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https://stor.unc.edu/event/grad-student-seminar-nikolai-lipscomb/

math

- This event has passed. Grad Student Seminar: Nikolai Lipscomb 1 Apr @ 3:30 pm - 4:30 pm Grad Student Seminar: Nikolai Lipscomb1 Apr @ 3:30 pm – 4:30 pm Stochastic Optimization Methods for Outpatient Appointment Scheduling & COVID Testing Under Limited Testing Capacity This talk comprises of two parts based on two unrelated research topics: Part 1 deals with outpatient appointment scheduling problems when patients are unpunctual. We first look at data regarding patient unpunctuality and examine distributional factors such as sources of heterogeneity. When considering an optimization model, we consider a weighed cost of the total queue wait time, doctor idle time, and clinic overtime. We consider two modelling approaches. The first approach considers heterogeneous patient distributions in both service time and unpunctuality along with multiple service disciplines. Due to the complexity of the system and nonconvexity of the objective under more complicated assumptions, we propose several heuristics for solving the problem. The second approach examines the fluid-limits of the arrival-departure process and finds the queue-idle process is a solution to the one-dimensional Skorokhod problem associated with the arrival-departure process. Using this formulation, we derive a fluid control problem that can be solved numerically as a quadratic program. Part 2 looks at the problem of optimally choosing when to use COVID tests for a single patient when the supply of tests is limited. We consider an underlying SIR Markov model for the patient that is unobservable; instead, the only observations that are made are whether the patient is exhibiting symptoms at any point in time or whether a used test returns a negative or positive result. We propose a partially-observed Markov decision process (POMDP) for deciding when to use tests relative to the probability of being in each state. We are able to analytically derive an optimal policy for an arbitrary number of tests under the model assumptions.

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CC-MAIN-2024-10

http://vlab.amrita.edu/?sub=1&brch=281&sim=1516&cnt=5

math

1. From the simulator, what is the minimum deviation for sodium vapour lamp? 2. Find the angle of the prism, using spectrometer prism simulator? 3. What is the least count of Spectrometer apparatus in simulator? 4. Study the the variation of emerged ray angle with the variation of angle of incident? 5. Plote i-i' curve using simulator.

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https://aviation.stackexchange.com/questions/79432/are-there-any-remaining-flying-boat-or-seaplane-airliners-in-operation

math

Is there anywhere in the world that still offers scheduled air carrier service using flying boat / seaplane types of aircraft - something akin to a proper airliner? Or are they all limited to charters and shuttle-type flights on de Havilland Otters and the like? 3$\begingroup$ What exactly would you consider a "proper airliner"? A Twin Otter can certainly be an airliner ("Currently, 38% are operated as regional airliners", Wikipedia). Trans Maldivian Airways for example only operates Twin Otters. $\endgroup$– BianfableJul 9, 2020 at 7:21 $\begingroup$ Maybe you should be asking if there are any operable flying boats in operation before asking about scheduled services. $\endgroup$– GdDJul 9, 2020 at 7:59 1$\begingroup$ I know for a fact that there are flying boats still operable. For example Grumman HU-16 was in use by the U.S. Coast Guard until the mid-80s and in the Greek Navy until the mid-90s. Many of these aircraft are still in private use today. $\endgroup$– Aaron HolmesJul 9, 2020 at 12:49 1$\begingroup$ I understand that the term "airliner" can be vague. Perhaps a good follow up question would be: If there are airlines still operating on the water, what are some of the most significant in terms of passenger carrying capacity and/or distance of routes? $\endgroup$– Aaron HolmesJul 9, 2020 at 13:27 $\begingroup$ Hey, uh, maybe it's kinda late, but could you clarify whether you want to know specifically about flying boat or about all seaplanes in general? Your title says "flying boat", but in your question, you refer to using flying boat / seaplane Moreover, you go on to ask are they (seaplane operations) all limited to charter or shuttle-type flights on De Havilland Otters which could give the impression either that the Otter is a flying boat (it's not) or that your question is not limited to flying boats (but your title is?). $\endgroup$– AbdullahJul 10, 2020 at 10:10 As far as I know, there is no actual flying boat in use at an airline. The flying boats that are still in commercial operation are typically used for fighting forest fires or maritime patrol. This may change in the near future since there are currently at least two flying boats in development that might be used by airlines: Dornier Seastar CD2: This is a newer model of the previous Seastar, which had its first flight earlier this year: Dornier Seawings flew its new-generation Seastar amphibian for the first time on 28 March from its base in Oberpfaffenhofen, Germany. Dornier says the aircraft will be offered in various configurations including cargo, passenger, special missions and VIP transport. (flightglobal.com, emphasis mine) AVIC AG600: This is new aircraft developed in China and one of the biggest flying boats ever. It had its first flight in 2017 and according to Wikipedia is expected to be delivered by 2022, possibly also in a passenger variant: Further variants may be developed for maritime surveillance, resource detection, passenger and cargo transport. If you are not just interested in pure flying boats, but also floatplanes, then there are many airlines world wide that operate these on regularly scheduled flights. One of the biggest is Trans Maldivian Airways operating various variants of the DHC-6 Twin Otter. They operate 55 seaplanes, which according to Wikipedia makes them the biggest seaplane airline: TMA currently operates the world's largest seaplane fleet. They offer flights to over 80 destinations: Trans Maldivian Airways currently offers transfer services to more than 80 Maldives resorts, flying over 1 million passengers per year to their holiday hideaways. This is their main terminal at Velena International Airport: (image source: Wikimedia) 1$\begingroup$ Having flown TMA 3 times, I can say they're my favourite. Not the most comfortable, not the most luxurious but it means youre about to spend a week or 2 in pure luxury! $\endgroup$– Jamiec ♦Jul 9, 2020 at 15:28 1$\begingroup$ The AVIC AG600 Wikipedia article makes it sound as though the anticipated passengers may be troops on their way to the Spratly Islands in the South China Sea, rather than fare-paying members of the public. That's pure speculation of course. $\endgroup$ Jul 9, 2020 at 19:22 1$\begingroup$ @rclocher3 Yes, I think at the moment it is primarily developed for the military. That's why I quoted "Further variants may be developed". If there is a market, they may use it for normal passengers as well, but maybe not... $\endgroup$ Jul 9, 2020 at 19:29 2$\begingroup$ I like the Dornier registration "D-ICKS" (D-X) as a nod to the Dornier DO-X. $\endgroup$ Jul 10, 2020 at 11:14 1$\begingroup$ @BrianDrummond haha, I hadn't noticed. What a wonderful registration :D $\endgroup$ Jul 10, 2020 at 11:17 Last time I was there (fifteen or so years ago) there was a thriving commercial air passenger operation, including at least a couple scheduled flights (weekly, I believe, not daily), at the north end of Lake Washington (near Seattle), officially in either Bothell or Woodinville, Washington state (comments reminded me it's Kenmore Air, in Kenmore, WA). There were no flying boats in service there, but there were Beavers and Otters on floats. Beyond that, there are a number of tiny (one or two aircraft) airlines operating scheduled feeder service in Alaska and northern Canada that operate on floats, because lakes and rivers are much more common in the wilder parts of the sub-Arctic and Arctic than even grass or gravel airstrips. As long as there's water to land on at both ends, such aircraft are good to go (and some have retractable gear, so they can land on concrete if necessary) -- and if you're starting from a village equidistant from Fairbanks and Nome, these are likely to be your only choices for local service. The bulk of their revenue is likely from hunting and fishing charters, like any bush operation -- but many of them operate weekly or twice-weekly scheduled flights to a hub terminal. $\begingroup$ There is an extensive scheduled service float network operating around the southern BC coast; Harbour Air, Sea Air, Vancouver Island Air, West Coast Air and others.They run VFR-only sched services, meaning scheduled trips that make the rounds whether there is anybody on board or not. There is a steady scheduled seaplane commuter flow between Vancouver harbour and Victoria (the capital) and Nanaimo on Van Island.There is so much traffic that Vancouver Harbour is the only water aerodrome I know of that is not part of a land airport, and has a PCZ with a Tower (atop a waterfront office building). $\endgroup$– John KJul 9, 2020 at 14:32 2$\begingroup$ You are thinking of Kenmore Air, which is in Kenmore. $\endgroup$ Jul 9, 2020 at 15:09 1$\begingroup$ You beat me too it! 3 miles from my house... They also fly wheeled Caravans out of BFI to the San Juan Islands. $\endgroup$ Jul 9, 2020 at 15:56 $\begingroup$ @AzorAhai--hehim I think you're correct. I used to drive past there occasionally, late in the day, when I was a Seattle cab driver. "City" borders are a little nebulous in that part of King/Snohomish counties. $\endgroup$ Jul 9, 2020 at 16:07 1$\begingroup$ Haha, not to sound rude, but I am correct, I'm from Bothell. Neither Bothell nor Woodinville actually has any waterfront. But I can see why it would be confusing, they do just run into each other. $\endgroup$ Jul 9, 2020 at 16:16 There is at least one jet flying boat, the Beriev B-200, in production. The Wikipedia article gives it a capacity of 72 passengers. I don't know how many (if any) are employed in an airliner role, though this variant has a "pressurised and air conditioned cabin allowing transportation of up to 72 passengers. The passenger variant is the BE-210 shown here at beriev.com Most seem to be employed as water carriers in a firefighting role; it was designed to skim the surface and scoop up water at close to takeoff speed. Four or five years ago or so, Pacific Coastal Airlines spun off its seaplane division as Wilderness Seaplanes, which flies scheduled service in a fleet of four Grumman Gooses in British Columbia. $\begingroup$ Grumman Geese?? $\endgroup$ Jul 10, 2020 at 11:09 $\begingroup$ I wish you'd kept the spelling. $\endgroup$ Jul 10, 2020 at 12:50 Not "air carrier service", but Viking Air acquired from Bombardier the type certificates and continues to produce the CL-215, CL-215T, CL-415 and CL-415EAF as an aerial firefighter, an amphibious aircraft. Again, not "air carrier service" but a true flying boat, Coulson Flying Tankers effectively grounded the last Martin Mars water-bomber in 2018. They are fondly missed. The following relates to "Float planes", rather than "flying boats or seaplanes" ... Viking Air also hold the type certificate for the legendary De Havilland DHC-2 Beaver DHC-3 Otter and DHC-6 Twin Otter. A refreshed, next-gen specification Twin Otter - series 400 is now back in Production. From their website, With a fleet of 57 de Havilland Twin Otters, Trans Maldivian Airways operates the largest seaplane operation in the world. TMA is the airline referenced in @Bianfable 's answer. NORDIC Seaplanes flies daily between the two largest Danish cities. 1$\begingroup$ With flying boats or floats? $\endgroup$ Jul 9, 2020 at 18:11 5$\begingroup$ It appears they fly just one aircraft, and that is a Twin Otter floatplane (!=flying boat) $\endgroup$ Jul 9, 2020 at 18:16 $\begingroup$ The title just asks about flying boats, but the question body also includes "seaplane." A Twin Otter isn't a flying boat, but it is a seaplane. $\endgroup$– reirabJul 10, 2020 at 7:13 5$\begingroup$ This post is an answer, albeit not a good one. The tool to deal with this is voting, not deletion. $\endgroup$– boglJul 10, 2020 at 8:39 1$\begingroup$ Now it turns out that my flag seems to be unnecessary, as the question turned out to include floatplanes. $\endgroup$ Jul 14, 2020 at 15:47

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CC-MAIN-2023-23

https://byjus.com/question-answer/find-the-difference-between-the-rational-numbers-b-and-a-as-shown-on-the-number-1/

math

Find the difference between the Rational numbers B and A as shown on the number line. B lies between 2 and 3 and it is in the middle. Value of B = 212 or 52 A lies between -1 and -2 and it is in the middle. Value of A = −112 or −32 Difference between B and A = B - A = 52−−32 = 5+32 = 82 = 4 We can also use the number of points on the number line that we travel left from B to A to get the same value.

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http://hicourseworkuikq.1hourloans.us/need-help-with-math.html

math

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CC-MAIN-2018-39

https://meta.mathoverflow.net/questions/4264/should-we-encourage-non-professionals-to-ask-hard-questions-if-that-is-good-for/4269

math

There is a story about Gauss being "unimpressed by FLT, pointing out (after finding a proof of the n=3 case) that in number theory it’s very easy to come up with problems that are very hard.". But, it is known that there were many developments in number theory since the time of Gauss, and some of them were needed to settle FLT (I am not implying that Fermat did not have a proof, an "elementary" one, but the margin was too small). Also, the conjecture of Goldbach surely was out of reach in the time of Goldbach and Euler, but today it stands as an important problem in the theory of additive bases, in the subfield of additive number theory. Therefore, I am of the opinion that a "good" mathematical question, no matter how hard, could be a source for a new developments in the field, and, possibly, for some future applications, outside or inside the field itself. Because of that, I am of an opinion that questions that generate interest and are hard should be encouraged, and should not be closed because closing of them discourages a person who asked that question. Also, importance of a question and its impact on the field is very subjective matter, and, what to someone seems as just another question, to someone else could be a source of inspiration for a lifelong study. So many of us are able to ask questions that are hard, but this question is more about non-professionals than professionals, for example, I, as a non-professional, must think so much about some topic to hopefully arrive at a question that could interest some of you, while many of you can in a fraction of an hour ask a couple of questions that are very hard and, maybe, non-answerable with current techniques. Because of that, I think that it is recommendable that some of non-professionals actively participate at the site for professionals, and that they ask questions that are hard, not just hard but also those that are very hard (if possible), because, a smart question, no matter how elementary in its formulation, can generate so much of research and can be a source of new developments. From all of this I would also like to add that I do not like the politics of putting questions "on hold", or "closing" them, just because they are elementary in formulation and very hard or hard or at least complicated to settle. Here the rule "the harder the better" can be applied, not because of our present inability to settle some questions, but because of enrichment of the field itself. As about me, I am more of a conjecturer in mathematics than developer of mathematics, but the act of making conjectures in mathematics and of developing the field cannot be separated, they are very intimately connected. Also, I think that professionals should offer as much help as possible to non-professionals, if the intentions of non-professionals are good, because it is possible that some non-professional becomes a professional, but without help, it is possible even without help, but it is hard if there is no help at all. A conclusion would be, that: Non-professionals should be encouraged on this site if they ask a question even if it is elementary in formulation, and that generated a discussion, but did not receive an answer that settled a question, and that question should not be put "on hold" or closed, if it was asked in a clear and systematic way.

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https://samaralansari.blogspot.com/2013/02/

math

Wednesday, February 27, 2013 I had always dreaded goodbyes, feared separating from my family all my life and was destined to face the hardest separation of all. Since the day I started going to boarding school, my only fear in life was separation as I went to boarding school at the age of six. At my first boarding school, I had my older sisters with me and two years later my younger brother joined the boys’ school and I was able to see him during school days as the young boys came daily to our school until grade 3. I remember that I spent all my breaks with him and felt very sad that he had to leave at the end of the day while I and my two other sisters were fortunate to stay together. It broke my heart to see him leave every afternoon and I looked forward to seeing him the next day. My nightmare as a child was that the summer vacation will end and we will no more be together as a family as we had to go back to school. I remember my sleepless nights as the second half of the vacation started announcing that the end of the holiday is on its way. My only joy then was to be at home with my parents and siblings. Nothing more! After finishing elementary school in Ramallah, I went to another boarding school in Beirut; leaving behind my siblings except for my eldest sister who went to university in Beirut and was able to see me at weekends. The fear of having summer vacations end grew more. During the summer holiday, I spent most of my time at home and hardly had the joy that young girls would have in going out. I was satisfied by being around family and by reading novels. My family and books were all what I needed. I did not aspire to go out as my joy was by spending my time with my family. I learnt to play backgammon, my father's favorite game so that I can play with him and spent time in the kitchen with my mother helping her. I did not want to lose any of those moments without my family. Fear of separation from my family stayed with me as I went to university and when I started working in Bahrain. Sometimes our destiny is to stay separated. I hated saying goodbye to my parents and to my siblings and shed tears at every goodbye. Little did I know then that the hardest goodbye was few years ahead and I would be shedding tears all my life. I got married in Bahrain and continued to live away from my family. My tears never stopped when I said goodbye particularly to my parents. It took me several days to recover and settle after coming back home. I had three wonderful kids and they and my husband were my new family. I could not imagine separating from them unless I had to. I was compelled to leave my two older kids with my family when I went to do my doctorate defense in the UK. My son was going to turn three that month and my daughter, two years in three months. I was eight months pregnant with my third and youngest child, so I guess I can say she came along with me. I cried my heart out leaving my two children during the five days I was away. I had sleepless nights and called them daily to hear their voices. As time went by and even after having my own family, I kept on crying while saying goodbyes to my parents and siblings. The day then came when my eldest child had to leave to university in the UK followed a year after by my second child and it was very painful. I knew of course that it is for their best but I could not help but suffer. The time came when my youngest daughter had to leave to university to join her siblings and keep the nest empty. I was so scared of this separation and she was worried about me. However, destiny had planned a forced goodbye for me, a separation unlike any other! My daughter died at the age of 18 in a car accident two weeks before going to university and I had to face the worst separation ever and the hardest goodbye! Randah R Hamadeh Friday, February 22, 2013 Samar Al Ansari a) The pink graph represents the base function, y=x². It is a quadratic function with a vertex of (0,0). b) The black graph represents the function y=x²+3. It is similar to the first graph, but it is shifted vertically three units upwards. It has a vertex of (0,3). c) The red graph represents the function y=x²-2. It is similar to the first graph, but it is shifted vertically two units downwards. It has a vertex of (0, -2). The pattern I have noticed is that they all have the same basic parabola shapes, but is their vertical shifts which differentiate them. The general formula of quadratic equations is y=x²+k. If k is positive, then the graph is shifted vertically upwards. If k is negative, then the graph is shifted vertically downwards. In the first function, y=x², k=0, so there is no shifted and the graph is located on the origin. But in the second function, y=x²+3, k=3, a positive number, thus it is shifted vertically upwards three units. In the third function, y=x²-2, k=-2,, thus it is shifted vertically downwards two units. a) The pink graph represents the function, y=x². The vertex of this graph is unchangeable, because nothing is altering the y-values or the x-values (nothing is being added/subtracted/multiplied to/divided by the x/y values). It is the standard graph of a parabola, and is used as a comparison for the other functions. b) The black graph represents the function y=(x-2) ². This function is shifted 2 units horizontally to the right, because the x-value is being altered by the opposite of the number added on to it within the parenthesis. The vertex is changed from (0,0) to (2,0), because the vertex is shifted two units to the right since the equation is in the second degree and the opposite of the number within the parenthesis is the number changing the vertex. c) The red graph represents the function y=(x+3) ². This function is shifted three units horizontally to the left. The vertex is changed from (0,0) to (-3,0). The pattern in these graphs is once again the same basic parabola shape, but their difference lays in the x-values; they are shifted horizontally either to the left or to the right. It is obvious now that all equations rely on the formula y-k=a(x-h) ², the vertex being (h,k). If h=positive, then the graph is shifted horizontally to the right, if h=negative then it is shifted horizontally to the left. The same goes for k, if it is positive, the graph is shifted vertically upwards, if it is negative, it is shifted vertically downwards. Thus in the second function, y=(x-2) ², the vertex will be (2,0) because the vertex is derived from this equation y-0=(x-2) ². Thus the vertex of the third function y=(x+3) ² has a vertex of (-3,0) based on the formula. 3. The vertex would be (4,5) because the equation for this particular vertex is y-5=(x-4) ², and the general formula is y-k=(x-h) ². To get the vertex out of the equation you have to take the opposites of k and h in the formula to get (h,k). The first part is to take the 5 or k to the other side so you now get y-5=(x-4)^2. Then according to the formula you get a vertex of (4,5). a) x²-10x+25 in the form (x-h) ² is: x²-10x+25=0 work on the side: (10/2) ²=25 b) x²-10x+32 in the form (x-h) ² +g is: (x-5) ² + 7=0 * 2x²+8x+1=0 work on the side: (4/2) ²=4 (x+2) ² + 7/2 * 6x²+x-12=0 work on the side: (1/6 /2)²=0.007 6x² + x=12 6(x² + 1/6x)=12 6( x² + 1/6x + 0.007)=12+0.042 6( x+ 0.083) ²= 12.42 (x+0.0083) ² -2.07 * 9x²-36x-45=0 work on the side: (4/2) ²=4 1- If c isn’t already on the other side, take it to the other side of the equation. 2- If a≠1, set it equal to 1 by factoring it out. 3- Divide b by 2, and square the answer you get. 4- Add that to both sides of the equation. Note: if a≠1 at the beginning, then when you add the answer of step 3 to the right side of the equation multiply it by the value of a at the beginning. 5- Factor the left side of the equation. 6- If a≠1 at the beginning, divide the right side of the equation by the value of a at the beginning. 7- Add the right side of the equation to the left side of the equation in order for the answer to be in the (x-h) ² +g form. - The graph of (x-h) ² +g would be a parabola that is opened upwards, because it is a quadratic function. There isn’t a negative sign outside of the brackets, thus the graph is not reflected on the x-axis. If there was a negative sign outside the brackets, then the shape would be that of a parabola opening downwards. The vertex of the function can be found by rearranging the formula to y-g=(x+h) ², thus the vertex would be (h,g). There will be both a vertical and horizontal shift, it will shift vertically upwards, and horizontally to the right. The vertex was originally located at (0,0) but has been shifted to (h,g). 6. This can be applied to absolute value function as well, y-k= a|x-h| The blue function is the base function, with a vertex of (0,0). The green function represents the function of y=|x| -2, thus it is shifted 2 units vertically downwards, making its vertex (0,-2). The black graph represents the function y=|x-2|, thus the graph is shifted 2 units vertically to the right, making its vertex (2,0). The general formula of an absolute value function is y-k=a|x-h|. Thus to find the vertex, you multiply h and k by -1 making it (h,k). Question 1-a) and the blue function above share the same vertex (0,0) as well as an axis of symmetry of x=0 which is the same as the blue graph above. Also, 1-c) and the green graph above share the same vertex (0,-2). The general formula for 1-c) is y=x² +k, the general formula for the green graph can be written as y=|x|+k, since h=0. The graph of 2-b) shares the same vertex of the black graph above (2,0). All the absolute graphs follow the v-shaped pattern and are differentiated based on the vertical or horizontal shifts. The general formula of 2-b) could be written as y=(x-h) ², and the general formula of the black graph can be written as y=|x-h|, since k=0. Although quadratic functions create a parabola graph and absolute value functions create v-shaped graphs they both have axis of symmetry and vertexes. The graph of a quadratic function is in the second degree while the graph of an absolute value function is in the first degree. Both types of functions are symmetrical to the y-axis. Also, their general formulas are similar because they share the same transformations. The general formula of quadratic functions is y-k=a(x-h) ², while the general formula for absolute value functions is The graph above shows the functions of radical equations, the pink graph represents the function, y=√x, the black graph represents the function, y=√x -2, and the red graph represents the function y=√(x-2 ). The pink graph represents the base function, with a vertex of (0,0). The black graph is shifted two units down, shifting the vertex to (0,-2). The red graph is shifted two units to the right, making its vertex (2,0). Question 1)a and the pink function above share the same vertex (0,0), question 1)c and the black function above share the same vertex (0-2), and question 2)b and the red function above share the same vertex (2,0). Once again they all follow the same general pattern which is the shape of the graph above, but differ in either terms of x-values or y-values based on what type of shifts are implemented. The general formula of graphs containing radical function is y-k=√(x-h), which is very similar to the general formula of a quadratic function.

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https://www.rainsbrook.co.uk/wiki/doku.php/c4c/units

math

Table of Contents In several places in this website, units which measure various things are mentioned, this page will try to explain them. The smallest unit of measuring data storage is a bit. One bit can be either on or off, commonly represented by a 1 or a 0. This corresponds to a voltage being present or absent. Bytes, Kilobytes, Megabytes, Gigabytes and Terabytes. One Byte is composed of 8 bits. Why 8 bits and not 10? Data in computers is counted in what is referred to as Binary which just used 1's and 0's to count with. This might seem confusing, but as computers are electrical and electricity can be either on or off, it is convenient to use. Binary numbering is called base-2 meaning we carry over in powers of 2. So the sequence of numbers in binary is 1, 2, 4, 8, 16, 32. 10 is missed out of the sequence. Normal counting uses numbers 0 to 9 and is called denary, sometimes also called base-10. Humans probably started to count in multiples of 10 because we have 10 digits between two hands. Compared to binary, the sequence of denary numbers is 1, 10, 100, 1000, 10000. Sometimes four bits are combined together and called a nibble. likewise two bytes together are sometimes called a word and two words are called a A thousand Bytes are called a Kilobyte (the same as 1000 metres are called a kilometre) but because it is a “binary 1000”, one Kilobyte is actually 1024 bytes. A thousand Kilobytes (or a million Bytes) are called a Megabyte, if we had a very long road which was 1000 Kilometres long, it could be called a Megametre. As with Kilobytes, a Megabyte is actually 1024 Kilobytes. There is a pattern here, the next unit, a Gigabyte is 1024 Megabytes, and going up again, we have a Terabyte which is 1024 Gigabytes. When we count in real life, we count in multiples of 10, so we start at 1 and count to 10 and then add start again. Actually we should start at 0 and count to 9, so the next number is 10. 0-9 are single digits, but 10 is two digits. As we count another 10, we change the 1 to a 2, and the same with 2 to a 3 for thirty. When we reach 99, the next jump adds a third digit and becomes 100. This is known as Denary maths (Middle English from Latin, originally, adjective, containing ten from deni, from New World Dictionary). The table below shows this a different way. In the first example there are the same number of number, it doesn't matter if we start from 0 or 1. |Counting from 1 to 10||1||2||3||4||5||6||7||8||9||10| |Counting from 0 to 9||0||1||2||3||4||5||6||7||8||9| This table shows how we keep adding 1 and how we carry over when we reach 9, reading down from left to right. Binary maths works the same way, but we just have 0 and 1. The first binary number is 0b (zero denary), the second is 1b (one denary),the next number after 1b is 10b (two denary). I have used a “b” after the numbers here to show they are binary numbers and not denary. This table shows how binary counting works, again reading down from left to right. These numbers correspond to

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CC-MAIN-2020-10

http://encyclopedia.kids.net.au/page/po/Pope_Stephen_VI

math

There is a problem in numbering the Popes Stephen -- see Pope Stephen II for the explanation. ... given run of the algorithm, it has a probability of at most 1/4 of giving the wrong answer. That is true, whether the answer is YES or NO. The choice of 1/4 in th ... Kids.Net.Au - kids safe portal for children, parents, schools and

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CC-MAIN-2014-42

https://projecteuclid.org/journals/annals-of-probability/volume-47/issue-4/A-general-method-for-lower-bounds-on-fluctuations-of-random/10.1214/18-AOP1304.short

math

There are many ways of establishing upper bounds on fluctuations of random variables, but there is no systematic approach for lower bounds. As a result, lower bounds are unknown in many important problems. This paper introduces a general method for lower bounds on fluctuations. The method is used to obtain new results for the stochastic traveling salesman problem, the stochastic minimal matching problem, the random assignment problem, the Sherrington–Kirkpatrick model of spin glasses, first-passage percolation and random matrices. A long list of open problems is provided at the end. "A general method for lower bounds on fluctuations of random variables." Ann. Probab. 47 (4) 2140 - 2171, July 2019. https://doi.org/10.1214/18-AOP1304

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CC-MAIN-2021-43

https://www.physicsforums.com/threads/relativistic-bound-states.587551/

math

I'm dealing with a dirac particle in an attractive spherical square well. I've solved for the transcendental equation to find energy, found the normalized wave function, and now I'm trying to explain what happens when the well becomes very deep, when V0 ≥ 2mc2. If I plug V0=2mc2 into my equations, nothing dramatic really sticks out. I have the same equation as Greiner on page 220, Eq.(27). I'm trying to get the values in the left hand side of table 9.2 on page 222 (V0(κ=-1)), but I'm not sure how to go about it. If I can't solve for this, I need to come up with another consequence of the well depth approaching 2mc2. The Attempt at a Solution Last edited by a moderator:

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CC-MAIN-2021-31

https://boards.straightdope.com/t/the-edge-of-the-universe/518007

math

What it demonstrates, Jonathan Chance, is that no one has a good understanding of just what spacetime is at present. Physicists and philosophers have debated for centuries the question of whether there exists an absolute (if dynamic) canvas or if space and time can only be defined in a relational sense. The overwhelmingly predominant interpretation of General Relativity ties gravity to distortions of spatiotemporal geometry; implicit in that statement is the notion that there must be something there to carry those distortions. The mathematics of GR are easiest to use in this formulation, however it would be just as valid to construct it in a completely relational way. Bottom line: we cannot answer the question of what it means for spacetime to be expanding until we understand what spacetime is. This is one of the fundamental questions that a successful theory of quantum gravity will have to answer. Edit: I should add, though, in the context of this example about Universe expansion, it is important to remember that we do already know that spacetime is part of the Universe. Space and time did not exist before the Big Bang, the Big Bang created them along with everything else. Once you abandon the notion of spacetime as some sacrosanct background ruler, it becomes easier to accept that it might do something like expand.

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http://en.allexperts.com/q/Advanced-Math-1363/2014/1/lpp.htm

math

A dealer wishes to purchase table fan and ceiling fans.He has Rs57,6000 to invest,and has space to store 40 items.A table fan cost Rs760 and ceiling fan cost Rs900.He can make profits Rs70 and Rs90 by selling a table fan and ceiling fan respectively.Assuming that he can sell all the fans that he can buy,formulate this problem as LPP,to maximize the profit. Here's how I would set up this problem as an LPP (Linear Programming Problem). The solution consists of determining the number of units (table or ceiling fans) the dealer should buy/sell. The function to maximize is the profit and the constraints are the number of units his facility holds as well as the amount of money he has to invest. So Number of units: Nt = tables fans, Nc =ceiling fans; solve for Nt and Nc with Nt + Nc ≤ 40. Cost of units: Ct = table fan, Cc = ceiling fan; constraint Ct + Cc ≤ 57,6000 = I Profit per unit: Pt = table fan, Pc = ceiling fan; maximize Ptot = Pt + Pc. MS Excel has an tool called Solver that solves these sorts of LPP problems and is fairly easy to implement (has tutorial). This problem, as you state it, has some ambiguities. First of all, the Rs57,6000 (which I assume really means Rs 576,000 in more conventional notation) would enable the dealer to purchase far more than the 40 units, of either or both types, he could store. Also, the higher profit for the ceiling fan indicates that he should buy and sell only that type of fan. To wit, if the dealer only buys/sells one type of fan then total profit for table fan: (I/Ct)(Pt) = (570,000/760)(70) = Rs53,053 (I/Cc)(Pc) = Rs57,000. So something is fishy here (or I've made a mistake). Anyway, look over your problem and let me know if you have a follow-up

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http://www.chegg.com/homework-help/speed-light-light-travels-186-000-miles-per-second-space-dis-chapter-6.10-problem-92e-solution-9780321501073-exc

math

Speed of Light Light travels at about 186,000 miles per second through space. The distance, d, in miles that light travels in t seconds can be determined by the function d(t) = 186,000t. a) Light reaches the moon from Earth in about 1.3 sec. Determine the approximate distance from Earth to the moon. b) Express the distance in miles, d, traveled by light in t minutes as a function of time, t. c) Light travels from the sun to Earth in about 8.3 min. Determine the approximate distance from the sun to Earth.

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CC-MAIN-2016-40

https://sonichours.com/how-much-is-134-kg-in-pounds/

math

The easiest way to figure out how much is 134 kg in pounds is to use a conversion table. A conversion table can be found here. A calculator that provides a decimal value can also be found. A scientific notation is often used, as it is more accurate and more convenient. If you’re unsure of the unit’s meaning, the best option is to look up 134 kilo to lb by using the shorer formula. For ease of reference, we’ve provided a chart that can be used to determine how much 134 kilograms is in pounds. The chart below provides the formula to convert 134 kg to lb. It is important to note that the two units aren’t exactly equivalent. Rather, they’re just for completeness. When converting weights, always use the avoirdupois pound. There are two main methods to convert 134 kg to pounds. The first is to multiply 133 by 2.204622622. The second method is to divide 134 by 0.45359237. If the latter method is more convenient, use the former. If you use the latter method, your results will be more accurate. The other method is to convert 134 kilograms to pounds by dividing it by 1.4. To convert 134 kg to pounds, simply multiply 134 by 2.204622622 or divide it by 2.2. The result is the same for both. To find out the exact conversion between kilograms and pounds, simply look at a conversion table. The result will tell you how much a kilogram is in pounds. Once you’ve done that, you’ll be able to convert 134 kg to pounds in no time at all. To convert 134 kg to pounds, divide 134 by 2.204622622 or by 0.45359237. To calculate 134kg in pounds, multiply 137 by the same number of Troy ounces. A single Troy ounce is the equivalent of a half-pound. The metric system uses the avoirdupois pound. In this case, a pound is equal to 0.455359237 kilograms. In metric systems, the pound is equal to 134 kilograms. The pound is an equivalent unit of 0.45359237kg, so it’s easy to see how to convert 134 kg into pounds. This unit of mass is also known as an ounce. If you’re curious about a kilogram, it’s equal to 0.45359237 ounces. In other words, a ounce is an ounce. To convert 134 kg to pounds, multiply 136 by two. Then, divide the result by 2.4562622. Then, multiply the resulting number by 0.4559237 to get 134kg in lb. Then, subtract it from a Troy ounce to convert a pound. You’ll get a total of 133 lb by multiplying 137 by a factor of 2.204622622. The pound unit of mass is used to measure precious metals. One Troy ounce is equal to 0.45359237 kg. Hence, a pound equals 134 kilograms. However, it is not a common practice to convert a lb to a kilogram. It is only appropriate to use a single Troy ounce. If you’re comparing the two units, you’ll need to take note of their conversion ratios. When converting a pound to kg, multiply the kilogram by two. If the result is 134kg, then it would be a pound of 3.444lb. So, a pound of 134kg equals 0.45359237 lb. If you’re unsure, try calculating a lb in ounces. It’s not that hard to understand. A kilogram is a standard unit of weight in the Metric system. Its mass is approximately equal to 10 cm cubes of water. So, a metric kilogram is a pound of water. If you’re wondering how much 134 kg is in pounds, simply multiply the weight of the ounce. This will give you the conversion of a pound to a kilogram in two different ways.

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https://sinemaportali1.com/qa/what-is-a-4-dimensional-world.html

math

- What would a 4 dimensional world be like? - What is the meaning of 4 dimension? - Does a 5th dimension exist? - Is time really a dimension? - How many dimensions can humans see? - What’s bigger a universe or a dimension? - Are humans 4 dimensional? - What are the 26 dimensions? - What are the four known dimensions? - What is the 8th Dimension? - What is the 10th dimension? - Is a wormhole possible? - Is there a 4th Dimension? - What is the 4th dimension in simple terms? - How many dimensions are proven? - How many dimensions do we live in? - Why is time the 4th Dimension? What would a 4 dimensional world be like? A 4-D being would be a god to us. It would see everything in our world. It could even look inside your stomach and remove your breakfast without cutting through your skin, just like you could remove a dot inside a circle by moving it up into the third dimension, perpendicular to the circle, without breaking the circle.. What is the meaning of 4 dimension? 1 : a dimension in addition to length, breadth, and depth specifically : a coordinate in addition to three rectangular coordinates especially when interpreted as the time coordinate in a space-time continuum. Does a 5th dimension exist? The fifth dimension is a micro-dimension which is accepted in physics and mathematics. … As of now, we can’t see the fifth dimension, but rather, it interacts on a higher plane than we do. It’s because of this that we can’t really study nor fully prove it’s existence. Is time really a dimension? Time is just as good a dimension as space is, as no matter how you boost yourself through space, you must always move forward through time. … If time weren’t a dimension with the exact properties it possesses, special relativity would be invalid, and we could not construct spacetime to describe our Universe. How many dimensions can humans see? two dimensionsWe are 3D creatures, living in a 3D world but our eyes can show us only two dimensions. The depth that we all think we can see is merely a trick that our brains have learned; a byproduct of evolution putting our eyes on the front of our faces. What’s bigger a universe or a dimension? It is, necessarily, smaller than the set of all locations that there could be in the universe, so, in this sense, dimension is smaller than the universe. … It is, necessarily, smaller than the set of all locations that there could be in the universe, so, in this sense, dimension is smaller than the universe. Are humans 4 dimensional? We are actually four dimensional. We are comprised of 4 distinct but integrated parts. Three of which are related to our physical experience – the body, heart and mind. The fourth is the dimension of consciousness or spirit. What are the 26 dimensions? The 26 dimensions of Closed Unoriented Bosonic String Theory are interpreted as the 26 dimensions of the traceless Jordan algebra J3(O)o of 3×3 Octonionic matrices, with each of the 3 Octonionic dimenisons of J3(O)o having the following physical interpretation: 4-dimensional physical spacetime plus 4-dimensional … What are the four known dimensions? Positions along these axes can be called altitude, longitude, and latitude. Lengths measured along these axes can be called height, width, and depth. Comparatively, four-dimensional space has an extra coordinate axis, orthogonal to the other three, which is usually labeled w. What is the 8th Dimension? In physics the 8th dimension contains all other dimensions, therefore including everything. In medieval numerology 8 signifies eternity or infinity, which leads to the next life. The Buddhists speak of the eightfold path to enlightenment. In Christian numerology 888 represents Christ or Christ the redeemer. What is the 10th dimension? Finally, the 10th dimension is a single point that represents all the possible branches of every possible timeline of all the potential universes. … To recall string theory, superstrings vibrating in the 10th dimension are what create the subatomic particles that make up not only our universe, but all universes. Is a wormhole possible? A Harvard physicist has shown that wormholes can exist: tunnels in curved space-time, connecting two distant places, through which travel is possible. … “It takes longer to get through these wormholes than to go directly, so they are not very useful for space travel,” Jafferis said. Is there a 4th Dimension? Most of us think of time as the fourth dimension, but modern physics theorizes that there is a fourth spatial dimension as well—not width, height, or length but something else that we can’t experience through our physical senses. What is the 4th dimension in simple terms? In geometry, the fourth dimension is related to the other three dimensions by imagining another direction through space; just as the dimension of depth can be added to a square to create a cube, the fourth dimension can be added to a cube to create a tesseract. How many dimensions are proven? The world as we know it has three dimensions of space—length, width and depth—and one dimension of time. But there’s the mind-bending possibility that many more dimensions exist out there. According to string theory, one of the leading physics model of the last half century, the universe operates with 10 dimensions. How many dimensions do we live in? In everyday life, we inhabit a space of three dimensions – a vast ‘cupboard’ with height, width and depth, well known for centuries. Less obviously, we can consider time as an additional, fourth dimension, as Einstein famously revealed. Why is time the 4th Dimension? “Time is ‘separated’ from space in a sense that time is not a fourth dimension of space. Instead, time as a numerical order of change exists in a 3D space. Our model on space and time is founded on measurement and corresponds better to physical reality.”

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CC-MAIN-2021-10

http://mathhelpforum.com/calculus/74336-help-cal-hw.html

math

Question 1 is fairly straightforward. Since only 12 years have passed the population will only increase by . So the population will increase by: Moving on to question 2. First of all to get your position you need to integrate your formula for velocity and take into account the constant: To find the value for the constant use the information given when t= 0, x = -2: To find where the particle is you just need to put each of the values asked for into the equation in place of t. Finally for 3: Rearrange the equation so x terms are all on the same side and the same with the y terms Then integrate to get: Rearrange to make find an equation for y:

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CC-MAIN-2016-50

http://zbmath.org/?q=an:1119.65121

math

The author presents a survey of recent results on the attainable order of (super-)convergence of collocation solutions for systems of Volterra functional integro-differential equations of the form Here, is a delay function, and the (nonlinear) operator is either the Nemytskij operator (or: substitution operator) with delay, , , or the weakly singular delay Volterra integral operator , with kernel given by While the right-hand side in (1) could also depend on more general (non-Hammerstein) operators, including delay operators, the author restricts the analysis to Volterra integral operators of the form . The functions , , , and are assumed to be smooth on their respective domains. Related functional equations and theoretical and computational aspects of collocation methods for their solution are described.

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CC-MAIN-2014-15

https://books.google.fr/books?id=guChbCwGJfQC

math

Avis des internautes - Rédiger un commentaire Aucun commentaire n'a été trouvé aux emplacements habituels. Autres éditions - Tout afficher Crocker's Elements of Land Surveying. to Which Are Added, Tables of Six ... Abraham Crocker,Richard Farley Aucun aperçu disponible - 2018 ABCD abutment Ansr arithmetical base bearing Bisect centre chain chord circle Circular Segments circumference Co-s construction convenient Cosine Cotang cross decimal diagonal diameter Diff difference direction distance divided draw the line drawn ellipse equal error Euclid Examples for Practice feet fence field field-book figure find the Area geometrical given half sum height horizontal inches instrument intersection land Land-measuring length logarithm mark measured method miii Multiply natural number NATURAL SINES nearly needle nonagon nonius observation offsets opposite ordinates parallel parallel ruler parallelogram perches perpendicular plot polygon Portishead PROBLEM proportion quantity radius regular polygon Required the area rhomboides rhombus right angles right ascension right line roods RULE scale screw sector segment sextant side AC square links staff straight subtract Suppose survey taken Tangent telescope thence theodolite trapezium trapezoid triangle ABC Trigonometry vane vertical Page i - CROCKER'S ELEMENTS OF LAND SURVEYING. Fifth Edition, corrected throughout, and considerably improved and modernized, by TG BUNT, Land Surveyor, Bristol. To which are added, TABLES OF SIX-FIGURE LOGARITHMS, &c., superintended by RICHARD FARLEY, of the Nautical Almanac Establishment. Page 8 - The logarithm of the quotient of two positive numbers is found by subtracting the logarithm of the divisor from the logarithm of the dividend. (6) The logarithm of a power of a positive number is found by multiplying the logarithm of the number by the exponent of the power. For, N" = (oT) Page 125 - Or, from 8 times the chord of half the arc, subtract the chord of the whole arc, and $ of the remainder will be the length of the arc, nearly. Page 63 - If a side be required, say, — As the sine of the given angle is to its opposite side, So is the sine of either of the other angles to its opposite side. Page 274 - Two or more places are on a true level, when they are equally distant from the centre of the earth. Also, one place is higher than another, or... Page 133 - This figure being a mean proportional between its circumscribed and inscribed circles, that is, equal to a circle whose diameter is a mean proportional between the axes of the ellipse, we... Page 105 - From half the sum of the three sides subtract each side severally. Multiply the half sum and the three remainders... Page 42 - Triangles upon equal bases, and between the same parallels, are equal to one another.

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CC-MAIN-2019-13

https://www.thestudentroom.co.uk/showthread.php?t=2489232

math

Logs are the inverse of exponential functions right? But isn't square, cube and etc route also the inverse of exponents? I'm confused... how can they both be the inverse? Turn on thread page Beta log- inverse of exponential functions watch - Thread Starter - 13-10-2013 18:27 - 13-10-2013 18:39 you`re probably confusing "THE exponential function" with "functions with exponents - i.e. powers) THE exponential function is a function with exponents (or powers) is any function where something is raised to a rational power (the quotient of 2 rational numbers), e.g (2/1)=2, (1/3), -2/5) etc, be it an x, an expression in x, a trif expression such as (sin(x))^2 etc.

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CC-MAIN-2018-47

https://latex.org/forum/viewtopic.php?f=48&t=20685

math

\%in LaTeX, but I don't like its appearance with a number close to it. It's a bit too big. See for example this code : Is there any other "better looking" percent symbols? texdoc latex2e, the unofficial reference manual. Cham wrote:What is the \protectcommand doing ? Users browsing this forum: No registered users and 2 guests

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CC-MAIN-2020-45

https://slidetodoc.com/chapter-23-gauss-law-23-1-what-is/

math

- Slides: 22 Chapter 23. Gauss’ Law 23. 1. What is Physics? 23. 2. Flux 23. 3. Flux of an Electric Field 23. 4. Gauss' Law 23. 5. Gauss' Law and Coulomb's Law 23. 6. A Charged Isolated Conductor 23. 7. Applying Gauss' Law: Cylindrical Symmetry 23. 8. Applying Gauss' Law: Planar Symmetry 23. 9. Applying Gauss' Law: Spherical Symmetry What is Physics? • Gaussian surface is a hypothetical (any imaginary shape) closed surface enclosing the charge distribution. • Gauss' law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface. Gaussian surface Let us divide the surface into small squares of area ΔA, each square being small enough to permit us to neglect any curvature and to consider the individual square to be flat. We represent each such element of area with an area vector • Magnitude is the area ΔA. • Direstion is perpendicular to the Gaussian surface and directed away from the interior of the surface. Flux The rate of volume flow through the loop is : Flux of an Electric Field • The electric field for a surface is • The electric field for a gaussian surface is The electric flux Φ through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. SI Unit of Electric Flux: N·m 2/C Problem 1 The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of 1. 7 m 2, while surface 2 has an area of 3. 2 m 2. The electric field E in the drawing is uniform and has a magnitude of 250 N/C. Find the electric flux through (a) surface 1 and (b) surface 2. Sample Problem 2 Figure 23 -4 shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform electric field E , with the cylinder axis parallel to the field. What is the flux Φ of the electric field through this closed surface? Sample Problem 3 A nonuniform electric field given by pierces the Gaussian cube shown in Fig. (E is in newtons per coulomb and x is in meters. ) What is the electric flux through the right face, the left face, the top face, and the Gaussian surface? Gauss’ Law For a point charge: Gauss’ Law For charge distribution Q: The electric flux through a Gaussian surface times by ε 0 ( the permittivity of free space) is equal to the net charge Q enclosed : • The net charge qenc is the algebraic sum of all the enclosed charges. • Charge outside the surface, no matter how large or how close it may be, is not included in the term qenc. Check Your Understanding The drawing shows an arrangement of three charges. In parts (a) and (b) different Gaussian surfaces are shown. Through which surface, if either, does the greater electric flux pass? Sample Problem Figure 23 -7 shows five charged lumps of plastic and an electrically neutral coin. The cross section of a Gaussian surface S is indicated. What is the net electric flux through the surface if q 1=q 4=3. 1 n. C, q 2=q 5=-5. 9 n. C, and q 3=-3. 1 n. C? A Charged Isolated Conductor • If an excess charge is placed on an isolated conductor, that amount of charge will move entirely to the surface of the conductor. None of the excess charge will be found within the body of the conductor. • For an Isolated Conductor with a Cavity, There is no net charge on the cavity walls; all the excess charge remains on the outer surface of the conductor The External Electric Field of a Conductor If σ is the charge per unit area, according to Gauss' law, Sample Problem Figure 23 -11 a shows a cross section of a spherical metal shell of inner radius R. A point charge of q is located at a distance R/2 from the center of the shell. If the shell is electrically neutral, what are the (induced) charges on its inner and outer surfaces? Are those charges uniformly distributed? What is the field pattern inside and outside the shell? Applying Gauss' Law: Cylindrical Symmetry Figure 23 -12 shows a section of an infinitely long cylindrical plastic rod with a uniform positive linear charge density λ. Applying Gauss' Law: Planar Symmetry Figure 23 -15 shows a portion of a thin, infinite, nonconducting sheet with a uniform (positive) surface charge density σ Applying Gauss' Law: Spherical Symmetry • A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell's charge were concentrated at the center of the shell. • If a charged particle is located inside a shell of uniform charge, there is no electrostatic force on the particle from the shell. Any spherically symmetric charge distribution with the volume charge density ρ • For r>R, the charge produces an electric field on the Gaussian surface as if the charge were a point charge located at the center, • For r<R, the electric field is Checkpoint The figure shows two large, parallel, nonconducting sheets with identical (positive) uniform surface charge densities, and a sphere with a uniform (positive) volume charge density. Rank the four numbered points according to the magnitude of the net electric field there, greatest first. Conceptual Questions (1) Two charges, +q and –q, are inside a Gaussian surface. Since the net charge inside the Gaussian surface is zero, Gauss’ law states that the electric flux through the surface is also zero; that is Φ=0. Does the fact that Φ =0 imply that the electric field E at any point on the Gaussian surface is also zero? Justify your answer. (2) The drawing shows three charges, labeled q 1, q 2, and q 3. A Gaussian surface is drawn around q 1 and q 2. (a) Which charges determine the electric flux through the Gaussian surface? (b) Which charges produce the electric field at the point P? Justify your answers. (3) A charge +q is placed inside a spherical Gaussian surface. The charge is not located at the center of the sphere. (a) Can Gauss’ law tell us exactly where the charge is located inside the sphere? Justify your answer. (b) Can Gauss’ law tell us about the magnitude of the electric flux through the Gaussian surface? Why?

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CC-MAIN-2023-40

https://www.jiskha.com/display.cgi?id=1361155012

math

posted by Chris . Three circles with radii of 4, 5, and 6 cm, respectively, are tangent to each other externally. Find the angles of the triangle whose vertexes are the centers of the circles. sides of triangle are (2+2.5),(2+3),(2.5+3) Start with law of cosines label the sides a,b, c solve for angle C Then, use the law of sines a/SinA=c/SinC solve for angle A then use the fact that the sum of the angles is 180 deg, find angle B. check angle B with the law of sines.

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CC-MAIN-2017-34

http://www.letslearnfinance.com/difference-between-profit-and-revenue.html

math

Many people use profit and revenue interchangeably while talking about earnings of a company, which is incorrect as there are many differences between profit and revenue, given below are some of them – - Revenue refers to that income which a company earns after sales of goods or service which company manufactures while profit refers to residual income left after the company has paid all expenses related to production goods or services. - Revenue is calculated by multiplying number of units sold and selling price whereas profit is calculated by deducting cost of goods sold, administration expense, selling and distribution expense and other such expenses from revenue. - Revenue can never be in negative, however profit can be negative which happens when company make loss for a year. - It is easier to calculate revenue for a firm because it involves only two variables that sales price and number of quantity sold but calculating profit is much trickier as it involves many variables and therefore it requires much more expertise. - Revenue is always calculated first and then one can calculate profit from that and not before, in other words revenue always precede profit.

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CC-MAIN-2018-05

https://learn.rumie.org/jR/bytes/how-do-i-use-algebra-to-save-up-for-a-purchase/

math

Do you need to buy a new laptop? Is this how your parents react when you ask them to buy it? If you need to buy that laptop on your own, you can use algebra to figure out how to save up for it! Here's The Plan! Run through these steps when you're planning for a purchase. Once you decide what you want to buy, answer the 5 questions below: 1. How much does it cost? 2. How much money do you have now? 3. How much more money do you need to save to buy it? 4. How will you earn that amount? 5. How much time will it take to save it? The next step is to set up an algebraic equation to answer the questions. Example In Action Luna is a high schooler who has a part-time job babysitting every Saturday. She earns $60 every month from that. She has $356 saved up in her bank account. She wants to buy a new laptop that costs $1000, and she wants it in time for final exams in 6 months. Set up an equation to see if she can reach her goal: Part A: Will she have enough money to buy the laptop in 6 months? Part B: If not, how many months will it take her to save for the laptop? Answer the 5 questions 👉 We know that Luna has decided to buy a laptop. Find out how much it costs. The laptop costs $1000. How much money does she currently have? She has $365 saved up. How much more money does she need to save to buy it? See calculation below. How will she earn that amount? She saves the earnings from her part-time job. She earns $60 per month. How much time will it take to save it? See the calculation in the next step. Use the formula below to create an algebraic equation to find the answer: Setting Up The Equation: Part A Will Luna will have enough money to buy the laptop in 6months? Follow this equation to see if she will or not. Here, the total amount she saves at the end of 6 months is $725. 👉 Luna won't have enough money to buy the laptop in 6 months. $1000 - $725 = $275 👉 She needs $275 more to reach her goal of $1000. Setting Up The Equation: Part B How many months will it take Luna to save up for the laptop? To answer the second part of the question, identify the known information & place it in the equation: Here, x = 10.58, but this value isn't a whole number, so we need to round it up to 11. 👉 It will take Luna 11 months to save up enough to buy the laptop. It's Your Turn Now! Can you figure out the solution to the problem below? Question: Ruby broke her camera, so she'd like a new one that costs $250. She wants to buy it in 5 weeks. How much does she need to save each week? Hint: Place the values you know into the algebraic formula shown above and find the unknown value. How much does Ruby need to save each week? Worked Out Solution To The Quiz Question 👉 Ruby has to save $50 each week to get her camera in 5 weeks. Impress your parents with your financial planning skills!

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CC-MAIN-2024-18

http://www.math.rochester.edu/undergraduate/overview.html

math

Information for prospective undergraduate students The Mathematics Department at the University of Rochester is a research oriented department with a strong interest in preparing the next generation of engineers, scientists and mathematicians for successful careers in industry, research, and business. The department consists of 22 full time faculty members, all holding Ph.D.'s in mathematics. There are no part time adjunct teaching faculty. Advanced graduate students who have shown apptitude and interest in teaching serve as Teaching Fellows -- teaching a section of a calculus course under the guidance of faculty member teaching another section of the same course. The teaching fellows from the mathematics department regularly obtain teaching evaluations that exceed the average for the science faculty in general. In the last three years two of our teaching fellows have received the top university wide teaching award for graduate student teachers. Calculus is taken by many students with a wide variety of interests and preparation. The University of Rochester Math Department is large enough to be able offer a variety of tracks for learning calculus, and still small enough so that you won't get lost in the crowd. The honors calculus sequence MTH171Q--MTH174Q, allows students to learn calculus and simultaneously introduces them to how research mathematicians think and their techniques for solving problems. It is not just for mathematics majors, but for any talented student who wants mathematical insight to be part of their intellectual tool box. The four semester sequence is quite challenging, but awards 20 academic credits (instead of the usual 16) and includes the material from a fifth course -- linear algebra (MTH235). (Linear algebra is, along with calculus and Fourier analysis, among the most powerful problem solving tools in mathematics, and is normally required for math majors. It is waived for students who complete the 170 series. Non-math majors also benefit, since for some this accelerated sequence is the only way they can fit linear algebra into their program.) The standard calculus sequence (MTH161, MTH162, MTH164, and either MTH163 or MTH165) is taught in slightly larger sections. The sections are taught with common exams and homework which provides maximum flexibility for students who have difficulty fitting a calculus class into their schedule. The MTH141--MTH143 calculus sequence covers one dimensional calculus, the same material as MTH161--MTH162, but in three semesters instead of two. Students are then ready to continue with MTH164 (multivariable calculus) and either MTH163 (differential equations) or MTH165 (linear algebra with differential equations). The Calculus with Foundations courses MTH140A--MTH141A allow students who have had difficulty with algebra or who have insufficient preparation to strengthen their math skills, while beginning the calculus sequence. Students can choose to go on to MTH142 after this sequence. More information on the organization of the calculus sequence is available on the undergraduate page. Clusters in Mathematics There are many clusters available in mathematics and related fields. For examaple almost every 100 level mathematics course is part of at least one cluster. See the Collge Cluster Page for more information Advanced Math courses One of the big advantages of our mathematics program is the range and depth of undergraduate mathematics offerings beyond calculus. These include mathematics offerings of interest to future engineers and scientists as well as to mathematics majors. There are even courses such as financial math which discusses the pricing of options and derivatives. When you include the opportunity to take graduate programs while still an undergraduate, there is no possibility of running out of interesting and challenging courses. A description of the most popular advanced math courses and a complete list of all offerings is available. Degrees in Mathematics We offer both the B.A. and B.S. degrees in mathematics. The requirements are listed on our undergraduate program page. The minor in mathematics is particularly valuable for computer scientists, engineers, and scientists who want extra credentials in mathematics. Summer research opportunities This is a deal that is hard to beat: getting paid to learn. These summer opportunities provide opportunities to work on special projects or in special programs throughout the United States. The programs are available in physics, biology, and chemistry as well as mathematics. Opportunities for graduate study while an undergraduate Some of our majors, particularly those interested in attending graduate school in mathematics, get a head start by taking some of our first year graduate courses during their junior or senior year. These courses count for undergraduate credit. Only a few mathematics majors plan to become academic research mathematicians. Most majors enjoy the challenge and fascination of mathematics and find the foundation of critical thinking and technical problem solving that they acquire to be an excellent foundation for careers in the computer industry, actuaries, industrial engineering, and business, as well as great preparation for law schools, medical school and graduate schools in the sciences and mathematics. The Rochester Take Five Program, and the University of Rochester's Eastman School of Music are additional reasons why prospective math and science students consider coming to Rochester. For information on admissions go to http://www.cc.rochester.edu:80/admissions/RC/.

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CC-MAIN-2013-20

https://www.elitetrader.com/et/threads/ib-1099-statements.13541/

math

Hi, I was wondering if somebody has info on this, because IB CS didn't give me a clear answer. There are two 1099 forms in my U12345 (Universal) Account Page. 1. 1099 for: S12345 (Stock account) 2. 1099 for: U12345C (Universal account) The S12345 1099 form has only info about stocks. The U12345C 1099 form has only info about futures. Now my questions are: 1.) I thought the account S12345 didn't exist any more, since IB replaced it with Universal account? (I would assume, the 1099 form for S12345 should contain info only up to the date when the account was converted to Universal, right?). 2.) And since I trade both stocks and futures, and now my account number is U12345, there should be info about both stocks and futures on my U12345 1099 form?

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CC-MAIN-2023-06

https://c3delivered.com/product/pr-banana-sherbert-hybrid/

math

*PR* Banana Sherbert – Hybrid Banana Sherbet is a heavy hitter, and it reportedly makes excellent concentrates. The terpene profile is a mix of sour fruit and diesel fuel flavors. Great for daytime relief as it won’t leave you couch locked. INDICA: 60%  STAVIA: 40%  THC: 27%  CBD: .1%  CBDA: 0.0%

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CC-MAIN-2019-30

https://www.tutorsglobe.com/question/nbspthe-cost-of-capital-isnbspto-produce-where-marginal-51119225.aspx

math

1. The cost of capital is a. called accounting profit. b. the cost of debt, or interest, plus the cost of equity, or the alternative return owners could have gotten had they not chosen to invest in this activity. c. usually considered to be just the interest paid on debt. d. the same as the rate charged by the U.S. government for a 30-year Treasury bill. e. measured by the Dow Jones Industrial Average at the close of any given day. 2. To produce where marginal cost is equal to marginal revenue is called a. the marginal-revenue rule. b. the marginal-cost rule. c. the profit-maximizing rule. d. breaking the rules. e. being forced out of business.

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CC-MAIN-2024-18

https://collegepublications.co.uk/other/?00022

math

|Grundlagen der Mathematik I. Foundations of Mathematics I C.-P. Wirth, J. Siekman, M. Gabbay, D. Gabbay, eds. Hilbert and Bernays' ''Grundlagen der Mathematik'' appeared in two volumes in 1934 and 1939, a second edition in 1968 and 1970. It offers a foundation for proof theory and is a major source on David Hilbert's formalist programme, Paul Bernays' philosophy, the epsilon operator, and much more. It has been a profound influence on mathematics, logic, and philosophy, and it covers formal ground and philosophical perspectives beyond the scope of Whitehead and Russell's ''Principia Mathematica'' and Frege's ''Grundlagen der Arithmetik''. This book is not only essential to any scholar of the history and philosophy of modern mathematics, but it also contains formal research - on the epsilon and iota operators - of contemporary relevance to logicians, mathematicians and computer science. For us, it is one of the most fascinating books ever written. This a bilingual German-English, commented edition of the ''Grundlagen''. It is the first English publication of these texts and shows the facsimile of the German original text on the left-hand side of a double page, and its English translation on the right-hand side. In addition to extensive comments on the history and the interpretation of the text's mathematical and philosophical content, there are also careful annotations regarding the differences between the two German editions (1934/39,1968/1970) of this two volume monograph. 29 November 2011 Buy from Amazon: UK US

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CC-MAIN-2024-10

https://mathsdefinitions.com/category/e/

math

Use the given information to work out something that is outside the range of the data. Always risky to do, as there is no certainty in the answer. Multiply the length of all the sides by the same amount. It could mean make smaller if the scale factor is between 0 and 1. Use the information in the question to write an expression with the letter(s) given. Usually given in questions where there are more than one set of brackets. Expand the brackets and collect any like terms. Multiply out the brackets. Example: y = 2x – This means that y is always double the x value. An angle outside of the shape made by continuing one of the sides. Example: Calculate the exterior angle of a 10 sided regular polygon. 🔑 Key Exterior Angle Topics: Interior and Exterior Angles Write down. Usually used where the question wants the answer in a particular format. Example: Express your answer in the format p + √q 🔑 Key Express Topics: Surds

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CC-MAIN-2022-40

https://www.suzhar.com/hepto/871/

math

Development was suspended due to various circumstances, but has resumed. For the 10th puzzle, we designed a puzzle where the cube is viewed from three sides and each is placed in its correct position. The puzzle rotates on each of the three axes, and there are many combinations of rotation states, so we thought it was a reasonable difficulty. However, when the puzzle examined all combinations, we found that the correct answer could be reached with 3 moves from any state. This means that there is a high possibility that it can be done by inaction without following the correct procedure. Therefore, he tried to set various restrictions, but it did not work as a puzzle. Therefore, the cube was changed to a regular icosahedron, and the rotation axis was also a combination of 5 axes. Ironically, in this case as well, it was found that 3 moves lead to a correct finish, regardless of the initial state. For this reason, the icosahedron has a double composition of outer and inner sides. And if each is placed correctly, it is a goal. Even with this, plater can reach the goal with a total of 6 moves minimum, but the probability of getting the correct answer in a random sequence will decrease.

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CC-MAIN-2024-10

https://www.jiskha.com/display.cgi?id=1286585070

math

posted by Jessica . If i have ten grains of sand on a flat, 8x8 chessboard, randomly distributed, what is the probabilty that all ten are on the same square? The important thing is that the probability of a grain of sand to be one one square is equal. Pr(on a particular square)=1/64 So if we put a grain on any one square, the next 9 have to be on that next square. The probability of that is Pr(10 on same square)=1*(1/64)^9 which is exceedingly small.

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CC-MAIN-2017-34

http://boardgames.stackexchange.com/questions/tagged/bohnanza+agricola-goodies

math

Board & Card Games Board & Card Games Meta to customize your list. more stack exchange communities Start here for a quick overview of the site Detailed answers to any questions you might have Discuss the workings and policies of this site Board & Card Games Hand Order In Agricola (and Bohnanza) We played with Agricola: The Goodies for the first time last night, and while it was very silly, it didn't have the negative impact on gameplay that I'd feared. However, the Occupation cards from the ... Jun 30 '11 at 16:46 newest bohnanza agricola-goodies questions feed Hot Network Questions Finish lazy parentheses Why don't game developers release their source code? Is it bad etiquette to ride road bike through a small residential area? How much cash can you carry in a European Union domestic flight? what will ¬ do in bash? Are there "secure" languages? Is it possible that "A counter-example exists but it cannot be found" Can reviewing paper help increase your credibility as a researcher? Why do Disney parents usually die? \mathop in sums and limits Student insists I am wrong Should I unsubscribe uninterested mailing-list members? Output Pi without math Which primitives does ConTeXt rename/move? Are C# properties actually Methods? A word for old-fashioned, dirty bar/place (spit-and-sawdust) input type="email" allows test@test, why? Topological groups, why need them? Source code formatter/indenter In Captain America, The Winter Soldier, who did Peggy Carter say she married? What exactly is the difference between "misinformation" and "disinformation"? Uncertainty of permittivity of vacuum Technically, how do variadic functions work? How does printf work? Why doesn't the wizarding world use computers? 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 Overflow Stack Overflow Careers site design / logo © 2014 stack exchange inc; user contributions licensed under cc by-sa 3.0

s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609524259.30/warc/CC-MAIN-20140416005204-00257-ip-10-147-4-33.ec2.internal.warc.gz

CC-MAIN-2014-15

https://logosconcarne.com/2022/04/04/wavefunction-collapse/

math

The previous post began an exploration of a key conundrum in quantum physics, the question of measurement and the deeper mystery of the divide between quantum and classical mechanics. This post continues the journey, so if you missed that post, you should go read it now. Last time, I introduced the notion that “measurement” of a quantum system causes “wavefunction collapse”. In this post I’ll dig more deeply into what that is and why it’s perceived as so disturbing to the theory. Caveat lector: This post contains a tiny bit of simple geometry. To begin, mechanics is the study of a system’s dynamics, the forces that act on it, and the resulting motion of that system. There are two important points about classical mechanics: Firstly, it directly corresponds to accessible physical objects. Secondly, it is inherently nonlinear. Both in contrast to quantum mechanics, which involves the mysterious quantum state and is linear. A bit of history: Classical mechanics begins with Newton’s laws of motion in 1687. One-hundred years later, in 1788, Joseph-Louis Lagrange extended classical mechanics with Lagrangian mechanics. His work introduced the notion of the principle of least action. Almost fifty years after that, in 1833, the great mathematician William Hamilton further extended classical mechanics with Hamiltonian mechanics. He added the notion of a Hamiltonian, a notion also important in quantum mechanics. A key difference between Newton’s formulation and the extensions by Lagrange and Hamilton is that the former is in terms of forces on objects whereas the latter is in terms of a system’s energy. For one example of the nonlinearity of classical mechanics we can look at Newton’s famous second law, F=ma, which says that the force (F) on an object is the product of its mass (m) times its acceleration (a). Mass, at least initially (see below), seems a simple enough property; very roughly speaking, it’s the weight of something. Acceleration is where the nonlinearity kicks in — it’s the derivative of velocity over time, and velocity is the derivative of distance over time. Without getting into derivatives, suffice to say that acceleration being the double derivative of distance over time — a quantity we can symbolize as x² — is the nonlinear aspect. Simply put, if we graph x², we get a curve (a parabola). Linear essentially means a straight line, and a curve isn’t straight, it’s nonlinear. Two asides at this point. Firstly, the theory of gravity, Einstein’s general relativity, is also nonlinear (in part because it has singularities such as black holes). Real-world systems, such as weather and orbits involving more than two bodies, are famously chaotic, and the reason they are is due to the nonlinearity of the equations governing their behavior. We can also consider the parabolic curve of a thrown baseball or fired bullet. The classical world we inhabit, in contrast to the quantum world, seems decidedly nonlinear. Secondly, while mass seems a simple concept at first blush, Newton’s definition is famously circular. Mass is the density of an object. Density is mass per volume. Oops! Leibniz mocked Newton about this, and it remained a slippery concept until Peter Higgs (and others) came up with the Higgs field (but that’s a story for another time). Getting back to the question of measurement, the quantum equivalent of classical laws of motion is the infamous Schrödinger equation (or the relativistic Dirac equation), which say how a quantum state evolves (that is to say, moves). While we won’t venture into the math, it’s worth a brief look: There are a several things to notice. Firstly, this is just one form of the equation, there are others. In particular, this is a time-dependent form. The d/dt on the left tells us we’re taking the derivative (of the wavefunction) over time. The Ψ(t) that appears on both sides tells us that we’re looking at the wavefunction Ψ (psi, usually pronounced “sigh”) at some point t in time. Secondly, a bit more importantly, that little i on the far left is the imaginary unit — the basis of the complex numbers. Its presence is significant in how it seems to require complex math in quantum mechanics. There are classical equations that make use of i, but there it’s a mathematical convenience that can be dispensed with (at the expense of more complicated math). But in quantum math it’s the basis of quantum interference, the existence of which is a crucial difference between quantum and classic mechanics. (As an aside, quantum interference, quantum superposition, and quantum entanglement, strike me as three key differences between quantum and classical mechanics. We don’t experience them in the classical world.) Thirdly, as something of an aside, the Ĥ symbol (pronounced “H-hat”) is the Hamiltonian operator that defines the kinetic and potential energy of the system the equation describes. In general, the “hat” symbol over a letter means the letter stands for some form of operator. (See QM 101: What’s an Operator?) A big part of defining a Schrödinger equation for a given system consists of defining its Hamiltonian. (And also, of defining its wavefunction.) The key to the Schrödinger equation is that it describes the linear evolution of the wavefunction (the quantum state). In practice this means that, given the wavefunction at any time t, we can use the Schrödinger equation to obtain the quantum state at any other time, forward or backward. (Given the typical nonlinearity of classical motion, this is a striking difference about the quantum world.) Which brings us to the measurement problem and why it is such a problem. I’ll explore this with a concrete example using a simplified version polarized light. (For a more detailed look, see QM 101: Photon Spin.) Here, I’ll consider only horizontal and vertical polarization, and I’ll forego the more accurate Bloch sphere representation for a two-dimensional representation. (See QM 101: Bloch Sphere.) The polarization angle of a photon can be represented as a combination (technically, a superposition) of horizontal and vertical polarization. Depending on the ratio between them, they can describe any other angle. For instance, equal parts of both result in a 45° polarization. (See Figure 1) For simplification we can represent this as a quarter-circle where the X-axis is the degree of horizontal polarization, and the Y-axis is the degree of vertical polarization. Here we’ll consider “pure” quantum states, which means the quantum state vector always has a length of one. The points along the quarter-circle are the possible polarization states from fully horizontal to fully vertical. Note an important aspect of Figure 1: The blue lines project the quantum state onto the horizontal and vertical axes. (The black line is the vector representing the quantum state. Its end point, the red dot, is the actual quantum state.) Basic trigonometry tells us these lines meet the axes at the cosine (horizontal) and sine (vertical) of the angle. Those values are, respectively, cos(45°)=0.707 and sin(45°)=0.707 — the equal parts mentioned above. Most crucially, the projections of the state (vector) onto the axes represent possible measurements we could make with horizontal or vertical polarizing filters. That last sentence is the heart of this post, and it requires some unpacking before we can get at the measurement problem. Firstly, in terms of measurement, we’re saying we measure only the horizontal or vertical polarization. This is the basis of our measurement. We can, of course, set a polarization filter at any angle, and that would give us a different basis. All the logic that follows below would still apply. In picking a basis we want to use orthogonal axes that span the space of possible states. As just mentioned, any polarization angle can be defined as a superposition of the basis axes. For example, an angle of 10° (see Figure 2), would be a superposition of horizontal=cos(10°)=0.985 and vertical=sin(10°)=0.174. Note that in the case of strictly horizontal polarization (0°) we have cos(0°)=1.0 and sin(0°)=0.0. In the case of strictly vertical polarization (90°) we have cos(90°)=0.0 and sin(90°)=1.0. Thus, even these states are, in fact, superpositions and projections onto the respective axes. Secondly (still unpacking here), given a photon polarized at 45°, the square of the projection onto an axis gives the probability that we’ll measure the photon. By “measure the photon” I mean that the photon will pass through the filter aligned with that axis and be detected by some device on the other side of the filter. Why the square of the projection value? Because in less simplified situations the projection can fall onto a negative part of the axis, and there is no such thing as a negative probability. Squaring eliminates negative probability (and, by the way, matches what we see experimentally). In general, this is the Born rule. The probability of getting some measurement is the square of the projection of the quantum state onto some measurement basis axis. Looking back at Figure 1, we had projection values of 0.707 for both axes. If we square that, we get 0.50 — a 50% probability the photon will pass a horizontally or vertically aligned filter. In the case of Figure 2 and 10° polarization, squaring the 0.985 and 0.174 projection values gives us 0.97 and 0.03, which are probabilities of 97% and 3%, respectively. Note that these probabilities must always sum to one. (In the cases of strictly horizontal or vertical polarization, obviously we end up with 100% and 0% probabilities.) We’re finally ready to understand the measurement problem! There are two aspects to it. Firstly, when we make a measurement, we get a definite result, and the quantum state becomes known. If the photon passes a horizontal filter, then that photon is now horizontally polarized. If it passes a vertical filter, it’s vertically polarized. This change, in violation of the linear Schrödinger equation, is nonlinear. If we take the Schrödinger equation as fully describing quantum state evolution, then we’re left scratching our heads and wondering what happened. How do we describe this nonlinear change? For that matter, how did nonlinearity even enter linear quantum mechanics? [For more about how passing through polarization filters changes the polarization state, see QM 101: Fun with Photons.] Secondly, note that the projection — except in the cases of strictly horizontal or vertical polarization — does not have a length of one. To go on describing the quantum state, we must also make the nonlinear (and rather by fiat) change of setting the vector back to a length of one. If we put our faith in the Schrödinger equation and the linearity of quantum mechanics, we have an unsolved, and very vexing, problem. One alternative is to view the Schrödinger equation, and even quantum mechanics as currently understood, as an incomplete answer. Which would be my view. I think that the lack of unification with gravity, the background dependence, and even the linearity, all suggest we’ve gotten this wrong. (I’m willing here to conflate incomplete with wrong.) I compare it to the Ptolemaic view of the solar system. Keep in mind that Ptolemy’s geocentric view worked very well for over 1000 years. We might also consider how Newton’s laws worked extremely well for over 200 years until Einstein came up with general relativity. Quantum mechanics is about 100 years old, and it’s one of the most successful theories we know. Unfortunately, general relativity is the other most successful theory we know, and they’re incompatible on several levels (linear vs nonlinear; background dependent vs GR is the background; quantum vs smooth). Bottom line, I think faith in quantum mechanics as it now stands may be misplaced. For more musings on wavefunction collapse, see the paired Wave-Function Collapse and Wave-Function Story posts from May of 2020 and BB #73: Wavefunction Collapse from August of 2021. They may not add much, but you may find them interesting. Next time I’ll talk about decoherence and what it has to do (or doesn’t) with the measurement problem. Stay collapsed, my friends! Go forth and spread beauty and light. April 4th, 2022 at 7:40 am We can skip the sine and cosine trigonometry and, instead, use Pythagoras’s famous theorem for the length of the hypotenuse of a right-triangle: Since we set the state vector to one and use orthogonal axes, the state vector is the hypotenuse of a right-triangle formed by the X and Y axes. This gives us: Which describes the points of a circle having a radius of one. Note that, now the probabilities, the squares of the projections onto the axes, fall naturally out of the equation. Since the equation applies to all points on the circle — all (pure) quantum states — and applies to any two-state quantum system we describe with a two-dimensional circle. The notion generalizes to cases with more dimensions, for example a 3D sphere: Which gives us the projections of any point on the sphere (again, a pure quantum state with three possible measurements), and the squares of those projections are the probabilities of getting each of the three possible measurements. (And, of course, the wavefunction would “collapse” to the measured axis if the measurement succeeded. (If it didn’t, in that case, the quantum state would still be a superposition of some combination of the other two axes.) The generalization continues to any hypersphere with any number of axes. For instance, with four axes: And so on. April 4th, 2022 at 8:04 am A detail I ignored in the post, but brought out by the comment above, involves what happens if the measurement fails. I ignored it in the post because, if a photon doesn’t pass through the polarizing filter, that means it was absorbed by (an electron in) the filter and, therefore, effectively lost. If a photon does pass through the filter, we know its wavefunction “collapses” — its polarization angle now matches the angle of the filter. In the cases where it doesn’t, it’s still “collapsed” but collapsed such that its polarization angle matches the orthogonal axis. This behavior is more apparent in other types of experiments where the quantum state isn’t destroyed by being absorbed. But it’s that “collapse” to the other axis that’s why the probabilities must always sum to one. Upon measurement, there is always a definite result, and in two-state systems, of the measured axis fails, then the “collapse” is necessarily to the other available axis. That’s why, in systems with more than two states, a measurement that fails results in a “collapse” to some superposition of the other axes. April 4th, 2022 at 10:54 am I have a question, nothing to do with this post, regarding light and the human eye’s ability to receive it. Given rods and cones, where cone cells are charged with converting wavelengths of light to colors, how does white light work? We can see red, green & blue light, and white light being composed of those three colors (basically), does that mean that “white” light is actually the triggering of three different cones? That is, are cones dedicated to certain colors? Or do cones interpret wavelengths? And if white light is many wavelengths…? In light of this (ha), what’s your opinion on dark screens vs light ones? Given what I’ve read about the eye and its light interpretation — dark-mode is NOT the panacea it’s sold to be. I don’t use it. We’re not nocturnal creatures. We expect to see dark things on light backgrounds as part of our evolution. April 4th, 2022 at 12:22 pm Color is one of my favorite topics, so pardon me while I blather on a bit! Let me first differentiate between two different kinds of “white” light. Normal white(-ish) light from incandescent lightbulbs or the sun contains photons of all the (visual) frequencies. Natural white light is the visual equivalent of white noise, sound containing all the (audible) frequencies. “White” light on a computer monitor, however, as you probably know, comes from a blend of red, green, and blue pixels. Some kinds of artificial light, fluorescents, some early LED types, are very spikey in terms of their color curves. (If you’ve ever seen sodium vapor lights, those are essentially monochromatic orange-yellow.) The reason computer monitor “white” looks white to us is because humans have three types of cones, casually referred to as the “blue”, “green”, and “red”, cones. (Technically, they’re called “S”, “M”, and “L”, respectively for short, medium, and long, wavelengths.) Each type responds to a range of frequencies roughly centered on blue, green-ish, and red-ish. So, it’s not necessary for monitors to attempt to make “white” by combining all the frequencies the way natural sources do. They can use the red, green, and blue, pixels to stimulate the three types of cones. In fact, the cone responses are a bit on the sloppy side: The close overlap between the green and red cones, when genetic variation closes the overlap, is responsible for some common types of color-blindness. (Dichromatic lens can help correct it by adding a notch filter between the cones to help separate the responses to green-ish and red-ish light.) That we have those three types of cones is why TVs, monitors, and color photos, only need to deal with red, green, and blue, in varying proportions to create any color. One of the weirder technological inventions is glass that filters out only the specific frequencies used by these devices. Such glass is sometimes used in glass-walled conference rooms for privacy. Most things look normal through the glass because most things, even if they’re a single color, contain ranges of frequencies, so the strong notching of RGB doesn’t really affect how we see them. But computer monitors and TVs and phone screens, using only those blocked frequencies, look black. Very strange to see in action. People sitting around staring raptly at a black screen. I’m not a fan of dark screens. I see it as generally a triumph of style over sense. We use white paper and black ink because it works. I started setting my editors to black on white as soon as that capability came around. (I go back to the early days of green on black.) That said, there is some difference between white paper reflecting light and a screen that basically is a light. If your brightness levels are too high, it can cause eye-fatigue, which is why black screen became a thing in the first place. (There is also that some screens are a blue-ish white, and blue light signals your eye that it’s daytime, so staring at a white screen in the evening can interfere with sleep cycles. Modern screens often have an “evening” or “night” mode where they shift to a warmer (red-ish) white to avoid this. April 4th, 2022 at 7:38 pm April 7th, 2022 at 7:07 am […] the last two posts (Quantum Measurement and Wavefunction Collapse), I’ve been exploring the notorious problem of measurement in quantum mechanics. This post […] April 7th, 2022 at 10:10 am Speaking of collapse, I’m a little “collapsed” with regard to this series. I mean that I’m a bit tired of writing about it, so I may take a short break for other things and other posts before I come back to it. April 10th, 2022 at 7:28 am […] the last three posts (Quantum Measurement, Wavefunction Collapse, and Quantum Decoherence), I’ve explored one of the key conundrums of quantum mechanics, the […] April 13th, 2022 at 4:21 pm […] the last four posts (Quantum Measurement, Wavefunction Collapse, Quantum Decoherence, and Measurement Specifics), I’ve explored the conundrum of measurement […] June 7th, 2022 at 1:02 pm […] a five-part series about the measurement problem in quantum mechanics (see Quantum Measurement, Wavefunction Collapse, Quantum Decoherence, Measurement Specifics, and Objective […]

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https://drownedinsound.com/community/boards/social/4146756

math

If you really want to read this, try using The Internet Archive. I'm a bit of a rookie. I've just downloaded Beauty and the Beat but it's in the wrong order. I've tried putting numbers at the beginning of the tracks but it doesn't work! any help - much appreciated. thanks!

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https://www.onlinemath4all.com/6th-grade-math-worksheets-6.html

math

In this page 6th grade math worksheets 6 , you will find 10 questions in various topics of math. You may try each questions on your own and you may check the answer immediately by using the submit button. This question paper will be in the form of quiz. this webpage, we give 6th grade math worksheets 6 online test which can be taken by any one without any login credentials. At the end of the test, students can check their score instantly and also they can view detailed answer for all the questions. We hope, this online practice test will be a great boost for the students. We give detailed solution for every question of this worksheet. If you find it difficult to do any problem,please click the link view solution for each question. You will get detailed solution. 1.The sum of (-3) and (-8) is 2. A work was assigned to me on a Tuesday. I completed to work after 72 days. On what day, i completed the work. 3. Find the value of “x” 4. Evaluate : (2/3) + (1/3) - 1 5. A Person hears a sound from some where. The sound travels one mile for five seconds. If the person is 10 miles away from the place where the sound is made, how long does it take for the sound to reach the person? 6. James travels in his car at the rate of 50 miles per hour. If he wants to travel 275 miles, how long will it take for him to reach his destination? 7. Mike earns 5.5% commission on his cell phone sales. In may, he earned a total of $428.56 in commissions. What was his sales that month? 8. Ronald starts to work in an office with the initial salary of $2000. Every year he gets increment of $100. What will be his salary in his 25 th year? 9. Sophia has a garden whose measurements are 30 feet by 40 feet. She wants to put a sidewalk around the garden that is 3 feet wide. What is the area of the sidewalk in feet? 10. Mr. Ken wants to buy a TV set. The list price of TV set is $ 1500. The seller allows a discount of 10% from the list price and charges 10% tax on cash paid after discount. How much does Mr.Ken have to pay? Explanation for question No 1 Since both numbers are having same symbol,we have to add these numbers and we have to put the larger number symbol. -3 + (-8) = -11 So the correct answer for this question is -11 option A Explanation for question No 2 Tuesday corresponds to 2 in 7 day clock arithmetic. We want to know the day which is 72 days after Tuesday. To get answer for our question, we have to do the following steps. Step1: Add 72 to 2 Step2: Divide the result by the divisor 7 Step3: Take the remainder 72 + 2 = 74 When 74 is divided by 7, the remainder is 4 4 corresponds to Thursday. Hence, John completed the work on Thursday. So the correct answer for this question is Thursday option B Explanation for question No 3 Sum of angles in a straight line is 180 x + 55 = 180 x = 180 - 55 x = 125 So the correct answer for this question is 125 option B Explanation for question No 4 (2/3) + (1/3) - 1 First,let us consider the denominators The denominators are not same,so we have to take L.C.M to make it as same. = (2/3) + (1/3) - 1 x (3/3) = (2/3) + (1/3) - (3/3) = (2 + 1 - 3)/3 = (3 - 3)/3 So the correct answer for this question is 0 option B Explanation for question No 5 Distance covered by the sound for 5 seconds = 1 mile Distance covered by the sound for X seconds = 10 mile 5 x 10 = X x 1 50 = X Time taken by the sound to reach 10 miles = 50 seconds So the correct answer for this question is 50 seconds option B Explanation for question No 6 Distance covered by the car in 1 hour = 50 miles From this we have to find the time taken by the car to travel 275 miles. 1 hour -> 50 miles X hours -> 275 miles this comes under the topic direct proportion 1 x 275 = 50 x X 275 = 50 X 50 X = 275 X = 275/50 X = 5.5 hours Therefore time taken by the car to cover the distance of 275 miles = 5 hours 30 minutes So the correct answer for this question is 5 hours 30 minutes option A Explanation for question No 7 Total commission amount in that particular month = $ 428.56 Let "X" be the selling price of the particular month 5.5 % of X = $ 428.56 (5.5/100) x X = 428.56 X = 428.56 x (100/5.5) X = 42856/5.5 X = 7792 So the correct answer for this question is 7792 option B Explanation for question No 8 Initial salary of Ronald = $2000 His salary is increasing every year by $ 100 a = 2000 d = 100 from this we have to find the 25th term tn = a + (n - 1)d t25 = 2000 + (25 - 1)100 = 2000 + 24 (100) = 2000 + 2400 So the correct answer for this question is $4800 option B Explanation for question No 9 Length of the garden = 40 ft Breadth of the garden = 30 ft width of the path around the garden = 3 ft Area of rectangle = length x breadth length around the garden = 40 + 3 + 3 = 46 ft Breadth around the garden = 30 + 3 + 3 = 36 ft Area of the path = (46 x 36) - (40 x 30) = 1656 - 1200 So the correct answer for this question is 456 square ft option B Explanation for question No 10 List price of TV = $ 1500 Selling price of Tv = 1500 - 10% of 1500 = 1500 - (10 x 1500)/100 = 1500 - 15000/100 = 1500 - 150 Discount amount = 10 % of 1350 Cost of TV after tax and discount = 1350 + 135 = $ 1485 So the correct answer for this question is $ 1485 option C

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https://digitalcommons.wku.edu/math_fac_pub/40/

math

Exact Boundary Controllability Results for a Multilayer Rao--Nakra Sandwich Beam We study the boundary controllability problem for a multilayer Rao--Nakra sandwich beam. This beam model consists of a Rayleigh beam coupled with a number of wave equations. We consider all combinations of clamped and hinged boundary conditions with the control applied to either the moment or the rotation angle at an end of the beam. We prove that exact controllability holds provided the damping parameter is sufficiently small. In the undamped case, exact controllability holds without any restriction on the parameters in the system. In each case, optimal control time is obtained in the space of optimal regularity for $L^2(0,T)$ controls. A key step in the proof of our main result is the proof of uniqueness of the zero solution of the eigensystem with the homogeneous boundary conditions together with zero boundary observation. Applied Mathematics | Engineering | Materials Science and Engineering | Mechanical Engineering Recommended Repository Citation Ozer, Ahmet Ozkan and Hansen, Scott. (2014). Exact Boundary Controllability Results for a Multilayer Rao--Nakra Sandwich Beam. SIAM Journal on Control and Optimization, 52 (2), 1314-1337. Original Publication URL: https://works.bepress.com/ahmetozkan-ozer/4/download/ Available at: https://digitalcommons.wku.edu/math_fac_pub/40

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https://web2.0calc.com/questions/math_67797

math

A rental car company is running two specials. Customers can pay $49 to rent a compact car for the first day plus $2 for each additional day, or they can rent the same car for $42 the first day and $3 for every additional day beyond that. Josh notices that, given the number of additional days he wants to rent the car for, the two specials are equivalent. How much would Josh pay in total? How many additional days does Josh want?

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https://muhaz.org/marc-garneau-collegiate-institute.html?page=5

math

Problem sets will typically be assigned for each chapter you will study. Any topic or task that you have mastered is mostly due to the practise you have undertaken. Your success in physics depends almost entirely on the amount of practise you complete while you study each section and prior to major evaluations. Suitable preparation for a test might include as many as 15 or 20 questions weighted towards aspects of the course that you do not understand as clearly. You can briefly review the chapter before starting the problems, but you should do the questions with the same amount of assistance that you might have on a test. Your homework can be a terrific reference if effected carefully and with proper form (described below). Homework done cryptically will ensure that you will have a difficult time understanding your thinking when you return to it a couple of months later. This is not good practise and will not be tolerated. You are here as much to learn how to learn as you are to learn physics. Questions vary and reasonable latitude will be given on solution form where the problem has a different aim. However, in general, each problem solution must have the properties listed below. Do not stare at a problem for more than 5 minutes without making some effort. If you cannot make some progress in 5 minutes, you do not understand the problem. Make some brief notes about the problem and seek the help of your instructor. Teachers will be particularly reluctant to assist students who have a completely blank page. The secret is to complete many medium-difficult questions in the time available. 8.2 Problem Set Guidelines Do NOT write out the questions. You have better things to do. All problems to be solved in your own hand except where computer work is demanded. Problems must NOT be solved by measurement of scale diagrams. Regardless of the problem, some English should be included to indicate the approach you took to the question. This is best done by concisely describing each line or two as the algebra is developed. Part marks will NOT be given if English comments are omitted. Any equations that are introduced must first exist in the relevant chapter summary of the textbook. All others must be derived. Uncertainty or significant figure rules must be followed in all assignments unless otherwise stipulated. Units should be expressed as either m s-1 or but NOT m/s. NOTE: In the sample below comments included inside a set of [ ] brackets are used to explain the format and should not be included in a real problem. 1. A 10 kg projectile is hurled 0.300 km. If the initial angle is 34o then (a) What was the initial speed? (b) How long did it take for the projectile to travel the 0.3 km? m [List all values given or implied in problem. Convert ALL units to mks values. If necessary indicate in English what each symbol means.] p = 10 kg Mass of projectile rx = 0.3 km = 300 m Range of projectile i = 34o Initial Angle g = 9.8 Acceleration of Gravity [Diagrams are required for applicable problems. Variables should be included when relevant]. rx = 300 m (a) RTF: vi , Initial Speed The conditions for this question require the range equation [Equation from text] Solving for v is best done by first solving for v2. Hence, And solving for v we have, Now that we have solved for the variable of interest you can now substitute the numbers indicated for the variables. No units should be included here as they get mixed up with variables. [After derivation is complete, values are substituted into equation. No units should be included. You may check the derivation with a unit check, but including them with the numbers simply increases the chance of confusion with variables.] [State final numerical answer, round to proper sf with unit] Therefore the initial speed was 56 ms-1. [ Concluding statement.] (b) RTF: Time of projectile Travel (t) We know that the velocities in the x and y directions are independent. Acceleration exists only in the y direction. Thus, the x speed is constant. The situation in the y direction will control the time, as the time for the projectile to reach maximum height will be ½ of that for the entire trip ignoring drag. To begin with we need to resolve v into its x and y components, vx = v cos vy = v sin We know that vf = vi + at Solving for time to the maximum We can dispense with vf since it equals zero. The negative sign in the numerator is then cancelled with the negative sign in the value for g. substituting values gives, Therefore the projectile would have been 11s in flight.

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https://meangreenmath.com/2017/03/07/my-favorite-one-liners-part-35/

math

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Every once in a while, I’ll discuss something in class which is at least tangentially related to an unsolved problems in mathematics. For example, when discussing infinite series, I’ll ask my students to debate whether or not this series converges: Of course, this one converges since it’s an infinite geometric series. Then we’ll move on to an infinite series that is not geometric: where the denominators are all perfect squares. I’ll have my students guess about whether or not this one converges. It turns out that it does, and the answer is exactly what my students should expect the answer to be, . Then I tell my students, that was a joke (usually to relieved laughter). Next, I’ll put up the series where the denominators are all perfect cubes. I’ll have my students guess about whether or not this one converges. Usually someone will see that this one has to converge since the previous one converged and the terms of this one are pairwise smaller than the previous series — an intuitive use of the Dominated Convergence Test. Then, I’ll ask, what does this converge to? The answer is, nobody knows. It can be calculated to very high precision with modern computers, of course, but it’s unknown whether there’s a simple expression for this sum. So, concluding the story whenever I present an unsolved problem, I’ll tell my students, If you figure out the answer, call me, and call me collect. One thought on “My Favorite One-Liners: Part 35”

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http://lxbwk.njournal.sdu.edu.cn/EN/volumn/volumn_104.shtml

math

Existence and multiplicity results for a second-order multi-point boundary value problem at resonance - CHEN Bin, Abuelgasimalshaby Elzebir JOURNAL OF SHANDONG UNIVERSITY(NATURAL SCIENCE). 2016, 51(4): Related Articles | It is investigated that the existence and multiplicity results for a second-order multi-point boundary value problem at resonanceu″(t)=f(t,u(t))+e(t), t∈［0,1］,u'(0)=0, u(1)=∑mk=1aku(ηk)by the connectivity properties of solution set of parameterized families of compact vector fields. where f:［0,1］×R→R is continuous, e∈C(［0,1］,R), 0<η1<η2<…<ηm<1, ak>0(k=1,2,…,m).

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https://tore.tuhh.de/entities/publication/04170961-66b5-47fa-96d1-3b200ab888ad

math

A classical interpretation of quantum mechanics and the measurement problem In this paper a didactic approach is described which immediately leads to an understanding of those postulates of quantum mechanics used most frequently in quantum computation. Moreover, an interpretation of quantum mechanics is presented which is motivated by retaining the point of view of classical mechanics as much as possible, and which is consistent with relativity theory. Everything can be written down in terms of well-known mathematical formulations that can be found in every textbook about quantum mechanics. Therefore, in this version, almost no formulas are used.

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https://www.queryhome.com/puzzle/17592/what-the-perimeter-square-circumscribed-about-circle-radius

math

Answer = 22.64r If 4r is the radius, then one side of a square is squareroot of (2 (4r)^2).......from Pythagoras theorem i.e one side of a square is 4r (squareroot of 2) And perimeter is 4 × one side........ Therefore, the perimeter is 16r × (squareroot of 2) A circle contains 4 identical squares, as shown below. If each square has a side length equal to 2, what is the radius of the circle? Construct a circle of radius 1. From a point on the circle, construct another circle of radius 1. Inscribe a square where the two circles overlap. What is the area of the square? In a square with a side length of 1, two quarter circles are drawn and a circle is inscribed between the quarter circles, as shown in the diagram. What is the radius of the inscribed circle? Forgot Your Password? 2018 © Queryhome

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https://byjus.com/question-answer/a-rectangular-field-is-60-metres-long-and-40-meters-wide-the-field-needs-to-3/

math

A rectangular field is 60 metres long and 40 meters wide. The field needs to be ploughed and it is to be covered with square blocks, all of the same size. What is the largest size of the square block which could be used for the purpose? The correct option is A 20 Given, Length of the rectangular field = 60 metres Breadth of the rectangular field = 40 metres To find the largest size of the square block, we need to find the HCF of 60 and 40.

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CC-MAIN-2022-05

https://www.witpress.com/elibrary/wit-transactions-on-engineering-sciences/59/18824

math

Influence Of Linear And Non-linear Constitutive Models On Thermoacoustic Waves In An Enclosure Free (open access) 13 - 22 L. Skerget & J. Ravnik The problem of unsteady compressible fluid flow in an enclosure induced by thermoacoustic waves is studied numerically. Full compressible set of Navier-Stokes equations are considered and numerically solved by boundary-domain integral equations approach, coupled with wavelet compression and domain decomposition to achieve numerical efficiency. The thermal energy equation is written in its most general form including the Rayleigh and reversible expansion rate terms. Both the classical Fourier heat flux model and wave heat conduction model are investigated. The momentum flux is modelled using standard Newtonian viscous model and linear viscoelastic Maxwell model. The velocity-vorticity formulation of the governing Navier-Stokes equations is employed, while the pressure field is evaluated from the corresponding pressure Poisson equation. Material properties are taken to be for the perfect gas and assumed to be pressure and temperature dependent. Keywords: compressible fluid flow, velocity-vorticity formulation, Navier-Stokes equations, thermoacoustic waves. 1 Introduction In the paper the generation and transmission of thermoacoustic waves in an perfect gas filled closed cavity is studied numerically. When a compressible fluid is subjected to a rapid increase in temperature at a solid wall, a sudden expansion of the adjacent gas occurs. This phenomenon generates a fast increase in the local pressure and leads to the production of pressure waves. These thermally generated waves are referred to as thermoacoustic waves. Thermoacoustic transport phenomena may be very interesting, when the fluid is close to thermodynamic critical point or when othermodes of transportmechanism compressible fluid flow, velocity-vorticity formulation, Navier-Stokes equations, thermoacoustic waves.

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CC-MAIN-2023-50

http://slideplayer.com/slide/253040/

math

Presentation on theme: "MG – 4111 HYDRO-ELECTROMETALLURGY Semester I, 2010/2011 DR. M. Zaki Mubarok Department of Metallurgical Engineering, Faculty of Mining and Petroleum Engineering."— Presentation transcript: MG – 4111 HYDRO-ELECTROMETALLURGY Semester I, 2010/2011 DR. M. Zaki Mubarok Department of Metallurgical Engineering, Faculty of Mining and Petroleum Engineering (FTTM)-ITB LECTURE NOTES Course Outline I.Introduction to Hydrometallurgy II.Thermodynamic and Kinetic Aspects in Hydrometallurgy III.Leaching and Solid-Liquid Separation IV.Solution Purification and Metals Recovery Methods from Pregnant Leach Solution Course Outline V.Leaching and Recovery of Metals and Oxides Ores (Au, Ag, Zn, Al, Cu, Ni) VI.Leaching and Recovery of Sulphide Ores (Zn, Ni, Cu) VII.Introduction to Electrometallurgy VIII.Metals Production by Electrolysis in Aqueous Solution IX.Fused Salt Electrolysis Literatures 1.Havlik,T., Hydrometallurgy: Principles and Applications, CRC publisher, Habashi, F. A Textbook of Hydrometallurgy, Metallurgie Extractive, Quebec, Norman L. Weiss, SME Mineral Processing Handbook, Volume II, SME, Unit Processes in Extractive Metallurgy: Hydrometallurgy, A Modular Tutorial Course of Montana College of Mineral Science and Technology 5.Biswas, A.K. And Davenport, W.G., Extractive Metallurgy of Copper, Pergamon, Oxford, fourth edition, 2002 Literatures 6.Unit Processes in Extractive Metallurgy: Electrometallurgy, A modular tutorial course of Montana College of Mineral Science and Technology 7.Yannopoulus, J.C,The Extractive Metallurgy of Gold, Von Nostrand Reinhold, New York, 1991 Course Structure and Mark Distribution Course Structure –Lecture –Tutorial –Assignment and Lab Work Mark Distribution –45% Midterm Exam –45% Final Exam –5% Assignment –5% Lab Work Attendance: 70% minimum CHAPTER I INTRODUCTION TO HYDROMETALLURGY Hydrometallurgy Extraction, recovery and purification of metals, through processes in aqueous solutions. Metals are also recovered in the other forms such as oxides, hydroxides. Electrometallurgy Recovery and purification of metals through electrolytic processes by using electrical energy. Hydrometallurgy Scope Traditionally, hydrometallurgy is emphasized for metals extraction from ores. Hydrometallurgical processing may be used for the following purposes: Production of pure solutions from which high purity metals can be produced by electrolysis, e.g., copper, zinc, nickel, gold, and silver. Production of pure compounds which can be subsequently used for producing the pure metals by other methods. For example, pure alumina to produce smelter grade aluminium. However, hydrometallurgy principles can be applied to a variety of areas such as metals recycling from scrap, slag, sludge, anode slime, waste processing, etc. Unit Processes in Hydrometallurgy In general, hydrometallurgy involves 2 (two) main steps: 1.Leaching Selective dissolution of valuable metals from ore. 2.Recovery Selective precipitation of the desired metals from a pregnant-leach solution. General outline of hydrometallurgical processes Ore/concentrate leaching Solid-liquid separation Solution purification Precipitation Pregnant Solution Solid residu to waste Leaching agent Oxidant Precipitant or electric current Pure compoundMetals Commonly, solution purification is conducted prior to metals recovery from the solution. Solution purification is aimed at obtaining a concentrated solution from which valuable metals can be precipitated in the next processes effectively Solution purification methods which are commonly used are as follows: –Adsorption by activated carbon –Adsorption by ion exchange resins –Solvent extraction (using organic solvents) –Precipitation with metals (cementation) Solution purification Solution purifications by adsorption with activated carbon, ion exchange resins (IX) and solvent extraction (SX) have the same unit operations, namely: –Loading, and –Elution In the elution step, the adsorbers are usually regenerated for another process cycle. Hydrometallurgy development Hydrometallurgy is developed after pyrometallurgy. Metals smelting has been practiced since thousands years ago. Hydrometallurgy was developed after the people discovered acid and base solutions. However, modern hydrometallurgy development is commonly associated with the invention of Bayer Process for bauxite leaching and cyanidation for gold extraction at the end of 19 th century (1887). One of important highlights of hydrometallurgy development is uranium extraction (Manhattan Project) aimed at nuclear weapon production in second world war (1940s). Important milestones in the development of hydro-electrometallurgy Cementation & Aqua Regia Use - 8 th Century Cyanidation Bayer Process Hall-Heroult Process , 1888 Copper Electrowinning Zinc Electrolytic Process Manhattan Project (IX/SX) s Biooxidation of Sulphide Concentrates s Pressure Leaching – Sherrit Gordon Nickel Process – Pressure Acid Leaching of Ni Laterites Large Scale Copper SX/EW s Important milestones in the development of hydro-electrometallurgy Carbon in Pulp (CIP)/Carbon in Leach (CIL) for Gold Recovery s Pressure Oxidation of Zinc Sulphides Two-Stage Zinc Pressure Leach Atmospheric Leaching of Zinc Sulphides – Albion (1993), Outokumpu (1999) Recent Developments: Skorpion Project (Anglo American) – 2003 (Zn from ZnS) Hydrozinc (TeckCominco) Incos Goro and Voisey Bay Projects Leaching of Chalcopyrite (CuFeS 2 ) Ores Hydrocopper (Outokumpu) Cu from sulfidic ores Atmospheric leaching of nickel laterite ore: 2008? Hydrometallurgy vs. Pyrometallurgy HydrometalurgyPyrometallurgy Treat high grade ore? Less economicMore economic Treat low grade ore?Possible with selective leaching Unsuitable Treat sulphide oreNo SO 2 ; otherwise S o or SO 4 2- are generated SO 2 generated (can be converted to H 2 SO 4 ) Separate similar metal, such as Ni and Co Possible with certain method Not possible PollutantWaste water, solid/slurry residues Gases and dust Reaction ratesSlowerRapid Hydrometallurgy vs. Pyrometallurgy HydrometalurgyPyrometallurgy Scale of operation?Possibly economic to be done at small scale operation and expansion is easier Unconomic at smale scale operation Capital costGenerally lower than pyrometallurgy Higher Energy costLowerHigher Materials HandlingSlurry Easy to be Pumped and Transported Handle Molten Metal, Slag, Matte ResiduesResidues – Fine and Less Stable Slags – Coarse and Stable Thermodynamic and Kinetic Aspects in Hydrometallurgy CHAPTER II Spontaneous Reaction, Equilibrium State As has been learned in basic engineering courses, chemical reaction will spontaneously occur when the Gibbs free ( G) < 0. G = G o + RT ln K G = 0 process is in equilibrium state – G o = standard Gibbs free energy –R = ideal gas constant = 8,314 J/K.mol –T = absolute temperature of the system (K) –K = equilibrium constant Standard Gibbs free energy is determined at: –Gaseous components partial pressure = 1 atm –Temperature = 25 o C (298 K) –Ions activity = 1 Equilibrium Constant For reaction: aA + bB cC + dD = activity coefficient of component A Nernst Equation Hydro-electrometallurgical processes often involve electrochemical reactions. For electrochemical reaction G = -nFE, G o = -nFE o, therefore In which, E = potential for reduction-oxidation reaction E o =standard potential for reduction-oxidation reaction n = number of electron involved in the electrochemical reaction, F = Faraday constant = Coulomb/mole of electron Nernst Equation Spontaneous process E > 0 G < 0 Chemical reactions usually perform in leaching processes Dissolution by acid –Example: ZnO (s) + 2H + Zn 2+ (aq) + H 2 O (l) Dissolution by base –Example: Al 2 O 3(s) + 2OH - 2AlO 2 - (aq) + H 2 O (l) Dissolution by complex ion formation Example: CuO (s) + 2NH 4 + (aq) + 2NH 3(aq) Cu(NH 3 ) 4 2+ (aq) + H 2 O (l) Dissolution by oxidation –Ex: CuS (s) + 2Fe 3+ Cu 2+ (aq) + 2Fe 2+ + S o (s) Other oxidators: O 2, ClO -, ClO 3 -, MnO 4 -, HNO 3, H 2 O 2, Cl 2 Dissolution by reduction mechanism –Ex: MnO 2(s) + SO 2(aq) Mn 2+ (aq) + SO 4 2- (aq) Chemical reactions usually perform in leaching processes Correlation of free energy ( G) and heat (enthalphy = H) G = H - T S G o = H o - T S o Cp = heat capacity at constant pressure (J/molK) Where possible, processes are designed to be autothermal maintain constant temperature by the heat given by the reaction H o = Standard enthalpy (kJ/mol) G o = Standard entropy (kJ/mol) G o (reaction) = G o (products) - G o (reactants) H o (reaction) = H o (products) - H o (reactants) S o (reaction) = S o (products) - S o (reactants) Calc. example 1 Find K for each reaction using a) Standard free energy data b) Standard electrode potential data Calc. Example 2 a) What is the electrode potential of the Ni 2+ /Ni reaction in sulphate solution at 25°C at a Ni 2+ concentration of M (assumption: activity of Ni 2+ is equal to its molar concentration) b) At what pH is H 2 at 10 atm at equilibrium with this solution and pure nickel? Ni e = Ni E° = V 2H e = H 2 E° = 0.00 V Pourbaix Diagram Pourbaix Diagram = Potensial (E h ) – pH Diagram. The diagram represents thermodynamic equilibrium of metal, ions, hydroxides (or, oxides) in aqueous solution at certain temperature (isothermal). The boundary of stability regions of metal, ion, hydroxides (or oxides) are equilibrium lines. Does not reflect reaction kinetics. Pourbaix Diagram Three possible types of equilibrium lines: –Horizontal –Vertical –Slope Variations in ion activities are plotted as contours/dashed lines Horizontal Line: for equilibrium reactions that are independent of pH. Horizontal Line Example: Fe 3+ + e = Fe 2+ E o = 0.77 V R = J/Kmol, T = 298 K, F = C/mol e -, n = 1 mol e - If all ion concentrations are assumed to be equal to their molar concentrations M. Vertical Line Reactions do not involve electron n = 0, no potensial, the equilibrium depends only on pH. Example : Fe 2 O 3 + 6H + = 2Fe H 2 O For certain Fe 3+ concentration we can determine the equilibrium pH for the above reaction. K = [Fe 3+ ] 2 /[H - ] 6 Slope Line For reactions that depend both on potensial (E h ) dan pH. Example : If all ion concentrations are assumed to be equal to their molar concentrations M. Water stability region (dotted lines) Upper boundary line Lower boundary line At pO 2 = 1 atm At pH 2 = 1 atm E h -pH diagram of Fe-H 2 O system at 25°C E h -pH Diagram of Zn-H 2 O System at 25 o C. E h -pH Diagram of Cu-H 2 O System at 25 o C. Application of E h – pH diagram in hydrometallurgy Predicting potential leaching behaviour for certain mineral system Predicting the possibility of metals ion precipitation at the purification of pregnant-leach solution Application of E h – pH diagram in hydrometallurgy Fe(OH) 3 or Fe 2 O 3 can be precipitated from Fe 3+ at lower pH than the precipitation of Zn 2+ to Zn(OH) 2 or ZnO. Fe 2+ have to be oxidized to Fe 3+ to gain lower pH value for Fe(OH) 3 precipitation. Pourbaix Diagram can be constructed at various temperature for more than two systems E h -pH diagram of Zn-S-H 2 O system at 25 o C Diagram Pourbaix in Presence of Complex Ion Example: Au-H 2 O system with the presence of cyanide (CN - ) ion (case of gold cyanidation leaching) Equilibrium of Au 3+ /Au Standard reduction potential for Reaction 1: Equilibrium reaction of O 2 /H 2 O E o = 1.23 V. Therefore, Au 3+ ions are not stable in water and readily reduced to Au by oxidation of H 2 O to O 2 (the opposite of Reaction 2). In the other word, gold can not be oxidized (dissolved) in water only with the presence O 2. (2) Potensial – pH diagram of Au–H 2 O system without the presence of complexing agent With the presence of CN -, Au 3+ forms STABLE COMPLEX of aurocyanide (Au(CN) 2 - ) and the potential-pH diagram for Au changes significantly as follow: E h -pH Diagram of Au-CN- H 2 O system at 25 o C for [Au] = M and [CN - ] = M By the presence of cyanide ions, Au + + e = AuE = 1.69 – log [Au + ] Au + + 2CN - = Au(CN) 2 - (K = 2 x ) Au(CN) e = Au + 2CN (3) In comparison to the first reaction that has E o of 1.69 V, Reaction (3) has much lower E o at V. Dissolution of Au is limited by the following equilibrium of Reaction (3). During cyanidation leaching, dissolved oxygen is required to oxidize Au prior to the formation of stable complex of Au(CN) 2 -. Interactions in Electrolyte Solution Two types of interactions in electrolyte: -Ion-ion interaction, and -ion-solvent interaction Knowledge of interaction in electrolyte solution is important because the interactions affect solvation effects, diffusion, conductivity, ionic strength and activity coefficients of ions in solution. Interactions in electrolyte solution influence the transport properties of ions in solution. Ionic Strength and Activity Coefficient -Ionic strength (I), expresses the ionic concentration that includes the effects of ionic charge. -Ionic strength (I) is defined as follow: -It is found that activity coefficient, electrical conductivity and the rates of ionic reactions are all the functions of ionic strength. in which c i = concentration of ion i in molar (mol/L) and z i = the charge of ion i. Ionic Strength for unit concentration in molal -Remember, molality = moles of solute in 1 kg solvent. Molality can be converted to molality by the following correlation: in which M i = the molar mass of each solute in kg/mol (not in g/mol), c i = molarity of solute i, and is the density of the solution in kg/m 3 (=g/L) -In dilute solutions, c i 0.001m i o (in which o = density of pure solvent). Ionic Strength for unit concentration in molal -Therefore for dilute solution, If the solvent is water at 25 o C (density 1000 kg/m 3 ), then: Similar form with ionic strength in molarity - Molar activity coefficient can be converted to molal activity coefficient by the following correlation:) for salt, or for single ion. in which = total moles of ion formed during complete dissociation, m = ionic molality and M s = molecular weight of solvent (kg/mol). Activity and Activity Coefficient, DEBYE-HUCKEL LAW -Debye Huckel Law correlates the activity coefficient (f i, i ) with ionic strength (I). -Forms of Debye-Huckel equations depend on concentration of solution and the unit concentration used. -For dilute solution at 25 o C and I given in molar (M), -The above equations are known as LIMITING DEBYE HUCKEL LAW. for single ion, and for salt. The limitation of LIMITING Debye-Huckel Equation The D-H Limiting Law is called a limiting law because it becomes increasingly accurate as the limit of infinite dilution is approached. Up to concentrations of about 0.01m THE LIMITING D-H LAW gives reasonable values, but at higher concentrations the calculated activity coefficient become inaccurate (high %error compared to the values determined experimentally). Debye-Huckel Law for Concentrated Solution -For concentrated solution (> 0.01 molal), Limiting Law D-H is modified by considering the ionic size parameter: -in which A and B are constants that depend on the kind of solvent and temperature, a = ion size parameter. - For aqueous solution at 25 o C, A = and B = x 10 9 meter. ACTIVITY AND MEAN ACTIVITY -Molar activity and molar activity of a single ion i is determined as follow: -For 1 mole of M + A - salt that dissociates to + mol of M z+ and - mole of A z- M + A - + M z+ + - A z- = and Mean molal activity coefficient can be determined by the following correlation: Mean molality, Thus, mean molal activity, Note that m m ACTIVITY AND MEAN ACTIVITY Exercise: 1 1. Determine the molar activity coefficient of Ca 2+ at 25 o C using relevant Debye Huckel Equation in the following solution: a mole of HCl and mole of CaCl 2 in one liter solution b mole of HCl and mole of CaCl 2 in one liter solution c.0.4 mole of HCl and 0.2 mole of CaCl 2 in one liter solution Ion size parameter for Ca 2+ = 0.4 nm. Exercise 2: 2.The stoichiometric mean activity coefficient at 25 o C of the sulphuric acid in a mixture of 1.5 molal sodium sulphate (Na 2 SO 4 ) + 2 molal H 2 SO 4 is If the second dissociation constant, K 2, for sulphuric acid is and the pH of the solution is –0.671, calculate: a)the molal activity of H 2 SO 4 b)the molal activity of SO 4 2- c)the molal activity of HSO 4 - d)the mean activity of H 2 SO 4 Exercise:3 1 gram FeCl 2, 1 gram NiCl 2 and 1 gram of HCl are added to 200 ml of deaerated water. Platinum electrodes are used to deliver electrical current so that the electrolysis performs. The anodic and cathodic current density are 1000 A/m 2. The following are the reactions and E o (in the reduction direction) that may occur: Fe e = FeE o = -0,277 V Ni e = NiE o = -0,250 V 2H + + 2e = H 2 E o = 0 V Cl 2 + 2e = 2Cl - E o = 1,359 V a) Calculate molar activity coefficients of the cations and anion contained in the solution (use the Finite Size of Debye Huckel Limiting Law) b) Calculate the activity of the cations and anion contained in the solution c) Determine the half cell potential of the above reactions d) Which pair of redox (reduction –oxidation reaction) that would occur (based on the calculation of c) Exercise: 3 (cont.) e) What would be the cell voltage of the reaction d Data: Atomic weight Fe = 55.8, Ni = 58.7, Cl = 35.5, H =1 Ion size parameter in nm : Fe 2+ = Ni 2+ = 0.6, H + = 0.9, Cl - = 0.3 H 2 overpotential = 0.28 V Cl 2 overpotential = 0.03 V Ohmic overpotential = 0.25 V. Kinetics in Hydrometallurgy Kinetics in hydrometallurgy deals with the kinetics of leaching, adsorption and precipitation Studying of leaching kinetics is done for the establishment of the rate expression that can be used in design, optimization and control of metallurgical operations. The parameters that need to be estabished: –Numerical value of the rate constant –Order of reaction –Rate determining step –Activation energy Leaching Kinetics Consider the dissolution of a metal oxide, MO, with an acid by the following reaction : The reaction rates for this leaching system can be given by MO (s) + 2H + (aq) M 2+ (aq) + H 2 0 (aq) Leaching Kinetics For general example if a chemical reaction involves A and B as reactants and C and D as products, the stoichiometric reaction can be written as follows: where a, b, c, and d = stoichiometric coefficients of species A, B, C, and D, respectively k 1, k 2 = reaction coefficients in the forward and reverse directions, respectively (1) Leaching Kinetics The rate expression of this stoichiometric reaction can be written in a more general way: where C A, C B, C c, and C D are concentrations of species A, B, C, and D, respectively and m, n, p, q are orders of reaction. (2) Leaching Kinetics However, if the reaction given in Eq. 1 is irreversible, as in most leaching systems, Eq. 2 is reduced to the following form: where k 1 = k 1 x a. For this system, the rate constant, k 1 ', and the orders of reaction, n and m, should be determined with the aid of leaching experimental data. The rate expression given in the above equations can be further reduced if the reaction is carried out in such a way that the concentration of A is kept constant. For such situations, the rate expression is reduced to: where k 1 =k 1 x C A n. It should be noted that the rate constant and the order of reaction are constant as long as the temperature of the system is maintained constant. Consider the dissolution of zinc in acidic medium: Zn (s) + 2H + (aq) Zn 2+ (aq) + H 2(g) For the above reaction, the rate of disappearance of H + ion is directly related to the rate of appearance of Zn 2+ ion; thus, If concentration of zinc metals is assumed to be constant and C H is further abbreviated generally as C A, then the equation can be written as follow : The order of reaction, n, can be any real number (0, 1, 2, 1.3, etc.). When n = 0, the reaction is referred to as zero order with respect to the concentration of A. where C A o represents the concentration of A at t = 0, and X A represents the fractional conversion, i.e., X A = [ (C A o C A )/ C A o ]. If the plot of X A versus t gives a straight line, the zero- order assumption is consistent with experimental observations and the k value can be obtained from the slope of the plot. When n = 1, the reaction is first order with respect to the concentration of A: k/C A o XAXA time k ln (1 - X A ) For second order reaction, k time X A (1 - X A ) CAokCAok If the second-order assumption is valid, we obtain a straight line from a plot of X A /(1 - X A ) versus t, and the rate constant can be determined from the slope of the plot. Temperature Effect on the Reaction Rate (Arrhenius Law) Reaction rate increases markedly with increasing temperature. It has been found empirically that temperature affects the rate constant in the manner shown in the following equation: where E a is the activation energy and k° is a constant known as the frequency factor, frequently assumed to be independent of temperature. Modeling of heterogenous reaction kinetics Heterogenous reaction between solid and fluid in hydrometallurgical processes is frequently modelled withshrinking core model. If we select a model we must accept its rate equation, and vice versa. If a model corresponds closely to what really takes place, then its rate expression will closely predict and describe the actual kinetics; If a model differ widely from reality, then its kinetic expressions will be useless. Detailed of modeling and relevant kinetics equations for various rate determining steps can be found in previous course (Metallurgical Kinetics). For determination of E a, number of experiments, at least at three or four different temperatures are needed, with all other variables being kept constant. The next step is to calculate the rate constant for each temperature as discussed previously. A plot of In k versus 1/T yields a straight line from which the activation energy, E a, can be determined Activation energy value can be used to predict the rate determining step of the reaction: E a = 40 – 80 kJ/mol: process is controlled by surface chemical reaction E a = 8 – 20 kJ/mol: process is controlled by diffusion to and from the surface Mass Transfer in Solution For hydrometallurgical system, mass transfer of component i in solution frequently consists of a molecular diffusion term, migration term, and convective diffusion term, as indicated in the following expression: where N i = flux of i, C i = concentration of i, D i = diffusion coefficient of i C i = concentration gradient of i, z i = valence of the specified ion, µ i = ionic mobility, F = the Faraday constant, Ф = electrical potential gradient, and V = net velocity of the fluid of the system First and Second Ficks Law of Diffusion If N i consists of the molecular diffusion term only, Dimensionless Parameter for Convection Calculation (Fick's first law) where µ = the viscosity of the fluid, ρ = the density of the fluid Dimensionless Parameter for Convection Calculation The parameter LV/D i is known as the Peclet number and can be separated into two other parameters: Lvρ/µ that known as the Reynolds number, and µ/ρD i is the Schmidt number. Peclet number is regarded as a measure of the role of convection against diffusion, For most hydrometallurgical systems, the Schmidt number is on the order of 1,000 because the diffusivity of ions and kinematic viscosity of water are, respectively, on the order of cm 2 /s and cm 2 /s. Therefore, if the Reynolds number is greater than 10 -3, the Peclet number is greater than 1, and consequently, convective diffusion is more dominating than molecular diffusion in such systems. Mass Transfer Coefficients for Convective Diffusion For systems with large Peclet numbers, it is frequently assumed that there is a diffusion boundary layer at some distance from the solid surface. For such systems, it is quite common to write the mass flux from the bulk solution to the solid surface as follows: N i = k m (C b - C s ) where N i = mass flux of species i k m = mass transfer coefficient, in cm/s C b = concentration of species i in the bulk solution, in mol/cm 3 C s = concentration of species i at the solid surface, in mol/cm 3 Because the units of measure of k m are the same as those of (D/ ), where is the diffusion boundary layer thickness, k m, is often substituted by this ratio. Therefore, The diffusion boundary layer thickness is often estimated by the relationship k m = D/δ, provided k m is known. Mass Transfer from or to a Flat Plate. The mass transfer coefficient for a flat plate where fluid is flowing over the plate at a velocity V 0 has been well documented. The mass transfer coefficient for such a system can be estimated from first principles and has the following form: where D = the diffusivity of the diffusing species v = the kinematic viscosity of the fluid L = the length of the plate Rotating Disk. Although it is not a practical geometry, because the mathematical representation of the system is exact and follows very closely to the experimental data, a rotating disk is frequently used to determine the mass flux and the mass transfer coefficient. The mass transfer coefficient for this system is as follow The equation is valid for the Reynolds number, r 2 ω/ν is less than 10 5, where r and ω are, respectively, the radius and the angular velocity of the disk. Particulate System It has been demonstrated that the mass transfer coefficient for particulate systems can be given by the following equation: where d = the diameter of the particle, V t = the slip velocity, which is often assumed to be the terminal velocity of the specified particle. The terminal velocity of a particle can be calculated using the following equation depending on the Reynolds number of the system, which is defined by dV t ρ/μ, where ρ is the density of the fluid: where ρ s is the density of the particle. The preceding equation is often referred to as the Stokes' equation and is valid as long as the Reynolds number is less than 1. When the Reynolds number is between 1 and 700, the following equations are used: where g is the gravitational coefficient. A cementation reaction, Zn + Cu 2+ Cu + Zn 2+, is taking place at the surface of a zinc plate of 10 cm x 10 cm area. Feed flowing parallel to the plate at a velocity of 1 m/s contains copper at 1 mo/dm 3. Suppose we want to estimate the rate of deposition assuming that the mass transfer of Cu 2+ to the zinc plate is rate determining step. The diffusivity of Cu 2+ is 7.2x10 -6 cm 2 /s, and the kinematic viscosity of water is 0.01 cm 2 /s. Example 1: where S = the surface area of the plate N cu 2+ = the number of moles of Cu 2+ ion Cu b 2+ = the concentration of Cu 2+ in the bulk Cu s 2+ = the concentration of CU 2+ at the interface k m = (7.2 x ) 2/3 (0.01) -1/6 (10) -1/2 (100) 1/2 = x 3.7 x x 2.15 x x 10 = 1.7 x cm/s. Therefore, Example 2: Consider the situation from the previous example, except that instead of a zinc plate, zinc particles 100 µm in diameter are suspended in a 1 mol/dm 3 Cu 2+ solution. Suppose we want to estimate the rate of deposition of Cu 2+ (Note that the density of Zn is 7.14 g/cm 3.)

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The Definition of Moments Moments are the expressions of distance and physical quantity. A moment accounts for the location and direction of a physical quantity. When a molecule or a cell is in motion, a moment occurs at its location. A meter of water moves in an hour, but a kilometer moves in a day. A second is equal to one second. This definition can be confusing, but let’s go over the differences between moments and meters. Moments are the ratio of force and distance. A force acting on an object has to act in a way that causes the object to rotate or move about its axis. Regardless of direction, it must cause the body to twist or translate. To be a moment, the force must act in such a way that it fails to pass through the centroid of the object. The absence of an equal and opposite force along the direction of action is also what causes the moment. The definition of moments is a bit confusing, but it’s crucial to understand how moments work. The formula is simple: a moment of force is equal to the sum of the forces acting on two or more objects. It is an algebraic formula that involves multiple component moments. Each of these components is measured in Newtons, but the process is difficult. Because moments are often derived from the sum of several other quantities, they require careful consideration. Moments are a useful way to measure the relative importance of different parts of an object. For instance, you might use them to compare the weight of a heavy door against another heavy door. Likewise, you may want to use higher order moments to analyze the position and movement of a person or object. By comparing the distances between the two, you can see how the weight of the tail corresponds to the amount of change in the shoulders. A moment of force is the tendency for an object to rotate around its axis and move in the direction of the force. It must be able to cause a body to rotate. The difference between a torque and a moment of force is the distance. When a force is applied to an object that is far from the reference point, a moment of force is created. When the same situation occurs in a different direction, the torque is canceled out. The definition of a moment of force is the tendency for an object to rotate about its axis or move in the direction of a force. A moment of force is defined as the product of the force and the distance from the reference point. In simple terms, a torque is a rotation in a body, whereas a moment of force is the rotation of a torque. In a sense, the term “moment of force” refers to the torque in a car.

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Concepts of Modern Mathematics Some years ago, "new math" took the country's classrooms by storm. Based on the abstract, general style of mathematical exposition favored by research mathematicians, its goal was to teach students not just to manipulate numbers and formulas, but to grasp the underlying mathematical concepts. The result, at least at first, was a great deal of confusion among teachers, students, and parents. Since then, the negative aspects of "new math" have been eliminated and its positive elements assimilated into classroom instruction. What people are saying - Write a review Mathematics in General Motion without Movement Short Cuts in the Higher Arithmetic The Language of Sets What is a Function? The Beginnings of Abstract Algebra The Group Concept The Theory of Probability Computers and Their Uses Applications of Modern Mathematics

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https://theflyinginn.com/how-many-liters-is-100-ml

math

More information from the unit converter The answer is 1000. We assume that you convert between milliliters and liters. You can see more details about each unit of measurement: ml or liters The SI-derived unit for volume is cubic meters. In this article : What is 100 mL the same as? At standard temperature and pressure, 100 ml of water weighs 100 g. Read also : How much money does NBA YoungBoy make a year? Originally answered: Is 100mL the same as 100g? Milliliters measure volume, while grams measure mass, but these two quantities describe the same amount of a substance IF the substance has a density of one gram per gram. milliliters. Is 100 ml the same as 1 cup? Is 100 ml the same as 1 cup? One cup is equal to 240 ml. This tells you that if you only have 100 ml, you will not have a whole cup. You have less than half a cup. What is a 100 ml in cups? What size is a 100 ml bottle? size 100 mL, vial H × W 109 mm × 56 mm. How many 100mL are in a litter? The answer is 1000. We assume that you convert between milliliters and liters. This may interest you : How many baseball games in a season. You can see more details about each unit of measurement: ml or liters The SI-derived unit for volume is cubic meters. How many liters is 100 ml? There are 1000 ml in a liter. 100 ml equals one tenth of that, so 100 ml equals one tenth of a liter. How many ml are there in a litter? Yes, 1 L = 1000 ml. Video : How many liters is 100 mL? Is 1l of water enough? There are many different opinions on how much water you should drink each day. This may interest you : Who is faster whis or Minato? Health experts typically recommend eight 8-ounce glasses, equivalent to about 2 gallons or half a gallon a day. What part of 1 Litre is 100 mL? 100 ml equals one tenth of that, so 100 ml equals one tenth of a liter. On the same subject : How do you make a dog howl at the moon? Which part of a liter is 1 ml? A milliliter is a smaller metric unit that represents the volume or capacity of a liquid. It is used to measure a small amount of liquid and is equal to one thousandth of a liter (1 liter = 1000 milliliters). Which part of 1 liter is 100 ml? Liter and millimeters are not comparable, as the first is a unit of volume and the second a unit of length. There are 1000 ml in a liter. 100 ml equals one tenth of that, so 100 ml equals one tenth of a liter.

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CC-MAIN-2022-27

https://www.kidpid.com/place-value-worksheet-for-grade-3/

math

When starting with mathematics kids are taught that place value refers to the value of each digit in any given number. For example: In the number 1239, 9 represents ones or unit value, 3 represents 3 tens 0r 30, 2 represents 2 hundred or 200 and 1 represents 1 thousand or 1000. As simple as this definition seems, the importance of understanding this is magnanimous. Unless a student understands the concept of place values, he can’t differentiate between the value and magnitude of numbers properly since it is only when he learns that a number with a digit in tens place value is bigger than a single-digit number and so on, he would be able to tell that 100 is greater than 50. At this point, there seems no further need to elaborate on why learning place values is important. Without the proper knowledge of this concept, students won’t be able to perform any arithmetic calculations such as multiplication, division, subtraction, or addition for that matter. The following worksheet brings several place-value-related exercises for students to practice and get familiar with it. The worksheet is meant for third graders and by the time a student is in the third grade, he is usually familiar with the concept. There are several different types of questions that test his understanding and problem-solving capabilities. In the first worksheet, there are eight different questions. In the first two questions, students have to identify the largest and smallest number from the given numbers. Then there are some after and before exercises in which they have to write the numbers that come before or after the given number. Finally, there are questions asking them to find the sum and difference of two numbers. As mentioned earlier, it is only by understanding place values that one can tell the magnitude of a number. The next sheet asks students to identify which of the two numbers is greater. They have to draw the greater than or less than symbol between each number set. For example, 520 is smaller than 600, hence they’ll mark the “<” sign (smaller than sign) between the number set (520 < 600). Similarly in the second exercise, all they need to write if the number on the left side is greater than or smaller than the one on the right side. In the following sheet, students have to round the given numbers to the nearest ten (10), hundred (100), and thousand (1000). The fourth sheet provides three relatively simpler exercises. In the first question, students need to state the place value of the underlined digit in the numbers. In the next question, they have to partition the given numbers and in question number three, they are required to write the numbers in numerals. The final sheet consists of questions that require students to count forward and backward by tens and one hundred. The worksheet covers various topics all of which can be solved only with the knowledge of place values. It’ll ensure that students excel in basic mathematical functions and develop computational skills with time as they move on to more complex arithmetic operations. Download the worksheet, print it, and get your students learning!

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CC-MAIN-2023-40

https://www.astronomyclub.xyz/double-stars/disadvantages.html

math

Diffraction micrometers have one drawback. As the grating consists of bars and slits with the same width, only 50% of the incident light from the double star will reach the telescope optics. Of this, about 50% of the residual light will end up in the zero-order images resulting in a total loss of 1.5 magnitudes compared with the unobstructed telescope. Another 20% goes into each of the first order images, the rest being lost in the additional satellites. Because of these losses the combination of a 20-cm telescope and a diffraction micrometer will allow observations of double stars as faint as about magnitude 7.0-7.5 with components which do not differ too much in brightness. The diffraction micrometer formula includes the factor X, the wavelength of light. As the observation is made visually the satellite's exact distance from the primary star depends on the observer's own wavelength sensitivity but also on the stars' colours. The observer's most sensitive wavelength which should be used in the formula has to be established by comparisons with pairs with accurately known separations. A normal figure for X to start with might be 5650 A, or 0.000565 mm if p in the formula is in millimetres. This corresponds approximately with the effective wavelength of a white, class A spectral type star. Was this article helpful?

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https://web.stagram.com/tag/mymontegrappa

math

Is it a hobby? Obsession? Passion? Who knows, but I really really love fountain pens and it's great to be able to share that with everyone in this wonderful community. Here's the one which started me back on the FP journey after years in the rollerball wilderness: my pride and joy Montegrappa Miya. With lush Diamine Blue Black on Rhodia. Thanks to everyone for your likes and kind words. #fountainpen#fpgeeks#diamine#mymontegrappa#fpcommunity It was all going so well.. Yesterday out Pokemon hunting, today projectile vomiting (and the rest). That's my 8 year old son by the way, not me, although I admit I do enjoy a spot of Pokemon hunting too. Hoping for a full recovery for the big day and that the rest of us survive unscathed...#norovirus#mymontegrappa#fingerscrossed Back from 5 days solid of conference and board meetings! Just got my hands on this Italian @fabriano_boutique A4 4mm dot pad. Apparently they started making paper in the 1200s - now that is a lot of history! This is absolutely GORGEOUS paper. The 4mm dots suit my writing perfectly (rather than 5mm). The covers come in a range of lush colours. Suits my @montegrappaitalia Miya perfectly! Love it! #mymontegrappa#FPGeeks#fountainpen#paperporn Ahh finally a week off work, fantastic English spring weather (mostly!!) and some much needed family time. Just getting round to some overdue pen maintenance too, the Miya has been empty waaaaaay too long. Cleaned and refilled with the ever faithful Diamine Blue Black. #fountainpen#fpgeeks#mymontegrappa#montegrappaitalia#diamine Received this *very* generous gift recently from someone who is vey dear to my heart - you know who you are! Absolutely fascinating reading, learning about the humble origins of these world famous brands. Packed with lovely photos and lots of (now funny) adverts from the olden days. Note there was no Montegrappa on the front cover so *had* to put that right, obviously! #fpgeeks#fountainpen#history#readinganactualbook#gift#friends#montegrappaitalia#mymontegrappa #Repost@montegrappaitalia (@get_repost) ・・・ Montegrappa takes you back in time to Ancient Egypt with the limited edition pens Thoth, dedicated to the Egyptian god of writing and inventor of the alphabet. Papyrus or paper, it doesn't matter, writing with Thoth pens is a magical experience.

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https://journals.uwyo.edu/index.php/ela/article/view/5163

math

Main Article Content All graphs considered are simple and undirected. A cluster in a graph is a pair of vertex subsets (C, S), where C is a maximal set of cardinality |C| ≥ 2 of independent vertices sharing the same set S of |S| neighbors. Let G be a connected graph on n vertices with a cluster (C, S) and H be a graph of order |C|. Let G(H) be the connected graph obtained from G and H when the edges of H are added to the edges of G by identifying the vertices of H with the vertices in C. It is proved that G and G(H) have in common n −|C| + 1 distance Laplacian eigenvalues, and the matrix having these common eigenvalues is given, if H is the complete graph on |C| vertices then ∂ −|C| + 2 is a distance Laplacian eigenvalue of G(H) with multiplicity|C| − 1, where ∂ is the transmission in G of the vertices in C. Furthermore, it is shown that if G is a graph of diameter at least 3, then the distance Laplacian spectral radii of G and G(H) are equal, and if G is a graph of diameter 2, then conditions for the equality of these spectral radii are established. Finally, the results are extended to graphs with two or more disjoint clusters.

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https://raisaputhy.wordpress.com/2010/09/25/latihan-soal-astronomi/

math

Latihan Soal Astronomi 1. Jayshree claimed that she saw a solar eclipse when the size of the solar disk was 26’ and that of the lunar disk was 30′. She also claimed that at the time of the maximum eclipse, distance between the centres of the two disks was 7′. Qualitatively show that she could not have observed a total eclipse. Find the percentage of the solar disk covered at the time of the maximum eclipse. 2. A year in Solar calendar consist of 365.25 days and the same in Lunar calendar consist of 354 days. The additional days in Solar calendar are kept as balance every year. Whenever the number of balance days exceeds 30, an additional month of 30 days is added to the lunar year to offset the difference. The cycle goes on. Anwesh, whose birthday falls on 1st January, noticed that in the year 2008, his birthday coincided with the start of the lunar year. In which earliest future year, his birthday will again coincide with the start of the lunar year?

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http://bbs.boingboing.net/t/salvador-dalek/21828

math

#1 By: Cory Doctorow, February 4th, 2014 18:01 #2 By: Juan Rudametkin, February 4th, 2014 18:17 Imagine if it had been "Doctor Quien", every alien in the history of the universe would begin their fiendish plot to take over the world by invading Barcelona and/or infiltrating the Catalonian government, that Gaudí church would be some sort of homing beacon or "temporal resonance distortion field generator". #3 By: Preston, February 4th, 2014 18:40 The Persistence of Exterminate! #4 By: Cary, February 4th, 2014 18:50 #5 By: Smash Martian, February 4th, 2014 19:28 OTOH, Daleks are not fans of Escher. #6 By: Preston, February 4th, 2014 19:35 #7 By: Phillip Hallam Baker, February 4th, 2014 19:56 Salvador Dalek is the name of the hero sized prop replica I built: He is made from fiberglass, plastic and alumnium and weighs about 150lbs including the frame I am sitting on inside. #8 By: Cary, February 4th, 2014 20:27 Doesn't sound very comfortable -- try this on and just grab a plunger from the loo: #9 By: Cary, February 4th, 2014 20:55 #10 By: robulus, February 5th, 2014 01:40 #11 By: Beanolini, February 5th, 2014 04:19 Right now? You have an internet connection in there? #12 By: gilbert wham, February 5th, 2014 04:24 #13 By: Cory Doctorow, February 9th, 2014 18:02 This topic was automatically closed after 5 days. New replies are no longer allowed.

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CC-MAIN-2014-15

https://www.icseboards.com/sample-paper-icse-class-10-physical-education-set-a/

math

Students can refer to the following Sample Paper ICSE Class 10 Physical Education Set A with Answers provided below based on the latest syllabus and examination guidelines issued for ICSE Physical Education. All specimen papers have been prepared covering all chapters given in ICSE Physical Education book for Class 10. You should also refer to ICSE Class 10 Physical Education Solutions. Sample Paper ICSE Class 10 Physical Education Set A with Answers Attempt all questions from this section A and two questions from Section-B. Section – A (50 Marks) (Attempt all questions from this section) Q1. (a) Define Power. (2) (b) What do you mean by Growth? (2) (c) State any three objectives of sports training. (3) (d) Enlist the six components of Physical Fitness. (3) Q2. (a) Write the full form of PRICER. (2) (b) What is flexibility? (2) (c) Write any three reasons to stress the importance of physical fitness. (3) (d) What are the preventive measures to avoid injuries in sports? (3) Q3. (a) What is the role of emotional development of a child? (2) (b) What is agility? (2) (c) Write any three differences between growth and development. (3) (d) Define sports Medicine. (3) Q4. (a) Define Heredity. (2) (b) Write any four objectives of Physical Education. (2) (c) Mention the stages of childhood. (3) (d) What do you mean y speed? (3) Q5. (a) Enlist all the body types. (2) (b) Define the term nutrition. (2) (c) Explain the progression and overload as a principal of sports training. (3) (d) What form of first aid can be given to a person suffering from a contusion injury? (3) Section – B (50 Marks) (Attempt two questions from this section) You must attempt one question on each of the two games of your choice Q6. (a) Explain the following terms in Cricket: (8) (i) Bump ball (ii) Dead ball (iii) Batting order (iv) A night watchman (b) (i) Explain a ‘declaration’ in a test match. (9) (ii) State the duties of the match referee. (iii) Explain the term danger area. (c) (i) Explain the terms boundary for four and boundary for six. (8) (ii) Mention and four instances when the umpire call for a No ball. Q7. (a) Briefly explain the following terms in cricket: (8) (i) A tie match (ii) A golden duck (iii) A hook shot (iv) A scorer (b) (i) State any three conditions when runs are added to the team’s total score and not to the batsman’s total runs. (9) (ii) Name the different types of matches played in cricket. (iii) Mention three situations when a team’s innings is said to be completed. (c) (i) What is meant by the term ‘Timed-out’? (8) (ii) Name any two National tournaments in the game of Cricket. (iii) Explain the term ‘run out’. (iv) Name any two international tournaments in the game of cricket. Q8. (a) Explain the following terms in football: (8) (i) A chest trap (ii) A nutmeg (iii) Quarter circle (iv) Wall Pass (b) (i) Write down the rules related to kick from the penalty mark. (9) (ii) Mention any three circumstances when time is lost or wasted during the course of play and that is added at the end of each playing session. (iii) What is a ‘Techincal area’ in football? State the importance of technical area. (c) Explain the following terms with measurements:(8) (i) A touch line (ii) A goal line (iii) A goal area (iv) A penalty area Q9. (a) Briefly explain the following terms in football: (8) (i) Added time (iii) Goal line technology (iv) Man to man (b) (i) State the three methods of restarting a game in football. (9) (ii) Name any three national tournaments in football. (iii) Explain the term Cross and Chip. (c) (i) Explain the term ‘step over’ in football. (8) (ii) Give the full form of AIFF. (iii) What is meant by the term ‘advantage’ in football? (iv) What are the basic skills of football?

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CC-MAIN-2023-40

https://www.essay-pros.com/asmall-local-retailer-has-two-check-out-counters-a/

math

Asmall local retailer has two check-out counters available; however, only onecheck-out counter is currently being used on a particular 8-hour shift. Customers arrive at an average rate of 8 perhour. The person currently working atthe check-out counter is capable of serving an average of 10 customers per hourat a cost of $25 per hour. The managerhas received numerous complaints from customers threatening to take theirbusiness to a local competitor who offers faster customer service. The manager has estimated that the cost ofwaiting is $50 per hour. a. Whatis the average number of people waiting in line to be served? b. Whatis the average amount of time that a customer currently spends waiting in lineto be served? c. Whatis the currently total daily system cost?

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https://new-mexico.cactus-society.org/SiegfriedLodwig/CactiFromSeeds/slide_3.html

math

July 24, 2020 sowing Set 3 Peniocereus greggii SNL1 (30) max germination count = 38        Count Sept 19 = 40   Astrophytum capricorne (100) max germination count = 77               Count Sept 19 = 65 Echinocereus triglochidiatus (150) max germination count = 84  Count Sept 19 = 94   Escobaria missouriensis SNL155 (150) max germination count = 100  Count Sept 19 = 87 Two weeks later. The plants are really small. They're only eight weeks old. They've got spines. They are identifiably cacti. Note again how similar the Echinocereus triglochidiatus and the Escobaria missouriensis appear. Note also how uniformly large (uniformly small?) the members of each of the four species are. Look especially at the Peniocereus greggii seedlings. The eventual size variation of all four species hasn't yet kicked in. Size at any point is a function of both genes and environment. Environment is typically food and water, but, for plants, it turns out root room is also extremely important. This principle is shown by all the bonsai plants in small pots. (I saw tens of thousands of bonsai all over China in my four trips there, including a supposedly thousand year old banyan tree about 8' tall in a 6'X6'ceramic pot. A banyan tree growing wild will be more than 50' tall and cover many acres(!), since the branches sent out roots and thereby grow to cover a lot of ground.) The key factor making those plants small is the pot size. There is no special skill required to make a bonsai plant. Small pots are all it takes. If you look for them, you can find lots of very old and very small cactus plants surviving in nature in some crack in a rock. The crack and rock are the bonsai pot. <<Prev Index Next>>

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https://essaywritekd.com/2021/05/05/math-problem-solving-calculator_t7/

math

Graphing calculator:a larger screen calculator that's capable of showing/drawing graphs and functions. therefore it is important that your math problem solving calculator math math problem solving calculator is right, or you could spend weeks math problem solving calculator trying to figure out what’s going wrong in your experiments. if you like this site about solving math problems, please let google know by clicking the 1 research proposal topics in psychology button i loved your post about the math problem solving. best tools to solve math’s problems. can do problem solving problems 1-5 will be no calculator. but some of them are limited to specific problems and sections of the subject. solving for math problem solving calculator x and figuring persuasive research essay topics out why. year 6 maths questions – assessment pack term solve math problems for me 1. i just started my masters in math and literacy critical paper example and was researching ways to improve problem solving in students with disabilities. if informative essay topics for high school you like this site about solving math problems, please let google know by clicking research paper hooks the 1 button year 6 how to begin a literary analysis essay addition and subtraction problem solving calculation maths mastery powerpoint. graphing calculator:a larger screen calculator that's capable of showing/drawing internship paper example graphs and functions. 2x 3=x 15. if you like this marketing assignment site about solving math problems, please let google know by clicking the 1 button solving math problems will help your child become more confident in the subject and help them develop several skills. scholarship essay introduction.

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CC-MAIN-2021-31

https://www.teacherspayteachers.com/Product/Parentheses-Practice-With-Dice-2542813

math

This freebie is a math mat to help students practice math problems with parentheses. Print the page out (card stock would be best) and laminate or put in a page protector to be used with a whiteboard marker. Provide students with three die. Three different colors would be ideal, but if you only have one color, that is fine. Students roll the die and place in the empty boxes on the math mat. If the die are different colors, you tell the students that they are to place certain colors in certain boxes. They would then rewrite the problem with numerals on the math mat. You can have the students either work out the problem on their math mat with a dry erase marker or they could copy it into a math journal or notebook and solve there.

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CC-MAIN-2017-34

https://www.heldermann.de/JLT/JLT15/JLT151/jlt15001.htm

math

Journal of Lie Theory 15 (2005), No. 1, 001--011 Copyright Heldermann Verlag 2005 Almost Transitive Actions on Spaces with the Rational Homotopy of Sphere Products Mathematisches Institut, Universität Würzburg, Am Hubland, 97074 Würzburg, Germany We determine the structure of transitive actions of compact Lie groups on spaces which have the dimension and the (rational) homotopy groups of a product S1 x Sm of spheres. These homogeneous spaces arise in several geometric contexts and may be considered as S1-bundles over certain spaces, e.g. over lens spaces and over certain quotients of Stiefel manifolds. Furthermore, we show that if a non-compact simply connected Lie group acts transitively on such a space, then the orbits of the maximal compact subgroups are all simply connected rational cohomology spheres of codimension one and hence classified. We obtain this by giving a short proof of the existence and structure of the natural bundle of Gorbatsevich under these much less general assumptions. In this special case the proof gets considerably shorter by the use of the homotopy properties of the spaces in question and a theorem of Mostert on the structure of orbit spaces of compact Lie groups on manifolds. [ Fulltext-pdf (176 KB)] for subscribers only.

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CC-MAIN-2023-40

http://suo.ieee.org/email/msg05875.html

math

SUO: Re: Logic & Programming Languages As usual, CSP had an answer to that question: >Where does random chance fit into the whole scheme of things? I know that >flipping a coin has a 50-50 chance of landing heads, so how can I logically >choose heads or tails? He said that reasoning about probability is not a special kind of reasoning. It is just logical reasoning about a different subject matter -- namely, probability. So in this case, you are correctly reasoning that the probability is 50%. Therefore, you can logically expect to get heads 50% of the time or tails 50% of the time.

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CC-MAIN-2013-20

http://crowdforgeeks.com/interview-questions/eicher-aptitude-placement-papers-eicher-aptitude-interview-questions-and-answers

math

Eicher Aptitude Placement Papers - Eicher Aptitude Interview Questions and Answers Q1. Ajay Bought 15 Kg Of Dal At The Rate Of Rs 14.50 Per Kg And 10 Kg At The Rate Of Rs 13 Per Kg. He Mixed The Two And Sold The Mixture At The Rate Of Rs 15 Per Kg. What Was His Total Gain In This Tracti Cost price of 25 kg = Rs. (15 x 14.50 + 10 x thirteen) = Rs. 347.50. Sell price of 25 kg = Rs. (25 x 15) = Rs. 375. Profit = Rs. (375 — 347.50) = Rs. 27.50. Q2. Amit, Raju And Ram Agree To Pay Their Total Electricity Bill In The Proportion 3 : 4 : Toatal bill paid by way of Amit, Raju and Ram = ( 50 + 55 +seventy five ) = Rs. 180 Let amount paid via Amit, Raju and Ram be Rs. 3x, 4x and 5x respectively. Therefore, (3x + 4x + 5x ) = a hundred and eighty 12x = one hundred eighty x = 15 Therefore, amount paid thru, Amit = Rs. Forty five Raju = Rs. 60 Ram = Rs. Seventy five But clearly as given within the query, Amit pays Rs. 50, Raju pays Rs. Fifty five and Ram will pay Rs. Eight@Hence, Amit pays Rs. 5 less than the actual amount to be paid. Hence he desires to pay Rs. Five to Raju settle the quantity. Q3. A Trader Mixes 26 Kg Of Rice At Rs. 20 Per Kg With 30 Kg Of Rice Of Other Variety At Rs. 36 Per Kg And Sells The Mixture At Rs. 30 Per Kg. His Profit Percent Is? C.P. Of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600. S.P. Of fifty six kg rice = Rs. (56 x 30) = Rs. 1680. Gain =(80/1600*100) % = 5%. Q4. A Man Buys An Item At Rs. 1200 And Sells It At The Loss Of 20 Percent. Then What Is The Selling Price Of That Item? Here normally undergo in mind, whilst ever x% loss, it me S.P. = (100 - x)% of C.P at the same time as ever x% earnings, it me S.P. = (a hundred + x)% of C.P So right here may be (100 - x)% of C.P. = 80% of 1200 = (80/one hundred) * 1200 Q5. A Shopkeeper Fixes The Marked Price Of An Item 35% Above Its Cost Price. The Percentage Of Discount Allowed To Gain eight% Is? Let the charge price = Rs one hundred then, Marked fee = Rs a hundred thirty five Required advantage = 8%, So Selling fee = Rs 108 Discount = one hundred thirty five - 108 = 27 Discount% = (27/one hundred thirty 5)*one hundred = 20%. Q6. Salaries Of Ram And Sham Are In The Ratio Of 4 : Assume precise salaries of Ram and Sham as 4x and 5x respectively. (4x + 5000)/= 50 (5x + 5000) 60 60 (4x + 5000) = 50 (5x + 5000) 10 x = 50,000 5x = 25, 000 Sham's present income = 5x + 5000 = 25,000 + 5000 Sham's gift income = Rs. 30,000. Q7. Find The Largest Number Of 4-digits Divisible By 12, 15 And 18? Required biggest extensive variety need to be divisible through the L.C.M. Of 12, 15 and 18 L.C.M. Of 12, 15 and 18 12 = 2 × 2 × 3 15 =five × three 18 = 2 × 3 × 3 L.C.M. = 180 Now divide 9999 with the aid of 100 eighty, we get the rest as ninety nine The required biggest quantity = (9999 – ninety nine) =9900 Number 9900 is exactly divisible via one hundred eighty. Q8. If Books Bought At Prices Ranging From Rs. Two hundred To Rs. 350 Are Sold At Prices Ranging From Rs. Three hundred To Rs. 425, What Is The Greatest Possible Profit That Might Be Made In Selling Eight Books ? Least Cost Price = Rs. (200 * 8) = Rs. 1600. Greatest Selling Price = Rs. (425 * 8) = Rs. 3400. Required income = Rs. (3400 - 1600) = Rs. 1800. Q9. A Man Buys Oranges At Rs 5 A Dozen And An Equal Number At Rs four A Dozen. He Sells Them At Rs 5.50 A Dozen And Makes A Profit Of Rs five Cost Price of dozen oranges Rs. (five + four) = Rs. 9. Sell fee of two dozen oranges = Rs. Eleven. If income is Rs 2, oranges bought = 2 dozen. If earnings is Rs. 50, oranges bought = (2/2) * 50 dozens = 50 dozens. Q10. A Wall Is four.Five Meters Long And 3.5 Meters High. Find The Number Of Maximum Sized Wallpaper Squares, If The Wall Has To Be Covered With Only The Square Wall Paper Pieces Of Same Size? Wall may be protected nice via the use of square sized wallpaper portions. Different sized squares aren't allowed. Length = four.5 m = 450 cm; Height = 3.5 m = 350 cm Maximum rectangular length feasible me HCF of 350 and 450 We can see that 350 and 450 can be divided via manner of 50. On dividing through manner of fifty, we get 7 and nine. Since we cannot divide similarly, HCF = 50 = length of aspect of square Number of squares =Wall region/=450 x 350/= 63 Square place=50 x 50. Q11. A Man, His Wife And Daughter Worked In A Graden. The Man Worked For 3 Days, His Wife For 2 Days And Daughter For four Days. The Ratio Of Daily Wages For Man To Women Is five : 4 And The Ratio For Man To Dau Assume that the every day wages of guy, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively. Multiply (no. Of days) with (assumed day by day wage) of all and sundry to calculate the value of x. [3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 100 and five [15x + 8x + 12x] = one hundred and 5 35x = 100 and five x = 3 Hence, man's every day wage = 5x = 5 x three = Rs. 15 Wife's every day income = 4x = 4 x 3 = Rs. 12 Daughter's each day wage = 3x = 3 x 3 = Rs. Nine. Q12. The Sum Of All three Digit Numbers Divisible By three Is? All 3 digit numbers divisible via three are : 102, one 0 5, 108, 111, ..., 999. This is an A.P. With first element 'a' as 102 and distinction 'd' as three. Let it incorporates n terms. Then, 102 + (n - 1) x3 = 999 102 + 3n-3 = 999 3n = 900 or n = 3 hundred Sum of AP = n/2 [2*a + (n-1)*d] Required sum = three hundred/2[2*102 + 299*3] = 165150. Q13. What Profit Percent Is Made By Selling An Article At A Certain Price, If By Selling At 2/third Of That Price, There Would Be A Loss Of 20%? SP2 = 2/three SP1 CP = 100 SP2 = eighty 2/3 SP1 = 80 SP1 = 120 one hundred --- 20 => 20%. Q14. Find The Largest Number To Divide All The Three Numbers Leaving The Remainders four, three, And 15 Respectively At The End? Here finest variety which can divide me the HCF Remainders are terrific so actually subtract remainders from numbers 17 - four = thirteen; forty two - three = 39; ninety 3 - 15 = seventy eight Now allow's find out HCF of thirteen, 39 and 78 By direct observation we are able to see that each one numbers are divisible via thirteen. ? HCF = thirteen = required best variety. Q15. 'a' Sold An Article To 'b' At A Profit Of 20%. 'b' Sold The Same Article To 'c' At A Loss Of 25% And 'c' Sold The Same Article To 'd' At A Profit Of 40%. If 'd' Paid Rs 252 For The Article, Then Find Let the item fees 'X' to A Cost charge of B = 1.2X Cost fee of C = zero.Seventy 5(1.2X) = 0.9X Cost charge of D = 1.Four(zero.9X) = 1.26X = 252 Amount paid by using A for the object = Rs. Two hundred. Q16. Find The Lcm Of Following Three Fractions:36/,48/,72/? Numerators = 36, forty eight and seventy two. Seventy two is largest quantity among them. Seventy two isn't divisible with the aid of 36 or forty eight Start with desk of 72. Seventy x 2 = one hundred forty 4 = divisible with the aid of seventy two, 36 and 48 ? LCM of numerators = a hundred and forty four Denominators = 225, a hundred and fifty and sixty five We can see that they can be divided with the resource of five. On dividing through 5 we get forty five, 30 and 13 We can not divide similarly. So, HCF = GCD = 5 LCM of fraction =a hundred and forty 4/five. Q17. The Speed Of A Car Increases By 2 Kms After Every One Hour. If The Distance Travelling In The First One Hour Was 35 Kms. What Was The Total Distance Travelled In 12 Hours? Total distance travelled in 12 hours =(35+37+39+.....Upto 12 terms) This is an A.P with first time period, a=35, amount of terms, Required distance = 12/2[2 x 35+12-1) x 2] = 552 kms. One hundred sixty@A Man Walking At The Rate Of five Km/hr Crosses A Bridge In 15 Minutes. The Length Of The Bridge (in Metres) Is? Q18. The Two Given Numbers A And B Are In The Ratio five:6 Such That Their Lcm Is 48 Let K be common element. So 2 numbers are 5K and 6K Also K is the greatest not unusual issue (HCF) as five and six have no other not unusual element ? 5K x 6K = 480 x K K = sixteen = HCF. Q19. The Profit Earned By Selling An Article For Rs. 832 Is Equal To The Loss Incurred When The Same Article Is Sold For Rs. Forty four Let C.P. = Rs. C. Then, 832 - C = C - 448 2C = 1280 => C = 640 Required S.P. = a hundred and fifty% of Rs. 640 = one hundred and fifty/100 x 640 = Rs. 960. Q20. If The Cost Price Is 25% Of Selling Price. Then What Is The Profit Percent? Let the S.P = one hundred then C.P. = 25 Profit = seventy five Profit% = (75/25) * a hundred = three hundred%. Q21. By Selling 45 Lemons For Rs 40, A Man Loses 20%. How Many Should He Sell For Rs 24 To Gain 20% In The Traction ? Let S.P. Of forty five lemons be Rs. X. Then, 80 : 40 = a hundred and twenty : x or x = 40×100 twenty/eighty= 60 For Rs.60, lemons offered = forty five For Rs.24, lemons supplied =4560×24= 18. Q22. Two Dice Are Tossed.The Probability That The Total Score Is A Prime Number Is? Clearly, n(S) = (6 x 6) = 36. Let E = Event that the sum is a top quantity. Then E= (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (three, 2), (3, 4), (4, 1), (four,three),(5, 2), (five, 6), (6, 1), (6, 5) n(E) = 15. P(E) = n(E)/n(S) = 15/36 = 5/12.

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https://doubtnut.com/question-answer/the-tangent-at-any-point-p-onthe-parabola-y24a-x-intersects-the-y-axis-at-qdot-then-tangent-to-the-c-39563

math

Ab clear karein apne doubts Whatsapp par bhi. Apna phone number register karein. Hurray! Now you can find solutions on Whatsapp easily. We will ping you shortly. The tangent at any point P onthe parabola y^2=4a x intersects the y-axis at Qdot Then tangent to the circumcircle of triangle P Q S(S is the focus) at Q a line parallel to x-axis a line parallel to y-axis (d) none of these Question from Class 11 Chapter Conic Sections Apne doubts clear karein ab Whatsapp par bhi. Try it now.

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http://quant.stackexchange.com/questions/tagged/financial-engineering+option-pricing

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Quantitative Finance Meta to customize your list. more stack exchange communities Start here for a quick overview of the site Detailed answers to any questions you might have Discuss the workings and policies of this site Better understanding of the Datar Mathews Method - Real Option Pricing in their paper "European Real Options: An intuitive algorithm for the Black and Scholes Formula" Datar and Mathews provide a proof in the appendix on page 50, which is not really clear to me. It's ... May 10 '11 at 10:00 newest financial-engineering option-pricing questions feed Podcast #57 – We Just Saw This On Florp Putting the Community back in Wiki Hot Network Questions Why is Earth's outer-core liquid? How can I roleplay a character more manipulative than myself? How to get top management support for security projects? Can you cut a smoker's tongue after it wraps around you? Paper rejected. Should I appeal against biased reviews? Continuous Learning when you have a family Is it common to say "late girlfriend"? Having students transcribe lectures Change the color of capital letters One large microcontroller, or lots of small microcontrollers? How do you choose treasure for encounters with creatures that drop treasure? Itemized list in scientific papers NTP - How are NTP servers so accurate PHP files browsable: is this a vulnerability? What's wrong with "stupider"? Does there exist an even polynomial that has a term with an odd power? Should academics allow students to submit and possibly publish weak papers? Is there a single word that means “hitting the target but missing the point”? ModernCV modifying the header - Contents in the next line Solving limit without L'Hôpital fortran code-algorithm for qr decomposition of non-square matrix Text Field Validation vs Prevention Why is Earth's inner core solid? Is there no radioactive decay between nuclear fusion and solid material formation? 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://www.wyzant.com/resources/answers/454359/suppose_your_friend_39_s_invest_15_000_in_an_account_paying_4_compounded_annually_what_will_the_balance_be_after_9_years

math

Suppose your friend's invest 15,000 in an account paying 4% compounded annually. What will the balance be after 9 years 1 Expert Answer Experienced Mathematics Tutor w/ Master's Degree in Math A = P(1 + r/n)nt A = 15000(1 + .04/1)1(9) A = 15000(1.04)9 A = 15000(1.423312) A = $21,349.68 Still looking for help? Get the right answer, fast. Find an Online Tutor Now Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

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https://mathematical-neuroscience.springeropen.com/articles/10.1186/2190-8567-3-15

math

- Open Access Gap Junctions, Dendrites and Resonances: A Recipe for Tuning Network Dynamics The Journal of Mathematical Neuroscience volume 3, Article number: 15 (2013) Gap junctions, also referred to as electrical synapses, are expressed along the entire central nervous system and are important in mediating various brain rhythms in both normal and pathological states. These connections can form between the dendritic trees of individual cells. Many dendrites express membrane channels that confer on them a form of sub-threshold resonant dynamics. To obtain insight into the modulatory role of gap junctions in tuning networks of resonant dendritic trees, we generalise the “sum-over-trips” formalism for calculating the response function of a single branching dendrite to a gap junctionally coupled network. Each cell in the network is modelled by a soma connected to an arbitrary structure of dendrites with resonant membrane. The network is treated as a single extended tree structure with dendro-dendritic gap junction coupling. We present the generalised “sum-over-trips” rules for constructing the network response function in terms of a set of coefficients defined at special branching, somatic and gap-junctional nodes. Applying this framework to a two-cell network, we construct compact closed form solutions for the network response function in the Laplace (frequency) domain and study how a preferred frequency in each soma depends on the location and strength of the gap junction. It has been known since the end of the nineteenth century and mainly from the work of Ramón y Cajal that neuronal cells have a distinctive structure, which is different to that of any other cell type. The most extended parts of many neurons are dendrites. Their complex branching formations receive and integrate thousands of inputs from other cells in a network, via both chemical and electrical synapses. The voltage-dependent properties of dendrites can be uncovered with the use of sharp micropipette electrodes and it has long been recognised that modelling is essential for the interpretation of intracellular recordings. In the late 1950s, the theoretical work of Wilfrid Rall on cable theory provided a significant insight into the role of dendrites in processing synaptic inputs (see the book of Segev et al. for a historical perspective on Rall’s work). Recent experimental and theoretical studies at a single cell level reinforce the fact that dendritic morphology and membrane properties play an important role in dendritic integration and firing patterns [3–5]. Coupling neuronal cells in a network adds an extra level of complexity to the generation of dynamic patterns. Electrical synapses, also known as gap junctions, are known to be important in mediating various brain rhythms in both normal [6, 7] and pathological [8–10] states. They are mechanical and electrically conductive links between adjacent nerve cells that are formed at fine gaps between the pre- and post-synaptic cells and permit direct electrical connections between them. Each gap junction contains numerous connexon hemi-channels, which cross the membranes of both cells. With a lumen diameter of about 1.2 to 2.0 nm, the pore of a gap junction channel is wide enough to allow ions and even medium-sized signalling molecules to flow from one cell to the next thereby connecting the two cells’ cytoplasm. Being first discovered at the giant motor synapses of the crayfish in the late 1950s, gap junctions are now known to be expressed in the majority of cell types in the brain . Without the need for receptors to recognise chemical messengers, gap junctions are much faster than chemical synapses at relaying signals. Earlier theoretical studies demonstrate that although neuronal gap junctions are able to synchronise network dynamics, they can also contribute toward the generation of many other dynamic patterns including anti-phase, phase-locked and bistable rhythms . However, such studies often ignore dendritic morphology and focus only on somato-somatic gap junctions. In the case of dendro-dendritic coupling, simulations of multi-compartmental models reveal that network dynamics can be tuned by the location of the gap junction on the dendritic tree [13, 14]. Here, we develop a more mathematical approach using the continuum cable description of a dendritic tree (either passive or resonant) that can compactly represent the response of an entire dendro-dendritic gap junction coupled neural network to injected current using a response function. This response function, often referred as a Green’s function, describes the voltage dynamics along a network structure in response to a delta-Dirac pulse applied at a given discrete location. Our work is based on the method for constructing the Green’s function of a single branched passive dendritic tree as originally proposed by Abbott et al. [15, 16] and generalised by Coombes et al. to treat resonant membrane (whereby subthreshold oscillatory behaviour is amplified for inputs at preferential frequencies determined by ionic currents such as ). This “sum-over-trips” method is built on the path integral formulation and calculates the Green’s function on an arbitrary dendritic geometry as a convergent infinite series solution. In Sect. 2, we introduce the network model for gap junction coupled neurons. Each neuron in the network comprises of a soma and a dendritic tree. Cellular membrane dynamics are modelled by an ‘LRC’ (resonant) circuit. In Sect. 3, we focus on an example of two unbranched dendritic cells, with no distinguished somatic node, with identical and heterogeneous sets of parameters and give the closed form solution for network response with a single gap junction. The complete “sum-over-trips” rules for the more general case of an arbitrary network geometry are also presented. In Sect. 4, we apply the formalism to a more realistic case of two coupled neurons, each with a soma and a branching structure. We introduce a method of ‘words’ to construct compact solutions for the Green’s function of this network and study how a preferred frequency in each soma depends on the location and strength of the gap junction. Finally, in Sect. 5, we consider possible extensions of the work in this paper. 2 The Model We consider a network of cells connected by gap junctions. The neural morphology of individual cells includes a branching dendritic structure and a lumped soma (see an illustrative example for two cells in Fig. 1a). We assume that the dendrites are not purely passive (i.e. modelled by the ‘RC’ circuit), but are resonant (i.e. modelled by the ‘LRC’ circuit shown in Fig. 1b). Many neurons exhibit resonances whereby subthreshold oscillatory behaviour is amplified for inputs at preferential frequencies, for example as seen in neurons of rat sensorimotor cortex . In this case, it is known that the non-linear ionic current is responsible, and in general it is believed that the presence of in dendrites can have a major impact on the integration of subthreshold synaptic activity . From a mathematical perspective, Mauro et al. have shown that a linearisation of channel kinetics (for currents such as ), about rest, may adequately describe the observed resonant dynamics. The resulting linear system has a membrane impedance that displays resonant-like behaviour due to the additional presence of inductances (which are determined by the choice of channel model). This circuit is described by the specific membrane capacitance C, the resistance across a unit area of passive membrane R and an inductance L in series with a resistance r. The transmembrane voltage on an individual branch i of each cell is then governed by the following set of equations: The constants and can be found in terms of the electrical parameters of the cell membrane as and , where is a diameter and is the specific cytoplasmic resistivity of branch i. The term models an external current applied to this branch. The dendritic structure of each cell is attached to an equipotential soma of the diameter modelled by the ‘LRC’ circuit with the parameters , , and . Moreover, individual branches of different cells can be connected by gap junctions with a coupling parameter . Equations (1)–(2) for each dendritic segment must be accompanied with additional equations describing the dynamics of voltage at two ends of a segment. If the proximal () or distal () end of a branch is a branching node point the continuity of the potential across a node and Kirchoff’s law of conservation of current are imposed. For example, boundary conditions for a node indicated in Fig. 1a take the form: where is the axial resistance on branch j. If a branch terminates at we either have a no-flux (a closed-end) boundary condition or a zero value (an open-end) boundary condition A lumped soma can be treated as a special node point with the somatic membrane voltage and the following set of equations, which imposes special boundary conditions on the proximal ends of branches connected to the soma: where the sum in Eq. (8) is over all branches connected to the soma. If the branches of two cells are coupled by a gap junction, the location of this coupling can be treated as a special node point on an extended branching structure. This gap-junctional (GJ) node requires the following set of boundary conditions (given here with an assumption that it is placed at ): where is the conductance of the gap junction and and ( and ) are two segments of branch m (branch n) connected at the gap junction (see Fig. 1a). The expressions in (10) reflect continuity of the potential across individual branches m and n, and Eqs. (11)–(12) enforce conservation of current. A whole network model can be viewed as an extended tree structure with each individual node belonging to one of the following categories: a terminal, a regular branching node, a somatic node or the GJ node. The voltage dynamics along the network structure are described by linear equations and, therefore, the model’s behaviour can be studied by constructing the network response function known as the Green’s function, . This function describes the voltage response at the location x on branch i in response to a delta-Dirac pulse applied to the location y on branch j at time (branches i and j can belong either to the same cell or to the two different cells). Knowing the Green’s function for the whole structure, it is easy to compute the voltage dynamics along the whole network for any form of an external input applied to branch j as where describes the initial conditions on branch k and the sum is over all branches of the tree. Multiple external stimuli can be tackled by simply adding new terms with additional inputs to Eq. (13). 3 The Green’s Function on a Network Earlier work of Coombes et al. demonstrated that the Green’s function for a single cell with resonant membrane can be constructed by generalising the “sum-over-trips” framework of Abbott et al. [15, 16] for passive dendrites. Here, we demonstrate how this framework can be extended to a network level starting with the simple case of two identical cells. 3.1 Two Simplified Identical Cells We consider the case of two identical cells coupled by a gap junction. Each cell consists of a single resonant dendrite of infinite length (see Fig. 2). A gap junction controlled by the parameter and located at divides two dendrites into four semi-infinite segments: , , , and . We assume that an external input is applied to segment . The Green’s function on each segment must satisfy the set of Eqs. (1)–(2) with the boundary conditions at the gap junction given by Eqs. (10)–(12). Introducing the Laplace transform with spectral parameter ω and assuming zero initial data, we can solve this model in the frequency domain: Solutions (14)–(16) are obtained using the “sum-over-trips” method where on each segment can be found as , and is the Laplace transform of the Green’s function for an infinite resonant cable. is the length of a path that starts at point x on one of the segments and ends at point y on segment . The trip coefficients which ensure that the boundary conditions at the gap junction hold are chosen according to the following rules (see Fig. 3): if the trip reflects along on the gap junction back onto the same dendrite. if the trip passes through the gap junction along the same dendrite. if the trip passes through the gap junction from one cell to another cell. Performing the numerical inverse Laplace transform () of Eqs. (14)–(16), we obtain the Green’s function in the time domain for each segment. These Green’s functions are plotted in Figs. 4a–c. For any arbitrary form of external input , the voltage response on each segment can be found by taking a convolution of the corresponding Green’s function with this stimulus. Using the Laplace representation of the Green’s function on each segment given by Eqs. (14)–(16) this can be computed as where . In Figs. 4d–f, we plot the voltage profiles on each segment in response to a rectangular pulse of strength and duration , where is the Heaviside step function. A response of the network model is characterised by the Green’s function and can be studied by introducing a power function defined as . Resonant dynamics of the model for a given pair of locations are directly linked with a value at which the function has its maximum. In Figs. 5a–c, we plot the voltage profiles on each segment in response to a chirp stimulus . These figures clearly demonstrate resonant behaviour of the system maximising the voltage responses for particular frequencies. In Fig. 5d, we plot the normalised power functions at the same locations. These power functions have their maximum at the value , the same for each segment and, therefore, the resonances (indicated by arrows) in Figs. 5a–c occur at the same time. We can also notice that the function decays to zero more rapidly than the functions and . This explains the rapid reduction of voltage amplitude straight after the resonance in Fig. 5c. Typical values of a unitary gap junction conductance are 10–550 pS giving (or ) for 1–10 gap junction channels per electrical connection, although these estimates may be conservative and conductances from a larger range could be considered, as for example corresponding to in . To demonstrate how the resistance of the gap junction affects the response function in a model of two identical cells, we plot Fig. 6 for (black curves, the case shown in Figs. 4a–c), MΩ (red curves) and (green curves). Low gap-junctional resistance significantly increases the amplitude of the Green’s function for Cell n and reduces the amplitude of the Green’s function for Cell m. Increasing the resistance reduces the response in Cell n and slightly increases the response in Cell m. In this model of two identical cells, the change of the resistance of the gap junction does not affect the resonant frequency , which is the same for each segment. Although solutions (14)–(16) are found for the case of the resonant membrane, an ‘LRC’ circuit can be naturally turned into a ‘RC’ circuit by using the limit which gives . In the case of purely passive membrane, it is possible to make extra progress and find analytical forms of the solutions in the time domain (see Appendix A). In Figs. 7a–c, we plot the Green’s functions for the model in Fig. 2 with passive (instead of resonant) membrane. Voltage responses on each segment in response to a rectangular pulse are shown in Figs. 7d–f. 3.2 Two Simplified Non-identical Cells Here, we consider a model in Fig. 2 with the assumption that the cells are non-identical. Then using the Laplace transform and solving the model (with zero initial data) in the frequency domain, we obtain where the parameters and are defined in terms of cells’ individual properties as Solutions (21)–(23) show that the trip coefficients depend on either or (see Fig. 8), which have the forms In Figs. 9a–c, we demonstrate how individual variations in cell parameters affect the voltage response in the system. For each set of the parameters, we plot the Green’s functions , , and obtained by taking the numerical inverse Laplace transform of (21)–(23). Black curves show the profiles for two identical cells. Dashed red curves are the Green’s functions for a case when is changed from to . This change affects the response in Cell n, but not in Cell m. Blue curves are plotted for a case when is changed from to . It has noticeable effect on both cells. Finally, green curves are plotted for the case instead of the original diameter showing changes in profiles in both cells. As the stimulus in these examples is applied to Cell m, any variations in the parameters of this cell have an immediate effect on the responses in Cell n. In contrast, Cell m seems to be mostly robust to variations in parameters in Cell n. Resonant properties of the cells’ responses can be studied by plotting the normalised power functions for each of the parameter sets (see Figs. 9d–f). The heterogeneity of the cells’ parameters leads to appearances of different values of (the maximum of the power function) for each cell. We can also notice that the power functions for Cell n are more localised around their peaks (Fig. 9f) in comparison to the power functions for Cell m (Figs. 9d, e) as it has been earlier observed in the case of two identical cells. 3.3 An Arbitrary Network Geometry Here, we consider a network of spatially-extended cells of arbitrary geometries. This network can be treated as a single extended tree structure which consists of individual branches (indexed by a finite non-repeating sequence ) and three types of nodes: a regular branching node, a somatic node and a GJ node (see Fig. 10). The Green’s function for the whole structure can be found by obtaining in the Laplace domain and then performing . We consider a general case when each branch of the network can have different biophysical parameters and is characterised by the function defined as where k labels an arbitrary branch of the network. Using the “sum-over-trips” formalism can be constructed as an infinite series expansion where and is the length of a path along the network structure that starts at the point on branch i and ends at the point on branch j. Note that the length of each branch of the network needs to be scaled by before is calculated for (29). It is also worth mentioning here that if all branches of a network have the same biophysical parameters, i.e. , the function defined by (19). The trip coefficients in (29) are chosen according to the following set of rules: For any branching node at which the trip passes from branch i to a different branch k, is multiplied by a factor . For any branching node at which the trip approaches a node and reflects off this node back along the same branch k, is multiplied by a factor . Here, the frequency dependent parameter is defined as where the sum is over all branches connected to the node. For every terminal which always reflects any trip, is multiplied by +1 for the closed-end boundary condition or by −1 for the open-end boundary condition. For the somatic node at which the trip passes through the soma from branch i to a different branch k, is multiplied by a factor . For the somatic node at which the trip approaches the soma and reflects off the soma back along the same branch k, is multiplied by a factor . Here, the frequency dependent parameter is defined as where the sum is over all branches connected to the soma. For the GJ node at which the trip passes through the gap junction from branch n to branch m, is multiplied by a factor . For the GJ node at which the trip passes through the gap junction from branch m to branch n, is multiplied by a factor . For the GJ node at which the trip approaches the gap junction, passes it and then continues along the same branch m, is multiplied by a factor . For the GJ node at which the trip approaches the gap junction, passes it and then continues along the same branch n, is multiplied by a factor . For the GJ node at which the trip approaches the gap junction and reflects off the gap junction back along the same branch m, is multiplied by a factor . For the GJ node at which the trip approaches the gap junction and reflects off the gap junction back along the same branch n, is multiplied by a factor . We refer the reader to Coombes et al. for a proof of rules for branching and somatic nodes. In Appendix B, we prove that the rules for generating the trip coefficients at the GJ node satisfy the gap-junctional boundary conditions. 4 Application: Two-Cell Network Here, we demonstrate how the “sum-over-trips” formalism can be applied to a two-cell network for obtaining insight into network response. As an example, we consider a model of two identical cells, each of which consists of a soma and N attached semi-infinite dendrites as shown in Fig. 11. The cells are coupled by a dendro-dendritic gap junction located at some distance away from their cell bodies. We assume that this network receives an input at the location . To study the dynamics of this network, we use the “sum-over-trips” framework and construct the Green’s functions and for Cell 1 and Cell 2, respectively. 4.1 Method of Words for Compact Solutions Here, we introduce a method which allows us to construct compact solution forms for the Green’s functions of this two-cell network. We describe this method in detail by constructing the Green’s function for Cell 2 when is placed between the soma and the gap-junction as shown in Fig. 12. Introducing points from 1 to 4 on this network, we associate letters with different directions as follows: From or from : letter A. From or from : letter B. From : letter W. From : letter Y. From : letter Z. Then the shortest trip which is a trip from is associated with the (ordered) word BZ consisting of one syllable. This and any trip associated with a word that starts with the letter B, i.e. from , and ends with the letter Z, i.e. from , will belong to class 1. Any trip associated with a word which starts from and ends with , will belong to class 2. The shortest trip in this class is associated with the word ABZ, consisting of two syllables, AB and BZ. We introduce the following table that associates individual coefficients in the “sum-over-trips” framework with the syllables: As the cells are identical in this network, the parameter is defined by (18) and where N is a number of dendrites attached to each soma and is given in (31). Then, using the table, it is easy to conclude that for example, the word BZ is associated with the coefficient and the word ABZ, consisting of the syllables AB and BZ, is associated with the coefficient . We also notice from the table that different coefficients are associated with the syllables BZ and YZ and, therefore, we need to introduce two additional classes. Class 3 will include the trips with the main skeleton and the associated word BWYZ. Class 4 will include the trips with the main skeleton and the associated word ABWYZ. Combining the skeleton structures (the shortest words) of the four classes, we have Any new word in each class can be formed by adding a combination of syllables AB and WY into the structure (33). Introducing such additions of n combinations of syllables consisting of k syllables AB and syllables WY by (both syllables AB and WY can take any position in this sequence of n syllables), class 3 and class 4 can be generalised as Similarly, we can generalise class 1 and class 2. However, to ensure that the words belong to class 1 and class 2, the syllable AB must be at the end of each word in combinations. This can be written as Using combinatorics, we can write which leads to Considering the possible trips in class 3 given by and substituting the expression for found in (38), we find Similarly, the possible trips in class 4 given by generate the trip coefficients Expressions (40) and (42) for coefficients in the trips belonging to class 3 and class 4 can now be used in the “sum-over-trips” expansion (29) to obtain where defined by (19). Similarly, we can show that the possible trips in class 1 and class 2 given by generate the terms Combining together (43) and (45), we obtain Terms in in (46) represent multiple trips in each of four classes and since there is a match in the length of trips among different classes, Eq. (46) can be simplified as Using this method of ‘words’, we can construct compact solution forms for the Green’s function for each of these two cells for any combinations of input, y, and output, x, locations. For example, placing in Cell 1 between its soma and the gap-junction we obtain 4.2 Network Dynamics To study the role of a gap-junction in this two-cell network model, we focus on the Green’s functions at the somas of these two cells in response to a stimulus at the location . Using Eqs. (47) and (48), we obtain the following somatic response functions: Resonant properties of each cell are analysed by studying a preferred frequency for each cell. This is defined as the frequency at which the corresponding power function, for Cell 1 and for Cell 2, reaches its maximum. This means that for each soma is simply a solution of one of the corresponding equations, and . In Fig. 13, we plot how this preferred frequency varies as a function of the distance of the gap-junction away from each soma. These plots are obtained for the case of a passive soma and resonant dendrites. In Fig. 14 we demonstrate how is affected if a resonant soma and passive dendrites are considered. Finally, in Fig. 15, we plot when each soma and dendritic branch is modelled with resonant membrane. All these three figures clearly demonstrate that the somatic response in each cell strongly depends on the location of the gap-junction and it is tuned to be maximised for different frequencies. This can be shown by applying a chirp stimulus at the location and plotting the somatic voltage for each cell (see Fig. 16). Resonances in each cell occur at different times as predicted by Fig. 15. All these figures are constructed for truncated series solutions (49), (50) when the index n increases up to 20, although it is possible to show that the solutions rapidly converge for a much smaller n such as . In this paper, we have generalised the “sum-over-trips” formalism for single dendritic trees to cover networks of gap-junction coupled resonant neurons. With the use of ideas from combinatorics, we have also introduced a so-called method of ‘words’ that allows for a compact representation of the Green’s function network response formulas. This has allowed us to determine that the position of a dendro-dendritic gap junction can be used to tune the preferred frequency at the cell body. Moreover we have been able to generate mathematical formula for this dependence without recourse to direct numerical simulations of the physical model. One clear prediction is that the preferred frequency increases with distance of the gap junction from the soma in a model with passive soma and resonant dendrites. In contrast for a system with a resonant soma and passive or resonant dendrite, the preferred frequency decreases as the gap junction is placed further away from the cell body. There are a number of natural extensions of the work in this paper. One is an application to more realistic network geometries or more than just two neurons, as may be found in retinal networks. Here, it would also be interesting to exploit any network symmetries (either arising from the identical nature of the cells, their shapes, or the topology of their coupling) to allow for the compact representation of network response (and further utilising the method of ‘words’). Another is to incorporate a model of an active soma whilst preserving some measure of analytical tractability. Schwemmer and Lewis have recently achieved this for a single unbranched cable model by coupling it to an integrate-and-fire soma model. The merger of our approach with theirs may pave the way for understanding spiking networks of gap junction coupled dendritic trees. Moreover, by using the techniques developed by them in (using weakly coupled oscillator theory) we may further shed light on the role of dendro-dendritic coupling in contributing to the robustness of phase-locking in oscillatory networks. Appendix A: Two Simplified Identical Cells with Passive Membrane and its inverse Laplace transform Then the Green’s function on each segment can be found in closed form as where is the Green’s function of the passive infinite dendritic cable, If an external stimulus has a form of a rectangular pulse with , the voltage response on each segment can also be found in closed form: These solutions generalise earlier results of Harris and Timofeeva applicable to a neural network, but with gap-junctional coupling at tip-to-tip contacts of two branches. Appendix B: Proof of the “Sum-over-Trips” Rules at the Gap Junction As the Green’s function is constructed in the Laplace domain, the rules at the gap-junction need to satisfy the following boundary conditions (after a length of each branch, say labelled by k, of the tree is re-scaled as , ): We prove here that the rules for generating the trip coefficients are consistent with these boundary conditions. Let X denote the distance away from the GJ node along the segment (see Fig. 17). The location of the stimulus , the segment number j and the variable ω are all considered to be arbitrary. Suppose that we sum all the trips starting from the GJ node itself and ending at point Y on branch j. We denote the result of summing over all trips that initially leave the GJ node along segment by , along segment by , along segment by and along segment by . Trips that start out from X and move away from the GJ node are identical to trips that start out from the GJ node itself along segment . The only difference is that the trips in the first case are shorter by the length X. We denote the sum of such shortened trips by . The argument −X means that a distance X has to be subtracted from the length of each trip summed to compute (and not that the trips start at the point −X). Trips that start out from X by moving toward the GJ node and then reflecting back along segment are also identical to trips that start out from the GJ node along segment except that these are longer by the length X. In addition, because of the reflection from the GJ node these trips pick up a factor according to the “sum-over-trips” rules. Therefore, the contribution to the solution from those trips is . Trips that start out from X by moving toward the GJ node and then continue moving along branch m, i.e. on segment , pick up a factor and the sum of such trips is given by . Finally, trips that start from X, move toward the GJ node and then leave the GJ node by moving out along segment or pick up a factor and contribute to the solution by the terms or . The full solution includes the contributions from all different types of trips we have been discussing. Thus, The functions in this formula consist of infinite sums over trips, but we do not need to know what they are to show that the solution satisfies the GJ node boundary conditions. At the GJ node, we have Considering the solution instead gives us the same expression as in (68) and, therefore, obeys the boundary condition (63). Similarly, we can show that which satisfies the boundary condition (64). To prove the boundary condition (65) we use Eq. (67) to find that Using the following properties for the term , , Eq. (70) can be simplified as Substituting (73) and (74) together with (68) and (69) in Eq. (65) gives us the right equality. Similarly, we can prove the boundary condition (66). Cajal R: Significación fisiológica de las expansiones protoplásmicas y nerviosas de la sustancia gris. Revista de ciencias medicas de Barcelona 1891, 22: 23. Segev I, Rinzel J, Shepherd GM (Eds): The Theoretical Foundations of Dendritic Function: Selected Papers of Wilfrid Rall with Commentaries. MIT Press, Cambridge; 1995. Mainen ZF, Sejnowski TJ: Influence of dendritic structure on firing pattern in model neocortical neurons. Nature 1996, 382: 363–366. 10.1038/382363a0 van Ooyen A, Duijnhouwer J, Remme MWH, van Pelt J: The effect of dendritic topology on firing patterns in model neurons. Network 2002, 13: 311–325. 10.1088/0954-898X/13/3/304 Spruston N, Stuart G, Häusser M: Dendritic integration. In Dendrites. Oxford University Press, New York; 2008. Hormuzdi SG, Filippov MA, Mitropoulou G, Monyer H, Bruzzone R: Electrical synapses: a dynamic signaling system that shapes the activity of neuronal networks. Biochim Biophys Acta 2004, 1662: 113–137. 10.1016/j.bbamem.2003.10.023 Bennet MVL, Zukin RS: Electrical coupling and neuronal synchronization in the mammalian brain. Neuron 2004, 41: 495–511. 10.1016/S0896-6273(04)00043-1 Carlen PL, Zhang FSL, Naus C, Kushnir M, Velazquez JLP: The role of gap junctions in seizures. Brains Res Rev 2000, 32: 235–241. 10.1016/S0165-0173(99)00084-3 Traub RD, Whittington MA, Buhl EH, LeBeau FEN, Bibbig A, Boyd S, Cross H, Baldeweg T: A possible role for gap junctions in generation of very fast EEG oscillations preceding the onset of, and perhaps initiating, seizures. Epilepsia 2001, 42(2):153–170. Nakase T, Naus CCG: Gap junctions and neurological disorders of the central nervous system. Biochim Biophys Acta, Biomembr 2004, 1662(1–2):149–158. Söhl G, Maxeiner S, Willecke K: Expression and functions of neuronal gap junctions. Nat Rev, Neurosci 2005, 6: 191–200. Bem T, Rinzel J: Short duty cycle destabilizes a Half-Center oscillator, but gap junctions can restabilize the anti-phase pattern. J Neurophysiol 2004, 91: 693–703. Traub RD, Kopell N, Bibbig A, Buhl EH, LeBeau FEN, Whittington MA: Gap junctions between interneuron dendrites can enhance synchrony of gamma oscillations in distributed networks. J Neurosci 2001, 21: 9478–9486. Saraga F, Ng L, Skinner FK: Distal gap junctions and active dendrites can tune network dynamics. J Neurophysiol 2006, 95: 1669–1682. 10.1152/jn.00662.2005 Abbott LF, Fahri E, Gutmann S: The path integral for dendritic trees. Biol Cybern 1991, 66: 49–60. 10.1007/BF00196452 Abbott LF: Simple diagrammatic rules for solving dendritic cable problems. Physica A 1992, 185: 343–356. 10.1016/0378-4371(92)90474-5 Coombes S, Timofeeva Y, Svensson CM, Lord GJ, Josic K, Cox SJ, Colbert CM: Branching dendrites with resonant membrane: a “sum-over-trips” approach. Biol Cybern 2007, 97: 137–149. 10.1007/s00422-007-0161-5 Hutcheon B, Miura RM, Puil E: Models of subthreshold membrane resonance in neocortical neurons. J Neurophysiol 1996, 76: 698–714. Magee JC: Dendritic hyperpolarization-activated currents modify the integrative properties of hippocampal CA1 pyramidal neurons. J Neurosci 1998, 18: 7613–7624. Mauro A, Conti F, Dodge F, Schor R: Subthreshold behavior and phenomenological impedance of the squid giant axon. J Gen Physiol 1970, 55: 497–523. 10.1085/jgp.55.4.497 Schwemmer MA, Lewis TJ: Bistability in a leaky integrate-and-fire neuron with a passive dendrite. SIAM J Appl Dyn Syst 2012, 11: 507–539. 10.1137/110847354 Schwemmer MA, Lewis TJ: The robustness of phase-locking in neurons with dendro-dendritic electrical coupling. J Math Biol 2012. 10.1007/s00285-012-0635-5 Harris J, Timofeeva Y: Intercellular calcium waves in the fire-diffuse-fire framework: Green’s function for gap-junctional coupling. Phys Rev E 2010., 82: Article ID 051910 Article ID 051910 YT would like to acknowledge the support provided by the BBSRC (BB/H011900) and the RCUK. DM would like to acknowledge the Complexity Science Doctoral Training Centre at the University of Warwick along with the funding provided by the EPSRC (EP/E501311). The authors declare that they have no competing interests. YT, SC and DM contributed equally. All authors read and approved the final manuscript. Authors’ original submitted files for images Below are the links to the authors’ original submitted files for images. About this article Cite this article Timofeeva, Y., Coombes, S. & Michieletto, D. Gap Junctions, Dendrites and Resonances: A Recipe for Tuning Network Dynamics. J. Math. Neurosc. 3, 15 (2013). https://doi.org/10.1186/2190-8567-3-15 - Gap junctions - Resonant membrane - Network dynamics

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https://chouprojects.com/imcosh-excel/

math

Do you feel overwhelmed by the numerous Excel formulae? IMCOSH is here to help! Our comprehensive guide provides easy, step-by-step instructions to help you get the most out of your spreadsheet work. Get ready to supercharge your data analysis and unleash the power of Excel. IMCOSH function in Excel IMCOSH is a mathematical function in Excel that calculates the hyperbolic cosine of a given angle. It is a part of the broader group of hyperbolic functions available in Excel alongside other popular functions such as SINH and TANH. IMCOSH function in Excel operates by taking a given angle and performing a specific mathematical operation on it to yield the hyperbolic cosine. It is commonly used in graphing and modeling functions and can be used in conjunction with other mathematical operations to create more complex formulas. It is worth noting that while hyperbolic functions might seem complex, they are relatively simple to understand when broken down into their constituent parts. With practice, users can readily understand and implement IMCOSH in their Excel spreadsheets. Interestingly, the origins of IMCOSH can be traced back to the 19th century, where hyperbolic functions first gained prominence amongst mathematicians. Today, IMCOSH, and its sister functions are essential tools for any Excel user working with mathematical models and data analysis. Explanation of IMCOSH Formula Understand IMCOSH formula! Here’s a guide, split into sections. Excel users – fresh or experienced – use this powerful formula to its fullest. Get ready to learn and use this formula to elevate your spreadsheet work! Definition, syntax, and usage – dive into IMCOSH! Definition of IMCOSH Function IMCOSH is an Excel formula that calculates the hyperbolic cosine of a complex number. It is a mathematical function used to analyze and manipulate data. IMCOSH accepts one argument which can be a real or imaginary number. The output is a complex number in polar coordinates. The IMCOSH function is a part of the larger family of Hyperbolic functions, commonly used in various scientific fields, including physics and engineering. These functions are defined as the analogs of trigonometric functions for hyperbolic angles instead of circular angles. IMCOSH expands the capabilities of Excel into more advanced mathematical computations, providing greater flexibility to its users. It can be used to model complex systems, perform statistical analysis on large data sets, and make predictions based on historical trends. One example usage of IMCOSH can be seen in electrical engineering, specifically in transmission line design. The function can help engineers determine how voltage fluctuations will occur along long distance power lines, ensuring efficient energy transmission with minimal loss. Source: Microsoft Support Documentation Get ready to channel your inner mathematician as we break down the syntax of the IMCOSH function – it’s like a puzzle you never knew you wanted to solve. Syntax of IMCOSH Function The IMCOSH Function syntax consists of a single argument, which is the angle in radians. It evaluates the hyperbolic cosine of a complex number in Excel. The function returns a complex number that satisfies the equation cosh(z) = (exp(z)+exp(-z))/2. The IMCOSH formula calculates the hyperbolic cosine of an imaginary number in Excel. To use the formula, enter “=IMCOSH(X)”, where “X” is the imaginary number for which you want to find the hyperbolic cosine. Excel has separate formulas for the real and imaginary parts of a complex number. Using them together allows us to work with complex numbers in Excel. Notably, using this function can simplify calculations that involve complex numbers because it eliminates the need to calculate sines and cosines independently. Don’t miss out on taking your Excel skills to the next level by mastering advanced functions such as IMCOSH. Understanding these formulas will save you time and effort while working with complex data sets. IMCOSH, the perfect formula for those who can’t cosh their way out of a math problem. How to use IMCOSH Function IMCOSH Function – Utilizing Excel Formulae Want to learn about using the IMCOSH function in Excel? Here’s a comprehensive guide to getting started! - Open your Excel spreadsheet and select the cell where you want the result to appear. - Start entering the formula with =IMCOSH(and add your value or reference to it, closing with - Press enter and see your result! - Alternatively, you can use the Function Wizard by going to Formulas > Insert Function > Engineering > IMCOSH. - Use this formula when working with inverse hyperbolic cosine trigonometric functions. Remember that IMCOSH is not available in all versions of Microsoft Office. Try upgrading for access to more formula tools. Here’s something interesting: The term ‘hyperbolic cosine’ was first coined by Vincenzo Riccati in 1763. Why do math teachers love IMCOSH? Because it’s cosine without the sin. Examples of IMCOSH Function Gaining an understanding of IMCOSH? Check out the “Examples of IMCOSH Function” section. It includes: - “Example 1: Simple calculation using IMCOSH” - “Example 2: Using IMCOSH with other Excel formulas” Exploring these sub-sections will help you use IMCOSH in different ways, and with other formulas. Example 1: Simple calculation using IMCOSH For this calculation, we will use the IMCOSH function. It is crucial for calculating hyperbolic cosine values in Excel. Please see the table below for Example 1: Simple calculation using IMCOSH. |Angle (degrees)||Hyperbolic Cosine Value| The table above shows an example where we need to calculate the hyperbolic cosine value of an angle in degrees using Excel’s IMCOSH function. The formula used is “IMCOSH (RADIANS (Angle))” which calculates the value of hyperbolic cosine in radians. It is crucial that the angle input be mentioned in radians, not degrees – hence why we had to use the RADIANS function first before plugging it into IMCOSH. Research has shown that using built-in Excel functions saves time and reduces errors (Source: Microsoft Excel Help). IMCOSH and other Excel formulas – a match made in spreadsheet heaven. Example 2: Using IMCOSH with other Excel formulas Using IMCOSH with Other Excel Formulas: An Informative Explanation Below is a table that showcases how IMCOSH can be used in conjunction with other Excel formulas to enhance data analysis and computation. The true and actual data presented below offer insight into how IMCOSH could be utilized in practical scenarios. |=IMCOSH(MAX(A2:A7),MIN(B2:B7))||Returns the hyperbolic cosine of the maximum value in cells A2 through A7 divided by the minimum value in cells B2 through B7.| |=IMCOSH(SUM(C2:C7),AVERAGE(D2:D7))||Returns the hyperbolic cosine of the sum of values in cells C2 through C7 divided by the average value of cells D2 through D7.| |=IMCOSH(IF(E2=”Yes”,F2,G2))||Returns the hyperbolic cosine of the value in cell F2 if cell E2 equals “Yes”, otherwise returns the hyperbolic cosine of the value in cell G2.| With an integration of functions like MAX, SUM, and IF within the IMCOSH formula, users can perform complex calculations with greater accuracy and efficiency. This feature enables analysts to quickly model large-scale datasets into actionable insights, transforming raw information into valuable business strategies. Don’t miss out on the benefits of incorporating IMCOSH into your data analysis toolbox. Boost your productivity and precision today! FAQs about Imcosh: Excel Formulae Explained What is IMCOSH: Excel Formulae Explained? IMCOSH: Excel Formulae Explained is a comprehensive guide to using the IMCOSH function in Microsoft Excel. The guide explains what the IMCOSH function is, how to use it, and includes numerous examples to help you understand how the function works. What is the IMCOSH Function? The IMCOSH function is an Excel function that calculates the hyperbolic cosine of a complex number. This function can be used to solve problems in calculus, physics, and engineering that involve complex numbers. How to Use the IMCOSH Function? To use the IMCOSH function in Excel, you need to enter the function name followed by the complex number you want to calculate the hyperbolic cosine of. For example, “=IMCOSH(2+3i)” would calculate the hyperbolic cosine of the complex number 2+3i. What are the Advantages of Using the IMCOSH Function? The IMCOSH function can help simplify complex calculations involving complex numbers. This function can save time by automating complex calculations that would otherwise need to be done manually. What are some of the common Applications of the IMCOSH Function? The IMCOSH function is commonly used in mathematical and engineering applications that involve complex numbers. This function is particularly useful in solving problems related to calculus, physics, and engineering. Where Can I Learn More About IMCOSH: Excel Formulae Explained? You can learn more about IMCOSH: Excel Formulae Explained by visiting the product website or by reading books or tutorials on Excel functions and formulas. The guide also includes numerous examples and tutorials to help you master the use of the IMCOSH function.

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https://rdsharmasolution.com/rd-sharma-class-10-solutions-chapter-11-constructions/

math

Here we are providing RD Sharma class 10 maths solutions for chapter 11 – constructions solved by expert maths teachers. We are offering step by step solutions for the questions given in class 10 maths textbook. Students can therefore, download the RD Sharma solutions that we are offering and use it to practice difficult maths problems. The solutions will prove useful to the students as they will further get a clear overview of the topic. The solution contains exercises solved by our experts and students can go through these as well to develop better math skills and prepare efficiently for the exams. Students can also use the math exercises to test their knowledge about the topics in chapter 11. The RD Sharma solutions are free and students can either view them online on the website or download the PDF. RD Sharma Class 10 Maths Solutions Chapter 11 Check out RD Sharma class 10 maths solutions for chapter 11 – constructions below. Several math exercises are also given in the following table. |Exercise 11.1||Exercise 11.2| The RD Sharma solutions for class 10 maths chapter 11 – constructions PDF is available for free here. For more RD Sharma maths solutions keep visiting us.

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https://ijim.srbiau.ac.ir/article_12653.html

math

Document Type : Research Paper Department of Mathematics, Islamic Azad University, Rasht Branch, Rasht, Iran. Department of Mathematics, Islamic Azad University, Hadishahr Branch, Hadishahr, Iran. Conventional data envelopment analysis (DEA) models normally assume all inputs and outputs are real valued and continuous. However in problems some inputs and outputs can only take integer values, also, both desirable and undesirable outputs can be generated . In this paper the effect of undesirable outputs in integer DEA model is discussed. The proposed model distinguishes weak disposability of outputs imposing non-uniform abatement factor.

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http://blog.rutwick.com/library/algebra-and-trigonometry-2-nd-edition

math

By James Stewart This most sensible promoting writer staff explains strategies easily and obviously, with no glossing over tricky issues. challenge fixing and mathematical modeling are brought early and bolstered all through, in order that whilst scholars end the path, they've got a great beginning within the rules of mathematical considering. This entire, calmly paced publication presents whole assurance of the functionality thought and integrates gigantic graphing calculator fabrics that support scholars increase perception into mathematical rules. The authors' awareness to element and readability, as in James Stewart's market-leading Calculus textual content, is what makes this article the marketplace chief. Read or Download Algebra and trigonometry, 2nd Edition PDF Similar popular & elementary books This Elibron Classics booklet is a facsimile reprint of a 1904 variation via Adam and Charles Black, London. Speedy Solvers for Mesh-Based Computations offers an alternate means of creating multi-frontal direct solver algorithms for mesh-based computations. It additionally describes tips to layout and enforce these algorithms. The book’s constitution follows these of the matrices, ranging from tri-diagonal matrices due to one-dimensional mesh-based tools, via multi-diagonal or block-diagonal matrices, and finishing with normal sparse matrices. Disk-Based Algorithms for giant facts is a manufactured from fresh advances within the components of massive facts, information analytics, and the underlying dossier platforms and information administration algorithms used to aid the garage and research of big facts collections. The booklet discusses difficult disks and their impression on facts administration, because harddrive Drives remain universal in huge facts clusters. - Pre-Calculus And Geometry - Arithmetic Theory of Elliptic Curves - Arithmetic Theory of Elliptic Curves - Mathematical Inequalities - Symmetric Designs: An Algebraic Approach Extra info for Algebra and trigonometry, 2nd Edition 1ab2 ϭ 1a # a # . . # a2 # 1b # b # . . # b2 ϭ a n b n 144424443 n factors 1442443 n factors 1442443 n factors Here we have used the Commutative and Associative Properties repeatedly. If n Յ 0, Law 4 can be proved using the definition of negative exponents. ■ You are asked to prove Laws 2 and 5 in Exercise 76. 24 CHAPTER P Prerequisites Example 3 Using Laws of Exponents (a) x 4x 7 ϭ x 4ϩ7 ϭ x 11 Law 1: a ma n ϭ a mϩn (b) y 4y Ϫ7 ϭ y 4Ϫ7 ϭ y Ϫ3 ϭ 1 y3 c9 ϭ c 9Ϫ5 ϭ c 4 c5 # (d) 1b 4 2 5 ϭ b 4 5 ϭ b 20 Law 2: a m/a n ϭ a mϪn (c) Law 3: (a m) n ϭ a mn (e) 13x2 3 ϭ 33x 3 ϭ 27x 3 Law 4: (ab)n ϭ a nb n x 5 x5 x5 (f) a b ϭ 5 ϭ 2 32 2 Example 4 Law 1: a ma n ϭ a mϩn Law 5: (a/b) n ϭ a n/b n ■ Simplifying Expressions with Exponents Simplify: x 3 y 2x 4 (b) a b a b z y (a) 12a 3b 2 2 13ab 4 2 3 Solution (a) 12a 3b 2 2 13ab 4 2 3 ϭ 12a 3b 2 2 333a 3 1b 4 2 3 4 ϭ 12a 3b 2 2 127a 3b 12 2 ϭ 122 1272a 3a 3b 2b 12 Law 4: (ab)n ϭ a nb n Law 3: (a m)n ϭ a mn Group factors with the same base Apago PDF ϭ 54a Enhancer b Law 1: a a 6 14 x 3 y 2x 4 x 3 1y 2 2 4x 4 (b) a b a b ϭ 3 z y y z4 ϭ x 3 y 8x 4 y 3 z4 ϭ 1x 3x 4 2 a ϭ x 7y 5 z4 m n ϭ a mϩn Laws 5 and 4 Law 3 y8 1 b y 3 z4 Group factors with the same base Laws 1 and 2 ■ When simplifying an expression, you will find that many different methods will lead to the same result; you should feel free to use any of the rules of exponents to arrive at your own method. Can you locate 15 by a similar method? What about 16? List some other irrational numbers that can be located this way. (a) Will the post office accept a package that is 6 in. wide, 8 in. deep, and 5 ft long? What about a package that measures 2 ft by 2 ft by 4 ft? (b) What is the greatest acceptable length for a package that has a square base measuring 9 in. by 9 in? œ∑2 _1 Apago PDF Enhancer L 21 0 1 1 2 5 ft=60 in. x 6 in. y 8 in. Discovery • Discussion 77. Signs of Numbers Let a, b, and c be real numbers such that a Ͼ 0, b Ͻ 0, and c Ͻ 0. 4 Rational Exponents and Radicals 31 (b) If b Ͼ 0, then 225b Ϫ 2b 3 ϭ 2252b Ϫ 2b 2 2b ϭ 52b Ϫ b2b Property 1: 1ab ϭ 1a1b Property 5, b Ͼ 0 ϭ 15 Ϫ b2 2b Distributive Property ■ Rational Exponents To define what is meant by a rational exponent or, equivalently, a fractional exponent such as a 1/3, we need to use radicals. In order to give meaning to the symbol a 1/n in a way that is consistent with the Laws of Exponents, we would have to have 1a 1/n 2 n ϭ a 11/n2n ϭ a 1 ϭ a So, by the definition of nth root, n a 1/n ϭ 1 a In general, we define rational exponents as follows.

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https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=154362

math

The flashcards below were created by user In a make-or-buy decison, the relevent costs are? 1. Manufacturing Costs that will be saved 2. Purchase Price 3. Opertunity Costs Manufacturing Costs include? 1. Direct Materials 2. Direct Labor 3. Variable Manufacturing Overhead 4. Fixed Manufacturing Overhead Annual Rate of Return Expected Annual Net Income / Average Investment Average Investment (Formula) The annual rate of return decision rule is? A project is acceptable if its rate of return is greater than managments rate of return Cash Back Period (Formula) Cash Flow (Formula) Net Present Value (Formula) Present Value of Net Cash Flows (Formula) The NPV decision rule is? Accept the project if NPV is zero or positive The IRR decision rule is? Acceot the project when IRR is equal to or greater than required rate of return Steps in determining IRR 1. Compute IRR factor: Capital Investment/Cash Flow=IRR Factor 2. Use the factor and the present value of an annuity of 1 table to find IRR Incremental analyis & Capital budgeting

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https://www.impan.pl/en/publishing-house/journals-and-series/fundamenta-mathematicae/online/113129/quasiconformal-and-geodesic-trees

math

Quasiconformal and geodesic trees Fundamenta Mathematicae MSC: Primary 30L10; Secondary 51F99. DOI: 10.4064/fm749-7-2019 Published online: 24 February 2020 A quasiconformal tree is a metric tree that is doubling and of bounded turning. We prove that every quasiconformal tree is quasisymmetrically equivalent to a geodesic tree with Hausdorff dimension arbitrarily close to 1.

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https://brainsanswer.com/question/304564

math

Because 0.75 is less than 1, the bigger x is, the smaller the whole thing will get. And the 4.1 in front is the starting point on the left side of the graph. Like on an actual graph with numbers: If x is 0, f(x) is 4.1 If x is 5, f(x) is 0.97 The graph is decreasing. This is an equation and graph for "decay"

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https://pintiparquet.es/category/chinalovecupid-inceleme-yorumlar-2

math

Herea€™s an example ERD utilizing Chen notation: Herea€™s a typical example of the symbols used with the Crowa€™s feet notation design. This is the design youra€™ll look at many on Database Star as ita€™s the main one Ia€™m more acquainted with. Ita€™s also known as a€?crowa€™s foota€™ due to the fact logo for a many commitment appears like the base of a crow with its three prongs. Herea€™s a good example ERD using Crowa€™s toes notation: Herea€™s a good example of the signs used with the Bachman notation preferences. Herea€™s an example ERD making use of Bachman notation: Herea€™s an example of the icons combined with the IDEF1X notation preferences. Herea€™s an illustration ERD utilizing IDEF1X notation: Herea€™s an example of the icons combined with the Barker notation design. Herea€™s an example ERD utilizing Bachman notation: Conceptual, sensible, bodily an Entity Relationship Diagram is pulled at three various amount: conceptual, reasonable, or physical. Each one of these amount have another standard of detail and so are used in yet another objective. Leta€™s discover some situations. Conceptual Information Model The conceptual facts unit demonstrates the business enterprise objects that exist when you look at the system as well as how they associate with both. Continuar leyendo “Herea€™s a good example of the signs used with the Chen notation design.”

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https://vladgkh.ru/how-to-solve-calculus-word-problems-616.html

math

Of course, cheating at math is a terrible way to learn, because the whole point isn't to know the answer to 2x 2 = 7x - 5, it's to understand the learn? Of course, cheating at math is a terrible way to learn, because the whole point isn't to know the answer to 2x 2 = 7x - 5, it's to understand the learn?Tags: Reality Tv Essay ThesisThesis Statement On Legalizing DrugsIn Content Essays AreSouthworth Resume Paper No WatermarkLogic And Critical Thinking QuestionsHesi A2 With Critical Thinking Study GuideIs There A Website That Will Write My Essay For MeFuture Tense EssayTitle For Dissertation It has trouble with word problems, but if you can write down a word problem in math notation it shouldn't be an issue. I also tried it on a weird fraction from an AP algebra exam, which it kind of failed at, but then I swiped over and it was showing me this graph, which included the correct answer: I love this app, not just because it would've helped 8th grade Paul out of a jam, but because it's such a computery use of computers. We’ve already found the relevant radius, $r = \sqrt\,.$To find the corresponding height, recall that in the Subproblem above we found that since the can must hold a volume of liquid, its height is related to its radius according to $$h = \dfrac\,. $$ Hence when $r = \sqrt\,,$ \[ \begin h &= \frac\,\frac \\[8px] &= \frac\,\frac \\[8px] &= \frac\,\frac \\[8px] &= 2^\frac \\[8px] h &= 2^\sqrt \quad \triangleleft \end \] The preceding expression for is correct, but we can gain a nice insight by noticing that $$2^ = 2 \cdot\frac$$ and so \[ \begin h &= 2^\sqrt \\[8px] &= 2 \cdot\frac\,\sqrt \\[8px] &= 2 \sqrt = 2r \end \] since recall that the ideal radius is $r = \sqrt\,.$ Hence the ideal height (height and radius) will minimize the cost of metal to construct the can? On about half the middle school science problems I tried, the app was able to identify the topic at question and show me additional resources about the concepts involved, but for others it was no more powerful than a simple web search. How To Solve Calculus Word Problems But for algebra this thing is I pointed it at 2x 2 = 7x - 5, which I wrote down at random, and it gave me a 10 step process that results in x = 7/5.Typical phrases that indicate an Optimization problem include: Before you can look for that max/min value, you first have to develop the function that you’re going to optimize.There are thus two distinct Stages to completely solve these problems—something most students don’t initially realize [Ref]. Now maximize or minimize the function you just developed.Notice, by the way, that so far in our solution we haven’t used any Calculus at all.That will always be the case when you solve an Optimization problem: you don’t use Calculus until you come to Stage II.Above, for instance, our equation for $A_\text$ has two variables, We can now make this substitution $h = \dfrac$ into the equation we developed earlier for the can’s total area: \[ \begin A_\text &= 2\pi r^2 2 \pi r h \[8px] &= 2\pi r^2 2 \pi r \left( \frac\right) \[8px] &= 2\pi r^2 2 \cancel \cancel \left(\frac\right) \[8px] &= 2\pi r^2 \frac \end \]We’re done with Step 3: we now have the function in terms of a single variable, , and we’ve dropped the subscript “total” from $A_\text$ since we no longer need it.This also concludes Stage I of our work: in these threes steps, we’ve developed the function we’re now going to minimize!In this problem, for instance, we want to minimize the cost of constructing the can, which means we want to use .So let’s write an equation for that total surface area:\begin A_\text &= A_\text A_\text A_\text \[8px] &= \pi r^2 2\pi r h \pi r^2 \[8px] &= 2\pi r^2 2 \pi r h \end That’s it; you’re done with Step 2!You'll see a button "View steps" and this takes you to the developer's site where you can purchase the full version of the solver (where you can see the steps). I was homeschooled (that's not the confession part), and in 8th grade my algebra textbook had the answers to half the problems in the back. That seems to be the premise behind app called Socratic. The app lets you take a picture of a problem (you can also type it in, but that's a little laborious), and it'll not only give you an answer, but the steps necessary to to arrive at that answer — and even detailed explanations of the steps and concepts if you need them.

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https://www.sanfoundry.com/mathematics-questions-answers-classification-polygons/

math

This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Classification of Polygons”. 1. Which of the following is correct? a) A polygon is a figure which comprises of straight lines b) A polygon is a figure which comprises of all types of curves c) A polygon is a closed figure made up of straight lines d) A polygon is a figure which comprises of straight lines and curves Explanation: The definition of a polygon states that ‘A simple closed curve made up of only line segments is called a polygon’. The other options which does not state the a polygon is a closed figure and made of straight lines are wrong. 2. Which of the following is a triangle? Explanation: There are four types of figures in the options, here we need to find which of the following is a triangle. The triangle is a polygon which have three sides which also form three angles with each other. 3. Which of the following can be classified as pentagon? Explanation: A polygon which has five sides and five angles is called a pentagon. One can find this by a small trick, a student should learn to break the word. For example, Hexagon: Hexa+Gon . As we know that hexa is six, it would be easy for the students to learn the names of the polygons. 4. All parallelograms are square. Explanation: All the squares are special cases of parallelogram. This is because parallelogram is a four-sided polygon with opposite sides parallel to each other. Moreover, a square is four-sided closed figure with all angles measuring 90° and all sides equal. So, the given statement is false. 5. If a polygon has its opposite sides equal and parallel to each other and all angles measuring 90°, identify the polygon. Explanation: If a rhombus has it’s opposites sides equal and parallel then it can be square or a rhombus but the question also states that the angle between the sides is 90°. Square is a special type of rhombus which has all the angles as 90°. 6. When a square is pushed from one corner keeping the base as it is, which polygon is formed? Explanation: If we select the upper corner of a square and push it the square will bend in the direction of the force so, the vertical edges would tilt in the direction of the force and this would lead to the formation of new polygon named as rhombus. 7. A seven sided figure is called _______ b) seven sided polygon Explanation: One can find this by a small trick, a student should learn to break the word. For example, Hexagon: Hexa+Gon. As we know that Hexa is six. Similarly, Seven is known as Hepta. ∴Heptagon. 8. A polygon with minimum number of sides is ________ Explanation: A polygon with minimum numbers of sides is a triangle. A single line does not have one side, an angle has two sides, but it is not closed. A triangle has three sides and its closed. Square has four sides and its closed. So, triangle is the polygon with minimum number of sides. 9. A parallelogram each of whose angles measures 90° is ___________ d) square as well as Rectangle Explanation: If a parallelogram has all it’s angle as 90° then it’s a special case of parallelogram and it can be either square or a rectangle. 10. If in a triangle two of it’s angles are 35° and 45° respectively. What is the measure of it’s third angle? Explanation: In a triangle the sum of all angles is 180°. So, now we can calculate the third angle. Angle 1 + Angle 2 + Angle 3 = 180° ∴ 35° + 45° + Angle 3 = 180° ∴ Angle 3 = 110°. 11. What is the general form of naming a polygon? c) n-gon (where n represents number of sides in a polygon) Explanation: One can find this by a small trick, a student should learn to break the word. For example Hexagon: Hexa+Gon now as we know that hexa is six. Similarly, if a polygon has ‘n’ sides then the polygon would be named as n-gon. Sanfoundry Global Education & Learning Series – Mathematics – Class 8. To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. - Practice Class 10 Mathematics MCQs - Buy Class 8 - Mathematics Books - Practice Class 9 Mathematics MCQs

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https://www.groundai.com/project/zassenhaus-conjecture-on-torsion-units-holds-for-operatornamepsl2p-with-p-a-fermat-or-mersenne-prime/

math

Zassenhaus Conjecture on torsion units holds for with a Fermat or Mersenne prime H.J. Zassenhaus conjectured that any unit of finite order in the integral group ring of a finite group is conjugate in the rational group algebra to an element of the form with . Though known for some series of solvable groups, the conjecture has been proved only for thirteen non-abelian simple groups. We prove the Zassenhaus Conjecture for the groups , where is a Fermat or Mersenne prime. This increases the list of non-abelian simple groups for which the conjecture is known by probably infinitely many, but at least by , groups. Our result is an easy consequence of known results and our main theorem which states that the Zassenhaus Conjecture holds for a unit in of order coprime with , for some prime power . 2010 Mathematics Subject Classification:16U60, 16S34 One of the most famous open problems regarding the unit group of an integral group ring of a finite group is the Zassenhaus Conjecture which was stated by H.J. Zassenhaus [Zas74]: Zassenhaus Conjecture111After this paper was submitted a metabelian counterexample to the Zassenhaus Conjecture was announced in [EM17]. Still no simple counterexample is known.: If is a finite group and is a unit of finite order in the integral group ring , then there exists a unit in the rational group algebra and an element such that . If for a given such and exist, one says that and are rationally conjugate. The Zassenhaus Conjecture found much attention and was proved for many series of solvable groups, e.g. for nilpotent groups [Wei91], groups possessing a normal Sylow subgroup with abelian complement [Her06] or cyclic-by-abelian groups [CMdR13]. Regarding non-solvable groups, however, the conjecture is only known for very few groups. The proofs of the results for solvable groups mentioned above often argue by induction on the order of the group. In this way one may assume that the conjecture holds for proper quotients of the original group. The first step in a similar argument for non-solvable groups should consist in proving the conjecture for simple groups. Although this has been studied by some authors, see e.g. [LP89, Her07, Her08, BKL08, Sal13, BM17, BC17], the conjecture is still only known for exactly thirteen non-abelian simple groups all being isomorphic to some for some particular small prime power (see [BM18, Theorem C] for an overview). Our aim in this paper is to extend this knowledge by proving the following theorem. Let for some prime power . Then any torsion unit of of order coprime with is rationally conjugate to an element of . We prove this result employing a variation of a well known method which uses characters of a finite group to obtain restrictions on the possible torsion units in . The idea of the method was introduced for ordinary characters by Luthar and Passi [LP89] and extended to Brauer characters by Hertweck [Her07]. Today this method is often called the HeLP (HertweckLutharPassi) Method. In fact to prove our results we do not use the HeLP Method in the classical sense, since this would imply too many case distinctions. For this reason we vary the method in a way suitable for the character theory of . Theorem 1.1 can be regarded as a generalization of [Mar16, Theorem 1]. Let be a Fermat or Mersenne prime. Then the Zassenhaus Conjecture holds for . This result increases the number of simple groups for which the Zassenhaus Conjecture is known from thirteen to sixty-two: The groups with or one of the four known Fermat primes different from 3 or one of the forty-nine known Mersenne primes different from 3 [Calb]. Actually, Theorem 1.2 proves the conjecture for probably infinitely many simple groups because, based on heuristic evidences, it has been conjectured that there are infinitely many Mersenne primes [Cala]. Lenstra, Pomerance and Wagstaff have proposed independently a conjecture on the growth of the number of Mersenne primes smaller than a given integer [Pom81, Wag83]. It has been shown in [dRS17] that a result as in Theorem 1.1 can not be achieved using solemnly the HeLP Method if the unit has order , where is coprime with and a prime bigger than . Looking on the orders of elements in , cf. Theorem 2.2, one should not expect a better result for the Zassenhaus Conjecture for when applying only this method. Thus, as so often in Arithmetics and Group Theory, the prime behaves very differently than the other primes. Let be a finite group. If , then denotes the order of , the cyclic group generated by is denoted by and denotes the conjugacy class of in . If is a ring then denotes the group ring of with coefficients in . Denote by the group of normalized units (i.e units of augmentation ) in . As mentioned above, we say that two elements of are rationally conjugate if they are conjugate in the units of . The main notion to study rational conjugacy of torsion units in are the so called partial augmentations. If is an element of a group ring , with each , then the partial augmentation of at is defined as The relevance of partial augmentations for the study of the Zassenhaus Conjecture is provided by a result of Marciniak, Ritter, Sehgal and Weiss. The following theorem states this result and collects some known information about partial augmentations. Let be a finite group and let be an element of order in . [MRSW87, Theorem 2.5] is rationally conjugate to element in if and only if for all and all divisors of . [JdR16, Proposition 1.5.1] (Berman-Higman Theorem) If then . [Her07, Theorem 2.3] If then divides . [Her07, Theorem 3.2] Let be a prime not dividing and let be a -Brauer character of associated to a modular representation for a suitable -modular system . Then extends to a -Brauer character defined on the -regular torsion units of , associated to the natural algebra homomorphism . Moreover, if are representatives of the -regular conjugacy classes of then We collect the group theoretical properties of and its integral group ring relevant for us. Let where for some prime and let . [Hup67, Hauptsatz 8.27] The following properties hold. The order of is . The orders of elements in are exactly and the divisors of and . Two cyclic subgroups of are conjugate in if and only if they have the same order. If with coprime with and multiple of then is conjugate in to an element of and the only elements of conjugate to in are and . In particular a conjugacy class of elements of order coprime with is a real conjugacy class. If is a torsion element of of order coprime with , is root of unity in an arbitrary field and is an -representation of then and have the same multiplicity as eigenvalues of . This follows from (1) and the formulas for multiplicities of eigenvalues of torsion units as presented in [Her07, Section 4]. [Mar16, Lemma 1.2] Let be a positive integer coprime with and let be an element of order . There exists a primitive -th root of unity in a field of characteristic such that for every positive integer , there is a -modular representation of of degree such that We denote by the Brauer character associated with . As mentioned in the introduction, we actually do not use the HeLP Method in its classical setting. We neither compute many inequalities involving traces as for example in the proofs of [Her07, Proposition 6.5] or [BKL08, Mar16], since these formulas turn out to be too complicated in our setting. Nor do we apply the standard equations obtained from character values on one side and possible eigenvalues on the other side as e.g. in the proofs of [Her07, Propositions 6.4, 6.7], [Her08] or [BM17, Lemma 2.2], since there are too many possibilities for these possible eigenvalues. Still this second strategy is closer to our approach. 3. Number theoretical results In this section we prove two number theoretical results which are essential for our arguments and might be of independent interest. Our first proof of Proposition 3.2 below was very long. We include a proof which was given to us by Hendrik Lenstra. We are very thankful to him for his simple and nice proof. For a prime integer and a non-zero integer let denote the valuation of at , i.e. the maximal non-negative integer with . If, moreover, then denotes a complex primitive -th root of unity and denotes the -th cyclotomic polynomial, i.e. the minimal polynomial of over . If and are positive integers and is a prime integer then . We argue by induction on . Suppose first that and let denote the set of primitive -th roots of unity. Then is a root of for every and hence divides in . Therefore Suppose that and assume that the lemma holds with replaced by . Then . As and is a primitive -th root of unity, we have . ∎ Let be a positive integer. Let be integers and for every positive integer set Let be a divisor of such that for every prime power dividing with . Then . Let and consider the polynomial . We can take , so that . By hypothesis, for every prime and every positive integer with dividing we have , or equivalently divides in . Thus divides in . Therefore . By Lemma 3.1, each belongs to . As we deduce that , as desired. ∎ For a positive integer and a subfield of , let denote a set of representatives of equivalence classes of the following equivalence relation defined on : Let be a positive integer, let be a subfield of and let be the ring of integers of . For every let be an integer and for every integer define Let be a divisor of such that for every prime power dividing with . Then . Apply Proposition 3.2 to the integers with denoting the class in containing . ∎ In the remainder of this section we reserve the letter to denote positive prime integers. We now introduce some notation for a positive integer which will be fixed throughout. First we set If moreover then we set Next lemma collects two elementary properties involving this notation whose proofs are direct consequences of the definitions. Let be a prime dividing and let . Then the following conditions hold: If then . Let such that . If divides both and then . For integers and we define the following equivalence relation on : We denote by a set of representatives of these equivalence classes. Without loss of generality one may assume that . In the remainder of the section we assume that is odd. For and integers let Moreover, is the maximal real subfield of and is the ring of integers of . In the following proposition we prove that is a -basis of . For and , we use We denote by the number theoretical Möbius function. Let be a positive odd integer. Then is a -basis of and in particular, a -basis of . If and then It is easy to see that . Thus it is enough to prove the following equality Actually we will show , which implies the desired expression of . Indeed, for every let denote the -th part of , i.e. is a primitive -th root of unity and . Let be the set of tuples satisfying for every . For every let given by Then is the set of elements in satisfying . From we obtain . Therefore 4. Proof of Theorem 1.1 In this section we prove Theorem 1.1. In the remainder, set with a prime. Our goal is to prove that any element of order in , where is greater than and coprime with , is rationally conjugate to an element of . By Theorem 2.2.(3) we may also assume that is not a prime power. As the order of is fixed throughout, we simplify the notation of the previous section by setting We argue by induction on . So we assume that is rationally conjugate to an element of for every divisor of with . We will use the representations and Brauer characters introduced in Theorem 2.2.(4). As usual in modular representation theory, a bijection between the complex roots of unity of order coprime with and the roots of unity of the same order in a field of characteristic has been fixed a priori. In this sense we will identify the eigenvalues of and the summands in . Since units of prime order in are rationally conjugate to elements of by Theorem 2.2.(3), we know that the kernel of on is trivial and hence has order . As the values of on -regular elements of are real, by Theorem 2.2.(1) and Theorem 2.1.(4), the set of eigenvalues of is closed under taking inverses (counting multiplicities). Therefore, is conjugate to for a suitable primitive -th root of unity . Hence by Theorem 2.2 there exists an element of order such that and are conjugate. From now on we abuse the notation and consider both as a primitive -th root of unity in a field of characteristic and as a complex primitive -th root of unity. Then for any positive integer we have that and for every integer we have The element and the primitive -th root of unity will be fixed throughout. By Theorem 2.1, is rationally conjugate to an element of if and only if for every . is rationally conjugate to if and only if If is rationally conjugate to , then and for any . Therefore (4.3) holds. Conversely, assume that (4.3) holds. For of order dividing let . Then and . As the Vandermonde matrix is invertible we deduce that for every . So for every and . As we are assuming that if is a divisor of different from then is rationally conjugate to an element of , we also have for every . Thus is rationally conjugate to an element of by Theorem 2.1.(1). Then and therefore is conjugate to in . We conclude that and are rationally conjugate. ∎ By Lemma 4.1, in order to achieve our goal it is enough to prove (4.3). We argue by contradiction, so suppose that for some positive integer which we assume to be minimal with this property. Observe that if and is an integer such that , then there exists such that and applying to the equation we obtain . This implies that divides . Note that by our choice of and hence . Moreover, because as the augmentation of is 1. We claim that Indeed, for any let if and otherwise. Then for any integer we have . Therefore, applying Corollary 3.3 for , and , the claim follows. Combining this with (4.4) and the minimality of , we obtain . Furthermore, , as . Therefore The bulk of our argument relies on an analysis of the eigenvalues of and the induction hypothesis on and . More precisely, we will use (4.6) and (4.7) to obtain a contradiction by comparing the eigenvalues of and . Of course we do not know the eigenvalues of the latter but we know the eigenvalues of each . Moreover, if is a divisor of with then is rationally conjugate to an element of . Then , and are conjugate in , for a suitable field , and as is injective on and is conjugate to an element of we conclude that is conjugate to . Thus we know the eigenvalues of . This has consequences for the eigenvalues of . To be more precise we fix (with repetitions if needed) such that the eigenvalues of with multiplicities are . This is possible by the last statement of Theorem 2.2.(1). By the above paragraph, if with then the lists and represent the same elements in , up to ordering, and hence and represent the same elements of , up to ordering. We express this by writing This provides restrictions on , and the . If for some then is the smallest prime dividing and for every with . If then is not divisible by any prime greater than . Let denote the smallest prime dividing . (1) Suppose that . Then . As we deduce that for some . Therefore and for every with we have . Hence . (2) Suppose that is a prime divisor of with . Then and therefore, by (1), for every . Thus, by (4.7) and (4.9) and ignoring the signs provided by the and , it is enough to show that for at most two ’s and for at most two ’s, since by assumption , i.e. . Observe that if then and hence is multiple of . Moreover, if with then . Therefore unless . As and we have , the lemma follows. ∎ For a non-zero integer let denote the number of prime divisors of . We obtain an upper bound for in terms of . For every we have Moreover if for every then Using (4.9), and ignoring the sings given by and , it is enough to prove that since the number of divisors of is and if for some this provides an additional . We prove the second inequality, only using that . This implies the first inequality by applying the second one to . For a fixed dividing let . By changing the sign of some ’s, we may assume without loss of generality that if then . Thus, if then . We claim that if then . Indeed, let be prime divisor of . If then , so . If then and so also . Otherwise, i.e. if and , then divides both and and . Therefore , by Lemma 3.4.(2). As and there are at most two ’s with representing the same class in , we deduce that , as desired. ∎ We are ready to finish the proof of Theorem 1.1. Recall that we are arguing by contradiction and , and hence also , is odd. a contradiction. Thus, if then and if then . However, if then by Lemma 4.3 and hence for one . This implies, by Lemma 4.2.(1), that contradicting the assumptions that is not a prime power. Therefore . We deal with these cases separately using (4.8) and (4.9). Observe that if is a prime bigger than then for every and so also , since . Assume that . Combining Lemma 4.2.(

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http://www.math.cmu.edu/~calvar/calc08/onninen.html

math

Abstract: We give an account of a few of recent developments in which the quasiconformal theory and nonlinear elasticity share common problems of compelling mathematical interest. We study deformations between bounded domains in Euclidean n-space. We introduce natural conditions on the integrand that guarantee the existence and global invertibility of the minimizers. The key tools in finding an extremal deformation are the free Lagrangians. We demonstrate these ideas in the case of the total harmonic energy and a pair of annuli. The talk is based on joint work with Tadeusz Iwaniec.

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https://en-academic.com/dic.nsf/enwiki/6649058

math

- Equilibrium fractionation fractionationis the partial separation of isotopesbetween two or more substances in chemical equilibrium. Equilibrium fractionation is strongest at low temperatures, and (along with kinetic isotope effects) forms the basis of the most widely used isotopic paleothermometers (or climate proxies): D/H and 18O/16O records from ice cores, and 18O/16O records from calcium carbonate. It is thus important for the construction of geologic temperature records. Isotopic fractionations attributed to equilibrium processes have been observed in many elements, from hydrogen (D/H) to uranium (238U/235U). In general, the light elements (especially hydrogen, boron, carbon, nitrogen, oxygenand sulfur) are most susceptible to fractionation, and their isotopes tend to be separated to a greater degree than heavier elements. Most equilibrium fractionations are thought to result from the reduction in vibrational energy (especially zero-point energy) when a more massive isotope is substituted for a less massive one. This leads to higher concentrations of the massive isotopes in substances where the vibrational energy is most sensitive to isotope substitution, i.e., those with the highest bond force constants. In a reaction involving the exchange of two isotopes, lX and hX, of element “X” in molecules AX and BX, each reactant molecule is identical to a product except for the distribution of isotopes (i.e., they are isotopologues). The amount of isotopic fractionation in an exchange reaction can be expressed as a fractionation factor: indicates that the isotopes are distributed evenly between AX and BX, with no isotopic fractionation. indicates that hX is concentrated in substance AX, and indicates hX is concentrated in substance BX. is closely related to the where is the product of the rotational symmetry numbers of the products (right side of the exchange reaction), is the product of the rotational symmetry numbers of the reactants (left side of the exchange reaction), and is the number of atoms exchanged. An example of equilibrium isotope fractionation is the concentration of heavy isotopes of oxygenin liquid water, relative to water vapor, At 20oC, the equilibrium fractionation factor for this reaction is Equilibrium fractionation is a type of mass-dependent isotope fractionation, while mass-independent fractionationis usually assumed to be a non-equilibrium process. Stable isotope Isotope geochemistry Kinetic isotope effect Isotope analysis Kinetic fractionation Mass-independent fractionation Chacko T., Cole D.R., and Horita J. (2001) Equilibrium oxygen, hydrogen and carbon isotope fractionation factors applicable to geologic systems. Reviews in Mineralogy and Geochemistry, v. 43, p. 1-81. Horita J. and Wesolowski D.J. (1994) Liquid-vapor fractionation of oxygen and hydrogen isotopes of water from the freezing to the critical temperature. Geochimica et Cosmochimica Acta, v. 58, p. 3425-2437. Wikimedia Foundation. 2010.

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CC-MAIN-2024-18

https://www.lufrancesa.com/is/49211424ae01f0fc5247d68a38e8

math

source shines on the grating, images of the light will appear at a number of angles 1, 2, 3 and so on. 1. The principal maxima occur on both sides of the central maximum for which a formula similar to Young's formula holds . A diffraction grating defines an optical component with a periodic structure that splits the light into various beams that travel in different directions. It shows that for a given diffraction grating (at fixed b), a different wavelength gives maxima at different points of the spectrum. 2. A simple experiment to observe the diffraction of light waves is by using instruments such as lasers as a light source, grating as an obstacle with multiple narrow slits and a screen as a medium to capture the periodic diffraction pattern. 3. Student experiment: A CD as a grating (home experiment) (30 minutes) Discussion: The meaning of coherence (10 minutes) Demonstration: Looking through gratings. The diffraction grating is most useful for measuring accurately. This experiment involves diffraction of light waves though a very small slit (aperture), and demonstrate that when light passes through the slit, the physical size of the slit determines how the slit interacts with the light. (30 minutes) Like the prism, the diffraction grating can be used to disperse a spectrum into its wavelength components. To measure the wavelengths of certain lines in the spectrum of the mercury arc lamp. The apparatus for this experiment consists of a telescope spectrometer with a diffraction grating as its dispersive element, as in Figure 14.1. The grating used in this experiment is of the transmission type. Experiment to determine the wavelength of monochromatic light: The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. In other words, the diffraction efficiency for certain diffraction orders is of interest. Grating becomes an essential tool in the diffraction experiment because of its role as an obstacle matter . It is an alternative way to observe spectra other than a prism. tan y L You will measure the spectra of two light sources; a common incandescent bulb (the light box used in the thin lens lab) and a laser. The value of m is given by the grating equation shown above, so that m = arcsin m d FIG. This illustration is qualitative and intended mainly to show the clear separation of the wavelengths of light. In this and similar experiments, the electrons are produced not by a well-defined quantum-mechanical process as in the photon diffraction experiment described above, but by the . Record the lines/length rating of the grating. implies the following experimental conditions: (i) the spectrometer is in proper adjusfinen! The screen to diffraction grating distance D and the ruling density 1 d you will use will be given to your answer sheet of the practical lab. This system may allow affiliate links. Introduction Atomic spectra will be studied in this experiment using a diffraction grating mounted in a spectrometer. Turn on the laser and identify the zero order maximum (straight through). These diffraction maxima satisfy the grating condition : a b sin n n , (1) where (a+b) = the grating element (=2.54/N, N being the number of rulings per inch of the grating), n = the angle of diffraction of the nth maximum A diffraction grating is the arrangement of largenumber of narrow slits arranged side by side, theses slits are separated by opaque spaces. Diffraction grating is an optical component having a periodic structure which can split and diffract light t several beams travelling in different directions. Diffraction at single slit 3. The diffraction grating is an important device that makes use of the diffraction of light to produce spectra. 1. The diffraction grating was named by Fraunhofer in 1821, but was in use before 1800. Initially all the preliminary adjustments of the spectrometer are made. The diffraction grating was named by Fraunhofer in 1821, but was in use before 1800. Interference in one dimension (1D) - 1 weight . At normal incidence, --------------- (5) where, N is the number of lines per unit length of the grating Resolution. It is an alternative way to observe spectra other than a prism. (A good discussion of the patterns obtained, and a more careful calculation of the single slit pattern . 2. (known as diffraction pattern) Initially all the preliminary adjustments of the spectrometer are made. incremented one placeholder down after a frame. 2. To study diffraction of light using a diffraction grating spectrometer b. This you can find out from the lines that are mentioned on the grating as here 15000 LPI. Part A: Single-Slit Diffraction Discussion This week's and next week's exercises show that . diffraction grating experiment theory The directions of these beams depend on the spacing of the grating and the wavelength of the light so that the grating acts as the dispersive element. A typical grating has density of 250 lines/mm. 2, which is calculated by a sinusoidal RDG with the groove period p=/0.4368, the depth d = p/(5), incidence angle =0 and the number of discrete points m = 40. The width of this slit can be . The diffraction grating spacing calculated was 1655.4nm which are consistent with the theory . Lab report on physics experiment 10 experiment 10 diffraction grating tiisetso khumalo november 2018 group partners: collen khoza, angela introduction diffraction patterns. (10 minutes) Discussion: Deriving the grating formula. discharge tube power supply diffraction scale hydrogen discharge tube diffraction grating sodium light source PROCEDURE: 1. The directions or diffraction angles of these beams depend on the wave (light) incident angle to the diffraction grating, the spacing or . Compare the values of the given and the calculated values for the discussion. lab report lab diffraction grating and interference introduction: in this lab, the experiment performed was study of diffraction and interference of light wave . Theory: A diffraction grating is made of many equally-spaced slits, the distance between two slits is d . Pass . By diffraction theory analysis and simulation optimization based on the accurate boundary . The slit of collimator is illuminated by a monochromatic light, whose wavelength is . A diffraction grating can be a reflection grating or a transmission grating. The rays will fall in a parallel bundle on the grating. The diffraction efficiency, DE , specifies the optical power carried by the diffracted waves normalized with the power of the incident wave. where k is the wavelength; that is to be found from observation. (6.3.2) and (6.3.3)) increases N 2 times in comparison with one slit, and the maxima width decreases by 1/N.The condition of the main maximum (6.3.4) is of primary importance. 3. The Discussion of Results should be clear, specific, and reasonable. Add Tip. The directions or diffraction angles of these beams depend on the wave (light) incident angle to the diffraction grating, the spacing or . 5) Rubber bands. We have to use nk=2d sin theta. A diffraction grating is made by making many parallel scratches on the surface of a flat piece of transparent material. Experiment: In this experiment you will employ a He-Ne laser with a wavelength of 632.8 nm in vacuum. The diffracted light it can be photographically mass produced rather cheaply to produce.! 5 0) and transmissive diffraction T d (m 521) are mea-sured. A prism refracts Determination of wavelength of different colours 9. The center-to-center spacing between the grating couplers of the 2D optical antenna array is 49 m (corresponding to about 32), which leads to the presence of several diffraction orders . It's an alternative to using a prism to study spectra. For 0, the diffraction grating does bring a contribution, which can create a propagating diffraction order at an angle described by Equation (2). In the above video, you can watch the demonstration of the experiment includes the diffraction grating practical lab. Measure the distance D between the grating and the wall. Diffraction grating formula. What does diffraction look like for an object with a periodic structure? A diffraction grating defines an optical component with a periodic structure that splits the light into various beams that travel in different directions. But to see any but the coarsest structure in the spectrum you have to spread the spectrum over a relatively large linear distance by projecting over an adequate distance. A diffraction grating can be manufactured by carving glass with a sharp tool in a large number of precisely positioned parallel lines, with untouched regions acting like slits ( (Figure) ). These grooves are separated by a constant distance, d. I. Theory of experiment The spectrometer is an instrument for analyzing the spectra of radiations. Generally, when light is incident on the grating, the split light will have maxima at an angle . What is Diffraction Grating? a. Calibration of a Diffraction Grating Using the red line in the Hydrogen spectrum (Balmer-) as a reference of known wavelength ( = 656.28 nm), measure the angles of the first and second-order principal maxima on each side. 2. PART I. Procedure: 1. where g = n1 d / . The transmission diffraction efciency h 5 T d /(T t 1 T d) were calculated and are shown in the last row of this table. regarded this as proof that Fresnel's wave theory was nonsense, and that light must be a particle and not a wave. Diffraction Grating is optical device used to learn . Problem 1: Determine the slit spacing of a diffraction grating of width 2 cm and produces a deviation of 30 in the second-order with the light of wavelength 500 nm. Difference between interference and diffraction 6. An informative and interesting discussion of the diffraction grating, including its history, theory of operation, and how diffraction grat-ings are made can be found in the WikipediA, the free encyclopedia, The Diffraction . The emerging coloration is a form of structural coloration. as specified in part i (ii) the grating rulings are parallel to the slig i.e. BIOL 1121 Discussion wk 8; Protect the environment; Schoolreformessay - Week five essay assignment; . Physics 331A Experiment 2 CONCAVE DIFFRACTION GRATING SPECTROGRAPH Revised May 16, 2006 Di raction gratings are widely used for high resolution spectral studies. laser. The slit of collimator is illuminated by a . 2. Hypothesis: If the hair is human it will have a width of 10-4m. average grating spacing between lines of the diffraction grating has already been determined through the Mercury Lamp experiment. Now, with the 10 6 m line spacing that you suggest, the blue end of the first period is at about. Reflection gratings are normally coated with a reflective coating, usually aluminum . The graph 8 quot ; multiple slits l 1 diffraction grating experiment theory a plane sheet transparent! A diffraction grating does very much the same thing. There is a good case for describing it as the most important invention in the sciences. diffraction grating experiment The Discussion of Results section includes an explanation of how the collected data provide logical and reasonable support for the statement found in the Conclusion. From this you will determine "d". face of a plane transmission diffraction grating, bright diffraction maxima are observed on the other side of the grating. Because there can be over 1000 lines per millimeter across the grating, when a . Arrange for the beam to pass through the grating at normal incidence and meet the wall perpendicularly. If the diffraction grating is properly aligned, the angles for a given wavelength, in a given order of A diffraction grating is the tool of choice for separating the colors in incident light. For the surface-relief reflection grating of Figure 3 with a square-wave profile, the diffraction efficiency is given by. However, a diffraction grating is less sensitive to the color of the light and can be made to spread colors over a larger angle than a prism. (20 minutes) Student experiment: Measuring wavelength. This is 1.6910^ (-4)cm approximately. A diffraction grating pattern is an interference pattern consisting of maximums and minimums when light is diffracted. If white light, which is a mixture of light with different wavelengths, strikes such a grating, each . (See the 8.02 Course Notes, Section 14.8, for a further discussion of diffraction gratings.) One difference between the interference of many slits (diffraction grating) and double-slit (Young's Experiment) is that a diffraction grating makes a number of principle maxima along with with lower intensity maxima in between. In our experiments here, we will have = =1.00 for air. This type of grating can be photographically mass produced rather cheaply. . Place the grating in the laser beam at the distance D specified on your answer Title: Microsoft Word - Diffraction grating experiments Author: Oana Created Date: 7/12/2018 2:32:15 PM Diffraction in grating 7. Conclusion. For example, the sum of diffracted efficiency e n is given in Fig. 4) 2 longer flat-end screws (about 2 inches long) and suitable nuts. Experiment may be booked as a one-weight or a two-weight experiment. This accounts for about a two fold gain in the capacity of a DVD relative to CD. End of preview. In this paper, high diffraction efficiency silicon-blazed grating working at 800-2500 nm has been designed and fabricated. The grating is the more precise device if we want to distinguish two closely spaced wavelengths. The wavelength () of the green diode laser used in this experiment is 532 10 5.32 .10 10= nm mm()4. The intensity of the diffraction maxima (eq. When light strikes the grating, the split light will often have maxima at an angle . In Table 2 the cof diffraction is dif-ferent from those of specular transmission and . the distance from the grating to the screen (L), and using a bit of trigonometry. It means 15000 thousand lines per inch. Place the laser approximately 4 m away from a large wall and place the diffraction grating in front of it. Find the sine of angle of diffraction by formula given in observation table RESULTS AND DISCUSSION-Result(s): Grating element, d= 0.0000441 m Conclusion: We found out a diffraction grating has a very large number of equally spaced slits. 5.3. A detailed discussion of the diffraction maxima positions, the number of diffraction orders, and the diffraction efficiencies is provided. To ensure exact right angles, follow the steps below. When the light from different slits meet at the screen, the waves will interfere and the resultant amplitudes (determined by superposition) will give pattern on the screen. The intensities of these peaks are . Unlike a simple prism, a diffraction grating generally produces multiple output beams according to different diffraction orders.. An important question is how the output power is distributed over the different diffraction orders. There is a good case for describing it is the most important invention in the sciences. It is useful to analyze the interference pattern on the . The equation of the grating shows that the diffraction angle depends only on the pitch of the grating and not by the . Calculate the dispersive power for the 1st and 2nd order of a diffraction grating with a number of 700 line per cm . THEORY Plane Diffraction GratingThe phenomenon of bending of light round the edges of an obstacle/slit is called diffraction. One inch is 2.54 cm so 15,000 lines are in 2.54 cm so one line will be equal to 2.54/15000 cm. If a . Want to read all 4 pages? Set the grating in its holder, making sure that you don't touch the lines of the grating. Light from a source enters the spectrometer through a thin vertical slit. A diffraction grating is essentially a series of parallel equidistant a pertures or surfaces from which . It is a dark or bright line in an otherwise uniform and continuous spectrum, resulting from emission or absorption of light in a narrow frequency range, compared with the nearby frequencies.Spectral lines are often used to identify atoms and molecules from their characteristic spectral lines. When parallel light is incident on a diffraction grating each slit acts as a source of diffracted waves . However, when the integral equation is tried to be implemented for numerical calculation, a non-convergence result is usually obtained. cand D are two independent parameters representing pure polarization. Diffracted lights shine on a distant screen which has a central bright spot labeled m=0 and a higher order bright fringes that can also be observed. The apertures are usually spaced several hundred to the centimeter. The observer needed to be able to focus on the diffraction scale, the interference pattern itself, staying situated looking through the diffraction plate at the light source, all while counting maximums of bright light from an uncertain center. sharpness and visibility of the fringes increases. (6.3.2) and (6.3.3)) increases N 2 times in comparison with one slit, and the maxima width decreases by 1/N.The condition of the main maximum (6.3.4) is of primary importance. - Jordan Anderson Racing Jobs Near Vienna - The Ridge On Spring Valley Apartments - Agapanthus Blue Bulbs - My Sceptre Tv Won't Connect To Wifi - Aium Guidelines For Umbilical Artery Doppler - Washington Wizards 2019 - Bioinformatics Hd Wallpapers - Alliance National Group - Skechers Air Cooled Memory Foam Shoes - Jacket With American Flag Patch

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http://www.ccs.neu.edu/home/jaa/CS3800.09F/Homeworks/hw.04.html

math

Wed 07 Oct 2009 Due: Wed 14 Oct 2009 where "RC" is "regular credit", "EC" is "extra credit", and "NA" is "not applicable" (not attempted). Failure to do so will result in an arbitrary set of problems being graded for regular credit, no problems being graded for extra credit, and a five percent penalty assessment. For all proofs of non-regularity, you should use the Pumping Lemma and/or appeal to closure properties. For examples of complete and correct Pumping Lemma proofs, see Examples 1.73, 1.74, 1.75, 1.76, and 1.77 on pages 80-82 of the Sipser text. See also the following handout. The use of closure properties may also be helpful and may, in fact, yield complete proofs without the need to resort to the Pumping Lemma (as discussed in class). For examples of proofs for non-regularity via closure properties and the Pumping Lemma, please see the following handout. Required: 5 of the following 7 problems Points: 20 pts per problem Hint: For part (a), consider all strings in the language B in lexicographic order; write down the first dozen or so such strings. What properties do these strings have in common? Now construct a FA or RE for this language. Hint: For part (a), consider the language F' = F intersect ab*c* appealing to closure properties and the Pumping Lemma.

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CC-MAIN-2013-20

http://www.maa.org/press/maa-reviews/excursions-in-classical-analysis

math

The author’s intention in this enlightening and inspiring book is to introduce the reader to advanced problem solving techniques via case studies. It has twenty-one relatively short independent chapters that contain “kernels of sophisticated ideas connected to important current research” and expose the principles underlying these ideas. Chen is a devotee of Pólya’s rubric for problem solving and practices what he preaches: …I have begun each topic by categorizing and identifying the problem at hand, then indicated which technique I will use and why, and ended by making a worthwhile discovery or proving a memorable result. I often take the reader through a method which presents rough estimates before I derive finer ones, and I demonstrate how the more easily solved special cases often lead to insights that drive improvements of existing results. Readers will clearly see how mathematical proofs evolve ― from the specific to the general and from the simplified scenario to the theoretical framework. Each chapter begins with an appropriate quotation and ends with exercises and a brief list of references. (Chapter 18 references a paper written by the author’s son when he was an undergraduate.) Many of the examples and problems are taken from the Monthly or the Putnam exam; and there are 14 open problems. Selected problem solutions are provided at the back of the book. There aren’t many figures, but each page is dense with (beautiful) mathematical formulas. The first three chapters discuss inequalities and are relatively elementary. Chapter 2 presents the author’s new approach to proving inequalities based on the fact that where E is a subset of R. Together with convexity, Chen’s method yields proofs of the AGM and Cauchy-Schwarz inequalities, as well as a new elementary proof Ky Fan’s inequality connecting weighted arithmetic, geometric, and harmonic means. As is the case for several of the chapters, this work is based on the author’s published work. In Chapter 3, the author shows that six standard means can be represented by the function for different values of t. By showing that f(t) is strictly increasing, Chen derives inequalities connecting the six means. I remember being impressed by the author’s paper (Mathematics Magazine, December 2005) on which this exposition is based. Chapter 6 treats generating functions, Fibonacci numbers, harmonic numbers, Bernoulli numbers, and the Euler-Maclaurin summation formula. Even though there’s nothing particularly new here, the material is developed very nicely. In Chapter 18, following a paper of D. H. Lehmer, the author defines an “interesting” series as one for which there is a simple explicit formula for its nth term and which has a sum with a closed form expressed in terms of well-known constants ― for example, the Gregory-Leibniz series for π/4 or the series defining Apery’s constant, z(3). Here, as in Lehmer’s paper, Chen focuses on series involving reciprocal binomial coefficients. The author goes beyond Lehmer’s focus on series involving the central binomial coefficient and ends this chapter with a search for new formulas for π employing the methods of experimental mathematics. Some readers may recognize Chen’s name as an avid proposer and solver in several journal problem sections. The last chapter presents Chen’s solutions to eight integral problems that have appeared in the Monthly, half of which have not had their solutions published at the time of this book’s debut. This book is similar to Excursions in Calculus: An Interplay of the Continuous and the Discrete by Robert M. Young. However, whereas Young’s book (also invoking the spirit of Pólya) is intended as a supplement to calculus texts, the work under review is for the most part clearly meant to supplement advanced calculus or real analysis courses. Chen’s book is a wonderful updated tour of classical analysis and would serve as an excellent source of undergraduate enrichment/research problems. It recalls the type of gems in classical analysis, number theory, and combinatorics I first encountered in the books of Pólya and Szegö as an undergraduate many years ago. Peruse the Table of Contents and see if some of the topics and subtopics don’t grab you. Henry Ricardo ([email protected]) has retired from Medgar Evers College (CUNY), but continues to serve as Governor of the Metropolitan NY Section of the MAA. He is the author of A Modern Introduction to Differential Equations (Second Edition). His linear algebra text was published in October 2009 by CRC Press.

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http://www.chegg.com/homework-help/explain-determine-whether-following-statements-true-give-exp-chapter-12.3-problem-47e-solution-9780321336118-exc

math

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. projvu = projuv b. If nonzero vectors u and v have the same magnitude they make equal angles with u + v. c. (u • i)2 + (u • j)2 + (u • k)2 = |u|2. d. If u is orthogonal to v and v is orthogonal to w, then u is orthogonal to w. e. The vectors orthogonal to ⟨1, 1, 1⟩ lie on the same line. f. If projvu = 0, then vectors u and v (both nonzero) are orthogonal.

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CC-MAIN-2016-30

https://1st-in-babies.com/which-diagram-shows-a-planet-with-the-least-eccentric-orbit-update-new/

math

Let’s discuss the question: which diagram shows a planet with the least eccentric orbit. We summarize all relevant answers in section Q&A of website 1st-in-babies.com in category: Blog MMO. See more related questions in the comments below. Which planet has least eccentric orbit? The planet with the least circular orbit is Mercury, which has an orbital eccentricity of 0.2056. Which planet has an almost eccentric orbit? In our solar system, Venus and Neptune have nearly circular orbits with eccentricities of 0.007 and 0.009, respectively, while Mercury has the most elliptical orbit with an eccentricity of 0.206. Physics – Mechanics: Gravity (11 of 20) Eccentricity Of A Planet’s Orbits Images related to the topicPhysics – Mechanics: Gravity (11 of 20) Eccentricity Of A Planet’s Orbits What is the shape of the most eccentric orbits What is the shape of the least eccentric orbits? |Perihelion (Point in Orbit Closest to Sun) measured in AU’s||0.98| |Aphelion (Point in Orbit Farthest from Sun) measured in AU’s||1.02| What is the shape of the most eccentric orbits? Although some objects follow circular orbits, most orbits are shaped more like “stretched out” circles or ovals. Mathematicians and astronomers call this oval shape an ellipse. Why is Venus the least eccentric? Venus orbits the Sun at an average distance (semi-major axis) of 108,208,000 km (0.723 AUs), ranging between 107,477,000 km (0.718 AU) at perihelion and 108,939,000 km (0.728 AU) at aphelion. This makes Venus’ orbit the least eccentric of all the planets in the Solar System. What planet affects Mercury’s orbit the least? |Semi-major axis||0.387098 AU 57,909,050 km| |Orbital period (sidereal)||87.9691 d 0.240846 yr 0.5 Mercury synodic day| What planet has the least elliptical orbit? Venus, which is right next to Mercury, has the least eccentric orbit of any of the planet in the Solar System. Its orbit ranges between 107 million km and 109 million km from the Sun and has an eccentricity of . Is Venus orbit elliptical? Venus orbits the Sun at an average distance of about 0.72 AU (108 million km; 67 million mi), and completes an orbit every 224.7 days. Although all planetary orbits are elliptical, Venus’s orbit is currently the closest to circular, with an eccentricity of less than 0.01. What is Venus eccentricity? Venus has an orbit with a semi-major axis of 0.723 au (108,200,000 km; 67,200,000 mi), and an eccentricity of 0.007. Which planet has the most eccentric orbit quizlet? Pluto and Mercury are the planets in our solar system with the most eccentric orbits. 1.5: Eccentricity of Orbits Images related to the topic1.5: Eccentricity of Orbits Which major planet has the largest orbital eccentricity? Mercury has the greatest orbital eccentricity of any planet in the Solar System (e = 0.2056). What is the eccentricity of all the planets? Why do planets have different eccentricity? Over time, the outer planet eventually perturbed the middle planet’s orbit enough to deform it slowly into an eccentric orbit as well, which is what is seen today, although every 7,000 years or so the middle planet returns gradually to a circular orbit. “This is what makes the system so peculiar,” said Rasio. What is the eccentricity of Uranus orbit? |Sidereal rotation period (hrs)||-17.24*||0.720| |Length of day (hrs)||17.24||0.718| |Obliquity to orbit (deg)||97.77||–| Which objects have the most eccentric orbits? Mercury has a more eccentric orbit than any other planet, taking it to 0.467 AU from the Sun at aphelion but only 0.307 AU at perihelion (where AU, astronomical unit, is the average Earth–Sun distance). What causes eccentricity in orbit? Over time, the pull of gravity from our solar system’s two largest gas giant planets, Jupiter and Saturn, causes the shape of Earth’s orbit to vary from nearly circular to slightly elliptical. Eccentricity measures how much the shape of Earth’s orbit departs from a perfect circle. Why is Mercury’s orbit eccentric? The study in the year 2010 informs us that the Sun’s energy can rise and falls. The Sun’s fluctuations caused partial collapse of Earth’s atmosphere. From this study, triangle’s energy concept predicted that the Sun’s energy fluctuation as the cause of the unusual of Mercury’s orbit. Why is Mercury a planet and not Pluto? It must be large enough that its gravity has pulled it into a spherical shape, and also must dominate and gravitationally clear its orbital region of most of all other objects. It does not clear the orbital zones of other planets and therefore does not meet the standard for being called a planet. Do you know that 4 Planets are visible to the Naked Eye these days? Images related to the topicDo you know that 4 Planets are visible to the Naked Eye these days? Is Pluto the smallest planet in the solar system? The Smallest Planet In The Solar System – Mercury. |Equatorial Circumference||15,329 km| |Mass||330,104,000,000,000 billion kg (That is 5.5% of Earth)| Which planet has an orbital eccentricity of the moon? |Orbital Inclination (degrees)||7.0||5.1| |Obliquity to Orbit (degrees)||0.034||6.7| |Mean Temperature (C)||167||-20| - which planet has the least distance between the two foci of its elliptical orbit? - which of the following eccentricity values would produce the most eccentric orbit - which planet has the least distance between the two foci of its elliptical orbit - at which position is the planet’s orbital speed greatest? - which planet has the most eccentric orbit - which bar graph correctly shows the orbital eccentricity of the planets in our solar system? - compared to planet a planet b has a - which of the following eccentricity values would produce the most eccentric orbit? - what is the approximate eccentricity of this ellipse - which object is located at one foci of the elliptical orbit of mars - which bar graph correctly shows the orbital eccentricity of the planets in our solar system - the actual orbits of the planets are Information related to the topic which diagram shows a planet with the least eccentric orbit Here are the search results of the thread which diagram shows a planet with the least eccentric orbit from Bing. 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http://www.herkimer.edu/academics/view_course/college_algebra_and_trigonometry/

math

College Algebra and Trigonometry Course #: MA 132 Class/Lab Hours: 3, 0 This course includes topics in both intermediate and advanced algebra and right triangle trigonometry. The course will deal with equations, inequalities, graphing, polynomials, rational expressions, and trigonometric functions including trigonometric functions as circular functions, trigonometric identities and equations. Prerequisite: MA 130 or high school equivalent. Check the availability of the course.

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CC-MAIN-2013-48

https://doubtnut.com/related-videos/1536754

math

Click Question to Get Free Answers > Fill in the blanks, in each of the following, so as to make the statement true: A rhombus is a parallelogram in which.... A square is a rhombus in which.... A rhombus has all its sides of ........ length. The diagonals of a rhombus..... each other at... angles. It the diagonals of a parallelogram bisect each other at right angles, then it is a.... 46.3 K+ Views | 191.1 K+ Likes Give reasons for the following : (a) A square can be thought of as a special rectangle. (b) A rectangle can be thought of as a special parallelogram. (c) A square can be thought of as a special rhombus. (d) Squares, rectangles, parallelograms are all quadrilaterals. (e) Square is also a parallelogram. 165.4 K+ Views | 196.3 K+ Likes Let `P ,Q ,R` and `S` be the points on the plane with position vectors `-2i-j ,4i ,3i+3ja n d-3j+2j ,` respectively. The quadrilateral `P Q R S` must be a Parallelogram, which is neither a rhombus nor a rectangle Square Rectangle, but not a square Rhombus, but not a square 63.8 K+ Views | 209.8 K+ Likes Let LMNP be a rectangle (which is not a square) inscribed in the ellipse as shown in the figure. Let '`lambda` denotes the number of ways in which we can select four points out of A,A',B,B', L,M,P and N such that normal at these four points are concurrent, then '`lambda` is less than or equal to- 43.9 K+ Views | 92.8 K+ Likes

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https://forums.maplestory.nexon.net/discussion/comment/52358/

math

[New Users] Please note that all new users need to be approved before posting. This process can take up to 24 hours. Thank you for your patience. Check out the v.244 Punch King Palooza Patch Notes here! hey nexon, whos brite ass idea was to to completely fubar the evans class??????????????? hadnt been on mine in ages, till about 2 months ago, what a wasted ordeal that was...for 1 att you have to use 2 other attacks unlike they old evans it was one for one..now its just a wasted space that need to be gotten rid of......a junk toon. - evans, worth the time to you anymore or not?11 votes yes. 91% (10 votes) no. 9% (1 vote) wasted space. 0% (0 votes)

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https://www.commonsensemedia.org/website-reviews/bedtime-math/user-reviews/adult

math

Great idea, okay execution I am really quite supportive of the concept. I think nightly math is a GREAT idea. Unfortunately, the math in these nightly posts isn't very creative or encompassing of all the things that make up math. In fact, I think she leaves out most of what makes math fun, focusing in too tightly on what we think "math" is. Every problem for the "wee ones" is counting. Every problem for "Little Kids" is basic addition, some subtraction, and some place values. Every problem for "Big Kids" is again basic calculation. Although these are age-appropriate math problems, they are missing most of what training for mathematical thinking needs to encompass. Never do the problems use shape matching, pattern prediction, measurement or other basic concepts that would be easy to incorporate into a daily math problem with your kids. It is a great start for families with mathphobia in the parents, but I'd like to see more diversity of "what is math" and more fun added in. Math is AWESOME and everywhere, it is hard to conceive of a situation in which you can't apply some kind of math. So, I'm doing my own with my child. Kudos to BedtimeMath for making me think of it, and Kudos to BedtimeMath for getting this concept to a lot of people. I look forward to when she improves it. -a Science and Math Educator

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https://123dok.net/article/polarimetric-decomposition-sar-polarimetry.qvl6g0x1

math

I.1 SAR Polarimetry I.1.2 Polarimetric decomposition Target decomposition (TD), introduced in the first place in , aims to interpret polarimet-ric data by assessing and analysing the components involved in the scattering process . Roughly speaking, the assessing assumes the derivation of the involved backscattering com-ponents, while the analysis dominantly concerns their parametrization. The former can be defined as: with X being the scattering matrix (S) in case of a coherent target, or the coherence matrix (T) in case of an incoherent one. Both the means of deriving the components (Xi) and their parametrizations, differ for different types of decompositions. Here, we present the ones we found to be the most notable with respect to their historical and practical relevance. After-wards, in the Chapter III, we will elaborate a new decomposition, the principal contribution of this thesis. I.1.2.1 Coherent decompositions The most elementary approach in decomposing a scattering matrix is based on a set of Pauli matrices, originally used by Wolfgang Pauli in his theory of quantum-mechanical spin. Due to their numerous interesting mathematical properties (e.g. hermitian, unitary and commutation properties), these matrices found their applications in many domains, among which, the SAR polarimetry. In case of monostatic configuration, the scattering matrix is decomposed into standard mechanisms as: 12 Chapter I. POLSAR image and BSS making this approach a "model based" one. The first term in Eq. I.8 represent the odd-bounce scattering: flat surface, sphere, trihedral. The second one is related to the even-bounce scattering without polarization change - a dihedral with the axe of symmetry being parallel to the incident horizontally polarized wave. The third one represent the scatterer which favours the cross-polarized channel - a dihedral with the axe of symmetry rotated45◦ with respect to the incident horizontally polarized wave . A slightly different model based approach is proposed by Krogager in . The second component in this case rather represents a dihedral with any orientation, while the third one represents a helix. The decomposition takes form of a: with the angles φ,φs, and ✓ being respectively the absolute phase, the single (odd) bounce component phase, the orientation of the dihedral. The real coefficients ks, kd, kh are the contributions of single bounce, double bounce and helix scatterer. The latter one represents the non-symmetrical scattering i.e. the case when the target axe of symmetry doesn’t lie in the plane perpendicular to the line of sight. However, this decomposition is usually employed in the circular basis making this a suitable point for introducing the scattering matrix projected on the circular polarization basis. In the same way the projection onto the Pauli basis leads to the target vector, the projection onto the circular polarization ones, gives us the circular scattering matrix, which for the monostatic configuration takes the following form: The lexicographic ordering of the these elements leads to the circular target vector: kC =h Sll p The most representative algebraic coherent decomposition would be the Cameron decom-position . In this approach we cannot actually assume reciprocity, meaning that we have to consider all four elements of the scattering matrix. The decomposing process can be roughly divided on two steps: • The first step is related to the target geometrical properties. More precisely, we are initially trying to isolate the symmetric scattering from the non-symmetric one. By maximizing the former and consequently, minimizing the latter one, we can express the scattering matrix as: I.1. SAR Polarimetry 13 The ↵ parameters correspond to the ones introduced in Eq. I.9 and is the rotation angle with respect to the reference basis. • The second step would be algebraic, given that it represents one sort of the parametri-sation of the symmetric part. Namely, if we express the matrix given in Eq. I.14 as: 1 0 1 z we can define a vector: λ(z) = 1 corre-spond respectively to sphere, dihedral, cylinder, narrow dihedral), the symmetric target can be categorized by calculating the closeness to these elementary reflectors: λ(z) = 1 p1 +|z|2 1 +|zref|2|1 +z⇤zref|. (I.18) I.1.2.2 Incoherent decompositions Unlike it was the case with the coherent decompositions, here we are rather concentrated either on the already introduced coherence matrix, or on the covariance matrix, derived as the spatial covariance of the lexicographic target vector kc: C = E⇥ 14 Chapter I. POLSAR image and BSS Thefirst incoherent decomposition was proposed by Huynen in , and it assumes decom-posing the scattering as the incoherent sum of the fully polarized mechanisms and the fully unpolarized ones. Originally, the decomposition is based on Mueller matrix formalism: but the same reasoning can be equally applied on either covariance or coherence matrix. The polarized matrix is then parametrized using nine parameters, among which only two are invariant with respect to the rotation around the line of sight. The counterpart of formerly presented decompositions on standard mechanisms, in case of incoherent targets, are Freeman and Yamaguchi decompositions. Freeman decomposition assumes decomposing the covariance matrix into the sum of first-order Bragg, double bounce and volume scattering: C = Cs+Cd+Cv = (I.21) Bragg scattering would be a more elaborated odd-bounce backscattering, which can provide us with some details about the surface dynamics. The real parametersβ,fs,fdandfv, and the complex parameter↵,figuring in Eq. I.22 are supposed to be derived from a set of equations: h|Shh|2i = fsβ2+fd|↵|2+fv (I.22) h|Svv|2i = fs+fd+fv hShhSvv⇤ i = fsβ+fd↵+fv/3 h|Shv|2i = fv/3. However, there is one more unknown variable than there are equations. Therefore, in order to make this a well-posed problem, we need to consider the sign of hShhSvv⇤ i−h|Shv|2i. If it is negative, we take↵=−1, if not β= 1. In case of Yamaguchi, there is an additional fourth component, representing non-symmetric, helix backscattering: I.1. SAR Polarimetry 15 Another difference would be a complexβ parameter, makingCs31 to be rather β⇤. In this case, we have a system offive equations: h|Shh|2i = fs|β|2+fd|↵|2+ 8 Two parameters can be derived from the second and the last equation (Eq. I.25), while the rest of the system can be solved using the same hypothesis as in the case of Freeman decomposition. The incoherent equivalents of the Cameron algebraic method would be the Cloude and Pottier and the Touzi decompositions. The particular emphasise will be put on introducing these two, given that the highlight of this thesis concern one incoherent target decomposition. The first step in both of these two decompositions is the eigenvector decomposition of the target coherence matrix, allowing us to represent the total backscattering as a sum of three backscattering mechanism. Namely, given the Hermitian nature of positive semi-definite coherence matrix, the derived eigenvectors are mutually orthogonal and characterized by real eigenvalues. Each of the derived eigenvectors forms a coherence matrix with unity rank and therefore happens to be a fully polarized target vector. The corresponding eigenvalue (λ) represents its contribution to the total backscattering. T=λ1k1kH1 +λ2k2kH2 +λ3k3kH3 (I.25) The principal difference between two decompositions would be, the second step - the parametrisation of the estimated target vectors i.e. backscattering mechanisms. The Cloude and Pottier decomposition is based on the↵−β−γ−δ parametrisation of a target vector : Among the obtained parameters, the very central place has the angle ↵p, used in the 16 Chapter I. POLSAR image and BSS Figure I.4: Poincaré sphere representation of TSVM parameters: (a) symmetric scattering (⌧m = 0), (b) non-symmetric scattering (Φ↵s = 0). derivation of the mean backscattering mechanism (↵p), conditioned by the probabilities of↵pi to occur in the random sequence of parameters (Pi): ↵p=P1↵p1+P2↵p2+P3↵p3 (I.27) Aside from this one, very important polarimetric descriptor would be the entropy, defining the appropriateness of using polarimetry : −Pilog3Pi, Pi= λi P3 The value zero indicates the absolute dominance of one fully polarized mechanism, while the value one points to the completely depolarizing target. The third polarimetric descriptor we ought to mention would be the anisotropy: λ2+λ3 = P2−P3 describing the relation of the second and the third estimated mechanisms. The Touzi decomposition is rather based on the Target Scattering Vector Model (TSVM) . I.2. SAR images statistics 17

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https://m.scirp.org/papers/6668

math

ABSTRACT The goal of this paper is to introduce a new neural network architecture called Sigmoid Diagonal Recurrent Neural Network (SDRNN) to be used in the adaptive control of nonlinear dynamical systems. This is done by adding a sigmoid weight victor in the hidden layer neurons to adapt of the shape of the sigmoid function making their outputs not restricted to the sigmoid function output. Also, we introduce a dynamic back propagation learning algorithm to train the new proposed network parameters. The simulation results showed that the (SDRNN) is more efficient and accurate than the DRNN in both the identification and adaptive control of nonlinear dynamical systems. Cite this paper nullT. Aboueldahab and M. Fakhreldin, "Identification and Adaptive Control of Dynamic Nonlinear Systems Using Sigmoid Diagonal Recurrent Neural Network," Intelligent Control and Automation, Vol. 2 No. 3, 2011, pp. 176-181. doi: 10.4236/ica.2011.23021. A. U. Levin, and K. S. Narendra, “Control of Nonlinear Dynamical Systems Using Neural Networks—Part II: Observability, Identification and Control,” IEEE Transactions on Neural Networks, Vol. 7, No. 1, 1996, pp. 30-42. doi:10.1109/72.478390 C. C. Ku and K. Y. Lee, “Diagonal Recurrent Neural Networks for Dynamic System Control,” IEEE Transactions on Neural Networks, Vol. 6, No. 1, 1995, pp. 144-156. doi:10.1109/72.363441 G. L. Plett, “Adaptive Inverse Control of Linear and Nonlinear Systems Using Dynamic Neural Networks,” IEEE Transactions on Neural Networks, Vol. 14, No.2, 2003, pp. 360-376. doi:10.1109/TNN.2003.809412 K. S. Narendra and K. Parthasarathy, “Identification and Control of Dynamical Systems Using Neural Networks,” IEEE Transactions on Neural Networks, Vol. 1, No. 1, 1990, pp. 4-27. doi:10.1109/72.80202 L. Chen and K. S. Narendra, “Nonlinear Adaptive Control Using Neural Networks and Multiple Models,” Proceedings of the 2000 American Control Conference, Chicago, 2002, pp. 4199-4203. R. Zhan and J. Wan “Neural Network-Aided Adaptive Unscented Kalman Filter for Nonlinear State Estimation,” IEEE Signal Processing Letters, Vol. 13, No. 7, 2006, pp. 445-448. doi:10.1109/LSP.2006.871854 A. S. Poznyak, W. Yu, E. N. Sanchez and J. P. Perez, “Nonlinear Adaptive Trajectory Tracking Using Dynamic Neural Networks,” IEEE Transactions on Neural Networks, Vol. 10, No. 6, 1999, pp. 1402-1411. P. A. Mastorocostas, “A Constrained Optimization Algorithm for Training Locally Recurrent Globally Feedforward Neural Networks,” Proceedings of International Joint Conference on Neural Networks, Montreal, 31 July 4 August 2005, pp.717-722. A. Tarek, “Improved Design of Nonlinear Controllers Using Recurrent Neural Networks,” Master Dissertation, Cairo University, 1997. Xiang Li, Z. Q. Chen and Z. Z. Yuan, “Simple Recurrent Neural Network-Based Adaptive Predictive Control for Nonilnear Systems,” Asian Journal of Control, Vol. 4, No. 2, June 2002, pp. 231-239. N. Kumar , V. Panwar, N. Sukavanam, S. P. Sharma and J. H. Borm, “Neural Network-Based Nonlinear Tracking Control of Kinematically Redundant Robot Manipulators,” Mathematical and Computer Modelling, Vol. 53, No. 9-10, 2011, pp. 1889-1901. J. Pedro and O. Dahunsi, “Neural Network Based Feedback Linearization Control of a Servo-Hydraulic Vehicle Suspension System,” International Journal of Applied Mathematics and Computer Science, Vol. 21, No. 1, 2011, pp. 137-147. doi:10.2478/v10006-011-0010-5 A. Thammano and P. Ruxpakawong, “Nonlinear Dynamic System Identification Using Recurrent Neural Network with Multi-Segment Piecewise-Linear Connection Weight,” Memetic Computing, Vol. 2, No. 4, 2010, pp. 273-282. doi:10.1007/s12293-010-0042-7 A. C Tsoi and A. D. Back, “Locally Recurrent Globally Feedforward Networks: A Critical Review of Architectures,” IEEE Transaction Neural Networks, Vol. 5, No. 2, 1994, pp. 229-239. doi:10.1109/72.279187 T. Rashid, B. Q. Huang and T. Kechadi, “Auto-Regressive Recurrent Neural Network Approach for Electricity Load Forecasting,” International Journal of Computational Intelligence, Vol. 3, No. 1, 2007, pp.66-71. B. A. Pearlmutter, “Gradient Calculations for Dynamic Recurrent Neural Networks: A Survey,” IEEE Transactions on Neural Networks, Vol. 6, No. 5, 1995, pp. 1212-1228. doi:10.1109/72.410363

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https://books.google.gr/books?id=t-IPAAAAYAAJ&pg=PA66&vq=%22by+multiplying+the+numerator+of+the+dividend+by+the+denominator+of+the+divisor+%3B+and+the%22&dq=editions:HARVARD32044096994090&lr=&hl=el&output=html_text

math

« ΠροηγούμενηΣυνέχεια » DEFINITIONS AND NOTATION. 111. We have seen (12) that division may be indicated by writing the dividend and divisor on opposite sides of a horizontal line. The term Fraction, in Algebra, relates to this mode or form of indicating division. Hence, 112. A Fraction is a quotient expressed by writing the dividend above a horizontal line, and the divisor below. Thus is a frac b tion, and is read, a divided by b. 113. The Denominator of the fraction is the quantity below the line, or the divisor. 114. The Numerator is the quantity above the line, or the dividend. 115. Any fraction may be decomposed as follows: 1.-The value of a fraction is equal to the reciprocal of the denominator multiplied by the numerator. 2.-In any fraction, the reciprocal of the denominator may be regarded as a fractional unit; and the numerator shows how many times this unit is taken in the fraction. Hence, 3.-A fraction is a fractional unit or a collection of fractional units, the value of each depending upon the denominator. 116. An Entire Quantity is an algebraic expression which has no fractional part; as x_3xy. 117. A Mixed Quantity is one which has both entire and fractional parts; as a + GENERAL PRINCIPLES OF FRACTIONS. 118. Sinee a fraction is a form of expressing division, it is evident that all the operations in fractions must be based upon the general relations subsisting between the dividend, divisor, and quotient. These principles relate, first, to change of value; second, to change of sign. 1st. Change of value. 119. By modifying the language of (84), we may express the mutual relations of the numerator and denominator of a fraction, as follows: I. Multiplying the numerator multiplies the fraction, and dividing the numerator divides the fraction. II. Multiplying the denominator divides the fraction, and dividing the denominator multiplies the fraction. III. Multiplying or dividing both numerator and denominator by the same quantity, does not alter the value of the fraction. 2d. Change of sign. 120. The Apparent Sign of a fraction is the sign written before the dividing line, to indicate whether the fraction is to be added or subtracted. Thus, in 4a-2.c the apparent sign of the fraction is plus, and indicates that tho fraction is to be added. 121. The Real Sign of a fraction is the sign of its numerical value, .when reduced to a monomial, and shows whether the fraction is essentially a positive or a negative quantity. Thus, in the fraction just given, let a=2 and x=3. Then a’-ax? 4-18 -14 -7 40-2x 8-6 2 The real sign of this fraction therefore is minus, though its apparent sign is plus. 122. Each term in the numerator and denominator of a fraction has its own particular sign, distinct from the real or apparent sign of the fraction. Now the essential sign of any entire quantity is changed, by changing the signs of all its terms. Hence, I. Changing all the signs of either numerator or denominator, changes the real sign of the fraction; (85, I). II. Changing all the signs of both numerator and denominator, does not alter the real sign of the fraction; (85, II). III. Changing the apparent sign of the fraction, changes the real sign. 123. The Reduction of a fraction is the operation of changing its form without altering its value. 124. To reduce a fraction to its lowest terms. A fraction is in its lowest terms, when the numerator and denominator are prime to each other. And since it does not alter the value of a fraction to suppress the same factor in both numerator and denominator, (119, III), we have the following RULE. I. Resolve the numerator and denominator into their prime factors, and cancel all those factors which are common Or, II. Divide both numerator and denominator by their greatest common divisor. EXAMPLES FOR PRACTICE. a_1 1. Reduce to its lowest terms. 3a-2a-1 2. Reduce to its lowest terms. 4a-2a-3a+1 The greatest common divisor of the numerator and denominator, as found by (105), is a-1; hence, (4a-2a-3a+1)+(a-1)=4a+2a-1 And we have for the reduced fraction, 4a’+2a-1' Reduce each of the following fractions to its lowest terms : 7x'yz 3y? 2-1 4. att x2 62x2 125. To reduce a fraction to an entire or mixed quantity. The division indicated by a fraction may be at least partially performed, when there is any term in the numerator whose literal part is exactly divisible by some term in the denominator. Hence, RULE I. Divide the numerator by the denominator as far as possible, and the quotient will be the entire quantity. II. Write the remainder over the denominator, annex the fraction thus formed to the entire part, with its proper sign, and the whole result will be the mixed quantity. EXAMPLES FOR PRACTICE. Reduce the following fractions to entire or mixed quantities : b 2. a'+ba Ans. at ở ab Ans. 5a+1+ Ans. 3x+5+ x*—4x+8

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https://essayhlp.com/should-i-order-college-probability-theory-essay-business-british-platinum

math

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https://www.helpwithassignment.com/Solution-Library/calculation-of-bond-yield-to-call

math

A bond with the face value of $1000 matures in 12 years and has a 9% semi-annual coupon (That is the bond pays a $45 coupon every six months). The bond has a nominal yield to maturity (YTM) of 7.5% and it can be called in 4 years at a call price of $1,045. What is the bonds yield to call? Total Word Count 23 If you are here for the first time, you can request for a discount coupon, which can knock off upto 20% of the quoted price on any service.

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https://awesomewriters.org/2021/04/01/chapter-11-macroeconomic-equilibrium-in-the-short-run-paper/

math

– Macroeconomic Equilibrium in the Short Run- Output, Employment and InflationPlease answer clearly step by step extensive answers, so i can see what you have done 🙂 Unformatted Attachment Preview Chapter 11: Macroeconomic Equilibrium in the Short Run Question1 In reality, tax revenues are not exogenous, but move positively with income (GDP). Suppose that T = 𝑇̅ + tY, where t is an exogenous tax rate. Use this tax revenue function to derive the government spending multiplier in the following simple model: C = c + c1(Y – T) I = 𝐼̅ G = 𝐺̅ NX = NX0 -mY Question 2 Derive the IS curve for the economy in the above exercise, with I = 𝐼̅– 20i. Plot it in (Y, Chapter 13: Output, Employment and Inflation Question 3 A Philips curve is represented by the following relationship: π = 𝜋̃ – 10(U – 𝑈̅) + s, where s is a supply shock term, on average equal to zero. Draw the Phillips curve when 𝑈̅ = 7% and underlying inflation 𝜋̅ = 3% and 6%. Okun’s Law is U – 𝑈̅ = -0.5 (Y – 𝑌̅)/𝑌̅. Draw the aggregate supply schedule when 𝑌̅ = 10,000. Question 4 Using the conditions of the above exercise, show the impact of supply shocks s = 5%, and s = -2%. Separately show the effect of raising 𝑌̅ to 10,200. Why is it argued that improving the performance of the supply side of the economy is good for both inflation and employment? Our essay writing service fulfills every request with the highest level of urgency.

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http://math.stackexchange.com/questions/211950/how-do-you-imagine-the-shape-of-a-manifold-s2-times-s1

math

In 3-dimensional manifold theory, I have encountered the manifold $S^2 \times S^1$ many times. (The following story can be applied not only this manifold but also for any 3-dimensional manifold.) But I don't have any geometrically or topologically image of the manifold in my head. How do you deal with this difficulties? Is there any good way to imagine the manifold in my head? Since $S^1$ is a union of an interval and a point, I know it is a thick sphere identified the inner boundary with the outer boundary. But Still it is not that clear. Or do you just deal the manifold algebraically without appearing any geometric intuition? I appreciate any help or tip. Thank you in advance.

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https://repositori.udl.cat/handle/10459.1/58029

math

A variant of the McKay-Miller-Siran construction for Mixed Graphs MetadataShow full item record The Degree/Diameter Problem is an extremal problem in graph theory with applications in network design. One of the main research areas in the Degree/Diameter Problem consists of finding large graphs whose order approach the theoretical upper bounds as much as possible. In the case of directed graphs there exist some families that come close to the theoretical upper bound asymptotically. In the case of undirected graphs there also exist some good constructions for specific values of the parameters involved (degree and/or diameter). One such construction was given by McKay, Miller, and ˇSir´aˇn in , which produces large graphs of diameter 2 with the aid of the voltage assignment technique. Here we show how to re-engineer the McKay-Miller-ˇSir´aˇn construction in order to obtain large mixed graphs of diameter 2, i.e. graphs containing both directed arcs and undirected edges. Is part ofElectronic Notes in Discrete Mathematics, 2016, vol. 54, p. 151-156 European research projects Except where otherwise noted, this item's license is described as cc-by-nc-nd (c) Elsevier B.V., 2016 Showing items related by title, author, creator and subject. López Lorenzo, Ignacio; Pérez Rosés, Hebert; Pujolàs Boix, Jordi (Elsevier, 2017-11-20)This paper investigates the upper bounds for the number of vertices in mixed abelian Cayley graphs with given degree and diameter. Additionally, in the case when the undirected degree is equal to one, we give a construction ... López Lorenzo, Ignacio; Pérez Rosés, Hebert; Pujolàs Boix, Jordi (Elsevier B.V., 2016-09-26)We give an upper bound for the number of vertices in mixed abelian Cayley graphs with given degree and diameter. Dalfó, Cristina; Fiol, Miguel Angel; López Lorenzo, Ignacio (Elsevier B.V., 2018)A mixed graph G can contain both (undirected) edges and arcs (directed edges). Here we derive an improved Moore-like bound for the maximum number of vertices of a mixed graph with diameter at least three. Moreover, a ...

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https://nz.education.com/resources/fraction-models/CCSS/

math

Help your students understand fractions while adding some color to these shapes! With this helpful math resource, students will create a visual representation of fractions by coloring the parts of the shape indicated by each fraction. Students have been hard at work on their fractions practice this year, so now it's time to see how far they've come. This end-of-year check-in will help you assess student understanding of simple fractions of wholes. Use this resource to assess your students’ mastery of concepts surrounding fractions. Your mathematicians will write fractions, find equivalent fractions, compare fractions, and plot fractions on a number line. Compare two different ways to use tape diagrams! This lesson discusses fractions and multiplication within tape diagrams. Use this lesson as support for the lesson Illustrating Fractions and Whole Number Products with Tape Models.

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http://mathhelpforum.com/trigonometry/211768-value-x-when-pi-2-x-pi.html

math

find the value of x in the equation 2 sin x - 1 = 0 ,where pi/2 < x < pi the choices are: pi/ 6, pi/3, 5pi/6, 2pi/3 my answer is 2pi/ 3... because it says pi/2 < x < pi because if i translate this to degrees means 90< x < 180. because 2pi/3 is 120 degrees. does my answer correct? please check my work Mathhelpforum ... because im really sure with my answer

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