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http://yzcourseworksunl.blogdasilvana.info/forced-convection-in-a-cross-flow-heat-exchanger-essay.html

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We assume that the cross flow is uniform and that forced convection is the primary mode of heat transfer the rod is heated internally by an electric resistance heater 1pre-lab section: theoretical analysis consider a heated cylinder subjected to a cross flow of air. M bahrami ensc 388 (f09) forced convection heat transfer 1 forced convection heat transfer convection is the mechanism of heat transfer through a fluid in the whereas in forced convection, the fluid is forced to flow over a surface or in a tube by external means such as a pump or fan. Estimate the flow rates of water required to attain the following reynold numbers for flow through the heat exchanger tube 100, 1000, 2000, 5000, 10000, 25000, 30000, 50000 page 6 of 15 experiment 8 – free & forced convection convection heat transfer doc 8 6 procedure caution for safety: 1. 90 numerical simulation of flow and forced convection heat transfer (7) (8) where vξ and vη are the velocity components in ξ and η directions, respectively, p is the pressure, t is the time, t. – a is cross-sectional area for flow – p is wetted perimeter – for a circular pipe where a = pd2/4 and p heat and mass transfer forced free (natural) 24 flow direction • flow direction depends on 08-forced convection twoppt. Force convection is a mechanism of heat transfer in which fluid motion is generated by an external source like a pump, fan, suction device, etc forced convection is often encountered by engineers designing or analyzing pipe flow, flow over a plate, heat exchanger and so on. Consider a heated cylinder subjected to a cross flow of air, as shown in fig 1 we assume that the cross flow is uniform and that forced convection is the primary mode of heat transfer. Forced and natural convection page 3 flow separation renders the analytical modelling of momentum, heat, and mass flows over bodies intractable except in very slender cases (blades and foils), and, although the numerical simulation using. Coupled conduction and convection where a (= wt) is the cross-sectional area of the fin, thus enabling the energy balance on the element to be written as: (3) the plate may be one of many making up an idealized plate heat exchanger, and the flow centerlines may be considered as axes of symmetry across which no energy is transferred. The aim of this lab is to determine the average convective heat transfer coefficient for forced convection of a fluid (air) past a copper tube, which is used as a heat transfer model. In this study, forced convection flow and heat transfer characteristics in semi-circular cross-sectioned micro-channel were studied numerically water, ethylene glycol (eg) and engine oil were used as working fluid. Convection heat transfer reading problems external flow: the flow engulfs the body with which it interacts thermally internal flow: the heat transfer surface surrounds and guides the convective stream forced convection: flow is induced by an external source such as a pump, compressor, fan, etc. The following table charts of typical convective convection heat transfer coefficients for fluids and specific applications. The enhancement of heat transfer characteristics for cross flow heat exchanger with using low integral finned tube has been experimentally studied in this paper. Forced convection heat transfer to incompressible power-law fluids from a heated elliptical cylinder in the steady, laminar cross-flow regime has been studied numerically. State heat transfer laboratory were to study the rates of heat transfer for different materials of varying sizes, to develop an understanding of the concepts of forced and free convection and to determine the heat transfer coefficients for several rods. Mayank vishwakarma et al (2013) an attempt made to decrease the pressure drop and to increase the heat transfer and the ratio of heat transfer and pressure drop in shell and tube type heat exchanger by tilting the baffle angle up to which we get the minimum pressure drop. Laminar fluid flow and heat transfer in an annulus with an externally enhanced inner tube ajay k agrawal department of mechanical engineering, clemson university, clemson, sc, usa university of michigan, dearborn, mi, usa laminar forced convection in a double-pipe heat exchanger is studied numerically in this study, an isothermal tube. 18 5 heat exchangers the general function of a heat exchanger is to transfer heat from one fluid to another the basic component of a heat exchanger can be viewed as a tube with one fluid running through it and another fluid flowing by on the outside. For forced water convection experiment, the testing facility (schematic shown in fig 3b) of prof px jiang j tian et al/international journal of heat and mass transfer 50 (2007) 2521–2536 2523. Convection heat transfer from tube banks in crossflow: analytical approach m m yovanovich, fellow aiaa w a khan j r culham microelectronics heat transfer laboratory department of mechanical engineering forced convection 2 steady, laminar, fully developed and 2-d flow 3 incompressible fluid with constant properties. In forced convection, the fluid is forced to flow by an external factor - eg wind in the atmosphere, a fan blowing air, water being pumped through a pipe typically heat transfer under forced convection conditions is higher than natural convection for the same fluid. Forced convection (in a cross flow heat exchanger) the aim of this lab is to determine the average convective heat transfer coefficient for forced convection of a fluid (air) past a copper tube, which is used as a heat transfer model. The heat transfer coefficient for convection is denoted by (h) and is measured in w/m^2k, this lab delves into the application of convection heat transfer and how it correlates to temperature, velocity, ect of the fluid in question.

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http://www.jiskha.com/display.cgi?id=1384148634

math

Are you sure it's 81.5 N on the surface of the moon? #1. Anyway, what we'll do is first get the mass of the man. Note that mass is constant, and in physics, it is different from the term 'weight' (though in other cases, they are interchangeable, like in chemistry). W = mg W = weight (in Newton) m = mass (in kg) g = acceleration due to gravity (on Earth) = 9.8 m/s^2 41 = m * 9.8 m = 4.18 kg On Moon, we'll solve for acceleration due to gravity of moon: 81.5 = 4.18 * g,moon g,moon = 19.5 m/s^2 How much lighter is he on the moon? If it means the mass, then his mass is constant and he is neither heavier nor lighter on the moon. #2. Given his weight on the moon, solve for the mass using the g,moon. 90 = m * 19.5 m = 4.62 kg Thus, his weight on the earth is W = 4.62 * 9.8 W = 45.3 N Hope this helps :3 Thank you so much Jai. You explained it very well:) Answer this Question Science - If Frankie the kangaroo climbed out of the crate and pushed the crate ... Physics - An 80 kg man is riding in an elevator from the first floor to the 34th... PHYSICS!!!! - An 80 kg man is riding in an elevator from the first floor to the ... physics....need help! - An 80 kg man is riding in an elevator from the first ... Calculus - Consider the transformation T:x=(41/40)u−(9/41)v , y=(9/41)u+(... PHYSICS - PLEASE HELP - The density of a 6 foot tall man that weighs 840. ... Science (simple machines) - The plank is 3.6 metres long. It is being used as a ... algebra - A man (we'll refer to him as man 1) starts from his home at 8 A.M. and... physics - If Frankie the kangaroo climbed out of the crate and pushed the crate ... Physics - A man lifts a 1.0 kg stone vertically with his hand at a constant ...

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https://pt.planetcalc.com/8401/

math

Explicit Runge–Kutta methods This online calculator implements several explicit Runge-Kutta methods so you can compare how they solve first degree differential equation with a given initial value. Artigos que descrevem esta calculadora Digits after the decimal point: 6 The file is very large. Browser slowdown may occur during loading and creation. Calculadora utilizadas por esta calculadora URL copiado para a área de transferência #differentiation #Runge-Kutta 3/8-rule fourth-order method Classic fourth-order method differentiation Explicit midpoint method Forward Euler Heun's method Heun's third-order method Kutta's third-order method Math Numerical differential equations Ralston's fourth-order method Ralston's method Ralston's third-order method Third-order Strong Stability Preserving Runge-Kutta (SSPRK3) PLANETCALC, Explicit Runge–Kutta methods

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http://slideplayer.com/slide/2433556/

math

Presentation on theme: "AXONOMETRIC PROJECTION"— Presentation transcript: 1 AXONOMETRIC PROJECTION C H A P T E R F I F T E E N 2 OBJECTIVES1. Sketch examples of an isometric cube, a dimetric cube, and atrimetric cube.2. Create an isometric drawing given a multiview drawing.3. Use the isometric axes to locate drawing points.4. Draw inclined and oblique surfaces in isometric.5. Use projection to create an axonometric drawing.6. Use offset measurements to show complex shapes in obliquedrawings.7. Add dimensions to oblique drawings.8. Describe why CAD software does not automatically createoblique drawings. 3 Axonometric projections Axonometric projections show all three principal dimensions using a singledrawing view, approximately as they appear to an observer. Pictorial drawings are also useful in developing design concepts. They can help you picture the relationships between design elements and quickly generate several solutions to a design problem.Axonometric projection(isometric shown)These projections are often called pictorial drawings because they look more like a picture than multiview drawings do. Because a pictorial drawing shows only the appearance of an object, it is not usually suitable for completely describing and dimensioning complex or detailed forms. 4 Projection Methods Reviewed The four principal types of projection 5 Types of Axonometric Projection Isometric projectionDimetric projectionThe degree of foreshortening of any line depends on its angle to the plane of projection. The greater the angle, the greaterthe foreshortening. If the degree of foreshortening is determined for each of the three edges of the cube that meet at one corner, scales can be easily constructed for measuring along these edges or any other edges parallel to themTrimetric projection 6 DIMETRIC PROJECTIONA dimetric projection is an axonometric projection of an object in which two of its axes make equal angles with the plane of projection, and the third axis makes either a smaller or a greater angle. 7 APPROXIMATE DIMETRIC DRAWING Approximate dimetric drawings, which closely resemble true dimetrics, can be constructed by substituting for the true angles.The resulting drawings will be accurate enough for all practical purposes. 8 TRIMETRIC PROJECTIONA trimetric projection is an axonometric projection of an object oriented so that no two axes make equal angles with the plane of projection.Because the three axes are foreshortened differently, each axiswill use measurement proportions different from the other two.Ellipses in Trimetric. (Method (b) courtesy of Professor H. E. Grant.) 9 TRIMETRIC ELLIPSESThe trimetric centerlines of a hole, or the end of a cylinder, become the conjugate diameters of an ellipse when drawn in trimetric.In constructions where the enclosing parallelogram for an ellipse is available or easily constructed, the major and minoraxes can be determined(Method (b) courtesy ofProfessor H. E. Grant.)(b)When you are creating a trimetric sketch of an ellipse, it works great to block in the trimetric rectangle that would enclose the ellipse and sketch the ellipse tangent to the midpoints of the rectangle. 10 AXONOMETRIC PROJECTION USING INTERSECTIONS Note that if the three orthographic projections, or in most cases any two of them, are given in their relative positions, the directions of the projections could be reversed so that the intersections of the projecting lines would determine the axonometric projection needed. 11 Use of an Enclosing Box to Create an Isometric Sketch using Intersections To draw an axonometric projection using intersections, it helps to make a sketch of the desired general appearance of the projection.Even for complex objects thesketch need not be complete, just an enclosing box. 12 COMPUTER GRAPHICSPictorial drawings of all sorts can be created using 3D CAD.The advantage of 3D CADis that once you makea 3D model of a partor assembly, you canchange the viewing directionat any time for orthographic,isometric, or perspective views.You can also apply different materials to the drawing objects and shade them to produce a high degree of realism in the pictorial view.(Courtesy ofRobert Kincaid.)(Courtesy of PTC) 13 OBLIQUE PROJECTIONSIn oblique projections, the projectors are parallel to each other but are not perpendicular to the plane of projection. 14 Directions of Projectors The directions of the projections BO, CO, DO, and so on, are independent of the angles the projectors make with the plane of projection. 15 ELLIPSES FOR OBLIQUE DRAWINGS It is not always possible to orient the view of an object so that all its rounded shapes are parallel to the plane of projection.Both cannot be simultaneously placed parallel to the plane of projection, so in the oblique projection, one of them must be viewed as an ellipse. 16 Alternative Four-Center Ellipses Normal four-center ellipses can be made only in equilateral parallelogram, so they cannot be used in an oblique drawing where the receding axis is foreshortened. Instead, use this alternative four-center ellipse to approximate ellipses in oblique drawings.Draw the ellipse on two centerlines. This is the same method as is sometimes used in isometric drawings, but in oblique drawings it appears slightly different according to the different angles of the receding lines… 17 OFFSET MEASUREMENTSCircles, circular arcs, and other curved or irregular lines can be drawn using offset measurements.Draw the offsets on the multiview drawing of the curve and then transfer them to the oblique drawing… 18 OBLIQUE DIMENSIONINGYou can dimension oblique drawings in a way similar to that used for isometric drawings.For the preferred unidirectional system of dimensioning, all dimension figures are horizontal and read from the bottom of the drawing. Use vertical lettering for all pictorial dimensioning. 19 COMPUTER GRAPHICSUsing CAD you can easily create oblique drawings by using a snap increment and drawing in much the same way as on grid paper. If necessary, adjust for the desired amount of foreshortening along the receding axis as well as the preferred direction of the axis.(Autodesk screen shots reprinted with the permission of Autodesk, Inc.) 20 PERSPECTIVE DRAWINGSC H A P T E R S I X T E E N 21 OBJECTIVES 1. Identify a drawing created using perspective projection. 2. List the differences between perspective projection andaxonometric projection.3. Create a drawing using multiview perspective.4. Describe three types of perspective5. Measure distances in perspective projection. 22 UNDERSTANDING PERSPECTIVES A perspective drawing involves four main elements:• The observer’s eye• The object being viewed• The plane of projection• The projectors from the observer’s eye to all points on the object 23 Rules to Learn for Perspective The following are some rules to learn for perspective:• All parallel lines that are not parallel to the picture plane vanish at a point.• If these lines are parallel to the ground, the vanishing point will beon the horizon.• Lines that are parallel to the picture plane, such as the vertical axis ofeach lamppost, remain parallel to one another and do not converge towarda vanishing point. 24 PERSPECTIVE FROM A MULTIVIEW PROJECTION It is possible to draw a perspective from a multiview projection,The upper portion of thedrawing shows the top view of the station point, the picture plane, the object, and the visual rays. The right-side viewshows the same station point, picture plane, object, and visual rays.In the front view, the picture plane coincides with the plane of the paper, and the perspective view is drawn on it. 25 NONROTATED SIDE VIEW METHOD FOR PERSPECTIVE The upper portion ofthe drawing shows the top views of the station point, picture plane, and the object. The lines SP-1, SP-2, SP-3, and SP-4are the top views of the visual rays.The perspective view is drawn on the picture plane where the front view would usually be located.The perspective view shows the intersectionof the ground plane with the picture plane. 26 POSITION OF THE STATION POINT The centerline of the cone of visual rays should be directed toward the approximatecenter, or center of interest, of the object.In two-point perspective,locating the station point (SP) in the plan view slightly to the left and not directly in front of the center of the object produces a better view, as if the object is seen at aglance without turning the head.The station point (SP) does not appear in the perspective view because the station point is in front of the picture plane. 27 LOCATION OF THE PICTURE PLANE The perspectives differ in size but not in proportion. The farther the plane is from the object, the smaller the perspective drawing will be. This distance controls the scale of the perspective. 28 BIRD’S-EYE VIEW OR WORM’S-EYE VIEW The horizon is level with the observer’s eye, so controlling the location for the horizon line controls whether the perspective view appears from above or below the object. The horizon line is defined by the observer’s point of view.To produce a perspective view that shows the objects as though viewed from above, place the object below the horizonline. To produce a perspective viewthat shows the object as thoughviewed from below, place the objectabove the horizon line. 29 ONE-POINT PERSPECTIVE In one-point perspective, orient the object so two sets of itsprincipal edges are parallel to the picture plane (essentially aflat surface parallel to the picture plane) and the third set is perpendicular to the picture plane. This third set of parallellines will converge toward a single vanishing point inperspective. 30 ONE-POINT PERSPECTIVE OF A CYLINDRICAL SHAPE A one point perspective representing a cylindrical machine part.The front surface of the cylinder isplaced in the picture plane. All circular shapes are parallel to the picture plane, and they project as circles and circular arcs in the perspective. The station point (SP) is located in front and to one side of the object. The horizon is placed above the ground line. The single vanishing point is on the horizon in the center of vision. 31 TWO-POINT PERSPECTIVE In two-point perspective, the object is oriented so that one set of parallel edges is vertical and has no vanishing point, whereas the two other sets have vanishing points. Two-point perspectives are often used to show buildings in an architectural drawing, or large structures in civil engineering, such as dams or bridges, especially for client presentation drawings.When multiview drawings are already available, tape their top (plan) and side (elevation) views in position, and usethem to construct the perspective. When you are finished, remove the taped portions. 32 THREE-POINT PERSPECTIVE In three point perspective, the object is placed so that none of its principal faces or edges are parallel to the picture plane. This means that each set of three parallel edges will have a separate vanishing point. The picture plane is approximately perpendicular to the centerlineof the cone of visual rays.Remember that to find the vanishingpoint of a line in any type of perspectiveyou draw a visual ray, or line, from thestation point parallel to that edge of theobject, then find the piercing point of thisray in the picture plane. 33 DIRECT MEASUREMENTS ALONG INCLINED LINES The method of direct measurements may also be applied to lines inclined to the picture plane (PP) and to the ground plane.Line XE, which pierces the picture plane (PP) at X. If you revolve the end of the house about a vertical axis XO into the picture plane (PP), line XE will be shown true length and tipped as shown at XY. This line XY may be used as themeasuring line for XE. Next find the corresponding measuring point MP. The line YE is the horizontal base of an isosceles triangle having its vertex at X, and a line drawn parallel to it through SP will determine MP 34 VANISHING POINTS OF INCLINED LINES To find the vanishing point of an inclined line, determine thepiercing point in the picture plane (PP) of a line drawn from the station point (SP) parallel to the given line. 35 CURVES AND CIRCLES IN PERSPECTIVE If a circle is parallel to the picture plane, its perspective is a circle. If the circle is inclined to the picture plane, its perspective drawing may be any one of the conic sections where the base of the cone is the givencircle, the vertex is the station point (SP),and the cutting plane is the picture plane (PP).The centerline of the cone of rays is usually approximately perpendicular to the picture plane, so the perspective will usually be an ellipse.A convenient method for determining the perspective of any planar curve… 36 THE PERSPECTIVE PLAN METHOD You can draw a perspective by first drawing the perspective ofthe plan of the object, then adding the vertical lines, and finally adding the connecting lines. 37 SHADINGShading pictorial drawings can be very effective in describing the shapes of objects in display drawings, patent drawings, and other pictorial drawings. Ordinary working drawings are not shaded.Methods of Shading 38 PERSPECTIVE VIEWS IN AUTOCAD AutoCAD software uses an interactive command calledDview (dynamic viewing) that you can use to show 3Dmodels and drawings in perspective. The Dview command uses a camera and target to create parallel and perspective views. You can use the camera option to select a new cameraposition with respect to the target point at which thecamera is aimed. The Dview distance option is used to create a perspective viewA Perspective View Created Using the Dview Command in AutoCAD. (Autodesk screen shots reprinted with the permission of Autodesk, Inc.)

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https://altenew.com/collections/kits-classes/class-theme_interactive-cards

math

Customer ReviewsWrite a review I'm only on day 3 of this class, but it is very good so far. Ashley Tucker is the teacher and in her videos she gives very clear, easy to understand directions and very complete PDFs. There are a total of 7 lessons and the price is very reasonable. I'm looking forward to the next 4. this class has many ways to make movement on cards, so far the ideas have been great

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http://blog.reino.co.jp/index.php/ebooks/cardinal-arithmetic

math

By Saharon Shelah Is the continuum speculation nonetheless open? If we interpret it as discovering the legislation of cardinal mathematics (or exponentiation, considering the fact that addition and multiplication have been classically solved), the speculation will be solved through the independence result of Godel, Cohen, and Easton, with a few remoted confident effects (like Gavin-Hajnal). so much mathematicians anticipate that purely extra independence effects stay to be proved. In Cardinal mathematics, in spite of the fact that, Saharon Shelah bargains an alternate view. by means of redefining the speculation, he will get new effects for the traditional cardinal mathematics, unearths new functions, extends older equipment utilizing general filters, and proves the lifestyles of Jonsson algebra. Researchers in set thought and similar components of mathematical common sense probably want to learn this provocative new method of an incredible subject. Read Online or Download Cardinal arithmetic PDF Best popular & elementary books Even though the hugely expected petascale pcs of the close to destiny will practice at an order of value swifter than today’s fastest supercomputer, the scaling up of algorithms and functions for this category of desktops is still a tricky problem. From scalable set of rules layout for enormous concurrency toperformance analyses and medical visualization, Petascale Computing: Algorithms and functions captures the cutting-edge in high-performance computing algorithms and functions. With an analogous layout and have units because the marketplace major Precalculus, 8/e, this concise textual content offers either scholars and teachers with sound, regularly established factors of the mathematical techniques. PRECALCULUS: A CONCISE direction is designed to supply an economical, one-semester replacement to the normal two-semester precalculus textual content. Algebra and Trigonometry Atomic correlations were studied in physics for over 50 years and often called collective results till lately once they got here to be famous as a resource of entanglement. this can be the 1st booklet that includes special and finished research of 2 presently generally studied matters of atomic and quantum physics―atomic correlations and their relatives to entanglement among atoms or atomic systems―along with the latest advancements in those fields. - On The Foundations of Combinatorial Theory: Combinatorial Geometries - Mathematical Inequalities - A Treatise on Solid Geometry - Schaum's outline of theory and problems of precalculus Extra info for Cardinal arithmetic D) Why should it cost $50 to serve zero cups of coffee? (b) Plot P against x. For what x-values is the graph of P below the x-axis? Above the x-axis? Interpret your results. (c) Interpret the slope and both intercepts of your graph in practical terms. 21. Owners of an inactive quarry in Australia have decided to resume production. They estimate that it will cost them $1000 per month to maintain and insure their equipment and that monthly salaries will be $3000. It costs $80 to mine a ton of rocks. 14 World 23. Make two tables, one comparing the radius of a circle to its area, the other comparing the radius of a circle to its circumference. Repeat parts (a), (b), and (c) from Problem 22, this time comparing radius with circumference, and radius with area. 24. Sri Lanka is an island that experienced approximately linear population growth from 1950 to 2000. 27 gives the population of these two countries, in millions. Which of these two countries is A and which is B? Explain. (b) What is the approximate rate of change of the linear function? What do the y-intercepts of the functions in Example 2 say about oxygen consumption? Often the y-intercept of a function is a starting value. e. x = 0). Since a person running on a treadmill must have a pulse, in this case it makes no sense to interpret the y-intercept this way. The formula for oxygen consumption is useful only for realistic values of the pulse. 3. Consider the following example. Example 3 We have $24 to spend on soda and chips for a party. A six-pack of soda costs $3 and a bag of chips costs $2.

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CC-MAIN-2021-04

https://www.physicsforums.com/threads/angular-velocity-of-gears.713244/

math

Gear C has radius 0.12 m and rotates with angular velocity 1.9 k rad/s. The connecting link rotates at angular velocity 1.5 k rad/s. Gear D has radius 0.05 m. Find the angular velocity of gear D (in rad/s). Note that gear C is pinned to ground and gear D is a planetary gear. Since gear c is pinned, the gear ratio ωc/ωd=rd/rc does not directly apply here right? I wonder if the correct solution would be ωd=ωc x rcd / rd. I got -2.125 rad/s. Thanks.

s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591578.1/warc/CC-MAIN-20180720100114-20180720120114-00171.warc.gz

CC-MAIN-2018-30

https://aviation.stackexchange.com/questions/39690/what-is-the-meaning-of-the-percentage-used-to-describe-the-thinness-of-this-glid

math

In a glider review for the Schempp-Hirth Discus, I encountered the following paragraph: The Discus 2 airfoil is thin (about 14.5 percent), incorporating studies by K.H. Horstmann and Dr. Wuerz (wing) and Luc Boermans (tail). After researching this a bit, I've been unable to find an explanation for what "14.5 percent" was referring to. I thought perhaps that it is the ratio (14.5/100) of the thickness of the wing to the width of the wing, but that is really just an educated guess. The Wikipedia page on airfoils has a section entitled Thin airfoil theory, with two bullets which also might qualify: (1) on a symmetric airfoil, the center of pressure and aerodynamic center lies exactly one quarter of the chord behind the leading edge (2) on a cambered airfoil, the aerodynamic center lies exactly one quarter of the chord behind the leading edge Is this "one quarter" the ratio that is mentioned in the article? I did see this post, but had trouble seeing how the 90% and 99% numbers mentioned in the answer are related.

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https://www.chemicalforums.com/index.php?topic=105493.msg372054

math

0 Members and 1 Guest are viewing this topic. That would mean all earlier dilutions are wrong as well, wouldn't it? But if that's the case then the last dilution is 1 in 5 relative to the penultimate dilution. No, I think you are misunderstanding something here. These dilutions are cumulative, each next one uses the previous solution.Quote from: Tyyyyuuu on September 28, 2020, 05:17:25 PMBut if that's the case then the last dilution is 1 in 5 relative to the penultimate dilution.No, it is 1.25 (240/192). 048 + 640 1 in 5 (not 1 in 10) Quote from: Tyyyyuuu on September 28, 2020, 06:28:09 PM048 + 640 1 in 5 (not 1 in 10)Why 640 and not 192? You have changed the rules for this one line. Your results are inconsistent. Show how you got 1 in 5 in the last line and explain how you got correct results for all previous lines. Page created in 0.071 seconds with 21 queries.

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https://www.authorea.com/doi/full/10.22541/au.170361383.35302659

math

A Novel Variant of Arithmetic Coding Using the Stern-Brocot Tree and Farey Addition AbstractIn arithmetic coding, the input data is mapped to a small subinterval within the interval [0, 1). In the conventional implementation, a rational number is selected from this, which can be represented as a fraction with a power of two in the denominator. Only the numerator of this fraction is saved in the compressed file because the power of the denominator corresponds to the number of bits in the numerator and can therefore be reconstructed from it. This paper introduces a new variant that instead uses the Stern-Brocot tree and the Farey addition to find the rational number that has the smallest possible denominator within the same interval. Now the denominator must be stored in the same way as the numerator, but the numerator and denominator together have only about as many binary digits as the number obtained using the conventional method. In around one in four cases, the result of the new approach is even shorter than the conventional binary number. The results show that this new method is not only interesting for theoretical reasons, but even has the potential to outperform established methods.

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https://www.socsci.uci.edu/newsevents/news/2013/2013-09-19-rin-speaks-at-two-international-conferences.php

math

Rin speaks at two international conferences - September 19, 2013 - Papers discuss theory of logic and computation Seventh year graduate student Ben Rin presented two papers at European conferences on logic and computability this summer. The first paper, called "On Set-Theoretic and Transfinite Analogues of Epistemic Arithmetic and Flagg Consistency," was given at a conference called "The Nature of Computation: Logic, Algorithms, Applications," hosted by the University of Milano-Bicocca from the 1st to the 5th of July. This paper addressed how ideas concerning epistemic arithmetic -- an attempt to use traditional arithmetic to model an alternative approach to logic -- can be extended to other domains of mathematics and logic. The second paper was presented at a workshop on the History and Philosophy of Infinity, hosted by Corpus Christi College at Cambridge University, from the 20th to the 23rd of July. This paper, called "Infinity and Recursion," discussed how ideas in the theory of computation extend to infinite sets.

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http://non-traditional-students.blogspot.com/2013/04/a-new-nontraditional-student.html

math

Hello, nontraditional students and everyone who helps them out. Here are the ten most popular postings here for this week. #1. Finding Scholarships and Grants #2. 10 Advantages of Face to Face Classes #3. The Writing Center - do you have one? #4. 10 Tips for Doing Well on Final Exams #5. Are you in a Nontrad Student Group? #6. Ivy League Schools are Beckoning to Nontraditional Students #7. Introduction to Law School for Nontraditional Students #8. Nontraditional Students Tips and Links #9. Final Project Excitement - Are you Excited? This can help. #10. Nontraditional Students in the Dorm - - Really?? AND here is a brand-new scholarship just for nontraditional students. I wrote about it on my Find Scholarships and Grants blog today. Here is that link too: From the Find Scholarships and Grants Blog: A New Scholarship for Nontraditional Students. Good luck on finals, and have a GREAT summer. Share your nontraditional journey as a comment here. Later! And here are more Nontrad links: The Nontrad site and blog Join Nontrads on Facebook Nontrads on Yahoo Nontrads on Twitter

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CC-MAIN-2017-17

https://slideplayer.com/slide/5928452/

math

Sound waves are made up of a series of regions of high and low pressure. The areas of high pressure are called compressions. The areas of low pressure are called rarefactions. Wave energy travels through the material. But a particular piece of material, only moves back a forth a short distance. Speed of Sound The speed of sound in the air is dependent on the amount of water vapor and temperature of the air. At normal room temperature, the speed of sound is about 343 m/s. As a general rule, if we can count the number of seconds between seeing and hearing a lightning bolt, we can approximate the distance to it. Every 5 seconds is equal to approximately 1 mile. Is this an accurate approximation? Sound Intensity Intensity = Sound Power/Area [W/m 2 ] The sound intensity drops as a function of the inverse square of the distance from the sound source. Compare the sound intensity at point 1 and 2 of a 12x10 -5 W sound traveling out of the speaker. (A 1 = 4 m 2 and A 2 = 12 m 2 ) Loudness The decibel is a measurement used to compare two sound intensities. I = 1x10 -12 W/m 2 is equal to 0 dB and considered the threshold for normal hearing. An increase of 10 dB roughly doubles the relative sound intensity. SoundW/m 2 dB Threshold of hearing10 -12 0 Whisper10 -10 20 Normal conversation3.2 x 10 -6 65 Garbage truck10 -3 90 Rock concert1120 Threshold of pain10130 Human Hearing A high pitched noise has a high vibration frequency, whereas a low pitched noise has a low vibration frequency. Human hearing ranges from 20 Hz to 20,000 Hz. Infrasonic is below this range. (earthquakes) Ultrasonic is above this range. (dogs) Natural Frequency All objects vibrate when they dropped or tapped against a surface. This natural frequency of an object depends on its elasticity and the shape of the object. When a forced vibration matches the frequency of an object a dramatic increase in amplitude occurs, this is called resonance. Doppler Effect Austrian physicist Christian Doppler identified frequency-shifting phenomena in 1842, and the “Doppler effect” was coined. In essence, a wave frequency, whether transverse or longitudinal, increases as a source and listener move toward one another. Similarly, the frequency decreases as a source and listener move away from one another.

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CC-MAIN-2019-51

https://www.brainkart.com/article/Analysis-and-Design-of-Sequential-Circuits_13572/

math

Analysis of Sequential Circuits The behaviour of a sequential circuit is determined from the inputs, the outputs and the states of its flip-flops. Both the output and the next state are a function of the inputs and the present Derive the state table and state diagram for the sequential circuit shown in Figure 7. STEP 1: First we derive the Boolean expressions for the inputs of each flip-flops in the schematic, in terms of external input Cnt and the flip-flop outputs Q1 and Q0. Since there are two D flip-flops in this example, we derive two expressions for D1 and D0: These Boolean expressions are called excitation equations since they represent the inputs to the flip-flops of the sequential circuit in the next clock cycle. STEP 2: Derive the next-state equations by converting these excitation equations into flip-flop characteristic equations. In the case of D flip-flops, Q(next) = D. Therefore the next state equal the excitation equations. Q0(next) = D0 = Cnt'*Q0 Q1(next) = D1 = Cnt'*Q1 + Cnt*Q1'*Q0 + Cnt*Q1*Q0' STEP 3: Now convert these next-state equations into tabular form called the next-state table. Each row is corresponding to a state of the sequential circuit and each column represents one set of input values. Since we have two flip-flops, the number of possible states is four - that is, Q1Q0 can be equal to 00, 01, 10, or 11. These are present states as shown in the For the next state part of the table, each entry defines the value of the sequential circuit in the next clock cycle after the rising edge of the Clk. Since this value depends on the present state and the value of the input signals, the next state table will contain one column for each assignment of binary values to the input signals. In this example, since there is only one input signal, Cnt, the next-state table shown has only two columns, corresponding to Cnt = 0 and Cnt = 1. Note that each entry in the next-state table indicates the values of the flip-flops in the next state if their value in the present state is in the row header and the input values in the column header. Each of these next-state values has been computed from the next-state equations in STEP 2. STEP 4: The state diagram is generated directly from the next-state table, shown in Figure 8. Each arc is labelled with the values of the input signals that cause the transition from the present state (the source of the arc) to the next state (the destination of the arc). In general, the number of states in a next-state table or a state diagram will equal 2m , where m is the number of flip-flops. Similarly, the number of arcs will equal 2m x 2k , where k is the number of binary input signals. Therefore, in the state diagram, there must be four states and eight transitions. Following these transition arcs, we can see that as long as Cnt = 1, the sequential circuit goes through the states in the following sequence: 0, 1, 2, 3, 0, 1, 2, .... On the other hand, when Cnt = 0, the circuit stays in its present state until Cnt changes to 1, at which the Since this sequence is characteristic of modulo-4 counting, we can conclude that the sequential circuit in Figure 7 is a modulo-4 counter with one control signal, Cnt, which enables counting when Cnt = 1 and disables it when Cnt = 0. To see how the states changes corresponding to the input signals Cnt, click on this image. Below, we show a timing diagram, representing four clock cycles, which enables us to observe the behaviour of the counter in In this timing diagram we have assumed that Cnt is asserted in clock cycle 0 at t0 and is disasserted in clock cycle 3 at time t4. We have also assumed that the counter is in state Q1Q0 = 00 in the clock cycle 0. Note that on the clock's rising edge, at t1, the counter will go to state Q1Q0 = 01 with a slight propagation delay; in cycle 2, after t2, to Q1Q0 = 10; and in cycle 3, after t3 to Q1Q0 = 11. Since Cnt becomes 0 at t4, we know that the counter will stay in state Q1Q0 = 11 in the next clock cycle. To see the timing behaviour of the circuit click on this image In Example 1.1 we demonstrated the analysis of a sequential circuit that has no outputs by developing a next-state table and state diagram which describes only the states and the transitions from one state to the next. In the next example we complicate our analysis by adding output signals, which means that we have to upgrade the next-state table and the state diagram to identify the value of output signals in each state. The input combinational logic in Figure 10 is the same as in Example 1.1, so the excitation and the next-state equations will be the same as in Example 1.1. In addition, however, we have computed the Output equation: Y = Q1Q0 As this equation shows, the output Y will equal to 1 when the counter is in state Q1Q0 = 11, and it will stay 1 as long as the counter stays in that state. Next-state and output table: Note that the counter will reach the state Q1Q0 = 11 only in the third clock cycle, so the output Y will equal 1 after Q0 changes to 1. Since counting is disabled in the third clock cycle, the counter will stay in the state Q1Q0 = 11 and Y will stay asserted in all succeeding clock cycles until counting is enabled again. Design of Sequential Circuits The design of a synchronous sequential circuit starts from a set of specifications and culminates in a logic diagram or a list of Boolean functions from which a logic diagram can be obtained. In contrast to a combinational logic, which is fully specified by a truth table, a sequential circuit requires a state table for its specification. The first step in the design of sequential circuits is to obtain a state table or an equivalence representation, such as a state diagram. A synchronous sequential circuit is made up of flip-flops and combinational gates. The design of the circuit consists of choosing the flip-flops and then finding the combinational structure which, together with the flip-flops, produces a circuit that fulfils the required specifications. The number of flip-flops is determined from the number of states needed in the The recommended steps for the design of sequential circuits are set out below.

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CC-MAIN-2024-10

https://www.groundai.com/project/paramagnetic-filaments-in-a-fast-precessing-field-planar-versus-helical-conformations/

math

Paramagnetic filaments in a fast precessing field: Planar versus helical conformations We examine analytically equilibrium conformations of elastic chains of paramagnetic beads in the presence of a precessing magnetic field. Conformations of these filaments are determined by minimizing their total energy, given in the harmonic approximation by the sum of the bending energy, quadratic in its curvature, and the magnetic dipolar interaction energy, quadratic in the projection of the vector tangent to the filament onto the precession axis. In particular, we analyze two families of open filaments with their ends aligned along the precession axis and described by segments of planar curves and helices. These configurations are characterized in terms of two parameters encoding their features such as their length, separation between their ends, as well as their bending and magnetic moduli, the latter being proportional to the magnitude and precession angle of the magnetic field. Based on energetic arguments, we present the set of parameter values for which each of these families of curves is probable to occur. pacs:87.16.Ka, 75.75.-c, 87.15.hp Flexible magnetic filaments can be synthesized by joining ferromagnetic or superparamagnetic beads with elastic linkers Wang et al. (2011); Tierno (2014). The combination of their elastic and magnetic properties gives rise to interesting phenomena, which have been a subject of active research since their introduction more than a decade ago Biswal and Gast (2003); Goubault et al. (2003). The mechanical properties of magnetic filaments have been extensively characterized. Their Young and bending moduli have been measured directly in bending and compression experiments performed with optical traps Biswal and Gast (2003) and indirectly from measurements of other quantities such as the separation between beads using Bragg diffraction Goubault et al. (2003), or more recently, via their thermal fluctuations Gerbal and Wang (2017). It has been found that the sole magnetic field can drive an Euler buckling instability in a free filament Cēbers (2005a), whose critical value has been determined theoretically and measured experimentally with a good agreement Cēbers and Javaitis (2004a); Gerbal et al. (2015). This kind of filaments exhibit diverse morphological features Cēbers and Cīrulis (2007), for instance, depending on their length, bending rigidity and magnetic field strength, they may adopt and shapes Goubault et al. (2003); Cēbers (2003) or configurations with more undulations Huang et al. (2016). Configurations of anchored superparamagnetic filaments with a free end may fold into loops, sheets and pillars for different combinations of their bending rigidity and the strength of the magnetic field Wei et al. (2016). In a precessing magnetic field, filaments with loads at their ends can adopt planar or helical configurations by changing the precessing angle and with a time-dependent precession free filaments can assemble into gels of diverse conformations Dempster et al. (2017). Due to their magnetic features, they have inspired the development of several applications. They can be used as micro-mechanical sensors used in the determination of force-extension laws at the micro-scale Biswal and Gast (2003); Goubault et al. (2003); Koenig et al. (2005). Furthermore, since they possess the interesting feature that their stiffness is tunable Cēbers (2005b), and conformational changes can be controlled through the magnetic field Cēbers and Cīrulis (2007); Huang et al. (2016) or the temperature Cerda et al. (2016), they can also be used as actuators Dempster et al. (2017) or grabbers Martinez-Pedrero et al. (2016). Their dynamics have also been studied in detail. In a magnetic field rotating on a plane, free filaments rotate rigidly synchronously or asynchronously depending on their length and whether the precession frequency is smaller or bigger than a critical value Biswal and Gast (2004); Cēbers and Javaitis (2004b). In the presence of an alternating magnetic field, magnetic filaments oscillate and displace, so they have been used to design self-propelled swimmers of controllable velocity and displacement direction Dreyfus et al. (2005); Belovs and Cēbers (2006); Roper et al. (2008); Belovs and Cēbers (2013). Further information about the properties and applications of these magnetic filaments can be found in the reviews Cēbers (2005b); Cēbers and Erglis (2016). In general, most of the previous works consider magnetic filaments with one or both free ends and in precessing fields at a fixed angle, typically precessing on a plane. In this paper we examine open superparamagnetic filaments in a magnetic field precessing at different constant angles, and with their ends held along the precession axis at a certain distance. Molecular dynamics simulations suggest that under these conditions filaments exhibit different behavior depending on the value of the precessing angle relative to a critical angle: if it is smaller, filaments bend but remain on a plane, whereas if it is larger, filaments explore the ambient space adopting helical structures Dempster et al. (2017). Here we present a detailed analytic description of these two families of filaments. We determine their equilibrium configurations by minimizing their ascribed total energy, which at quadratic order has two contributions, the bending energy and the magnetic energy due to dipolar interactions between the beads Cēbers (2003); Goubault et al. (2003). In principle, the behavior of these magnetic filaments depends on their intrinsic properties: length and bending modulus; as well as of their extrinsic properties: separation between their ends and magnetic modulus, which depends on the magnetic field parameters (magnitude and precession angle) and can be positive, negative or even vanish. However, it is possible to characterize their equilibrium configurations in terms of just two parameters capturing all of their characteristics: the boundary separation and the ratio of the magnetic to bending moduli, both scaled with powers of the total length of the filament so as to adimensionalize them. We discuss the forces required to hold the boundaries of the filaments and the behavior of their total energy as a function of these two parameters. Although both families are critical points of the total energy regardless of the precession angle, by comparing their total energies we investigate their plausibility in each precession regime, determining the parameter values for which each family is more likely to take place. This paper is organized as follows. We begin in Sec. II with the framework that we employ to describe superparamagnetic filaments, to this end we define their energy and we express the corresponding Euler-Lagrange (EL) equations, that their equilibrium configurations must satisfy, in terms of the stresses on the filaments. In Sect. III we specialize this framework to the case of planar curves, which in Sec. IV is applied to examine the family of vertical planar filaments (their ends are fixed and aligned with the precession axis) in the perturbative and non-linear regimes. In Sec. V we do the respective analysis of the family of helices. In Sec. VI we compare the total energy of both families of filaments with same parameters to assess their possible physical realization. We close with our conclusions and discussion of future work in Sec. VII. Some derivations and calculations used or discussed in the main text are presented in the appendices. Ii Energy and stresses The magnetic filament is described by the curve , parametrized by arc length in three-dimensional Euclidean space and passing trough the center of the beads. Geometric quantities of the curve are expressed in terms of the Frenet-Serret (FS) frame adapted to the curve, denoted by , see Fig. 1. The rotation of the FS frame along the curve is given by the FS formula where is the Darboux vector; and are the FS curvature and torsion, quantifying how the curve bends in the osculating and normal planes, respectively Kreyszig (1991). We consider paramagnetic filaments in the presence of a magnetic field precessing at an angle about a direction we choose as the axis, see Fig. 2(a). The total energy density of the paramagnetic filament is the sum of the bending and magnetic energies ,111We disregard the weight of the filament so we do not include gravitational effects; nor we include the magnetic dipole induced by the neighbors, whose influence would rescale the magnetic modulus Zhang and Widom (1995); Cēbers (2003) with quadratic in the curvature Kratky and Porod (1949); Kamien (2002), and given by the time-averaged dipolar interactions between nearest neighbors induced by the magnetic field (a derivation of is presented in Appendix A) Goubault et al. (2003); Cēbers (2003), where is the projection of the tangent vector onto the precession axis; is the bending modulus (with units of force times squared length); is the magnetic modulus (with units of force) defined by with the vacuum permeability, the magnitude of the magnetic dipoles, the separation between their centers and the precession angle. As shown below, configurations of the filaments depend sensitively on the sign of , determined in turn by . vanishes at the so-called “magic” angle , so in this case the leading order of the magnetic energy will be the quadrupolar term, which is of short range so filaments behave mostly as elastic curves Osterman et al. (2009). In the regime , which will be termed as regime , see Fig. 2(b), the magnetic modulus is positive, (the magnetic dipolar interactions are attractive), and from Eq. (2) we see that in order to minimize the filaments will tend to align with the precession axis to maximize Martin et al. (2000). By contrast in the regime , termed as regime , the magnetic modulus becomes negative, (the magnetic dipolar interactions are repulsive), and is minimized when vanishes, so the filaments will tend to lie on the plane orthogonal to the precession axis Martin et al. (2000). To reduce the material parameters space we rescale all quantities by the bending modulus and denote the rescaled quantity by an overbar. In particular, the rescaled quantity possesses units of inverse squared length, so the inverse of its square root, , provides the characteristic length scale at which buckling occurs. The dimensionless parameter is known as the magnetoelastic parameter. This parameter quantifies the ratio of magnetic to bending energies: bending and magnetic energy scale as and , so their ratio scale as . Below, we use as a parameter to characterize the conformations of the filaments. Typical experimental values of these paramagnetic filaments222For beads with diameter , magnetic susceptibility in a magnetic field , the magnitude of the induced dipole is . are , , , so , and , Biswal and Gast (2003); Goubault et al. (2003); Biswal and Gast (2004). The total bending and magnetic energies of the filament are given by the line integrals of the corresponding energy densities Thus the total energy is , but since the filament is inextensible, we consider the effective energy where is a Lagrange multiplier fixing total length which acts as an intrinsic line tension. The change of the energy under a deformation of the curve is given by Capovilla et al. (2002); Guven and Vázquez-Montejo (2012); Guven et al. (2014) In the first term, which represents the response of the energy to a deformation in the bulk, is the force vector, given by the sum of the bending and magnetic forces, , defined by Langer and Singer (1996); Capovilla et al. (2002); Guven and Vázquez-Montejo (2012); Guven et al. (2014); Dempster et al. (2017) where and . is the force exerted by the line element at on the neighboring line element at , so () represents compression (tension). From the force balance at the boundaries follows that is the external force on the filament Langer and Singer (1996). We see in Eq. (7) that the magnitudes of the bending and magnetic forces scale as and . Thus, the magnetoelastic parameter also quantifies the ratio of magnetic to bending forces, , Cēbers (2003). If the filaments are immersed in a medium of viscosity , we have from the balance of bending and viscous forces that the characteristic bending relaxation time is , Powers (2010), whereas the characteristic magnetic relaxation time is Dempster et al. (2017). The ratio of bending to magnetic relaxation times is the magnetoelastic parameter . For a filament of length , bending and magnetic moduli and , in water () we have and . Therefore, in order to be legitimate, the use of the time-averaged magnetic energy density given in Eq. (2) is justified if the precessing period is less than a millisecond, (frequency Dempster et al. (2017)), so that the characteristic relaxation times are much larger, . The second term in Eq. (6) contains quantities arising after integration by parts and is given by the total derivative of so it represents the change of the energy due to boundary deformations. Stationarity of the energy implies that in equilibrium the force vector is conserved along the filament, , a consequence of the translational invariance of the total energy. By contrast, the torque vector, , with , is not conserved: ,333In this expression we have used the identity , Capovilla et al. (2002). while the first term vanishes in equilibrium, the second term, representing a torque per unit length due to the magnetic field, , does not vanish in general. However, the component of the torque along the precession axis, , is conserved on account of the rotational symmetry of the energy about such direction: , which vanishes in equilibrium. Spanning the derivative of in terms of the two normals as ,444The projection onto the tangent vanishes identically due to the reparametrization invariance of the energy Capovilla et al. (2002). so the normal projections of the conservation law provide the Euler-Lagrange (EL) equations satisfied by equilibrium configurations, which read Dempster et al. (2017) In solving these equations, the Lagrange multiplier is determined from boundary or periodicity conditions. The squared magnitude of the force vector is constant on account of the conservation law of . This constant corresponds to the first Casimir of the Euclidean group and provides a first integral of the EL equations.555EL Eq. (9b) can be written as . Thus, the scalar quantity , corresponding to the second Casimir in the case of Euler Elastica, is not conserved in equilibrium because the magnetic field breaks the rotational invariance of the energy and introduces a source of stresses. Below we analyze solutions of two families of curves satisfying these equations with their ends held along the precession axis: curves lying on a plane passing through the precession axis and helices whose axis is parallel to the precession axis. Iii Planar cuves Let us consider curves on a plane, say -, so the embedding functions are and the tangent vector is . Since the curve lies on a plane, it has vanishing torsion, , and the EL equation associated with deformations along reduces to whereas the EL corresponding to deformations along is satisfied identically, because vanishes on account of the orthogonality of the binormal vector to the plane of the curve, i.e. . The force vector, defined in Eq. (7), lies on the osculating plane of the curve Projecting onto the FS basis we obtain Differentiating Eq. (13a) with respect to and using the FS formula we obtain Eq. (13b), whereas differentiation of Eq. (13b) reproduces the EL Eq. (11). Therefore Eq. (13a) provides a second integral of the EL Eq. (11), which permit us to express the difference between the bending and magnetic energy densities as the sum of the tangential component of the force and the constant . Moreover, this relation can be used to eliminate the curvature in favor of the projections of the tangent vector, for instance, the total energy density can be recast as In terms of the tangent and normal vectors are and , whereas the FS curvature is . Expressing Eq. (13a) in terms of , it reduces to a quadrature for : where we have defined which in the mechanical analogy would represent the corresponding equation of motion, see Appendix B. We consider filaments with their boundaries fixed, but not the tangents. Thus the variation vanishes at the boundaries (which we set at ), i.e. , and from Eq. (8) we have that the stationarity of the energy at the boundaries, , imply the vanishing of the curvature at those points. Therefore the appropriate boundary conditions (BC) for equilibrium configurations is In consequence, the intrinsic torque vanishes at the ends and only the torque coupling position and force contributes. Furthermore, the quadrature implies that the maximum value of the angle, say , occurs at the boundaries, i.e. . Thus at the turning points the “kinetic” energy vanishes and the “potential” energy is equal to the “total energy” Audoly and Pomeau (2010), which determines the Lagrange multiplier in terms of the angle : Once has been determined as a function of , the coordinates are obtained by integrating the tangential components In the next section we apply these results to the case of filaments aligned with the precession axis. Iv Vertical filaments Here we consider a curve resulting from a deformation of a straight filament lying along the precession axis, chosen as the axis, such that the end points remain along this axis (see Fig. 3). In consequence the force is also along the precession axis: and . Thus, the potential reduces to , which has period and left-right symmetry . We set the mid point of the curve at from where arc length is measured, being positive (negative) above (below) the axis, i.e. with . We denote the height of the boundary by so that the height difference is . We characterize the curves by the height difference rescaled with the total length To gain some insight about how the magnetic field modifies the behavior of the filaments, we first solve the quadrature (15) in the regime of small deviations from a vertical straight line. iv.1 Perturbative regime We consider a small perturbation of a straight line with and we expand the constants perturbatively as , .666 and are constants, however, we are interested in determining the corrections as functions of a small parameter, determined below, required by deviations from a straight line. At quadratic order, the quadrature describes a harmonic motion Only for 777If , then and the curve is a straight line. the quadratic potential is positive and bounded solutions are possible, given by If the filament develops half periods,888For instance the filament shown in Fig. 3 completes only one half-period (). the wave number is given by We see that the magnetic field modifies the minimum force required to trigger an Euler buckling instability. Furthermore, unlike the purely elastic case where the force on the filaments is always compressive, the magnetic contribution enables the force to be either tensile or compressive depending on the value of the magnetoelastic parameter relative to the squared number of half-periods: for (precession regime or ) the force is positive, , so the filament is under compression, whereas for (precession regime ) the force becomes negative, , and the filament is under tension. In the particular case with , free filaments with are possible Cēbers (2005a). The coordinates can be obtained by integrating Eq. (20), obtaining Evaluating the second expression at the boundaries we determine the amplitude in terms of the scaled height difference defined in (21): The total energy of the filament is , where is the scaled energy of the original straight vertical state and the second order correction is Since increases linearly with the magnitude of the force it can be either positive or negative. Let us now look at the stability of these states. To lowest order, the differential operator of the second variation of the energy, (derived in Appendix C, Eq. (78)), reads The two trivial zero modes (with vanishing eigenvalues), , with constant, are associated to the translational invariance of the energy: at lowest order they correspond to infinitesimal vertical and horizontal translations respectively. To analyze the eigenmodes of we use the basis , which in order to preserve the periodicity of the original curves should have wave numbers , , so the corresponding eigenvalues for each case are The two non-trivial zero modes with correspond to infinitesimal rotations in the plane, but for finite rotations such eigenmodes will not be zero modes, because the energy is only invariant under rotations about the precession axis (in the - plane). The eigenvalues corresponding to states with are shown in Fig. 4. We see that, like in the purely elastic case, only the eigenvalues of the ground state are all positive, so it is the only stable state in the perturbative regime. Therefore, any excited state would decay recursively to the next intermediate state with the more negative eigenvalue until the ground state is reached, Guven et al. (2012). As we will see below, comparison of the total energy of successive states leads suggests that the state is still the ground state in the non-linear regime. iv.2 Non-linear regime Now, we describe the behavior of the filaments in the non-linear regime, i.e. deformations far from straight configurations under the influence of a magnetic field (for comparison purposes, the case of elastic curves, is reviewed in Appendix D). If (), the quadrature (15) can be recast as Integrating the quadrature twice, we obtain the coordinates in terms of elliptic functions in each regime, (details are provided in Appendix E for the interested reader): where the constants and are defined by , and are the sine, cosine, and delta Jacobi elliptic functions; is the Jacobi amplitude; is the incomplete elliptic integral of the third kind Abramowitz and Stegun (1965); Gradshteyn and Ryzhik (2007). Recall is the buckling characteristic length. Like in the perturbative case, if the filament possesses half periods, the wave number is given by where is the complete elliptic integral of the first kind Abramowitz and Stegun (1965); Gradshteyn and Ryzhik (2007). The last relation permits us to express constants and , defined in Eq. (33), in terms of the modulus and the magnetoelastic parameter . The FS curvature of the filament is given by where is the maximum value of the curvature. The condition (18) of vanishing curvature at the boundaries determines, . Evaluating expressions (32) for at the boundaries and using the identities and , we get the following equation for the scaled boundary separation , defined in Eq. (21) To determine , these equations are solved numerically for given values of , , and .999Alternatively, one could extend the method employed in Ref. Hu et al. (2013) for the purely elastic case, in which case would be expanded as a series in and and the coefficients would be determined from Eq. (36). This completes the determination of all parameters of the curve. States with in regime are plotted for different values of and in Figs. (5) and (6). Corresponding states with in regime are plotted in Figs (7) and (8). In these sequences we choose initial states with different boundary angles , specifically (top rows with ), (middle rows with ) and (bottom rows with ).101010For smaller separations , the boundaries, and also upper and bottom segments of the filaments for strong magnetic fields, get close and in consequence our nearest neighbors approximation is no longer valid. In both regimes we observe that for relative small absolute values of the magnetoelastic parameter, , the filaments behave mainly as elastic curves (shown with dashed black lines in the plots with ), adopting and shapes for and , respectively. As is increased, we observe deviations from elastic behavior depending on the regime: in regime , at , filaments begin to elongate along the precession axis and squeezing inwards along the orthogonal direction, and for a large value, , they form thin vertical hairpins connected by straight segments aligned with the precession axis; in regime the converse behavior is observed, at they begin to stretch outwards and orthogonally to the precession axis, and at they are mostly straightened and with the filament’s horizontal extremum farthest from the precession axis. The magnitude of the forces in these planar curves is given by111111It can be positive or negative depending on the sign of the constant satisfying Eq. (36). is plotted for states in Fig 9. For vertical curves with , the force is linear in the magnetoelastic parameter, , as found in the perturbative analysis. We see that in regime is positive for all values of and , indicating that the filaments are under compression, as is usual for elastic curves bent under compression. By contrast, there are regions in regime where becomes negative in which case filaments are under tension, reflecting the fact that they tend to lie orthogonally to the precession axis. The bending and magnetic energy densities in terms of arc length read The bending energy is positive, increasing with (); the magnetic energy is positive in regime and negative in regime , thus the total energy density , can be positive or negative in regime , but it is strictly positive in regime . is shown with a color scale for states in Figs. (5)-(8). For filaments in regime , we see that initially is concentrated in the extremum and low in the boundaries, but as is increased, the high-energy regions migrates towards the hairpins near the boundaries and low-energy regions move to the extremum where straight segments (minimizing both energies) are developed. In regime , high-energy regions always occur at the extremum where the curvature concentrates, whereas low-energy regions correspond to the straight segments near to the boundaries. Moreover, the former regions become more localized and the latter regions more spread as is increased. In the calculation of the total energy, although integration of is simple, integration of is rather complicated because it involves . However, we can integrate expression (14) for the total energy density, where was replaced in favor of by means of the quadrature, (13a), obtaining the following expressions of the total energy for each case (details are presented in Appendix E): where the constant is defined by The total energy of states are plotted in Figs. 10(a) and 10(b). As found in the perturbative regime, regardless of , straight lines with have scaled total energy . We see that is negative almost everywhere (except in a small fringe of values in the vicinity of ) in regime and it is positive everywhere in regime . Values of and for which are shown with a solid black line, and the energies of elastic curves () are shown with a dashed black line. The total energy of the filaments increases as augments for any value of and . To show this, in Fig. 10(c), we plot the energy difference between states and , , where we see that everywhere on the parameter space , result that can be verified for states with higher , i.e, . Thus is the ground state among planar configurations for all parameter values. However, as we will see below this does not hold in general when non-planar configurations are considered, in particular the energy may be lowered for some values of and if filaments adopt helical configurations, which we examine in the next section. Here we demonstrate that helices are also critical points of the total energy. Recall that a helix is characterized by its radius and pitch , with the pitch angle defined by (see Fig. 11). The helix can be parametrized in cylindrical coordinates by the azimuthal angle as, A helical segment is specified by the total azimuthal angle . We consider helices completing full turns, so and the distance between the end points is . Arc length is proportional to , , so total length is proportional to , . From these relations follows that . Inverting to get and in terms of , , and , we get The FS basis adapted to the helix is whereas the FS curvature and torsion are given by The sign of the torsion determines the chirality of the helix, () corresponds to right (left) handed helices. The degenerate cases and correspond to circles on the plane - and to vertical lines, respectively. For helices, the Darboux vector is along the helical axis . Since and are constant and , the EL Eq. (9a) is satisfied if is constant, taking the value and the EL Eq. (9b) vanishes identically. Hence, helices satisfying Eq. (45) minimize the total energy . Using these expressions for , , and in Eq. (7) for , we find that the scaled force required to hold the helix is linear in the separation of the end points and directed along the helical axis, The magnitude of the force is plotted for states as a function of and in Fig. 12. In this plot the line over () represents the scaled force required in the Euler buckling instability of a straight line, with the elastic term four times larger as compared with the scaled force required in the planar case, . Like the case of planar curves, the force can be tensile or compressive depending on value of the magnetoelastic parameter relative to the total azimuthal angle: if () the magnitude of the axial force is positive (negative), (), and the helix is under compression (tension). For circles with () and configurations with , there is no vertical force, . The difference of the force between successive states and , is independent of and positive for any value of , and it increases with . The torque vector has two components, one introduced by the magnetic field and another one of elastic character The magnitude of the azimuthal torque is linear in the magnetoelastic parameter, so it vanishes for elastic curves with and its direction is reversed when changing from regime to regime . For a given value of it vanishes for circles and lines with , respectively, and is maximum for helices of maximum torsion with . The magnitude of the axial torque is proportional to the total azimuthal angle and increases as the boundary points are approached, so it is maximal for circles and vanishing for vertical filaments. The total scaled energy of the helices is harmonic in the separation of the end points Dempster et al. (2017)

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CC-MAIN-2020-10

https://download.cnet.com/Mixed-Numbers/3000-20415_4-75754141.html

math

The download button opens the iTunes App Store, where you may continue the download process. You must have iTunes installed with an active iTunes account in order to download and install the software. This download may not be available in some countries. Convert mixed number into improper fractions with this Common Core standards aligned iPad app. Learning Objective Students will learn to: Interpret fractions as the division of the numerator by the denominator Convert a mixed number into an improper fraction Blank page to create and solve your own examples in a group settingGrade Level(s)4 5Common Core Standards5.NF.3 Interpret a fraction as division of the numerator by the denominator (a/b = a * b). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers, e.g., by using visual fraction models or equations to represent the problem. For example, interpret 3/4 as the result of dividing 3 by 4, noting that 3/4 multiplied by 4 equals 3, and that when 3 wholes are shared equally among 4 people each person has a share of size 3/4. If 9 people want to share a 50-pound sack of rice equally by weight, how many pounds of rice should each person get? Between what two whole numbers does your answer lie?5.NF.2 Solve word problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators, e.g., by using visual fraction models or equations to represent the problem. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2/5 + 1/2 = 3/7, by observing that 3/7 < 1/2.

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CC-MAIN-2018-30

https://projecteuclid.org/euclid.aop/1176996942

math

The Annals of Probability - Ann. Probab. - Volume 1, Number 3 (1973), 484-487. A Note on Fine's Axioms for Qualitative Probability Fine gives axioms on a binary relation $\precsim$ on a field of events, with $A \precsim B$ interpreted as "$A$ is (subjectively) no more probable than $B$," sufficient to guarantee the existence of an order-preserving probability measure and an additive order-preserving probability measure. It is noted that one of Fine's axioms, that the order topology have a countable base, can be replaced by the more appealing axiom that there is a countable order-dense subset. Ann. Probab., Volume 1, Number 3 (1973), 484-487. First available in Project Euclid: 19 April 2007 Permanent link to this document Digital Object Identifier Mathematical Reviews number (MathSciNet) Zentralblatt MATH identifier Roberts, Fred S. A Note on Fine's Axioms for Qualitative Probability. Ann. Probab. 1 (1973), no. 3, 484--487. doi:10.1214/aop/1176996942. https://projecteuclid.org/euclid.aop/1176996942 - Correction: Fred S. Roberts. Correction Note: Correction to "A Note on Fine's Axioms for Qualitative Probability". Ann. Probab., Vol. 2, Iss. 1 (1974), 182.

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CC-MAIN-2019-43

https://bhs.fcschools.net/students/high-school-course-information-guide/mathematics

math

A preparatory course for Fundamentals of Algebra, this course acquaints the student with different number systems through mathematical language consisting of symbols and new concepts dealing with sets. Primary goals are to teach students to deal with variables and polynomials through language and application as well as to solve equations. This course is a preparatory course for math I. Students are introduced to algebra, geometry, and other mathematical topics that are integrated in a format that connects mathematics to students’ lives and the world of work. Math I provides students the opportunity to study concepts of algebra, geometry, functions, number and operations, statistics and modeling throughout the course. These concepts include expressions in the real number system, creating and reasoning with equations and inequalities, interpreting and building simple functions, expressing geometric properties and interpreting categorical and quantitative data. The final exam is the North Carolina End-of-Course Test based on the Common Core Math 1 Standards. This course continues a progression of the standards established in Math I. In addition to these standards, Math II includes: polynomials, congruence and similarity of figures, trigonometry with triangles, modeling with geometry, probability, making inferences and justifying conclusions. Progresses from the standards learned in Math I and Math II. In addition to these standards, Math III extends to include algebraic concepts such as: the complex number system, inverse functions, trigonometric functions and the unit circle. Math III also includes the geometric concepts of conics and circles. Designed for those students who have potential for outstanding performance in mathematics, Math III Honors is an accelerated, expanded, and demanding course. Students will work with real, irrational and imaginary numbers, solving systems of equations, problem solving with logarithms, conic sections and polynomials. NC Math 4 focuses on functions and statistical thinking, continuing the study of algebra, functions, trigonometry and statistical concepts previously experienced in NC Math 1-3. The course is designed to be a capstone to introductory statistical concepts. Additionally, the course intentionally integrates concepts from algebra and functions to demonstrate the close relationship between algebraic reasoning as applied to the characteristics and behaviors of more complex functions. This is a survey course of various topics that will prepare the student for calculus and college-level mathematics courses. Emphasis is placed on functions, logarithms, and exponential systems of equations. Graphing calculators will be used on a regular basis. This course satisfies the 4th math requirement for the public universities in the UNC system. Math I, Math II with A or B recommended. Calculus focuses on the solution of problems which cannot be solved by algebra or trigonometry. Finding the slope of the tangent to a curve, areas of planar and spatial surfaces, the volume of solids, and the mathematics of speed and acceleration are examples. The ability to construct and interpret graphs is a necessary component of many solutions. Calculus is intended for students with a high aptitude in mathematics who intend to pursue fields related to mathematics, physics, and engineering while in college. Graphing calculators are used on a regular basis. This is a college level course and is offered for students who will take the Calculus AB AP exam. Course is intended to be challenging and demanding. Calculus AB is primarily concerned with developing the students’ understanding of the concepts of calculus and providing experience with its methods and applications. The course emphasizes a multi-representational approach to calculus, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. The connections among these representations are also important. This is a college level course and is offered for students who will take the Calculus BC AP exam. Calculus BC is an extension of Calculus AB rather than an enhancement; common topics require a similar depth of understanding. The course is intended to be challenging and demanding. Calculus BC is primarily concerned with developing the students’ understanding of the concepts of calculus and providing experience with its methods and applications. The course emphasizes a multi-representational approach to calculus, with concepts, results, and problems being expressed graphically, numerically, analytically, and verbally. The connections among these representations are also important. Advanced Placement Statistics introduces students to the major concepts and tools for collecting, analyzing, and drawing conclusions from data. Students will observe patterns and departures from patterns, decide what and how to measure, produce models using probability and simulation, and confirm models. Appropriate technology, from manipulatives to calculators and application software, should be used regularly for instruction and assessment.

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CC-MAIN-2023-50

https://blog.wolfram.com/2020/10/

math

The pandemic has postponed or canceled a lot of things this year, but luckily learning isn’t one of them. Check out these picks for new Wolfram Language books that will help you explore new software, calculus, engineering and more from the comfort of home. Date Archive: 2020 October Our annual Wolfram Technology Conference took place October 6–9, and along with it the 10th annual Wolfram Innovator Award Ceremony. This year, Stephen Wolfram recognized 13 outstanding individuals from around the globe for their significant work using the Wolfram Language across fields and disciplines. Although this year’s Wolfram Technology Conference was virtual, that didn’t stop us from running the ninth annual One-Liner Competition, where attendees vie to produce the most amazing results they can with 128 or fewer characters of Wolfram Language code. Here are the winners, including an audio game, a hands-free 3D viewer and code that makes up countries. Blockchain was integrated into the Wolfram Language in 2018 with the release of Version 11.3, featuring a set of functions that is constantly improved and expanded upon by our team. Currently supporting a seamless connection to the Bitcoin, Ethereum, ARK and bloxberg mainnets, testnets and devnets, Wolfram introduced to the distributed ledger technology (DLT) space its philosophy of injecting computational intelligence everywhere through Wolfram Blockchain Labs, with the mission of enabling blockchain-based commerce and business model innovation. Earth has experienced five major extinctions since life first appeared almost four billion years ago. The sixth is happening right now; the current extinction rate is between one hundred and one thousand times greater than what it was before 1800. Despite the alarming extinction rate, it’s easier than ever to document biodiversity with the help of the Wolfram Language. Using the monarch butterfly as an example, I will explore the new biodiversity data access functions in the Wolfram Function Repository and how they can help you join a community of thousands of citizen scientists from iNaturalist in preserving biodiversity.

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CC-MAIN-2024-10

https://web2.0calc.com/questions/geometry_92201

math

How many units are in the sum of the lengths of the two longest altitudes in a triangle with sides 8, 12, and 14? We calculate the area by Heron's formula, s = (8 + 12 + 14)/2 = 17 Area = sqrt(s(s - 8)(s - 12)(s - 14)) = 3 sqrt(255) Altitude length corresponding to a side x is 2(area)/x. Use the shortest side lengths as x, and you will get the longest altitude in result. You can substitute area = 3 sqrt(255) and x = 8, x = 12 to get the length of the two longest altitudes.

s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00646.warc.gz

CC-MAIN-2023-06

http://umj-old.imath.kiev.ua/article/?lang=en&article=8005

math

On a representation of smooth invariants of Coxeter groups in terms of anisotropic spaces A smooth function invariant under the action of the Coxeter group can be represented as a function of basic invariants. We propose to describe the latter in terms of special anisotropic spaces, which enables us to obtain more precise estimates of its smoothness. English version (Springer): Ukrainian Mathematical Journal 49 (1997), no. 4, pp 661-664. Citation Example: Gokhman A. O. On a representation of smooth invariants of Coxeter groups in terms of anisotropic spaces // Ukr. Mat. Zh. - 1997. - 49, № 4. - pp. 597–600.

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CC-MAIN-2021-04

https://physics.stackexchange.com/questions/292521/explain-the-origin-of-magnetic-fields-in-layman-terms

math

Electromagnetism always seemed so weird to me. It feels like it should be 2 different things entirely. At first I thought magnetism was just applied electric charge, and that a magnetic dipole was just an electric dipole on a net neutral macroscopic object. However I learned this was false when I learned a magnetic field can be created by an electron, which has electric monoploe. Electric fields I at least understand the origins of. When there are 2 charges in space, they apply a force on one another. The net effect of those forces on any one charge at any given point in space can be represented mathematically with the electric field. But what force of attraction/repulsion, in that case, is a magnetic field representing? It's apparently still caused by the same electromagnetic force, but it follows entirely different rules, so different that a partical with an electric monopole can have a magnetic dipole. I understand a magnetic field is created when an electric charge moves, as well as by the particle spin of a charged particle. But is that all we've got on its origins? Does it just show up when electric charges move and that's that? Why is it considered the same fundamental force as electric attraction/repulsion when it follows wildly different rules? CAN it be derived from the rules of electric charges? Please understand, when you explain it, that you're talking to an undergraduate chemical engineer here, not a physics major. Please try to explain it while keeping physics terminology and math beyond high school level to an absolute minimum. Edit: Once again, I want to stress that I know next to nothing about wave functions, the concept of symmetry, and angular momentum is something I've only kind of learned about and only kind of understand at an intuitive level. (My basic understanding of angular momentum is that it is a constant change in linear momentum as governed by a force perpendicular to the velocity of the object at any given time. I understand this could be technically wrong). Not saying I'm incapable of understanding these concepts if you explain them to me in layman's terms. I'm not dumb. I just know next to none of the vocab.

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CC-MAIN-2021-21

http://www.jiskha.com/display.cgi?id=1360626918

math

Posted by mythreyee on Monday, February 11, 2013 at 6:55pm. enzo and eric are a sharing a pie. if enzo eats 1/3 of the pie and eric eats 1/5 of the pie, is there more than 1/2 of the pie remaining. 3rd grade math - minecraft, Monday, February 11, 2013 at 6:57pm 1/3 + 1/5 = once you get your answer. turn into decimals and compare with 0.50 since 1/2 is 0.50. 3rd grade math - JJ, Monday, February 11, 2013 at 6:58pm You have to add 1/3 and 1/5. You need to get a common denominator of 15. after you add the fractions, subtract from 1 for 1 whole pie. (You can write 1 as 15/15). compare your answer with 1/2 to see if it is larger or smaller. (It is always good to be sure the denominators are the same which makes it easier to compare. 3rd grade math - Ms. Sue, Monday, February 11, 2013 at 6:59pm 1/3 = 5/15 1/5 = 3/15 5/15 + 3/15 = 8/15 = 0.5333 They ate 0.533 of the pie. 3rd grade math - minecraft, Monday, February 11, 2013 at 7:00pm compare and get your answer is which is yes there is more then one half because 0.533>0.50 3rd grade math - Ms. Sue, Monday, February 11, 2013 at 7:02pm There is less than half a pie remaining. Reread the problem and your answer, minecraft. 3rd grade math - minecraft, Monday, February 11, 2013 at 7:03pm i knew i was right at first Answer This Question More Related Questions - 3rd grade math - Enzo and Eric are sharing a pie. If Enzo eats 1/3 of the pie ... - Math - Eric eats 1/6 of an apple pie and Jack eats 1/4 of the apple pie. How ... - Math - Hands on : Multiplication patterns with 10,100,and 1,000 1/ Rebecca has 4... - math(help) - please help me and answers these i feel clueless!with this math 2.... - 6 th grade math - 2 students are learning how to bake apple and blueberry pies. ... - Math - Daniel made a chocolate pie, a cream pie, and an apple pie that were the ... - physics - The motion of a particle is described by x = 10 sin (piet +pie/2 ). At... - Calculus grade 12 - solve for (<_ = less than or equal to / pie = pie sign... - math - evaluate the expressions below and leave them in radical form. 1. sin(pie... - Math - Sabrina and John cut a pie into 4 slices. Sabrina eats 2 slices. John ...

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CC-MAIN-2017-09

http://www.patentgenius.com/patent/7050957.html

math

Projection electron beam lithography apparatus and method employing an estimator ||Projection electron beam lithography apparatus and method employing an estimator ||May 23, 2006 ||August 29, 2001 ||Stanton; Stuart T. (Bridgewater, NJ) ||Agere Systems Inc. (Allentown, PA)| ||Homere; Jean R. |Attorney Or Agent: ||430/30; 700/30; 703/13; 703/2; 703/3; 703/4; 708/300; 708/313 |Field Of Search: ||703/3; 703/4; 703/13; 703/2; 700/30; 708/300; 708/313; 430/30 |U.S Patent Documents: ||5719796; 6177218; 6243158; 6285971; 6631299; 6925478; 2002/0120656 |Foreign Patent Documents: ||Stanton et al., "Initial Wafer Heating Analysis for a SCALPEL Lithography System", Microelectronic Engineering, vol. 46, Issues 1-4, May 1999,pp. 235-238. cited by examiner. ||A process and method for projection beam lithography which utilizes an estimator, such as a Kalman filter to control electron beam placement. The Kalman filter receives predictive information from a model and measurement information from a projection electron beam lithography tool and compensates for factors which cause beam placement error such as wafer heating and beam drift. The process and method may also utilize an adaptive Kalman filter to control electron beam placement. The adaptive Kalman filter receives predictive information from a number of models and measurement information from a projection electron beam lithography tool and compensates for factors which cause beam placement error such as heating and beam drift. The Kalman filter may be implemented such that real-time process control may be achieved. ||What is claimed is: 1. A projection electron lithography system, comprising: a lithography tool for emitting a beam of electrons and producing measurement information; and a processorincluding, a plurality of different predictive models for producing predictive information, and an adaptive estimator that iteratively selects a best predictive model from said plurality of different predictive models and controls placement of said beamof electrons based on said predictive information from said best predictive model and said measurement information from said lithography tool, said adaptive estimator employing a tunable strength parameter to determine an optimal adaptation weightingcriterion. 2. The system of claim 1, wherein said adaptive estimator compensates for beating and beam drift effects. 3. The system of claim 1, wherein said adaptive estimator employs least-squares based linear matrix algebra. 4. The system of claim 1, wherein said system is a SCALPEL system. 5. The system of claim 1, wherein said adaptive estimator is an adaptive Kalman filter. 6. The system of claim 1, wherein said adaptive estimator is an adaptive Kalman filter and each of said plurality of different predictive models is partitioned into wafer scale components and die scale components, said adaptive Kalman filteronly employed for wafer scale components. 7. The system of claim 1, wherein said plurality of different predictive models differ due to a single parameter that varies in each of said plurality of different predictive models. 8. The system of claim 1, wherein said plurality of different predictive models includes three or more models. 9. The system of claim 1 wherein said plurality of different predictive models are only directed to producing said predictive information for corrections associated with a die scale. 10. A computer implemented process for controlling projection electron lithography, comprising: emitting a beam of electrons; producing measurement information on said emitting step; producing predictive information related to the projectionelectron lithography process based on a plurality of different predictive models; iteratively selecting one of said plurality of different predictive models until a best predictive model from said plurality of different predictive models emerges; andcontrolling placement of the beam of electrons based on selected predictive information from said best predictive model and said measurement information, wherein said controlling includes determining an optimal adaptation weighting criterion employing atunable strength parameter. 11. The process of claim 10, wherein said controlling step employs an adaptive Kalman filter. 12. The process of claim 10, wherein said controlling step compensates for heating and beam drift effects. 13. The process of claim 10, wherein said process is a SCALPEL process. 14. The process of claim 10, wherein said controlling step is implemented as an adaptive Kalman filter and each of said plurality of different predictive models is partitioned into wafer scale components and die scale components, said adaptiveKalman filter only employed for wafer scale components. 15. The process of claim 10, wherein said plurality of different predictive models differ due to a single parameter that varies in each of said plurality of different predictive models. This invention relates to the field of projection electron beam lithography and in particular, to projection electron beam lithography employing an estimator. In projection electron beam lithography, precise control of the placement of the electron beam is required in order to ensure that the image is constructed without distortion and aligned to a prior process level. Precise control of the electronbeam placement is difficult because electron beam placement depends on many factors. One of these factors is a wafer distortion response to the heating action of a projection electron beam lithography beam, ranging up to many hundreds of nanometers, depending on conditions. Correction schemes include a model-based predictor forsub-field center placement adjustment. The algorithm implemented by the model-based predictor controls the writing of a matched dynamic distortion with an accuracy of about 1% or better for the largest, long-length-scale effects of approximately 500 nm. Other factors in addition to a predictable heating response, such as beam drift and wafer-to-chuck contact variation, also affect placement accuracy. Their effect may be either random or very difficult to correctly model. As stated above, wafer-to-chuck contact may have an effect on the response that requires enhancement to a basic predictive model. Modeling and experiments have both demonstrated the desirable result that good thermal contact to the chuck(.about.150 W/m.sup.2K) can lower the accumulated size of the wafer-heating response by a factor of roughly 10, thus enlarging the fractional correction error tolerance similarly. However, there are several factors, such as wafer-flatness, particletolerance, frictional contact, and pulling-force that may remain variable or random despite efforts in chuck design. Realistically, the chuck design process can only reduce frictional influences on the heating response to a form ofchuck-coordinate-system drift that is slow and indistinguishable from beam drift. Since important parameters in the predictive model may be variable from wafer to wafer, prediction alone is not sufficient for full correction of beam placement. Further, it is difficult to perform the complex model computation required to determine correct beam placement in a short period of time. The only alternative to prediction is measurement. The obvious primary measurement of beam placement involves an alignment mark sensing process. The use of a re-alignment strategy, or some variation of local alignment, is a common approach todealing with drift in many other electron beam lithography applications, such as mask-making and direct-writing. This often involves time-consuming actions like extra stage motions that detract from throughput, but this can be a tolerable situation whenmaking relatively few high-value exposures. In the area of production wafer-level lithography using SCALPEL, throughput is a concern even without the use of local alignment or complex re-alignment strategies. Hence, re-alignment is not a suitable correction strategy for a high-throughputSCALPEL tool. Based on the above, it is clear that an enhancement to the predictive models used for beam placement correction is desirable, making use of alignment mark sensing and efficient computation. SUMMARY OF THE INVENTION The method and apparatus of the present invention include an estimator that integrates a predictive model and a measurement capability, both subject to substantial noise sources, plus measurement sampling limitations. The estimator works in realtime with only historical data. In one exemplary embodiment, the estimator is a Kalman filter, which may be a least-squares based optimum estimation algorithm for the states of time-dependent systems, using linear matrix algebra. In the present invention, the Kalman filter is used to correct for wafer heating, beam drift and/or other errors in a projection electron beam lithography system, such as for example, SCALPEL. By using a Kalman filter, real time process controlis obtained using a greater amount of information than could be used if conventional modeling/process control and measurement techniques were used. The method and apparatus of the present invention may also employ an adaptive Kalman filter (A-KF) correction for wafer heating, beam drift and/or other errors. The adaptive Kalman filter correction may be based on a numerical response-modelinterface that allows efficient integration of relatively slow but infrequent pre-calculation results, and allows real-time adaptive Kalman filter functionality. An adaptive Kalman filter is particularly effective when a model parameter uncertainty problem is superimposed on a more elementary state noise problem. The two types of unknown system response can both be handled using only one measurement datasequence, but are distinguishable in terms of their statistical behavior. In SCALPEL, an example of an uncertain parameter is wafer-to-chuck thermal contact, which should be a nearly-fixed quantity on length scales of interest, during each waferexposure. The effect of wafer-to-chuck thermal contact on the response of the system is momentarily stable and non-random for any one execution of the Kalman filter, even if poorly known. This is in contrast to the lumped beam drift and frictionalchuck-coordinate-system drifts that may be more like a random-walk effect, and hence most readily treated as a band-limited state noise. In a preferred embodiment, the control algorithm which performs the predictive model can be partitioned into global (wafer scale) and local (die scale) components. A pure-predictor would suffice for the local problem since the main noise anduncertainty terms do not act on this scale and the errors are inherently smaller. The use of an adaptive Kalman filter only for the global part of the problem would be very efficient. The method and apparatus of the present invention may also employ a multi-model adaptation corrector, which provides a best estimate that converges on the correct unknown model parameter choice. The behavior of the Kalman filter is very good for scenarios that are realistic or somewhat pessimistic in key parameters pertaining to SCALPEL operation, including a slow beam drift of typically 40 nm and a 15 nm 3-sigma one-site alignmentnoise. Adaptation in a multi-model form is effective at handling the problem of at least a factor of two thermal contact parameter uncertainty. Combined errors on the order of 50 nm in predicting responses that are well over 100 nm can be reduced to 10 nm or better, in a case of low contact and thermal dissipation to the chuck. With some optimization and the benefit of maximum chuckthermal contact, error budget requirements of nominally 5 nm can also be met. BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 illustrates a projection electron beam lithography system in one exemplary embodiment of the present invention. FIG. 2 illustrates the Kalman filter of FIG. 1 in one exemplary embodiment of the present invention. FIG. 3 illustrates the steps of multi-model adaptation in one exemplary embodiment of the present invention. FIGS. 4a and 4b illustrate a weight-determining function in one exemplary embodiment of the present invention. FIGS. 5a and 5b illustrate the response of a nominally tuned adaptation scheme based on residual curves and multi-model execution. DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENT FIG. 1 illustrates a projection electron beam lithography system 10 in one exemplary embodiment of the present invention. As illustrated, the system 10 includes a processor 12 (either with or without external memory) and a projection electronbeam lithography tool 14. In a preferred embodiment, the projection electron beam lithography tool 14 is a SCALPEL tool. A predictive model 16 and a Kalman filter 18 are both implemented in processor 12. The Kalman filter 18 receives predictions fromthe predictive model 16 and measurements from the projection electron beam lithography tool 14 and controls placement of an electron beam output from the projection electron beam lithography tool 14 as described in more detail below. A Kalman filter 18 is a recursive algorithm using linear matrix algebra to make an optimal estimate of the state of a system, given a combination of state and measurement noises. The most common form of the optimization is least-squares, whichis readily formulated in linear matrix algebra form and is optimum for Gaussian noise, but the algorithm can be more general as well. Nonlinear systems also can be linearized in order to make use of the linear algebra form of the filter. The essence of the Kalman filter 18 is to use one or more models 16 to describe the statistical behavior of both the measurement noise and the physical system state noise, so that this information can be used to determine the weighting in thecombination of prediction and measurement. This is referred to as "propagating the noise or error covariance", which is an ingredient in one of the two major recursive steps of the filter illustrated by FIG. 2. As illustrated in FIG. 2, predictionsfrom the model 16 and measurements from the tool 14 are recursively processed. By propagating the error covariance, an update of the Kalman gain (K) can be made. This quantity determines the weighting in the filter 18; 0 for pure prediction and 1 forpure measurement. The other major step is propagating the predictive model 16 iteratively based on a starting value from the estimate made in the previous step. This process continues iteratively in a loop. This estimate updating process is notnecessarily smooth since the quality of measurement information can change abruptly even if the system state cannot. "Tuning" the Kalman filter 18 may entail making adjustments in the proposed error/noise statistics model 16 in order to better match "reality". Variants of the Kalman filter 18 allow this to be done adaptively during the course of the filter 18operation, but it is also common to tune by trial and error as a series of experiments or simulations are performed. In the SCALPEL heating response application, the tuning is motivated by a need to estimate the required sub-field position adjustmentfor exposures in a sequence, hence reducing the worst error that occurs at any time in the exposure for an ensemble of wafer exposures. Using the Kalman filter 18, prediction alone is good enough in early stages when state errors have not accumulatedyet. This is due to the band-limited nature of the beam drift and the action of errors in thermal contact. A Kalman filter 18 usually uses differential equations of the system state expressed in state-space matrix form. However, the description below uses a common alternative notation, namely "discrete form" notation, which expresses the result atstep k+1 caused by propagation forward from step k. This is appropriate for a discrete measurement process, such as the SCALPEL process. Note that the steps modeled are absolutely not limited to those where measurements are made. The Kalman filter 18naturally deals with this by assigning non-measurement steps with a very large measurement covariance, resulting in the gain (K) being set to zero for those times. So the model 16 can naturally interpolate the state estimate in closely-spaced stepsbetween relatively sparse measurements. The five basic matrix equations are: 1) State prediction update: X(k+1/k)=.PHI.(k+1,k)X(k/k)+.PSI.(k+1,k)U(k) 2) Covariance prediction update: P(k+1/k)=.PHI.(k+1,k)P(k/k).PHI..sup.T(k+1,k)+Q*(k+1) WithQ*(k+1)=.GAMMA.(k+1,k)Qd(k).PHI..sup.T(k+1,k) 3) Gain computation: K(k+1)=P(k+1/k)H.sup.T(k+1)[H(k+1)P(k+1/k)H.sup.T(k+1)+R(k+1)].sup.-1 4) Estimation update: X(k+1/k+1)=[I-K(k+1)H(k+1)]X(k+1/k)+K(k+1)Z(k+1) 5) Covariance update:P(k+1/k+1)=[I-K(k+1)H(k+1)]P(k+1/k) These five equations correspond to a state-space representation of the propagation of state X and process of measurement Z, including noise, given by: X(k+1)=.PHI.(k+1,k)X(k)+.GAMMA.(k+1,k)Wd(k)+.PSI.(k+1,k)U(k); andZ(k+1)+H(k+1)X(k+1)+V(k+1). In all of the above equations, k is a step counter. The use of (n/m), such as (k+1/k), designates value in step n if given the value in step This is distinct from (k+1,k) which designates that the matrix value is sensitive to both the prior andpresent step count in general. Two examples clarify this notation: X(k+1/k) is the pure prediction update of the state vector X and X(k+1/k+1) is the update of the estimate of state X including measurement. In the state equations, the values are defined as: .PHI.=state propagator model from differential equations which also propagates the state covariance; U=input term for state, which can be generalized as we will discuss later, except that it doesnot propagate the state covariance; .PSI.=matrix which translates input to state form; Wd=state noise in raw form; .GAMMA.=matrix that translates state noise into state form; V=measurement noise in raw measurement form; and H=matrix that translates thestate into measurement form. The other quantities in the filter equations 1) 5) are: P=state covariance matrix, standard definition with terms in the form .sigma..sub.i .sigma..sub.j; has a starting value but is later generated by the filter 18; Qd=covariance matrix of statenoise Wd, in a form like P; nominally an assumed constant, or may be a sequence; generally subject to tuning; R=covariance matrix for measurement, similar to Qd; usually derived from measurement process modeling or experiments; may be tuned; K=calculatedKalman gain representing weight of measurement in estimate; and I=Identity matrix. Further, .sup.T refers to the transpose operation, and .sup.-1 refers to matrix inversion. As indicated by equation 3), K is computed completely from the propagation of measurement covariance and state noise covariance, which includes initial errors and added state noise. These can be done ahead of time in a situation that is notadaptively tuned and when the covariance model is stable. Similarly, as indicated by equation 4), K acts as a weight on the use of measurement in the estimation update, and a term of the form "I-K" is the converse weight of predictive update. Equations 1) 5) do not consider time-correlated noise (also known as "non-white" or "colored" noise) in any category. Equations 1) 5) assume that each new time step gives independent new random noise terms. The entire Kalman filter 18 equation set 1) 5) above can be modified to deal with correlated noises, although there may be a different process for measurement than there is for state. In the case of SCALPEL, measurement by alignment is expectedto have no time-correlation in the sense that information at each site has an error with no dependence on prior measurements. However, the state noise of drift clearly cannot be a white noise. Therefore, the state noise may be considered colored andthe Kalman filter 18 may be modified accordingly. The basic form of the equation 1) 5) stays the same except that a few elemental vectors and matrices should be augmented, meaning that new vectors and matrices are composed from old vectors and matrices with terms attached that represent atime-correlation or filter model. One such example is a one-step filter function with variable time constant t0, in the form: .PHI.wf=Exp[-(t.sub.k+1-t.sub.k)/t0]. One-step colored noise (Wdco) at step k+1 is generated from a new white random noisevalue (Wdwf) plus a fixed residual amount of the last noise value at k determined by the filter function: Wdco(k+1)=.PHI.wf(k+1,k)Wdco(k)+Wdwf(k) Augmentation processes are well-known. Below the equation changes are shown symbolically as extended vectors or groupings of matrices of the same dimensions to form larger matrices, where: X[X Wdco].sup.T H[H 0] .GAMMA. replaced by .GAMMA.aw=[0I].sup.T The original .GAMMA. is integrated with the state propagator: .PHI..times..PHI..function..kappa..kappa..GAMMA..function..PHI..times..tim- es..function. ##EQU00001## .PSI..times..PSI..times. ##EQU00001.2## .times..times. ##EQU00001.3## Q[U 0].sup.T Qd. terms in form .sigma..sup.2 become .sigma..sup.2[1-Exp[-2.DELTA.t/t0]] "0" represents a matrix of zeroes. For the purpose of running Monte-Carlo simulations of the application of a Kalman filter 18 to a specific model, it is typical to only provide a white-noise generator. Either the truth model is propagated in an augmented fashion to obtainfiltered noise, or the filter is applied a-priori (as shown here) to a time-series of random elements of the noise matrix. The use of the model 16 can be totally consistent by design, or the effect of an erroneous assumption about the time-correlationcan also be simulated. The Kalman filter 18 described assumes a singular "good" model 16 exists and that physical effects are appropriately modeled as additive random noise. This accurately describes the beam drift effects in SCALPEL. A different problem occurs ifthe model 16 is not fully known, so an assumed model leads to poorer filter performance than an ideal one would achieve. In general, there are known system model identification procedures that can be used to "learn" what a model should be. Particularlyin the absence of state noise, there are many non-Kalman filter approaches to using real-time measurements to converge on the right model and iteratively best-fit a measurement sequence. However, the same limited data may be subject to both noise andparameter uncertainty, as in SCALPEL. For this situation, an adaptive Kalman filter 18 implemented in a multi-model form is a powerful tool. In general, it is possible for one noise model to actually be the net effect of many more. It is not always obvious which type of disturbance is best treated as a "noise" versus an "uncertain parameter". In all cases, the Kalman filter 18equations must still have only one linear-additive noise vector in the state. The ability of the Kalman filter 18 to rapidly and efficiently perform real-time estimation depends on the linearity of the matrix formulation. Therefore, a multiplicativenoise or a product of two model components having noise must be linearized. However, if two disturbances are distinguishable because their statistical natures are very different, then one disturbance may be deemed to be a parameter that is momentarily fixed relative to another that varies more rapidly. In general,adaptation schemes can be applied sequentially to attempt to choose this parameter at any time as this parameter may evolve. In this case, time-correlation is the trait that distinguishes one from another even though both may have a stochastic nature. A multi-model adaptive Kalman filter 180 may be used to discern the best model 160. A set 160 of N assumed models 161, 162, 163 . . . are continuously tested to see if one emerges as a "better" model than the rest. This is a particularly goodapproach when only one unknown parameter really matters, such as chuck thermal contact. As each of N filters 181, 182, 183, . . . are run in parallel, each defines an optimal estimate for the same measurement sequence but using a different model 161,162, 163, . . . . Usually the models 161, 162, 163, . . . are basically the same, and a single parameter is varied N times in some series of steps. In the event that the response of the model 160 to the unknown parameter is continuous and not too severe, a limited number of models may be used in combination with a scheme that interpolates to determine a weighted combination of "best discretemodels". Obviously, the more models needed (N) and the more parameters not known (M), the less efficient the process may be since a total of NxM models must be run. One issue is what criterion can be used to guide the "adaptation", which is the process of selecting the correct model or weighted combination of models in real-time. Publications exist on this topic, with various ideas depending on the natureof the problem. The common thread is analysis of the "residual", which is the historic record of differences between the estimate and the measurement. Therefore, in addition to the use of multiple filters 181, 182, 183, . . . , the other practicalfacet of a multi-model adaptation approach is a certain amount of historic book-keeping. The steps in multi-model adaptation are illustrated in FIG. 3. First, an initial model is selected, then several models and filters are run. A minimum is foundfor a key criterion at 200 and a revised model is selected at 210. The adapted estimate is output and looped back to the different model 161, 162, 163, . . . . In the case of the SCALPEL responses, it may be reasonable to consider the unknown thermal contact parameter to be nearly fixed in the whole time-frame of one wafer exposure, then changed but fixed again for a second wafer exposure. For any oneassumed parameter model, if the assumption is relatively bad the Kalman filter 180 behavior will be relatively bad, which will lead to a residual which is "large" in some key criterion. The prediction will diverge from reality and the filter willdefault to an estimate dominated by measurement (K.about.1), but directly limited by measurement noise and not much helped by the model 160. Therefore, the model that reduces some criterion composed from the historic residual should be the "best model" and the Kalman filter 180 should transition from an initial assumption to the selection of this model. In general, this occursgradually since the measurements are noisy, but a large enough amount of data will eventually establish a trend. Effectiveness in many real systems is based on the time-growth of the response associated with the uncertain parameter, such that tolerablylittle error accumulates in the time required to converge on the correct model. The specific length of the history considered and the specific criterion designed to make a selection depend on many factors, such as the duration one would expect theparameter to be nominally fixed, or the ultimate application where the best estimate is needed at a "singular end-event" time instead of all times. Of course, the real state is not known for real situations, but should be known in a Monte-Carlo adaptive Kalman filter simulation, which is a common filter development method. Adaptation criterion and model-selection methods are described below. A decision criterion is based on the history of residuals, where the residual is the vector difference between the measurement and the estimate for the whole state at eachstep, for each model acting in parallel. The momentary position error radius at each step is of interest in the SCALPEL problem. Therefore, the position error radius can be formed from appropriate residual components at each step, and a simple averageerror radius over some history length can be calculated for each model 161, 162, 163, . . . . This average could consider a length of time either shorter than or up to the total time of the system propagation or the full length of the history at eachstep. This average error radius is the best criterion for adaptation in the SCALPEL case. In running an adaptative Kalman filter, the average error radius is calculated for each model number at each time step. As the system propagates, a clear minimum inside the assumed model range occurs, and this almost always corresponds to theselection of the correct model used to generate a truth simulation, unless the state noise effects are overwhelmingly large. The plot is a visual representation of the data that is analyzed at every step to form an adaptation scheme. The correct or "best" model occurs at the model number having the lowest residual radius error over some characteristic averaging time. Essentially, the strength of the minimum within the available model set is used as the selection criterion. The minimum should be both pronounced and sustained. Simulations or trials can be used to determine if the range of models assumed isappropriate to make sure that a minimum can eventually be found. Analysis of the position and strength of this minimum is aided by using a normalized contrast criterion ranging from 0 to 1 to compare the maximum and minimum values of this residual radius error across the model set as a function of timecontrast(k)=[Max-Min]/[Max+Min]@step k where Max and Min refer to the averaged error radius of each model. To translate these fairly small contrast values into a criterion for selecting a given model, it may be useful to use a second weight-determining function. The second weight-determining function should be a smooth function that translates thisbasic contrast evaluation in a simple way, over a normalized range of 0 to 1. The specific function chosen is not important as long as tuning of the parameters is done in simulations. FIGS. 4a and 4b illustrate a function(Adaptweight=1-Exp[-(contrast/strength)^2]) that can be made to saturate the weight versus contrast relationship depending on a single strength parameter (with examples shown for strength=0.2 and 0.5). Therefore, the process of developing an adaptive filter entails tuning the strength parameter to determine the weighting of adaptation. This weight can be considered to be similar to an "outside loop" version of the Kalman gain (K) that goesfrom 0 to 1 as the measurement data provides enough information to select a best model. A distinction is that this weight operates on a whole history of residual data from action of the set of filters, while the K in each filter operates only one stepat a time and within its own assumptions. Although the present invention has been described above as the implementation of a Kalman filter 18 or a multi-model adaptive Kalman filter 180 in a projection electron lithography method or apparatus, other additions or refinements may bepossible including: using the weight to interpolate between discrete models and allow selection of a best model that combines two near-minimal residual models; using a "no-turning-back" scheme where the weight is not allowed to go back down in theunusual event that a longer history of measurements does not continue to converge on a stronger minimum residual (this option makes sense if there must be a singular fixed model and state noise is relatively small, but tuning can become complex if statenoise is large, namely the measurements must counter both noise and parameter uncertainty problems); replacing the starting-assumption model at some threshold weight value with the last adapted model; smoothing of the adaptation process, which may yielda smoother result but not necessarily a better one, and is subject to tuning. FIGS. 5a and 5b illustrate the response of a nominally tuned adaptation scheme based on averaged error radius curves and multi-model execution. Note that in FIG. 5b, the starting assumption is model 190 6, but the truth model is model #4, bothof which lie inside a range from a low at #1 to a high at #9. The weight of adaptation in FIG. 5a rises sharply at about 1/4 the time into the sequence and is locked at its last high value. The model selection oscillates slightly after the assumedmodel is rejected, and then it converges close to the true model. In a preferred embodiment, more than three models are used, and in a more preferred embodiment, five models are used. The SCALPEL wafer-heating response requires a complex heat transfer and elastic strain model based on partial differential equations and boundary conditions, with mixed cylindrical and Cartesian coordinate systems used for key features. Theresponse cannot be simplified by treating only certain dominant modes of the response. The response can be almost arbitrarily complex and variable with several parameters. The dynamic distortion process should be corrected to a few nanometers accuracyat times corresponding to unique sub-field locations throughout the exposure, corresponding to roughly one million model steps in about 2 minutes, or a step rate of 8333 Hz. In each step, a full history-dependent snapshot of an extended system modelwould have to be executed. The likelihood of obtaining even one adequately fast and accurate real-time model is poor, and running an array of models for adaptation may be impractical. However, the Kalman filtering described above is an inherently numerical approach to propagating system state estimates based on differential equations. The Kalman filtering described above is also inherently linear in the way it incrementallyadds a new prediction to the prior estimation of the state using a predictive model. Therefore, it is natural to substitute a sequence of numbers in the matrix positions for what would otherwise be a discretely propagated function-based model. If thenumbers exist a-priori, the linear matrix algebra can be very fast because the differential equations have been effectively solved before-hand. A remaining issue is the speed of the a-priori number generation processor. Since this process is not done in real-time during the one or two minute exposure time, presumably much more time could be taken. However, throughput requirements onthe exposure tool require that such a calculation does not add significant time to the batch process time of many wafers, for example 30 wafers exposed in an hour. The up front calculations have to be some combination of fast and/or done in parallel toother necessary lithography tool functions. Since high throughput is usually associated with repetitive exposure batches, the up-front model variations should be limited to occasions when the pattern (mask) is changed or significant conditions (exposure current or resist dose) mightchange. If at least 25 wafers are run with the expectation of completing them in about an hour, spending one minute overall on computation is acceptable but spending 25 minutes in repetitive computation is not. As stated earlier, the main distinction of each wafer exposure in a batch is likely to be chuck thermal contact and beam drift. However, due to the linearity in the combination of basic elements of the Kalman filter 18, there is nothing aboutthe operation of the Kalman filter 18 that would "feed back" a required change to the basis predictive model. They are uncoupled, and it is well known that many elements of a Kalman filter 18 can be pre-computed and stored to minimize the real-timecomputation burden. This is also true for adaptive Kalman filtering 180 as well, assuming that a whole array of models exists for the full time. In fact, this may be a reason to implement the multi-model adaptation scheme, instead of a scheme thatminimizes the number of models used as the unknown parameter is discerned. If number sequences are chosen for the model, the predictive model and Kalman filter can be decoupled entirely to allow any good model technique to be used for any up-front calculation. A remaining issue in implementing the Kalman filter isdeciding what position the model-result sequence should take in the Kalman filter equations. It is tempting to just substitute the number sequence for the whole predictive step to give X(k+1/k), but this is incorrect. The general reason why it isincorrect is because the .PHI. component of the state space predictor also propagates the state error covariance that makes the filter work. Therefore these substitutions must be consistent and careful. For SCALPEL wafer heating and beam drift response, the nature of the system actually simplifies the model integration problem. The "model" of beam drift propagation may only require the state-noise band-limit filter function. This is consistentwith the idea that the electron beam is a system with negligible inertia. Further, drift noise is instantly and fully added to the position state, and the modified state has no effect on incremental propagation to the next state. Therefore, given the fact that the (D matrix is augmented with this filter function already, the simplest answer is to use a "null" basis state propagation model with the pre-calculation treated as "input", given by: .PHI.=0 U=[X.sub.u, 0,y.sub.u,0].sup.T The x and y entries in U are a sequence of pre-calculated predicted sub-field center responses at known times. The use of a state vector comprised of position and velocity is continued. This approach has been shown to work adequately by simulation. However, other methods are possible. For example, it may be possible to propagate the state noise covariance with a simple, approximate model that has some basic physicalsensibility. As described above, the present invention is directed to a method and apparatus that implements a Kalman filter 18 or an adaptive Kalman filter 180 correction scheme for wafer heating and beam drift in projection electron beam lithography, suchas SCALPEL. The Kalman filter is based on a numerical response model interface that allows efficient integration of relatively slow but infrequent pre-calculation results, and allows real-time adaptive Kalman filter functionality. The present inventiondemonstrates the feasibility of a die-center correction for the critical "global" part of the correction scheme. The local part can be done by pure prediction since the errors are smaller and less subject to effects of drift and chuck contactuncertainty. The adaptive Kalman filter 180 behavior is very good for a scenario that is realistic or somewhat pessimistic in key parameters, including a slow beam drift of typically 40 nm and a 15 nm 3-sigma one-site alignment noise. Adaptation in amulti-model form is effective at handling the problem of at least a factor of two thermal contact parameter uncertainty, in a scenario where the contact is a great deal lower than what we know is possible, hence giving relatively large responses. Combined errors on the order of 50 nm in predicting responses that are well over 100 nm can be reduced to 10 nm or better. With some optimization of the corrector and the benefit of maximum chuck thermal contact, it is likely that error budgetrequirements of nominally 5 nm will be met. Although the various embodiments of the Kalman filter described above may be used to correct for wafer heating, beam drift and/or other errors in a SCALPEL or other projection electron beam lithography system, the present invention is not limitedto correction of these errors. Other correctable errors may include errors related to the current at the wafer, the thickness of the wafer, thermal response parameters (which may include heat capacity, heat conductivity, thermal expansion coefficient,Young's modulus, or Poisson's ratio of Si), wafer-to-chuck frictional contact, wafer-to-chuck thermal contact, wafer initial temperature profile, and/or beam drift (which may be related to charging, stray fields, electronics, and/or thermal factors). It is noted that the functional blocks in FIGS. 1 3 representing the Kalman filter 18,180 and model 16,160 may be implemented in hardware and/or software. The hardware/software implementations may include a combination of processor(s) andarticle(s) of manufacture. The article(s) of manufacture may further include storage media and executable computer program(s). The executable computer program(s) may include the instructions to perform the described operations. The computer executableprogram(s) may also be provided as part of externally supplied propagated signal(s). In an exemplary implementation of the numerical integration approach described above, the real-time operation of a die-by-die Kalman filter, using pre-existing numerical model results, only took 14 seconds to run on a 400 MHz PC runningnoncompiled and relatively-slow Mathematical.RTM. 3.0 by Wolfram Research Inc. Champaign, Ill., with many extra plotting and data output steps. This is easily fast enough for real-time use if die exposures take at least 1 second. This result isexpected because the recursive part of the Kalman filter is mainly linear matrix algebra. Equivalent compiled code runs should be much faster for real tool implementation. Other control system development and simulation software, such as MatLab.RTM.,by the MathWorks Inc., Natick, Mass. could also be used, as could any of the C-family of languages. Although the estimator described above is a Kalman filter, any number of other estimators such as simple observers, full order observers, reduced order observers, trackers, or other estimation techniques known to one of ordinary skill in the artor combinations thereof, are also contemplated by the present application. Still further, although the statistical technique utilized above is a least squares technique, other techniques, such as variance, (linear or not), general optimal, maximumlikelihood, maximum a-posteriori, weighted leased squares, or other techniques known to one of ordinary skill in the art or combinations thereof, are also contemplated by the present application. The invention being thus described, it will be obvious that the same may be varied in many ways. Such variations are not to be regarded as a departure from the spirit and scope of the invention, and all such modifications as would be obvious toone skilled in the art are intended to be included within the scope of the following claims. * * * * * ||Randomly Featured Patents

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CC-MAIN-2019-04

https://byjus.com/question-answer/in-chloroplast-the-highest-number-of-protons-are-found-ininter-membrane-spaceantenna-complexstromalumen-of-thylakoids/

math

In chloroplast, the highest number of protons are found in Lumen of thylakoids The correct option is D Lumen of thylakoids During photosynthetic electron transport, protons (H$$^+$$) accumulate in the thylakoid space. These protons are released into thylakoid space after each split of water molecule during photooxidation and after each electron transport between PQH$$_2$$ to cytochrome-f. Increase in the number of protons in the thylakoid space results in an increase in a proton gradient. Hence, the correct answer is 'Lumen of thylakoids'.

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https://awm-math.org/awards/noether-lectures/noether-lectures-2022/

math

2022 Lecturer: Marianna Csörnyei The Kakeya needle problem for rectifiable sets A planar set admits the “Kakeya property” if it can be moved continuously to any other position covering arbitrary small area during the movement. It was known for more than 100 years that line segments have this property, but until recently there were only very few other known examples. In the talk we will study two variants of this problem, the geometric and the analytic version. In the classical, geometric version, we find all connected closed sets with the Kakeya property. In the analytic version, where we are allowed to delete a null set at each time moment, we will show that every rectifiable set admits the Kakeya property, moreover, they can be moved to any other position covering not only arbitrary small but zero area. Marianna Csörnyei is a Professor of Mathematics at The University of Chicago. She received her Ph.D. from the Loránd Eötvös University in Budapest in 1999. Before joining the faculty at the University of Chicago, she was a Research Fellow at the University College London during 1999 – 2003, a member of the Institute for Advanced Study in Princeton in 2003/2004, and a Professor of Mathematics at the University College London during 2004 – 2011. Csörnyei has made significant contributions to several areas of Mathematical Analysis, including Geometric Measure Theory, Functional Analysis and Real Analysis. While she was still an undergraduate, she established a reputation as a brilliant problem solver. (Even before that, in her high school years, she won a gold medal at the International Mathematical Olympiad.) Later she worked on deep innovative long-term projects. She is known for example for her results concerning various versions of the Kakeya needle problem and for her work on the structure of Lebesgue null sets in Euclidean spaces. The latter work is connected to problems in partial differential equations and the calculus of variations, questions concerning the possibility/impossibility of strengthening the Rademacher Theorem about the almost everywhere differentiability of Lipschitz functions, as well as to some combinatorial problems. Csörnyei was an invited speaker at the 2010 International Congress of Mathematicians and has given lectures at distinguished institutions around the world. She won the Whitehead Prize from the London Mathematical Society in 2002 and the Philip Leverhulme Prize in 2008. In 2019 she was elected an External Member of the Hungarian Academy of Sciences.

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https://study.com/academy/lesson/trapezoid-definition-properties-formulas.html

math

Definition of a Trapezoid A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term 'trapezium,' from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning 'little table.' There are a few special trapezoids that are worth mentioning. In an isosceles trapezoid, the legs have the same length and the base angles have the same measure. In a right trapezoid, two adjacent angles are right angles. If the trapezoid has no sides of equal measure, it is called a scalene trapezoid. A parallelogram is a trapezoid with two sets of parallel sides. There is actually some controversy over whether a parallelogram is a trapezoid. One group states that the definition of a trapezoid is having only one set of parallel sides, which would exclude the parallelogram because it has two sets of parallel sides. The other, more mainstream group, states that the definition of a trapezoid is having at least one set of parallel sides, which includes the parallelogram. For our discussions, because it is the more widely accepted view, we will consider a parallelogram to be a trapezoid. Properties of a Trapezoid The formula for the perimeter of a trapezoid is P = (a + b + c + d). To find the perimeter of a trapezoid, just add the lengths of all four sides together. The formula for the area of a trapezoid is A = (1/2)(h)(a + b), where: - h = height (This is the perpendicular height, not the length of the legs.) - a = the short base - b = the long base An isosceles trapezoid has special properties that do not apply to any of the other trapezoids: - Opposite sides of an isosceles trapezoid are the same length (congruent). - The angles on either side of the bases are the same size or measure (also congruent). - The diagonals are congruent. - Adjacent angles (next to each other) along the sides are supplementary. This means that their measures add up to 180 degrees. Let's try a couple of practice problems to test your newfound trapezoid knowledge. Feel free to pause the video at any point to work through the problems yourself. 1.) Find the perimeter and area of the following trapezoid: To find the perimeter, simply add all four sides together. P = 12mm + 14mm + 18mm + 13mm = 57mm To find the area, use the formula A = (1/2)(h)(a + b). A = (1/2)(11mm)(12mm + 18mm) = 165mm^2 2.) Find the area of the following trapezoid: Again, use the area formula A = (1/2)(h)(a + b). A = (1/2)(6ft)(9ft + 4ft) A = 39ft^2 A trapezoid is a 2-dimensional figure with four sides. In order for it to be classified as a trapezoid, it must have at least one set of parallel sides. Trapezoids play a key role in architecture and also can be found in numerous everyday items. Take a look at the glass you are drinking from at your next meal. From the side, it's probably shaped like a trapezoid. Review the video lesson and its corresponding transcript so that you can: - Define a trapezoid and identify its properties - Illustrate several special trapezoids - Point out the properties of isosceles trapezoids - Find the perimeter and area of a trapezoid To unlock this lesson you must be a Study.com Member. Create your account Practice Trapezoid Questions 1. Given the area of a trapezoid, whose parallel sides are 11 and 13 units respectively, is 36 square units, find the height of this trapezoid or the perpendicular distance between its parallel sides. A sketch of this trapezoid is presented below: 2. First, we have an isosceles trapezoid. The area of this trapezoid is 100 cm^2. The height or the perpendicular distance between the two parallel sides of this trapezoid is 5 cm. One of the parallel sides is 15 cm. What is the length of the other parallel side? 1. The formula for the area of a trapezoid is A = (1/2) x (a + b) x h, where a and b are the lengths of the two parallel sides, and h is the perpendicular distance between the two parallel sides. Then, substituting in the formula gives us: 36 = (1/2) x (11 + 13) x h = (1/2) x 24 x h = 12 x h Then, dividing both sides by 12 yields: 36/12 = 12 x h/12 or 3 = h Hence, the height of this trapezoid is 3 units. 2. Again, the formula for the area of a trapezoid is A = (1/2) x (a + b) x h, where a and b are the lengths of the two parallel sides, and h is the perpendicular distance between the two parallel sides. Then, substituting in the formula gives us: 100 = (1/2) x (a + 15) x 5 Multiplying both sides of the above equation by 2, and then distributing on the right-hand sides yields: 200 = 5a + 75 Subtracting 75 from both sides of the above equation leads to: 200 - 75 = 5a or 125 = 5a Dividing both sides by 5 finally gives us: 125/5 = 5a/5 or 25 = a The length of the other parallel side is 25 cm. Register to view this lesson Unlock Your Education See for yourself why 30 million people use Study.com Become a Study.com member and start learning now.Become a Member Already a member? Log InBack

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CC-MAIN-2022-21

https://www.blueasanorange.com/post/asking-questions

math

There are two types of questions at the core of my approach to language learning. First, the questions that students can ask at any time, without feeling that they should know better. As we teachers always say, there are no silly questions, because they truly aren't! There are only answers to be provided, so that the language point becomes clear and students can use it. Then there are questions I am pushing my students to practise every week, so that they can initiate conversation in the language they are learning, as opposed to waiting for others to talk to them. Each and every one of my students prepares three questions for every lesson. These don't have to be about me. In fact, I actively encourage my students to prepare questions they’re very likely to use in the future, when travelling or talking to their friends/family. In other words, I’m looking, as always, at the practical application of language learning, as opposed to the theory of it all. Speaking during a lesson equals rehearsing for when you're going to speak. My job as a teacher, is to make sure that my students are clear about the language points they're using, about the logic of the language they're studying, and they have confidence in both and therefore in themselves. I create a safe space for them to try different ways of saying what they want to say, so they can have a go on their own and trust themselves to figure out a way to have a conversation. More about my teaching approach Don't hesitate to send me your questions #1. Is it too late to learn a language? #2. Not knowing the word is not the end of the conversation #3. Is learning vocabulary lists a good idea? #4. Should I use bilingual books? #5. Practising all four skills equally #6. On the importance of making mistakes #7. You know more than you think you do #8. Can everyone learn a language? #9. With or without subtitles? #10. Should I use a dictionary? #11. What is the main obstacle when learning a language? #12. There's no such thing as perfect sentences #13. Learning French when English is your mother tongue #15. Language learning does not mean translating #16. The right tools to learn a language #18. Leaving your linguistic comfort zone behind

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https://minimalsurfaces.blog/home/repository/tori/hoffman-karcher-tori/

math

The planar middle end of the Costa surface can be deformed into a catenoidal end. This gives a 1-parameter family of embedded minimal tori, as first announced in a 1987 paper by William Meeks and David Hoffman, described in Celso Costa’s Classification paper and proven to exist and be embedded in the the Finite Total Curvature paper by David Hoffman and Hermann Karcher. With changing parameter, the separation of the two catenoidal necks becomes more pronounced. Costa has shown that these are the only embedded, 3-ended minimal tori of finite total curvature. It is an open problem whether there are any other embedded minimal tori of finite total curvature (for infinite curvature, there is the genus one helicoid). It is also an open problem to come up with a conceptual proof of Costa’s result. D. Hoffman and W.H. Meeks III: Properties of properly embedded minimal surfaces of finite topology, Bull. Amer. Math. Soc. 17 (1987), 296-300. C. J. Costa: Classification of complete minimal surfaces in R³ with total curvature 12π, Inventiones mathematicae 105 (1991), 273–303. D. Hoffman, H. Karcher: Complete embedded minimal surfaces of finite total curvature, in Geometry V, Encyclopedia Math. Sci. 90 Springer, (1997) 5-93. - Costa Surface - The Genus One Helicoid

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CC-MAIN-2023-14

https://gmatclub.com/reviews/comments/e-gmat-gmat-live-prep-345350035

math

You have found 1 out of 13 Christmas decorations! Find them all And get a super prize from GMAT Club: |All Reviews > e-GMAT > e-GMAT GMAT Live Prep > Review Comments| Joined: Jun 23, 2014 700 Q50 V35 In my first attempt, I signed up for a bangalore based test prep company and studies for more than 7 months and got Q40. The problem was that the did not teach the basic concepts very clearly and instead started teaching to solve complex problems. Also they taught shortcuts for each problem and they took a different approach to solve each problem What I found at the end was that the shortcuts and the approaches worked only for the problems from their course and when i took the actual test, i could not solve most of the problems and got Q40. Then I found e-GMAT course. Initially I used only verbal live prep and i was my accuracy increased in verbal after using e-GMAT course. Then I thought i will just spend extra $150 and see how this quant live prep is going to be. Later I realized that that was the best thing i did. I am out of college for almost 7-8 years now and this Quant course was very helpful to teach me all math concepts required for GMAT exam. I digested all the concepts very clearly and had a standard approach for each question type. Even though it took little longer to apply the standard approach for each problem, it helped me improve my accuracy. Its certain that we cannot answer all the questions correct in GMAT, but the eGMAT concepts with enough practice tests will be sufficient to score Q 50 / 51. After using this math course for 3 months, i got Q-50 in my second attempt.

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CC-MAIN-2021-10

https://www.rediff.com/cricket/report/ipl-2011-stats-dhoni-badrinath-csk/20110513.htm

math

Statistical highlights of Thursday's Indian Premier League match between Chennai Super Kings and Delhi Daredevils. # Chennai Super Kings have won eight and lost four of 12 matches played in the IPL 2011. With 16 points, they are placed at the top of the points table. # Delhi Daredevils is the second team after Deccan Chargers to lose eight matches in the present edition of the IPL. # Chennai have won four and lost three out of seven played against Delhi -- winning percentage is 57.14. # Suresh Raina (14 off 10 balls) is the first batsman to complete 1700 runs in the IPL, totaling 1709 at an average of 36.36, including twelve fifties, in 58 games. # Raina has taken his tally of sixes in the IPL to 69, emulating Yusuf Pathan's tally. Adam Gilchrist (71) has hit more sixes than Raina. # Mahendra Singh Dhoni (63 not out off 31 balls) posted his seventh half-century in the IPL. # Dhoni has hit 46 sixes in the IPL -- the second best tally for Chennai, behind Raina's 69. # Dhoni has aggregated 1295 runs at an average of 39.24 in 55 matches -- the second highest for Chennai Super Kings in the IPL. # Subramaniam Badrinath (55 off 43 balls) became the first batsman to register five fifties in the IPL 2011. # Badrinath averages 87.25 in the IPL 4, totaling 349 in 12 innings at a strike rate of 133.71. # Badrinath has registered 10 fifties in the IPL â his runs' tally being 1074 (ave 35.80) in 58 matches, at a strike rate of 125.17. # Badrinath and Dhoni were involved in a partnership of 96 for the fourth wicket -- Chennai's highest against Delhi for any wicket in the IPL. # Pathan has aggregated 750 at an average of 24.19 in 54 matches in the IPL. # Venugopal Rao (30 off 20 balls) has scored 320 runs at an average of 29.09 in 12 matches in IPL 2011. # For Chennai, three bowlers have bagged 14 wickets each in IPL 4 -- Doug Bollinger (ave 14.92), Albie Morkel (21.57) and Ravichandran Ashwin (21.78) # Dhoni has got the Man of the Match award for the sixth time in the IPL -- a tally exceeded by Yusuf Pathan (10) and Sachin Tendulkar (7). # Naman Ojha (25 off 14 balls, including two sixes) has 50 sixes in Twenty20 -- 51 in 51 innings. # The three top wicket-takers for Delhi in the IPL 4 so far are Morne Morkel (12), Ajit Agarkar (8) and Irfan Pathan (8). # Irfan Pathan (44 not out off 34 balls) registered his highest innings in IPL 2011.

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CC-MAIN-2020-45

https://snucongo.org/triangles-pqr-and-xyz-are-similar/

math

If two triangles are similar then their equivalent anglesare equal and corresponding sides space proportional. You are watching: Triangles pqr and xyz are similar Here, the two triangles XYZ and also PQR room similar. So, ∠X = ∠P, ∠Y = ∠Q, ∠Z = ∠R and (fracXYPQ) = (fracYZQR) = (fracXZPR). ∆XYZ is comparable to ∆PQR. We write ∆XYZ ∼ ∆PQR (the prize ‘∼’ means ‘similar to‘.) Sides opposite come equal angle in similar triangles are known as matching sides and also they room proportional. Here indigenous the given numbers ∠X = ∠P, ∠Y = ∠Q and also ∠Z = ∠R. Therefore, XY and also PQ are corresponding sides together they room opposite come ∠Z and ∠R respectively. Similarly indigenous the offered figure, YZ and also QR are a pair of equivalent sides. XZ and PR are likewise a pair of matching sides. Thus, (fracXYPQ) = (fracYZQR) = (fracXZPR), as corresponding sides of similar triangles are proportional. Angles opposite come proportional sides in similar triangles are known as matching angles. If ∆XYZ ∼ ∆PQR and also (fracXYPQ) = (fracYZQR) = (fracXZPR) climate ∠X = ∠P as they space opposite to equivalent sides YZ and QR respectively. Similarly native the offered figure, ∠Y = ∠Q and ∠Z = ∠R. Congruency and Similarity of Triangles: Congruency is a certain case of similarity. In both the cases, three angles that one triangle room equal come the three equivalent angles that the other triangle. However in similar triangles the matching sides are proportional, when in congruent triangles the corresponding sides room equal. ∆XYZ ∼ ∆TUV. Therefore, (fracXYTU) = (fracYZUV) = (fracXZTV)= k, where k is the consistent of proportionality or the scale element of sizetransformation. ∆XYZ ≅ ∆PQR. Here, (fracXYPQ) = (fracYZQR) = (fracXZPR)= 1. Therefore, in congruent triangle the consistent ofproportionality in between the equivalent sides is same to one. Thus,congruent triangles have the very same shape and size while comparable triangles havethe same shape however not have to the very same size. Congruent triangle are always similar, but comparable trianglesare no necessarily congruent. Note: Triangles the are comparable to the same triangle aresimilar to every other. Here, ∆XYZ ∼ ∆PQR and also ∆ABC ∼ ∆PQR. Therefore, ∆XYZ ∼ ∆ABC. 9th class Math From Similar Triangles to home PAGE New! CommentsHave your say about what you just read! leaving me a comment in the box below. Ask a inquiry or answer a Question. See more: How Many Orbitals Are In The 4P Sublevel ? The Orbitron: 4P Atomic Orbitals Didn"t uncover what you were looking for? Or want to know more informationabout Math only Math.Use this Google search to find what you need. Share this page: those this?

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CC-MAIN-2022-27

https://statshelponline.com/partial-least-squares-regression-assignment-help-21532

math

Partial Least Squares Regression Assignment Help For example, you might approximate (i.e., forecast) an individual’s weight as a function of the individual’s height and gender. You might utilize direct regression to approximate the particular regression coefficients from a sample of information, determining height, weight, and observing the topics’ gender. For lots of information analysis issues, price quotes of the direct relationships in between variables are sufficient to explain the observed information, and to make affordable forecasts for brand-new observations (see Numerous Regression or General Step-by-step Regression for extra information. PLS regression is mostly utilized in the chemical, food, plastic, and drug markets. In PLS regression, the focus is on establishing predictive designs.The algorithm minimizes the number of predictors utilizing a method comparable to primary elements analysis to draw out a set of parts that explains optimum connection in between the predictors and action variables. Minitab then carries out least-squares regression on these uncorrelated parts. Partial Least Squares regression (PLS) is a fast, ideal and effective regression technique based upon covariance. It is advised in cases of regression where the variety of explanatory variables is high, and where it is most likely that the explanatory variables are associated.An excellent benefit of PLS regression over traditional regression are the readily available charts that explain the information structure. It can be relationships amongst the reliant variables or explanatory variables, as well as in between reliant and explanatory variables. This example reveals how to use Partial Least Squares Regression (PLSR) and Principal Elements Regression (PCR), and talks about the efficiency of the 2 approaches. PLSR and PCR are both techniques to design a reaction variable when there are a big number of predictor variables, and those predictors are extremely associated or even collinear. Both approaches build brand-new predictor variables, understood as elements, as direct mixes of the initial predictor variables, however they build those parts in various methods.The Partial Least Squares Regression treatment approximates partial least squares (PLS, likewise called “forecast to hidden structure”) regression designs. PLS is a predictive method that is an alternative to regular least squares (OLS) regression, canonical connection, or structural formula modeling, and it is especially beneficial when predictor variables are extremely associated or when the variety of predictors goes beyond the variety of cases. PLS integrates functions of primary elements analysis and numerous regressions. It initially draws out a set of hidden aspects that discuss as much of the covariance as possible in between the reliant and independent variables. A regression action anticipates worths of the reliant variables utilizing the decay of the independent variables.X is an n-by-p matrix of predictor variables, with rows corresponding to columns and observations to variables. XL is a p-by-comp matrix of predictor loadings, where each row consists of coefficients that specify a direct mix of PLS parts that approximate the initial predictor variables. YL is an m-by-comp matrix of reaction loadings, where each row consists of coefficients that specify a direct mix of PLS elements that approximate the initial action variables. We construct connections in between envelopes, a just recently proposed context for effective evaluation in multivariate stats, and multivariate partial least squares (PLS) regression. In specific, we develop an envelope as the nucleus of both Univariate and multivariate PLS, which unlocks to pursuing the exact same objectives as PLS however utilizing various envelope estimators. It is argued that a likelihood-based envelope estimator is less conscious the variety of PLS parts that are picked which it surpasses PLS in forecast and evaluation. The derivation of analytical residential or commercial properties for Partial Least Squares regression can be a difficult job. In this work, we study the intrinsic intricacy of Partial Least Squares Regression.We reveal that the Degrees of Liberty depend on the cob linearity of the predictor variables: The lower the co linearity is, the greater the Degrees of Liberty are. In specific, they are usually greater than the ignorant technique that specifies the Degrees of Liberty as the number of elements. Even more, we highlight how the Degrees of Liberty technique can be utilized for the contrast of various regression approaches.The approach is based on partial least squares regression, which breaks down the thermo graphic PT information series gotten throughout the cooling routine into a set of hidden variables. The regression approach is used to speculative PT information from a carbon fiber-reinforced composite with simulated flaws. The speaker utilizes customer rankings for 24 kinds of bread to show ways to utilize PLS to recognize item credit to assist brand-new bread formula and style procedures. He utilizes leave-one-out cross-validation and demonstrates how to analyze and analyze Root Mean Press; NIPALS Fit x and y ratings for a single hidden element; Diagnostic Plots; and VIP vs. Coefficients Plots. He utilizes the Forecast Profiler to optimize desirability. This paper provides a brand-new strategy for predictive heart movement modeling and correction, which utilizes partial least squares regression to extract intrinsic relationships in between three-dimensional (3-D) heart contortion due to respiration and numerous one-dimensional real-time quantifiable surface area strength traces at chest or abdominal area. In spite of the reality that these surface area strength traces can be highly paired with each other however inadequately associated with respiratory-induced heart contortion, we show how they can be utilized to precisely forecast heart movement through the extraction of hidden variables of both the input and output of the design. You might utilize direct regression to approximate the particular regression coefficients from a sample of information, determining height, weight, and observing the topics’ gender. For lots of information analysis issues, price quotes of the direct relationships in between variables are sufficient to explain the observed information, and to make affordable forecasts for brand-new observations (see Numerous Regression or General Step-by-step Regression for extra information.

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CC-MAIN-2018-30

https://chrome.google.com/webstore/detail/guzinta-math-order-and-ab/cdhgapbgpjjieblpijgjjfckbepofblk

math

Math lesson for 6th grade for teaching inequalities. Video instruction, worked examples, guided practice, and instructor notes. For one-on-one or whole-class mathematics instruction, use this Guzinta Math lesson as a supplement for teaching about comparing and ordering rational numbers, estimating the magnitude of fractions, and absolute value--key concepts in 6th grade. The app includes detailed instructor notes for each guided practice problem, with notes about extending discussion around each worked example and problem. In this lesson, students have the opportunity to practice estimating the magnitude of fractions using a blank number line. Instructors and/or students can turn on and use the new canvas feature (with improved writing functionality) in the app to write all over the screen--take notes, highlight key terms and ideas, show work, etc.

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CC-MAIN-2017-17

https://somme2016.org/recommendations/what-is-a-logical-argument-called/

math

What is a logical argument called? The building blocks of a logical argument are propositions, also called statements. A proposition is a statement which is either true or false. For example, “Washington D.C. is the capital of the United States.” There are three stages to creating a logical argument: Premise, inference, and conclusion. What is example of logical argument? Example. The argument “All cats are mammals and a tiger is a cat, so a tiger is a mammal” is a valid deductive argument. Both the premises are true. To see that the premises must logically lead to the conclusion, one approach would be use a Venn diagram. What is the best definition of a logical argument quizlet? An argument is a set of statements, some of which provide support for another statement. An argument is a set of assertions. A. True or False: According to Sutherland, a ‘subject’ is a topic to be discussed. What is the nature of logical argument? A logical argument is a process of creating a new statement from the existing statements. It comes to a conclusion from a set of premises by means of logical implications via logical inference. An argument can be valid or invalid. An argument can have more than one premises but one conclusion. What is an example of a sound argument? A sound argument is one that is not only valid, but begins with premises that are actually true. The example given about toasters is valid, but not sound. However, the following argument is both valid and sound: In some states, no felons are eligible voters, that is, eligible to vote. What is the best definition of an argument? 1 : a reason or the reasoning given for or against a matter under discussion — compare evidence, proof. 2 : the act or process of arguing, reasoning, or discussing especially : oral argument. Which is the best definition of argument as used in a logic class? What is a sound argument? A sound argument is a valid argument that has true premises. A cogent argument is a strong non-deductive argument that has true premises. And we defined an argument as being strong if it’s a non-deductive argument in which the premises succeed in providing strong support for the conclusion. How many types of logical reasoning are there? These two types include logical reasoning and analytical reasoning. What makes a logical argument sound? A deductive argument is sound if and only if it is both valid, and all of its premises are actually true. In effect, an argument is valid if the truth of the premises logically guarantees the truth of the conclusion. What are the elements of a logical argument? Premises – A logical argument must have premises. A good, complete argument clearly states those premises and identifies them as premises. Premises are declarative statements known as propositions from which the conclusion is concluded. They are the assumptions. What are some examples of bad arguments? Divorce Over Sodomy Image via A Hot Mama. Example one: man divorces his wife. What are examples of logical arguments? Though arguments do use inferences that are indisputably purely logical (such as syllogisms), other kinds of inferences are almost always used in practical arguments. For example, arguments commonly deal with causality, probability and statistics or even specialized areas such as economics. What are the styles of argument? Standard types. There are several kinds of arguments in logic, the best-known of which are “deductive” and “inductive.”. An argument has one or more premises but only one conclusion. Each premise and the conclusion are truth bearers or “truth-candidates”, each capable of being either true or false (but not both).

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CC-MAIN-2023-40

https://www.jiskha.com/display.cgi?id=1301753259

math

posted by Mark A 33kg traffic light hangs from a vertical beam by two wires. The wire on the left side of the light has an angle of 53 degrees to the beam and the right side wire has an angle of 37 degrees with the beam. Calculate the tension in each of the wires. Would the tension just be 33kg(sin53) + 33kg(cosin53) for the left side, and 33kg(sin37) + 33kg(cosine37) for the right side? Call the left side tension T1 and the right side tension T2. Vertical and horizontal equilibrium equations are: Mg = T1 cos53 + T2 cos37 0 = T1 sin53 - T2 sin37 Mg = T1 cos 53 + T1*cos37*(sin53/sin37) T1 = Mg/(0.6918 + 1.0598) = 0.5709 M*g You set up the equations or did the algebra incorrectly, and you also omitted the g factor that you need to get the tension in Newtons. Thanks , I knew this just didn't seem right.

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CC-MAIN-2018-17

http://www.pitt.edu/~super1/lecture/lec0821/016.htm

math

will talk about how to analyze the hazard period data and control data to estimate the effect of exposure on the outcome. Since there are 2 types of control data, we will also have 2 types of analyses. If the control data are from a comparable control period, the analysis will be the same as that for a matched case-control study. However, instead of concordant and discordant pairs of study subjects in a matched case-control study, here we will have concordant and discordant pairs of exposures from the hazard period and control period. That is, the pairs are made up of two intervals(hazard period and control period) for each subject, and the pairs are either concordant or discordant for exposure. After arranging the data in a 2 by 2 table as shown in the slide, we can just use the standard method for matched case-control to get the odds ratio.

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CC-MAIN-2018-05

https://www.shaalaa.com/question-bank-solutions/every-quadratic-equation-has-at-least-two-roots-nature-of-roots-of-a-quadratic-equation_269732

math

True or False Every quadratic equation has at least two roots. This statement is False. For example, a quadratic equation x2 – 4x + 4 = 0 has only one root which is 2. Concept: Nature of Roots of a Quadratic Equation Is there an error in this question or solution?

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CC-MAIN-2023-14

https://biology.stackexchange.com/questions/93668/is-cloning-exact-or-almost-similar-to-parent

math

I was studying "Reproduction in Organism" as an interest of my own and there was a line offspring produced through single parent are exact copies of their parent. Also, they are genetically identical. So, in here, it says, "genetically identical" (I assume that identical here implies exactly same) but then the next line quite confused me..... The term clone is used to describe such morphologically and genetically similar individuals. Here, the term "genetically similar" is used. So, clones (offspring produced through asexual reproduction- involvement of one parent) are "genetically exact" to their parents or genetically similar? If "genetically exact", Then how is it possible since, there is 'No 100% efficient DNA Replication' that can be achieved?

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CC-MAIN-2021-10

https://blablawriting.net/integers-negative-and-non-negative-numbers-and-absolute-value-essay

math

Integers: Negative and Non-negative Numbers and Absolute Value - Pages: 2 - Word count: 422 - Category: Values A limited time offer! Get a custom sample essay written according to your requirements urgent 3h delivery guaranteedOrder Now Integers are the first numbers that we learn to use. Along with their usefulness in everyday life, integers are building blocks from which all others numbers are derived. The integers are all the whole numbers including zero, all negative and all the positive numbers Basics of integers * Whole numbers greater than zero are called positive integers. These numbers are to the right of zero on the number line. * Whole numbers less than zero are called negative integers. These numbers are to the left of zero on the number line. * The integer zero is neutral. It is neither positive nor negative. * The sign of an integer is either positive (+) or negative (-), except zero, which has no sign. * Two integers are opposites if they are each the same distance away from zero, but on opposite sides of the number line. One will have a positive sign, the other a negative sign. In the number line above, +3 and -3 are labelled as opposites. * We compare integers just as we compare whole numbers. For any two numbers graphed on a number line, the number to the right is the greater number and the number to the left is the smaller number * The absolute value of a number is the number’s distance from 0 on the number line. Addition and subtraction of integers To Add Two Numbers With the Same Sign · Step 1. Find the sum of their absolute values. · Step 2. Use their common sign as the sign of the sum. E.g. 12 + 3 = 15 To Add Two Numbers with Different Signs 1. Find the absolute value of each integer. 2. Subtract the smaller number from the larger number you get in Step 1. 3. The result from Step 2 takes the sign of the integer with the greater absolute Value. E.g. -3 + 5 = 5 – 3 = 2 The only thing you need to remember is you can rearrange a subtraction to an addition (only in most cases) Multiplying and dividing integers The product of two numbers having different (opposite) signs is a negative number E.g. 4 x -3= -12 The product of two numbers having the same sign is a positive number. E.g. 4 x 3 =12 The quotient of two integers having the same sign is positive. E.g. -24 /-3 = 8 The quotient of two integers having opposite (different) signs is negative. E.g. -24/3 = -8

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CC-MAIN-2023-50

https://blog.recessedlighting.com/recessed-lighting-placement/

math

Recessed lighting placement in the ceiling is calculated based on the surface, object, or area to be illuminated. In this post, I’ll explain how to manually calculate the placement for your lights, and then give you a calculator that does it all for you! The formula for spacing recessed lights is the distance between the lights is always double what it is at the ends. This formula, combined with the number of lights, the layout, and the dimensions of the room or surface, is what’s used to calculate the placement of recessed lighting in the ceiling. Note: Before you can calculate the placement, you need to know the layout for your lights. Calculating Placement for General and Task Lighting - Determine the desired “area” that is to be illuminated (It may be the whole room, a section of a larger room, or a work surface). - Measure the length of the area and write down your answer. - Divide the length of the area by twice (2x) the number of lights to be placed in that row and write down your answer. This will be the distance from the wall to your first light in that row. - Double (2x) your answer from the previous step and write down your answer. This will be the distance between the rest of the lights in that row. So remember, the distance between your lights is always twice (2x) the distance from the wall to the first light.[divider] - Now do the same for the width of the area: Measure the width of the area and write down your answer. - Divide the width of the area by twice (2x) the number of lights to be placed in that row and write down your answer. This will be the distance from the wall to your first light in that row. - Double (2x) your answer from the previous step and write down your answer. This will be the distance between the rest of the lights in that row. Example: Room with 6 recessed lights Calculating Placement for Accent Lighting Accent lighting is somewhat of a broad term, so I’d like to clarify that this formula is specifically for calculating the placement of recessed lights that use adjustable trims for the purpose of highlighting a painting or object on the wall. The optimal aiming angle to minimize glare is 30-degrees from the ceiling, so that will be the starting point. Don’t worry if the light cannot be placed at exactly 30-degrees. There may be something in the ceiling like framing or an air duct that prevents you from installing a light there. In this case you can just place the light as close to the ideal location as possible. Most accent trims have plenty of adjustment to compensate for various placements. By using a trigonometric formula, we can calculate the ideal placement for your fixture(s). In the diagram to the right, you will notice that the accent light, the wall, and the ceiling form a right triangle. This type of triangle is known as a “30-60-90” right triangle. Since we know all three angles of the triangle, can measure the distance of side b (from the center of your object on the wall to the ceiling), we only need to solve for a. This will be the distance away from the wall to place your accent light(s) to achieve the desired 60-degree aiming angle. The rule for this type of triangle is that the sides always have a ratio of 1 : 2 : √3. Using this rule and the known side of the triangle (side b), we can use the following formula to solve for side a: Side a=(Side b√3)/3. As promised, there is a picture light calculator on this page that does it all for you!

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CC-MAIN-2023-50

https://gulpmatrix.com/ludwig-boltzmanns-constant/

math

Boltzmann constant which is denoted by the symbol “k,” is a fundamental physical constant that relates the average kinetic energy of particles in a gas to the temperature of that gas. It was first introduced by the Austrian physicist LudwigBoltzmann in the late 19th century as a means of explaining the different behavior of gases.. One of its applications is in the ideal gas law, this law states that pressure of an ideal gas is directly proportional to its temperature and also to the number of particles present in a given volume. The ideal gas law can be written as; PV = nRT Where P is represented as the pressure, V is represented as the volume, n represented as the number of particles, R is represented as the ideal gas constant, and T is represented as the temperature. The Boltzmann constant can be expressed in terms of the ideal gas constant as; k = R/NA Where NA is the Avogadro’s number, which is the number of particles present in one mole of a substance. The value of Boltzmann constant is approximately 1.38 x 10^-23 joules per kelvin (J/K). This figure may seem like a relatively small number, but it is actually very important in determining the behavior of gases at different temperatures. For example, when the temperature of a gas increases, the average kinetic energy of the particles also increases, leading to an increase in pressure. Boltzmann constant plays an important role in the study of thermodynamics. Thermodynamics is the study of the relationship between heat and work and the second law of thermodynamics states that the total entropy (a measure of the disorder or randomness of a system) of a closed system will always increase over time. Boltzmann constant is used in calculating the change that has been noticed in the entropy for a system undergoing a change in temperature. In addition to its importance in the study of gases and thermodynamics,Boltzmann constant has also been used in different fields like the field of statistical mechanics, which is the study of the behavior of large numbers of particles in a system. It is used in calculating the probability of a particle being in a particular energy state, and can also help to explain why certain substances behave in the way they do at the molecular level. Overall, Boltzmann constant is a fundamental physical constant that plays a crucial role in our understanding of the behavior of gases and their relationships with heat, work, and energy. Its applications are vast and varied, and it continues to be a very important tool in many areas of science and engineering. What is Boltzmann constant? Boltzmann constant is defined as the fundamental physical constant that relates the average kinetic energy of particles in a gas to the temperature of the gas. It was named after the Austrian physicist Ludwig Boltzmann, who developed the concept of statistical mechanics. It is denoted by “k” and has a value of approximately 1.38 x 10^-23 J/K. What is the relationship between Boltzmann constant and temperature? Boltzmann constant relates the average kinetic energy of particles in a gas to the temperature of the gas. The average kinetic energy of the particles in a gas is directly proportional to the temperature of the gas, according to the equation: E = (3/2)kT where E is the average kinetic energy of the particles, k is represented as Boltzmann constant, and T is the temperature of the gas. This equation is known as the equipartition theorem. What is Boltzmann constant used for? It is used in calculating thermodynamic properties of gases, such as pressure, volume, and internal energy of a gas. It is also used in calculating the entropy of a system, which is the measure of the amount of thermal energy that is unavailable to do work. In addition, Boltzmann constant is used in calculating the rate at which chemical reactions occur and the rate at which particles in a gas collide with each other. Is Boltzmann constant an exact value? Boltzmann constant is known for its high accuracy. However, like all physical constants, the value of the Boltzmann constant is not an exact value, rather it is a measured value that is subject to uncertainty. Its current value is determined by the International Committee for Weights and Measures (CIPM), is k = 1.380 649 x 10^-23 J/K, with an uncertainty of about 0.000 000 083 x 10^-23 J/K. Is Boltzmann constant the same for all gases? Yes, Boltzmann constant is the same for all gases, regardless of the specific properties of the gas and this is because Boltzmann constant relates the average kinetic energy of particles in a gas to the temperature of the gas, and the kinetic energy of a particle depends only on its mass and velocity, which are independent of the type of gas in which the particle is found.

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CC-MAIN-2023-06

https://news.ycombinator.com/item?id=12420362

math

1. TOC and TOP ask different questions. 2. The disparity of the computational complexity involved in the two classes of models is so great, that it is objective proof that they represent two distinctly different things, and therefore comparing them directly is meaningless. That a jet and a bicycle require vastly different amounts of energy to power is conclusive and objective proof that completely different tradeoffs must be involved in choosing them.

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CC-MAIN-2019-26

https://study.com/academy/topic/working-with-exponential-logarithmic-functions.html

math

About This Chapter Working with Exponential & Logarithmic Functions - Chapter Summary High school students need to read and understand a great deal of information both before they graduate and as they move on in college or the professional world. This chapter provides students with focused training on developing their ability to work with exponential and logarithmic functions. Instead of an endless chapter of information, the chapter has been divided into short, individualized lessons. Students have the option to review all of the lessons in order, but they can also pick and choose lesson topics in which they need the most assistance. The chapter menu provides a list of all lessons available, and since each lesson is only focused on a single topic, students can easily find what they need. After completing this chapter, students will be ready to: - Understand and describe the transformation of exponential functions - Define and use the natural base e - Write inverses of logarithmic functions - Define and identify basic and shifted graphs of logarithmic functions - Use the change-of-base formula for logarithms - Calculate rate and exponential growth as used in the population dynamics problem - Analyze end behavior of exponential and logarithmic functions 1. Transformation of Exponential Functions: Examples & Summary A transformation within an exponential function involves different changes to a graph. Explore more about transformations, the basic exponential function, the three types of changes, and examples of each. 2. Using the Natural Base e: Definition & Overview The natural base e is an irrational number used in calculations and graphing. Learn more about the definition of the natural base e, explore the ways it is used in logarithms and inverse functions, discover how it appears graphically, and apply it to calculating compounds. 3. Writing the Inverse of Logarithmic Functions Logarithmic functions can be notated in reverse, where the expressions are communicated the same way but are written inversely. Learn this concept through a set of examples, and discover how to solve for the inverse as well. 4. Basic Graphs & Shifted Graphs of Logarithmic Functions: Definition & Examples Logarithmic functions can appear on graphs and be shifted by altering their equation. Learn how graphs can be flipped and shifted, both horizontally and vertically, through the examples provided in this lesson. 5. Using the Change-of-Base Formula for Logarithms: Definition & Example For logarithms, using the change-of-base formula is essential when trying to calculate a log of a base that is not the standard base 10. Learn more about this formula for logarithms, including the bases, how to use the formula, and an example of it in practice. 6. Calculating Rate and Exponential Growth: The Population Dynamics Problem Calculating rate and exponential growth is a method that can be used to solve the population dynamics problem. Learn what the population dynamics problem is about, and study the population growth formula and how its also used in other rate problems. 7. Behavior of Exponential and Logarithmic Functions Exponential and logarithmic functions are the inverse of one another, the variable being in the exponent in the former and serving as the argument in the latter. Discover the definitions and behavior of these related functions. Earning College Credit Did you know… We have over 220 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. To learn more, visit our Earning Credit Page Other chapters within the High School Precalculus: Homeschool Curriculum course - Graphing with Functions Review - Working With Inequalities Review - Absolute Value Equations Review - Working with Complex Numbers Review - Quadratic Functions Review - Solving & Graphing Quadratic Functions - Polynomial Functions Review - Working with Polynomial Functions - Graphing Piecewise Functions - Graph Symmetry - Rate of Change - Rational Functions & Difference Quotients - Rational Expressions & Equations Review - Graphing Rational Functions - Exponential & Logarithmic Functions Review - Trigonometric Functions & Graphs Review - Solving Trigonometric Identities - Trigonometric Graphs & Applications - Understanding Vectors, Matrices & Determinants - Working with Matrices & Determinants - Mathematical Sequences & Series Review - Sets in Algebra - Analytic Geometry & Conic Sections Review - Polar Coordinates and Parameterizations - Precalculus: Homeschool Assignments & Projects - Homeschool Resources for Precalculus in High School

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CC-MAIN-2022-33

https://kasmana.people.cofc.edu/MATHFICT/mfview.php?callnumber=mf559

math

Garmt de Vries The members of the Gun Club want to use a giant cannon's recoil to change the Earth's rotation axis, so they can exploit the presumed coalfields at the North Pole. An unfortunate side effect is that many regions will be submerged in the sea or end up at unpleasant altitudes. The world holds its breath as the gun is fired, but nothing happens. It turns out that Maston, who did all the calculations, forgot three zeros in the beginning, which led to a result that was wrong by a factor 1018. The novel contains a supplementary chapter by Alain Badoureau, a mine engineer who helped Verne with all the math. In this extra chapter, all the calculations are rigorously presented. Garmt de Vries has written to me with other suggestions of works by Jules Verne to add to this database. Some of them fall below the borderline of what I consider to be mathematical fiction. (I'm afraid that I have to be somewhat restrictive about what is included, or else the list will become too long to be useful!) Sans Dessus Dessous is included not because of the appendix containing computations, but because mathematics is truly essential to the plot and because it says something about mathematics. Here are some other suggestions from de Vries: Garmt de Vries - Voyage au centre de la Terre (1864, English: Journey to the centre of the Earth) starts with the deciphering of a cryptogram, using a simple transposition. See chapters 2--5. - Mathias Sandorf (1885) also starts out with a message encrypted by transposition, but slightly more complicated, with the use of a grid. See vol. 1, ch. 4. - La jangada (1881, English: Eight Hundred Leagues on the Amazon) is all about cryptography. The first lines of the novel consist of the cryptogram to be solved. In the end, a man's life depends on the information in this document, and several chapters are spent trying to decipher it. The system used is a Vigenere system, where each letter of the clear text can be represented by different letters in the cryptotext, depending on its position. The solution is found in the nick of time, when the name of the document's author is found out. This yields the key to the cryptogram, and the protagonist's life is saved. When this novel was first serialised before appearing as a volume, a math student apparently solved the riddle before the necessary clue was given or even hinted at. Jules Verne went to see the student, who explained how he had done it. Verne was much impressed. The method used by the student is not known. It may have been something like Kasiski counting, or using a probable word approach. - Aventures de trois Russes et de trois Anglais (1872, English: Meridiana) is about an Anglo-Russian team of astronomers who set out to measure a meridian in southern Africa. Chapter 4 gives all the gory details of trigonometry, along with a history of the metre, and why it is important to measure a meridian. One of the characters is an absent-minded mathematician, who wanders off into the wild for days and is almost eaten by crocodiles, because he is verifying the logarithmic tables of James Wolston (ch. 9). - In Hector Servadac (1877, English: Off on a comet), a group of people is taken away by a comet that collides with the Earth. One of them is a French astronomer, who is often compleely absorbed in his calculations. At one point, he gives a lecture on how to calculate the mass, density, etc. of their comet. See vol.2 , ch. 8. - Mirifiques aventures de Maître Antifer (1894) is based on a geometrical problem. Three characters inherit an immense treasure, which is buried on an island. Each time they receive a longitude that they have to combine with a latitude in the possession of another heir to find the location of a new island. In the end, they have visited three islands, and find a document that has become illegible. There are some traces of text: "it suffices... circumference... pole...". The location of the last island is the centre of the circle through all three islands. It is a bit silly that they determine this final location using only a globe and a ruler, but the idea is nice. In this section, you've listed many works by Verne involving Math. I think the following two also belong to that list: a. From the Earth to the Moon: where the author discusses details of possible geometric ways of communication with extraterrestrials, mathematical considerations of the trip to the moon, orbital recalculations, etc b. Mysterious Island: where geometry and a simple apparatus is used to fix the castaway's longitude and latitude.

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CC-MAIN-2024-10

https://math.stackexchange.com/questions/92986/weil-and-cartier-divisors-on-a-curve

math

I'm trying to understand the relationship between Weil divisors and Cartier divisors, and I would like to see why these are the same in the simple case where $X$ is a nonsingular projective curve over an algebraically closed field. What is the most concrete way of explaining the equivalence between the two sorts of divisors in this situation? In particular, if $P \in X$ is a closed point thought of as a prime Weil divisor, what is the Cartier divisor corresponding to $P$? What about the invertible sheaf corresponding to $P$? 1) The Cartier divisor corresponding to the Weil divisor $1.P$ is given by the pair $s=\lbrace (U,z),(V, 1) \rbrace$ described as follows: $\bullet$ $U$ is an open neigbourhood of $P$ and $z$ is a regular function on $U$ whose sole zero is $P$ with multiplicity one. $\bullet \bullet$ $V=X\setminus \lbrace P\rbrace $ and of course $1$ is the constant function equal to $1$ on $V$. (This pair determines a section $s\in \Gamma(X,\mathcal K^* _X/\mathcal O^*_X)$ if you unravel what it means to be a section of a quotient sheaf.) 2) The invertible sheaf corresponding to $P$ is the sheaf denoted by $\mathcal O(P)$. Its $k$-vector space of sections $\Gamma(W,\mathcal O(P))$ over an open subset $W\subset X$ consists of those rational functions $f\in Rat(W)=Rat(X)$ regular on $W$ except perhaps at $P$, where $f$ is allowed a pole of order at most $1$.

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CC-MAIN-2020-10

https://www.hindawi.com/journals/aaa/2014/465782/

math

Research Article | Open Access Jian Wang, Yong Wang, "A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Manifolds with Boundary", Abstract and Applied Analysis, vol. 2014, Article ID 465782, 18 pages, 2014. https://doi.org/10.1155/2014/465782 A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Manifolds with Boundary We give a brute-force proof of the Kastler-Kalau-Walze type theorem for 7-dimensional manifolds with boundary. The noncommutative residue found in [1, 2] plays a prominent role in noncommutative geometry. For one-dimensional manifolds, the noncommutative residue was discovered by Adler in connection with geometric aspects of nonlinear partial differential equations. For arbitrary closed compact -dimensional manifolds, the noncommutative residue was introduced by Wodzicki in using the theory of zeta functions of elliptic pseudodifferential operators. In , Connes used the noncommutative residue to derive a conformal 4-dimensional Polyakov action analogy. Furthermore, Connes made a challenging observation that the noncommutative residue of the square of the inverse of the Dirac operator was proportional to the Einstein-Hilbert action in . Let be the scalar curvature and let Wres denote the noncommutative residue. Then, the Kastler-Kalau-Walze theorem gives an operator-theoretic explanation of the gravitational action and says that, for a 4-dimensional closed spin manifold, there exists a constant , such that In , Kastler gave a brute-force proof of this theorem. In , Kalau and Walze proved this theorem in the normal coordinates system simultaneously. And then, Ackermann proved that the Wodzicki residue in turn is essentially the second coefficient of the heat kernel expansion of in . On the other hand, Fedosov et al. defined a noncommutative residue on Boutet de Monvel’s algebra and proved that it was a unique continuous trace in . In , Schrohe gave the relation between the Dixmier trace and the noncommutative residue for manifolds with boundary. For an oriented spin manifold with boundary , by the composition formula in Boutet de Monvel’s algebra and the definition of , should be the sum of two terms from interior and boundary of , where is an element in Boutet de Monvel’s algebra . It is well known that the gravitational action for manifolds with boundary is also the sum of two terms from interior and boundary of . Considering the Kastler-Kalau-Walze theorem for manifolds without boundary, then the term from interior is proportional to gravitational action from interior, so it is natural to hope to get the gravitational action for manifolds with boundary by computing . Based on the motivation, Wang proved a Kastler-Kalau-Walze type theorem for 4-dimensional spin manifolds with boundary where is the canonical volume of . Furthermore, Wang found a Kastler-Kalau-Walze type theorem for higher dimensional manifolds with boundary and generalized the definition of lower-dimensional volumes in to manifolds with boundary. For 5-dimensional spin manifolds with boundary , Wang got and for 6-dimensional spin manifolds with boundary, In order to get the boundary term, we computed the lower-dimensional volume for 6-dimensional spin manifolds with boundary associated with and in and obtained the volume with the boundary term where is the extrinsic curvature. In , Wang proved a Kastler-Kalau-Walze type theorem for general form perturbations and the conformal perturbations of Dirac operators for compact manifolds with or without boundary. Let be 4-dimensional compact manifolds with the boundary and let be a general differential form on , from Theorem 10 in ; then Recently, we computed for -dimensional spin manifolds with boundary in case of . In the present paper, we will restrict our attention to the case of . We compute for 7-dimensional manifolds with boundary. Our main result is as follows. Main Theorem. The following identity for 7-dimensional manifolds with boundary holds: where and are, respectively, scalar curvatures on and . Compared with the previous results, up to the extrinsic curvature, the scalar curvature on and the scalar curvature on appear in the boundary term. This case essentially makes the whole calculations more difficult, and the boundary term is the sum of fifteen terms. As in computations of the boundary term, we will consider some new traces of multiplication of Clifford elements. And the inverse 4-order symbol of the Dirac operator and higher derivatives of -1-order and -3-order symbols of the Dirac operators will be extensively used. This paper is organized as follows. In Section 2, we define lower-dimensional volumes of compact Riemannian manifolds with boundary. In Section 3, for 7-dimensional spin manifolds with boundary and the associated Dirac operators, we compute and get a Kastler-Kalau-Walze type theorem in this case. 2. Lower-Dimensional Volumes of Spin Manifolds with Boundary In this section, we consider an -dimensional oriented Riemannian manifold with boundary equipped with a fixed spin structure. We assume that the metric on has the following form near the boundary: where is the metric on . Let be a collar neighborhood of which is diffeomorphic . By the definition of and , there exists such that and for some sufficiently small . Then, there exists a metric on which has the form on such that . We fix a metric on the such that . Let us give the expression of Dirac operators near the boundary. Set and , where are orthonormal basis of . Let denote the Levi-Civita connection about . In the local coordinates and the fixed orthonormal frame , the connection matrix is defined by The Dirac operator is defined by By Lemma 6.1 in and Propositions 2.2 and 2.4 in , we have the following lemma. Lemma 1. Let and be a Riemannian manifold with the metric . For vector fields and in , then Denote ; then we obtain the following lemma. Lemma 2. The following identity holds: Others are zeros. By Lemma 2, we have the following definition. Definition 3. The following identity holds in the coordinates near the boundary: To define the lower-dimensional volume, some basic facts and formulae about Boutet de Monvel’s calculus which can be found in Section 2 in are needed. Denote by the Fourier transformation and (similarly, define , where denotes the Schwartz space and We define and which are orthogonal to each other. We have the following property: iff which has an analytic extension to the lower (upper) complex half-plane such that, for all nonnegative integers , as , . Let be the space of all polynomials and let ; . Denote by , respectively, the projection on . For calculations, we take rational functions having no poles on the real axis} ( is a dense set in the topology of ). Then, on , where is a Jordan close curve which included surrounding all the singularities of in the upper half-plane and . Similarly, define on : So, . For , and for , . Let be an -dimensional compact oriented manifold with boundary . Denote by Boutet de Monvel’s algebra; we recall the main theorem in . Theorem 4 (Fedosov-Golse-Leichtnam-Schrohe). Let and be connected, let , and let , and denote by , , and the local symbols of , and , respectively. Define Then, (a) , for any ; (b) it is a unique continuous trace on . Let and be nonnegative integers and let . Then, by Section 2.1 of , we have the following definition. Definition 5. Lower-dimensional volumes of spin manifolds with boundary are defined by Denote by the -order symbol of an operator . An application of (2.1.4) in shows that where and the sum is taken over , , . 3. A Kastler-Kalau-Walze Type Theorem for 7-Dimensional Spin Manifolds with Boundary In this section, we compute the lower-dimensional volume for 7-dimensional compact manifolds with boundary and get a Kastler-Kalau-Walze type formula in this case. From now on, we always assume that carries a spin structure so that the spinor bundle and the Dirac operator are defined on . The following proposition is the key of the computation of lower-dimensional volumes of spin manifolds with boundary. Proposition 6 (see ). The following identity holds: Nextly, for 7-dimensional spin manifolds with boundary, we compute . By Proposition 6, for 7-dimensional compact manifolds with boundary, we have Recall the Dirac operator of Definition 3. Write By the composition formula of pseudodifferential operators, then we have Thus, we get Lemma 7. Consider the symbol of the Dirac operator where Since is a global form on , so for any fixed point , we can choose the normal coordinates of in (not in ) and compute in the coordinates and the metric . The dual metric of on is . Write and ; then, Let be an orthonormal frame field in about which is parallel along geodesics and ; then, is the orthonormal frame field in about . Locally, . Let be the orthonormal basis of . Take a spin frame field such that , where is a double covering; then, is an orthonormal frame of . In the following, since the global form is independent of the choice of the local frame, we can compute in the frame . Let be the canonical basis of and let be the Clifford action. By , then then, we have in the above frame. By Lemma 2.2 in , we have the following. Lemma 8. With the metric on near the boundary, where . Then, the following lemma is introduced. Lemma 9. The following identity holds: Proof. From Lemma 5.7 in , we have Then, we obtain . Lemma 10. Let be the metric on 7-dimensional spin manifolds near the boundary; then, Proof. From Lemma 2.3 in , we have Let , , and . Then, When , , Similarly, when , , or , , . When , . On the other hand, from definitions (10) and (11), then When , , Similarly, when , , . When , , . When , Lemma 11. When , When , Next, we can compute (see formula (23) for definition of ). Since the sum is taken over , , then we have that is the sum of the following fifteen cases. Case 2. Consider , , , , and . From (23), we have By Lemma 7, a simple computation shows By (18) and the Cauchy integral formula, then Similarly, we obtain From (51) and (53), we get Note that ; then, from (50), (54), and direct computations, we obtain Therefore, where is the canonical volume of .

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CC-MAIN-2020-45

https://brainly.com/question/194292

math

First of all, you would have to set that expression equal to zero and factor. 2x^2+3x-9=0 would factor out to (2x-3)(x+3)=0. Now, you would have to set EACH expression to zero and solve for "x." Your two answers should be x=3/2 and x= -3. Those are your x-intercepts. Okay, two points on the same axis won't do you any good, so you need a y-intercept. To find the y-intercept, plug in "0" for all x's in the equation" 2x^2+3x-9=0, and that would come out as just 9. So, with your two x-intercepts and one y-intercept, you should be able to approximate the curve of the quadratic. Hope this helped!

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CC-MAIN-2017-04

https://iosappspy.com/with-lock-land-free-you-can-create-the-perfect-ios-lock-screen-home-screen-wallpaper/

math

The #1 Ranked (in more than 60 countries) Best Wallpaper app with Ultimate features& beautiful and cool HD wallpapers at your fingertips! Download NOW for FREE! Please go to settings in the app and download in-app purchase items!!!! Lock Land is your one-stop resource to save time and money and to get all the best Skins ,Themes,Screens,Shelves,Frames and many more.. If you enjoy using Lock Lands,would you mind taking a moment to rate it ?Don’t forget to leave a comment.. # Compatible with iOS 8, iOS 7 & iOS 6 # Optimal for iPhone 6 Plus,6,5S, 5C, 5, 4S, 4, 3GS & iPod Touch 5, 4, 3 & iPad Retina # Special Designs for Christmas,Bible God screens,Different locks like locklab # New and Popular galleries with Category wise # Wallpapers made to fit both home screen & lock screen # Powerful Text Editing # Powerful Photo Editing # Ultimate designs # Unique Themes Ideas # Full Screen Preview feature:There’s no need to save each design to see how it looks on your screen! # Easy to use! # Share your images on Facebook, Twitter, Tumblr, Instagram or email # Add calendars and frames to your Lock Screens # 3D and 2D shelves! # Unique frames! # Endless customization possibilities!

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CC-MAIN-2021-39

http://meessaywlut.gvu-edu.us/math-lcm.html

math

Lcm 2 3 5 360 hcf 2 3 6 ∴ a b 120 18 2160 lcm hcf 360 6 2160 ∴ a b lcm hcf the proof is not difficult: let the hcf of two numbers a,. The least common multiple (lcm) of two or more numbers is the smallest number which can be exactly divided by each of the given number. The least common multiple, or the lowest common multiple, is the smallest common multiple of two integers home basic math factors and multiples. Lcd stands for least common denominator and lcm stands for least common multiple the least common multiple is the smallest number that. Addition games - minko's milkshake shoppe - math games fun4thebrain common multiple (lcm) find the lcm of the numbers to win the snowball fight. Other articles where least common multiple is discussed: arithmetic: fundamental theory:of the numbers, called their least common multiple (lcm. In this lesson you will learn how to find the gcf and lcm of two whole numbers by using their prime factors. Whenever lcm becomes greater than the mod, the above mentioned property gets destroyed you must try to find lcm in terms of products of. Time - least common multiple (lcm) - solved math problems, problem solving and knowledge review problems count: 12. This page should be merged with least common multiple and/or greatest common least common multiple (abbreviated lcm) of two natural numbers is the. This calculator will find the highest common factor and lowest common denominator of any two numbers. The math worksheet sitecom on-line math worksheet generator least common multiple number of problems 10 20 multiple worksheets create 1, 2, 3, 4. Calculate the lcm least common multiple of 2 or more numbers find the for example, lcm (2,3) = 6 and lcm (6,10) = 30 the math forum: lcm, gcf. The class and function templates in provide run-time and compile-time evaluation of the greatest common divisor (gcd) or. Since we are in a field and the least common multiple is only determined up to a unit, it is correct to either return zero or one note that fraction fields of unique. Quiz theme/title: finding the lcm description/instructions practice finding the lcm of a set of numbers group: math math quizzes topic: math. Find the least common multiple of 72, 108 and 21 multiply all by 100 first 720 = 6, 6, 5, 4 108 = 6, 6, 3 210 = 6, 5, 7 lcm x 100 = 6 x 6 x 5 x 4. Quick way to find the lcm for any pair of numbers 7 september 2010 josh rappaport mental math, uncategorized 7 comments finding the lcm, gcf,. The least common multiple (lcm) of two or more numbers is the smallest number (not counting 0) which is a multiple of all of the numbers. Calculator to find greatest common divisor (highest common factor) and lowest common multiple. Sal finds the lcm (least common multiple) of 12 and 36, and of 12 and 18 he shows how to do that using least common multiple common core math: 6ns. In arithmetic and number theory, the least common multiple, lowest common multiple, or smallest common multiple of two integers a and b, usually denoted by . Lcm is found under math → num, arrow down to #8: lcm( from the home screen, choose lcm( followed by two numbers separated by a comma note: lcm (. How to find the lowest common multiple (lcm) of two (or more) numbers in math grade 7 math questions are presented along with detailed. This lesson will teach you 3 methods for finding the least common multiple(lcm) of two whole numbers we will start with a definition of the. Get the answer to lcm of 8,12 with the cymath math problem solver - a free math equation solver and math solving app for calculus and algebra.

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CC-MAIN-2018-43

https://nrich.maths.org/public/leg.php?code=31&cl=2&cldcmpid=32

math

This challenge extends the Plants investigation so now four or more children are involved. The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"? Place six toy ladybirds into the box so that there are two ladybirds in every column and every row. First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items. Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column? Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this? Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers? Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code? Find the values of the nine letters in the sum: FOOT + BALL = GAME Can you use the information to find out which cards I have used? Make your own double-sided magic square. But can you complete both sides once you've made the pieces? Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square. Using the statements, can you work out how many of each type of rabbit there are in these pens? Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total. Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions. Using 3 rods of integer lengths, none longer than 10 units and not using any rod more than once, you can measure all the lengths in whole units from 1 to 10 units. How many ways can you do this? Can you find all the ways to get 15 at the top of this triangle of numbers? Can you substitute numbers for the letters in these sums? Who said that adding couldn't be fun? Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon? Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done? Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number. Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make? What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates A group of children are using measuring cylinders but they lose the labels. Can you help relabel them? Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes? This is an adding game for two players. There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places. This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards. A game for 2 players. Practises subtraction or other maths What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? Exactly 195 digits have been used to number the pages in a book. How many pages does the book have? Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? If you take a three by three square on a 1-10 addition square and multiply the diagonally opposite numbers together, what is the difference between these products. Why? An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore. There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it? There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the Use your addition and subtraction skills, combined with some strategic thinking, to beat your partner at this game. This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . . Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers? Can you arrange 5 different digits (from 0 - 9) in the cross in the Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens? This dice train has been made using specific rules. How many different trains can you make? If each of these three shapes has a value, can you find the totals of the combinations? Perhaps you can use the shapes to make the Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the Can you explain the strategy for winning this game with any target? This challenge focuses on finding the sum and difference of pairs of two-digit numbers.

s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917123549.87/warc/CC-MAIN-20170423031203-00162-ip-10-145-167-34.ec2.internal.warc.gz

CC-MAIN-2017-17

http://www.jiskha.com/display.cgi?id=1245900271

math

The definition of a composite function: f º g (x) means f (g(x) ), which tells you to work out g(x) first, and then fill that answer into f. See (Broken Link Removed) In this case, f(x) = 6/x-1 and g(x) = 1+3/x f º g (x) and the composite function would be? There is probably a mis-interpretation of the parentheses: f(x) = 6/(x-1) and g(x) = 1+3/x f º g (x) If this is the case, the answer is (A). This is what I thought as well. Thanks! Algebra - I am getting ready to start an Algebra class for the first time and I ... SHAY - THE MATH QUESTION IS IT ALGEBRA , PRE ALGEBRA' OR GEOMETRY? Quadriatic ... Algebra 1A - How is algebra a useful tool? what concepts investigated in algebra... Algebra - In an interview of 50 math majors, 12 liked calculus and geometry 18 ... 7th grade - There are at least 5 more than twice as many students taking algebra... algebra 1 - at a certain high school,350 students are taking algebra. the ratio ... Algebra-still need some help - Homework Help Forum: Algebra Posted by Jena on ... math - 8. In an interview of 50 math majors, 12 liked calculus and geometry 18 ... Algebra - The average mark on a test in an algebra class is 80. If the two ... pre-algebra - kk my teach told me to describe and draw things that are about pre...

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CC-MAIN-2013-48

https://www.macewan.ca/wcm/Registrar/FeesandFinancialInformation/TuitionFees/OtherFees/index.htm

math

|Challenge Exam / Prior Learning Assessment ||$40| |Course Audit ||50% of course tuition, and course material and special fees | |Deferred Examination ||$50 for single exam; $100 for 2+ exams | |Extension to Incomplete Grade ||$50 | |Field Placement / Practicum Identification Card ||$10 | |Locker Rental ||$20 per term | |Locker Contents Reclaiming ||$25 | |Non-Program International Student Service ||$50 | |Photo Identification Card ||$10 | |Photo ID Card Replacement ||$10 | |Reassessment of Final Exam ||$20 | |Replacement T2202A Tax Receipt ||$20 | |Returned Cheque (NSF) ||$40 | |Skills Appraisal / Skills Appraisal Rewrite ||$50 | |Tuition Payment Plan ||$50 | The continuing/returning student deposit is required to confirm enrolment in second and subsequent years in the following program. This deposit will be allocated towards your tuition. Continuing/returning student deposits are non-refundable. You must pay the deposit even if you have applied for student loans, scholarships, bursaries, or sponsored payments. Students with money still owing from previous terms must pay this balance in addition to the deposit amount. You must pay your continuing/returning student deposit on or before August 1 each year. If your deposit is not received by this date, your enrolment will be cancelled. |Bachelor of Physical Education transfer ||$2501|

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CC-MAIN-2018-09

https://hardyramanujan.wordpress.com/2011/03/08/euclidean-division-of-an-angle/

math

Back in my school days, probably in the year 1997, I first read about a famous and ancient geometrical problem, the problem of trisecting an angle using only a compass and an unmarked ruler. It took me some time to understand that a general method for trisecting all angles is impossible arbitrary angle . Of course with a marked ruler, we can always trisect an arbitrary angle, and moreover, in case of some special angles such as , or some non-constructible angles such as , we can trisect even with an unmarked ruler. I found some approximate trisection methods, each of which gave a dominant term and an error terms that was much smaller than the dominant term. At that time, I thought is it possible to find a method that gives an approximate division of an arbitrary angle into an arbitrary ratio. In formal terms, this problem can be stated as: Problem: Let be positive integer and be any positive real. Given an angle and a line that divides it in the ratio find a method with finite number of steps to construct an line that approximately divides the angle in the ratio using only a compass and an unmarked ruler. Solution: We shall use the notation to denote the angle . In the diagram let be an acute angle and let be given. - With O as center and radius OC, draw an arc to cut OB at D. - Join CD and draw DJ perpendicular to OB. - Let OI intersect DJ at I and the arc CD at K. - Produce DK to intersect OC at G. - Let OI intersect CD at E. - Extent GE to intersect OB at H. - Join IH to intersect GD at F. - Draw a line OA through F. . - Repeating this method times, to construct a line that approximately divides in the ratio . As increases, the lines OC and DG tend to become parallel to each other and therefore the paper size required for construction increases. To overcome this problem, we can consider an obtuse angle as the sum of a right angle and an acute angle and apply our method on each part separately and then add up the final angle resulting form each of the two parts.

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CC-MAIN-2018-26

https://tohoku.pure.elsevier.com/ja/publications/vortex-induced-vibration-energy-extraction-modeling-via-forced-ve

math

Vortex induced vibrations (VIV) for energy extraction have been revisited in last years by both marine power and wind power communities. Even though vortex induced vibrations have been focus of research for many years, energy extraction from vortex induced vibrations is relevantly new field which needs more detailed investigation and modeling. To this end, there has been recent experimental and modeling parametric studies where VIV was modeled by solution of one-degree-of-freedom ordinary differential equation spring system where engineering modeling of vortex induced vibration for energy extraction has been investigated based on a spring system with the forces defined from forced oscillation experiments where full coupling of free oscillations were not taken into account. Herein a Computational Fluid Dynamics (CFD) modeling of a circular cylinder will be studied to compare forced and free vibrations in the context of vortex-induced energy extraction. The model is essentially solved by partial differential isothermal incompressible Navier-Stokes equations to model fully mathematical model of the fluid-structure interaction of vortex induced vibration. The comparison between forced and free oscillation response studies of this paper will serve to improve the scientific knowledge where vortex induced vibration modeling are comparatively more limited. The preliminary results are presented herein for forced and free oscillations for the Reynolds number regimes Re = 100 and Re = 3800 in two dimensions for combinations of amplitudes and frequency of oscillations in the context of energy extraction modeling.

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CC-MAIN-2022-40

https://wiki.math.wisc.edu/index.php?title=PDE_Geometric_Analysis_seminar&direction=prev&oldid=3972

math

PDE Geometric Analysis seminar The seminar will be held in room B115 of Van Vleck Hall on Mondays from 3:30pm - 4:30pm, unless indicated otherwise. Seminar Schedule Spring 2012 Yao Yao (UCLA) Degenerate diffusion with nonlocal aggregation: behavior of solutions The Patlak-Keller-Segel (PKS) equation models the collective motion of cells which are attracted by a self-emitted chemical substance. While the global well-posedness and finite-time blow up criteria are well known, the asymptotic behaviors of solutions are not completely clear. In this talk I will present some results on the asymptotic behavior of solutions when there is global existence. The key tools used in the paper are maximum-principle type arguments as well as estimates on mass concentration of solutions. This is a joint work with Inwon Kim. Xuan Hien Nguyen (Iowa State) Gluing constructions for solitons and self-shrinkers under mean curvature flow In the 1990s, Kapouleas and Traizet constructed new examples of minimal surfaces by desingularizing the intersection of existing ones with Scherk surfaces. Using this idea, one can find new examples of self-translating solutions for the mean curvature flow asymptotic at infinity to a finite family of grim reaper cylinders in general position. Recently, it has been shown that it is possible to desingularize the intersection of a sphere and a plane to obtain a family of self-shrinkers under mean curvature flow. I will discuss the main steps and difficulties for these gluing constructions, as well as open problems. Nestor Guillen (UCLA) We consider the Monge-Kantorovich problem, which consists in transporting a given measure into another "target" measure in a way that minimizes the total cost of moving each unit of mass to its new location. When the transport cost is given by the square of the distance between two points, the optimal map is given by a convex potential which solves the Monge-Ampère equation, in general, the solution is given by what is called a c-convex potential. In recent work with Jun Kitagawa, we prove local Holder estimates of optimal transport maps for more general cost functions satisfying a "synthetic" MTW condition, in particular, the proof does not really use the C^4 assumption made in all previous works. A similar result was recently obtained by Figalli, Kim and McCann using different methods and assuming strict convexity of the target. Charles Smart (MIT) PDE methods for the Abelian sandpile Abstract: The Abelian sandpile growth model is a deterministic diffusion process for chips placed on the $d$-dimensional integer lattice. One of the most striking features of the sandpile is that it appears to produce terminal configurations converging to a peculiar lattice. One of the most striking features of the sandpile is that it appears to produce terminal configurations converging to a peculiar fractal limit when begun from increasingly large stacks of chips at the origin. This behavior defied explanation for many years until viscosity solution theory offered a new perspective. This is joint work with Lionel Levine and Wesley Pegden. Vlad Vicol (University of Chicago) Title: Shape dependent maximum principles and applications Abstract: We present a non-linear lower bound for the fractional Laplacian, when evaluated at extrema of a function. Applications to the global well-posedness of active scalar equations arising in fluid dynamics are discussed. This is joint work with P. Constantin. Jiahong Wu (Oklahoma State) "The 2D Boussinesq equations with partial dissipation" The Boussinesq equations concerned here model geophysical flows such as atmospheric fronts and ocean circulations. Mathematically the 2D Boussinesq equations serve as a lower-dimensional model of the 3D hydrodynamics equations. In fact, the 2D Boussinesq equations retain some key features of the 3D Euler and the Navier-Stokes equations such as the vortex stretching mechanism. The global regularity problem on the 2D Boussinesq equations with partial dissipation has attracted considerable attention in the last few years. In this talk we will summarize recent results on various cases of partial dissipation, present the work of Cao and Wu on the 2D Boussinesq equations with vertical dissipation and vertical thermal diffusion, and explain the work of Chae and Wu on the critical Boussinesq equations with a logarithmically singular velocity. Joana Oliveira dos Santos Amorim (Universit\'e Paris Dauphine) "A geometric look on Aubry-Mather theory and a theorem of Birkhoff" Given a Tonelli Hamiltonian $H:T^*M \lto \Rm$ in the cotangent bundle of a compact manifold $M$, we can study its dynamic using the Aubry and Ma\~n\'e sets defined by Mather. In this talk we will explain their importance and give a new geometric definition which allows us to understand their property of symplectic invariance. Moreover, using this geometric definition, we will show that an exact Lipchitz Lagrangian manifold isotopic to a graph which is invariant by the flow of a Tonelli Hamiltonian is itself a graph. This result, in its smooth form, was a conjecture of Birkhoff. Gui-Qiang Chen (Oxford) "Nonlinear Partial Differential Equations of Mixed Type" Many nonlinear partial differential equations arising in mechanics and geometry naturally are of mixed hyperbolic-elliptic type. The solution of some longstanding fundamental problems in these areas greatly requires a deep understanding of such nonlinear partial differential equations of mixed type. Important examples include shock reflection-diffraction problems in fluid mechanics (the Euler equations), isometric embedding problems in in differential geometry (the Gauss-Codazzi-Ricci equations), among many others. In this talk we will present natural connections of nonlinear partial differential equations with these longstanding problems and will discuss some recent developments in the analysis of these nonlinear equations through the examples with emphasis on identifying/developing mathematical approaches, ideas, and techniques to deal with the mixed-type problems. Further trends, perspectives, and open problems in this direction will also be addressed. This talk will be based mainly on the joint work correspondingly with Mikhail Feldman, Marshall Slemrod, as well as Myoungjean Bae and Dehua Wang. Jacob Glenn-Levin (UT Austin) We consider the Boussinesq equations, which may be thought of as inhomogeneous, incompressible Euler equations, where the inhomogeneous term is a scalar quantity, typically density or temperature, governed by a convection-diffusion equation. I will discuss local- and global-in-time well-posedness results for the incompressible 2D Boussinesq equations, assuming the density equation has nonzero diffusion and that the initial data belongs in a Besov-type space.

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http://www.chegg.com/homework-help/questions-and-answers/suppose-nolan-ryan-stands-surface-moon-throws-baseball-horizontally-baseball-high-speed-do-q2158645

math

0 pts endedThis question is closed. No points were awarded. Suppose that Nolan Ryan stands on the surface of the moon and throws a baseball horizontally. If the baseball has a high enough speed and does not strike any mountain, it can orbit around the moon and, after completing the orbit, strike Nolan from behind. The mass of the moon is mm = 7.35×1022 kg, and its radius is rm = 1740 km. The gravitation constant G = 6.67×10-11 Nm2/kg2. Find the speed at which Nolan must throw the ball for such a circular orbit. How long (in hours) does the ball take to complete one orbit?

s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394021585790/warc/CC-MAIN-20140305121305-00044-ip-10-183-142-35.ec2.internal.warc.gz

CC-MAIN-2014-10

https://cloudutil.player.fm/series/the-chris-rose-rotation-mlb-players-podcast/austin-riley-been-playing-playoff-ball-since-the-mets-series

math

Archived series ("Inactive feed" status) When? This feed was archived on January 10, 2023 11:46 (). Last successful fetch was on December 08, 2022 23:07 () Why? Inactive feed status. Our servers were unable to retrieve a valid podcast feed for a sustained period. What now? You might be able to find a more up-to-date version using the search function. This series will no longer be checked for updates. If you believe this to be in error, please check if the publisher's feed link below is valid and contact support to request the feed be restored or if you have any other concerns about this. Manage episode 343589078 series 2886556 Use code ROSE for $20 off your first SeatGeek order https://seatgeek.onelink.me/RrnK/ROSE Braves third baseman Austin Riley talks with Chris Rose about adjusting to playing baseball and being a dad, signing a 10-year extension, facing the Phillies in the divisional round, the Mets/Padres series, walking the red carpet at the All-Star game and more!

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CC-MAIN-2023-06

https://www.jiskha.com/display.cgi?id=1217280343

math

posted by Diane . This is a 5 part question; (a-e)The question reads: Suppose that a market is described by the following supply & demand equations: Qs=2P & Qd=300-P a) Solve for the equalibrium price & quantity. (I think I understand this process.) b)Suppose that a tax of T is placed on buyers so the new demand equation is: Qd=300-(P+T). Solve for new equalib. What happens to the price received by seller, the price paid by buyers, & qty sold? c)Tax revenue is T x Q. Use your answer to part (b) to solve for tax revenue as a function of T. d) Graph the DWL e) Gov't levies a tax on good of $200/unit. Is this a good policy? Why/Why not? Can you propose a better policy? a) set Qs=Qd and solve for P. b) Same, set Qs=Qd and solve for P. Buyer pays P+T, seller gets P. (I get P=100-T/3) c) Let P^ and Q^ be the equilibrium price and quantity. TR=T*Q^. From supply Q^=2P^. Substitute. TR=T*2P^ = T*2*(100-T/3)= 200T-2T^2/3 d) dwl is dead weight loss, and is represented by the little triangle below demand above supply and to the right of the equilibrium Q. e) In general, whether a tax is a good policy or not depends on how it compares to the other policy choices. However, this $200 tax is a BAD policy because the government could raise the same amount of money with a lower tax rate. Use calculus on the equation from c) to find the revenue maximizing tax rate. TR' = 200 - 4T/3. You have a maxima at T=150.

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CC-MAIN-2017-34

https://www.hackmath.net/en/m

math

MathematicsIs there a bad grade of mathematics for certification or exam? Do you want to become more accustomed to it and transform the boring subject into fun exploration? Want to feel the joy of understanding and computing an example? Do you want train in thinking, looking for problems solutions? We consider mathematics to be a bearer of common sense, without which it is difficult in the next career to achieve success. Whether as a manager, programmer, judge, mason or electrician, you will make decisions every day, the quality of which will depend on your wisdom, maturity, concentration. The smartest and better decisions you make in life, the greater the potential for success in your personal and working life. This will help you with our portal with examples of mathematics with their solutions. If you are looking for math material, you are getting ready for a prize, a graduate, a monitor, a nationwide knowledge test is also right here. In addition to examples, we also offer online math calculators. - Let x Let x represent one quantity. State what that quantity represents. Express the second quantity in terms of x. The length of the rectangle is 4 inches less than 8 times the width. - AMSL and skiing Tomas skis from point A (3200m above sea level to place B. Hill has 20% descent. The horizontal distance between the start and finish is 2,5km. At what altitude is point B? - Turtles 2 A box turtle hibernates in the sand at 11 5/8. A spotted turtle hibernates at 11 16/25 feet. Which turtle is deeper? - Age problems A) Alex is 3 times as old as he was 2 years ago. How old is he now? b) Casey was twice as old as his sister 3 years ago. Now he is 5 years older than his sister. How old is Casey? c) Jessica is 4 years younger than Jennifer now. In 10 years, Jessica wi - Tank of fuel A 14.5-gallon tank of fuel is 3/4 full. How many more gallons will it take to fill up the tank? - Depth angles At the top of the mountain stands a castle, which has a tower 30 meters high. We see the crossroad in the valley from the top of the tower and heel at depth angles of 32° 50 'and 30° 10'. How high is the top of the mountain above the crossroad - Area of a rectangle Calculate the area of a rectangle with a diagonal of u = 12.5cm and a width of b = 3.5cm. Use the Pythagorean theorem. next math problems »

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CC-MAIN-2019-43

https://www.beautetrade.com/buyers/skin-toner/

math

Hi how much is it per box? If order 20 box how much ya? I need 50pcs as sample to check I am interested in knowing the price and MOQ. Whats the price for 50-100? whats the prices? pls send me your quote I want to get 100 pieces Shipping to canada, what is the cost? Whats the price? Please give me your quote How much please Do you have more ? please issue a quote. Thank you.

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CC-MAIN-2022-05

http://ineedacprclass.com/essay-on/Igtu/166111

math

Business and Management Submitted By zainabnaz External Factor Evaluation (EFE) matrix method is a strategic-management tool often used for assessment of current business conditions. The EFE matrix is a good tool to visualize and prioritize the opportunities and threats that a business is facing. The EFE matrix is very similar to the IFE matrix. The major difference between the EFE matrix and the IFE matrix is the type of factors that are included in the model. While the IFE matrix deals with internal factors, the EFE matrix is concerned solely with external factors. External factors assessed in the EFE matrix are the ones that are subjected to the will of social, economic, political, legal, and other external forces. List factors: The first step is to gather a list of external factors. Divide factors into two groups: opportunities and threats. Assign weights: Assign a weight to each factor. The value of each weight should be between 0 and 1 (or alternatively between 10 and 100 if you use the 10 to 100 scale). Zero means the factor is not important. One or hundred means that the factor is the most influential and critical one. The total value of all weights together should equal 1 or 100. Rate factors: Assign a rating to each factor. Rating should be between 1 and 4. Rating indicates how effective the firm’s current strategies respond to the factor. 1 = the response is poor. 2 = the response is below average. 3 = above average. 4 = superior. Weights are industry-specific. Ratings are company-specific. Multiply weights by ratings: Multiply each factor weight with its rating. This will calculate the weighted scorefor each factor. Total all weighted scores: Add all weighted scores for each factor. This will calculate the total weighted score for the company. EXTERNAL FACTOR EVALUATION MATRIX OF SAMSUNG Internal Factor Evaluation (IFE) matrix is a strategic management tool…...

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CC-MAIN-2018-51

https://quatr.us/math/percents-fractions-decimals.htm

math

Percent comes from the Latin words “per cent”. That means “out of a hundred,” like there are 100 cents in a dollar. So “ten per cent” means ten out of every hundred. A dime is ten percent of a dollar. A dollar is ten percent of ten dollars. If we say that 50% (fifty percent) of people are women, that means that fifty out of a hundred people are women, or half of the people. If we say that 100% of people have hearts, that means that all of them have hearts. To calculate a percentage for yourself, you need to understand how to multiply fractions and simplify equations. Suppose you have sixty-five seeds, and ten of them have gone moldy. Say you need to know what percentage of the seeds went bad. First write your problem as a fraction: 10/65: ten out of sixty-five seeds went bad. Now you need to know how much that would be out of a hundred, so set it up as an equation: To solve this equation and find out what x is, you’ll need to find a common denominator for your two fractions. Just use 6500 (65×100). Remember to multiply the numerators by the same amount as the denominators! So you get: Percentages can also be more than 100 percent. If you planted ten good seeds, and they grew into a hundred peas, you’d have 1000% of your original seeds back, or ten times as much. If you weigh 100 pounds, and your little brother weighs 50 pounds, you weigh 200% as much as he does, or twice as much.

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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=cheb&paperid=368&option_lang=eng

math

This article is cited in 1 scientific paper (total in 1 paper) Anatoly Alekseevich Karatsuba G. I. Arkhipovab, V. N. Chubarikova a Lomonosov Moscow State University b Steklov Mathematical Institute of Russian Academy of Sciences The authors set themselves two main objectives: to characterize the main stages of the life of the outstanding Russian mathematician, Honored Scientist of Russia, Professor of Moscow State University named after M. V. Lomonosov, head of the department of number theory Steklov Mathematical Institute, Doctor of Physical and Mathematical Sciences Anatolii Alexeevich Karatsuba and give a brief analysis of his scientific work, which has had a significant impact on the development of analytic number theory. Sufficiently detailed description of the research of Professor A. A. Karatsuba and his disciples on the analytic theory of numbers, which are allocated following the Karatsuba, three main areas: - trigonometric sums and trigonometric integrals; - the Riemann zeta function; - Dirichlet characters. A. A. Karatsuba, being a disciple of Professor N. M. Korobov, led the scientific schools and seminars on analytic number theory at Moscow State University named after M. V. Lomonosov. Among his many students defended their dissertations, with seven of them later became Doctor of Physical and Mathematical Sciences. Anatoly Alekseevich has published 158 scientific papers, including 4 monographs and a classic textbook on analytic number theory, was the translator of a number of fundamental scientific monographs. He was a member of the editorial board of the journal “Mathematical notes” and a member of the program committees of several international conferences on algebra and number theory. Bibliography: 58 titles. trigonometric sums, trigonometric integrals, Riemann zeta function, Dirichlet characters. PDF file (305 kB) G. I. Arkhipov, V. N. Chubarikov, “Anatoly Alekseevich Karatsuba”, Chebyshevskii Sb., 16:1 (2015), 32–51 Citation in format AMSBIB \by G.~I.~Arkhipov, V.~N.~Chubarikov \paper Anatoly Alekseevich Karatsuba \jour Chebyshevskii Sb. Citing articles on Google Scholar: Related articles on Google Scholar: This publication is cited in the following articles: M. A. Korolev, “On Anatolii Alekseevich Karatsuba's works written in the 1990s and 2000s”, Proc. Steklov Inst. Math., 299 (2017), 1–43 |Number of views:|

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CC-MAIN-2021-25

https://www.physicsforums.com/threads/calculating-resultant-forces-given-fbd.382251/

math

A truck of mass 8000 kg is accelerating at 5.0 m/s2 up a hill of inclination 4 degrees to the horizontal The friction forces on the truck add up to 5500 N. What force does the truck exert against the road in the direction of travel? What force does the road exert on the truck in the direction of travel? With the truck's engine working just as hard, what would be the acceleration of the truck down the hill? The Attempt at a Solution For the force the truck is exerting i used F=ma so F=(8000)(5)=40 000N I just don't believe it's that simple, do i need to break down the incline into x and y components? In which case I am not given enough information..

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CC-MAIN-2022-49

http://www.solutioninn.com/the-following-is-a-list-of-important-abbreviations-used-in

math

Question: The following is a list of important abbreviations used in The following is a list of important abbreviations used in the chapter. These abbreviations also are used widely in business. For each abbreviation, give the full designation. The first one is an example. Relevant QuestionsLearning which items belong in each cash flow statement category is an important first step in understanding their meaning. Use a letter to mark each item in the following list as a cash flow from Operating, Investing, or ...Items from the 2013 income statement, statement of retained earnings, and balance sheet of Electronic Arts, Inc., are listed below in alphabetical order. Solve for the missing amounts, and explain whether the company was ...Oakley, Inc., reported the following items in its financial statements. For each item, indicate (1) the type of account (A = asset, L = liability, SE = stockholders’ equity, R = revenue, E = expense, D = dividend) and (2) ...Ken Young and Kim Sherwood organized Reader Direct as a corporation; each contributed $ 49,000 cash to start the business and received 4,000 shares of stock. The store completed its first year of operations on December 31, ...Assume that you are the president of High Power Corporation. At the end of the first year of operations (December 31), the following financial data for the company are available: Accounts Payable………………………. ... Post your question

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CC-MAIN-2017-39

https://www.yourdictionary.com/algebraical

math

Such is the basis of the algebraical or modern analytical geometry. Thus, if x= horned and y = sheep, then the successive acts of election represented by x and y, if performed on unity, give the whole of the class horned sheep. Boole showed that elective symbols of this kind obey the same primary laws of combination as algebraical symbols, whence it followed that they could be added, subtracted, multiplied and even divided, almost exactly in the same manner as numbers. The only other algebraical symbol is A for minus; plus being expressed by merely writing terms one after another. In his youth he went to the continent and taught mathematics at Paris, where he published or edited, between the years 1612 and 1619, various geometrical and algebraical tracts, which are conspicuous for their ingenuity and elegance. His earliest publications, beginning with A Syllabus of Plane Algebraical Geometry (1860) and The Formulae of Plane Trigonometry (1861), were exclusively mathematical; but late in the year 1865 he published, under the pseudonym of "Lewis Carroll," Alice's Adventures in Wonderland, a work that was the outcome of his keen sympathy with the imagination of children and their sense of fun.

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CC-MAIN-2021-31

https://oregonbeachroofinspection.com/

math

How to solve quadratics by factoring There are a few different methods that can be used to solve quadratics, but one of the most common is by factoring. To factor a quadratic, one must first determine the factors of the constant term and then the factors of the leading coefficient. Once these are determined, they can be used to write the quadratic in factored form. How can we solve quadratics by factoring From there, the roots of the quadratic can be determined by setting each factor equal to 0 and solving. There are a few different methods that can be used to solve quadratics by factoring. One method is to factor the quadratic equation into two linear equations. Another method is to use the quadratic formula. To factor a quadratic equation, one must first determine the greatest common factor of the terms. Then, the terms are divided by the greatest common factor and the equation is rewritten. The next step is to find two factors of the leading coefficient that add To solve a quadratic equation by factoring, one must first determine the factors of the equation. The factors of the equation must then be set equal to zero. The solutions to the equation will be the values of x that make the equation equal to zero. There are a few different methods that can be used to solve quadratics, but one of the most straightforward is by factoring. To do this, start by setting the equation equal to zero and then factoring the quadratic equation. This will usually result in two linear equations that can be solved for the roots of the original equation. The first step in solving a quadratic equation by factoring is to determine the greatest common factor (GCF) of the coefficients. The GCF is the largest integer that will divide evenly into all of the coefficients. Once the GCF is determined, it can be used to factor the quadratic equation. The GCF can be factored out of the coefficients, and the resulting terms can be factored using the distributive property. The resulting factors can then be set equal We will support you with math difficulties an amazing and very helpful app to get know and understand math better. it explains everything. love it, but the only thing is that you need to pay a monthly fee if you want a further explanation Phenomenal app. It can almost always find a problem in your text book so it will always give the correct answer + it will show the work or show you how to do it. 1st time reviewing an app but I had too just because it's so helpful. 15/10 a must have No ads btw

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https://boardgames.stackexchange.com/questions/15573/may-i-play-only-one-tenacious-z-or-all-of-them-from-discard-pile-in-my-turn

math

The card text of "Tenacious Z" says: Special: During your turn you may play this card from your discard pile as an extra minion. You may only use the ability of one Tenacious Z each turn. I'm not sure if that means "once for each card" or "one time each turn, no matter which card it is". So may I play 1 or 3 Tenacious Z from discard pile when I have 3 of them in the discard pile?

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CC-MAIN-2024-10

https://www.onlinemath4all.com/hcf-and-lcm-word-problems.html

math

Problem 1 : A merchant has 120 liters and 180 liters of two kinds of oil. He wants to sell the oil by filling the two kinds in tins of equal volumes. Find the greatest volume of such a tin. The given two quantities 120 and 180 can be divided by 10, 20,... exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,... liters. But, the target of the question is, the volume of oil filled in tins must be greatest. So, we have to find the largest number which exactly divides 120 and 180. That is the highest common factor (HCF) of (120, 180). HCF (120, 180) = 60 liters The 1st kind 120 liters is sold in 2 tins of of volume 60 liters in each tin. The 2nd kind 180 liters is sold in 3 tins of volume 60 liters in each tin. Hence, the greatest volume of the tin is 60 liters. Problem 2 : Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m can be covered completely. We require the least number of square tiles. So, each tile must be of maximum dimension. To get the maximum dimension of the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is the highest common factor (HCF) of (16.58, 8.32). To convert meters into centimeters, we have to multiply by 100. 16.58 ⋅ 100 = 1658 cm 8.32 ⋅ 100 = 832 cm HCF (1658, 832) = 2 cm Hence the side of the square tile is 2 cm. Required no. of tiles : = (Area of the floor)/(Area of a square tile) = (1658 ⋅ 832)/(2 ⋅ 2) Hence, the least number of square tiles required is 344,864. Problem 3 : A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing. For the least possible number of casks of equal size, the size of each cask must be of the greatest volume. To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is the highest common factor (HCF) of (403, 434, 465). HCF (403, 434, 465) = 31 liters Each cask must be of the volume 31 liters. Required number casks : = 403/31 + 434/31 + 465/31 = 13 + 14 + 15 Hence, the least possible number of casks of equal size required is 42. Problem 4 : Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ? (excluding the one at start) For example, let the two bells toll after every 3 and 4 seconds respectively. Then the first bell tolls after every 3, 6, 9, 12 seconds... Like this, the second bell tolls after every 4, 8, 12 seconds... So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 is the least common multiple (LCM) of 3 and 4. The same thing happened in our problem. To find the time, when they will all toll together, we have to find the LCM of (2, 4, 8, 6, 10, 12). LCM (2, 4, 8, 6, 10, 12) is 120 That is, 120 seconds or 2 minutes. So, after every two minutes, all the bell will toll together. For example, in 10 minutes, they toll together : 10/2 = 5 times That is, after 2, 4, 6, 8, 10 minutes. It does not include the one at the start. Similarly, in 30 minutes, they toll together : = 15 times (excluding one at the start). Problem 5 : The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hours, when will they again change simultaneously ? For example, let the two signals change after every 3 secs and 4 secs respectively. Then the first signal changes after 3, 6, 9, 12 seconds... Like this, the second signal changes after 4, 8, 12 seconds... So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 is the least common multiple (LCM) of 3 and 4. The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the LCM of (48, 72, 108). LCM (48, 72, 108) = 432 seconds or 7 min 12 sec So, after every 7 min 12 sec, all the signals will change simultaneously. At 8:20:00 hours, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hours. Hence, three signals will change simultaneously at 8:27:12 seconds. Problem 6 : Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square. To answer this question, we have to find the least number which is exactly divisible by the given numbers 15, 20 and 25. That is the least common multiple of (15, 20, 25). LCM (15, 20, 25) = 300 So, we need 300 soldiers such that they stand in rows of 15, 20 , 25. But, it has to form a perfect square (as per the question). To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square. To make 300 as perfect square, we have to multiply 300 by 3. Then, it is 900 which is a perfect square. Hence, the least number of soldiers required is 900. Kindly mail your feedback to email@example.com We always appreciate your feedback.

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CC-MAIN-2023-40

https://www.zeroing.com/viewtopic/0wyv4.php?7f5ce6=probability-distribution-formula

math

probability distribution formula Let’s suppose a coin was tossed twice and we have to show the probability distribution of showing heads. A probability distribution can be compiled like the table below, which shows the probability of getting any particular number on one roll: Probability Distribution Table 1.4 Unlock Content Binomial Probability Distribution. The probability density function (PDF) is: The cumulative distribution function (CDF) is: Notation. A discrete probability distribution is a table (or a formula) listing all possible values that a discrete variable can take on, together with the associated probabilities.. x = Normal random variable. Where, μ μ = Mean σ σ = Standard Distribution. Events A and B are independent iff. In this article, we will mainly be focusing on probability formula and examples. `sigma=sqrt(V(X)` is called the standard deviation of the probability distribution. Correlation . P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75. Conditional Probability. Distribution Function Definitions. Poisson Distribution Formula – Example #1. If mean(μ μ) = 0 and standard deviation(σ σ) = 1, then this distribution is known to be normal distribution. of heads selected will be – 0 or 1 or 2 and the probability of such event could be calculated by using the following formula: Calculation of probability of an event can be done as follows, Using the Formula, Probability of selecting 0 Head = No of Possibility of Event / No of Total Possibility 1. These are normally plotted as straight horizontal lines. Probability is a wonderfully usable and applicable field of mathematics. i.e. The normal distribution is the only distribution whose cumulants beyond the first two (i.e., other than the mean and variance) are zero. The formula for normal probability distribution is as stated. It would be the probability that the coin flip experiment results in zero heads plus the probability that the experiment results in one head. One of the most important parts of a probability distribution is the definition of the function as every other parameter just revolves around it. Theory of probability began in the 17th century in France by two mathematicians Blaise Pascal and Pierre de Fermat. The area under each curve is `1`. In probability theory and statistics, if in a discrete probability distribution, the number of successes in a series of independent and identically disseminated Bernoulli trials before a particularised number of failures happens, then it is termed as the negative binomial distribution. Probability rules. 1 – p = Probability of failure. p (x) = 1 2 π σ 2 −−−−√ e (x − μ) 2 2 σ 2 p(x)=12πσ2e(x−μ)22σ2. Geary has shown, assuming that the mean and variance are finite, that the normal distribution is the only distribution where the mean and variance calculated from a set of independent draws are independent of each other. The average number of yearly accidents happen at a Railway station platform during train movement is 7. Uniform Distribution Formula. The concept of probability distribution formula is very important as it basically estimates the expected outcome on the basis of all the possible outcomes for a given range of data. Counting rules. The function f(x) is called a probability density function for the continuous random variable X where the total area under the curve bounded by the x-axis is equal to `1`. Term Description; ξ : location parameter: θ: scale parameter: e: base of the natural logarithm: v: Euler constant (~0.57722) t-distribution. P(A | B) = P(B | A) ⋅ P(A) / P(B) Independent Events. For a number p in the closed interval [0,1], the inverse cumulative distribution function (ICDF) of a random variable X determines, where possible, a value x such that the probability of X ≤ x is greater than or equal to p.

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CC-MAIN-2023-06

https://www.medbroadcast.com/channel/infection/health-tools

math

Not sure if your child has an ear infection? There might be symptoms you're overlooking. Take this quiz to get more information. What do you know about genital herpes? Test yourself with this quiz. 1. Genital herpes is a contagious viral infection that affects _______ people. a) 1 in 4000 b) 1 in 400 c) 1 in 40 d) 1 in 14 e) 1 in 4 2. This percentage of people have genital herpes but are unaware of their disease: a) 20% b) 30% c) 50% d) 60% e) 80% 3. Answer true or false to the following statements. 1. Exposure to HIV always leads to infection. true false 2. HIV is a disease. true false 3. AIDS stands for "acquired infectious disease syndrome." true false 4.

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CC-MAIN-2022-27

http://www.e-booksdirectory.com/details.php?ebook=6834

math

Lectures On Some Fixed Point Theorems Of Functional Analysis by F.F. Bonsall Publisher: Tata Institute Of Fundamental Research 1962 Number of pages: 147 The book is concerned with the application of a variety of methods to both non-linear (fixed point) problems and linear (eigenvalue) problems in infinite dimensional spaces. A wide choice of techniques is available for linear problems, and I have usually chosen to use those that give something more than existence theorems. Download or read it online for free here: by N.P. Landsman - arXiv A graduate-level introduction to C*-algebras, Hilbert C*-modules, vector bundles, and induced representations of groups and C*-algebras, with applications to quantization theory, phase space localization, and configuration space localization. by Gerald Teschl - University of Vienna This free manuscript provides a brief introduction to Functional Analysis. The text covers basic Hilbert and Banach space theory including Lebesgue spaces and their duals (no knowledge about Lebesgue integration is assumed). by J. Cigler, V. Losert, P.W. Michor - Marcel Dekker Inc This book is the final outgrowth of a sequence of seminars about functors on categories of Banach spaces (held 1971 - 1975) and several doctoral dissertations. It has been written for readers with a general background in functional analysis. by Ville Turunen - Aalto TKK In this book you will learn something about functional analytic framework of topology. And you will get an access to more advanced literature on non-commutative geometry, a quite recent topic in mathematics and mathematical physics.

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CC-MAIN-2020-05

https://www.coursehero.com/tutors-problems/Accounting/9659833-WCP9-Waterways-Corporation-is-preparing-its-budget-for-the-coming-year/

math

WCP9 Waterways Corporation is preparing its budget for the coming year, 2014. The first step is to plan for the first quarter of that coming year. Waterways gathered the following information from the managers. Unit sales for November 2013 112,500 Unit sales for December 2013 102,100 Expected unit sales for January 2014 113,000 Expected unit sales for February 2014 112,500 Expected unit sales for March 2014 116,000 Expected unit sales for April 2014 125,000 Expected unit sales for May 2014 137,500 Unit selling price $12 Waterways likes to keep 10% of the next month’s unit sales in ending inventory. All sales are on account. 85% of the Accounts Receivable are collected in the month of sale, and 15% of the Accounts Receivable are collected in the month after sale. Accounts receivable on December 31, 2013, totaled $183,780. Direct Materials Item Amount Used per Unit Inventory, Dec. 31 Metal 1 lb @ 58¢ per lb. 5,177.5 lbs Plastic 12 oz @ 6¢ per oz 3,883.125 lbs Rubber 4 oz @ 5¢ per oz 1,294.375 lbs 2 lbs per unit 10,355.0 lbs Metal, plastic, and rubber together are 75¢ per pound per unit. Waterways likes to keep 5% of the materials needed for the next month in its ending inventory. Payment for materials is made within 15 days. 50% is paid in the month of purchase, and 50% is paid in the month after purchase. Accounts Payable on December 31, 2013, totaled $120,595. Raw Materials on December 31, 2013, totaled 11,295 pounds. Direct Labor Labor requires 12 minutes per unit for completion and is paid at a rate of $8 per hour. Manufacturing Overhead Indirect materials 30¢ per labor hour Indirect labor 50¢ per labor hour Utilities 45¢ per labor hour Maintenance 25¢ per labor hour Salaries $42,000 per month Depreciation $16,800 per month Property taxes $ 2,675 per month Insurance $ 1,200 per month Janitorial $ 1,300 per month Selling and Administrative Variable selling and administrative cost per unit is $1.60. Advertising $15,000 a month Insurance $ 1,400 a month Salaries $72,000 a month Depreciation $ 2,500 a month Other fixed costs $ 3,000 a month Waterways Continuing Problem 11 Other Information The Cash balance on December 31, 2013, totaled $100,500, but management has decided it would like to maintain a cash balance of at least $800,000 beginning on January 31, 2014. Dividends are paid each month at the rate of $2.50 per share for 5,000 shares outstanding. The company has an open line of credit with Romney’s Bank. The terms of the agreement requires borrowing to be in $1,000 increments at 8% interest. Waterways borrows on the first day of the month and repays on the last day of the month. A $500,000 equipment purchase is planned for February. Instructions For the first quarter of 2014, do the following. (a) Prepare a sales budget. (b) Prepare a production budget. (c) Prepare a direct materials budget. (Round to nearest dollar) (d) Prepare a direct labor budget. (For calculations, round to the nearest hour.) (e) Prepare a manufacturing overhead budget. (Round amounts to the nearest dollar.) (f) Prepare a selling and administrative budget. (g) Prepare a schedule for expected cash collections from customers. (h) Prepare a schedule for expected payments for materials purchases. (Round totals to nearest dollar) (i) Prepare a cash budget. The answer to this question... View the full answer

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CC-MAIN-2019-04

http://blog.voltagp.com/2011/02/2010-season-highlights-part-2-of-3.html

math

Sunday, February 20, 2011 2010 Season Highlights (Part 2 of 3) More action from both indoor and outdoor racing. Posted by Volta Grand Prix at 5:42 PM Labels: kart racing, karting, volta gp, volta grand prix, voltagp Subscribe to: Post Comments (Atom) When you could have} been dealt your two palms, you have to resolve whether or not to take further cards or persist with the cards you could have} in your possession. If you benefit from the pleasure of enjoying in} blackjack in a real casino, take a look at|check out} the live blackjack video games obtainable by way of video streaming. These let you play blackjack with actual dealers in actual time, 안전한 카지노사이트 giving you all of the fun of a live casino experience from the consolation of your personal residence. A participant who splits Aces is often solely allowed to obtain a single additional card on each hand.ReplyDelete

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https://www.teacherspayteachers.com/Product/One-Dozen-Practice-Battles-3536892

math

Great for teams wanting to practice what the real battles are like! Here are twelve Oregon Battle of the Books battles. Each battle has eight In Which Book questions and eight Content questions, just like the ones from State OBOB. There is one question for each of this year’s sixteen books in each battle (grades 3-5). The questions range from easy to medium-hard. Each battle can be used over and over again by different teams. HOW TO BATTLE: - You need a moderator (the person who asks the questions and determines if answers are correct) and a timer/score keeper. - Two teams. Each team sits huddled together. Each team will decide which child is their “spokes person,” the person who answers the questions for the team. - Decide which team will answer the odd numbered questions (1, 3, 5, etc.), and which team will answer the even numbered questions (2, 4, 6, etc.) by a coin toss. - The moderator asks the odd team the first question. - When the question has been asked, the timekeeper begins timing 15 seconds. During this time the team may whisper among themselves to come up with the right answer. - At the end of 15 seconds the timer says “time.” Then the spokes person says the answer. It is very important to know that once time has been called, or when the spokes person begins to speak (if they know the answer in advance of time being called) all talk among the team must end. - If the answer is wrong, or they do not give an answer, the other team has a chance to “steal” the points. In this case the question is asked again, and the timer times 15 seconds again, etc. - Each question is worth 5 points. If it is a 2 part question, the first correct answer is worth 3 points and the second correct part is worth 2 points. A team can steal the second part of a question answered incorrectly or not answered by the other team. All In Which Book questions 2 part questions, with the answers being the title and the author of a book. Some Ccontent questions are also 2 part questions.

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CC-MAIN-2018-43

http://www.fact-archive.com/encyclopedia/Geodesy

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Geodesy, also called geodetics, is the scientific discipline that deals with the measurement and representation of the earth, its gravitational field and geodynamic phenomena (polar motion, earth tides, and crustal motion) in three-dimensional time varying space. Geodesy is primarily concerned with positioning and the gravity field and geometrical aspects of their temporal variations, although it can also include the study of the Earth's magnetic field. Wolfgang Torge quotes in his 2001 textbook Geodesy (3rd edition) Friedrich Robert Helmert as defining geodesy as "the science of the measurement and mapping of the earth's surface." As Torge also remarks, the shape of the earth is to a large extent the result of its gravity field. This applies to the solid surface (orogeny; few mountains are higher than 10 km, few deep sea trenches deeper than that). It affects similarly the liquid surface (dynamic sea surface topography) and the earth's atmosphere. For this reason, the study of the Earth's gravity field is seen as a part of geodesy, called physical geodesy. The figure of the Earth Primitive ideas about the figure of the Earth, still found in young children, hold the Earth to be flat, and the heavens a physical dome spanning over it. Already the ancient Greeks were aware of the spherical shape of the Earth. Lunar eclipses, e.g., always have a circular edge of appox. three times the radius of the lunar disc; as these always happen when the Earth is between Sun and Moon, it suggests that the object casting the shadow is the Earth and must be spherical (and four times the size of the Moon, the lunar and solar discs being the same size). Also an astronomical event like a lunar eclipse which happened high in the sky in one end of the Mediterranean world, was close to the horizon in the other end, also suggesting curvature of the Earth's surface. Finally, Eratosthenes determined a remarkably accurate value for the radius of the Earth at around 200 BC. The Renaissance brought the invention of the telescope and the theodolite, making possible triangulation and grade measurement . Of the latter especially should be mentioned the expedition by the French Academy of Sciences to determine the flattening of the Earth. One expedition was sent to Lapland as far North as possible under Pierre Louis Maupertuis (1736-37), the other under Pierre Bouguer was sent to Peru, near the equator (1735-44). At the time there were two competing theories on the precise figure of the Earth: Isaac Newton had calculated that, based on his theory of gravitation, the Earth should be flattened at the poles to a ratio of 1:230. On the other hand the astronomer Jean Dominique Cassini held the view that the Earth was elongated at the poles. Measuring the length, in linear units, of a degree of change in north-south direction of the astronomical vertical, at two widely differing latitudes would settle the issue: on a flattened Earth the length of a degree grows toward the poles. The flattening found by comparing the results of the two grade measurement expeditions confirmed that the Earth was flattened, the ratio found being 1:210. Thus the next approximation to the true figure of the Earth after the sphere became the flattened, biaxial ellipsoid of revolution . In South America Bouguer noticed, as did George Everest in India, that the astronomical vertical tended to be "pulled" in the direction of large mountain ranges, obviously due to the gravitational attraction of these huge piles of rock. As this vertical is everywhere perpendicular to the idealized surface of mean sea level, or the geoid, this means that the figure of the Earth is even more irregular than an ellipsoid of revolution. Thus the study of the "undulations of the geoid" became the next great undertaking in the science of studying the figure of the Earth. Geoid and reference ellipsoid The geoid is essentially the figure of the Earth abstracted from its topographic features. It is an idealized equilibrium surface of sea water, the mean sea level surface in the absence of currents, air pressure variations etc. and continued under the continental masses. The geoid, unlike the ellipsoid, is irrgular and too complicated to serve as the computational surface on which to solve geometrical problems like point positioning. The geometrical separation between it and the reference ellipsoid is called the geoidal undulation. It varies globally between 110 m. A reference ellipsoid, customarily chosen to be the same size (volume) as the geoid, is described by its semi-major axis (equatorial radius) a and flattening f. The quantity f = (a - b) / a, where b is the semi-minor axis (polar radius) is a purely geometrical one. The mechanical ellipticity of the earth (dynamical flattening) is determined by observation and differs from the geometrical because the earth is not of uniform density. The 1980 Geodetic Reference System (GRS80) posited a 6,378,137 m semi-major axis and a 1:298.257 flattening. This system was adopted at the XVII General Assembly of the International Union of Geodesy and Geophysics (IUGG). It is essentially the basis for geodetic positioning by the Global Positioning System and is thus in extremely widespread use also outside the geodetic community. The numerous other systems which have been used by diverse countries for their maps and charts are gradually dropping out of use as more and more countries move to global, geocentric reference systems using the GRS80 reference ellipsoid. Co-ordinate systems in space The locations of points in three-dimensional space are most conveniently described by three cartesian or rectangular coordinates, X,Y and Z. Since the advent of satellite positioning, such coordinate sytems are typically geocentric: the Z axis is aligned with the Earth's (conventional or instantaneous) rotation axis. Before the satellite geodesy era, the coordinate systems associated with geodetic datums attempted to be geocentric, but their origins differed from the geocentre by hundreds of metres, due to regional deviations in the direction of the plumbline (vertical). These regional geodetic datums, such as ED50 (European Datum 1950) or NAD83 (North American Datum 1983) have ellipsoids associated with them that are regional 'best fits' to the geoids within their areas of validity, minimising the deflections of the vertical over these areas. It is only because GPS satellites orbit about the geocentre, that this point becomes naturally the origin of a coordinate system defined by satellite geodetic means, as the satellite positions in space are themselves computed in such a system. Geocentric co-ordinate systems used in geodesy can be divided naturally into two classes: - Inertial reference systems, where the co-ordinate axes retain their orientation relative to the fixed stars, or equivalently, to the rotation axes of ideal gyroscopes; the X axis points to the vernal equinox - Co-rotating, also ECEF ("Earth Centred, Earth Fixed"), where the axes are attached to the solid body of the Earth. The X axis lies within the Greenwich observatory's meridian plane. The co-ordinate transformation between these two systems is described to good approximation by (apparent) sidereal time. A more accurate description takes also length-of-day variations and polar motion into account, phenomena currently closely monitored by geodesists. Co-ordinate systems in the plane - Plano-polar, in which points in a plane are defined by a distance s from a specified point along a ray having a specified direction α with respect to a base line or axis; - Rectangular, points are defined by distances from two perpendicular axes called x and y. It is geodetic practice -- contrary to the mathematical convention -- to let the x axis point to the North and the y axis to the East. Rectangular co-ordinates in the plane can be used intuitively with respect to one's current location, in which case the x axis will point to the local North. More formally, such co-ordinates can be obtained from three-dimensional co-ordinates using the artifice of a map projection. It is not possible to map the curved surface of the Earth onto a flat map surface without deformation. The compromise most often chosen -- called a conformal projection -- preserves angles and length ratios, so that small circles are mapped as small circles and small squares as squares. An example of such a projection is UTM (Universal Transverse Mercator). Within the map plane, we have rectangular co-ordinates x and y. In this case the North direction used for reference is the map North, not the local North. The difference between the two is called meridian convergence. It is easy enough to "translate" between polar and rectangular co-ordinates in the plane: let, as above, direction and distance be α and s respectively, then we have The reverse translation is slightly more tricky. In geodesy, point or terrain heights are "above sea level", an irregular, physically defined surface. Therefore a height should ideally not be referred to as a co-ordinate. It is more like a physical quantity, and though it can be tempting to treat height as the vertical coordinate z, in addition to the horizontal co-ordinates x and y, and though this actually is a good approximation of physical reality in small areas, it becomes quickly invalid in larger areas. Heights come in the following variants: - Orthometric height s - Normal height s - Geopotential number s Each have their advantages and disadvantages. Both orthometric and normal heights are heights in metres above sea level, which geopotential numbers are measures of potential energy (unit: m2s - 2) and not metric. Orthometric and normal heights differ in the precise way in which mean sea level is conceptually continued under the continental masses. The reference surface for orthometric heights is the geoid, an equipotential surface approximating mean sea level. None of these heights are in any way related to geodetic or ellipsoidial heights, which express the height of a point above the reference ellipsoid. Satellite positioning receivers typically provide ellipsoidal heights, unless they are fitted with special conversion software based on a model of the geoid. Because geodetic point c-oordinates (and heights) are always obtained in a system that has been constructed itself using real observations, we have to introduce the concept of a geodetic datum: a physical realization of a co-ordinate system used for describing point locations. The realization is the result of choosing conventional co-ordinate values for one or more datum points. In the case of height datums, it suffices to choose one datum point: the reference bench mark, typically a tide gauge at the shore. Thus we have vertical datums like the NAP (Normaal Amsterdams Peil), the North American Vertical Datum 1988 (NAVD88), the Kronstadt datum, the Trieste datum, etc. In case of plane or spatial coordinates, we typically need several datum points. A regional, ellipsoidal datum like ED50 can be fixed by prescribing the undulation of the geoid and the deflection of the vertical in one datum point, in this case the Helmert Tower in Potsdam. However, an overdetermined ensemble of datum points can also be used. Changing the coordinates of a point set referring to one datum, to make them refer to another datum, is called a datum transformation. In the case of vertical datums, this consists of simply adding a constant shift to all height values. In the case of plane or spatial coordinates, datum transformation takes the form of a similarity or Helmert transformation, consisting of a rotation and scaling operation in addition to a simple translation. In the plane, a Helmert transformation has four parameters, in space, seven. A note on terminology In the abstract, a co-ordinate system as used in mathematics and geodesy is, e.g., in ISO terminology, referred to as a coordinate system. International geodetic organizations like the IERS (International Earth Rotation and Reference Systems Service) speak of a reference system. When these co-ordinates are realized by choosing datum points and fixing a geodetic datum, ISO uses the terminology coordinate reference system, while IERS speaks of a reference frame. A datum transformation again is referred to by ISO as a coordinate transformation. (ISO 19111: Spatial referencing by coordinates). Point positioning is the determination of the coordinates of a point on land, at sea, or in space with respect to a coordinate system. Point position is solved by compution from measurements linking the known positions of terrestrial or extraterrestrial points with the unknown terrestrial position. This may involve transformations between or among astronomical and terrestrial coordinate systems. The known points used for point positioning can be, e.g., triangulation points of a higher order network, or GPS satellites. Traditionally, a hierarchy of networks has been built to allow point positioning within a country. Highest in the hierarchy were triangulation networks. These were densified into networks of traverse s (polygons), into which local mapping surveying measurements, usually with measuring tape, corner prism and the familiar red and white poles, are tied. Nowadays all but special measurements (e.g., underground or high precision engineering measurements) are performed with GPS. The higher order networks are measured with static GPS , using differential measurement to determine vectors between terrestrial points. These vectors are then adjusted in traditional network fashion. A global polyhedron of permanently operating GPS stations under the auspices of the IERS is used to define a single global, geocentric reference frame which serves as the "zeroth order" global reference to which national measurements are attached. For surveying mappings, frequently Real Time Kinematic GPS is employed, tying in the unknown points with known terrestrial points close by in real time. One purpose of point positioning is the provision of known points for mapping measurements, also known as (horizontal and vertical) control. In every country, thousands of such known points exist in the terrain and are documented by the national mapping agencies. Constructors and surveyors involved in real estate will use these to tie their local measurements to. In geometric geodesy we formulate two standard problems: the geodetic principal problem and the geodetic inverse problem. - Geodetic principal problem (also: first geodetic problem) - Given a point (in terms of its coordinates) and the direction (azimuth) and distance from that point to a second point, determine (the co-ordinates of) that second point. - Geodetic inverse problem (also: second geodetic problem) - Given two points, determine the azimuth and length of the line (straight line, great circle or geodesic) that connects them. In the case of plane geometry (valid for small areas on the Earth's surface) the solutions to both problems reduce to simple trigonometry. On the sphere, the solution is significantly more complex, e.g., in the inverse problem the azimuths will differ between the two end points of the connecting great circle arc. On the ellipsoid of revolution , closed solutions do not exist; series expansions have been traditionally used that converge rapidly. In the general case, the solution is called the geodesic for the surface considered. It may be nonexistent or non-unique. The differential equations for the geodesic can be solved numerically, e.g., in MatLab(TM). Geodetic observational concepts Here we define some basic observational concepts, like angles and coordinates, defined in geodesy (and astronomy as well), mostly from the viewpoint of the local observer. - The plumbline or vertical is the direction of local gravity, or the line that results by following it. It is slightly curved. - The zenith is the point on the celestial sphere where the direction of the gravity vector in a point, extended upwards, intersects it. More correct is to call it a <direction> rather than a point. - The nadir is the opposite point (or rather, direction), where the direction of gravity extended downward intersects the (invisible) celestial sphere. - The celestial horizon is a plane perpendicular to a point's gravity vector. - Azimuth is the direction angle within the plane of the horizon, typically counted clockwise from the North (in geodesy) or South (in astronomy and France). - Elevation is the angular height of an object above the horizon, Alternatively zenith distance, being equal to 90 degrees minus elevation. - Local topocentric co-ordinates are azimut (direction angle within the plane of the horizon) and elevation angle (or zenith angle) as well as distance if known. - The North celestial pole is the extension of the Earth's (precessing and nutating) instantaneous spin axis extended Northward to intersect the celestial sphere. (Similarly for the South celestial pole.) - The celestial equator is the intersection of the (instantaneous) Earth equatorial plane with the celestial sphere. - A meridian plane is any plane perpendicular to the celestial equator and containing the celestial poles. - The local meridian is the plane containing the direction to the zenith and the direction to the celestial pole. Geodetic observing instruments The level is used for determining height differences and height reference systems, commonly referred to mean sea level. The traditional spirit level produces these practically most useful heights above sea level directly; the more economical use of GPS instruments for height determination requires precise knowledge of the figure of the geoid, as GPS only gives heights above the GRS80 reference ellipsoid. As geoid knowledge accumulates, one may expect use of GPS heighting to spread. The theodolite is used to measure horizontal and vertical angles to target points. These angles are referred to the local vertical. The tacheometer additionally determines, electronically or electro-optically, the distance to target, and is highly automated in its operations. The method of free station position is widely used. For local detail surveys, tacheometers are commonly employed although the old-fashioned rectangular technique using angle prism and steel tape is still an inexpensive alternative. More and more, also real time kinematic (RTK) GPS techniques are used. Data collected is tagged and recorded digitally for entry into a Geographic Information System (GIS) data base. Geodetic GPS receivers produce directly three-dimensional co-ordinates in a geocentric co-ordinate frame. Such a frame is, e.g., WGS84, or the frames that are regularly produced and published by the International Earth Rotation Service (IERS). GPS receivers have almost completely replaced terrestrial instruments for large-scale base network surveys. For planet-wide geodetic surveys, previously impossible, we can still mention satellite laser and Very Long Baseline Interferometer (VLBI) techniques. All these techniques also serve to monitor Earth rotation irregularities as well as plate tectonic motions. Gravity is measured using gravimeters . Common field gravimeters are spring based and referred to a relative. Absolute gravimeters, which nowadays can also be used in the field, are based directly on measuring the acceleration of free fall of a reflecting prism in a vacuum tube. Gravity surveys over large areas can serve to establish the figure of the geoid over these areas. Units and measures on the ellipsoid Geographical latitude and longitude are stated in the units degree, minute of arc, and second of arc. They are angles, not metric measures, and describe the direction of the local normal to the reference ellipsoid of revolution. This is approximately the same as the direction of the plumbline, i.e., local gravity, which is also the normal to the geoid surface. For this reason, astronomical position determination, measuring the direction of the plumbline by astronomical means, works fairly well provided an ellipsoidal model of the figure of the Earth is used. A geographic mile, defined as one minute of arc on the equator, equals 1,855.32571922 m. A nautical mile is one minute of astronomical latitude. The radius of curvature of the ellipsoid varies with latitude, being the longest at the pole and shortest at the equator as is the nautical mile. A metre was originally defined as the 40 millionth part of the length of a meridian. This means that a kilometre is equal to (1/40,000) * 360 * 60 meridional minutes of arc, which equals 0.54 nautical miles. Similarly a nautical mile is on average 1/0.54 = 1.85185... km. - The Geodesy Page. - Welcome to Geodesy

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https://methods.sagepub.com/Reference/the-sage-encyclopedia-of-educational-research-measurement-and-evaluation/i18967.xml

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Researchers are often faced with the task of making statements about entire populations. However, including every member of a population into a [Page 1520]study is often not possible and simply not feasible. Thus, subsets of the population (samples) must be chosen to represent the population. If samples are collected properly, precise statements can be made about a population, with a fairly high degree of confidence, from relatively small samples. Numerous techniques have been developed to ensure that the subset, or sample, is representative of the overall population so generalizations can be made. Simple random sampling is a probability method of selecting a subset, or sample, from a larger population in such a manner that every element (individual member of the population whose characteristics are to ... Looks like you do not have access to this content.

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https://www.srmek.xyz/2019/06/chanchal-ghosh-maths-trick-bengali-pdf-download.html

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Chanchal Ghosh Maths Trick Bengali PDF Download Hello Readers, here you can find Chanchal Ghosh Maths tricks for PDF. Are you searching for the pdf formula of Bengali Maths Tricks? Then here’s the correct location for you. Mathematics is the most significant topic for any type of competitive examination, such as wbcs, ssc, upsc, rrb, bank, rail, main etc. Most learners are scared of mathematics. But if you get a better outcome on the Competitive Exam, Today freegk.in shares the link of Chanchal Ghosh Maths to download pdf book tricks. Most Bengal students search the pdf book of Chanchal Ghosh Maths on the Internet. Here we update some portion of the pdf book on Chanchal Ghosh math tricks. You can readily download this Chanchal Ghosh pdf book from our website. Chanchal Ghosh math books are one of West Bengal’s most famous books. So readers download this Chanchal Ghosh pdf book as quickly as possible.

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CC-MAIN-2023-40

https://www.econstor.eu/handle/10419/93630

math

I establish that inflation risk is priced in the cross section of stock returns: Stocks that have low returns during inflationary times command a risk premium. I estimate a market price of inflation risk that is comparable in magnitude to the price of risk for the aggregate market. Inflation is therefore a key determinant of risk in the cross section of stocks. The inflation premium cannot be explained by either the Fama-French factors or industry effects. Instead, I argue the premium arises because high inflation lowers expectations of future real consumption growth. To formalize and test this hypothesis, I develop a consumption-based general equilibrium model. The model generates a price of inflation risk consistent with my empirical estimates, while simultaneously matching the joint dynamics of consumption and inflation, the aggregate equity premium, and the level and slope of the yield curve. My model suggests that the costs of inflation are significant: A representative agent would be willing to give up 1.5 percent of lifetime consumption to eliminate all inflation risk.

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https://a2group5.medium.com/simply-supported-beams-with-uniform-distributed-load-udl-11c0245ba9d8?source=post_internal_links---------1----------------------------

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Authors of this Blog are Atharva Dandapur, Sonal Dangare, Tejas Dani, Arjun Desai. WHAT ARE BEAMS? Beams are important types of structural elements that play a key role in creating a safe load path to transfer the weight and forces on a structure to the foundations and into the ground. They are horizontal members which carry perpendicular loads in their longitudinal direction. They are used to support the weight of the floor, roofs of the building and transfer the load to the vertical load bearing member of the structure. Following are the loads acting on a BEAM: b. Varied by length c. Single point TYPES OF BEAMS: - Beams are classified according to various features such as: According to the End Support Conditions: · Simply Supported Beam · Cantilever beam · Continuous beam · Fixed end beam · Overhanging beam · Double overhanging beam Ø According to the Cross Section: According to Geometry · Straight beam · Curved Beam · Tapered Beam According to Casting Condition · In-situ casting beam · Precast concrete beam · Pre-stressed concrete beam SIMPLY SUPPORTED BEAMS: - The simply supported beam is one of the simplest structures. It features only two supports, one at each end. One is a pinned support and the other is a roller support. With this configuration, the beam is inhibited from any vertical movement at both ends whereas it is allowed to rotate freely. Due to the roller support it is also allowed to expand or contract axially, though free horizontal movement is prevented by the other support. Removing any of the supports inserting an internal hinge, would render the simply supported beam to a mechanism, that is body the moves without restriction in one or more directions. Obviously this is unwanted for a load carrying structure. Therefore, the simply supported beam offers no redundancy in terms of supports, and if a local failure occurs the whole structure would collapse. These type of structures that offer no redundancy are called critical or determinant structures. To the contrary, a structure that features more supports than required to restrict its free movements is called redundant or indeterminate structure. FINITE DIFFERENCE METHOD: - In numerical analysis, finite-difference methods (FDM) are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences. Here , spatial domain and time interval are discretized, or broken into a finite number of steps, and the value of the solution at these discrete points is approximated by solving algebraic equations containing finite differences and values from nearby points. Finite difference methods convert ordinary differential equations (ODE) or partial differential equations (PDE), which may be solved by matrix algebra techniques. Today, FDM are one of the most common approaches to the numerical solution of PDE, along with finite element methods. We solve second-order ordinary differential equations of the form: Basic equation: The differential equation that governs the deflection of a simply supported beam under uniformly distributed load is: · x = Location along the beam (in) · E = Young’s modulus of elasticity of beam (psi) · I = Second moment of area (in⁴) · Q = Uniform loading intensity (lb/in) · L = Length of beam (in) 2. Solution of equation 3. Conversion of equation to Matrix 4. Applying boundary conditions 5. Find Error The deflection ‘y’ in a simply supported beam with a uniform load ‘q’ and a tensile axial load T is given by: · x = location along the beam (in) · T = tension applied (lbs) · E = young’s modulus of elasticity of beam (psi) · I = second moment of area (in⁴) · q = uniform load intensity (lb/in) · L = length of beam (in) T = 7200lbs, q = 5400lbs/in, L = 75in, E = 30Msi , I = 120 in⁴ 1. Find the deflection of beam at x=50”. Use a step size of △x = 25” and the approximate the derivatives by central divided difference approximation. 2. Find the relative true error in the calculation of y (50). d²y ∕ dx²-Ty / El = qx(L -X)/2EL d²y ∕ dx² -7200y/(30*10⁶)*(120) = (5400)*x*(75-x) / 30*10⁶*2*120 d²y ∕ dx² -2*10⁶y = 7.5*10^-2x*(75-x) Approximating the derivative at node i by CENTRAL DIVIDED DIFFERENCE approximation d² y/dx² =y(i+1) -2yi+ y(i-1) 1/(∆x)² Since △x=25, we have 4 nodes 1/((∆x)² ) y(i+1) -2yi+y(i-1) -2∗10⁶ yi=7.5∗10^(-7) xi∗(75-xi) Equations at nodes: Node 1- y1=0 …..(from simply supported boundary condition at x=0) Node 2 — (y3 — 2y2+ y1)/(25)² -2∗10^(-6) y2=7.5∗10^(-7) x2∗(75-x2 ) — 0.0016y1–0.003202y2+0.0016y3=9.375∗10^(-4) Node 3 — (y4 — 2y3+ y2)/(25)² -2∗10^(-6) y3=7.5∗10^(-7) x3∗(75-x3 )- 0.0016y2–0.003202y3+0.0016y4=9.375∗10^(-4) Node 4 — y4=0 …..(from simply supported boundary condition at x=75) Solving the equations we get, y(50)= y(x2) ≈ y2=-0.58521 yh= k1 e⁰.0014142x+ k2 e^(-0.0014142x) ……homogeneous part yp=Ax²+Bx+C …..particular part (d² yp)/dx² -2∗10⁶ yp=7.5∗10^(-7) x∗(75-x) (d² (Ax²+Bx+C ))/dx² -2∗10⁶ (Ax²+Bx+C )=7.5∗10^(-7) x∗(75-x) Equating the terms we get, Solving the equations, Applying the following boundary conditions, To calculate the RELATIVE TRUE ERROR, The true error is given by, Et = Exact Value — Approximate Value ; Et = 0.05320 The Relative Error is given by, MATLAB CODE: - A common usage is for things like solving Differential Equations numerically, and approximating derivatives for root finding and numerical optimization schemes. They are generally used to solve Partial Differential Equation (PDE) like- heat, diffusion, flow, and electromagnetics. From the graph we can see that as the nodes increases the value of error decreases. As the number of nodes increases the deflection between the true value and the approximate value decreases. Finite difference method is easy, simple but errors are more, thus to reduce these errors we need to increase the step size. Beam Deflection problems are accurately solved by Finite Difference Method.

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https://social.librem.one/@petrisch/106993799293081616

math

@petrisch that looks very good @petrisch this is really great I'm sorry but... this image seems really noisy. I've looked at images at the given address. I've seen some are way off in White Balance. Does anybody know how is done the WB? Is it post-process or by an ISP?

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https://www.purplemath.com/casstown_precalculus_tutors.php

math

...In fact, some of my research publications have focused on students' understanding of optimization problems, related rates problems, and marginal change (i.e., marginal cost, marginal revenue, and marginal profit). Please feel free to contact me with any questions or to schedule your first less... (read more) ...I am Enthusiastic and Encouraging yet Honest about what it takes to get to your goal. I have worked with many students who came in testing from 14 to 30 on ACT Math with great success and score gains! I have tutored and worked with elementary aged students, ages 6 - 14, for many years in a variety of situations and settings, and have served students K-8 from over 6 school districts. (read more) ...I have tutored several subjects that involve using linear algebra, including finite math, engineering calculus, and elementary linear algebra. As a student, I have taken linear algebra as well as two abstract algebra classes. As a math education, I have taken many proof-based math classes. (read more) ...I also made myself available to the students for one-on-one help, which was tantamount to tutoring. I also taught college algebra (which is equivalent to precalculus) as well as remedial math at Central State for a semester. As a graduate student at Clarkson University, I tutored Physics at the same time I taught physics recitations. (read more)

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CC-MAIN-2022-05

https://www.jiskha.com/questions/1384156/A-self-supporting-ladder-placed-so-that-the-angle-between-the-ladder-and-the-ground

math

A self-supporting ladder placed so that the angle between the ladder and the ground is 75.5° can be used to test whether the ladder can support a certain load. The diagram represents a ladder placed next to a window that is 20 feet above the ground. What is the length of the ladder? Round to the nearest tenth. I don't see any diagram but by a simple trig ratio, we have cos 75.5 = 20/L L = 20/cos75.5 = appr 79.9 ft Wow, a ladder to go up about 8 stories high??? posted by Reiny

s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221210408.15/warc/CC-MAIN-20180816015316-20180816035316-00620.warc.gz

CC-MAIN-2018-34

http://mathhelpforum.com/pre-calculus/97463-distance-point-line-hyperbola-asymptotes-problem.html

math

I've been struggling with this for a while now and to no avail. Here's the question. A hyperbola of the form has asymptotes with equations and passes through the point (a,0). Find the equation of the hyperbola in terms of x, y, a and m. A point P on this hyperbola is equidistant from one of its asymptotes and the x-axis. Prove that, for all values of m, P lies on the curve with equation. I hate these type of questions as I always end up with a pile of messy algebra and ton of screwed up paper. Here's my attempt this time. I got the first part ok, the equation of the hyperbola is For the second part I decided to pick an arbitrary point on the hyperbola and find its distance from the asymptote and the x-axis and set them equal, then eliminate m. So my general point is . Now the distance of this point to the x-axis is simply . For the moment I chose the point to be in the first quadrant so the equation of the corresponding asymptote is . Wolfram says that the distance between the point and the asymptote is where the line is So for my point and line this becomes which I set equal to the distance from the y-axis to obtain It is here where I have no idea how to eliminate m to get the required equation or even if I've been doing it wrong the whole time. Please can someone help, thanks!! I also had trouble with messy algebra, but I got there in the end, using a more geometric/trigonometric approach. The asymptote makes an angle with the x-axis, where . If a point is equidistant from these two lines, then it must lie on their bisector, namely the line , where . The formula for tan of twice an angle then tells you that . Therefore . If the point (x,y) lies on the line then , and so , from which . If the point (x,y) also lies on the hyperbola then , from which . Substitute the formula for m from the previous paragraph into that, and you get the result. That's a pretty smart way! I got there in the end by saying that the point be (x,y) and using their relationship only at the end, it went something like this Now I square again because this remove any ambiguity as to which asymptote we are referring to, so we now have Now, I used the fact that x and y lie on a hyperbola to get So I got there in the end, but thank you for all your help I don't know what it is about these type of questions (conic section/locus ones) but they seem to knock me dead with insane algebra every single time. Are they generally quite tricky or am I being really dumb! :P

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CC-MAIN-2016-50

https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-4-section-4-4-the-trapezoid-exercises-page-213/40b

math

Work Step by Step DK is the median because it is the center step on the ladder and its length is equal to the average of the top and bottom steps. DK=AH+3(.125) DK=2.375 DK=.5(AH+GN) DK=.5(2+2.75) DK=.5(4.75) DK=2.375 You can help us out by revising, improving and updating this answer.Update this answer After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00470.warc.gz

CC-MAIN-2022-27

https://bz.standardtoday.co.uk/812-electrical-resistance.html

math

We are searching data for your request: Upon completion, a link will appear to access the found materials. We know that materials present degrees of difficulty for the passage of electric current. This degree of difficulty is called electrical resistance. Even metals, which are generally good conductors, have resistance. Resistance unit of measure is ohm ( ). Devices that are used in an electrical circuit are called resistors. Resistors are used in a circuit to increase or decrease the intensity of the electrical current that flows through it. We can compare the electrical resistance to those barriers we encounter on the running tracks for the obstacle course. The more obstacles the slower is the average speed of the runners. In a circuit it happens the same way: the more electrical resistance, the less current that flows through the conductor wire. The most common application of resistors is to convert electrical energy to thermal energy. This is because the electrons moving in the resistor collide with the crystalline lattice that forms it, generating heat. This phenomenon is called the joule effect in our daily lives: in electric showers, irons, electric stoves, etc. Note that all of these appliances "provide heat." The incandescent lamp itself converts more electrical energy into thermal energy than light energy, the latter being its great purpose: 85% of the energy it consumes is transformed into heat. In contrast, fluorescent lamps, considered “cold light bulbs”, have a much smaller share of electrical energy converted to heat and are therefore economical. Ohm's First Law It was observed experimentally in some resistors, which The current set in a circuit is directly proportional to the applied voltage and inversely proportional to the resistance of the circuit devices and the wires that connected them. That is: the higher the generator voltage, the higher the current and the higher the resistance, the lower the current. This relationship is expressed mathematically by: on what: U it's the tension R is the resistance i is the current Let's look at an example: A small lamp has a voltage of 12 V. Knowing that its resistance is determine the current through the lamp. We know that . We have to:

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CC-MAIN-2022-33

https://forum.openwrt.org/t/how-to-configure-a-cantenna/73919

math

I connected a homemade 2.4GHz cantenna instead of one of the 4 antennas of my Netgear Nighthawk X4S R7800. I would like to tell to R7800 to use the cantenna as a source of the Wi-Fi signal (to be "redistributed" therefore around the house through the 3 remaining antennas). I have already flashed OpenWrt on Netgear, but I have no idea how to proceed. Would anyone know how to help me? Thanks in advance for your help!

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https://appcdstage.collegedunia.com/exams/iit-jam/mathematical-statistics-syllabus

math

IIT JAM 2022 Syllabus for Mathematical Statics is divided into two parts i.e. Mathematics and Statistics. The weightage of the mathematics section is 40% and of the statistics section, it is 60%. Candidates who are planning to appear for Mathematical Statistics can download the complete IIT JAM 2022 Syllabus for Mathematical Statics from the official website. Candidates need to analyze section-wise syllabus rigorously before appearing in the exam. Candidates should also note that the question paper will be divided into 3 sections and three types of questions will be asked in the paper i.e. MCQ, MSQ, and NAT. Read the article to know more about IIT JAM section-wise syllabus of mathematical statistics, exam pattern, best books, and preparation tips. Check IIT JAM 2022 Exam Pattern Read the complete article for detailed Mathematical Statics Syllabus, Preparation Tips and Important Books. IIT JAM Syllabus for Mathematical Statics There are two sections such as Mathematics and Statistics in IIT JAM Syllabus. Section-wise details for Mathematics and Statistics Syllabus for IIT JAM are tabulated below. IIT JAM Mathematics Syllabus |Sequences and Series||All the areas included in this section are Convergence of real numbers sequences, Comparison, root and ratio tests for convergence of series of real numbers.| |Differential Calculus||In this unit, all the subtopics covered are Limits, continuity and differentiability of functions of one and two variables. indeterminate forms, maxima and minima of functions of one and two variables. Apart from that, there are some theorems such as Rolle's theorem, mean value theorems, Taylor's theorem, etc.| |Integral Calculus||Under this section all the properties are Fundamental theorems of integral calculus; Double and triple integrals; applications of definite integrals, arc lengths, areas and volumes, etc| |Matrices||From Rank, inverse of a matrix; Systems of linear equations; Linear transformations, eigenvalues and eigenvectors to Cayley-Hamilton theorem; symmetric; skew-symmetric and orthogonal matrices are explained in this section.| IIT JAM Statistics Syllabus |Probability||This section covers the Axiomatic definition of probability and properties, conditional probability, and multiplication rule. It also covers the Theorem of total probability along with Bayes’ theorem and independence of events.| |Random Variables||There are various sub topics such as Probability mass function, density function and cumulative distribution functions along with distribution of a function of a random variable. Mathematical expectation, moments and moment generating function. Chebyshev's inequality in this unit.| |Standard Distributions||This section includes Binomial, negative binomial, geometric, Poisson and hypergeometric, uniform, exponential, gamma, beta, and normal distributions. Apart from that, there are Poisson and normal approximations of a binomial distribution.| |Joint Distributions||In this part, all the areas are Joint, marginal and conditional distributions; Distribution of functions of random variables; Joint moment generating function. It also covers Product moments, correlation, simple linear regression; Independence of random variables.| |Sampling distributions and Limit Theorems||In the first part, there are Chi-square along with t and F distributions, and their properties. And in the second part, all the areas are the Weak law of large numbers and the Central limit theorem that defines i.i.d. with finite variance case only.| |Estimation||This section consists of Unbiasedness, consistency, and efficiency of estimators, details of the method of moments as well as the method of maximum likelihood. Besides that, there is Sufficiency, factorization theorem also included. From Completeness, Rao-Blackwell and Lehmann-Scheffe theorems, uniformly minimum variance unbiased estimators to Rao-Cramer inequality; Confidence intervals for the parameters of univariate normal; two independents normal, and one parameter exponential distributions are there.| |Testing of Hypotheses||Here are some of the primary concepts and applications of Neyman-Pearson Lemma for testing simple and composite hypotheses along with Likelihood ratio tests for parameters of the univariate normal distribution, etc.| What are the Valuable Tips to Prepare IIT JAM Syllabus for Mathematics and Statistics? Some of the most salient principles for taking preparation for the subject Mathematics and Statistics of IIT JAM Syllabus are given below. - As the preparation depends on the primary concepts of the syllabus, so, candidate need to well aware of the entire syllabus - Make a study plan with all the subject matter of the syllabus for some months - Need to improve problem-solving skills as this section is based on the that - Centre of attention must be in probability, variable and also in distribution - Practice all the theorems and formula - Note down all the formulas one by one and keep practicing daily with the time table - Make important notes from all the theorems and mark them in bold - It is a must to solve all the questions from the previous year paper - Start giving the mock test as many times as possible and do not forget to solve the sample papers - It is a good practice to revise all the areas from the syllabus to remember Download IIT JAM Practice Papers IIT JAM Mathematics and Statistics Exam Pattern IIT JAM Mathematics and Statistics Exam Pattern is listed below in a table format: |Total Duration of the Test||3 hours| |Total number of questions||60| |Number of Section||3 (Sec A, B and C)| |Question Type for the Test||MCQ, MSQ and NAT| |Marks in Total||100| Section Wise Marks Division |Section||Number of Questions in Total||Question Wise Marks Distribution||Marks in Total| IIT JAM Statistics and Mathematics Books for Preparation Below are some recommended books in a table for preparation for the examination: |Mathematical Statistics||IIT-JAM: M.Sc. Mathematical Statistics Joint Admission Test||Anand Kumar| |Mathematics||A Complete Resource Manual M.Sc Mathematics Entrance examination||Suraj Singh and Reshmi Gupta| |Mathematical Statistics||Fundamentals of Mathematical Statistics||S.C. Gupta and V K Kapoor| |Mathematical Statistics||Introduction to Mathematical Statistics||Hogg| |Statistics||An Introduction to Probability and Statistics||Vijay K. Rohatgi and A.K. Md. Ehsanes Saleh| Also Check: IIT JAM Paper Analysis IIT JAM Syllabus for Mathematics and Statistics FAQ Ques. Are there any changes in IIT JAM Syllabus for Mathematics and Statistics? Ans. No, there are no changes in the Mathematics and Statistics Syllabus for IIT JAM. All the sections will remain the same as the previous year. Ques. What should I know as the basic mathematical concepts from IIT JAM Mathematics and Statistics Syllabus? Ans. Many sections fall under this section such as functions, maxima, and minima, vectors, matrices, integrals, determinants, etc. Ques. Which part is the most difficult part between Statistics and Mathematics in IIT JAM Syllabus? Ans. If you study in a proper way no part will be difficult for you to score well. Ques. Which section is the scoring one between Statistics and Mathematics in IIT JAM Syllabus? Ans. As per the expert advice, you can score maximum from both the section Mathematics and Statistics but the syllabus covers 60% area of Statistics and 40% of Mathematics. Ques. Can I give less importance to the Central limit theorem from the Statistics and Mathematics Syllabus in IIT JAM? Ans. No, you cannot avoid this section as this is one of the most important parts of Statistics. *The article might have information for the previous academic years, which will be updated soon subject to the notification issued by the University/College.

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CC-MAIN-2022-05

https://www.anobii.com/books/An_Introduction_to_Probability_Models/9780125984560/009bc1891b67f0d661

math

The seventh edition of the successful Introduction to Probability Models introduces elementary probability theory and the stochastic processes and is particularly well-suited to those applying probability theory to the study of phenomena in engineeri The seventh edition of the successful Introduction to Probability Models introduces elementary probability theory and the stochastic processes and is particularly well-suited to those applying probability theory to the study of phenomena in engineering, management science, the physical and social sciences, and operations research. Skillfully organized, Introduction to Probability Models covers all essential topics. Sheldon Ross, a talented and prolific textbook author, distinguishes this carefully and substantially revised book by his effort to develop in students an intuitive, and therefore lasting, grasp of probability theory. The seventh edition includes many new examples and exercises, with the majority of the new exercises being less demanding of the student. In addition, the text introduces stochastic processes, stressing applications, in an easily understood manner. There is a comprehensive introduction to the applied models of probability that stresses intuition. Both students and professors will agree that this is the most solid and widely used text for probability theory. * Provides a detailed coverage of the Markov Chain Monte Carlo methods and Markov Chain covertimes * Gives a thorough presentation of k-record values and the surprising Ignatov's theorem * Includes examples relating to: "Random walks to circles," "The matching rounds problem," "The best prize problem" and many more * Contains a comprehensive appendix with the answers to approximately 100 exercises from throughout the text * Accompanied by a complete instructor's solutions manual with step-by-step solutions to all exercises NEW TO THIS EDITION * Includes many new and easier examples and exercises * Offers new material on utilizing probabilistic method in combinatorial optimization problems * Includes new material on suspended animation reliability models * Contains new material on random algorithms and cycles of random permutations ...Continua Nascondi

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CC-MAIN-2020-40

https://vanessahodgkinson.com/math-solve-638

math

Help me solve this math problem free Apps can be a great way to help students with their algebra. Let's try the best Help me solve this math problem free. Our website can help me with math work. The Best Help me solve this math problem free Math can be a challenging subject for many students. But there is help available in the form of Help me solve this math problem free. You'll start by plugging in numbers into a few simple equations, starting with easy ones like addition and multiplication. Then, when you come across a harder one that you can't figure out, the app will ask you to simplify it. As you work through each problem, your confidence will grow as you work out more complex equations from simpler ones. A good way to use solvers is by starting with an easier problem first. Eventually, you'll get used to the way they work and be able to tackle more difficult problems on your own. They can be really helpful for anyone who finds calculus intimidating! Linear equations are mathematical equations that have one variable in terms of the other. For example, if you have a 2x2 table, an equation could be written as 2 + 2 = 4. This equation could be used to put together the pieces of the puzzle by adding or subtracting the corresponding numbers. If you have a 3x3 table, an equation could be written as 3 + 3 = 6. An important thing to remember about linear equations is that they are always true (assuming they make sense). As you can see in the examples above, this means that if you add or subtract variables, you will always get the same answer. The only way to get a different result is if there is a typo or some other mistake in your math. Math word problem solvers are a great way to practice math skills, such as addition and subtraction. Math word problem solvers can be used in a number of ways — for example, to help students learn how to write mathematical equations. They can also be used to practice sequencing and sequencing order, as well as numerusing and number sense. There are many different ways of solving math word problems. One way is to use the four operations. For example, if you are asked to add 5 + 3 + 1, you could solve this using addition by saying "5 plus 3 equals 8." Another way is to use the inverse operation (subtracting). If you are asked to subtract 2 - 1, you could solve this using subtraction by saying "2 minus 1 equals 1." You can also use zero-to-one and one-to-zero visual cues when solving math word problems. Finally, you can use the strategy of breaking down the problem into smaller pieces and then solving each piece separately. Online math tutoring is a great way to help people who are having difficulties in math. It is a very good way to improve one's understanding of math concepts and work on weaknesses in math. For this reason, it is becoming increasingly popular among parents and students. There are many different reasons why you may need to take a math tutor. You may be struggling with the course material or you may simply need help making sure that you are getting the most out of your time in class. Either way, taking advantage of online math tutoring is an excellent way to get the help that you need. When looking for online math tutoring, you should keep several things in mind. First, make sure that the person you are working with is highly qualified. You want someone who can help explain concepts clearly and who will be able to teach you what you need to know. Second, make sure that they have experience with the type of material that you are working on. Finally, make sure that they are flexible enough to meet your needs when it comes to scheduling and location. Calc solvers are applications that solve linear and non-linear mathematical problems, such as finding the solution to a differential equation. Solvers of this type can be used to solve many different types of problems and give an accurate answer. There are two main types of calculators in modern computing: handheld calculators and desktop computers. Handheld calculators are very common in classrooms because they are easy to use, but desktop computers are more powerful and allow for more complex calculations. Calc solvers fall into the category of software, which means they can be downloaded from the Internet. Because there are so many different solver programs available, it is important to choose one that fits your needs. One of the main advantages of using a calc solver is that it does not require any programming knowledge or tools. Another advantage is that it can be set up and used quickly, allowing you to get your answer quickly. However, there are also disadvantages to using a calc solver. A major disadvantage is that they are very expensive compared to hand-held calculators, making them out of reach for many people.

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CC-MAIN-2022-49

https://news.met.police.uk/tag/west-london

math

News View all 86 hits News • Apr 11, 2021 12:30 BST Officers have issued more than 30 Covid fines after being called to a large house party in Mayfair. News • Apr 08, 2021 06:00 BST Five brave mothers who lost their sons to knife crime have shared their emotional stories as part of a new campaign supported by the Metropolitan Police Service and the independent charity, Crimestoppers. Blog posts 3 hits Blog posts • Mar 20, 2021 08:00 GMT BLOG by PC Suzanne Stanbrook on UN Anti-Racism Day Stephen Lawrence Inquiry 20 years on – a perspective from Neil Basu, the Met’s most senior BAME police officer Blog posts • Feb 23, 2019 09:00 GMT The professionalism and the standards we now see are a result of a just campaign by the Lawrence family and the officers they inspired to do the right thing.

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CC-MAIN-2021-17

https://greatminds.org/math/blog/eureka/post/how-far-away-is-the-moon

math

Imagine staring up at the moon on a spring evening and wondering, “how far away is the moon?” What would you do to figure out how far away it is? Perhaps take out your smart phone to Google it? How is it then that the Greeks, who posed the same question 2,200 years ago, found the answer without the aid of Google, NASA, or any modern technology? The Greeks were able to determine a reasonably accurate distance from the earth to the moon using careful observations, measurements, and a clever geometric model of the situation. In the Eureka curriculum, when presented with this same question,“students would use modeling or mathematical pictures — Eureka uses a short list of such flexible models throughout its 14 grades — to bridge the gap between the word problem and the abstract operations needed for the right answer" (Center for Digital Education). IDENTIFYING KEY INFORMATION The Greek astronomer and mathematician, Aristarchus of Samos, calculated the distance from the earth to the moon with only the following information: The earth’s diameter was approximately 8,000 miles (another ingenious calculation, made by Eratosthenes). Experimentation showed that in order to block out the sun with a spherical object, the ratio of the length of the object’s shadow to the object’s diameter must be 108:1. For example, a one-inch marble can just block out the sun at 108 inches from the eye. If it were any further, the marble would appear smaller than the sun and a ring of light would be visible around the marble. If the marble were any closer than 108 inches, it would completely block out the sun. This phenomenon is true regardless of the size of the obstructing object and is also true in order to block out the moon, which has the same apparent size in the sky as the sun. It can be concluded, then, that the moon is 108 moon diameters away from the earth since it just blocks out the sun during a solar eclipse. Shadows cast by planetary bodies are conical in shape and are assumed to be similar. In a two dimensional view, then, the shadows cast by the moon and the earth are similar isosceles triangles: The next piece of information was a result of close observation of lunar eclipses. During a total eclipse, it was observed that the moon traveled a distance equal to two and one half moon diameters while passing through the earth’s shadow. Check out an animation of the moon passing through the earth’s shadow HERE. FORMULATING A MODEL With these pieces of information, a model was developed. In this model, the moon and its shadow is reversed so that one side of the shadow aligned with one side of the earth’s shadow and is positioned so that the moon’s shadow tapered just as it reached the earth. Let the moon diameter be equal to 1 unit. Then AF is 1 unit, and FD must be 2.5 units, as determined by the lunar eclipse observations. We have now established that the diameter of the moon, AF, has a length of approximately 2,300 miles. AB is 108 times this length. Then, based on the information we began with, the distance between the earth and the moon is (2,300 x 108) miles, or approximately 248,000 miles. Modern-day calculations show that the distance from the earth to the moon varies between 225,622 miles and 252,088 miles due to the Earth’s orbit being elliptical versus circular. Still, Aristarchus managed to make an incredible estimation, considering the tools he had at his disposal and the magnitude of the distances he was working with.

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CC-MAIN-2018-47

https://www.reference.com/math/many-edges-pentagonal-pyramid-d00b6cd0bc42efa2

math

A pentagonal pyramid is characterized by six faces, six vertices and 10 edges. This three-dimensional polyhedron is a type of pyramid containing five triangular lateral faces and one pentagonal face, which is the base of the pyramid. Pyramids are solid geometric figures made up of planar surfaces called "faces" that are bounded by line segments referred to as "edges" or "sides." The points where the edges of the faces meet are called "vertices" or "corner points." The faces are all triangular polygons, which are two-dimensional shapes that are completely enclosed. The intersection point of the lateral faces is called the apex of the pyramid, located above the base. Pyramids are named based on the shape of their bases. The Swiss mathematician Leonhard Euler devised a formula for computing the number of faces, vertices and edges for most three-dimensional polyhedrons. Known as Euler's formula, this mathematical equation is given as F + V - E = 2, where "F" indicates faces, "V" denotes vertices and "E" represents edges. The sum of the faces and vertices minus the number of edges is always equal to 2. The formula for calculating the number of edges can then be expressed by the derived equation F + V - 2 = E. By substituting the correct values for a pentagonal pyramid such that 6 + 6 - 2 = E, the number of edges is equal to 10.

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http://addinbreathin.blogspot.com/2013/05/

math

orang sibuk pasal keputusan pru13, aku sibuk2 siapkan assingments tertunggak.. biarlah..biarlah hati dan suasana mula reda baru aku berbicara.. Definition — a precise and unambiguous description of the meaning of a mathematical term. It characterizes the meaning of a word by giving all the properties and only those properties that must be true. Theorem — a mathematical statement that is proved using rigorous mathematical reasoning. In a mathematical paper, the term theorem is often reserved for the most important results. Lemma — a minor result whose sole purpose is to help in proving a theorem. It is a stepping stone on the path to proving a theorem. Very occasionally lemmas can take on a life of their own (Zorn’s lemma, Urysohn’s lemma, Burnside’s lemma, Sperner’s lemma). Corollary — a result in which the (usually short) proof relies heavily on a given theorem (we often say that “this is a corollary of Theorem A”). Proposition — a proved and often interesting result, but generally less important than a theorem. esok quiz kut!

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http://openstudy.com/updates/56699242e4b0b955825def5b

math

. Trained shepherding dogs take the following path to gather the sheep in the field and bring them in for sheering. Let’s consider the lead dog’s motion. First he runs to the edge of the pasture (120, -200)m the he starts working the sheep. 90m at an angle of 160º, then 50m at an angle of 170º, then finally 165m at an angle of 60º. What is the dogs distance of running and what is the displacement? Stacey Warren - Expert brainly.com Hey! We 've verified this expert answer for you, click below to unlock the details :) At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat. I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help! can I send the drawing on your email, because its not posting here screen grab it using gyazo then post a link its much easier Not the answer you are looking for? Search for more explanations. try this for size displacement in x will be \(120 + 90 \cos (160) + 50 \cos(170) + 165 \cos(60)\) in y, \(-200 + 90 \sin (160) + 50 \sin(170) + 165 \sin(60)\) Convert all to rectangular (120, -200) (120,-200) 90(sin70* more than north 90*) = 84.6 left 90(cos70*) = 30.8 up (-84.6,30.8) 50(sin80* more than north 90*) = 49.2 left 50(cos80* ) = 8.7 up (-49.2,8.7) 165(sin60*) = 142.9 up 165(cos60*) = 82.5 right (82.5,142.9) Add them up……. The question asks for the displacement of running – that would be the hypoteneuse of these rectangular coordinates…….sqrt(68.7^2 + 17.6^2) = 70.9m Also asked for distance and that would be all the lengths added up 233.23 (the hyp of (120,-200) Total distance = 538.23 Is this right? looks good. displacement numbers agree haven't checked the actual distance number but that is the easy one and method looks good

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https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:kinetics/x2eef969c74e0d802:concentration-changes-over-time/v/half-life-of-a-first-order-reaction

math

The half-life of a reaction is the time required for a reactant to reach one-half its initial concentration or pressure. For a first-order reaction, the half-life is independent of concentration and constant over time. Created by Jay. Want to join the conversation? - So, to clarify, the main point here is that no matter the initial concentration of a reactant, it will take the same amount of time for half of the reactant to be disappear?(11 votes) - I have a little confusion about first order reactions that produce products being dependent upon the concentration of the reactant, (i.e. if you double the reactant in a first order reaction, you double the amount of product produced), while the half life decay of a first order reaction that produces a product (i.e. half the initial concentration of the reactant) is not dependent upon the initial concentration. Could you please explain these two differences when you get a chance? By the way, these videos have been extremely helpful and I appreciate all the hard work you and Khan Academy have put into making these materials available to anyone, anytime, anyplace.(10 votes) - In earlier videos we see the rate law for a first-order reaction R=k[A], where [A] is the concentration of the reactant. If we were to increase or decrease this value, we see that R (the rate of the reaction) would increase or decrease as well. When dealing with half-life, however, we are working with k (the rate constant). While the rate of reaction is measured in units molar/second, a rate constant for a first-order reaction is 1/(second).(4 votes) - Why would we need to know about the half-time ? and where does the symbol t 1/2come from ?(4 votes) - when you know the half life of a rxn, then you can determine the amount of product formed and we all know time is 't' 1/2 is half, so t1/2 is just a symbol for half life.(6 votes) - Why is the function of half life exponential. And the function in the previous video is a straight line. If both are first-order reaction.(3 votes) - If you graph half life data you get an exponential decay curve. It’s kind of the definition of it. If you graph something that starts at 100 and decays by half every 1 minute, 50 by minute 2, 25 by 3, 12.5 by 4, 6.25 by 5 etc. you’ll see. Go back to the previous video and look at the label on the y axis, then compare with the y axis on this one. A quick method for working out the reaction order is to plot [A] vs t, ln[A] vs t and 1/[A] vs t, one of them will give a straight line which tells you the order See: https://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/IntegratedRateLaws.html(3 votes) - Does half life increase or decrease(1 vote) - Is the time you get from the half life always seconds?(1 vote) - why is there e in the half life formula, i thought that half life was calculated using .5 as the exponential base?(1 vote) - any exponential decay can be written with any base you want (check your properties of logarithms) e is common but 1/2 is also good for half-life. e is common because it is much easier to deal with when you want to manipulate these equations with higher math, like calculus. When you study calculus you will see why.(3 votes) - At05:23he said that k is a constant. I know that k is a constant, but it has a different value depending on the ordo of the reaction (like if the ordo is one, the unit of k would be 1/s) Is that alright? Or am I get it wrong? Can someone help me?(1 vote) - So each chemical reaction has it's own rate constant, which indeed is constant, at a specific temperature. So a reaction like A -> B will have a rate constant associated with the rate law of the reaction, but we have to specify the temperature of the reaction. This is because the rate constant for the same reaction at say 300K will be different in value than the rate constant at 400K. Now this is juts talking about the value of a rate constant, but what I think what you mean are the units of the rate constant the same? In which case then no the units of the rate constant will be different depending on the overall order of the reaction. So a first order reaction's rate constant will indeed be using units of 1/s (or s^(-1)) while the rate for a second order reaction will be in units of 1/(M*s) (or s^(-1)*M^(-1)). So a first order reaction and a second order reaction can have the same numerical value for their rate constants, but it's inaccurate to say that they are the same since they are using different units. I think I might be getting hung up on your use of different value as opposed to different unit, but I hope that made sense.(3 votes) - How can we tell if a reaction is first order?(1 vote) - We can tell what order a reaction is graphically by plotting the reactant's concentration versus time and seeing if it produces a linear curve. We essentially rearrange a reaction order's integrated rate to resemble a linear equation of the form: y=mx+b. For a zeroth order reaction: [A] = -kt +[A]0, the y variable is the reactant concentration or [A], the x variable is time or t. If plotting [A] versus t yields a straight line then the reaction is zeroth order. Additionally the slope, m, will be -k and the y-intercept, b, will be the initial concentration of the reactant. So if the reaction is truly zeroth order then it'll form a straight line modeled on the integrated rate law. This same logic follows for first, second, etc. order reactions too. If we plot [A] versus t and we do not find a straight line, then it's not zeroth order and will follow another order's integrated rate law. For a first order reaction: ln([A]) = -kt + ln([A]0), the y variable is now ln([A]) and the x variable is still time. If we tried plotting ln([A]) versus time and get a straight line now, then it's first order. The slope will be -k and the y-intercept will be ln([A]0). And if it's not first order, then it could second order which uses: 1/[A] = kt + 1/[A]0, with y being 1/[A] and x being time again. The slope would be k and the y-intercept would be 1/[A]0. If 1/[A] versus time produces a straight line, it's second order. If not we keep repeating this process for other reaction orders until we find a straight line. It's a nice trick in chemistry of fitting data to a straight line to see if there's a relationship between variables. Hope that helps.(2 votes) - How can half life be achieved, while a chemical reaction is actually going on?(1 vote) - The half life is the time required for half of a reactant to disappear. We can't measure it unless the reaction has actually gone on long enough.(2 votes) - [Voiceover] Here we have one form of the integrated rate law for a first order reaction. And we're gonna keep going with the math here, so we eventually will talk about the half-life. So over here on the left, the natural log of the concentration of A at time t minus the natural log of the initial concentration of A. That's the same thing as the natural log of the concentration of A over the initial concentration of A. So that's just the log property. And this is equal to negative kt, where k is your rate constant. Next, we need to get rid of our natural logs. So we're going to exponentiate both sides. So we're gonna take e to both sides here and that's gonna get rid of our natural log. So now, on the left side we have our concentration over the initial concentration. On the right side we have e to the negative kt. So we're gonna multiply both sides by the initial concentration of A. So we get that our concentration of A at time t is equal to the initial concentration of A times e to the negative kt. And now, it's a little bit easier to think about the graph. We can put the concentration of A on the y-axis and we can put time on the x-axis. And this is in the form of an exponential decay. So down here I've graphed an exponential decay graph, just to show you what it looks like here. Let's think about this point on our graph. So that's when time is equal to zero. So when time is equal to zero, what is the concentration? So you would just plug in time is equal to zero into here. So you would have your concentration is equal to the initial concentration times e to the zero. And e to the 0 is of course one. So this is one. So our initial concentration is obviously this point right here, time is equal to zero. This is obviously our initial concentration. So I'll write that in here. So that's this point. And as time approaches infinity, as time goes to infinity, your concentration of A goes to zero. So as you go out here, obviously your concentration of A is going to approach zero. So that's the idea of an exponential decay graph. Next, let's think about half-life. So over here is our definition of half-life. It's the time it takes for the concentration of a reactant to decrease to half of its initial concentration. So if the initial concentration, if this is the initial concentration here, what would be the concentration after half of it has reacted? We would get our initial concentration divided by two. So we're gonna plug this in for our concentration and then the symbol for half-life is t 1/2. So t 1/2. So we're gonna plug this in for time. So when the time is equal to the half-life, your concentration is half of your initial concentration. So let's plug those in and solve for the half-life. So on the left side we would have our initial concentration divided by two. And then this would be equal to the initial concentration of A times e to the negative k and then this would be the half life, so we plug in t 1/2 here. And so now we're just gonna solve for t 1/2. We're gonna find the half-life for a first order reaction. Let's get some more space down here. And we can immediately cancel out our initial concentration of A. So now we have 1/2 is equal to this is e to the negative kt 1/2. Alright, next, we need to get rid of our e here. So we can take the natural log of both sides. So we can take the natural log of 1/2 this is equal to the natural log of e to the negative kt to the 1/2. And so that gets rid of our e. So now we have the natural log of 1/2 is equal to negative kt 1/2. So we're just solving for t 1/2, cause t 1/2 is our half-life. So our half-life, t 1/2, would be equal to this would be negative natural log of 1/2 divided by k. Let's get out the calculator and let's find out what natural log of 1/2 is. So let's get some space over here. So natural log of .5 is equal to negative .693. So we have the negative of that, so we get a positive value here for our half life. So our half-life is equal to, let me rewrite this here, so our half-life, t 1/2, is equal to .693 divided by k, where k is our rate constant. So here is your half-life for a first order reaction. Now let's think about this. If k is a constant, obviously .693 is a constant. And so your half-life is constant. Your half-life of a first order reaction is independent of the initial concentration of A. So you're gonna get the same half-life. And let's think about that for an example. Let's go back up here to our graph and let's think about half-life. So lets' say we're starting with some initial concentration, let me go ahead and change colors here, so we can think about it. I'm just going to represent our initial concentration here with eight dots. So that's our, let's say we have eight particles we're starting out here. So obviously this is a theoretical reaction. So we're gonna wait until we've lost half of our reactant. Alright, so we've lost half of our reactant, obviously we'd be left with four. We'd be left with four here. So, where would that be on our graph? Well this point right here is out initial concentration. This is concentration here, so we'd go half that. So that would be right here on our graph. So we'd go over here and we find this point. Then we drop down to here on our x-axis. And let's just put in some times. Let's say that this is 10 seconds and 20 seconds and 30 seconds and 40 and so on. So we can see that it took, this took 10 seconds for us to decrease the concentration of our reactant by half. And so the first half-life is 10 seconds. So let me write that in here, so t 1/2 is equal to 10 seconds. Again, a made up reaction, just to think about the idea of half-life. And then, let's say, now we have four. How long does it take for half of that to react? So if half of that reacts, we're left with two particles. And on our graph, let's see, that would be right here. This point would be half of this, so we go over to here, and then we drop down. And how long did it take for us to decrease the concentration of our reactant by half? Once again, 10 seconds. So this time here would be 10 seconds. So this half-life is 10 seconds. We could do it again. So we lose half of our reactant again. And we go over to here on our graph and we drop down to here. How long did it take to go from two particles to one particle? Once again, it took 10 seconds. So the half-life is once again 10 seconds. So your half-life is independent of the initial concentration. So it didn't matter if we started with eight particles or four or two. Our half-life was always 10 seconds. And so, this is the idea of half-life for a first order reaction.

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https://www.dissertations.se/dissertation/3ea5c907cd/

math

Topics in multifractal measures, nonparametrics and biostatistics Abstract: This thesis consists of four papers. The first two papers, which comprise the main part of the thesis, deal with an unexpected connection between kernel density estimators and dimension spectra for multifractal measures. The third paper presents a fully automated expert system for the diagnosis of pulmonary embolism from ventilation/perfusion scintigraphy. The final paper concerns statistical properties of the parameters of the operational model of pharmacological agonism, a widely applied model for dose-response curves in pharmacology. In the first paper kernel density estimators for singular distributions are studied. The density estimator f is a function of the sample size and the bandwidth. It was found that the integral of H(f), where H is a suitable “magnifying” functional, diverges as the sample increases to infinity and the bandwidth goes to 0. In the second paper it is shown that, for a particular choice of H, the velocity with which the integral of H(f) diverges depends on the q:th generalized Hentschel-Procaccia dimension of the measure from which the sample is drawn. This gives a new way to estimate dimension spectra for multifractal measures. An alternative kernel-based method that gives the correlation integral as a special case is also studied, which enables the estimation of the correlation dimension. The classic way of estimating generalized fractal dimensions with the aid of grids gives the generalized Rényi dimension. For q>-1 this is proved to be equivalent to the generalized Hentschel-Procaccia dimension. For q<-1 the Rényi dimension may depend on the choice of grid and thus be different from the uniquely defined Hentschel-Procaccia dimension. Examples of such measures are given. This dissertation MIGHT be available in PDF-format. Check this page to see if it is available for download.

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https://teacollection.kidizen.com/items/tea-mix-match-baby-outfit-bundle-0-3m-11703575

math

Mix and match options with these baby sets! Two knit pants, one long sleeved bodysuit and one dress. Gently used, no holes, stains or pilling. Slight cracking on graphic. Listing cautiously as good since it is baby. See all pics. Sales final. Bundle with more Tea here > #teamk#teacollection#bundles#ohbaby#outfit

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https://leofinance.io/@steemseph/re-antisocialist-72cwez

math

RE: CVR Diary 5 Predicting the Most EPIC Haul EVER for me. I don't have an answer, but you may have answered a question of mine. My question: why is it so easy for them to round up the change? Those pennies would amount to a lot after a year. Maybe subsidized is the reason for that lazy rounding I enjoy. (Maybe because there might have been a flirt vibe or curious vibe with the cashier gal... maybe... not sure... maybe... ) Anyway, not sure if it's subsidized. Some say Orange County doesn't have CRV and then other say OC does have it... looking now... https://en.wikipedia.org/wiki/California_Redemption_Value seems like a yes. Posted Using LeoFinance Beta

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