand required = 115/0.167 = 688 Kgs or 14.98 cft Aggregate required = 209/0.167 = 1251 kgs or 29.96 cft
## Quantity of Cement, Sand & Water required for Plastering
Apr 13, 2018 · The Quantity of cement required = 574Kgs Calculated Quantity of Sand (Fine aggregate) required = 2560Kgs Quantity of Water required = 627 litres. Related Articles : Quantitiy of Cement, Sand, Aggregate and water required for different grades of concrete. For Instant updates Join our Whatsapp Broadcast.
## Concrete Mix Design Just Got Easier
Source: Standard Practice for Selecting Proportions for Normal, Heavyweight, and Mass Concrete (ACI 211.1-91) Step 3: Mixing Water and Air Content. Now you get the first estimation of the amount of water required to obtain the appropriate workability for your mix, based on the slump flow and aggregate size.
## Concrete Mix Ratio Calculator | On Site Concrete Calculator
Concrete Mix Ratio Calculator. An online on site concrete calculator to calculate the concrete mix ratio. It plays a major role in creating a strong, durable concrete block. The major materials needed in the preparation of concrete blocks are portland cement, sand, aggregate (stone), and water.
## On-Site concrete calculator
Water - this is needed to chemically react with the cement and to make the concrete workable. Aggregate - The fines are sand and gravel or crushed stone is the coarse aggregate (10, 20 & 40mm) in most mixes. Admixtures in Concrete Accelerators - added to the concrete to reduce overall setting times and gain early strength. Calcium chloride is ...
## how much reinforcement requerd for 1m3 rcc
material qty required for 1m3 rcc m20 grade – Crusher South Africa Posts Related to material qty required for 1m3 rcc m20 grade. how much water recuired for 1cu.m concrete. … reinforcement required for 1 cubic feet concrete in … »More detailed
## Actual Concrete Mix Ratios For 3000, 3500, 4000, and 4500 psi
These are the actual concrete mix ratios for 3000, 3500, 4000, and 4500 psi concrete that I use to pour concrete floors, patios, pool decks and more. I'll show you the actual concrete batch plant ticket with the cement, sand, and aggregate break downs for the yards we used.
## How much sand cement water to make 1m3 of strong concrete?
Hi can anyone tell me how many kilograms of cement how much sand m3 water etc to make one cubic meter 1m3 of concrete. I will be using a standard cement mixer.
## Concrete 1 cubic meter volume to liters converter
To convert concrete measuring units can be useful when building with concrete and where handling of concrete is required. How much of concrete amount is from cubic meters ( m3 ) into corresponding liters ( L ) unit. Exchange between multiple concrete mass/weight and volume measures with instantly calculated unit values for this masonry material.
## What does 1m3 mean?
What does 1m3 mean? The transporting job requires talking in cubic meters (m3). For every removal is important to know how many m3 is. What means 1 m3.
## How much of water is required to produce 1 m3 of concrete (25
Jun 15, 2006 · You will be requiring about 270 litres of water to obtain 1m3 of finished concrete, if you are using crushed aggregate. However it is a very rough estimate, for more precise assessment, the following data are required:-1. Maximum Nominal size of aggregate. 2. Type of aggregate. 3. Degree of workability. 4. Degree of QC. 5.
## Concrete 1 cubic meter volume to cubic feet converter
To convert concrete measuring units can be useful when building with concrete and where handling of concrete is required. How much of concrete amount is from cubic meters ( m3 ) into corresponding cubic feet ( cu ft - ft3 ) unit. Exchange between multiple concrete mass/weight and volume measures with instantly calculated unit values for this ...
## Quantities of Materials Per Cubic Meter of Concrete Mix
Quantities of Materials Per Cubic Meter of Concrete and Mortar Mix Proportions Quantity of materials such as cement, sand, coarse aggregates and water required per cubic meter of concrete and mortar for different mix proportions varies with the mix design of the concrete and mortar respectively.
## How To Calculate Cement, Sand, & Aggregate Quantity In 1 Cubic metre Concrete
Consider volume of concrete = 1m3 Dry Volume of Concrete = 1 x 1.54 = 1.54 m3 (For Dry Volume Multiply By 1.54) Now we start calculation for find Cement, Sand and Aggregate quality in 1 cubic meter concrete CALCULATION FOR CEMENT QUANTITY 3
## How to Calculate Cement, Sand and Aggregate Quantity in
Oct 16, 2016 · How to Calculate Cement, Sand and Aggregate Quantity in Concrete Today In this video I am going to teach you, How to Calculate Cement, Sand and Aggregate Quantity in Concrete Theoretically At Site.
## Concrete slab calculation
So for 0.15 of a cubic metre you would need 52.5kg cement, 131.25kg sand, 157.50kg aggregate (20-5mm size) and 27 litres of water. Cement usually comes in 25kg bags these days so it should be OK to use 2 x 25kg bags (50kg) for a 0.15m3 batch. Try not to make the mix any wetter than is needed to ensure full compaction as this will reduce the ...
## Concrete 1 liter volume to cubic meters converter
To convert concrete measuring units can be useful when building with concrete and where handling of concrete is required. How much of concrete amount is from liters ( L ) into corresponding cubic meters ( m3 ) unit. Exchange between multiple concrete mass/weight and volume measures with instantly calculated unit values for this masonry material.
## Adding Water to Concrete| Concrete Construction Magazine
The cement (and cementitious materials like fly ash) in the concrete needs water to hydrate and form calcium-silicate-hydrate (C-S-H) which is the crystalline glue that holds the concrete together. The water is chemically bound (consumed) during the reaction with the cement at approximately 25 pounds of water to every 100 pounds of cement ... | 3,069 | 13,408 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2021-04 | latest | en | 0.922345 |
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# Reflecting the community having an interest in the
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[ Reflecting the community having an interest in the preservation of historical buildings], the council allocated seven thousand dollars for the restoration of the courthouse.
(A) Reflecting the community having an interest in the preservation of historical buildings
(B) Reflecting the interest in preserving historical buildings that the community had
(C) Reflecting the interest of the community in preserving historical buildings
(D) As a reflection of the community having -interest in preserving historical buildings
(E) As a reflection of the interest in the preservation of historical buildings that the
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C correctly conveys the reflection of the interest of the community
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http://bluecatmotors.com/epub/complex-manifolds-without-potential-theory-with-an-appendix-on-the-geometry-of | 1,513,297,806,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551501.53/warc/CC-MAIN-20171215001700-20171215021700-00362.warc.gz | 36,313,352 | 9,146 | > > Complex Manifolds without Potential Theory: with an appendix by Shiing-shen Chern
# Complex Manifolds without Potential Theory: with an appendix by Shiing-shen Chern
By Shiing-shen Chern
From the experiences of the second one edition: "The new tools of advanced manifold conception are very helpful instruments for investigations in algebraic geometry, complicated functionality conception, differential operators etc. The differential geometrical tools of this concept have been constructed basically lower than the effect of Professor S.-S. Chern's works. the current ebook is a moment edition... it could possibly function an advent to, and a survey of, this concept and relies at the author's lectures held on the college of California and at a summer season seminar of the Canadian Mathematical Congress....
The textual content is illustrated through many examples... The booklet is warmly advised to every person drawn to advanced differential geometry." #Acta Scientiarum Mathematicarum, forty-one, 3-4#
Read or Download Complex Manifolds without Potential Theory: with an appendix on the geometry of characteristic classes PDF
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Additional resources for Complex Manifolds without Potential Theory: with an appendix on the geometry of characteristic classes
Example text
E 1\ Ai' ... 39) o , + where e is a (qxq)-matrix of I-forms. trices of the form is a one-form. 38). The invariant polynomials constitute a link between the local properties of a connection and its global properties. that a family of matrices the adjoint type in local frame field {~s} is a tensorial matrix of k-forms of M , if such a matrix s Ai defined in is a form of degree M If P(Al, ... 20). 29). form of degree R2j ~ H2 j(M,C) (B) dimension in W: Let Let E M be a complex vector bundle with fiber + r!
Y of is a given real- runs over all real-valued l-forms 52 in M. 1 -l - . 2 l It suffices to prove that As previously let COO r-forms and d. Also let and Cpq definition Cr APq i2H~1,1)(M,Z) = 0 . Ar be the sheaf of germs of complex-valued be the sub sheaf of the germs of be the sheaf of germs of be the sub sheaf of germs of CO =C and COO o. APq Ar closed under Coo-forms of type (p,q) closed under We have the diagram a. +l Or k' C - > Aorl ----+ k tive. Moreover both horizontal sequences are exact.
43) Thus, n'g n, gn the difference of two connection matrices, is a tensorial matrix of l-forms of the adjoint type. 44) Then Wt = W We put o + tn , < t < 1 . is a connection matrix depending on the parameter w t =1 The curvature matrix of the connection wt which reduces to wand for t =0 and t , respectively. •. ,Ar ) be a symmetric invariant polynomial. PCA) = PCA, ... 46) ----- QCB,A) rPCB,A, ... ) t , we get - PCrl) A2 = ... Let 43 This proves (B). 2). 50). are two COO-sections of the bundle, H(~,n) E is a COO-function having properties corresponding to A complex vector bundle with an hermitian structure is called an hermitian vector bundle. | 899 | 3,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-51 | latest | en | 0.909681 |
https://wiki.gis.com/wiki/index.php/Frequency | 1,696,093,159,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510697.51/warc/CC-MAIN-20230930145921-20230930175921-00403.warc.gz | 669,339,412 | 14,074 | # Frequency
Sinusoidal waves of various frequencies; the bottom waves have higher frequencies than those above. The horizontal axis represents time.
Frequency is the number of occurrences of a repeating event per unit time. It is also referred to as temporal frequency. The period is the duration of one cycle in a repeating event, so the period is the reciprocal of the frequency.
## Definitions and units
For cyclical processes, such as rotation, oscillations, or waves, frequency is defined as a number of cycles, or periods, per unit time. In physics and engineering disciplines, such as optics, acoustics, and radio, frequency is usually denoted by a Latin letter f or by a Greek letter ν (nu).
In SI units, the unit of frequency is hertz (Hz), named after the German physicist Heinrich Hertz. For example, 1 Hz means that an event repeats once per second, 2 Hz is twice per second, and so on. This unit was originally called a cycle per second (cps), which is still sometimes used. Heart rate and musical tempo are measured in beats per minute (BPM). Frequency of rotation is often expressed as a number of revolutions per minute (rpm). BPM and rpm values must be divided by 60 to obtain the corresponding value in Hz: thus, 60 BPM translates into 1 Hz.
The period is usually denoted as T, and is the reciprocal of the frequency f:
$T = \frac{1}{f}.$
The SI unit for period is the second.
## Measurement
### By timing
To calculate the frequency of an event, the number of occurrences of the event within a fixed time interval are counted, and then divided by the length of the time interval.
In experimental work (for example, calculating the frequency of an oscillating pendulum) it is generally more accurate to measure the time taken for a fixed number of occurrences, rather than the number of occurrences within a fixed time. The latter method introduces — if N is the number of counted occurrences — a random error between zero and one count, so on average half a count, causing a biased underestimation of f by ½ f / (N + ½)[citation needed] in its expected value. In the first method, which does not suffer this particular error, frequency is still calculated by dividing the number of occurrences by the time interval; however it is the number of occurrences that is fixed, not the time interval.
### By stroboscope effect, or frequency beats
In case when the frequency is so high that counting is difficult or impossible with the available means, another method is used, based on a source (such as a laser, a tuning fork, or a waveform generator) of a known reference frequency f0, that must be tunable or very close to the measured frequency f. Both the observed frequency and the reference frequency are simultaneously produced, and frequency beats are observed at a much lower frequency Δf, which can be measured by counting. This is sometimes referred to as a stroboscope effect. The unknown frequency is then found from $f=f_0\pm \Delta f$.
## Frequency of waves
Frequency has an inverse relationship to the concept of wavelength, simply, frequency is inversely proportional to wavelength λ (lambda). The frequency f is equal to the speed v of the wave divided by the wavelength λ of the wave:
$f = \frac{v}{\lambda}.$
In the special case of electromagnetic waves moving through a vacuum, then v = c , where c is the speed of light in a vacuum, and this expression becomes:
$f = \frac{c}{\lambda}.$
When waves from a monochromatic source travel from one medium to another, their frequency remains exactly the same — only their wavelength and speed change.
## Examples
### Physics of light
Radiant energy is energy which is propagated in the form of electromagnetic waves. Most people think of natural sunlight or electrical light, when considering this form of energy. The type of light which we perceive through our optical sensors (eyes) is classified as white light, and is composed of a range of colors (red, orange, yellow, green, blue, indigo, violet) over a range of wavelengths, or frequencies.
Visible (white) light is only a small fraction of the entire spectrum of electromagnetic radiation. At the short end of that wavelength scale is ultraviolet (UV) light from the sun, which cannot be seen. At the longer end of that spectrum is infrared (IR) light, which is used for night vision and other heat-seeking devices. At even shorter wavelengths than UV are X-rays and Gamma-rays. At longer wavelengths than IR are microwaves,radio waves, electromagnetic waves in megahertz and kHz range, as well as natural waves with frequencies in the millihertz and microhertz range. A 2 millihertz wave has a wavelength approximately equal to the distance from the earth to the sun. A microhertz wave would extend 0.0317 light years. A nanohertz wave would extend 31.6881 light years.
Complete spectrum of electromagnetic radiation with the visible portion highlighted
Electromagnetic radiation is classified according to the frequency (or wavelength) of the light wave. This includes (in order of increasing frequency): natural electromagnetic waves, radio waves, microwaves, terahertz radiation, infrared (IR) radiation, visible light, ultraviolet (UV) radiation, X-rays and gamma rays. Of these, natural electromagnetic waves have the longest wavelengths and gamma rays have the shortest. A small window of frequencies, called visible spectrum or light, is sensed by the eye of various organisms, with variations of the limits of this narrow spectrum.
### Physics of sound
Sound is vibration transmitted through a solid, liquid, or gas; particularly, sound means those vibrations composed of frequencies capable of being detected by ears. For humans, hearing is limited to frequencies between about 20 Hz and 20,000 Hz (20 kHz), with the upper limit generally decreasing with age. Other species have a different range of hearing. For example, dogs can perceive vibrations higher than 20 kHz. As a signal perceived by one of the major senses, sound is used by many species for detecting danger, navigation, predation, and communication.
The mechanical vibrations that can be interpreted as sound are able to travel through all forms of matter: gases, liquids, solids, and plasmas. The matter that supports the sound is called the medium. Sound cannot travel through vacuum.
#### Longitudinal and transverse waves
Sinusoidal waves of various frequencies; the bottom waves have higher frequencies than those above. The horizontal axis represents time.
Sound is transmitted through gases, plasma, and liquids as waves, also called compression waves. Through solids, however, it can be transmitted as both longitudinal and transverse waves. Longitudinal sound waves are waves of alternating pressure deviations from the equilibrium pressure, causing local regions of compression and rarefaction, while transverse waves in solids, are waves of alternating shear stress.
Matter in the medium is periodically displaced by a sound wave, and thus oscillates. The energy carried by the sound wave converts back and forth between the potential energy of the extra compression (in case of longitudinal waves) or lateral displacement strain (in case of transverse waves) of the matter and the kinetic energy of the oscillations of the medium.
#### Sound wave properties
Sound waves are characterized by the generic properties of waves, which are frequency, wavelength, period, amplitude, intensity, speed, and direction (sometimes speed and direction are combined as a velocity vector, or wavelength and direction are combined as a wave vector).
Transverse waves, also known as shear waves, have an additional property of polarization.
Sound characteristics can depend on the type of sound waves (longitudinal versus transverse) as well as on the physical properties of the transmission medium.
Whenever the pitch of the soundwave is affected by some kind of change, the distance between the sound wave maxima also changes, resulting in a change of frequency. When the loudness of a soundwave changes, so does the amount of compression in airwave that is travelling through it, which in turn can be defined as amplitude.
In music and acoustics, the frequency of the pitch A above middle C on a piano is usually defined as 440 Hz, that is, 440 cycles per second () and known as concert pitch, to which an orchestra tunes.
### Other examples
In Europe, Africa, Australia, Southern South America, most of Asia, and Russia, the frequency of the alternating current in household electrical outlets is 50 Hz (close to the tone G), whereas in North America and Northern South America, the frequency of the alternating current is 60 Hz (between the tones B♭ and B — that is, a minor third above the European frequency). The frequency of the 'hum' in an audio recording can show where the recording was made — in countries utilizing the European, or the American grid frequency.
## Period versus frequency
As a matter of convenience, longer and slower waves, such as ocean surface waves, tend to be described by wave period rather than frequency. Short and fast waves, like audio and radio, are usually described by their frequency instead of period. These commonly used conversions are listed below:
Frequency Period (time) 1 mHz (10-3) 1 Hz (100) 1 kHz (103) 1 MHz (106) 1 GHz (109) 1 THz (1012) 1 ks (103) 1 s (100) 1 ms (10-3) 1 µs (10-6) 1 ns (10-9) 1 ps (10-12)
## Other types of frequency
• Angular frequency ω is defined as the rate of change in the orientation angle (during rotation), or in the phase of a sinusoidal waveform (e.g. in oscillations and waves):
$\omega=2\pi f\,$.
• Spatial frequency is analogous to temporal frequency, but the time axis is replaced by one or more spatial displacement axes.
• Wavenumber is the spatial analogue of angular frequency. In case of more than one spacial dimension, wavenumber is a vector quantity.
• Frequency converter
• Frequency range or frequency band
• Letter frequencies
• Natural frequency
• Negative frequency
• Periodicity
• Pitch (music)
• Rate (mathematics)
• Wavelength
## References
• Giancoli, D.C. (1988), Physics for Scientists and Engineers (2nd ed.), Prentice Hall, ISBN 013669201X | 2,185 | 10,204 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.939833 |
https://www.oilngasprocess.com/instrumentation/proportional-control.html | 1,720,879,133,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514494.35/warc/CC-MAIN-20240713114822-20240713144822-00892.warc.gz | 784,118,276 | 12,193 | # Proportional Control
(Controller output can go directly to a valve or to the setpoint of another controller. In the following discussions, it is assumed that controllers send their output directly to a valve.)
Figure 300-7 shows the relationship between valve position and error that is characteristic of proportional control: The valve position changes in exact proportion to the amount of error, not to its rate or duration. The response is almost instantaneous, and the valve returns to its initial value when the error returns to zero.
Control Algorithm
The linear relationship between the setpoint deviation (error) and the valve position (controller output) for proportional action can be expressed as follows:
O = KcE
(Eq. 300-1)
where:
O = Controller output
Kc = Controller Gain = DOutput / DError
E = Error = (Setpoint – Measurement)
This equation is called the control algorithm. The gain, Kc, is also called the controller sensitivity. It represents the proportionality constant between the control valve position and controller error.
Proportional Band
Another way of characterizing a proportional controller is to describe its proportional band. The proportional band is the percent change in value of the controlled variable necessary to cause full travel of the final control element. The proportional band, PB, is related to its gain as follows:
Kc= 100/PB
(Eq. 300-2)
Both proportional band and gain are expressions of proportionality. Manufacturers may call their adjustments gain, sensitivity, or proportional band. Figure 300-8 shows the relationship between valve opening and proportional bands of different percentages. High percentage proportional bands (wide bands) have a less sensitive response than low percentage proportional bands (narrow bands).
Bias
Bias is the amount of output from a proportional controller when the error is zero. Equation 300-1 implies that when the error is zero, controller output is zero. The valve is either fully open or fully closed and provides no throttling action. Adding a bias provides this throttling action. Equation 300-1 then becomes:
O = KcE + B
(Eq. 300-3)
where:
B = Bias (percent of full output)
Typically, manufacturers set the bias at 50%. To prevent a process bump, the operator is sometimes allowed to adjust the bias before putting the controller in automatic. Figure 300-9 shows controller output versus error at different proportional bands with a 50% bias. At zero error, the controller output is 50% of full range for any proportional band.
Offset
A controller’s error is the difference between its setpoint and measurement. In a proportional-only controller, a change in setpoint or load introduces a permanent error called offset (see Figure 300-10). It is impossible for a proportional-only controller to return the measurement exactly to its setpoint, because proportional output only changes in response to a change in the error, not to the error’s duration.
Assume that a proportional-only controller controls the outlet temperature of a furnace and that the temperature is at the setpoint. If the feed rate to the furnace increases, more fuel will be needed. This disturbance represents a load change to the furnace. To get more fuel, the fuel valve must be opened more. As is suggested by Equation 300-3, the only way that the valve can be at some value other than its starting point is for an error to exist. Thus, the proportional controller alone cannot return the outlet temperature to its setpoint. As mentioned, some controllers allow the operator to adjust the bias until the value of E (the error, or offset) is zero.
Offset is determined by the proportional band value for the controller and the change in valve position that occurs when a disturbance takes place:
The proportional-only controller is the easiest continuous controller to tune. It provides rapid response and is relatively stable. If offset can be tolerated (loose control), proportional-only control can be used. | 808 | 3,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.910342 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-7-dividing-polynomials-exercise-set-page-398/61 | 1,531,926,451,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590199.42/warc/CC-MAIN-20180718135047-20180718155047-00452.warc.gz | 883,890,012 | 13,587 | ## Algebra: A Combined Approach (4th Edition)
c. $\frac{a}{7}+1$
$\frac{a+7}{7}$ $=a(\frac{1}{7})+7(\frac{1}{7})$ $=\frac{a}{7}+1$ c. $\frac{a}{7}+1$ | 76 | 150 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-30 | latest | en | 0.383484 |
http://www.buenastareas.com/ensayos/Problemas-Resueltos-Fisica/3307776.html | 1,508,832,957,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828189.71/warc/CC-MAIN-20171024071819-20171024091819-00352.warc.gz | 391,065,512 | 14,893 | # Problemas resueltos fisica
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Problemas Resueltos Fisica Libro E. Tippens
Chapter 10. Uniform Circular Motion
Centripetal Acceleration
10-1. A ball is attached to the end of a 1.5 m string and it swings in a circle witha constant speed of 8 m/s. What is the centripetal acceleration?
[pic] ac = 42.7 m/s2
10-2. What are the period and frequency of rotation for the ball in Problem 10-1?
[pic]; T =1.18 s
[pic]; f = 0.849 rev/s
10-3. A drive pulley 6-cm in diameter is set to rotate at 9 rev/s. What is the centripetal acceleration of a point on the edge of the pulley? What wouldbe the linear speed of a belt around the pulley? [ R = (0.06 m/2) = 0.03 m ]
[pic]; ac = 95.9 m/s2
[pic]; v = 1.70 m/s
10-4. An object revolvesin a circle of diameter 3 m at a frequency of 6 rev/s. What is the period of revolution, the linear speed, and the centripetal acceleration? [ R = (3 m/2) = 1.5 m ]
[pic]; T = 0.167 s ;[pic]; v = 56.5 m/s
[pic]; ac = 2130 m/s2
10-5. A car moves around a curve 50 m in radius and undergoes a centripetal acceleration of 2 m/s2. What is its constant speed?
[pic];v = 10.0 m/s
10-6. A 1500-kg car moves at a constant speed of 22 m/s along a circular track. The centripetal acceleration is 6 m/s2. What is the radius of the track and the centripetal force onthe car?
[pic]; R = 80.7 m
[pic]; Fc = 9000 N
10-7. An airplane dives along a curved path of radius R and velocity v. The centripetal acceleration is 20 m/s2. If both the velocityand the radius are doubled, what will be the new acceleration?
[pic];
a2 = 2a1 = 2(20 m/s2; a = 40 m/s2
Centripetal Force
10-8. A 20-kg child riding a loop-the-loop at the Fair moves at 16m/s through a track of radius 16 m. What is the resultant force on the child?
[pic]; Fc = 320 N
10-9. A 3-kg rock, attached to a 2-m cord, swings in a horizontal circle so that it makes... | 669 | 1,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-43 | latest | en | 0.77101 |
https://math.stackexchange.com/questions/4141610/how-can-i-prove-that-set-t-is-countably-infinite-countable-and-uncountable-sets | 1,652,918,720,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00172.warc.gz | 452,731,069 | 68,540 | # How can I prove that set T is countably infinite? Countable and uncountable sets; rational numbers
Let S be the collection of all non-vertical lines in the 2-dimensional plane $$R^2$$ passing through the origin. We can index the collection $$S$$ using $$R$$ as the index set, as follows.
For each $$i∈R$$, define $$L_i={\{(x,y) ∈ R^2 | y = ix}\}$$ Note that $$i$$ is simply the slope of the line $$L_i$$. We can then write $$S={\{L_i | i∈R}\}$$
1. What is $$⋃_{i∈R} L_i$$ ?
2. Explain why $$S$$ is uncountable.
3. We shall call a line $$L_i∈_S$$ special if at least one point on the line $$L_i$$ other than the origin has rational numbers for both coordinates (i.e., there is at least one point $$(x,y)≠(0,0)$$ on $$L_i$$ such that $$x∈Q$$ and $$y∈Q$$). Let $$T=\{L_i\in S\ |\ L_i\text{ is special}\}$$. Prove that $$T$$ is countably infinite.
$$R$$ denotes set of real numbers and $$Q$$ denotes set of rational numbers.
1. $$⋃_{i∈R} L_i$$ is the union of all the lines with slope $$i \in R$$.
2. $$S$$ is uncountable because the set $$R$$ of real numbers is uncountable since $$S$$ is indexed using $$R$$ as the index set.
3. I am not sure how to prove this. What i have in mind is, since $$T$$ is a special set with a special line that has at least one point with rational numbers for both its coordinates, should I show that $$T$$ is countable because the set of rational numbers $$Q$$ is countable? Please help.
Also are my answers to 1 and 2 correct? I somewhat feel they are but would love to hear from you all.
• The rationals are countably infinite. So if we scatter a representation of rationals around a plane and then require lines to pass through them and the origin, there cannot be more lines than there are rationals.
– Kaz
May 17, 2021 at 15:50
• “$i$” is not a good letter to represent an arbitrary real number. Did somebody tell you to use that notation? May 17, 2021 at 17:16
1. Your answer doesn't make sense. The set $$\bigcup_{i\in\Bbb R}L_i$$ is a subset of $$\Bbb R^2$$ and you should say which set this is. It's $$\Bbb R^2\setminus\{(0,y)\mid y\in\Bbb R\setminus\{0\}\}$$.
2. By that argument, if $$L_i=\{(0,0)\}$$, then the set $$\{L_i\mid i\in\Bbb R\}$$ would be uncountable too, but, it fact, it consists of a single element. Your set is uncountable because, for each $$i\in\Bbb R$$, $$(1,i)\in L_i$$, but $$(1,i)\notin L_j$$ when $$j\ne i$$. So, when $$i\ne j$$, $$L_i\ne L_j$$, and therefore the map $$i\mapsto L_i$$ is injective. So, since $$\Bbb R$$ is uncountable, the set os all $$L_i$$'s is uncountable too.
3. The set $$T$$ is countable because $$\Bbb Q^2\setminus\{(0,0)\}$$ is countable. So, only countably many lines can contain points of $$\Bbb Q^2\setminus\{(0,0)\}$$.
Yeah, your solutions are (almost) correct. However, the notion of being "indexed" using an index set is not clearly defined. It should be defined as follows:
A set is $$S$$ indexed by $$R$$ if there is a bijective mapping between the index set $$R$$ and the set $$S$$, i.e. for $$i\neq j\in R$$ we have $$L_i\neq L_j\in S$$. The important point is the condition of $$i\neq j$$ implying $$L_i\neq L_j$$ because otherwise, one can just take the set of one line $$\{L\}$$ and assign this line to all real numbers. In that case we have also a mapping $$\mathbb{R} \to \{L\}$$, but that is not bijective and certainly $$\{L\}$$ is not uncountable infinite.
1. It is clear that the union of those lines is the union of all lines, that's basically the definition. Therefore I would argue that $$\bigcup_{i\in\mathbb{R}} L_i = \mathbb{R}^2 \setminus (\{0\} \times \mathbb{R}\setminus \{0\}) = (\mathbb{R}\setminus\{0\} \times \mathbb{R}) \cup \{(0,0)\}$$ as the union of those lines cover almost full $$\mathbb{R}^2$$ and only the $$x$$-axis (without $$(0,0)$$) is not included in that union.
2. See above paragraph about the indexing of those lines.
3. You can answer the third question similar to the second. If there is at least one rational point on a line, that point has a rational slope. Therefore, $$T$$ can be indexed by $$\mathbb{Q}$$ and is indeed countably infinite.
• I disagree with "all fine" for 2. I'm not downvoting because you address this in your remark, but "for $i\ne j$ we have $L_i\ne L_j$" is a crucial piece of the argument and is missing in the OP. May 17, 2021 at 17:53
• Okay, I will make it more clear in the whole answer. For me the meaning of "indexes" is as I described it, so I didn't see any problems with that until José answered May 17, 2021 at 17:55
• I hope it's clearer now May 17, 2021 at 18:02
Regarding 3., you're also right. $$T$$ is infinite countable because $$\mathbb Q$$ is infinite countable and because when a line passing through the origin also passes through a point having both rational coordinates, then the slope of such a line is also rational.
1.Union of such lines gives $$R^2$$ -{Y axis} .(0,0) is included. Since Y- axis is vertical is not included in it. For any other point in $$R^2$$ We can find a line passing through it and origin.
3.A line having rational numbers for both coordinates other than origin say $$(r_1,r_2)$$ will have a rational slope $$\frac{r_1}{r_2}$$. Thus we can create a bijection between rationals and such lines. So they are countable. | 1,624 | 5,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 69, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2022-21 | latest | en | 0.932438 |
https://en.1answer.info/6d617468-7a32373930383637 | 1,540,210,127,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515041.68/warc/CC-MAIN-20181022113301-20181022134801-00420.warc.gz | 670,048,031 | 16,754 | Probability of drawing the Jack of Hearts?
geocalc33 05/22/2018. 6 answers, 4.954 views
You have a standard deck of cards and randomly take one card away without looking at it and set it aside. What is the probability that you draw the Jack of Hearts from the pile now containing 51 cards? I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it.
fleablood 05/22/2018.
The Hard Way
The probability that the first card is not the Jack of Hearts is $\frac {51}{52}$ so the probability that the first card is not the Jack of Hearts and the second card is the Jack of Hearts is $\frac {51}{52}\times \frac 1{51}$.
The probability that the first card is the Jack of Hearts is $\frac 1{52}$ so the probability that the first card is the Jack of Hearts and the second card is the Jack of Hearts is $\frac 1{52}\times 0$.
So the total probability that the second card is Jack of Hearts is:
The probability that the second card is after the first card is not +
The probability that the second card is after the first card already was $$= \frac {51}{52}\times \frac 1{51} + \frac 1{52}\times 0 = \frac 1{52} + 0 = \frac 1{52}$$
That was the hard way.
The Easy Way
The probability that any specific card is any specific value is $\frac 1{52}$. It doesn't matter if it is the first card, the last card, or the 13th card. So the probability that the second card is the Jack of Hearts is $\frac 1{52}$. Picking the first card and not looking at, just going directly to the second card, putting the second card in an envelope and setting the rest of the cards on fire, won't make any difference; all that matters is the second card has a one in $52$ chance of being the Jack of Hearts.
Any thing else just wouldn't make any sense.
The thing is throwing in red herrings like "what about the first card?" doesn't change things and if you actually do try to take everything into account, the result, albeit complicated, will come out to be the same.
Graham Kemp 05/22/2018.
I'm confused by this question because if the card you removed from the pile was the Jack of Hearts then the probability would be zero so I'm not sure how to calculate it.
Well, to go the long way, you need to use the Law of Total Probability. Letting $X$ be the card you took away, and $Y$ the card you subsequently draw from the remaining deck.
$$\mathsf P(Y=\mathrm J\heartsuit) = \mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)~\mathsf P(X\neq \mathrm J\heartsuit)+0\cdot\mathsf P(X=\mathrm J\heartsuit)$$
Well, now, $\mathsf P(X\neq\mathrm J\heartsuit) = 51/52$ is the probability that the card taken is one from the 51 not-jack-of-hearts.
Also $\mathsf P(Y=\mathrm J\heartsuit\mid X\neq \mathrm J\heartsuit)$ is the probability that the subsequent selection is the Jack of Hearts when given that that card is among the 51 remaining cards.
Alternatively, the short way is to consider : When given a shuffled deck of 52 standard cards, what is the probability that the second from the top is the Jack of Hearts?
NicNic8 05/22/2018.
The probability is 1/52.
It doesn't matter that one card was moved somewhere else.
Philip C 05/22/2018.
I think the reason you're getting a bit confused is that you're conflating two different probabilities:
1. Your level of belief that you will choose the Jack of Hearts.
2. Whether or not you will choose the Jack of Hearts.
The first case depends entirely on your knowledge of the state of the deck, and won't be altered until you learn something new about the deck. If you take a random card from the deck and don't look at it, you've learned nothing and your level of belief remains at $1/52$.
The second case depends entirely on the actual, true state of your deck and can be altered if the deck is altered: if the random card you discard is the Jack of Hearts, the true chance of you subsequently randomly selecting that card is zero, otherwise it is $1/51$ as you said.
Incidentally, the second case is analagous to removing a random card and looking at it.
Paul Evans 05/23/2018.
Through extension: You randomly remove 51 cards from a shuffled deck and put them in an envelope. What is the probability the card left behind is the Jack of Hearts?
CiaPan 05/24/2018.
Suppose you don't remove the first card from the pile, but just remember its position, and choose the second card from any of remaining $51$ positions.
This is equivalent to the initial problem, isn't it?
Now, that in turn is equivalent to randomly chosing two positions in a pile and taking one card from the second chosen position.
But in all possible ordered pairs of distinct numbers from the set $\{1,2,\dots 52\}$, each number has exactly the same probability of appearing as the second one in a pair.
This implies you can drop choosing the first position at all — just choose randomly the 'second' position and see if it is the Jack of Hearts there.
This, however, is a simple random selection of a single element from the set, consequently the probability of taking the specific one is $1$ over the number of elements: $\frac 1{52}$. | 1,293 | 5,171 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-43 | latest | en | 0.94987 |
https://web2.0calc.com/questions/how-to-simplify-sin-and-cos | 1,547,956,111,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583690495.59/warc/CC-MAIN-20190120021730-20190120043730-00436.warc.gz | 683,578,460 | 6,320 | +0
# How to simplify? Sin and cos
0
623
4
Show that this: 6 sin2(x) * cosx - 3cosx
Is equal to: -3 cos (x) cos (2x)
Aug 5, 2017
#1
+27342
+2
Show that 6 sin2(x) * cosx - 3cosx = -3 cos (x) cos (2x)
Now cos(2x) = 1 - 2sin2(x) so the RHS of the above becomes:
-3 cos (x) cos (2x) → -3 cos (x) (1 - 2sin2(x)) → -3 cos (x) + 6cos(x)*sin2(x) which equals the LHS of the original.
Aug 5, 2017
#2
0
I looked up the relationship between cos and sin, where I found this: https://www.afit.edu/KNEEBOARD/images/halfangle2.gif
Which, I believe, is exactly the rule you used. At what age are you expected to know this, because I've never heard of such a thing.
Guest Aug 5, 2017
#4
+27342
+1
If you are being asked to tackle the question you specified then you should already have come across the half-angle trigonometric formulae. Such questions don't make sense otherwise!
Alan Aug 5, 2017
#3
+94544
+1
6 sin^2(x) * cosx - 3cosx = -3 cos (x) cos (2x)
We can work on both sides and show that they are equal.....
First, factor out cos x on both sides and write cos 2x as 1 - 2sin^2 x
cos x [ 6sin^2 x - 3 ] = cos x [ -3 (1 - 2sin^2 x) ]
Simplify
cos x [ 6sin^2 x - 3 ] = cos x [ 6sin^2 x - 3 ]
Aug 5, 2017
edited by CPhill Aug 5, 2017 | 502 | 1,256 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-04 | latest | en | 0.811656 |
https://www.numwords.com/words-to-number/en/5176 | 1,561,552,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00167.warc.gz | 824,072,711 | 1,852 | NumWords.com
# How to write Five thousand one hundred seventy-six in numbers in English?
We can write Five thousand one hundred seventy-six equal to 5176 in numbers in English
< Five thousand one hundred seventy-five :||: Five thousand one hundred seventy-seven >
Ten thousand three hundred fifty-two = 10352 = 5176 × 2
Fifteen thousand five hundred twenty-eight = 15528 = 5176 × 3
Twenty thousand seven hundred four = 20704 = 5176 × 4
Twenty-five thousand eight hundred eighty = 25880 = 5176 × 5
Thirty-one thousand fifty-six = 31056 = 5176 × 6
Thirty-six thousand two hundred thirty-two = 36232 = 5176 × 7
Forty-one thousand four hundred eight = 41408 = 5176 × 8
Forty-six thousand five hundred eighty-four = 46584 = 5176 × 9
Fifty-one thousand seven hundred sixty = 51760 = 5176 × 10
Fifty-six thousand nine hundred thirty-six = 56936 = 5176 × 11
Sixty-two thousand one hundred twelve = 62112 = 5176 × 12
Sixty-seven thousand two hundred eighty-eight = 67288 = 5176 × 13
Seventy-two thousand four hundred sixty-four = 72464 = 5176 × 14
Seventy-seven thousand six hundred forty = 77640 = 5176 × 15
Eighty-two thousand eight hundred sixteen = 82816 = 5176 × 16
Eighty-seven thousand nine hundred ninety-two = 87992 = 5176 × 17
Ninety-three thousand one hundred sixty-eight = 93168 = 5176 × 18
Ninety-eight thousand three hundred forty-four = 98344 = 5176 × 19
One hundred three thousand five hundred twenty = 103520 = 5176 × 20
Sitemap | 425 | 1,441 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2019-26 | latest | en | 0.802523 |
https://socratic.org/questions/how-do-you-prove-corresponding-angles-are-equal#203251 | 1,723,669,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641121834.91/warc/CC-MAIN-20240814190250-20240814220250-00651.warc.gz | 420,099,053 | 6,242 | # How do you prove corresponding angles are equal?
Dec 24, 2015
Suppose you have two parallel lines cut by a transversal.
Due to the straight angle (linear pair) theorem, we know that
{(mangle2+mangle3=180˚),(mangle5+mangle6=180˚):}
Thought the transitive property, we can say that
$m \angle 2 + m \angle 3 = m \angle 5 + m \angle 6 \textcolor{w h i t e}{\times \times}$ (1)
Though the alternate interior angles theorem, we know that
$m \angle 3 = m \angle 5$
Use substitution in (1):
$m \angle 2 + m \angle 3 = m \angle 3 + m \angle 6$
Subtract $m \angle 3$ from both sides of the equation
$m \angle 2 = m \angle 6$
$\therefore \angle 2 \cong \angle 6$
Thus $\angle 2$ and $\angle 6$ are corresponding angles and have proven to be congruent. | 243 | 756 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.779828 |
https://the360report.com/types-of-numbers-chart/ | 1,680,023,946,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948868.90/warc/CC-MAIN-20230328170730-20230328200730-00794.warc.gz | 648,363,911 | 14,007 | # Types of numbers chart
### What are the five types of numbers?
Types of numbers
• Natural Numbers (N), (also called positive integers, counting numbers, or natural numbers); They are the numbers {1, 2, 3, 4, 5, …}
• Whole Numbers (W).
• Integers (Z).
• Rational numbers (Q).
• Real numbers (R), (also called measuring numbers or measurement numbers).
Lesson Summary
### What is classification of numbers?
There are 5 classifications of real numbers: rational, irrational, integer, whole, and natural/counting.
### Is square root of 7 a real number?
How do we know that √7 is irrational? For a start, 7 is a prime number, so its only positive integer factors are 1 and 7 .
### What is the square of 7?
List of Perfect Squares
### What are the first 20 square numbers?
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 2, 5, 8, 10, 13, 17, 18, 20, 25, 26, 29, 32, 50, 65, 85, 125, 130, 145, 170, 185, 200,
### What is the square of 1 to 30?
Square, Cube, Square Root and Cubic Root for Numbers Ranging 0 – 100
### What are the 6 square numbers?
Informally: When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.
Perfect Square:
### What is the square of19 20?
In radical form: √x. In exponential form: (x) Largest Square Root: √20 = 4.4721.
Square Root 1 to 20 for Non-Perfect Squares.
### What is the square of 37?
What is the square root of 37? The square root of 37 is ± 6.082.
### What are all the real square roots of 100?
The square root of 100 is 10.
### How do I calculate square root?
The product property of square roots states that for any given numbers a and b, Sqrt(a × b) = Sqrt(a) × Sqrt(b). Because of this property, we can now take the square roots of our perfect square factors and multiply them together to get our answer. In our example, we would take the square roots of 25 and 16.
### How do you calculate a square?
For a square or rectangular room, you will first need to measure the length and then the width of the room. Then multiply the length and width. Length x Width = Area. So, if your room measures 11 feet wide x 15 feet long, your total area will be 165 square feet.
### Is square root of 14 a SURD?
The square root of 14 is expressed as √14 in the radical form and as (14)½ or (14)0.5 in the exponent form.
Square Root of 14 in radical form: √14.
### Is 14 a perfect square?
In mathematics, a square is a product of a whole number with itself. For instance, the product of a number 2 by itself is 4. In this case, 4 is termed as a perfect square. A square of a number is denoted as n × n.
Example 1.
### Is Square Root 36 a SURD?
The square root of 36 is 6. It is the positive solution of the equation x2 = 36. The number 36 is a perfect square.
Square Root of 36 in radical form: √36.
### Is square root of 17 a SURD?
1 Answer. √17 is not simplifiable and is irrational.
### Is 16 a SURD?
In general:
To simplify a surd, write the number under the root sign as the product of two factors, one of which is the largest perfect square. Note that the factor 16 is the largest perfect square. Recall that the numbers 1, 4, 9, 16, 25, 36, 49, are perfect squares.
Types of numbers chart
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Scroll to top | 994 | 3,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-14 | latest | en | 0.863752 |
https://www.physicsforums.com/threads/show-these-functions-are-2-pi-periodic.874096/ | 1,618,973,456,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039503725.80/warc/CC-MAIN-20210421004512-20210421034512-00200.warc.gz | 1,029,139,220 | 18,545 | # Show these functions are 2 pi periodic
Member warned about posting without the HW template
g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) )
show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ?
½( f(t)+f(-t) ) = g(t)
h(t+2π)=½( f(t+2π)-f(t-2π) )
½( f(t)-f(-t) ) = h(t) is this correct?
can anybody show me some similar examples please?
this is from a fourier series question paper.
thanks
LCKurtz
Homework Helper
Gold Member
Of course, this problem is false unless ##f(t) = f(t+2\pi)##, which you haven't told us.
g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) )
show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ?
It doesn't. If you put ##t+2\pi## in for ##t## in ##f(-t)## you get ##f(-(t+2\pi)) = f(-t -2\pi)##. Do you see why the numerator becomes ##f(t)+f(-t)##?
Delta2
RUber
Homework Helper
An example would be the exponential definition of the cosine function.
##\cos(x) = \frac{1}{2} (e^{ix} + e^{-ix}) ##
This is analogous to your problem for ##g(x)## with ##f(x) = e^{ix}##.
##\cos(x+2\pi) = \frac{1}{2} (e^{i(x+2\pi)} + e^{-i(x+2\pi)})\\
= \frac{1}{2} (e^{ix}e^{i2\pi} +e^{-ix}e^{-i2\pi})\\
= \frac{1}{2} (e^{ix} + e^{-ix})\\
= \cos(x) ##
Delta2
Homework Helper
Gold Member
Hint: prove that if a function has period T, then ##f(x-T)=f(x)## as long as f is defined in x-T.
Last edited:
haruspex
Homework Helper
Gold Member
2020 Award
Of course, this problem is false unless ##f(t) = f(t+2\pi)##,
or maybe the sum of a periodic function and an odd function.
LCKurtz
Homework Helper
Gold Member
Of course, this problem is false unless ##f(t) = f(t+2\pi)##, which you haven't told us.
or maybe the sum of a periodic function and an odd function.
Well, I assumed, which was also unstated, that problem was to prove that if ##f## has period ##2\pi##, then when you write ##f = g + h## with ##g## even and ##h## odd, that ##g## and ##h## have period ##2\pi##. In that context I'm not sure what you are suggesting.
And, annoyingly enough and a pet peeve of mine, the OP hasn't returned to the thread to clarify anything
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Are you writing the 2022 Waec Exams? Are you searching on How to Get Waec 2022 Economics Correct Answer? Are you looking for legit and sure Questions answers especially for Economics?
If yes, congratulations, for your search ends here as NAIJACLASS is the Number one Trusted Exam Specialists, An expert in all Exams Sector; we assist candidates to pass various examinations with ease.
======================
“”ECONOMICS EXAM””
=====================
ECONOMICS OBJ
01-10: DBCDCDBCCD
11-20: BAABACDBAB
21-30: BCBDBCCACC
31-40: CBBDDBCDBA
41-50: BBCBDCCACC
COMPLETED
======+++++===========
*INSTRUCTIONS TO FOLLOW*
You’re To Answer Four Questions In All, One Question Only From Section A And Any Other Three From Section B
(1a)
GROSS DOMESTIC PRODUCT (GDP)
Wages and salaries = 250
Income from self employment =120
Rent = 25
Interest = 10
Royalties = 2
Profit and dividend = 35
GDP = 442
(1b)
GROSS NATIONAL PRODUCT (GNP)
GNP = GDP + Net factor income
GNP = 442+ income received from Abroad – Income paid Abroad
GNP = 442+(50-75)
GNP = 442+(-25)
GNP = 442 – 25
GNP = 417
(1c)
NET NATIONAL PRODUCT (NNP)
NNP = GNP – Depreciation
NNP = 417 – 3
NNP = 414
=============
NUMBER 1 a
NUMBER 1 b and c
NUMBER 2
(2ai)
X is demand curve
Y is supply curve
Z is supply curve
(2aii)
The supply curve shift from Y to Z
(2bi)
Before the introduction of subsidy
\$15×20 = \$300
(2bii)
After the introduction of subsidy
\$10×40 = \$400
(2c)
Percentage increase = \$400-\$300 = \$100
100/300 × 100/1 = 33.3%
(2di)
% change in quantity demanded/ % change in price = 100/33.3 = 3.0
% change in quantity demanded= change/old × 100/1
= 20/20 × 100/1
= 100%
% change in price = change/old × 100/1
= 5/15 × 100/1 = 33.3%
(2dii)
It is elastic because the coefficient is greater than 1
========================
(8a)
Economic integration is a form of economic cooperation among countries with the aim of achieving higher economic growth and development. This economic cooperation includes the reduction or elimination of trade barriers and the coordination of monetary and fiscal policies.
(8b)
(PICK ANY THREE)
(i) Language barrier: Member states are polarized into English, French and Portuguese languages. This is seen as a barrier to the unity of the peoples of the sub-region;
(ii) Fear of domination : Some of the smaller nations have great fear that big nations in the Community might dominate them. Hence, there is mutual suspicion among them.
(iii) Influence of foreign powers and ex-colonial masters: Member states depend on them economically.
(iv) Political Instability in member states: This results in different approach to ECOWAS issues by leaders of member states.
(v) Inadequate finance: Many members states do not fulfill their financial obligation as at and when due. This is crippling the effective operation
of ECOWAS.
(8c)
(PICK ANY THREE)
(i) Road construction between big cities: The highways Lagos-Abidjan, Nouakchott-Lagos have made commuting much easier that it was before. The road network Elubo – Alflao – Lagos is one of the achievements.
(ii) Relationship establishment between Anglophone and Francophone: The relations between the Anglophone and Francophone have been stabilized. It has been done through the ECOWAS passport. It has significantly eased the movement of people within those areas.
(iii) Provision of Telephone network for the member states: Nowadays, interconnection is available for all countries of the African Union.
(iv) No strict borders for nations and trades: ECOWAS has enabled West Africans to easily move among the West African countries. Also, the trade barrier has been removed. People, goods and services travel around the countries with ease.
(v) Peace and security throughout the sub-region: The state of peace is kept by ECOWAS Monitoring Groups. The zones of conflicts are not so troublesome compared to what it used to be.
=======================
(3a)
Limited liability is a type of legal structure for an organization where a corporate loss will not exceed the amount invested in a partnership or limited liability company.
(3b)
-PUBLIC JOINT-STOCK COMPANY-
(i) It implies a company that is listed on a recognized stock exchange and whose shares are traded openly by the public.
(ii) The minimum number of members are 7
(iii) The maximum number of members are unlimited
(iv) The minimum number of directors are 3.
-PRIVATE JOINT-STOCK COMPANY-
(i) It refers to the company which is not listed on a stock exchange and the shares are held privately by the members concerned.
(ii) Minimum number of members are 2
(iii) Maximum number of members are 200, except in case of one person company
(iv) Minimum number of directors are 2.
(3c)
(Pick Three Only)
(i) Personal savings
(ii) Retained profits
(iii) Working capital
(iv) Sale of assets
(v) Bank loans.
===================
(4a)
(i)Labour force: Labour force can be defined as the total number of persons available to supply the labour for the production of economic goods and services. In other words, it is the total number of people of working age in a country who are able and willing by law to work.
(ii)Efficiency of labour: Efficiency of labour may be defined as the ability of labour to increase output without increasing the quantity of labour. Increase in efficiency is usually expressed in terms of increase in output of labour within a shorter period of time without any fall in the quality of goods and services produced.
(4b)
(PICK ANY FIVE)
(i)Age structure of the population: The structure of a country’s population is a significant determinant of the size of the labour force. The lower the dependant people, the higher the supply of labour and vice versa. In otherword, the labour force will increase in a country with a greater number of its people between the ages of 18 and 65 years.
(ii)Role of women in the society: In some societies, women are usually prevented from engaging in gainful employment because of religious belief, social and cultural factors and this affects the size of labour force.
(iii)Number of working hours and working days: The number of working hours per day and the number of working days in a week or a year also helps to determine the supply of labour.
(iv) The number of disabled: When the number of disabled persons is high especially within the working population, the supply for labour will be low.
(v) The number of people unwilling to work: There are certain number of able bodied people who are also between the age bracket of 18 and 65 years but are unwilling to work. If their population is high, it will affect the size of supply of labour
(vi) Migration: The rate of migration can also affect the size of labour force. If the rate at which the working population leaves a country is higher than the rate at which people come in, it will lead to reduction in the supply of labour.
(vii)Trade union activities: The activities of trade union may also affect the supply of supply. For example, when a long period of training is imposed on a certain trade, this may discourage people from engaging in such trade or profession leading to a reduction in supply of labour.
(viii)Government policies: Certain government policies can affect the supply of labour. E.g. specific laws are made to exclude children and women from working in ministries. This can reduce the supply of labour to that area or field.
=============================6a i.) Specific tax
A specific tax is a fixed amount of tax placed on a particular good. It is also referred to as a per-unit tax, and the tax will depend on the quantity sold (not price).
ii.) Value-added tax (VAT) : is a flat tax levied on an item. It is similar to a sales tax in some respects, except that with a sales tax, the full amount owed to the government is paid by the consumer at the point of sale. With a VAT, portions of the tax amount are paid by different parties to a transaction
=========================
(5a)
Demand schedule is the table that shows the relationship between quantity demanded and price of goods.
Or
It can be defined as the list showing the number of quantities of a commodity demanded at various prices.
(5b)
(i) Effective demand: this is the type of demand which is supported with the ability and willingness to pay.
(ii). Composite Demand: Demand is composite when a commodity can be used for more than one purpose
(iii)Derived Demand: It occurs when a commodity is wanted not for the satisfaction it yields but for the reason that they are useful in producing another commodity.
(7a)
MORTGAGE BANK
(i) A mortgage bank is a financial institution that specializes in granting loans to individual and corporate bodies for building purposes. Such loans are repaid by installments and can be spread over several years. On the other hand.
MERCHANT BANK
Merchant bank is a financial institution that provides specialized services like acceptance of bills of exchange, corporate finance; portfolio management, equipment leasing and acceptance of deposits
(7aii)
COMMERCIAL BANK
A commercial banks are financial institutions that per-form the services of holding people’s money and ac-counts and using such money to make loans and other financial services available to customers. The loans are usually for short and medium terms.
DEVELOPMENT BANK
Development bank is a financial institution set up to pro-vide long term loans to groups of individuals and governments for development projects. They provide financial assistance in high risk, low profit and long gestation period investments which are unattractive to commercial banks.
(7b)
[PICK ANY FOUR]
(i) Acceptance of deposit:
Customers money can be kept in any of the three different types of commercial bank accounts-savings, current or demand deposit and time or fixed deposit accounts.
(ii) Lending of money:
Commercial banks make avail-able loans, overdraft and discount bills of exchange for their customers.
(iii) Commercial banks provide:
facilities for domestic and foreign remittances. Such transfers can be done telegraphically or by cable, or by ordinary sail or through traveller’s cheques.
(iv) Commercial banks provide:
facilities for the safe keeping of valuables
(v) Trust services for individuals and organizations:
Trust services include the management of trust funds.
(vi) Agency services:
The banks act as agents for their customers in the purchase and sale of securities.
(vii) Money creating function:
Deposits received can be given out as credit to customers which in turn create further deposits.
(viii) Commercial banks offer advisory services to customers.
=========================
NUMBER 2
==========================
1 Comment
1. obinna says:
thanks for ur help pls correct answer GOD BLESS U FOR ME AMEN ? | 2,434 | 10,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.830963 |
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MTTSNG Discussion and analysis of MTTSNGs.
08-30-2011, 09:50 PM #2 Grindation veteran Join Date: Aug 2010 Location: At the top Posts: 2,077 Re: Risk of Ruin Simulator and Calculator 1st. will test tomorrow. looking forward to it
08-31-2011, 12:40 AM #4 Camzace old hand Join Date: Jun 2010 Posts: 1,351 Re: Risk of Ruin Simulator and Calculator This is very cool. Will try it out right now. Am familiar with R but probably will have forgotten how to use it.
11-19-2011, 12:06 PM #5 self journeyman Join Date: Apr 2011 Location: Stars Posts: 323 Re: Risk of Ruin Simulator and Calculator I have used your function for 6-max Turbo Single Table Tournaments. I input this command in R: Code: RoR.sim(prizes=c(8.94, 3.20, -3.50, -3.50, -3.50, -3.50), probs=c(.1538, .2906, .2051, .1282, .1538, .0685), sims=1000, startBR=40*3.5) The results: Results Sims 1000.0000000 ROI % 0.1029406 Traditional Ruin Prob. 0.2689182 Simulated Ruin Prob. 0.0130000 Simulated Win Prob. 0.9870000 How can we explain the big gap between Traditional RoR and Simulated RoR? 1% Sim RoR seems too low for BR = 40 BIs, while Trad RoR is pretty high @ 26%. I have used ev++ calculator and used same BR and ROI with a Standard Deviation of 1.7 and got a 6.28% RoR, which is in between the two results above. Another thing is HEM2 shows my Std Dev @ 5.37 which brings about crazy results for RoR. I couldn't find a typical Std Dev, so I started this thread. I am very interested in this matters, mainly because of my challenge here. Thanks for your work Sherman, I appreciate it!
11-19-2011, 12:53 PM #6 Sherman Carpal \'Tunnel Join Date: Jun 2005 Location: Psychology Department Posts: 7,764 Re: Risk of Ruin Simulator and Calculator Can you tell me a bit about the payout structure? I've actually changed some pieces of this function and created another function that may be of some use here. The new functions are improved. I need the payout structure and the BI (it looks like \$3.50 is the BI). But I need the payout structure without the BIs subtracted. If I knew what the rake was (is it 10%?) I could do it myself, but I don't know what it is.
11-20-2011, 01:26 AM #7
self
journeyman
Join Date: Apr 2011
Location: Stars
Posts: 323
Re: Risk of Ruin Simulator and Calculator
Quote:
Originally Posted by Sherman Can you tell me a bit about the payout structure? I've actually changed some pieces of this function and created another function that may be of some use here. The new functions are improved. I need the payout structure and the BI (it looks like \$3.50 is the BI). But I need the payout structure without the BIs subtracted. If I knew what the rake was (is it 10%?) I could do it myself, but I don't know what it is.
Yes, buy-in is \$3.50 (\$3.19+0.31 rake, which is ~9.71%)
1st pays \$12.45 and 2nd is \$6.69
11-20-2011, 10:58 AM #8
Sherman
Carpal \'Tunnel
Join Date: Jun 2005
Location: Psychology Department
Posts: 7,764
Re: Risk of Ruin Simulator and Calculator
Quote:
Originally Posted by self Yes, buy-in is \$3.50 (\$3.19+0.31 rake, which is ~9.71%) 1st pays \$12.45 and 2nd is \$6.69 Thanks for the reply.
Ok. I've got some info for you.
Here are the results:
Code:
Results
Sims 1000.0000000
ROI % 10.2549714
SD 4.6981403
Iter. Ruin % 0.8533218
Sim. Ruin % 0.7000000
Sim. Win % 99.3000000
First, note that the Ruin % are in percentage. So 1.05 is 1.05%. Not 105%. But basically all three methods indicate that you have around a 1% Risk of Ruin (assuming all of your information is accurate).
There are three ruin % here including the Traditional Ruin % and the Simulated Ruin %. They are the same as before. The third Ruin % is the Iter. Ruin % is done through the use of iterative methods. I have written a sub-function that calls it:
Code:
RoR.iter <- function(prizes, probs, BI, BR) {
if(sum(prizes < 0) > 0) {stop("Prizes must not be below 0.")}
prizes.BI <- prizes/BI
BR.BI <- BR/BI
ROI.BI <- sum((prizes.BI-1)*probs)
ifelse(ROI.BI < 0, R <- 1, R <- uniroot(function(x) sum(probs*x^prizes.BI) - x, c(0,.9999999), tol=1e-70)\$root)
RoR <- R^BR.BI
return(RoR)
}
The key here is finding the value R because Rnumber of BIs in bankroll is the Risk of Ruin. Finding the value of R is complicated and iterative methods are used to find it (i.e. "try a value and see if it works, if not adjust and find a better fitting value, stop when the fit is good."). Iterative methods are great unless there is more than one solution. For most of the problems we are talking about there is only one best solution though.
The advantage of the RoR.iter() is that it never goes over 100% (like the traditional RoR sometimes does). It is well behaved.
In any case, the new RoR.sim() function includes this sub-function:
Code:
RoR.sim <- function(prizes, probs, BI, BR, win=5*BR, sims=1000) {
if(BR < BI) {stop("BR must be higher than BI.")}
if(sum(prizes < 0) > 0) {stop("Prizes must not be below 0.")}
if(sum(probs) != 1) {stop("Finish probabilities do not total 1.00. Please adjust input probabilities.")}
if(length(prizes) != length(probs)) {stop("prizes and probs must be the same length (have the same number of elements).")}
ROI <- sum((prizes-BI)*probs) / BI
VAR <- sum((prizes-BI)^2 * probs) - ROI^2
if(sims==FALSE) {
iterRoR <- RoR.iter(prizes, probs, BI, BR)
colnames(out) <- "Results"
return(out)
}
res <- rep(NA, sims)
for (i in 1:sims) {
simBR <- BR
if(i == .25*sims) {print("25% of sims complete.")}
if(i == .5*sims) {print("50% of sims complete.")}
if(i == .75*sims) {print("75% of sims complete.")}
while(simBR >= BI & simBR < win) {
simBR <- simBR + sample((prizes-BI), 1, T, prob=probs)
}
res[i] <- simBR
}
iterRoR <- RoR.iter(prizes, probs, BI, BR)
out <- rbind(sims, BI, BR/BI, ROI*100, sqrt(VAR), tradRoR*100, iterRoR*100, (sum(res < BI) / sims)*100, (sum(res >= win) / sims)*100)
rownames(out) <- c("Sims", "BuyIn", "BR (in Buyins)", "ROI %", "SD", "Trad. Ruin %", "Iter. Ruin %", "Sim. Ruin %", "Sim. Win %")
colnames(out) <- "Results"
return(out)
}
So you have to run the RoR.iter() function first (to store it in R) for the RoR.sim() function to work.
But here is the code I used with the functions for your data:
Code:
RoR.sim(prizes=c(12.45, 6.69, 0, 0, 0 , 0), probs=c(.1538, .2906, .2051, .1282, .1538, .0685), BI=3.5, sims=1000, BR=40*3.5, win=1000)
Notice now that you list "prizes" instead of "profits." I think this makes the function more user-friendly. The downside is that you have to explicitly list the BI now, but that is okay in my opinion.
Here is another function I built to examine situations where you are going to play X amount of MTTs and you have a particular finish distribution. This function assumes you can always afford to play another (so it isn't a risk of ruin calculator). It is really useful for understanding variability in MTTs. It draws pretty graphs as well.
Code:
MTT.sim <- function(prizes, probs, BI, games=1000, sims=1000, CI=.95, plots=500, ignore.prob=F) {
if(sum(prizes < 0) > 0) {stop("Prizes must not be below 0.")}
if(ignore.prob==F) {
if(sum(probs) != 1) {stop("Finish probabilities do not total 1.00. Please adjust input probabilities. Or set ignore.prob==T.")}
}
if(length(prizes) != length(probs)) {stop("prizes and probs must be the same length (have the same number of elements).")}
if(CI >= 1.0 | CI <= 0) {stop("CI must be between .00 and 1.00.")}
LL <- 1-CI / 2
UL <- 1-LL
profits <- prizes - BI
profits.BI <- profits / BI
ROI <- sum(profits*probs) / BI
ROI.BI <- sum(profits.BI*probs)
ITM <- sum(probs[prizes>0])
VAR <- sum(profits^2 * probs) - ROI^2
VAR.BI <- sum(profits.BI^2 * probs) - ROI.BI^2
res <- matrix(nrow=games, ncol=sims)
cumres <- matrix(nrow=games, ncol=sims)
dswings <- rep(NA, sims)
uswings <- rep(NA, sims)
for(i in 1:sims) {
res[,i] <- sample(profits, games, T, prob=probs)
cumres[,i] <- cumsum(res[,i])
dswings[i] <- max(rle(res[,i])\$lengths[rle(res[,i])\$values<0])
uswings[i] <- max(rle(res[,i])\$lengths[rle(res[,i])\$values>0])
}
cumres.BI <- cumres / BI
op <- par(mfrow=c(2,2), font.main=1)
plot(cumsum(rep(ROI*BI,games)), type="l", lty=5, lwd=4, xlim=c(0,games),
ylim=c(min(cumres) - 3*BI, max(cumres) + 3*BI), main=paste(plots, " Simulated Profits"),
xlab="Tournaments", ylab="Profits (\$)")
for(j in 1:plots) {
lines(cumres[,j], col=sample(sims, sims, T))
}
lines(cumsum(rep(ROI*BI,games)), type="l", lty=5, lwd=4)
legend("topleft", legend="Expected Line", col="black", lty=5, bty="n", lwd=4)
plot(density(cumres[games,]), main="Distribution of Profits", ylab="Density", xlab="Profits (\$)", col="green")
polygon(density(cumres[games,]), col="green")
plot(cumsum(rep(ROI.BI,games)), type="l", lty=5, lwd=4, xlim=c(0,games),
ylim=c(min(cumres.BI) - 3, max(cumres.BI) + 3), main=paste(plots, " Simulated Profits"),
for(j in 1:plots) {
lines(cumres.BI[,j], col=sample(sims, sims, T))
}
lines(cumsum(rep(ROI.BI,games)), type="l", lty=5, lwd=4)
legend("topleft", legend="Expected Line", col="black", lty=5, bty="n", lwd=4)
plot(density(cumres.BI[games,]), main="Distribution of Profits", ylab="Density", xlab="Profits (BuyIns)", col="green")
polygon(density(cumres.BI[games,]), col="green")
out.reg <- rbind(sims, games, BI, ITM*100, ROI*100, sqrt(VAR),
min(cumres[games,]), quantile(cumres[games,],UL), mean(cumres[games,]),
median(cumres[games,]), quantile(cumres[games,],LL), max(cumres[games,]),
max(dswings), max(uswings), ((sum(cumres[games,] < 0))/sims)*100)
out.BI <- rbind(sims, games, 1, ITM*100, ROI.BI*100, sqrt(VAR.BI),
min(cumres.BI[games,]), quantile(cumres.BI[games,],UL), mean(cumres.BI[games,]),
median(cumres.BI[games,]), quantile(cumres.BI[games,],LL), max(cumres.BI[games,]),
max(dswings), max(uswings), ((sum(cumres.BI[games,] < 0))/sims)*100)
out <- cbind(out.reg, out.BI)
colnames(out) <- c("Results \$", "Results BIs")
rownames(out) <- c("Simulations", "Tournies per Sim", "BuyIn", "ITM %", "ROI %", "SD", "Worst Profit",
"CI Lowerbound", "Avg. Profit", "Median Profit", "CI Upperbound", "Best Profit",
"Longest OOTM Streak", "Longest ITM Streak", "% Finishes with Loss")
return(out)
}
Running it on your data I get the following results (1000 simulations of playing 1000 \$3.50 STTs):
Code:
MTT.sim(prizes=c(12.45, 6.69, 0, 0, 0 , 0), probs=c(.1538, .2906, .2051, .1282, .1538, .0685), BI=3.5)
Results \$ Results BIs
Simulations 1000.00000 1000.000000
Tournies per Sim 1000.00000 1000.000000
ITM % 44.44000 44.440000
ROI % 10.25497 10.254971
SD 4.69814 1.338724
Worst Profit -89.51000 -25.574286
CI Lowerbound 354.61000 101.317143
Avg. Profit 364.47523 104.135780
Median Profit 365.20000 104.342857
CI Upperbound 373.57000 106.734286
Best Profit 815.08000 232.880000
Longest OOTM Streak 25.00000 25.000000
Longest ITM Streak 12.00000 12.000000
% Finishes with Loss 0.20000 0.200000
11-25-2011, 02:00 PM #9 arrtvandelay banned Join Date: Nov 2010 Location: Brazil Posts: 437 Re: Risk of Ruin Simulator and Calculator bump
07-03-2013, 04:23 PM #10 rer newbie Join Date: Jun 2013 Posts: 16 Re: Risk of Ruin Simulator and Calculator cool!
11-18-2013, 08:31 PM #11 onlinepokerwiz banned Join Date: Sep 2013 Posts: 380 Re: Risk of Ruin Simulator and Calculator how can i run this for a 180s SNG?
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Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top | 4,527 | 15,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.871691 |
https://totheinnovation.com/serialize-and-deserialize-bst-leetcode-solution/ | 1,717,038,227,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059418.31/warc/CC-MAIN-20240530021529-20240530051529-00362.warc.gz | 497,037,559 | 36,878 | # Serialize and Deserialize BST LeetCode Solution
Here, We see Serialize and Deserialize BST LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
# List of all LeetCode Solution
## Problem Statement
Serialization is converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary search tree. There is no restriction on how your serialization/deserialization algorithm should work. You need to ensure that a binary search tree can be serialized to a string, and this string can be deserialized to the original tree structure.
The encoded string should be as compact as possible.
Example 1:
Input: root = [2,1,3]
Output: [2,1,3]
Example 2:
Input: root = []
Output: []
## Serialize and Deserialize BST LeetCode SolutionC++
``````class Codec {
public:
void serializehelper(TreeNode* root, string& s){
if(root==nullptr) return;
s+=to_string(root->val) + ",";
serializehelper(root->left, s);
serializehelper(root->right, s);
}
string serialize(TreeNode* root) {
if(root==nullptr) return "";
string s="";
serializehelper(root, s);
return s;
}
int convertStringtoInt(string& data, int& pos){
pos=data.find(',');
int value=stoi(data.substr(0, pos));
return value;
}
TreeNode* deserializehelper(string& data, int min, int max) {
if(data.size()==0) return nullptr;
int pos=0;
int value = convertStringtoInt(data, pos);
if (value < min || value > max) return nullptr;
TreeNode* tnode = new TreeNode(value);
data=data.substr(pos+1);
tnode->left=deserializehelper(data, min, tnode->val);
tnode->right=deserializehelper(data, tnode->val, max);
return tnode;
}
TreeNode* deserialize(string data) {
if(data=="") return nullptr;
return deserializehelper(data, INT_MIN, INT_MAX);
}
};```Code language: PHP (php)```
## Serialize and Deserialize BST LeetCode SolutionJava
``````class Codec {
public String serialize(TreeNode root) {
StringBuilder res = new StringBuilder();
helpserialize(root,res);
return res.toString();
}
private void helpserialize(TreeNode root, StringBuilder res){
if(root == null){
res.append("x,");
return ;
}
res.append(root.val);
res.append(',');
helpserialize(root.left, res);
helpserialize(root.right, res);
}
public TreeNode deserialize(String data) {
return helpdeserialize(q);
}
private TreeNode helpdeserialize(Deque<String> q){
String res = q.remove();
if(res.equals("x")){
return null;
}
TreeNode root = new TreeNode(Integer.parseInt(res));
root.left = helpdeserialize(q);
root.right = helpdeserialize(q);
return root;
}
}```Code language: JavaScript (javascript)```
## Serialize and Deserialize BST SolutionJavaScript
``````var serialize = function(root) {
let preorder = [];
let dfs = function(node) {
if (node==null) return;
preorder.push(node.val);
dfs(node.left);
dfs(node.right);
}
dfs(root);
return preorder.join(',');
};
var deserialize = function(data) {
if (data == '') return null;
let preorder = data.split(',');
let recur = function(lower, upper) {
if (Number(preorder[0]) < lower || Number(preorder[0]) > upper) return null;
if (preorder.length == 0) return null;
let root = new TreeNode(preorder.shift());
root.left = recur(lower, root.val);
root.right = recur(root.val, upper);
return root;
}
return recur(-Infinity, Infinity);
};```Code language: JavaScript (javascript)```
## Serialize and Deserialize BST SolutionPython
``````class Codec:
def serialize(self, root):
path_of_preorder = []
def helper( node ):
if node:
path_of_preorder.append( node.val )
helper( node.left )
helper( node.right )
helper( root )
return '#'.join( map(str, path_of_preorder) )
def deserialize(self, data):
if not data:
return None
node_values = deque( int(value) for value in data.split('#') )
def helper( lower_bound, upper_bound):
if node_values and lower_bound < node_values[0] < upper_bound:
root_value = node_values.popleft()
root_node = TreeNode( root_value )
root_node.left = helper( lower_bound, root_value )
root_node.right = helper( root_value, upper_bound )
return root_node
return helper( float('-inf'), float('inf')) ``````
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https://usethinkscript.com/threads/adding-addorder-to-existing-parabolicsar-script.7701/ | 1,726,867,442,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00217.warc.gz | 548,370,836 | 18,982 | fernandesj
New member
VIP
Hello All, Newbie here. I am trying to use the Floating PL study to see how this works (for /ES) and I would like to add the AddOrder functionality to the ParabolicSAR study which I have pasted below. I know I have no idea what I am doing with thinkscript
but what I would like to do is
1. when the state goes from short to long or from init to long I would like to buy one futures contract.
2. when the state goes from long to short I would like to sell (to close) one futures contract
3. when the initial state is short (no action at this time)
I hope someone can help straighten me out.
### Code Below ###
Ruby:
``````input accelerationFactor = 0.02;
input accelerationLimit = 0.2;
input quantity = 1;
assert(accelerationFactor > 0, "'acceleration factor' must be positive: " + accelerationFactor);
assert(accelerationLimit >= accelerationFactor, "'acceleration limit' (" + accelerationLimit + ") must be greater than or equal to 'acceleration factor' (" + accelerationFactor + ")");
def state = {default init, long, short};
def extreme;
def SAR;
def acc;
switch (state[1]) {
case init:
state = state.long;
acc = accelerationFactor;
extreme = high;
SAR = low;
case short:
if (SAR[1] < high)
then {
state = state.long;
acc = accelerationFactor;
extreme = high;
SAR = extreme[1];
} else {
state = state.short;
if (low < extreme[1])
then {
acc = min(acc[1] + accelerationFactor, accelerationLimit);
extreme = low;
} else {
acc = acc[1];
extreme = extreme[1];
}
SAR = max(max(high, high[1]), SAR[1] + acc * (extreme - SAR[1]));
}
case long:
if (SAR[1] > low)
then {
state = state.short;
acc = accelerationFactor;
extreme = low;
SAR = extreme[1];
} else {
state = state.long;
if (high > extreme[1])
then {
acc = min(acc[1] + accelerationFactor, accelerationLimit);
extreme = high;
} else {
acc = acc[1];
extreme = extreme[1];
}
SAR = min(min(low, low[1]), SAR[1] + acc * (extreme - SAR[1]));
}
}
plot parSAR = SAR;
parSAR.SetPaintingStrategy(PaintingStrategy.POINTS);
parSAR.SetDefaultColor(GetColor(5));
Last edited by a moderator:
Solution
Hello All, Newbie here. I am trying to use the Floating PL study to see how this works (for /ES) and I would like to add the AddOrder functionality to the ParabolicSAR study which I have pasted below. I know I have no idea what I am doing with thinkscript
but what I would like to do is
1. when the state goes from short to long or from init to long I would like to buy one futures contract.
2. when the state goes from long to short I would like to sell (to close) one futures contract
3. when the initial state is short (no action at this time)
I hope someone can help straighten me out.
See if this helps. This has to be copied and pasted in the Strategy Tab.
Ruby:
``````#
# TD Ameritrade IP Company, Inc. (c) 2008-2021
#...``````
Hello All, Newbie here. I am trying to use the Floating PL study to see how this works (for /ES) and I would like to add the AddOrder functionality to the ParabolicSAR study which I have pasted below. I know I have no idea what I am doing with thinkscript
but what I would like to do is
1. when the state goes from short to long or from init to long I would like to buy one futures contract.
2. when the state goes from long to short I would like to sell (to close) one futures contract
3. when the initial state is short (no action at this time)
I hope someone can help straighten me out.
See if this helps. This has to be copied and pasted in the Strategy Tab.
Ruby:
``````#
# TD Ameritrade IP Company, Inc. (c) 2008-2021
#
input accelerationFactor = 0.02;
input accelerationLimit = 0.2;
Assert(accelerationFactor > 0, "'acceleration factor' must be positive: " + accelerationFactor);
Assert(accelerationLimit >= accelerationFactor, "'acceleration limit' (" + accelerationLimit + ") must be greater than or equal to 'acceleration factor' (" + accelerationFactor + ")");
def state = {default init, long, short};
def extreme;
def SAR;
def acc;
switch (state[1]) {
case init:
state = state.long;
acc = accelerationFactor;
extreme = high;
SAR = low;
case short:
if (SAR[1] < high)
then {
state = state.long;
acc = accelerationFactor;
extreme = high;
SAR = extreme[1];
} else {
state = state.short;
if (low < extreme[1])
then {
acc = Min(acc[1] + accelerationFactor, accelerationLimit);
extreme = low;
} else {
acc = acc[1];
extreme = extreme[1];
}
SAR = Max(Max(high, high[1]), SAR[1] + acc * (extreme - SAR[1]));
}
case long:
if (SAR[1] > low)
then {
state = state.short;
acc = accelerationFactor;
extreme = low;
SAR = extreme[1];
} else {
state = state.long;
if (high > extreme[1])
then {
acc = Min(acc[1] + accelerationFactor, accelerationLimit);
extreme = high;
} else {
acc = acc[1];
extreme = extreme[1];
}
SAR = Min(Min(low, low[1]), SAR[1] + acc * (extreme - SAR[1]));
}
}
plot parSAR = SAR;
parSAR.SetPaintingStrategy(PaintingStrategy.POINTS);
parSAR.SetDefaultColor(GetColor(5));
#1. when the state goes from short to long or from init to long I would like to buy one futures contract.
def buy = if state[1] == state.short and state == state.long or state[1] == state.init and state == state.long then 1 else 0;
#2. when the state goes from long to short I would like to sell (to close) one futures contract
def sell = if state[1] == state.long and state == state.short then 1 else 0;
AddOrder(type = OrderType.SELL_TO_CLOSE, condition = sell, name = "SAR Sell Strategy", arrowColor = Color.YELLOW);
#3. when the initial state is short (no action at this time)``````
Not the exact question you're looking for?
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TheInfoList
OR:
In
mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in modern mathematics ...
, a sheaf is a tool for systematically tracking data (such as sets, abelian groups, rings) attached to the open sets of a topological space and defined locally with regard to them. For example, for each open set, the data could be the ring of continuous functions defined on that open set. Such data is well behaved in that it can be restricted to smaller open sets, and also the data assigned to an open set is equivalent to all collections of compatible data assigned to collections of smaller open sets covering the original open set (intuitively, every piece of data is the sum of its parts). The field of mathematics that studies sheaves is called sheaf theory. Sheaves are understood conceptually as general and abstract objects. Their correct definition is rather technical. They are specifically defined as sheaves of sets or as sheaves of rings, for example, depending on the type of data assigned to the open sets. There are also maps (or morphisms) from one sheaf to another; sheaves (of a specific type, such as sheaves of abelian groups) with their morphisms on a fixed topological space form a category. On the other hand, to each continuous map there is associated both a direct image functor, taking sheaves and their morphisms on the domain to sheaves and morphisms on the codomain, and an inverse image functor operating in the opposite direction. These functors, and certain variants of them, are essential parts of sheaf theory. Due to their general nature and versatility, sheaves have several applications in topology and especially in algebraic and
differential geometry Differential geometry is a mathematical discipline that studies the geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics Mathematics is an area of knowledge that includes the topics of numbers, form ...
. First, geometric structures such as that of a differentiable manifold or a scheme can be expressed in terms of a sheaf of rings on the space. In such contexts, several geometric constructions such as vector bundles or divisors are naturally specified in terms of sheaves. Second, sheaves provide the framework for a very general cohomology theory, which encompasses also the "usual" topological cohomology theories such as singular cohomology. Especially in algebraic geometry and the theory of complex manifolds, sheaf cohomology provides a powerful link between topological and geometric properties of spaces. Sheaves also provide the basis for the theory of ''D''-modules, which provide applications to the theory of differential equations. In addition, generalisations of sheaves to more general settings than topological spaces, such as Grothendieck topology, have provided applications to
mathematical logic Mathematical logic is the study of formal logic within mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantiti ...
and to
number theory Number theory (or arithmetic or higher arithmetic in older usage) is a branch of pure mathematics devoted primarily to the study of the integer An integer is the number zero (), a positive natural number In mathematics, the natura ...
.
Definitions and examples
In many mathematical branches, several structures defined on a topological space $X$ (e.g., a differentiable manifold) can be naturally ''localised'' or ''restricted'' to open
subset In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented i ...
s $U \subset X$: typical examples include continuous real-valued or complex-valued functions, $n$-times differentiable (real-valued or complex-valued) functions, bounded real-valued functions, vector fields, and sections of any vector bundle on the space. The ability to restrict data to smaller open subsets gives rise to the concept of presheaves. Roughly speaking, sheaves are then those presheaves, where local data can be glued to global data.
Presheaves
Let $X$ be a topological space. A ''presheaf of sets'' $F$ on $X$ consists of the following data: *For each open set $U$ of $X$, a set $F\left(U\right)$. This set is also denoted $\Gamma\left(U, F\right)$. The elements in this set are called the ''sections'' of $F$ over $U$. The sections of $F$ over $X$ are called the ''global sections'' of $F$. *For each inclusion of open sets $V \subseteq U$, a function $\operatorname_ \colon F\left(U\right) \rightarrow F\left(V\right)$. In view of many of the examples below, the morphisms $\text_$ are called ''restriction morphisms''. If $s \in F\left(U\right)$, then its restriction $\text_\left(s\right)$ is often denoted $s, _V$ by analogy with restriction of functions. The restriction morphisms are required to satisfy two additional ( functorial) properties: *For every open set $U$ of $X$, the restriction morphism $\operatorname_ \colon F\left(U\right) \rightarrow F\left(U\right)$ is the identity morphism on $F\left(U\right)$. *If we have three open sets $W \subseteq V \subseteq U$, then the composite $\text_\circ\text_=\text_$ Informally, the second axiom says it doesn't matter whether we restrict to ''W'' in one step or restrict first to ''V'', then to ''W''. A concise functorial reformulation of this definition is given further below. Many examples of presheaves come from different classes of functions: to any ''$U$'', one can assign the set $C^0\left(U\right)$ of continuous real-valued functions on ''$U$''. The restriction maps are then just given by restricting a continuous function on ''$U$'' to a smaller open subset ''$V$'', which again is a continuous function. The two presheaf axioms are immediately checked, thereby giving an example of a presheaf. This can be extended to a sheaf of holomorphic functions $\mathcal\left(-\right)$ and a sheaf of smooth functions $C^\infty\left(-\right)$. Another common class of examples is assigning to $U$ the set of constant real-valued functions on $U$. This presheaf is called the ''constant presheaf'' associated to $\mathbb$ and is denoted $\underline^$.
Sheaves
Given a presheaf, a natural question to ask is to what extent its sections over an open set ''$U$'' are specified by their restrictions to smaller open sets $U_i$ of an open cover $\mathcal = \_$ of ''$U$''. A ''sheaf'' is a presheaf that satisfies both of the following two additional axioms: # (''Locality'') Suppose $U$ is an open set, $\_$ is an open cover of $U$, and $s, t \in F\left(U\right)$ are sections. If $s, _ = t, _$ for all $i \in I$, then $s = t$. # ( ''Gluing'') Suppose $U$ is an open set, $\_$ is an open cover of $U$, and $\_$ is a family of sections. If all pairs of sections agree on the overlap of their domains, that is, if $s_i, _ = s_j, _$ for all $i, j \in I$, then there exists a section $s \in F\left(U\right)$ such that $s, _ = s_i$ for all $i \in I$. The section ''$s$'' whose existence is guaranteed by axiom 2 is called the ''gluing'', ''concatenation'', or ''collation'' of the sections ''s''''i''. By axiom 1 it is unique. Sections ''$s_i$'' and ''$s_j$'' satisfying the agreement precondition of axiom 2 are often called ''compatible''; thus axioms 1 and 2 together state that ''any collection of pairwise compatible sections can be uniquely glued together''. A ''separated presheaf'', or ''monopresheaf'', is a presheaf satisfying axiom 1. The presheaf consisting of continuous functions mentioned above is a sheaf. This assertion reduces to checking that, given continuous functions $f_i : U_i \to \R$ which agree on the intersections $U_i \cap U_j$, there is a unique continuous function $f: U \to \R$ whose restriction equals the $f_i$. By contrast, the constant presheaf is usually ''not'' a sheaf as it fails to satisfy the locality axiom on the empty set (this is explained in more detail at constant sheaf). Presheaves and sheaves are typically denoted by capital letters, $F$ being particularly common, presumably for the French word for sheaf, ''faisceau''. Use of calligraphic letters such as $\mathcal$ is also common. It can be shown that to specify a sheaf, it is enough to specify its restriction to the open sets of a basis for the topology of the underlying space. Moreover, it can also be shown that it is enough to verify the sheaf axioms above relative to the open sets of a covering. This observation is used to construct another example which is crucial in algebraic geometry, namely quasi-coherent sheaves. Here the topological space in question is the spectrum of a commutative ring $R$, whose points are the prime ideals $p$ in $R$. The open sets $D_f := \$ form a basis for the Zariski topology on this space. Given an $R$-module $M$, there is a sheaf, denoted by $\tilde M$ on the Spec $R$, that satisfies : the localization of $M$ at $f$.
Further examples
Sheaf of sections of a continuous map
Any continuous map $f:Y\to X$ of topological spaces determines a sheaf $\Gamma\left(Y/X\right)$ on $X$ by setting :$\Gamma\left(Y/X\right)\left(U\right) = \.$ Any such $s$ is commonly called a section of ''$f$'', and this example is the reason why the elements in $F\left(U\right)$ are generally called sections. This construction is especially important when $f$ is the projection of a fiber bundle onto its base space. For example, the sheaves of smooth functions are the sheaves of sections of the trivial bundle. Another example: the sheaf of sections of :$\C \stackrel \to \C\setminus \$ is the sheaf which assigns to any ''$U$'' the set of branches of the complex logarithm on ''$U$''. Given a point $x$ and an abelian group $S$, the skyscraper sheaf $S_x$ is defined as follows: if $U$ is an open set containing $x$, then $S_x\left(U\right)=S$. If $U$ does not contain $x$, then $S_x\left(U\right)=0$, the trivial group. The restriction maps are either the identity on $S$, if both open sets contain $x$, or the zero map otherwise.
Sheaves on manifolds
On an $n$-dimensional $C^k$-manifold $M$, there are a number of important sheaves, such as the sheaf of $j$-times continuously differentiable functions $\mathcal^j_M$ (with $j \leq k$). Its sections on some open $U$ are the $C^j$-functions $U \to \R$. For $j = k$, this sheaf is called the ''structure sheaf'' and is denoted $\mathcal_M$. The nonzero $C^k$ functions also form a sheaf, denoted $\mathcal_X^\times$.
Differential form In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in mo ...
s (of degree $p$) also form a sheaf $\Omega^p_M$. In all these examples, the restriction morphisms are given by restricting functions or forms. The assignment sending $U$ to the compactly supported functions on $U$ is not a sheaf, since there is, in general, no way to preserve this property by passing to a smaller open subset. Instead, this forms a cosheaf, a dual concept where the restriction maps go in the opposite direction than with sheaves. However, taking the dual of these vector spaces does give a sheaf, the sheaf of distributions.
Presheaves that are not sheaves
In addition to the constant presheaf mentioned above, which is usually not a sheaf, there are further examples of presheaves that are not sheaves: * Let $X$ be the two-point topological space $\$ with the discrete topology. Define a presheaf $F$ as follows: $F(\varnothing) = \,\ F(\) = \R,\ F(\) = \R,\ F(\) = \R\times\R\times\R$The restriction map $F\left(\\right) \to F\left(\\right)$ is the projection of $\R \times\R\times\R$ onto its first coordinate, and the restriction map $F\left(\\right) \to F\left(\\right)$ is the projection of $\R \times\R\times\R$ onto its second coordinate. $F$ is a presheaf that is not separated: a global section is determined by three numbers, but the values of that section over $\$ and $\$ determine only two of those numbers. So while we can glue any two sections over $\$ and $\$, we cannot glue them uniquely. * Let $X = \R$ be the real line, and let $F\left(U\right)$ be the set of bounded continuous functions on $U$. This is not a sheaf because it is not always possible to glue. For example, let $U_i$ be the set of all $x$ such that
Motivating sheaves from complex analytic spaces and algebraic geometry
One of the historical motivations for sheaves have come from studying complex manifolds, complex analytic geometry, and scheme theory from
algebraic geometry Algebraic geometry is a branch of mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. ...
. This is because in all of the previous cases, we consider a topological space $X$ together with a structure sheaf $\mathcal$ giving it the structure of a complex manifold, complex analytic space, or scheme. This perspective of equipping a topological space with a sheaf is essential to the theory of locally ringed spaces (see below).
Technical challenges with complex manifolds
One of the main historical motivations for introducing sheaves was constructing a device which keeps track of holomorphic functions on complex manifolds. For example, on a compact complex manifold $X$ (like complex projective space or the vanishing locus of a
homogeneous polynomial In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in ...
), the ''only'' holomorphic functions
$f:X \to \C$
are the constant functions. This means there could exist two compact complex manifolds $X,X\text{'}$ which are not isomorphic, but nevertheless their ring of global holomorphic functions, denoted $\mathcal\left(X\right), \mathcal\left(X\text{'}\right)$, are isomorphic. Contrast this with smooth manifolds where every manifold $M$ can be embedded inside some $\R^n$, hence its ring of smooth functions $C^\infty\left(M\right)$ comes from restricting the smooth functions from $C^\infty\left(\R^n\right)$. Another complexity when considering the ring of holomorphic functions on a complex manifold $X$ is given a small enough open set $U \subset X$, the holomorphic functions will be isomorphic to $\mathcal\left(U\right) \cong \mathcal\left(\C^n\right)$. Sheaves are a direct tool for dealing with this complexity since they make it possible to keep track of the holomorphic structure on the underlying topological space of $X$ on arbitrary open subsets $U \subset X$. This means as $U$ becomes more complex topologically, the ring $\mathcal\left(U\right)$ can be expressed from gluing the $\mathcal\left(U_i\right)$. Note that sometimes this sheaf is denoted $\mathcal\left(-\right)$ or just $\mathcal$, or even $\mathcal_X$ when we want to emphasize the space the structure sheaf is associated to.
Tracking submanifolds with sheaves
Another common example of sheaves can be constructed by considering a complex submanifold $Y \hookrightarrow X$. There is an associated sheaf $\mathcal_Y$ which takes an open subset $U \subset X$ and gives the ring of holomorphic functions on $U \cap Y$. This kind of formalism was found to be extremely powerful and motivates a lot of homological algebra such as sheaf cohomology since an intersection theory can be built using these kinds of sheaves from the Serre intersection formula.
Operations with sheaves
Morphisms
Morphisms of sheaves are, roughly speaking, analogous to functions between them. In contrast to a function between sets, which have no additional structure, morphisms of sheaves are those functions which preserve the structure inherent in the sheaves. This idea is made precise in the following definition. Let $F$ and $G$ be two sheaves on $X$. A '' morphism'' $\varphi:G\to F$ consists of a morphism $\varphi_U:G\left(U\right)\to F\left(U\right)$ for each open set $U$ of $X$, subject to the condition that this morphism is compatible with restrictions. In other words, for every open subset $V$ of an open set $U$, the following diagram is commutative. :$\begin G\left(U\right) & \xrightarrow & F\left(U\right)\\ r_\Biggl\downarrow & & \Biggl\downarrow r_\\ G\left(V\right) & \xrightarrow\left[\right] & F\left(V\right) \end$ For example, taking the derivative gives a morphism of sheaves on $\R$: $\mathcal O^n_ \to \mathcal O^_.$ Indeed, given an ($n$-times continuously differentiable) function $f : U \to \R$ (with $U$ in $\R$ open), the restriction (to a smaller open subset $V$) of its derivative equals the derivative of $f, _V$. With this notion of morphism, sheaves on a fixed topological space $X$ form a category. The general categorical notions of
mono- Numeral or number prefixes are prefixes derived from numerals or occasionally other number A number is a mathematical object used to count, measure, and label. The original examples are the natural numbers 1, 2, 3, 4, and so forth. ...
, epi- and isomorphisms can therefore be applied to sheaves. A sheaf morphism $\varphi$ is an isomorphism (resp. monomorphism) if and only if each $\varphi_U$ is a bijection (resp. injective map). Moreover, a morphism of sheaves $\varphi$ is an isomorphism if and only if there exists an open cover $\$ such that $\varphi, _$ are isomorphisms of sheaves for all $\alpha$. This statement, which also holds for monomorphisms, but does not hold for presheaves, is another instance of the idea that sheaves are of a local nature. The corresponding statements do not hold for epimorphisms (of sheaves), and their failure is measured by sheaf cohomology.
Stalks of a sheaf
The ''stalk'' $\mathcal_x$ of a sheaf $\mathcal$ captures the properties of a sheaf "around" a point $x\in X$, generalizing the germs of functions. Here, "around" means that, conceptually speaking, one looks at smaller and smaller neighborhoods of the point. Of course, no single neighborhood will be small enough, which requires considering a limit of some sort. More precisely, the stalk is defined by :$\mathcal_x = \varinjlim_ \mathcal\left(U\right),$ the direct limit being over all open subsets of $X$ containing the given point $x$. In other words, an element of the stalk is given by a section over some open neighborhood of $x$, and two such sections are considered equivalent if their restrictions agree on a smaller neighborhood. The natural morphism $F\left(U\right)\to F_x$ takes a section $x$ in $F\left(U\right)$ to its ''germ'' at $x$. This generalises the usual definition of a germ. In many situations, knowing the stalks of a sheaf is enough to control the sheaf itself. For example, whether or not a morphism of sheaves is a monomorphism, epimorphism, or isomorphism can be tested on the stalks. In this sense, a sheaf is determined by its stalks, which are a local data. By contrast, the ''global'' information present in a sheaf, i.e., the ''global sections'', i.e., the sections $\mathcal F\left(X\right)$ on the whole space $X$, typically carry less information. For example, for a compact complex manifold $X$, the global sections of the sheaf of holomorphic functions are just $\C$, since any holomorphic function :$X \to \C$ is constant by Liouville's theorem.
Turning a presheaf into a sheaf
It is frequently useful to take the data contained in a presheaf and to express it as a sheaf. It turns out that there is a best possible way to do this. It takes a presheaf $F$ and produces a new sheaf $aF$ called the ''sheafification'' or ''sheaf associated to the presheaf'' $F$. For example, the sheafification of the constant presheaf (see above) is called the '' constant sheaf''. Despite its name, its sections are ''locally'' constant functions. The sheaf $aF$ can be constructed using the étalé space of $F$, namely as the sheaf of sections of the map :$\mathrm\left(F\right) \to X.$ Another construction of the sheaf $aF$ proceeds by means of a functor $L$ from presheaves to presheaves that gradually improves the properties of a presheaf: for any presheaf $F$, $LF$ is a separated presheaf, and for any separated presheaf $F$, $LF$ is a sheaf. The associated sheaf $aF$ is given by $LLF$. The idea that the sheaf $aF$ is the best possible approximation to $F$ by a sheaf is made precise using the following universal property: there is a natural morphism of presheaves $i\colon F\to aF$ so that for any sheaf $G$ and any morphism of presheaves $f\colon F\to G$, there is a unique morphism of sheaves $\tilde f \colon aF \rightarrow G$ such that $f = \tilde f i$. In fact $a$ is the left adjoint functor to the inclusion functor (or forgetful functor) from the category of sheaves to the category of presheaves, and $i$ is the unit of the adjunction. In this way, the category of sheaves turns into a Giraud subcategory of presheaves. This categorical situation is the reason why the sheafification functor appears in constructing cokernels of sheaf morphisms or tensor products of sheaves, but not for kernels, say.
Subsheaves, quotient sheaves
If $K$ is a subsheaf of a sheaf $F$ of abelian groups, then the quotient sheaf $Q$ is the sheaf associated to the presheaf $U \mapsto F\left(U\right)/K\left(U\right)$; in other words, the quotient sheaf fits into an exact sequence of sheaves of abelian groups; :$0 \to K \to F \to Q \to 0.$ (this is also called a sheaf extension.) Let $F,G$ be sheaves of abelian groups. The set $\operatorname\left(F, G\right)$ of morphisms of sheaves from $F$ to $G$ forms an abelian group (by the abelian group structure of $G$). The sheaf hom of $F$ and $G$, denoted by, :$\mathcal\left(F, G\right)$ is the sheaf of abelian groups $U \mapsto \operatorname\left(F, _U, G, _U\right)$ where $F, _U$ is the sheaf on $U$ given by $\left(F, _U\right)\left(V\right) = F\left(V\right)$ (note sheafification is not needed here). The direct sum of $F$ and $G$ is the sheaf given by $U \mapsto F\left(U\right) \oplus G\left(U\right)$, and the tensor product of $F$ and $G$ is the sheaf associated to the presheaf $U \mapsto F\left(U\right) \otimes G\left(U\right)$. All of these operations extend to sheaves of modules over a sheaf of rings $A$; the above is the special case when $A$ is the constant sheaf $\underline$.
Basic functoriality
Since the data of a (pre-)sheaf depends on the open subsets of the base space, sheaves on different topological spaces are unrelated to each other in the sense that there are no morphisms between them. However, given a continuous map $f:X\to Y$ between two topological spaces, pushforward and pullback relate sheaves on $X$ to those on $Y$ and vice versa.
Direct image
The pushforward (also known as direct image) of a sheaf $\mathcal$ on $X$ is the sheaf defined by :$\left(f_* \mathcal F\right)\left(V\right) = \mathcal F\left(f^\left(V\right)\right).$ Here $V$ is an open subset of $Y$, so that its preimage is open in $X$ by the continuity of $f$. This construction recovers the skyscraper sheaf $S_x$ mentioned above: :$S_x = i_* \left(S\right)$ where $i: \ \to X$ is the inclusion, and $S$ is regarded as a sheaf on the singleton (by $S\left(\\right)=S, S\left(\emptyset\right) = \emptyset$. For a map between locally compact spaces, the direct image with compact support is a subsheaf of the direct image. By definition, $\left(f_! \mathcal F\right)\left(V\right)$ consists of those $f \in \mathcal F\left(f^\left(V\right)\right)$ whose support is proper map over $V$. If $f$ is proper itself, then $f_! \mathcal F = f_* \mathcal F$, but in general they disagree.
Inverse image
The pullback or inverse image goes the other way: it produces a sheaf on $X$, denoted $f^ \mathcal G$ out of a sheaf $\mathcal G$ on $Y$. If $f$ is the inclusion of an open subset, then the inverse image is just a restriction, i.e., it is given by $\left(f^ \mathcal G\right)\left(U\right) = \mathcal G\left(U\right)$ for an open $U$ in $X$. A sheaf $F$ (on some space $X$) is called locally constant if $X= \bigcup_ U_i$ by some open subsets $U_i$ such that the restriction of $F$ to all these open subsets is constant. One a wide range of topological spaces $X$, such sheaves are equivalent to representations of the fundamental group $\pi_1\left(X\right)$. For general maps $f$, the definition of $f^ \mathcal G$ is more involved; it is detailed at inverse image functor. The stalk is an essential special case of the pullback in view of a natural identification, where $i$ is as above: :$\mathcal G_x = i^\mathcal\left(\\right).$ More generally, stalks satisfy $\left(f^ \mathcal G\right)_x = \mathcal G_$.
Extension by zero
For the inclusion $j : U \to X$ of an open subset, the extension by zero of a sheaf of abelian groups on $U$ is defined as :$\left(j_! \mathcal F\right)\left(V\right) = \mathcal F\left(V\right)$ if $V \subset U$ and $\left(j_! \mathcal F\right)\left(V\right) = 0$ otherwise. For a sheaf $\mathcal G$ on $X$, this construction is in a sense complementary to $i_*$, where $i$ is the inclusion of the complement of $U$: :$\left(j_! j^* \mathcal G\right)_x = \mathcal G_x$ for $x$ in $U$, and the stalk is zero otherwise, while :$\left(i_* i^* \mathcal G\right)_x = 0$ for $x$ in $U$, and equals $\mathcal G_x$ otherwise. These functors are therefore useful in reducing sheaf-theoretic questions on $X$ to ones on the strata of a stratification, i.e., a decomposition of $X$ into smaller, locally closed subsets.
Complements
Sheaves in more general categories
In addition to (pre-)sheaves as introduced above, where $\mathcal F\left(U\right)$ is merely a set, it is in many cases important to keep track of additional structure on these sections. For example, the sections of the sheaf of continuous functions naturally form a real
vector space In mathematics and physics Physics is the natural science that studies matter, its fundamental constituents, its motion and behavior through space and time, and the related entities of energy and force. "Physical science is th ...
, and restriction is a
linear map In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in ...
between these vector spaces. Presheaves with values in an arbitrary category $C$ are defined by first considering the category of open sets on $X$ to be the posetal category $O\left(X\right)$ whose objects are the open sets of $X$ and whose morphisms are inclusions. Then a $C$-valued presheaf on $X$ is the same as a contravariant functor from $O\left(X\right)$ to $C$. Morphisms in this category of functors, also known as natural transformations, are the same as the morphisms defined above, as can be seen by unraveling the definitions. If the target category $C$ admits all limits, a $C$-valued presheaf is a sheaf if the following diagram is an equalizer for every open cover $\mathcal = \_$ of any open set ''$U$'': :$F\left(U\right) \rightarrow \prod_ F\left(U_i\right) \prod_ F\left(U_i \cap U_j\right).$ Here the first map is the product of the restriction maps :$\operatorname_ \colon F\left(U\right) \rightarrow F\left(U_i\right)$ and the pair of arrows the products of the two sets of restrictions :$\operatorname_ \colon F\left(U_i\right) \rightarrow F\left(U_i \cap U_j\right)$ and :$\operatorname_ \colon F\left(U_j\right) \rightarrow F\left(U_i \cap U_j\right).$ If $C$ is an abelian category, this condition can also be rephrased by requiring that there is an exact sequence :$0 \to F\left(U\right) \to \prod_i F\left(U_i\right) \xrightarrow \prod_ F\left(U_i \cap U_j\right).$ A particular case of this sheaf condition occurs for $U$ being the empty set, and the index set $I$ also being empty. In this case, the sheaf condition requires $\mathcal F\left(\emptyset\right)$ to be the terminal object in $C$.
Ringed spaces and sheaves of modules
In several geometrical disciplines, including
algebraic geometry Algebraic geometry is a branch of mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. ...
and
differential geometry Differential geometry is a mathematical discipline that studies the geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics Mathematics is an area of knowledge that includes the topics of numbers, form ...
, the spaces come along with a natural sheaf of rings, often called the structure sheaf and denoted by $\mathcal_X$. Such a pair $\left(X, \mathcal O_X\right)$ is called a '' ringed space''. Many types of spaces can be defined as certain types of ringed spaces. Commonly, all the stalks $\mathcal O_$ of the structure sheaf are local rings, in which case the pair is called a ''locally ringed space''. For example, an $n$-dimensional $C^k$ manifold $M$ is a locally ringed space whose structure sheaf consists of $C^k$-functions on the open subsets of $M$. The property of being a ''locally'' ringed space translates into the fact that such a function, which is nonzero at a point $x$, is also non-zero on a sufficiently small open neighborhood of $x$. Some authors actually ''define'' real (or complex) manifolds to be locally ringed spaces that are locally isomorphic to the pair consisting of an open subset of $\R^n$ (resp. $\C^n$) together with the sheaf of $C^k$ (resp. holomorphic) functions. Similarly, schemes, the foundational notion of spaces in algebraic geometry, are locally ringed spaces that are locally isomorphic to the spectrum of a ring. Given a ringed space, a ''sheaf of modules'' is a sheaf $\mathcal$ such that on every open set $U$ of $X$, $\mathcal\left(U\right)$ is an $\mathcal_X\left(U\right)$-module and for every inclusion of open sets $V\subseteq U$, the restriction map $\mathcal\left(U\right) \to \mathcal\left(V\right)$ is compatible with the restriction map $\mathcal\left(U\right) \to \mathcal\left(V\right)$: the restriction of ''fs'' is the restriction of $f$ times that of $s$ for any $f$ in $\mathcal\left(U\right)$ and $s$ in $\mathcal\left(U\right)$. Most important geometric objects are sheaves of modules. For example, there is a one-to-one correspondence between vector bundles and locally free sheaves of $\mathcal_X$-modules. This paradigm applies to real vector bundles, complex vector bundles, or vector bundles in algebraic geometry (where $\mathcal O$ consists of smooth functions, holomorphic functions, or regular functions, respectively). Sheaves of solutions to differential equations are $D$-modules, that is, modules over the sheaf of differential operators. On any topological space, modules over the constant sheaf $\underline$ are the same as sheaves of abelian groups in the sense above. There is a different inverse image functor for sheaves of modules over sheaves of rings. This functor is usually denoted $f^*$ and it is distinct from $f^$. See inverse image functor.
Finiteness conditions for sheaves of modules
Finiteness conditions for module over commutative rings give rise to similar finiteness conditions for sheaves of modules: $\mathcal$ is called ''finitely generated'' (resp. ''finitely presented'') if, for every point $x$ of $X$, there exists an open neighborhood $U$ of $x$, a natural number $n$ (possibly depending on $U$), and a surjective morphism of sheaves $\mathcal_X^n, _U \to \mathcal, _U$ (respectively, in addition a natural number $m$, and an exact sequence $\mathcal_X^m, _U \to \mathcal_X^n, _U \to \mathcal, _U \to 0$.) Paralleling the notion of a coherent module, $\mathcal$ is called a '' coherent sheaf'' if it is of finite type and if, for every open set $U$ and every morphism of sheaves $\phi : \mathcal_X^n \to \mathcal$ (not necessarily surjective), the kernel of $\phi$ is of finite type. $\mathcal_X$ is ''coherent'' if it is coherent as a module over itself. Like for modules, coherence is in general a strictly stronger condition than finite presentation. The Oka coherence theorem states that the sheaf of holomorphic functions on a complex manifold is coherent.
The étalé space of a sheaf
In the examples above it was noted that some sheaves occur naturally as sheaves of sections. In fact, all sheaves of sets can be represented as sheaves of sections of a topological space called the ''étalé space'', from the French word étalé , meaning roughly "spread out". If $F \in \text\left(X\right)$ is a sheaf over $X$, then the étalé space of $F$ is a topological space $E$ together with a local homeomorphism $\pi: E \to X$ such that the sheaf of sections $\Gamma\left(\pi, -\right)$ of $\pi$ is $F$. The space ''$E$'' is usually very strange, and even if the sheaf ''$F$'' arises from a natural topological situation, ''$E$'' may not have any clear topological interpretation. For example, if ''$F$'' is the sheaf of sections of a continuous function $f: Y \to X$, then $E=Y$ if and only if $f$ is a local homeomorphism. The étalé space ''$E$'' is constructed from the stalks of ''$F$'' over ''$X$''. As a set, it is their disjoint union and ''$\pi$'' is the obvious map that takes the value $x$ on the stalk of $F$ over $x \in X$. The topology of ''$E$'' is defined as follows. For each element $s \in F\left(U\right)$ and each $x \in U$, we get a germ of $s$ at $x$, denoted or $s_x$. These germs determine points of ''$E$''. For any $U$ and $s \in F\left(U\right)$, the union of these points (for all $x \in U$) is declared to be open in ''$E$''. Notice that each stalk has the discrete topology as subspace topology. Two morphisms between sheaves determine a continuous map of the corresponding étalé spaces that is compatible with the projection maps (in the sense that every germ is mapped to a germ over the same point). This makes the construction into a functor. The construction above determines an equivalence of categories between the category of sheaves of sets on ''$X$'' and the category of étalé spaces over ''$X$''. The construction of an étalé space can also be applied to a presheaf, in which case the sheaf of sections of the étalé space recovers the sheaf associated to the given presheaf. This construction makes all sheaves into representable functors on certain categories of topological spaces. As above, let ''$F$'' be a sheaf on ''$X$'', let ''$E$'' be its étalé space, and let $\pi:E \to X$ be the natural projection. Consider the overcategory $\text/X$ of topological spaces over $X$, that is, the category of topological spaces together with fixed continuous maps to $X$. Every object of this category is a continuous map $f:Y\to X$, and a morphism from $Y\to X$ to $Z\to X$ is a continuous map $Y\to Z$ that commutes with the two maps to $X$. There is a functor
$\Gamma:\text/X \to \text$
sending an object $f:Y\to X$ to $f^ F\left(Y\right)$. For example, if $i: U \hookrightarrow X$ is the inclusion of an open subset, then
$\Gamma\left(i\right) = f^ F\left(U\right) = F\left(U\right) = \Gamma\left(F, U\right)$
and for the inclusion of a point $i : \\hookrightarrow X$, then
$\Gamma\left(i\right) = f^ F\left(\\right) = F, _x$
is the stalk of $F$ at $x$. There is a natural isomorphism
$\left(f^F\right)\left(Y\right) \cong \operatorname_\left(f, \pi\right)$,
which shows that $\pi: E \to X$ (for the étalé space) represents the functor $\Gamma$. ''$E$'' is constructed so that the projection map ''$\pi$'' is a covering map. In algebraic geometry, the natural analog of a covering map is called an étale morphism. Despite its similarity to "étalé", the word étale has a different meaning in French. It is possible to turn $E$ into a scheme and ''$\pi$'' into a morphism of schemes in such a way that ''$\pi$'' retains the same universal property, but ''$\pi$'' is ''not'' in general an étale morphism because it is not quasi-finite. It is, however, formally étale. The definition of sheaves by étalé spaces is older than the definition given earlier in the article. It is still common in some areas of mathematics such as
mathematical analysis Analysis is the branch of mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These top ...
.
Sheaf cohomology
In contexts, where the open set $U$ is fixed, and the sheaf is regarded as a variable, the set $F\left(U\right)$ is also often denoted $\Gamma\left(U, F\right).$ As was noted above, this functor does not preserve epimorphisms. Instead, an epimorphism of sheaves $\mathcal F \to \mathcal G$ is a map with the following property: for any section $g \in \mathcal G\left(U\right)$ there is a covering $\mathcal = \_$ where
$U = \bigcup_ U_i$
of open subsets, such that the restriction $g, _$ are in the image of $\mathcal F\left(U_i\right)$. However, $g$ itself need not be in the image of $\mathcal F\left(U\right)$. A concrete example of this phenomenon is the exponential map :$\mathcal O \stackrel \to \mathcal O^\times$ between the sheaf of holomorphic functions and non-zero holomorphic functions. This map is an epimorphism, which amounts to saying that any non-zero holomorphic function $g$ (on some open subset in $\C$, say), admits a complex logarithm ''locally'', i.e., after restricting $g$ to appropriate open subsets. However, $g$ need not have a logarithm globally. Sheaf cohomology captures this phenomenon. More precisely, for an exact sequence of sheaves of abelian groups :$0 \to \mathcal F_1 \to \mathcal F_2 \to \mathcal F_3 \to 0,$ (i.e., an epimorphism $\mathcal F_2 \to \mathcal F_3$ whose kernel is $\mathcal F_1$), there is a long exact sequence$0 \to \Gamma(U, \mathcal F_1) \to \Gamma(U, \mathcal F_2) \to \Gamma(U, \mathcal F_3) \to H^1(U, \mathcal F_1) \to H^1(U, \mathcal F_2) \to H^1(U, \mathcal F_3) \to H^2(U, \mathcal F_1) \to \dots$By means of this sequence, the first cohomology group $H^1\left(U, \mathcal F_1\right)$ is a measure for the non-surjectivity of the map between sections of $\mathcal F_2$ and $\mathcal F_3$. There are several different ways of constructing sheaf cohomology. introduced them by defining sheaf cohomology as the derived functor of $\Gamma$. This method is theoretically satisfactory, but, being based on injective resolutions, of little use in concrete computations. Godement resolutions are another general, but practically inaccessible approach.
Computing sheaf cohomology
Especially in the context of sheaves on manifolds, sheaf cohomology can often be computed using resolutions by soft sheaves, fine sheaves, and flabby sheaves (also known as ''flasque sheaves'' from the French ''flasque'' meaning flabby). For example, a partition of unity argument shows that the sheaf of smooth functions on a manifold is soft. The higher cohomology groups $H^i\left(U, \mathcal F\right)$ for $i > 0$ vanish for soft sheaves, which gives a way of computing cohomology of other sheaves. For example, the de Rham complex is a resolution of the constant sheaf $\underline$ on any smooth manifold, so the sheaf cohomology of $\underline$ is equal to its de Rham cohomology. A different approach is by Čech cohomology. Čech cohomology was the first cohomology theory developed for sheaves and it is well-suited to concrete calculations, such as computing the coherent sheaf cohomology of complex projective space $\mathbb^n$. It relates sections on open subsets of the space to cohomology classes on the space. In most cases, Čech cohomology computes the same cohomology groups as the derived functor cohomology. However, for some pathological spaces, Čech cohomology will give the correct $H^1$ but incorrect higher cohomology groups. To get around this, Jean-Louis Verdier developed hypercoverings. Hypercoverings not only give the correct higher cohomology groups but also allow the open subsets mentioned above to be replaced by certain morphisms from another space. This flexibility is necessary in some applications, such as the construction of Pierre Deligne's mixed Hodge structures. Many other coherent sheaf cohomology groups are found using an embedding $i:X \hookrightarrow Y$ of a space $X$ into a space with known cohomology, such as $\mathbb^n$, or some weighted projective space. In this way, the known sheaf cohomology groups on these ambient spaces can be related to the sheaves $i_*\mathcal$, giving $H^i\left(Y,i_*\mathcal\right) \cong H^i\left(X,\mathcal\right)$. For example, computing the coherent sheaf cohomology of projective plane curves is easily found. One big theorem in this space is the Hodge decomposition found using a spectral sequence associated to sheaf cohomology groups, proved by Deligne. Essentially, the $E_1$-page with terms
$E_1^ = H^p\left(X,\Omega^q_X\right)$
the sheaf cohomology of a smooth projective variety $X$, degenerates, meaning $E_1 = E_\infty$. This gives the canonical Hodge structure on the cohomology groups $H^k\left(X,\mathbb\right)$. It was later found these cohomology groups can be easily explicitly computed using Griffiths residues. See Jacobian ideal. These kinds of theorems lead to one of the deepest theorems about the cohomology of algebraic varieties, the decomposition theorem, paving the path for Mixed Hodge modules. Another clean approach to the computation of some cohomology groups is the Borel–Bott–Weil theorem, which identifies the cohomology groups of some line bundles on flag manifolds with irreducible representations of Lie groups. This theorem can be used, for example, to easily compute the cohomology groups of all line bundles on projective space and grassmann manifolds. In many cases there is a duality theory for sheaves that generalizes
Poincaré duality In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in mod ...
. See Grothendieck duality and Verdier duality.
Derived categories of sheaves
The derived category of the category of sheaves of, say, abelian groups on some space ''X'', denoted here as $D\left(X\right)$, is the conceptual haven for sheaf cohomology, by virtue of the following relation: : The adjunction between $f^$, which is the left adjoint of $f_*$ (already on the level of sheaves of abelian groups) gives rise to an adjunction :$f^ : D\left(Y\right) \rightleftarrows D\left(X\right) : R f_*$ (for $f: X \to Y$), where $Rf_*$ is the derived functor. This latter functor encompasses the notion of sheaf cohomology since $H^n\left(X, \mathcal F\right) = R^n f_* \mathcal F$ for $f: X \to \$. Like $f_*$, the direct image with compact support $f_!$ can also be derived. By virtue of the following isomorphism $R f_! F$ parametrizes the cohomology with compact support of the fibers of $f$: :$\left(R^i f_! F\right)_y = H^i_c\left(f^\left(y\right), F\right).$ This isomorphism is an example of a base change theorem. There is another adjunction :$Rf_! : D\left(X\right) \rightleftarrows D\left(Y\right) : f^!.$ Unlike all the functors considered above, the twisted (or exceptional) inverse image functor $f^!$ is in general only defined on the level of derived categories, i.e., the functor is not obtained as the derived functor of some functor between abelian categories. If $f: X \to \$ and ''X'' is a smooth orientable manifold of dimension ''n'', then : This computation, and the compatibility of the functors with duality (see Verdier duality) can be used to obtain a high-brow explanation of
Poincaré duality In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in mod ...
. In the context of quasi-coherent sheaves on schemes, there is a similar duality known as coherent duality. Perverse sheaves are certain objects in $D\left(X\right)$, i.e., complexes of sheaves (but not in general sheaves proper). They are an important tool to study the geometry of singularities.
Derived categories of coherent sheaves and the Grothendieck group
Another important application of derived categories of sheaves is with the derived category of coherent sheaves on a scheme $X$ denoted $D_\left(X\right)$. This was used by Grothendieck in his development of intersection theory using derived categories and K-theory, that the intersection product of subschemes $Y_1, Y_2$ is represented in K-theory as
where $\mathcal_$ are coherent sheaves defined by the $\mathcal_X$-modules given by their structure sheaves.
Sites and topoi
André Weil's Weil conjectures stated that there was a cohomology theory for algebraic varieties over finite fields that would give an analogue of the
Riemann hypothesis In mathematics, the Riemann hypothesis is the conjecture that the Riemann zeta function The Riemann zeta function or Euler–Riemann zeta function, denoted by the Greek letter ( zeta), is a mathematical function of a complex varia ...
. The cohomology of a complex manifold can be defined as the sheaf cohomology of the locally constant sheaf $\underline$ in the Euclidean topology, which suggests defining a Weil cohomology theory in positive characteristic as the sheaf cohomology of a constant sheaf. But the only classical topology on such a variety is the Zariski topology, and the Zariski topology has very few open sets, so few that the cohomology of any Zariski-constant sheaf on an irreducible variety vanishes (except in degree zero). Alexandre Grothendieck solved this problem by introducing Grothendieck topologies, which axiomatize the notion of ''covering''. Grothendieck's insight was that the definition of a sheaf depends only on the open sets of a topological space, not on the individual points. Once he had axiomatized the notion of covering, open sets could be replaced by other objects. A presheaf takes each one of these objects to data, just as before, and a sheaf is a presheaf that satisfies the gluing axiom with respect to our new notion of covering. This allowed Grothendieck to define étale cohomology and ℓ-adic cohomology, which eventually were used to prove the Weil conjectures. A category with a Grothendieck topology is called a ''site''. A category of sheaves on a site is called a ''topos'' or a ''Grothendieck topos''. The notion of a topos was later abstracted by William Lawvere and Miles Tierney to define an elementary topos, which has connections to
mathematical logic Mathematical logic is the study of formal logic within mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantiti ...
.
History
The first origins of sheaf theory are hard to pin down – they may be co-extensive with the idea of analytic continuation. It took about 15 years for a recognisable, free-standing theory of sheaves to emerge from the foundational work on cohomology. * 1936 Eduard Čech introduces the '' nerve'' construction, for associating a
simplicial complex In mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in mo ...
to an open covering. * 1938 Hassler Whitney gives a 'modern' definition of cohomology, summarizing the work since J. W. Alexander and Kolmogorov first defined '' cochains''. * 1943 Norman Steenrod publishes on homology ''with local coefficients''. * 1945 Jean Leray publishes work carried out as a
prisoner of war A prisoner of war (POW) is a person who is held captive by a belligerent power during or immediately after an armed conflict. The earliest recorded usage of the phrase "prisoner of war" dates back to 1610. Belligerents hold prisoners of ...
, motivated by proving fixed-point theorems for application to PDE theory; it is the start of sheaf theory and spectral sequences. * 1947 Henri Cartan reproves the de Rham theorem by sheaf methods, in correspondence with André Weil (see De Rham–Weil theorem). Leray gives a sheaf definition in his courses via closed sets (the later ''carapaces''). * 1948 The Cartan seminar writes up sheaf theory for the first time. * 1950 The "second edition" sheaf theory from the Cartan seminar: the sheaf space (''espace étalé'') definition is used, with stalkwise structure. Supports are introduced, and cohomology with supports. Continuous mappings give rise to spectral sequences. At the same time Kiyoshi Oka introduces an idea (adjacent to that) of a sheaf of ideals, in several complex variables. * 1951 The Cartan seminar proves theorems A and B, based on Oka's work. * 1953 The finiteness theorem for coherent sheaves in the analytic theory is proved by Cartan and Jean-Pierre Serre, as is Serre duality. * 1954 Serre's paper '' Faisceaux algébriques cohérents'' (published in 1955) introduces sheaves into
algebraic geometry Algebraic geometry is a branch of mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. ...
. These ideas are immediately exploited by Friedrich Hirzebruch, who writes a major 1956 book on topological methods. * 1955 Alexander Grothendieck in lectures in Kansas defines abelian category and ''presheaf'', and by using injective resolutions allows direct use of sheaf cohomology on all topological spaces, as derived functors. * 1956 Oscar Zariski's report '' Algebraic sheaf theory'' * 1957 Grothendieck's ''Tohoku'' paper rewrites homological algebra; he proves Grothendieck duality (i.e., Serre duality for possibly singular algebraic varieties). * 1957 onwards: Grothendieck extends sheaf theory in line with the needs of algebraic geometry, introducing: schemes and general sheaves on them, local cohomology, derived categories (with Verdier), and Grothendieck topologies. There emerges also his influential schematic idea of ' six operations' in homological algebra. * 1958 Roger Godement's book on sheaf theory is published. At around this time Mikio Sato proposes his hyperfunctions, which will turn out to have sheaf-theoretic nature. At this point sheaves had become a mainstream part of mathematics, with use by no means restricted to
algebraic topology Algebraic topology is a branch of mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. Th ...
. It was later discovered that the logic in categories of sheaves is
intuitionistic logic Intuitionistic logic, sometimes more generally called constructive logic, refers to systems of symbolic logic that differ from the systems used for classical logic by more closely mirroring the notion of constructive proof. In particular, system ...
(this observation is now often referred to as Kripke–Joyal semantics, but probably should be attributed to a number of authors). | 13,082 | 51,704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 514, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-06 | latest | en | 0.951589 |
https://citizensactionnetworks.com/qa/question-how-fast-should-a-teenage-girl-run-a-mile.html | 1,603,193,546,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872686.18/warc/CC-MAIN-20201020105000-20201020135000-00541.warc.gz | 269,943,129 | 9,113 | Question: How Fast Should A Teenage Girl Run A Mile?
What is the average time for a teenager to run a mile?
about 6 to 3 minutesBut on average, most teens run a mile in about 6 to 3 minutes..
What is the average mile time for a 13 year old female?
I think for a 13 year old girl 10 minutes is quick. Most girls ive seen run the mile are around 14-15 minutes.
What is the average mile time for a 14 year old?
A noncompetitive, relatively in-shape runner usually completes one mile in about 9 to 10 minutes, on average. If you’re new to running, you might run one mile in closer to 12 to 15 minutes as you build up endurance.
How many minutes does it take to drive 1 mile?
Time to cover 1 mile = t = ? x = (1 hour / 60 miles) * mile = 1/60 hour = (1/60) * 60 minutes = 1 minute. Thus, it takes 1 minute to travel 1 mile, with the speed of 60 miles per hour.
What happens if you run a mile everyday?
According to medical science, if you run a mile every day, you have: 42% lower risk of esophageal cancer, 27% lower risk of liver cancer, 26% lower risk of lung cancer, 23% lower risk of kidney cancer, 16% lower risk of colon cancer, and 10% lower risk of breast cancer.
What is a good mile time for a high school girl?
Anywhere from 4 to 14 minutes, depending on your height, weight, nutrition, and overall fitness level. But as Stephen wrote, any time is a good time, especially if you’re getting faster over time. A typical walking pace is 20 minutes/mile, just for reference.
How fast should a 14 year old female run a mile?
It depends. It’s like asking what the average weight of a 14-year old is. If some girls are athletic and like running they may run a mile in 5 minutes. But if every 14-year old girl in the school ran a mile the average would probably drop to about a 9:00 minute time for a mile.
What is an average mile time for a 13 year old?
around 5:35- 5:50So to answer your question the average mile time for an athletic 13 year old boy would probably be around 5:35- 5:50. For a runner a competitive time would be around 5:00–5:30. This track season I’ll be a sophomore and I hope to go 4:50.
How fast can a 14 year old run?
The average under-17 girl cancomplete a 100-meter sprint in 12.8 to 13.1 seconds and a 200-meter sprint in 26.4 to 27 seconds.
Is a 10 minute mile slow?
A 9-minute mile for a man and 10:30 for a woman are signs of moderate fitness; men who can’t run better than a 10-minute mile, and women slower than 12 minutes, fall into the low-fitness category. … Over all, he said, a 10-minute mile for a middle-aged man and a 12-minute mile for a woman suggest a good level of fitness.
How long should a 13 year old run a mile?
The International Association of Athletics Federations recommends that 13-year-olds run no more than 10K at a time — approximately 6.2 miles. At a fast pace of six minutes per mile, this works out to just over 37 minutes, but at a slower pace of 11 minutes per mile, it works out to about an hour and eight minutes.
Is 7 minutes a good mile time?
No. From my running experience in High School, I understand that the greater runners run usually close to 5 minute mile times or under, sometimes way under. A 7 minute mile time probably means you are somewhat fit, but it doesn’t make you a great runner. … So no, this is a good time, but it’s not a great time.
How fast can a 13 year old Run mph?
For a 12–13yr old, the top speed usually is between 12 and 16.4mph. But don’t take my word for it because it is not always the case. Yup, you can also get some of them reaching speeds of up to 17 mph with a very small number of them just shy of 18 mph.
How fast should I run a 5k for my age?
Running a 5K is a fairly achievable feat that’s ideal for people who are just getting into running or who simply want to run a more manageable distance….Average by age and sex.Age groupMenWomen50 to 5434:3041:2055 to 5937:3345:1860 to 6440:3345:4965 to 9942:5950:138 more rows•Aug 9, 2019
How long should it take a 13 year old to run 5k?
Average 5k Pace by Age and SexAgeMenWomen1111:13:2911:42:071210:20:1911:20:491310:22:1311:34:271409:28:2411:31:5873 more rows
What is the fastest mile time by a high schooler?
3:53.43The time is the fastest outdoor mile by a high school boy since 2001, when Alan Webb ran the still-standing national record of 3:53.43. Only Webb, Jim Ryun (3:55.3), and Drew Hunter (3:57.81 indoors) have ever run faster than Brown as a prep.
How fast should a 14 year old run 100m?
Age 13: 16–17 seconds. Age 15: 14,5–16 seconds. Age 17: 14–15 seconds. Age 19: 13,5–14 seconds.
Who is the fastest 14 year old?
Malaysian teenager Muhammad Azeem runs 100m in 10.63; minister Syed Saddiq hails 14-year-old as ‘our very own Usain Bolt’ | 1,323 | 4,734 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-45 | latest | en | 0.946246 |
https://www.enotes.com/homework-help/y-x-4-cos-x-use-product-rule-quotient-rule-find-521317 | 1,669,572,423,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00756.warc.gz | 806,177,333 | 18,541 | # `y = (x^4)/(cos(x))` Use the Product Rule or the Quotient Rule to find the derivative of the function.
`y=x^4/cosx`
Using the quotient rule for evaluating the derivative of the function,
`y'=(cosxd/dx(x^4)-x^4d/dxcosx)/(cosx)^2`
`y'=(cos(x)*(4x^3)-x^4(-sin(x)))/(cos^2(x))`
`y'=(4x^3cos(x)+x^4sin(x))/(cos^2(x))`
`y'=(x^3(4cos(x)+xsin(x)))/(cos^2(x))`
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`y=x^4/cosx`
Using the quotient rule for evaluating the derivative of the function,
`y'=(cosxd/dx(x^4)-x^4d/dxcosx)/(cosx)^2`
`y'=(cos(x)*(4x^3)-x^4(-sin(x)))/(cos^2(x))`
`y'=(4x^3cos(x)+x^4sin(x))/(cos^2(x))`
`y'=(x^3(4cos(x)+xsin(x)))/(cos^2(x))`
Approved by eNotes Editorial Team
You need to evaluate the derivative of the given function and since the function is a quotient of two functions, then you must use the quotient rule, such that:
`f'(x) = ((x^4)'(cos x) - (x^4)(cos x)')/((cos x)^2)`
`f'(x) = (4x^3*cos x + x^4*sin x)/((cos x)^2)`
Factoring out `x^3` yields:
`f'(x) = x^3(4x*cos x + x*sin x)/((cos x)^2)`
Hence, evaluating the derivative of the function, using the product rule, yields `f'(x) = x^3(4x*cos x + x*sin x)/((cos x)^2).`
Approved by eNotes Editorial Team | 477 | 1,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-49 | latest | en | 0.653347 |
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