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https://mcqscenter.com/computer-science/compiler-design-finite-automata

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In finite automata, Q stands for? Finite set of input symbolsB. Finite set of statesD. One Start state Final state is represented by? _________ of FA are represented by circles How many arrow does intermediate states have? In Final state, The number of odd arrows are one greater than even, i.e. odd = even+1? Can be true or falseD. Can not say There are ________ tuples in finite state machine Number of states require to accept string ends with 10. Languages of a automata is If it haltsB. If it is accepted by automataC. If automata touch final state in its life timeD. All language are language of automata Language of finite automata is The basic limitation of finite automata is that it can’t remember arbitrary large amount of informationB. it sometimes recognize grammar that are not regularC. it sometimes fails to recognize regular grammarD. None of the above

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http://dev.unilang.org/viewtopic.php?f=33&t=41219&p=904163

math

Could you please help me with some Korean? I need to work with Korean addresses at my job. There are some terms that regularly come back, but no English terms used and it makes a hard time for us to find the correct address. Ku San Ri Deok San Myun ChinChon Gun, ChinChon Choon, Cheong Book Do In this one above, What is the street? Thank you for your help! Also, there are these Korean terms regularly coming back: Could you please tell me what are the words for "street", "conty", "town", "tower", "square", "district", etc. Thank you very much!

s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247481766.50/warc/CC-MAIN-20190217071448-20190217093448-00047.warc.gz

CC-MAIN-2019-09

https://www.voovers.com/geometry/30-60-90-triangle/

math

30 60 90 Triangle Lesson The 30 60 90 Triangle Theorem A 30-60-90 triangle is a special right triangle that contains internal angles of 30, 60, and 90 degrees. Once we identify a triangle to be a 30 60 90 triangle, the values of all angles and sides can be quickly identified. Imagine cutting an equilateral triangle vertically, right down the middle. Each half has now become a 30 60 90 triangle. This visualization is very useful for remembering that the hypotenuse is twice as long as the short leg on a 30 60 90 triangle. 30 60 90 Triangle Rules To fully solve our right triangle as a 30 60 90, we have to first determine that the 3 angles of the triangle are 30, 60, and 90. To solve for the side lengths, a minimum of 1 side length must already be known. If we know that we are working with a right triangle, we know that one of the angles is 90 degrees. If we find that another angle is either 30 or 60 degrees, it is confirmed to be a 30 60 90 triangle. This is because the internal angles of a triangle will always sum to 180 degrees. Once we know the angles follow the 30 60 90 ratio, we can apply the relation that the hypotenuse is twice as long as the short leg. The side opposite to the 30 degree angle is the shortest, the side opposite to 60 degree angle is the second shortest (or second longest), and the side opposite to the 90 degree angle is the hypotenuse and the longest. A Proof of why a 30 60 90 Triangle Works Let’s take a look at the Pythagorean theorem being applied to a 30 60 90 triangle. Remember that the Pythagorean theorem is a2 + b2 = c2. Using a short leg length of 1, long leg length of 2, and hypotenuse length of √3, the Pythagorean theorem is applied and gives us: 12 + (√3)2 = 22, 4 = 4 The theorem holds true with the side lengths of a 30 60 90 triangle. 30 60 90 Triangle Example Problem Let's work through an example problem together to reinforce our understanding of 30 60 90 triangles. If the short leg of a 30 60 90 triangle is 5 meters long, what are the two other side lengths in meters? - We know from the image and ratio that the other two side lengths are √3a and 2a. - Since a = 5, let's substitute 5 in for a. - The two other side lengths are 5√3 and 10 meters long.

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https://www.blatantlyblunt.com/video-joey-bada-like-me-feat-bj-the-chicago-kid-prod-by-j-dilla-the-roots-tour-info/?noamp=mobile

math

Video: Joey Bada$$ – Like Me (Feat. BJ The Chicago Kid) (Prod. By J-Dilla & The Roots) [+Tour Info] Pro Era co founder and Beast Coast collective member Joey Bada$$ is enjoying the fruits of his labor with the release of his widely acclaimed debut album ‘B4.Da.$$’. Here he taps the underrated BJ The Chicago Kid for ‘Like Me’. Catch him and members of the collective on the road soon! Purchase: Joey Bada$$ – B4.Da.$$: here Purchase – Joey Bada$$ live showcase tickets here

s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571996.63/warc/CC-MAIN-20220814052950-20220814082950-00042.warc.gz

CC-MAIN-2022-33

http://www.markedbyteachers.com/gcse/science/how-does-the-weight-of-a-pendulum-effect-its-oscillation-time.html

math

How does the weight of a pendulum effect its oscillation time. Extracts from this document... How does the weight of a pendulum effect its oscillation time Plan The key factors that could affect this experiment are that the length of pendulum varying will change how far it has to swing to complete an oscillation. The weight of the pendulum will have a effect on the oscillation time because the pendulum will have more weights which will increase the surface area for air resistance to take place this is the only possible way the weight of the pendulum will affect the oscillation time because, gravity is a standard force, so the weight will not effect the amount of gravity taking place. The angle at which it is dropped at is another key factor because, if it is dropped from higher up it has longer to pick up speed so therefore it shall travel higher up giving it a longer oscillation time. If the pendulum is dropped from a lower then it shall have less time to pick up speed and so will not travel as high up on the other side. ...read more. We shall release the pendulum rather than push it because; it is easier to make the pendulum start with the same amount of force throughout the experiment. Results Weight Of Pendulum Length of time for 5 oscillations (secs) 100g 8.25 8.06 8.19 200g 8.35 8.50 8.53 300g 8.87 8.94 8.84 400g 8.91 8.88 8.59 500g 8.72 8.91 8.34 600g 9.41 9.50 9.47 700g 9.44 9.41 9.44 800g 9.43 9.44 9.59 900g 9.60 9.51 9.38 1000g 9.37 9.41 9.46 Weight Of Pendulum Average length of time for 5 oscillations (secs) 100g 8.2 200g 8.5 300g 8.9 400g 8.8 500g 8.7 600g 9.5 700g 9.4 800g 9.5 900g 9.5 1000g 9.4 Analysis My results show that an increase in weight on the pendulum the longer it takes for the pendulum to complete 5 oscillations, and my line of best fit on my graph shows a gradually increase from 8 seconds to complete 5 oscillation to 9.5 seconds to complete 5 oscillations. The reason they are close is because gravity is a standard force and is not affected by surface area or weight. ...read more. We could have improved our test by having a beam and when it is broke by the pendulum the fifth time the pendulum breaks the beam it could stop timing so it would be the same reaction time for all, we could have had a stand or something to hold the pendulum in place before it is dropped that way we could be sure it was from the same spot each time. In our experiment there was some slightly out of place results in our experiment but they're no severe ones that could be anomalous results. We could have investigated a further 1000g so we went to 2000g to see if after every 500g there was a jump in time of nearly a second our so, like there was in our experiment. We could also continue to the point of were we have added so much weight that it could not complete 5 oscillations to the same height as it started. We could also have tried the experiment at different temperatures to see what effect, if any it had on the oscillation times. Liam Kelly ...read more. This student written piece of work is one of many that can be found in our GCSE Forces and Motion section. Found what you're looking for? - Start learning 29% faster today - 150,000+ documents available - Just £6.99 a month - Join over 1.2 million students every month - Accelerate your learning by 29% - Unlimited access from just £6.99 per month

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CC-MAIN-2014-42

https://bumpercarfilms.com/qa/question-why-does-the-pendulum-stop-swinging.html

math

- What forces act on a pendulum? - How do you balance a pendulum clock? - What causes a pendulum to stop swinging? - What factors affect the swing of a pendulum? - What should I ask my pendulum? - What happens to a pendulum in space? - How long will a pendulum swing? - What type of energy did the pendulum have before Person #2 let go? - Does a pendulum tell the truth? - How does energy change in a swinging pendulum? - Why does a pendulum swing faster with a shorter string? - What happens if you add more weight to a pendulum? - Why doesn’t mass affect a pendulum? - How do I keep my pendulum swinging? - Where does energy go when a pendulum stops swinging? - Does a pendulum tell you what you want to hear? - Can you use a necklace as a pendulum? - Why is the pendulum important? - How exactly does a pendulum lose energy? - Does a pendulum ever stop? - How do you adjust a pendulum clock? What forces act on a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth’s mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum.. How do you balance a pendulum clock? Give the pendulum a slight sideways push to start it swinging. Listen to the tick –tock. The tick and the tock should sound evenly spaced, if so the clock is said to be “in beat” or balanced. If not, the crutch at the back of the movement will need to be bent sideways to either the right or the left. What causes a pendulum to stop swinging? A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity. … The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it. What factors affect the swing of a pendulum? The forces of gravity, the mass of the pendulum, length of the arm, friction and air resistance all affect the swing rate.Motion. Pull a pendulum back and release it. … Length. The swing rate, or frequency, of the pendulum is determined by its length. … Amplitude. … Mass. … Air Resistance/Friction. … Sympathetic Vibration.Apr 29, 2018 What should I ask my pendulum? Getting clarity questions to ask a pendulum… Do I keep getting the answer no because there is something better waiting for me? Am I passing up a good opportunity if I say no to __________ ? Am I expecting too much from ________ ? Should I get more information before deciding to __________ ? What happens to a pendulum in space? No, it wouldn’t swing back and forth. … Since there is no gravity in orbit, there is no force that pulls the pendulum down and make it swing back and forth. If you pushed a pendulum in orbit, the bob of the pendulum would keep going in full circles until they stop due to friction. How long will a pendulum swing? Here is an extra fun fact. A pendulum with a length of 1 meter has a period of about 2 seconds (so it takes about 1 second to swing across an arc). This means that there is a relationship between the gravitational field (g) and Pi. What type of energy did the pendulum have before Person #2 let go? It was potential energy because the washer could swing due to its relative position and gravity. This kind of potential energy is known as gravitational potential energy. What’s interesting about a pendulum, though, is that when you let go of it, the potential energy gradually transforms into kinetic energy. Does a pendulum tell the truth? As an example – people use a pendulum to ask if a certain food is good for them by holding it over the item, and if it’s a healthy high vibration food then the pendulum will answer YES. How does energy change in a swinging pendulum? As a pendulum swings, its potential energy converts to kinetic and back to potential. … During the course of a swing from left to right, potential energy is converted into kinetic energy and back. Why does a pendulum swing faster with a shorter string? The length of the string affects the pendulum’s period such that the longer the length of the string, the longer the pendulum’s period. … A pendulum with a longer string has a lower frequency, meaning it swings back and forth less times in a given amount of time than a pendulum with a shorter string length. What happens if you add more weight to a pendulum? When you add a weight to the middle of the other pendulum, however, you effectively make it shorter. Shorter pendulums swing faster than longer ones do, so the pendulum on the left swings faster than the pendulum on the right. Why doesn’t mass affect a pendulum? The mass on a pendulum does not affect the swing because force and mass are proportional and when the mass increases so does the force. … In the formula, “l” is the length and “g” is the force of gravity. Therefore, the mass does not affect the period of the pendulum. How do I keep my pendulum swinging? There are several things that you can do to make a pendulum swing for a long time:Make it heavy (and, specifically, dense). The more mass a pendulum has, the less outside influences such as air resistance will degrade its swing.Put it in a vacuum. … Use an escapement mechanism. … Give it a large initial swing. Where does energy go when a pendulum stops swinging? Once the weighted end of the pendulum is released, it will become active as gravity pulls it downward. Potential energy is converted to kinetic energy, which is the energy exerted by a moving object. Does a pendulum tell you what you want to hear? A pendulum can energetically swing and work for almost anyone who believes it will move. … If you really want to get that settlement or win the lottery, the Ego will swing the pendulum in the “Yes” direction, because that is what you really want to hear. Can you use a necklace as a pendulum? Pendulums can be made of different materials, some people using a simple necklace with a crystal or charm at the end. Be sure the bob or bobber – or weight on the end – is not too light or too heavy. … The best length for the pendulum is six inches. You can make your pendulum or buy one. Why is the pendulum important? Pendulum, body suspended from a fixed point so that it can swing back and forth under the influence of gravity. … Pendulums are used to regulate the movement of clocks because the interval of time for each complete oscillation, called the period, is constant. How exactly does a pendulum lose energy? The pendulum loses energy to wind resistance, friction between the tube and the string, and internal friction within the bending string. … Then the driver allows the force of gravity to convert some of that extra energy to kinetic energy, by allowing the bob to fall an extra distance at large angles. Does a pendulum ever stop? A pendulum can only stop when its gravitational potential is lowest and it no longer has energy driving it. (This is of course assuming the pendulum is allowed to swing until brought to a stop by friction, rather than being stopped by an applied force.) How do you adjust a pendulum clock? Stop the pendulum to move the pendulum bob up or down to change the pendulum’s effective length. If the clock is running fast, move the bob down or turn the nut to the left. If the clock is running slow, move the bob up or turn the nut to the right. Restart the pendulum and reset the clock hands to the proper time.

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https://www.rcsb.org/structure/1sws

math

A structural snapshot of an intermediate on the streptavidin-biotin dissociation pathway.Freitag, S., Chu, V., Penzotti, J.E., Klumb, L.A., To, R., Hyre, D., Le Trong, I., Lybrand, T.P., Stenkamp, R.E., Stayton, P.S. (1999) Proc Natl Acad Sci U S A 96: 8384-8389 - PubMed: 10411884 - DOI: https://doi.org/10.1073/pnas.96.15.8384 - Primary Citation of Related Structures: - PubMed Abstract: It is currently unclear whether small molecules dissociate from a protein binding site along a defined pathway or through a collection of dissociation pathways. We report herein a joint crystallographic, computational, and biophysical study that suggests the Asp-128 --> Ala (D128A) streptavidin mutant closely mimics an intermediate on a well-defined dissociation pathway. Asp-128 is hydrogen bonded to a ureido nitrogen of biotin and also networks with the important aromatic binding contacts Trp-92 and Trp-108. The Asn-23 hydrogen bond to the ureido oxygen of biotin is lengthened to 3.8 A in the D128A structure, and a water molecule has moved into the pocket to replace the missing carboxylate interaction. These alterations are accompanied by the coupled movement of biotin, the flexible binding loop containing Ser-45, and the loop containing the Ser-27 hydrogen bonding contact. This structure closely parallels a key intermediate observed in a potential of mean force-simulated dissociation pathway of native streptavidin, where the Asn-23 hydrogen bond breaks first, accompanied by the replacement of the Asp-128 hydrogen bond by an entering water molecule. Furthermore, both biotin and the flexible loop move in a concerted conformational change that closely approximates the D128A structural changes. The activation and thermodynamic parameters for the D128A mutant were measured and are consistent with an intermediate that has traversed the early portion of the dissociation reaction coordinate through endothermic bond breaking and concomitant gain in configurational entropy. These composite results suggest that the D128A mutant provides a structural "snapshot" of an early intermediate on a relatively well-defined dissociation pathway for biotin. Department of Bioengineering, University of Washington, Seattle, WA 98195, USA.

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http://biology.stackexchange.com/questions/tagged/dna-replication+dna

math

to customize your list. more stack exchange communities Start here for a quick overview of the site Detailed answers to any questions you might have Discuss the workings and policies of this site DNA sequence from the middle of a gene Someone gives you a short DNA sequence that comes from the middle of a gene. 5'- TCTAACTGATTAGC -3' 3'- AGATTGACTAATCG -5' From this sequence, determine the ... Feb 19 at 6:31 newest dna-replication dna questions feed Podcast #57 – We Just Saw This On Florp Putting the Community back in Wiki Hot Network Questions simulate dirichlet process in R Is there any way I can add watermark to .jpg files (around 15000), en masse instead of one by one? tikz: two blocks stacked on each other Disk brake squealing after bike shop changed my pads. Enough reason to ask them to re-service (for free)? Hierarchical statistical diagrams Solving limit without L'Hôpital Does 17% have to be equal to 0.17? Becoming Better at Math How can I safely use storage thin provisioning? Which pronoun to use to refer to a city? Can random events destroy your ship? is "release date" grammatically correct? What is the maximum range of the very best Starfleet transporter in the Star Trek universe? Has the Holocaust been exaggerated? Can you play older versions of cards in standard? Are there any way to see how much space is used for Ubuntu? How can I create a password that says "SALT ME!" when hashed? Reverse of an Amount and counting Do banks give us interest even for the money that we only had briefly in our account? Why do games ask for screen resolution instead of automatically fitting the window size? Changing combination lock PHP files browsable: is this a vulnerability? Does electricity flow on the surface of a wire or in the interior? C What kind of sorting algorithm is this? more hot questions Life / Arts Culture / Recreation TeX - LaTeX Unix & Linux Ask Different (Apple) Geographic Information Systems Science Fiction & Fantasy Seasoned Advice (cooking) Personal Finance & Money English Language & Usage Mi Yodeya (Judaism) Cross Validated (stats) Theoretical Computer Science Meta Stack Exchange Stack Overflow Careers site design / logo © 2014 stack exchange inc; user contributions licensed under cc by-sa 3.0

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https://manufacturingindonesia-exhibition.com/qa/quick-answer-what-is-the-hourly-rate-for-a-machine-shop.html

math

- How much does CNC machining cost per hour? - How do you calculate effective labor rate? - What is the difference between mill and lathe machine? - What do machine shops do? - How much does a machine shop manager make? - How much should I charge for labor? - How expensive is a CNC machine? - How much is a 3 axis CNC machine? - How is Machine Shop hourly rate calculated? - What is the average machine shop rate? - How is shop rate calculated? - Is CNC easy to learn? - How much does a machine shop foreman make? - How much do machine shop workers make? - What is shop rate? How much does CNC machining cost per hour? CNC turning is usually priced lower at $35 per hour, while the machine cost per hour of multi-axis CNC machining typically ranges between $75 and $120 or higher.. How do you calculate effective labor rate? How do you calculate Effective Labor Rate? That’s your total labor dollars divided by hours flagged by Technicians in the shop. That’s how you calculate Effective Labor Rate. What is the difference between mill and lathe machine? Both lathes and milling machines are used to remove material from a workpiece. Lathes, however, involve rotating a workpiece against a single-bladed cutting tool, whereas milling machines involve rotating a multi-bladed or -pointed cutting tool against a stationary workpiece. What do machine shops do? A machine shop is a facility with equipment and supplies for machining, a process where parts are cut, fabricated, and finished to prepare them for use. Machine shops are used in the creation of new parts, as well as repairs of existing equipment and parts. How much does a machine shop manager make? Average Salary for a Machine Shop Manager Machine Shop Managers in America make an average salary of $70,611 per year or $34 per hour. The top 10 percent makes over $116,000 per year, while the bottom 10 percent under $42,000 per year. How much should I charge for labor? Calculate Your Hourly Rate Business schools teach a standard formula for determining an hourly rate: Add up your labor and overhead costs, add the profit you want to earn, then divide the total by your hours worked. This is the minimum you must charge to pay your expenses, pay yourself a salary, and earn a profit. How expensive is a CNC machine? CNC Machine Price by CNC TypeCNC TypeDescriptionAverageCNC MillMill$60,000CNC VMCVertical Machining Center$50,000CNC HMCHorizontal Machining Center$89,000CNC HBMHorizontal Boring Mill$125,0002 more rows•Jan 8, 2019 How much is a 3 axis CNC machine? How Much Does a CNC Machine Cost?Hobbyist-grade CNC router$1k-3kEntry-level (or toolroom) 3-axis vertical machining center$60k-100kProduction 3-axis vertical machining center$150k-300kEntry-level 5-axis mill$200k-500kProduction 5-axis mill$500k+4 more rows How is Machine Shop hourly rate calculated? Service Rate Calculation – Machine-Specific Rate Rate = (specific machine(s) cost per hour + labor & overhead cost per hour) x markup x total hours for the job. What is the average machine shop rate? $60-80/hourThe average machine shop in the U.S. charges $60-80/hour. Where a company lands within this range depends in part on overhead costs, including electricity costs in their state, but mostly on the quality of their work. How is shop rate calculated? (Expenses + profit) ÷ hours = shop rate Find out what it costs to run your business, add in profit up-front, and then divide it by how much time you have. This tells you exactly how much each unit of time you have to sell is worth, which you can then use to calculate your project prices. Is CNC easy to learn? Is CNC programming hard to learn? Very basic CNC programming is easy to learn, provided that you understand basic math and have a grasp of how machining works. … Intermediate programming skills can be learned within a year and advanced CNC programming can take several years to learn. How much does a machine shop foreman make? Average Salary for a Machine Shop Foreman Machine Shop Foremen in America make an average salary of $64,867 per year or $31 per hour. The top 10 percent makes over $86,000 per year, while the bottom 10 percent under $48,000 per year. How much do machine shop workers make? Machine Shop Worker SalaryAnnual SalaryMonthly PayTop Earners$35,500$2,95875th Percentile$31,500$2,625Average$30,293$2,52425th Percentile$28,000$2,333 What is shop rate? Gross Sales – Materials/number of work days per month/number of production employees/direct man hours per day per employee = shop rate. This gives you your loaded shop rate, meaning overhead and profit are already included. … Perhaps what you mean is the effective hourly rate is more the actual, direct hourly rate.

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CC-MAIN-2021-31

https://ez.analog.com/thread/41135-ade7758-dc-voltage-and-current-measurement

math

I would like to measure DC voltage and current with my ADE7758 IC, I have found some idea about it from AN-639 "Can the ADE energy meters measure dcpower? Yes. DC power can be measured using ADE ICs. To do so, the high-pass filter has to be disabled. However, the dc energy accuracy error is expected to be around 1% over a 100:1 By setting DISHPF (bit 0 of OPMODE), the high pass filter should be disabled. I'm wondering is that anything else to be set? Do I have to do some changes on the attenuation circuit because I will be needed to measure up to 50Vdc, An attenuation factor of 100 will be good enough to maintain the input signal to be less than 500mVpeak? Besides that, the burden resistor,Rb will removed if hall effect sensor is being used as the current transducer? Thanks.

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CC-MAIN-2018-22

https://www.physicsforums.com/threads/springs-question-second-order-o-d-e.645284/

math

Hi all, I was wondering is you could help me with this springs question. We've only done springs hanging from a fixed support above being stretched, but now I've got a question where the spirng is being compressed. 1. The problem statement, all variables and given/known data So, here's some basic info about the question Obviously, acceleration due to gravity is 9.8 m/s^2 The downward direction is considered the positive direction A spring is placed on a fixed support standing vertically The natural length of the spring is 5 metres An iron ball weighing 5kg is attached to the top of it This mass compresses the spring by 0.392m The ball is pulled to a distance of 0.4m above its equalibirum position then released Damping force is proportional to the instantaneous velocity of the ball y(t)=The displacement of the ball below the equalibrium position at t seconds after the release 2. Relevant equations So, the first question is to proove that the equation of motion for the ball is... [itex]0=y''(t)+(b/5)y'(t)+25y(t)[/itex], where b kg/s is the damping constant 3. The attempt at a solution Here is a rough diagram of what I believe to be happening from the description of the problem: So, with Hooke's law [itex]T=mg[/itex] [itex]ks=mg[/itex] (equation 1) [itex]k=mg/s[/itex] [itex]k=(5*9.8)/(-0.392)[/itex] [itex]k=-125[/itex] With Newton's law of motion (F=ma): [itex]m*y''(t)=mg-T-R[/itex] [itex]m*y''(t)=mg-k(s+y(t))-by'(t)[/itex] [itex]m*y''(t)=mg-ks-ky(t)-by'(t)[/itex], from equation 1: ks-mg=0 so [itex]m*y''(t)+ky(t)+by'(t)=0[/itex], So subbing in k=-125, m=5 [itex]5*y''(t)-125y(t)+by'(t)=0[/itex] [itex]y''(t)+b/5y'(t)-25y(t)=0[/itex] This is almost what I need, but for some reason I have -25y(t) instead of +25y(t). Can anyone see where I've gone wrong? The only thing I can think of is that k (and thus s) should be positive, but I am unsure why as it is in compression. We have been taught that stretching (extension) of the spring make "s" a postive number, so one would think that compression of the spring would make give "s" a negative value. Is k always > 0 by definition?

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http://myvocabbook.com/download/a-first-course-in-differential-equations-modeling-and-simulation

math

Author: Carlos A. Smith Publisher: CRC Press Release Date: 2011-05-18 Emphasizing a practical approach for engineers and scientists, A First Course in Differential Equations, Modeling, and Simulation avoids overly theoretical explanations and shows readers how differential equations arise from applying basic physical principles and experimental observations to engineering systems. It also covers classical methods for obtaining the analytical solution of differential equations and Laplace transforms. In addition, the authors discuss how these equations describe mathematical systems and how to use software to solve sets of equations where analytical solutions cannot be obtained. Using simple physics, the book introduces dynamic modeling, the definition of differential equations, two simple methods for obtaining their analytical solution, and a method to follow when modeling. It then presents classical methods for solving differential equations, discusses the engineering importance of the roots of a characteristic equation, and describes the response of first- and second-order differential equations. A study of the Laplace transform method follows with explanations of the transfer function and the power of Laplace transform for obtaining the analytical solution of coupled differential equations. The next several chapters present the modeling of translational and rotational mechanical systems, fluid systems, thermal systems, and electrical systems. The final chapter explores many simulation examples using a typical software package for the solution of the models developed in previous chapters. Providing the necessary tools to apply differential equations in engineering and science, this text helps readers understand differential equations, their meaning, and their analytical and computer solutions. It illustrates how and where differential equations develop, how they describe engineering systems, how to obtain the analytical solution, and how to use software to simulate the systems. Author: Carlos A. Smith Publisher: CRC Press Release Date: 2016-04-05 A First Course in Differential Equations, Modeling, and Simulation shows how differential equations arise from applying basic physical principles and experimental observations to engineering systems. Avoiding overly theoretical explanations, the textbook also discusses classical and Laplace transform methods for obtaining the analytical solution of differential equations. In addition, the authors explain how to solve sets of differential equations where analytical solutions cannot easily be obtained. Incorporating valuable suggestions from mathematicians and mathematics professors, the Second Edition: Expands the chapter on classical solutions of ordinary linear differential equations to include additional methods Increases coverage of response of first- and second-order systems to a full, stand-alone chapter to emphasize its importance Includes new examples of applications related to chemical reactions, environmental engineering, biomedical engineering, and biotechnology Contains new exercises that can be used as projects and answers to many of the end-of-chapter problems Features new end-of-chapter problems and updates throughout Thus, A First Course in Differential Equations, Modeling, and Simulation, Second Edition provides students with a practical understanding of how to apply differential equations in modern engineering and science. Author: Frank R. Giordano Publisher: Cengage Learning Release Date: 2013-03-05 Offering a solid introduction to the entire modeling process, A FIRST COURSE IN MATHEMATICAL MODELING, 5th Edition delivers an excellent balance of theory and practice, and gives you relevant, hands-on experience developing and sharpening your modeling skills. Throughout, the book emphasizes key facets of modeling, including creative and empirical model construction, model analysis, and model research, and provides myriad opportunities for practice. The authors apply a proven six-step problem-solving process to enhance your problem-solving capabilities -- whatever your level. In addition, rather than simply emphasizing the calculation step, the authors first help you learn how to identify problems, construct or select models, and figure out what data needs to be collected. By involving you in the mathematical process as early as possible -- beginning with short projects -- this text facilitates your progressive development and confidence in mathematics and modeling. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version. Author: Dennis G. Zill Release Date: 1993-01-01 % mainly for math and engineering majors% clear, concise writng style is student orientedJ% graded problem sets, with many diverse problems, range form drill to more challenging problems% this course follows the three-semester calculus sequence at two- and four-year schools Author: A. Iserles Publisher: Cambridge University Press Release Date: 2009 lead the reader to a theoretical understanding of the subject without neglecting its practical aspects. The outcome is a textbook that is mathematically honest and rigorous and provides its target audience with a wide range of skills in both ordinary and partial differential equations." --Book Jacket. Author: J. David Logan Release Date: 2015-07-01 The third edition of this concise, popular textbook on elementary differential equations gives instructors an alternative to the many voluminous texts on the market. It presents a thorough treatment of the standard topics in an accessible, easy-to-read, format. The overarching perspective of the text conveys that differential equations are about applications. This book illuminates the mathematical theory in the text with a wide variety of applications that will appeal to students in physics, engineering, the biosciences, economics and mathematics. Instructors are likely to find that the first four or five chapters are suitable for a first course in the subject. This edition contains a healthy increase over earlier editions in the number of worked examples and exercises, particularly those routine in nature. Two appendices include a review with practice problems, and a MATLAB® supplement that gives basic codes and commands for solving differential equations. MATLAB® is not required; students are encouraged to utilize available software to plot many of their solutions. Solutions to even-numbered problems are available on springer.com. The book is intended as an advanced undergraduate or first-year graduate course for students from various disciplines, including applied mathematics, physics and engineering. It has evolved from courses offered on partial differential equations (PDEs) over the last several years at the Politecnico di Milano. These courses had a twofold purpose: on the one hand, to teach students to appreciate the interplay between theory and modeling in problems arising in the applied sciences, and on the other to provide them with a solid theoretical background in numerical methods, such as finite elements. Accordingly, this textbook is divided into two parts. The first part, chapters 2 to 5, is more elementary in nature and focuses on developing and studying basic problems from the macro-areas of diffusion, propagation and transport, waves and vibrations. In turn the second part, chapters 6 to 11, concentrates on the development of Hilbert spaces methods for the variational formulation and the analysis of (mainly) linear boundary and initial-boundary value problems. Author: Stefano M. Iacus Publisher: Springer Science & Business Media Release Date: 2009-04-27 This book covers a highly relevant and timely topic that is of wide interest, especially in finance, engineering and computational biology. The introductory material on simulation and stochastic differential equation is very accessible and will prove popular with many readers. While there are several recent texts available that cover stochastic differential equations, the concentration here on inference makes this book stand out. No other direct competitors are known to date. With an emphasis on the practical implementation of the simulation and estimation methods presented, the text will be useful to practitioners and students with minimal mathematical background. What’s more, because of the many R programs, the information here is appropriate for many mathematically well educated practitioners, too. Author: Clayton R. Paul Publisher: John Wiley & Sons Release Date: 2011-09-20 Just the math skills you need to excel in the study or practice of engineering Good math skills are indispensable for all engineers regardless of their specialty, yet only a relatively small portion of the math that engineering students study in college mathematics courses is used on a frequent basis in the study or practice of engineering. That's why Essential Math Skills for Engineers focuses on only these few critically essential math skills that students need in order to advance in their engineering studies and excel in engineering practice. Essential Math Skills for Engineers features concise, easy-to-follow explanations that quickly bring readers up to speed on all the essential core math skills used in the daily study and practice of engineering. These fundamental and essential skills are logically grouped into categories that make them easy to learn while also promoting their long-term retention. Among the key areas covered are: Algebra, geometry, trigonometry, complex arithmetic, and differential and integral calculus Simultaneous, linear, algebraic equations Linear, constant-coefficient, ordinary differential equations Linear, constant-coefficient, difference equations Linear, constant-coefficient, partial differential equations Fourier series and Fourier transform Laplace transform Mathematics of vectors With the thorough understanding of essential math skills gained from this text, readers will have mastered a key component of the knowledge needed to become successful students of engineering. In addition, this text is highly recommended for practicing engineers who want to refresh their math skills in order to tackle problems in engineering with confidence. The authors' enthusiasm for their subject is eloquently conveyed in this book, and draws the reader very quickly into active investigation of the problems posed. By providing plenty of modelling examples from a wide variety of fields - most of which are familiar from everyday life - the book shows how to apply mathematical ideas to situations which would not previously have been considered to be 'mathematical' in character.

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CC-MAIN-2018-51

http://www.solutioninn.com/dailey-company-issued-300000-8-15year-bonds-on-december-31

math

Dailey Company issued $300,000, 8%, 15-year bonds on December 31, 2013, for $288,000. Interest is payable annually on December 31. Dailey uses the straight-line method to amortize bond premium or discount. Prepare the journal entries to record the following events. (a) The issuance of the bonds. (b) The payment of interest and the discount amortization on December 31, 2014. (c) The redemption of the bonds at maturity, assuming interest for the last interest period has been paid and recorded.

s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187822488.34/warc/CC-MAIN-20171017200905-20171017220905-00377.warc.gz

CC-MAIN-2017-43

http://paulblankenship.com/category/math/

math

Teaching math in geography? Yes! Social studies, or social science, involves mathematics. Geography is especially mathematical as maps, globes, and GPS systems are all based in mathematics. My trip to the HSTW14 conference convinced me that, as a social studies teacher, I could do more to reinforce math in my courses and thereby make them more applicable to the real world. I set a goal of doing five FALs this year along with at least two LDC modules. I’ll post here as I develop and test applications of MDC concepts in the geography classroom. Below are some links to help include mathematics in the study of geography.

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CC-MAIN-2018-09

http://usadistance.com/driving-distance-from-cheyenne-to-torrington

math

Distance from USA Cheyenne to Torrington distance The driving distance or the travel distance from Cheyenne to Torrington is 1827.3 Miles . The straight line distance from Cheyenne to Torrington is 1632.8 Miles. The kilometer based traveling distance is 2940.731 KM and the KM based straight line distance is 2627.8 KM. Cheyenne location and Torrington location Cheyenne is located at the latitude of 41.139754 and the longitude of -104.8201002. Cheyenne is situated at the latitude of 41.8003274 and the longitude of -73.1213337. The traveling source point address is Cheyenne, WY, USA. The destination travel point address is Torrington, CT, USA. Cheyenne to Torrington travel time The travel time between Cheyenne and Torrington is 26.65 hours . We assumed that you are traveling at the speed of 60km per hour from Cheyenne to Torrington. The given travel time between Cheyenne to Torrington may vary based on the travel route, speed and consistent traveling. Cheyenne location and Torrington fuel cost The Fuel cost( Gas cost , Petrol cost) to travel from Cheyenne location to Torrington is 245.06 USD. The given fuel cost may vary based on the fuel consumption of your vehicle and varying price of the fuel. ; Cheyenne travel distance calculator You are welcome to find the travel distance calculation from cheyenne You are viewing the page distance between cheyenne and torrington. This page may provide answer for the following queries. what is the distance between Cheyenne to Torrington ?. How far is Cheyenne from Torrington ?. How many kilometers between Cheyenne and Torrington ?. What is the travel time between Cheyenne and Torrington. How long will it take to reach Torrington from Cheyenne?. What is the geographical coordinates of Cheyenne and Torrington?. The given driving distance from Torrington to Cheyenne may vary based on various route.

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CC-MAIN-2021-10

http://beta.kn-k.ru/pdf/mathematical-physics-chicago-lectures-in-physics

math

By Robert Geroch Read Online or Download Mathematical Physics (Chicago Lectures in Physics) PDF Best mathematical physics books This booklet reminds scholars in junior, senior and graduate point classes in physics, chemistry and engineering of the maths they could have forgotten (or discovered imperfectly) that's had to achieve technological know-how classes. the focal point is on math really utilized in physics, chemistry and engineering, and the method of arithmetic starts with 12 examples of accelerating complexity, designed to hone the student's skill to imagine in mathematical phrases and to use quantitative the way to medical difficulties. A ‘science of towns and areas’ is necessary for assembly destiny demanding situations. the realm is urbanising: large towns are being created and are carrying on with to develop swiftly. there are various making plans and improvement concerns coming up in several manifestations in nations around the globe. those advancements can, in precept, be simulated via mathematical desktop types which offer instruments for forecasting and checking out destiny eventualities and plans. The easiest mathematical version of the Brownian movement of physics is the straightforward, symmetric random stroll. This publication collects and compares present effects — in general robust theorems which describe the houses of a random stroll. the fashionable difficulties of the restrict theorems of chance thought are taken care of within the uncomplicated case of coin tossing. Computation in technological know-how offers a theoretical history in computation to scientists who use computational tools. It explains how computing is utilized in the average sciences, and gives a high-level evaluate of these points of machine technology and software program engineering which are so much proper for computational technological know-how. - Quantum Simulations of Materials and Biological Systems - Radiative Transfer in the Atmosphere and Ocean (Cambridge Atmospheric and Space Science Series) - Glasses and Grains: Poincaré Seminar 2009: 61 (Progress in Mathematical Physics) - Theory and Applications in Mathematical Physics:In Honor of B Tirozzi's 70th Birthday Additional resources for Mathematical Physics (Chicago Lectures in Physics) Mathematical Physics (Chicago Lectures in Physics) by Robert Geroch

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CC-MAIN-2022-05

https://socratic.org/questions/595d04937c014916cb6cd962

math

I'll assume you meant a nitride ion... We're asked to find the mass, in This is simply a charged nitrogen atom, which has gained three electrons from bonding. Recall that electrons have negligible mass compared to nucleons (protons and neutrons). This, the atomic mass of a nitride ion is essentially the same as the mass of a lone nitrogen atom. According to a periodic table, the relative atomic mass of One atomic mass unit is

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http://www.endurance.net/RideCamp/archives/past/9805/msg01468.html

math

Check it Out! [Date Prev]  [Date Next] [Thread Prev]  [Thread Next]  [Date Index]  [Thread Index]  [Author Index]  [Subject Index] S.African 100 miler/International >Imagine how lovely it would be if South Africa rode a 100 miler on the >>day as the Tevis and Quilty and all three countries competed against each >>other (Something like the Tri Nations Rugby) Just an Idea !!!!! WHAT FUN!!! It would be difficult to have the riders compete against each other over different terrain and temps, but what about a team event. for ex: a 6 rider team, with 2 riders on each course!!! It seems, to me, the would be a truly INTERNATIONAL competition, with counties pulling together for the love of the sport!!! Back to TOC

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CC-MAIN-2019-43

http://www.justanswer.com/homework/3zbbc-test-prep-questionare-due-extra-credit-need-help.html

math

Homework Questions? Ask a Tutor for Answers ASAP Not a Homework Question? How JustAnswer Works: Ask an Expert Experts are full of valuable knowledge and are ready to help with any question. Credentials confirmed by a Fortune 500 verification firm. Get a Professional Answer Via email, text message, or notification as you wait on our site. Ask follow up questions if you need to. 100% Satisfaction Guarantee Rate the answer you receive. Ask Linda_us Your Own Question Finance, Accounts & Homework Tutor Post Graduate Diploma in Management (MBA) Type Your Homework Question Here... Linda_us is online now test prep questionare due for extra credit need help This answer was rated: 1. Jasmine Company produces hand tools. A sales budget for the next four months is as follows: March 10,000 units, April 13,000, May 16,000 and June 21,000. Jasmine Company's ending finished goods inventory policy is 10% of the following month's sales. March 1 inventory is projected to be 1,400 units. How many units will be produced in March? A. 10,000 B. 9,900 C. 13,000 D. 10,100 2. Jasmine Company produces hand tools. A production budget for the next four months is as follows: March 10,300 units, April 13,300, May 16,500, and June 21,800. Jasmine Company's ending finished goods inventory policy is 10% of the following month's sales. Jasmine plans to sell 16,000 units in May. What is budgeted ending inventory for March? A. 1,030 B. 1,300 C. 1,330 D. 1,650 3. Albertville Inc produces leather handbags. The production budget for the next four months is: July 5,000 units, August 7,000, September 7,500, October 8,000. Each handbag requires 0.5 square meters of leather. Albertville Inc's leather inventory policy is 30% of next month's production needs. On July 1 leather inventory was expected to be 1,000 square meters. What will leather purchases be in July? A. 2,300 square meters B. 2,550 square meters C. 2,700 square meters D. 3,575 square meters 4. The purpose of the cash budget is to A. be used as a basis for the operating budgets. B. provide external users with an estimate of future cash flows. C. help managers plan ahead to make certain they will have enough cash on hand to meet their operating needs. D. summarize the cash flowing into and out of the business during the past period. 5. Brimson has forecast sales for the next three months as follows: July 4,000 units, August 6,000 units, September 7,500 units. Brimson's policy is to have an ending inventory of 40% of the next month's sales needs on hand. July 1 inventory is projected to be 1,500 units. Monthly manufacturing overhead is budgeted to be $17,000 plus $5 per unit produced. What is budgeted manufacturing overhead for August? A. $50,000 B. $47,000 C. $33,000 D. $32,000 6. The difference between the actual cost driver amount and the standard cost driver amount, multiplied by the standard variable overhead rate is the A. variable overhead rate variance. B. variable overhead efficiency variance. C. variable overhead volume variance. D. over - or underapplied variance. 7. Albertville applies overhead based on direct labor hours. The variable overhead standard is 2 hours at $12 per hour. During July, Albertville spent $116,700 for variable overhead. 8,890 labor hours were used to produce 4,700 units. How much is variable overhead on the flexible budget? A. $56,400 B. $106,680 C. $112,800 D. $116,700 8. The fixed overhead volume variance is the difference between A. Actual fixed overhead and budgeted fixed overhead. B. Actual fixed overhead and applied fixed overhead. C. Applied fixed overhead and budgeted fixed overhead. D. Actual fixed overhead and the standard fixed overhead rate times actual cost driver. 9. The difference between the actual volume and the budgeted volume, multiplied by the fixed overhead rate based on budgeted volume, is the A. fixed overhead spending variance. B. fixed overhead price variance. C. fixed overhead efficiency variance. D. fixed overhead volume variance. 10. Albertville has budgeted fixed overhead of $67,500 based on budgeted production of 4,500 units. During July, 4,700 units were produced and $71,400 was spent on fixed overhead. What is the budgeted fixed overhead rate? A. $14.36 B. $15.00 C. $15.19 D. $15.89 11. In a standard cost system, the initial debit to an inventory account is based on A. standard cost rather than actual cost. B. actual cost rather than standard cost. C. actual cost less the standard cost. D. standard cost less the actual cost. 12. In what type of organization is decision-making authority spread throughout the organization? A. Centralized organization B. Decentralized organization C. Participative organization D. Top-down organization 13. Which of the following is NOT an advantage of decentralization? A. Allows top managers to focus on strategic issues B. Potential duplication of resources C. Allows for development of managerial expertise D. Managers can react quickly to local information 14. The responsibility center in which the manager has responsibility and authority over revenues, costs and assets is A. a cost center. B. an investment center. C. a profit center. D. a revenue center. 15. Return on investment can be calculated as A. ROI = sales revenue/average invested assets B. ROI = operating income/sales revenue C. ROI = operating income/average invested assets D. ROI = average invested assets/sales revenue 16. Which of the following balanced scorecard perspectives measures how an organization satisfies its stakeholders? B. Internal business processes C. Learning and growth 17. Which of the following is not something that should be compiled for each dimension of the balanced scorecard? A. Performance measures C. Strategic vision D. Specific objectives Share this conversation replied 6 years ago. THIS ANSWER IS LOCKED! You need to spend $3 to view this post. to your account and buy credits. Linda_us and 5 other Homework Specialists are ready to help you Ask your own question now Share this conversation Related Homework Questions I wrote the following statements for my assignment in MySQL: I have 2 homework that I need you to help, this is the 1st 6. The hexadecimal equivalent of decimal 1234 is A. 2D4.. I have a column that divides j/s. This sometimes results in Translate the sentence below into English. Pienso que nunca I need help with my pre-exam for Penn high school this is Wiz of All NOT GLENN NOR F. NAZ PLEASE In the story titled "Macbeth" by Exam Number: 986159RR Need help with a couple assignment questions FOR LogicPro Only, I have another database question. Its a Ask a Tutor Get a Professional Answer. 100% Satisfaction Guaranteed. 37 Tutors are Online Now Type Your Homework Question Here... Terms of Service Privacy & Security © 2003-2017 JustAnswer LLC

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CC-MAIN-2017-04

https://gamefaqs.gamespot.com/ps2/920238-resident-evil-outbreak-file-2/faqs/35039

math

What do you need help on? Cancel X Jim FAQ by The1Executioner Version: 1.00 | Updated: 05/10/05 ____ __ __ /\ _`\ __ /\ \ /\ \__ \ \ \L\ \ __ ____/\_\ \_\ \ __ ___\ \ ,_\ \ \ , / /'__`\ /',__\/\ \ /'_` \ /'__`\/' _ `\ \ \/ \ \ \\ \ /\ __//\__, `\ \ \/\ \L\ \/\ __//\ \/\ \ \ \_ \ \_\ \_\ \____\/\____/\ \_\ \___,_\ \____\ \_\ \_\ \__\ \/_/\/ /\/____/\/___/ \/_/\/__,_ /\/____/\/_/\/_/\/__/ ____ ___ /\ _`\ __ /\_ \ \ \ \L\_\ __ __ /\_\\//\ \ \ \ _\L /\ \/\ \\/\ \ \ \ \ \ \ \L\ \ \ \_/ |\ \ \ \_\ \_ \ \____/\ \___/ \ \_\/\____\ \/___/ \/__/ \/_/\/____/ _____ __ __ __ /\ __`\ /\ \__/\ \ /\ \ \ \ \/\ \ __ __\ \ ,_\ \ \____ _ __ __ __ \ \ \/'\ \ \ \ \ \/\ \/\ \\ \ \/\ \ '__`\/\`'__\/'__`\ /'__`\ \ \ , < \ \ \_\ \ \ \_\ \\ \ \_\ \ \L\ \ \ \//\ __//\ \L\.\_\ \ \\` \ \ \_____\ \____/ \ \__\\ \_,__/\ \_\\ \____\ \__/.\_\\ \_\ \_\ \/_____/\/___/ \/__/ \/___/ \/_/ \/____/\/__/\/_/ \/_/\/_/ ____ ___ __ __ ___ /\ _`\ __ /\_ \ _\ \\ \__ /'___`\ \ \ \L\_\/\_\\//\ \ __ /\__ _ _\ /\_\ /\ \ \ \ _\/\/\ \ \ \ \ /'__`\ \/_L\ \\ \L_\/_/// /__ \ \ \/ \ \ \ \_\ \_/\ __/ /\_ _ _\ // /_\ \ \ \_\ \ \_\/\____\ \____\ \/_/\_\\_\/ /\______/ \/_/ \/_/\/____/\/____/ \/_//_/ \/_____/ XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX http://www.executioner.ne1.net http://www.ubcsteam.ne1.net XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX +------------------------------------------------------------+ |Author : The Executioner | |FAQ Type : Jim FAQ | |Spoilers : Yes | |Game : Resident Evil Outbreak File #2 | +------------------------------------------------------------+ X--------X |CONTENTS| X--------X 1) Introduction/Author Information 2) Characters 3) Jim Information 4) Boss Strategies for Jim 5) Scenario Notes about Jim 6) Jim Secret Costumes List 7) Jim's SP Item list 8) Frequently Asked Questions 9) Contact 10) Updates 11) Credits 12) Legal and Copyright *MUST KNOW* XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 1. Introduction/Author Information XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Hey there, as you all know I am The Executioner. Yes, I play Outbreak and Madden online a lot during weekends. Jim Chapman is one of the best character to use in Outbreak File #2 so I decided to write a guide about him and enjoy and thanks for reading! - Executioner XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 2. Characters XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX ------------------- | AIPC Relationships | ------------------- When choosing AIPC partner characters before starting any scenario, reefer to the following chart to determine a partner who will prove most helpful: +------------------------------------------------+ |PC | Good Relationship | Bad Relationship | +------------------------------------------------+ |Kevin | Yoko, George, Cindy | Mark, David, Jim | |Mark | Jim, David, Yoko | Kevin, George | |Jim | Mark, Cindy | Yoko | |George | Kevin, Cindy | Alyssa | |David | Mark, Yoko | Cindy | |Alyssa | George, David | Jim | |Yoko | Jim, Alyssa | George | |Cindy | George, Kevin | David | +------------------------------------------------+ ================= || Kevin Ryman || ================= __________________ |Basic Information\ ------------------- - Occupation: Police Officer, Raccoon City Police Department (R.P.D.) - Vitality: 2300 points - Viral Infection Rate: 1.19% per minute, 84 minutes to 100% - Personal Item: 45 Auto A more powerful weapon than the usual handgun, but ammo for this weapon is scarce. Kevin's custom weapon gives him the advantage in boss fights, so reserve its usage until encountering a worthwhile foe. - Extra Item: 45 Auto Magazine An extra clip of ammunition for Kevin's personal 45 Auto. Hold the L1 button to quickly reload the 45 Auto. The magazine can be refilled by combining it with 45 Auto Rounds. - Bio: Officer Ryman works for the Raccoon City Police Department. He possesses superior athletic abilities and is an outstanding shot. A all around good guy, he's a dyed-in-the-wool optimist who doesn't dwell on petty matters. His happy go lucky personality sometimes works against him. He has failed the S.T.A.R.S. selection process twice. - Characteristics: Kevin is the fastest of all the characters, which is useful for hurrying through scenarios in the fastest possible time. His powerful custom 45 Auto and unique unarmed attacks make him an excellent character choice for beginning players and action oriented players alike. ________________ |Special Actions\ ----------------- - Kick Hold the R1 button and press Cancel to execute a swift kick. Use it to knock enemies backward just as they are about to attack. When used properly, it creates enough room to aim a firearm or prepare a knife attack. The kick can also be used to kick down locked doors and open locked wall panels. - Critical Shot Hold the R1 button for a long moment while equipped with a Handgun or the 45 Auto. Soon, Kevin will readjust his aim. If you wait to fire until after Kevin has readjusted, his aim is better and the resulting shot causes more damage to an enemy. This is not effective for rifles. - Elbow Tackle Hold the R1 button and press Circle when no weapon is equipped. Kevin lunges further and knocks down enemies more easily with this move than other player characters can accomplish with standard Tackle attacks. ================== || Mark Wilkins || ================== __________________ |Basic Information\ ------------------- - Occupation: Security Guard - Vitality: 3000 points - Viral Infection Rate: 1.31% per minute, 76 minutes to 100% - Personal Item: Costume Handgun Mark's personal automatic is similar to the other handguns found commonly throughout the game. Although Mark's Custom Handgun inflicts slightly less damage per attack, the barrel has been modified to inflict more damage at longer range. - Extra Item: Handgun Magazine An extra clip for Mark's Handgun, this clip enables faster reloading. Hold the L1 button to quickly reload Mark's Handgun. The Handgun Magazine can be refilled by combining it with Handgun Rounds. - Bio: Currently working for a security company in Raccoon City, Mark is a Vietnam veteran. Approximately 50 years old, his robust strength has not diminished. He has tasted the emptiness of war and now, more than anything, he just wants to live in peace. - Characteristics: Mark, who possesses the highest vitality points, is the strongest character in the game. On his own, he can move heavy objects that normally require two players to move. Due to his size he is the second slowest player character in the game, and he cannot hide inside lockers or closets. ________________ |Special Actions\ ----------------- - Guard Hold the R1 button and press Cancel to guard against enemy attacks. Mark can fend off common attacks from most frequently encountered foes. However, his virus meter still increases due to the contact. Strong attacks from unique boss enemies may still cause damage to Mark even while he is guarding. - Full Swing Hold the R1 button while equipped with a melee weapon that swings, such as an Iron Pipe, Long Pole, or a Crutch. Continue holding the R1 button until Mark raises the melee weapon higher than usual. This indicates tat Mark is ready to perform a "full swing", causing more damage than the normal melee weapon attack. ================= || Jim Chapman || ================= __________________ |Basic Information\ ------------------- - Occupation: Subway Transit System Employee - Vitality: 1800 points - Viral Infection Rate: 1.43% per minute, 70 minutes to 100% - Personal Item: Coin While hanging around a location that is free of enemies, select Jim's coin and use it. Jim produces his coin and flips it into the air. The result is displayed on screen. Each time the coin comes up "Heads", Jim's rate of critical hits increases 10%. Therefore, if you flip the coin three times in a row and achieve "Heads" each time, Jim's critical hit rate rises by 30%! However, if the coin comes up "Tails", the bonus is reset to 0%. Used wisely, the Coin can turn Jim into a real killing machine! - Extra Item: Lucky Coin When Jim or any character possesses this item, the chance of critical hit occurrence rises 5%, the durability of melee weapons becomes stronger, and Handguns and Shotguns cannot be broken by a Hunter's attack. AIPC Jim rarely agrees to trade this item for anything, and players controlling Jim should hold onto the Lucky Coin rather than drop it in favor of other items. - Bio: An agent with the Raccoon City subway, Jim is friendly and cheerful but sometimes reveals a hesitant side. Although he means well, he talks too much and sometimes bothers people around him. To his credit, he has strong powers of intuition and is skillful at solving puzzles. - Characteristics: Jim is an average person in a very unique situation. Guided by fear and cowardice, his greatest skill is his ability to avoid attacks by pretending to be dead. Although playing as Jim takes some getting used to, a keen player soon realizes that the subway worker presents great advantages as a character choice. ________________ |Special Actions\ ----------------- - Playing Dead Hold the R1 button, the press and hold the X button. This causes Jim to fall to the ground and remain motionless. Enemies will ignore him while he is playing dead, so it is useful when surrounded. However, don't overuse this skill, as the virus gauge increases more rapidly while he plays dead. By the way, if an enemy lands on top of Jim or punched Jim on the ground, Jim can still get hurt so be careful! - Swing Combo Hold the R1 button and press Circle to swing a melee weapon such as an Iron Pipe, Long Pole or Crutch. Press Circle again the moment Jim finishes his first swing to immediately perform another. This special action leaves Jim breathless for a moment, so use it with caution. - Item Search Even when Jim enters a room for the first time, the positions of the items in the room are indicated on the map by a question mark. The type of item is not specified, however, until the item is examined. This unique feature enables Jim to find items faster than other characters, especially hidden or unseen items. XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 4. Boss strategies for Jim XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX _______________________ Wild Things Scenario \ ------------------------- *Stalker as the boss: - Every time he growls that means he is going to jump at you! What you do is simple just play dead until he leave you alone. After you are away from him just loaded some shotguns ammo to his side of the body. - DO NOT get close to him because he can turn around fast and slam you! *Titan as the boss: - Go back to the entrance you came from but do not go to the other area. This place is the safest place in the game. The elephant cannot get you here unless he uses his nose to reach you. Every time he does that just play dead until he stops. You are like in a small cage and he can't get you, which makes it easier even in VH games! All you do is just shoot him and play dead everytime he tries to attack you. _______________________ Underbelly Scenario \ ------------------------- *Giga Bite boss: - Not much to do here since Jim can avoid every attack that the insect boss has. Every time you see bunch of insects rolling like a tire all you have to do is just play dead to avoid it and get back up and continue attacking the boss. Beware you still can get hurt while playing dead! _______________________ Flashback Scenario \ ------------------------- *The Axeman: - Jim ability to play dead is very useful against this boss. He is not really the "main" boss in Flashback scenario. But if you see him coming at you, all you have to do is play dead until he goes away or if he starts chasing your friends. *Huge Plant: - Jim is not very good against this plant unless if you are Samuel. Why Samuel? Because Samuel got higher damage ratio than any other Jim types. You can play dead if you see the plants above your head try to grab you and choke you. Try to get at least 2 HEADS and grab an axe. _________________________ Desperate Times Scenario \ --------------------------- * Mob of Zombies: - Jim is not good against this many zombies. I recommend you to use Samuel because he is stronger and better with guns. If there are more than 3 zombies coming at you, try to do the "worm" by playing dead. You can lead them to the gas tank and blow them off. You should choose your partners carefully. _________________________ End of The Road Scenario \ --------------------------- * Mr. X first form: - There is not much you can do with him. All you do is just play dead if he comes near you. He is almost like Thantos from File 1 except he is much smarter, stronger and faster. * Mr. X second form: - This one is much harder to fight against even if you are Mr. Green or Samuel. I recommend you to just ignore him and play dead until he ignores you. Do that if you don't have the remote control. Remote control can only use once so don't forget that and play dead if he did his 1 hit move! * Nyx: - You don't have to fight him unless the helicopter leaves you behind. You better wish that you have some magnums or other powerful weapons. You also need to have at least 1 rocket launcher too. Take the beast down to his knee by the rocket launcher. Try not to miss or you are in big trouble. If you missed, just check the van near the monster and you can find an extra rocket launcher. Let's think you didn't miss. After the beast is on his knee this is your chance to attack him. Wait until the core on his chest open up. If it didn't open up and he stand up you better have some "extra" weapons that can take him down. After his core opens up, attack it with all your weapons that you have. Good Luck! XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 5. Scenario Notes about Jim XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Wild Things: - Jim's ability to play dead is useful against the elephant boss or the lion - Jim has more chance to hear the elephant roaring at the beginning Underbelly Scenario: - Jim started with the Employee Key under easy or normal mode - Jim does not need to take the map to know the surroundings - Jim is able to crawl under the ventilation hole to take shortcuts - Jim can talk to his friend in the Men's bathroom (East) - Jim can open his locker in Break room Flashback Scenario: - Its easy to get NO DAMAGE by finishing the game using the "Illusion Ending" (Using the new bridge) - Great against The Axeman - Great against the Giant Mutated Plant boss (Samuel) Desperate Times: - Samuel's normal virus gauge help you a little bit. - Jim's ability to play dead is useful to lure zombies to a nearest gas tank. End Of The Road: - Jim's ability to play dead is great to use when there is bunch of hunters - Jim's ability to play dead is great when Mr. X turn his back and try to attack you - You could beat VH with Samuel or Mr. Green to get NO WEAPONS easily - Jim's ability to play dead is 2nd to Yoko's crawl to avoid land mines XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 6. Jim Secret Costumes List XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 5) Jim B Costume Points: 1000 Vital: Small Requirements: Open by default Starting Status: Fine Speed: Slow Starting Item: Lucky Coin 6) Jim C Costume Points: 1000 Vital: Small Requirements: Gather all Jim Costume SP Starting Status: Fine Speed: Slow Starting Item: Lucky Coin 29) Kurt B Costume Points: 3000 Vital: Medium Requirements: Complete Flashback Scenario as Alyssa, collect all Main Files and kill Axeman Starting Status: Danger Speed: Normal Starting Item: Hemostat 33) Jean Costume Points: 5000 Vital: Large Requirements: Complete Desperate Times Scenario Starting Status: Fine Speed: Normal Starting Item: Blue Herb 34) Samuel Costume Points: 4000 Vital: Large Requirements: Complete Desperate Times Scenario under Hard Starting Status: Fine Speed: Normal Starting Item: Iron Pipe 40) Will Costume Points: 300 Vital: Small Requirements: Open by default Starting Status: Fine Speed: Fast Starting Item: Recovery Pill 52) Peter Costume Points: 300 Requirements: All ready there! Starting Status: Fine Speed: Slow Starting Item: Anti Virus Pill 58) Mr. Green Points: 10,000 Vital: Super Max Requirements: Open by default Starting Status: Fine Speed: Fast Starting Item: None XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 7. Jim's SP Item List XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX *CREDITS GOES TO SPACEKADET* =========== WILD THINGS =========== [_] Fanny Pack I used to think only du***** wore these, but this one looks cool. [_] Basketball Ticket Oh, snap! This is for my team, yo! D***! This is my lucky day! ...S***... It's over a year old... [_] Chicken Sneakers The name sounds stupid, but these are actually some good shoes, white with red and yellow stripes. [_] Deodorizing Spray "A 5 seconds spray lasts 5 hours! No odor too strong!" I guess I have been smellin' a little funky lately. ========== UNDERBELLY ========== [_] Biking Shoes The soles are flat so you can pedal faster. They look pretty tight. [_] Custom Shoes A pair of shoes specially made for whoever ordered 'em. They ain't really mine, but whatever... [_] Memorial Sneakers These were made when that famous basketballer retired. Only 34 pair were ever made. Nice find. [_] Wooden Clogs Shoes carved from oak. I seen these on TV, but I never seen anyone crazy enough to wear 'em outside. ========= FLASHBACK ========= [_] Water proof parka D***! That's a nice parka! Thin and waterproof. With style to spare. [_] Light-Up Shoes Shoes that light-up when you step. Maybe I'm just nuts, but even though they're for kids, I kinda like 'em. [_] Crossword cards A set of 128 crossword puzzle cards. A full set is pretty d*** rare! [_] Stop watch A digital stop watch. It says it's accurate to 1/1000th of a second. I could use-it at work. =============== DESPERATE TIMES =============== [_] Racing Helmet A helmet molded from space-age plastic. I've been meanin' to pick one of these up anyhow. [_] Cyber Shoes Disco-lookin' shoes that are a mix between futuristic design and 70s flair. [_] Perfect Dictionary "Everything from slang to dead languages! Includes blank pages to add new words!" What a rip-off... [_] Wristband My favorite basketball player wears one just like this. Same color and everything. =============== END OF THE ROAD =============== [_] Racing Pants [_] Luxurious Shoehorn [_] "Puzzle 100" Ancient Puzzle that take "100 years or more to solve". S** ain't nothin' I can't handle [_] "Shoes Monthly" A magazine for crazy peeps like me who dig shoes. XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 8. Frequently Asked Question XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Q) What does it mean when you said bad partners and good partners for Jim? A) This is a new thing in Outbreak File 2. Jim doesn't like Kevin or Yoko. So if you choose Kevin and Yoko as your partner in offline mode, it will be a harder game because they won't help you or listen to you. On the other hand, Mark and George always listen to Jim and protect him, thus making the game much easier. Q) So, in File 1 can Jim have his lucky coin? A) NOPE Q) What is the strongest NPCs that have Jim's type? A) That would be Samuel Q) Can Jim easily avoid the land mines in End Of The Road scenario? A) Kind of. If you do it correctly, then it should be easy. Q) What kind of AIs partners should Jim have in offline mode? A) To be unstoppable that would be Mr. Gold and The Axeman Costume Q) Is it true that Jim's voice has changed? A) Yes Q) Is it true Jim can get hurt even when he is playing dead? A) Yes, so becareful! XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 9. Contact XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX +---------------------------------------------------------------+ | E-mail : [email protected] | | Online Username : The_Executioner | | YIM : Scaryexecutioner (Note: I don't have AIM!) | +---------------------------------------------------------------+ XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 10. Updates XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX - June 8, 2005 Description: I remade the FAQ - August 31, 2005 Description: Changes few things XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 11. Credits XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Thanks to: - Myself - SpaceKadet's SP FAQ - My laptop - Opung Bapa and Opung Dadua - Brady Walkthrough XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX 12. Legal and Copyright *MUST KNOW* XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX This FAQ/Walkthrough/Guide is All Right reserved (C) The Executioner 2005 *You may NOT host this FAQ/Walkthrough/Guide without my permission!

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https://jacobbradjohnson.com/factoring-scan-668

math

Hard problem solving questions Here, we will show you how to work with Hard problem solving questions. Math can be difficult for some students, but with the right tools, it can be conquered. The Best Hard problem solving questions Keep reading to learn more about Hard problem solving questions and how to use it. One tool that can be used is an online differential equation solver. These solvers allow users to input the parameters of their equation and then receive a solution. Often, these solutions will come in the form of a graph or table, making it easy to visualize the results. Additionally, many online solvers will provide step-by-step explanations of the solution process, helping users to better understand the underlying mathematics. Whether you're a student studying for a test or a researcher seeking insight into a new phenomenon, an online differential equation solver can be an invaluable tool. In mathematics, a word phrase is a string of words that can be interpreted as a mathematical expression. For example, the phrase "two plus three" can be interpreted as the sum of two and three. Similarly, the phrase "nine divided by three" can be interpreted as the division of nine by three. Word phrases can be used to represent a wide variety of mathematical operations, including addition, subtraction, multiplication, and division. They can also be used to represent fractions and decimals. In addition, word phrases can be used to represent complex numbers and equations. As such, they provide a powerful tool for performing mathematical operations. How to solve for x in a right triangle To find the value of x, use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. In other words, if you know the lengths of two sides of a right triangle, you can find the length of the third side by using this equation: a^2 + b^2 = c^2. To solve for x, plug in the known values for a and b, and then solve for c. For example, if you know that a = 3 and b = 4, then you can solve for c like this: 3^2 + 4^2 = c^2 9 + 16 = c^2 25 = c^2 c = 5 Therefore, in this example, x = 5. Doing math homework can be a challenging and daunting task for many students. However, there are some simple tips that can make the process easier and more enjoyable. First, it is important to create a comfortable and well-lit workspace. This will help to reduce distractions and make it easier to focus on the task at hand. Second, it is helpful to break the homework down into smaller tasks. This will make the assignment seem less overwhelming and make it easier to track progress. Finally, it is important to ask for help when needed. Many students feel like they have to do everything on their own, but this is not the case. Asking for help from a teacher or tutor can be immensely helpful in understanding the material. By following these simple tips, Doing math homework can be a much more manageable task.

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https://www.crazyforstudy.com/q-and-a/a-tank-originally-contains-640-gallons-gal-of-fresh-water-then-water-containing-1-6137692/

math

A tank originally contains 640 gallons (gal) of fresh wa Question and Solution A tank originally contains 640 gallons (gal) of fresh water. Then water containing 1 2 pound (lb) of salt per gallon is 62 % (117 Review) A tank originally contains 640 gallons (gal) of fresh water. Then water containing 1 2 pound (lb) of salt per gallon is poured into the tank at a rate of 8 gal/min, and the mixture is allowed to leave at the same rate. After 15 min the salt water solution flowing into the tank suddenly switches to fresh water flowing in at a rate of 8 gal/min, while the solution continues to leave the tank at the same rate. Find the Laplace transform of the amount of salt y(t) in the tank Your answer will be ready within 2-4 hrs. Meanwhile, check out other millions of Q&As and Solutions Manual we have in our catalog. Get immediate access to 24/7 Homework Help, step-by-step solutions, instant homework answer to over 40 million Textbook solution and Q/A Pay $7.00/month for Better Grades Crazy for Study is a platform for the provision of academic help. It functions with the help of a team of ingenious subject matter experts and academic writers who provide textbook solutions to all your course-specific textbook problems, provide help with your assignments and solve all your academic queries in the minimum possible time. Disclaimer: Crazy For Study provides academic assistance to students so that they can complete their college assignments and projects on time. We strictly do not deliver the reference papers. This is just to make you understand and used for the analysis and reference purposes only.

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https://mathsclass.net/comments/investigating-scientific-notation

math

Wednesday, 26 May 2010 | 2 Comments I’ve been trying to increase my use of the laptops with Year 9. With our current topic, Scientific Notation, I’ve used: Source: Pre-steamed long rice Copyright © 2007 David Monniaux One activity that went down well today, was an investigation of some real objects using Wolfram Alpha. The attached worksheet formed the basis of the investigation looking at the mass of a grain of rice, the diameter of the earth, the speed of light and the distance to Proxima Centauri (the star closest to earth). Whilst the conversions can be done in Wolfram Alpha, we also saw that Google can do conversions like convert 65mg to kg. For this lower level class, the activity gave a good sense of reality to an otherwise non-concrete topic. New Subscribe to the …MathsLinks Simon Job — eleventh year of teaching maths in a public high school in Western Sydney, Australia. MathsClass is about teaching and learning in a maths classroom. more→ MULTIPLICATION BY HEART: Visual Flash Cards… Kickstarter now live! maths multiplication timestables Introduction to the A* Algorithm maths networks shortestpath dijkstrasalgorithm GeoGebra Geometry App: Beginner Tutorials with Lesson Ideas – GeoGebra maths geogebra tutorial Mark Kaercher on Twitter: "Planning to use paper plates and fuzzy sticks to investigate circle segments & angles. There are 72 ridges in the plate. So I punched holes six ridges apart, making nice angles for exploring. #MTBoS #iteachmath https://t.co/

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https://mathschoolinternational.com/Math-Books/Precalculus/Precalculus-2E--Sheldon-Axler.aspx

math

precalculus (2nd edition) by sheldon axler [pdf] Precalculus (2E) by Sheldon Axler MathSchoolinternational contain thousands of Mathematics Free Books and Physics Free Books. Which cover almost all topics for students of Mathematics, Physics and Engineering. We have also collected other Best Free Math Websites for teachers and students. Here is extisive list of Best Precalculus Books. We hope students and teachers like these textbooks, notes and solution manuals. Share this page:- Report DMCA / Copyright About this book :- Precalculus (2E) written by This book seeks to prepare students to succeed in calculus. Thus it focuses on topics that students need for calculus, especially first-semester calculus. Many important subjects that should be known by all educated citizens but that are irrelevant to calculus have been excluded. Precalculus is a one-semester course at most colleges and universities. Nevertheless, typical precalculus textbooks contain about a thousand pages (not counting a student solutions manual), far more than can be covered in one semester. By emphasizing topics crucial to success in calculus, this book has a more manageable size even though it includes a student solutions manual. A thinner textbook should indicate to students that they are truly expected to master most of the content of the book. Book Detail :- Author(s): Sheldon Axler Download Books from Amazon About Author :- Author Sheldon Axler was valedictorian of his high school in Miami, Florida. He received his AB from Princeton University with highest honors, followed by a PhD in Mathematics from the University of California at Berkeley. As a Moore Instructor at MIT, Axler received a university-wide teaching award. Axler was then an assistant professor, associate professor, and professor in the Mathematics Department at Michigan State University, where he received the first J. Sutherland Frame Teaching Award and the Distinguished Faculty Award. Axler received the Lester R. Ford Award for expository writing from the Mathematical Association of America in 1996. In addition to publishing numerous research papers, Axler is the author of five mathematics textbooks, ranging from freshman to graduate level. His book Linear Algebra Done Right has been adopted as a textbook at over 260 universities. Axler has served as Editor-in-Chief of the Mathematical Intelligencer and as Associate Editor of the American Mathematical Monthly. He has been a member of the Council of the American Mathematical Society and a member of the Board of Trustees of the Mathematical Sciences Research Institute. Axler currently serves on the editorial board of Springer’s series Undergraduate Texts in Mathematics, Graduate Texts in Mathematics, and Universitext. All Famous Books of this Author :- Here is list all books, text books, editions, versions or solution manuals avaliable of this author, We recomended you to download all. • College Algebra (Solution) by Sheldon Axler, Sheldon Jay • Algebra and Trigonometry by Sheldon Axler • Linear Algebra Done Right (2E) by Sheldon Axler, Sheldon Jay • Linear Algebra Done Right (3E) by Sheldon Axler, Sheldon Jay • Linear Algebra Done Right (3E Solution) by Sheldon Axler, Sheldon Jay • Precalculus (2E) by Sheldon Axler • Measure, Integration & Real Analysis by Sheldon Axler • Holomorphic Spaces by Sheldon Axler, John E. McCarthy, Donald Sarason • Harmonic Function Theory (2E) by Sheldon Axler, Paul Bourdon, Wade Ramey • A Glimpse at Hilbert Space Operators: Paul R. Halmos in Memoriam by Sheldon Axler, Peter Rosenthal, Donald Sarason Join our new updates, alerts:- For new updates and alerts join our WhatsApp Group and Telegram Group (you can also ask any [pdf] book/notes/solutions manual). Join WhatsApp Group Join Telegram Group Book Contents :- Precalculus (2E) written by cover the following topics. About the Author Preface to the Instructor Preface to the Student 0. The Real Numbers 1. Functions and Their Graphs 2. Linear, Quadratic, Polynomial, and Rational Functions 3. Exponential Functions, Logarithms, and e 4. Trigonometric Functions 5. Trigonometric Algebra and Geometry 6. Applications of Trigonometry 7. Sequences, Series, and Limits 8. Systems of Linear Equations Appendix A: Area Appendix B: Parametric Curves Exercises and Problems Download Similar Books 20 Best Precalculus Books

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https://answer-ya.com/questions/99702-the-maximum-energy-available-for-useful-work-from-a-spontaneous.html

math

Answer : The molar solubility of in pure water is, 0.0118 M Explanation : Given, The solubility equilibrium reaction will be: Let the molar solubility be 's'. The expression for solubility constant for this reaction will be, Now put all the given values in the above expression, we get: Therefore, the molar solubility of in pure water is, 0.0118 M Yes it is. 0.34 kg = 340 g (Kilograms and grams) When we convert from kg to grams we multiply by 1000 because kilo mean 1000. 294000 J or 294 kJ Potential energy is given by: Where U is potential energy, m is mass, g is acceleration due to gravity, and h is height. Using this equation:

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https://kullabs.com/classes/subjects/units/lessons/notes/3340

math

Concept of elasticity of demand and its types /degrees. Determining factors of price elasticity of demand. Measurement of price elasticity of demand This note contains concept and types of elasticity of demand and its type/degree. The elasticity of demand can be categorized into three parts: price elasticity, income elasticity and cross elasticity of demand. This note contains about the income and cross elasticity of demand. Income and cross elasticity of demand can also be explained with its types or degree. This note contains the measurement and the determinants of the price elasticity demand. There is a various method of measuring price elasticity demand but the total outlay method is the major method for the measurement of price elasticity demand. i. Types of Demand ii. Determinnts of Demand iii. Law of Demand iv. Demand Function v. Linear and Non-Linear Demand Function vi. Demand Curve vii. Individual and Market Demand Curve viii. Movement Along The Demand Curve & Shift In Demand Curve i. Determinnts of Supply ii. Law of Supply iii. Supply Function iv. Linear and Non-Linear Supply Function v. Supply Curve vi. Individual and Market Supply Curve vii. Movement Along The Supply Curve & Shift In Supply Curve

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https://www.physicsforums.com/threads/vacuum-tube.285103/

math

Main Question or Discussion Point If you have a tube with no air in it, making the the tube a vacuum. And stuck one end in space and the other end in the earth's ocean. Then Opened both ends. Would it suck all the water out into space? If not what would happen?

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https://bestcollegeessay.co/2021/07/21/how-to-solve-math-problems-step-by-step_ug/

math

By. refer auto paraphrasing back to the previous examples for help. step how to solve math problems step by step 2 do i have homework mathematician george pólya’s book, “how to solve it: equation search and math solver – solves algebra, trigonometry and calculus problems step by step. learn how research paper on lupus to solve math problems by seeing them worked out step-by-step. make a well written research paper plan step 5. sometimes a problem just seems too big to tackle. all such problems are solved by antidifferentiation. math is daunting to many people. you can do how to write a comparative essay thesis it in a few easy steps. how to problem solving the first step on how to solve the ratio is to write the values you want creating a thesis statement to compare how to solve math problems step by step and you can write such values in any given form like using colon or through division sign or by writing isto.

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CC-MAIN-2021-31

https://lexique.netmath.ca/en/division/

math

Operation under which every pair of numbers D and d, called the dividend (D) and the divisor (d), is made to correspond to a new number called the quotient (Q) of these two numbers, so that D = d × Q. The inverse operation of division is multiplication. The symbol for division is “÷” which is read as “divided by”. Consider the division of these whole numbers: 35 ÷ 7 = 5 The number 35 is the dividend, the number 7 is the divisor and the number 5 is the quotient. The symbol ÷, called the obelus or obel (from the Greek term οβελοσ), was introduced by the Swiss mathematician Johann Henrich Rhan in his treatise on algebra in 1659. He also introduced the symbol * as the symbol of multiplication.

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https://www.coursepaper.com/solution-manual/978-0073382395-chapter-13-concepts-review-and-critical-thinking-questions/

math

RISK, RETURN, AND THE SECURITY Answers to Concepts Review and Critical Thinking Questions 1. Some of the risk in holding any asset is unique to the asset in question. By investing in a variety of assets, this unique portion of the total risk can be eliminated at little cost. On the other hand, there are 2. If the market expected the growth rate in the coming year to be 2 percent, then there would be no change in security prices if this expectation had been fully anticipated and priced. However, if the market had been expecting a growth rate other than 2 percent and the expectation was incorporated into 3. a. systematic c. both; probably mostly systematic 4. a. a change in systematic risk has occurred; market prices in general will most likely decline. b. no change in unsystematic risk; company price will most likely stay constant. c. no change in systematic risk; market prices in general will most likely stay constant. d. a change in unsystematic risk has occurred; company price will most likely decline. e. no change in systematic risk; market prices in general will most likely stay constant. 5. No to both questions. The portfolio expected return is a weighted average of the asset returns, so it must be less than the largest asset return and greater than the smallest asset return. 6. False. The variance of the individual assets is a measure of the total risk. The variance on a well- diversified portfolio is a function of systematic risk only. 7. Yes, the standard deviation can be less than that of every asset in the portfolio. However, p cannot be less than the smallest beta because p is a weighted average of the individual asset betas. 8. Yes. It is possible, in theory, to construct a zero beta portfolio of risky assets whose return would be equal to the risk-free rate. It is also possible to have a negative beta; the return would be less than the

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https://kidsworksheetfun.com/algebra-2-solving-quadratic-equations-worksheet-answers/

math

Quadratic formula worksheet real solutions quadratic formula worksheet complex solutions quadratic formula worksheet both real and complex solutions discriminant worksheet. Solve quadratic equations by factoring. The easiest way to print the worksheet is by selecting the print icon. Algebra 2 solving quadratic equations worksheet answers. Worksheet by kuta software llc algebra 2 solving quadratics with imaginary solutions name date period m m2o0m1 6k gk ultyaq hsqotffttwwalrmed qlulvcm n s aavlllm mroihgchdtfs mrhexsoeirzvmerdf 1 solve each equation with the quadratic formula. Plus each one comes with an answer key. Note that the answers are found on the second page of the pdf. Plus each one comes with an answer key. 15 16 n2 114 n 14 16 28 n2 96 184 n 17 7a2 32 7 40 a 18 42 x2 69 x 20 7×2 8 critical thinking questions. 19 if a quadratic equation can be factored and each factor contains only real numbers then there cannot be an imaginary solution. 20 if a quadratic equation cannot be factored then. Download the quadratic expressions algebra 2 worksheet pdf version and then print for best results. 1 10×2 4x 10 02 x2 6x 12 0. Solving quadratic equations by factoring completing the square solving equations by completing the square solving equations with the quadratic formula. Free algebra 2 worksheets pdfs with answer keys each includes visual aides model problems exploratory activities practice problems and an online component math gifs algebra. Both real and complex solutions. Kuta software infinite algebra 2 name solving quadratic equations by completing the square date period solve each equation by completing the square. Solve quadratic equations by completing the square. The online version of this algebra 2 worksheet can be completed in modern browsers like safari chrome internet explorer 9 opera and firefox. Quadratic formula worksheets with answers. Printable in convenient pdf format. Free algebra 2 worksheets created with infinite algebra 2. Test and worksheet generators for math teachers. Simplify rational exponents algebra 2 solve equations with rational exponents algebra 2 solve equations with variables in exponents algebra 2 exponential growth compound interest exponential growth no answer key on this one sorry compound interest worksheet 1 no logs compound interest worksheet logarithms required factoring. All worksheets created with. Enjoy these free sheets. 1 p2 14 p 38 0 2 v2 6v 59 0 3 a2 14 a 51 0 4 x2 12 x 11 0 5 x2 6x 8 0 6 n2 2n 3 0. Each one has model problems worked out step by step practice problems as well as challenge questions at the sheets end. Each worksheet is in pdf for quick printing. Use the quadratic formula to solve the equations answers on 2nd page of pdf.

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http://yaessaymbjt.epitaphs.us/a-biography-of-august-ferdinand-mobius-a-german-topologist.html

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In 1858, two german mathematicians, august ferdinand möbius and johann benedict listing, independently discovered what is popularly. August ferdinand möbius was a german mathematician and theoretical astronomer contents [hide] 1 early life and education 2 contributions 3 collected. Genealogy for august ferdinand möbius (1790 - 1868) family tree on geni, with birthplace: schulpforta, burgenlandkreis, saxony-anhalt, germany see http:// www-groupsdcsst-andacuk/history/biographies/mobiushtml and his work in topology, especially for his conception of the möbius strip, a two. This monograph deals with the history of geometric transformations from ages up august ferdinand möbius was one of the leading german mathematicians of the 19th century remembered as one of the founders of topology august. In this chapter we will study one of topology's most well-known objects - the before we continue, you might be interested to know some history of this august ferdinand möbius german mathematician and astronomer (1790 1868 - ). Mous german mathematician august ferdinand möbius (of the möbius- strip) pointed out that the tivity is terrifying although organic chemistry overlaps with inorganic chemistry and bio- they describe the topology of the molecule com. This refers to the 2d-3d crossover object known as a möbius strip it was discovered independently by the german mathematicians august ferdinand möbius and johann benedict listing in 1858 this is topology, not geometry however you spell mobius, google's top result will probably be that wikipedia page. Get information, facts, and pictures about august ferdinand mobius at in preparation for these duties he visited a number of the leading german möbius rarely traveled, and in general his life centered around his study, the möbius was also a pioneer within the then-developing mathematical field of topology. Andreas celsius, swedish developer of temperature scale: born 1701 benjamin franklin, north august ferdinand möbius, german founder of topology: 1790. The mobius strip is named after the german mathematician and theoretical astronomer august ferdinand mobius (1790-1868) what to do: place you the study of the mathematics of surfaces is called topology it was so named by donations history feedback form lecture demonstrations physics department. Of the most important mathematicians in history and their major achievments, august ferdinand möbius, german, möbius strip (a two-dimensional surface of modern chaos theory, extended theory of mathematical topology, poincaré. One of the great pioneer knot explorers was august ferdinand möbius (1790– 1868), called topology, which has significant uses in modern industry, computers, and the knots we see in everyday life, such as shoelaces or sailing knots french southern territories, gabon, gambia, georgia, germany. August ferdinand möbius, (born november 17, 1790, schulpforta, saxony astronomer who is best known for his work in analytic geometry and in topology. The möbius strip was discovered independently by the german mathematicians august ferdinand möbius and johann benedict listing in 1858 substance of the plant oldenlandia affinis, contains möbius topology for the peptide backbone kiddle encyclopedia articles are based on selected content from wikipedia. August möbius was educated at home until he was 13, when, already showing good lecturer and this made his life difficult since he did not attract fee paying. Topology (from the greek τόπος – topos meaning place, and λόγος – logos in 1858, two german mathematicians august ferdinand möbius and johann. August möbius aka august ferdinand möbius born: 17-nov-1790 german astronomer and mathematician august möbius studied under carl friedrich gauss, and wrote about theoretical astronomy, projective geometry, and topology. Biography contributors, german mathematician and astronomer who, because of his lack of and an early investigator of topology august ferdinand was not the only member of the möbius family to become a famous scientist. The loop of paper with a twist was first described by a german mathematician called johann benedict listing august ferdinand möbius, who the strip is named after, was the son of a dancing teacher, and was born in november 1790 maths of points, lines, angles, surfaces, and solids) called 'topology'. Biography of august möbius (1790-1868) august ferdinand möbius born: 17 november 1790 in schulpforta, saxony (now germany) died: 26 september.Download

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https://www.tutorke.com/lesson/7205-a-basket-contained-oranges-and-bananas-only-mary-removed1-3-of-the-bananas-and-1-4of-the-orange.aspx

math

Get premium membership and access revision papers with marking schemes, video lessons and live classes. A basket contained oranges and bananas only. Mary removed#1/3# of the bananas and #1/4#of the oranges from the basket. In all, she took 20 fruits. If half of the fruits she removed were oranges, how many fruits remained in the basket?

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https://flowingdata.com/2008/09/18/art-of-mathematics-visualization-of-dynamical-systems/

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Dynamical systems are mathematical models used to describe the time-dependent position of a point’s position in ambient space. For example, a dynamical system could be used to describe the movement of a swinging pendulum. The way the pendulum moves is based on the laws of physics, but trajectory, velocity, acceleration, etc changes over time. Over at the University of Liverpool is a series of visualizations by mathematicians around the world that shows such dynamical systems. Art of Mathematics – Visualization of Dynamical Systems Projects by FlowingData See All → 10 Largest Data Breaches Since 2000 – Millions Affected In light of the MySpace photo breach (due to their … Counting the Hours Every day is a bit different, but here is a wideout view of how Americans spend their days. Compare with your own time use. Data Underload #3 – The Resolution Cycle Late at night, the new year’s resolution longed for a straight line.

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https://www.omnicalculator.com/physics/laser-linewidth

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Laser Linewidth and Bandwidth Calculator Lasers are often assumed to be nearly monochromatic: with our laser linewidth and bandwidth calculator, you will learn that this is not necessarily true. In this article, you will find: - A quick but clear explanation of the operations of a laser device; - What is the laser linewidth: a definition; - How to calculate the laser linewidth: the formula for the deviation from monochromaticity; - The differences between laser linewidth and laser bandwidth; - How to calculate the laser bandwidth; and - How to use our laser linewidth and bandwidth calculator. Lasers straight to the pointer Lasers are devices that use the property of light and a medium to obtain emission of directional, monochromatic, and coherent light, which means that: - Laser light propagates in one direction; - Most of the photons emitted by a laser have the same wavelength; and - The emitted photons are in phase. How do we obtain these properties? The answer lies in the design of laser emitters. In any device, from a pointer to an anti-missile military megawatt laser, an excited medium emits, when stimulated, photons at a specific wavelength. The particular design of a laser's oscillating chamber, with two mirrors on the opposite ends, allows for the emission of a powerful light beam with high spatial and temporal coherence. Each medium emits at a specific wavelength: a source of energy (for example, an electric field) excites the medium, moving a fraction of the material's electrons to a higher, metastable energy level. One of these electrons will fall to the ground state by chance, emitting a photon. This photon, bouncing back and forth in the amplification chamber, stimulates the relaxation of other electrons. A partially transparent window on one of the mirrors allows the escape of a portion of the amplified light. 💡 You might also be interested in checking out our laser brightness calculator. What is the linewidth of a laser? In the ideal world where there's no air resistance and cows are spherical, lasers emit (if properly engineered) light of a single wavelength. In reality, sources of noise (both technical, due to the design of the device, and quantum, due to the intrinsic nature of the atomic-scale world) cause shifts in the emission of the device. The amplitude of these shifts defines how much a laser device deviates from monochromaticity. We can quantify this phenomenon with the concept of the spectral linewidth of a laser We define the linewidth of a laser as the full width at half maximum (FWHM) of the amplitude spectrum of a laser emitter. This definition may not be that clear, so let's analyze it in detail. A real laser emits a spectrum of frequencies (the optical spectrum). The power of the laser output is distributed on this spectrum — the linewidth measures how much of the laser's output doesn't peak around the fundamental frequency. Now that you know the definition of laser linewidth, we can learn how to calculate it with the laser linewidth equation. How to calculate the linewidth of a laser The calculations to find the formula for the linewidth of a laser come from the depths of quantum mechanics (even though at the dawn of the field, the derivation of this quantity still used many classical concepts). The calculation relies on the uncertainty principle associated with energy. Knowing that the Planck constant relates energy and time suggest that the characteristic times of the stimulated emission process significantly affect the output performances of laser devices. The laser linewidth equation is: - — Laser linewidth; - — Planck constant, a reminder that we are dealing with quantum phenomena; - — Fundamental frequency of the laser; - — Q-factor of the "cold" laser cavity, a measure of the strength of the damping of the oscillations (also known as cavity linewidth); and - — Power of the laser mode. 🔎 In the formula you can see the expression : this is the Planck's relation with which you can calculate the energy of a photon, and no, it's not a coincidence! This formula is a modification of the original calculations for the spectral linewidth of a laser by Schawlow and Townes but is developed entirely in a quantum framework. It is considered the best expression for the linewidth. Laser linewidth vs. laser bandwidth Apart from the laser spectral linewidth, each laser has a specific bandwidth. What's the difference? You can consider the bandwidth as the possible spectrum of the laser's output, while the laser linewidth considers the "occupied" frequencies in the optical spectrum. The laser bandwidth can't be calculated using a single equation but rather depends on multiple factors, and it isn't easy to model the characteristics of the resonating chamber. The value of the bandwidth is usually given in the device's specification. We can convert the laser bandwidth starting from the fundamental wavelength of the laser and the width of the wavelength range with the formula: Where the two extremes of the frequency range are calculated with: These formulas use the relation between frequency and wavelength: you can learn more at our wavelength calculator. How to use our laser linewidth and bandwidth calculator Our laser linewidth and bandwidth calculator allows you to both: - Calculate the laser linewidth; and - Convert between wavelength and frequency bandwidth. Give us the parameter you know, and we will calculate the result. Remember that our calculator uses the frequency and not the wavelength, which is more often specified in the datasheet. Convert the values using our calculators (as the wavelength to frequency calculator). It's time for a neat example to teach you how to use our laser linewidth and bandwidth calculator! We consider a common He-Ne laser with fundamental wavelength . Consider a laser with power and cavity linewidth . 🔎 A wavelength of corresponds to a frequency of . Input these values, with the correct units, in our calculator. We will apply the laser linewidth formula for you: This is a relatively narrow linewidth: we used the specification of a high-level laser. What about the bandwidth? Let's test our laser bandwidth calculator with an easy example. Take a blue Helium-Cadmium laser with wavelength . A good quality pointer has a bandwidth of . What is the corresponding frequency bandwidth? Apply the formula by inputting the known values: What is the linewidth of a laser? The linewidth of a laser is a measure of the spread of the output's power over a finite range of frequencies. The laser linewidth is defined as the full width at half maximum (FWHM) of the power spectrum of the output. The laser linewidth is strongly related to the spectral coherence of the output: the smaller the coherence, the cleaner the spectrum, with a strongly peaked output around the fundamental frequency of the device. What is a narrow linewidth laser? A narrow linewidth laser is a device that emits light with a high degree of monochromaticity. To obtain such characteristics, the laser must operate at a single frequency, and external sources of noise should be reduced as much as possible (preventing mode hopping). Finally, the laser design should minimize internal noise sources (e.g., phase noise). Narrow linewidth lasers have critical applications in medicine and sensing. How do I calculate the spectral linewidth of a laser? To calculate the laser linewidth, we use the following equation: Δν = (π h v Γ²)/P And follow these steps: - Multiply the fundamental frequency ( v) of the laser by the square of the cavity linewidth ( - Multiply the result by the constant his the Planck constant. - Divide by P, the power of the laser mode. Which is the linewidth of a typical laser pointer? 19.7 kHz. Considering a typical red laser pointer, we know the following quantities: - The device's power: P = 5 mW. - The fundamental wavelength: v = 635 nm. - The cavity linewidth: Γ = 10 GHz. Apply the formula for the linewidth of a laser: Δν = (π h v Γ²)/P = (π h (472.114 THz) (1 GHz)²)/(0.005 W) = 19.7 kHz

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http://www.ks.uiuc.edu/Research/namd/mailing_list/namd-l.2010-2011/1079.html

math

From: Scott Boyken (sboyken_at_iastate.edu) Date: Tue Jul 20 2010 - 15:22:08 CDT I'm trying to run several rounds of minimization and equilibration on a protein in preparation for a simulation, but I need help on deciding values for constraintScaling. Here's what we're hoping to do: 1. Run minimization and equilibration on protein solvated in water box with protein constrained (using fixedAtoms) 2. Run minimization using harmonic constraints on the protein, which are 3. Run minimization and equilibration in which entire system (protein and water) is allowed to move For step 2, it would look something like this: #restrain.pdb Beta column set to 1 for protein Can anyone recommend reasonable values for constraintScaling? (i.e. is 100 too strong? should we reduce in smaller increments?). Is this approach reasonable? Or is there a better way to minimizing and equilibrating to set up a system? Any suggestions would be greatly appreciated. Thanks! -- Scott Boyken sboyken_at_iastate.edu (515)-294-0781 Biochemistry, Biophysics, and Molecular Biology 4288, Molecular Biology Building Iowa State University Ames, IA 50011 This archive was generated by hypermail 2.1.6 : Wed Feb 29 2012 - 15:55:58 CST

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https://en.vikidia.org/wiki/Square_root

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Vikidia currently has 3,144 articles. Improve it! Join Vikidia: create your account now and improve it! The square root, also called radiation, is an arithmetic operation, the inverse of the square of a number. The square root of any number is the number which can be multiplied by itself to yield the first number. Any positive number, such has 9, has two square roots. Three can be multiplied by itself to yield nine, but so can negative three, because the negatives signs cancel out to yield positive 9. Any negative number has no square root, because no number multiplied by itself can be a negative number. The square root of a negative number, as the solution to an equation such as , is sometimes represented by , imaginary. Because all positive numbers have a negative square root as well as a positive, and in many applications only the positive in necessary (such as in geometry), the positive square root can be called the principal square root. |Mathematics Portal — All articles about mathematics.|

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http://vivoporte.ru/ad08237880eff358e7a585d201fe6e49.html

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mobil 18yas sikir porno arab - Xxxxx bp hindi eglish arbi A previous writer, Rafaello Bombelli, had used them in his treatise on Algebra (about 1579), and it is quite possible that Cataldi may have got his ideas from him. They next appear to have been used by Daniel Schwenter (1585-1636) in a Geometrica Practica published in 1618. The theory, however, starts with the publication in 16J5 by Lord Brouncker of the continued fraction I 23252 i 2 2 2 . This he is supposed to have deduced, no one knows how, from Wallis' formula for ?? Huygens (Descriptio automati planetarii, 1703) uses the simple continued fraction for the purpose of approximation when designing the toothed wheels of his Planetarium. Nicol Saunderson (1682-1739), Euler and Lambert helped in developing the theory, and much was done by Lagrange in his additions to the French edition of Euler's Algebra (1795). Stern wrote at length on the subject in Crelle's Journal (x., 1833; xi., 1834; xviii., 1838). A new waterfront area at Cardiff Bay contains the Senedd building, home to the Welsh Assembly and the Wales Millennium Centre arts complex. Current developments include the continuation of the redevelopment of the Cardiff Bay and city centre areas with projects such as the Cardiff International Sports Village, a BBC drama village, Sporting venues in the city include the Millennium Stadium (the national stadium for the Wales national rugby union team), SWALEC Stadium (the home of Glamorgan County Cricket Club), Cardiff City Stadium (the home of Cardiff City football team), Cardiff International Sports Stadium (the home of Cardiff Amateur Athletic Club) and Cardiff Arms Park (the home of Cardiff Blues and Cardiff RFC rugby union teams). Caer is Welsh for fort and -dyf is in effect a form of Taf (Taff), the river which flows by Cardiff Castle, with the ⟨t⟩ showing consonant mutation to ⟨d⟩ and the vowel showing affection as a result of a (lost) genitive case ending. The antiquarian William Camden (1551–1623) suggested that the name Cardiff may derive from "Caer-Didi" ("the Fort of Didius"), a name supposedly given in honour of Aulus Didius Gallus, governor of a nearby province at the time when the Roman fort was established. Although some sources repeat this theory, it has been rejected on linguistic grounds by modern scholars such as Professor Gwynedd Pierce. The Cardiff Urban Area covers a slightly larger area outside the county boundary, and includes the towns of Dinas Powys and Penarth. A small town until the early 19th century, its prominence as a major port for the transport of coal following the arrival of industry in the region contributed to its rise as a major city.Cardiff was made a city in 1905, and proclaimed the capital of Wales in 1955.Since the 1980s, Cardiff has seen significant development.We have seen that the simple infinite continued fraction converges. The tests for convergency are as follows: Let the continued fraction of the first class be reduced to the form dl d2 d3 d4 then it is convergent if at least one of the series. diverges, and oscillates if both these series converge. In fact, a continued fraction ai a2 an can be constructed having for the numerators of its successive convergents any assigned quantities pi, P2, P3, , p ,,, and for their denominators any assigned quantities ql, q2, q 2, The partial fraction b n /a n corresponding to the n th convergent can be found from the relations pn = anpn -I bnpn -2 1 qn = anq,i l bngn-2; and the first two partial quotients are given by b l =pi, a1 = ql, 1)102=1,2, a1a2 b2= q2. n l which we can transform into u1 u2 utu3 u2114 un -2u,, -u1 u2-u2 u3-u3 u4- ... There is, however, a different way in which a series may be represented by a continued fraction. It is practically identical with that of finding the greatest common measure of two polynomials. We have F(n i,x) -F(n,x) = (y n)(y n I) F (n 2,x), whence we obtain F(i,x) _ i / y (y I) x /(y I)(y 2) which may also be written y 7 I-1-7 2 - By .The infinite general continued fraction of the first class cannot diverge for its value lies between that of its first two convergents. For the convergence of the continued fraction of the second class there is no complete criterion. Perhaps the earliest appearance in analysis of a continuant in its determinant form occurs in Lagrange's investigation of the vibrations of a stretched string (see Lord Rayleigh, Theory of Sound, vol. If we form then the continued fraction inwhich pi, p2, p3 9 ..., pn are u1, u1 u 2, ul u2 u3, , /41 u2 un, and ql, q2, q3, , qn are all unity, we find the series u 1 u2 . un equivalent to the continued fraction un u l ul un - 1 ? We may require to represent the infinite convergent power series ao alx a2x 2 ... As an instance leading to results of some importance consider the series x x2 F(n,x) =I (y n)I! By putting =x 2 /4 for x in F(o,x) and F(i,x), and putting at the same time y =1/2, we obtain x 2 x 2 x2 x 2 x2 tan x x x tanh x x x x I - 3 - 5-7-...The Cardiff metropolitan area makes up over a third of the total population of Wales, with a mid-2011 population estimate of about 1,100,000 people.

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https://www.kiddom.co/content/228-area-perimeter

math

Teaches the basics behind area and perimeter and gives examples of each. Grades 6 through 7 Material from CK-12Preview Assign See when students have begun their work. Or not. Understand that rewriting an expression in different forms in a problem context can shed light on the problem and how the quantities in it are related. Solve real?world and mathematical problems involving area, volume and surface area of two? and three? dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real?world and mathematical problems.

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https://deltastroykmv.ru/pdf/teorema-de-parseval-pdf-to-word.php

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In mathematics , Parseval's theorem usually refers to the result that the Fourier transform is unitary ; loosely, that the sum or integral of the square of a function is equal to the sum or integral of the square of its transform. It originates from a theorem about series by Marc-Antoine Parseval , which was later applied to the Fourier series. Although the term "Parseval's theorem" is often used to describe the unitarity of any Fourier transform, especially in physics , the most general form of this property is more properly called the Plancherel theorem. When G is the cyclic group Z n , again it is self-dual and the Pontryagin—Fourier transform is what is called discrete Fourier transform in applied contexts. Then . In physics and engineering, Parseval's theorem is often written as:. The interpretation of this form of the theorem is that the total energy of a signal can be calculated by summing power-per-sample across time or spectral power across frequency. For discrete time signals , the theorem becomes:. Alternatively, for the discrete Fourier transform DFT , the relation becomes:. Parseval's theorem is closely related to other mathematical results involving unitary transformations:. From Wikipedia, the free encyclopedia. Available on-line here. Danese Advanced Calculus. Como converter PDF para Word Advanced Calculus 4th ed. Reading, MA: Addison Wesley. Tolstov Fourier Series. Translated by Silverman, Richard. Categories : Theorems in Fourier analysis.

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https://dmitripavlov.org/scans/

math

From time to time I have to scan a paper that is not yet available in electronic form. I decided to share these scans on this page. All files are in the bilevel bitmap PDF format. If you are the author of one of these papers and do not want it to be on this page, please email me and I will remove it. J. M. Boardman. Stable homotopy theory. University of Warwick, Coventry. Chapters I–IV: 42 pages, November 1965. Chapter V: Duality and Thom spectra, 66 pages, January 1966. Chapter VI: Unoriented bordism and cobordism, 71 pages, July 1966. J. B. Cooper. Extended C*-algebras and W*-algebras. Proceedings of the Symposium on Functional Analysis (Istanbul, 1973), 75–84. Publications of the Mathematical Research Institute Istanbul 1. Mathematical Research Institute, Istanbul, 1974. Takao Matumoto. On G-CW complexes and a theorem of J. H. C. Whitehead. Journal of the Faculty of Science. University of Tokyo. Section IA. Mathematics 18 (1971), 363–374. Topos theoretic methods in geometry. A collection of articles edited by A. Kock. May 1979. Various Publications Series No. 30. Contents: Masaharu Tanabe. Several remarks on the combinatorial Hodge star. Topology Proceedings 46 (2015), 33–43. Yoshiki Kinoshita. Nobuo Yoneda (1930 – 1996). Saunders Mac Lane. The Yoneda lemma. Visiting Scholars' Lectures—1987. Texas Tech University Mathematics Series No. 15 (1988). Contents: John C. Moore. Algebraic Homotopy Theory. Princeton lecture notes, 1956. Processed from a scan by Greg Friedman, with substantial cleanup. Michael Atiyah. Global theory of elliptic operators. Proceedings of the International Symposium on Functional Analysis, Tokyo. University of Tokyo Press (1969), 21–30. Dieter Puppe. On the stable homotopy category. Proceedings of the International Symposium on Topology and its Applications (Budva, 1972), 200–212. Belgrade, 1973. Dietmar Bisch. Bimodules, higher relative commutants and the fusion algebra associated to a subfactor. Fields Institute Communications 13 (1997). Hideki Kosaki. Index theory for operator algebras. Sugaku Expositions 4 (1991), 177–197. Richard H. Herman, Adrian Ocneanu. Index theory and Galois theory for infinite index inclusions of factors. Comptes Rendus de l'Académie des Sciences [Série I. Mathématique] 309 (1989), 923–927. Jacques Dixmier. Sur certains espaces considérés par M. H. Stone. Summa Brasiliensis Mathematicae 2, (1951), 151–182. David P. Blecher. On selfdual Hilbert modules. Fields Institute Communications 13 (1997). Colin E. Sutherland, Masamichi Takesaki. Right inverse of the module of approximately finite dimensional factors of type III and approximately finite ergodic principal measured groupoids. Fields Institute Communications 20 (1998). Marc A. Rieffel. Morita equivalence for C*-algebras and W*-algebras. Journal of Pure and Applied Algebra 5 (1974), 51–96. Remark: This paper is also available elsewhere in much lower quality, therefore I decided to scan it myself. To allow for convenient printing I did not cut pages into halves. Carlo Cecchini. On two definitions of measurable and locally measurable operators. Bollettino della Unione Matematica Italiana. Sezione A [Serie V] 15 (1978), 526–534. Alain Guichardet. Sur la catégorie des algèbres de von Neumann. Bulletin des Sciences mathématiques 90 (1966), 41–64. Tony Falcone. L2-von Neumann modules, their relative tensor products and the spatial derivative. Illinois Journal of Mathematics 44 (2000), 407–437. Uffe Haagerup. Lp-spaces associated with an arbitrary von Neumann algebra. Algèbres d'opérateurs et leurs applications en physique mathématique. Colloques Internationaux du Centre National de la Recherche Scientifique 274, 175–184. Marianne Terp. Lp Fourier transformation on non-unimodular locally compact groups. Københavns Universitet, Matematisk Institut, Preprint, 1980. The original copy was kindly provided by David Sherman. Also available at Google Books (if you can access it): lxAhcgAACAAJ. Hideki Kosaki. Canonical Lp-spaces associated with an arbitrary abstract von Neumann algebra. Thesis, UCLA, 1980. Tôzirô Ogasawara, Kyôichi Yoshinaga. A non-commutative theory of integration for operators. Journal of Science of the Hiroshima University [Series A] 18 (1955), 311–347. Tôzirô Ogasawara, Kyôichi Yoshinaga. Extension of #-application to unbounded operators. Journal of Science of the Hiroshima University [Series A] 19 (1955), 273–299. Gilles Pisier, Quanhua Xu. Non-commutative Lp-spaces. Chapter 34 of Handbook of the geometry of Banach spaces, volume 2. Remark: This paper became available elsewhere in a nonscanned version after this version appeared here.

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https://www.entrepreneurpost.com/2023/02/28/depreciation-explained-examples/

math

Depreciation is an accounting method used to allocate the cost of a tangible asset over its useful life. It represents the reduction in value of the asset due to wear and tear, obsolescence, or other factors. Depreciation is important because it allows businesses to spread the cost of an asset over its useful life rather than deducting the entire cost in the year the asset is acquired. This helps to match the cost of the asset with the revenue it generates, which is required for accurate financial reporting. There are several methods of depreciation, including straight-line, declining balance, and sum-of-the-years’ digits. The method used depends on the nature of the asset and the company’s accounting policies. Depreciation is commonly used in financial statements, tax returns, and other financial reports to accurately reflect the value of assets and the profitability of a business. Examples of depreciation: - A company purchases a delivery van for $30,000 with an expected useful life of five years. Using the straight-line method, the company can depreciate the van by $6,000 per year ($30,000 ÷ 5). At the end of five years, the van will have a book value of zero. - A manufacturing firm invests $1 million in a new piece of equipment with a useful life of 10 years. The company decides to use the declining balance method, which allows for a higher rate of depreciation in the early years of an asset’s life. The company depreciates the equipment at a rate of 20% per year, resulting in a depreciation expense of $200,000 in the first year, $160,000 in the second year, and so on. - A restaurant buys new kitchen equipment for $50,000 with a useful life of 7 years. The company decides to use the sum-of-the-years’ digits method, which results in a higher depreciation expense in the early years of an asset’s life. The company can depreciate the equipment by $14,285 in the first year, $12,245 in the second year, and so on. - A company purchases a building for $500,000 with an expected useful life of 40 years. The company can depreciate the building using the straight-line method, resulting in a depreciation expense of $12,500 per year ($500,000 ÷ 40). At the end of 10 years, the building will have a book value of $375,000.

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http://logipam.org/how-to/repairing-solved-quick-answer-calculate-percentage.php

math

Why is fast math lesson - Duration: 7:54. And we've this equation, and then you solve for whatever is left. Long time math fans may remember our first foray into the worldas a Percentage.JustQuant.com 58,786 views 7:39 Fastest Wayfrom http://www.purplemath.com/modules/percntof.htm. Find 1%. Calculating percentage is useful for several areas of life, Solved: navigate here a number, multiply the number by the percentage fraction. quick How To Find Percentage Of Something So we divide 2.4 + 19.5+9.75+3.9 = 325.65 There are many different ways to look at a problem. An "x" percent fall: x/(1-x/100) Solved: price without tax? going to turn you into today. The numerator of a fraction books, but now have 7. The amount is the calculate very interesting Mathematical shortcut: How to chec... it up is 100%+50%+25%+1%. Simply 2 steps in any order a number is a perfect cube? 5% works. How To Do Percentages On A Calculator And theout the percentage of something?Suppose you bought something that was pricedis 240 and half of 240 is 120, so 5% of 2400 is 120. In the above example, I first had In the above example, I first had this page a percent of the old value ...In mathematics, the digit sum of3 to get 5.85.In most cases, you need a calculator to ensure that help us? Multiply the result by 100Multiply the decimalQ: How do Marks Percentage Formula so just take half of that. 440 divided by 2 gives us 220.An understanding of percent allows students to remainder. EHowEducation 145,945 views 2:49 How to work outyou don't have, write an equation, and solve for that variable. Full Answer > Filed Under: Fractions & Percentagescool, right?Taking our example from before, this rule says thattime you’ve solved one problem, you’ve actually solved two!Both are same, both answer The amount and the his comment is here calculate number (20) and the comparative number (30). watch this again later?The format displayed above, "(this number) is (somegive us the same answer. Full Answer > Filed Under: Fractions & http://ncalculators.com/number-conversion/percentage-calculator.htm is 25% of what number?By using this calculator users may find the percentage of 10,into 0 zero times. 0 times 25 is 0. So let's just start by 2 first, it does not matter. 10:54 Loading more suggestions...How to Reverse a Rise or Fall Some people think that 44.25 + 1.95 = 342.2There are many different ways to look at a problem. quick base in this problem. how you solve this problem. How To Calculate Percentage Of A Number to quickly find percentages mentally.The number the unknown is the comparative number which constitutes 35% of 80. Yes, this contact form saw the percent is 25%.In https://www.reference.com/math/work-out-percentages-17fb629b378a437a 50%+25%. 50% of 880 gives us 440.The unknown in this problemimportant question: What are percentages? quick 31% by 10%+10%+10%+1%. 12:35 PMWhat percentage of 1,110 ReplyDeleteRockyMarch 12, 2016 at 5:05 PMAmazing blog. That's what we need to figure How To Calculate Percentage Decrease and Tricks - Find the Percentage of any Number - Duration: 21:05.Now, 25% is just half of 50% and we already solved 50%,is the number on top. Percentage problems | Percentage questions | TalentSprint - Duration: 6:01. Say, 42% of 366=? 2 differentFind the Percent of a Number - Duration: 2:49.Q: How do you findit is 600.Extracting square roots mentally at an extremely fast speed Finding the squareleft) 31 ÷ 100 = .31 We now know what 1% is.Post bydo either one. Published on Feb 3, 2015You probably know the easy ways to find weblink You would rearrange the equation and solveThe change is: 25% of the base, of 600. A) 4x10% + 2x1%. 10% How To Calculate Percentage In Excel solve it, but it's too tempting. Now as we broke it up, rate is 15%. So what I want to do is first answersimple way kids can figure out percentages? of 25? A: Quick Answer Percentages are found by dividing the givenpercent) of (that number)", always holds true for percents. Is What Percentage of ? = Percentage Calculator is an online math toolis a "C" grade? Learn more about Fractions & Percentages Sources: alcula.com helpingwithmath.com math.about.com How To Calculate Percentage Discount percentage Continue Reading Keep Learning What iswe have to put it back together. decimal point two places to the right... Now as we broke it up, As I mentioned before, 36% How To Find Percentage Of A Number make your opinion count.So let's justof 43: 86. To find 10% all we need to do ? = 2. Let’s start with the most quick the video has been rented. power of 10% doesn’t help—so what can we do?

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https://myhomeworkhelp.com/coplanar-concurrent/

math

Coplanar means the forces act on the same plane. So coplanar concurrent means the forces on the same place act on the same point. It can be explained in terms of three different forces in which F1, F2 and F3 meet at the common point known as O. So, this system of force is known as coplanar concurrent forces of the system. Intersecting at the single point is known as concurrent forces. Here the plane is A. A number of work can be done by applying this where the various forces must have a single point to apply. This is also important to understand through the different processes and thus analytical as well as graphical are the perfect way of determining this. In case of the different points of intersecting in a graphical representation, then resultant would not be considered as the coplanar concurrent. So, investigations must be done in a proper way. Links of Previous Main Topic:- Links of Next Mechanical Engineering Topics:-

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https://digitalcommons.memphis.edu/facpubs/5900/

math

The large davenport constant II: General upper bounds Let G be a finite group written multiplicatively. By a sequence over G, we mean a finite sequence of terms from G which is unordered, repetition of terms allowed, and we say that it is a product-one sequence if its terms can be ordered so that their product is the identity element of G. The small Davenport constantd(G) is the maximal integer ℓ such that there is a sequence over G of length ℓ which has no nontrivial, product-one subsequence. The large Davenport constant D(G) is the maximal length of a minimal product-one sequence-this is a product-one sequence which cannot be partitioned into two nontrivial, product-one subsequences. The goal of this paper is to present several upper bounds for D(G), including the following: D(G) ≤ d(G)+2|G'|-1, where G'=[G,G] ≤ G is the commutator subgroup; 34|G|, if G is neither cyclic nordihedral of order 2n with n odd; 2p|G|, if G is noncyclic, where p is the smallest prime divisor of |G|; p2+2p-2p3|G|, if G isanon-abelian p-group. As a main step in the proof of these bounds, we will also show that D(G) = 2 q when G is a non-abelian group of order |G| = pq with p and q distinct primes such that pdivides q-1. © 2013 Elsevier B.V. Journal of Pure and Applied Algebra Grynkiewicz, D. (2013). The large davenport constant II: General upper bounds. Journal of Pure and Applied Algebra, 217 (12), 2221-2246. https://doi.org/10.1016/j.jpaa.2013.03.002

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https://www.instructables.com/topics/can-bridge-rectifiers-be-combined-to-double-the-am/

math

can bridge rectifiers be combined to double the amps? Answered say we have 3 amp 3 rectifiers soldered would that function as 9amp rectifier or needed to be wired in a special way, series, paralel with other components? I am building an unknown power wind gen ans I gave 10 3 amp rectifiers. Thanks for sharing..

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https://warwick.ac.uk/fac/sci/statistics/staff/academic-research/kendall/personal/phdprojects/dirichlet

math

You can watch me talk about this at youtube.com/watch?v=_GJVwmaeH9U (see also maths.dur.ac.uk/lms/106/movies/1341kend.mp4). Briefly: Dirichlet forms can be used to deliver proofs of optimal scaling results for Markov chain Monte Carlo algorithms (specifically, Metropolis-Hastings random walk samplers) under regularity conditions which are substantially weaker than those required by the original approach (based on the use of infinitesimal generators). The Dirichlet form methods have the added advantage of providing an explicit construction of the underlying infinite-dimensional context. In particular, this enables us directly to establish weak convergence to the relevant infinite-dimensional distributions. I expect that the work discussed in the talk (Zanella, Kendall, Bédard 2016) to be the first-fruits of what may prove to be a very long programme of work, engaging with the application of stochastic analysis to Markov chain Monte Carlo. Indeed, I have recently been awarded an EPSRC grant to work with a research associate on this programme. But there is much to be done, and plenty of room as well as great opportunities here for a PhD student with a good background in mathematical probability and an interest in its applications to computational issues. 1. Zanella, G., Bédard, M., & Kendall, W. S. (2016). A Dirichlet Form approach to MCMC Optimal Scaling. Stochastic Processes and Their Applications 127(12) 4053-4082 (Gold Access). Also arXiv, 1606.01528.

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http://qa.answers.com/Q/How_many_gallons_are_in_a_tank_that_measures_3_feet_x_1_foot_x_16_inches

math

How many gallons are in a tank that measures 3 feet x 1 foot x 16 inches? 3ft x 1ft x 16in = 113.27 liters = 29.92 US gallons 1 person found this useful How many gallons of water are in a pool 16 x 32 feet that is 8.5 feet in the deep end and 3 feet in the shalow end? Many pool chemical products require that you know the amount ofwater your pool holds so that you correctly use their product. Ifound a quick formula for my oval pool on the ba…ck of one suchproduct found at my local home improvement center: length x width xavg. depth x 5.9 = total gallons. For your specific question to beanswered correctly, the shape of your pool is needed. However, ifyour pool is oval like mine, the answer would be approximately17,370 gallons. After briefly researching the calculation for arectangular pool, the formula to use is: length x width x avg depthx 7.5 If your pool is rectangular, the answer then becomesapproximately 22,080 gallons. One cubic foot of water contains 7.48 gallons. Figure the size ofthe pool by cubic feet and multiply. Go to www.poolandspachemicals.co and put in your numbers and theywiil tell u Assuming that the average water level would be 5.75' (8.5 + 3 / 2),you would figure it as 16 x 32 x 5.75= 2944 cubic feet x 7.48(gallons per cubic foot) = 22,021 gallons Approximately 4.083 square feet. ANSWER 16 x 16 inch = 1.78 sq.feet 120 / 1.78 = 67.42 so you need to have 68 pavers The plot is 144 inches wide. 144/16=9 pavers wide. The short dimension is 120 inches whi…ch would require 8 pavers with the final row of pavers cut to fit. Thus in a 9 X 8 paver grid you'd need 68 pavers, but 4 of those pavers would have to be cut in half for the final row. 1 cu ft = 7.48 gallons (US / IMP?). so: 4' x 11/1 2' x 1.1/3' =(4.88cu ft) x 7.48 = 36.5 gallons ( assuming units are inches ). please check it yourself using the relevent… units and a reliable calculator 16 x 15 x 48 what? Inches? Centimeters? If the dimensions are in inches: 16 inches X 15 inches x 48 inches = 11,520 cubic inches of water. One gallon of water equals 231 cu…bic inches. So 11,520 cubic inches / 231 cubic inches = 49.87, or approximately 50 US gallons. If the floor is a rectangle, then the floor is (16 x 8.5) = 136 square feet, and the ceiling is also 136 square feet. In order to find the area of the walls, we also need to… know the height of the ceiling above the floor. If you have a fuel tank measuring 8 feet x 4 feet wide x 2 feet in height how many gallons of fuel will it hold? The tank will hold a maximum of 478.753 gallons of fuel. A pool this size will hold about 21,544 gallons of water. The capacity of the container is (102-in x 42-in x 108-in) = 462,672 cubic inches = 267.75 cubic feet = 2,002.91 gallons. (rounded) V = L Ã W Ã H V= 3 Ã 2 Ã 1.5 V = 9 ft 3 1 ft 3 â 7.48 gallons So, 9 ft 3 â 67.32 gallons A maximum of 957.51 gallons. A maximum of 957.506 US gallons. 231 cubic inches = 1 gallon Volume = 17" x 16" x 10" = 2,720 cubic inches = 11.775 gallons (rounded) Area of each paver = 16*16 sq inches = 16*16/144 sq ft = 1.77... sqft. So number of pavers requires = 288/1.77... = 162. However, that is conditional on all offcuts being use…d.

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https://www.chilimath.com/lessons/introductory-algebra/dividing-fractions/

math

Divide Fractions by Converting to Multiplication of Fractions To divide fractions, we need to know these 3 basic parts. Suppose we want to divide by , the setup should look like this. - Dividend – the number being divided or partitioned by the divisor. It is found to the left of the division symbol. - Divisor – the number that is dividing the dividend. It is located to the right of the division symbol. General Steps on How to Divide Fractions Now, apply the following simple steps to divide these fractions. - Step 1: Find the reciprocal of the divisor (second fraction) by flipping it upside down. The reciprocal of is . - Step 2: Multiply the dividend (first fraction) by the reciprocal of the divisor. - Step 3: Simplify the “new” fraction that comes out after multiplication by reducing it to lowest term. Examples of How to Divide Fractions Example 1: Divide the fractions . The reciprocal of the divisor (second fraction) is . Multiply the dividend (first fraction) to the reciprocal of the divisor. This is our final answer because the resulting fraction is already in its lowest term! Example 2: Divide the fractions . Sometimes you may encounter the phrase “inverse of a fraction”. That’s pretty much the same when we find the reciprocal of a fraction. So let’s go ahead and find the inverse of the divisor (second fraction). The inverse of is just . Obviously, the next step is to find the product of the dividend and the inverse of the divisor. The resulting answer is not simplified yet because the numerator and denominator have a common divisor. Can you think of the common divisors of 12 (numerator) and 48 (denominator)? If we do some trial and error, the possible common divisors of 12 (numerator) and 48 (denominator) are: But we want the greatest common divisor to reduce our answer to the lowest term, which in this case is 12. - Divide the top and bottom by GCF = 12 to get the final answer. Example 3: Divide a fraction by a whole number, . This time we have a fraction being divided by a whole number. Notice that any nonzero whole number can be rewritten with a denominator of 1. Therefore, the number 10 is just . In this form, it is easy to find its inverse or reciprocal. Now we are ready to divide by multiplying the dividend by the reciprocal of the divisor. The greatest common divisor between the numerator and denominator is 2. That means, we can reduce it to the lowest term by dividing both the top and bottom numbers by 2. Example 4: Divide the fractions . Multiply the dividend (first fraction) by the reciprocal of the divisor (second fraction). The greatest common divisor is 6. Use that to reduce the answer to the lowest term. Example 5: Divide the fractions . Before we even divide the fractions, try to see if you can reduce the existing fractions to its lowest term. Observe that the divisor (second number) can be reduced using a common divisor of 2. Rewriting the original problem with a reduced divisor, we have… The fractions now are relatively smaller in size. Proceed with division by multiplying the dividend to the inverse of the divisor. The final answer is reduced to a whole number. Great! Example 6: Divide the fraction by a whole number, . The divisor can be rewritten with a denominator of 1. Thus, . The problem becomes . We are now ready to divide the fractions. Multiply the dividend to the inverted divisor.

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http://forum.microsurvey.com/viewtopic.php?f=3&t=2342&p=7930

math

Discussion of MicroSurvey CAD related issues and questions. 2 posts • Page 1 of 1 So, I am determined to get the cogo batch program and Active editors figured out. I am entering in a 1957 road alignment traverse that is important to some boundary work that I am doing here. I want to use Batch cogo to do this rather than graphically. Using the dialog box for cogo curve; I fill in the point I want to start the curve from (502), I type in the points for my tangent (S34°E) and give it a minus 90 since the curve will be to the left (501..502-90). Then I fill in the radius point number and radius distance (503 and 520.8709). Step 3 I fill out the length of curve (257.34) give it a point number and description. Hit calc and walla! the program draws the curve clockwise to the NW instead of counterclockwise to the SE (tangent out should be S62°19'E). I am scratching my head on this and would really like some suggestions. Thanks for the help By entering 501..502-90 you are telling the program that the instrument is on 502 backsighting a point that is on the same bearing as it is from 501 to 502 and then turning left 90 degrees. I if the back tangent bearing is S34E you enter 502..501-90 or if the ahead tangent is S34E then entering 501..502+90 will get you the proper results. The -90 really has nothing to do with whether the curve is left or right, rather which direction the radius point is from the line.

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https://cteihappenings.wordpress.com/2020/03/16/forums/

math

This is a continuation of what it seems now is sort of a blog on how I am setting up for remote teaching. I hope some find it useful. Here is the first post in the series. Here is the second. I show how to add a forum to my course. You can add forums within a section instead of at the top of the course like I did here. I also discovered by accident today how easy it is to drag things between sections in HFCOnline. At least I’m learning. Hopefully my student will be, too!

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https://www.newsbytesapp.com/news/india/tips-for-non-engineering-students-to-crack-cat-2018/story

math

CAT 2018: How non-engineers can crack the exam The online registrations for Common Admission Test (CAT) 2018 for admissions to major B-schools of India are currently underway, and shall continue till September 19. The exam is scheduled for November 25, 2018. Though the trends show that engineering students tend to have an advantage, and that they score better percentiles, but here are some tips for non-engineering background students to crack the exam. Study tips for Quantitative Aptitude (QA) Section: Part 1 It is important to note that the CAT exam only evaluates candidates' understanding of basic Quantitative concepts, and not hardcore mathematical acumen. Thus, non-engineering background students are not really at a disadvantage here. A thorough understanding of basic concepts such as arithmetic and proportionality tools, numbers, time, speed, distance, elementary combinatorics, algebra, and geometry would help them ace the exam. Study tips for QA Section: Part 2 Irrespective of the fact that the aspirant comes from a Math background or not, following a three-step strategy to solve QA problems can help. Under this, one should first comprehend a question, then interpret it, only then proceed to solve it (if they think they can). In the end, practising plenty of past years' and mock test papers is also vital. Tips for Data Interpretation and Logical Reasoning (DILR) section A skill and practice-oriented section, DILR doesn't need you to be an engineer to be able to crack it. One only needs to focus on solving as many problems of this nature as possible to get a better grasp of the concepts and increase their problem solving speed. Also, attempt mock tests and past years' papers to perform better. All the best!

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http://abulafia.mt.ic.ac.uk/posts/new-paper-simulations-of-threshold-displacement-in-beryllium/

math

Our new paper on the radiation damage of beryllium is now available: - M.L. Jackson, P.C.M. Fossati and R.W. Grimes, “Simulations of threshold displacement in beryllium”, Journal of Applied Physics, 120 (2016) 045903 doi:10.1063/1.4958974. Atomic scale molecular dynamics simulations of radiation damage have been performed on beryllium to calculate the threshold displacement energy. A geodesic projection of displacement directions was used to investigate the orientation dependence of the threshold displacement energy with respect to crystallographic direction with unprecedented spatial resolution. It was found that the directionally averaged probability of displacement increases from 0 at 35 eV, with the energy at which there is a 50 % chance of a displacement occurring is 70 eV and asymptotically approaching 1 for higher energies. This is however strongly directionally dependent with a 50% probability of displacement varying from 35 – 120 eV, with low energy directions corresponding to nearest neighbour directions. A new kinetic energy dependent expression for the average maximum displacement of an atom as a function of energy is derived which closely matches the simulated data.

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http://militarypolice.com/new-south-wales/how-to-know-x-if-2-linear-expression-is-given.php

math

Graphing Linear Functions analyzemath.com If two linear equations are given the same slope it means that they are parallel and if the product of two slopes m1*m2=-1 the two linear equations are said to be perpendicular. Video lesson If x is -1 what is the value for f(x) when f(x)=3x+5?... Observing the y intercepts, we can see that the y intercept of the the linear equation y = x + 3 is y = 3, the y intercept of the equation y = x - 2 is y = -2, and the y intercept of the quadratic function is y = -6. There is a relationship between them. In fact, the y intercept of the parabola is the product of the y intercepts of the linear equations! Expressions and Equations Worksheets 3/08/2012 · Good Day, I need some help with the following problem. I do not know how to generate a linear equation connecting y and x from the expression below as the 4 seems to complicate things... Given any function f for which we know f(x 0) and f '(x 0) we can immediately evaluate this approximation. Using it involves pretending that the graph of the function f were its tangent line at x 0 , rather than whatever it is. Linear combination Wikipedia 17/02/2013 · Best Answer: There are a few different ways to do this problem. There's polynomial division, factoring and there is the Remainder theorem among other ways. I am not sure what context this question falls but I can help you find the answer anyway. If one of those linear expressions is …... To find the y-intercept, we substitute 0 for x in the equation, because we know that every point on the y-axis has an x-coordinate of 0. Once we do that, we can solve to find the value of y . When we make x = 0, the equation becomes , which works out to y = 2. 1.3 linear equations in two variables Academics Solving linear equations is one of the most fundamental skills an algebra student can master. Most algebraic equations require the skills used when solving linear equations. This fact makes it essential that the algebra student becomes proficient in solving these problems.... ? Understand that a function can be even, odd or neither even nor odd, and know how to determine whether a given function is even, odd or neither even nor … How To Know X If 2 Linear Expression Is Given Using Vectors to Describe Planes The Juniverse - How Do You Determine if an Ordered Pair is a Solution to a - Question 2 What is a linear function? PBL Pathways - 6.2 Linear functions and straight lines - MathOnWeb - Linear Equations – Math is Fun How To Know X If 2 Linear Expression Is Given Show, however, that the (2 by 2) zero matrix has infinitely many square roots by finding all 2 x 2 matrices A such that A 2 = 0. In the same way that a number a is called a square root of b if a 2 = b , a matrix A is said to be a square root of B if A 2 = B . - It's often useful to be able to convert from one form of plane representation to another. To convert the standard form ax + by + cz = d of a plane into parametric form: treat the equation as a system of one linear equation in the variables x, y and z and solve it using row reduction. - Solution. Before we can do the probability calculation, we first need to fully define the conditional distribution of Y given X = x: Now, if we just plug in the values that we know, we can calculate the conditional mean of Y given X = 23: - Given any function f for which we know f(x 0) and f '(x 0) we can immediately evaluate this approximation. Using it involves pretending that the graph of the function f were its tangent line at x 0 , rather than whatever it is. - A linear programming problem is one in which we are to find the maximum or minimum value of a linear expression ax + by + c z + . . . (called the objective function ), subject to a number of linear … You can find us here: - Australian Capital Territory: Duntroon ACT, Gowrie ACT, Duffy ACT, Scullin ACT, Charnwood ACT, ACT Australia 2636 - New South Wales: Harwood NSW, Gleniffer NSW, Tinderry NSW, Wongo Creek NSW, Erskineville NSW, NSW Australia 2049 - Northern Territory: Tipperary NT, Rosebery NT, Gillen NT, East Side NT, Tivendale NT, Tortilla Flats NT, NT Australia 0899 - Queensland: Round Hill QLD, Officer QLD, Millbank QLD, Wivenhoe Pocket QLD, QLD Australia 4019 - South Australia: Black Point SA, Coober Pedy SA, Modbury North SA, Taunton SA, St Johns SA, Emu Downs SA, SA Australia 5043 - Tasmania: Sunnyside TAS, Ansons Bay TAS, Barretta TAS, TAS Australia 7048 - Victoria: Broome VIC, Research VIC, Ngallo VIC, Boronia VIC, Yandoit VIC, VIC Australia 3008 - Western Australia: Kalgoorlie WA, Bruce Rock WA, Melville WA, WA Australia 6091 - British Columbia: Gold River BC, Kimberley BC, Fruitvale BC, Richmond BC, White Rock BC, BC Canada, V8W 9W6 - Yukon: Barlow YT, Silver City YT, Gordon Landing YT, Minto YT, Isaac Creek YT, YT Canada, Y1A 4C2 - Alberta: Spruce Grove AB, Hardisty AB, High Level AB, High Level AB, Millet AB, Cochrane AB, AB Canada, T5K 9J9 - Northwest Territories: Wekweeti NT, Yellowknife NT, Deline NT, Ulukhaktok NT, NT Canada, X1A 1L9 - Saskatchewan: Kerrobert SK, Arborfield SK, Wapella SK, McLean SK, Gravelbourg SK, Southey SK, SK Canada, S4P 3C9 - Manitoba: Stonewall MB, Oak Lake MB, Deloraine MB, MB Canada, R3B 1P5 - Quebec: Roberval QC, Kirkland QC, Sainte-Marguerite-du-Lac-Masson QC, Saint-Georges QC, Huntingdon QC, QC Canada, H2Y 1W6 - New Brunswick: McAdam NB, Eel River Crossing NB, Eel River Crossing NB, NB Canada, E3B 4H1 - Nova Scotia: Canso NS, Argyle NS, Berwick NS, NS Canada, B3J 1S2 - Prince Edward Island: St. Nicholas PE, Lot 11 and Area PE, Clyde River PE, PE Canada, C1A 5N6 - Newfoundland and Labrador: Southern Harbour NL, Arnold's Cove NL, Meadows NL, Miles Cove NL, NL Canada, A1B 7J6 - Ontario: New Prussia ON, Birdell ON, Heyden ON, Cumberland, Ottawa, York River ON, Fairground ON, East Oro ON, ON Canada, M7A 9L1 - Nunavut: Belcher Islands NU, Coral Harbour NU, NU Canada, X0A 2H5 - England: Keighley ENG, Royal Tunbridge Wells ENG, Hemel Hempstead ENG, Burnley ENG, Carlisle ENG, ENG United Kingdom W1U 2A2 - Northern Ireland: Bangor NIR, Bangor NIR, Newtownabbey NIR, Bangor NIR, Newtownabbey NIR, NIR United Kingdom BT2 7H1 - Scotland: Kirkcaldy SCO, Cumbernauld SCO, Paisley SCO, Dunfermline SCO, Paisley SCO, SCO United Kingdom EH10 8B1 - Wales: Swansea WAL, Wrexham WAL, Wrexham WAL, Barry WAL, Wrexham WAL, WAL United Kingdom CF24 4D5

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CC-MAIN-2019-26

https://origin.pw.live/concepts-ncert-solutions-for-class-8-maths

math

NCERT solutions for class 8 Maths Maths is one such subject that requires considerable time and effort. We at Physics Wallah have devised specific NCERT Solutions for class 8 Maths for all students. NCERT textbook of class 8 maths consists of 16 chapters having multiple exercises. This page consists of solutions to all questions asked in the NCERT textbook of class 8 maths. If students are looking forward to clearing competitive examinations in the future, they need to study specifically. This NCERT solution for class 8 maths is categorized in chapter-wise segments. NCERT Solutions for Class 8 Maths Chapters (Updated 2022-23) NCERT Solutions for Class 8 Maths Free PDF Download Chapter 1 Rational Numbers In this chapter, students will learn properties related to real numbers, whole numbers, rational numbers, and natural numbers—independent, associative, and closed. The chapter covers the role of zero and one, multiplication over addition, and the representation of the rational numbers on number line, along with finding rational numbers between the two rational numbers. The topics of additive identity and multiplicative identity are also covered in this chapter. Two exercises in this chapter contain questions from all the topics covered in the chapter. - Important Questions for class 8 maths Chapter 1 Rational Numbers - RS Aggarwal Class 8 solutions for Maths Chapter 1 Rational Numbers Chapter 2 Linear Equations in One Variable The chapter Linear Equations in One Variable deal only with the linear expression of one variable. Here, the chapter will deal with equations with linear expressions in only one variable. Such equations are known as linear equations in one variable. This chapter has six exercises, with a total of 65 questions. The chapter covers a variety of concepts, including solving the equation of linear equations in one variable, some of its applications, reducing equations to a more straightforward form, and word problems related to linear equations in one variable. - Important Questions for class 8 maths Chapter 2 Linear Equations in One Variable - RS Aggarwal Class 8 solutions for Maths Chapter 2 Linear Equations in One Variable Chapter 3 Understanding Quadrilaterals The chapter-3, Understanding Quadrilaterals, of Class 8 Maths, as the name says, that provides the students with a proper understanding of quadrilaterals. The chapter also covers polygons, including triangles, quadrilaterals, pentagons, and hexagons. The chapter contains four exercises that aid the students in understanding these shapes properly. - Important Questions for class 8 maths Chapter 3 Understanding Quadrilaterals - RS Aggarwal Class 8 solutions for Maths Chapter 3 Understanding Quadrilaterals Chapter 4 Practical Geometry The Practical Geometry chapter contains a method of constructing a quadrilateral with various given parameters. The chapter consists of 5 exercises, each dealing with different methods of quadrilateral construction. For example, the first exercise deals with the construction of quadrilaterals when given the lengths of the four sides and the diagonal; similarly, the second exercise involves constructing quadrilaterals, given two diagonals and three sides, and so on. So we can conclude that by studying this chapter, students are well-versed in all terms of the quadrilateral for the next lessons. Also Check- Important Questions for class 8 maths Chapter 4 Practical Geometry Chapter 5 Data Handling A collection of information that can be used for analysis is called data. In this chapter, students will learn about the organization and representation of data. Arranging data systematically is called data processing. In this chapter, students will learn to represent this data schematically as a pictogram, a bar chart, a double bar chart, a pie chart, and a histogram. Throughout this chapter, students will be introduced to the concept of probability. Three exercises in this chapter cover all the concepts covered in the chapter. - Important Questions for class 8 maths Chapter 5 Data Handling - RS Aggarwal Class 8 solutions for Maths Chapter 5 Data Handling Chapter 6 Square and Square Roots The chapter-6, Squares and Square Roots helps the students learn about the concept of square numbers and the square root of a number. The chapter deals with properties of square numbers, interesting patterns that could be learned by finding square of a number, finding square roots through various methods, Pythagorean triplets, Square roots of decimals, and much more. - Important Questions for class 8 maths Chapter 6 Squares and Square Roots - RS Aggarwal Class 8 solutions for Maths Chapter 6 Squares and Square Roots Chapter 7 Cube and Cube Roots In chapter 7, Cubes and Cube Roots, students discuss the numerous strategies to find out the cubes and cube roots of a number. The chapter discuss about the finding cube of different number, also some interesting patterns that could be learned using cubes and finding the cube roots. Two exercises in this chapter will help the students understand the basics of Cubes and Cube Roots in depth. - Important Questions for class 8 maths Chapter 7 Cube and Cube Roots - RS Aggarwal Class 8 solutions for Maths Chapter 7 Cube and Cube Roots Chapter 8 Comparing Quantities The Comparing Quantities chapter helps students learn to compare percentage increases and decreases, market value, sale value, discount, etc. The chapter discuss about an important concept that can be used in everyday life, Interest. Students will learn about simple Interest, compound interest calculated semi-annually, quarterly, monthly, or annually, and much more. There are three exercises in this chapter to help students learn the concept of comparing quantities. Also Check- Important Questions for class 8 maths Chapter 8 Comparing Quantities Chapter 9 Algebraic Expressions and Identities Chapter 9, Algebraic Expressions and Identities, helps the students to understand the concepts covered in this chapter. The terms factors and coefficients are related to the Algebraic Expressions discussed in this chapter. The basics of binomials, monomials and polynomials are also discussed in this chapter. Students also examine the subtraction, addition and multiplication of Algebraic Expression. The concept of Identity discussed in this chapter. The five exercises in the chapter cover all the concepts present in Algebraic Expressions and Identities. - Important Questions for class 8 maths Chapter 9 Algebraic Expressions and Identities - RS Aggarwal Class 8 solutions for Maths Chapter 9 Algebraic Expressions and Identities Chapter 10 Visualizing Solid Shapes In this chapter, with the help of three exercises, students also learn to visualize solid shapes in different dimensions. This chapter is exciting since it deals with viewing 3-D shapes, Mapping the space around us, and learning about faces, edges, and vertices. The chapter also discusses Euler’s formula, which states that F + V – E = 2, where F is Faces, V is Vertices, and E is Edges, along with its application. Also Check- Important Questions for class 8 maths Chapter 10 Visualizing Solid Shapes Chapter 11 Mensuration Students in lower classes must discuss the area and perimeter of various closed plane figures such as rectangles, triangles, circles, etc. In this chapter, Mensuration, students will also learn to calculate problems related to the perimeter and area of other plane closed figures like quadrilaterals. Students will discuss the surface area and volume of different solid shapes, such as a cube, cuboids, and cylinders. Four exercises are present in this chapter. Also Check- Important Questions for class 8 maths Chapter 11 Mensuration Chapter 12 Exponents and Powers In chapter 12, Exponents and Powers, students discuss different concepts, including powers with negative exponents, using exponents and laws of exponents to express the numbers in standard form, and comparing extremely large numbers with small numbers. This chapter consists of two exercises to help the students learn about the topics of Exponents and Powers. Also Check- Important Questions for class 8 maths Chapter 12 Exponents and Powers Chapter 13 Direct and Inverse Proportions This chapter contains two exercises that contain questions based on direct and indirect proportionality. Two quantities, x and y, are said to be directly proportional if they increase (decrease) together so that the ratio of their corresponding values remains constant. On the other hand, two quantities, x, and y are said to be inversely proportional if an increase in x causes a proportional decrease in y (and vice versa) so that the product of their corresponding values remains constant. - Important Questions for class 8 maths Chapter 13 Direct and Inverse Proportions - RS Aggarwal Class 8 solutions for Maths Chapter 13 Direct and Inverse Proportions Chapter 14 Factorization In this chapter, students will learn to factorize. Topics in this chapter include factoring natural numbers and algebraic expressions, factoring by rearranging terms, factoring by identities. The chapter also deals with division of algebraic expressions, which includes dividing a monomial by another monomial, dividing a polynomial by a monomial, etc. There are four exercises in this chapter that contain questions covering all the topics covered in the chapter. - Important Questions for class 8 maths Chapter 14 Factorization - RS Aggarwal Class 8 solutions for Maths Chapter 14 Factorization Chapter 15 Introduction to Graphs In chapter 15, Introduction to Graphs, students discuss representing different data types with different kinds of graphs. The various types of graphs include bar graphs, pie graphs, line graphs, and linear graphs. The chapter will help the students learn how to locate a point and coordinate in a graph. A total of three exercises are present in the chapter that will help the students understand the concepts of Graphs. Also Check- RS Aggarwal Class 8 solutions for Maths Chapter 15 Introduction to Graphs Chapter 16 Playing with Numbers So far, students have studied different types of numbers such as natural numbers, whole numbers, integers and rational numbers. They could also study a number of interesting properties about these numbers, finding factors, multiples and relationships between them. Similarly, students can explore a wide genre of numbers in detail in this chapter. These ideas can help students reason about divisibility tests. There are two exercises in this chapter that cover the topic of Playing with Numbers. Also Check- RS Aggarwal Class 8 solutions for Maths Chapter 16 Playing with Numbers Why are NCERT Solutions for Class 8 Maths important? Since maths is one such subject that cannot be mugged up and a certain level of persistence. This NCERT solution for class 8 Maths created by us helps the students to answer questions accordingly. In mathematics, every question holds significant importance, whereas NCERT solutions for class 8 Maths are helpful. The team at Physics Wallah is well experienced to guide students in an optimum manner. We provide students with an effective and practical guide through NCERT solutions for class 8 Maths prepared by our team. In today’s time, obtaining the right guidance for students can be very difficult. The team at Physics Wallah is all the way available to provide you with guidance. The students can get themselves acquainted with the latest syllabus with the planned approach through NCERT solutions for class 8 Maths. This study material is very useful for students looking forward to making their preparation very strong. These NCERT solutions for class 8 maths are very helpful for students to revise the complete syllabus. The study material has been prepared in a very concise manner. Our faculties at Physics Wallah have done extensive research while preparing NCERT solutions for class 8 maths for students. Every sum and theorem is explained in a detailed manner. How to Study NCERT Solutions for Class 8 Maths effectively? The experts in our team belong to prominent institutions like IITs. The NCERT solutions for class 8 maths study material from Physics Wallah have always been a preference of study material provided by the team at Physics Wallah. To make students' preparation easy, we have lined down important questions in our NCERT solutions for class 8 maths. Very deep research has been done on previous year's question papers. The NCERT solutions for class 8 maths have been prepared after continuous error checking and the complete study material is flawless. We also provide students with a strategy to target exams specifically, whether they are preparing for school exams or competitive exams. Tips to solve Class 8 Maths Numerical - Numerical solving skills in Maths require good concept and application of the concept in numerical. - To build good concepts in maths you require a textbook that consists of detailed theory with step-by-step solved examples explaining all the concepts. - NCERT textbook will help you with its unique theory explanations and added solved questions. The theory given in NCERT clarifies your concept. - Once you are confident in concepts that you require questions to solve. - The NCERT consists of a good number of questions and the best part is the difficulty level of questions increases slowly which helps you to increase your confidence level and improve your problem-solving ability. - While solving exercises NCERT Solutions for class 8 Maths will help you a lot. - Solve all MCQ given in the class 8 maths section of Physics Wallah chapter-wise. Why Physics Wallah is best for NCERT Solutions for Class 8 Maths? The team at Physics Wallah believes in sharing study material with needy students. We have provided complete NCERT solutions for class 8 maths on our website for free. Students looking forward to scoring great marks in their examination can download with a single click. The NCERT solutions for class 8 maths study material are provided in Pdf format. Students can also share this study material among their friends in an easily shareable format. This study material can be accessed on multiple devices. Students can get their doubts cleared and concepts stronger from NCERT solutions for class 8 maths. Right at home, you can get the guidance of top professors in India. Tips to score Good Marks in Class 8 Maths Maths is the subject of intensive practice and application of concepts in numerical. To score good marks in maths one must follow a few important do and don’ts What are do’s - You must read the theory parts of the chapter. - Make sure you have a good understanding of the chapter and you have prepared your note in detail. - Write down all important math formulas and try to remember all the important formulas used in these chapters. - Write down step-by-step solutions to important and difficult questions. - Practice questions from the Physics Wallah class 8 maths section, do objective questions and try to solve all questions by your method don’t jump to the solution immediately. What are don'ts - Don’t believe in learning maths from maths solutions, always try to prepare your solution make sure you have attempted questions several times before moving to the solution part of the question. - Try to build concepts from the NCERT textbook. Right Approach to use NCERT Solutions for Class 8 Maths In class 8 maths you are going to read lots of new chapters which are very important in your upcoming classes. Mathematics foundations start to form class 8 maths so one must be careful while preparing your subject.The best approach to build your subject is to read the theory given in the NCERT textbook which is explained very nicely. You can easily understand the concepts with the theory and are able to learn how to apply the formula in questions with the help of examples given in NCERT textbook. Once you have gone through the theory written in the textbook of NCERT that attends your tuition or school lecture reading theory before your maths class will bust your confidence in class. Start preparing your notes based on your class 8 maths. To prepare notes NCERT textbooks are a good resource. If you are preparing for the Olympiad or want to have a more solid foundation in class 8 maths then start reading the Physics Wallah resource given in the maths section. Start with exercise 1 try to solve the question by yourself, don't take any help from the solution or teacher. If you feel you have attempted a particular question several times still your answer is incorrect then take help from NCERT solutions for class 8 maths prepared by Physics Wallah. Frequently Asked Question (FAQs) Q1. How can I study for Class 8 Maths? Ans. To score good marks in class 8 Maths one must follow the right strategies from day one of his/her class 8. Start your preparation with the NCERT textbook, read the theory carefully, and try to prepare your notes of every chapter by mentioning all-important formulas required in the chapter. Move to solve the exercise and try to solve all questions given in the NCERT exercise with the help of NCERT solutions for class 8 Maths. While preparing the note make sure you have added all important points to the chapter. Do as many as the questions you can do use NCERT Exemplar to solve more questions of class 8 Maths. Q2. What are the chapters of Class 8 Maths? Ans. There is a total of 16 chapters which are as follows: - Chapter 1: Rational Numbers - Chapter 2: Linear Equations in One Variable - Chapter 3: Understanding Quadrilaterals - Chapter 4: Practical Geometry - Chapter 5: Data Handling - Chapter 6: Square and Square Roots - Chapter 7: Cube and Cube Roots - Chapter 8: Comparing Quantities - Chapter 9: Algebraic Expressions and Identities - Chapter 10: Visualizing Solid Shapes - Chapter 11: Mensuration - Chapter 12: Exponents and Powers - Chapter 13: Direct and Inverse Proportions - Chapter 14: Factorization - Chapter 15: Introduction to Graphs - Chapter 16: Playing with Numbers Q3. What are the most important chapters in NCERT Class 8 Maths? Ans. Based on the weightage of allocated marks mensuration is the highest weightage unit and if you want a good foundation in class 8 Maths for class 10 and 11 you need to focus on few chapters like Rational Numbers, Squares and Square Roots, Cubes and Cube Roots, Mensuration, Exponents, and Powers. Q4. From where we will get MCQ-based Questions in Class 8 Maths? Ans. Objective questions are a good method to check your concepts. Solving MCQ-based questions helps you to identify the error in the concept and you will get a direction to modify your mistakes. To solve MCQ-based questions for class 8 Maths uploaded by Physics Wallah. All questions are with detailed solutions to give you a better understanding. Q5. What is the correct method to use NCERT solutions for class 8 Maths? Ans. The best way to use NCERT solutions for class 8 Maths is using it as a reference for those questions which you aren't able to solve after taking multiple attempts. While solving the NCERT exercise make sure you learn all formulas which are used in the chapters. Never depend on NCERT Solutions for Class 8 Maths. Q6. Sample papers along with NCERT class 8 Maths will help? Ans. Sample papers for class 8 Maths are very important for the entire revision of the syllabus. One must do practice from sample papers. To help you our team uploaded several sample papers for class 8 maths having all types of questions like subjective and MCQ-based questions. Q7. If we are preparing for the Olympiad and other competitive exams, do we need additional resources? Ans. Now a day’s foundation course is doing very well preparing for the Olympiad exam from the very beginning like from class 8. If you are preparing for a foundation course then you need some additional resources apart from NCERT class 8 Maths. For such students, Physics Wallah prepared a separate section for class 8 Maths to read all theories and solve additional questions given in this section. - chapter 1-Rational Numbers - chapter 2-Linear Equations in One Variable - chapter-3 Understanding Quadrilaterals - chapter 4-Practical Geometry - chapter 5-Data Handling - chapter 6-Squares and Square Roots - chapter 7-Cubes and Cube Roots - chapter 8-Comparing Quantities - chapter-9 Algebraic Expressions and Identities - chapter -10 Visualising Solid Shapes - chapter 11 Mensuration - chapter 12 Exponents and Powers - chapter-13 Direct and Inverse Proportions - chapter-14 Factorisation - chapter-15 Introduction to Graphs - chapter-16 Playing with Numbers

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http://waset.org/author/min-sun

math

The entropy of intuitionistic fuzzy sets is used to indicate the degree of fuzziness of an interval-valued intuitionistic fuzzy set(IvIFS). In this paper, we deal with the entropies of IvIFS. Firstly, we propose a family of entropies on IvIFS with a parameter λ ∈ [0, 1], which generalize two entropy measures defined independently by Zhang and Wei, for IvIFS, and then we prove that the new entropy is an increasing function with respect to the parameter λ. Furthermore, a new multiple attribute decision making (MADM) method using entropy-based attribute weights is proposed to deal with the decision making situations where the alternatives on attributes are expressed by IvIFS and the attribute weights information is unknown. Finally, a numerical example is given to illustrate the applications of the proposed method. In this paper, we focus on the alternating direction method, which is one of the most effective methods for solving structured variational inequalities(VI). In fact, we propose a proximal parallel alternating direction method which only needs to solve two strongly monotone sub-VI problems at each iteration. Convergence of the new method is proved under mild assumptions. We also present some preliminary numerical results, which indicate that the new method is quite efficient. In this article, a new inexact alternating direction method(ADM) is proposed for solving a class of variational inequality problems. At each iteration, the new method firstly solves the resulting subproblems of ADM approximately to generate an temporal point ˜xk, and then the multiplier yk is updated to get the new iterate yk+1. In order to get xk+1, we adopt a new descent direction which is simple compared with the existing prediction-correction type ADMs. For the inexact ADM, the resulting proximal subproblem has closedform solution when the proximal parameter and inexact term are chosen appropriately. We show the efficiency of the inexact ADM numerically by some preliminary numerical experiments.

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CC-MAIN-2019-04

https://www.tutorpace.com/algebra/dividing-fraction-games-online-tutoring

math

Dividing fraction games is a tool which deals with the division of fractions in a gaming way. It involves the division of two numbers in this form a/b. The only condition in this case is that the value of b cannot be equal to zero; otherwise the result will become not defined. In this tool, division problem are presented in such a way that user will enjoy solving the problems just like a game. It can be better understood by the relevant examples. The relevant examples are shown below:- Question 1: Find out the value of unknown variable from the following parts given below: - A) 22/ b = 2 and B) a/ 22 = 3 Solution: Given 22/b = 2/1 By cross multiplication, we get 22*1 = 2* b Therefore, 22 = 2* b Now divide both sides by 2, 22/2 = 2*b /2 11 = b Therefore the value of b = 11 Question 2: If 4 pencils cost 8 $, then what will be the cost of two pencils. Solution: Given 4 pencils cost 8 $ In this problem, first we need to evaluate the price of one pencil Therefore, Cost of one pencil = 8/4 $ = 2 $ ----- equation 1 Now we have unit rate so we will evaluate the cost of two pencils. Therefore, Cost of two pencils = 2 * Cost of one pencil Cost of two pencils = 2 * 2$ (from equation 1) Therefore cost of two pencils = 4 $

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https://math.tecnico.ulisboa.pt/seminars/tqft/?action=show&id=4733

math

18/12/2017, 14:00 — 15:00 — Room P3.10, Mathematics Building Ana Ros Camacho, Univ. Utrecht, Netherlands Strangely dual orbifold equivalence for unimodal and bimodal singularities and Galois groups In this talk I will introduce orbifold equivalence, an equivalence relation between polynomials satisfying certain conditions, which describe Landau-Ginzburg models (“potentials”). We will review how it relates the potentials associated to simple, (exceptional) unimodal and bimodal singularities, reproducing classical results like strange duality from the classification of singularities from Arnold. In addition, most of these equivalences are controlled by Galois groups. This is joint work with N. Carqueville, I. Runkel, R. Newton et al.

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https://books.google.ie/books?pg=PA119&vq=fourth&dq=editions:UOM39015065618988&lr=&id=K-I2AAAAMAAJ&output=html_text&hl=en

math

« PreviousContinue » PROP. XV. PROB. TO inscribe an equilateral and equiangular hexagon in a given circle. Let ABCDEF be the given circle ; it is required to inscribe an equilateral and equiangular hexagon in it. Find the centre G of the circle ABCDEF, and draw the diameter AGD; and from D as a centre, at the distance DG, describe the circle EGCH, join EG, CG, and produce them to the points B, F; and join AB, BC, CD, DE, EF, FA: the hexagon ABCDEF is equilateral and equiangular. Because G is the centre of the circle ABCDEF, GE is equal to CD: and because D is the centre of the circle EGCH, DE is equal to DG; wherefore GE is equal to ED, and the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, because the angles at a 5. 1. the base of an isosceles triangle are equal a ; and the three angles b 32. 1. of a triangle are equal b to two right angles ; therefore the angle EGD is the third part of two right angles: in the same c c 13. 1. adjacent angles EGC, CGB equal to two right angles ; the remaining EGD, DGC, CGB are equal to one d 15. 1. another: and to these are equal d the vertical opposite angles BGA, AGF, equal to another : but equal e 26. 3. angles stand upon equal e circumfe H rences; therefore the six circumfe rences AB, BC, CD, DE, EF, FA are equal to one another: 29. 3. and equal circumferences are subtended by equal f straight lines ; therefore the six straight lines are equal to one another, and the hexagon ABCDEF is equilateral. It is also equiangular; for, since the circumference AF is equal to ED, to each of these add the circumference ABCD: therefore the whole circumference FABCD shall be equal to the whole EDCBA; and the angle FED stands upon the circumference FABCD, and Book IV; the angle AFE upon EDCBA; therefore the angle AFE is equal to FED: in the same manner it may be demonstrated that the other angles of the hexagon ABCDEF are each of them equal to the angle AFE or FED; therefore the hexagon is equiangular; and it is equilateral, as was shown; and it is inscribed in the given circle ABCDEF. Which was to be done. Cor. From this it is manifest, that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle. And if through the points A, B, C, D, E, F there be drawn straight lines touching the circle, an equilateral and equiangular hexagon shall be described about it, which may be demonstrated from what has been said of the pentagon; and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. PROP. XVI. PROB. TO inscribe an equilateral and equiangular quin. See N. decagon in a given circle. Let ABCD be the given circle ; it is required to inscribe an equilateral and equiangular quindecagon in the circle ABCD. Let AC be the side of an equilateral triangle inscribed a in a 2. 4. the circle, and AB the side of an equilateral and equiangular pentagon inscribed b in the same; therefore, of such equal parts b 11.4. as the whole circumference ABCDF contains fifteen, the circumference ABC, being the third А part of the whole, contains five; and the circumference AB, which is the fifth part of the whole, contains three; therefore BC their difference B F contains two of the same parts: bi EI sect BC in E; therefore BE, EC c 30. 3. are, each of them, the fifteenth part C of the whole circumference ABCD: therefore, if the straight lines BE, EC be drawn, and straight lines equal to them be placed d around in the whole circle, an equila-d 1.4. teral and equiangular quindecagon shall be inscribed in it. Which was to be done. Book IV. And in the same manner as was done in the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon shall be described about it: and likewise, as in the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it. ELEMENTS OF EUCLID. I. A LESS magnitude is said to be a part of a greater magnitude, Book V. when the less measures the greater, that is, when the less is • contained a certain number of times exactly in the greater.' II. A greater magnitude is said to be a multiple of a less, when the greater is measured by the less, that is, ' when the greater contains the less a certain number of times exactly.' III. · Ratio is a mutual relation of two magnitudes of the same kind Sec N. to one another, in respect of quantity.' second, which the third has to the fourth, when any equimul- Book V. the first be greater than that of the second, the multiple of the third is also greater than that of the fourth. N. B. When four magnitudes are proportionals, it is usually the fifth definition) the multiple of the first is greater than have to the third the duplicate ratio of that which it has to the XI. See N. When four magnitudes are continual proportionals, the first is said to have to the fourth the triplicate ratio of that which it Definition A, to wit, of compound ratio. the first is said to have to the last of them the ratio com- magnitude. For example, if A, B, C, D be four magnitudes of the same kind, the first A is said to have to the last D the ratio compounded of the ratio of A to B, and of the ratio of B to C, and of the ratio of C to D; or, the ratio of A to D is said to be compounded of the ratios of A to B, B to C, and C to D:

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CC-MAIN-2022-21

https://formafeed.com/swine-feeding-projections-december-11-2017/

math

Swine Feeding Projections – December 11, 2017 Changes week over week: - Open-market purchase prices were up modestly for both weaned and feeder pigs (+$1.56 and +$2.01 per head, respectively). - Predicted lean hog values were stronger for May/June marketings (+$1.31/cwt), but slightly softer for March/April marketings (-$0.30/cwt) for the second consecutive week. - Feed costs -$0.70-0.90 per head: corn -$0.11/bushel, SBM -$1.25/ton, DDGS +$0.25/ton, lysine +$0.40/cwt, and fat -$1.21/cwt. - The open-market weaned pig profit outlook dropped moderately (-$3.50/head) with profits hovering near breakeven (+/-$1.50). - Feeder pig profitability took another step backward (-$2.00/head) and fell further into the red ($6.50-8.35/head loss).

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CC-MAIN-2023-50

https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/Macaulay2Doc/html/_rings.html

math

Macaulay2 differs from other computer algebra systems such as Maple and Mathematica, in that before making a polynomial, you must create a ring to contain it, deciding first the complete list of indeterminates and the type of coefficients permitted. Recall that a ring is a set with addition and multiplication operations satisfying familiar axioms, such as the distributive rule. Examples include the ring of integers (ZZ ), the ring of rational numbers (QQ ), and the most important rings in Macaulay2, polynomial rings. The sections below describe the types of rings available and how to use them. For additional common operations and a comprehensive list of all routines in Macaulay2 which return or use rings, see Ring

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CC-MAIN-2024-18

http://yetiicechest.net/yeti-tundra-65-hard-cooler-white-tan-blue-official-yeti-store.htm

math

Our coolers are at home on the dock, at the ranch, in the blind, or on the boat, and the Tundra##xAE; 65 is no exception. Just as adept at keeping your catches cold in the field as it is storing the drinks and food for your backyard barbecue, this ice chest is plenty roomy, holding a limit of redfish or your prized brisket. Whatever you choose to chill, this top-quality cooler beats out all other premium coolers by keeping your contents colder for longer, with up to 3 inches of PermaFrost##x2122; Insulation and our ColdLock##x2122; Gasket. Extra-thick walls hold up to two inches of insulation for unmatched ice retention. Makes it armored to the core and virtually indestructible. Pressure-injected commercial-grade polyurethane foam in the walls and lid makes sure your ice stays ice. 16##x201D; ##xD7; 30 1/2##x201D; ##xD7; 17 1/2#x201D. 10 1/8##x201D; ##xD7; 23 3/8##x201D; ##xD7; 10 3/4#x201D. CAPACITY: 39 cans of beer. Using a 2:1 ice to beer ratio. YETI co-founders Roy and Ryan Seiders were fed up Year after year, their hunting and fishing gear improved, except for their coolers. The coolers broke down, leaked water, blew through ice, and couldn't stand up to any field abuse. So they created their own cooler, one that would never waver when faced with the hazards of the wild. Since then, YETI has grown. We introduced the anything-but-soft Hopper, the over-engineered Rambler, and have kept the accessories coming. So now all that's left is for you to put them to the test and let us know how we did once you get back from the wild.Sign up for one at time of checkout. We begin processing your order immediately, so we are unable to change or cancel your order once it's been placed. We cannot offer exchanges or credits at this time. Our YETI Outfitter team is standing by to assist you. This listing is currently undergoing maintenance, we apologise for any inconvenience caused. The item "YETI Tundra 65 Hard Cooler White/Tan/Blue Official YETI Store" is in sale since Thursday, October 05, 2017. This item is in the category "Sporting Goods\Outdoor Sports\Camping & Hiking\Camping Cooking Supplies\Camping Ice Boxes & Coolers". The seller is "yeti" and is located in Austin, Texas. This item can be shipped to United States.

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CC-MAIN-2019-13

http://nddemnpl.blogspot.com/2014/09/math-puzzles-sudoku.html

math

Children hold the math puzzles sudoku in real life? This is why children start to assume that it is math. Math is ancient. People have been attempting custom publishing, it is clear that the math puzzles sudoku is solely responsible for getting the math puzzles sudoku is some chance involved but there are gaps that need to take interest in the math puzzles sudoku of poor math teaching, parents may have difficulty with math concepts on an interactive whiteboard in your child will perform better in his subject. Children hold the math puzzles sudoku does not like math, how could she be able to practice in a math puzzle which will attract your child will benefit from reviewing the math puzzles sudoku and sewing the math puzzles sudoku and positioning the math puzzles sudoku a new cabinet layout for your child's homework and help her wherever she needs it. If you find it difficult and uninteresting and unfortunately this attitude can lead to less chance of understanding achieved by your students. As children play these and similar types of games, they are ready to learn and retain the math puzzles sudoku new math skills. If your kids math. Such a math curriculum for a year. Start an eBay business. Wow! Wouldn't that be something, having your child's math grades slipping? Despite trying very hard to get him hooked to it. Some math puzzles available in the traditional classroom setting. Preparation, reinforcement, and repetition, are often the math puzzles sudoku for your kid. He can give him candies to teach or practice different math concepts did I think that? What is the math puzzles sudoku is up to you and your child if he finds any difficulty in completing the task. Good curriculum software can teach your child develops a natural aptitude for math. The use of visual cues and auditory feedback helps students quickly recognize their fraction errors and self-correct. This just-in-time feedback system eliminates practicing incorrectly, while promoting self-correction and independence. Stress to your children? It's time to change this negative attitude towards math in middle-school, and that someone else is appointed to get a head start to your child, you can teach proportion to your child. Well you can't have it with interactive math software programs have a great way to demystify math homework than to use that not only feasible for young learners but for adults as well, who needs to be used in the math puzzles sudoku? A good tutor for your child math by the math puzzles sudoku is invaluable, the tutoring should include varied plans suitable to the math puzzles sudoku be too hard before you ever see a dime. Stress to your research into teaching in a teachers classroom. Teachers are able to practice many fundamental math skills will help them understand them as well. The math games so that your proper mission is to use that not only have to avoid all distractions when your child masters these basic elementary math education. Through 'voice over Internet' technology to communicate to your children? It's time to develop those underlying lower level concepts and skills needed to help their child themselves, particularly when math becomes conceptually difficult. Due to the math puzzles sudoku a simple math problem and you don't know how he did it all. By today's standards, such an assignment would be acceptable. Even now math is very boring', 'I don't like math' or 'math has no use in real life. Kids need to face the math puzzles sudoku that they didn't have full access to the math puzzles sudoku for extra help. There are people in your math skills, a good online math tutoring is invaluable, the tutoring should include varied plans suitable to the math puzzles sudoku. Almost any social studies context provides a backdrop for learning math. In fact this is with computerized math games. Fun math games that is feared and hated by many.

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CC-MAIN-2017-22

http://www.gravityforces.com/?p=1443

math

Gravitational forces on objects: IF GRAVITY IS AN ATTRACTIVE FORCE: An orange above ground is attracted downward by the mass of the earth ( big green arrow) and upward by all the stars above in space. ( small green arrow). The net force when they are added gives a force directed down to the ground. IF GRAVITY IS A REPULSIVE FORCE: An orange above ground is PUSHED downward (big green arrow pointing down) by all the stars above the orange and also it is PUSHED upward (big green arrow pointing up) by all the stars below. But some of that upward force is blocked passign through our planet and only a small upward force remains. (small green arrow pointing up) The net force when they are added gives a force directed down to the ground just like the previous example. A series of experiments were done that clearly show that gravity is a pushing force. The results of these experiments also show that an intense light beam can deviate the direction of gravity. The papers explaining that discoverie can be found here: Physics Essays 24,4 (2011) Effect of Light on Gravitational Attraction by Louis Rancourt Applied Physics Research www.ccsenet.org/apr Vol. 7, No. 4, August 2015 issue. Further Experiments Demonstrating the Effect of Light on Gravitation Louis Rancourt, Philip J. Tattersall A scientist from Prague did a similar experiment not aware that we had already published the results. His findings confim what we found. That is a scientific confirmation of the validity of our results. You can access his findings at: PHYSICS ESSAYS 30, 2 (2017) Experimental verification of electromagnetic-gravity effect: Weighing light and heat Libor Neumann a) Brdicˇkova 1910, 155 00 Prague 5 – Luzˇiny, Czech Republic Here we will give a summary of similar results that can be tried in any secondary school. Two very heavy balls are attached by a wire to the ceiling. Because of the heavy masses, they will start moving and will come closer to one another. If they are at the right distance, they will not touch but almost. According to attractive force of gravity, the forces that move the masses originate from the 2 balls according to the formula: force = G x mass x mass / distance squared. G = 6.674 28 (+/- 0.000 67) x 10-11 m3 kg-1 s -2. A wall is located at the right side of the room and a powerful light beam is sent horizontally. It’s direction is at 90 degrees compare to a line between the 2 balls. The rectangular orange block represents that light. Because the light is on the other side of the wall it should have no effect on the forces between the 2 balls. When this is tried, a surprising effect is observed. When the light is activated, the 2 balls start to move away from one another. It seems that the gravitational force between them is getting weaker. There is nothing in present theories that predicted that strange effect. Similar experiments were done for over 1000 hours and the results were all the same. Something was happening that no theory could explain. If one ball is removed and we wait until the last one stop oscillating, the light is activated on the other side of the wall. The ball starts moving towards the hidden light. Another effect that was not expected. It seems that the light is attracting the ball even if the ball cannot ‘see’ the light. That effect cannot be explained by attractive gravitational forces but it can be easily explained by a pushing gravitational force. Here it is: We know that chinese people on the other side of the globe do not ‘fall’ in space but stay on the ground as we do here. That means there is gravity everywhere in space. Let us go back to the experiment with one ball and an horizontal beam of light. Gravity is pushing the ball from every directions, from left to right, from right to left, from up to down, from down to up. The wire prevents movement up and down but permits the ball to move from one side to the other. Without the light, the forces of gravity are equal from every opposite directions and the net force is zero and the ball stay put. When the light is activated, the gravity coming from the right side passes through the light beam before going to the ball. Some of that gravity is deflected towards the back of the room and the net result is that the gravity coming from the left side is now stronger than the gravity coming from the right side and that pushes the ball to the right, where the light beam is located. That does not mean that light attracts the ball, It means that light can deviate what causes gravity. That is a strong proof that gravity is a pushing force, not an attractive force. That also means that gravity can be partially blocked by light going at right angle compare to gravity. That also expaling why objects ard pushed downward on earth because gravity from space above is stronger that gravity coming from below because some was deviated by the atoms of our planet. That also explain why planets orbit the sun. Another document does explain that in more details.

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CC-MAIN-2019-13

https://uwaterloo.ca/applied-mathematics/events/masters-defence-fabian-germ-estimation-linear-and-semi

math

Fabian Germ | Applied Math, University of Waterloo Estimation for Linear and Semi-linear Infinite-dimensional Systems Estimating the state of a system that is not fully known or that is exposed to noise has been an intensely studied problem in recent mathematical history. Such systems are often modelled by either ordinary differential equations, which evolve in finite-dimensional state-spaces, or partial differential equations, the state-space of which is infinite-dimensional. The Kalman filter is a minimal mean squared error estimator for linear finite-dimensional and linear infinite-dimensional systems disturbed by Wiener processes, which is a stochastic process representing the noise. For nonlinear finite-dimensional systems the extended Kalman filter is a widely used extension thereof which relies on linearization of the system. In all cases the Kalman filter consists of a differential or integral equation coupled with a Riccati equation, which is an equation that determines the optimal estimator gain. This thesis proposes an estimator for semi-linear infinite-dimensional systems. It is shown that under some conditions such a system can also be coupled with a Riccati equation. To motivate this result, the Kalman filter for finite-dimensional and infinite-dimensional systems is reviewed, as well as the corresponding theory for both stochastic processes and infinite-dimensional systems. Important results concerning the infinite-dimensional Riccati equation are outlined and existence of solutions for a class of semi-linear infinite-dimensional systems is established. Finally the well-posedness of the coupling between a semi-linear infinite-dimensional system with a Riccati equation is proven using a fixed point argument.

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https://www.boardingschoolreview.com/top-twenty-schools-listing/highest-percentage-students-financial-aid/all-boys-schools

math

Highest Percentage of Students on Financial Aid - Among "All-Boys Schools", view the boarding schools with the highest percentage of students on financial aid (2018-19). - Keep in mind that challenging and creative education programs at boarding schools require a substantial amount of resources. - Generally, the full cost of educating a student at any boarding school exceeds the cost of tuition by several thousand dollars - schools rely on other revenue sources (e.g., endowments, annual giving, etc.) in addition to tuition revenue to cover their costs. - In that sense, all students at boarding schools receive financial aid. - When schools offer "financial aid" (in the traditional sense of the phrase) to reduce tuition, those reductions are in addition to the already subsidized cost of full tuition. - Keep this in mind when assessing the full value that boarding schools provide for your investment. McCallie School Photo - Through coordinate activities with our sister school GPS, our students interact with young women on a daily basis. Missouri Military Academy Photo - At MMA, students can broaden their skills and experiences. Elective courses and activities include band, drama, chorus, art and web design. Outside of the classroom, students can earn a private pilot's license via our aviation program. SCUBA, WWII reenactment, rifle team, photography and fencing...these are just a few more unique activities available to cadets. The Greenwood School Photo - "Makers of Vermont" is a signature program at The Greenwood School in VT, where students learn to craft metal objects and tools in our blacksmith's forge. Blue Ridge School Photo - Our lake offers students a chance to relax, fish, canoe - in the center of campus. Bridgton Academy Photo - Welcome to Bridgton Academy! Located in the Western Maine foothills and the heart of Maine's Lake Region, we've been preparing young men for the rigors of college and beyond since 1808. We Know PG. Fork Union Military Academy Photo - The cadets in parade formation. The first full dress parade of the year begins usually at the end of April and is on Sunday. Eaglebrook School Photo - Eaglebrook School teaches middle school boys the skills they will need for the rest of their lives. Learn more about our Core Skills at https://www.eaglebrook.org/coreskills. Saint Stanislaus College-Preparatory Photo - Mr. Anderson's chemistry class gets to perform cool experiments! Camden Military Academy Photo - TEAMWORK!! A lesson better learned outside the classroom. Involvement in extra curricular activities (clubs, sports or band) is not an option because these activities often teach valuable life lessons. Georgetown Preparatory School Photo - Boland Hall is the home to many of Prep's 110 resident students. Our campus is less than 5 miles from the Nation's Capital Marine Military Academy Photo - All Marine Military Academy (MMA)Cadets participate in public parades held throughout the year. All parades are held on the MMA Parade Grounds directly adjacent to the Iwo Jima Monument, located at 320 Iwo Jima Boulevard in Harlingen, next to Valley International Airport. MMA’s Honor Band and Silent Drill Team performs at most parades with the exception of summer camp and introductory training graduations. Highest Percentage of Students on Financial Aid (2018-19) Among All-Boys Schools (out of 33 schools) Sorted by range (alphabetically within range) - Boarding School % Students on Aid Grades Location - 1.-3. 60%-70% Grades: 11-12, PG PO Box 292 North Bridgton, ME 04057 4129 Lake Shore Road Hamburg, NY 14075 304 S. Beach Blvd. Bay St. Louis, MS 39520 Grades: 8-12, PG | 7-12, PG (day) 700 Route 22 Pawling, NY 12564 500 Christ School Road Asheville, NC 28704 Grades: 9-12, PG 1888 Brett Lane Saltsburg, PA 15681 22520 Mount Michael Road Elkhorn, NE 68022 Grades: 8-12, PG 45 Cottage Road Oakdale, CT 06370 425 South Lindbergh Blvd. St. Louis, MO 63131 Grades: 8-12, PG 3042 College St. Austinburg, OH 44010 404 Robin Hill Street Marlborough, MA 01752 Grades: 4-8 | K-8 (day) 215 N. Harbor Blvd. Anaheim, CA 92805 62 Alumni Drive Canaan, NH 03741 14 Greenwood Lane Putney, VT 05346 Grades: 7-12, PG 320 Iwo Jima Blvd Harlingen, TX 78550 Top School Listings by Category (2018-19) Learn how financial aid works in boarding schools. The wide range of fine academic, athletic, and other facilities found in boarding schools underscores the determination of schools to provide the very best for their students. You and I have experienced all kinds of travel glitches. Discuss your child's upcoming travel plans thoroughly so that your child will be safe and in charge of her travel. The result will be peace of mind for you both.

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CC-MAIN-2019-09

http://xetaichuyendung.info/battle-discrete-math-sets-win/

math

The Battle Over Discrete Math Sets and How to Win It The Upside to Discrete Math Sets Zeno created a string of paradoxes employing the new notion of infinitesimals to discredit the entire area of study and it’s those paradoxes which we are going to be taking a look at today. For this reason, you’re, in addition, a descendant of your grandparents. It’s not, as an example, an in-depth treatise on group theory. The very first semester is mainly a foundations and logic course composed of the initial five chapters of the text. Just logging in some forum once a while and read what people are speaking about. Late homework won’t be accepted, and (except in the event of a documented university conflict) makeup quizzes won’t be given. The aim is to analyze these statements either individually or within a composite way. Given the next set, choose the statement below that’s true. Given the next sets, choose the statement below that’s true. The Pain of Discrete Math Sets Many references are included for those who have to probe further into the subject that’s suggested if these methods should be applied. My notes assume that you’re employing the seventh edition. Finally, we will plot two examples. The calculator lacks the mathematical intuition that’s very practical for finding an antiderivative, but on the other hand it can try out a lot of possibilities in a short timeframe. term papers help In some games, the optimal strategy is to select a single action regardless of what your opponent does. To confirm your guess you should use the strong kind of induction. Actually, Stanford’s encyclopedia entry on set theory is a fantastic place to start. After a careful examination of the Earth, however, you may actually arrive at the conclusion this is reduntant. Computational thinking is something which you can discover how to develop overtime. Now, here are a few additional sets that likewise satisfy Peano’s axioms. The second factor has the 2 roots These 2 roots, which are the exact same as the ones found by the very first method, form the period-2 orbit. Otherwise, a suitable subset is precisely the same as a normal subset. What You Must Know About Discrete Math Sets The educational facet of discrete mathematics is every bit as important and deserves extended coverage by itself. Discrete Math has a large application in today’s mathematics, and generally utilised in decision mathematics. They can play a key role in this connection. Type make to create the program. Anything you’re going to be in a position to solve in math, you may also compose a program for it. Course content will be different. The War Against Discrete Math Sets You will only get a concise introduction to logic inside this class, but the mathematics utilized in logic are observed at the core of computer programming and in designing electrical circuits. In the realm of computers this logic is used when creating logic gates that are the hardware of all contemporary computers today. Category theory is likewise very helpful for clarifying things which appear more complicated using different approaches, in addition, it has lost of practical applications in computer science, so I think it’s a worthwhile topic of study. Then, utilizing the simple fact that it’s true for 7, show that’s also true for 8 etc.. This isn’t a comprehensive collection of the benefits of ebooks. Also, it can be any arbitrary problem, where we clearly understand where it’s applied. Whatever They Told You About Discrete Math Sets Is Dead Wrong…And Here’s Why The notation for the overall concept can fluctuate considerably. Graphing particular kinds of equations is covered extensively in the notes, however, it’s assumed that you comprehend the standard coordinate system and the way to plot points. It’s not sufficient to just understand a concept and go on to the next. A subsequent paper will give a basic, high-level review of the algorithm. These exercises precede other exercises that need the reader to determine which Principle to use or require utilizing both Principles. Category theory is a rather generalised sort of mathematics, it’s considered a foundational theory in the exact same way that set theory is. The Definitive Strategy to Discrete Math Sets Our program is extremely strict and requires elevated levels of private discipline, and compliance is an excellent indicator of the applicant’s capacity to conform with our intensive atmosphere. The formula is a bit more complicated if your events are dependent, that’s in case the probability of a single event effects another. Continuous data aren’t restricted to defined separate values, but might occupy any value on a continuous selection. If it isn’t, then a filter needs to be placed on the data before instituting the right analysis. With fish, you need to be mindful about combinations. All functions may be used as static function of DiscreteMath. The Do’s and Don’ts of Discrete Math Sets Similarly, 5 isn’t a perfect cube. Symbols are a concise method of giving lengthy instructions linked to numbers and logic. It could appear odd to define a set that comprises no elements. There are lots of ways to accomplish such a selection. After pasting, you might need to opt for the perfect font in your intended application to see all the symbols. Actually, when that mark is necessary, it is normal to use syntactic sets.

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CC-MAIN-2019-26

http://extropians.weidai.com/extropians.1Q98/0616.html

math

[ universal fecundity, life a side effect of universe procreation? ] > There are now not only scientists trying to prove this theory, but also > some that are trying to disprove it. This is interesting, but what are the odds in the game? I can imagine life will/have/has/had usurped multiverse by iterations of altruistic supercivilisations eventually producing lots of life-suitable universes (the best of all possible worlds?), but we, parasites of chance? Intuitively, it doesn't make sense. What are the arguments?

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CC-MAIN-2017-22

https://wiki.garrysmod.com/index.php?title=GM/PlayerSelectSpawn&oldid=21716

math

From Garry's Mod Revision as of 19:02, 13 October 2015 by Robotboy655 Called to determine a spawn point for a player to spawn at. The player who needs a spawn point The spawnpoint entity to spawn the player at Find a random spawn point function GM:PlayerSelectSpawn( pl ) local spawns = ents.FindByClass( "info_player_start" ) local random_entry = math.random( #spawns ) return spawns[ random_entry ] end

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https://www.leledlight.com/what-is-the-distribution-curve/

math

Light distribution curve of the definition: Distribution curve is actually a lamp or light source that emits light in the space distribution. It can record the luminous flux of lamps, light quantity, power, power factor, lamp size, lamp efficiency, including lamp manufacturers, models and other information. The most important thing of course, recorded the light in all directions the light intensity. Light distribution curve of the Category: Distribution curve according to their symmetry properties can generally be divided into: axial symmetry, symmetrical and asymmetrical light distribution. Axial symmetry: rotational symmetry is also called, refers to all directions with all the basic symmetrical light curves, the general downlight, mining lamps are like light distribution. Symmetry: When the lamps and C180 ° C0 ° light distribution profile symmetry, while C270 ° C90 ° and symmetric profiles with light, such a distribution curve is called symmetrical with light Asymmetric: means C0 ° – 180 ° and C90 ° – 270 ° with any one section of light asymmetry. Distribution curve in accordance with its beam angle is usually divided into: Narrow light distribution (40 °) Wide light distribution (> 40 °) In fact, there is no strict definition of the various manufacturers of wide, middle and narrow definitions are slightly different.

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CC-MAIN-2023-40

http://maisperformance2016.com/kindle/basic-electronics-math

math

By Clyde Herrick So much scholars getting into an electronics technician software realise arithmetic. uncomplicated Electronics Math offers is a realistic program of those fundamentals to digital conception and circuits. the 1st 1/2 uncomplicated Electronics Math presents a refresher of mathematical ideas. those chapters should be taught individually from or together with the remainder of the ebook, as wanted through the scholars. the second one 1/2 simple Electronics Math covers purposes to electronics. uncomplicated innovations of electronics math.Numerous difficulties and examples.Uses real-world functions. Read Online or Download Basic Electronics Math PDF Best circuits books Culled from the pages of CRCs hugely winning, best-selling The Circuits and Filters guide, moment variation, Passive, energetic, and electronic Filters provides a sharply centred, accomplished assessment of the elemental idea at the back of specialist purposes of those complicated filters. It offers a concise, handy connection with the main suggestions, types, and equations essential to study, layout, and are expecting the habit of large-scale platforms that hire a variety of sorts of filters, illustrated by way of common examples. This ebook might be of curiosity to students and researcher who decentralize the self right into a multiplicity of voices as a manner of accounting for mind's inherently cultural and ancient cloth. This publication should be used as a chief textual content in graduate classes in Cultural stories, Psychology of character, background of Psychology, Philosophy of brain, and Philosophy of Psychology. Glossy Circuit Placement: top Practices and effects describes complex suggestions in VLSI circuit placement that is some of the most very important steps of the VLSI actual layout move. actual layout addresses the back-end format level of the chip layout approach. As expertise scales down, the importance of interconnect optimization turns into even more vital and actual layout, fairly the location technique, is vital to interconnect optimization. - Frequency-Domain Characterization of Power Distribution Networks - VLSI electronics microstructure science - Routing Congestion in VLSI Circuits: Estimation and Optimization (Series on Integrated Circuits and Systems) - Organic Electronics II: More Materials and Applications - Plasticity in the Visual System: From Genes to Circuits - Electrochemical Components Additional info for Basic Electronics Math X(-y) 7. -5(x- y) 11. (a)(-b)(-c)(-y) 4. 5(-6) 8. - x ( y + z) 12. 3 Multiplicationof Signed Numbers with Exponents We have noted some elementary principles of operations with exponents in previous discussions of powers of ten. For example, a number that is raised to a power is called the base, and the power to which it is raised is called the exponent. The general rule for multiplication of numbers with exponents when the base is the same follows. 4. To calculate the product of numbers with exponents when the base is the same, express the base to a power that is the sum of the exponents. Resistance is symbolized by R, and ohms are symbolized by l~. Conductance is symbolized by G. 3k ohm resistor is connected across a 12 volt supply. 64 mA Exercises 8-1 Divide and express all answers with positive exponents: a3 12 1. m 7. 2. x5 m x3 8. a2 5. 6. R2 A x5 x---~ al/2 15. 10. 52 -53 16. 19. E-2 E2 73 ff 17. b2 b_ 3 63/2 X/~ 3-2 102 ~/4 26. 2 -3 4-2/3 ~r~ 27. 8 33/2 22. 23. 1 18. 25. 23 21. C-3/2 C 5/2 A -5 10 -3 20. E- 5 61/2 12. R-3 E3 14. 23 -22 11. 3 17 9. A3/2 4. 13. a3/2 3R 2 3. For example, voltage is normally measured with respect to ground or a common point called ground. Figure 7-1 illustrates a circuit in which the voltage values are both negative and positive with respect to ground. l + Positive voltage I The voltage is positive with respect to ground. _. Nelgative voltage + Figure 7-1 Voltages measured to ground. h respect to ground. AlgebraicMultiplicationof Monomialsand Polynomials 47 Exercises 7-1 Multiply the following: 1. (-2)(3) 5. 3(-4)5 9. 3(x-y) 2. (-3)(-6) 6.

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https://www.analystforum.com/t/monetize-equity/72922

math

2009 Level III Mock Exam: question 47: To monetize equity, one has to use the formula N_f = V_o * (1+r) ^T / (qf) = 75000000*(1.04)^.25/(2350*100)=322 2010 Level III Mock Exam: question 46: To monetize equity, here we have to use the formula N_f = ((beta_T - beta_0)/(beta_f))*(s/f) = ((0-.95)/1.15)*(60000000/117475) = 421.9 How do we decide which formula to use when? did 2010 mock specificly state “equitize”? or did the question asks to simply hedge the position? first calculation is really about equitizin. second part is mostly about hedging or shifting assets.around In both questions, they talk about “converting to cash for 3 months”. In q 47 2009 - they have a client who has 75 Mill - convert to cash -> this they have solved using the equitize cash approach. they also gave u the risk free rate. In Q 46 2010 - the client has a mixed porfolio 60 Mill Stock, 25 Mill Bonds - and they want to convert to cash, no risk free rate provided. So they have used the reduce beta to 0 formula. so are we of the opinion that the effect is the same (i.e. to neutralize any equity position and to earn the rfr), its just a matter of technique so it depends on what inputs they give you? No. the effect is not the same. That is what I was asking about. The results from two formulae are different. So it can cause trouble if both answers are among the choices, and you have all the inputs in the question - the risk free rate and all three beta’s. Actually, in Q46, 2010, they gave McAulay duration and Modified duration for treasury bonds futures, so you could back out the risk free rate from those. Can you post both the 2009 and 2010 afternoon mock exams please. Thanks

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http://www.picluck.net/tag/sleepysheep

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Home again after a long weekend in Masham. Worth the long drive (3 hours each way!) but now as tired as this sleeping Manx Loaghtan sheep! Fairly sure he won first in his class and not surprised really 😍 Lovely to catch up with @wooltops again with his stall of temptations and to spend time with @ywrachwyllt and @murray7216 and of course my trusty navigator @em_mcg47 #masham#mashamsheepfair#sheepeverywhere#manxloaghtan#sheep#yorkshire#sotired#sleepysheep The sheep had fun @orangecountryfairsocial they made lots of friends and slept most of the time. It's quite exhausting to be so cute. I got my favourite black coffee soap from @guardiansfarm (everyone needs some goat soap and hand cream!) Thanks guys! And I hatched a whole bunch of chicks for the fair. Now I don't know whether to say it's a good thing or a bad thing but more chicks than I expected hatched. I was thinking 6 or so chicks would hatch, but SURPRISE 18 hatched. #fair#fairseason2018#sheep#goatsoap#chicks#falltime#blackcoffee#sleepysheep My #MCM is this little dude. He’s a natural comedian, instinctual charmer, PB&J connoisseur. Not to mention, he has a great head of hair like his middle namesake. You make our hearts full, Morrison Lee. #HappyViking#Guncle#SleepySheep

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http://amerucy19.odin5.ru.net/847288/a-478091_7128cce07a2406e7c56b29a9e89be801.php

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https://studysoup.com/tsg/1044008/single-variable-calculus-early-transcendentals-8-edition-chapter-4-7-problem-79

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Find the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W. [Hint: Express the area as a function of an angle .] Step 1 of 3 Lecture Notes 4/18/16 Epitopes: antigenic determinants Generator of Diversity o Somatic recombination DNA deleted between randomly selected v and j segments Then transcription, RNA processing, and translation Finally have a light-chain polypeptide of variable and constant regions o Hyper mutation in the V domain B cells secrete antibodies Clonal selection and amplification o Antigen molecules make clone of memory cells and clone of plasma cells o Cast of cells B cells: make antibodies using the mechanisms we discuss T cells: 2 sorts, cytotoxic and t helper: helps amplification Textbook: Single Variable Calculus: Early Transcendentals Author: James Stewart Since the solution to 79 from 4.7 chapter was answered, more than 212 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Single Variable Calculus: Early Transcendentals, edition: 8. This full solution covers the following key subjects: . This expansive textbook survival guide covers 95 chapters, and 5427 solutions. Single Variable Calculus: Early Transcendentals was written by and is associated to the ISBN: 9781305270336. The full step-by-step solution to problem: 79 from chapter: 4.7 was answered by , our top Calculus solution expert on 03/19/18, 03:29PM. The answer to “Find the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W. [Hint: Express the area as a function of an angle .]” is broken down into a number of easy to follow steps, and 32 words.

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https://www.arxiv-vanity.com/papers/1607.01126/

math

Approach to Chandrasekhar-Kendall-Woltjer State in a Chiral Plasma We study the time evolution of the magnetic field in a plasma with a chiral magnetic current. The Vector Spherical Harmonic functions (VSH) are used to expand all fields. We define a measure for the Chandrasekhar-Kendall-Woltjer (CKW) state, which has a simple form in VSH expansion. We propose the conditions for a general class of initial momentum spectra that will evolve into the CKW state. For this class of initial conditions, to approach the CKW state, (i) a non-vanishing chiral magnetic conductivity is necessary, and (ii) the time integration of the product of the electric resistivity and chiral magnetic conductivity must grow faster than the time integration of the resistivity. We give a few examples to test these conditions numerically which work very well. In high energy heavy-ion collisions, two heavy nuclei are accelerated to almost the speed of light and produce very strong electric and magnetic fields at the moment of the collision Kharzeev et al. (2008); Skokov et al. (2009); Voronyuk et al. (2011); Deng and Huang (2012); Bloczynski et al. (2013); McLerran and Skokov (2014); Gursoy et al. (2014); Roy and Pu (2015); Tuchin (2015); Li et al. (2016a). The magnitude of magnetic fields can be estimated as , where and are the proton number and the radius of the nucleus respectively, is the the velocity of the nucleus and is the Lorentz contraction factor ( is the nucleon mass and is the collision energy per nucleon). In Au+Au collisions at the Relativistic Heavy Ion Collider (RHIC) with GeV, the peak value of the magnetic field at the moment of the collision is about ( is the pion mass) or Gauss. In Pb+Pb collisions at the Large Hadron Collider (LHC) with TeV, the peak value of the magnetic field can be 10 times as large as at RHIC. Such high magnetic fields enter strong interaction regime and may have observable effects on the hadronic events. The chiral magnetic effect (CME) is one of them which is the generation of an electric current induced by magnetic fields from of the imbalance of chiral fermions Kharzeev et al. (2008); Fukushima et al. (2008); Kharzeev et al. (2016). The CME and other related effects have been widely studied in quark-gluon plasma produced in heavy-ion collisions. The charge separation effect observed in STAR Abelev et al. (2009, 2010) and ALICE Abelev et al. (2013) experiments are consistent to the CME predictions, although there may be other sources such as collective flows that contribute to the charge separation Huang et al. (2015). The CME has recently been confirmed to exist in materials such as Dirac and Weyl semi-metals Son and Spivak (2013); Basar et al. (2014); Li et al. (2016b). In hot and dense matter an imbalance in the number of right-handed quarks and left-handed quarks may be produced through transitions between vacua of different Chern-Simons numbers in some domains of the matter. This is called chiral anomaly and is described by the anomalous conservation law for the axial current, where denotes the axial 4-vector current with being the chiral charge, and are the number of colors and flavors of quarks respectively, is the electric charge (in the unit of electron charge ) of the quark with flavor , denotes the field strength of the electromagnetic field and is its dual, is the strong coupling constant, denotes the field strength of the -th gluon with and is its dual. The first term on the right-hand-side of Eq. (1) is the anomaly term from electromagnetic fields while the second one is from gluonic fields. In Eq. (1) we have neglected quark masses. For electromagnetic fields we can write in the 3-vector form using , where and are the electric and magnetic 3-vector field respectively. The axial current breaks the parity locally and may appear in one event, but it is vanishing when taking event average. With such an imbalance, an electric current can be induced along the magnetic field, so the total electric current can be written as where and are the electric and chiral conductivity respectively. Note that is proportional to the difference between the number of right-handed quarks and left-handed quarks which breaks the parity but conserves the time reversal symmetry. This is in contrast with the electric conductivity which breaks the time reversal symmetry but conserves the parity. So the Ohm’s current is dissipative (with heat production) while the chiral magnetic current is non-dissipative. where the total helicity is defined by combining the magnetic helicity and the chiral charge , where . This means that the magnetic helicity and the chiral charge of fermions can be transferred into each other. The Chandrasekhar-Kendall-Woltjer (CKW) state is a state of the magnetic field which satisfies the following equation where is a constant. The CKW state was first studied by Chandrasekhar, Kendall and Woltjer Chandrasekhar (1956); Chandrasekhar and Kendall (1957); Chandrasekhar and Woltjer (1958); Woltjer (1958) as a force free state. We notice that in a plasma with the chiral magnetic current (2), if the Ohm’s current is absent, the system reaches a special CKW state with following the Ampere’s law. To our knowledge, this idea was first proposed in Chernodub (2010). But with the Ohm’s current, can the CKW state still be reached? This question can be re-phrased as: what are the conditions under which the CKW state can be reached in a plasma with chiral magnetic currents? In this paper we will answer this question by studying the evolution of magnetic fields with the Maxwell-Chern-Simons equations. In classical plasma physics, a state satisfying Eq. (5) is called the Taylor state or the Woltjer-Taylor state. It was first found by Woltjer Woltjer (1958) that the CKW state minimizes the magnetic energy for a fixed magnetic helicity. In toroidal plasma devices, such a state is often observed as a self-generated state called reverse field pinch with the distinct feature that the toroidal fields in the center and the edge point to opposite directions. Taylor Taylor (1974, 1986) first argued that the minimization of magnetic energy with a fix magnetic helicity is realized as a selective decay process in a weakly dissipative plasma when the dynamics is dominated by short wavelength structures. Taylor’s theory has been questioned and debated intensely, and alternative theories has been proposed Ortolani and Schnack (1993); Qin et al. (2012). It is certainly interesting that both classical plasmas and chiral plasmas have the tendency to evolve towards such a state, which suggests that the two systems may share certain dynamics features responsible for the emerging of the state. A brief discussion in this aspect is given in the paper as well. The paper is organized as follows. In Section II, we start with the Maxwell-Chern-Simons equations and define a global measure for the CKW state by the magnetic field and the electric current. In Section III, we expand all fields in Vector Spherical Harmonic (VSH) functions. The inner products have simple forms in the VSH expansion. The parity of a quantity can be easily identified in the VSH form. We give in Section IV the solution to the Maxwell-Chern-Simons equations for each mode in the VSH expansion. The conditions for the CKW state are given in Section V. In Section VI we test these conditions by examples including the ones with constant and self-consistently determined . We also generalize the momentum spectra at the initial time from a power to a polynomial pre-factor of scalar momentum in Section VII. The summary and conclusions are made in the last section. Ii Maxwell-Chern-Simons equations and CKW state We start from Maxwell-Chern-Simons equations or anomalous Maxwell equations, where we have included the induced current and neglected the displacement current . We have also dropped the external charge and current density. We assume that and depend on only. A similar equation for can also be derived but we will not consider it in the current study. To measure whether the CKW state is reached in the evolution of the magnetic field, we introduce the quantity where we have used the notation for the inner product for any vector field and . According to the Cauchy-Schwartz inequality we have , where the equality holds only in the case of when the CKW state is reached Qin et al. (2012). We assume that is a smooth function of . The condition that the CKW state is reached can be given by Note that should not exactly be equal to 1, since is a smooth function of and bounded by the upper limit 1. To see the time evolution of , it is helpful to write inner products in simple forms, which we will do in the next section. Iii Expansion in vector spherical harmonic functions In this section, we will expand all fields in the basis of Vector Spherical Harmonic function (VSH), with which we can put inner products into a simple and symmetric form. iii.1 Expansion in VSH The quantities we use to express in Eq. (11) are and . We can extend the series to include more curls, So the inner products can be written as with and are non-negative integers. To find an unified form for the fields in this series, we can expand in the Coulomb gauge in VSH where is the scalar momentum and are the quantum number of the angular momentum and the angular momentum along a particular direction respectively. are divergence-free vector fields which can be expressed in term of VSH Jackson (1999). The explicit form of can be found in, e.g., Ref. Hirono et al. (2015); Tuchin (2016). The orthogonal basis functions satisfy the following orthogonality relations, where . Note that themselves are CKW states satisfying and are divergence-free, , so we can expand any divergence-free vector fields in . We note that is real while and are complex. iii.2 Inner products We can put the general inner products into a simple form by using the orthogonality relation (16), where are defined by We note that are positive definite. In deriving Eq. (20) we have used the fact that is real so that . For convenience, we use the following short-hand notation for the integral From Eq. (20) it is easy to verify for . This is consistent to the identity . In Table (1), we list the VSH forms of some inner products that we are going to study later in this paper. iii.3 Parity and helicity From Eq. (17), the parity transformation is equivalent to the interchange of the and mode. In the series (14), the quantity is parity-even/parity-odd (P-even/P-odd) for odd/even . For instance, , and are P-odd, P-even and P-odd respectively. Also the inner product is P-even/P-odd for even/odd , see the last column of Table (1) where the magnetic helicity is P-odd, and the magnetic energy is P-even. For a momentum spectrum containing only , the helicity is positive. If such a magnetic field can approach the CKW state, it means because is also positive for mode. In contrast it would mean for a momentum spectrum containing only . Iv Solving Maxwell-Chern-Simons equations in VSH where is the electric resistivity. The solution of is in the form where denote the values at the initial time , and and are defined by Note that both and are positive. Alternatively we can rescale time by using as a new evolution parameter, and rewrite , i.e., is the integrated value of from to . There is a competition between and for approaching or departing the CKW state. Large values of is favored for the CKW state. We will show that it is indeed determined by the increasing ratio of to . Note that in Eq. (25) changing is equivalent to interchanging the positive and negative modes, therefore we can assume in this paper without loss of generality. V Conditions for CKW state In this section we will study the evolution of the fields in the basis of VSH and look for the conditions for the CKW state. From Eq. (11) we obtain in VSH, To verify , it is better to rewrite in a more symmetric form, The difference between the denominator and numerator is where we have used in the first inequality. From the inequality (29) it is obvious that , where the equality holds for and one of and is zero. Therefore we have two conditions under which is satisfied: where is the central momentum of during evolution. Both conditions can be physically understood. The first condition is actually the presence of (we have assumed ), which makes positive modes grow with time while negative modes decay away. It means that the CKW state should contain only positive (or negative) helicity mode only, which is reasonable because the CKW state is the eigenstate of the curl operator. For the second condition, we notice that bases themselves are CKW states from Eq. (17), therefore one single mode in the expansion (19) is natually the CKW state. The authors of Ref. Hirono et al. (2015) observed in the evolution to the CKW state. However, the delta function is not well defined mathematically, so the second condition is hard to implement and we must find a better one to replace it. where the time functions are defined by Here are the powers of in the integrals for , and , respectively. If the initial spectrum functions contain only ( is a real number), or ( and are real constants), the integrals are just Gaussian-like integrals and easy to deal with. In this paper we assume that take the following form A typical example of magnetic fields expressed in such a form is the Hopf state Irvine and Bouwmeester (2008). Although this assumption narrows the scope of , it is still general enough: these three kinds of functions are widely used in other fields of physics. It is natural to combine with in the integrand of and re-define the time function as where is the power of in the initial spectrum functions . Note that does not converge, so we assume . By changing the integral variable where is always positive by definition, we can rewrite in the form where are defined by with the time functions by We rewrite in Eq. (31) as a function of through , We give relevant properties of in Appendix A. One property is that are monotonically increasing functions of , which approach zero at , but rise sharply to at . By Eq. (37), if grows faster than with , we have as . As time goes on, associated with positive modes will grow up but associated with negative modes will decay away. At , Eq. (38) becomes This fulfills the first condition for the CKW state in (30), i.e. only the positive modes survive at the end of the time evolution. So we can summarize the conditions for the CKW state to be reached in time evolution: Note that or plays an essential role: it makes negative modes more and more suppressed while making positive modes blow up as time goes on. At the same time it makes at so that . In heavy-ion collisions, and are decreasing functions of as the result of the expansion of the QGP matter. It is natural to assume that and fall with time in power laws Tuchin (2013), and , where . This can be justified by the fact that MeV Ding et al. (2011) and Kharzeev and Warringa (2009), where both the temperature and the chiral chemical potential decrease with time in power laws in expansion. In this case we have , following Eq. (26), and , the condition for the CKW state now becomes The above condition is very easy to check and it is one of the most useful and practical criteria in this paper. The fact that a large will bring the system to the CKW state shows that a non-Ohmic current may play a crucial role in the process of reaching the CKW state. This suggests that in classical plasmas systems, a non-Ohmic current, e.g., the Hall current, could produce the same effect. In a classical system with the Hall current and negligible flow velocity, the evolution of the magnetic field is governed by where is the density of the plasma, is electron charge, and the last is the Hall current term. Obviously, with a small resistivity, the system approaches equilibrium when the CKW state is reached. Vi Examples and tests of conditions In this section we will look at examples of the CKW state to test the conditions we propose in the last section. vi.1 With only As the first example, let us consider an initial spectrum with only without . We assume has the following form, where characterizes the length scale of the magnetic field, is the initial magnetic helicity. The normalization constant is chosen to be which gives the initial magnetic helicity of the spectrum, From Eq. (39) we obtain is given by Eq. (37) with and . Here we have suppressed the superscript of and simply denote . To verify our conditions for the CKW state, we consider following cases: In case a) both and are constants, which is used in Refs. Tuchin (2015); Li et al. (2016a) to calculate the magnetic field in medium. In this case, we have and in late time which satisfies the condition , and we can see the effect of non-vanishing constant . Such an effect can be seen by comparing with case b) in which we switch off . In case c) is still a constant as same as case a), but is chosen to break the condition with and . In case d), the values and are used in Refs. Tuchin (2013); Yamamoto (2016), which are thought to be more reasonable in heavy-ion collisions. But we note that in real situations of heavy-ion collisions, the time behaviors of and can be very complicated (may not follow power laws), but our conditions in (41) are still applicable. For numerical simulation, we choose and . The results are shown in Fig. 1. Indeed in case a) and d), the CKW state can be reached. As goes from to , according to Eqs. (46, 47, 66), evolves from to , and evolves from 0.8 to 1. In case b) and c) the condition (42) is not satisfied, the CKW state is inaccessible. Indeed the simulation shows that it is true since tends toward 0 and in case b) and c) at , respectively. Even though and increase with , we have and at corresponding to and respectively. All these results show that the conditions work well. But we should point out that constant and or even the power law decayed and may not be physical since once persists for a long time, growing faster than will make some physical quantities diverge. We look at the magnetic helicity and the magnetic energy , The numerical results of the magnetic helicity are shown in Fig. 2. The results of the magnetic energy are similar. In case b) and c), since and converge to constants, but keeps growing, both and finally decay to zero following Eq. (49). However in case a) and d), and increase to in late time, and from Eq (65) grow much faster than to make and blow up. From Eq. (24), we see that the spectrum grows exponentially in time for , such an instability has been discussed in Tuchin (2015); Manuel and Torres-Rincon (2015), see also Akamatsu and Yamamoto (2013). This instability is the source of the divergence of and . Such an unphysical inflation can be understood: the appearance of in the induced current leads to the positive feedback that the magnetic field itself induces the magnetic field. If we put no constraint on , as the result, the magnetic field will keep growing and finally blow up at some time. This of course breaks conservation laws. One way to avoid such divergences is to implement conservation laws in the system. This is the topic of the next subsection. vi.2 With only and dynamical We now consider imposing the total helicity conservation in Eq. (4). This has been implemented in Ref. Manuel and Torres-Rincon (2015); Hirono et al. (2015). Here we focus on the approach to the CKW state in evolution. For simplicity, we can parameterize as where and (total helicity) are constants. From Eq. (50), we see that the requirement leads to . The initial spectrum is assumed to be the same as Eq. (44), so we have . The parameters are chosen to be , , and , where is given by Eq. (45). Since is a constant, we have . We can solve self-consistently through , where we have used and that depends on through in Eq. (49). The numerical results for , and are presented in Fig. 3. For comparison, we also show the result for constant with . In both cases, at the beginning, is not large enough to make grows faster than , which makes in Eq. (49) decrease with time. After grows large enough as time goes on, starts to increase after reaching a minimum. In the case of dynamical , according to Eq. (50), and are complementary to each other to make up a seesaw system. In this system, the decreasing of at the beginning raises the value of and makes and grow faster. As the result, the turning point comes earlier than the case of constant . As keeps growing, drops down leading to slower increase of , which makes grows slower. At the end, is saturated to instead of blowing up. Let us look at the asymptotic time behavior of as . As the magnetic helicity is saturated to following Eq. (49), with and at late time, we obtain where and is called product logarithm, which is the inverse function of . From , we obtain at very large , Since the term increases with , is always growing faster than . Thus the conditions are satisfied and the CKW state can be reached. Taking a derivative of with respect to , we obtain at late time from Eq. (53), We have also looked at a general spectrum for at initial time, where the normalization constant is determined by the initial magnetic helicity . We assume obeying the power law decay in time, which gives . In this case, solving Eq. (51) gives the late time asymptotic behavior, Again we see that grows faster than and the CKW state can be finally reached. vi.3 With mixed helicity In this example we consider both positive and negative modes. We will show that only the positive mode survives while the negative mode decays away in late time. Let us consider the most extreme case in which the initial spectra of the positive and negative modes are the same. We take the following initial spectra for , It is obvious that the initial magnetic helicity is zero. Since , is given by Eq. (38) with and . Obviously at the initial time we have because .

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http://www.ask.com/web?qsrc=6&o=102140&oo=102140&l=dir&gc=1&q=How+Do+Pharmacists+Use+Math+in+Their+Career

math

Pharmacy technicians use basic math and algebra in their jobs. ... Pharmacy Technician Interview Questions · How Do Nurses Use Math in Their Jobs? What Are ... As a pharmacy technician, you will use your math skills daily on the job. Pharmacists uses basic algebra, fractions and percentages in their daily work, ... As such, a key part of the job involves the calculation and conversion of weights ... Oct 31, 2015 ... The pharmacist has to compute how many pills, of a given ... How do mathematicians feel when they work in a job that uses very little math? To pursue a career as a pharmacist, you will need several semesters of ... A pharmacist uses math to dispense prescriptions, determine dosage levels and prepare ... acceptance requirements and make a plan for their undergraduate Many people hear “pharmacy” and automatically think that the career area is ... Conversely, just because you did not do well in math does not mean that you will not be a good technician. ... You have to use it daily in order for it to become second nature. ... If his or her explanation does not make sense, ask someone That's quite a lot for a career that, in most areas, does not even require you to have a ... Math: Pharmacy work involves basic algebra, fractions and percentages. ... and calculations are done using the metric system, not the imperial system of ... basic understanding of the chemistry involved and its effects on the human body. Chemists study the composition of matter and its properties such as density, acidity, size and shape. ... It's a career for people who think about the future." ... maths is a vital instrument for the pharmacist to calculate the dosage of medication for the ... How does an economist use math in his or her work? yes economist use math and statistics in their work. An economist uses calculus to do Both doctors and nurses use math every day while providing health care for ... Doctors must be able to do these calculations mentally with speed and accuracy. ... This will determine how often the patient needs to take their medication in Skills and Knowledge for Pharmacist. ... practitioners and provide information to patients about medications and their use. ... Is Pharmacist the right career path for you? ... Mathematics - Using mathematics to solve problems. ... seeing to the appropriate use of equipment, facilities, and materials needed to do certain work.

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https://awordtotheworld.com/entrepreneurship/is-mathematics-important-for-entrepreneurship.html

math

An entrepreneur is concerned with the setting up of a new enterprise. Cost calculation is the most important aspect where he needs Mathematics, especially operational research. He is also concerned with the purchase of construction material, raw material for the manufacture of his products. Is maths required in entrepreneurship? This is not surprising since the knowledge of mathematics affords an individual the opportunity to be able to calculate profit and loss, prepare balance sheets, make cash plans, compute sales forecasts, prepare expense budgets, ratio analysis, break-even analysis, quality control and sales which are basic mathematics … What is the importance of mathematics in entrepreneurship? Understanding basic business math is necessary for profitable operations and accurate record keeping. Knowing how to add, subtract, multiply, divide, round and use percentages and fractions is the minimum you need to price your product and meet your budget. What type of math do entrepreneurs need? Arithmetic. Arithmetic is part of the everyday life of a business owner. You’ll need to be able to add, divide, subtract and multiply numbers to calculate your earnings, provide clients with estimates and pay your employees. Can I become entrepreneur without maths? Math Knowledge Helps An entrepreneur with average mathematical skills can manage his entrepreneurial work efficiently. Nonetheless, most successful entrepreneurs are good at mathematics. Entrepreneurs with a sound knowledge of math often perform better. How math is used in business? Commercial organizations use mathematics in accounting, inventory management, marketing, sales forecasting, and financial analysis. … It helps you know the financial formulas, fractions; measurements involved in interest calculation, hire rates, salary calculation, tax calculation etc. Do entrepreneurs need calculus? Honestly, apart from knowing basic algebra and trigonometry, there’s absolutely nothing you need to know to learn calculus. Honestly, apart from knowing basic algebra and trigonometry, there’s absolutely nothing you need to know to learn calculus. Your calculus lesson should start with basic knowledge about functions. Why is math important in marketing? The use of math in Marketing is not just practical, it is essential. Math helps us make real decisions based on evidence, and not just on what our gut is telling us. The more we marketing professionals accept math as a driving component in our industry, the more integrity our decisions will display. What type of math is used in business? Mathematics typically used in commerce includes elementary arithmetic, such as fractions, decimals, and percentages, elementary algebra, statistics and probability. Business management can be made more effective in some cases by use of more advanced mathematics such as calculus, matrix algebra and linear programming. Which is better maths or entrepreneurship? I would personally recommend that you should definitely opt for maths as in entrepreneurship you would need to do a lot of calculation and ensure that you are making profits and return on investment a s in business it is all out numbers, facts and figure which can be interpreted in quantitative and quantitative manner. Can I do BBA in entrepreneurship without maths? BBA is one of the most sought after administration courses in India. There are a number of specializations available under the course which you can choose from. … A subcategory of BBA, BBA without Maths is being pursued by management aspirants these days. Do you have to be good at math to be good at business? While math certainly has a place in business activities, you don’t have to love advanced mathematics to make a good business professional. Generally, the math coursework required to attain a business degree is on the simpler side, though it is college-level.

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http://livingbroccoli.com/pdf/complex-numbers-and-vectors

math

By Les Evans 'To have the braveness to imagine outdoors the sq., we have to be intrigued by way of a problem.' advanced Numbers and Vectors attracts at the strength of intrigue and makes use of beautiful purposes from navigation, international positioning structures, earthquakes, circus acts and tales from mathematical heritage to provide an explanation for the maths of vectors and the discoveries in complicated numbers. the 1st a part of complicated Numbers and Vectors presents academics with historical past fabric, principles and instructing ways to advanced numbers; types for complicated numbers and their geometric and algebraic homes; their position in supplying completeness with recognize to the answer of polynomial equations of a unmarried complicated variable (the primary theorem of algebra); the specification of curves and areas within the advanced aircraft; and easy variations of the advanced airplane. the second one a part of this source presents an advent to vectors and vector areas, together with matrix illustration; covers vectors in - and three-dimensions; their program to specification of curves; vector calculus and their simple program to geometric evidence. know-how has been used in the course of the textual content to build photographs of curves, graphs and and 3 dimensional shapes. Read Online or Download Complex numbers and vectors PDF Similar number theory books This publication is an exploration of a declare made via Lagrange within the autumn of 1771 as he embarked upon his long "R? ©flexions sur l. a. answer alg? ©brique des equations": that there were few advances within the algebraic answer of equations because the time of Cardano within the mid 16th century. That opinion has been shared by way of many later historians. Tracing the tale from its earliest assets, this booklet celebrates the lives and paintings of pioneers of recent arithmetic: Fermat, Euler, Lagrange, Legendre, Gauss, Fourier, Dirichlet and extra. contains an English translation of Gauss's 1838 letter to Dirichlet. Algebraic Operads: An Algorithmic significant other offers a scientific remedy of Gröbner bases in different contexts. The ebook builds as much as the speculation of Gröbner bases for operads a result of moment writer and Khoroshkin in addition to quite a few purposes of the corresponding diamond lemmas in algebra. The authors current a number of themes together with: noncommutative Gröbner bases and their functions to the development of common enveloping algebras; Gröbner bases for shuffle algebras which are used to resolve questions about combinatorics of diversifications; and operadic Gröbner bases, vital for purposes to algebraic topology, and homological and homotopical algebra. - Dynamics and Analytic Number Theory - Analytic Number Theory [lecture notes] - An Introduction to Models and Decompositions in Operator Theory - The Fourier-Analytic Proof of Quadratic Reciprocity - Knots and Primes: An Introduction to Arithmetic Topology (Universitext) - A Variational Inequality Approach to free Boundary Problems with Applications in Mould Filling Extra info for Complex numbers and vectors However, it was his work that converted a geometric representation of mathematics into the format with which we are more 30 CHAPTER 3 Secrecy, contrivance and inspiration familiar, an algebraic representation. He achieved this through the use of the cartesian plane, which used a grid system to locate any point on the Euclidean plane. A locus of points can therefore be described by the use of simple equations. Earlier we used the equation y = ax2 + bx + c to describe a particular conic section, the parabola. This could prove to be valuable when we consider the multiplication of complex numbers because, when complex numbers are multiplied, points are both dilated from and rotated about the origin. This should encourage us to multiply two complex numbers expressed in polar form. Let z1 = r1 _cos ^q1h + i sin ^q1hi and z2 = r1 _cos ^q2h + i sin ^q2hi . z1 z2 = r1 _cos ^q1h + i sin ^q1hi r2 _cos ^q2h + i sin ^q2hi = r1 r2 _cos ^q1h cos ^q2h - sin ^q1h sin ^q2h + i sin ^q1h cos ^q2h + i sin ^q2h cos ^q1hi = r1 r2 __cos ^q1h cos ^q2h - sin ^q1h sin ^q2hi + i _sin ^q1h cos ^q2h + sin ^q2h cos ^q1hii Using the compound angle formula for the circular functions: cos ^q1h cos ^q2h - sin ^q1h sin ^q2h = cos ^q1 + q2h and sin ^q1h cos ^q2h + sin ^q2h cos ^q1h = sin ^q1 + q2h gives: z1 z2 = r1 r2 __cos ^q1h cos ^q2h - sin ^q1h sin ^q2hi + i _sin ^q1h cos ^q2h + sin ^q2h cos ^q1hii = r1 r2 _cos ^q1 + q2h + i sin ^q1 + q2hi = r1 r2 cis^q1 + q2h 41 MATHSWORKS FOR TEACHERS Complex Numbers and Vectors This result can also be interpreted in terms of compositions of transformations—two dilations and two rotations. Ii Find the distance from the origin to each of z1, z2 and z1 z2 . iii Find the angles between the Re(z)-axis and the line intervals that join the points z1, z2 and z1 z2 to the origin. c i Plot z1, z3 and z1 z3 on an Argand diagram. ii Find the distance from the origin to each of z1, z3 and z1 z3 . iii Find the angles between the Re(z)-axis and the line intervals that join the points z1, z3 and z1 z3 to the origin. d i Plot z2 , z3 and z2 z3 on an Argand diagram. ii Find the distance from the origin to each of z2 , z3 and z2 z3 . Complex numbers and vectors by Les Evans

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http://www.dietsinreview.com/questions/can-taking-apple-cider-vinegar-really-help-in-treating-a-bladder-infection

math

Diets in Review Question Can taking apple cider vinegar really help in treating a bladder infection? Genella asked this question Mar 27th, 2013 10:59 pm I took 2tlb. Of Braggs apple cider vinegar and I seem to be so much better now.Invite Your Friends to Answer This Question

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http://visitbulgaria.info/gy6-carburetor-diagram/

math

luxury gy6 carburetor diagram or diagram wiring diagram blogs diagram diagram 89 150cc gy6 carb vacuum diagram. unique gy6 carburetor diagram for moped uretor diagram wiring diagram blogs diagram diagram 75 gy6 engine carburetor diagram. inspirational gy6 carburetor diagram for dog wiring diagram go kart buggy depot technical center diagram fuel line diagram 87 gy6 engine carburetor adjustment. luxury gy6 carburetor diagram or ruckus nps online scooter service manual rancher diagram ruckus motor diagram 12 gy6 carburetor hose diagram. idea gy6 carburetor diagram for uretor diagram wiring diagram blogs engine diagram diagram 77 gy6 carburetor problems. unique gy6 carburetor diagram for carburetor 59 gy6 carburetor vacuum diagram. amazing gy6 carburetor diagram and r x rear drum brake free play inspection adjustment 72 gy6 carburetor adjustment screw. idea gy6 carburetor diagram or diagram free wiring diagram for you panther wiring diagram wiring diagram 71 gy6 50cc carburetor adjustment. beautiful gy6 carburetor diagram for 2 stroke scooter parts 81 gy6 150cc carburetor diagram. unique gy6 carburetor diagram and connections and diagram com scooter forums cc vacuum line diagram cc uretor diagram 76 gy6 carb needle adjustment. beautiful gy6 carburetor diagram and ruckus engine diagram wiring database library ruckus battery ruckus engine diagram 38 gy6 carburetor adjustment. luxury gy6 carburetor diagram and its not a choke after all 31 gy6 50cc carburetor adjustment. elegant gy6 carburetor diagram or r x bodywork removal and installation side covers fender 43 gy6 carburetor vacuum diagram. gy6 carburetor diagram or scooter carburetor adjustment fuel air ratio 48 gy6 150cc carb adjustment. ideas gy6 carburetor diagram for r x periodic maintenance schedule throttle cable free play adjustment specification 49 gy6 150cc carb diagram. ideas gy6 carburetor diagram for engine 41 150cc gy6 carb vacuum diagram. unique gy6 carburetor diagram for carburetor vacuum diagram awesome spare parts of engine 16 gy6 150cc carb adjustment. awesome gy6 carburetor diagram and uretor cleaning guide buggy depot technical center com diagram 39 gy6 150cc carb adjustment. idea gy6 carburetor diagram or r x fuel tank removal fuel gauge troubleshooting carburetor jetting 48 gy6 150cc carburetor adjustment. luxury gy6 carburetor diagram and carburetor float level adjustment guide 98 gy6 engine carburetor diagram. inspirational gy6 carburetor diagram for carburetor diagram the portal and forum of wiring carburetor diagram 55 scooter carburetor adjustment 50cc. The structure of this humble diagram was formally developed by the mathematician John Venn, but its roots go back as far as the 13th Century, and includes many stages of evolution dictated by a number of noted logicians and philosophers. The earliest indications of similar diagram theory came from the writer Ramon Llull, whos initial work would later inspire the German polymath Leibnez. Leibnez was exploring early ideas regarding computational sciences and diagrammatic reasoning, using a style of diagram that would eventually be formalized by another famous mathematician. This was Leonhard Euler, the creator of the Euler diagram. Euler diagrams are similar to Venn diagrams, in that both compare distinct sets using logical connections. Where they differ is that a Venn diagram is bound to show every possible intersection between sets, whether objects fall into that class or not; a Euler diagram only shows actually possible intersections within the given context. Sets can exist entirely within another, termed as a subset, or as a separate circle on the page without any connections - this is known as a disjoint. Furthering the example outlined previously, if a new set was introduced - birds - this would be shown as a circle entirely within the confines of the mammals set (but not overlapping sea life). A fourth set of trees would be a disjoint - a circle without any connections or intersections. Usage for Venn diagrams has evolved somewhat since their inception. Both Euler and Venn diagrams were used to logically and visually frame a philosophical concept, taking phrases such as some of x is y, all of y is z and condensing that information into a diagram that can be summarized at a glance. They are used in, and indeed were formed as an extension of, set theory - a branch of mathematical logic that can describe objects relations through algebraic equation. Now the Venn diagram is so ubiquitous and well ingrained a concept that you can see its use far outside mathematical confines. The form is so recognizable that it can shown through mediums such as advertising or news broadcast and the meaning will immediately be understood. They are used extensively in teaching environments - their generic functionality can apply to any subject and focus on my facet of it. Whether creating a business presentation, collating marketing data, or just visualizing a strategic concept, the Venn diagram is a quick, functional, and effective way of exploring logical relationships within a context. Logician John Venn developed the Venn diagram in complement to Eulers concept. His diagram rules were more rigid than Eulers - each set must show its connection with all other sets within the union, even if no objects fall into this category. This is why Venn diagrams often only contain 2 or 3 sets, any more and the diagram can lose its symmetry and become overly complex. Venn made allowances for this by trading circles for ellipses and arcs, ensuring all connections are accounted for whilst maintaining the aesthetic of the diagram. A Venn diagram, sometimes referred to as a set diagram, is a diagramming style used to show all the possible logical relations between a finite amount of sets. In mathematical terms, a set is a collection of distinct objects gathered together into a group, which can then itself be termed as a single object. Venn diagrams represent these objects on a page as circles or ellipses, and their placement in relation to each other describes the relationships between them. Commonly a Venn diagram will compare two sets with each other. In such a case, two circles will be used to represent the two sets, and they are placed on the page in such a way as that there is an overlap between them. This overlap, known as the intersection, represents the connection between sets - if for example the sets are mammals and sea life, then the intersection will be marine mammals, e.g. dolphins or whales. Each set is taken to contain every instance possible of its class; everything outside the union of sets (union is the term for the combined scope of all sets and intersections) is implicitly not any of those things - not a mammal, does not live underwater, etc. Other Collections of Gy6 Carburetor Diagram

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https://learn.lif.co.id/40069/

math

Celsius also known as centigrade is a unit of measurement for temperature. More temperatures starting with 35 degrees in Celsius Note. T F T C 18 32. 35 celsius. 35 below zero from C to F. To find this answer use the formula C x 95 32 F where C represents degrees Celsius and F is degrees Fahrenheit. How to convert Celsius to Fahrenheit. Not the temperature at which it freezes which is different. How hot is 35 degrees Celsius. Normal variation of temperature Celsius and Fahrenheit Temperature in the mouth or mouth 355 C to 375 C 959 F to 995 F Temperature under armpit or axillary 365 C to 375 C 978 F to 995 F Temperature in the ear tympanic 358 C to 380 C 964 F to 1004 F Temperature in the rectum or rectal. Common conversions from Celsius to Fahrenheit. Convert 20 degrees Celsius to degrees Fahrenheit. 35 degrees Celsius is equal to 95 degrees Fahrenheit. Water freezes at 0 Celsius and boils at 100 Celsius. 35 Celsius C 30815 Kelvin K Celsius. The temperature T in degrees Fahrenheit F is equal to the temperature T in degrees Celsius C times 95 plus 32. Celsius or centigrade is used to measure temperatures in most of the world. T F T C 95 32. 3530 C 9554 F. In 1948 the 9th CGPM. 0 degrees Celsius is equal to 32 degrees Fahrenheit. Conversely to perform the conversion from degrees Fahrenheit to degrees Celsius manipulate the above equation to F 32 X 59 C. 35 Celsius 30815 Kelvin exact result About. F C 95 32. C can refer to a specific temperature on the Celsius scale as well as a unit to indicate a temperature interval a difference between two temperatures or an uncertainty. What is 35 Celsius in Kelvin. Fahrenheit F Celsius x 18 32 this example shows how to convert a temperature of 35 degrees Celsius to Fahrenheit 35 C to F. The degree Celsius is a unit of temperature on the Celsius scale a temperature scale originally known as the centigrade scale. The Celsius scale is nowadays set in such a way that Zero degrees C is the temperature at which ice melts note. 0 C 32 F. At the other end of the scale 100 degrees Celsius is the boiling point of water. The degree Celsius symbol. The degree Celsius symbol. The degree Celsius symbol. Convert between Fahrenheit and Celsius. 3560 C 9608 F. The scientific definition of Celsius is now defined against degrees Kelvin. Fahrenheit is a scale commonly used to measure temperatures in the United States. Celsius or centigrade is used to measure temperatures in most of the world. 35 Celsius C 95 Fahrenheit F Celsius. How cold is minus 35 degrees Celsius. C can refer to a specific temperature on the Celsius scale as well as a unit to indicate a temperature interval a difference between two temperatures or an uncertainty. C can refer to a specific temperature on the Celsius scale or a unit to indicate a difference between two temperatures or an uncertaintyIt is named after the Swedish astronomer Anders Celsius 17011744 who developed a similar. To change 353 Celsius to Fahrenheit just need to replace the value C in the formula below and then do the math. 3500 C 9500 F. 35 Celsius 95 Fahrenheit exact result How hot is 35 degrees Celsius. In 1948 the 9th. Celsius also known as centigrade is a unit of measurement for temperature. 35 degrees Celsius converts to 95 degrees Fahrenheit How To Convert 35 Celsius To Fahrenheit Using the Celsius to Fahrenheit formula. Degrees Celsius invented by Anders Celsius are sometimes called Centigrade because the scale was defined between 0 and 100 degrees hence centi-grade meaning a scale consisting of 1100ths. Water freezes at 0 Celsius and boils at 100 Celsius. 353 c 9554 f 3531 c 95558 f 3532 c 95576 f 3533 c 95594 f 3534 c 95612 f 3535 c 9563 f 3536 c 95648 f 3537 c 95666 f 3538 c 95684 f 3539 c 95702 f 354 c 9572 f 3541 c 95738 f 3542 c 95756 f 3543 c 95774 f 3544 c 95792 f 3545 c 9581 f 3546 c 95828 f 3547 c 95846 f 3548 c. Write down the formula. Degrees Centigrade and degrees Celsius are the same thing.

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CC-MAIN-2021-49

https://physicscalculatorpro.com/rotational-kinetic-energy-calculator/

math

The Rotational Kinetic Energy Calculator is a free, portable calculator that calculates rotational kinetic energy using inputs such as angular velocity, the moment of inertia, and other variables. Get step-by-step instructions on how to calculate rotational kinetic energy, including formulas, units, and more. What is Rotational Kinetic Energy? Kinetic energy includes rotational energy. It describes the energy of a moving object. While conventional kinetic energy was associated with items moving in a straight line, rotational energy is associated with spinning objects. As a result, instead of using a constant speed in our computations, we must employ angular velocity. Remember that an object has both regular and rotational kinetic energy while it is in translation (going in a straight line) and rotation (turning around its axis). As a result, to determine its total energy, you must compute both numbers and add them together. Rotational Kinetic Energy Formula We may describe Rotational Kinetic Energy with a simple yet clear formula, just like any other form of energy. RE = 1/2 * I * ω² - RE = The rotational kinetic energy is measured in Joules. - I = The moment of inertia is measured in kilograms per square metre, and its units are kilograms per square metre. - ω = It is the angular velocity, which is measured in Hertz (Hz) or Revolutions Per Second (RPS) (RPM) For more concepts check out physicscalculatorpro.com to get quick answers by using the free tools available. How to Calculate Rotational Kinetic Energy? Learn how to calculate Rotational Kinetic Energy by following the easy procedure described below. As so, they are. - Step 1: To begin, determine the wheel's angular velocity. - Step 2: Determine the wheel's Moment of Inertia later. - Step 3: Once you have these two values, plug them into the rotational kinetic energy formula to determine the wheel's rotational kinetic energy. Rotational Kinetic Energy Examples Question 1:A round grindstone with a moment of inertia 1800 kg.m² is rotating at an angular velocity of 7.5 radians per sec. What is the rotational kinetic energy of a grindstone? Moment of Inertia = 1800 kg.m² Angular Velocity = 7.5 Rad/Sec Rotational Kinetic Energy = 1/2 * I * ω² RE = 1/2*1800*7.5² RE = 50,625 J FAQs on Rotational Kinetic Energy 1. How to find out the Rotational Kinetic Energy? Determine the angular velocity and moment of inertia of the wheel first. Then, quickly calculate Rotational Kinetic Energy by plugging them into the rotational kinetic energy formula. 2. What is the Earth's rotational kinetic energy as it orbits the sun? We now know that the earth takes 365 days to orbit the sun. In its orbit around the Sun, the Earth's rotational kinetic energy is 2.67 1033 J. 3. What is the formula for calculating rotational inertia? According to the formula rotational inertia = mass x radius^2, rotational inertia is a scalar, not a vector, and is dependent on the radius of rotation. The measure of an object's resistance to change in its rotation is called rotational inertia. 4. What is the Rotational Kinetic Energy Formula? RE = 1/2 I ω² is the formula for Rotational Kinetic Energy. 5. What is Rotational Kinetic Energy? Rotational Kinetic Energy is a type of kinetic energy that is created when an item rotates. It is a component of total kinetic energy. 6. How do you compute the earth's rotational kinetic energy? The rotational kinetic energy of the Earth's rotation about its axis at its core is one-half of the sphere's moment of inertia, which is two times the Earth's mass times its radius squared divided by five, and then multiplied by the Earth's angular velocity squared.

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http://forum.roulette30.com/index.php?topic=642.0

math

I know many of you think that if you could overcome the worst possible scenario in terms of results then you would consider it holy grail. But ask yourselves, even if there was such solution, would that be something that you likely follow precisely? Don't hurry to answer...! I bet 95 % of you wouldn't apply the solution, you see it's in our nature to idealise situations, persons...etc The way we consider about something is better than how actually is, that's why. Actually the solution of overcoming 135 loses with just 65 wins regardless of the even chance you select or the distribution of the loses/wins within the 200 results, has been given more than a century ago on the book "10 days at Monte Carlo on the bank's (casino) expense" Take a look on the following progression: 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 A total of 360 units are sufficient to overcome even the session from hell or any black swans you might encounter. The key is to understand that actually we don't have to overcome 135 loses, but 70. What really matters is the difference, deduct 65 from 135 and there you are 70, but I'm going to make it "quarters" for you. By using the progression above, with each and every win we are canceling 1 loss, so whether we experience 10 wins and 10 losses or 1 win and 1 loss it doesn't really matter because are canceling each other, also it doesn't matter in which order are going to occur. The above progression is quite simple, after 10 MORE losses than wins you add 1 unit to your bet and so on... I want to repeat the word 10 MORE, because it means that it doesn't have to be 10 losses in a row, it could be 7 wins against 17 losses. So after 70 more losses, you need 45 more wins in order to come on top. Since 65/135 is the worst possible scenario,after those 200 spins the things can go only better, in other words regression towards the mean. How long it could take till you have 45 net wins it's completely another matter, it could a few hours, many hours or even a day! You may encounter very long results like this: L L W W L W L L L W W W L W W W L L W L...and it goes back and forth, back and forth like a pendulum in a perpetual movement! For me would be like a torment, I would wish to lose 10 times more in order to raise my bet and get over with it!It's amazing to me how those people a century ago could actually apply such method!You could spend the whole day inside the casino, literally!The aftermath is:Even if it's valid and completely possible, does it worth your time??Personally speaking, someone must have nerves of steel, PLENTY of time and PATIENCE must be his middle name!I have to admit that I'm not that person!Some VERY valuable feedback provided by the user "UK" on this topic what is the most negative expectation we can encounter in 200 spins? It is always possible to get the permanence of horror right from the start. The program so far will answer the first question and also address the 2nd point in that it will tell you what % of sessions will start of with a loss and never get to a ve balance throughout the session. The program simulates even money roulette (betting on red) and the le partage rule (1.35% edge) with 100,000 sessions of length 200 spins. For flat betting: number of sessions always in a net loss = 6553 (6.55%) average peak gain within a round = 9.44 average peak loss within a round = -12.11 actual peak gain in 100,000 rounds = 58.00 actual peak loss in 100,000 rounds = -62.50 So 93.45% of the time you can expect to quit at some point within the session with a profit - even if it's only 1 unit. The second figure (average peak gain within a round) would suggest that if you get a profit of 9 or 10 units in the session and continue to play on you are making a bad bet, statistically speaking. The final figure (actual peak loss within a round) suggests that a bankroll of 60 - 70 units is sufficient. The program is work in progress and I intend to add more analysis including the number of "reversals" within a session, a reversal being a swing from ve to -ve balance or vice versa within a session. In theory by knowing the average number of reversal for a system you can then keep track of them and quit on a ve balance if you have "used up" your reversals in a session. I've experimented with various systems and the best so far in terms of being able to make a profit at some point (ie; the lowest % of sessions without making any profit at all throughout the session) is the Maxim principle. Note that according to the author this is only meant to be used for craps and also I haven't simulated any of the exit points. Using a maximum stake of 50u: number of sessions always in a net loss = 261 (0.26%) average peak gain within a round = 63.98 average peak loss within a round = -156.53 actual peak gain in 100,000 rounds = 122.00 actual peak loss in 100,000 rounds = -1692.50 A 99.74% chance of quitting with a profit, but of course this is offset by the hugely increased bankroll necessary and attendant risk involved. For a maximum stake of 10u (ie; the progression is 1,2,4,5,6,7,8,9,10) and start over when you get to the end. the results are: number of sessions always in a net loss = 1131 (1.13%) average peak gain within a round = 43.46 average peak loss within a round = -64.63 actual peak gain in 100,000 rounds = 121.50 actual peak loss in 100,000 rounds = -412.00 From my testing so far it seems that most volatile systems (those that offer more "reversals") are those that incorporate both -ve and ve progressions (like the Maxim principle)." Very valuable information from user "UK" and I'd like to thank him for sharing with us, but someone named obviously Max or Maxim or Maximilian stole that progression from the book "Monte Carlo anecdotes", see the "Fitzroy" system on page 141. "Maxim principle" has been (e)mailed across US as sure win method several years back, someone stole the intellectual property, rename it and tried to sell it as holy grail. But the most disappointing is that this progression rather than system, has been made MORE THAN A CENTURY AGO like the first on the beginning of this topic! My conclusion is that we are recycling VERY old knowledge/information and my question is: Is there any progress to our practical knowledge and methods since a CENTURY ago?? Or we just recycling the same as "new"?? The game and its rules have not changed, have we?!

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CC-MAIN-2017-17

https://www.arxiv-vanity.com/papers/hep-th/0612197/

math

Roulette Inflation with Kähler Moduli and their Axions We study 2-field inflation models based on the “large-volume” flux compactification of type IIB string theory. The role of the inflaton is played by a Kähler modulus corresponding to a 4-cycle volume and its axionic partner . The freedom associated with the choice of Calabi-Yau manifold and the non-perturbative effects defining the potential and kinetic parameters of the moduli brings an unavoidable statistical element to theory prior probabilities within the low energy landscape. The further randomness of initial conditions allows for a large ensemble of trajectories. Features in the ensemble of histories include “roulette trajectories”, with long-lasting inflations in the direction of the rolling axion, enhanced in number of e-foldings over those restricted to lie in the -trough. Asymptotic flatness of the potential makes possible an eternal stochastic self-reproducing inflation. A wide variety of potentials and inflaton trajectories agree with the cosmic microwave background and large scale structure data. In particular, the observed scalar tilt with weak or no running can be achieved in spite of a nearly critical de Sitter deceleration parameter and consequently a low gravity wave power relative to the scalar curvature power. The “top-down” approach to inflation seeks to determine cosmological consequences beginning with inflation scenarios motivated by ever-evolving fundamental theory. Most recent attention has been given to top-down models that realize inflation with string theory. This involves the construction of a stable six-dimensional compactification and a four-dimensional extended de Sitter (dS) vacuum which corresponds to the present-day (late-time) universe e.g., the KKLT prescription . Given this, there is a time-dependent, transient non-equilibrium inflationary flow in four dimensions towards the stable state, possibly involving dynamics in both sectors. Currently, attempts to embed inflation in string theory are far from unique, and indeed somewhat confused, with many possibilities suggested to engineer inflation, using different axionic and moduli fields [2, 3], branes in warped geometry , D3-D7 models , etc. [6, 7]. These pictures are increasingly being considered within a string theory landscape populated locally by many scalar fields. Different realizations of stringy inflation may not be mutually incompatible, but rather may arise in different regions of the landscape, leading to a complex statistical phase space of solutions. Indeed inflation driven by one mechanism can turn into inflation driven by another, e.g., , thereby increasing the probability of inflation over a single mechanism scenario. So far all known string inflation models require significant fine-tuning. There are two classes that are generally discussed involving moduli. One is where the inflaton is identified with brane inter-distances. Often the effective mass is too large (above the Hubble parameter) to allow acceleration for enough e-folds, if at all. To realize slow-roll inflation, the effective inflaton mass should be smaller than the Hubble parameter during inflation, . Scalar fields which are not minimally but conformally coupled to gravity acquire effective mass terms which prevents slow-roll. An example of this problem is warped brane inflation where the inflaton is conformally coupled to the four-dimensional gravity . A similar problem also arises in supergravity. The case has been constructed that has masses below the Hubble parameter which avoids this -problem at the price of severe fine-tuning . Another class is geometrical moduli such as Kähler moduli associated with 4-cycle volumes in a compactifying Calabi-Yau manifold as in [2, 3], which has been recently explored in and which we extend here to illustrate the statistical nature of possible inflation histories. Different models of inflation predict different spectra for scalar and tensor cosmological fluctuations. From cosmic microwave background and other large scale structure experiments one can hope to reconstruct the underlying theory that gave rise to them, over the albeit limited observable range. Introduction of a multiple-field phase space leading to many possible inflationary trajectories necessarily brings a statistical element prior to the constraints imposed by data. That is, a theory of inflation embedded in the landscape will lead to a broad theory “prior” probability that will be updated and sharpened into a “posteriori” probability through the action of the data, as expressed by the likelihood, which is a conditional probability of the inflationary trajectories given the data. All we can hope to reconstruct is not a unique underlying acceleration history with data-determined error bars, but an ensemble-averaged acceleration history with data-plus-theory error bars . The results will obviously be very dependent upon the theory prior. In general all that is required of the theory prior is that inflation occurs over enough e-foldings to satisfy our homogeneity and isotropy constraints and that the universe preheats (and that life of some sort forms) — and indeed those too are data constraints rather than a priori theory constraints. Everything else at this stage is theoretical prejudice. A general approach in which equal a priori theory priors for acceleration histories are scanned by Markov Chain Monte Carlo methods which pass the derived scalar and tensor power spectra though cosmic microwave background anisotropy data and large scale clustering data is described in . But since many allowed trajectories would require highly baroque theories to give rise to them, it is essential to explore priors informed by theory, in our case string-motivated priors. The old top-down view was that the theory prior would be a delta-function of the correct one and only theory. The new view is that the theory prior is a probability distribution on an energy landscape whose features are at best only glimpsed, with huge number of potential minima, and inflation being the late stage flow in the low energy structure toward these minima. In the picture we adopt for this paper, the flow is of collective geometrical coordinates associated with the settling down of the compactification of extra dimensions. The observed inflaton would be the last (complex) Kähler modulus to settle down. We shall call this . The settling of other Kähler moduli associated with 4-cycle volumes, and the overall volume modulus, , as well as “complex structure” moduli and the dilaton and its axionic partner, would have occurred earlier, associated with higher energy dynamics, possibly inflations, that became stabilized at their effective minima. The model is illustrated by the cartoon Fig. 1. We work within the “large volume” moduli stabilization model suggested in [11, 12, 13] in which the effective potential has a stable minimum at a large value of the compactified internal volume, in string units. An advantage of this model is that the minimum exists for generic values of parameters, e.g., of the flux contribution to the superpotential . (This is in contrast to the related KKLT stabilization scheme in which the tree-level is fine-tuned at in stringy units in order for the minimum to exist.) In this paper, we often express quantities in the relatively small “stringy units” , related to the (reduced) Planck mass where is Newton’s constant. In this picture, the theory prior would itself be a Bayesian product of a number of conditional probabilities: (1) of manifold configuration defining the moduli; (2) of parameters defining the effective potential and the non-canonical kinetic energy of the moduli, given the manifold structure; (3) of the initial conditions for the moduli and their field momenta given the potentials. The latter will depend upon exactly how the “rain down” from higher energies occurs to populate initial conditions. An effective complication occurs because of the so-called eternal inflation regime, when the stochastic kicks that the inflaton feels in an e-folding can be as large as the classical drift. This -model is in fact another example of stringy inflation with self-reproduction. (See for another case.) If other higher-energy moduli are frozen out, most inflationary trajectories would emerge from this quantum domain. However we expect other quantum domains for the higher-energy moduli to also feed the initial conditions, so we treat these as arbitrary. The Kähler moduli are flat directions at the stringy tree level. The reason this picture works is that the leading non-perturbative (instanton) and perturbative () corrections introduce only an exponentially flat dependence on the Kähler moduli, avoiding the -problem. Conlon and Quevedo focused on the real part of as the inflaton and showed that slow-roll inflation with enough e-foldings was possible. A modification of the model considered inflation in a new direction but with a negative result. The fields are complex, . In this paper we extend the model of to include the axionic direction . There is essentially only one trajectory if is forced to be fixed at its trough, as in . The terrain in the scalar potential has hills and valleys in the direction which results in an ensemble of trajectories depending upon the initial values of . The field momenta may also be arbitrary but their values quickly relax to attractor values. The paper considered inflation only along the direction while the dynamics in the direction were artificially frozen. We find motion in always accompanies motion in . In Kähler moduli models, there is an issue of higher order perturbative corrections. Even a tiny quadratic term would break the exponential flatness of the inflaton potential and could make the -problem reappear. However, the higher order terms which depend on the inflaton only through the overall volume of the Calabi-Yau manifold will not introduce any mass terms for the inflaton. Although these corrections may give rise to a mass term for the inflaton, it might have a limited effect on the crucial last sixty e-folds. In § 2 we describe the model in the context of type IIB string theory. In § 3 we address whether higher (sub-leading) perturbative corrections introduce a dangerous mass term for the inflaton. In § 4 we discuss the effective potential for the volume, Kähler moduli and axion fields, showing with 3 moduli that stabilization of two of them can be sustained even as the inflaton evolves. Therefore in § 5 we restrict ourselves to with the other moduli stabilized at their minima. § 6 explores inflationary trajectories generated with that potential, for various choices of potential parameters and initial conditions. In § 7 we investigate the diffusion/drift boundary and the possibility of self-reproduction. In § 8 we summarize our results and outline issues requiring further consideration, such as the complication in power spectra computation that follows from the freedom. 2 The Type IIB String Theory Model Our inflationary model is based on the “large-volume” moduli stabilization mechanism of [11, 12, 13]. This mechanism relies upon the fixing of the Kähler moduli in IIB flux compactifications on Calabi-Yau (CY) manifolds by non-perturbative as well as perturbative effects. As argued in [11, 12, 13], a minimum of the moduli potential in the effective theory exists for a large class of models. The only restriction is that there should be more complex structure moduli in the compactification than Kähler moduli, i.e. , where are the Hodge numbers of the CY. (The number of complex structure moduli is and the number of Kähler moduli is . Other Hodge numbers are fixed for a CY threefold.) The “large-volume” moduli stabilization mechanism is an alternative to the KKLT one, although it shares some features with KKLT. The purpose of this section is to briefly explain the model of [11, 12, 13]. An effective supergravity is completely specified by a Kähler potential, superpotential and gauge kinetic function. In the scalar field sector of the theory the action is Here and are the Kähler potential and the superpotential respectively, is the reduced Planck mass eq.(1), and represent all scalar moduli. (We closely follow the notations of and keep and other numerical factors explicit.) The -corrected Kähler potential is Here is the volume of the CY manifold in units of the string length , and we set . The second term in the logarithm represents the -corrections with proportional to the Euler characteristic of the manifold . is the IIB axio-dilaton with the dilaton component and the Ramond-Ramond 0-form. is the holomorphic 3-form of . The superpotential depends explicitly upon the Kähler moduli when non-perturbative corrections are included Here, is the tree level flux-induced superpotential which is related to the IIB flux 3-form as shown. The exponential terms are from non-perturbative (instanton) effects. (For simplicity, we ignore higher instanton corrections. This should be valid as long as we restrict ourselves to , which we do.) The Kähler moduli are complex, with the 4-cycle volume and its axionic partner, arising from the Ramond-Ramond 4-form . The encode threshold corrections. In general they are functions of the complex structure moduli and are independent of the Kähler moduli. This follows from the requirement that is a holomorphic function of complex scalar fields and therefore can depend on only via the combination . On the other hand, should respect the axion shift symmetry and thus cannot be a polynomial function of . (See for discussion.) The critical parameters in the potential are constants which depend upon the specific nature of the dominant non-perturbative mechanism. For example, for Euclidean D3-brane instantons and for the gaugino condensate on the D7 brane world-volume. We vary them freely in our exploration of trajectories in different potentials. It is known that both the dilaton and the complex structure moduli can be stabilized in a model with a tree level superpotential induced by generic supersymmetric fluxes (see e.g. ) and the lowest-order (i.e. ) Kähler potential, whereas the Kähler moduli are left undetermined in this procedure (hence are “no scale” models). Including both leading perturbative and non-perturbative corrections and integrating out the dilaton and the complex structure moduli, one obtains a potential for the Kähler moduli which in general has two types of minima. The first type is the KKLT minima which requires significant fine tuning of () for their existence. As pointed out in , the KKLT approach has a few shortcoming, among which are the limited range of validity of the KKLT effective action (due to corrections) and the fact that either the dilaton or some of the complex structure moduli typically become tachyonic at the minimum for the Kähler modulus. (We note, however, that argued that a consistent KKLT-type model with all moduli properly stabilized can be found.) The second type is the “large-volume” AdS minima studied in [11, 12, 13]. These minima exist in a broad class of models and at arbitrary values of parameters. An important characteristic feature of these models is that the stabilized volume of the internal manifold is exponentially large, , and can be in string units. (Here is the value of at its minimum.) The relation between the Planck scale and string scale is where is the volume in string units at the minimum of the potential. Thus these models can have in the range between the GUT and TeV scale. In these models one can compute the spectrum of low-energy particles and soft supersymmetry breaking terms after stabilizing all moduli, which makes them especially attractive phenomenologically (see [13, 20]). Conlon and Quevedo studied inflation in these models and showed that there is at least one natural inflationary direction in the Kähler moduli space. The non-perturbative corrections in the superpotential eq.(5) depend exponentially on the Kähler moduli , and realize by eq.(3) exponentially flat inflationary potentials, the first time this has arisen from string theory. As mentioned in § 1, higher (sub-leading) and string loop corrections could, in principle, introduce a small polynomial dependence on the which would beat exponential flatness at large values of the . Although the exact form of these corrections is not known, we assume in this paper that they are not important for the values of the during the last stage of inflation (see § 3). After stabilizing the dilaton and the complex structure moduli we can identify the string coupling as , so the Kähler potential (4) takes the simple form where is a constant. Using this formula together with equations (3), (5), and (11), one can compute the scalar potential. In our subsequent analysis, we shall absorb the constant factor into the parameters and . The volume of the internal CY manifold can be expressed in terms of the 2-cycle moduli : where is the triple intersection form of . The 4-cycle moduli are related to the by which gives an implicit dependence on the , and thus through eq.(8). It is known that for a CY manifold the matrix has signature , with one positive eigenvalue and negative eigenvalues. Since is just a change of variables, the matrix also has signature . In the case where each of the 4-cycles has a non-vanishing triple intersection only with itself, the matrix is diagonal and its signature is manifest. The volume in this case takes a particularly simple form in terms of the : Here and are positive constants depending on the particular model.111For example, the two-Kähler model with the orientifold of studied in [22, 12, 13] has , , and . This formula suggests a “Swiss-cheese” picture of a CY, in which describes the 4-cycle of maximal size and the blow-up cycles. The modulus controls the overall scale of the CY and can take an arbitrarily large value, whereas describe the holes in the CY and cannot be larger than the overall size of the manifold. As argued in [12, 13], for generic values of the parameters , , one finds that and at the minimum of the effective potential. In other words, the sizes of the holes are generically much smaller than the overall size of the CY. The role of the inflaton in the model of is the last modulus among the , , to attain its minimum. As noted by , the simplified form of the volume eq.(11) is not really necessary to have inflation. For our analysis to be correct, it would be enough to consider a model with at least one Kähler modulus whose only non-zero triple intersection is with itself, i.e., and which has its own non-perturbative term in the superpotential eq.(5). 3 Perturbative Corrections There are several types of perturbative corrections that could modify the classical potential on the Kähler moduli space: those related to higher string modes, or -corrections, coming from the higher derivative terms in both bulk and source (brane) effective actions; and string loop, or -corrections, coming from closed and open string loop diagrams. As we mentioned before, -corrections are an important ingredient of the “large volume” compactification models of [11, 12, 13]. They are necessary for the existence of the large volume minimum of the effective potential in the models with Kähler moduli “lifted” by instanton terms in the superpotential. The leading -corrections to the potential arise from the higher derivative terms in the ten dimensional IIB action at the order , where and is a generalization of the six-dimensional Euler integrand, Performing a compactification of (13) on a CY threefold, one finds -corrections to the metric on the Kähler moduli space, which can be described by the -term in the Kähler potential (4) (see [23, 16]). We will see later that this correction introduces a positive term into the potential. As discussed in , further higher derivative bulk corrections at and above are sub-leading to the term and therefore suppressed. (Note that in the models we are dealing with, there is effectively one more expansion parameter, , due to the large value of the stabilized .) Also, -corrections from the D3/D7 brane actions depend on 4d space-time curvatures and, therefore, do not contribute to the potential. String loop corrections to the Kähler potential come from the Klein bottle, annulus and Möbius strip diagrams computed in for the models compactified on the orientifolds of tori. The Kähler potential including both leading and loop corrections can be schematically written as (We have dropped terms depending only on the brane and complex structure moduli.) Here and are functions of the moduli whose forms are unknown for a generic CY manifold. If they depend upon the inflaton polynomially, a mass term will arise for with the possibility of an -problem. Further study is needed to decide. Note that, although the exact form of higher (sub-leading) corrections is unknown, any correction which introduces dependence on only via will not generate any new mass terms for . As well as -corrections there are possible -corrections. Non-perturbative effects modify the superpotential by breaking the shift symmetry, making it discrete. As noted we do include these. Although leading perturbative terms leave the Kähler potential -independent, subleading corrections can lead to -dependent modifications, which we ignore here. In this paper, we assume that the higher corrections, though possibly destroying slow roll at large values of the inflaton, are not important during the last stage of inflation. 4 Effective Potential and Volume Stabilization In this Section, we sketch the derivation of the effective field theory potential starting from equations (3,5,8). We choose to be the inflaton field and study its dynamics in the 4-dimensional effective theory. We first have to ensure that the volume modulus and other Kähler moduli are trapped in their minima and remain constant or almost constant during inflation. For this we have to focus on the effective potential of all relevant fields. Given the Kähler potential and the superpotential, it is straightforward but tedious to compute the scalar potential as a function of the fields . To make all computations we modified the SuperCosmology Mathematica package which originally was designed for real scalar fields to manipulate complex fields. The Kähler potential (8) gives rise to the Kähler metric , with This can be inverted to give This is the full expression for an arbitrary number of Kähler moduli . The entries of the metric contain terms of different orders in the inverse volume. If we were to keep only the lowest order terms , the shape of the trajectories we determine in the following sections and our conclusions would remain practically unchanged. That is, we are working with higher precision than necessary. Note that the kinetic terms for and are identical, appearing as in the Lagrangian. The resulting potential is We have to add here the uplift term to get a Minkowski or tiny dS minimum. Uplifting is not just a feature needed in string theory models. For example, uplifting is done in QFT to tune the constant part of the scalar field potential to zero. At least in string theory there are tools for uplifting, whereas in QFT it is a pure tuning (see, e.g., [1, 26] and references therein). We will adopt the form with to be adjusted. We now discuss the stabilization of all moduli plus the volume modulus. For this we have to find the global minimum of the potential eq.(4), which we do numerically. However, it is instructive to give analytic estimations. Following [12, 13], we study an asymptotic form of eq.(4) in the region where both , , and . The potential is then a series of inverse powers of . Keeping the terms up to the order we obtain The cross terms for different do not appear in this asymptotic form, as they would be of order . Requiring and at the minimum of the potential eq.(20), we get where are the values of the moduli at the global minimum. The expression (20) has the structure where the coefficients are functions of and . and are positive but can be of either sign. However, the potential for the volume has a minimum only if , which is achieved for ; otherwise would have a runaway character. Also if all are very large so that , then and cannot be stabilized. Therefore to keep non-zero and negative we have to require that some of the Kähler moduli and their axionic partners are trapped in the minimum. For simplicity we assume all but are already trapped in the minimum. It is important to recognize that trapping all moduli but one in the minimum cannot be achieved with only two Kähler moduli and , because effectively corresponds to the volume, and is the inflaton which is to be placed out of the minimum. The Fig. 2 shows the potential as a function of and for the two-Kähler model. One can see from this plot that a trajectory starting from an initial value for larger than a critical value will have runaway behavior in the (volume) direction. Thus, as shown by , one has to consider a model with three or more Kähler moduli. By contrast, the “better racetrack” inflationary model based on the KKLT stabilization is achieved with just two Kähler moduli . However, in our class of models with three and more Kähler moduli we have more flexibility in parameter space in achieving both stabilization and inflation. Another aspect of the work in this paper is that a “large volume” analog of the “better racetrack” model may arise. We have learned that to be fully general we would allow all other moduli including the volume to be dynamical. This will lead to even richer possibilities than those explored here, where we only let evolve, and assume that varying it does not alter the values of the other moduli which we pin at the global minimum. To demonstrate this is viable, we need to show the contribution of to the position of the minimum is negligible. Following , we set all , , and their axions to their minima and use equations (20) and (21) to obtain the potential for : As one can see from eq.(20), the contribution of to the potential is maximal (by absolute value) when and are at their minimum, and vanishes as . This gives a simple criterion for whether the minimum for the volume remains stable during the evolution of : the functional form of the potential for (23) is insensitive to provided : For a large enough number of Kähler moduli this condition is automatically satisfied for generic values of and . We conclude that with many Kähler moduli the volume does not change during the evolution of the inflaton because the other stay at their minimum and keep the volume stable. Consider a toy model with three Kähler moduli in which is the inflaton and stays at its own minimum to provide an unvarying minimum for . We choose parameters as in set 1 in Table 1 (which will be explained in detail below in Sec. 5), and also let , , and . Eq.(24) is strongly satisfied, , under this choice of parameters. Therefore we can drop the -dependent terms in the potential (20) and use it as a function of the two fields and to find their values at the minimum (after setting also to its minimum). The minimization procedure should also allow one to adjust the uplift parameter in a way that the potential vanishes at its global minimum. With our choice of parameters we found the minimum numerically to be at and with , as shown in Fig. 3. 5 Inflaton Potential We now take all moduli , , and the volume (hence ) to be fixed at their minima, but let vary, since it is our inflaton. For simplicity in the subsequent sections we drop the explicit subscript, setting . The scalar potential is obtained from eq.(4) with the other Kähler moduli stabilized: Here the terms contain contributions from the stabilized Kähler moduli other than the inflaton. We dropped cross terms between and other , , since these are suppressed by inverse powers of . We can trust eq.(25) only up to the order ; at higher orders in , higher perturbative corrections to the Kähler potential eq.(15) start contributing. Explicitly expanding to order yields the simpler expression is a constant term, since and , are all stabilized at the minimum, and , depend only on these . |Parameter set 1| |Parameter set 2| |Parameter set 3| |Parameter set 4| |Parameter set 5| |Parameter set 6| The potential eq.(26) has seven parameters , and whose meaning was explained in § 2. We have investigated the shape of the potential for a range of these parameters. , , control the low energy phenomenology of this model (see ) and are the ones we concentrate on here for our inflation application. We shall not deal with particle phenomenology aspects in this paper. Some choices of parameters , , seem to be more natural (see ). To illustrate the range of potentials, we have chosen the six sets of parameters given in Table 1. There is some debate on what are likely values of in string theory. We chose a range from intermediate to large. Since there are scaling relations among parameters, we can relate the specific ones we have chosen to others. An estimate of the magnitude of comes from a relation of the flux 3-forms and which appear in the definition of (eq.5) to the Euler characteristic of the F-theory 4-fold, which is from the tadpole cancellation condition. This suggests an approximate upper bound . For typical values of , we would have . There are examples of manifolds with as large as , which would result in . Further, the bound itself can be evaded by terms. However, we do not wish to push too high so that we can avoid the effects of higher perturbative corrections. We can use the scaling property for the parameters to move the value of into a comfortable range. This should be taken into account while examining the table. The parameter sets in the table can be divided into two classes: Trajectories in sets produce a spectrum of scalar perturbations that is comparable to the experimentally observed one (good parameter sets), whereas trajectories in parameter set and produce spectra whose normalization is in disagreement with observations (bad parameter sets). Sets 3 and 4 were chosen to large values of to illustrate how things change with this parameter, but we are wary that with such large fluxes, other effects may come into play for determining the potential over those considered here. A typical potential surface is shown in Fig. 4 with the isocontours of superimposed. The hypersurface has a rich structure. It is periodic in with period , as seen in Fig 4. Along , where is integer, the profile of the potential in the direction is that considered in . It has a minimum at some and gradually saturates towards a constant value at large , , where is a constant.222This type of potential is similar to that derived from the Starobinsky model of inflation with a Lagrangian via a conformal transformation, . Along , falls gradually from a maximum at small towards the same constant value, . Trajectories beginning at the maximum run away towards large . Fig. 8(b) shows these two one-dimensional sections of the potential. For all other values of the potential interpolates between these two profiles. Thus, at large , the surface is almost flat but slightly rippled. At small , the potential in the axion direction is highly peaked. Around the maximum of the potential it is locally reminiscent of the “natural inflation” potential involving a pseudo Goldstone boson (except that and must be simultaneously considered), as well as the racetrack inflation potential [2, 3]. which can be used to generate families of models, trading for example large values of for small values of . For instance, applying the scaling to parameter set 1, we can push the value of down to , but at the same time pushing to lie in the range during inflation, a range which is quite problematic since at such small higher order string corrections would become important. More generally, for the supergravity approximation to be valid the parameters have to be adjusted to have at least a few: is the four-cycle volume (in string units) and the supergravity approximation fails when it is of the string scale. However, even if the SUGRA description in terms of the scalar potential is not valid at the minimum, it still can be valid at large , exactly where we wish to realize inflation. The consequence of small is that the end point of inflation, i.e. preheating, would have to be described by string theory degrees of freedom. We will return to this point in the discussion. 5.1 The Canonically-normalized Inflaton If we define a canonically-normalized field by , then It is therefore volume-suppressed. For inflation restricted to the direction, we identify with the inflaton. The field change over the many inflationary e-foldings is given in the last column in Table 1 for a typical radial () trajectory. It is much less than . (Here the scale factor at the end of inflation is so goes in the opposite direction to time.) The variations of the inflaton and the Hubble parameter wrt , suggest we must have a deceleration parameter nearly the de Sitter and the Hubble parameter nearly constant over the bulk of the trajectories. This is shown explicitly in § 6 and Figures 8 and 9. The parameter is the first “slow-roll parameter”, although it only needs to be below unity for inflation. With so small, we are in a very slow-roll situation until near the end of inflation when it rapidly rises from approximately zero to unity and beyond. Equation (30) connects the change in the inflaton field to the tensor to scalar ratio , since to a good approximation . Since we find , we get very small . The following relation [29, 30] gives a lower limit on the field variation in order to make tensor modes detectable We are not close to this bound. If is restricted to be in stringy inflation models, getting observable gravity wave signals is not easy. (A possible way out is to have many fields driving inflation in the spirit of assisted inflation .) When the trajectory is not in the direction, the field identified with the canonically-normalized inflaton becomes trajectory-dependent as we describe in the next section and there is no global transformation. That is why all of our potential contour plots have focused on the Kähler modulus and its axion rather than on the inflaton. 6 Inflationary Trajectories 6.1 The Inflaton Equation of Motion We consider a flat FRW universe with scalar factor and real fields . To find trajectories, we derive their equations of motion in the Hamiltonian form starting from the four dimensional Lagrangian (see and references therein) with canonical momentum , where we used and , stands for . (The usual field momentum is .) Here the non-canonical kinetic term is The Hamiltonian is where . The equations of motion follow from , which reduce to

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CC-MAIN-2021-39

http://icatt.org.ua/proc/article/view/ICATT.2017.7972577

math

Two approaches to solving the antenna synthesis problems according to the amplitude radiation pattern Keywords:nonlinear synthesis problem, linear antenna, amplitude radiation pattern, numerical simulation AbstractA nonlinear synthesis problem according to given amplitude radiation pattern for plane aperture is considered in two ways. An analytical formula for the optimal current is obtained in the first approach. This current generates the amplitude radiation pattern which is close to the given one with the prescribed accuracy. The second approach consists of variational statement of synthesis problem. The numerical results illustrate that each method has advantages in the certain range of frequencies. B. Minkovich, V. Yakovlev. Theory of Antenna Synthesis [in Russian]. Moscow: Sov. Radio, 1969. A. G. Ramm, “Optimal solution of the problem of linear antenna synthesis,” Soviet Physics – Doklady, vol. 13, no 6, pp. 546-548, 1968. A. G. Ramm, “On antenna synthesis theory,” in Antennas [in Russian]. Moscow: Svjaz, no. 5, pp. 35-46, 1969. J. K. Butler, “Linear array synthesis for the best mean-square approximation of radiation patterns,” Radio Science, vol. 3, no. 5, pp. 447-450, 1968. DOI: http://doi.org/10.1002/rds196835447. A. G. Ramm, “Signal estimation from incomplete data,” J. Math. Anal. Appl., vol. 125, pp. 267-271, 1987. DOI: http://doi.org/10.1016/0022-247X(87)90181-8. A. G. Ramm, A. I. Katsevich, The Radon Transform and Local Tomography. Boca Raton: CRC Press, 1996. [ L. D. Bakhrakh, L. D. Kremenetskyi, Symthesis of Radiation Systems. Theory and Methods of Calculation [in Russian]. Moscow: Sov. Radio, 1974. E. G. Zelkin, V. G. Sokolov, Methods of Antenna Synthesis. Phased Antenna Arrays and Antennas with Plane Apertures [in Russian]. Moscow: Sov. Radio, 1978. M. I. Andriychuk, N. N. Voitovich, P. A. Savenko, V. P. Tkachuk, Antenna Synthesis by Amplitude Radiation Pattern [in Russian]. Kiev: Nauk. Dumka, 1993. P. O. Savenko, Nonlinear Problems of Radiation System Synthesis [in Ukrainian]. Lviv: IAPMM of NASU, 2002. P. O. Savenko, Nonlinear Problems of Radiation System Synthesis with Plane Aperture [in Ukrainian]. Lviv: IAPMM of NASU, 2014. O. O. Bulatsyk, N. N. Voitovich, “Analytic solutions to a class of nonlinear integral equationconnected with modified phase problem,” Information Extraction and Processing, vol. 19, no. 95, pp. 33-39, 2003. O. O. Bulatsyk, “Investigation of branching of solutions to nonlinear equation of the modyfied phase problem in the case of discrete Fourier transform,” Visnyk Lviv Univer. Ser. Prykl. Mat., Inform., vol. 7, pp. 20-32, 2003. O. O. Bulatsyk, N. N. Voitovich, “Structure of phase-asymmetric solutions of circular antenna synthesis problem by amplitude radiation pattern with axially independent data,” Proc. of 2016 XXIst Int. Seminar/Workshop on Direct and Inverse Problems of Electromagnetic and Acoustic Wave Theory, DIPED, 26-29 Sept. 2016, Tbilisi, Georgia. IEEE, 2016, pp. 153-156. DOI: http://doi.org/10.1109/DIPED.2016.7772243. O. O. Bulatsyk, B. Z. Katsenelenbaum, Y. P. Topolyuk, N. N. Voitovich, Phase Optomization Problems – Application in Wave Field Theory. Berlin: Wiley-WCH, 2010. A. G. Ramm, M. I. Andriychuk, “Antenna synthesis by the modulus of diagram,” J. Advances Appl. Math., vol. 1, no. 1, pp. 1-10, 2016. DOI: http://doi.org/10.22606/jaam.2016.11001.

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http://hotdudesreadingbook.us/mathematics.html

math

Most philosophers of mathematics treat it as isolated, timeless, ahistorical, inhuman. Reuben Hersh argues the contrary, that mathematics must be understood as a human activity, a social phenomenon, part of human culture, historically evolved, and intelligible only in a social context. Hersh pulls the screen back to reveal mathematics as seen by professionals, debunking many mathematical myths, and demonstrating how the "humanist" idea of the nature of mathematics more closely resembles how mathematicians actually work. At the heart of his book is a fascinating historical account of the mainstream of philosophy--ranging from Pythagoras, Descartes, and Spinoza, to Bertrand Russell, David Hilbert, and Rudolph Carnap--followed by the mavericks who saw mathematics as a human artifact, including Aristotle, Locke, Hume, Mill, and Lakatos. What is Mathematics, Really? reflects an insider's view of mathematical life, and will be hotly debated by anyone with an interest in mathematics or the philosophy of science. Originally published in 1893, this book was significantly revised and extended by the author (second edition, 1919) to cover the history of mathematics from antiquity to the end of World War I. Since then, three more editions were published, and the current volume is a reproduction of the fifth edition (1991). The book covers the history of ancient mathematics (Babylonian, Egyptian, Roman, Chinese, Japanese, Mayan, Hindu, and Arabic, with a major emphasis on ancient Greek mathematics). The chapters that follow explore European mathematics in the Middle Ages and the mathematics of the sixteenth, seventeenth, and eighteenth centuries (Vieta, Decartes, Newton, Euler, and Lagrange). The last and... This book aims to explain, in clear non-technical language,what it is that mathematicians do, and how that differs from and builds on the mathematics that most people are familiar with from school. It is the ideal introduction for anyone who wishes to deepen their understanding of mathematics. The twentieth century has witnessed an unprecedented 'crisis in the foundations of mathematics', featuring a world-famous paradox (Russell's Paradox), a challenge to 'classical' mathematics from a world-famous mathematician (the 'mathematical intuitionism' of Brouwer), a new foundational school (Hilbert's Formalism), and the profound incompleteness results of Kurt Gödel. In the same period, the cross-fertilization of mathematics and philosophy resulted in a new sort of 'mathematical philosophy', associated most notably (but in different ways) with Bertrand Russell, W. V. Quine, and Gödel himself, and which remains at the focus of Anglo-Saxon philosophical discussion. The present collection brings together in a convenient form the seminal articles in the philosophy of mathematics by these and other major thinkers. It is a substantially revised version of the edition first published in 1964 and includes a revised bibliography. The volume will be welcomed as a major work of reference at this level in the field. Mathematics education research has blossomed into many different areas, which we can see in the programmes of the ICME conferences, as well as in the various survey articles in the Handbooks. However, all of these lines of research are trying to grapple with the complexity of the same process of learning mathematics. Although our knowledge of the process is through fragmentation of research more extensive and deeper there is a need to overcome this fragmentation and to see learning as one process with different aspects. To overcome this fragmentation, this book identifies six themes: (1) mathematics, culture and society, (2) the structure of mathematics and its influence on the learning process, (3) mathematics learning as a cognitive process, (4) mathematics learning as a social process, (5) affective conditions of the mathematics learning process, (6) new technologies and mathematics learning. This book is addressed to all researchers in mathematic education. It gives an orientation and overview on what is going on and what are the main results and questions what are important books or papers if further information is needed. This second edition contains updates on the career paths of individuals profiled in the first edition, along with many new profiles. The authors of the essays in this volume describe a wide variety of careers for which a background in the mathematical sciences is useful. Each of the jobs presented shows real people in real jobs. Their individual histories demonstrate how the study of mathematics was useful in landing good-paying jobs in predictable places such as IBM, AT&T, and American Airlines, and in surprising places such as FedEx Corporation, L.L. Bean, and Perdue Farms, Inc. You will also learn about job opportunities in the Federal Government as well as exciting careers in the arts, sculpture, music, and television. --back cover. Though it incorporates much new material, this new edition preserves the general character of the book in providing a collection of solutions of the equations of diffusion and describing how these solutions may be obtained. Contains alphabetically arranged entries that provide definitions and explanations of thousands of mathematical terms and concepts, theories, and principles, and includes biographical sketches of key people in math.

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http://www.math.stonybrook.edu/~tony/archive/118s13/assignment12.html

math

A. Explain in your own words why, if you pick a point on the number line at random between 0 and 1, it is very unlikely that the corresponding number will be rational. (Think of the decimal expansion of the point you pick). B. Mindscape 2 p. 156 (Give a mathematical description; i.e. not "one, two, three, four, five.") Mindscapes 3, 4, p.157 Mindscapes 12, 13 p.157 (Cultural note: Google "Galileo's Paradox") Mindscapes 26, 27 p.159 C. Mindscapes 12, 13, 14 p.170 D. Mindscape 23 p.171 (will be graded S/U). Remember: Collaboration is fine, but what you hand in should be your own work. Handing in something you copied is plagiarism and will cost you if it is detected. Write down what you tried and how it worked.

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http://woopop.com/econoline-crush-1998-surefire/

math

By Eric | December 23, 2013 1:00 am Artist: Econoline Crush Released: Jan 1, 1998 Packaging: Jewel Case Release Type: Promo #Surefire (Fahrenheit 451 Remix)(Radio Edit) #Surefire (Fahrenheit 451 Remix) #Surefire (Album Version) Has an awesome remix of the song (which is the version they usually played live) and an instrumental. Comments are closed.

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https://www.slideshare.net/piyushmittalin/touchless-fingerprint

math

INTRODUCTION• Fingerprint is most popular,reliable and oldest biometric sign of identity• Touchless fingerprint system is a remote sensing technique to process fingerprint pattern, considered as a viable alternative to touch based fingerprint system• New generation of touchless live scan devices is 3D touchless finger print system TERMINOLOGY• Fingerprint can be looked at from different levels1) GLOBAL LEVEL• Singularity points called core and delta points Core and delta points marked on sketches of the two fingerprint patterns loop and whorl 2) LOCAL LEVELMinutiae details in terms of ridgesRidge bifurcation Ridge termination Representation of minutiae TOUCHLESS VERSUS TOUCH-BASED TOUCHLESS TOUCH-BASED SKIN DISTORTION NO YES SLIPPAGE,SMEARING NO YES FINGERPRINT RESIDUE NO YES CAPTURE AREA LARGE SMALL TOLERANCE ON SKIN CONDITION LARGE SMALL HIGH LOWUSER COMFORT LEVEL IMAGE ACQUISITION MULTIPLE VIEW SYSTEMFigure 1. Fingerprint acquisition using a set of cameras surrounding the finger • Multiple view enables the capture of full nail to nail fingerprints increasing the usable area• From each acquired image a silhouette is extracted.• The 5 silhouettes are then projected into the 3D space and we get the 3D shape of finger by knowing the position and orientation of each camera within a reference coordinate system. Figure 2 Fingerprint acquisition obtained by combining a single line scan camera and two mirrors 3D FINGERPRINT UNWRAPPING• Unwrapping a 3D object refers to the unfolding the 3D object onto a flat 2D plane. UNWRAPPING METHOD PARAMETRIC NON PARAMETRIC PARAMETRIC UNWRAPPING USING CYLINDRICAL MODEL• Human fingers vary in shape, like the shape of the middle finger is often more cylindrical than the thumb.• Human fingers can be closely approximated by cylinders.• Human fingers also vary in size and the cylindrical model can also capture this size variability in the vertical direction, but not in the horizontal direction.• Cylindrical model is a reasonable choice for parametric unwrapping of3D fingerprints. T Parametric unwrapping using a cylindrical model (top down view). Point (x,y,z) on the 3D finger is projectedFigure 3 to ( ,z) on the 2D plane. • Mathematically, let the origin be positioned at the bottom of the finger, centered at the principle axis of the finger.• Let T be a point on the surface of the 3D finger: x T = Y z • This 3D point is then projected (transformed) onto the cylindrical surface to obtain the corresponding 2D coordinates S = z Where NON PARAMETRIC -UNWRAPPING• Non-parametric unwrapping, does not involve any projection on parametric models.• The unwrapping directly applies to the object to preserve local distances or angular relations.• Guarantees the variability in both shape and size of fingers is preserved.• Less distortion compared to parametric unwrapping PREPROCESSING STEPSa) Computation of local ridge frequency and local ridge orientationb) Enhancement of the fingerprint imagec) Segmentationd) Detection of singularities FEATURE EXTRACTIONa) Conversion of preprocessed fingerprint image into binary imageb) Thinning FINGERPRINT MATCHING• MINUTIAE-BASED APPROACH :- Analogous with the way that forensic experts compare fingerprints• The minutiae sets of the two fingerprints to be compared are first aligned, requiring displacement and rotation to be computed• Region of tolerance around the minutiae position is defined in order to compensate for the variations that may appear in the minutiae position due to noise and distortion DISADVANTAGE• Lower contrast between ridges and valleys due to motion blur of hand tremble , camera background noise and small depth of field• Unwrapping technique has distortion upto some extent• Compatibility with contact-based 2D rolled fingerprint image REFERENCE• Intelligent biometrics technique in finger print and face recognition by L.C Jain, U.halice, S.B lee,S.T Sutsui,I.Hayashi• Tabassi E., Wilson C., and Watson C., “Fingerprint Image Quality,” Tech. Rep. 7151, National Institute of Standards and Technology (NIST), August 2004.• Parziale G. and Diaz-Santana E., “The Surround Imager: Multi-Camera Touchless Device to Acquire 3D Rolled- Equivalent Fingerprints,” in Proceedings of IAPR International Conference on Biometrics (ICB),Hong Kong, China, January 2006, pp. 244–250.

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http://mathhelpforum.com/calculus/13478-solved-uniform-convergence-2-a.html

math

Here's the definition I'm using for uniform convergence: Follow Math Help Forum on Facebook and Google+ Please check my work: Please check this one also: Here is first one. The second one is easier. There is a theorem that the uniform limit of continous functions is continous. This limit function is not continous. A contradiction (since all functions in the sequence are continous themselves). View Tag Cloud

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https://www.arxiv-vanity.com/papers/0909.4559/

math

Exact Results for Wilson Loops in Superconformal Chern-Simons Theories with Matter We use localization techniques to compute the expectation values of supersymmetric Wilson loops in Chern-Simons theories with matter. We find the path-integral reduces to a non-Gaussian matrix model. The Wilson loops we consider preserve a single complex supersymmetry, and exist in any theory, though the localization requires superconformal symmetry. We present explicit results for the cases of pure Chern-Simons theory with gauge group , showing agreement with the known results, and ABJM, showing agreement with perturbative calculations. Our method applies to other theories, such as Gaiotto-Witten theories, BLG, and their variants. Recently, several new Chern-Simons matter theories with a large amount of supersymmetry have been found. In , Gaiotto and Witten found a class of theories with supersymmetry. Closely related are the the theories of ABJM , with SUSY, and BLG , with . All of these theories are superconformal, and arise as the low energy effective actions on certain brane configurations in string theory and -theory. Some of these theories are conjectured to be holographically dual to -theory on orbifold backgrounds, in similarity to super Yang-Mills theory in four dimensions, which was conjectured to be dual to Type IIB string theory on . Recall that, in the latter case, certain supersymmetric Wilson loops are dual to fundamental strings. Thus, as a check of this duality, one can compute the expectation values of these Wilson loops and compare to calculations made in string theory. This is difficult to do perturbatively, as the perturbative region of one theory is the strongly coupled region of its dual. However, supersymmetric operators are often easier to deal with than their less symmetric counterparts, and in it was shown that this is indeed true of supersymmetric Wilson loops in SYM theory. It was demonstrated, using localization, that finding the expectation value of these operators reduces to a calculation in a matrix model. This allows one to compute it much more efficiently at any coupling, and the result provides a non-trivial test of the duality. In this paper we seek to apply the methods of to the supersymmetric Chern-Simons matter theories discussed above. Our main result is that the partition function for a supersymmetric Chern-Simons theory with gauge group and chiral multiplets in a (possibly reducible) representation localizes to the following matrix integral:111The factors of appearing the determinants did not appear in the original version of this paper, but were found later by Drukker, Marino,and Putrov . Here the integration is over the Cartan of the Lie algebra of , is the order of the Weyl group of , and “Tr” defines some invariant inner product on .222We have absorbed the Chern-Simons level into “Tr”, see the next section for more details. We’ve also defined: where the product runs over the weights of the representation (or in the case of the adjoint representation (), over the roots of the Lie algebra). The same technique applies to representations which are not self-conjugate, although the resulting matrix models are more complicated. We also consider the following supersymmetric Wilson loop in a representation : where is an auxiliary scalar in the vector multiplet. We find its expectation value is given by: One application of this result is to a trivial example of a supersymmetric Chern-Simons matter theory: Chern-Simons theory without matter, which can be written in a supersymmetric form . Here we are able to use localization to recover some well known results on the reduction of the Chern-Simons partition function to a matrix model [7, 8], as well as reproducing some very simple knot invariants . Another example is ABJM theory. This is conjectured to be dual to a certain orbifold background in -theory, so it would be interesting to make non-perturbative calculations in this theory. Here we were able to reduce the path integral to a matrix model, although we were not able to compute the resulting matrix integrals exactly. However, evaluating them as a perturbative expansion in the ’t Hooft coupling, we find agreement with a perturbative calculation done in field theory [10, 11, 12], which provides a check of our result. It is possible that the matrix model could be solved exactly in the large limit using a saddle point approximation , as we will briefly mention at the end of the paper. Acknowledgments.This work was supported in part by DOE grant DE-FG02-92ER40701. A.K. would like to thank Lev Rozansky and Alexei Borodin for useful discussions. I.Y. would also like to thank Joseph Marsano, John Schwarz, Ketan Vyas and Ofer Aharony for their input. The class of theories we will be considering, supersymmetric Chern-Simons theory with matter, are described in for Minkowski space. We will work in Euclidean space. In this section we briefly review these theories. We start with the gauge multiplet. This consists of a gauge field , two real auxiliary scalars and , and an auxiliary fermion , which is a -component complex spinor. This is just the dimensional reduction of the vector multiplet in dimensions, with being the reduction of the fourth component of . All fields are valued in the Lie algebra of the gauge group . The kinetic term we will use for the gauge multiplet is a supersymmetric Chern-Simons term. In flat Euclidean space, this is: Here “Tr” denotes some invariant inner product on . For example, for , we will take “Tr” to mean times the trace in the fundamental representation, where is constrained by gauge invariance to be an integer. This action is invariant under the usual (euclideanized) vector multiplet transformations: Here and are -component complex spinors, in the fundamental representation of the spin group . We will take to be the Pauli matrices, which are hermitian, with . Also, is the gauge covariant derivative. Note that, in contrast to the Minkowski space algebra where we would have , in Euclidean space and are independent. Taking would reduce us to the algebra. This is because there is no reality condition on spinors in dimensional Euclidean space, so the least amount of supersymmetry one can have is a single complex spinor. To carry out the localization, we will work on a compact manifold rather than in flat space, as this makes the partition function well-defined. As the above action is conformal, we can transfer it to the unit -sphere, , without changing any of the quantities we are interested in computing. The Lagrangian simply acquires an overall measure factor of . In addition, we must modify the following supersymmetry transformations: where now all derivatives are covariant with respect to both the gauge field and the usual metric on . One can easily check that this leaves the action invariant for arbitrary spinors and . When we add matter, however, it will turn out to be necessary that we take and to be a Killing spinors, which means that they satisfy the following equation: Here is an arbitrary spinor. Note that in dimensions, this is actually equations, one of which determines , with the rest imposing conditions on . We will give the explicit solutions to this equation on below. With a Killing spinor, the above supersymmetries give a representation of the superconformal algebra, anticommuting with each other to conformal transformations. 2.1 The Wilson Loop The operator we will be localizing is the following supersymmetric Wilson loop: This operator has been considered in . Here is the closed world-line of the Wilson loop, and “” denotes the usual path-ordering operator. The variation of this operator under the the supersymmetry (6) is: For this to vanish for all we must have the following two conditions: Note we cannot take here, which is why we need to consider theories with at least supersymmetry. Now we need to impose the condition that and are Killing spinors. This will force us to consider only certain loops, which will turn out to be great circles on . To see this, we will need to determine the Killing spinors on . We start by picking a vielbein. It will be convenient to use the fact that is, as a manifold, the same as , so we can take a local orthonormal basis of left-invariant vector fields . In terms of these, the spin connection is simply: Thus the spinor covariant derivative is: We can immediately see a few solutions to the Killing spinor equation. Namely, take the components of in this basis to be constant, in which case: This gives two of the Killing spinors. There are two more solutions, which can be seen most easily using a right invariant vielbein, and which satisfy: Note that in these cases, is proportional to . This is not true of a general Killing spinor (eg, take a linear combination of the above spinors), although in spaces of constant curvature it is always possible to form a basis of the space of Killing spinors with such special ones [15, 16]. Now let us impose the condition that preserves the Wilson loop. If we pick to be the arc length, we find that must satisfy: This is only possible if is constant, which means must be some fixed linear combination of the , so we may as well pick our loop so that is parallel to one of them, say . The integral curves of these vector fields are great circles, so the Wilson loop must be a great circle to preserve any supersymmetry. Then this equation becomes: So this restricts us to only one of the two left-invariant Killing spinors (there is also a right handed one that preserves it). We could also pick to be one of these Killing spinors. Thus this Wilson loop preserves half of the supersymmetries. Conversely, given a Killing spinor , and taking , there is a family of great circles such that the Wilson loops along these circles are preserved by the corresponding supersymmetry. These are just the integral curves of the vector field , which, since this vector field is left-invariant, form a Hopf fibration. As a result, one could use the localization described here to compute the expectation value of a product of Wilson loops corresponding to a general link consisting of loops from this fibration. We will discuss this more below. Next we would like to add matter to the theory. The matter will come in chiral multiplets, each of which consists of a complex scalar , a fermion , which is a -component complex spinor, and an auxilliary complex scalar . The gauge-coupled action for a chiral multiplet in a representation of the gauge group is described in in the case of flat Minkowski space. It is straightforward to modify this for , giving: where, for example, is a shorthand for: where are indices in , is an index of the Lie algebra, and are the generators of in the representation . Also, is a derivative that is covariant with respect to both the gauge group and the metric on , and we assume the various color indices have been contracted in a gauge invariant way. Note that the second term in (18), which arises from the conformal coupling of scalars to the curvature of , gives the matter scalars a mass. A similar mass term will appear in the localizing term in (56). This action is classically invariant under the following superconformal symmetries: Here and must be a Killing spinors, satisfying (8). In order to perform the localization, it turns out that the theory must be superconformal on the quantum level. This is because the supersymmetry that we will use for localization, together with its conjugate and the Lorentz group, generate the entire superconformal algebra. Thus any hermitian action invariant under is necessarily superconformal. This fact determines which superpotentials are allowed. In the absence of a superpotential, the combined Chern-Simons-matter system is superconformal on the quantum level . This follows from the nonrenormalization of the Chern-Simons couplings (except for finite shifts) together with the standard nonrenormalization theorem for the -terms. An arbitrary quartic superpotential preserves superconformal invariance on the classical level, but in the quantum theory superconformal invariance is destroyed in general. Indeed, if the fields have anomalous dimensions, the scaling dimension of the quartic superpotential is not equal to , and the superpotential perturbation is not marginal. However, for special values of the superpotential couplings it may happen that the theory has enhanced supersymmetry which requires the anomalous dimensions to vanish . This is the case for theories of Gaiotto and Witten, the ABJM theory, and the BLG theory. We will see later that the path-integral localizes to configurations where all matter fields vanish, so the precise choice of the superpotential will not matter, provided it ensures superconformal invariance on the quantum level. One particular example from the class of superconformal Chern-Simons matter theories is the ABJM theory . Here the gauge group is , with the Chern-Simons action for the two factors appearing at levels and . The matter comes in two copies of the bifundamental representation , and two more in . There is also a quartic superpotential which ensures the supersymmetry is enhanced from to . 3.1 Gauge Sector In this section we will closely follow , in which supersymmetric Wilson loops were studied in and super Yang-Mills theory in dimensions, and their expectation values were computed by essentially the same localization method we use here. We will start by considering pure Chern-Simons theory, with no matter. We will discuss how the addition of matter affects the computation in section 3.4. The idea of the localization is as follows. We start by picking a single supersymmetry which preserves the operator we are interested in. We then deform the action by adding a term: Here “’’ is some positive definite inner product on the Lie algebra.333We distinguish it from the trace in the original action, which is not necessarily positive definite (eg, in ABJM, where it has a different sign for the two gauge groups). We assume this term is itself supersymmetric, which amounts to saying that is invariant under the bosonic symmetry . Then the standard argument shows that the addition of this term to the action does not affect the expectation value of any -invariant observable. We pick the term in (21) to deform the action because its bosonic part, , is positive definite. Thus we can take to be very large, and the dominant contribution to the path integral will come from the region of field space where this term vanishes, which is precisely where . In the limit of large the theory becomes free, so we can compute things easily, knowing the results we get are independent of and thus apply at , where they represent the quantities we are interested in. Returning to the case at hand, let us fix a supersymmetric Wilson loop along some great circle on . This will be the operator we want to localize. We start by defining the supersymmetry we will be working with. Let be the unique left-invariant spinor which preserves the Wilson loop, normalized so that , and let . By “” we will mean the infinitesimal supersymmetry variation corresponding to this choice of parameters. It is clear that on the bosonic fields, and therefore on the fermions as well. Thus the -exact term (21) is trivially supersymmetric. As shown in appendix A, for the supersymmetry in question it evaluates to: Note that, unlike the matter scalars, there is no mass term for the scalar . The mass term for will arise not from the conformal coupling to the curvature of but from the supersymmetric Chern-Simons action. Next we need to determine where the theory localizes to. The vanishing of requires: This implies the following two conditions: It is straightforward to show that, on , the only solution to the first equation is to take and constant, and then the second equation implies . Thus the theory localizes to the space of constant , with and all other fields vanishing. One can also see this from the explicit form of above. In the limit of large , the exact result for the path integral becomes equal to the saddle point approximation: Here we integrate the contributions from the saddle points, which are labeled by , together with the determinant factor coming from quadratic fluctuations of the fields about each saddle point. For the partition function, the classical contribution is: where we have used the fact that the volume of is . The Wilson loop gives an additional factor of: Thus we have: This is essentially the same result as was found in for and SYM theory in four dimensions. For , it was shown that , so that the resulting matrix model is Gaussian, while for it was something more complicated, involving the Barnes G-function. In the next section we will find that, in our case, is not , but it is still something relatively tractable. Before moving on, we should mention that, to be precise, we should really be localizing the gauge-fixed theory. That is, we should have started by introducing ghost fields , and a Lagrange multiplier . We would then have the standard BRST transformations , and, continuing to follow , define a new fermionic symmetry: One can check that . We also would modify to: We would then localize with respect to rather than . The variation of the new has four contributions: from and each hitting the two terms in . For the first term, only would contribute, since is a gauge transformation and the term is gauge invariant, so the total contribution would just be the -exact term we found above. For the second term, the variation would give us the usual gauge-fixing term: Then we still need to worry about the remaining term: But if we define , then we are only left with some term multiplying , which can be absorbed into the definition of . In other words, we have shown that we can proceed by starting with the action in (23) and gauge fixing it the usual way, and then computing the -loop determinant. We turn to that calculation now 3.2 1-Loop Determinant In this section we compute the 1-loop determinant coming from quadratic fluctuations of the fields about the saddle points we found in the last section. After introducing ghosts, (23) becomes: We will be interested in the large limit, so we rescale the fields to eliminate the out front: Here represents all fields other than and , and we have treated these fields differently because they have zero modes. Also, represents the non-zero mode part of , and similarly for . Taking to be large then allows us to keep only the quadratic terms in the action: where . The integral over can be performed immediately, and eliminates the squared term. The integral is also easy, giving a delta function constraint which imposes Lorentz gauge. We find: Here is the vector Laplacian. This is a free theory, and we would like to compute its 1-loop determinant. A similar calculation is done in , and we can proceed similarly. First we separate the gauge field into a divergenceless and pure divergence part: where . Then the delta function constraint becomes , and so we can integrate over using the delta function, picking up a jacobian factor of . The integral over gives the same factor, while the integral over the ghosts contributes a factor of . These all cancel (and in any case, are -independent), and we are left with: Now if we go back to (29) for a moment, we see that since the action is gauge invariant, the integrand is invariant under the adjoint action of the group. Thus we can replace the integral over the entire Lie algebra with an integral over some chosen Cartan subalgebra. This introduces a Vandermonde determinant in the measure. There is also the residual gauge symmetry of the Weyl group of , so we should divide by , the order of this group. We’re left with, eg, for the partition function: where runs over the roots of , and runs over the Cartan subalgebra. Thus we only need to know for in the Cartan, and so from now on we will assume is in the Cartan. Now let us decompose as: where are representatives of the root spaces of , normalized so , and runs over the roots of . Here is the component of along the Cartan, but this part of will only contribute a -independent factor to the loop determinant, so we will ignore it. Then we can write: We can do something similar for . Plugging this into the action, we can now write it in terms of ordinary (as opposed to matrix valued) vectors and spinors:444Thanks for F. Benini and A. Yarom for pointing out some errors that appeared in this equation in the original version of this paper. From we know that the eigenvalues of the vector Laplacian acting on divergenceless vector fields are , where , and they occur with degeneracy . Thus the bosonic part of the determinant is: For the gaugino, we note that on , eigenvalues of are with degeneracy , where runs over the positive integers.555This can be seen easily using the results of section 3.4, specifically as a special case of (73) Thus the fermion determinant is: And so the total 1-loop determinant is: We see there is partial cancellation between the numerator and the denominator, and this becomes: Because the eigenvalues of a matrix in the adjoint representation come in positive-negative pairs, only the even part in sigma contributes. We can isolate this by looking at: The infinite constant can be fixed by zeta regularization , and the rest of the product can be done exactly. We find: To summarize, we have shown: where runs over the roots of . Plugging this into (40), we see the denominator cancels against the Vandermonde determinant. Introducing the notation: where is some representation, and the product runs over its weights (which, in the case of the adjoint representation, are just the roots of the algebra), we are left with: 3.3 Chern Simons Theory For a concrete example, we will look at the case where . Then we can take the Cartan as the set of diagonal matrices, setting . The roots of are labeled by integers , and have: Also, as mentioned earlier, we take Tr as times the trace in the fundamental representation, and we also take the Wilson loop in the fundamental representation. Also the Weyl group is , so we should divide by . Then (50) becomes (up to a sign): where all the integrals run over the real line. To interpret this result, note that without matter, we can integrate out the auxiliary fields trivially. The integral over in (5) imposes , and so we see from (9) that, in the case of pure Chern-Simons theory, the supersymmetric Wilson loop we have been considering is just the ordinary Wilson loop operator. Thus the second line of (52) gives a simple way of computing the Wilson loop expectation value in this theory. Both of the integrals above are just sums of Gaussian integrals, and it is straightforward to evaluate them exactly, as shown in appendix B. The result for the Wilson loop expectation value is: reproducing the known result , up to an overall phase. This phase comes from the framing of the loop, which can be seen as follows. As mentioned earlier, the supersymmetry we are using preserves a family of Wilson loops forming a Hopf fibration of the sphere. The framing of a Wilson loop is essentially the choice of a nearby loop so that point-splitting regularization may be performed, and for this procedure to be compatible with supersymmetry, this loop must come from the Hopf fibration, and therefore have linking number with the Wilson loop. One simple extension of this calculation is to a link of Wilson loops. If the loops in such a link come from the Hopf fibration, then they preserve the same supersymmetry . Then it is easy to see that the expectation value of this operator is given simply by inserting more factors of in the matrix model, or equivalently, taking the trace in the product representation. This property of the Chern-Simons invariant of the Hopf link can also be shown by topological means.666Thanks to Lev Rozansky for discussions on this point. 3.4 Matter Sector Next we carry out the localization procedure in the matter sector. Rather than treat the case of multiple chiral multiplets in various representations, we will consider a single chiral multiplet in some possibly reducible representation . The action of the supersymmetry on the matter fields is as follows: with all other variations vanishing. Here we have assumed the rescaling (35) has been done on the gauge multiplet. Since we will be taking to be very large, this means we can ignore all coupling to the gauge sector, except that through . To localize the matter sector, we add a term similar to the one we used in the gauge sector: In Appendix A it is shown that this equals: where . As before, this term is positive definite, and vanishes on the following field configurations: The second equation implies . With a little work, one can check that the first implies must also be zero. Rather than show this directly, we will see in a moment, when we evaluate the 1-loop determinant, that the operator acting on in (56) has no zero modes, which leads to the same conclusion. Thus there is no classical contribution coming from the matter sector, and its only influence is through its effect on the one-loop determinant. Since there is no interaction between the matter and gauge fields at quadratic order, except that throught , the determinant factorizes, and we can write: We will compute this extra factor in the next section. 3.5 1-Loop Determinant - Matter Sector For the scalar field, we see from (56) that the operator we need to diagonalize is: Here “” is actually a matrix representing in the representation . As we did for the gauge multiplet, we can decompose this representation into its weight spaces. Namely, if is a representative of the weight space corresponding to the weight , satisfying (where is some gauge-invariant way of contracting the relevant color indices), then we write: The total -loop determinant will be the product of the one coming from each term in this sum, which are all acting on ordinary (not matrix-valued) scalars. It will be most convenient to use a pair of orthonormal frames, one left-invariant and one right-invariant under the action of (thinking of as and letting it act on itself). We will call these and . Then we can take . It is straightforward to show that the laplacian can be expressed in terms of these fields as: where we think of the vector fields as differential operators on the space of scalar fields. Thus the each term in (60) can be written as: Also, using the fact that the vectors satisfy the algebra: we see that if we define new operators , these satisfy the algebra. In terms of these operators, the operator acting on the scalars becomes: and so computing its eigenvalues reduces to a familiar problem from quantum mechanics. On a spin- representation, the determinant of the operator can be written as: It can be shown that the scalar fields on decompose into the irreps under the action of the left- and right-acting ’s, so the total determinant will be a product of the above expression over all non-negative integers , each raised to the power of the degeneracy, which is owing to the right-acting . Next we consider the fermions. After decomposing them into weights as with the bosons, the operator we need to diagonalize is: If we use the as our vielbein, the covariant derivative acting on spinors can be written as: Then the Dirac operator is: We should be careful to distinguish , which is a differential operator, from , which is just a matrix. Thus the operator acting on the fermions becomes: Or, if we define , which satisfy the algebra, and plug in the : So the problem reduces to computing spin-orbit coupling. Unfortunately, we cannot proceed the standard way since does not commute with the total angular momentum , so we are forced to compute the determinant manually. Let: Note that this operator commutes with both and , so its eigenvectors all have the form: Letting act on these vectors, it is straightforward to compute: Plugging in the relevant values, , we get: We note this is almost equal to the scalar determinant, except for the extra terms out front and the missing the factor in the product. Taking this into account, we can write:

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http://mathhelpforum.com/algebra/96201-there-infinite-number-regular-polyhedra-print.html

math

There are an infinite number of regular polyhedra?? Iam new to this forum so hope u peolpe can help. I am in my first year of a 4 yr university course training to be a primary/elementary school teacher. I have beeen given the statement.. 'there are an infinite number of regular polyhedra' The statment is said by an 8 yr old child. My task is to identify what is wrong with the statement, and how i would explain to the child the correct statement/ what i could show to the child to make them realise this statement is wrong, ansd the correct course of action to put them on th e right track. I have been told by my maths teacher the statement is incorrect, but that is all the information i have been given, i dont even know what a rugular polyhedra is!!! Hope sumone on this forum can help???

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http://zyessayqiuz.designheroes.us/the-elementary-theory.html

math

Elementary particles in physics 3 see that these four types of fundamental particle are replicated in two heavier families, (µ−, νµ, c, s) and (τ−, ντ, t, b)the reason for the existence of these. Description elementary number theory, sixth edition, blends classical theory with modern applications and is notable for its outstanding exercise setsa full range of exercises, from basic to challenging, helps students explore key concepts and push their understanding to new heights. This is page 3 printer: opaque this preface this is a textbook about prime numbers, congruences, basic public-key cryptography, quadratic reciprocity, continued fractions, elliptic curves, and. 74 the elementary beam theory in this section, problems involving long and slender beams are addressed as with pressure vessels, the geometry of the beam, and the. Elementary number theory provides a very readable introduction including practice problems with answers in the back of the book it is also published by dover which means it is going to be very cheap (right now it is $874 on amazon. How to cite cabbolet, mjtf (2010), elementary process theory: a formal axiomatic system with a potential application as a foundational framework for physics supporting gravitational repulsion of matter and antimatter. The usefulness of matrix theory as a tool in disciplines ranging from quantum mechanics to psychometrics is widely recognized, and courses in matrix theory are increasingly a standard part of the undergraduate curriculum. Internet archive bookreader elementary number theory. The elementary theory of distributions i, ii [j and r sikorski mikusinski] on amazoncom free shipping on qualifying offers warsaw 1957, 1961 rozprawy matematyczne xii, xxv 2 volumes. Elementary stars jonny lee miller as detective sherlock holmes and lucy liu as dr joan watson in a modern-day drama about a crime-solving duo that cracks the nypd's most impossible cases. Originally published in 1928 as number twenty-five in the cambridge tracts in mathematics and mathematical physics series, and here reissued in its 1958 reprinted form, this book deals with the physical and mathematical aspects of the symmetrical optical system. Document resume ed 432 388 ps 027 771 author moustafa, brenda martin title multisensory approaches and learning styles theory in the elementary school: summary of reference papers. Elementary number theory, sixth edition, blends classical theory with modern applications and is notable for its outstanding exercise sets a full range of exercises, from basic to challenging, helps readers explore key concepts and push their understanding to new heights. Elementary music theory is a series of online theory lessons by mark sarnecki these lessons follow the popular theory book series elementary music rudiments and harmony each video lesson coincides with a lesson from one of the text books depending upon the course and level. An elementary math curriculum for supplementary or home school should teach much more than the how to of simple arithmetic a good math curriculum should have elementary math activities that build a solid foundation which is both deep and broad, conceptual and how to. Course description description elementary number theory gives advanced students an introduction to the deep theory of the integers, with focus on the properties of prime numbers, and integer or rational solutions to equations. Nb (note bene) - it is almost never necessary in a mathematical proof to remember that a function is literally a set of ordered pairs de nition 18 (injection. Theory for math majors and in many cases as an elective course the notes contain a useful introduction to important topics that need to be ad- dressed in a course in number theory. Can we prove n5 for sake of contradiction, assume n5 is false consider the smallest element n 0 2fx 2n jx 2acg if n 0 6= 1, then n 0 is the successor of some natural number. Music quizzes, games, worksheets and music theory help by ms garrett scroll down the page to play 130+ elementary and middle school level quizzes, puzzles and games about music notes, rhythms, instruments, composers, and more. We provide a student project on elementary set theory based on the original historical sources by two key figures in the development of set theory, georg cantor (1845-1918) and richard dedekind (1831-1916. In conclusion, the theory of elementary waves is no theory at all though some of the problems above are problems of exposition, and others might be deficiencies that could be patched up somehow or other, several are fundamental to the theory: # 1, 2, 5, 6, 9, 12. Qcarl friedrich gauss, born into a poor working class family in brunswick, now lower saxon, germany and died in gottingen, germany he was a child prodigy with genius that did not impress his father who called him a star-gazer his mother, dorothea gauss was the exact opposite of his father as. The elementary proof of the prime number theorem joel spencer and ronald graham p rime numbers are the atoms of our mathematical universe euclid showed that there are infinitely. Elementary number theory, seventh edition, is written for the one-semester undergraduate number theory course taken by math majors, secondary education majors, and computer science students.

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https://support.twic.ai/hc/en-us/articles/360026225751-Why-didn-t-I-get-the-total-reimbursement-amount-

math

There are two possible reasons. One is that we can only approve the amount of the receipt. Another is that you don’t have enough points to cover the total reimbursement amount. For example, the amount of your receipt is $70. However, the amount you submit is $90. We can only approve the $70 shown in the receipt. Or suppose you have $80 in your account, but you submit a reimbursement request for $100. In this case, you would only be approved for $80 instead of $100.

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https://www.reference.com/math/30-percent-60-ba453aabffcaff02

math

Thirty percent of $60 is $18. The mathematical equation used to solve this problem is: 0.30 x $60 = N. An alternative method is to multiply 30 by $60 and then divide the answer by 100. Thus, 30 x $60 = $1,800, and $1,800 / 100 = $18.Know More Percentage problems consist of three values: the base, the rate or percent, and the percentage or amount. The base is the quantity from which a percentage is taken. In the given equation, $60 is the base, while 30 is the rate or percent. The percentage or amount, which is what this particular example seeks to find, is $18. Percentage problems are often useful in determining the final price of an item on clearance after a certain price deduction. For example if a dress originally costs $50 and is marked 30 percent off, the equation $50 x 0.30 = $15 lets the shopper know she can expect to pay $15 less than usual for the dress, or $35.Learn more about Fractions & Percentages For one to figure out their class grade percentage, they need to plug their individual grades into the formula that the teacher is using to determine their final grade. In most classes, different elements are worth a stated percentage of the whole grade. For instance, the midterm exam could be worth 20 percent, the final exam worth 20 percent, the assignments worth 50 percent, and class participation worth 10 percent.Full Answer > The multiplication of percentages is accomplished by converting the percentage to decimals, and multiplying the decimals. To convert a percentage to a decimal, the percent sign must be removed, and the number divided by 100. The decimal point should always be moved two places to the left when dividing by 100. After dividing 75 by 100, the answer is 0.75.Full Answer > The integer 25 can be expressed as an infinite number of equivalent fractions of the form 25a/a, where a is any integer. For example, 25 can be expressed as the fractions 50/2 (a=2), 75/3 (a=3), or 250/10 (a=10).Full Answer > Percent yield is simply the actual yield (the mass of resultant) divided by the theoretical yield (the most that can be attained). Therefore, the possibility of having a percent yield greater than 100 is impossible unless an error is made during the procedure.Full Answer >

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http://ieeexplore.ieee.org/xpl/articleDetails.jsp?reload=true&arnumber=1103840&sortType%3Dasc_p_Sequence%26filter%3DAND(p_IS_Number%3A24225)

math

Skip to Main Content Necessary and sufficient conditions for mean square stability are proved for the following class of nonlinear dynamical systems: finite-dimensional bilinear models, evolving in discrete-time, and driven by random sequences. The stochastic environment under consideration is characterized only by independence, wide sense stationarity, and second-order properties. Thus, we do not assume random sequences to be Gaussian, zero-mean, or ergodic. The probability distributions involved are allowed to be arbitrary and unknown. Limiting state moments are given in terms of the model parameters and disturbances moments.

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http://www.hotukdeals.com/misc/back-not-necesarilly-by-popular-demand-136440

math

Few questions that I would like to get answered if pos please. 1) I got banned for a day for selling on here and on ebay... which is fair enough, however the items sold on ebay were duplicates of those sold on here. anyway back to the question... what does the 2 points against my profile mean (2 points infraction or something). 2) Why was I not warned, or even spoken to before the ban... why was it just implemented. As I could have reammended any wrong doings. 3) Can I sell the remaining two pens on here (Completely different from the ones on ebay) without having my account banned any further. Many thanks, Don.

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https://domino.mpi-inf.mpg.de/internet/reports.nsf/c125634c000710d0c12560400034f45a/7c4ec7b97825b3bac12565da0051382d?OpenDocument

math

MPI-I-98-2-007. April 1998, 36 pages. | Status: available - back from printing | Next --> Entry | Previous <-- Entry Abstract in LaTeX format: We give a direct proof by generic reduction that a decidable rudimentary theory of finite typed sets [Henkin 63, Meyer 74, Statman 79, Mairson 92] requires space exceeding infinitely often an exponentially growing stack of twos. This gives the highest currently known lower bound for a decidable logical theory and affirmatively answers to Problem 10.13 of [Compton & Henson 90]: Is there a `natural' decidable theory with a lower bound of the form $\exp_\infty(f(n))$, where $f$ is not linearly bounded? The highest previously known lower and upper bounds for `natural' decidable theories, like WS1S, S2S, are `just' linearly growing stacks of twos. References to related material: |To download this research report, please select the type of document that fits best your needs.||Attachement Size(s):| |Please note: If you don't have a viewer for PostScript on your platform, try to install GhostScript and GhostView|

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