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http://au.metamath.org/mpegif/cdleme10.html | 1,532,056,821,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591481.75/warc/CC-MAIN-20180720022026-20180720042026-00638.warc.gz | 29,258,064 | 9,161 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > cdleme10 Structured version Unicode version
Theorem cdleme10 31051
Description: Part of proof of Lemma E in [Crawley] p. 113, 2nd paragraph on p. 114. represents s2. In their notation, we prove s s2 = s r. (Contributed by NM, 9-Jun-2012.)
Hypotheses
Ref Expression
cdleme10.l
cdleme10.j
cdleme10.m
cdleme10.a
cdleme10.h
cdleme10.d
Assertion
Ref Expression
cdleme10
Proof of Theorem cdleme10
StepHypRef Expression
1 cdleme10.d . . 3
21oveq2i 6092 . 2
3 simp1l 981 . . . 4
4 simp3l 985 . . . 4
5 simp2 958 . . . . 5
6 eqid 2436 . . . . . 6
7 cdleme10.j . . . . . 6
8 cdleme10.a . . . . . 6
96, 7, 8hlatjcl 30164 . . . . 5
103, 5, 4, 9syl3anc 1184 . . . 4
11 simp1r 982 . . . . 5
12 cdleme10.h . . . . . 6
136, 12lhpbase 30795 . . . . 5
1411, 13syl 16 . . . 4
15 hllat 30161 . . . . . 6
163, 15syl 16 . . . . 5
176, 8atbase 30087 . . . . . 6
18173ad2ant2 979 . . . . 5
196, 8atbase 30087 . . . . . 6
204, 19syl 16 . . . . 5
21 cdleme10.l . . . . . 6
226, 21, 7latlej2 14490 . . . . 5
2316, 18, 20, 22syl3anc 1184 . . . 4
24 cdleme10.m . . . . 5
256, 21, 7, 24, 8atmod3i1 30661 . . . 4
263, 4, 10, 14, 23, 25syl131anc 1197 . . 3
276, 7latjcom 14488 . . . . 5
2816, 18, 20, 27syl3anc 1184 . . . 4
29 eqid 2436 . . . . . 6
3021, 7, 29, 8, 12lhpjat2 30818 . . . . 5
31303adant2 976 . . . 4
3228, 31oveq12d 6099 . . 3
33 hlol 30159 . . . . 5
343, 33syl 16 . . . 4
356, 7latjcl 14479 . . . . 5
3616, 20, 18, 35syl3anc 1184 . . . 4
376, 24, 29olm11 30025 . . . 4
3834, 36, 37syl2anc 643 . . 3
3926, 32, 383eqtrd 2472 . 2
402, 39syl5eq 2480 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wa 359 w3a 936 wceq 1652 wcel 1725 class class class wbr 4212 cfv 5454 (class class class)co 6081 cbs 13469 cple 13536 cjn 14401 cmee 14402 cp1 14467 clat 14474 col 29972 catm 30061 chlt 30148 clh 30781 This theorem is referenced by: cdleme10tN 31055 cdleme20aN 31106 cdleme20g 31112 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 ax-rep 4320 ax-sep 4330 ax-nul 4338 ax-pow 4377 ax-pr 4403 ax-un 4701 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2285 df-mo 2286 df-clab 2423 df-cleq 2429 df-clel 2432 df-nfc 2561 df-ne 2601 df-nel 2602 df-ral 2710 df-rex 2711 df-reu 2712 df-rab 2714 df-v 2958 df-sbc 3162 df-csb 3252 df-dif 3323 df-un 3325 df-in 3327 df-ss 3334 df-nul 3629 df-if 3740 df-pw 3801 df-sn 3820 df-pr 3821 df-op 3823 df-uni 4016 df-iun 4095 df-iin 4096 df-br 4213 df-opab 4267 df-mpt 4268 df-id 4498 df-xp 4884 df-rel 4885 df-cnv 4886 df-co 4887 df-dm 4888 df-rn 4889 df-res 4890 df-ima 4891 df-iota 5418 df-fun 5456 df-fn 5457 df-f 5458 df-f1 5459 df-fo 5460 df-f1o 5461 df-fv 5462 df-ov 6084 df-oprab 6085 df-mpt2 6086 df-1st 6349 df-2nd 6350 df-undef 6543 df-riota 6549 df-poset 14403 df-plt 14415 df-lub 14431 df-glb 14432 df-join 14433 df-meet 14434 df-p0 14468 df-p1 14469 df-lat 14475 df-clat 14537 df-oposet 29974 df-ol 29976 df-oml 29977 df-covers 30064 df-ats 30065 df-atl 30096 df-cvlat 30120 df-hlat 30149 df-psubsp 30300 df-pmap 30301 df-padd 30593 df-lhyp 30785
Copyright terms: Public domain W3C validator | 1,900 | 3,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-30 | latest | en | 0.089685 |
http://www.jiskha.com/members/profile/posts.cgi?name=jeffrey | 1,481,208,317,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541529.0/warc/CC-MAIN-20161202170901-00046-ip-10-31-129-80.ec2.internal.warc.gz | 537,478,556 | 8,715 | Thursday
December 8, 2016
# Posts by jeffrey
Total # Posts: 62
math geometry
The area of a rectangular wall of a barn is 80 square feet. The length is 12 feet longer than twice its width. Find the length and width of the barn.
September 15, 2016
ALGEBRA
A SPINNER WITH FOUR EQUAL SECTIONS NUMBERED 3,5,7 AND 8 IS SPUN TWICE. WHAT IS THE PRBABILITY THAT AN ODD NUMBER IS SPUN THE FIRST TIME AND AN EVEN NUMBER IS PUN THE SECOND? IS THE ANSWER 3/4? IS IT THAT EASY OR NOT?
May 23, 2016
life orientation
this it put or result people's life in danger because of insufficient of natural resource
March 31, 2016
Math
An automobile purchased for \$31,000 is worth \$2800 after 6 years. Assuming that the car's value depreciated steadily from year to year, what was it worth at the end of the third year?
January 13, 2016
Math
Repeat the following procedure for the four given numbers. Add 7 . Double the result. Subtract 4 . Divide by 2. Subtract the original selected number. The 1st number is 1 . The result is 5 . The 2nd number is 4 . The result is 5 . The 3rd number is 7 . The result is 5 . The ...
January 10, 2016
Algebra
Got it! Thank you Steve!
August 22, 2015
Algebra
I am working on a review worksheet and there are 2 problems I can't remember how to do. Each question has 6 other questions, so helping me with these will remind me how to do the others! Thank you!! 1.2(x+5) = 1.6(2x+5) and Solve for y a(y+c) = b(y-c) Thank you again! Jeffrey
August 22, 2015
Geometry
The volume of a rectangular prism is 960 cubic inches. If the dimensions of the base are doubled and the height remains the same to create a new prism, what will be the volume of the new rectangular prism in cubic inches?
July 19, 2015
Trigonometry
A building is 50 feet high. At a distance away from the building, an observer notices that the angle of elevation to the top of the building is 41 degrees. How far is the observer from the base of the building?
June 1, 2015
physics
1. An electric pump, used to obtain water from 20 metres below the ground, is marked 5000W. a) If the pump operates as rated, how much energy is used to pump water every second? b) Describe the energy conversions that take place when the pump is used. c) What is the maximum ...
April 6, 2015
Circles
2 circles have different radius touch at a point and at the point has 1 common tangent and another common tangent touching the circles at different points other than the first
January 15, 2015
Graphing
The line l1 passes through the points (3,-3) and (-5,2). The line is the graph of the equation Ax + By = C, where A, B, and C are integers with greatest common divisor 1, and A is positive. Find A + B+ C.
December 31, 2014
College physics
When a 0.15 kg baseball is hit, it approaches the bat with a speed of 20 m/s and undergoes an elastic collision. (a) What is the impulse delivered to the bat by the ball? (b) If the baseball is in contact with the bat for 1.3 ms, what is the average force exerted by the bat on...
October 24, 2014
Math
If 2 1/3 feet of silver ribbon cost \$.21, how much will 9 feet cost
June 13, 2014
math
Michael and mateo begin running around a circular track from the starting point at the same time. it takes Michael 48 seconds one lap around the track. it takes mateo 60 seconds to complete one lap around the track how many minutes does it take before the two boys will meet at...
December 15, 2013
Statistics
Thank you, but I'm still confused. So is there not enough info to answer this question? Thank you for your time.
October 28, 2013
Statistics
A city ballot includes a local initiative that would legalize gambling. The issue is hotly contested, and two groups decide to conduct polls to predict the outcome. The local newspaper finds that 53% of 1200 randomly selected voters plan to vote "yes," while a ...
October 28, 2013
Statistics
A city ballot includes a local initiative that would legalize gambling. The issue is hotly contested, and two groups decide to conduct polls to predict the outcome. The local newspaper finds that 53% of 1200 randomly selected voters plan to vote "yes," while a ...
October 27, 2013
Statistics
In 1998 a San Diego Reproductive clinic reported 49 live births to 207 women under the age of 40 who had previously been unable to get pregnant. Using the 90% confidence interval calculated above, would it be misleading for the clinic to report a success rate of 25%? A) Yes, ...
October 27, 2013
Math estimation problem
It is 100 for estimation if yo are dividing.
October 14, 2013
calculus
The function is x^3+6x^2/x^2+36.
May 29, 2013
calculus
For the function r(x)=x3+6x2x2−36, find the following. a. the y - intercept b. the x - intercepts c. all horizontal asymptotes. y = d. all vertical asymptotes. x =
May 29, 2013
math x
8. The Chicago Public School system has about 680 schools. About 130 of the schools will be closed. If five schools are chosen at random find the probability that at least 3 of them will be closed.
April 3, 2013
chemistry
thanks, sorry, 10ml of the unknown mixture was titrated methyl orange was used it took 7.5ml of 1M HCl (.0074Mol ?) so this means .00074Mol CH3NH2 / ml ?
August 24, 2012
chemistry
Q: Calculate the unknown molar concentration of CH3NH2 in MeOH titrated against a known molar concentration of HCl ? Could you please answer this, there are lots of examples around calculation PH of two known's but not so much of unknown concentration. thankyou.
August 24, 2012
pre algebra
the ratio of boys to girls in an art class is 3.5. there are 12 boys in the class. how many girls are in the class?
October 4, 2011
Sociology
The socialization process ends with the onset of puberty.
September 16, 2011
calculus
A spherical balloon is being inflated so that its volume is increasing at the rate of 200 cm3/min. At what rate is the radius increasing when the radius is 15 cm.
July 31, 2011
math
A balloon rises at a rate of 3 meters per second from a point on the ground 30 meters from an observer. Find the rate of change of the angle of elevation (in radians per second or rad/sec) of the balloon from the observer when the balloon is 30 meters above the ground.
July 25, 2011
math
A right circular cylinder is to be designed to hold 750 cm3 of processed milk, and to use a minimal amount of material in its construction. Find the dimensions for the container.
July 25, 2011
math
A cardboard box manufacturer wishes to make open boxes from rectangular pieces of cardboard with dimensions 40 cm by 60 cm by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the cut-out square so that the box has the largest...
July 25, 2011
physics
static friciton is greater than sliding friction prove
July 19, 2011
Science
dilute
May 19, 2011
7 Mathematics
A team of 4 golfers scored 69,73,70, and 74 on the first round on a par 72 course. They reduced their team score by 3 on the second round. a) How many strokes above or below par was the team score on the first round? b) How many strokes about or below par was the team score on...
May 16, 2011
Probability as a Fraction
In a bag there are 7 red tiles, 3 blue tiles, and 2 purple tiles. How many and what color tiles would you have to add to the bag so that the probability of picking red is 1/3?
April 24, 2011
Chemistry
April 17, 2011
math
Given the following equation x,y data pairs find the least squares equation for these data. Use 3 decimal places. Y equals what + what x(3,3)(4,5)(6,8,)
January 19, 2011
math
given the following equations x,y data pairs find the least squares equation fo these data. use 3 decimal places. Y equals what + what x (3,3)(4,5)(6,8)
January 19, 2011
Technology
Would the president of a technologies workers union be considered a stakeholder for a telecommunication company?
January 6, 2011
algebra
6P4 = 6!/(6-4)! =6*5*4*3*2*1/2! =6*5*4*3*2*1/2*1 =720/2 =360. (/)-represents division. (*)-represents multiplication. Thanks. I hope it was super helpful.
December 14, 2010
Physics
A lumberjack (mass = 98 kg) is standing at rest on one end of a floating log (mass = 245 kg) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of +2.7 m/s relative to the shore, and then hops onto an identical floating log that is ...
November 14, 2010
thesis statement
whats a good thesis statement on cooking
August 30, 2010
algebra
thank you
August 25, 2010
algebra
this tells me to rewrite the expression using the distributive property and then simplify plz help me and show me the work here are the questions. 15(f+1/3) 16(3b-0.25) (c-4)d
August 25, 2010
CHEMISTRY
A 31.43 mL volume of 0.108 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 4.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
August 25, 2010
JJ
A 30.84 mL volume of 0.128 M NaOH is required to reach the phnolphthalein endpoint in the titration of a 5.441 g sample of vinegar. Calculate % acetic acid in the vinegar.
August 25, 2010
CHEMISTRY
Assuming the density of a 5% acetic acid solution is 1.0 g/mL,determine the volume of the acetic acid solution necessary toneutralize 25.0 mL of 0.10M NaOH.
August 25, 2010
stats
Among the contestants in a competition are 44 women and 26 men. if 5 winners are randomly selected, what is the probability that they are all men?
July 14, 2010
Pre algebra
y/2 - 5y/6 + 1/3=1/2
May 8, 2010
CHEMISTRY
In some solid calcium carbonate samples, calcium bicarbonate, Ca(HCO3)2, is also present. Write a balanced euation for its reaction with hydrochloric acid.
March 8, 2010
algebra
I need help!!! 3/2,1/2, 1/6,1/8,... I need to determine if this sequence is arithmetic, geometric, or neither, then I need to find the ninth term
April 19, 2009
algebra
I need help!!! 3/2,1/2, 1/6,1/8,... I need to determine if this sequence is arithmetic, geometric, or neither, then I need to find the ninth term
April 19, 2009
debate
resolved that capital punishment in the philippines in affirmative side
March 17, 2009
social studies
how does enginering use math
October 26, 2008
Chemistry- Stochiometry
The conversion factor for deriving the number of moles of salicylic acid, C7H6O3, from a given number of grams of salicylic acid is...
February 28, 2008
Honors Geometry
Okay. I figured the problem out now. Thanks for your help!
January 18, 2008
Honors Geometry
Okay, the ITT is If a triangle has two congruent sides, then the angles oppositeof those sides are congruent
January 18, 2008
Honors Geometry
hmm.. okay, but i'm not really seeing how that goes with the ITT
January 18, 2008
Honors Geometry
I'm stuck on these 2 problems. My Geometry teacher won't help me. What is the inverse of the Isosceles Triangle Theorem (ITT)? What is the contrapositive of the Isosceles Triangle Theorem (ITT)?
January 18, 2008
chemistry
Could someone help me with this please? In chemistry, we did a lab in which he put brands of commerical vinegar in a microplate. We put in 10 drops of each vinegar in each plate thinger then we added 1 drop of phenolphthalein and then put in as many drops of NaOH as it took to...
May 14, 2007
Social Studies
Is there a desert in Italy? http://www.google.com/search?q=map+italy&rls=com.microsoft:en-us:IE-SearchBox&ie=UTF-8&oe=UTF-8&sourceid=ie7&rlz=1I7SUNA Nope. =)
April 23, 2007
Economics
Determine whether each of the following would cause a shift in the aggregate demand curve, the aggregate supply curve, neither, or both. Which curve shifts, and in which direction? What happens to aggregate output and the price level in each case? a. The price level changes b...
September 6, 2006
1. Pages:
2. 1 | 3,273 | 11,656 | {"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} | 3.578125 | 4 | CC-MAIN-2016-50 | longest | en | 0.910067 |
https://physics.stackexchange.com/questions/424051/is-the-net-power-of-an-open-system-always-equal-to-zero | 1,580,169,692,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00192.warc.gz | 598,275,247 | 30,687 | # Is the net power of an open system always equal to zero?
Are the following statements, line of reasoning all correct? I'm particularly interested in the last bullet.
• For a closed system energy is conserved and constant.
• For an open system energy is not necessarily conserved but is
constant. In other words if we account for the work done on the
system, the work done by the system (energy losses and gains due to
work) and the energy stored in the system, then the net energy is
constant.
• The power is the time rate of change of energy.
• Therefore the net power (over any interval of time) for an open system is equal to zero.
Any caveats?
• Would it be better to say "isolated" instead of "closed" system? In thermodynamic lingo, closed usually just means no mass transfer; but doesn't restrict energy. – JMac Aug 21 '18 at 19:12
• what is the difference between conserved and constant? – Adam Aug 21 '18 at 20:13
• If by net power you mean mechanical power (turning a turbine for example) then it is not necessarily zero for an open system. – Deep Aug 22 '18 at 6:25 | 266 | 1,085 | {"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-2020-05 | latest | en | 0.922494 |
https://www.thepokerbank.com/strategy/concepts/rem/maximize/ | 1,722,660,920,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640356078.1/warc/CC-MAIN-20240803025153-20240803055153-00513.warc.gz | 814,581,212 | 6,966 | The REM Process - Maximize
REM Process: Range > Equity > Maximize
The final (and arguably most important) step in the REM process is "Maximize". This involves making the optimum play to get the most value from your hand.
Unfortunately, the art of maximizing your profits from every hand you play is out of the scope of one article. However, what I will be able to do is give you some groundwork for making optimum plays at the table.
Maximizing value on a basic level.
On a basic level, optimum play is to get as much money in the middle with the best hand and save your money when you do not.
When you have the hand with the greatest equity you want to get as much money in to the pot as possible. When you don't have good equity in the hand you want to see future cards as cheaply as possible, which generally involves checking and folding.
You want to play as closely as you can to the fundamental theorem of poker to help you maximize your winnings from each hand you play.
The fundamental theorem of poker.
“Every time you play a hand differently from the way you would have played it if you could see all your opponents' cards, they gain; and every time you play your hand the same way you would have played it if you could see all their cards, they lose.”
So if we knew what our opponent was holding at all times, we would be able to make the optimum play on every single betting round. When maximizing value from our hands, we want to try and make the same play that we would make if we could see our opponent's holecards.
One problem.
That's all well and good, but there is a hell of a lot of middle ground where we will not know if we are ahead of behind in a hand. Therefore, we just have to make the best play we can with the limited information we have from the range and equity sections of the REM process.
Betting and raising.
Whenever you bet or raise, ask yourself this question:
Do I want my opponent to fold or do I want them to call?
• If you want them to call, you are betting for value.
• If you want them to fold, you are bluffing.
That's all there is to it. If you do not know whether you want your opponent to do either, you should not be betting.
Far too many players make bets with no reasoning behind why they are making them. The chances are that if you do not know why you are betting, a large number of your bets are going to be -EV.
Betting for value.
A super-common example for you. We raise in the CO and the BB calls.
Our hand: A K
Flop: A 9 7
As we all should know, on this flop we should bet out. Why? Because we have good equity against our opponent's range, which is likely to be a weaker ace or lower pocket pair. We are betting for value in the hope that our opponent will call with a weaker ace or a flush draw because we have greater equity in the hand (even if they may have a better hand like 2 pair or a set).
The equity of our AdKc against villain's range. Found using PokerStove.
Yep that's right, we actually want our opponent to call with a flush draw. If the pot is \$10 and we bet \$8, they are getting terrible odds to chase their flush. Therefore, if they call they are making a mistake and we are gaining from this mistake. If they fold on the other hand, they are making the correct play and we gain nothing.
"Betting to protect your hand" is very bad alternate terminology for "betting for value". What are we protecting our hand from? Players calling with bad odds and worse hands? According to the fundamental theorem of poker we want our opponents to call with bad odds and worse hands, as we gain nothing otherwise.
Try your best to get used to the idea of either betting for value or bluffing. "Protecting your hand" is not the right way to explain your bets - "betting for value" is.
Bluffing.
Our hand: A Q
Flop: T 2 7
Let's say we raised preflop in the CO and the button calls. Our opponent then bets out 3/4 of the pot on this flop. With our overcards and nut flush draw, we reraise. However, is this reraise for value or as a bluff? In other words, do we want our opponent to call of to fold?
Well, first things first, this all depends on our opponents range. If we figure that our opponent is the type of player to only ever bet out on this sort of flop with top pair or better, our equity in the hand is as follows:
The equity of our AsQs against villain's range. Found using PokerStove.
As you can see, we do not have the dominant equity in the hand. So in this situation, by raising we are hoping that our opponent will fold their better hand. Therefore we are bluffing. Simple as that.
Sure, it's technically a semi-bluff because it's not all that bad if our opponent calls this raise, but based on the equity we have right now our raise is a bluff. Hopefully this helps to illustrate the difference between bluffing and betting for value.
Evaluation of "maximize" in the REM process.
I only really covered the north-facing side of the tip of the ice berg in this article. I was going to cover more on maximizing value with checking and calling, but in all honesty I would have just been rehashing some pretty basic strategy to reinforce the same point I put across in the betting and raising section.
Maximizing value in poker is a skill that you will be perfecting for life. A lot of the strategy you read all boils down to making the best decisions possible and getting the most from your hands. As long as you always think about every decision you make, you will continue to improve your game and make optimum plays more frequently.
Poker will always be a game of broken information - one big jigsaw puzzle with a few pieces missing. We just have to try to put it together and do the best we can with the information we have at hand - and then make the best play we can.
Go back to the overview of The REM Process.
Go back to the awesome Texas Hold'em Strategy. | 1,323 | 5,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.976849 |
https://crackmyproctoredexam.com/how-you-can-use-reason-to-learn/ | 1,720,923,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00289.warc.gz | 158,919,782 | 22,103 | # How You Can Use Reason to Learn
Two different types of logical reasoning are commonly distinguished as well as formal deduction and inductive reasoning. Given a particular condition or proposition, one can infer the other or a result from the condition or proposition. In inductive reasoning, one deduces something from its antecedent or prior condition. Inductive reasoning is more precise than deductive reasoning, because it takes into account a second and not just a first antecedent. The latter deductive reasoning does not take into account a second antecedent, which is not needed in inductive reasoning.
At an intuitive level, intuition is a way to represent what a person knows without having to put together an argument. The logic used at an intuitive level can be very simple; for example, when someone says, “Bill’s car runs on gasoline.” Someone else might have said, “Bill’s car is powered by an engine that uses gas.”
This example shows that intuition is a way to express information at an intuitive level. Logic, on the other hand, is the process of applying the information to a particular situation and its consequences.
Logic is also commonly used on an intuitive level because of its directness. When people use logic, they are using their “gut feeling,” which is essentially an intuitive form of reasoning. They do not need to analyze what they feel because they know instinctively. When logic is used on an intuitive level, it is easier for people to find the best solution in a complicated set of circumstances.
Logical reasoning is different from inductive reasoning, because it takes into account both the premises and the consequent of a given proposition. Logical reasoning can also be called deduction, since it relies on a rule that can be deduced from all of the premises. As a rule, one will usually say, “This rule tells us that A is true.” The deduction can be done from any of the premises. The most common deduction is the one that involves A, B, and C.
Logical reasoning is very important when deciding on a course of action. For example, if we want to write a song about our experiences, we would likely use logic.
However, it is also important that people are not too much logical about things; they should follow their instinct and intuition. Sometimes logic can help them avoid being too much logical.
To summarize, logic is not always right and logical arguments can be bad. However, logic can help us find the best solution when it is a useful part of how we reason.
A simple example that uses logic: “Bill and Bob love animals; therefore, Bill and Bob must get a pet.” This is a logical argument that shows how logical reasoning can help a person to decide on the best solution to a problem.
The problem with this type of logical argument is that the conclusion of this argument is too obvious. For example, “Bill and Bob must get a dog because dogs are cute and look good with a collar.” is not a good conclusion.
However, when people use intuition alone without reasoning, they often do not understand that they are not as logical as they might think. They may make mistakes, especially if they are new to a particular situation.
The main thing to remember about logic and intuition is that they are both equally useful. Logic and intuition can be used together in conjunction with logic to help people find the best solution. Using logic when appropriate is great for solving problems and in many cases will lead to the right solution. | 699 | 3,495 | {"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-2024-30 | latest | en | 0.963463 |
http://www.mathisfunforum.com/post.php?tid=20018&qid=284645 | 1,398,252,286,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223202457.0/warc/CC-MAIN-20140423032002-00490-ip-10-147-4-33.ec2.internal.warc.gz | 731,848,063 | 8,144 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## Topic review (newest first)
rogerwaters
2013-11-05 03:57:07
What is the answer to number 5?
D or A? Or something else?
bob bundy
2013-09-24 16:53:29
hi demha,
Well done for that result!
Bob
bobbym
2013-09-24 06:21:31
Hi;
Very good result.
demha
2013-09-24 03:44:27
Hi Bobbym,
I actually just wanted help to do the equation my self, not have someone else do it for me but thank you very much!
And Bob I thank you too for your help!
Sent in the lesson and got 20/20.
bob bundy
2013-09-24 03:13:49
I have a rectangle, with a length of 7 and a width of 4:
15. What is the perimeter? - Answer: E
A 31
B 26
C 58
D 35
E 22
F 59
Correct.
16. What is the area? - Answer: A
A 28
B 65
C 79
D 13
E 10
F 19
Correct.
This rectangle just became the base of a regular prism, with a height of 6:
17. What is the lateral area? - Answer: F
A 119
B 873
C 256
D 349
E 332
F 132
Correct.
18. What is the total surface area? - Answer: D
A 119
B 873
C 256
D 188
E 332
F 132
Correct.
19. What is the volume? - Answer: E
A 651
B 327
C 395
D 221
E 168
F 342
Correct.
20. What is the area of the largest rectangular side? - Answer: D
A 65
B 32
C 36
D 42
E 16
F 34
Correct.
Well done.
Bob
bob bundy
2013-09-24 03:05:16
hi demha
Firstly Q6-10 are all correct. Well done!
Q11 and Q12 and Q14 are also correct.
I'll look at the last section in a new post
Bob
bobbym
2013-09-24 01:39:35
Hi;
For 13) you indicate that you need some help.
You have:
h = 4
b = 6
l = 8
V = ( 4 x 6 x 8 ) / 2 = 96
Answer is E.
demha
2013-09-24 01:31:11
I have a rectangle, with a length of 7 and a width of 4:
15. What is the perimeter? - Answer: E
A 31
B 26
C 58
D 35
E 22
F 59
16. What is the area? - Answer: A
A 28
B 65
C 79
D 13
E 10
F 19
This rectangle just became the base of a regular prism, with a height of 6:
17. What is the lateral area? - Answer: F
A 119
B 873
C 256
D 349
E 332
F 132
18. What is the total surface area? - Answer: D
A 119
B 873
C 256
D 188
E 332
F 132
19. What is the volume? - Answer: E
A 651
B 327
C 395
D 221
E 168
F 342
20. What is the area of the largest rectangular side? - Answer: D
A 65
B 32
C 36
D 42
E 16
F 34
demha
2013-09-24 01:25:11
Last 10 to go!
I have an isosceles triangle with a height of 4 and a base of 6:
11. What is the area? - Answer: C
A 19
B 35
C 12
D 16
E 22
F 54
This triangle just became the base of a regular prism, with a height of 8:
12. What is the lateral area? - Answer: D
A 105
B 28
C 35
D 128
E 56
F 12
13. What is the volume? **need help solving
A 95
B 54
C 32
D 23
E 96
F 10
14. What is the area of the largest rectangular side? - Answer: D
A 95
B 54
C 32
D 48
E 96
F 10
demha
2013-09-24 01:13:44
The shape is a pyramid with:
6. a rectangular base with a width of 4
A Lateral area: 120; Volume: 98
B Lateral area: 192; Volume: 76
C Lateral area: 240; Volume: 51
D Lateral area: 176; Volume: 33
E Lateral area: 288; Volume: 75
F Lateral area: 100; Volume: 64
Volume:
a = 6 x 4
a = 24
v = 1/3 x 24 x 8
v = 64
Lateral Area:
6 + 4 + 6 + 4 = 20
20 x .50 = 10
10 x 10 = 100
Answer: F
7. a square base
A Lateral area: 120; Volume: 96
B Lateral area: 90; Volume: 64
C Lateral area: 176; Volume: 144
D Lateral area: 192; Volume: 35
E Lateral area: 288; Volume: 288
F Lateral area: 240; Volume: 75
Volume:
a = 6 x 6
a = 36
v = 1/3 x 36 x 8
v = 96
Lateral Area:
6 + 6 + 6 + 6 = 24
24 x .50 = 12
12 x 10 = 120
Answer: A
8. a rectangular base with a width of 3
A Lateral area: 192; Volume: 144
B Lateral area: 90; Volume: 48
C Lateral area: 276; Volume: 64
D Lateral area: 176; Volume: 144
E Lateral area: 92; Volume: 96
F Lateral area: 62; Volume: 24
Volume:
a = 6 x 3
a = 18
v = 1/3 x 18 x 8
v = 48
Lateral Area:
3 + 6 + 3 + 6 = 18
18 x .50 = 9
9 x 10 = 90
Answer: B
9. a rectangular base with a width of 5
A Lateral area: 100; Volume: 48
B Lateral area: 240; Volume: 112
C Lateral area: 176; Volume: 96
D Lateral area: 110; Volume: 80
E Lateral area: 288; Volume: 144
F Lateral area: 90; Volume: 64
Volume:
a = 5 x 6
a = 30
v = 1/3 x 30 x 8
v = 80
Lateral Area:
6 + 5 + 6 + 5 = 22
22 x .50 = 11
11 x 10 = 110
Answer: D
10. a rectangular base with a width of 7
A Lateral area: 240; Volume: 64
B Lateral area: 188; Volume: 96
C Lateral area: 176; Volume: 144
D Lateral area: 130; Volume: 112
E Lateral area: 144; Volume: 215
F Lateral area: 100; Volume: 128
Volume:
a = 6 x 7
a = 42
v = 1/3 x 42 x 8
v = 112
Lateral Area:
7 + 6 + 7 + 6 = 26
26 x .50 = 13
13 x 10 = 130
Answer: D
bob bundy
2013-09-23 18:27:30
hi demha,
Glad you enjoyed our discussion about these problems. We have identified a big error with the work that Compu High are teaching you. The teaching page is available as a 'demo' so I have been able to read it all. The volume calculations are fully correct but the slant height section is WRONG.
Because I am a teacher of maths, I cannot just let you do these questions without showing you the correction. I'll do that first. Then you'll want to get their answers so you get full marks. That's OK. I have used their method and Q6 to Q10 all have a 'correct' answer. So hopefully, at the end, you will understand the work properly and be able to get full marks. I would just add, this is not the only criticism I have of this on-line course but I won't burden you with all my objections. Let's get on.
Correction: If you buy some wire so you can make a wire-frame version of these pyramids, you could proceed as follows. (see diagram)
(i) Cut two lengths a and two widths b. Join them to make the base rectangle.
(ii) Join the diagonals to find the middle and make a temporary wire for the height h. This fixes the vertex.
At this point the five points of the pyramid are all fixed.
(iii) You can cut four wires to join the base to the vertex and the model is complete.
It is not necessary is say how long the red lines for the slant heights are because they are already there, whatever size they need to be to join the midpoints of the base to the vertex. Just choosing a length for these red lines is not an option. Their lengths can be calculated using Pythagoras' theorem. Furthermore, with a rectangular base the two slant heights will not be the same, so giving a single slant height value is only possible where the base is a regular polygon. A rectangle is not a regular polygon.
This is why I have had such a problem with the slant height values given. The pyramids described cannot exist because those slant height values won't work!
Sorry for the rant, but I felt I had to get that off my chest. I feel better. If that's leaving you worried and confused I apologise, but I have to take account of the possibility that other members of the forum may be reading this thread and I don't want them being taught incorrectly.
So now, back to the problems. You can, of course, ignore all I have just said, and concentrate on getting the 'right' answers. (Anyway, once you have a volume, you'll know which answer to choose, without having to calculate the lateral areas at all. This is another criticism whoops, I'd better move on!)
The formula for the volume is
You have done Q6 volume correctly.
Their formula for lateral area is
So what you need to do is
(i) Add a + b + a + b to get the perimeter (distance around the base).
(ii) Calculate 1/2 of this and multiply by the slant height.
You can be fairly confident you have the right result if you can choose a single multi-choice answer that fits both the volume and area you have worked out.
Good luck.
Bob
anonimnystefy
2013-09-23 09:47:29
It truly is. The way you two tumble down those problems is amazing!
demha
2013-09-23 09:10:44
It's interesting to see two math geniuses at work
The shape is a pyramid with:
6. a rectangular base with a width of 4
A Lateral area: 120; Volume: 98
B Lateral area: 192; Volume: 76
C Lateral area: 240; Volume: 51
D Lateral area: 176; Volume: 33
E Lateral area: 288; Volume: 75
F Lateral area: 100; Volume: 64
So this is what I did. First I got the area of the rectangle:
a = 6 x 4
a = 24
v = 1/3 x 24 x 8
v = 64
That's what I'm getting for the volume.
As for the lateral area, would I have to do:
2 x 6 x 8
2 x 4 x 8
Get the answers for these, add them together to get the perimeter and then move on?
anonimnystefy
2013-09-23 06:00:13
Huh? I'm not sure what to make of your reply. I don't know how you will proceed when their questions are incorrect.
bob bundy
2013-09-23 05:58:17
Thank you so much. It was driving me mad because I wasn't getting the right volume ... but, somehow, just having you say that has re-aligned my brain correctly to answer A too. Oh joy! Just got to square these formulas with the warped non Euclidean space that they occupy and I'm there. Many, many thanks for your input. I think I'm OK to proceed now.
Bob
## Board footer
Powered by FluxBB | 3,108 | 9,048 | {"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-2014-15 | latest | en | 0.809824 |
https://reference.wolframcloud.com/language/ref/CumulantGeneratingFunction.html | 1,709,240,259,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00583.warc.gz | 462,226,132 | 14,209 | # CumulantGeneratingFunction
CumulantGeneratingFunction[dist,t]
gives the cumulant-generating function for the distribution dist as a function of the variable t.
CumulantGeneratingFunction[dist,{t1,t2,}]
gives the cumulant-generating function for the multivariate distribution dist as a function of the variables t1, t2, .
# Examples
open allclose all
## Basic Examples(3)
Compute a cumulant-generating function (cgf) for a continuous univariate distribution:
The cgf for a univariate discrete distribution:
The cgf for a multivariate distribution:
## Scope(5)
Compute the cgf for a formula distribution:
Find the cgf for a function of random variates:
Compute the cgf for data distribution:
Find the cgf for a truncated distribution:
Find the cgf for the slice distribution of a random process:
## Applications(5)
The cumulant-generating function of a difference of two independent random variables is equal to the sum of their cumulant-generating functions with oppositive sign arguments:
Illustrate the central limit theorem:
Find the cumulant-generating function for the standardized random variate:
Find the moment-generating function for the sum of standardized random variates rescaled by :
Find the large limit:
Compare with the moment-generating function of a standard normal distribution:
Find the Esscher premium for insuring against losses following GammaDistribution:
Compare with the definition:
Construct a BernsteinChernoff bound for the survival function :
Large approximation of the bound:
Construct Daniel's saddle point approximation to PDF of VarianceGammaDistribution:
Find the saddle point associated with the argument of probability density function :
Select the solution that is valid for all real , including the origin:
The approximation is constructed using the cumulant-generating function at the saddle point:
Find the normalization constant:
Compare the approximation to the exact density:
## Properties & Relations(3)
Exponential of CumulantGeneratingFunction gives MomentGeneratingFunction:
CumulantGeneratingFunction is an exponential generating function for the sequence of cumulants:
Use CumulantGeneratingFunction directly:
Cumulant is equivalent to :
Use SeriesCoefficient formulation:
## Possible Issues(2)
For some distributions with long tails, cumulants of only several low orders are defined:
Correspondingly, CumulantGeneratingFunction is undefined:
CumulantGeneratingFunction is not always known in closed form:
Use Cumulant to find cumulants directly:
Wolfram Research (2010), CumulantGeneratingFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html.
#### Text
Wolfram Research (2010), CumulantGeneratingFunction, Wolfram Language function, https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html.
#### CMS
Wolfram Language. 2010. "CumulantGeneratingFunction." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html.
#### APA
Wolfram Language. (2010). CumulantGeneratingFunction. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html
#### BibTeX
@misc{reference.wolfram_2023_cumulantgeneratingfunction, author="Wolfram Research", title="{CumulantGeneratingFunction}", year="2010", howpublished="\url{https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html}", note=[Accessed: 29-February-2024 ]}
#### BibLaTeX
@online{reference.wolfram_2023_cumulantgeneratingfunction, organization={Wolfram Research}, title={CumulantGeneratingFunction}, year={2010}, url={https://reference.wolfram.com/language/ref/CumulantGeneratingFunction.html}, note=[Accessed: 29-February-2024 ]} | 801 | 3,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2024-10 | latest | en | 0.747211 |
https://kmmiles.com/63-991-miles-in-km | 1,686,308,054,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00127.warc.gz | 360,106,039 | 5,897 | kmmiles.com
Search
# 63.991 miles in km
## Result
63.991 miles equals 102.9615 km
You can also convert 63.991 mph to km.
## Conversion formula
Multiply the amount of miles by the conversion factor to get the result in km:
63.991 mi × 1.609 = 102.9615 km
## How to convert 63.991 miles to km?
The conversion factor from miles to km is 1.609, which means that 1 miles is equal to 1.609 km:
1 mi = 1.609 km
To convert 63.991 miles into km we have to multiply 63.991 by the conversion factor in order to get the amount from miles to km. We can also form a proportion to calculate the result:
1 mi → 1.609 km
63.991 mi → L(km)
Solve the above proportion to obtain the length L in km:
L(km) = 63.991 mi × 1.609 km
L(km) = 102.9615 km
The final result is:
63.991 mi → 102.9615 km
We conclude that 63.991 miles is equivalent to 102.9615 km:
63.991 miles = 102.9615 km
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case sixty-three point nine nine one miles is approximately one hundred two point nine six two km:
63.991 miles ≅ 102.962 km
## Conversion table
For quick reference purposes, below is the miles to kilometers conversion table:
miles (mi) kilometers (km)
64.991 miles 104.570519 km
65.991 miles 106.179519 km
66.991 miles 107.788519 km
67.991 miles 109.397519 km
68.991 miles 111.006519 km
69.991 miles 112.615519 km
70.991 miles 114.224519 km
71.991 miles 115.833519 km
72.991 miles 117.442519 km
73.991 miles 119.051519 km
## Units definitions
The units involved in this conversion are miles and kilometers. This is how they are defined:
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada.
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world. | 650 | 2,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.863581 |
https://exceljet.net/formulas/get-days-before-a-date | 1,695,427,671,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506429.78/warc/CC-MAIN-20230922234442-20230923024442-00521.warc.gz | 283,776,036 | 12,116 | ## Summary
To calculate the number of days before a certain date, you can use subtraction and the TODAY function. In the example, D5 contains this formula:
``````=B4-TODAY()
```
```
## Generic formula
``=date-TODAY()``
## Explanation
In Excel, dates are simply serial numbers. In the standard date system for windows, based on the year 1900, where January 1, 1900 is the number 1. Dates are valid through 9999, which is serial number 2,958,465. This means that January 1, 2050 is the serial number 54,789.
In the example, the date is March 9, 2016, which is the serial number 42,438. So:
``````= B4-TODAY()
= January 1 2050 - April 27, 2014
= 54,789 - 42,438
= 12,351
``````
This means there are 13,033 days before January 1, 2050, when counting from March 9, 2016.
### Without TODAY
Note: you don't need to use the TODAY function. In the second example, the formula in D6 is:
``````=B6-C6
``````
### Concatenating with text
In the third example, the same basic formula is used along with concatenation operator (&) to embed the calculated days in a simple text message:
``````="Just "& B6-C6 &" days left!"
```
```
Since there are 15 days between December 10, 2014 and December 25, 2014, the result is this message: Just 15 days left!
### Workdays only
To calculate workdays between dates, you can use the NETWORKDAYS function as explained here.
Author
### Dave Bruns
Hi - I'm Dave Bruns, and I run Exceljet with my wife, Lisa. Our goal is to help you work faster in Excel. We create short videos, and clear examples of formulas, functions, pivot tables, conditional formatting, and charts. | 458 | 1,612 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2023-40 | longest | en | 0.921951 |
https://convertoctopus.com/257-ounces-to-kilograms | 1,618,483,610,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00212.warc.gz | 295,441,367 | 7,763 | ## Conversion formula
The conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:
1 oz = 0.028349523125 kg
To convert 257 ounces into kilograms we have to multiply 257 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:
1 oz → 0.028349523125 kg
257 oz → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 257 oz × 0.028349523125 kg
M(kg) = 7.285827443125 kg
The final result is:
257 oz → 7.285827443125 kg
We conclude that 257 ounces is equivalent to 7.285827443125 kilograms:
257 ounces = 7.285827443125 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.13725277023183 × 257 ounces.
Another way is saying that 257 ounces is equal to 1 ÷ 0.13725277023183 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred fifty-seven ounces is approximately seven point two eight six kilograms:
257 oz ≅ 7.286 kg
An alternative is also that one kilogram is approximately zero point one three seven times two hundred fifty-seven ounces.
## Conversion table
### ounces to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms
ounces (oz) kilograms (kg)
258 ounces 7.314 kilograms
259 ounces 7.343 kilograms
260 ounces 7.371 kilograms
261 ounces 7.399 kilograms
262 ounces 7.428 kilograms
263 ounces 7.456 kilograms
264 ounces 7.484 kilograms
265 ounces 7.513 kilograms
266 ounces 7.541 kilograms
267 ounces 7.569 kilograms | 456 | 1,783 | {"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-2021-17 | latest | en | 0.731536 |
http://mathhelpforum.com/calculus/52985-stationary-pt-curve.html | 1,524,548,217,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946564.73/warc/CC-MAIN-20180424041828-20180424061828-00638.warc.gz | 205,873,980 | 11,560 | # Thread: stationary pt. of a curve
1. ## stationary pt. of a curve
Help me with this.
Can you tell me what is your value of x.Thanks
i dunno where went wrong..
2. Can't read the text in the figure.
To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.
3. Originally Posted by maybeline9216
Help me with this. Thanks
Fortunately I have a degree in Egyptology majoring in heiroglyphics. The curve is $\displaystyle y = 3 e^{-x/2} + e^{x/2}$.
Solve $\displaystyle \frac{dy}{dx} = 0$ for x and substitute the value into y:
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
$\displaystyle \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3$.
Therefore $\displaystyle y = 3 e^{-(\ln 3) /2} + e^{(\ln 3)/2} = 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$.
4. Originally Posted by ticbol
Can't read the text in the figure.
To get the stationary point, find the dy/dx and then set that to zero. The x there will give the stationary point.
Click the attached file and then click it again to get a better view
5. Originally Posted by maybeline9216
Click the attached file and then click it again to get a better view
Hmmpph .... So in fact only my mathematical skills were required.
6. $\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
can you tell me how to get the value of x from here.
Its because i tried to use the substitution method to get x
For eg. Let x=e^x/2
But i got a different answer.Can you show me another method?
P.S.My dy/dx is correct
7. Originally Posted by mr fantastic
Fortunately I have a degree in Egyptology majoring in heiroglyphics.
Wow! What is that?? lols
8. Originally Posted by maybeline9216
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
can you tell me how to get the value of x from here.
Its because i tried to use the substitution method to get x
For eg. Let x=e^x/2
But i got a different answer.Can you show me another method?
P.S.My dy/dx is correct
Originally Posted by Mr Fantastic
[snip]
$\displaystyle 0 = -\frac{3}{2} e^{-x/2} + \frac{1}{2} e^{x/2}$
Mr F edit: Multiply both sides by $\displaystyle {\color{red}2 e^{x/2}}$:
$\displaystyle \Rightarrow 0 = -3 + e^x \Rightarrow x = \ln 3$.
[snip]
..
9. Thanks!!that's so much easier!
10. $\displaystyle 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$
Sorry.... i have another qn, how do u get this(see above) ?
11. Originally Posted by maybeline9216
$\displaystyle 3 = 3 e^{\ln \left(1/\sqrt{3}\right)} + e^{\ln \sqrt{3}} = \frac{3}{\sqrt{3}} + \sqrt{3} = 2 \sqrt{3}$
Sorry.... i have another qn, how do u get this(see above) ?
$\displaystyle e^{\ln A} = A \, ....$ | 938 | 2,791 | {"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} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.877016 |
https://www.teacherspayteachers.com/Product/Fractions-Decimals-Game-Converting-Tenths-BUMP-2757481 | 1,493,159,683,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120881.99/warc/CC-MAIN-20170423031200-00631-ip-10-145-167-34.ec2.internal.warc.gz | 957,389,613 | 26,679 | Total:
\$0.00
# Fractions & Decimals Game - Converting Tenths BUMP
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7.36 MB | 12 pages
### PRODUCT DESCRIPTION
Fractions & Decimals Games Players throw a die and move their counter around the outside track. When they land on an outside space they convert the fraction or decimal they land on into the place a counter on the answer. This is a game students really enjoy.
Included in this download are 5 full color boards - fractions as tenths to decimals, fractions as tenths to percentages, fractions as tenths to symbolic representations, decimals as tenths to symbolic representations & decimals as tenths to percentages.
- 5 low color boards
- 1 set of rules and teaching notes
What You Need:
2 Players
1 die numbered 1 to 6
1 Bump Fractions and Decimals Board
7 Counters of different colors per player
Dice Mat (optional)
How to Play:
1. Add up how many letters you have in your first and last names. The student with the least letters goes first.
2. Players choose a monster inside a space on the board and place one of their counters on it. This is their starting position.
3. Player One throws the die onto the Dice Mat and moves their counter that many spaces forward.
4. Player One then converts the fraction they have landed on to a decimal and places one of the remaining counters over the decimal in the center of the board.
5. Player Two repeats steps 3 & 4.
The 'Bump' Rule :
- If a number is already covered by a single opposition counter a player can 'BUMP' the counter off the board and replace it with their own.
The 'Locked In' Rule:
- If a number is covered is covered by your piece and you throw that number, A player can place a second counter on top of the one already on the board. This means the piece is 'LOCKED IN' and it cannot be 'BUMPED' off the board.
6. Play continues in this turnabout fashion of throwing 2 dice on one turn and one die on the next.
How to Win:
The first player to use up all their counters wins the game.
The Dice Mat:
- The Dice Mat is a piece of felt or A4 paper. It is used to keep down noise, confine the 'throws' to a small area and or to eliminate dice related 'silliness' before it occurs. hehehe
- If any of the dice roll off the mat the throw is disallowed.
Conduct Number Talks Before and After the Game:
A 'Number Talk' is a 5 to 15 minute class discussion which focuses on the efficient use of number strategies to solve problems. The teacher and students verbally communicate problem solving strategies to explore and expose the group to new ways of thinking. The talks serve to deepen mathematical understandings and develop computational fluency. In this case the Number Talk would be held in relation to the students understandings of what fractions and decimals actually are and how to convert between the two.
A Possible Number Talks:
- A fraction can be read as 2 out of 3. Use concrete materials to demonstrate this. Use the PowerPoint file to show the board where MAB material is colored to show fractions. Tell your partner which graphic shows 4 out of 10. Which graphic has 4 tenths covered. Which graphic is 60% covered.
Possible Journal Reflections:
- Write 2 'I wonder' questions you still have as a result of today's Math session.
- I enjoyed playing this Math game today because______________
- I did not enjoy playing this Math game today because___________
- While playing the Math game today I figured out ________
- On a scale from 1 to 10 I rated my Math work a ______ today because________
Enjoy!
If You Like the Look of This Game, CHECK OUT MY MEGA BUNDLE! - The package includes ALL my current Math and Literacy games PLUS every resource I create for the next YEAR for FREE!!! You only have to use 7 of the games to get your money's worth :-)
Plus... Special thanks goes to Ron Leishman for creating the monsters used in this game.
You're a Legend!
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12
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Teaching Duration
Lifelong Tool
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\$3.60 | 1,027 | 4,333 | {"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-2017-17 | longest | en | 0.924943 |
http://us.metamath.org/mpeuni/clublem.html | 1,643,187,105,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00191.warc.gz | 73,159,892 | 4,150 | Mathbox for Richard Penner < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > clublem Structured version Visualization version GIF version
Theorem clublem 37736
Description: If a superset 𝑌 of 𝑋 possesses the property parameterized in 𝑥 in 𝜓, then 𝑌 is a superset of the closure of that property for the set 𝑋. (Contributed by RP, 23-Jul-2020.)
Hypotheses
Ref Expression
clublem.y (𝜑𝑌 ∈ V)
clublem.sub (𝑥 = 𝑌 → (𝜓𝜒))
clublem.sup (𝜑𝑋𝑌)
clublem.maj (𝜑𝜒)
Assertion
Ref Expression
clublem (𝜑 {𝑥 ∣ (𝑋𝑥𝜓)} ⊆ 𝑌)
Distinct variable groups: 𝜒,𝑥 𝑥,𝑋 𝑥,𝑌
Allowed substitution hints: 𝜑(𝑥) 𝜓(𝑥)
Proof of Theorem clublem
StepHypRef Expression
1 clublem.sup . . 3 (𝜑𝑋𝑌)
2 clublem.maj . . 3 (𝜑𝜒)
3 clublem.y . . . . 5 (𝜑𝑌 ∈ V)
43a1d 25 . . . 4 (𝜑 → ((𝑋𝑌𝜒) → 𝑌 ∈ V))
5 clublem.sub . . . . . 6 (𝑥 = 𝑌 → (𝜓𝜒))
65cleq2lem 37733 . . . . 5 (𝑥 = 𝑌 → ((𝑋𝑥𝜓) ↔ (𝑋𝑌𝜒)))
76elab3g 3351 . . . 4 (((𝑋𝑌𝜒) → 𝑌 ∈ V) → (𝑌 ∈ {𝑥 ∣ (𝑋𝑥𝜓)} ↔ (𝑋𝑌𝜒)))
84, 7syl 17 . . 3 (𝜑 → (𝑌 ∈ {𝑥 ∣ (𝑋𝑥𝜓)} ↔ (𝑋𝑌𝜒)))
91, 2, 8mpbir2and 956 . 2 (𝜑𝑌 ∈ {𝑥 ∣ (𝑋𝑥𝜓)})
10 intss1 4483 . 2 (𝑌 ∈ {𝑥 ∣ (𝑋𝑥𝜓)} → {𝑥 ∣ (𝑋𝑥𝜓)} ⊆ 𝑌)
119, 10syl 17 1 (𝜑 {𝑥 ∣ (𝑋𝑥𝜓)} ⊆ 𝑌)
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 196 ∧ wa 384 = wceq 1481 ∈ wcel 1988 {cab 2606 Vcvv 3195 ⊆ wss 3567 ∩ cint 4466 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1720 ax-4 1735 ax-5 1837 ax-6 1886 ax-7 1933 ax-9 1997 ax-10 2017 ax-11 2032 ax-12 2045 ax-13 2244 ax-ext 2600 This theorem depends on definitions: df-bi 197 df-or 385 df-an 386 df-tru 1484 df-ex 1703 df-nf 1708 df-sb 1879 df-clab 2607 df-cleq 2613 df-clel 2616 df-nfc 2751 df-v 3197 df-in 3574 df-ss 3581 df-int 4467 This theorem is referenced by: mptrcllem 37739 trclubgNEW 37744
Copyright terms: Public domain W3C validator | 1,060 | 1,855 | {"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} | 3.109375 | 3 | CC-MAIN-2022-05 | latest | en | 0.242145 |
https://www.simplyfreshers.com/aricent-placement-paper-2008-orissa-pattern-interview/ | 1,725,732,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00655.warc.gz | 971,983,658 | 10,209 | Company: Aricent Group
Hi friends ,I am GURU from university college of Engineering,burla,orissa .ARICENT came to our college on 4th sep 2008 It had 4 stages in selection procedure.
1.written test
2.critical incident form filling
3.GI(group interview)
4.tech interview+critical incident interview
1.WRITTEN TEST
This time the pattern was different .This test was consisting of 4 sections .each paper will b given 1 after another there was sectional cut-off. so b careful.
1st was Analytical ability test including puzzle test+ statement- assumption+ logical reasoning+ statement- conclusions+ figure series etc. puzzle section was quite difficult n time taking n others easy. 30 ques were there.
2nd section was Aptitude test having 20 ques. which was very very easy. i managed to do them within 10 minutes.
3rd was Data structure which had 20 ques regarding time complexity of sorting algorithm like quick sort , merge sort also time complexity if an array consists of some particular no of elements through binary search(O(log2 n=? if n=10).and ques from linked list, tree( which tree is used in binary sorting). This DS section was very very easy i took only 5 minutes to do all d ques. so friends read DS very carefully , that ll help u both in written test as well as in interview.
In 4th test one has 2 opt for c or c++ or java as u strong accordingly.i chose c .there ll be 30 ques. so practising test your c skill by kanetkar ll help u a lot. One thing keep in mind that there is sectional cut-off n a good marks obtained in this round will help u much more in the selection procedure.21 out of 120 qualified for the next round. fortunately i was one of them. after the written test , all the remaining stages are open to the selected candidates.
2.CRITICAL INCIDENT FORM FILLING
After written test , we had PPT. then all candidates selected through written test were given a form having around 7 critical situation based ques and u have to answer all the ques within 30 mins. like:———-
1.give a recent experience where u had some proposal but u r enforced to change ur opinion while working in a team
2.give your one recent situation while working in a team dat u have managed to do some tasks at a time
3.while working in a group , how did u manage yourself to accomplish a task where u have no preliminary idea about that .
4. how did u make ur team members understand a task, while they had no knowledge …………….etc.
3.GROUP INTERVIEW(GI)
Here i had 12 members in my group and as i said this is not an elimination round. our topic was :”PACKAGED FOOD IS A GOOD IDEA “(as compared to the mid-day meals or hot meals delivered to school going children)
We were divided in to two groups . don’t b worried u ll b given a sheet having some facts about d topic n u have 2 present your own views n not to oppose or to support others
4.TECH-INTERVIEW + CRITICAL-INTERVIEW (TI+CI)
strong>
Here u ll b asked about some technical ques as well as from critical incident form u filled up. There were 4 panels and i was d first to b interviewed in panel no.-1. my interview was @6.30pm n like this.
1.introduce yourself briefly
2.what are storage classes used in c n their advantages , disadvantages n their uses
3.do u know fibonacci series n implement it 1.using main() as recursive 2.using user defined recursive function with parameter passing 3.without parameter passing (he just wanted to know whether you have sound programming skill or not)
4.different stages of c programming
5.function used for dynamic memory allocation n their differences
6.what is data structure n ADT (abstact data type)
7.given an expression n told 2 draw d tree. then 2 find out post-order n pre-order n explanation
8.implementing doubly linked list and adding n deleting nodes at some given positions
9.difference between c n c++
11.about new n delete operator in comparison with malloc(), calloc() and free() functions
12.why malloc() function isn’t used in c++ . instead we r using new operator
13.difference between structure and class in c++
14.difference between DBMS n RDBMS (told abt E.M.CODD’S 12 rules n key concept)
15.abt INF and 2NF
16.some simple queries
17.abt page replacement algorithm
18.what is page fault n thrashing
19.what is synchronization n different tools
20.how to implement semaphor n monitor
21.what is deadlock n different mechanism to avoid it
22.which c/c++ complier u r using and it’s effect on size of data types. what do u mean by a 32-bits os
23.what is computer network n distinguish from distributed system
24.what is an IP, it’s function n d layer on which it is used
25.diffeent version of IP addresses n explain IPV4 in details with classes
26.abt OSI n TCP/IP network model
27.application of queue n stack
28.abt router,bridge ,switch ……………………………………….etc.
29.1-2 situation based ques
30.What is endianness. how can u know your computer is either little or big one. write a code to find it.
31.where extern n static variables are stored.
32.write program for binary search and derive its time complexity.
33.what is constructor and why it is mainly used.
34.what is binary search tree and it’s uses
35.hav u any ques to me
Definitely it was too long .My interview took around 1.30 hrs n almost technical. he was very cool, helping in nature n gave some hints when not finding any idea.8 out of 21 got selected ……..THAT DAY WAS FOR ME AND LUCKILY I WAS ONE OF THEM …..So my dear friends b cool, confident n do not try to bluff them .if u rn’t finding any point , simply tell i don’t know or i m not getting any point. so have faith on yourself n god .WHEN UR DAY WILL COME , NO ONE CAN STOP U, EVEN NOT U. BUT KEEP IN MIND THAT HARD WORK ALWAYS PAYS , BCOZ GOOD LUCK COMES TO IT
For technical, have depth concept on c, data structure , os , n/w and some fundamental concept on c++ , dbms and microprocessor .i was from computer science . for electronics student depth concept about DEC, MP & MC n N/W is essential. | 1,427 | 5,944 | {"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-2024-38 | latest | en | 0.976034 |
https://isabelle.in.tum.de/repos/isabelle/diff/339a8b3c4791/src/HOL/Enum.thy | 1,606,264,633,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141177607.13/warc/CC-MAIN-20201124224124-20201125014124-00245.warc.gz | 336,007,265 | 3,170 | src/HOL/Enum.thy
changeset 45140 339a8b3c4791 parent 45119 055c6ff9c5c3 child 45141 b2eb87bd541b
```--- a/src/HOL/Enum.thy Thu Oct 13 23:02:59 2011 +0200
+++ b/src/HOL/Enum.thy Thu Oct 13 23:27:46 2011 +0200
@@ -758,69 +758,13 @@
unfolding enum_the_def by (auto split: list.split)
qed
+
subsection {* An executable card operator on finite types *}
-lemma
- [code]: "card R = length (filter R enum)"
-by (simp add: distinct_length_filter[OF enum_distinct] enum_UNIV Collect_def)
-
-subsection {* An executable (reflexive) transitive closure on finite relations *}
-
-text {* Definitions could be moved to Transitive Closure theory if they are of more general use. *}
-
-definition ntrancl :: "('a * 'a => bool) => nat => ('a * 'a => bool)"
-where
- [code del]: "ntrancl R n = (UN i : {i. 0 < i & i <= (Suc n)}. R ^^ i)"
-
-lemma [code]:
- "ntrancl (R :: 'a * 'a => bool) 0 = R"
-proof
- show "R <= ntrancl R 0"
- unfolding ntrancl_def by fastforce
-next
- {
- fix i have "(0 < i & i <= Suc 0) = (i = 1)" by auto
- }
- from this show "ntrancl R 0 <= R"
- unfolding ntrancl_def by auto
-qed
-
lemma [code]:
- "ntrancl (R :: 'a * 'a => bool) (Suc n) = (ntrancl R n) O (Id Un R)"
-proof
- {
- fix a b
- assume "(a, b) : ntrancl R (Suc n)"
- from this obtain i where "0 < i" "i <= Suc (Suc n)" "(a, b) : R ^^ i"
- unfolding ntrancl_def by auto
- have "(a, b) : ntrancl R n O (Id Un R)"
- proof (cases "i = 1")
- case True
- from this `(a, b) : R ^^ i` show ?thesis
- unfolding ntrancl_def by auto
- next
- case False
- from this `0 < i` obtain j where j: "i = Suc j" "0 < j"
- by (cases i) auto
- from this `(a, b) : R ^^ i` obtain c where c1: "(a, c) : R ^^ j" and c2:"(c, b) : R"
- by auto
- from c1 j `i <= Suc (Suc n)` have "(a, c): ntrancl R n"
- unfolding ntrancl_def by fastforce
- from this c2 show ?thesis by fastforce
- qed
- }
- from this show "ntrancl R (Suc n) <= ntrancl R n O (Id Un R)" by auto
-next
- show "ntrancl R n O (Id Un R) <= ntrancl R (Suc n)"
- unfolding ntrancl_def by fastforce
-qed
+ "card R = length (filter R enum)"
+ by (simp add: distinct_length_filter [OF enum_distinct] enum_UNIV Collect_def)
-lemma [code]: "trancl (R :: ('a :: finite) * 'a => bool) = ntrancl R (card R - 1)"
-by (cases "card R") (auto simp add: trancl_finite_eq_rel_pow rel_pow_empty ntrancl_def)
-
-(* a copy of Nitpick.rtrancl_unfold, should be moved to Transitive_Closure *)
-lemma [code]: "r^* = (r^+)^="
-by simp
subsection {* Closing up *}
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Browsing for any effective and successful GPA Calculator?
GPA stands for quality issue standard and it’s yet one more metric you’ll must sustain keep track of of in very high university, higher education, and over and above. Students oftentimes get perplexed on all things GPA – the right way to estimate and raise GPA; differing types of GPA, weighted vs. non-weighted, cumulative vs. semester GPA etc., and there are not any smart useful resource presented over the internet that will help pupils with all their problems. We have stepped in to help you out which has a GPA information.
That’s this tutorial for?
Any substantial faculty or university pupil along with a grade letter or 4-point scale dependent grade reporting strategy. Some academic techniques site extra weight and significance on the GPA calculations than other individuals, nevertheless it is an critical piece of academic conditions in American schools particularly.
### Just what is GPA?
I continue to keep listening to about GPA and exactly how some pupils have got a GPA of three or 4, and so on. What do these numbers suggest and how should I estimate my GPA?
Grade Issue Typical (GPA) is a summary statistic that signifies a student’s ordinary operation inside their studies through a stated time period of time, these kinds of as a single semester, a tutorial yr, or a complete educational overall performance at an institution. Becoming numerical, GPAs are often calculated to two decimals. These are utilised as indicators of even if amounts of university student effectiveness fulfill some mounted criterion, and for sorting groups of students into rank buy.
Whereas GPA scores are universally comprehended, grading scales vary significantly across establishments and nations. Conversion tables are normally obtainable for evaluating grades and GPAs in countries and internationally.
When an entire study software is arranged being a selection of “units”, every single period of time of your time presents increase to its possess GPA. The commonest examine time period for your program is 1 semester, commonly 12-15 months of class. If a full-time pupil enrolls in four programs in a very certain semester, the GPA is calculated as the typical general performance around those people four programs.
### Why is GPA so really important?
Schools use this variety to measure your complete overall performance in school and do a comparison of you to definitely other students.
Your GPA is very important for your long run due to the fact: | 491 | 2,586 | {"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} | 2.859375 | 3 | CC-MAIN-2020-34 | latest | en | 0.942928 |
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2007-01-23, 08:10 #100
Citrix
Jun 2003
5×317 Posts
Quote:
Originally Posted by paulunderwood We are at nearly 3.1 million bit tests and it is not the end of January yet That's about 80% towards a million digits. We are due a prime, but, fingers crossed, that prime will occur at just over a million digits But who knows? Join us and chance your arm.
IF you find a million digit prime soon (which I hope you do), will you still continue to LLR to 5 M?
2007-01-23, 16:22 #101
paulunderwood
Sep 2002
Database er0rr
3,947 Posts
Quote:
IF you find a million digit prime soon (which I hope you do), will you still continue to LLR to 5 M?
Hardly, No. Personally, the main goal is a 1 million plus digit prime. The effort would have to be redoubled to find the following prime, most likely and that could be greater than 5M bits So it just a final push now. Help us
Last fiddled with by paulunderwood on 2007-01-23 at 16:28
2007-01-23, 21:15 #102 Citrix Jun 2003 5×317 Posts Will help soon! (As soon as I can free up a computer).
2007-02-12, 12:01 #103
paulunderwood
Sep 2002
Database er0rr
1111011010112 Posts
Thomas wrote:
Quote:
(Just as a note: The next FFT length change is at n=3121158. Tests will take about 20% longer above that n-value.)
Thanks for the reminder. This new FFT size will last until n=3719191, i.e. for the rest of the year. 32-bit athlons can continue at the old length until n=3171158.
That is now 5 hours per test on a Pentium 4 at 3 GHz. Quicker on a Core 2 Duo.
Last fiddled with by paulunderwood on 2007-02-12 at 12:04
2007-02-12, 13:55 #104
Thomas11
Feb 2003
111011110112 Posts
Quote:
Originally Posted by paulunderwood 32-bit athlons can continue at the old length until n=3171158.
The above holds for version 3.5 (or below) of LLR. For LLR 3.6 and higher the 32bit Athlons will use the old length until n=3180158.
The make the things even more complicated: We found LLR 3.5 slightly faster than the newer versions on the Athlons (a fraction of a percent). So the Athlon testers might use version 3.5 up to n=3171158 and then switch to version 3.6/3.6.2/3.7...
2007-02-26, 19:12 #105 paulunderwood Sep 2002 Database er0rr 3,947 Posts Welcome back James! I have asked Xyzzy to upload the next set of LLR input files, covering 3.15M to 3.3M bits, which will take us very close to 1 million digits (3.32M). Tests take nearly 6 hours each on a Pentium 4 at 2.66GHz. An average file will take 2 weeks -- with 58 numbers. This will be the case until about 3.7M bits. Thanks to Geoff for the new sieve, which Carlos and Thomas used to get the least divisor up to 245 trillion. We are due to find a prime If we find a prime now it will be 12th largest known. Good luck all Last fiddled with by paulunderwood on 2007-02-26 at 19:25
2007-02-27, 01:00 #106 SB2 Jul 2003 1628 Posts Thanks Paul, it's good to be back.
2007-03-22, 01:10 #107 paulunderwood Sep 2002 Database er0rr F6B16 Posts I have just completed the last sub-3 million bit 321 candidate. Great work people. Our next goal is to reach 1 million decimal digits by the end of May. After reaching this mile stone I hope we find another prime quicky. In the long term, it would be great to get to 3.6 million by the end of this year and then a 4 million bits target for the end of 2008 will be easily attainable. Our next prime is due at 4.1 million bits. However we do not know where it actually is. It could be much smaller or much larger. Good luck all with finding our next 321 prime!
2007-05-03, 10:05 #108 paulunderwood Sep 2002 Database er0rr 3,947 Posts It looks as though we will miss the end-of-May target for all candidates below one million digits to be booked out, but we should easily get there by the end of June For all 0
2007-06-02, 21:33 #109 paulunderwood Sep 2002 Database er0rr 394710 Posts Thanks to Xyzzy for uploading the latest 321 input files ready for LLR crunching, covering from n=3.3M to 3.45M. We are also grateful to Thomas for further sieving these to 271.2 trillion. This batch will take us over 1 million decimal digits (3.32M) Grab yourself some and maybe you will find a mega-prime Also we have covered all n from 0 to 3M with a factor or an LLR residue. After some more analysis of errors, we will make all public... Last fiddled with by paulunderwood on 2007-06-02 at 22:09
2007-07-11, 09:34 #110 paulunderwood Sep 2002 Database er0rr 3,947 Posts We are now testing candidates for a prime number with over 1 million digits, a so called mega-prime Thanks to all who have helped over the last 4 years to get us here. Join us! Last fiddled with by paulunderwood on 2007-07-11 at 09:35
Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Lounge 137 2021-12-01 18:27 paulunderwood 3*2^n-1 Search 2 2004-05-17 22:13 rogue Factoring 13 2004-05-01 14:48 Xyzzy Hardware 0 2003-08-21 00:06 Xyzzy Lounge 4 2003-06-28 13:37
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Tue Dec 7 20:47:05 UTC 2021 up 137 days, 15:16, 1 user, load averages: 1.51, 1.43, 1.40 | 1,588 | 5,192 | {"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-2021-49 | latest | en | 0.893851 |
https://ohbug.com/uva/11030/ | 1,723,008,524,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00094.warc.gz | 349,624,035 | 1,977 | # Predator II
Oh No!!! The predator has entered the room again. But this time it is a different kind of room. The room is a square of size 10000 × 10000. There are several compartments of different shapes and sizes, situated strictly inside the room. The walls of different compartments don’t overlap, but a compartment can be completely inside that of another. This time the predator has learned to hop over walls, but it can jump over at most one wall at a time. Given the starting and ending coordinates of the predator, and the positions of the compartments, your job is to find out the minimum number of hops required for the predator to reach his destination from the start. The starting and ending positions are never on the boundary of any compartment. Input The first line of input is an integer (T ≤ 20), that indicates the number of test cases. Each case starts with an integer (n ≤ 20), that determines the number of compartments in the room. The next n lines give the positions of the compartments. The compartments are simple polygons. The description of each compartment starts with an integers (S ≤ 10), that gives the number of sides of the polygon, followed by S pairs of x, y coordinates in order. Next there is an integer Q that determines the number of queries for this scenario. Each of the next Q lines contains 4 integers x1,y1,x2,y2. (x1,y1) is the starting position and (x2, y2) is the ending position. The lower left and upper right coordinates of the room are (0, 0) and (10000, 10000) respectively. Output For each case, output the case number. Then for each query, output the minimum number of hops required. Illustration: The following diagram depicts the first sample input. • Pink spot a starting position • Black spot a target • The three hops are shown by three broad line segments. Sample Input 2 3 411515515 422424424 3 7 7 10 10 7 10 1 3389 1 4 1 1 10 1 10 10 1 10 2
2/2 2 2 100 100 100 100 2 2 Sample Output Case 1: 3 Case 2: 1 1 | 490 | 1,970 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.901427 |
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Math Central Quandaries & Queries
My son is working on a hexagonal pyramid. The base is 4 and the slant height is 6. We know that the answer is 48 times the sqrt of 2. We don't know how they got it. We attached the drawing of it. Thanks Angie
Hi Angie,
The volume of a pyramid is 1/3 × (the area of the base) × (the height) so you need to find the area of the base and the height.
You can find the area of the base using the technique Stephen used in his response to an earlier problem.
o find the height I added a line to your diagram as well as some labels.
The height of the pyramid is the length of the side AB of the right triangle ABC. |AC| = 6 and |BC| you found in determining the area of the hexagonal base. Thus you can use Pythagoras theorem to find |AB|.
I hope this helps,
Penny
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https://financial-dictionary.thefreedictionary.com/returnable | 1,627,262,859,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151972.40/warc/CC-MAIN-20210726000859-20210726030859-00599.warc.gz | 269,132,247 | 13,216 | # Return
(redirected from returnable)
Also found in: Dictionary, Thesaurus, Medical, Legal, Encyclopedia.
## Return
The change in the value of a portfolio over an evaluation period, including any distributions made from the portfolio during that period.
## Rate of Return
In securities, the amount of revenue an investment generates over a given period of time as a percentage of the amount of capital invested. The rate of return shows the amount of time it will take to recover one's investment. For example, if one invests \$1,000 and receives \$150 in the first year of the investment, the rate of return is 15%, and the investor will recover his/her initial \$1,000 in six years and eight months. Different investors have different required rates of return at different levels of risk.
See yield.
## Return.
Your return is the profit or loss you have on your investments, including income and change in value.
Return can be expressed as a percentage and is calculated by adding the income and the change in value and then dividing by the initial principal or investment amount. You can find the annualized return by dividing the percentage return by the number of years you have held the investment.
For example, if you bought a stock that paid no dividends at \$25 a share and sold it for \$30 a share, your return would be \$5. If you bought on January 3, and sold it the following January 4, that would be a 20% annual percentage return, or the \$5 return divided by your \$25 investment.
But if you held the stock for five years before selling for \$30 a share, your annualized return would be 4%, because the 20% gain is divided by five years rather than one year.
Percentage return and annual percentage return allow you to compare the return provided by different investments or investments you have held for different periods of time.
References in periodicals archive ?
Due to the factor that returnable transport packaging (RTP) is environmentally friendly and cost-efficient, the market is growing at a significant rate.
Regional analysis of returnable transport packaging market is presented for following market segments:
South Africa has one of the most efficient returnable bottle systems in the world.
Returnable transit products are used in many industrial sectors such as automotive and play a vital role enclosing and securely protecting products in strong, rigid structures.
By the end of the month, more than half of the students had been able to convince their parents to buy only returnable bottles when that was an option.
The Oregon Grocery Association has successfully resisted proposals to lengthen the list of returnable containers in the past, and in 1996 even persuaded voters to reject an initiative calling for a broader Bottle Bill.
"We also found modern glass recycling processes mean it is more environmentally friendly to use single glass recycling bottles than to transport, clean and refill returnable bottles.
Pamela Lowell's RETURNABLE GIRL (0761453172, \$16.99) tells of Ronnie, whose mother to Alaska when after becoming involved with a good-for-nothing boyfriend.
As part of the revised terms retailers will receive a 46% discount on orders of 10 or more units shipped on a returnable basis and non-returnable discounts will range from 46% to 52%, Publishers Weekly reported.
A new polypropylene copolymer is said to provide the toughness and stiffness required for returnable containers, trays, nestable totes, and pallets.
Returnable plastic container (RPC) use remains virtually unchanged, at about 8% of retail volume, or 4% of total produce volume including both retail and foodservice channels.
Those who knowingly deliver or dispose returnable containers or tires would face a fine of up to \$500.
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level 1
4 points · 4 months ago
Circonus DataScientist here. I had the percentile aggregation discussion with John Rauser DataScientist@Snapchat a while ago on Twitter: https://twitter.com/heinrichhartman/status/748562001392111617
He wrote up a 10pager with some simulations where percentile aggregation works: http://rpubs.com/jrauser/percentiles
I found some practical examples where stuff goes sideways: https://twitter.com/heinrichhartman/status/750089215946264577
An in the end we agreed that:
Yep. It's easy to find real world examples where what seems like a good idea fails. https://twitter.com/jrauser/status/750140187225632768
There is also research literature about this topic in the "Big Data" context. It runs under the name "mergable summaries" in particular:
Agarwal, et. al. PODS’12 - Mergable Summaries - https://www.cs.utah.edu/~jeffp/papers/merge-summ.pdf
contains a section about Quantiles. The issue has nothing to do with rounding errors of floating point numbers. Let me try to outline the basic problem:
If you are averaging percentiles from equally sized populations that were sampled from the same distribution, you will get results that are close to the true percentile of the distribution. This follows from the law of large numbers. This is the good case, and happens e.g. if you are:
1. Averaging percentiles from a fleet of equally loaded web servers
2. Averaging percentiles over time of a single consistently loaded service.
Problems arise if the distributions of the individual samples are very different, e.g.
(A) Say one web-server hangs and serves only one request with latency 60sec => p99 = 60s The remaining 3 web-servers are running fine and are handling 10K requests each with p99 = 1ms. The true p99 of the overall sample (30Kx1ms + 1x50s) is 1ms The averaged p99 is (60s + 1ms + 1ms) / 3 ~= 20s
(B) Another common case is when load drops over night, and a search engine is querying expensive endpoints. Over the day your p99 can be well behaved, say 1ms. Over night, you have basically no traffic but p99 is 10s b/c the queries you get from the search engine are so expensive. The true p99 over the day will still be ~1ms. The averaged p99 will be ~5s.
Now there are ways to do better, by remembering more stuff e.g. the count along with the p99. If you go down that route, you end up with t-digests and other clever ways to calculate approximate percentiles over streams. Those methods tend to be rather complicated to understand, and often give you only probabilistic bounds of the errors under some model assumptions on your data. So you can say that "With 99.99% probability your p99 is lower than 10s." ... which might be good enough.
One of these "mergable enhacements" that we found to be working well in practice are sparsely-encoded log-linear histograms. With them you can get arbitrary percentiles (no up-front choice) and deterministic error bounds (5% worst-case with a single sample). In addition they are pretty efficient (ns insertion time) and straight forward to implement.
Now, there might be ways to be more space efficient or more performant if you are only ever interested in a single percentile. But these kind of histograms are efficient enough for all applications that I have seen (which involve measuring syscall latencies in kernel) and encode enough information to calculate almost any summary statistic in a mergable way.
Google is also known to use log-linear Histograms for Monitoring of their APIs. John Banning talked about this at Monitorama PDX 2016 (https://vimeo.com/173607638 @ 7:08) I hear that they have some regret for not to have fixed the bin sizes upfront, but essentially they are using stuff like http://www.hdrhistogram.org/ or our https://github.com/circonus-labs/libcircllhist.
The main reason that this is not adopted more widely, is that it's not trivial to store Histograms in a TSDB that was designed for purely numeric data (one float at a time). You can try to store the counts of a a small number of fixed buckets as metrics, but this will only get a small value range covered at pretty low resolution. With sparse encoding you can get away with using a set of 46K buckets that cover all data you will ever see in praxis pretty comfortably (https://support.circonus.com/solution/articles/6000066608-histogram-bin-count).
So yes. Percentile aggregation is subtle. There are ways to do it bad (plain averaging) there are ways to do it better (t-digest), and there is a pragmatic approach that's reasonably simple and efficient but requires you to put histograms into your TSDB.
level 2
1 point · 4 months ago
Thank you for the detailed answer! This is what I was looking for and as the blog post points out, averaging can work, but falls down in a lot of scenarios.
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Be exelent to eachother | 1,221 | 5,158 | {"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} | 2.859375 | 3 | CC-MAIN-2018-51 | longest | en | 0.906945 |
https://gateway.ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/R%C3%B6ssler_attractor.html | 1,642,582,472,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301264.36/warc/CC-MAIN-20220119064554-20220119094554-00659.warc.gz | 305,408,691 | 16,835 | # Rössler attractor
The Rössler attractor
Rössler attractor as a stereogram with , ,
The Rössler attractor /ˈrɒslər/ is the attractor for the Rössler system, a system of three non-linear ordinary differential equations originally studied by Otto Rössler.[1][2] These differential equations define a continuous-time dynamical system that exhibits chaotic dynamics associated with the fractal properties of the attractor.[3]
Some properties of the Rössler system can be deduced via linear methods such as eigenvectors, but the main features of the system require non-linear methods such as Poincaré maps and bifurcation diagrams. The original Rössler paper states the Rössler attractor was intended to behave similarly to the Lorenz attractor, but also be easier to analyze qualitatively.[1] An orbit within the attractor follows an outward spiral close to the plane around an unstable fixed point. Once the graph spirals out enough, a second fixed point influences the graph, causing a rise and twist in the -dimension. In the time domain, it becomes apparent that although each variable is oscillating within a fixed range of values, the oscillations are chaotic. This attractor has some similarities to the Lorenz attractor, but is simpler and has only one manifold. Otto Rössler designed the Rössler attractor in 1976,[1] but the originally theoretical equations were later found to be useful in modeling equilibrium in chemical reactions.
## Definition
The defining equations of the Rössler system are:[3]
Otto E. Rössler studied the chaotic attractor with , , and , though properties of , , and have been more commonly used since. Another line of the parameter space was investigated using the topological analysis. It corresponds to , , and was chosen as the bifurcation parameter.[4] How Rössler discovered this set of equations was investigated by Letellier and Messager.[5]
## An analysis
plane of Rössler attractor with , ,
Some of the Rössler attractor's elegance is due to two of its equations being linear; setting , allows examination of the behavior on the plane
The stability in the plane can then be found by calculating the eigenvalues of the Jacobian , which are . From this, we can see that when , the eigenvalues are complex and both have a positive real component, making the origin unstable with an outwards spiral on the plane. Now consider the plane behavior within the context of this range for . So long as is smaller than , the term will keep the orbit close to the plane. As the orbit approaches greater than , the -values begin to climb. As climbs, though, the in the equation for stops the growth in .
### Fixed points
In order to find the fixed points, the three Rössler equations are set to zero and the (,,) coordinates of each fixed point were determined by solving the resulting equations. This yields the general equations of each of the fixed point coordinates:
Which in turn can be used to show the actual fixed points for a given set of parameter values:
As shown in the general plots of the Rössler Attractor above, one of these fixed points resides in the center of the attractor loop and the other lies comparatively removed from the attractor.
### Eigenvalues and eigenvectors
The stability of each of these fixed points can be analyzed by determining their respective eigenvalues and eigenvectors. Beginning with the Jacobian:
the eigenvalues can be determined by solving the following cubic:
For the centrally located fixed point, Rössler’s original parameter values of a=0.2, b=0.2, and c=5.7 yield eigenvalues of:
The magnitude of a negative eigenvalue characterizes the level of attraction along the corresponding eigenvector. Similarly the magnitude of a positive eigenvalue characterizes the level of repulsion along the corresponding eigenvector.
The eigenvectors corresponding to these eigenvalues are:
Examination of central fixed point eigenvectors: The blue line corresponds to the standard Rössler attractor generated with , , and .
Rössler attractor with , ,
These eigenvectors have several interesting implications. First, the two eigenvalue/eigenvector pairs ( and ) are responsible for the steady outward slide that occurs in the main disk of the attractor. The last eigenvalue/eigenvector pair is attracting along an axis that runs through the center of the manifold and accounts for the z motion that occurs within the attractor. This effect is roughly demonstrated with the figure below.
The figure examines the central fixed point eigenvectors. The blue line corresponds to the standard Rössler attractor generated with , , and . The red dot in the center of this attractor is . The red line intersecting that fixed point is an illustration of the repulsing plane generated by and . The green line is an illustration of the attracting . The magenta line is generated by stepping backwards through time from a point on the attracting eigenvector which is slightly above – it illustrates the behavior of points that become completely dominated by that vector. Note that the magenta line nearly touches the plane of the attractor before being pulled upwards into the fixed point; this suggests that the general appearance and behavior of the Rössler attractor is largely a product of the interaction between the attracting and the repelling and plane. Specifically it implies that a sequence generated from the Rössler equations will begin to loop around , start being pulled upwards into the vector, creating the upward arm of a curve that bends slightly inward toward the vector before being pushed outward again as it is pulled back towards the repelling plane.
For the outlier fixed point, Rössler’s original parameter values of , , and yield eigenvalues of:
The eigenvectors corresponding to these eigenvalues are:
Although these eigenvalues and eigenvectors exist in the Rössler attractor, their influence is confined to iterations of the Rössler system whose initial conditions are in the general vicinity of this outlier fixed point. Except in those cases where the initial conditions lie on the attracting plane generated by and , this influence effectively involves pushing the resulting system towards the general Rössler attractor. As the resulting sequence approaches the central fixed point and the attractor itself, the influence of this distant fixed point (and its eigenvectors) will wane.
### Poincaré map
The Poincaré map is constructed by plotting the value of the function every time it passes through a set plane in a specific direction. An example would be plotting the value every time it passes through the plane where is changing from negative to positive, commonly done when studying the Lorenz attractor. In the case of the Rössler attractor, the plane is uninteresting, as the map always crosses the plane at due to the nature of the Rössler equations. In the plane for , , , the Poincaré map shows the upswing in values as increases, as is to be expected due to the upswing and twist section of the Rössler plot. The number of points in this specific Poincaré plot is infinite, but when a different value is used, the number of points can vary. For example, with a value of 4, there is only one point on the Poincaré map, because the function yields a periodic orbit of period one, or if the value is set to 12.8, there would be six points corresponding to a period six orbit.
### Mapping local maxima
In the original paper on the Lorenz Attractor,[6] Edward Lorenz analyzed the local maxima of against the immediately preceding local maxima. When visualized, the plot resembled the tent map, implying that similar analysis can be used between the map and attractor. For the Rössler attractor, when the local maximum is plotted against the next local maximum, , the resulting plot (shown here for , , ) is unimodal, resembling a skewed Hénon map. Knowing that the Rössler attractor can be used to create a pseudo 1-d map, it then follows to use similar analysis methods. The bifurcation diagram is specifically a useful analysis method.
### Variation of parameters
Rössler attractor's behavior is largely a factor of the values of its constant parameters , , and . In general, varying each parameter has a comparable effect by causing the system to converge toward a periodic orbit, fixed point, or escape towards infinity, however the specific ranges and behaviors induced vary substantially for each parameter. Periodic orbits, or "unit cycles," of the Rössler system are defined by the number of loops around the central point that occur before the loops series begins to repeat itself.
Bifurcation diagrams are a common tool for analyzing the behavior of dynamical systems, of which the Rössler attractor is one. They are created by running the equations of the system, holding all but one of the variables constant and varying the last one. Then, a graph is plotted of the points that a particular value for the changed variable visits after transient factors have been neutralised. Chaotic regions are indicated by filled-in regions of the plot.
#### Varying a
Here, is fixed at 0.2, is fixed at 5.7 and changes. Numerical examination of the attractor's behavior over changing suggests it has a disproportional influence over the attractor's behavior. The results of the analysis are:
• : Converges to the centrally located fixed point
• : Unit cycle of period 1
• : Standard parameter value selected by Rössler, chaotic
• : Chaotic attractor, significantly more Möbius strip-like (folding over itself).
• : Similar to .3, but increasingly chaotic
• : Similar to .35, but increasingly chaotic.
#### Varying b
Bifurcation diagram for the Rössler attractor for varying
Here, is fixed at 0.2, is fixed at 5.7 and changes. As shown in the accompanying diagram, as approaches 0 the attractor approaches infinity (note the upswing for very small values of . Comparative to the other parameters, varying generates a greater range when period-3 and period-6 orbits will occur. In contrast to and , higher values of converge to period-1, not to a chaotic state.
#### Varying c
Bifurcation diagram for the Rössler attractor for varying
Here, and changes. The bifurcation diagram reveals that low values of are periodic, but quickly become chaotic as increases. This pattern repeats itself as increases – there are sections of periodicity interspersed with periods of chaos, and the trend is towards higher-period orbits as increases. For example, the period one orbit only appears for values of around 4 and is never found again in the bifurcation diagram. The same phenomenon is seen with period three; until , period three orbits can be found, but thereafter, they do not appear.
A graphical illustration of the changing attractor over a range of values illustrates the general behavior seen for all of these parameter analyses – the frequent transitions between periodicity and aperiodicity.
Varying c
c = 4, period 1
c = 6, period 2
c = 8.5, period 4
c = 8.7, period 8
c = 9, chaotic
c = 12, period 3
c = 12.6, period 6
c = 13, chaotic
c = 18, chaotic
The above set of images illustrates the variations in the post-transient Rössler system as is varied over a range of values. These images were generated with .
• , period-1 orbit.
• , period-2 orbit.
• , period-4 orbit.
• , period-8 orbit.
• , sparse chaotic attractor.
• , period-3 orbit.
• , period-6 orbit.
• , sparse chaotic attractor.
• , filled-in chaotic attractor.
The banding evident in the Rössler attractor is similar to a Cantor set rotated about its midpoint. Additionally, the half-twist that occurs in the Rössler attractor only affects a part of the attractor. Rössler showed that his attractor was in fact the combination of a "normal band" and a Möbius strip.[7]
## References
1. Rössler, O. E. (1976), "An Equation for Continuous Chaos", Physics Letters, 57A (5): 397–398, doi:10.1016/0375-9601(76)90101-8.
2. Rössler, O. E. (1979), "An Equation for Hyperchaos", Physics Letters, 71A (2,3): 155–157, doi:10.1016/0375-9601(79)90150-6.
3. Peitgen, Heinz-Otto; Jürgens, Hartmut; Saupe, Dietmar (2004), "12.3 The Rössler Attractor", Chaos and Fractals: New Frontiers of Science, Springer, pp. 636–646.
4. Letellier, C.; P. Dutertre; B. Maheu (1995). "Unstable periodic orbits and templates of the Rössler system: toward a systematic topological characterization". Chaos. 5 (1): 272–281. doi:10.1063/1.166076.
5. Letellier, C.; V. Messager (2010). "Influences on Otto E. Rössler's earliest paper on chaos". International Journal of Bifurcation & Chaos. 20 (11): 3585–3616. doi:10.1142/s0218127410027854.
6. Lorenz, E. N. (1963), "Deterministic nonperiodic flow", J. Atmos. Sci., 20 (2): 130–141, Bibcode:1963JAtS...20..130L, doi:10.1175/1520-0469(1963)020<0130:DNF>2.0.CO;2.
7. Rössler, Otto E. (1976). "Chaotic behavior in simple reaction system". Zeitschrift für Naturforschung A. 31: 259–264. | 3,049 | 12,901 | {"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-2022-05 | latest | en | 0.923918 |
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This date is established independent of stratigraphy and chronology. Certain unstable isotopes of trace radioactive elements in both organic and inorganic materials decay into stable isotopes. The remaining atoms have exactly the same decay probability, so in another half-life, one half of the remaining atoms will decay. Relative Dating Techniques Explained.
For this reason, many archaeologists prefer to use samples from short-lived plants for radiocarbon dating. Hardest Math Problem in the World. Neanderthals Were Inbreeding.
## Relative Vs. Absolute Dating The Ultimate Face-off
1. Potassium gradually decays to the stable isotope argon, which is a gas.
2. Unlike observation-based relative dating, most absolute methods require some of the find to be destroyed by heat or other means.
3. The half-life is a measure of the probability that a given atom will decay in a certain time.
The smallest of these rock units that can be matched to a specific time interval is called a bed. When dendrochronology can be used, it provides the most accurate dates of any technique. These layers are often different colours or made from different textured sediment.
The nucleus of every radioactive element such as radium and uranium spontaneously disintegrates over time, transforming itself into the nucleus of an atom of a different element. The longer the exposure to the radiation, the more electrons that are bumped into an excited state, and the more light that is emitted upon heating. Radiogenic Isotope Geology. Fossils are found in between these layers and thus can be estimated to be a similar age to the rocks that they around. The rate at which this process occurs is called the half-life.
The origins of human beings according to ancient Sumerian texts. With death, the uptake of carbon stops. Research where the fossil was found. The worst candidates are bits of wood that have been saturated with sea water, since sea water contains dissolved atmospheric carbon dioxide that may throw off the results.
## Dating methods in Archaeology. Are they accurate
Both methods date rock instead of organic material. In recent years, a few of these methods have undergone continual refinement as scientists strive to develop the most accurate dating techniques possible. Although certain dating techniques are accurate only within certain age ranges, whenever possible, scientists attempt to use multiple methods to date specimens. Depositional rates of sediments have also been employed as a dating method, but only recently has absolute dating been made possible through the use of radioactive isotopes.
## 8.4 Isotopic Dating Methods
Facts about Albert Einstein. Interesting Facts About Hurricanes. There are well over labs worldwide that do radiocarbon dating.
As long as the plant is alive, the relative amount ratio of carbon to carbon remains constant at about one carbon atom for every one trillion carbon atoms. Archaeologists can then use this information to determine the relative ages of some sites and layers within sites. Beds that are related are grouped together into members, dating in and members are grouped into formations.
## Dating Techniques
The particular radioisotope used to determine the age of an object depends on the type of object and its age. Potassium-argon dating relies on the fact that when volcanic rocks are heated to extremely high temperatures, they release any argon gas trapped in them. Radiocarbon dating is used to date charcoal, wood, and other biological materials.
Dendrochronology, also known as tree-ring dating, is the earliest form of absolute dating. Canon of Kings Lists of kings Limmu. Stratigraphy and Seriation. Protactinium begins to accumulate via the decay of U after the organism dies.
This means that no matter how many atoms are in a sample, approximately one-half will decay in one half-life. In relative dating, mostly the common sense principles are applied, and it is told that which artifact or object is older than the other one. These same Greek pottery styles could be associated with monuments in Greece whose construction dates were fairly well known. The greater the number of fissures in a rock, the older the fossil is likely to be. Glaciology Hydrogeology Marine geology.
Scientists can determine how many years have passed since a ceramic was fired by heating it in the laboratory and measuring how much light is given off. This method is not widely used in archaeology, since most archaeological deposits are not associated with volcanic activity. To evaluate the exact age, both the chemical and physical properties of the object are looked keenly. This is a radiometric technique since it is based on radioactive decay. Scientific dating techniques have had a huge impact on archaeology.
Before the advent of absolute dating methods, nearly all dating was relative. For this reason, and because some of the amino acid racimization dates have disagreed with dates achieved by other methods, the technique is no longer widely used. At first, there were not many methods of dating were available, but now with advancement in the technology, we mainly have two types of techniques to ascertain ages of ancient belongings. That way, dates reported in magazine articles and books do not have to be adjusted as the years pass.
## Would you like to take a short survey
• Over time, certain kinds of rocks and organic material, such as coral and teeth, are very good at trapping electrons from sunlight and cosmic rays pummeling Earth.
• The main relative dating method is stratigraphy.
• This isotope of uranium spontaneously undergoes fission.
Pollen that ends up in lakebeds or peat bogs is the most likely to be preserved, but pollen may also become fossilized in arid conditions if the soil is acidic or cool. When the organism dies, however, its body stops incorporating new carbon. Volcanic rocks can be dated by measuring the amount of argon in them. They then use that absolute date to establish a relative age for fossils and artifacts in relation to that layer.
Absolute dating methods are used to determine an actual date in years for the age of an object. This absolute dating method is also known as dendrochronology. Dendrochronology is a dating technique that makes use of tree growth rings. In other words, we can say that in relative dating the archaeologist determines that which of the two fossil or the artifacts are older. Since all of the trees in a region experience the same climate variations, they will have similar growth patterns and similar tree ring patterns.
Excavations, in combination with surveys, may yield maps of a ruin or collections of artifacts. In Search of the Trojan War. In other words, we can say that the age in relative dating is ascertained by witnessing the layers of deposition or the rocks. When radiometric techniques are applied to metamorphic rocks, the results normally tell us the date of metamorphism, not the date when the parent rock formed. As long as an organism is alive, the supply of carbon is replenished.
Lunisolar Solar Lunar Astronomical year numbering. The absolute dating method utilizing tree ring growth is known as dendrochronology. Absolute dating methods are carried out in a laboratory.
Because of this limitation, other dating techniques are often used along with radioactive dating to ensure accuracy. It is distinguished from other forms of inquiry by its method of study, excavation. Radiometric dating is another crucial technique through which the exact age can be obtained. Here we come to the question of how accurate the dates are that we currently have regarding the history of the human race and our planet. In the case of a daughter excess, edmonton a larger amount of the daughter is initially deposited than the parent.
Archaeologists are seeking an accurate dating technique, but this method is yet to be found. Fluorine is found naturally in ground water. As climates change over time, the plants that grow in a region change as well. Article Info This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Thus, measuring the ratio of D to L in a sample enables one to estimate how long ago the specimen died.
Dating techniques Dating techniques are procedures used by scientists to determine the age of an object or a series of events. The Seven Wonders of the Ancient World are seven awe-inspiring monuments of classical antiquity that reflect the skill and ingenuity of their creators. Samples that were heated or irradiated at some time may yield by radioactive dating an age less than the true age of the object. If we dated a number of individual grains in the sedimentary rock, we would likely get a range of different dates, miss dating all older than the age of the rock.
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https://prof.clontz.org/classes/2017/06/ma227/standards/s06/ | 1,632,277,037,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00392.warc.gz | 497,389,989 | 3,890 | # MA 227 Standard S06
Linearization
At the end of the course, each student should be able to…
• S06: Lineariz. Compute the linearization of a two-variable real-valued function at a point and use it for approximation.
## S06: Linearization
• Differentiable functions $$f(x,y)$$ may be approximated near a point $$P_0$$ by the linearization $$L(x,y)$$ defined by the tangent plane at that point.
• Its equation is $$L(x,y)=f(P_0)+f_x(P_0)(x-x_0)+f_y(P_0)(y-y_0)$$.
• Thus if $$<x,y>\approx P_0$$, then $$L(x,y)\approx f(x,y)$$.
### Textbook References
• University Calculus: Early Transcendentals (3rd Ed)
• 13.6 (exercises 25-30) | 200 | 635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-39 | latest | en | 0.758407 |
https://community.zemax.com/people-pointers-9/how-is-m2-m-squared-calculated-in-physical-optics-propagation-popd-364 | 1,718,635,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00409.warc.gz | 160,402,665 | 36,347 | # How is M2 (M squared) calculated in Physical Optics Propagation (POPD)?
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• Zemax Staff
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Many users might have questions how exactly M2 is calculated and let’s discover this in this post! We will first explain its equation and then show how you can reproduce it in OpticStudio.
The only way to get M2 is by using POPD operand in merit function. To calculated M2, the Data parameter must be either 25 (X direction) or 26 (Y direction). To use POPD operand, first define the settings on the POP analysis feature as desired, then press Save on the settings box.
More information about POPD can be found in Help file > The Optimize Tab (sequential ui mode) > Automatic Optimization Group > Merit Function Editor (automatic optimization group) > Optimization Operands by Category > Physical Optics Propagation (POP) Results
Now let’s start to see how it’s calculated. As can be found in Help file, we use the following equation to calculate:
where Wx(0) is the minimum beam size of the real beam and W(z) is the beam size at a large z position.
For the Wx(0), we use the beam width (POPD with Data = 23 or 24) at the position of the pilot beam's waist.
For the Wx(z) with z->inf, we use the beam width (POPD with Data = 23 or 24) at a large z. For example, we can use 1e5*Rayleigh Length for large z.
Let’s open the following example file to test.
{Zemax}\Samples\Sequential\Objectives\Double Gauss 28 degree field.zmx
Then we open POP analysis, click Reset and Save. Now any POPD you set in the merit function will use the settings you just saved.
As shown below, we can see the M2 is calculated as 3.992486 in both X and Y directions. It’s not surprised value at X and Y are same because this is an rotationally-symmetrical system.
Let’s check the pilot beam position and Rayleigh range using Data = 5 and 6.
Now add two surfaces in LDE to get position of pilot beam waist position and far field. The far field is calculated as 1e5 times of Rayleigh range. This can be any number as long as it’s large compared to Rayleigh range.
The w(0) and w(z) is then calculated by using POPD with Data = 23 at surface 13 and surface 14.
Then we can calculate the M2 as shown below. M2 is at the final line. Note there is a little error if we compare line 28 below and line 2 above. This is because in POPD we use a more efficient way to calculate this. However, the concept is still same and we can see the difference very small between 3.992361 and 3.992554.
The final file is attached! | 644 | 2,544 | {"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-2024-26 | latest | en | 0.867308 |
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Determine your monthly student loan payments on income-driven repayment plans — even the newest.
Updated
Edited by Des Toups
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If you’re struggling to afford federal student loan payments, you may be able to lower them with an income-driven repayment plan. Your new monthly payment will be capped at 5%, 10%, 15% or 20% of your discretionary income, depending on the plan. Your eligibility will depend on the type of federal loan you have.
Use the top calculator to estimate payments under existing income-driven repayment plans.
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## Discretionary income explained
Discretionary income matters for federal student loans because the Education Department uses it to calculate payments for income-based repayment and other income-driven plans. By accounting for your necessities, discretionary income helps determine how much you could reasonably pay each month. If yours is low enough, your payment may be reduced to \$0 a month.
Of course, people have different needs — or things they consider needs. The government isn’t going to have borrowers submit receipts and defend their spending choices. Instead, it uses a standardized discretionary income definition to make things as fair as possible.
## How is discretionary income calculated?
To calculate discretionary income for most student loan repayment plans, the Education Department:
• Finds the correct federal poverty guideline for your location and family size.
• Multiplies that number by 1.5.
Income-Contingent Repayment, which sets payments at 20% of discretionary income, uses 100% of the poverty line instead of 150%.
Adjusted gross income is the amount you pay taxes on. You’ll find it on your most recent tax return on Line 37 if you filed Form 1040; Line 21 on 1040A; or Line 4 on Form 1040-EZ.
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Ivica Nakić, Zoran Tomljanović, Ninoslav Truhar
Optimal Direct Velocity Feedback
Applied Mathematics and Computation 225 (2013) ; 590-600
Preliminary version appeared as Preprint 3/2013, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: We present a novel approach to the problem of Direct Velocity Feedback (DVF) optimization of vibrational structures, which treats simultaneously small as well as large gains. For that purpose, we use two different approaches. The first one is based on the gains optimization using the Lyapunov equation. In the scope of this approach we present a new formula for the optimal gain and we present a relative error for modal approximation. In addition, we present a new formula for the solution of the corresponding Lyapunov equation for the case with multiple undamped eigenfrequencies, which is a generalization of existing formulae. The second approach studies the behavior of the eigenvalues of the corresponding quadratic eigenvalue problem. Since this approach leads to the parametric eigenvalue problem we consider small and large gains separately. For the small gains, which are connected to a modal damping approximation, we present a standard approach based on Gerschgorin discs. For the large gains we present a new approach which allows us to approximate all eigenvalues very accurately and efficiently.
Ivana Kuzmanović, Zoran Tomljanović, Ninoslav Truhar
Optimization of material with modal damping
Applied Mathematics and Computation 218 (2012); 7326-7338
Preliminary version appeared as Preprint 9/2010, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: This paper considers optimal parameters for modal damping $D=Mf_1(M^{-1}K;\alpha_1,\dots,\alpha_k)+Kf_2(K^{-1}M;\alpha_1,\dots,\alpha_k)$ in mechanical systems described by the equation $M\ddot{x}+D\dot{x}+Kx=0$, where matrices $M$ and $K$ are mass and stiffness matrices, respectively. Different models of proportional and generalized proportional damping are considered and optimal parameters with respect to different optimization criteria related to the solution of the corresponding Lyapunov equation are given. Also, some specific example problems are compared with respect to the optimal and estimated parameters.
Peter Benner, Zoran Tomljanović, Ninoslav Truhar
Optimal Damping of Selected Eigenfrequencies Using Dimension Reduction
Numerical Linear Algebra with Applications, 20 (2013) , 1; 1-17 , Published online in Wiley Online Library. DOI: 10.1002/nla.833
Preliminary version appeared as Preprint 1/2011, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: We consider a mathematical model of a linear vibrational system described by the second-order differential equation $M \ddot{x} + D \dot{x} + Kx = 0$, where $M$ and $K$ are positive definite matrices, representing mass and stiffness, respectively. The damping matrix $D$ is positive semidefinite. We are interested in finding an optimal damping matrix which will damp a certain (critical) part of the undamped eigenfrequencies. For this we use an optimization criterion based on minimization of the average total energy of the system. This is equivalent to the minimization of the trace of the solution of the corresponding Lyapunov equation $A X+ X A^T =-GG^T$, where $A$ is the matrix obtained from linearizing the second-order differential equation and $G$ depends on the critical part of the eigenfrequencies to be damped. The main result is the efficient approximation and corresponding error bound for the trace of the solution of the Lyapunov equation obtained by dimension reduction, which includes the influence of the right-hand side $G G^T$ and allows us to control the accuracy of the trace approximation. This trace approximation yields a much accelerated optimization algorithm for determining the optimal damping.
Peter Benner, Zoran Tomljanović, Ninoslav Truhar
Dimension reduction for damping optimization in linear vibrating system
Journal of Applied Mathematics and Mechanics (ZAMM), 91 (3), (2011); 179-191; DOI: 10.1002/zamm.201000077.
Editor's choice
of this ZAMM issue!
Preliminary version appeared as Preprint 2/2010, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: We consider a mathematical model of a linear vibrational system described by the second-order differential equation $M \ddot{;x}; + D \dot{;x}; + Kx = 0$, where $M$ and $K$ are positive definite matrices, called mass and stiffness, espectively. We consider the case where the damping matrix $D$ is positive semidefinite. The main problem considered in the paper is the construction of efficient algorithm for calculating an optimal damping. As optimization criterion we use the minimization of the average total energy of the system which is equivalent to the minimization of the trace of the solution of the corresponding Lyapunov equation $A X+ X A^T =-I$, where $A$ is the matrix obtained from linearizing the second-order differential equation. Finding the optimal $D$ such that the trace of $X$ is minimal is a very demanding problem, caused by the large number of trace calculations, which are required for bigger matrix dimensions. We propose a dimension reduction to accelerate the optimization process. We will present an approximation of the solution of the structured Lyapunov equation and a corresponding error bound for the approximation. Our algorithm for efficient approximation of the optimal damping is based on this approximation.
Zoran Tomljanović, Ninoslav Truhar, Krešimir Veselić
Optimizing a damped system - a case study
International journal of computer mathematics, 88 (7), (2011); 1533 - 1545
Preliminary version appeared as Preprint 6/2008, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: We consider a second order damped-vibrational system described by the equation $M \ddot{;x}; + C(v) \dot{;x}; + K x = 0$, where $M, C(v), K$ are real, symmetric matrices of order $n$. We assume that the undamped eigenfrequencies (eigenvalues of $(\lambda^2 M + K) x = 0$) $\omega_1, \omega_2, \ldots, \omega_n \,$, are multiple in the sense that $\omega_1 = \omega_2$, $\omega_3 = \omega_4$, \ldots, $\omega_{;n-1}; = \omega_n$, or are given in close pairs $\omega_1 \approx \omega_2$, $\omega_3 \approx \omega_4$, \ldots, $\omega_{;n-1}; \approx \omega_n$. We present a formula which gives the solution of the corresponding phase space Lyapunov equation, which then allows us to calculate the first and second derivatives of the trace of the solution, with no extra cost. This one can serve for the efficient trace minimization.
Ninoslav Truhar, Zoran Tomljanović, Ren-Cang Li
Analysis of the solution of the Sylvester equation using low-rank ADI with exact shifts
Systems and Control Letters, 59 (2010); 248-257
Preliminary version appeared as Preprint 4/2008, Department of Mathematics, J. J. Strossmayer University of Osijek. (*.PDF)
ABSTRACT: The solution to a general Sylvester equation $AX+XB = GF^*$ with a low rank right- hand side is analyzed quantitatively through Low-rank Alternating-Directional- Implicit method (LR-ADI) with exact shifts. New bounds and perturbation bounds on X are obtained. A distinguished feature of these bounds is that they reflect the interplay between the eigenvalue decompositions of A and B and the right-hand side factors G and F. Numerical examples suggest that because of this inclusion of details, new perturbation bounds are much sharper than the existing ones.
Ninoslav Truhar , Zoran Tomljanović
Estimation of optimal damping for mechanical vibrating systems
International Journal of Applied Mathematics and Mechanics , 5 (2009) , 5; 14-26.
Preliminary version appeared as Preprint 2/2008, Department of Mathematics, J. J. Strossmayer University of Osijek. ( *.PDF)
ABSTRACT:This paper is concerned with the efficient algorithm for dampers' and viscosity optimization in mechanical systems. Our algorithm optimize simultaneously the dampers' positions and their viscosities. For the criterion for optimization we use minimization of the average total energy of the system which can be done by the minimization of the trace of the solution of the corresponding Lyapunov equation. Efficiency of the algorithm is obtained by new heuristics for finding the optimal dampers' positions and for the approximation of the trace of the solution of the Lyapunov equation.
### Papers published at proceedings of international conferences:
Peter Benner, Zoran Tomljanović, Ninoslav Truhar
Damping Optimization for Linear Vibrating Systems Using Dimension Reduction (*.PDF)
In J. Náprstek, J. Horáček, M. Okrouhlík, B. Marvalová, F. Verhulst, and J.T. Sawicki (Eds.),
Vibration Problems ICOVP 2011. The 10th International Conference on Vibration Problems,, Springer Proceedings in Physics, Vol. 139, Springer-Verlag, pp. 297-305, 2011. [ISBN: 978-94-007-2068-8]
ABSTRACT: We consider a mathematical model of a linear vibrational system de- scribed by the second-order system of differential equations $M \ddot{x} + D \dot{x} + Kx = 0$, where M, K and D are positive definite matrices, called mass, stiffness and damp- ing, respectively. We are interested in finding an optimal damping matrix which will damp a certain part of the undamped eigenfrequencies. For this we use a minimiza- tion criterion which minimizes the average total energy of the system. This is equiv- alent to the minimization of the trace of the solution of a corresponding Lyapunov equation. In this paper we consider an algorithm for the efficient optimization of the damping positions based on dimension reduction techniques. Numerical results illustrate the efficiency of our approach.
### Profesional papers
Zoran Tomljanović
Horner's algorithm and applications
Osječki matematički list , Vol. 7, No. 2 (2008) 99-106
ABSTRACT: In this paper we study Horner's algorithm which is efficient for calculating value of a polynom at a given point. Horner's algorithm we can easily expand to the algorithm with which we can obtain Taylor's expansion around given point. Also, this paper includes illustrative examples and applications of this algorithm. | 2,453 | 10,257 | {"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} | 2.546875 | 3 | CC-MAIN-2023-14 | latest | en | 0.868095 |
https://playtaptales.com/numbers/duononagintaseptingentillion/ | 1,670,120,546,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00759.warc.gz | 490,458,809 | 4,272 | ## Duononagintaseptingentillion
A Duononagintaseptingentillion (1 Duononagintaseptingentillion) is 10 to the power of 2379 (10^2379). This is a stupendously astronomical number!
## How many zeros in a Duononagintaseptingentillion?
There are 2,379 zeros in a Duononagintaseptingentillion.
## What's before Duononagintaseptingentillion?
A Unnonagintaseptingentillion is smaller than a Duononagintaseptingentillion.
## What's after Duononagintaseptingentillion?
A Trenonagintaseptingentillion is larger than a Duononagintaseptingentillion.
## Duononagintaseptingentillionaire
A Duononagintaseptingentillionaire is someone whos assets, net worth or wealth is 1 or more Duononagintaseptingentillion. It is unlikely anyone will ever be a true Duononagintaseptingentillionaire. If you want to be a Duononagintaseptingentillionaire, play Tap Tales!
## Is Duononagintaseptingentillion the largest number?
Duononagintaseptingentillion is not the largest number. Infinity best describes the largest possible number - if there even is one! We cannot comprehend what the largest number actually is.
## Duononagintaseptingentillion written out
Duononagintaseptingentillion is written out as:
1,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
## Big Numbers
This is just one of many really big numbers! | 1,961 | 4,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.665151 |
http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8860 | 1,455,295,633,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701164289.84/warc/CC-MAIN-20160205193924-00262-ip-10-236-182-209.ec2.internal.warc.gz | 141,901,111 | 28,614 | # How can I really motivate the Zariski topology on a scheme?
First of all, I am aware of the questions about the Zariski topology asked here and I am also aware of the discussion at the Secret Blogging Seminar. But I could not find an answer to a question that bugged me right from my first steps in algebraic geometry: how can I really motivate the Zariski topology on a scheme?
For example in classical algebraic geometry over an algebraically closed field I can define the Zariski topology as the coarsest $T_1$-topology such that all polynomial functions are continuous. I think that this is a great definition when I say that I am working with polynomials and want to make my algebraic set into a local ringed space. But what can I say in the general case of an affine scheme?
Of course I can say that I want to have a fully faithful functor from rings into local ringed spaces and this construction works, but this is not a motivation.
For example for the prime spectrum itself, all motivations I came across so far are as follows: well, over an algebraically closed field we can identify the points with maximal ideals, but in general inverse images of maximal ideals are not maximal ideals, so let's just take prime ideals and...wow, it works. But now that I know that one gets the prime spectrum from the corresponding functor (one can of course also start with a functor) by imposing an equivalence relation on geometric points (which I find very geometric!), I finally found a great motivation for this. What is left is the Zariski topology, and so far I just came across similar strange motivations as above...
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Can you explain what you mean by imposing an equivalence relation on geometric points? My understanding is that this only makes sense for algebras over a field and presumably you want to talk about commutative rings in general. – Qiaochu Yuan Dec 8 '09 at 18:18
Let k be a commutative ring and A a k-algebra. Let t:A -> K, t':A -> K' be two geometric points (i.e. K,K' are fields). Say that t and t' are equivalent if and only if there exists a third geometric point s:A -> L and k-algebra morphisms f:K -> L, f':K' -> L such that s = ft = f't'. By taking kernels of the morphisms t you get a bijection between the equivalence classes of this relation and the prime ideals of A. And this is my first motivation, why I take the spectrum of A! (I think of this as compressing the corresponding zero set functor into something which still contains all I want). – user717 Dec 8 '09 at 18:48
I'm not sure I understand. If t, t' are not required to be surjective then their kernels are already prime ideals and not maximal ideals. If t, t' are required to be surjective then it is not possible to recover the prime ideals of A from its maximal ideals; consider the case A = k[[x, y]], k a field. Am I misunderstanding something? – Qiaochu Yuan Dec 8 '09 at 19:08
Yes, I think you misunderstand something. It's precisely my point that the kernels are prime ideals, because this is what gives you the bijection between equivalence classes (wrt the relation described above) of geometric points and prime ideals! you can read about this (probably in exactly the same words) in the 1971 edition of EGA, introduction, nr. 13.! – user717 Dec 8 '09 at 23:33
I just realized that normally 'geometric point' means that the field K is algebraically closed. I used the terminology of the introduction to EGA and there 'geometric point' is what I was talking about above...Perhaps this was the source of your confusion. Sorry. – user717 Dec 9 '09 at 14:24
In the category of sets there is no such thing as the initial local ring into which R maps, i.e. a local ring L and a map f:R-->L such that any map from R into a local ring factors through f.
But a ring R is a ring object in the topos of Sets. Now if you are willing to let the topos vary in which it should live, such a "free local ring on R" does exist: It is the ring object in the topos of sheaves on Spec(R) which is given by the structure sheaf of Spec(R). So the space you were wondering about is part of the solution of forming a free local ring over a given ring (you can reconstruct the space from the sheaf topos, so you could really say that it "is" the space).
Edit: I rephrase that less sloppily, in response to Lars' comment. So the universal property is about pairs (Topos, ring object in it). A map (T,R)-->(T',R') between such is a pair
(adjunction $f_*:T \leftrightarrow T':f^*$ , morphism of ring objects $f^*R'\rightarrow R$).
Note that by convention the direction of the map is the geometric direction, the one corresponding to the direction of a map topological spaces - in my "universal local ring" picture I was stressing the algebraic direction, which is given by $f^*$.
Now for a ring R there is a map $(Sh(Spec(R)), O_{Spec(R)})\rightarrow(Set,R)$: $f^* R$ is the constant sheaf with value R on Spec(R), the map $f^* R \rightarrow O_{Spec(R)}$ is given by the inclusion of R into its localisations which occur in the definition of $O_{Spec(R)}$. This is the terminal map (T,L)-->(Set,R) from pairs with L a local ring. For a simple example you might want to work out how such a map factors, if the domain pair happens to be of the form (Set,L).
This universal property of course determines the pair up to equivalence. It thus also determines the topos half of the pair up to equivalence, and thus also the space Spec(R) up to homeomorphism.(end of edit)
An even nicer reformulation of this is the following (even more high brow, but to me it gives the true and most illuminating justification for the Zariski topology, since it singles out just the space Spec(R)):
A ring R, i.e. a ring in the topos of sets, is the same as a topos morphism from the topos of sets into the classifying topos T of rings (by definition of classifying topos). There also is a classifying topos of local rings with a map to T (which is given by forgetting that the universal local ring is local). If you form the pullback (in an appropriate topos sense) of these two maps you get the topos of sheaves on Spec(R) (i.e. morally the space Spec(R)). The map from this into the classifying topos of local rings is what corresponds to the structure sheaf.
Isn't that nice? See Monique Hakim's "Schemas relatifs et Topos anelles" for all this (the original reference, free of logic), or alternatively Moerdijk/MacLane's "Sheaves in Geometry and Logic" (with logic and formal languages).
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This is a cool answer, but it seems completely contrary to what the OP was asking for. – Harry Gindi Feb 6 '10 at 0:41
In your first construction, would the "free local ring on R" also exist in the topos of sheaves on wrt the, say, fppf or etale topology on Spec R. If so, then this argument would not "motivate" the Zariski topology, since there is still an (a priori) arbitrary choice involved. Now if the Zariski sheaves on Spec R were in some sense a minimal topos with the property that the "free local ring on R" existed... – Lars Feb 6 '10 at 8:49
Ok, I was sloppy. Zariski sheaves are the "minimal" topos in the sense now explained above. Intuitively you give the ring R just enough space to spread out into a collection of local rings. – Peter Arndt Feb 6 '10 at 14:13
Thanks for the edit...this is pretty neat. – Lars Feb 6 '10 at 16:50
And by the way, in the same fashion the etale topos gives you a universal Henselian ring. I haven't heard of any such thing for fppf, though. – Peter Arndt Feb 6 '10 at 17:27
Here is what Eisenbud and Harris ('The Geometry of Schemes') have to say on this (page 10):
[comments by myself are in square brackets]
"By using regular functions, we make Spec R [R being a arbitrary comm. ring with 1] into a topological space; the topology is called the Zariski topology. The closed sets are defined as follows.
For each subset S ⊂ R, let
V (S) = {x ∈ Spec R | f (x) = 0 for all f ∈ S} = {[p] ∈ Spec R | p ⊃ S}.
The impulse behind this definition is to make each f ∈ R behave as much like a continuous function as possible. Of course the fields κ(x) [the residual fields at x ∈ Spec R] have no topology, and since they vary with x the usual notion of continuity makes no sense. But at least they all contain an element called zero, so one can speak of the locus of points in Spec R on which f is zero; and if f is to be like a continuous function, this locus should be closed. Since intersections of closed sets must be closed, we are led immediately to the definition above: V (S) is just the intersection of the loci where the elements of S vanish."
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If you buy into the idea that you want a topological model for your ring, then right away it becomes sensible to ask that any map Ring -> Top be functorial. Of course, m-Spec -- which is already classically motivated -- doesn't lend itself to this, simply because there isn't an obvious way to use a ring homomorphism $f : A \to B$ to move maximal ideals around.
Such a map can move around ideals, both by extension and contraction, and this is a good first thing to investigate. Your choice of whether or not you want to push ideals forward or pull them back will determine if your functor should be co- or contravariant.
To decide between these two, look at the initial and terminal objects in Ring, as well as the initial and terminal objects in Top. The ring {0,1} has a single ideal, hence its topological space has (at-most) a single point, hence should probably be sent to the final object in Top. The 0 ring has no ideals, hence its topological space has no points, hence should be sent to the initial object in Top. Your hand has thus been forced: you need a contravariant functor, hence contraction is the thing to look at.
Now observe that $f^{-1}(\mathfrak m)$ need not be maximal, even if $\mathfrak m$ is, but it will be prime. You're thus immediately led to seeing if you can put a topology on Spec the same way you did for m-Spec. It works, and you move on.
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The way I think about contravariance is that one should always associate with an ideal I the homomorphism A \to A/I. Then it's the most natural thing in the world to take a homomorphism B \to A and compose it with a homomorphism A \to A/I. The reason prime ideals pull back to prime ideals is that if A/I is an integral domain then B \to A/I is still a map with image an integral domain; however, a subring of a field need not be a field, so maximal ideals don't pull back. – Qiaochu Yuan Dec 8 '09 at 18:10
ring {0,1} = k? – user2035 Dec 8 '09 at 18:14
Contravariance is also sensible in light of the ideal correspondence for quotients and localization, which suggest that the topological spaces associated to A/p and A_p should both be subspaces of the space for A, a fact which runs in the opposite direction of the canonical maps A->A/p and A->A_p. – Tim Carstens Dec 8 '09 at 18:15
Here is my favourite way to motivate the Zariski topology: it is the coarsest topology which makes the functions defined (below) by ring elements "continuous" in the following sense:
Given a classical variety $V$ over $\mathbb{C}$ and a "regular function" $f:V\to\mathbb{C}$, one can identify the value
$f(x)$ with the the image of $f$ in $A_V/m_x$, where $A_V$ is the coordinate ring of $V$ and $m_x$ is the maximal ideal at $x$. This perpsective has the advantage of generalizing to any ring, if you allow the target field to vary from point to point:
First, motivate working with primes instead of maximal ideals because primes pull back under ring maps, and because non-maximal primes act like "generic points" in classical algebraic geometry.
Next, at each prime ideal of a ring $p\triangleleft A$, you get a domain $A/p$ (which people often like to think of as living inside a residue field, $k(p):=Frac(A/p)$). Then an element of the ring $a\in A$ defines a function $f_a$ on $Spec(A)$ taking values in various domains (or fields): $f_a(p):=image_{A/p}(a)$.
All domains/fields have the element $0$ in common, so it makes sense to talk about the vanishing set
$f_a^{-1}(0)=V(a):=$ {$p\in Spec(A) | a\in p$}, and these sets form a base for the closed sets of the Zariski topology.
Moreover, the finite unions and arbitrary intersections we need turn out to be extremely manageable because of the definition of primes, in a way that is intuitively meaningful in the context above: For any collection of basic closed sets $V(a)$ with $a$ ranging over a set $E\subset Spec(A)$, we get
• $\bigcap_{a\in E} V(a) = V(E) :=$ {$p\in Spec(A) | E \subset P$}. These are the primes where
"all of $E$ vanishes" in the residue domain/field.
• $V(a)\cup V(b) = V(ab)$, the primes where $a$ and $b$ "both vanish".
-
This is nice. Besides, this viewpoint allows to say every element of my ring is giving rise to a continuous function from Spec(A) to the corresponding domain/field. – Csar Lozano Huerta Dec 10 '09 at 8:43
This looks like the easiest and most natural way to motivate Zariski topology. – Fei YE Feb 6 '10 at 19:11
So, let's accept for now that considering the points of an affine scheme as a set with no topological structure makes sense.
Then each element of your ring either vanishes at a point (i.e. lies in the ideal) or doesn't. In any topology on this set that's compatible with the idea that ring elements are sections of some bundle on it, this zero set had better be closed.
By general topology nonsense, there is a unique coarsest topology where these are closed sets, the one that uses their complements as a basis (this exists because they cover, and the intersection of the sets for a and b is the set for ab). This is the Zariski topology.
-
Hm, does this make sense? Do you mean morphisms into arbitrary local ringed spaces (whether this makes sense or not)? – user717 Dec 8 '09 at 17:40
So for example, in my motivation for the Zariski topology on varieties over an algebraically closed field k, you first have to equip the field k with the Zariski topology to make this definition well-defined. So, you need some "initial data". – user717 Dec 8 '09 at 17:45
Sorry if I am getting on your nerves :) But I don't see why this is well-defined. Because "scheme" already carries a topology (the Zariski topology) and for "coarsest" to make sense you somehow need a (non trivial) set of topologies on your scheme. "My" case above makes sense because you just consider morphisms into one fixed space already carrying a topology and so you consider all topologies on the domain satisfying a condition and then you take the coarsest one. – user717 Dec 8 '09 at 18:53
Note to passersby: comments above apply to an earlier version of the answer. – Ben Webster Dec 8 '09 at 19:24
I don't think that you have to motivate the Zariski topology as anything other than a correct description of something that can be defined without it.
Suppose from the beginning that you are interested in commutative rings $R$, in general. Suppose that you would like to interpret the reverse category of ring homomorphisms as "geometry". After all, many other kinds of geometry are reverse categories of ring homomorphisms, for certain kinds of rings and homomorphisms. For example, any smooth map $M \to N$ between smooth manifolds is equivalent to an algebra homomorphism $C^\infty(N) \to C^\infty(M)$ with certain favorable properties. To keep things as general as possible, and in a strictly algebraic setting, let's call any ring homomorphism $R \to S$ a geometric map in the opposite direction. Let's call the map $\text{Spec}(S) \to \text{Spec}(R)$, where for the moment "Spec" doesn't mean anything other than reversing arrows. Then this is the category of affine schemes with scheme morphisms as the morphisms.
Having taken the plunge to call this an abstract "geometry", we can try to fill in a tangible geometry. Certainly maximal ideals should be called points in this "geometry", given the motivating example of polynomial rings and their ring homomorphisms. (Proposition: A homomorphism between polynomial rings is equivalent to a polynomial map between affine spaces.) Should we perhaps stop at the maximal ideals? That would be nice, if it were consistent. However, having committed ourselves to all ring homomorphisms between all rings as "geometry", it isn't consistent. For example, $\mathbb{Z} \to \mathbb{Q}$ is an important ring homomorphism. However, the inverse of maximal ideal $\{0\}$ in the target is a prime ideal which is not maximal. As this example suggests, prime ideals are the smallest viable extension of the maximal ideals in the contravariant geometry of ring homomorphisms.
They aren't the only viable extension. We could have taken all ideals instead of just the prime ideals. As far as I know, another Grothendieck and another Zariski would have defined the points of an affine scheme using all ideals instead of just the prime ideals. In any case, the prime ideals work; the maximal ideals don't.
Okay, what about topology. I think that the Zariski topology is still, as you say, the coarsest $T_1$ topology available to be able to call regular maps, by definition the maps on prime ideals induced by ring homomorphisms, continuous.
To summarize, the framework is a ruse to study all rings and ring homomorphisms in a geometric language.
-
You define points to be prime ideals and not just any ideals because you want to be able to localize at points, i.e. if $R$ is supposed to be the set of functions over the space of points $SpecR$, you want to be allowed to speak of the "local functions" around each point. – Qfwfq Mar 8 '11 at 14:19
Fields are characterized in the category of rings by the property that an epimorphism whose source is a field must be an isomorphism. The prime ideals of a ring can be identified with the epimorphisms from that ring to fields. From a categorical perspective, the prime spectrum therefore seems a more natural object than the max. spectrum.
I don't have a good explanation for the Zariski topology. It is the coarsest topology in which the maps induced by homomorphisms of rings are continuous and the origin in $\mathbf{A}^1$ is closed. This is not very satisfying, though.
The real reason the Zariski topology is important is because descent works for lots of things. For example, a module defined Zariski locally over a ring can be glued to give a module over the original ring. I wonder if the Zariski topology is the finest topology (i.e., finest Grothendieck topology defined by covers by subfunctors subobjects) in which descent works for quasi-coherent sheaves. Does anyone know?
-
Isn't there faithfully flat descent for quasi-coherent sheaves? – Dinakar Muthiah Dec 9 '09 at 7:47
Maybe you should ask your last question as a new question. It is very interesting, but Grothendieck topologies may not be part of the original question here. – Konrad Voelkel Dec 9 '09 at 15:03
Dinakar: yes, but the fppf topology is not defined by subobjects (I changed the wording above to emphasize this). The induced topology defined by subobjects is the Zariski topology (since an fppf embedding is an open embedding in the Zariski topology). – Jonathan Wise Dec 9 '09 at 18:03
Ah, I see your distinction. – Dinakar Muthiah Dec 10 '09 at 0:00
think there are two questions here: (1) why study the prime spectrum, and (2) why think of it in terms of the axioms for a topology. (1) has been pretty well handled by other commenters. And a number of them point out that (2) isn't really especially useful.
Part of the problem, according to a very interesting article I read by Grothendieck (maybe where he introduces dessins d'enfants?), is that the axioms for topological spaces are "wrong". Alas, he doesn't know what the right axioms are; he's just sure that the field of general topology should never have existed. From that point of view, discovering that the prime spectrum has a topology automatically isn't that interesting. (This guy has a contrary viewpoint on the field of general topology, but unsurprisingly I find Grothendieck more convincing.)
-
If you (or somebody else) can find that article, that would be awesome. – Kevin H. Lin Feb 6 '10 at 7:50
to Kevin: I heard from my advisor, he said the definition of grothendieck topology(not pretopology) "is equivalent to" all the possible existing topology in various branches of mathematics – Shizhuo Zhang Feb 6 '10 at 12:41
Steven Gubkin found the article; see mathoverflow.net/questions/14634/what-is-a-topological-space – Allen Knutson Feb 8 '10 at 15:07
This article, and all of Grothendeick's writing, have just recently been removed from the Grothendieck Circle's webpage, apparently on Grothendieck's request (Wikipedia says the request was made in January 2010 in a letter from Grothendieck to Illusie). Too bad. – Dan Ramras Jun 17 '10 at 5:00
Grothendieck's article can be found here: dl.dropbox.com/u/1963234/EsquisseEng.pdf – Gunnar Þór Magnússon Nov 14 '10 at 1:50
This is probably a standard question, so allow me write down (what I think) the standard answer. Viewing any ring $R$ as a ring of functions (allowing nilpotents and all that) on the prime spectrum $Spec R$, you naturally want all such functions (elements of $R$) to be continuous, thus it needs that, for any $f \in R$, $V(f) = [p \in Spec\; R: f(p) = 0 ] = [p \in Spec \; R: f \; mod \; p = 0] = [p \in Spec \; R: f \in p]$ (I used [ ] to denote a set...don't know why { } doesnt work)) being a closed set, where $p$ is a prime ideal and the 'value' of a function $f \in R$ at a point $p$ is the image of the residual of $f$ mod $p$ in the field of fractions of $R/p$ (which is an integral domain). I think there's no difficulty in showing that the field of fraction of $R/p$ is isomorphic to the residual field $k(p)$ of the local ring $O_p$, coinciding with the other definition of the 'value' of function $f \in R$.
Now any ideal $I$ of $R$ is generated by its elements, and so in order for a bunch of functions to be continuous, it needs to have the closed set $V(I)=[p \in Spec \; R: I \subset p]$.
-
Hm, what do you really mean by continuity of f? I mean, f is in general not a proper function into some topological space!? – user717 Dec 8 '09 at 16:50
It's really a motivation for Zariski topology on an affine scheme. If you look at affine varieties over a algebraically closed field $k$, then as you say you can see why Zariski is motivated; similar here, although the only difference is that the residual fields are different (and can be of different characteristics), unlike a unique field $k$ for the "functions" to take values. These are heuristics anyway, you may not need to take the motivations too seriously - otherwise why bother a formal theory. – user2148 Dec 8 '09 at 16:57
But when your "functions" aren't functions, then this motivation is just wrong, or am I wrong? Of course, I know about this intuition, but for exactly this reason I don't know what I should do with it... – user717 Dec 8 '09 at 17:42
I don't think so. If you want to think of topological spaces in terms of their ring of functions, then for continuous real-valued functions into R a topological space has the initial topology if and only if it's completely regular, and this is equivalent to the zero sets of continuous functions being a basis for the closed sets. So the Zariski topology makes Spec R behave as if it were completely regular in the sense that elements of R separate points and closed sets. – Qiaochu Yuan Dec 8 '09 at 17:49
You can take a look at James Milne's notes on AG. It uses the ringed space to define an affine variety over an algebraically closed field, and the structure sheaf indeed consists of regular functions (from the variety to the filed) on Zariski open subsets. The motivation from this example is clear enough, and the functions here are indeed functions. – user2148 Dec 8 '09 at 17:52
Here's an idea related to Tim Carstens' answer. As in Ben's answer we start from the point of view that it makes sense to think of $\text{Spec } R$ as a set. Given an ideal $I$ we have a homomorphism $R \to R/I$ and by the correspondence theorem the prime ideals of $R/I$ are precisely the prime ideals of $R$ containing $I$, so we get an injection $\text{Spec } R/I \to \text{Spec } R$. In any reasonable assignment of a topology to the spectrum this injection should be an embedding.
The Zariski topology accomplishes this by the simple requirement that the above map be both closed and continuous. Why is being closed a reasonable requirement? Well, an embedding of a compact Hausdorff spaces into another is always closed, so if we think of the relationship between $\text{Spec } R$ and $R$ as analogous to the relationship between a compact Hausdorff space $X$ and its C*-algebra then this is a natural requirement. (This is similar to the comment I made about completely regular spaces, so if you don't like that reasoning you probably won't like this argument either.)
This is perhaps a little unsatisfying until it's shown that there are no other reasonable ways to turn the map $\text{Spec } R/I \to \text{Spec } R$ into an embedding (certainly the discrete topology works, but I wouldn't call that reasonable), but hopefully somebody else has some insight here.
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# Current Loop simulation in HFSS
Status
Not open for further replies.
#### lokeyh
current loop simulator
Hi,
Has anyone done any simulation on the current loop in HFSS? Is it possible to upload the file?
Thanks.
#### e_m_c
##### Full Member level 3
calculating current in hfss through a loop
What file do you mean?
#### lokeyh
e_m_c said:
What file do you mean?
The HFSS file for current loop simulation.
#### e_m_c
##### Full Member level 3
I don't have such projects actually. But probably I can do it in MoM. What do you need to calculate? Field radiated by impressed current in free space?
#### lokeyh
I want to simulate the magnetic field (x-, y-, and z-component) at a given position for a current loop.
#### e_m_c
##### Full Member level 3
If you can publish more information - current loop parameters (geometry, values), field probe poisition and frequency range you are interested in it would be possible to discuss solution more effectively.
#### lokeyh
e_m_c said:
If you can publish more information - current loop parameters (geometry, values), field probe poisition and frequency range you are interested in it would be possible to discuss solution more effectively.
The loop has a radius of 10cm and a sinusoidal current (100 Hz) with r.m.s 1A will pass through the loop.
The loop is situated at the origin and I want to find the magnitude of the x-component of the point (0<= x <= 50cm, y = 20cm, z = 20cm) in HFSS.
I have simulations in HFSS but I cannot get the magnetic information I required. So I am asking if anyone has done any previous simulation.
Hopes this clarifies the problem.
Thanks.
#### e_m_c
##### Full Member level 3
OK, and the loop is situated in origin in XY or YZ or XZ plane?
#### lokeyh
It is on the XY plane.
#### ksugahar
##### Member level 1
do not place current source in series. you will get an inaccurate result.
#### e_m_c
##### Full Member level 3
I have simulated this in EMC Studio. Please see results in attached pdf file. Is this what you have looked for?
neverquick
Points: 2
### neverquick
Points: 2
#### lokeyh
ksugahar said:
do not place current source in series. you will get an inaccurate result.
What do you mean by this? Thanks.
e_m_c said:
I have simulated this in EMC Studio. Please see results in attached pdf file. Is this what you have looked for?
Hi, this is the result I am looking for. Thanks. But it would be good to have it in HFSS as that is the only software available to me.
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• # Fundamental Economics Concepts and Optimisation Techniques
LEARNING OUTCOMES
By the end of this topic, you should be able to:
1.
Apply marginal analysis concepts in making managerial decisions;
2.
Apply mathematics, in particular, differential calculus to find optimal
solutions to economics and management problems at the firm level;
3.
Compute present values of future cash flows; and
4.
Identify risks and uncertainties associated with business decisions.
• ### INTRODUCTION
An economic problem involves tradeoffs and opportunity costs. An economic problem can be illustrated in the pricing decisions by Malaysia Airlines on whether to offer discounts for unsold business class seats, when it is unsure whether the seats will ever be sold. This topic provides selected basic tools often used in managerial economics. These include concepts of marginal analysis, net present value and the tradeoff between risk and returns.
In this topic, we will focus on learning the mathematical tools to find the best value. The tools are called optimisation techniques. Depending on the problem, the highest value or lowest value is the optimum. In golf, for instance, the lowest
TOPIC 2
FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
13
number is the best but in bowling, the highest number is the best. Finding the least cost input combinations to produce a given output, the most profitable output, or the maximum net present value for a given investment; are all optimisation problems.
2.1
### MARGINAL ANALYSIS
Marginal analysis is the most useful concept in economic decision-making. Since resources are scarce and we cannot have everything that we want, choices must be made. The concept of opportunity cost reminds us that every time we make a choice, something else must be given up. Economics provides us with a set of tools that can help us make better choices. Most of the time, the best decision is made by weighing the marginal benefits against the marginal costs. We will carry out the decision as long as the marginal benefit is greater than the marginal cost. The best is when marginal benefit equals marginal cost.
• ### 2.1.1 What is Marginal Return or Revenue?
Marginal return (or marginal benefit or marginal revenue) is the change in total benefits derived from doing an activity and is also known as the additional benefits received when one more unit of the activity is produced. Benefits can be expressed in terms of units of utility or satisfaction, or they can be expressed in monetary values (for example, Ringgit Malaysia).
In microeconomics, Marginal Revenue (MR) is the extra revenue that an additional unit of product will bring to the firm. It can also be described as the change in total revenue/change in number of units sold.
More formally, marginal revenue is equal to the change in total revenue over the change in quantity when the change in quantity is equal to one unit (or the change in output in the bracket where the change in revenue has occurred).
• ### 2.1.2 What is Marginal Cost?
In economics and finance, marginal cost is the change in total cost that arises when the quantity produced changes by one unit. Mathematically, the marginal cost (MC) function is expressed as the derivative of the total cost (TC) function with respect to quantity (Q). Note that the marginal cost may change with volume and so at each level of production, the marginal cost is the cost of the next unit produced.
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FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
MC
dTC
dQ
or
MC
/
TC
Q
In general terms, marginal cost at each level of production includes any additional costs required to produce the next unit. If producing additional vehicles requires, for example, building a new factory, the marginal cost of those extra vehicles includes the cost of the new factory. In practice, the analysis is segregated into short and long-run cases, and over the longest run, all costs are marginal. At each level of production and time period being considered, marginal costs include all costs which vary with the level of production and other costs are considered fixed costs (http://www.wikipedia.org).
In a nutshell, marginal cost is the change in total cost of doing an activity and is also known as the additional costs incurred when one more unit of the activity is produced.
SELF-CHECK 2.1
1.
When do you use „marginal analysis?‰
2.
Your firm is operating a long-distance express bus service from
Kuala Lumpur to Kota Bharu. During the festive season, you
obtain licences to offer additional trips. Using the concepts of
marginal benefits and marginal costs, how would you decide on
the optimal number of additional trips to offer?
2.2
### RELATIONSHIPS AMONG TOTAL PROFIT, AVERAGE PROFIT AND MARGINAL PROFIT
The relationships among total profit, average profit and marginal profit are discussed below:
Total Profit, π(Q), is Total Revenue (TR) minus Total Cost (TC). Total profits are maximised at the output level (Q) where marginal revenue equals marginal cost. Another way to state the rule is that marginal revenue minus marginal cost must be zero for total profit to be a maximum.
Average profit is total profit divided by quantity: A π (Q) = π(Q)/Q. Marginal profit is the profit attributable to the last unit of output:
M π (Q) = Δπ(Q)/ΔQ.
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15
Tables, graphs and/or algebraic expressions are usually used to show the relationships among total profit, average profit and marginal profit. The relationships are shown in Table 2.1 and Figure 2.1.
Table 2.1: Quantity (Q), profit (), average profit and marginal profit.
TR TC (Q) M (Q) A (Q) 0 20 0 20 - - 34 1 26 8 28 8 2 34 66 32 24 16 3 44 96 52 20 17.33 4 56 124 68 16 17 150 5 70 80 12 16 174 6 86 88 8 14.67 7 104 196 92 4 13.14 8 124 216 92 0 11.5 9 146 234 88 4 9.78 10 250 170 80 8 8
In Table 2.1, total profit rises up to a maximum. Marginal profit is the slope of the total profit curve. The slope of total profit is zero at its maximum, because the slope of the horizontal line is zero. Hence, marginal profit is zero at maximum profit. The decision rule for maximising profits is to expand output until marginal profit is zero.
16
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TECHNIQUES
Figure 2.1: Total profit, average profit and marginal profit. Source: McGuigan, J. R., Moyer, R. C., & Harris, F. H. (2005) Managerial economics:
Applications, strategy and tactics (10th ed.). Mason, Ohio: South-Western.
In algebra, profit (called the dependent variable) depends on the level of output (the independent variable). The highest profits occur where ∆π(Q) = 0. You see that a quick method to find maximum profits uses calculus: marginal profit is the derivative of total profit. Therefore, local maximum profits occur at the quantity where the derivative of total profits with respect to output equals zero.
SELF-CHECK 2.2
Discuss and show the relationships among total profit, average profit
and marginal profit. Make sure you understand the use of symbols
representing all the terms.
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17
2.3
### TYPES OF OPTIMISATION TECHNIQUES
There are two types of optimisation techniques:
Unconstrained Optimisation is a relatively simple calculus problem that can be solved using differentiation, such as finding the quantity that maximises profit in the function π (Q) = 16Q - Q². [The answer is Q = 8.]
Constrained Optimisation involves one or more constraints. This happens when there are inequalities, for instance, when you must spend less than or equal to your budget allocation. Lagrangian multipliers are used to solve these problems.
• ### 2.3.1 Calculus in Managerial Economics
We study differentiation and the rules for differentiating functions because these methods can be used to find optimal solutions to the various kinds of maximising and minimising problems in managerial economics.
The rules for differentiating functions are summarised below:
Name Function Derivative Example Constant Y = c dY/dX = 0 Y = 5 Functions dY/dX = 0 A Line Y = cX dY/dX = c Y = 5X dY/dX = 5 Power Y = cX b dY/dX = bcX b-1 Y = 5X 2 Functions dY/dX = 10X Sum of Y = G(X) + H(X) dY/dX = dG/dX + dH/dX Y = 5X + 5X 2 Functions dY/dX = 5 + 10X Product of Y = G(X) x H(X) dY/dX = (dG/dX)H + (dH/dX)G Y = (5X)(5X 2 ) Two Functions dY/dX = 5(5X 2 ) + (10X)(5X) = 75X 2
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• ### 2.3.2 Optimisation in One Variable Case
MAXIMISATION PROBLEM: Profit maximisation assumes that there is some output level that is the most profitable. A profit function might look like an arch, rising to a peak and then declining at even larger outputs. A firm might sell huge amounts at very low prices but discover that profits are low or negative.
MINIMISATION PROBLEM: Cost minimisation assumes that there is a least cost point to produce. An average cost curve might have a U-shape. At the least cost point, the slope of the cost function is zero.
The first order condition for an optimum is that the derivative at that point is zero. To determine whether that optimum is either maximum or minimum, you must find the second derivative, which is the derivative of the first derivative.
The second order condition states that:
If the second derivative is negative, then itÊs a maximum If the second derivative is positive, then itÊs a minimum
Examples:
• 1. π = 100Q - Q 2
First derivative: d π /dQ = 100 -2Q To find the optimum point, set the first derivative as equals 0.
dπ /dQ = 100 -2Q = 0
Therefore Q = 50 To determine whether Q = 50 is a maximum or minimum, find the second derivative. The second derivative is -2, thus implying that Q =50 is a MAXIMUM.
• 2. π = 50 + 5X 2 dπ/dX = 10X = 0 Therefore X = 0 The second derivative is 0, implying X = 0 is a MINIMUM.
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FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
19
### 2.3.3 Optimisation in More-than-One Variable Case
Economic relationships usually involve several independent variables. For example: the production of shoes requires materials, machine and labour; while the demand for shoes depends on price and income. In this case, we use Partial Differentiation. A partial derivative is like a controlled experiment – it holds the „other‰ variables constant. Assume quantity, Q = f (P, I). If price is increased, holding the disposable income constant then the partial derivative of Q with respect to price is Q/P, holding income constant. Similarly, the partial derivative of Q with respect to income is Q/I, holding price constant.
Example:
Assume Sales is a function of advertising in newspapers and magazines (X, Y) and it is given in the following relationship:
Max S = 200X 100Y 10X 2 20Y 2 + 20XY
To find the optimum, differentiate the above with respect to X and Y and set them equal to zero:
S/X = 200 – 2 – 0X + 20Y= 0 S/Y = 100 – 40Y + 20X = 0
Then, solve for X & Y and Sales:
• 200 – 20X + 20Y= 0
• 100 – 40Y + 20X = 0
Adding them, the 20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15
Plug into one of them: 200 20X + 300 = 0,
hence X = 25
To find Sales, plug into the equation: S = 200X + 100Y 10X2 20Y2 + 20XY =
3,250
20
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TECHNIQUES
SELF-CHECK 2.3
1.
Find the derivatives of the following:
(i)
TC = 50 + 100Q 6Q 2 + .5Q 3
(ii) ATC = 50/Q + 100 6Q + .5Q 2
(iii)
(iv)
MC = 100 12Q + 1.5Q 2
Q = 50 .75P
(v)
Q = .4X 1.5
2.
Given Y = -2X 2 + 20X 20,
(i)
Find the optimal value of X.
(ii) Is this a maximum or a minimum?
3.
Suppose Q = 2000 + 15Y 5.5P, find the optimal values of Y, P
and Q.
2.4
### CONCEPT OF NET PRESENT VALUE
To find managerial decision rules that maximise shareholdersÊ wealth over a long period of time, you must consider the present value of the benefits as well as the present value of the costs. The net present value is the difference between the present values of all the benefits and costs. When the net present value is positive, then the decision improves shareholdersÊ wealth.
Present value recognises that a dollar received in the future is worth less than a dollar in hand today, because a dollar today could be invested to earn a return. To compare monies in the future with today, the future dollars must be discounted by a present value interest factor, PVIF = l/(l+i), where it is the interest compensation for postponing receiving cash by one period. For dollars received in n periods, the discount factor is PVIF n = [l/(l+i)] n
.
What will be the present value of \$500 to be received 10 years from today if the discount rate is 6%?
PV = \$500 {1/(1+.06) 10 } = \$500 (1/1.791) = \$500 (.558) = \$279
TOPIC 2
FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
21
Net Present Value, NPV = Present value of future returns minus initial outlay. This is for the simple example of a single cost today yielding a benefit or stream of benefits in the future. For the more general case, NPV = Present value of all cash flows (both positive and negative ones). Example: A project with an initial cash outlay of \$40,000 with the following cash flows for five years. The firm has a 12% required rate of return. Year In-flows Out-flows Net Cash flows (NCF) Initial outlay 40,000 -40,000 1 50,000 36,000 14,000 2 51,000 38,000 13,000 3 51,000 38,000 13,000 4 52,000 40,000 12,000 5 52,000 41,000 11,000 The present value of the NCFÊs is \$47,678. Subtracting the initial cash outlay of \$40,000 leaves an NPV of \$7,678. NPV>0, therefore we accept. NPV Rule: Carry out all projects that have a positive net present value. By doing this, the manager will maximise shareholder wealth.
Some investments may increase NPV, but at the same time, they may increase risk. Whether the extra risk is acceptable depends on what is the acceptable rate of return for that risk.
SELF-CHECK 2.4
1.
What is Net Present Value (NPV)?
2.
What is the rule when you use NPV?
2.5
RISK
• ### 2.5.1 What is Risk?
Uncertainty about a situation can often indicate risk, which is the possibility of loss, damage or any other undesirable event. Most people desire low risk, which would translate to a high probability of success, profit or some form of gain.
22
TOPIC 2
FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
Risk
event.
is the possibility of loss, damage or any other undesirable
For example, if sale for next month is above a certain level (a desirable event), then these orders will reduce inventory and if there is a delay in shipping orders (an undesirable event) which means losing orders, then that possibility of delay is a risk to the firm.
An investment decision is risk-free when the expected dollar returns and initial investment are certain. But most managerial decisions involve uncertainties. There is always a possibility that cash flows will fall below the expected level and sometimes there is also the possibility that the cash flows will be negative (a loss).
• ### 2.5.2 How to Measure Risk?
Variability in the outcomes of an investment is a measure of risk. The bigger the variability of the possible outcomes, the higher is the risk of the investment. Variability can be described using probability distributions.
n
In any distributions, the sum of the probabilities, p j , must equal one p j = 1.
j=1
This assures that all possible outcomes, r j , have been exhausted and each outcome is discrete. Probabilities can be thought of as the percentage likelihood that each outcome, or state of nature, occuring:
• (a) Expected Value is the weighted average of the possible outcomes:
^ n r j=1 ____________ σ = √ Σ (r j -
= p j r j = 1.
j=1
• (b) Standard Deviation, σ measures the dispersion of outcomes around its expected value.
r^) 2 p j
The expected values and standard deviations of two projects, with differing cash flows and differing probability distributions, can be compared. If two projects have the same expected value, you may wish to select the one with the lower standard deviation. Or if two projects have the same standard deviations, you may wish to select the one with the higher expected value. If one project has a higher expected value and a higher standard deviation, then the choice depends on a tradeoff between risk and return.
TOPIC 2
FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
23
• ### 2.5.3 Risk and Return
We face risk every day. Driving a car or riding a motorcycle, crossing a busy street, even eating your favourite meals, involves some form of risk. With investment decisions, balancing risk and return can be very tricky. Investors want to maximize their return, while at the same time want to minimize risk. Unfortunately, for most investments, the higher the return, the higher is the risk associated with it.
Some investments are certainly more "risky" than others but no investment is risk-free. Avoiding risk by not investing at all can be the riskiest move of all. Try to keep your money under your pillow! You will not earn anything from it; you will definitely lose due to inflation or worse, somebody might steal it from you. Trying to avoid risk is like standing at the curb, never setting your foot into the street to get to the other side. You will neve r be able to get to your destination if you do not accept some risk. In investing, just like crossing that street, you carefully consider the situation, accept a comfortable level of risk and proceed to where you are going. Risk can never be eliminated but it can be managed.
In an investment decision, we can divide the required return into two parts, the risk- free return and a risk premium.
Required Return = Risk-free Return + Risk Premium
The greater the risk, the greater must be the risk premium as a reward for accepting that risk.
Two mutually exclusive projects with different risks can be compared using the NPV rule. You can discount the riskier project with a higher required return (because of its higher risk premium). The different discount rates adjust for risk. The project with a higher risk-adjusted NPV should be selected.
SELF-CHECK 2.5
1.
In your own words, define „risk‰.
2.
How do you measure risk?
3.
How do you select projects based on the expected values and
standard deviations?
4.
What is the relationship between risk and return?
24
TOPIC 2
FUNDAMENTAL ECONOMIS CONCEPT AND OPTIMISATION TECHNIQUES
ACTIVITY 2.1 | 4,702 | 18,533 | {"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} | 3.1875 | 3 | CC-MAIN-2020-05 | latest | en | 0.896075 |
https://physics.stackexchange.com/questions/659895/mwi-and-quantum-probability | 1,723,728,456,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00684.warc.gz | 365,907,368 | 42,139 | # MWI and quantum probability
One thing that I cannot wrap my head around, and which never seems to be addressed in the videos and articles I've read about the Many Worlds Interpretation, is how "branching" is supposed to account for the probabilistic outcomes we see in quantum experiments.
In the classical interpretation, the wavefunction of a particle exists in a superposition of possible outcomes until it is measured, after which decoherence kicks in and the particle settles into one if its possible outcomes. The "more likely” outcomes – corresponding to the square modulus of the wavefunction – occur more frequently. And yet any one particular outcome is essentially random.
In the MWI, we are in one fixed branch of the universal wavefunction. Whatever the measurement of the particle could be, will be, in some branch – just not necessarily our branch of reality.
So how is there to be any reconciling of the probability we see in experiments with this interpretation? If a quantum event is measured and outcome A is observed to have probability $$1/3$$, and probability B has probability $$2/3$$, then what does this mean in terms of the "branches" in the MWI?
The idea of "branching" or "branches" is fundamentally discrete. Does this mean there would be exactly 1 branch where outcome A happens, and exactly 2 branches where outcome B happens? What about for more complicated probabilities, like an outcome of $$12214/36537$$? And what about an irrational probably? Like $$1/\sqrt2$$ or $$1/\pi$$?
Are we to believe that there are no irrational quantum probabilities? That the density of the rational numbers in the reals is good enough? Or can branching somehow be made into a continuous phenomenon? And what would that even mean?
• Popular science sources that talk about "branches" are being inaccurate. Wavefunctions have a domain. This is $\mathbb{R}^3$ if you're talking about a single particle's position and something much larger if you're talking about the whole universe. If $x$ is the universe in which you perform the above experiment and see outcome A the first time, while $y$ is the universe in which you do the same and see outcome B the first time, it must be that $\psi(y) = 2\psi(x)$. But there are still no branches. Commented Aug 17, 2021 at 19:11
• There's no problem with having an infinite number of branches. See my answer here: physics.stackexchange.com/a/536580/123208 Commented Aug 17, 2021 at 21:40
I'm not clear on what videos and articles you have been reading but there is a large literature on probability in the MWI and on branching.
According to the MWI the universe you see around you is part of a much larger and more complex structure called the multiverse, which looks a bit like a collection of classical universes to some approximation. A universe is a structure within the multiverse in which information about a particular outcome of an event flows freely. For example, there is another universe in which I typed the letter 'a' at the end of this sentence: b. We're not in that universe but that universe exists and the people in that universe can see the letter 'a' at the end of the sentence and discuss it and check that I typed the letter 'a' and so on. In this universe I typed the letter 'b' instead and the people in the 'a' universe can't see the letter 'b' at the end of that sentence. So those two outcomes are in different universes.
For more detail see:
https://arxiv.org/abs/quant-ph/0104033
https://arxiv.org/abs/quant-ph/0107144
The parallel universe approximation breaks down in various kinds of experiments like single particle interference experiments and the EPR experiment. The approximation holds when information about a particular outcome has been copied from one system into another: this suppresses interference and that process goes by various names like decoherence and quantum darwinism:
https://arxiv.org/abs/1212.3245
The outcome of an experiment is discrete and is basically an eigenstate of some observable. Before the state of a system is decohered it will look a bit like this: $$\sum a_j|j\rangle$$.
The probabilities of doing a measurement on that state don't come from counting the number of branches, i.e. - the number of $$j$$ labels. Rather, the gist of the explanation goes like this. The probability of an outcome is a function of the state that satisfies particular properties required by decision theory, like if a person uses those probabilities you can't make him take a series of bets that will make him lower the expectation value of his winnings, the probability of an outcome can't be changed by a later measurement and some other assumptions and from this you can explain that the probability of the $$j$$th outcome will be $$|a_j|^2$$:
https://arxiv.org/abs/0906.2718
https://arxiv.org/abs/quant-ph/0303050
https://arxiv.org/abs/quant-ph/9906015
https://arxiv.org/abs/1508.02048
So if you are offered a bet such that if the outcome of the measurement if $$j$$ you will win £1 and otherwise you get nothing the probability of that outcome is $$1/\pi$$, then if somebody offers to let you place that bet for less then £$$1/\pi$$ you would be rational to take the bet, otherwise you wouldn't be rational to take it. Another way of thinking about it is that each branch has a certain thickness compared to the branches from which it is a descendant and the probability is the thickness of the branch that is relevant to betting that is rational in the decision theoretic sense.
There is another approach to deriving the probability rule called envariance that I think is less satisfactory but I'll give you a link for the sake of completeness:
https://arxiv.org/abs/quant-ph/0405161 | 1,274 | 5,712 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-33 | latest | en | 0.947577 |
https://www.zackvision.com/weblog/2013/page/2/ | 1,534,798,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217006.78/warc/CC-MAIN-20180820195652-20180820215652-00351.warc.gz | 1,017,339,764 | 41,978 | ## 1,000 Miles Progress
One of my new year resolutions this year was to do 1,000 miles in 2013.
Running a thousand miles in a year requires 2.74 miles/day or 19.19 miles/week. That is beyond my ability. But I do some cycling on the weekends. So I decided to count my cycling miles at one-third.
Let’s see how I have done since January 1.
Running: 4.26 + 4.05 + 5.1 + 4.19 + 4.04 + 3.22 + 4.11 + 4.2 + 6.04 + 4.31 + 4.25 + 6.16 + 4.2 + 4.29 + 6.33 + 4.07 = 72.82 miles
Cycling: 29.5 + 31.5 + 40.3 + 28.9 = 130.2 miles
Adding them up, we get 72.82 + 130.2/3 = 116.22 miles.
Today is February 6, the 37th day of the year. So 10% of the year is gone. To keep my pace for the 1,000 miles, I needed to do 100 miles till noon today. Thus I am on track.
## Len Foote Hike Inn
In October, we usually head to North Georgia to hike, enjoy the nice weather and see some fall colors. This time we booked a room at the Len Foote Hike Inn which can only be reached by hiking 5 miles from the Amicalola Falls State Park.
Here is our hike path from Amicalola Falls to the Inn.
On our return the next day, we decided to take the Appalachian Approach trail which connects Amicalola Falls to Springer Mountain.
It was an easy introduction for our trip to LeConte Lodge in the spring.
## Hot Chocolate 5K
When chocolate’s on offer, of course, I’ll run. And that’s how I registered for the Hot Chocolate 5K.
On the morning of January 13, the “Hot” referred to the weather since it was 60F early in the morning.
It was slow going to get to the race area due to traffic. About 18,000 people were taking part and traffic was completely stopped for miles.
After the race, we got some chocolate fondue with a banana and other dips. It was yummy.
Here’s a map of the route, along with elevation and my heart rate and speed graphs.
Hot Chocolate 5K [urldisplaymode=none;gpxspeedchart=show;gpxheartratechart=show]
This was my first race with a timing snafu. Initially, the results online showed my chip time as 36 minutes. I knew that was definitely wrong. My Wahoo Fitness iPhone app had clocked me at 32:59 and I had started that before my corral even started running and had stopped it a little after crossing the finish line. So I sent them a correction request. My time has now been changed to 33:33, which is still about a minute too long. But I am going to let that stand.
Based on the new time of 33min 33sec, I was ranked 2169th/9083 overall, 756th/1732 among men, and 95th/269 among men aged 40-44. (Yes, men were only 19% of the 5K participants.)
Now, I am waiting for my first 10K! | 753 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-34 | longest | en | 0.956302 |
http://www.instructables.com/id/How-to-make-your-own-personalized-rubiks-cube/CXODVYAG4D61LRE | 1,369,401,856,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704662229/warc/CC-MAIN-20130516114422-00045-ip-10-60-113-184.ec2.internal.warc.gz | 543,195,352 | 28,974 | # How to make your own personalized rubik's cube
9 Steps
Okay, so I was playing portal for the sixth time in a row and while doing that I was looking at the rubik's cube I got from one of my colleagues for secret santa. Now the rubik's cube I got was a so called Sudoku rubik's cube and it was way to hard to solve, so I thought why not make it into the companion cube from portal?
Now after that thought I got an even better idea for an instructable. Let's face it sometimes we forget our girlfriend's birthday, that of our parents, somebody's christmas present... and we have to come up with something that is makeable in one evening and with easy materials. Well here it is! The personalized rubik's cube!
During this instructable I'll be making the companion cube but I will also show you how you can add your own family pictures, favorite game, tv-show... anything you want to be on the rubik's cube.
Now let's go to step 1 What will you need.
Note: English is not my first language so if you encounter any spelling mistakes in this instructable please leave a comment and I will remove the mistakes.
Note2: Excuse me if some of these steps in the tutorial are to "Well duuuuh we are not idiots". I'm used to teaching 8-9 year olds so it kind of grows on you.
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## Step 1: What you will need:
- A rubik's cube (in my case the sudoku version but maybe you want to make one of those?)
- Photoshop CS4 (you can get a trail version here: www.adobe.com/nl/products/photoshop/photoshop/)
- if you don't want to use photoshop, paint can also work but it's not coverd in this instructable.
- A sharp knife
- A metal ruler
- Double sided tape
- Normal tape (optional)
- piece of wood to cut on/ cuttingboard (so you don't ruin your desk)
- scissors (not shown in picture)
Note: Don't worry if you can't use photoshop CS4, were not going to do anything hard with it and the steps are explained using screenshots
1-40 of 43 Next »
DVecc1217 says: Sep 5, 2011. 1:55 AM
This is informative and all but wouldn't it be much simpler to just buy the Rubik's custom sticker kit from the official website, download the blank template, add your photos and then just print it? You don't need to go through the trouble of laminating it and all and could just stick them on when it's done printing.
AwesomeSmiley3707 says: Apr 26, 2010. 8:03 PM
Hey, I just wanted to point out that instead of spending money on photoshop
or wasting time and patience on paint, you can also use GIMP. Its a free, photoshop-like program that has most of the functions as photoshop, and ITS FREE!!! You can easily download it off the internet and it is completely safe. Just search GIMP on Google. I just wanted to recommend that also, BTW, great tut. . .
MichelMoermans (author) in reply to AwesomeSmiley3707Apr 27, 2010. 1:00 PM
Yeah, alot of people have recommended GIMP to me in the past but since I already have photoshop there is no point for me to learn it.
And because I can't work with it I didn't give any instructions on GIMP.
So if visitors want to follow the instructions I just advice them to get the photoshop trial version instead of letting them try on their own with GIMP.
But I'm sure that the people who know how to work with GIMP will use that program instead of photoshop already ;)
So thank you for the suggestion and comment :)
AwesomeSmiley3707 in reply to MichelMoermansApr 27, 2010. 4:20 PM
oh, ok.
MichelMoermans (author) in reply to AwesomeSmiley3707Apr 28, 2010. 12:12 AM
But thank you now people will know there is an alternative if they want to keep using drawing programs.
bakumaster117 in reply to MichelMoermansSep 23, 2010. 11:44 AM
AwesomeSmiley3707 in reply to MichelMoermansApr 28, 2010. 4:41 PM
ok, cool
scott rox says: Aug 24, 2010. 2:21 AM
I like your concept but i'm gonna leave all squares black and spray-paint coloured dots in the center
Hunter O. says: Jul 21, 2010. 5:05 PM
Sudokubes are practically impossible. I tried taking it apart and it broke...
scott rox in reply to Hunter O.Aug 24, 2010. 2:16 AM
Rofl me too. i was doing the above by myself and when i took apart my cube it snapped :( so i i had to use my expensive one
krickerd says: Jul 25, 2010. 3:19 PM
Not sure what I'm looking at here. Try using the macro setting on your camera and make sure the green light comes on telling you it's in focus.
santy22 says: Apr 17, 2010. 3:08 PM
But a rubik with all same sides is ALWAYS solved... so the companion cube is kinda not a good idea.
seabananers says: Apr 10, 2010. 12:39 PM
i wish i could get photo shop so i could do this with pics of friends/family
MichelMoermans (author) in reply to seabananersApr 10, 2010. 1:06 PM
You can get a trial version at the adobe website.
It's good for 30 days and will give you enough time to make all the templates with pics of family and friends you want :)
Also you can save them at that moment as a .jpeg this way even when your trial expires you can still print them with windows picture viewer :)
seabananers in reply to MichelMoermansApr 11, 2010. 10:46 AM
thx i'll bug my parens for it -----
kijiyuuki says: Mar 15, 2010. 10:33 PM
hey i like your idea and im making one myself, only im making one so that it has album art all over it, its going to be albums by the band tool, if i ever get a pic of it ill post it for you to see
MichelMoermans (author) in reply to kijiyuukiMar 16, 2010. 10:26 AM
Okzy I love to see it :D
I love it that you took a different concept to it. That was the reason I put that step in :)
mettaurlover says: Feb 19, 2010. 8:25 AM
COMPANION CUBE!!!!!
Duct Tape Dude says: Dec 26, 2009. 6:40 PM
I made a solvable comanion rubiks cube by painting (in ms paint...duh)
the hearts a different color...1 is red, others are pink, purple, blue, green, and yellow
MichelMoermans (author) in reply to Duct Tape DudeDec 27, 2009. 2:52 AM
Well done. I would like to see a picture if you have it :)
How did you handle the squares around the heart? Did you make the little pink stripes change color to?
Duct Tape Dude in reply to MichelMoermansDec 27, 2009. 6:53 AM
Yeah...forgot to tell about that,its still super easy (WOW You were up early! I just woke up!!!)
MichelMoermans (author) in reply to Duct Tape DudeDec 27, 2009. 7:16 AM
Early bird get's the worm ;)
Well if you have the file saved I would be delighted to add it to the instructable and I'm sure other members would like to see it as well. Ofcourse you will be properly credited.
Let me know what you think :)
Duct Tape Dude in reply to MichelMoermansDec 27, 2009. 7:29 AM
you can post as soon as i can get some pics...my camera need some batteries -_-
MichelMoermans (author) in reply to Duct Tape DudeDec 27, 2009. 8:05 AM
Okay, thanks alot :)
Duct Tape Dude in reply to MichelMoermansFeb 13, 2010. 4:05 PM
Uhh I took pics...but then dropped my camera -_- and now its broken. But it looks just like it sounds...I guess I could 'shopp some of your pics...if you reall want to see it...
Duct Tape Dude in reply to Duct Tape DudeFeb 13, 2010. 4:08 PM
Oh and I changed colors to all pink just so I can say " I know my Companion Cube loves me, because no matter which way I turn her she always stays solved!"
MichelMoermans (author) in reply to Duct Tape DudeFeb 14, 2010. 3:53 AM
:D That's one of the best lines I have heard about the companion cube rubik's cube :D
If you have the time to shop it, sure I would love to see it :)
Duct Tape Dude in reply to MichelMoermansFeb 14, 2010. 9:00 PM
I whipped up a template for the public in Gimp...it's not the best in the world(about 3.5 out of 5) but you can "profesionalize" the hearts in Photoshop if you want...I'd give you the pic...but I can't post comment pics :P can you help?
MichelMoermans (author) in reply to Duct Tape DudeFeb 15, 2010. 9:16 AM
Normally in your reply screen you see in the left corner on the bottem (below the red bar with the be nice policy) an option "add images" click on that one and yo should be fine :)
spydog4 says: Jan 14, 2010. 9:22 AM
i know intrucables people don't believe in buying things, but they do sell a kit on rubiks website that lets you print your cube
MichelMoermans (author) in reply to spydog4Jan 14, 2010. 10:14 AM
Why buy something that is cheaper and funner to make yourself? :)
spydog4 in reply to MichelMoermansJan 15, 2010. 1:41 PM
im not saying i would buy it just putting that out there
Ward_Nox says: Dec 24, 2009. 9:32 AM
ths is a neat and useful ible well done
one thing you MIGHT want to try is using Decoupage to attach the new squares
or paring that use sticker paper in your printer
MichelMoermans (author) in reply to Ward_NoxDec 24, 2009. 11:56 AM
Thank you. And yes you can use sticker paper, I didn't try that because I didn't have any laying around and because the sticker paper is thicker then the double sided tape I used. And the thicker your stickers are the easier they go off during use. But I guess if you remove the original it could work as well. If you try with sticker paper let me know how it turned out and I'll add it to the instructable :)
Oh and if you liked it, please rate :)
LuminousObject says: Dec 24, 2009. 7:35 AM
Dude, that's sweet! Your English was fine too. It's better than some of the people who have English as their first language!
Stephen D. Alverez says: Dec 24, 2009. 4:55 AM
WOW! some of these comments of yours are from 3:10 AM, thats early as all get out! JEEZ!
MichelMoermans (author) in reply to Stephen D. AlverezDec 24, 2009. 5:07 AM
What can I say, I'm commited to project feedback ;)
Nah to tell you the truth it was around 11.10 AM I think. I live in Belgium Europe. The time is GMT+1 here.
thanks for viewing :)
mangoicecream says: Dec 24, 2009. 1:55 AM
Fabulous! Thank.you!!! =)
MichelMoermans (author) in reply to mangoicecreamDec 24, 2009. 3:12 AM
Thanks for viewing :)
If you decide to make one please post a picture, I'd love to see it :)
Flashflint says: Dec 23, 2009. 2:40 PM
What kind of knife is shown in the last picture?
1-40 of 43 Next » | 2,837 | 10,013 | {"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-2013-20 | latest | en | 0.955644 |
http://www.chegg.com/homework-help/questions-and-answers/cheers-q19539 | 1,371,718,227,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711005985/warc/CC-MAIN-20130516133005-00028-ip-10-60-113-184.ec2.internal.warc.gz | 368,300,202 | 8,556 | ## Signals an Systems
Cheers
• 1). the responses due to x1(t) and x2(t) are givenby
y1(t) =
. y2(t)=
the out put due to weighted sum of inputs is
now the weighted sum of outputs is
+
for a s ystem to be linear the below condition shouldbe satisfied ay1(T) + b y2(t) = y3(T).
but for this problem this is not vfarified so the systemis not a linear one. its a nonlinearsystem.
2) the given system is causalsystem because for all inputs the out put is depending on the present inputs only.
take t = 0,1,2 seconds then y(1)=2x2(1)+3x(1).
y(0) = 2x2(0)+3x(0).
a system can be called as causal if its output depends on thepast prsent values of inputs only not on the futureinputs,
3) y(t) = T2 x2 (t) + 3 x(t)
the out put due to delayed input is
y(t,T) = T(x(t-T)) = 2x2(t-T)+3x(t-T).
the delayed output is
y(t-T) = 2x2(t-T)+3x(t-T).
y(t-T) = y(t,T). so the systemis timeinvariant
4)from the impulse response h(n) we can conclude thegiven system is memory or memory less,
if h(n) is equal to zero mens it is withoutmemory
if h(n) is not equal to zero systemis a memoryone. | 367 | 1,109 | {"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} | 3.078125 | 3 | CC-MAIN-2013-20 | longest | en | 0.658481 |
https://www.airmilescalculator.com/distance/ric-to-rno/ | 1,620,547,199,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988961.17/warc/CC-MAIN-20210509062621-20210509092621-00434.warc.gz | 629,276,000 | 83,693 | # Distance between Richmond, VA (RIC) and Reno, NV (RNO)
Flight distance from Richmond to Reno (Richmond International Airport – Reno–Tahoe International Airport) is 2284 miles / 3675 kilometers / 1984 nautical miles. Estimated flight time is 4 hours 49 minutes.
Driving distance from Richmond (RIC) to Reno (RNO) is 2669 miles / 4295 kilometers and travel time by car is about 43 hours 52 minutes.
## Map of flight path and driving directions from Richmond to Reno.
Shortest flight path between Richmond International Airport (RIC) and Reno–Tahoe International Airport (RNO).
## How far is Reno from Richmond?
There are several ways to calculate distances between Richmond and Reno. Here are two common methods:
Vincenty's formula (applied above)
• 2283.559 miles
• 3675.033 kilometers
• 1984.359 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
• 2278.084 miles
• 3666.220 kilometers
• 1979.601 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 Richmond International Airport
City: Richmond, VA
Country: United States
IATA Code: RIC
ICAO Code: KRIC
Coordinates: 37°30′18″N, 77°19′10″W
B Reno–Tahoe International Airport
City: Reno, NV
Country: United States
IATA Code: RNO
ICAO Code: KRNO
Coordinates: 39°29′56″N, 119°46′4″W
## Time difference and current local times
The time difference between Richmond and Reno is 3 hours. Reno is 3 hours behind Richmond.
EDT
PDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 250 kg (551 pounds).
## Frequent Flyer Miles Calculator
Richmond (RIC) → Reno (RNO).
Distance:
2284
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
2284
Round trip? | 501 | 1,955 | {"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-2021-21 | latest | en | 0.821837 |
http://matise.com.br/blog/4qns6bx.php?c44f62=introduction-to-percolation-theory-pdf | 1,653,295,576,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662556725.76/warc/CC-MAIN-20220523071517-20220523101517-00282.warc.gz | 40,207,738 | 5,535 | # introduction to percolation theory pdf
lecture, computer simulation and experimental measurements) to percolation theory is discussed. PDF | An integrated approach (i.e. Once percolation theory is defined, we explore applications to the renormalization group, computer simulations of potts models, and randomly punctured conducting sheets. this chapter we give a short introduction to percolation theory and describe one ap-plication to composites. Percolation Theory Isaac Brodsky This essay describes percolation theory. Percolation theory, the theory of the properties of classical particles interacting with a random medium, is of wide applicability and provides a simple picture exhibiting critical behaviour, the features of which are well understood and amenable to detailed calculation. Resistor networks, from which resistors are removed at random, provide the natural generalization of the lattice models for which percolation thresholds and percolation probabilities have previously been considered. 3922 Lectures: 9:25-10:40 AM T. A good knowledge of linear algebra (including eigenvalues, bilinear forms, euclidean spaces, and tensor products of vector spaces) is presupposed, as well as some. We start with the structural properties of site percolation clusters and their substructures and report on other percolation systems after that. Introduction to Sets. Learning and Generalization Early machine learning algorithms aimed to learn representations of simple functions. Imagine a very large lattice of empty sites. Ethan Lewis. PDF Introduction to Percolation Theory. Fisher and Kadanoff – helped to develop percolation theory and understand the percolation as a critical phenomena 9Fractal concept (Mandelbrot, 1977) – new tools (fractal geometry) together with computer development pushed forward the percolation theory 9Still – many open questions exist ! Kesten, combined with work of T. Harris from the 1960’s, proved the “obvious result” that the critical probability for bond percolation on Z2 is 1/2. At random, a site could be occupied with probability p or unoccupied with probability 1-p. spectacular result in percolation theory. Extensions of percolation theory to treat transport are described. Introduction to Bernoulli percolation Hugo Duminil-Copin∗ October 7, 2018 Contents 1 Phase transition in Bernoulli percolation 2 2 Everyone’s toolbox 4
Seashore Paspalum Shade Tolerance, Differentiated Instruction Chemistry Lesson Plans, Sophisticated Bohemian Decor, Cross Classic Century Ballpoint Pen Price, How To Open Pontiac Solstice Trunk, Detroit Lake News, Land For Sale Elko, Ga, Deccan Plateau Map, How To Stop Escapism, Custodes Psychic Awakening, Di Potassium Peroxodisulphate, Creation Kit Rotate Camera, Dinhe Traditional Dance Pdf, Stranger Things Quotes Friends, Watering Rings For Trees, Missouri Waterfowl Season 2020-2021, How To Keep A Guy Interested Without Sleeping With Him, Daffodil International University Logo, Webscan Hp Envy 4520, Mercedes Gle Coupe 2019 Price, Dessert Recipes With Eggs And Milk, All Female Characters In Mahabharata In Telugu, 347 Volt Led High Bay Lights, Land Raider Proteus Size, Best Counterfeit Pen, Suncast Lawn Edging Installation, | 708 | 3,218 | {"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-2022-21 | latest | en | 0.909737 |
http://www.onemathematicalcat.org/algebra_book/online_problems/basic_concepts.htm | 1,579,800,521,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00468.warc.gz | 251,673,210 | 9,308 | Basic Concepts Involved in Factoring Trinomials
BASIC CONCEPTS INVOLVED IN FACTORING TRINOMIALS
Here, you will practice the basic concepts involved in factoring trinomials of the form $\,x^2 + bx + c\,$.
These trinomials have an $\,x^2\,$ term with a coefficient of $\,1\,$, an $\,x\,$ term, and a constant term.
Recall that factoring is the process of taking a sum (things added)
and rewriting it as a product (things multiplied).
Observe that for all real numbers $\,f\,$, $\,g\,$ and $\,x\,$: $$(x+f)(x+g) = \overset{\text{First}}{\overbrace{\strut\ x^2\ }} + \overset{\text{Outer}}{\overbrace{\strut\ gx\ }} + \overset{\text{Inner}}{\overbrace{\strut\ fx\ }} + \overset{\text{Last}}{\overbrace{\strut\ fg\ }} = x^2 + (f+g)x + fg$$
Now, think of going ‘backwards’:
from $\,x^2 + (f+g)x + fg\,$
back to the factored form $\,(x+f)(x+g)\,$.
We'd need two numbers that add together to give the coefficient of the $\,x\,$ term,
and that multiply together to give the constant term.
This gives the following result, which is the primary tool used in factoring trinomials:
KEY TOOL FOR FACTORING TRINOMIALS
(where the coefficient of the squared term is $\,1\,$)
To factor a trinomial of the form $\,x^2 + bx + c\,$,
start by finding two numbers, $\,f\,$ and $\,g\,$, that
• add together to give $\,b\,$ (the coefficient of the $\,x\,$ term); and
• multiply together to give $\,c\,$ (the constant term).
Then: $$\,x^2 + bx + c \ \ =\ \ x^2 + (\overset{= b}{\overbrace{f+g}})x + \overset{= c}{\overbrace{\ fg\ }} \ \ =\ \ (x + f)(x + g)\,$$
For example, to factor $\,x^2 + 5x + 6\,$,
we must find two numbers that add to $\,5\,$ and multiply to $\,6\,$.
The numbers $\,2\,$ and $\,3\,$ work, since $\,2+3 = 5\,$ and $2\cdot 3 = 6\,$.
Thus: $$x^2 + 5x + 6 \ \ =\ \ x^2 + (2 + 3)x + (2\cdot 3) \ \ =\ \ (x+2)(x+3)$$
(FOIL it out to check!)
When everything in sight is positive and coefficients are small,
then it may be easy to come up with the ‘numbers that work’.
For example, it may not be too hard for you to find numbers that add to $\,5\,$ and multiply to $\,6\,$.
However, bring some negative numbers into the picture and make coefficients bigger,
and things can get considerably tricker.
Fortunately, there are some key ideas that will help you find the ‘numbers that work’ (if they exist),
and the purpose of this web exercise is to give you practice with these ideas.
KEY IDEAS FOR FINDING THE ‘NUMBERS THAT WORK’:
• If two numbers multiply to give a POSITIVE number,
then they must both be positive or they must both be negative.
That is, two numbers that multiply to a positive number must have the same sign.
• If two numbers multiply to give a NEGATIVE number,
then one must be positive, and the other must be negative.
That is, two numbers that multiply to a negative number must have different signs.
• When you add two numbers that have the same sign,
For example, to mentally compute the sum $\,(-2) + (-5)\,$,
in your head you would compute $\,2 + 5\,$, and then assign a negative sign to your answer.
Think of the number line interpretation of this fact.
Start at zero.
You must walk in only one direction (both numbers have the same sign).
Each time you walk, you're getting farther from zero.
Your final distance from zero is the sum of the individual distances you walk.
• When you add two numbers that have different signs,
For example, to mentally compute $\,-5 + 2\,$,
you would think $\,5 - 2\,$, and then make the answer negative.
Think of the number line interpretation of this fact.
Start at zero.
You must walk in both directions (the numbers have different signs).
So, you'll be doing some back-tracking (some overlapping).
Your final distance from zero is the difference of the individual distances you walk.
• If two numbers have the same sign and their sum is positive,
then both numbers must be positive.
For example, if two numbers have the same sign and add to $\,10\,$,
then they must both be positive.
Think of the number line interpretation of this fact.
Start at zero.
You need to walk in only one direction (both numbers have the same sign).
You need to end up to the right of zero (the sum is positive).
So, you must walk to the right both times (both numbers must be positive).
• If two numbers have the same sign and their sum is negative,
then both numbers must be negative.
For example, if two numbers have the same sign and add to $\,-10\,$,
then they must both be negative.
Think of the number line interpretation of this fact.
Start at zero.
You need to walk in only one direction (both numbers have the same sign).
You need to end up to the left of zero (the sum is negative).
So, you must walk to the left both times (both numbers must be negative).
• If two numbers have different signs and their sum is positive,
then the bigger number must be positive.
(Remember that ‘bigger’ means farther away from zero.)
For example, if two numbers have different signs and add to $\,10\,$,
then the bigger number must be positive.
(Like $\,12 + (-2) = 10\$: the numbers being added are $\,12\,$ and $\,-2\,$;
$12\,$ is bigger because it is farther from zero.)
Think of the number line interpretation of this fact.
Start at zero.
You must walk in both directions (the numbers have different signs).
You need to end up to the right of zero (the sum is positive).
So, you must walk farther to the right (the bigger number must be positive).
• If two numbers have different signs and their sum is negative,
then the bigger number must be negative.
(Remember that ‘bigger’ means farther away from zero.)
For example, if two numbers have different signs and add to $\,-10\,$,
then the bigger number must be negative.
(Like $\,-12 + 2 = -10\$: the numbers being added are $\,-12\,$ and $\,2\,$;
$-12\,$ is bigger because it is farther from zero.)
Think of the number line interpretation of this fact.
Start at zero.
You must walk in both directions (the numbers have different signs).
You need to end up to the left of zero (the sum is negative).
So, you must walk farther to the left (the bigger number must be negative).
EXAMPLES:
Question: Suppose two numbers multiply to $\,36\,$.
Then, the numbers have (choose one):
THE SAME SIGN DIFFERENT SIGNS
Question: Suppose two numbers multiply to $\,-36\,$.
Then, the numbers have (choose one):
THE SAME SIGN DIFFERENT SIGNS
Question: Suppose two numbers have the same sign, and they add to $\,10\,$.
Then, the numbers are (choose one):
BOTH POSITIVE BOTH NEGATIVE
Question: Suppose two numbers have the same sign, and they add to $\,-10\,$.
Then, the numbers are (choose one):
BOTH POSITIVE BOTH NEGATIVE
Question: When you add two numbers that have the same sign,
Question: When you add two numbers that have different signs,
Question: Suppose two numbers have different signs, and they add to $\,10\,$.
Then, the bigger number is (choose one):
POSITIVE NEGATIVE
Question: Suppose two numbers have different signs, and they add to $\,-10\,$.
Then, the bigger number is (choose one):
POSITIVE NEGATIVE
Master the ideas from this section
Factoring Trinomials (coefficient of $\,x^2\,$ term is $\,1\,$) | 1,904 | 7,109 | {"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} | 4.96875 | 5 | CC-MAIN-2020-05 | latest | en | 0.814948 |
http://www.numbersaplenty.com/132000101 | 1,590,567,483,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347392142.20/warc/CC-MAIN-20200527075559-20200527105559-00079.warc.gz | 195,355,484 | 3,467 | Search a number
132000101 is a prime number
BaseRepresentation
bin1111101111000…
…10100101100101
3100012101022002122
413313202211211
5232243000401
621033115325
73161660642
oct767424545
9305338078
10132000101
11685686a2
1238258b45
1321468c8c
1413760dc9
15b8c621b
hex7de2965
132000101 has 2 divisors, whose sum is σ = 132000102. Its totient is φ = 132000100.
The previous prime is 132000091. The next prime is 132000103. The reversal of 132000101 is 101000231.
Adding to 132000101 its reverse (101000231), we get a palindrome (233000332).
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 125104225 + 6895876 = 11185^2 + 2626^2 .
It is an emirp because it is prime and its reverse (101000231) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-132000101 is a prime.
It is a Sophie Germain prime.
Together with 132000103, it forms a pair of twin primes.
It is a Chen prime.
It is a Curzon number.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (132000103) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 66000050 + 66000051.
It is an arithmetic number, because the mean of its divisors is an integer number (66000051).
Almost surely, 2132000101 is an apocalyptic number.
It is an amenable number.
132000101 is a deficient number, since it is larger than the sum of its proper divisors (1).
132000101 is an equidigital number, since it uses as much as digits as its factorization.
132000101 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6, while the sum is 8.
The square root of 132000101 is about 11489.1296885360. The cubic root of 132000101 is about 509.1644668285.
The spelling of 132000101 in words is "one hundred thirty-two million, one hundred one", and thus it is an aban number. | 591 | 1,980 | {"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} | 3.375 | 3 | CC-MAIN-2020-24 | latest | en | 0.868352 |
http://www.johnamcnamara.com/grade7/grade8/mar14.html | 1,660,926,529,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573744.90/warc/CC-MAIN-20220819161440-20220819191440-00249.warc.gz | 74,373,538 | 3,165 | Week of March 14
1. A person spends everything they have in their pockets during a visit to five stores.
In each store, the person spends \$10.00 more than half what they had when they entered the store. How much money did the person have when they set off from home?
2. Rachel has to go on a trip 400 miles from home. She drives 200 miles, using 14 gallons of gasoline per 100 miles. As the price of gas is pretty high, she wants to keep her travel costs down and so is trying to lower her fuel consumption. How many gallons of gas must Rachel use over the remaining 200 miles so that her average fuel consumption for the whole journey works out at 10 gallons per 100 miles?
3. What is behind this sequence of numbers?
7893 1512 10 0
4. On average, 800 hens lay 800 eggs in 8 days. How many eggs will 400 hens lay in 4 days?
5. What number has to come next for this sequence to be logical?
one three five four
seven five four ? | 261 | 963 | {"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-2022-33 | latest | en | 0.956308 |
http://www.cartalk.com/content/perfect-square-dance?answer | 1,464,636,919,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051054181.34/warc/CC-MAIN-20160524005054-00183-ip-10-185-217-139.ec2.internal.warc.gz | 404,598,521 | 10,902 | # The Perfect Square Dance!
RAY: Here's the answer. We know the sum of each pair has to be a perfect square. And we know that 4, 9, 16, and 25 actually are the only perfect squares available to us.
So let's look at dancer 18. Dancer 18 must be paired up with dancer number 7 because that adds up to 25. Dancer 17 must be paired up with dancer number 8. And 16 must be paired up with 9. All those three pairs of dancers have 25 as their total.
So who's left? We have 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 14, and 15. Well let's look at our hostess Sally and her possible dance partners. Remember she's number 1. So she could be dancing with number 3, or number 8. Or she could dance with 15.
But she can't dance with 8 because 8's already been taken by 17. If she chooses 3, then dancer number 6 must pair up with 10. And now there's no one left for dancer number 15, because 15's only choice would have been either 1 or 10.
So the only way it's going to work is that Sally has to dance with 15. 10 gets paired with 6, 13 with 3, and then you can figure out the rest.
TOM: Wow.
RAY: Who's our winner this week?
TOM: The winner this week is Jerry Darmond from Madison, Wisconsin, and for having his answer selected at random from among all the correct answers that we got, Jerry's going to get a \$26 gift certificate to the Shameless Commerce Division at cartalk.com, with which he can get our new book, Ask Click and Clack.
RAY: Congratulations, Jerry!
Support for Car Talk is provided by: | 417 | 1,495 | {"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} | 3.21875 | 3 | CC-MAIN-2016-22 | longest | en | 0.963215 |
http://forum.allaboutcircuits.com/threads/circuit-elements.105753/ | 1,484,570,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00371-ip-10-171-10-70.ec2.internal.warc.gz | 111,427,798 | 17,112 | # circuit elements
Discussion in 'General Electronics Chat' started by kamarul amin, Jan 14, 2015.
1. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
guys,, how can we know the differences between the ideal dependant sources and ideal independant source???
2. ### #12 Expert
Nov 30, 2010
16,655
7,293
I'm a bit blurry about this, too. How can an ideal source be dependent? Great sex but no money?
3. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
hmmm i dont know just learnt this topic about couple days ago but still cant understand it well. this is my first semester undergraduate
4. ### #12 Expert
Nov 30, 2010
16,655
7,293
OK! Good start. Now we know you have some education. If you tell us some more about where these sources come from and what they are about, we can probably get you some serious help.
5. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
hm i'm not really sure about this source but it is the branches for active element. active element is the element that can generate energy. that is what i know so far. hehe
6. ### studiot AAC Fanatic!
Nov 9, 2007
5,005
515
Well there is no mystery here.
Circuits are considered to be assembled from ideal elements for the purpose of analysis.
So the word ideal is really almost redundant.
It just means that, unlike real sources there are no loads that could cause it to fail.
So your question comes down to what is the difference between a dependent source and an independent one?
Well an independent source provides its voltage or current (Direct or Alternating) regardless of the conditions in the rest of the circuit.
A 9volt PP3 battery is a 9volt independent Direct voltage source, for the purposes of most circuit analysis.
A dependent source is a source whose value of current or voltage depends upon the voltage or current somewhere else in the circuit.
So dependent voltage source may supply a voltage equal to say 5 times the voltage at some other point in the circuit.
A dependent current source may supply a (constant) current equal to 3 times the current in some other branch of the circuit.
Last edited: Jan 14, 2015
kamarul amin and #12 like this.
7. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
V1 is an ideal (independent) voltage source which happens to be 2Vpp 10Hz Sine wave offset by +1V that creates V(x). It is independent because we get to specify what it is.
E1 is is an ideal dependent voltage source which producesV(y) which is -3*V(x). It is dependent on the voltage at node x.
Get it?
kamarul amin and #12 like this.
8. ### ErnieM AAC Fanatic!
Apr 24, 2011
7,433
1,623
First off, when you use the term "ideal" you are into the realm of fantasy as you cannot build an "ideal" anything.
Now fantasy may be a useful thing especially when it is a mathematical construct. An ideal source simply means a mathematical model where there are no parasitic parts, such as a voltage source that can put out an infinite current and still maintain the same voltage across itself.
An "ideal independent source" is simply something that supplies (sources) voltage or current and it does not depend on any other parameters. One example is a fixed voltage source, or a fixed current source.
An "ideal dependant source" does depend on some other parameter, such as current source that supplies a current of some multiple of anther current. One popular model of a transistor places a dependent current source between the collector and emitter, where the current is (beta) times the base current.
You know the difference by how the source is written. I would need to see what you are looking at to comment further.
kamarul amin and #12 like this.
9. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
know i get understand it better than before. still having a little bit confuse on dependent source. thank you bro
10. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
yeah you make me understand about them now. hehe. thank you bro
it is said that there are four type dependent source, right??
i use this book ---> fundamentals of electric circuit by Charles K. Alexander/ Mathhew N. O. sadiku (fifth edition)
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11. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
the graph make me nervous lol. i'm dont learnt this graph yet i think, because the lecturer just taught about the dependent and independent source.
12. ### studiot AAC Fanatic!
Nov 9, 2007
5,005
515
To tell an independent source froma dependent one;
An independent current source will be labelled something like 5amps or whatever on the circuit diagram.
An independent sources has a fixed value.
A dependent current source will not have a specified current it will be labelled something like 5i2 so that if i2 changes so does the dependent current so that is always 5 times i2.
A dependent source does not have a fixed value.
Last edited: Jan 14, 2015
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13. ### WBahn Moderator
Mar 31, 2012
18,064
4,904
You have two possible outputs for a dependent source -- voltage or current.
You have two possible controlling inputs for a dependent source -- voltage or current.
You thus have four possible combinations for a dependent source:
Voltage-controlled voltage source (VCVS)
Current-controlled voltage source (CCVS)
Voltage-controlled current source (VCCS)
Current-controlled current source (CCCS)
These are the main types. You could also have models that have different control parameters, such as a light-controlled source or a temperature-controlled source.
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14. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
great!! thats what i saw in the example, some of the values just like 7i,5i, so it is dependent source. now you make me understand it better. thank you bro
15. ### kamarul amin Thread Starter Member
Dec 2, 2014
62
3
great explanation. thank you for helping me | 1,449 | 5,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-04 | latest | en | 0.902086 |
http://www.bbc.co.uk/schools/gcsebitesize/maths/algebra/graphshirev1.shtml | 1,394,293,625,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1393999655040/warc/CC-MAIN-20140305060735-00025-ip-10-183-142-35.ec2.internal.warc.gz | 245,573,248 | 15,651 | Print
Maths
Parallel and perpendicular lines - Higher
Page:
1. Next
On the higher paper you may be asked questions about parallel and perpendicular lines on graphs.
# Parallel and perpendicular lines
On a graph, parallel lines have the same gradient.
For example, y = 2x + 3 and y = 2x - 4 are parallel because they both have a gradient of 2.
Remember that perpendicular lines will always cross at right angles.
In this diagram, the lines y = 2x + 3 and y = -1/2 x -1 cross at right angles.
The gradients of these lines are 2 and -1/2.
The product of the gradients is 2 x -1/2= -1.
You can work out whether 2 lines are perpendicular by multiplying their gradients. The product of the gradient of perpendicular lines will always be -1.
If lines are perpendicular, M1× M2 = − 1
In exams, you will often be asked to find the equation of a line that is perpendicular to a given line. To do this, you will need to work out the gradient of one line before finding the gradient and equation of the other.
## Example
Find the perpendicular line to 4y - 3x = 8 through the point (0, 2).
• Re-arrange the equation 4y - 3x = 8 in the form y = mx + c
• y = 3/4x +2
Now we need to work out the gradient of the 2nd line. Remember that when 2 lines are perpendicular the product of their gradients is -1. Let's call the gradient of the second line m.
• 3/4m = -1
• m = -4/3
In the question we are told that the line passes through the point (0, 2). This means that the line crosses the y axis at +2.
So the equation of the line that is perpendicular to 4y - 3x = 8 is y = - 4/3x + 2
## Finding the gradient of a line between two points
To find the gradient of a line we need to know how many it goes up, for every one across.
### Example
Find the gradient of the line joining (1,3) to (4,9).
As we go from (1,3) to (4,9) the y value increases by 6, and the x value increases by 3. So the line goes 6 up for 3 across. So this line has a gradient of 6/3 = 2.
Use this technique to answer the following question:
Question
Line A goes through the points (4, 9) and (1, 3). Find the perpendicular line through the point (2, 0).
First, find the gradient of line A. We know that the line passes through the points (1, 3) and (4, 9). To calculate the gradient, find the difference in the y coordinates and the difference in the x coordinates, and then divide the y value by the x value.
9 - 3/4 - 1 = 6/3 = 2.
The gradient of line A is 2.
We know that the product of the gradients of perpendicular lines is -1. If we call the gradient of line B m, then:
• m × 2 = -1
• m = -1/2
A straight line always has the equation y= mx + c
m is always the gradient so we know this equation is
y = -1/2x + c
Line B passes through the point (2, 0). To find out the value of c in the equation
y = -1/2x + c
we substitute the values in the equation with x = 2 and y = 0.
• 0 = -1/2 × 2 + c
• 0 = -1 + c
• c = 1
The equation of line B is therefore y = -1/2x + 1
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# Problem 3100. Rumis Scorer 2
Solution 2153937
Submitted on 8 Mar 2020 by Nikolaos Nikolaou
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
clear board board(:,:,1) = ['rrbrbbr';'rbrbbrb';'bbrbbrb';'rbrrrbr';'bbrbbbr';'bbbrrbr']; board(:,:,2) = ['brbbbrb';'rrrbbbr';'bbrbrbr';'brbrbrr';'rbrrrbr';'brbbbbr']; %blue wins 22 to 20 assert(isequal(rumis_scorer2(board),'b'))
winner = 'b'
2 Pass
clear board board(:,:,1) = ['ppbrpbr';'rbrpbrb';'bbrbbrb';'rbrrppr';'bbrpbbr';'bbprrbr']; board(:,:,2) = ['brbpbrb';'rrpbbpr';'pprbrbr';'brprbrr';'rbrrpbr';'brppbbr']; %red wins 18 to 15 (b) to 9 (p) assert(isequal(rumis_scorer2(board),'r'))
winner = 'r'
3 Pass
clear board board(:,:,1) = ['brbrb';'rrrbb';'bggob';'ggoro';'bbrbr']; board(:,:,2) = ['bobrb';'roggr';'bogor';'bgoor';'rgrrr']; board(:,:,3) = ['bobrg';'rrbgr';'rooor';'brbrg';'rgrrr']; board(:,:,4) = ['borbg';'rrbbr';'bbrbr';'bboog';'rgrog']; board(:,:,5) = ['robbr';'bbror';'borbr';'rrobb';'boobg']; %blue wins 10 to 8 (r) to 6 (o) to 1 (g) assert(isequal(rumis_scorer2(board),'b'))
winner = 'b'
4 Pass
clear board board(:,:,1) = ['rrbrryr';'rbrrryb';'brrrbyb';'rbrbyyr';'rbrbryr';'brrrybr']; board(:,:,2) = ['brbybrb';'rbybbbr';'bbrbrbr';'brbryrr';'rbrrbbr';'brbbybr']; board(:,:,3) = ['byyybry';'ybbbyyr';'bbrbyyr';'brbyyrr';'rbyyybr';'brybybr']; %yellow wins 16 to 15 (b) to 11 (r) assert(isequal(rumis_scorer2(board),'y'))
winner = 'y'
5 Pass
clear board board(:,:,1) = ['prbrby';'pprbby';'bggoyb';'gpyoro';'bppbyr']; board(:,:,2) = ['pobrby';'rogygr';'bogoyr';'bgyoor';'rgyyyr']; board(:,:,3) = ['ppbrgy';'rpybgr';'roooyr';'brybrg';'rgryrr']; board(:,:,4) = ['borbgy';'rpbybr';'bbrbry';'bbyoog';'rgrpog']; board(:,:,5) = ['robbry';'bprory';'borbry';'rroybb';'booybg']; board(:,:,6) = ['oobyry';'bpryor';'bopbyr';'rooppb';'boobyg']; %orange wins 8 to 7 (b) to 5 (r,y) to 4 (p) to 1 (g) assert(isequal(rumis_scorer2(board),'o'))
winner = 'o'
6 Pass
ind = randi(5); switch ind case 1 clear board board(:,:,1) = ['rrbrbbr';'rbrbbrb';'bbrbbrb';'rbrrrbr';'bbrbbbr';'bbbrrbr']; board(:,:,2) = ['brbbbrb';'rrrbbbr';'bbrbrbr';'brbrbrr';'rbrrrbr';'brbbbbr']; assert(isequal(rumis_scorer2(board),'b')) case 2 clear board board(:,:,1) = ['ppbrpbr';'rbrpbrb';'bbrbbrb';'rbrrppr';'bbrpbbr';'bbprrbr']; board(:,:,2) = ['brbpbrb';'rrpbbpr';'pprbrbr';'brprbrr';'rbrrpbr';'brppbbr']; assert(isequal(rumis_scorer2(board),'r')) case 3 clear board board(:,:,1) = ['brbrb';'rrrbb';'bggob';'ggoro';'bbrbr']; board(:,:,2) = ['bobrb';'roggr';'bogor';'bgoor';'rgrrr']; board(:,:,3) = ['bobrg';'rrbgr';'rooor';'brbrg';'rgrrr']; board(:,:,4) = ['borbg';'rrbbr';'bbrbr';'bboog';'rgrog']; board(:,:,5) = ['robbr';'bbror';'borbr';'rrobb';'boobg']; assert(isequal(rumis_scorer2(board),'b')) case 4 clear board board(:,:,1) = ['rrbrryr';'rbrrryb';'brrrbyb';'rbrbyyr';'rbrbryr';'brrrybr']; board(:,:,2) = ['brbybrb';'rbybbbr';'bbrbrbr';'brbryrr';'rbrrbbr';'brbbybr']; board(:,:,3) = ['byyybry';'ybbbyyr';'bbrbyyr';'brbyyrr';'rbyyybr';'brybybr']; assert(isequal(rumis_scorer2(board),'y')) case 5 clear board board(:,:,1) = ['prbrby';'pprbby';'bggoyb';'gpyoro';'bppbyr']; board(:,:,2) = ['pobrby';'rogygr';'bogoyr';'bgyoor';'rgyyyr']; board(:,:,3) = ['ppbrgy';'rpybgr';'roooyr';'brybrg';'rgryrr']; board(:,:,4) = ['borbgy';'rpbybr';'bbrbry';'bbyoog';'rgrpog']; board(:,:,5) = ['robbry';'bprory';'borbry';'rroybb';'booybg']; board(:,:,6) = ['oobyry';'bpryor';'bopbyr';'rooppb';'boobyg']; assert(isequal(rumis_scorer2(board),'o')) end
winner = 'b'
7 Pass
ind = randi(5); switch ind case 1 clear board board(:,:,1) = ['rrbrbbr';'rbrbbrb';'bbrbbrb';'rbrrrbr';'bbrbbbr';'bbbrrbr']; board(:,:,2) = ['brbbbrb';'rrrbbbr';'bbrbrbr';'brbrbrr';'rbrrrbr';'brbbbbr']; assert(isequal(rumis_scorer2(board),'b')) case 2 clear board board(:,:,1) = ['ppbrpbr';'rbrpbrb';'bbrbbrb';'rbrrppr';'bbrpbbr';'bbprrbr']; board(:,:,2) = ['brbpbrb';'rrpbbpr';'pprbrbr';'brprbrr';'rbrrpbr';'brppbbr']; assert(isequal(rumis_scorer2(board),'r')) case 3 clear board board(:,:,1) = ['brbrb';'rrrbb';'bggob';'ggoro';'bbrbr']; board(:,:,2) = ['bobrb';'roggr';'bogor';'bgoor';'rgrrr']; board(:,:,3) = ['bobrg';'rrbgr';'rooor';'brbrg';'rgrrr']; board(:,:,4) = ['borbg';'rrbbr';'bbrbr';'bboog';'rgrog']; board(:,:,5) = ['robbr';'bbror';'borbr';'rrobb';'boobg']; assert(isequal(rumis_scorer2(board),'b')) case 4 clear board board(:,:,1) = ['rrbrryr';'rbrrryb';'brrrbyb';'rbrbyyr';'rbrbryr';'brrrybr']; board(:,:,2) = ['brbybrb';'rbybbbr';'bbrbrbr';'brbryrr';'rbrrbbr';'brbbybr']; board(:,:,3) = ['byyybry';'ybbbyyr';'bbrbyyr';'brbyyrr';'rbyyybr';'brybybr']; assert(isequal(rumis_scorer2(board),'y')) case 5 clear board board(:,:,1) = ['prbrby';'pprbby';'bggoyb';'gpyoro';'bppbyr']; board(:,:,2) = ['pobrby';'rogygr';'bogoyr';'bgyoor';'rgyyyr']; board(:,:,3) = ['ppbrgy';'rpybgr';'roooyr';'brybrg';'rgryrr']; board(:,:,4) = ['borbgy';'rpbybr';'bbrbry';'bbyoog';'rgrpog']; board(:,:,5) = ['robbry';'bprory';'borbry';'rroybb';'booybg']; board(:,:,6) = ['oobyry';'bpryor';'bopbyr';'rooppb';'boobyg']; assert(isequal(rumis_scorer2(board),'o')) end
winner = 'b'
8 Pass
ind = randi(5); switch ind case 1 clear board board(:,:,1) = ['rrbrbbr';'rbrbbrb';'bbrbbrb';'rbrrrbr';'bbrbbbr';'bbbrrbr']; board(:,:,2) = ['brbbbrb';'rrrbbbr';'bbrbrbr';'brbrbrr';'rbrrrbr';'brbbbbr']; assert(isequal(rumis_scorer2(board),'b')) case 2 clear board board(:,:,1) = ['ppbrpbr';'rbrpbrb';'bbrbbrb';'rbrrppr';'bbrpbbr';'bbprrbr']; board(:,:,2) = ['brbpbrb';'rrpbbpr';'pprbrbr';'brprbrr';'rbrrpbr';'brppbbr']; assert(isequal(rumis_scorer2(board),'r')) case 3 clear board board(:,:,1) = ['brbrb';'rrrbb';'bggob';'ggoro';'bbrbr']; board(:,:,2) = ['bobrb';'roggr';'bogor';'bgoor';'rgrrr']; board(:,:,3) = ['bobrg';'rrbgr';'rooor';'brbrg';'rgrrr']; board(:,:,4) = ['borbg';'rrbbr';'bbrbr';'bboog';'rgrog']; board(:,:,5) = ['robbr';'bbror';'borbr';'rrobb';'boobg']; assert(isequal(rumis_scorer2(board),'b')) case 4 clear board board(:,:,1) = ['rrbrryr';'rbrrryb';'brrrbyb';'rbrbyyr';'rbrbryr';'brrrybr']; board(:,:,2) = ['brbybrb';'rbybbbr';'bbrbrbr';'brbryrr';'rbrrbbr';'brbbybr']; board(:,:,3) = ['byyybry';'ybbbyyr';'bbrbyyr';'brbyyrr';'rbyyybr';'brybybr']; assert(isequal(rumis_scorer2(board),'y')) case 5 clear board board(:,:,1) = ['prbrby';'pprbby';'bggoyb';'gpyoro';'bppbyr']; board(:,:,2) = ['pobrby';'rogygr';'bogoyr';'bgyoor';'rgyyyr']; board(:,:,3) = ['ppbrgy';'rpybgr';'roooyr';'brybrg';'rgryrr']; board(:,:,4) = ['borbgy';'rpbybr';'bbrbry';'bbyoog';'rgrpog']; board(:,:,5) = ['robbry';'bprory';'borbry';'rroybb';'booybg']; board(:,:,6) = ['oobyry';'bpryor';'bopbyr';'rooppb';'boobyg']; assert(isequal(rumis_scorer2(board),'o')) end
winner = 'b' | 3,140 | 6,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2020-40 | latest | en | 0.31588 |
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Suppose we are minimizing $f(x)$. The first order necessary condition of $x^*$ being local minmum is: $$\nabla f(x^*)= \mathbf{0}.$$ For sufficiency, we check if also $\nabla^2f(x^*) \succ 0$, i.e., ... | 468 | 1,915 | {"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} | 2.90625 | 3 | CC-MAIN-2021-39 | latest | en | 0.863405 |
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# Least Common Multiples and Greatest Common Factors - PowerPoint PPT Presentation
Least Common Multiples and Greatest Common Factors. Least Common Multiple (LCM). The least common multiple is the smallest number that is common between two lists of multiples. EXAMPLE: Find the LCM of 12 and 18. The multiples of 12: 12 x 1 = 12 12 x 2 =24 12 x 3 = 36 12 x 4 = 48
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Presentation Transcript
### Least Common Multiples and Greatest Common Factors
• The least common multiple is the smallest number that is common between two lists of multiples.
EXAMPLE: Find the LCM of 12 and 18
• The multiples of 12:
• 12 x 1 = 12
• 12 x 2 =24
• 12 x 3 = 36
• 12 x 4 = 48
• 12 x 5 =60
• The multiples of 18:
• 18 x 1 = 18
• 18 x 2 = 36
• 18 x 3 = 54
• 18 x 4 = 72
• 18 x 5 = 90
18, 36, 54, 72, 90
The first number you see in both lists is 36.
The least common multiple of 12 and 18 is 36.
Example 2: Find the LCM of 9 and 10
9, 18, 27, 36, 45, 54, 63, 72
81, 90, 99
10, 20, 30, 40, 50, 60, 70, 80
90, 100, 110
If you don’t see a common multiple, make each list go further.
The LCM of 9 and 10 is 90
Example 3:Find the LCM of 4 and 12
4, 8, 12, 16
12, 24, 36
Example 4:Find the LCM of 5 and 8
5, 10, 15, 20, 25, 30
35, 40
8, 16, 24, 32, 40, 48
Example 5:Find the LCM of 6 and 20
6, 12, 18, 24, 30, 36
42, 48, 54, 60
20, 40, 60, 80, 100, 120
• The greatest common factor is the largest factor that two numbers share.
• Let’s find the GCF of 12 and 42. First, we need to make a list of factors for each number.
42
1 x 12
1 x 42
Factors of 12:
1, 2, 3, 4, 6,12
2 x 21
2 x 6
3 x 14
3 x 4
4 x ??
4 x 3
Factors of 42:
1, 2, 3, 6, 7, 14, 21, 42
5 x ??
6 x 7
7 x 6
Common Factors: 1, 2, 3, 6
Greatest Common Factor: 6
18
27
Factors of 18:
1, 2, 3, 6, 9, 18
1 x 18
1 x 27
2 x ?
2 x 9
3 x 9
3 x 6
Factors of 27:
1, 3, 9, 27
4 x ?
5 x ?
4 x ?
6 x ?
5 x ?
7 x ?
Common Factors: 1, 3, 9
8 x ?
6 x 3
9 x 3
GCF: 9
Factors of 48:
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
48
60
1 x 48
1 x 60
2 x 30
2 x 24
3 x 20
3 x 16
4 x 15
Factors of 60:
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
4 x 12
5 x 12
6 x 8
6 x 10
Common Factors: 1, 2, 3, 4, 6, 12
GCF: 12 | 1,163 | 2,873 | {"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} | 4.75 | 5 | CC-MAIN-2017-39 | latest | en | 0.848427 |
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Apr 18 comment How to prove that $\binom{X+L-1}{L-1} \ge (X-L\times N)^{L-1}$? Take $X = 0$ to get a counterexample when $L$ is odd Mar 28 awarded Popular Question Jan 23 answered If $\mathbb{RP}^n$ can be immersed in $\mathbb{R}^{n+1}$, how do I see that $n$ must be of the form $2^r - 1$ or $2^r - 2$? Jan 18 revised Find all triples of non-negative Integers $a,b,c$ such that $a!b!=a!+b!+c!$ deleted 7 characters in body Jan 18 answered Find all triples of non-negative Integers $a,b,c$ such that $a!b!=a!+b!+c!$ Jan 16 awarded Yearling Jan 14 comment A subset of $[0,1]\times[0,1]$ containing at most one point from each horizontal and vertical section whose boundary is $[0,1]\times[0,1]$ the map $\pi: \mathbb{R}^{2} \to [0,1)\times [0,1)$ by $\pi(x,y) = (\{x\},\{y\}).$ Jan 13 answered Determinant of the following $(n\times n)$ matrix Jan 13 answered How to proof that two lines in cube are perpendicular, without use of vectors Jan 13 comment A subset of $[0,1]\times[0,1]$ containing at most one point from each horizontal and vertical section whose boundary is $[0,1]\times[0,1]$ You really only need to show that the projection of a line with irrational slope is dense in the unit square, which follows from a standard argument. Jan 13 answered Finding the number of sequences with $0 \leq a_m \leq 3m$ Jan 13 comment Geometrical proof by induction gave a proof, then looked at answers; since David seems to have chosen explicitly not to give a full solution, I removed mine Jan 8 comment Tiling a Rectangle with integer length horizontal/vertical strips @JMoravitz - Yes, I've seen this result before and the article you linked, but I'm not sure how it would help. Thanks, though Jan 8 comment Tiling a Rectangle with integer length horizontal/vertical strips @ArchisWelankar - no, it holds in any case. Jan 8 asked Tiling a Rectangle with integer length horizontal/vertical strips Jun 22 revised Diagonals of $2n$-gon bisecting area implies? edited tags Jun 21 asked Diagonals of $2n$-gon bisecting area implies? Jun 18 awarded Tumbleweed Jun 17 comment Fundamental polygon square $abab$ He identified all of the vertices here, though, so it should only have one vertex. Also I lied about the homology; it has the same homology as that of the klein bottle Jun 17 comment Fundamental polygon square $abab$ you could look at the euler characteristic, which is zero, so if it's actually a surface, then it would be either the klein bottle or torus. unless i'm mistaken, calculating the homology (using the cellular boundary maps) makes it clear that this isn't the case, so it's not a surface. | 747 | 2,661 | {"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} | 3.484375 | 3 | CC-MAIN-2016-18 | latest | en | 0.868795 |
https://www.liveabout.com/are-edge-balls-out-of-bounds-3174114 | 1,653,755,468,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016949.77/warc/CC-MAIN-20220528154416-20220528184416-00202.warc.gz | 967,904,711 | 40,880 | # Are Edge Balls Out of Bounds?
## In or Out?
N writes:
Greg,
I searched a question on google to determine who gets the point when a ball "deflects" from the table by hitting an edge. I believe that you are incorrect in asserting that either player can receive the point based on the circumstances. I assert that the player who hits the ball resulting in the deflection NEVER receives the point.
The official rules defining the table state:
The playing surface shall be uniformly dark colored and matt, but with a white side line, 2cm wide, along each 2.74m edge and a white end line, 2cm wide, along each 1.525m edge.
These lines are referred to as "boundary lines" and therefore any ball landing outside and beyond their limits are "out of bounds". Since the boundary lines are intentionally drawn a finite distance away from the edge of the table, any ball which can deflect due to striking the edge is out of bounds and the point should go to the receiving player.
Hi N,
Thanks for sharing your opinion - and while I do agree that your theory would make the process of judging edge balls much easier, I am afraid that I still do not believe that it is correct.
I've never come across the argument that the perimeter lines are drawn intentionally a little distance away from the edge of the table. The ITTF Technical Leaflet regarding tables only mentions that there shall be 20mm lines around the perimeter of the playing surface to ensure that its limits are clearly visible, with a tolerance on width of all lines of +- 1mm. This may be the cause of any gaps that you see. This is on page 7 of the leaflet, which you can find on the ITTF website (this is a .pdf file).
Furthermore, on page 15 of the ITTF Handbook for Match Officials (this is also a .pdf file), clear mention is made of the procedure for dealing with edge balls. As you can see, the ITTF do not agree with your intepretation, since they give guidelines on how to determine whether an edge ball is the server or receiver's point.
12.2 Edge Balls
12.2.1 It is necessary to decide whether a ball which touches the edge of the table makes contact on or below the playing surface, and the path of the ball before and after it touches the table can help the umpire or assistant umpire to arrive at the correct decision. If the ball first passed over the playing surface the return is good, but if it touched while it is still rising from below the level of the playing surface it almost certainly touched the side.
12.2.2 The main difficulty arises when a ball arrives from outside, and above the level of, the playing surface, and here the best guide is the direction of the ball after contact with the table. There is no infallible guide but, if, after touching the edge, the ball travels upwards, it is reasonable to assume that it touched the playing surface but, if it continues downwards, it is more likely to have touched the side.
12.2.3 The assistant umpire is solely responsible for edge ball decisions at the side of the table nearest to him. If he believes that the ball touched the side he should call “side”, and the umpire must award a point to the opponent(s) of the last striker. Only the umpire can decide on edge balls at the ends and at the side nearest to him.
So although I do think that your method would simply edge balls to a large degree, I don't think that is how the ITTF intend edge balls to be handled.
Greg | 750 | 3,420 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-21 | longest | en | 0.942921 |
http://www.studymode.com/essays/Quantitative-Techniques-530995.html | 1,490,686,241,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189680.78/warc/CC-MAIN-20170322212949-00050-ip-10-233-31-227.ec2.internal.warc.gz | 718,007,781 | 19,262 | # Quantitative Techniques
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Assignment No. 1
QUANTITATIVE TECHNIQUES (5564)
Executive MBA/MPA (Col)
ZAHID NAZIR
Roll.No. AB523655 Semester:Autumn 2008
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Question 1
a). Explain the use of Quantitative Techniques in Business and Management? Marks: 10 b). What are limitations of Statistics? Marks: 10
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a). QUATITATIVE TECHNIQUES
Quantitative techniques refers to the group of statistical and operations research techniques. All these techniques require preliminary knowledge of certain topics in mathematics. Quantitative Techniques
Statistical Techniques Techniques
Operations Research
USE OF QUANTITATIVE TECHNIQUES IN BUSINESS AND MANAGMENT
Due to increasing complexity in business and industry, decision making based on intuition has become highly questionable especially when the decision involves the choice among several courses of action each of which can achieve several management actions. So there is need for training the people who can manage a system efficiently and creatively. Quantitative Techniques now have a major role in effective decision making in various functional areas of management i.e. marketing, finance, production and personnel. These techniques are also widely used in planning, transportation, public health, communication, military, agriculture etc. Quantitative techniques are also used extensively as an aid in business decision making. Some of the areas where quantitative techniques can be used are:
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MANAGEMENT i). Marketing • Analysis of marketing research information • Statistical records for building and maintaining an extensive market • Sales forecasting • • • • iii). • • • • • iv). • • • • ECONOMICS • Measurement of gross national product and input-output analysis • Determination of business cycle, long term growth and seasonal fluctuations • Comparison of market prices, cost and profits of individual firms • Analysis of population, land economies and economic geography Page 4 of 29
ii).
Production Production planning, control and analysis Evaluation of machine performance Quality control requirement (to analyze the data/trends) Inventory control measures Finance, Accounting and Investment Financial forecast, budget preparation Financial investment decisions Selection of securities Auditing function Credit policies, credit risk and delinquent accounts Personnel Labour turn over rate Employment trends Performance appraisal Wage rates and incentive plans
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• Operational studies of public utilities • Formulation of appropriate economic policies and evaluation of their effect RESEARCH & DEVELOPMENT • Development of new product lines • Optimal use of resources • Evaluation of existing products NATURAL SCIENCE • Diagnosis of disease based on data like temperature, pulse rate, blood pressure etc. • Judging the efficiency of a particular drug for curing a certain disease • Study of plant life In the competitive and dynamic business world, firms/companies who likes to succeed and survive are those which are capable of maximizing the use of tools of management like quantitative techniques.
b) LIMITATIONS OF STATISTICS
Statistics with all its wide application in every sphere of human activity has its own limitations. Some of them are given below. 1. STATISTICS IS NOT SUITABLE TO THE STUDY OF QUALITATIVE PHENOMENON: Since statistics is basically a science and deals with a set of numerical data, it is applicable to the study of only these subjects of enquiry, which can be expressed in terms of quantitative measurements. As a matter of fact, qualitative phenomenon like honesty, poverty, beauty, intelligence etc, cannot be expressed numerically and any Page 5 of 29
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statistical analysis cannot be directly applied on these qualitative phenomenons. Nevertheless, statistical techniques may be applied indirectly by first reducing the qualitative expressions to accurate quantitative terms. For... | 797 | 4,043 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-13 | latest | en | 0.867945 |
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# Abc costing
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### Abc costing
1. 1. Activity-Based Costing and Activity-Based Management
2. 2. Learning Objective 1 <ul><li>Explain undercosting </li></ul><ul><li>and overcosting of </li></ul><ul><li>products and services. </li></ul>
3. 3. Undercosting and Overcosting Example Tariq, Nasir, and Hamid order separate items for lunch. Tariq’s order amounts to Rs 14 Nasir consumed 30 Hamid’s order is 16 Total Rs 60 What is the average cost per lunch?
4. 4. Undercosting and Overcosting Example Rs 60 ÷ 3 = Rs 20 Tariq and Hamid are overcosted. Nasir is undercosted.
5. 5. Learning Objective 2 <ul><li>Present three guidelines for </li></ul><ul><li>refining a costing system. </li></ul>
6. 6. Existing Single Indirect- Cost Pool System Example Super Corporation manufactures a normal lens (NL) and a complex lens (CL). Super currently uses a single indirect-cost rate job costing system. Cost objects: 80,000 (NL) and 20,000 (CL).
7. 7. Existing Single Indirect- Cost Pool System Example Normal Lenses (NL) Direct materials Rs 1,520,000 Direct mfg. labor 800,000 Total direct costs Rs 2,320,000 Direct cost per unit: Rs 2,320,000 ÷ 80,000 = Rs29
8. 8. Existing Single Indirect- Cost Pool System Example Complex Lenses (CL) Direct materials Rs 920,000 Direct mfg. labor 260,000 Total direct costs Rs1,180,000 Direct cost per unit: Rs 1,180,000 ÷ 20,000 = Rs 59
9. 9. Existing Single Indirect- Cost Pool System Example All Indirect Costs Rs 2,900,000 50,000 Direct Manufacturing Labor-Hours INDIRECT-COST POLL INDIRECT COST-ALLOCATION BASE Rs 58 per Direct Manufacturing Labor-Hour
10. 10. Existing Single Indirect- Cost Pool System Example Indirect Costs Direct Costs COST OBJECT: NL AND CL LENSES DIRECT COSTS Direct Materials Direct Manufacturing Labor
11. 11. Existing Single Indirect- Cost Pool System Example Super uses 36,000 direct manufacturing labor-hours to make NL and 14,000 direct manufacturing labor-hours to make CL. How much indirect costs are allocated to each product?
12. 12. Existing Single Indirect- Cost Pool System Example NL: 36,000 × Rs 58 = Rs 2,088,000 CL: 14,000 × Rs 58 = Rs 812,000 What is the total cost of normal lenses? Direct costs Rs 2,320,000 + Allocated costs Rs 2,088,000 = Rs 4,408,000 What is the cost per unit? Rs 4,408,000 ÷ 80,000 = Rs 55.10
13. 13. Existing Single Indirect- Cost Pool System Example What is the total cost of complex lenses? Direct costs Rs 1,180,000 + Allocated costs Rs 812,000 = Rs 1,992,000 What is the cost per unit? Rs 1,992,000 ÷ 20,000 = Rs 99.60
14. 14. Existing Single Indirect- Cost Pool System Example Normal lenses sell for Rs 60 each and complex lenses for Rs 142 each. Normal Complex Revenue Rs 60.00 Rs 142.00 Cost 55.10 99.60 Income Rs 4.90 Rs 42.40 Margin 8.2% 29.9%
15. 15. Refining a Costing System Direct-cost tracing Indirect-cost pools Cost-allocation basis
16. 16. Refining a Costing System 1. Design of Products and Process The Design Department designs the molds and defines processes needed (details of the manufacturing operations).
17. 17. Refining a Costing System 2. Manufacturing Operations Lenses are molded, finished, cleaned, and inspected. 3. Shipping and Distribution Finished lenses are packed and sent to the various customers.
18. 18. Learning Objective 3 <ul><li>Distinguish between the </li></ul><ul><li>traditional and the </li></ul><ul><li>activity-based costing </li></ul><ul><li>approaches to designing </li></ul><ul><li>a costing system. </li></ul>
19. 19. Activity-Based Costing System Fundamental Cost Objects Activities Costs of Activities Assignment to Other Cost Objects <ul><li>Cost of: </li></ul><ul><li>Product </li></ul><ul><li>Service </li></ul><ul><li>Customer </li></ul>
20. 20. Activity-Based Costing System A cross-functional team at Super Corporation identified key activities: Design products and processes. Set up molding machine. Operate machines to manufacture lenses. Maintain and clean the molds.
21. 21. Activity-Based Costing System Set up batches of finished lenses for shipment. Distribute lenses to customers. Administer and manage all processes.
22. 22. Activity-Based Costing System No. of Setup Hours Lenses NL Lenses CL Lenses Other Cost Allocation Base Product Cost Objects No. of Shipments Parts- Square feet Setup Design Shipping Activity Indirect Cost Pool
23. 23. Activity-Based Costing System NL CL Quantity produced 80,000 20,000 No. produced/batch 250 50 Number of batches 320 400 Setup time per batch 2 hours 5 hours Total setup-hours 640 2,000 Total setup costs are Rs 409,200.
24. 24. Activity-Based Costing System What is the setup cost per setup-hour? Rs 409,200 ÷ 2,640 hours = Rs 155 What is the setup cost per direct manufacturing labor-hour? Rs 409,200 ÷ 50,000 = Rs 8.184
25. 25. Activity-Based Costing System Allocation using direct labor-hours: NL: Rs 8.184 × 36,000 = Rs 294,624 CL: Rs8.184 × 14,000 = Rs114,576 Total Rs409,200 Allocation using setup-hours: NL: Rs155 × 640 = Rs 99,200 CL: Rs155 × 2,000 = Rs310,000 Total Rs409,200
26. 26. Learning Objective 4 <ul><li>Describe a four-part </li></ul><ul><li>cost hierarchy. </li></ul>
27. 27. Cost Hierarchies A cost hierarchy is a categorization of costs into different cost pools. Cost drivers bases (cost-allocation bases) Degrees of difficulty in determining cause-and-effect relationships
28. 28. Cost Hierarchies ABC systems commonly use a four-part cost hierarchy to identify cost-allocation bases: 1. Output unit-level costs 2. Batch-level costs 3. Product-sustaining costs 4. Facility-sustaining costs
29. 29. Output Unit-Level Costs These are resources sacrificed on activities performed on each individual unit of product or service. Energy Machine depreciation Repairs
30. 30. Batch-Level Costs These are resources sacrificed on activities that are related to a group of units of product(s) or service(s) rather than to each individual unit of product or service. Setup-hours Procurement costs
31. 31. Product-Sustaining Costs These are often called service-sustaining costs and are resources sacrificed on activities undertaken to support individual products or services. Design costs Engineering costs
32. 32. Facility-Sustaining Costs These are resources sacrificed on activities that cannot be traced to individual products or services but support the organization as a whole. General administration – rent – building security
33. 33. Learning Objective 5 <ul><li>Cost products or services using </li></ul><ul><li>activity-based costing. </li></ul>
34. 34. Implementing Activity-Based Costing Identify cost objects. NL CL Identify the direct costs of the products. Direct material Direct labor Mold cleaning and maintenance Step 1 Step 2
35. 35. Implementing Activity-Based Costing Cleaning and maintenance costs of Rs360,000 are direct batch-level costs. Why? Because these costs consist of workers’ wages for cleaning molds after each batch of lenses is run.
36. 36. Implementing Activity-Based Costing <ul><li>Normal Lenses (NL) </li></ul><ul><li>Cost Hierarchy </li></ul><ul><li>Description Category </li></ul><ul><li>Direct materials Unit-level Rs1,520,000 </li></ul><ul><li>Direct mfg. labor Unit-level 800,000 </li></ul><ul><li>Cleaning and maint. Batch-level 160,000 </li></ul><ul><li>Total direct costs Rs2,480,000 </li></ul>
37. 37. Implementing Activity-Based Costing <ul><li>Complex Lenses (CL) </li></ul><ul><li>Cost Hierarchy </li></ul><ul><li>Description Category </li></ul><ul><li>Direct materials Unit-level Rs 920,000 </li></ul><ul><li>Direct mfg. labor Unit-level 260,000 </li></ul><ul><li>Cleaning and maint. Batch-level 200,000 </li></ul><ul><li>Total direct costs Rs1,380,000 </li></ul>
38. 38. Implementing Activity-Based Costing Select the cost-allocation bases to use for allocating indirect costs to the products. (1) (2) (3) Activity Cost Hierarchy Total Costs Design Product-sustaining Rs450,000 Setups Batch-level Rs409,200 Operations Unit-level Rs637,500 Step 3
39. 39. Implementing Activity-Based Costing Identify the indirect costs associated with each cost-allocation base . Overhead costs incurred are assigned to activities, to the extent possible, on the basis of a cause-and-effect relationship. Step 4
40. 40. Implementing Activity-Based Costing Compute the rate per unit. (1) (5) NL CL Total Setup-hours: 640 2,000 2,640 Step 5 Rs409,200 ÷ 2,640 = Rs155
41. 41. Implementing Activity-Based Costing Compute the indirect costs allocated to the products. NL: Rs155 × 640 = Rs 99,200 CL: Rs155 × 2,000 = 310,000 Total Rs409,200 Step 6
42. 42. Implementing Activity-Based Costing Compute the costs of the products. NL and CL would show three direct cost categories. Step 7 1. Direct materials 2. Direct manufacturing labor 3. Cleaning and maintenance
43. 43. Implementing Activity-Based Costing NL and CL would show six indirect cost pools. 1. Design 2. Molding machine setups 3. Manufacturing operations 4. Shipment setup 5. Distribution 6. Administration
44. 44. Learning Objective 6 <ul><li>Use activity-based </li></ul><ul><li>costing systems for </li></ul><ul><li>activity-based management. </li></ul>
45. 45. Activity-Based Management ABM describes management decisions that use activity-based costing information to satisfy customers and improve profits. Product pricing and mix decisions Cost reduction and process improvement decisions Design decisions
46. 46. Product Pricing and Mix Decisions ABC gives management insight into the cost structures for making and selling diverse products. It provides more accurate product cost information and more detailed information on costs of activities and the drivers of those costs.
47. 47. Cost Reduction and Process Improvement Decisions Manufacturing and distribution personnel use ABC systems to focus on cost-reduction efforts. Managers set cost-reduction targets in terms of reducing the cost per unit of the cost-allocation base.
48. 48. Design Decisions Management can identify and evaluate new designs to improve performance by evaluating how product and process designs affect activities and costs. Companies can work with their customers to evaluate the costs and prices of alternative designs.
49. 49. Learning Objective 7 <ul><li>Compare activity-based costing </li></ul><ul><li>systems and department- </li></ul><ul><li>costing systems. </li></ul>
50. 50. ABC and Department Indirect-Cost Rates Many companies have evolved their costing system from using a single cost pool to using separate indirect-cost rates for each department: Design Manufacturing Distribution
51. 51. ABC and Department Indirect-Cost Rates Why? Because the cost drivers of resources in each department or sub department differ from the single, company-wide, cost-allocation base. ABC systems are a further refinement of department costing systems.
52. 52. Learning Objective 8 <ul><li>Evaluate the costs and benefits </li></ul><ul><li>of implementing activity-based </li></ul><ul><li>costing systems. </li></ul>
53. 53. Benefits of ABC Systems Significant amounts of indirect costs are allocated using only one or two cost pools. All or most costs are identified as output unit-level costs. Products make diverse demands on resources because of differences in volume, process steps, batch size, or complexity.
54. 54. Benefits of ABC Systems Products that a company is well-suited to make and sell show small profits while products for which a company is less suited show large profits. Complex products appear to be very profitable and simple products appear to be losing money.
55. 55. Benefits of ABC Systems Operations staff have significant disagreements with the accounting staff about the costs of manufacturing and marketing products and services.
56. 56. Limitations of ABC Systems The main limitations of ABC are the measurements necessary to implement the system. ABC systems require management to estimate costs of activity pools and to identify and measure cost drivers for these pools.
57. 57. Limitations of ABC Systems Activity-cost rates also need to be updated regularly. Very detailed ABC systems are costly to operate and difficult to understand.
58. 58. ABC In Service and Merchandising Companies The general approach to ABC in the service and merchandising areas is very similar to the approach in manufacturing. Costs are divided into homogeneous cost pools and classified as output unit-level, batch-level, product- or service-sustaining, and facility-sustaining costs.
59. 59. ABC In Service and Merchandising Companies The cost pools correspond to key activities. Costs are allocated to products or customers using activity drivers or cost-allocation bases that have a cause-and-effect relationship with the cost in the cost pool.
60. 60. ATTENTION STUDENTS JOIN KHALID AZIZ FOR ACCOUNTING OF ICMAP 1,2,3..CA.. B,C & D PIPFA,MBA,BBA B.COM,M.COM MA-ECONOMICS(STATS) CONTACT:0322-3385752
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https://www.integers.co/questions-answers/what-is-the-place-value-chart-for-the-number-2.html | 1,623,495,374,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487582767.0/warc/CC-MAIN-20210612103920-20210612133920-00345.warc.gz | 748,984,731 | 4,622 | # Q: What is the place value chart for the number 2?
A:
Ones
2
Here are the values of each number:
Number x Multiplier = Value
2x12
Add up all of the values and you get 2.
## What is a place value chart?
A place value chart is used to help understand the value of each digit based on the place or position.
### Billions to Hundred Thousandths Place Value Chart
The number 2 is written as two.
### More Examples
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### Explore more about the number 2:
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## Problem Statement
#### For example:
``````Let’s say the array ‘A’ = [1, 3, 1, 5], then if we select index ‘2’, the sum of the prefix is ‘1’, and the sum of the suffix is 1 + 5 = 6. Since the sum is not the same, hence index ‘2’ is not a beautiful index.
``````
Detailed explanation ( Input/output format, Notes, Constraints, Images )
##### Sample Input-1
``````2
3
1 1 1
3
1 2 3
``````
##### Sample Output-1
``````2
-1
``````
##### Explanation for Sample Input 1:
``````For test case 1:
Index ‘2’ is the leftmost beautiful index. The left sum is 1 and the right sum is also 1.
For test case 2:
No index is beautiful.
``````
##### Sample Input -2
``````2
6
1 7 3 6 5 6
3
2 1 -1
``````
##### Sample Output -2
``````4
1
``````
Auto
Console | 492 | 1,599 | {"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-2023-06 | latest | en | 0.775258 |
https://www.imaginary.org/program/the-sphere-of-the-earth | 1,627,667,814,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00314.warc.gz | 874,350,451 | 11,657 | # The Sphere of the Earth
program
Program (zip, Windows installer). See below documentation and other platforms
## Credits
This program has been developed by Daniel Ramos, MMACA (Museu de Matemàtiques de Catalunya). "An Album of Map Projections", 1989, Snyder, John P.; Voxland, Philip M. Credit: U.S. Geological Survey Department of the Interior/USGS
## Contributors
Programming and DesignDaniel Ramos
This exhibit explores the science of cartography and the geometry of the sphere. The geometric properties of the sphere and the plane are essentially different, and no map can faithfully represent the Earth without distortion. This module goes through some of these properties, comparing different map projections and trying to get a feeling of what “distortion” means and why is there such impossibility for a “perfect map”.
Studying the problem of representing the spherical surface of the Earth onto a flat map is the subject of cartography, and has been an important mathematical problem along History (navigation, position, frontiers, land ownership…). An essential theorem in Geometry (Gauss’ Egregium theorem) ensures that there is no perfect map, that is, there is no way of representing the Earth keeping distances at scale. However, this is exactly what makes cartography a discipline: developing several different maps that try to solve well enough the problem of representing the Earth.
We present six different map projections that are to be compared with a physical globe. Although all maps are created at nominal scale 1:1 of the globe, the distortion is apparent. We propose several activities with the maps and some tools that explore the properties of each map.
On a second part, we use the program “The sphere of the Earth”, that displays the Tissot indicatrix for each map. This indicatrix is a mathematical graphical tool that helps to understand the inherent distortion of a map. When moving the mouse over a map, an ellipse is drawn around the pointer; this ellipse represents actually a real circle, but the distortion of the map makes it appear in this shape. The inspection of this ellipses at different points tells us a lot of information about the projection properties.
Files to download include the six maps at poster size, the script programs that generate them, the program displaying Tissot indicatrices, a technical manual and a set of proposed activities.
This exhibit was awarded the first prize in the “Mathematics of Planet Earth 2013” competition. | 508 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2021-31 | longest | en | 0.897962 |
https://proofwiki.org/wiki/Immediate_Successor_in_Toset_is_Unique | 1,585,819,099,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506870.41/warc/CC-MAIN-20200402080824-20200402110824-00517.warc.gz | 649,783,206 | 10,180 | # Immediate Successor in Toset is Unique
It has been suggested that this article or section be renamed: "unique" implies existence. suggest something with "at most one" One may discuss this suggestion on the talk page.
## Theorem
Let $\left({S, \preceq}\right)$ be a toset.
Let $a \in S$.
Then $a$ has at most one immediate successor.
## Proof
Let $b, b' \in S$ be immediate successors of $a$.
Because $\preceq$ is a total ordering, WLOG:
$b \preceq b'$
By virtue of $b'$ being a immediate successor of $a$:
$\neg \exists c \in S: a \prec c \prec b'$
However, since $b$ is also an immediate successor:
$a \prec b$
Hence, it cannot be the case that $b \prec b'$.
Since $b \preceq b'$, it follows that $b = b'$.
Hence the result.
$\blacksquare$ | 226 | 760 | {"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} | 3.53125 | 4 | CC-MAIN-2020-16 | latest | en | 0.868396 |
https://mr-mathematics.com/product/coordinates-in-the-first-quadrant/ | 1,716,747,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058972.57/warc/CC-MAIN-20240526170211-20240526200211-00630.warc.gz | 350,001,908 | 24,609 | # Coordinates in the First Quadrant
#### Resources
Click an icon to access the resources.
# Coordinates in the First Quadrant
Students learn how to plot and read coordinate pairs in the first quadrant.
As learning progresses students use 2D shapes to plot vertices and mid-points.
##### Differentiated Learning Objectives
• All students should be able to plot and read coordinates in the first quadrant.
• Most students should be able to plot and read coordinates in the first quadrant and determine the midpoint of a line segment.
• Some students should be able to apply their knowledge of geometrical properties to solve problems in a grid.
##### Schemes of Work
Patterns and Sequences
SKU: 207
### Mr Mathematics Blog
#### Planes of Symmetry in 3D Shapes
Planes of Symmetry in 3D Shapes for Key Stage 3/GCSE students.
Use isometric paper for hands-on learning and enhanced understanding.
#### GCSE Trigonometry Skills & SOH CAH TOA Techniques
Master GCSE Math: Get key SOH-CAH-TOA tips, solve triangles accurately, and tackle area tasks. Ideal for students targeting grades 4-5.
#### Regions in the Complex Plane
Explore Regions in the Complex Plane with A-Level Further Maths: inequalities, Argand diagrams, and geometric interpretations. | 272 | 1,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.857797 |
http://www.let.rug.nl/~gosse/termpedia2/termpedia.php?language=dutch_general&density=7&link_color=000000&termpedia_system=perl_db&url=http%3A%2F%2Fen.wikipedia.org%2Fwiki%2FTopology | 1,558,569,061,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256980.46/warc/CC-MAIN-20190522223411-20190523005411-00377.warc.gz | 301,725,728 | 42,650 | # Topology
Möbius strips, which have only one surface and one edge, are a kind of object studied in topology.
In mathematics, topology (from the Greek τόπος, place, and λόγος, study) is concerned with the properties of space that are preserved under continuous deformations, such as stretching, twisting, crumpling and bending, but not tearing or gluing.
An n-dimensional topological space is a space (not necessarily Euclidean) with certain properties of connectedness and compactness.[1]
The space may be continuous (like all points on a rubber sheet), or discrete (like the set of integers). It can be open (like the set of points inside a circle) or closed (like the set of points inside a circle, together with the points on the circle).
Topology developed as a field of study out of geometry and set theory, through analysis of concepts such as space, dimension, and transformation.[2] Such ideas go back to Gottfried Leibniz, who in the 17th century envisioned the geometria situs (Greek-Latin for "geometry of place") and analysis situs (Greek-Latin for "picking apart of place"). Leonhard Euler's Seven Bridges of Königsberg Problem and Polyhedron Formula are arguably the field's first theorems. The term topology was introduced by Johann Benedict Listing in the 19th century, although it was not until the first decades of the 20th century that the idea of a topological space was developed. By the middle of the 20th century, topology had become a major branch of mathematics.
A three-dimensional depiction of a thickened trefoil knot, the simplest non-trivial knot
## History
The Seven Bridges of Königsberg was a problem solved by Euler.
Topology, as a well-defined mathematical discipline, originates in the early part of the twentieth century, but some isolated results can be traced back several centuries.[3] Among these are certain questions in geometry investigated by Leonhard Euler. His 1736 paper on the Seven Bridges of Königsberg is regarded as one of the first practical applications of topology.[3] On 14 November 1750, Euler wrote to a friend that he had realised the importance of the edges of a polyhedron. This led to his polyhedron formula, VE + F = 2 (where V, E, and F respectively indicate the number of vertices, edges, and faces of the polyhedron). Some authorities regard this analysis as the first theorem, signalling the birth of topology.[4][5]
Further contributions were made by Augustin-Louis Cauchy, Ludwig Schläfli, Johann Benedict Listing, Bernhard Riemann and Enrico Betti.[6] Listing introduced the term "Topologie" in Vorstudien zur Topologie, written in his native German, in 1847, having used the word for ten years in correspondence before its first appearance in print.[7] The English form "topology" was used in 1883 in Listing's obituary in the journal Nature to distinguish "qualitative geometry from the ordinary geometry in which quantitative relations chiefly are treated".[8] The term "topologist" in the sense of a specialist in topology was used in 1905 in the magazine Spectator.[citation needed]
Their work was corrected, consolidated and greatly extended by Henri Poincaré. In 1895, he published his ground-breaking paper on Analysis Situs, which introduced the concepts now known as homotopy and homology, which are now considered part of algebraic topology.[6]
Topological characteristics of closed 2-manifolds[6]
Manifold Euler num Orientability Betti numbers Torsion coefficient (1-dim)
b0 b1 b2
Sphere 2 Orientable 1 0 1 none
Torus 0 Orientable 1 2 1 none
2-holed torus −2 Orientable 1 4 1 none
g-holed torus (genus g) 2 − 2g Orientable 1 2g 1 none
Projective plane 1 Non-orientable 1 0 0 2
Klein bottle 0 Non-orientable 1 1 0 2
Sphere with c cross-caps (c > 0) 2 − c Non-orientable 1 c − 1 0 2
2-Manifold with g holes
and c cross-caps (c > 0)
2 − (2g + c) Non-orientable 1 (2g + c) − 1 0 2
Unifying the work on function spaces of Georg Cantor, Vito Volterra, Cesare Arzelà, Jacques Hadamard, Giulio Ascoli and others, Maurice Fréchet introduced the metric space in 1906.[9] A metric space is now considered a special case of a general topological space, with any given topological space potentially giving rise to many distinct metric spaces. In 1914, Felix Hausdorff coined the term "topological space" and gave the definition for what is now called a Hausdorff space.[10] Currently, a topological space is a slight generalization of Hausdorff spaces, given in 1922 by Kazimierz Kuratowski.[11]
Modern topology depends strongly on the ideas of set theory, developed by Georg Cantor in the later part of the 19th century. In addition to establishing the basic ideas of set theory, Cantor considered point sets in Euclidean space as part of his study of Fourier series. For further developments, see point-set topology and algebraic topology.
## Introduction
Topology can be formally defined as "the study of qualitative properties of certain objects (called topological spaces) that are invariant under a certain kind of transformation (called a continuous map), especially those properties that are invariant under a certain kind of invertible transformation (called homeomorphisms)."
Topology is also used to refer to a structure imposed upon a set X, a structure that essentially characterizes the set X as a topological space by taking proper care of properties such as convergence, connectedness and continuity, upon transformation.
Topological spaces show up naturally in almost every branch of mathematics. This has made topology one of the great unifying ideas of mathematics.
The motivating insight behind topology is that some geometric problems depend not on the exact shape of the objects involved, but rather on the way they are put together. For example, the square and the circle have many properties in common: they are both one dimensional objects (from a topological point of view) and both separate the plane into two parts, the part inside and the part outside.
In one of the first papers in topology, Leonhard Euler demonstrated that it was impossible to find a route through the town of Königsberg (now Kaliningrad) that would cross each of its seven bridges exactly once. This result did not depend on the lengths of the bridges or on their distance from one another, but only on connectivity properties: which bridges connect to which islands or riverbanks. This Seven Bridges of Königsberg problem led to the branch of mathematics known as graph theory.
A continuous deformation (a type of homeomorphism) of a mug into a doughnut (torus) and a cow into a sphere
Similarly, the hairy ball theorem of algebraic topology says that "one cannot comb the hair flat on a hairy ball without creating a cowlick." This fact is immediately convincing to most people, even though they might not recognize the more formal statement of the theorem, that there is no nonvanishing continuous tangent vector field on the sphere. As with the Bridges of Königsberg, the result does not depend on the shape of the sphere; it applies to any kind of smooth blob, as long as it has no holes.
To deal with these problems that do not rely on the exact shape of the objects, one must be clear about just what properties these problems do rely on. From this need arises the notion of homeomorphism. The impossibility of crossing each bridge just once applies to any arrangement of bridges homeomorphic to those in Königsberg, and the hairy ball theorem applies to any space homeomorphic to a sphere.
Intuitively, two spaces are homeomorphic if one can be deformed into the other without cutting or gluing. A traditional joke is that a topologist cannot distinguish a coffee mug from a doughnut, since a sufficiently pliable doughnut could be reshaped to a coffee cup by creating a dimple and progressively enlarging it, while shrinking the hole into a handle.[12]
Homeomorphism can be considered the most basic topological equivalence. Another is homotopy equivalence. This is harder to describe without getting technical, but the essential notion is that two objects are homotopy equivalent if they both result from "squishing" some larger object.
Equivalence classes of the English (i.e., Latin) alphabet (sans-serif)
Homeomorphism Homotopy equivalence
An introductory exercise is to classify the uppercase letters of the English alphabet according to homeomorphism and homotopy equivalence. The result depends on the font used, and on whether the strokes making up the letters have some thickness or are ideal curves with no thickness. The figures here use the sans-serif Myriad font and are assumed to consist of ideal curves without thickness. Homotopy equivalence is a coarser relationship than homeomorphism; a homotopy equivalence class can contain several homeomorphism classes. The simple case of homotopy equivalence described above can be used here to show two letters are homotopy equivalent. For example, O fits inside P and the tail of the P can be squished to the "hole" part.
Homeomorphism classes are:
• no holes corresponding with C, G, I, J, L, M, N, S, U, V, W, and Z;
• no holes and three tails corresponding with E, F, T, and Y;
• no holes and four tails corresponding with X;
• one hole and no tail corresponding with D and O;
• one hole and one tail corresponding with P and Q;
• one hole and two tails corresponding with A and R;
• two holes and no tail corresponding with B; and
• a bar with four tails corresponding with H and K; the "bar" on the K is almost too short to see.
Homotopy classes are larger, because the tails can be squished down to a point. They are:
• one hole,
• two holes, and
• no holes.
To classify the letters correctly, we must show that two letters in the same class are equivalent and two letters in different classes are not equivalent. In the case of homeomorphism, this can be done by selecting points and showing their removal disconnects the letters differently. For example, X and Y are not homeomorphic because removing the center point of the X leaves four pieces; whatever point in Y corresponds to this point, its removal can leave at most three pieces. The case of homotopy equivalence is harder and requires a more elaborate argument showing an algebraic invariant, such as the fundamental group, is different on the supposedly differing classes.
Letter topology has practical relevance in stencil typography. For instance, Braggadocio font stencils are made of one connected piece of material.
## Concepts
### Topologies on sets
The term topology also refers to a specific mathematical idea central to the area of mathematics called topology. Informally, a topology tells how elements of a set relate spatially to each other. The same set can have different topologies. For instance, the real line, the complex plane, and the Cantor set can be thought of as the same set with different topologies.
Formally, let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:
1. Both the empty set and X are elements of τ.
2. Any union of elements of τ is an element of τ.
3. Any intersection of finitely many elements of τ is an element of τ.
If τ is a topology on X, then the pair (X, τ) is called a topological space. The notation Xτ may be used to denote a set X endowed with the particular topology τ.
The members of τ are called open sets in X. A subset of X is said to be closed if its complement is in τ (i.e., its complement is open). A subset of X may be open, closed, both (a clopen set), or neither. The empty set and X itself are always both closed and open. An open subset of X which contains a point x is called a neighborhood of x.
### Continuous functions and homeomorphisms
A function or map from one topological space to another is called continuous if the inverse image of any open set is open. If the function maps the real numbers to the real numbers (both spaces with the standard topology), then this definition of continuous is equivalent to the definition of continuous in calculus. If a continuous function is one-to-one and onto, and if the inverse of the function is also continuous, then the function is called a homeomorphism and the domain of the function is said to be homeomorphic to the range. Another way of saying this is that the function has a natural extension to the topology. If two spaces are homeomorphic, they have identical topological properties, and are considered topologically the same. The cube and the sphere are homeomorphic, as are the coffee cup and the doughnut. But the circle is not homeomorphic to the doughnut.
### Manifolds
While topological spaces can be extremely varied and exotic, many areas of topology focus on the more familiar class of spaces known as manifolds. A manifold is a topological space that resembles Euclidean space near each point. More precisely, each point of an n-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension n. Lines and circles, but not figure eights, are one-dimensional manifolds. Two-dimensional manifolds are also called surfaces, although not all surfaces are manifolds. Examples include the plane, the sphere, and the torus, which can all be realized without self-intersection in three dimensions, and the Klein bottle and real projective plane, which cannot (that is, all their realizations are surfaces that are not manifolds).
## Topics
### General topology
General topology is the branch of topology dealing with the basic set-theoretic definitions and constructions used in topology.[13][14] It is the foundation of most other branches of topology, including differential topology, geometric topology, and algebraic topology. Another name for general topology is point-set topology.
The basic object of study is topological spaces, which are sets equipped with a topology, that is, a family of subsets, called open sets, which is closed under finite intersections and (finite or infinite) unions. The fundamental concepts of topology, such as continuity, compactness, and connectedness, can be defined in terms of open sets. Intuitively, continuous functions take nearby points to nearby points. Compact sets are those that can be covered by finitely many sets of arbitrarily small size. Connected sets are sets that cannot be divided into two pieces that are far apart. The words nearby, arbitrarily small, and far apart can all be made precise by using open sets. Several topologies can be defined on a given space. Changing a topology consists of changing the collection of open sets, and this changes which functions are continuous, and which subsets are compact or connected.
Metric spaces are an important class of topological spaces where distances between any two points are defined by a function called a metric. In a metric space, an open set is a union of open disks, where an open disk of radius r centered at x is the set of the points whose distance to x is less than d. Many common spaces are topological space whose topology can be defined by a metric. This is the case of the real line, the complex plane, real and complex vector spaces and Euclidean spaces. Having a metric simplifies many proofs.
### Algebraic topology
Algebraic topology is a branch of mathematics that uses tools from algebra to study topological spaces.[15] The basic goal is to find algebraic invariants that classify topological spaces up to homeomorphism, though usually most classify up to homotopy equivalence.
The most important of these invariants are homotopy groups, homology, and cohomology.
Although algebraic topology primarily uses algebra to study topological problems, using topology to solve algebraic problems is sometimes also possible. Algebraic topology, for example, allows for a convenient proof that any subgroup of a free group is again a free group.
### Differential topology
Differential topology is the field dealing with differentiable functions on differentiable manifolds.[16] It is closely related to differential geometry and together they make up the geometric theory of differentiable manifolds.
More specifically, differential topology considers the properties and structures that require only a smooth structure on a manifold to be defined. Smooth manifolds are 'softer' than manifolds with extra geometric structures, which can act as obstructions to certain types of equivalences and deformations that exist in differential topology. For instance, volume and Riemannian curvature are invariants that can distinguish different geometric structures on the same smooth manifold—that is, one can smoothly "flatten out" certain manifolds, but it might require distorting the space and affecting the curvature or volume.
### Geometric topology
Geometric topology is a branch of topology that primarily focuses on low-dimensional manifolds (i.e. spaces of dimensions 2, 3, and 4) and their interaction with geometry, but it also includes some higher-dimensional topology.[17][18] Some examples of topics in geometric topology are orientability, handle decompositions, local flatness, crumpling and the planar and higher-dimensional Schönflies theorem.
In high-dimensional topology, characteristic classes are a basic invariant, and surgery theory is a key theory.
Low-dimensional topology is strongly geometric, as reflected in the uniformization theorem in 2 dimensions – every surface admits a constant curvature metric; geometrically, it has one of 3 possible geometries: positive curvature/spherical, zero curvature/flat, and negative curvature/hyperbolic – and the geometrization conjecture (now theorem) in 3 dimensions – every 3-manifold can be cut into pieces, each of which has one of eight possible geometries.
2-dimensional topology can be studied as complex geometry in one variable (Riemann surfaces are complex curves) – by the uniformization theorem every conformal class of metrics is equivalent to a unique complex one, and 4-dimensional topology can be studied from the point of view of complex geometry in two variables (complex surfaces), though not every 4-manifold admits a complex structure.
### Generalizations
Occasionally, one needs to use the tools of topology but a "set of points" is not available. In pointless topology one considers instead the lattice of open sets as the basic notion of the theory,[19] while Grothendieck topologies are structures defined on arbitrary categories that allow the definition of sheaves on those categories, and with that the definition of general cohomology theories.[20]
## Applications
### Biology
Knot theory, a branch of topology, is used in biology to study the effects of certain enzymes on DNA. These enzymes cut, twist, and reconnect the DNA, causing knotting with observable effects such as slower electrophoresis.[21] Topology is also used in evolutionary biology to represent the relationship between phenotype and genotype.[22] Phenotypic forms that appear quite different can be separated by only a few mutations depending on how genetic changes map to phenotypic changes during development. In neuroscience, topological quantities like the Euler characteristic and Betti number have been used to measure the complexity of patterns of activity in neural networks.
### Computer science
Topological data analysis uses techniques from algebraic topology to determine the large scale structure of a set (for instance, determining if a cloud of points is spherical or toroidal). The main method used by topological data analysis is:
1. Replace a set of data points with a family of simplicial complexes, indexed by a proximity parameter.
2. Analyse these topological complexes via algebraic topology – specifically, via the theory of persistent homology.[23]
3. Encode the persistent homology of a data set in the form of a parameterized version of a Betti number, which is called a barcode.[23]
### Physics
Topology is relevant to physics in areas such as condensed matter physics,[24] quantum field theory and physical cosmology.
The topological dependence of mechanical properties in solids is of interest in disciplines of mechanical engineering and materials science. Electrical and mechanical properties depend on the arrangement and network structures of molecules and elementary units in materials.[25] The compressive strength of crumpled topologies is studied in attempts to understand the high strength to weight of such structures that are mostly empty space.[26] Topology is of further significance in Contact mechanics where the dependence of stiffness and friction on the dimensionality of surface structures is the subject of interest with applications in multi-body physics.
A topological quantum field theory (or topological field theory or TQFT) is a quantum field theory that computes topological invariants.
Although TQFTs were invented by physicists, they are also of mathematical interest, being related to, among other things, knot theory, the theory of four-manifolds in algebraic topology, and to the theory of moduli spaces in algebraic geometry. Donaldson, Jones, Witten, and Kontsevich have all won Fields Medals for work related to topological field theory.
The topological classification of Calabi-Yau manifolds has important implications in string theory, as different manifolds can sustain different kinds of strings.[27]
In cosmology, topology can be used to describe the overall shape of the universe.[28] This area of research is commonly known as spacetime topology.
### Robotics
The possible positions of a robot can be described by a manifold called configuration space.[29] In the area of motion planning, one finds paths between two points in configuration space. These paths represent a motion of the robot's joints and other parts into the desired pose.[30]
### Games and puzzles
Tanglement puzzles are based on topological aspects of the puzzle's shapes and components.[31][32][33][34]
### Fiber art
In order to create a continuous join of pieces in a modular construction, it is necessary to create an unbroken path in an order which surrounds each piece and traverses each edge only once. This process is an application of the Eulerian path.[35]
## References
### Citations
1. ^ "the definition of topology".
2. ^ Bruner, Robert (2000). "What is Topology? A short and idiosyncratic answer".
3. ^ a b Croom 1989, p. 7
4. ^ Richeson 2008, p. 63
5. ^ Aleksandrov 1969, p. 204
6. ^ a b c Richeson (2008)
7. ^ Listing, Johann Benedict, "Vorstudien zur Topologie", Vandenhoeck und Ruprecht, Göttingen, p. 67, 1848
8. ^ Tait, Peter Guthrie, "Johann Benedict Listing (obituary)", Nature 27, 1 February 1883, pp. 316–317
9. ^ Fréchet, Maurice (1906). Sur quelques points du calcul fonctionnel. PhD dissertation. OCLC 8897542.
10. ^ Hausdorff, Felix, "Grundzüge der Mengenlehre", Leipzig: Veit. In (Hausdorff Werke, II (2002), 91–576)
11. ^ Croom 1989, p. 129
12. ^ Hubbard, John H.; West, Beverly H. (1995). Differential Equations: A Dynamical Systems Approach. Part II: Higher-Dimensional Systems. Texts in Applied Mathematics. 18. Springer. p. 204. ISBN 978-0-387-94377-0.
13. ^ Munkres, James R. Topology. Vol. 2. Upper Saddle River: Prentice Hall, 2000.
14. ^ Adams, Colin Conrad, and Robert David Franzosa. Introduction to topology: pure and applied. Pearson Prentice Hall, 2008.
15. ^ Allen Hatcher, Algebraic topology. (2002) Cambridge University Press, xii+544 pp. ISBN 0-521-79160-X, 0-521-79540-0.
16. ^ Lee, John M. (2006). Introduction to Smooth Manifolds. Springer-Verlag. ISBN 978-0-387-95448-6.
17. ^ Budney, Ryan (2011). "What is geometric topology?". mathoverflow.net. Retrieved 29 December 2013.
18. ^ R.B. Sher and R.J. Daverman (2002), Handbook of Geometric Topology, North-Holland. ISBN 0-444-82432-4
19. ^ Johnstone, Peter T. (1983). "The point of pointless topology". Bulletin of the American Mathematical Society. 8 (1): 41–53. doi:10.1090/s0273-0979-1983-15080-2.
20. ^ Artin, Michael (1962). Grothendieck topologies. Cambridge, MA: Harvard University, Dept. of Mathematics. Zbl 0208.48701.
21. ^ Adams, Colin (2004). The Knot Book: An Elementary Introduction to the Mathematical Theory of Knots. American Mathematical Society. ISBN 978-0-8218-3678-1
22. ^ Stadler, Bärbel M.R.; Stadler, Peter F.; Wagner, Günter P.; Fontana, Walter (2001). "The Topology of the Possible: Formal Spaces Underlying Patterns of Evolutionary Change". Journal of Theoretical Biology. 213 (2): 241–274. CiteSeerX 10.1.1.63.7808. doi:10.1006/jtbi.2001.2423. PMID 11894994.
23. ^ a b Gunnar Carlsson (April 2009). "Topology and data" (PDF). Bulletin (New Series) of the American Mathematical Society. 46 (2): 255–308. doi:10.1090/S0273-0979-09-01249-X.
24. ^ "The Nobel Prize in Physics 2016". Nobel Foundation. 4 October 2016. Retrieved 12 October 2016.
25. ^ Stephenson, C.; et., al. (2017). "Topological properties of a self-assembled electrical network via ab initio calculation". Sci. Rep. 7: 41621. Bibcode:2017NatSR...741621S. doi:10.1038/srep41621. PMC 5290745. PMID 28155863.
26. ^ Cambou, Anne Dominique; Narayanan, Menon (2011). "Three-dimensional structure of a sheet crumpled into a ball". Proceedings of the National Academy of Sciences. 108 (36): 14741–14745. arXiv:1203.5826. Bibcode:2011PNAS..10814741C. doi:10.1073/pnas.1019192108. PMC 3169141. PMID 21873249.
27. ^ Yau, S. & Nadis, S.; The Shape of Inner Space, Basic Books, 2010.
28. ^ The Shape of Space: How to Visualize Surfaces and Three-dimensional Manifolds 2nd ed (Marcel Dekker, 1985, ISBN 0-8247-7437-X)
29. ^ John J. Craig, Introduction to Robotics: Mechanics and Control, 3rd Ed. Prentice-Hall, 2004
30. ^ Michael Farber, Invitation to Topological Robotics, European Mathematical Society, 2008
31. ^ https://math.stackexchange.com How to reason about disentanglement puzzles.
32. ^ Horak, Mathew (2006). "Disentangling Topological Puzzles by Using Knot Theory". Mathematics Magazine. 79 (5): 368–375. doi:10.2307/27642974. JSTOR 27642974..
33. ^ http://sma.epfl.ch/Notes.pdf A Topological Puzzle, Inta Bertuccioni, December 2003.
34. ^ https://www.futilitycloset.com/the-figure-8-puzzle The Figure Eight Puzzle, Science and Math, June 2012.
35. ^ Eckman, Edie: Connect the shapes crochet motifs: creative techniques for joining motifs of all shapes. ©2012 Storey Publishing
### Bibliography
• Aleksandrov, P.S. (1969) [1956], "Chapter XVIII Topology", in Aleksandrov, A.D.; Kolmogorov, A.N.; Lavrent'ev, M.A. (eds.), Mathematics / Its Content, Methods and Meaning (2nd ed.), The M.I.T. Press
• Croom, Fred H. (1989), Principles of Topology, Saunders College Publishing, ISBN 978-0-03-029804-2
• Richeson, D. (2008), Euler's Gem: The Polyhedron Formula and the Birth of Topology, Princeton University Press | 6,226 | 26,711 | {"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} | 3.078125 | 3 | CC-MAIN-2019-22 | latest | en | 0.945827 |
http://www.brookings.edu/research/opinions/2010/10/22-political-momentum-wolfers?rssid=wolfersj&utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+BrookingsRSS%2Fexperts%2Fwolfersj+%28Brookings+Experts+-+Justin+Wolfers%29 | 1,368,920,342,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382989/warc/CC-MAIN-20130516092622-00030-ip-10-60-113-184.ec2.internal.warc.gz | 354,969,763 | 18,267 | The New York Times
# Detecting Political Momentum Is Harder Than You Think
Over at FiveThirtyEight.com, Nate Silver has a post attempting to debunk the idea that there is momentum in political campaigns. But I think he’s wrong. And his post provides a fun opportunity for a simple statistics lesson on the difficulty of discovering momentum.
Here’s what Nate does: He compares the change in polls between successive periods. If there were momentum, he argues, a rise in the polls this month would make it more likely there is another rise next month, while a fall in this month’s polls would likely yield another fall next month. That is, he believes that momentum will cause successive changes in a time series to be positively correlated. Instead, he finds the opposite to be true. (His graph is here.)
But this isn’t a fair test of the momentum hypothesis. Here’s the problem: Imagine that there is some measurement error in the poll taken this month — perhaps the pollsters just interviewed too many Democrats, or too many Republicans. If there’s any kind of measurement error, this could drive Nate’s findings. There are two cases to consider:
If the estimate for this month is too high, it will make the change between this month and last month look too large, and it will make the change from this month to next month look too small (or perhaps negative).
Or, if the estimate for this month is too low, it will make the change between this month and last month look too small, and it will make the change from this month to next month look too large.
Either way, if there’s any kind of noise in your estimate for this month, you get a negative correlation between adjoining changes in polls. But in my example, this negative correlation is driven by measurement error in the polls, not by the presence or absence of momentum.
The problem is that he’s using the same measurement — this month’s poll — in constructing both of the variables he’s analyzing. And this is likely responsible for much of the (negative) correlation he observes. So perhaps this negative correlation is in fact disguising true momentum in political races.
Perhaps a simple example will help. I’m going to use data on the black unemployment rate to make my point, just because these are the data I have handy:
It’s crystal clear that there’s substantial momentum here: When black unemployment has been rising for a few months, it usually continues to rise; when it has been falling, it usually continues to fall for a few more months.
But let’s see what happens if we perform Nate Silver’s test, analyzing successive monthly changes in this black unemployment rate?
Instead of finding positive momentum, we find that this month’s change is negatively correlated with next month’s change in unemployment! This would lead Nate’s approach to (wrongly) conclude there’s no momentum. The reason is that the black unemployment rate is measured with error, and by construction, these errors cause this month’s change and last month’s change to move in opposite directions.
There’s a simple solution: Analyze changes where you don’t use the same measure — this month’s unemployment rate — in constructing both your dependent and independent variables. I try this alternative test in the next graph, showing the change in the black unemployment rate between next month and this month versus the change between last month and the previous month:
And now this analysis shows what was obvious from the first graph: Yes, there is momentum in the black unemployment rate.
Is this what is going on with Nate’s analysis of polling data? I don’t know for sure, because I don’t have his database to test this idea on. But I’m willing to bet it is. Nate: Here’s my proposal. Re-run your analysis, but instead of analyzing the relationship between changes over periods A and B as a function of changes between periods B and C, analyze them as a function of changes between periods C and D. I’m confident that you’ll find less evidence of negative correlation; you may even find evidence of a positive correlation.
In fact, let’s bet a fancy dinner on the outcome — I reckon you’ll find that non-overlapping changes in polling are in fact positively related. That is, there probably is momentum in political races.
Do we have a bet? | 866 | 4,308 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2013-20 | latest | en | 0.931708 |
https://fr.mathworks.com/matlabcentral/answers/480966-i-am-trying-to-solve-for-an-integral-using-the-relation-y-n-ny-n-1-1-e-and-that-y-1-1-2-e-be | 1,571,263,371,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986670928.29/warc/CC-MAIN-20191016213112-20191017000612-00287.warc.gz | 511,933,257 | 22,327 | ## I am trying to solve for an integral using the relation y(n)=ny(n-1)-(1/e) and that y(1)=1-(2/e). Below is my function but I am not sure what to do about the 'y(n-1)' part.
### Emily Gallagher (view profile)
on 18 Sep 2019 at 14:08
Latest activity Edited by John D'Errico
### John D'Errico (view profile)
on 18 Sep 2019 at 15:50
### John D'Errico (view profile)
I am trying to solve for an integral using the relation y(n)=ny(n-1)-(1/e) and that y(1)=1-(2/e). Below is my function but I am not sure what to do about the 'y(n-1)' part.
% This function gives us the integral of x^n*exp^(-x) over [0,1].
function [y] = integral(n)
%INTEGRAL
if n == 1
y(n) = 1 - (2./ exp(1));
return;
end
y(n) = n .* y(n-1) - (1 ./exp(1));
end
### John D'Errico (view profile)
on 18 Sep 2019 at 14:46
Edited by John D'Errico
### John D'Errico (view profile)
on 18 Sep 2019 at 15:50
First, you might recognize that 1/e = 1/exp(1) = exp(-1).
So if you are going to compute an exponential, thus exp(1), then why not save time and just compute exp(-1) in the first place?????
Next, do NOT call a function integral!!!!!!!!!!!!! If you do, then the existing, VERY useful function named integral will no longer be accessible. And then your next frenzied question will be about why integral does not work anymore. So call your function something like myintegral instead.
Next, your real question is what do you do with y(n-1). But you already have a formula for y(n). This is a basic recurrence relation. Just call the function recursively!
By the way, you don't need a return statement in the if statement, if your code will just exit the function anyway.
Finally, you set the result to y(n). But your function takes a value for n, and then returns a SCALAR value for y_n. However, if you try to stuff it into y(n), then you will crete the VECTOR y, that is n elements long. All of the other elements will be zero, except the nth element of that vector.
function y = myint1(n)
%INTEGRAL
if n == 1
y = 1 - 2*exp(-1);
end
y = n .* myint1(n-1) - exp(-1);
end
Of course, this is slightly inefficient if n would be at all large, because it will require you to call the function integral n times.
Or, perhaps, what you really want is to return the value for all values of y(n) up to but not exceeding n. Then it would be efficient to just write it as a loop.
function y = myint2(n)
%INTEGRAL
y = zeros(1,n);
expminus1 = exp(-1);
y(1) = 1 - 2*expminus1;
for m = 2:n
y(m) = m.*y(m-1) -expminus1;
end
That code would return a vector y, for all values of up to n.
With n==20, lets see how the latter code works.
format long g
>> y
y =
Columns 1 through 5
0.264241117657115 0.160602794142788 0.113928941256923 0.0878363238562483 0.0713021781097991
Columns 6 through 10
0.0599336274873523 0.0516559512400239 0.0453681687487486 0.0404340775672951 0.0364613345015088
Columns 11 through 15
0.0331952383451544 0.03046341897041 0.0281450054438883 0.0261506350429941 0.024380084473469
Columns 16 through 20
0.0222019104040609 0.00955303569759369 -0.195924798614756 -4.0904506148518 -82.1768917382075
What is interesting is that it starts to yield negative numbers for n as large as 20. Even 18 seems to be a problem. (By the way, note that I actually need to use the true functino integral here to compare results!)
m = 1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
1 0.264241117657115 0.264241117657115
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
2 0.160602794142788 0.160602794142788
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
3 0.113928941256923 0.113928941256923
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
4 0.0878363238562483 0.0878363238562491
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
5 0.0713021781097991 0.0713021781098032
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
6 0.0599336274873523 0.0599336274873766
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
7 0.0516559512400239 0.0516559512401941
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
8 0.0453681687487486 0.0453681687501108
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
9 0.0404340775672951 0.040434077579555
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
10 0.0364613345015088 0.0364613346241073
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
11 0.0331952383451544 0.0331952396937377
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
12 0.03046341897041 0.0304634351534098
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
13 0.0281450054438883 0.0281452158228847
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
14 0.0261506350429941 0.026153580348944
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
15 0.024380084473469 0.0244242640627179
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
16 0.0222019104040609 0.0229087838320443
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
17 0.00955303569759369 0.0215698839733108
>> m = m+1;[m,y(m),integral(@(x) kernel(x,m),0,1)]
ans =
18 -0.195924798614756 0.0203784703481514
Look carefully at what you see here. The last few digits start to go bad as n gets larger, and it does not even require you to go very far. Even at m==4, we see some deviation. And one thing that we should recognize is that the integral in question should NEVER yield a negative number, since all terms are positive, over the interval [0,1]. So how could the integral be negative?
What is happening? Why does this fail?
You need to see that for larger values of n, we are multiplying by n in this recursive relation. And we are also subtracting two numbers.
n .* myint1(n-1) - exp(-1);
So what happens is any tiny error in y(n-1) gets amplified, because you just multiplied it by n. Think of it like the factorial function. nd what is factorial(18), or a number that large?
factorial(18)
ans =
6.402373705728e+15
So, if there is an error in the least significant bit of y(1), by only one bit in that decimal place, then it gets multiplied (AMPLIFIED!) by something on the order of 6e15 by the time you get to y(18).
Hmm. Can we fix this? Well, yes. There is a trick we can use, and it is one that is often of value in problems like this. The ide is to start out past the END, and work backwards.
That is, if we wish to compute the value of y(n), start at n+20. We have this recursion:
y(n) = n .* y(n-1) - exp(-1);
Rewrite it as:
y(n-1) = (y(n) + exp(-1))/n
Now we can write it in a loop, going backwards.
function yn = myint3(n)
%INTEGRAL, starting past the top...
yn = .1; % Anything even close will suffice here.
N = n + 20;
expminus1 = exp(-1);
for m = N:-1:(n+1)
yn = (yn + expminus1)/m;
end
Does this work?
m=1,integral(@(x) kernel(x,m),0,1),myint3(m)
m =
1
ans =
0.264241117657115
ans =
0.264241117657115
>> m=18,integral(@(x) kernel(x,m),0,1),myint3(m)
m =
18
ans =
0.0203784703481514
ans =
0.0203784703481514
>> m=30,integral(@(x) kernel(x,m),0,1),myint3(m)
m =
30
ans =
0.0122495029578589
ans =
0.0122495029578589
As you see, now this code with the backwards loop works for virtually any value of n, whereas when you use the forward loop, it starts to fail very early. We can even compare it to a symbolic integral.
m = 18
m =
18
>> int(X.^m*exp(-X),0,1)
ans =
6402373705728000 - 17403456103284421*exp(-1)
>> vpa(ans)
ans =
0.020378470348151434104822774864287
>> myint3(18)
ans =
0.0203784703481514
Again, it yields essentially full precision results. Why did it work? Think about the difference in terms of how I formulated the recurrence. Dividing by large numbers will decrease any noise, until it all gets killed off, even if I started with complete crap for my initial guess at y(n+20).
This is a good trick to remember. (You may forget it by tomorrow, but perhaps not, or perhaps it will ring a bell in some later class.) | 2,661 | 7,762 | {"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} | 3.859375 | 4 | CC-MAIN-2019-43 | latest | en | 0.803927 |
https://www.geeksforgeeks.org/find-position-given-number-among-numbers-made-4-7/?ref=rp | 1,656,376,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00363.warc.gz | 842,542,564 | 27,682 | Find position of the given number among the numbers made of 4 and 7
• Difficulty Level : Medium
• Last Updated : 08 Jun, 2022
Consider a series of numbers composed of only digits 4 and 7. The first few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. etc. Given a number constructed by 4, 7 digits only, we need to find the position of this number in this series.
Examples:
```Input : 7
Output : pos = 2
Input : 444
Output : pos = 7```
It is reverse of the following article :
Find n-th element in a series with only 2 digits (4 and 7) allowed | Set 2 (log(n) method)
``` ""
/ \
1(4) 2(7)
/ \ / \
3(44) 4(47) 5(74) 6(77)
/ \ / \ / \ / \```
The idea is based on the fact that all even positioned numbers have 7 as the last digit and all odd positioned numbers have 4 as the last digit.
If the number is 4 then it is the left node of the tree, then it corresponds to (pos*2)+1. Else right child node(7) corresponds to (pos*2)+2.
C++
`// C++ program to find position of a number``// in a series of numbers with 4 and 7 as the``// only digits.``#include ``#include ``using` `namespace` `std;` `int` `findpos(string n)``{`` ``int` `i = 0, pos = 0;`` ``while` `(n[i] != ``'\0'``) {` ` ``// check all digit position`` ``switch` `(n[i])`` ``{` ` ``// if number is left then pos*2+1`` ``case` `'4'``:`` ``pos = pos * 2 + 1;`` ``break``;` ` ``// if number is right then pos*2+2`` ``case` `'7'``:`` ``pos = pos * 2 + 2;`` ``break``;`` ``}`` ``i++;`` ``}`` ``return` `pos;``}` `// Driver code``int` `main()``{`` ``// given a number which is constructed`` ``// by 4 and 7 digit only`` ``string n = ``"774"``;`` ``cout << findpos(n);`` ``return` `0;``}`
Java
`// java program to find position of a number``// in a series of numbers with 4 and 7 as the``// only digits.``import` `java.util.*;` `class` `GFG {`` ` ` ``static` `int` `findpos(String n)`` ``{`` ` ` ``int` `k = ``0``, pos = ``0``, i = ``0``;`` ``while` `(k != n.length()) {` ` ``// check all digit position`` ``switch` `(n.charAt(i)) {` ` ``// if number is left then pos*2+1`` ``case` `'4'``:`` ``pos = pos * ``2` `+ ``1``;`` ``break``;` ` ``// if number is right then pos*2+2`` ``case` `'7'``:`` ``pos = pos * ``2` `+ ``2``;`` ``break``;`` ``}`` ` ` ``i++;`` ``k++;`` ``}`` ` ` ``return` `pos;`` ``}` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ` ` ``// given a number which is constructed`` ``// by 4 and 7 digit only`` ``String n = ``"774"``;`` ` ` ``System.out.println(findpos(n));`` ``}``}` `// This code is contributed by Sam007.`
Python3
`# python program to find position``# of a number in a series of``# numbers with 4 and 7 as the``# only digits.``def` `findpos(n):`` ``i ``=` `0`` ``j ``=` `len``(n)`` ``pos ``=` `0`` ``while` `(i
C#
`// C# program to find position of``// a number in a series of numbers``// with 4 and 7 as the only digits.``using` `System;` `class` `GFG``{`` ``static` `int` `findpos(String n)`` ``{`` ` ` ``int` `k = 0, pos = 0, i = 0;`` ``while` `(k != n.Length) {` ` ``// check all digit position`` ``switch` `(n[i]) {` ` ``// if number is left then pos*2+1`` ``case` `'4'``:`` ``pos = pos * 2 + 1;`` ``break``;` ` ``// if number is right then pos*2+2`` ``case` `'7'``:`` ``pos = pos * 2 + 2;`` ``break``;`` ``}`` ` ` ``i++;`` ``k++;`` ``}`` ` ` ``return` `pos;`` ``}` ` ``// Driver code`` ``static` `void` `Main()`` ``{`` ` ` ``// given a number which is constructed`` ``// by 4 and 7 digit only`` ``String n = ``"774"``;`` ` ` ``Console.Write(findpos(n));`` ``}`` ` `}` `// This code is contributed by Sam007`
PHP
``
Javascript
``
Output:
`13`
Time Complexity: O(n), where n represents the size of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
This article is contributed by Devanshu Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 1,614 | 4,827 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.69228 |
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https://www.interviewcake.com/question/java/find-duplicate-optimize-for-space | 1,680,108,789,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949009.11/warc/CC-MAIN-20230329151629-20230329181629-00538.warc.gz | 918,860,686 | 42,727 | ### You only have free questions left (including this one).
But it doesn't have to end here! Sign up for the 7-day coding interview crash course and you'll get a free Interview Cake problem every week.
Find a duplicate, Space Edition™.
We have an array of integers, where:
1. The integers are in the range 1..n
2. The array has a length of n+1
It follows that our array has at least one integer which appears at least twice. But it may have several duplicates, and each duplicate may appear more than twice.
Write a method which finds an integer that appears more than once in our array. Don't modify the input! (If there are multiple duplicates, you only need to find one of them.)
We're going to run this method on our new, super-hip MacBook Pro With Retina Display™. Thing is, the damn thing came with the RAM soldered right to the motherboard, so we can't upgrade our RAM. So we need to optimize for space!
We can do this in space.
We can do this in less than time while keeping space.
We can do this in time and space.
We can do this without modifying the input.
Most algorithms double something or cut something in half. How can we rule out half of the numbers each time we iterate through the array?
Actually, we don't support password-based login. Never have. Just the OAuth methods above. Why?
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2. It lets us avoid storing passwords that hackers could access and use to try to log into our users' email or bank accounts.
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Actually, we don't support password-based login. Never have. Just the OAuth methods above. Why?
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space and time.
Tricky as this solution is, we can actually do even better, getting our runtime down to while keeping our space cost at . The solution is NUTS; it's probably outside the scope of what most interviewers would expect. But for the curious...here it is!
This method always returns one duplicate, but there may be several duplicates. Write a method that returns all duplicates.
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Reset editor | 624 | 2,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-14 | longest | en | 0.935011 |
http://www.cgl.uwaterloo.ca/wmcowan/teaching/cs251/s08/notes/l26.html | 1,516,794,327,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00658.warc.gz | 399,582,398 | 2,296 | # Lecture 26 - Multi-cycle Control
## Practical Details
1. Assignment 6
# Multi-cycle Control
## The Basic Idea
1. You have already put in the
• datapaths
• MUXes
• control signals
2. Work out the control signal values for each state.
• Note that different instruction types mean that there is more than one state per step (cycle).
3. Make a finite-state machine
This is done in detail in both the notes and in the text-book. Sometimes the details are a little sketchy, but you should be able to fill in the blanks by this point. This is what you have to do in order to master the material.
Assignment 6, which asks you to give datapaths and control cycles for a much simpler will give you some indication how well you are doing.
## Control Signals
### One-bit
Signal name 0 1 Comment 1 2 3R 3M 3B 4R 4M 5 RegDst write to rt write to rd MUX2 X X X X X 1 X 0 RegWrite write to register X X X X X 1 X 1 ALUSrcA PC register A MUX4 0 0 1 1 1 MemRead read from memory 1 1 MemWrite write to memory X MemToReg data from ALU data from memory MUX3 0 1 IorD address from PC address from ALU MUX1 0 1 IRWrite latch instruction register 1 PCWrite write PC 1 CondPCWrite write PC if ALUout = 0 1
### Two-bit
Signal Value Effect 1 2 3E 3M 3B 4 5 ALUOp 00 add S S S S 01 subtract S 10 determine by funct ALUSrcB 00 input from register B S 01 input is 4 S 10 input is immediate field S S 11 input is shifted immediate field S PCSource 00 PC from ALUout S 01 PC from ALUOut S 10 PC gets jump target from immediate
#### Implementation
Either
1. Combinational logic: fast, small, inflexible
• input: last state x opcode
• output: this state x control signals
2. Programmable logic array (PLA): fast, big, more flexible
3. Read-only memory (could be flash): slower, very big, yet more flexible
• last-state x opcode can be 20 or 30 bits
• an address this big makes a very big ROM
4. Micro-code: faster or slower, small, the most flexible
## Micro-code
Anything we can make with a finite state machine we can make with a program.
This has re-entered CS hardware with a new name
• V(ery)L(ong)I(nstruction)W(ord)
## Exceptions
What do we do when
1. an external event needs to be accepted by the processor
2. something screwy happens, like an address fault | 637 | 2,257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2018-05 | latest | en | 0.858919 |
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Topic: brain buster problem
Replies: 0
Search Thread: Advanced Search
Robert L Strawderman Posts: 9 Registered: 12/18/04
brain buster problem
Posted: Jul 17, 1996 5:53 PM
Plain Text Reply
Here's a tough problem -- any ideas?
Is it possible to construct a (dependent) sequence of rv's
{t_i, i=1..n} such that for all i,j,k=1..n
1. E[t_i] = 0
2. E[t^2_i] = 1
3. E[t_i t_j] = 0 i!=j
4. E[t_i t_j t_k] = 0 i!=j!=k
5. E[t^2_i t_j] = 0 i < j
6. E[t^2_i t_j] = 1 i >= j
Generating a sequence of rv's which satisfies 1-5 is rather
trivial - its 6 that appears to be the killer. Note that for 6
to be satisfied, its sufficient if E[t^2_i | t_j] = t^2_j and
E[t^3_j]=1. However, constructing such a sequence seems difficult
if one wishes to preserve 3.
The t_i's can be discrete, continuous, or a mix.
Any ideas would be appreciated....
***************************************************************************
Robert Strawderman, Sc.D. Email: strawder@umich.edu
Department of Biostatistics Office: (313) 936 - 1002
University of Michigan Fax: (313) 763 - 2215
1420 Washington Heights
Ann Arbor, MI 48109-2029 Web: http://www.sph.umich.edu/~strawder/
***************************************************************************
© The Math Forum at NCTM 1994-2018. All Rights Reserved. | 445 | 1,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2018-17 | longest | en | 0.809777 |
https://www.jiskha.com/questions/1126265/Four-students-worked-to-find-an-estimate-for-square-root-22-Who-is-closest-to-finding | 1,561,380,878,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999482.38/warc/CC-MAIN-20190624104413-20190624130413-00030.warc.gz | 768,159,254 | 5,093 | # Math Help!
Four students worked to find an estimate for square root 22. Who is closest to finding the true estimate?
A. Tiffany: "Use square root 16 and square root 25 to estimate."
B. Todd: "I use square root 9 and square root 16."
C. Trenton: "It should be square root 21 and square root 22."
D. Fredrica: "Use square root 25 and square root 36 to get the esimate."
Thank you
1. 👍 0
2. 👎 0
3. 👁 389
1. Yes, A.
1. 👍 0
2. 👎 0
2. go back and see how I did the earlier one with linear interpolation. It clearly works best when you pick perfect squares above and below the desired answer.
1. 👍 0
2. 👎 0
posted by Damon
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More Similar Questions | 989 | 3,473 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-26 | latest | en | 0.8844 |
https://studymoose.com/gcse-maths-coursework-round-round-new-essay | 1,632,158,519,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057083.64/warc/CC-MAIN-20210920161518-20210920191518-00444.warc.gz | 593,926,107 | 28,830 | # GCSE Maths Coursework: Round and Round
Categories: CourseworkMath
I have been asked to investigate the equation –
n–>(+1) –>(?2) –>
* ?
(a) (b)
I will do this by first of all, changing the first number (a) to find out if that it has any relevance to the answer that comes out of the equation at the end.
Then I am going to change the (b) number to find out weather that has anything to do with the outcome of the final number, I will also be looking for patterns and sequences in the answers.
Investigation 1
6–>(+1)–>(?2)–> = 3.5
2.25
1.625
1.3125
1.15625
1.078125
1.0390625
1.01953125
1.009765625
1.004882813
1.002441406
1.001220703
1.000610352
1.000305176
As you can see, when the sum is entered in to a graphical calculator, the numbers that come out are as above. The numbers are decending from 3.5 to 1.000305176, that is the most that I have done down to so far. I predict that the numbers will eventually stop ay the number 1 as near the end there is three 0’s and before that there were two 0’s and before that there was one 0, so I estimate that there will eventually be nine 0’s and the number will be finally 1.
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000000000.
Investigation 1:carried on
1.000152588
1.000076294
1.000038147
1.000019074
1.00009537
1.000047685
1.000023843
1.000011922
1.000005961
1.000029805
1.000014903 here an extra “0” is added
1.000007452
1.000003726
1.000001863
1.000000932
1.000000466
1.000000233
1.000000177
1.000000089 here one extra “0” is added.
1.000000045
1.000000023
1.000000012
1.000000006 here one more “0” is added
1.000000003
1.000000002
1.000000001
1.000000001
1.000000001
At this point, it is the furthest that the number is been to the number one, I was almost right in my prediction.
I will now try a change in the sequence, I am going do the sequence as followed;
6–>(+2)–>(?2)–> = 4
3
2.5
2.25
2.125
2.0625
2.03125
2.016625
2.0078125
2.00390625
2.001953125
2.000976563
2.000488281
Now that I have changed the “+1” to a “+2” I can see that the answer goes up.
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When the (a) number is “+1” the number eventually rounds down, close to 1, when the (a) number is “+2” the number will eventually also round down to the (a) number which is now “2”.
I can now see quite clearly, that if the (a) number in the sequence is for example, “20” the number, when divided by two and after been passed through the sequence, the answer will be “20”.
I am now going to put my method to practice:
6 –>(+20)–>(?2)–> = 13
16.5
18.25
19.125
19.5625
19.78125
19.890623
19.9453125
19.97263625
19.98632813
19.99316406 19.99658203
19.99829102
19.99914551
Investigation 2
I think that I have found a pattern in the numbering, I think that the higher the “a” number, the longer it takes to get to the final answer.eg. When the “a”number is 6, it takes only a short time to get to the final answer, but with the “a” number as 20, it takes a lot longer to get the final answer.
I will show this now:
High number
6 –>(+20)–>(?2)–> = 13 19.99998665
16.5 19.99999333
18.25 19.77777667
19.125 19.88888834
19.126 19.94444417
19.5625 19.97222209
19.78125 19.98611105
19.890623 19.99305553
19.9453125 19.99652777
19.97263625 19.99826389
19.98632813 19.99913195
19.99316406 19.99956598 19.99658203 19.99978299
19.99829102 19.9998915
19.99914551 19.9994575
19.99957276 19.99972875
19.99978638 19.99986438
19.99989319 19.99993219
19.9999466 19.9999661
19.9999733 19.99998305
Lower number
6–>(+1)–>(?2)–> = 3.5
2.25
1.625
1.3125
1.15625
1.078125
1.0390625
1.01953125
1.009765625
1.004882813
1.002441406
1.001220703
1.000610352
1.000305176
1.000152588
1.000076294
1.000038147
1.000019074
1.00009537
1.000047685
1.000023843
1.000011922
1.000005961
1.000029805
1.000014903
1.000007452
1.000003726
1.000001863
1.000000932
1.000000466
1.000000233
1.000000177
1.000000089.
1.000000045
1.000000023
1.000000012
1.000000006
1.000000003
1.000000002
1.000000001
1.000000001
1.000000001
As you can see, when the “a” number is high, it takes a longer time to get to the final answer then when the number is lower.
When the “a”number was +1, it took only 42 attempts to get to the closest that it would get to 1, but when the “a”number was +20, after 40 attempts, it was still not even close to becoming 20. However, my analysis that the “a” number decides how long the answer takes to find could be wrong, so I am going to carry out one more investigation to certify my prediction and give me a clearer view.
Investigation 3
6–> (+10)–>(?2)–> = 89 9.999969483
9.5 9.999984742
9.75 9.999992371
9.875 9.999996186
9.9375 9.999998093
9.96875 9.999999047
9.984375 9.999999524
9.9921875 9.999999762
9.99609375 9.999999881
9.998046875 9.999999941
9.999023438 9.999999971
9.999511719 9.999999986
9.99975586 9.999999995
9.99987793 9.999999997
9.999938965 9.999999999
10
It took 31 attempt to get to the final number, 10, but this shows me that the size of the number, does not decide how long it takes to find the final answer.
Investigation 4
I am now going to change some more numbers in the sequence, I will change the “n” number in the sequence to see if that has any relevance to the final number given.
18–>(+1)–>(?2)–> = 9.5
5.25
3.125
2.0625
1.53125
1.265625
1.1328125
1.06640625
1.033203125
1.016601563
1.008300782
I now believe that it is not the “a” number that decides how long it takes to find the answer, but the “n” number.
I will now see how long it takes to get to the final answer when the “n “number is 1.
1–>(+1)–>(?2)–> = 1
1
1
it takes only one attempt to find the answer in this sequence, but I will now try with 2 as the “n” number.
2–>(+1)–>(?2)–> = 1.5
1.25
1.125
1.0625
1.03125
1.015625
1.0078125
1.00390625
1.001953125
1.000976563
1.000488282
1.000244141
1.000122171
1.000061086
1.000030543
1.000015272
1.000007636
1.000003818
1.000001909
1.000000955
1.000000478
1.000000239
1.000000165
1.000000083
1.000000042
1.000000021
1.000000011
1.000000006
1.000000003
1.000000002
1.000000001
1.000000001 this is the lowest that the number will go.
It took 32 attempts to get the final answer and so this shows me that not the “n” number or the “a” number have anything to do with how long the equation takes to solve.
Now I will investigate what the “b” number has to do with the equation. I will do this by changing the number, higher and lower, looking for patterns and rhythms.
Investigation 5: Changing the “b” value
6–>(+1)–>(?3)–> = 2.333333333
1.111111111
0.703703703
0.567901234
0.522633744
0.507544581
0.50251486
0.75125743
0.583752476
0.791876238
0.895938119
0.631979373
0.543993124
0.514664374
As you can see, when the “b” number has been changed, it alters the whole answer completely. There are no patterns and the number is not heading in a pattern that I can see.
Now that I have done all my working, I think that I have found a solution.
To see if this solution works, I am going to predict something:
I predict that 6 ? (+1)? (?5) = 1/4
With my equation: +x ?y ? x
?
y-1
I predict that the number will eventually round down to 0.25 (1/4).
1.4 1.2 0.44 0.288 0.2576 0.25152 0.250304 0.2500608 0.25001216 0.250002432 0.25000486 0.250000486 0.250000097 0.250000019 0.250000003 0.250000000 = 1/4
My prediction was correct. I will now see if it works on other sequences.
I predict that
6 ?(+3) ? (?2) = 3
will eventually end up at the number three:
4.5
3.75
3.375
3.1875
3.09375
3.046875
3.0224375
3.01171875
3.005835375
3.002917688
3.001458844
3.000729422
3.000364711
3.000182356
3.000091178
3.000045589
I was correct, it does eventually end up at 3.
This means that my equation works.
X? (a) ?(?b) = a
?
(b-1)
e.g. 1?(+2)?(?2)=
2
?
2-1 =
2
?
1 =
2
So, I have worked out that the final rule is +x ?y = x
?
y-1
GCSE Maths Coursework: Round and Round. (2020, Jun 02). Retrieved from https://studymoose.com/gcse-maths-coursework-round-round-new-essay
👋 Hi! I’m your smart assistant Amy!
Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes. | 3,041 | 8,531 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2021-39 | latest | en | 0.910188 |
https://www.coursehero.com/file/p7h1kf3/so-the-general-solution-is-n-k-t-n-0-k-e-Dk-2-t-11241-where-n-0-k-n-k-t-0-to-be/ | 1,638,662,974,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363125.46/warc/CC-MAIN-20211204215252-20211205005252-00486.warc.gz | 795,418,789 | 58,430 | # So the general solution is n k t n 0 k e dk 2 t 11241
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, so the general solution is ˜ n ( k, t ) = ˜ n 0 ( k ) e - Dk 2 t . (11.241) where ˜ n 0 ( k ) ˜ n ( k, t = 0), to be determined by the initial conditions. The answer (i.e. general solution) comes via an inverse FT: n ( x, t ) = Z -∞ dk 2 π ˜ n ( k, t ) e ikx = Z -∞ dk 2 π ˜ n 0 ( k ) e ikx - Dk 2 t . (11.242) 71
n(x,t) x increasing time Figure 11.18: Variation of concentration with distance x at various diffusion times. SPECIFIC EXAMPLE: We add a small drop of ink to a large tank of water (assumed 1- dimensional). We want to find the density of ink as a function of space and time, n ( x, t ). Initially, all the ink ( S particles) is concentrated at one point (call it the origin): n ( x, t = 0) = S δ ( x ) (11.243) implying (using the sifting property of the Dirac delta function), ˜ n 0 ( k ) ˜ n ( k, 0) = Z -∞ dx n ( x, t = 0) e - ikx = Z -∞ dx δ ( x ) e - ikx = S. (11.244) Putting this into Eqn. (11.242) we get: n ( x, t ) = Z -∞ dk 2 π ˜ n ( k, t ) e ikx = Z -∞ dk 2 π ˜ n 0 ( k ) e ikx - Dk 2 t = Z -∞ dk 2 π S e ikx - Dk 2 t = S 2 π 2 Dt e - x 2 / (4 Dt ) . (11.245) (we used the ‘completing the square’ trick that we previously used to FT the Gaussian). Compare this with the usual expression for a Gaussian, 1 2 πσ exp - x 2 2 σ 2 (11.246) and identify the width σ with 2 Dt . So, the ink spreads out with concentration described by a normalized Gaussian centred on the origin with width σ = 2 Dt . The important features are: normalized: there are always S particles in total at every value of t centred on the origin: where we placed the initial drop width σ = 2 Dt : gets broader as time increases 72
σ t : characteristic of random walk (‘stochastic’) process σ D : if we increase the diffusion constant D , the ink spreads out more quickly. The solution n ( x, t ) is sketched for various t in Fig. 11.18. FOURIER ANALYSIS: LECTURE 18 11.3 Fourier solution of the wave equation One is used to thinking of solutions to the wave equation being sinusoidal, but they don’t have to be. We can use Fourier Transforms to show this rather elegantly, applying a partial FT ( x k , but keeping t as is). The wave equation is c 2 2 u ( x, t ) ∂x 2 = 2 u ( x, t ) ∂t 2 (11.247) where c is the wave speed. We Fourier Transform w.r.t. x to get ˜ u ( k, t ) (note the arguments), remembering that the FT of 2 /∂x 2 is - k 2 : - c 2 k 2 ˜ u ( k, t ) = 2 ˜ u ( k, t ) ∂t 2 . (11.248) This is a harmonic equation for ˜ u ( k, t ), with solution ˜ u ( k, t ) = Ae - ikct + Be ikct (11.249) However, because the derivatives are partial derivatives, the ‘constants’ A and B can be functions of k . Let us write these arbitrary functions as ˜ f ( k ) and ˜ g ( k ), i.e. ˜ u ( k, t ) = ˜ f ( k ) e - ikct + ˜ g ( k ) e ikct . (11.250) We now invert the transform, to give u ( x, t ) = Z -∞ dk 2 π h ˜ f ( k ) e - ikct + ˜ g ( k ) e ikct i e ikx = Z -∞ dk 2 π ˜ f ( k ) e ik ( x - ct ) + Z -∞ dk 2 π ˜ g ( k ) e ik ( x + ct ) = f ( x - ct ) + g ( x + ct ) and f and g are arbitrary functions. | 1,116 | 3,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-49 | latest | en | 0.79308 |
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http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/121673 | 1,539,786,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511175.9/warc/CC-MAIN-20181017132258-20181017153758-00168.warc.gz | 40,870,750 | 3,571 | Below is my solution, formatted to use Jannis' testing classes.
The solution basically recursively calls clean? on the full array of
words, then on the first half and on the second half, then on the
first quarter, the second quarter, the third quarter, etc. Whenever
clean? returns true, it stops dividing that section of the array.
(This description is breadth-first, but the algorithm actually runs
depth-first. Breadth-first is easier to explain, and the final result
is the same.)
I then noticed that we don't really need to make all these calls to
clean?, because we can sometimes deduce what the results will be
without calling clean?. For example, if we call clean? on [0...8]
and the result is false, then we call clean? on [0...4] and the result
is true, then we know that clean? on [4...8] must return false, so we
don't need to make this call.
The ordinary recursion is done by the method findBanned(aWords). When
the above deduction tells us that clean? will return false, we instead
call the method findBannedThereIsOne(aWords). In the above example we
would call findBannedThereIsOne(aWords[4...8]), and it will skip
calling clean? on aWords[4...8].
(Yes, findBannedThereIsOne is an awkward name, but at least it's
better than findBannedAndWeKnowThereIsAtLeastOne)
When findBannedThereIsOne is called with an array with only one
element, it doesn't need to call clean? at all, because we know this
must be a banned word.
So that findBanned and findBannedThereIsOne only need to deal with
arrays whose size is a power of two, the main routine,
runYourAlgorithm, divides the input array into chunks of these sizes.
As I'm typing this, it occurs to me that this code might actually run
faster if runYourAlgorithm didn't do this. I'll give this a try, but
for now I'll post the version that does the chunking.
runYourAlgorithm also handles the case of an empty array, so the other
two routines don't need to handle it.
Here are the results with Jannis' testing classes. For the times in
seconds, keep in mind that these tests were run on an ancient 500MHz
Pentium III with other work going on at the same time.
Runs: 10
Words: 121430
Banned Words: 3
Average
Seconds: 0.5561
Mails: 76
Verified: 10/10
Rebuilt Banned Words
Runs: 10
Words: 121430
Banned Words: 126
Average
Seconds: 254.4113
Mails: 1948
Verified: 10/10
Rebuilt Banned Words
Runs: 10
Words: 3930
Banned Words: 29
Average
Seconds: 12.5187
Mails: 327
Verified: 10/10
Rebuilt Banned Words
Runs: 10
Words: 125282
Banned Words: 80
Average
Seconds: 139.9058
Mails: 1319
Verified: 10/10
Rebuilt Banned Words
Runs: 10
Words: 127367
Banned Words: 10
Average
Seconds: 1.5328
Mails: 210
Verified: 10/10
Rebuilt Banned Words
Wayne Vucenic
No Bugs Software
"Ruby and C++ Contract Programming in Silicon Valley"
======================================
class YourAlgorithm < RQ9Algorithm
def run()
runYourAlgorithm(@words)
end
# Returns an array containing all banned words from aWords
def runYourAlgorithm(aWords)
return [] if aWords.empty?
# Compute the largest power of two <= aWords.size
powerOfTwo = 1
powerOfTwo *= 2 while powerOfTwo * 2 <= aWords.size
# To simplify the logic in findBanned, we always call it with an
# array whose size is a power of two.
aBanned = findBanned(aWords[0...powerOfTwo])
# If we didn't process all of aWords, recursively do the rest
if powerOfTwo < aWords.size
aBanned += runYourAlgorithm(aWords[powerOfTwo..-1])
end
aBanned
end
# Returns an array containing all banned words from aWords
# aWords.size is a power of 2, and is > 0
def findBanned(aWords)
if aWords.size == 1
@filter.clean?(aWords[0]) ? [] : aWords
elsif @filter.clean?(aWords.join(' '))
[]
else
iSplit = aWords.size / 2
if @filter.clean?(aWords[0...iSplit].join(' '))
# There is at least one banned word in 0..-1, but not in 0...iSplit,
# so there must be one in iSplit..-1
findBannedThereIsOne(aWords[iSplit..-1])
else
# From the test above we know there is a banned word in 0...iSplit
findBannedThereIsOne(aWords[0...iSplit]) +
findBanned(aWords[iSplit..-1])
end
end
end
# Returns an array containing all banned words from aWords
# aWords.size is a power of 2, and is > 0
# Our caller has determined there is at least one banned word in aWords
def findBannedThereIsOne(aWords)
if aWords.size == 1
# Since we know there is at least one banned word, and since there is
# only one word in the array, we know this word is banned without
# having to call clean?
aWords
else
iSplit = aWords.size / 2
if @filter.clean?(aWords[0...iSplit].join(' '))
# There is at least one banned word in 0..-1, but not in 0...iSplit,
# so there must be one in iSplit..-1
findBannedThereIsOne(aWords[iSplit..-1])
else
# From the test above we know there is a banned word in 0...iSplit
findBannedThereIsOne(aWords[0...iSplit]) +
findBanned(aWords[iSplit..-1])
end
end
end
end | 1,441 | 5,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-43 | latest | en | 0.916764 |
https://brainor.com/math-worksheets/grade-1/number-counting/1-100/ | 1,725,855,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00726.warc.gz | 128,249,215 | 45,693 | # 30 Counting Worksheets 1-100 | Free Printable
In the following discussion, we’re diving into counting up to 100 through various engaging activities. These counting worksheets 1-100 will help you to visualize and understand the number system and count them correctly. With our free worksheets, students from first grade will learn and practice basic counting abilities.
Welcome to the world of numbers! Get ready to explore big numbers and see how they come together in exciting ways
## Worksheet for Counting 1-100 and Filling Half-Full Number Chart
In this worksheet, the chart is half-filled and half-empty for students to fill in. Students will count from 1 to 100 and fill in the gaps to complete the chart.
## Worksheet for Counting 1-100 and Filling Quarter-Full Number Chart
In this worksheet, the chart is quarter-filled. Students have to count from 1 to 100 and fill in the gaps to complete the chart.
## Worksheet for Counting 1-100 and Filling Empty Number Chart
In this worksheet, the chart is almost empty. Students will count from 1 to 100 and fill in the gaps to complete the chart.
## Worksheet for Counting Blocks to 1-100
In the first worksheet, students will count the blocks and write the answers in the boxes. In the second worksheet, they have to count tens, and in the third worksheet, they will count tens and ones and write the answers.
## Worksheet for Counting 1-100 and Completing the Sentences
In this worksheet, students will count the objects carefully to complete the sentences correctly.
## Count and Fill in the Gaps in Number Lines 1-100
Using these worksheets, students will count and fill in the gaps in number lines 1 to 100.
## Worksheet for following the path by counting 1-100
Using these worksheets, students will count 1-100 and color to follow the path to go to the park.
## Worksheet for Counting 1-100 Backward and Filling Number Chart
In this worksheet, students will count backward and fill in the gaps to complete the chart.
## Worksheet for Counting Objects from 1-100
In this worksheet, students have to count the objects and write their answers carefully to complete the task.
So today, we’ve discussed the counting worksheets with the number range from 1-100. To make our discussion more interactive, we have gone through some fun and engaging activities like counting in a number chart, counting with blocks, completing sentences after counting, counting and filling gaps, object counting, and backward counting. Download our free worksheets and continue to explore the realm of numbers. | 552 | 2,565 | {"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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.902807 |
http://www.ramblingteacher.com/2015/07/problems-that-do-not-compute.html | 1,726,360,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00102.warc.gz | 46,055,781 | 19,661 | ### Problems that do not compute
Today, my task was to convince my students that there are problems which computers cannot solve in a practical time, say a year or a few years! It was meant to be an informal introduction to the concept of the limits of computation and we did it by exploring two specific problems: The towers of Hanoi and the Travelling Salesman.
### The Towers of Hanoi
Legend has it that three tall towers stand vertically outside of a temple. The priests of that temple are tasked with transferring the 64 golden disks from one tower to another one. The conditions are:
1. Only one disk can be moved at any one time
2. No disk can rest on top of one that is smaller than itself
When all 64 disks have been moved, the world would end!
At first, we played the game with 1, 2, 3, 4 and then 5 disks, each time recording the minimum number of moves required. The video below shows our attempt with 3 disks:
We now had to deduce a general formula for the number of moves required to transfer any number of disks. To do this, consider the table below, to which I have added a third column to help the reader (a luxury which the students didn't have!):
Number of moves m = 2 to the nth power - 1. For 64 disks, we need roughly 16 x 1 billion x 1 billion moves! If the priests could make a move every 10 seconds, “it would take them well over five trillion years to get the job done.” (Harel, Algorithmics: The Spirit of Computing, 3rd Ed, p. 182, emphasis mine)
Assuming that a computer can make a move each microsecond, it would take around 1,000 years to work this problem out!
That was our first taste of an "intractable" problem, one that cannot be solved in a realistic time frame. We went on to explore another problem. If you found this post interesting and would like to know how we approached the Travelling Salesman Problem, let me know and I will try to report on that too. | 449 | 1,904 | {"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-2024-38 | latest | en | 0.966175 |
http://deleeuwpdx.net/pubfolders/gower/gower.html | 1,582,855,019,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146940.95/warc/CC-MAIN-20200228012313-20200228042313-00031.warc.gz | 41,948,533 | 232,230 | Note: This is a working paper which may be expanded/updated frequently. All suggestions for improvement are welcome. The directory gifi.stat.ucla.edu/gower has a pdf version, the complete Rmd file with all code chunks, and the bib file.
# 1 Theory
In full-dimensional scaling (De Leeuw (1993), De Leeuw, Groenen, and Mair (2016)) or FDS we minimize a loss function defined as $$\sigma(C):=\sum_{i=1}^n\sum_{j=1}^nw_{ij}(\delta_{ij}-d_{ij}(C))^2$$ over all positive semi-definite (psd) matrices $$C$$. Here $$w_{ij}$$ are known non-negative weights and $$\delta_{ij}$$ are known non-negative dissimilarities. To prevent complete triviality we assume $$w_{ij}\delta_{ij}>0$$ for at least one pair $$i,j$$. The distances $$d_{ij}(C)$$ are defined by $$d_{ij}(C):=\sqrt{c_{ii}+c_{jj}-2c_{ij}}.$$
It follows that $$\sigma$$ is convex on the convex cone of psd matrices, and thus all local minimizers of the FDS problem are global. Suppose $$\mathcal{I}$$ is the set of all pairs $$(i,j)$$ with $$w_{ij}\delta_{ij}>0$$. Then (De Leeuw (1984)) at a local minimum $$d_{ij}(C)>0$$ for all $$(i,j)\in\mathcal{I}$$. Thus $$\sigma$$ is differentiable in a neighborhood of local minima.
We need some additional definitions. We use the Loewner order, i.e. $$A\gtrsim B$$ means that $$A-B$$ is psd. Define the matrices $A_{ij}:=(e_i-e_j)(e_i-e_j)',$ with $$e_i$$ and $$e_j$$ unit vectors, i.e. vectors with one element equal to one and the rest equal to zero. Note that $$d_{ij}(C)=\sqrt{\mathbf{tr}\ A_{ij}C}$$. Using the matrices $$A_{ij}$$ we define \begin{align} V&:=\sum_{j=1}^n\sum_{j=1}^n w_{ij}A_{ij},\label{E:vdef}\\ B(C)&:=\sum_{j=1}^n\sum_{j=1}^n w_{ij}\frac{\delta_{ij}}{d_{ij}(C)}A_{ij}.\label{E:bdef} \end{align}
If we assume, for convenience, that $$\sum_{j=1}^n\sum_{j=1}^n w_{ij}\delta_{ij}^2=1$$, then $$\sigma(C)=1-2\mathbf{tr}\ B(C)C+\mathbf{tr}\ VC$$.
From convex analysis (Rockafellar (1970), theorem 31.4) the necessary and sufficient conditions for $$C$$ to be a minimizer are \begin{align} C&\gtrsim 0,\label{E:psd}\\ B(C)&\lesssim V,\label{E:pos}\\ \mathbf{tr}\ (B(C)-V)C&=0.\label{E:ort} \end{align}
In general we do not have uniqueness. Think of the case in which $$\mathcal{I}$$ has only one element, say $$(1,2)$$. Then any psd matrix $$C$$ with $$c_{11}+c_{22}-2c_{12}=\delta_{12}^2$$ is a minimizer with stress equal to zero. The local minimum is unique if and only if $$C=0$$ is the only psd solution of the homogeneous linear system $$d_{ij}^2(C)=\mathbf{tr}\ A_{ij}C=0$$ with $$(i,j)\in\mathcal{I}$$. Let us assume that this is the case, and that the global minimizer $$C$$ is of rank $$p$$. Then $$p$$ is called the Gower rank of the pair $$(W,\Delta)$$. If the minimum is not unique the Gower rank is the minimum rank of the minimizers.
There is a clear analogy with classical MDS, in which the corresponding quantity $$p^\star$$ is the number of positive eigenvalues of $$-\frac12 J\Delta^{(2)}J$$, the doubly-centered matrix of squared dissimilarities. In fact, the Gower conjecture is that the Gower rank $$p$$ is less than or equal to $$p^\star$$.
If $$C=XX'$$ is a full rank decomposition, with $$X$$ of dimensions $$n\times p$$, then we can write $$\eqref{E:ort}$$ as $$V^+B(C)X=X$$, using superscript $$+$$ for the Moore-Penrose inverse. Thus $$X$$ are eigenvectors of $$V^+B(C)$$ with eigenvalues equal to one. Condition $$\eqref{E:pos}$$ says that all eigenvalues of $$V^+B(C)$$ are less than or equal to one, and $$X$$ corresponds with the $$p$$ largest eigenvalues.
# 2 Algorithm
Multidimensional scaling (MDS) algorithms such as smacof (De Leeuw and Mair (2009)) parametrize the MDS problem in terms of an $$n\times p$$ configuration $$X$$. Thus $$\sigma(X):=\sum_{i=1}^n\sum_{j=1}^nw_{ij}(\delta_{ij}-d_{ij}(X))^2$$
with $$d_{ij}(X)=\sqrt{\mathbf{tr}\ A_{ij}XX'}$$. An MDS(p) algorithm minimizes $$\sigma$$ over all $$n\times p$$ configurations.
An iterative MDS(p) algorithm converges to a configuration with $$B(X)X=VX$$, where $$B(X)$$ is defined by $$\eqref{E:bdef}$$ with $$C=XX'$$. Thus $$C=XX'$$ solves $$\eqref{E:psd}$$ and $$\eqref{E:ort}$$, but the eigenvalues of $$V^+B(X)$$ can be larger than one, and consequently $$C$$ does not solve the FDS problem. But conversely if the MDS(p) solution $$X$$ gives a matrix $$V^+B(X)$$ with $$p$$ eigenvalues equal to one, then $$p$$ is the Gower rank and $$C=XX'$$ solves the FDS problem. This also implies that $$X$$ gives the global minimum of the MDS(q) problem for all $$q\geq p$$.
# 3 First Example
The following asymmetric matrix is taken from Wang and Chin (2010), table 5. Wong and Beasley (1990) constructed artificial data comparing seven hypothetical departments in a university in terms of three input variables (number of academic staff, academic staff salaries, support staff salaries) and three output variables (number of undergraduates, number of postgraduate students, number of resaerch papers). The data are used to construct a cross-efficiency matrix, which uses linear programming to compute input and output weights that measure efficiency of a department relative to another department. This produces cross efficiencies between zero and one, which can be interpreted as similarities, so we actually use one minus the data from Wang and Chin (2010).
## a b b d d f g
## A 0.0000 0.0188 0.2310 0.3589 0.0618 0.0000 0.0000
## B 0.0781 0.0000 0.2281 0.2987 0.1010 0.0000 0.0000
## C 0.0000 0.1490 0.0000 0.5458 0.5050 0.0000 0.7059
## D 0.3125 0.0000 0.2651 0.1803 0.2350 0.0494 0.0000
## E 0.0000 0.1539 0.3349 0.5865 0.0000 0.0896 0.0000
## F 0.0000 0.0188 0.2310 0.3589 0.0618 0.0000 0.0000
## G 0.0000 0.0188 0.2310 0.3589 0.0618 0.0000 0.0000
It was observed by Shahin Ashkiani, PhD candidate at Autonomous University of Barcelona, that metric unfolding solutions for this matrix exhibit some curious behaviour (personal communication, June 4, 2016). Increasing the dimensionality of the solution does not decrease the minimum stress. The MDS(p) solution has the same minimum stress for all $$p\geq 2$$. This indicates the Gower rank of these data is two.
Although metric unfolding can be more efficiently done by smacofRect(), we use smacofSym() on the $$14\times 14$$ augmented matrix, with the data in the off-diagonal submatrices and with zero weights for the diagonal submatrices. This corresponds more directly with the notation we use in this paper. We start with a random configuration and iterate until difference between successive stresses is less that 1e-15.
## ndim = 1 | iterations = 5 | stress = 0.5805246365
## ndim = 2 | iterations = 136 | stress = 0.4067723531
## ndim = 3 | iterations = 129 | stress = 0.4067723531
## ndim = 4 | iterations = 144 | stress = 0.4067723531
## ndim = 5 | iterations = 119 | stress = 0.4067723531
## ndim = 6 | iterations = 142 | stress = 0.4067723531
We can also illustrate this by showing the singular values of the configuration computed by smacofSym().
## [1] 2.724878
## [1] 2.574972 1.997666
## [1] 2.5749717 1.9976662 0.0000001
## [1] 2.574972 1.997666 0.000000 0.000000
## [1] 2.5749712 1.9976669 0.0000005 0.0000000 0.0000000
## [1] 2.574972 1.997666 0.000000 0.000000 0.000000 0.000000
# 4 Second Example
There is a more interesting example in Wang and Chin (2010), analyzing cross efficiences computed from data of Tofallis (1997). In this case the decision making units are 14 airlines. The input variables are aircraft capacity, operating cost, and non-flight assets. The output variables are passenger kilometres and non-passenger revenue. The cross-efficiences are
## a b c d e f g h i j k l
## A 0.13 0.55 0.38 0.13 0.15 0.53 0.19 0.21 0.30 0.25 0.13 0.23
## B 0.83 0.66 0.95 0.83 0.83 0.98 0.75 0.73 0.72 0.79 0.83 0.80
## C 0.12 0.81 0.05 0.12 0.12 0.31 0.28 0.32 0.38 0.22 0.12 0.19
## D 0.04 0.57 0.30 0.04 0.06 0.30 0.18 0.21 0.30 0.19 0.04 0.17
## E 0.03 0.63 0.00 0.03 0.00 0.00 0.23 0.26 0.22 0.00 0.03 0.00
## F 0.12 0.89 0.04 0.12 0.12 0.02 0.34 0.39 0.49 0.28 0.12 0.25
## G 0.08 0.22 0.52 0.08 0.12 0.66 0.00 0.00 0.16 0.22 0.08 0.20
## H 0.22 0.39 0.48 0.22 0.23 0.71 0.15 0.14 0.18 0.25 0.22 0.24
## I 0.21 0.27 0.49 0.21 0.21 0.73 0.12 0.09 0.05 0.16 0.21 0.16
## J 0.22 0.36 0.35 0.22 0.17 0.64 0.22 0.21 0.00 0.00 0.22 0.03
## K 0.00 0.00 0.57 0.00 0.00 0.56 0.00 0.00 0.00 0.00 0.00 0.00
## L 0.05 0.37 0.25 0.05 0.04 0.56 0.06 0.06 0.00 0.00 0.05 0.00
## M 0.00 0.57 0.00 0.00 0.00 0.54 0.00 0.00 0.00 0.02 0.00 0.00
## N 0.00 0.77 0.00 0.00 0.00 0.00 0.22 0.27 0.35 0.14 0.00 0.12
## m n
## A 0.13 0.13
## B 0.83 0.83
## C 0.12 0.12
## D 0.04 0.04
## E 0.03 0.03
## F 0.12 0.12
## G 0.08 0.08
## H 0.22 0.22
## I 0.21 0.21
## J 0.22 0.22
## K 0.00 0.00
## L 0.05 0.05
## M 0.00 0.00
## N 0.00 0.00
We repeat the same analysis as in the first example. In this case we conclude the Gower rank is equal to three.
## ndim = 1 | iterations = 7 | stress = 0.4984859586
## ndim = 2 | iterations = 291 | stress = 0.2403664444
## ndim = 3 | iterations = 1340 | stress = 0.2386715692
## ndim = 4 | iterations = 1784 | stress = 0.2386715692
## ndim = 5 | iterations = 1870 | stress = 0.2386715692
## ndim = 6 | iterations = 1920 | stress = 0.2386715692
The singular values for the second example are
## [1] 4.487714
## [1] 4.247787 2.580708
## [1] 4.151090 2.632157 1.042903
## [1] 4.151090 2.632154 1.042909 0.000020
## [1] 4.151090 2.632154 1.042909 0.000020 0.000000
## [1] 4.151090 2.632154 1.042909 0.000020 0.000000 0.000000
# 5 Third Example
The third example was already analyzed in De Leeuw, Groenen, and Mair (2016). The Ekman (1954) color data give similarities between 14 colors. We transform to dissimilarities by using the transformation $$(1-x)^3$$, and we repeat the analysis from the previous sections. Note that this is not an unfolding example, all weights are equal to one. The Gower rank turns out to be two, i.e. for this transformation the two-dimensional smacof solution is the global minimizer of stress in $$p\geq 2$$ dimensions.
## ndim = 1 | iterations = 4 | stress = 0.3893229700
## ndim = 2 | iterations = 47 | stress = 0.1049991045
## ndim = 3 | iterations = 174 | stress = 0.1049991045
## ndim = 4 | iterations = 174 | stress = 0.1049991045
## ndim = 5 | iterations = 173 | stress = 0.1049991045
## ndim = 6 | iterations = 175 | stress = 0.1049991045
The singular values for the third example are
## [1] 2.348357
## [1] 2.060186 1.477827
## [1] 2.0601856 1.4778274 0.0000007
## [1] 2.0601856 1.4778274 0.0000007 0.0000000
## [1] 2.0601856 1.4778274 0.0000007 0.0000000 0.0000000
## [1] 2.0601856 1.4778274 0.0000007 0.0000000 0.0000000 0.0000000
# 6 NEWS
001 06/06/16
• First release
002 08/06/16
• Third Example
003 08/06/16
# References
De Leeuw, J. 1984. “Differentiability of Kruskal’s Stress at a Local Minimum.” Psychometrika 49: 111–13. http://www.stat.ucla.edu/~deleeuw/janspubs/1984/articles/deleeuw_A_84f.pdf.
———. 1993. “Fitting Distances by Least Squares.” Preprint Series 130. Los Angeles, CA: UCLA Department of Statistics. http://www.stat.ucla.edu/~deleeuw/janspubs/1993/reports/deleeuw_R_93c.pdf.
De Leeuw, J., and P. Mair. 2009. “Multidimensional Scaling Using Majorization: SMACOF in R.” Journal of Statistical Software 31 (3): 1–30. http://www.stat.ucla.edu/~deleeuw/janspubs/2009/articles/deleeuw_mair_A_09c.pdf.
De Leeuw, J., P. Groenen, and P. Mair. 2016. “Full-Dimensional Scaling.” doi:10.13140/RG.2.1.1038.4407.
Ekman, G. 1954. “Dimensions of Color Vision.” Journal of Psychology 38: 467–74.
Rockafellar, R.T. 1970. Convex Analysis. Princeton University Press.
Tofallis, C. 1997. “Input Efficiency Profiling: an Application to Airlines.” Computers and Operations Research 24: 253–58.
Wang, Y.-M., and K.-S. Chin. 2010. “A Neutral DEA Model for Cross-efficiency Evaluation and its Extension.” Expert Systems with Applications 37: 3666–75.
Wong, Y.-H. B., and J.E. Beasley. 1990. “Restricting Weight Flexibility in Data Envelopment Analysis.” The Journal of the Operational Research Society 41: 829–35. | 4,842 | 12,362 | {"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": 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": 2, "equation": 3, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2020-10 | latest | en | 0.617966 |
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## Nested IF statements with several variables - convert from excel to DAX
I'm currently trying to convert existing excel formulas I have into PowerQuery (which i've only just started using).
A lot of it is straightforward, but i've come unstuck with nested IF statements.
My formula in excel is:
IF([@ED]=TRUE,"high financial stress",
IF(AND([@LWM1]>2,[@FC]>2,[@LWM2]>2,[@ED]=FALSE,[@R]=FALSE),"no financial stress",
IF(AND([@LWM1]=1,(OR([@FC]<3)),(OR([@LWM2]<3)),[@R]=TRUE),"high financial stress","financial stress")))
And i need to convert this to DAX for a custom column.
Any help would be hugely appreciated
1 ACCEPTED SOLUTION
Accepted Solutions
Community Support Team
## Re: Nested IF statements with several variables - convert from excel to DAX
You modify your formula using DAX below:
```Result =
IF (
[@ED] = TRUE,
"high financial stress",
IF (
( [@LWM1] > 2 )
&& ( [@FC] > 2 )
&& ( [@LWM2] > 2 )
&& ( [@ED] = FALSE )
&& ( [@R] = FALSE ),
"no financial stress",
IF (
( [@LWM1] = 1 )
&& ( ( [@FC] < 3 )
|| ( [@LWM2] < 3 )
|| ( [@R] = TRUE ) ),
"high financial stress",
"financial stress"
)
)
)
```
Community Support Team _ Jimmy Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Community Support Team
## Re: Nested IF statements with several variables - convert from excel to DAX
You modify your formula using DAX below:
```Result =
IF (
[@ED] = TRUE,
"high financial stress",
IF (
( [@LWM1] > 2 )
&& ( [@FC] > 2 )
&& ( [@LWM2] > 2 )
&& ( [@ED] = FALSE )
&& ( [@R] = FALSE ),
"no financial stress",
IF (
( [@LWM1] = 1 )
&& ( ( [@FC] < 3 )
|| ( [@LWM2] < 3 )
|| ( [@R] = TRUE ) ),
"high financial stress",
"financial stress"
)
)
)
```
Community Support Team _ Jimmy Tao
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly. | 567 | 1,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.858926 |
https://www.coursehero.com/file/6235169/Lesson-25-Printable-PPT/ | 1,496,022,281,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612003.36/warc/CC-MAIN-20170528235437-20170529015437-00130.warc.gz | 1,056,919,946 | 25,282 | Lesson_2.5_Printable_PPT
# Lesson_2.5_Printable_PPT - Electric Potential Comparing...
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Unformatted text preview: Electric Potential Comparing gravitational and electric fields. You are familiar with gravitational potential energy. Gravity does egative Gravity does positive An external force does work against gravity to raise an object. The work done by gravity on the object is negative negative work work This work done by the external force gives the object potential energy ( U ) When the object is allowed to fall, gravity does positive work on the object The net work on the object in the Potential energy is a measure of the negative work done by gravity closed path is zero. A force field where the net work done in a closed path is zero is called a conservative force field . Potential energy in a conservative force field is the negative work done by the conservative force. U F ds = - ⋅ ∫ Here we study the motion of electric charges in an electric field. Since there are two types of electric charges, we will take the positive charge as the standard test charge . positive st charge Like gravitational field, electric field is a conservative force field. A positive test charge is placed in a field produced by a positive charge. An external force has to do work to move the test charge from position A to position B increasing its potential energy. The electric field does negative work on the test harge increasing its When the test charge is allowed to move, it falls to low potential in the direction of the electric field when the electric field does positive work on it. charge increasing its U The potential decreases in the direction of the field. The potential energy of the test charge at infinity is zero. When the test charge is moved opposite to the direction of the field, the gain in potential energy is a measure of the negative work done by the filed. U F ds = - ⋅ ∫ Consider an electric field E generated by a positive charge . Since E varies with the distance from the charge, the electric field has different values at different points. + + + q o F = E q o E F = -E q o A test charge q o placed in the field experiences a force F = E q o . To move the test charge in the direction of increasing potential, an equal opposite external force is needed. F = - Eq o + + + q o F = E q o E F = -E q o When the test charge is moved a distance dr opposite to the direction of the field, its increase in potential energy is: dr dU = - Eq o dr F = - Eq o The increase in potential nergy per unit charge is: This is a measure of the potential difference between the two points and it is the work done on a unit test charge to move it through a distance dr . energy per unit charge is: o dU q = - E dr + + + q o F = E q o E F = -E q o dr o dU q = - E dr We will use V to represent potential at a point and ∆ V the potential difference between two points separated by small distance ∆ r ....
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## This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.
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Lesson_2.5_Printable_PPT - Electric Potential Comparing...
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Ask a homework question - tutors are online | 841 | 3,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2017-22 | longest | en | 0.917857 |
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# length conversion
## Amount: 1 Japanese mō (毛, 毫) of length Equals: 0.000100 Japanese kanejaku (曲尺) in length
Converting Japanese mō to Japanese kanejaku value in the length units scale.
TOGGLE : from Japanese kanejaku into Japanese mō in the other way around.
## length from Japanese mō to Japanese kanejaku Conversion Results:
### Enter a New Japanese mō Amount of length to Convert From
* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
* Precision is how many numbers after decimal point (1 - 9)
Enter Amount :
Decimal Precision :
CONVERT : between other length measuring units - complete list.
Conversion calculator for webmasters.
## Length, Distance, Height & Depth units
Distance in the metric sense from any two A to Z points (interchangeable with Z and A), also applies to physical lengths, depths, heights or simply farness. Tool with multiple distance, depth and length measurement units.
Convert length measuring units between Japanese mō (毛, 毫) and Japanese kanejaku (曲尺) but in the other reverse direction from Japanese kanejaku into Japanese mō.
conversion result for length: From Symbol Equals Result To Symbol 1 Japanese mō 毛, 毫 = 0.000100 Japanese kanejaku 曲尺
# Converter type: length units
This online length from 毛, 毫 into 曲尺 converter is a handy tool not just for certified or experienced professionals.
First unit: Japanese mō (毛, 毫) is used for measuring length.
Second: Japanese kanejaku (曲尺) is unit of length.
## 0.000100 曲尺 is converted to 1 of what?
The Japanese kanejaku unit number 0.000100 曲尺 converts to 1 毛, 毫, one Japanese mō. It is the EQUAL length value of 1 Japanese mō but in the Japanese kanejaku length unit alternative.
How to convert 2 Japanese mō (毛, 毫) into Japanese kanejaku (曲尺)? Is there a calculation formula?
First divide the two units variables. Then multiply the result by 2 - for example:
0.0001 * 2 (or divide it by / 0.5)
QUESTION:
1 毛, 毫 = ? 曲尺
1 毛, 毫 = 0.000100 曲尺
## Other applications for this length calculator ...
With the above mentioned two-units calculating service it provides, this length converter proved to be useful also as a teaching tool:
1. in practicing Japanese mō and Japanese kanejaku ( 毛, 毫 vs. 曲尺 ) values exchange.
2. for conversion factors training exercises between unit pairs.
3. work with length's values and properties.
International unit symbols for these two length measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for Japanese mō is:
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for Japanese kanejaku is:
### One Japanese mō of length converted to Japanese kanejaku equals to 0.000100 曲尺
How many Japanese kanejaku of length are in 1 Japanese mō? The answer is: The change of 1 毛, 毫 ( Japanese mō ) unit of length measure equals = to 0.000100 曲尺 ( Japanese kanejaku ) as the equivalent measure for the same length type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solution. If there is an exact known measure in 毛, 毫 - Japanese mō for length amount, the rule is that the Japanese mō number gets converted into 曲尺 - Japanese kanejaku or any other length unit absolutely exactly.
Conversion for how many Japanese kanejaku ( 曲尺 ) of length are contained in a Japanese mō ( 1 毛, 毫 ). Or, how much in Japanese kanejaku of length is in 1 Japanese mō? To link to this length Japanese mō to Japanese kanejaku online converter simply cut and paste the following.
The link to this tool will appear as: length from Japanese mō (毛, 毫) to Japanese kanejaku (曲尺) conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting. | 1,036 | 3,978 | {"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-2018-26 | latest | en | 0.814989 |
https://ctarchery.org/and-pdf/1370-pythagoras-theorem-questions-and-answers-pdf-128-670.php | 1,652,976,630,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00271.warc.gz | 237,193,169 | 10,152 | # Pythagoras Theorem Questions And Answers Pdf
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All questions and answers from the Mathematics Solutions Book of Class 7 Math Chapter 14 are provided here for you for free. In the figures below, find the value of ' x '. Hence, side LN is the hypotenuse.
Example 1 — The hypotenuse of a right triangle is 1 inch longer than the longer leg. The shorter leg is 7 inches shorter than the longer leg. Find the length of the hypotenuse. Example 2 —Carrie works dues south of her apartment. Her friend Sarah works due east of the apartment.
## More Challenging Pythagorean Theorem Problems - Answers
Here are some questions which can be answered using Pythagoras' Theorem. You can earn a trophy if you get at least 14 questions correct. Each time you finish a question click the 'Check' button lower down the page to see if you got it right! What is the length of the longest side of a right angled triangle if the two shorter sides are 5cm and 12cm? A rectangular swimming pool is 23m wide and 44m long. Calculate the length of a diagonal in metres to 1 decimal place. A ladder is 7m long.
Measurement and Geometry : Module 15 Year : PDF Version of module. Is there a simple relationship between the length of the sides of a triangle? Apart from the fact that the sum of any two sides is greater than the third, there is, in general, no simple relationship between the three sides of a triangle. Among the set of all triangles, there is a special class, known as right-angled triangles or right triangles that contain a right angle. The longest side in a right-angled triangle is called the hypotenuse. The word is connected with a Greek word meaning to stretch because the ancient Egyptians discovered that if you take a piece of rope, mark off 3 units, then 4 units and then 5 units, this can be stretched to form a triangle that contains a right angle.
## Pythagorean Theorem Practice Worksheets with Answers PDF
Practice the questions given in the worksheet on Pythagorean Theorem. We know, in a right angled triangle the square of the hypotenuse is equal to the sum of the squares of its remaining two sides. The side of the triangle are of length 7. Is this triangle a right triangle? If so, which side is the hypotenuse? A tree broke from a point but did not separate. Its top touched the ground at a distance of 24 m from its base.
These solutions for Pythagoras Theorem are extremely popular among Class 10 students for Math Pythagoras Theorem Solutions come handy for quickly completing your homework and preparing for exams. Identify, with reason, which of the following are Pythagorean triplets. We know that, In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided. In the given figure. For finding AB and BC with the help of information given in the figure, complete following activity. It is given that ABCD is a square. In the given figure, M is the midpoint of QR.
I had no time to compete my dissertation, but my friend the pythagorean theorem homework help recommended this website. How can we use the pythagorean theorem to figure how the distance from the catcher at home plate to throw out the the player that is trying to steal 2nd base. We provide perfect homework help to students all around the globe. The pythagorean theorem assignment answers pdf. The egyptians used a form of the pythagorean theorem to lay out their fields and the greeks borrowed it from the egyptians.
## Geometry: Pythagorean Theorem
Левый крайний Джорджтауна, подавая угловой, отправил мяч в аут, и трибуны негодующе загудели. Защитники поспешили на свою половину поля. - А ты? - спросил Беккер. - Что предпочитаешь .
Пожалуй, дело кончится тем, что его выставят на улицу. Клушар продолжал бушевать: - И этот полицейский из вашего города тоже хорош. Заставил меня сесть на мотоцикл.
ГЛАВА 36 Ручное отключение. Сьюзан отказывалась что-либо понимать. Она была абсолютно уверена, что не вводила такой команды - во всяком случае, намеренно. Подумала, что, может быть, спутала последовательность нажатия клавиш.
Отчаянным движением он развернул Сьюзан так, чтобы она оказалась выше его, и начал спускаться.
### Pythagoras' Theorem Exercise
Как только получит денежки, так и улетит. Беккер почувствовал тошноту. Это какая-то глупая шутка.
Самое место, где толкнуть колечко: богатые туристы и все такое прочее. Как только получит денежки, так и улетит. Беккер почувствовал тошноту. Это какая-то глупая шутка. Он не находил слов. - Ты знаешь ее фамилию.
ГЛАВА 22 Дэвид Беккер быстро подошел к койке и посмотрел на спящего старика. Правое запястье в гипсе. На вид за шестьдесят, может быть, около семидесяти. Белоснежные волосы аккуратно зачесаны набок, в центре лба темно-красный рубец, тянущийся к правому глазу. Ничего себе маленькая шишка, - подумал Беккер, вспомнив слова лейтенанта.
Конечно. Алгоритм, не подающийся грубой силе, никогда не устареет, какими бы мощными ни стали компьютеры, взламывающие шифры. Когда-нибудь он станет мировым стандартом. Сьюзан глубоко вздохнула. - Да поможет нам Бог, - прошептала .
Но вместо признаков срыва Фонтейн обнаружил подготовительную работу над беспрецедентной разведывательной операцией, которую только можно было себе представить. Неудивительно, что Стратмор просиживает штаны на работе. Если он сумеет реализовать свой замысел, это стократно компенсирует провал Попрыгунчика. Фонтейн пришел к выводу, что Стратмор в полном порядке, что он трудится на сто десять процентов, все так же хитер, умен и в высшей степени лоялен, впрочем - как .
Если бы возникла проблема, он тут же позвонил бы. Мидж долго молчала. Джабба услышал в трубке вздох - но не мог сказать, вздох ли это облегчения. - Итак, ты уверен, что врет моя статистика.
Dov Переведя взгляд на рабочий кабинет Стратмора, она поняла, что больше не может ждать, пусть даже помешает его разговору по телефону. Она встала и направилась к двери. Хейл внезапно почувствовал беспокойство - скорее всего из-за необычного поведения Сьюзан.
- Я позвоню Стратмору и попрошу прислать нам письменное подтверждение. - Нет, - сказала Мидж, - игнорируя сарказм, прозвучавший в его словах. - Стратмор уже солгал нам. - Она окинула Бринкерхоффа оценивающим взглядом.
При первых же признаках беды он тут же поднял бы тревогу - а в этих стенах сие означает, что он позвонил бы. - Джабба сунул в рот кусочек сыра моцарелла. - Кроме всего прочего, вирус просто не может проникнуть в ТРАНСТЕКСТ. Сквозь строй - лучший антивирусный фильтр из всех, что я придумал. Через эту сеть ни один комар не пролетит.
Сейчас ему надо было совершить давно уже откладываемую прогулку в туалетную комнату.
Халохот ударился сначала о внешнюю стену и только затем о ступени, после чего, кувыркаясь, полетел головой. Пистолет выпал из его рук и звонко ударился о камень. Халохот пролетел пять полных витков спирали и замер.
Стратмор медленно повернулся к Сьюзан. Тоже неподвижная, она стояла у дверей шифровалки. Стратмор посмотрел на ее залитое слезами лицо, и ему показалось, что вся она засветилась в сиянии дневного света. Ангел, подумал. Ему захотелось увидеть ее глаза, он надеялся найти в них избавление.
Ничего не выйдет, - пробормотал. В разделе Служба сопровождения в справочнике было только три строчки; впрочем, ничего иного все равно не оставалось. Беккер знал лишь, что немец был с рыжеволосой спутницей, а в Испании это само по себе большая редкость.
Немецкий акцент и просьба снять девушку на ночь - это же очевидная подстава. Интересно, что они еще придумают. Телефон на столе громко зазвонил. Сеньор Ролдан поднял трубку с обычной для него самоуверенностью.
Мгновение спустя компьютер подал звуковой сигнал. СЛЕДОПЫТ ОТОЗВАН Хейл улыбнулся. Компьютер только что отдал ее Следопыту команду самоуничтожиться раньше времени, так что ей не удастся найти то, что она ищет. Помня, что не должен оставлять следов, Хейл вошел в систему регистрации действий и удалил все свои команды, после чего вновь ввел личный пароль Сьюзан. Монитор погас.
Наверху включились огнетушители. ТРАНСТЕКСТ стонал. Выли сирены.
- Я воспользуюсь вашим лифтом.
#### Basic welsh a grammar and workbook pdf
15.06.2021 at 21:55
09.02.2021 at 00:48
#### Cisco bgp-4 command and configuration handbook pdf
09.02.2021 at 00:48
1. Percy G. 16.05.2021 at 23:46
The following diagram gives the formula for the Pythagorean Theorem, scroll down the page for more examples and solutions that use the Pythagorean Theorem.
2. Zara A. 17.05.2021 at 15:57
3. Jennifer H. 17.05.2021 at 18:32
Welcome to our Pythagoras' Theorem Questions area.
4. DorothГ©e G. 19.05.2021 at 13:28
All english idioms with their meanings pdf topology and functional analysis pdf
5. AurГ©lie A. 20.05.2021 at 05:19
Learn html5 and css3 with w3schools pdf free pdf cellular and molecular immunology abbas 7th edition download | 2,686 | 9,047 | {"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} | 4.21875 | 4 | CC-MAIN-2022-21 | latest | en | 0.891885 |
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posted by .
kyle's credit bill is \$360. kyle sends a check to the credit card company for \$70, charges another \$109 in merchandise, and then pays off another \$231 of the bill. how much does kyle owe the company?
• algebra -
That depends on the time period between these transactions. Credit cards charge interest on the unpaid balance each month.
• algebra -
-360+70-109+231
kyle owes =168\$ +interest
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Started by March 28, 2001
Hello, I have some questions on sampling the bandpass signals. If I have a analog band pass signal with the frequncy range Fmin to Fmax,is it possible to sample with 2*(Fmax-Fmin) instead of The Nyquist frequncy ,2*Fmax? Is it sufficient to use a bandpass filter to reconstruct the analog signal from the sampled signal? If I use a filtering operation on the sampled signal,is it equivalent to the same operation on a signal sampled with Nyquist frequncy? Is there any difference in sampling narrowband pass signals and wide band pass signals? And can some one point out the resources on this topic which give the details of system design and mathematical details? Thanks in advance. Regards Kiran
Kiran > Hello, > I have some questions on sampling the bandpass signals. > > If I have a analog band pass signal with the frequncy range Fmin to > Fmax,is it possible to sample with 2*(Fmax-Fmin) instead of The > Nyquist frequncy ,2*Fmax? > Yes, this will translate the down to baseband. Normally people perform complex sampling (in-phase and quadrature) to ensure an accurate representation. > Is it sufficient to use a bandpass filter to reconstruct the analog > signal from the sampled signal? > Once you sample the data, it no longer has the original freq content. In order to reconstruct the signal, you would need to bandshift the data. > If I use a filtering operation on the sampled signal,is it equivalent > to the same operation on a signal sampled with Nyquist frequncy? > No, the frequency content has been shifted. However, it is usually easier to design an equivalent filter that can be applied to the basebanded signal. > Is there any difference in sampling narrowband pass signals and wide > band pass signals? > Don't think so. > And can some one point out the resources on this topic which give the > details of system design and mathematical details? > You probably want a book on communication theory and one on spectral analysis. Ex. An Introduction to Random Signals and Communication Theory, B.P. Lathi, Int'l Textbook Company Discrete-Time Signal Processing, Alan Oppenheim & Ronald Schafer, Prentice Hall > Thanks in advance. > > Regards > Kiran Bill Zimmerman Good mathematics promotes Intellectual Economy by helping us understand more while obliging us to memorize less without violating . Einstein's Dictum: "Everything should be as simple as possible, but no simpler." - W. Kahan ManTech (954)929-8604 Voice One Oakwood Blvd. Suite 180 (954)925-0205 FAX Hollywood, FL 33020 | 582 | 2,591 | {"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-2024-22 | latest | en | 0.91597 |
http://puzzle.queryhome.com/610/how-many-toffees-you-can-get-in-rs-15 | 1,500,835,222,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424586.47/warc/CC-MAIN-20170723182550-20170723202550-00034.warc.gz | 261,139,398 | 30,830 | # How many toffee’s you can get in Rs.15?
+1 vote
543 views
Cost of one toffee is Rs.1 and as a special promotion scheme you get 1 toffee free for three wrappers.
How many toffee’s you can get in Rs.15?
posted May 2, 2014
## 1 Solution
Step 1: 15 from 15 rupees
Step 2: 5 from 15 wrappers
Step 3: 1 from three (2 wrappers remaining)
Step 4: 1 (2 wrappers remaining + 1 wrapper in step 3 gets added)
Total: 22
solution May 2, 2014
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There are seven thieves, They steal diamonds from a diamond merchant and run away in jungle. While running, night sets in and they decide to rest in the jungle When everybody’s sleeping, two of the best friends get up and decide to distribute the diamonds among themselves and run away. So they start distributing but find that one diamond was extra. So they decide to wake up 3rd one and divide the diamonds again …..only to their surprise they still find one diamond extra. So they decide to wake up fourth one. Again one diamond is spare. 5th woken up……still one extra. 6th still one extra. Now they wake up 7th and diamonds are distributed equally.
How many minimum diamonds they steal? | 304 | 1,154 | {"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} | 4.0625 | 4 | CC-MAIN-2017-30 | longest | en | 0.9327 |
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If it takes work W to stretch a Hooke’s-law spring (F = kx) a distance d from its unstressed length, determine the extra work required to stretch it an additional distance d (Hint: draw a graph and give answer in terms of W!).
I don't understand why the answer is not 2W since Force is proportional to x, or even how to begin using a graph to disprove why the answer is not 2W. Any help would be greatly appreciated.
-
Even without a graph, the answer is straightforward. The potential energy stored in a spring is proportional to the square of the difference b/w stretched and unstretched length,
$$V = \frac{1}{2}kx^2$$
Thus, the work required to stretch it (mind you, you have to do this slowly so that the Kinetic Energy isn't changed) would be:
$$W = \frac{1}{2}k(2d)^2 - \frac{1}{2}k(d)^2 = \frac{3}{2}kd^2 = 3W_0$$
This is just straightforward conservation of energy.
-
Ah, how did I miss something so easy? Thanks! – Joe Feb 16 '12 at 1:49
@Joe You might also want to see this question – pewfly Feb 16 '12 at 2:13 | 302 | 1,084 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2016-30 | latest | en | 0.942345 |
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## K.CC.A
Know number names and the count sequence.
Check out this Trial Lesson on K.CC.A Counting and Cardinality worksheets for Kindergarten!
Count Forward Starting From Any Number
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• K.CC.A
## Number Worksheet: Counting With Peapods
Help your child learn multiplication and develop their place value skills in a fun way with this number worksheet: Counting with Peapods! Each pod has 10 peas, and kids can simply count each one and add a zero to the ones place to get the answer. It's a great, colorful way to learn!
Number Worksheet: Counting With Peapods
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## Counting Fun Worksheet
Math and fun can go hand-in-hand with this free PDF worksheet! Bright, colourful illustrations will activate your child's brain, as they count the red and yellow flowers. How many can you see? Count together and write down the total. Make learning and counting fun for your little ones!
Counting Fun Worksheet
Worksheet
## Counting In The Neighborhood Part 2 Worksheet
Download this free worksheet to help your child build number sense and math reasoning skills! It uses pictures of objects they know, letting your child count and match the right numeral to the objects to count past 10. Let them feel successful in mastering numbers.
Counting In The Neighborhood Part 2 Worksheet
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## Holiday Counting Worksheet
Let's spread some holiday cheer! Santa needs help counting stockings hung up for gifts - can your little one lend a hand? Find the missing numbers on the worksheet and have them check the boxes for the correct answers. Time to get counting!
Holiday Counting Worksheet
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## Peter Pan Worksheet
Kindergarteners can join Peter Pan on an adventure to Neverland! They can help him by coloring squares to create a path, strengthening their shape identification skills in a fun maze. Let the journey begin!
Peter Pan Worksheet
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## Learning to Write 1 Worksheet
Learning to write numbers can be easy and fun! Get this new number worksheet to help your child excel. Practice counting, tracing and writing the number “one”, then circle the fish with “1” on them. Revise spelling and have fun! Get more tracing numbers worksheets at Kids Academy and enjoy the learning process.
Learning to Write 1 Worksheet
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## Kindergarten Number Tracing: Counting in Charts Worksheet
Little learners can find big numbers tricky! Help them practice counting and build early math skills, like place value and addition, with this kindergarten number tracing PDF worksheet. Count the dots to solve the problem and trace each answer!
Kindergarten Number Tracing: Counting in Charts Worksheet
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## Counting Forward On the Farm Worksheet
Help your child practice counting with this fun worksheet. Have them help a farmer feed the right animal by counting on from a starting point. This activity will help them build number line thinking and refine fine motor skills.
Counting Forward On the Farm Worksheet
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## Counting Numbers Worksheet For Kindergarten
This worksheet is a great way to get your child practicing foundational math skills. With bright pictures to count, counting numbers just got a lot more fun! Get them ready for future math success and help them hone their early math skills.
Counting Numbers Worksheet For Kindergarten
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## Smart Shopping: Trade Tens for a Hundred Worksheet
Use this worksheet to test counting skills: each bag has 10 pears. Ask your child how many pears are in 10 bags. Guide them as they count through the printout and check the answer below. See if they got it right!
Smart Shopping: Trade Tens for a Hundred Worksheet
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## Counting in the Neighborhood Part1 Worksheet
Counting using one-to-one representation is a great way to develop a child's number sense. Use objects, pictures and images to help counting beyond the fingers. This worksheet uses everyday objects to practice counting. It encourages children to find the correct number and count forward. Give it a try!
Counting in the Neighborhood Part1 Worksheet
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## Sweet Counting - Part 1 Worksheet
Help your child learn numbers by counting cakes! Download this fun worksheet and have them fill in the numbers missing in the line of cakes. They'll have fun helping the cook and be learning at the same time!
Sweet Counting - Part 1 Worksheet
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## Check. Checkmate or Stalemate? Worksheet
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Check. Checkmate or Stalemate? Worksheet
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## Number Puzzles Worksheet
In this tracing paper, your child must identify the missing numbers in the puzzle. Guide them as they trace the line for the correct number to complete the worksheet. This activity will help boost their problem-solving skills.
Number Puzzles Worksheet
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## Sidewalk Counting Worksheet
Counting is a vital math skill, helping kids with addition and improving their speed when solving math problems. Let them practice with this fun sidewalk counting worksheet: have them take a walk and fill in the next number in the sequence as they count.
Sidewalk Counting Worksheet
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## Number Tracing Worksheet For Kindergarten
Help your child develop addition skills and improve number recognition with this easy-to-use worksheet. Have them count the dots and add to solve each problem. Then, trace over the number to complete each row! It's a fun and simple way for kindergarteners to practice their math skills.
Number Tracing Worksheet For Kindergarten
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## What Is a Stalemate? Worksheet
Chess is an ideal way to develop logic, strategizing, problem-solving and more! Even kids can learn it. With this worksheet, children can learn about draws and stalemates. Sample boards will help them decide which team, black or white, is at a stalemate.
What Is a Stalemate? Worksheet
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## Learn to Write the Number 2 Worksheet
Teaching numbers should be fun! Ask your kids to count the two chicks, circle the groups of two among the animals, then trace and write the number and word. Get more fun worksheets from Kids Academy.
Learn to Write the Number 2 Worksheet
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## Number Tracing Worksheet
Kids learn math with counters and place value! This handy worksheet helps form a foundation for math skills and number sense! Have your child count the dots and trace the answers to the addition problems. It's a great way to support their math skills!
Number Tracing Worksheet
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## Trace And Write Number 0 Worksheet
Let’s learn numbers together! Start with zero: practice writing, counting, matching spots. Tracing lines will help. Then move on to the collection of number worksheets. Enjoy learning with Kids Academy!
Trace And Write Number 0 Worksheet
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## Find Stalemate! Worksheet
Teach your kids Chess and give them more than a hobby - equip them with critical thinking, rationale and problem-solving skills! This free worksheet helps them understand stalemates by examining various sample boards. By providing visual discernment and the joy of the game, your child's logical skills will soar!
Find Stalemate! Worksheet
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## Kindergarten Number Tracing: Medow Flowers Worksheet
Take your child to the garden to count flower petals and trace numbers with this fun kindergarten PDF worksheet. They'll learn to count and write, all while enjoying the sweet scent of the flowers.
Kindergarten Number Tracing: Medow Flowers Worksheet
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Learning Skills | 1,600 | 7,762 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.854444 |
https://www.edaboard.com/threads/how-to-use-%E2%88%AB-integral-to-calculate-heat-dissipation.278784/ | 1,709,587,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00431.warc.gz | 764,209,001 | 20,787 | Continue to Site
# How to use ∫ (integral) to calculate heat dissipation
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##### Newbie level 4
Hi;
I am going to build a coffee cup warmer plate using resistors - as an experiment.
Instead of using a USB port from my computer I am going to use a 5V, 700mA cell phone charger.
Using Ohms Law I have calculated that I there will be 7.14$\Omega$ of resistance so I have ordered 3 x 2.4$\Omega$ 5W resistors (Yageo - through-hole) and also 3 x .12$\Omega$ resistors (Yageo - through-hole) in case they are all at the bottom end of the 5% tolerance for each resistor ((3x2.4) + (3x.12) x.95) = 7.182$\Omega$ ... just slightly larger than the 7.14$\Omega$ required for the circuit.
I would like to try to determine ahead of time what will be the heat dissipated from the resistors so I can 1) learn how to estimate heating of circuit components, and 2) know if I will need to use a larger or smaller supply (and resistor values).
I have not done any calculus before and I am wondering if someone can assist me with a quick tutorial on how to use the ∫ (integral) symbol to calculate the heat that would be dissipated from resistors.
I have this forumla W= ∫ V(t) I(t) dt ... with a t2 at the top of the ∫ symbol and a t1 at the bottom of the ∫ symbol.
Here's an image of the formula:
The values that I am using are:
5V,
.7A,
7.14$\Omega$
100% duty cycle - always on.
Thanks for any assistance.
EDIT: For clarification, I would like to try to determine the temperature that the resistors will achieve.
Last edited:
When you have any two of voltage, current and resistance constant, you don't need to use calculus.
P = VI = (V^2)/R = (I^2)R = [(5 * 5)/7.14] W = 3.5W
You may need to use a power supply with larger capacity. 700mA is the maximum rating of the power supply and you're operating it at/near maximum ratings.
Hope this helps.
Tahmid.
Points: 2
What I am specifically looking to figure out is how power (W) through the circuit relates to the temperature that the resistors will generate as the purpose of this little experiment is to create a heating / warming plate for my coffee / tea cup.
It's a simple / somewhat mindless experiment but it helps me in starting to learn about electronics.
I have numerous power supplies around and can use a more powerful supply for the project - that's no prob.
If the coffee cup did not loose heat then you would not have to build your heater, so you must find out at what rate the coffee cup looses heat. 1. Find out the typical volume of its contents (300 ml?). Then find out the temperature of the poured hot water (90 deg C?), work out when the coffee is too cold (50 degrees C?) and the time its takes to get to this temperature. Now you know you have a volume of water, which drops so many degrees so you can work out how many calories have been lost/ persecond, multiply by 4.2 and this should give you a figure of watts. Which I hope is less then 3.5, but I suspect is more like 15 W.
Frank
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Not open for further replies. | 779 | 3,036 | {"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} | 3.890625 | 4 | CC-MAIN-2024-10 | latest | en | 0.912289 |
https://qaalot.com/what-does-10-penny-nail-mean/ | 1,716,894,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00863.warc.gz | 404,639,593 | 19,447 | In the US, the size of a nail is designated by its penny dimension, written with a quantity and the abbreviation d for penny; for instance, 10d for a tenpenny nail. A bigger quantity signifies an extended nail, proven within the desk under.
Equally, what does Penny imply for nails?
The time period “penny“, used with nails, was an English measurement initially which means value per 100. It now means nail size and is abbreviated “d” from the outdated Roman “penny” the denarius. Beneath the unique measurement, 6d nails price 6 pence per 100. A 60d nail, being a lot heavier, price 60 pence per 100.
Secondly, what’s the penny dimension of a nail? Instance: 100 3-1/2” nails = 16 pennies, or denarius. To at the present time, 3-1/2” nails are 16d (penny) nails. The opposite is that one thousand eight penny nails weighed 8 kilos, however this does not precisely clarify the size.
The origin of the “d” image on nails.
(*10*)
Penny (d) = Inches
8d 2-1/2”
9d 2-3/4”
10d 3”
12d 3-1/4”
Folks additionally ask, how large is a ten penny nail?
The ‘denarius’ was an outdated Roman coin and because the centuries handed, the abbreviation ‘d’ ultimately grew to become related to the outdated British penny. What concerning the expression “10 penny nails“?
NAIL SIZE CHART. Be aware: The ‘d’ means ‘penny‘. For instance, a ’10d nail‘ is a ’10 penny nail‘.
(*10*)
Dimension Inches Cm
100d 10.00 25.400
How lengthy is a 40 penny nail?
Nail Measurement Information: Second to 16d 4d nails measure 1.5 inches and 5d nails measure 1.75. A 6d nail is 2 inches lengthy and an 8d nail is 2.5 inches lengthy. 12d nails are 3.25 inches lengthy and 16d nails (16 penny) are 3.5 inches lengthy.
Associated Query Solutions
## What does 16d imply?
For historic causes, nails are bought each by a quantity adopted by d and (much less confusingly) by size. The “d” stands for penny, so 8d refers to an 8-penny nail, 16d to a 16-penny nail and so forth. It is a technique to point out nail size, as you may see within the desk under.
## What sort of nails do you employ for framing?
Galvanized and Vinyl Sinkers
When toenailing, which implies to drive nails at an angle to safe a butt joint, 8d nails are finest. Inside framing nails have a vinyl coating to make them simpler to drive. They’re often called vinyl sinkers. When doing exterior framing, framers use galvanized nails, that are rust resistant.
## How lengthy is a penny?
The coin is 0.75 inches (19.05 mm) in diameter and 0.0598 inches (1.52 mm) in thickness. Its weight has different, relying upon the composition of metals utilized in its manufacturing (see additional under).
## What is a 16d sinker nail?
16D Metal Coated Sinker Nails (30 lb. Pack) are glorious on the whole building, carpentry and framing functions. The nails are coated for easy driving and agency holding energy. For basic building, carpentry and framing use. Fabricated from metal.
## What dimension is a 60 penny nail?
#2 x 6 in. 60Penny Sizzling-Galvanized Metal Frequent Nails (50 lb.
## What is a 7d nail?
Our 7d nails for bee hives are specifically designed for a protracted life in outside situations. Trace: Use a 1 1/4″ nail on every of the highest finish corners of your hive our bodies and supers to maintain the wooden from splitting.
## The place did the time period penny come from?
As for the British penny, it acquired its identify from the Previous English penning, which in flip is believed to derive from the German pfennig. The time period nickel has not all the time been the identify for the US’ five-cent coin. You see, the half disme (pronounced like dime), because it was initially referred to, wasn’t manufactured from nickel.
## What number of cm is a penny?
The Lincoln penny, in its present design since 1909, has a diameter of 19.05 mm (0.75 in) and an space of two.850 sq. centimeters. It is about one-and-one-tenth instances as large as a Dime. In different phrases, 2.850 sq. centimeters is 1.1310 instances the dimensions of a Dime, and the dimensions of a Dime is 0.88420 instances that quantity.
## How lengthy ought to nails be?
The rule of thumb with nails 2 factors should you get the pun is that you simply ought to select a nail that’s 3 times as lengthy because the thickness of the fabric you might be fastening. If you wish to maintain 1/2″ drywall to a stud wall, the size of the nails ought to be at the very least 1 1/2″.
## How lengthy is a 10d nail?
For years, the constructing code outlined a 10d widespread nail as 0.148 inches in diameter and three inches lengthy, and that definition occurred in a single location within the code.
## How do you learn nail sizes?
An 8d nail, for instance, price 8 pennies for 100. Right this moment, the penny system refers particularly to nail size. A Second nail is 1 inch lengthy, for instance, whereas a 16d nail is 3 1/2 inches lengthy. Every increased quantity within the penny system represents a 1/4-inch size enhance, as much as a 12d nail (3 1/4 inches lengthy).
## What number of kilos is 3 Inch Nails?
For 3 inch nails:360 (nail amount per minute) x 60 (minutes) x8 (hours) / 148 (148 numbers 3 inch nails is 1 kg)= 1167 kg.
## How large is a 16 gauge nail?
16gauge end nailer
Like 15-gauge nailers, most 16gauge weapons shoot nails as much as 2-1/2 in. lengthy and are appropriate for thick trim. The principle benefit of a 16gauge gun is that it is smaller and lighter.
## What is a Tenpenny?
Definition of tenpenny. : amounting to, price, or costing 10 pennies.
## What is the shear power of a 16 penny nail?
Is there a distinction in power?
(*10*)
SHEAR
* 16d widespread nail .162” 134 lb.
* 16d sinker .148” 112 lb
* 16d field (nail gun) .131” 93 lb.
* #6 screw .138 70 lb.
## How thick is an 8 penny nail?
Customary Nail Sizes/Nail Measurements
(*10*)
Nail Dimension Shank Diameter
8D 10 0.134
10D 9 0.148
12D 9 0.148
16D 8 0.165
## What are ending nails?
Ending nails are small nails with very small heads. They’re normally used for “last touches” resembling molding or trim. They normally are sunken into the floor and go away a small gap that may be crammed in with putty and painted over for a easy end and invisible nail and nail gap. | 1,682 | 6,208 | {"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-2024-22 | latest | en | 0.922137 |
https://math.answers.com/math-and-arithmetic/What_is_the_total_value_of_what_a_household_owns_minus_what_it_owes | 1,722,895,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00447.warc.gz | 312,596,404 | 47,411 | 0
# What is the total value of what a household owns minus what it owes?
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I would say the average household has 4 credit cards? One for the mom, one for the dad, and a business credit card for each.That is wrong the average household has eight credit cards, and owes more than 8,000 on them. About 50 percent report having problems making the minimum payments on their bill.
-1 | 500 | 2,095 | {"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} | 3.21875 | 3 | CC-MAIN-2024-33 | latest | en | 0.946681 |
https://slideplayer.com/slide/7288389/ | 1,713,220,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00446.warc.gz | 492,327,793 | 22,795 | # Calculating with Significant Figures
## Presentation on theme: "Calculating with Significant Figures"— Presentation transcript:
Calculating with Significant Figures
When we do math with these numbers, always round to the number of significant figures represented by the most uncertain number. There are rules, depending on the operations you perform.
Calculating with Significant Figures: Multiplication & Division
To determine the number of significant figures in your answer, look for the term with the smallest number of significant figures, because that is the least accurate measurement:
Multiplication Example:
4.56 has three significant figures and 1.4 has two significant figures, therefore round off to two significant figures in your answer = 6.4
Division Example: Example: = Since 298 has the least number of significant figures (3), we round the answer to
Multiplication & Division Practice
67.90 ÷ 2 5600 ÷0.368 884.00÷76. ÷ 14 x 2,096 x 1.3 47,249 x 38,000 x 536 x
Calculating with Significant Figures: Addition & Subtraction
To determine the number of significant figures in your answer, find the term with the smallest number of decimal places. Use that many decimal places for your significant figures in your answer.
Addition Example: Example: Since 18.0 has just one decimal place, we will round off the answer to one decimal place = 31.1
– , 400. – 1.43 ,
Rounding Off Once you have determined how many significant figures is in your answer, there are a few rules for rounding off: Round down if the digit to be removed is less than rounded to two significant figures becomes 1.3 Round up if the digit to be removed is 5 or greater Rounding to two significant figures, 1.36 becomes 1.4 and becomes 3.2. If you are removing a string of numbers, only look at the first number to the right Rounding to two significant figures becomes 4.3. In a series of calculations, keep the extra digits until your final result, then round.
Scientific Notation The mass of a proton = 1.67 x 10-27 grams
What does this mean???
Review: What are the following #s?
101 102 103 104 100 =10 =10x10=100 =10x10x10=1,000 = 10,000 = 1 10-1 10-2 10-3 10-4 = .1 = .01 = .001 = .0001
Scientific Notation uses a number multiplied by a power of ten:
2000 = 2x103 0.004 = 4x10-3
Rules for Scientific Notation:
Add decimal point if it is missing: Move decimal point so there is ONE non-zero number to the left of it: Exponent is the number of places the decimal point was shifted: x103 Exponent can be positive or negative: = 6.3x10-3
Express in Scientific Notation
4001 32,560,000 78,941,000,000 | 661 | 2,599 | {"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} | 4.78125 | 5 | CC-MAIN-2024-18 | latest | en | 0.912238 |
https://www.teacherspayteachers.com/Product/Times-Table-Reverse-Flash-Cards-3335559 | 1,537,930,404,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00324.warc.gz | 894,293,994 | 18,646 | # Times Table - Reverse Flash Cards
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This flash card set is one of a kind! Children can finally understand what multiplication means through colorful visual cards that they will enjoy practicing times table. You don't need to tell the children; they will quickly see that multiplication is only a repeated addition when they practice times table with these cards.
Students are instructed to look at a card only for a second and try to remember the number of columns and rows that are shaded and the product as well. Blue line markers dividing the columns and rows half way enable children to notice this quite quickly even after glancing for a second. This practice strengthens their number sense in an amazing way; children will know instinctively 8 is 3 more than 5 or 2 less than 10 and that 7 is 2 more than 5 and 3 less than 10. This becomes a fun game for the children even if they are working on their own. They focus intensely like you've never seen them before and the result is incredible. Product (answer) is given in bold and unique fonts making it less intimidating to memorize the multiplication facts. The visual ques and repetition make memorizing times table a piece of cake.
Even children with severe ADHD love practicing times table with these cards because it's fun! And they will know the facts in order and in mixed order in no time. They won't have to count by numbers to get the answer they need. This will increase their times table fluency by many folds. I have used these cards with several 6th graders, yes 6th graders, who had the hardest time memorizing times table and they managed to memorize 4 and 6 times table within 10 minutes both in order and in mixed order as well. Their retention level was amazing even a few days later.
This can be used in math centers and also for one on one tutoring. Try it today.
Recommendation: For 2x3=6 card, encourage students to say ""two, three, six" instead of saying "two times three equals six". This simply takes too long and is not effective. They can either say it out loud and check their answers immediately or write the facts quickly on a blank worksheet.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 522 | 2,441 | {"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} | 3.40625 | 3 | CC-MAIN-2018-39 | latest | en | 0.962016 |
http://www.stata.com/statalist/archive/2009-07/msg00766.html | 1,448,483,317,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445679.7/warc/CC-MAIN-20151124205405-00050-ip-10-71-132-137.ec2.internal.warc.gz | 695,914,199 | 6,462 | # Re: st: gllamm with pweights
From sjsamuels@gmail.com To statalist@hsphsun2.harvard.edu Subject Re: st: gllamm with pweights Date Fri, 17 Jul 2009 14:53:40 -0400
```--
It looks like a three-level model is the most appropriate for you.
However compute scaled sampling weights only for the "census tract"
level"; "state", the highest level, gets a weight of 1. You can
use Korn and Graubard's method D referenced on page 814 of
Rabe-Hesketh and Skrondal, 2006, p 814, or one of the others. (Be
sure to cite the original sources, not just Chantala.) Scaling the
weight for "tracts" is needed to properly estimate the between-tract
component of variance.
You do not need the -cluster()- option in -gllamm-.
Good luck!
-Steve
Rabe-Hesketh, S. & Skrondal, A. (2006). Multilevel modelling of
complex survey data. Journal of the Royal Statistical Society: Series
A (Statistics in Society), 169(4), 805-827.
On Fri, Jul 17, 2009 at 12:38 PM, Kanter, Rebecca<rkante8r@jhsph.edu> wrote:
> Hi Steve and list,
>
> The original survey design is a multi-stage stratified design. The PSU is essentially the equivalent of a U.S. census tract (the probability that one of these tracts was selected was proportional to the number of households within it and the number of tracts selected corresponded to the sample size in the strata within the state) ...from which households are selected (with probability proportional to size). For each census tract selected six "blocks" are selected with probability proportional to the number of houses in each block; within each chosen block 6 households are selected via systematic random sampling and then individuals within the household via simple random sampling.
>
> I would just use the original individual survey pweights for the gllamm, but the pweight command for the gllamm does not work unless weights for all levels are specified.
>
> Thus, I go back to a previous suggestion on this thread...should I just set the pweight for L2 just equal to a constant 1.
>
> Or do I need to use the method by Chantala (as my advisor reminded me for example that while I am taking into account the urban and rural area of each state within the country that I only have a "sample" of x number of rural or urban tracts of the total number of rural or urban tracts within each state)?
>
> Or what?
>
> Thanks so much!
> Rebecca
> ___________________________________________
> Rebecca M. Kanter
> PhD Candidate
> Johns Hopkins Bloomberg School of Public Health
> Department of International Health
> Center for Human Nutrition
> ________________________________________
> From: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] On Behalf Of sjsamuels@gmail.com [sjsamuels@gmail.com]
> Sent: Friday, July 17, 2009 10:27 AM
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: gllamm with pweights
>
> Rebecca, I didn't follow the original thread, so I apologize. As
> there was no sampling of your level-two units, you do not need
> sampling weights for them, nor, therefore, the weights computed by
> Chantala's code. We could have been more helpful if you had
> described the original design. What was it, and what were the PSUs?
> The PSUs are the units which should be designated as clusters in
> -gllamm-. They need not be part of the two-level model, but might be
> interesting as units in a three-level model.
>
> You can use the original sampling weights, but perhaps you have
> enough information to post-stratify the weights for individuals, for
> example by gender and state. This is less necessary if gender is a
> predictor for your multi-level model.
>
> Good luck
>
> Steve
>
>
>
>
> On Thu, Jul 16, 2009 at 3:08 PM, Kanter, Rebecca<rkanter@jhsph.edu> wrote:
>> Thanks Steven, these resources are a big help.
>>
>> I am now trying to apply this method to my 2 level model (L1 = individual L2 = urban or rural part of state they live in; 64 units based on 32 states).
>>
>> In the method by Chantala et al, if I am interpreting this correctly...the PSU takes on a new meaning here (from the original complex survey design)...
>>
>> whereby PSU_wtj = 1 / Pr(urstate j selected) --> so if I am including all urban and rural parts of states (i.e. all 64 units that in turn make up the 32 states in a country) then is 1 for every urstate ?
>>
>> Furthermore, then, if FSU_wt i|j = 1 / Pr(person i selected / urstate j selected) then is FSU_wt i|j = 1 / Pr ( (1 / total number of people in urstate j) / 1) as in their example with schools = j each "student selected from school j will have a sampling weight equal to the number of students within school j represented by that student."?
>>
>> And in the end the original survey individual pweight is not used?
>>
>> Thanks so much for all your help,
>> Rebecca
>>
>>
>> ___________________________________________
>> Rebecca M. Kanter
>> PhD Candidate
>> Johns Hopkins Bloomberg School of Public Health
>> Department of International Health
>> Center for Human Nutrition
>> ________________________________________
>> From: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu] On Behalf Of sjsamuels@gmail.com [sjsamuels@gmail.com]
>> Sent: Thursday, July 16, 2009 12:24 PM
>> To: statalist@hsphsun2.harvard.edu
>> Subject: Re: st: gllamm with pweights
>>
>> --
>>
>> Also, see: http://www.stata.com/meeting/4nasug/Chantala.ppt and
>> http://www.cpc.unc.edu/restools/data_analysis/ml_sampling_weights.
>> These contain links to the Stata program -pwigls- which will scale the
>> weights. Rabe-Hesketh and Skrondal (2006), the second citation that
>> Stas listed, compute the "Method 1" weights by hand and illustrate an
>> analysis in GLLAMM.
>>
>> Rabe-Hesketh, S. & Skrondal, A. (2006). Multilevel modelling of
>> complex survey data. Journal of the Royal Statistical Society: Series
>> A (Statistics in Society), 169(4), 805-827.
>>
>> On Thu, Jul 16, 2009 at 12:28 PM, Stas Kolenikov<skolenik@gmail.com> wrote:
>>> Oh, I see. With 64 second level units, you are in a much better shape.
>>> I would probably have an urban/rural dummy as an explanatory variables
>>> for those second levels with -feq- option.
>>>
>>> If you sum up the weights, you are using the weights twice. And that's
>>> hardly a great idea: you are overcompensating for unequal
>>> probabilities of selection, if there were any. Were these
>>> states/ruran/urban areas selected via a sampling procedure? Or what
>>> you have is a complete list? In the latter case, you surely would need
>>> to specify unit weights at the second level.
>>>
>>> On the issue of weights in multilevel models, see:
>>> http://www.citeulike.org/user/ctacmo/article/711637,
>>> http://www.citeulike.org/user/ctacmo/article/850244,
>>> http://www.citeulike.org/user/ctacmo/article/3158754. There's probably
>>> more by now, but I am not tracking this literature very closely.
>>>
>>> On Thu, Jul 16, 2009 at 11:18 AM, Kanter, Rebecca<rkanter@jhsph.edu> wrote:
>>>> Hi Stan and statalist,
>>>>
>>>> Regarding my second level it is more than 2 values...as there are 32 states in the country...that makes 64 values (or areas/clusters that i illustrate via one variable called urstate...e.g. if urstate=1 it is the urban area of the 1st state and if urstate=33 it is the rural area of the 1st state and so on) if one divides each state into its urban and rural areas, respectively. Each one I want to take its own intercept and slopes etc to better account and visualize the urban and rural differences in the country....
>>>>
>>>> Thus, is it better to sum the individual weights per urstate (1-64) or let all weights for this second level equal one and keep my individual pweights as is for the individual level (level 1)?
>>>>
>> ________________
>>>> From: owner-statalist@hsphsun2.harvard.edu [owner-statalist@hsphsun2.harvard.edu]
>>>> On Wed, Jul 15, 2009 at 5:37 PM, Kanter, Rebecca<rkanter@jhsph.edu> wrote:
>>>>> Hi,
>>>>>
>>>>> I am running 2 level multi-level models using gllamm. Level one is individuals and Level two is either the urban or rural part of the country's state (i.e. urstate).
>>>>>
>>>>> I would like to use the survey pweights I have...I only have pweights for the individual level (adul_sr), but it seems that you need pweights for all levels specified in gllamm (?) so this is what I did to create pweights for urstate based on these weights:
>>>>>
>>>>>
>>>>> then I merged them to the rest of my dataset
>>>>>
>>>>> and made this weight for the gllamm:
>>>>>
>>>>> *MLM-level pweights
>>>>>
>>>>> Then ran the most basic random-intercept only (around urstate) in gllamm and get the follow error below and am assuming it is a pweight problem but I do not know where the problem is coming from so if anyone has insight that would be much appreciated. Thanks so much!
>>>>>
>>>>> (note: diettag==1 & exwt==1 is the subpopulation i want to look at for this series of models)
>>>>>
>>>>>
>>>>>
>>>>> Convergence not achieved: try with more quadrature points
>>>>>
>>>>>
>>>>
>> Steven Samuels
>> sjsamuels@gmail.com
>> 18 Cantine's Island
>> Saugerties NY 12477
>> USA
>> 845-246-0774
>>
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>>
>> *
>> * For searches and help try:
>> * http://www.stata.com/help.cgi?search
>> * http://www.stata.com/support/statalist/faq
>> * http://www.ats.ucla.edu/stat/stata/
>>
>
>
>
> --
> Steven Samuels
> sjsamuels@gmail.com
> 18 Cantine's Island
> Saugerties NY 12477
> USA
> 845-246-0774
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
>
--
Steven Samuels
sjsamuels@gmail.com
18 Cantine's Island
Saugerties NY 12477
USA
845-246-0774
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 2,793 | 10,205 | {"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-2015-48 | longest | en | 0.907761 |
https://www.ultimate-guitar.com/columns/music_theory/reading_scales.html | 1,454,998,303,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701156520.89/warc/CC-MAIN-20160205193916-00264-ip-10-236-182-209.ec2.internal.warc.gz | 882,332,974 | 18,927 | author: guitar98765432 date: 01/12/2006 category: music theory
### Thanks for subscribing! Check your email soon for some great stories from UG
I like this 2916 votes: 45 views: 1,425
Reading scale tab is actually very easy. First, I'll show you a scale tab:
e|--0--|-----|-----|--0--| B|-----|--0--|--0--|-----| G|--0--|-----|-----|--0--| D|-----|--0--|--0--|-----| A|--0--|-----|-----|--0--| E|-----|--0--|--0--|-----|
Don't worry, its not as hard as it looks. This is the same scale in standard tab:
e|-1--4---------------------------| B|-------2--3---------------------| G|------------1--4----------------| D|-----------------2--3-----------| A|----------------------1--4------| E|---------------------------2--3-|
Scale tab shows you what fret to press down in a differen way then tab does. On the high E string it tells you to press down the 1st fret then the 4th fret. After you are done with the Hgh E string then you can go to the B strin and play the 2nd fret then the 3rd fret and so on. This should make it easier:
|-1st-|-2nd-|-3rd-|-4th-| <--fret number-- e|--0--|-----|-----|--0--| B|-----|--0--|--0--|-----| G|--0--|-----|-----|--0--| D|-----|--0--|--0--|-----| A|--0--|-----|-----|--0--| E|-----|--0--|--0--|-----|
Now you're asking yourself why don't they just number the frets in the fiirst place? Good question. The reason is scales are moveable. You can start on the high e string at the 4th fret, the 5th fret, any fret you want. If this was the tab for the scale:
e|--1--4-------------------------------------| B|--------2--3-------------------------------| G|--------------1--4-------------------------| D|--------------------2--3-------------------| A|--------------------------1--4-------------| E|--------------------------------2--3-------|
Then this is also true, since scales are moveable:
e|--5--8-------------------------------------| B|--------6--7-------------------------------| G|--------------5--8-------------------------| D|--------------------6--7-------------------| A|--------------------------5--8-------------| E|--------------------------------6--7-------|
You can start on any fret you want. Once you learn the scale foward then you can play it backwards, too this is the same scale, only backwards:
e|--------------------------------5--8-------| B|--------------------------6--7-------------| G|--------------------5--8-------------------| D|--------------6--7-------------------------| A|--------5--8-------------------------------| E|--6--7-------------------------------------|
If you are still confused, then this should help. This is theslace tab again, with the order you are supposed to play it in:
e|--1--|-----|-----|--2--| B|-----|--3--|--4--|-----| G|--5--|-----|-----|--6--| D|-----|--7--|--8--|-----| A|--9--|-----|-----|--10-| E|-----|--11-|--12-|-----|
Once you learn a scale then you will start memorizing it and eventually start playing it faster and faster, wich will help your ability to create and play solos. If you have any corrections, questions, or comments, please feel free to email me at mike98765432@aol.com.
Comments BIU:) Only "https" links are allowed for pictures, otherwise they won't appear | 839 | 3,176 | {"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-2016-07 | latest | en | 0.49825 |
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### Topic: how can i tell if device is rated for 5v or 3.3v ? (Read 912 times)previous topic - next topic
#### Southpark
#15
##### Sep 10, 2018, 11:24 amLast Edit: Sep 10, 2018, 11:30 am by Southpark
- a guy said to me that Hall sensors (such as the ones in the encoder) require 4.7K pull-up resistors . but for now everything seems to be working without those , should i add them anyway in my printed circuit design ?
One way to find out is to measure the DC voltage at one of the hall sensor output pins (relative to the gnd pin) for a case where the hall sensor supply voltage is 3.3V. As you rotate the disk, the voltage will probably switch from 0V to 3.3V or vice versa. Then do the same thing for a hall sensor supply voltage of 5V. If you get sensor output voltages that switch between 0V and 5V, then chances are that your hall sensor can work with either 3.3V or 5V. This means that you would be free to choose what voltage you want to operate with, by choosing your hall sensor supply voltage.
Those pin-out sheets are at least helpful in that they tell you what the wires are, and provide some clues about voltage levels. Otherwise, it's a bit of fail on their part on providing adequate details. It's probably a fail on their suppliers.
The other thing is ..... maybe you've thought about it already ---- such as what sort of measurements are planned for those encoders. Relatively constant angular velocity measurements?
#### amine2
#16
##### Oct 18, 2018, 04:41 pm
Flippant answer to the original question: Power it from 7V, if it blows up its a 3.3V part, if it works its a
5V part (probably a somewhat damaged 5V part, but hey).
Slightly less flippant answer: Test a bunch of them at a range of voltages and plot the results. In other
words do the work of characterizing the parts in detail.
Yes thank you very much Sir ! i am sorry for the late response
it's all about the melons .
#### amine2
#17
##### Oct 18, 2018, 04:42 pm
One way to find out is to measure the DC voltage at one of the hall sensor output pins (relative to the gnd pin) for a case where the hall sensor supply voltage is 3.3V. As you rotate the disk, the voltage will probably switch from 0V to 3.3V or vice versa. Then do the same thing for a hall sensor supply voltage of 5V. If you get sensor output voltages that switch between 0V and 5V, then chances are that your hall sensor can work with either 3.3V or 5V. This means that you would be free to choose what voltage you want to operate with, by choosing your hall sensor supply voltage.
Those pin-out sheets are at least helpful in that they tell you what the wires are, and provide some clues about voltage levels. Otherwise, it's a bit of fail on their part on providing adequate details. It's probably a fail on their suppliers.
The other thing is ..... maybe you've thought about it already ---- such as what sort of measurements are planned for those encoders. Relatively constant angular velocity measurements?
Yes that actually did work !
it turned out that encoders were equally rated for 3.3v or 5v .
it's all about the melons .
#### amine2
#18
##### Oct 18, 2018, 04:45 pm
Of course you realise that to do a real hand, you arguably need twin motors for each finger - the superficialis and profundus for each finger, not to mention the lumbricals and interossei.
Cannot live long and prosper without.
Well yeah in the human body that's how it works xD the superficialis and profundus are for finger precision , while when it comes to extension and flexion it depends on the muscle group ! either the flexors or the extenders . that's mainly because muscle can only apply force when they contract . there for a single muscle can only move the joint in one direction ! though Motors can turn in both directions . i am using a 4 Bar linking mechanism . the motor motor flexes and extends the finger . in other designs they use an elastic substance to extend the finger , and a motor to flex it !
it's all about the melons .
#### amine2
#19
##### Oct 18, 2018, 04:57 pmLast Edit: Oct 18, 2018, 04:59 pm by amine2
Hello again since its the same project i didn't want to make an other thread .
ima just post my question here .
it is time now to make a printed circuit for the hand . and we're gunna get the made in china . the making doesn't take more than 2 days but the delivery takes more than a month ! that's why i wanna make sure that everything is done correctly to avoid a major delay of an other month after the circuit comes in !
the PCB house has two parameters in the making process that i can't quite grasp :
- the copper weight and the surface finish , i have no idea how to pick the right ones . i even tried searching online without any hope .
- the clearance for everything on the board is 0.2mm which is around 6mil . even with the +12V line the clearance is also 6mil . that wouldn't cause any problems right ?
-for a copper weight of 1oz , the width of the 12V line is 1.7mm . i just wanted something more than enough to avoid problems , that would work right ?
when it comes to the 5v line , i used used the standard width of 0.4mm .
- one last thing , i used many connectors on this board , the thing is : two of them are not on the edge of the board , they are inside it as you can see , i marked them with green arrows in this photo :
the connectors will still snap correctly right ?
Thank you very much for your attention .
it's all about the melons .
#### srnet
#20
##### Oct 18, 2018, 08:44 pmLast Edit: Oct 18, 2018, 08:54 pm by srnet
the connectors will still snap correctly right ?
How can anyone know, without knowing exactly which connectors they are, and the other exact components that are nearby ?
Print out the board, at 1:1 scale, stick the printout to a bit of foam and actually check that all the actual components that you are using will fit, simple.
\$50SAT is now Silent (but probably still running)
http://www.50dollarsat.info/
http://www.loratracker.uk/
#### srnet
#21
##### Oct 18, 2018, 08:52 pmLast Edit: Oct 18, 2018, 08:53 pm by srnet
- the copper weight and the surface finish , i have no idea how to pick the right ones . i even tried searching online without any hope
Where were you searching ?
If you type 'HASL with lead'
Into Google is already knows you are asking about PCB surface finishes, and suggests several explanatory links.
For the full answers however, you should tell us if this is DIY project or something you intend to sell.
\$50SAT is now Silent (but probably still running)
http://www.50dollarsat.info/
http://www.loratracker.uk/
#### amine2
#22
##### Oct 20, 2018, 03:31 pm
Hello , thank you very much for your Attention Srnet , i am sorry for the lack of detail .
this is the exact connector that i have :
connector
you can ignore all the nearby components , my question is whether or not it can slide on the surface . because the connector is in the middle of the board and not on any of the edges .
- i am not intending to sell this , neither is it for a DIY project . we are constructing these hands for local amputees . so it must be long lasting and the design has to be as reliable as it can be .
- what about the clearance and the routing width sir ?
it's all about the melons .
#### MarkT
#23
##### Oct 20, 2018, 06:09 pm
Maybe go for gold plating and conformal coating for reliability - will be more robust to getting
damp and less likely to corrode if it does. Certainly all connectors must be gold plated sort
for long term reliability in harsh environments.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### srnet
#24
##### Oct 20, 2018, 08:44 pm
my question is whether or not it can slide on the surface . because the connector is in the middle of the board and not on any of the edges .
They will (normally) fit flush, but it depends on the wires used and other stuff nearby.
But as I am sure you were going to do a trial component fit (see post #20) you can check at that time if the wires are intefering with other stuff.
\$50SAT is now Silent (but probably still running)
http://www.50dollarsat.info/
http://www.loratracker.uk/
#### amine2
#25
##### Oct 23, 2018, 12:04 am
Thank you very much guys , this is what i finally went with for the design .
- 2oz copper is an extreme , and this design can do well with just 1oz
.
- i went with HASL with lead , it seems to have sufficient endurance . also went with that to avoid the higher temperatures and lack of compatibility of some parts with ROHS surface finishes .
- a 0.2mm clearance is more than enough for this application , even near the 12v+ rail .
- i ramped the 12V line width to 2.1mm just for extra safety .
- the connector will fit just fine .
if you guys have any remarks i would like to hear them .
thank you .
it's all about the melons .
Go Up | 2,231 | 8,857 | {"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-2019-18 | latest | en | 0.929427 |
http://studylib.net/doc/18129970/electricity-lesson-1-how-are-electricity-and-magnetism-re.. | 1,519,297,635,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814101.27/warc/CC-MAIN-20180222101209-20180222121209-00128.warc.gz | 334,581,083 | 13,787 | ```Electricity
Lesson 1 How Are Electricity and Magnetism Related?
Electricity
Recall that atoms contain charged particles.
A proton has a p_____________________ charge.
An electron has a n_____________________ charge.
An atom can have m_________ electrons than protons.
This gives the atom a n________________ charge.
An atom with more protons than electrons has a p_____________ charge.
Atoms with o________________ charges attract each other.
In an atom, electrons move around o________________ the nucleus.
In some elements, electrons can also move from one atom to another.
This m________________ of e________________ p________________
e________________.
Electricity(DEFINE):___________________________________________________
__________________________________________________________________
Lamps change electricity into l________________ energy.
Electricity changes to motion, or m________________ energy.
Electricity can also change to h________________ energy.
What do a doorbell and a radio have in common?
They both use electricity to produce s________________ energy.
MAIN IDEA AND DETAILS: What are four kinds of energy that electricity
can be changed into?
________________________________
________________________________
________________________________
________________________________
Electricity and Magnetism
An electromagnet:____________________________________________________
_________________________________________________________________.
When electricity flows through a wire, the wire has a m______________
f______________ around it.
A single wire doesn’t have a strong magnetic field.
Winding the wire into a c_____________ makes the magnetic field stronger. Each
t________________ of the wire makes the field s________________.
Electromagnets are useful because they can be turned on and off.
You have seen that e________________ can produce m________________.
This also works the other way around—a m________________ can produce
e________________.
MAIN IDEA AND DETAILS: What are some ways in which electromagnets
are used?
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________
__________________________________________________
Electric Motors
You know that e________________ can be changed to m________________
e________________.
The motion of an electric device is produced by an e________________
m________________.
The electric motor is the opposite of a generator that produces electricity in an
energy station.
A motor contains a c________________ of w________________ that can
s________________ inside the m________________ f________________ of a
permanent magnet.
When the motor is switched on, electricity produces a magnetic field in the coil of
wire.
The c________________ becomes an e________________.
The poles of this electromagnet are attracted and repelled by the poles of the
permanent magnet.
This causes the electromagnet to spin.
The motion of the spinning coil can turn the blades of the blender or the fan.
MAIN IDEA AND DETAILS: What are the two main parts of a motor?
_________________________________
_________________________________
Lesson 2 What Are Static and Current Electricity?
Static Electricity
Most objects have no charge. The atoms making up the matter are
n______________________. They have equal numbers of protons and electrons.
But when one object rubs against another, electrons move from atoms of one
object to atoms of the other object. The numbers of protons and electrons in the
atoms are no longer equal. The objects become either p_____________________
or n______________________ charged. The buildup of charges on an object is
called s______________________ e______________________.
O______________________ charges a______________________ each
other.
When objects with opposite charges get close, e______________________
sometimes j________________ from the n______________________ object to the
p______________________ object. This evens out the charges, and the objects
become neutral.
CAUSE AND EFFECT What causes an object to build up a static charge?
__________________________________________________________________
__________________________________________________________________
Current Electricity
For electricity to be a useful source of energy, it must be a
s_______________ f____________ of charges. If electrons have a path to follow,
they will move in a steady flow instead of building up a static charge. This
f____________ of e___________________ is called an e_________________
c__________________ . Electricity that flows in this way is a kind of kinetic
energy called current electricity. To keep the charges flowing, a
c__________________ supply of e________________ is needed.
Electrical pressure forces current through the wires, giving the electrons
energy. This e______________________ p______________________ is
measured in v_________________.
The rate at which electric current flows is measured in a_________. The
combination of volts and amps can be dangerous, which is why many objects that
use electricity have warning labels.
The amount of electrical energy a device uses each second is measured
in w_____________. A label on a hair dryer, light bulb, or clock shows how many
watts it uses. One kilowatt is equal to 1,000 watts.
CAUSE AND EFFECT What causes an electric current to keep moving?
__________________________________________________________________
__________________________________________________________________
Conductors and Insulators
Electricity moves more e______________ through some kinds of matter than
others. A material through which electricity moves well is a c______________ .
Most metals are conductors. The electrons of m_____________ are held loosely by
the atoms. This makes it e____________ for the electrons to move between atoms,
causing current to flow.
P______________________ doesn’t conduct electricity well. Its electrons are
n__________ free to move between atoms. A material that conducts electricity
p______________________ is an i______________________ . Wood, glass, and
rubber are also insulators.
Insulators are important because they p______________________ you from
the electric current in the wire. If the layer of plastic on a wire peels off or cracks,
the w______________ s_______________ be r_______________. If you touch a
bare wire that is conducting current, the current will flow through you and
c______________________ h____________ you. Also, wires get warm when
they carry electricity. A bare wire that touches paper or cloth could start a fire.
CAUSE AND EFFECT What causes a metal to be a good conductor?
__________________________________________________________________
__________________________________________________________________
Lesson 3: What are Electric Circuits?
Series Circuits
The wires give the e__________________ a path to follow. The path an
electric current follows is called an electric circuit.
An electric circuit needs t______ things for current to flow. First, it needs a
s__________________ of current, or electrons.
Second, the circuit has to be c__________________. If there is a
b__________________ in the circuit, the current won’t flow.
A battery, wires, and light bulbs can form a complete circuit.
A s__________________ controls the flow of current by
o__________________ and c__________________ the circuit. When the switch is
on, the circuit is complete. The light comes on. When you turn the switch off, a
piece of metal inside the switch moves. This b__________________ the flow of
current, and the light goes out.
In a series circuit, the current has only o__________________ path to
follow. The parts are connected one after the other in a s__________________
l__________________. Removing any part of the circuit breaks the circuit, and
current s__________________ flowing.
MAIN IDEA AND DETAILS What two things are needed in order for
current to flow in a circuit?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
Parallel Circuits
Instead of giving the current only one path, you can make a path for
e__________________ d__________________ in the circuit. A circuit that has
m__________________ t__________________ one path for current to follow is
called a p__________________ c__________________.
In a parallel circuit, if one device is turned o______ or r________________,
current stops flowing along the loop f_______ t__________
d_________________. But current c__________________ to f________________
through the rest of the circuit. The other devices don’t stop working. This makes a
parallel circuit m____________ p_________________ to use than a series circuit.
A parallel circuit has another advantage over a series circuit. You can
connect m_______________ d________________ in a parallel circuit. If you
connect too many light bulbs in a series circuit, they all become dimmer. That’s
why old signs wired in series circuits didn’t have many bulbs. But in a parallel
circuit, adding more bulbs doesn’t change how bright they are.
MAIN IDEA AND DETAILS What are two advantages that parallel
circuits have over series circuits? ____________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
Drawing Circuits
You have learned that d__________________ that use
e__________________ have c__________________ inside them. Engineers
design these circuits. Some circuits, such as those in a blender or a lamp, are
simple. Some, such as those in a car or an airplane, are m__________________
c__________________.
When engineers work on a new design, they don’t begin by connecting a lot
of wires. Instead, they draw diagrams of the circuits they will use. Circuit diagrams
use s__________________ to show the p__________________ of a
c__________________. Each part of a circuit has a d__________________
s__________________.
A diagram of the series circuit you built in the Investigate would have
symbols for the b__________________, the w__________________, and the
t__________________ lights. These symbols would be connected in a
s__________________ l__________________.
MAIN IDEA AND DETAILS What do you see on a circuit diagram?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
``` | 1,920 | 10,959 | {"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-2018-09 | longest | en | 0.782136 |
https://avatest.org/tag/phys311/ | 1,713,676,950,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00498.warc.gz | 105,116,805 | 17,353 | Posted on Categories:Theoretical mechanics, 物理代写, 理论力学
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## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Angular Momentum
The angular momentum of a particle is defined as
$$\boldsymbol{L}{O}=\boldsymbol{r} \times \boldsymbol{P}=\boldsymbol{r} \times(m v) .$$ Similar to the concept of moment, it can also be decomposed into three components in the three axes, i.e., $$\boldsymbol{L}{O}=L_{x} \boldsymbol{i}+L_{y} \boldsymbol{j}+L_{z} \boldsymbol{k} .$$
The derivative of the angular momentum is
\begin{aligned} \dot{\boldsymbol{L}}{O} &=\dot{\boldsymbol{r}} \times m v+\boldsymbol{r} \times m \dot{\boldsymbol{v}} \ &=\boldsymbol{v} \times m \boldsymbol{v}+\boldsymbol{r} \times m a \ &=\boldsymbol{r} \times \boldsymbol{F} \ &=M{O}(\boldsymbol{F}) . \end{aligned}
## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy
“Work” is defined as the cross product of the force and the corresponding displacement, which is formulated as
\begin{aligned} W &=\boldsymbol{F} \cdot \boldsymbol{r} \ &=F s \cos \varphi, \end{aligned}
where $\boldsymbol{F}$ is a constant force, $s$ is the magnitude of the displacement, and $\varphi$ is the angle between the lines of the applied force and the displacement, as shown in Fig. 13.1. This means that the work represents the actual contribution of the force to the displacement, and it can be understood that the displacement times the projection quantity of the force along the line of displacement. Notably, the work is a scalar, whose unit is $\mathrm{J}$, which is in memory of the great scientist Joule.
For an arbitrary force, the work should be expressed via integration
$$W=\int_{C} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{r}$$
\begin{aligned} &=\int_{C} F \cos \varphi \mathrm{d} s \ &=\int_{C} F v \cos \varphi \mathrm{d} t . \end{aligned}
## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Angular Momentum
$$\boldsymbol{L} O=\boldsymbol{r} \times \boldsymbol{P}=\boldsymbol{r} \times(m v) .$$
$$\boldsymbol{L} O=L_{x} \boldsymbol{i}+L_{y} \boldsymbol{j}+L_{z} \boldsymbol{k} .$$
$$\dot{\boldsymbol{L}} O=\dot{\boldsymbol{r}} \times m v+\boldsymbol{r} \times m \dot{\boldsymbol{v}} \quad=\boldsymbol{v} \times m \boldsymbol{v}+\boldsymbol{r} \times m a=\boldsymbol{r} \times \boldsymbol{F} \quad=M O(\boldsymbol{F}) .$$
## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy
“功”定义为力和相应位移的叉积，公式为
$$W=\boldsymbol{F} \cdot \boldsymbol{r} \quad=F s \cos \varphi,$$
$$\begin{gathered} W=\int_{C} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{r} \ =\int_{C} F \cos \varphi \mathrm{d} s \quad=\int_{C} F v \cos \varphi \mathrm{d} t . \end{gathered}$$
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,429 | 3,446 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-18 | latest | en | 0.588664 |
https://www.jiskha.com/display.cgi?id=1302153242 | 1,502,936,078,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00012.warc.gz | 939,335,192 | 3,926 | # Physics
posted by .
A 65kg boy and his 40kg sister, both wearing roller blades, face each other at rest. The girl pushes the boy hard, sending him backwards with a velocity of 2.9m/s towards the west. Ignore friction. What is the minimum amount of chemical energy is converted into mechanical energy in the girl’s muscles?
• Physics -
It is easy to calculate how much kinetic energy must be given to both brother and sister. The girl acquires an equal momentum in the opposite direction, with velocity 2.9*64/40 = 4.64 m/s.
Calculate and add the kinetic energies of both brother and sister.
Chemical (food) energy cannot be converted to mechanical energy with 100% efficiency. The efficiency varies from person to person, but is around 20%. They may want you to assume 100% efficiency for the minimum food energy required, but the real number is much higher.
## Similar Questions
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More Similar Questions | 755 | 2,858 | {"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} | 2.859375 | 3 | CC-MAIN-2017-34 | latest | en | 0.87202 |
https://www.physicsforums.com/threads/oniel-world-gravity.834395/ | 1,513,398,006,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948581053.56/warc/CC-MAIN-20171216030243-20171216052243-00786.warc.gz | 770,209,475 | 15,772 | # ONiel world gravity?
1. Sep 24, 2015
### darthsasquatch
Im doing some research on ONiel cylinder worlds and the use of rotation to imitate gravity. My question is, if you jump in a ONiel cylinder (full of air) is it just the rotating air which exerts force to bring you back to the inner surface? Expecting the cylinder has too little mass to create earth like gravity.
2. Sep 25, 2015
### 256bits
Air, or vacuum, doesn't matter. You would "fall" back to the surface in either case. With air though there may be some air currents, or wind, if you prefer, but any such effect upon the imitation gravity would not be taking that into account, just as much one does not consider the minimal effects of air currents on earth.
3. Sep 25, 2015
### darthsasquatch
Thank you for responding. Im unclear of the mechanics, so just by placing a rotating cylinder around an object floating in space, that object would be attracted to the surface? Where does the force come from?
4. Sep 25, 2015
### 256bits
No, that will not work.
Basically, all objects on the cylinder surface travel at the same velocity. If you jump, that velocity stays with you. Since there is no force acting upon you, you will travel in a straight line( assuming the jump upwards is not too great in strength ) and then bump back into the cylinder. It will feel as if the cylinder has attracted you to it.
5. Sep 25, 2015
### darthsasquatch
Right, it ocurred to me after my relpy that just jumping in the Y does not change the X the surface was giving you. So if a person could momentarly negate the cylinders spin, say by jumping against its rotation, they could "hover" over the rotating surface?
6. Sep 25, 2015
### A.T.
Yes, running against the spin will reduce the apparent gravity, and when it's down to zero you can just hover (ignoring the effect of air).
7. Sep 25, 2015
### darthsasquatch
Great! Thank you for clearing this up for me. Now i can plug more numbers into a calculator i found for these type of stations, and figure out how feasable such an attempt would be. | 509 | 2,064 | {"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} | 3.328125 | 3 | CC-MAIN-2017-51 | longest | en | 0.945596 |
https://www.clutchprep.com/physics/practice-problems/50021/a-5-kg-mass-hangs-from-a-rope-wrapped-around-the-surface-of-a-cylindrical-pulley | 1,643,452,472,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304883.8/warc/CC-MAIN-20220129092458-20220129122458-00513.warc.gz | 748,868,573 | 19,335 | # Problem: A 5 kg mass hangs from a rope wrapped around the surface of a cylindrical pulley. The pulley has a mass of 20 kg and a radius of 0.8 m. If the 5 kg mass is suddenly released, falling down, what speed will it have after it has fallen 3.2 m?
###### FREE Expert Solution
84% (400 ratings)
###### Problem Details
A 5 kg mass hangs from a rope wrapped around the surface of a cylindrical pulley. The pulley has a mass of 20 kg and a radius of 0.8 m. If the 5 kg mass is suddenly released, falling down, what speed will it have after it has fallen 3.2 m? | 153 | 561 | {"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-2022-05 | latest | en | 0.884341 |
https://kr.mathworks.com/matlabcentral/profile/authors/2507048-informaton | 1,582,482,733,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00107.warc.gz | 439,127,830 | 21,695 | Community Profile
# Informaton
377 2014 이후 총 참여 횟수
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https://gravity.wikia.org/wiki/Wind_tunnel | 1,555,922,255,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578548241.22/warc/CC-MAIN-20190422075601-20190422101601-00168.warc.gz | 425,691,452 | 37,335 | ## FANDOM
771 Pages
A wind tunnel is a research tool used in aerodynamic research. It is used to study the effects of air moving past solid objects.
## Theory of operationEdit
Wind tunnels were first proposed as a means of studying vehicles (primarily airplanes) in free flight. The wind tunnel was envisioned as a means of reversing the usual paradigm: instead of the air's standing still and the aircraft moving at speed through it, the same effect would be obtained if the aircraft stood still and the air moved at speed past it. In that way a stationary observer could study the aircraft in action, and could measure the aerodynamic forces being imposed on the aircraft.
Later, wind tunnel study came into its own: the effects of wind on manmade structures or objects needed to be studied, when buildings became tall enough to present large surfaces to the wind, and the resulting forces had to be resisted by the building's internal structure. Determining such forces was required before building codes could specify the required strength of such buildings.
Still later, wind-tunnel testing was applied to automobiles, not so much to determine aerodynamic forces per se but more to determine ways to reduce the power required to move the vehicle on roadways at a given speed. In these studies, the interaction between the road and the vehicle plays a significant role, and this interaction must be taken into consideration when interpreting the test results. In an actual situation the roadway is moving relative to the vehicle but the air is stationary relative to the roadway, but in the wind tunnel the air is moving relative to the roadway, while the roadway is stationary relative to the test vehicle. Some automotive-test wind tunnels have incorporated moving belts under the test vehicle in an effort to approximate the actual condition.
## Measurement of aerodynamic forcesEdit
Ways that air velocity and pressures are measured in wind tunnels:
• air velocity through the test section (called the throat) is determined by Bernoulli's principle. Measurement of the dynamic pressure, the static pressure, and (for compressible flow only) the temperature rise in the airflow
• direction of airflow around a model can be determined by tufts of yarn attached to the aerodynamic surfaces
• direction of airflow approaching an aerodynamic surface can be visualized by mounting threads in the airflow ahead of and aft of the test model
• dye, smoke, or bubbles of liquid can be introduced into the airflow upstream of the test model, and their path around the model can be photographed (see particle image velocimetry)
• pressures on the test model are usually measured with beam balances, connected to the test model with beams or strings or cables
• pressure distributions across the test model have historically been measured by drilling many small holes along the airflow path, and using multi-tube manometers to measure the pressure at each hole
• pressure distributions can more conveniently be measured by the use of pressure-sensitive paint, in which higher local pressure is indicated by lowered fluorescence of the paint at that point
• pressure distributions can also be conveniently measured by the use of pressure-sensitive pressure belts, a recent development in which multiple ultra-miniaturized pressure sensor modules are integrated into a flexible strip. The strip is attached to the aerodynamic surface with tape, and it sends signals depicting the pressure distribution along its surface.[1]
• pressure distributions on a test model can also be determined by performing a wake survey, in which either a single pitot tube is used to obtain multiple readings downstream of the test model, or a multiple-tube manometer is mounted downstream and all its readings are taken (often by photograph).
## History of wind tunnelsEdit
English military engineer and mathematician Benjamin Robins (1707–1751) invented a whirling arm apparatus to determine drag and did some of the first experiments in aviation theory.
Sir George Cayley (1773-1857) also used a whirling arm to measure the drag and lift of various airfoils. His whirling arm was 5 feet long and attained top speeds between 10 and 20 feet per second.
However, the whirling arm does not produce a reliable flow of air impacting the test shape at a normal incidence. Centrifugal forces and the fact that the object is moving in its own wake mean that detailed examination of the airflow is difficult. Francis Herbert Wenham (1824-1908), a Council Member of the Aeronautical Society of Great Britain, addressed these issues by inventing, designing and operating the first enclosed wind tunnel in 1871. Once this breakthrough had been achieved, detailed technical data was rapidly extracted by the use of this tool. Wenham and his colleague Browning are credited with many fundamental discoveries, including the measurement of l/d ratios, and the revelation of the beneficial effects of a high aspect ratio.
Carl Rickard Nyberg used a wind tunnel when designing his Flugan from 1897 and onwards.
In a classic set of experiments, the Englishman Osborne Reynolds (1842-1912) of the University of Manchester demonstrated that the airflow pattern over a scale model would be the same for the full-scale vehicle if a certain flow parameter were the same in both cases. This factor, now known as the Reynolds Number, is a basic parameter in the description of all fluid-flow situations, including the shapes of flow patterns, the ease of heat transfer, and the onset of turbulence. This comprises the central scientific justification for the use of models in wind tunnels to simulate real-life phenomena. However, there are limitations on conditions in which dynamic similarity is based upon the Reynolds number alone.
The Wright brothers' use of a simple wind tunnel in 1901 to study the effects of airflow over various shapes while developing their Wright Flyer was in some ways revolutionary.[2] It can be seen from the above, however, that they were simply using the accepted technology of the day, though this was not yet a common technology in America.
Subsequent use of wind tunnels proliferated as the science of aerodynamics and discipline of aeronautical engineering were established and air travel and power were developed.
Wind tunnels were often limited in the volume and speed of airflow which could be delivered.
The wind tunnel used by German scientists at Peenemünde prior to and during WWII is an interesting example of the difficulties associated with extending the useful range of large wind tunnels. It used some large natural caves which were increased in size by excavation and then sealed to store large volumes of air which could then be routed through the wind tunnels. This innovative approach allowed lab research in high-speed regimes and greatly accelerated the rate of advance of Germany's aeronautical engineering efforts. By the end of the war, Germany had at least three different supersonic wind tunnels, with one capable of Mach 4.4 (heated) airflows.[3]
Later research into airflows near or above the speed of sound used a related approach. Metal pressure chambers were used to store high-pressure air which was then accelerated through a nozzle designed to provide supersonic flow. The observation or instrumentation chamber ("test section") was then placed at the proper location in the throat or nozzle for the desired airspeed.
For limited applications, Computational fluid dynamics (CFD) can augment or possibly replace the use of wind tunnels. For example, the experimental rocket plane SpaceShipOne was designed without any use of wind tunnels. However, on one test, flight threads were attached to the surface of the wings, performing a wind tunnel type of test during an actual flight in order to refine the computational model. It should be noted that, for situations where external turbulent flow is present, CFD is not practical due to limitations in present day computing resources. For example, an area that is still much too complex for the use of CFD is determining the effects of flow on and around structures, bridges, terrain, etc.
The most effective way to simulative external turbulent flow is through the use of a boundary layer wind tunnel.
There are many applications for boundary layer wind tunnel modeling. For example, understanding the impact of wind on high-rise buildings, factories, bridges, etc. can help building designers construct a structure that stands up to wind effects in the most efficient manner possible. Another significant application for boundary layer wind tunnel modeling is for understanding exhaust gas dispersion patterns for hospitals, laboratories, and other emitting sources. Other examples of boundary layer wind tunnel applications are assessments of pedestrian comfort and snow drifting. Wind tunnel modeling is accepted as a method for aiding in Green building design. For instance, the use of boundary layer wind tunnel modeling can be used as a credit for Leadership in Energy and Environmental Design (LEED) certification through the U.S. Green Building Council.
Wind tunnel tests in a boundary layer wind tunnel allow for the natural drag of the earth's surface to be simulated. For accuracy, it is important to simulate the mean wind speed profile and turbulence effects within the atmospheric boundary layer. Most codes and standards recognize that wind tunnel testing can produce reliable information for designers, especially when their projects are in complex terrain or on exposed sites.
## How it worksEdit
Air is blown or sucked through a duct equipped with a viewing port and instrumentation where models or geometrical shapes are mounted for study. Typically the air is moved through the tunnel using a series of fans. For very large wind tunnels several meters in diameter, a single large fan is not practical, and so instead an array of multiple fans are used in parallel to provide sufficient airflow. Due to the sheer volume and speed of air movement required, the fans may be powered by stationary turbofan engines rather than electric motors.
The airflow created by the fans that is entering the tunnel is itself highly turbulent due to the fan blade motion (when the fan is blowing air into the test section - when it is sucking air out of the test section downstream, the fan-blade turbulence is not a factor), and so is not directly useful for accurate measurements. The air moving through the tunnel needs to be relatively turbulence-free and laminar. To correct this problem, closely-spaced vertical and horizontal air vanes are used to smooth out the turbulent airflow before reaching the subject of the testing.
Due to the effects of viscosity, the cross-section of a wind tunnel is typically circular rather than square, because there will be greater flow constriction in the corners of a square tunnel that can make the flow turbulent. A circular tunnel provides a smoother flow.
The inside facing of the tunnel is typically as smooth as possible, to reduce surface drag and turbulence that could impact the accuracy of the testing. Even smooth walls induce some drag into the airflow, and so the object being tested is usually kept near the center of the tunnel, with an empty buffer zone between the object and the tunnel walls. There are correction factors to relate wind tunnel test results to open-air results.
Lighting is usually recessed into the circular walls of the tunnel and shines in through windows. If the light were mounted on the inside surface of the tunnel in a conventional manner, the light bulb would generate turbulence as the air blows around it. Similarly, observation is usually done through transparent portholes into the tunnel. Rather than simply being flat discs, these lighting and observation windows may be curved to match the cross-section of the tunnel and further reduce turbulence around the window.
Various techniques are used to study the actual airflow around the geometry and compare it with theoretical results, which must also take into account the Reynolds number and Mach number for the regime of operation.
### Pressure measurements Edit
Pressure across the surfaces of the model can be measured if the model includes pressure taps. This can be useful for pressure-dominated phenomena, but this only accounts for normal forces on the body.
### Force and moment measurements Edit
With the model mounted on a force balance, one can measure lift, drag, lateral forces, yaw, roll, and pitching moments over a range of angle of attack. This allows one to produce common curves such as lift coefficient versus angle of attack (shown).
Note that the force balance itself creates drag and potential turbulence that will affect the model and introduce errors into the measurements. The supporting structures are therefore typically smoothly shaped to minimize turbulence.
### Flow visualization Edit
Because air is transparent it is difficult to directly observe the air movement itself. Instead, a smoke particulate or a fine mist of liquid is sprayed into the tunnel just ahead of the device being tested. The particulate is sufficiently low mass to stay suspended in the air without falling to the floor of the tunnel, and is light enough to easily move with the airflow.
If the air movement in the tunnel is sufficiently non-turbulent, a particle stream released into the airflow will not break up as the air moves along, but stays together as a sharp thin line. Multiple particle streams released from a grid of many nozzles can provide a dynamic three-dimensional shape of the airflow around the object being tested. As with the force balance, these injection pipes and nozzles need to be shaped in a manner that minimizes the introduction of turbulent airflow into the airstream.
High-speed turbulence and vortices can be difficult to see directly, but strobe lights and film cameras or high-speed digital cameras can help to capture events that are a blur to the naked eye.
High-speed cameras are also required when the subject of the test is itself moving at high speed, such as an airplane propeller. The camera can capture stop-motion images of how the blade cuts through the particulate streams and how vortices are generated along the trailing edges of the moving blade.
## Wind tunnel classification Edit
There are many different kinds of wind tunnels, an overview is given in the figure below:
## List of wind tunnels Edit
The aerodynamic principles of the wind tunnel work equally on watercraft, except the water is more viscous and so imposes a greater forces on the object being tested. A looping flume is typically used for underwater aquadynamic testing. The interaction between 2 different types of fluids means that pure windtunnel testing is only partly relevant. However, a similar sort of research is done in a towing tank
### Low-speed Oversize Liquid Testing Edit
Air is not always the best test medium to study small-scale aerodynamic principles, due to the speed of the air flow and airfoil movement. A study of fruit fly wings designed to understand how the wings produce lift was performed using a large tank of mineral oil and wings 100 times larger than actual size, in order to slow down the wing beats and make the vortices generated by the insect wings easier to see and understand.[4]
### Wind Tunnel Testing for Wind Engineering Edit
In Wind Engineering, Wind Tunnel Tests are often used to measure the velocity around, and forces or pressures upon structures. Usually very tall buildings, buildings with unusual or complicated shapes (such as a tall building with a parabolic or a hyperbolic shape), cable suspension bridges or cable stayed bridges are analysed in specialized atmospheric boundary layer wind tunnels. These feature a long upwind section to accurately represent the wind speed and turbulence profile acting on the structure. Wind tunnel tests provide the necessary design pressure measurements for use in the dynamic analysis of the structure. | 3,044 | 16,074 | {"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-2019-18 | latest | en | 0.960163 |
http://en.allexperts.com/q/Algebra-2061/2016/3/binary-operation.htm | 1,490,401,615,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188717.24/warc/CC-MAIN-20170322212948-00456-ip-10-233-31-227.ec2.internal.warc.gz | 119,322,443 | 6,648 | You are here:
# Algebra/Binary... Operation
Question
Mathematics is my favorite subject, but the following question makes me feel that mathematics is pointles and grey:
A binary operation * is defined on the set R of real numbers by a * b= a + b + ab whare a, b belongs to R
(i) calculate 5*(-2)*5
(ii) find the identity element of R under the operation *
(* symbolizes binary operation; that is how it is used in my book.)
HELP ME SOLVE IT!
Since a * b = a + b + ab, this says we only take two numbers at once.
Since the 1st two numbers are 5 and -2, take a=5 and b=-2.
This gives a*b = 5 -2 -10 = 3-10= -7.
We then use -7 and 5 as a and b.
This says that -7 * 5 = -7 + 5 - 35. = -37.
Algebra
Volunteer
#### Scott A Wilson
##### Expertise
Any algebraic question you've got. That includes question that are linear, quadratic, exponential, etc.
##### Experience
I have solved story problems, linear equations, parabolic equations. I have also solved some 3rd order equations and equations with multiple variables.
Publications
Documents at Boeing in assistance on the manufacturiing floor.
Education/Credentials
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Past/Present Clients
Students in a wide variety of areas since the 80's; over 1,000 of them have been in algebra. | 379 | 1,430 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-13 | latest | en | 0.935213 |
https://www.physicsforums.com/threads/can-someone-help-me-with-these-problems.87376/ | 1,669,518,320,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00086.warc.gz | 1,004,174,992 | 14,308 | # Can someone help me with these problems?
thugg
i can't remember how to do this at all, but can someone just post a simple step by step solution for this? thanks
http://img275.imageshack.us/img275/2392/ss2mx.jpg [Broken]
also having trouble with this one
http://img374.imageshack.us/img374/1146/dfsd4mu.jpg [Broken]
Last edited by a moderator:
Homework Helper
Q1: resolve T1 into horizontal and vertical components. Then equate all the horizontal forces and all the vertical forces.
Homework Helper
Q2: let T1 be the tension in the angled rope/string and T2 be the tension in the horizontal rope/string. Use the same strategy as for Q1 to solve for T1 and T2. Hence get coefft of friction
thugg
hmm still kinda confused, any chance you could show me step by step how to solve it
like how do you resolve T1 and T2 to get values.....like i know the 20 has components of x = 0, and y = -20 right? but how do u get values with the 30 degree angle?
Last edited:
teclo
thugg said:
hmm still kinda confused, any chance you could show me step by step how to solve it
like how do you resolve T1 and T2 to get values.....like i know the 20 has components of x = 0, and y = -20 right? but how do u get values with the 30 degree angle?
draw out the angles and use trig.
T1+T2=(0,+20)
use the sin and cos function to resolve the two tensions into vectors. you should have 2 sets of equations to solve for T1 or T2, once you know one the other is only a matter of substitution. | 387 | 1,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-49 | latest | en | 0.91376 |
http://caldwellcobeekeepers.org/e10n0c3w/ | 1,611,310,235,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703529179.46/warc/CC-MAIN-20210122082356-20210122112356-00245.warc.gz | 19,043,810 | 19,194 | # Math Addition And Subtraction Worksheets
Published at Tuesday, September 29th 2020. by in Addition Worksheets.
Play a magnetic fish game with cardboard fish with a paper-clip and a piece of dowel and string with a magnet on the end as a fishing rod. Count the fish in the pond. When one gets caught subtraction how many are left? Division can be as simple as a sharing exercise. "There are 4 people here and I have 8 counters. Let us see how many we will get each". Use play dough or counters or blocks to make groups of items. Talk about what happens when you put groups together (multiplication). Make the terminology you use simple. This age group need simple language instead of mathematical terms. These activities are laying the foundations for further learning.
If the materials do not specifically indicate "brain-based," determine if they are at least brain-friendly. This would mean that you are looking for lots of color, material interesting to the child, many varied activities-especially involving movement, and using several of the senses. I saw one company whose worksheets included the instruction to "say the number out loud as you..." This is very good! Speaking out loud is very important for learning to occur. Ideally, all worksheets should include this instruction. If you can not find any that do, then you need to add that instruction yourself.
Are you the parent of a toddler? If you are, you may be looking to prepare your child for preschool from home. If you are, you will soon find that there are a number of different approaches that you can take. For instance, you can prepare your child for social interaction by setting up play dates with other children, you can have arts and crafts sessions, and so much more. Preschool places a relatively large focus on education; therefore, you may want to do the same. This is easy with preschool worksheets.
Interactive math games for first graders allow young students to play their way to a deeper understanding of numerical concepts. Addressing addition, subtraction and other first grade math skills through games helps make learning fun and expands a child has academic experience beyond the traditional classroom setting. By merging print materials with technology, both teachers and parents can aid children in becoming more proficient with the concepts they will need to be successful in school and in daily life.
In a growing move amongst home-schoolers to look at online courses, one subject area lends itself towards a bit more hesitation from the group. Home-schoolers want to like online courses because of the flexibility of them, but with regard to math, they are just not so sure about the validity of online math. There is reason for this, but many students are having good success with online math programs, and slowly but surely, the homeschooling community is coming around. Home-schoolers tend to shy away from online math due to the perception that math is better learned with a real person giving instruction and students following along in their textbooks. Many students learn well this way, but online math courses operate on a different philosophy. They presume that students can learn to understand material with information, practice, and feedback, and in essence, can become their own teachers. This is a far more effective method of instruction in the long run, and while it does take some adjustment, many programs make this method very viable for students of all abilities.
NEVER use "skill and drill" worksheets. These are the worksheets just made up of columns of problems. There are better materials out there, so do not resort to skill and drill. The very worst problem of skill and drill worksheets is the greatly increased chance of a practiced mistake. The same problem will likely appear several times on the same sheet. A wrong answer once means a wrong answer several times; and a practiced mistake takes hundreds of correct repetitions to fix. This danger alone is important enough to never use any worksheet. I am quite serious about how difficult it is to repair a practiced mistake. Learning is hard enough. Re-learning is much more difficult. | 807 | 4,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-04 | latest | en | 0.956928 |
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