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https://handwiki.org/wiki/Physics:Giant_magnetoresistance	[ "# Physics:Giant magnetoresistance\n\nGiant magnetoresistance (GMR) is a quantum mechanical magnetoresistance effect observed in multilayers composed of alternating ferromagnetic and non-magnetic conductive layers. The 2007 Nobel Prize in Physics was awarded to Albert Fert and Peter Grünberg for the discovery of GMR.\n\nThe effect is observed as a significant change in the electrical resistance depending on whether the magnetization of adjacent ferromagnetic layers are in a parallel or an antiparallel alignment. The overall resistance is relatively low for parallel alignment and relatively high for antiparallel alignment. The magnetization direction can be controlled, for example, by applying an external magnetic field. The effect is based on the dependence of electron scattering on the spin orientation.\n\nThe main application of GMR is magnetic field sensors, which are used to read data in hard disk drives, biosensors, microelectromechanical systems (MEMS) and other devices. GMR multilayer structures are also used in magnetoresistive random-access memory (MRAM) as cells that store one bit of information.\n\nIn literature, the term giant magnetoresistance is sometimes confused with colossal magnetoresistance of ferromagnetic and antiferromagnetic semiconductors, which is not related to the multilayer structure.", null, "The founding results of Albert Fert and Peter Grünberg (1988): change in the resistance of Fe/Cr superlattices at 4.2 K in external magnetic field H. The current and magnetic field were parallel to the axis. The arrow to the right shows maximum resistance change. Hs is saturation field.[note 1]\n\n## Formulation\n\nMagnetoresistance is the dependence of the electrical resistance of a sample on the strength of an external magnetic field. Numerically, it is characterized by the value\n\n$\\displaystyle{ \\delta_H = \\frac{R(H)-R(0)}{R(0)} }$\n\nwhere R(H) is the resistance of the sample in a magnetic field H, and R(0) corresponds to H = 0. Alternative forms of this expression may use electrical resistivity instead of resistance, a different sign for δH, and are sometimes normalized by R(H) rather than R(0).\n\nThe term \"giant magnetoresistance\" indicates that the value δH for multilayer structures significantly exceeds the anisotropic magnetoresistance, which has a typical value within a few percent.\n\n## History\n\nGMR was discovered in 1988 independently by the groups of Albert Fert of the University of Paris-Sud, France, and Peter Grünberg of Forschungszentrum Jülich, Germany. The practical significance of this experimental discovery was recognized by the Nobel Prize in Physics awarded to Fert and Grünberg in 2007.\n\n### Early steps\n\nThe first mathematical model describing the effect of magnetization on the mobility of charge carriers in solids, related to the spin of those carriers, was reported in 1936. Experimental evidence of the potential enhancement of δH has been known since the 1960s. By the late 1980s, the anisotropic magnetoresistance had been well explored, but the corresponding value of δH did not exceed a few percent. The enhancement of δH became possible with the advent of sample preparation techniques such as molecular beam epitaxy, which allows manufacturing multilayer thin films with a thickness of several nanometers.\n\n## Experiment and its interpretation\n\nFert and Grünberg studied electrical resistance of structures incorporating ferromagnetic and non-ferromagnetic materials. In particular, Fert worked on multilayer films, and Grünberg in 1986 discovered the antiferromagnetic exchange interaction in Fe/Cr films.\n\nThe GMR discovery work was carried out by the two groups on slightly different samples. The Fert group used (001)Fe/(001) Cr superlattices wherein the Fe and Cr layers were deposited in a high vacuum on a (001) GaAs substrate kept at 20 °C and the magnetoresistance measurements were taken at low temperature (typically 4.2 K). The Grünberg work was performed on multilayers of Fe and Cr on (110) GaAs at room temperature.\n\nIn Fe/Cr multilayers with 3-nm-thick iron layers, increasing the thickness of the non-magnetic Cr layers from 0.9 to 3 nm weakened the antiferromagnetic coupling between the Fe layers and reduced the demagnetization field, which also decreased when the sample was heated from 4.2 K to room temperature. Changing the thickness of the non-magnetic layers led to a significant reduction of the residual magnetization in the hysteresis loop. Electrical resistance changed by up to 50% with the external magnetic field at 4.2 K. Fert named the new effect giant magnetoresistance, to highlight its difference with the anisotropic magnetoresistance. The Grünberg experiment made the same discovery but the effect was less pronounced (3% compared to 50%) due to the samples being at room temperature rather than low temperature.\n\nThe discoverers suggested that the effect is based on spin-dependent scattering of electrons in the superlattice, particularly on the dependence of resistance of the layers on the relative orientations of magnetization and electron spins. The theory of GMR for different directions of the current was developed in the next few years. In 1989, Camley and Barnaś calculated the \"current in plane\" (CIP) geometry, where the current flows along the layers, in the classical approximation, whereas Levy et al. used the quantum formalism. The theory of the GMR for the current perpendicular to the layers (current perpendicular to the plane or CPP geometry), known as the Valet-Fert theory, was reported in 1993. Applications favor the CPP geometry because it provides a greater magnetoresistance ratio (δH), thus resulting in a greater device sensitivity.\n\n## Theory\n\n### Fundamentals\n\n#### Spin-dependent scattering", null, "Electronic density of states (DOS) in magnetic and non-magnetic metals. 1: the structure of two ferromagnetic and one non-magnetic layers (arrows indicate the direction of magnetization). 2: splitting of DOS for electrons with different spin directions for each layer (arrows indicate the spin direction). F: Fermi level. The magnetic moment is antiparallel to the direction of total spin at the Fermi level.\n\nIn magnetically ordered materials, the electrical resistance is crucially affected by scattering of electrons on the magnetic sublattice of the crystal, which is formed by crystallographically equivalent atoms with nonzero magnetic moments. Scattering depends on the relative orientations of the electron spins and those magnetic moments: it is weakest when they are parallel and strongest when they are antiparallel; it is relatively strong in the paramagnetic state, in which the magnetic moments of the atoms have random orientations.\n\nFor good conductors such as gold or copper, the Fermi level lies within the sp band, and the d band is completely filled. In ferromagnets, the dependence of electron-atom scattering on the orientation of their magnetic moments is related to the filling of the band responsible for the magnetic properties of the metal, e.g., 3d band for iron, nickel or cobalt. The d band of ferromagnets is split, as it contains a different number of electrons with spins directed up and down. Therefore, the density of electronic states at the Fermi level is also different for spins pointing in opposite directions. The Fermi level for majority-spin electrons is located within the sp band, and their transport is similar in ferromagnets and non-magnetic metals. For minority-spin electrons the sp and d bands are hybridized, and the Fermi level lies within the d band. The hybridized spd band has a high density of states, which results in stronger scattering and thus shorter mean free path λ for minority-spin than majority-spin electrons. In cobalt-doped nickel, the ratio λ can reach 20.\n\nAccording to the Drude theory, the conductivity is proportional to λ, which ranges from several to several tens of nanometers in thin metal films. Electrons \"remember\" the direction of spin within the so-called spin relaxation length (or spin diffusion length), which can significantly exceed the mean free path. Spin-dependent transport refers to the dependence of electrical conductivity on the spin direction of the charge carriers. In ferromagnets, it occurs due to electron transitions between the unsplit 4s and split 3d bands.\n\nIn some materials, the interaction between electrons and atoms is the weakest when their magnetic moments are antiparallel rather than parallel. A combination of both types of materials can result in a so-called inverse GMR effect.\n\n#### CIP and CPP geometries", null, "Spin valves in the reading head of a sensor in the CIP (left) and CPP (right) geometries. Red: leads providing current to the sensor, green and yellow: ferromagnetic and non-magnetic layers. V: potential difference.\n\nElectric current can be passed through magnetic superlattices in two ways. In the current in plane (CIP) geometry, the current flows along the layers, and the electrodes are located on one side of the structure. In the current perpendicular to plane (CPP) configuration, the current is passed perpendicular to the layers, and the electrodes are located on different sides of the superlattice. The CPP geometry results in more than twice higher GMR, but is more difficult to realize in practice than the CIP configuration.\n\n#### Carrier transport through a magnetic superlattice", null, "Spin valve based on the GMR effect. FM: ferromagnetic layer (arrows indicate the direction of magnetization), NM: non-magnetic layer. Electrons with spins up and down scatter differently in the valve.\n\nMagnetic ordering differs in superlattices with ferromagnetic and antiferromagnetic interaction between the layers. In the former case, the magnetization directions are the same in different ferromagnetic layers in the absence of applied magnetic field, whereas in the latter case, opposite directions alternate in the multilayer. Electrons traveling through the ferromagnetic superlattice interact with it much weaker when their spin directions are opposite to the magnetization of the lattice than when they are parallel to it. Such anisotropy is not observed for the antiferromagnetic superlattice; as a result, it scatters electrons stronger than the ferromagnetic superlattice and exhibits a higher electrical resistance.\n\nApplications of the GMR effect require dynamic switching between the parallel and antiparallel magnetization of the layers in a superlattice. In first approximation, the energy density of the interaction between two ferromagnetic layers separated by a non-magnetic layer is proportional to the scalar product of their magnetizations:\n\n$\\displaystyle{ w = - J (\\mathbf M_1 \\cdot \\mathbf M_2). }$\n\nThe coefficient J is an oscillatory function of the thickness of the non-magnetic layer ds; therefore J can change its magnitude and sign. If the ds value corresponds to the antiparallel state then an external field can switch the superlattice from the antiparallel state (high resistance) to the parallel state (low resistance). The total resistance of the structure can be written as\n\n$\\displaystyle{ R = R_0 + \\Delta R \\sin^2 \\frac{\\theta}{2}, }$\n\nwhere R0 is the resistance of ferromagnetic superlattice, ΔR is the GMR increment and θ is the angle between the magnetizations of adjacent layers.\n\n### Mathematical description\n\nThe GMR phenomenon can be described using two spin-related conductivity channels corresponding to the conduction of electrons, for which the resistance is minimum or maximum. The relation between them is often defined in terms of the coefficient of the spin anisotropy β. This coefficient can be defined using the minimum and maximum of the specific electrical resistivity ρ for the spin-polarized current in the form\n\n$\\displaystyle{ \\rho_{F\\pm}=\\frac{2\\rho_F}{1\\pm\\beta}, }$\n\nwhere ρF is the average resistivity of the ferromagnet.\n\n#### Resistor model for CIP and CPP structures\n\nIf scattering of charge carriers at the interface between the ferromagnetic and non-magnetic metal is small, and the direction of the electron spins persists long enough, it is convenient to consider a model in which the total resistance of the sample is a combination of the resistances of the magnetic and non-magnetic layers.\n\nIn this model, there are two conduction channels for electrons with various spin directions relative to the magnetization of the layers. Therefore, the equivalent circuit of the GMR structure consists of two parallel connections corresponding to each of the channels. In this case, the GMR can be expressed as\n\n$\\displaystyle{ \\delta_H = \\frac{\\Delta R}{R}=\\frac{R_{\\uparrow\\downarrow}-R_{\\uparrow\\uparrow}}{R_{\\uparrow\\uparrow}}=\\frac{(\\rho_{F+}-\\rho_{F-})^2}{(2\\rho_{F+}+\\chi\\rho_N)(2\\rho_{F-}+\\chi\\rho_N)}. }$\n\nHere the subscript of R denote collinear and oppositely oriented magnetization in layers, χ = b/a is the thickness ratio of the magnetic and non-magnetic layers, and ρN is the resistivity of non-magnetic metal. This expression is applicable for both CIP and CPP structures. Under the condition $\\displaystyle{ \\chi\\rho_N \\ll \\rho_{F\\pm} }$ this relationship can be simplified using the coefficient of the spin asymmetry\n\n$\\displaystyle{ \\delta_H = \\frac{\\beta^2}{1-\\beta^2}. }$\n\nSuch a device, with resistance depending on the orientation of electron spin, is called a spin valve. It is \"open\", if the magnetizations of its layers are parallel, and \"closed\" otherwise.\n\n#### Valet-Fert model\n\nIn 1993, Thierry Valet and Albert Fert presented a model for the giant magnetoresistance in the CPP geometry, based on the Boltzmann equations. In this model the chemical potential inside the magnetic layer is split into two functions, corresponding to electrons with spins parallel and antiparallel to the magnetization of the layer. If the non-magnetic layer is sufficiently thin then in the external field E0 the amendments to the electrochemical potential and the field inside the sample will take the form\n\n$\\displaystyle{ \\Delta\\mu = \\frac{\\beta}{1-\\beta^2}eE_0\\ell_se^{z/\\ell_s}, }$\n$\\displaystyle{ \\Delta E = \\frac{\\beta^2}{1-\\beta^2}eE_0\\ell_se^{z/\\ell_s}, }$\n\nwhere s is the average length of spin relaxation, and the z coordinate is measured from the boundary between the magnetic and non-magnetic layers (z < 0 corresponds to the ferromagnetic). Thus electrons with a larger chemical potential will accumulate at the boundary of the ferromagnet. This can be represented by the potential of spin accumulation VAS or by the so-called interface resistance (inherent to the boundary between a ferromagnet and non-magnetic material)\n\n$\\displaystyle{ R_i= \\frac{\\beta(\\mu_{\\uparrow\\downarrow}-\\mu_{\\uparrow\\uparrow})}{2ej} = \\frac{\\beta^2\\ell_{sN}\\rho_N}{1+(1-\\beta^2)\\ell_{sN}\\rho_N/(\\ell_{sF}\\rho_F)}, }$\n\nwhere j is current density in the sample, sN and sF are the length of the spin relaxation in a non-magnetic and magnetic materials, respectively.\n\n## Device preparation\n\n### Materials and experimental data\n\nMany combinations of materials exhibit GMR, and the most common are the following:\n\n• FeCr\n• Co10Cu90: δH = 40% at room temperature\n• Co95Fe5/Cu: δH = 110% at room temperature.\n\nThe magnetoresistance depends on many parameters such as the geometry of the device (CIP or CPP), its temperature, and the thicknesses of ferromagnetic and non-magnetic layers. At a temperature of 4.2 K and a thickness of cobalt layers of 1.5 nm, increasing the thickness of copper layers dCu from 1 to 10 nm decreased δH from 80 to 10% in the CIP geometry. Meanwhile, in the CPP geometry the maximum of δH (125%) was observed for dCu = 2.5 nm, and increasing dCu to 10 nm reduced δH to 60% in an oscillating manner.\n\nWhen a Co(1.2 nm)/Cu(1.1 nm) superlattice was heated from near zero to 300 K, its δH decreased from 40 to 20% in the CIP geometry, and from 100 to 55% in the CPP geometry.\n\nThe non-magnetic layers can be non-metallic. For example, δH up to 40% was demonstrated for organic layers at 11 K. Graphene spin valves of various designs exhibited δH of about 12% at 7 K and 10% at 300 K, far below the theoretical limit of 109%.\n\nThe GMR effect can be enhanced by spin filters that select electrons with a certain spin orientation; they are made of metals such as cobalt. For a filter of thickness t the change in conductivity ΔG can be expressed as\n\n$\\displaystyle{ \\Delta G = \\Delta G_{SV} + \\Delta G_f (1 - e^{\\beta t/\\lambda}), }$\n\nwhere ΔGSV is change in the conductivity of the spin valve without the filter, ΔGf is the maximum increase in conductivity with the filter, and β is a parameter of the filter material.\n\n### Types of GMR\n\nGMR is often classed by the type of devices which exhibit the effect.\n\n#### Films\n\n##### Antiferromagnetic superlattices\n\nGMR in films was first observed by Fert and Grünberg in a study of superlattices composed of ferromagnetic and non-magnetic layers. The thickness of the non-magnetic layers was chosen such that the interaction between the layers was antiferromagnetic and the magnetization in adjacent magnetic layers was antiparallel. Then an external magnetic field could make the magnetization vectors parallel thereby affecting the electrical resistance of the structure.\n\nMagnetic layers in such structures interact through antiferromagnetic coupling, which results in the oscillating dependence of the GMR on the thickness of the non-magnetic layer. In the first magnetic field sensors using antiferromagnetic superlattices, the saturation field was very large, up to tens of thousands of oersteds, due to the strong antiferromagnetic interaction between their layers (made of chromium, iron or cobalt) and the strong anisotropy fields in them. Therefore, the sensitivity of the devices was very low. The use of permalloy for the magnetic and silver for the non-magnetic layers lowered the saturation field to tens of oersteds.\n\n##### Spin valves using exchange bias\n\nIn the most successful spin valves the GMR effect originates from exchange bias. They comprise a sensitive layer, \"fixed\" layer and an antiferromagnetic layer. The last layer freezes the magnetization direction in the \"fixed\" layer. The sensitive and antiferromagnetic layers are made thin to reduce the resistance of the structure. The valve reacts to the external magnetic field by changing the magnetization direction in the sensitive layer relatively to the \"fixed\" layer.\n\nThe main difference of these spin valves from other multilayer GMR devices is the monotonic dependence of the amplitude of the effect on the thickness dN of the non-magnetic layers:\n\n$\\displaystyle{ \\delta_H(d_N) = \\delta_{H0} \\frac{\\exp\\left(-d_N/\\lambda_N\\right)}{1 + d_N/d_0}, }$\n\nwhere δH0 is a normalization constant, λN is the mean free path of electrons in the non-magnetic material, d0 is effective thickness that includes interaction between layers. The dependence on the thickness of the ferromagnetic layer can be given as:\n\n$\\displaystyle{ \\delta_H(d_F) = \\delta_{H1} \\frac{1 - \\exp\\left(-d_F/\\lambda_F\\right)}{1 + d_F/d_0}. }$\n\nThe parameters have the same meaning as in the previous equation, but they now refer to the ferromagnetic layer.\n\n##### Non-interacting multilayers (pseudospin valves)\n\nGMR can also be observed in the absence of antiferromagnetic coupling layers. In this case, the magnetoresistance results from the differences in the coercive forces (for example, it is smaller for permalloy than cobalt). In multilayers such as permalloy/Cu/Co/Cu the external magnetic field switches the direction of saturation magnetization to parallel in strong fields and to antiparallel in weak fields. Such systems exhibit a lower saturation field and a larger δH than superlattices with antiferromagnetic coupling. A similar effect is observed in Co/Cu structures. The existence of these structures means that GMR does not require interlayer coupling, and can originate from a distribution of the magnetic moments that can be controlled by an external field.\n\n##### Inverse GMR effect\n\nIn the inverse GMR, the resistance is minimum for the antiparallel orientation of the magnetization in the layers. Inverse GMR is observed when the magnetic layers are composed of different materials, such as NiCr/Cu/Co/Cu. The resistivity for electrons with opposite spins can be written as $\\displaystyle{ \\rho_{\\uparrow,\\downarrow}=\\frac{2\\rho_F}{1\\pm\\beta} }$; it has different values, i.e. different coefficients β, for spin-up and spin-down electrons. If the NiCr layer is not too thin, its contribution may exceed that of the Co layer, resulting in inverse GMR. Note that the GMR inversion depends on the sign of the product of the coefficients β in adjacent ferromagnetic layers, but not on the signs of individual coefficients.\n\nInverse GMR is also observed if NiCr alloy is replaced by vanadium-doped nickel, but not for doping of nickel with iron, cobalt, manganese, gold or copper.\n\n#### GMR in granular structures\n\nGMR in granular alloys of ferromagnetic and non-magnetic metals was discovered in 1992 and subsequently explained by the spin-dependent scattering of charge carriers at the surface and in the bulk of the grains. The grains form ferromagnetic clusters about 10 nm in diameter embedded in a non-magnetic metal, forming a kind of superlattice. A necessary condition for the GMR effect in such structures is poor mutual solubility in its components (e.g., cobalt and copper). Their properties strongly depend on the measurement and annealing temperature. They can also exhibit inverse GMR.\n\n## Applications\n\n### Spin-valve sensors\n\n#### General principle\n\nOne of the main applications of GMR materials is in magnetic field sensors, e.g., in hard disk drives and biosensors, as well as detectors of oscillations in MEMS. A typical GMR-based sensor consists of seven layers:\n\n1. Silicon substrate,\n2. Binder layer,\n3. Sensing (non-fixed) layer,\n4. Non-magnetic layer,\n5. Fixed layer,\n6. Antiferromagnetic (Pinning) layer,\n7. Protective layer.\n\nThe binder and protective layers are often made of tantalum, and a typical non-magnetic material is copper. In the sensing layer, magnetization can be reoriented by the external magnetic field; it is typically made of NiFe or cobalt alloys. FeMn or NiMn can be used for the antiferromagnetic layer. The fixed layer is made of a magnetic material such as cobalt. Such a sensor has an asymmetric hysteresis loop owing to the presence of the magnetically hard, fixed layer.\n\nSpin valves may exhibit anisotropic magnetoresistance, which leads to an asymmetry in the sensitivity curve.\n\n#### Hard disk drives\n\nIn hard disk drives (HDDs), information is encoded using magnetic domains, and a change in the direction of their magnetization is associated with the logical level 1 while no change represents a logical 0. There are two recording methods: longitudinal and perpendicular.\n\nIn the longitudinal method, the magnetization is normal to the surface. A transition region (domain walls) is formed between domains, in which the magnetic field exits the material. If the domain wall is located at the interface of two north-pole domains then the field is directed outward, and for two south-pole domains it is directed inward. To read the direction of the magnetic field above the domain wall, the magnetization direction is fixed normal to the surface in the antiferromagnetic layer and parallel to the surface in the sensing layer. Changing the direction of the external magnetic field deflects the magnetization in the sensing layer. When the field tends to align the magnetizations in the sensing and fixed layers, the electrical resistance of the sensor decreases, and vice versa.\n\n### Magnetic RAM", null, "The use of a spin valve in MRAM. 1: spin valve as a memory cell (arrows indicate the presence of ferromagnetic layers), 2: row line, 3: column line. Ellipses with arrows denote the magnetic field lines around the row and column lines when electric current flows through them.\n\nA cell of magnetoresistive random-access memory (MRAM) has a structure similar to the spin-valve sensor. The value of the stored bits can be encoded via the magnetization direction in the sensor layer; it is read by measuring the resistance of the structure. The advantages of this technology are independence of power supply (the information is preserved when the power is switched off owing to the potential barrier for reorienting the magnetization), low power consumption and high speed.\n\nIn a typical GMR-based storage unit, a CIP structure is located between two wires oriented perpendicular to each other. These conductors are called lines of rows and columns. Pulses of electric current passing through the lines generate a vortex magnetic field, which affects the GMR structure. The field lines have ellipsoid shapes, and the field direction (clockwise or counterclockwise) is determined by the direction of the current in the line. In the GMR structure, the magnetization is oriented along the line.\n\nThe direction of the field produced by the line of the column is almost parallel to the magnetic moments, and it can not reorient them. Line of the row is perpendicular, and regardless of the magnitude of the field can rotate the magnetization by only 90 °. With the simultaneous passage of pulses along the row and column lines, of the total magnetic field at the location of the GMR structure will be directed at an acute angle with respect to one point and an obtuse to others. If the value of the field exceeds some critical value, the latter changes its direction.\n\nThere are several storage and reading methods for the described cell. In one method, the information is stored in the sensing layer; it is read via resistance measurement and is erased upon reading. In another scheme, the information is kept in the fixed layer, which requires higher recording currents compared to reading currents.\n\nTunnel magnetoresistance (TMR) is an extension of spin-valve GMR, in which the electrons travel with their spins oriented perpendicularly to the layers across a thin insulating tunnel barrier (replacing the non-ferromagnetic spacer). This allows to achieve a larger impedance, a larger magnetoresistance value (~10x at room temperature) and a negligible temperature dependence. TMR has now replaced GMR in MRAMs and disk drives, in particular for high area densities and perpendicular recording.\n\n### Other applications\n\nMagnetoresistive insulators for contactless signal transmission between two electrically isolated parts of electrical circuits were first demonstrated in 1997 as an alternative to opto-isolators. A Wheatstone bridge of four identical GMR devices is insensitive to a uniform magnetic field and reacts only when the field directions are antiparallel in the neighboring arms of the bridge. Such devices were reported in 2003 and may be used as rectifiers with a linear frequency response." ]	[ null, "https://upload.wikimedia.org/wikipedia/commons/thumb/7/73/GMR.svg/340px-GMR.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/2/2d/Electron_density_in_magnets.svg/300px-Electron_density_in_magnets.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/Spin-valve_CIP_CPP.svg/300px-Spin-valve_CIP_CPP.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/9/93/Spin-valve_GMR.svg/300px-Spin-valve_GMR.svg.png", null, "https://upload.wikimedia.org/wikipedia/commons/thumb/a/ad/Spin-valve_in_MRAM.svg/300px-Spin-valve_in_MRAM.svg.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8043828,"math_prob":0.93292457,"size":35957,"snap":"2021-31-2021-39","text_gpt3_token_len":9531,"char_repetition_ratio":0.16766334,"word_repetition_ratio":0.031835936,"special_character_ratio":0.26067248,"punctuation_ratio":0.16920598,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97324145,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,6,null,3,null,1,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-03T23:32:42Z\",\"WARC-Record-ID\":\"<urn:uuid:ef8e304d-fb71-4e72-add6-6f4ceb847bca>\",\"Content-Length\":\"148965\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3ad31474-866c-4660-87ff-a5ff26854da1>\",\"WARC-Concurrent-To\":\"<urn:uuid:fe7eb32f-d55a-4ae9-8920-878f76674ba1>\",\"WARC-IP-Address\":\"69.197.158.244\",\"WARC-Target-URI\":\"https://handwiki.org/wiki/Physics:Giant_magnetoresistance\",\"WARC-Payload-Digest\":\"sha1:NCGTCZEO4HPG5IL4MQ33EJB6MLVP5EU5\",\"WARC-Block-Digest\":\"sha1:K3PNAPONF3GV6AXMY4UAFZQ2M6MX3P2B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154486.47_warc_CC-MAIN-20210803222541-20210804012541-00124.warc.gz\"}"}
https://contents.driverguide.com/content.php?id=1202325&path=data2.cab%2FApp_MyLogo%2FMyLogo.ini	[ "# MyLogo.ini Driver File Contents (AsusUpdt_V71304.zip)\n\nDriver Package File Name: AsusUpdt_V71304.zip\nFile Size: 8.8 MB\n\n```[SUPPORT MB]\nNUMBER = 22\nMB(1)= CUSL2-CB<R>\nMB(2)= P4B<R>\nMB(3)= P4B-E<R>\nMB(4)= P4B266<R>\nMB(5)= P4B266-E<R>\nMB(6)= P4B266-C<R>\nMB(7)= P4S333<R>\nMB(8)= A7V333<R>\nMB(9)= P4S133<R>\nMB(10)=A7S333<R>\nMB(11)=P4B533<R>\nMB(12)=P4B533-E<R>\nMB(13)=P4B533-V<R>\nMB(14)=P4T533<R>\nMB(15)=P4T533-C<R>\nMB(16)=P4S533<R>\nMB(17)=P4S533-E<R>\nMB(18)=P4SE\nMB(19)=A7V8X<R>\nMB(20)=P4S8X<R>\nMB(21)=P4G8X<R>\nMB(22)=P4PE<R>```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.544178,"math_prob":0.99486333,"size":699,"snap":"2020-45-2020-50","text_gpt3_token_len":317,"char_repetition_ratio":0.21007194,"word_repetition_ratio":0.0,"special_character_ratio":0.48640916,"punctuation_ratio":0.06535948,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9996692,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T10:21:43Z\",\"WARC-Record-ID\":\"<urn:uuid:e336203a-aaa5-4ef2-9975-9d21fb49c118>\",\"Content-Length\":\"20386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:840887a3-0abe-4320-ab26-504dd65fbe77>\",\"WARC-Concurrent-To\":\"<urn:uuid:da1076f8-585c-440e-80a0-077b8c3c5273>\",\"WARC-IP-Address\":\"184.173.96.94\",\"WARC-Target-URI\":\"https://contents.driverguide.com/content.php?id=1202325&path=data2.cab%2FApp_MyLogo%2FMyLogo.ini\",\"WARC-Payload-Digest\":\"sha1:HVXVAOIUXDBYVI6IZZZZYJRFZ3SQCI3M\",\"WARC-Block-Digest\":\"sha1:D566WBZPJJMQ6RJEO4ZLEADCDYPYIHKD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141187753.32_warc_CC-MAIN-20201126084625-20201126114625-00555.warc.gz\"}"}
https://www.gaageegoo.info/2015/01/behaviour-of-gases10th-science-notes.html	[ "# Behaviour of Gases–10th Science Notes and Question Bank\n\n### Text Book Question and Answers\n\n1. Define Charles law.\n\n“At constant pressure, the volume of a fixed mass of a gas is directly proportional to its absolute temperature”– This is Charles law.\n\n2. Caution “store in cool and dry place” is printed on medicine bottles. Why?\n\nWhen the bottle is exposed, due to heat, it expands as the gas inside the bottle expands as temperature rises. To increase the volume the bottle explodes.\n\n3. Define Boyle’s law.\n\n“At constant temperature, the volume of a given mass of dry gas is inversely proportional to its pressure” – This is Boyles law.\n\n4. What do you mean by Diffusion?\n\nRandom movement of gaseous molecules from the region of higher concentration to the region of lower concentration is called diffusion.\n\n5. Define rate of diffusion?\n\nThe volume of gas diffusing per unit time is called rate of diffusion.\n\n6. What is the relation between diffusion and mass?\n\nThe rate of diffusion is inversely proportional to square root of mass\n\nr = K\n\nWhere m = mass, r = rate of diffusion, v = volume.\n\n7. The volume of the gas found to be at a pressure of 2500 Pa. when the pressure was decreased by 500 Pa the gas occupied a volume of 2400 litres. Calculate the initial volume occupied by gas before the change was done at constant temperature?\n\nData:\n\nP1 = 2500,\n\nP2 = 2500-500 = 2000\n\nV2 = 2400\n\nV1 =?\n\nSolution:\n\nAccording to Boyle’s law\n\nP1 V1 = P2 V2 (at constant temperature)\n\n2500 x V1 = 2000 x 2400\n\nV1 = 1920 litres\n\n8. About 6 litres of oxygen gas is collected at 300K. If for a particular use, the volume has to be reduced to 1/3 of its original volume, find the gas has to be cooled? (pressure is kept constant)\n\nData:\n\nV1 = 6 litre\n\nV2 = 1/3 x 6 = 2 litre\n\nT1 = 300K,\n\nT2 =?\n\nSolution:\n\nAccording to Charles law (at constant pressure)\n\nV1/T1 = V2/T2\n\n6/300 = 2/ T2\n\nT2 = 100K.\n\nThe gas has to be cooled to 100k,\n\nII. For each of the following statements or questions four alternatives are given. Choose the most appropriate alternative:\n\n1. The expression which do not says relation between volume and temperature at constant temperature is -------\n\na. V = KT b. T = KV c. V/T = K d. V = K/T\n\n2. The temperature which is also called Absolute zero is ------\n\na.00C b. 00F c. -273 0C d. -273 0K\n\n3. The expression for relation between volume and pressure at constant temperature is\n\na. P/V = K b. T = PV c. V/ P = K d. PV = K\n\nIII. Fill in the blanks with right answer\n\n1. Under constant pressure , per degree rise in temperature , the volume of gas is increased by ------- of the original volume at 00C. ( 1 / 273)\n\n2. The S.I unit of temperature is --------- (kelvin)\n\n3. The volume of gas diffusing per unit time is --------- ( rate of diffusion )\n\nIV. Match the followings\n\n A B C 1.Charle’s law i)Volume of gas diffused / Time(3) a) TV= PK 2. Boyle’s law ii) constant volume b) V = KT (1) 3. Rate of diffusion iii) constant temperature (2) c) PV = K (2) iv) Constant pressure (1) d) r = k √v / √m (3)\n\n1. List the main features of gases.\n\na) Gases are highly compressible.\n\nb) Gases excerts pressure equally in all directions\n\nc) Gases mix evenly in all proportions without any mechanical aid\n\nd) Gases are least dense than other two states of matter.\n\n2. Give the formula for Charles law.\n\nVolume α temperature\n\nV α T\n\nV = KT here V= volume T= temperature K = proportionality constant\n\n3. List the applications of Charles law.\n\na) Soda bottles should store in cool and dry place.\n\nb) Air balloons pops out during hot summer than in winter.\n\n4. Give the formula for Boyle’s law.\n\nV α 1 / P\n\nV α K / P\n\nPV =K here V= volume , P= Pressure , K = proportionality constant\n\n5. List the applications of Boyle’s law.\n\na)Popping of balloon when squeezed due to increase in the pressure.\n\nb) Deep sea fishes die when suddenly they brought to surface due to low pressure.\n\nc) Scuba driver’s life is under threat when he suddenly comes to surface quickly.\n\n6. State Graham’s law of diffusion.\n\n“The rate of diffusion of gas is inversely proportional to the square root of its density”- This is called Graham’s law of diffusion\n\n7. Give the formula for Graham’s law of diffusion\n\nr α 1 / √d\n\nr = K / √d\n\nK = r x √d\n\nhere r = rate of diffusion, d = density of gas, K = proportionality constant\n\nYou Might Like\n\n1.", null, "" ]	[ null, "https://www.blogger.com/img/blogger_logo_round_35.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87531954,"math_prob":0.9921317,"size":4489,"snap":"2023-40-2023-50","text_gpt3_token_len":1204,"char_repetition_ratio":0.15117057,"word_repetition_ratio":0.01627907,"special_character_ratio":0.29026508,"punctuation_ratio":0.111834966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99657714,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T08:29:37Z\",\"WARC-Record-ID\":\"<urn:uuid:eb2f103a-89ec-4ca0-825c-05835ff6f341>\",\"Content-Length\":\"148386\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e0a11da-f4c8-4d38-83e9-a4ba76ca5c02>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a235007-1b25-4bd6-9506-7cf0d1d53f1f>\",\"WARC-IP-Address\":\"172.253.115.121\",\"WARC-Target-URI\":\"https://www.gaageegoo.info/2015/01/behaviour-of-gases10th-science-notes.html\",\"WARC-Payload-Digest\":\"sha1:BNBXOP4ZIXWZGVL2FW7CYFJVAKBEPBM2\",\"WARC-Block-Digest\":\"sha1:FNCQO5EVNREB4QVFGGLNP5SMIPV3IHHB\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100739.50_warc_CC-MAIN-20231208081124-20231208111124-00462.warc.gz\"}"}
https://calculus7.com/how-do-you-calculate-the-volume-of-a-tank-in-gallons-and-litres/	[ "# How do you calculate the volume of a tank in Gallons and Litres\n\n### How Can I Find Out How Much Water My Fish Tank Holds\n\nRectangular Tanks\n\nThe measurements are 20 inches by 12 inches by 12 inches, i.e Length 12 inches, Width 12 inches, height 20 inches\n\nthe volume is 2,880 cubic inches.\n\nFinally, divide the cubic volume in inches by 231, as a gallon is equal to 231 cubic inches.\n\nA 2,880 cubic-inch tank is 12.47 gallons" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8082949,"math_prob":0.97052383,"size":352,"snap":"2022-27-2022-33","text_gpt3_token_len":96,"char_repetition_ratio":0.20114942,"word_repetition_ratio":0.0,"special_character_ratio":0.30397728,"punctuation_ratio":0.13414635,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97920924,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T00:39:50Z\",\"WARC-Record-ID\":\"<urn:uuid:8ec0755b-af12-41cc-8267-9dcd5b1402f7>\",\"Content-Length\":\"57708\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bb43407b-e1e9-4fbb-8bac-ed21924f9ec0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3007160d-3319-4fb8-8661-1db26864c946>\",\"WARC-IP-Address\":\"104.21.11.128\",\"WARC-Target-URI\":\"https://calculus7.com/how-do-you-calculate-the-volume-of-a-tank-in-gallons-and-litres/\",\"WARC-Payload-Digest\":\"sha1:HDB2NVJQJF5JSCIZA5XOKGGZQ5BPEON3\",\"WARC-Block-Digest\":\"sha1:WOFIWX7JRIBI5RVCTFYXEJJONFA44Y6B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103646990.40_warc_CC-MAIN-20220630001553-20220630031553-00786.warc.gz\"}"}
https://mathoverflow.net/questions/342926/coinciding-weak-operator-and-strong-operator-topologies	[ "# Coinciding weak operator and strong operator topologies\n\nLet $$A \\subset \\mathcal{B}(H)$$ a subalgebra, not necessarily a $$$$-algebra. In Murphy's book 'C-algebras and Operator Theory', in Remark 4.2.1 you can find a proof of the failure of strong compactness for the ball of $$\\mathcal{B}(H)$$:\n\nIf the ball is strongly compact, then the identity map of the ball with the relative strong topology to the ball with the relative weak operator topology is a continuous bijection from a compact space to a Hausdorff one, and therefore a homeomorphism, so the relative strong operator and the relative weak operator topologies coincide. But the involution operator is weakly continuous but not strongly continuous restricted to the ball, and it shows that the unit ball can't be strongly compact.\n\nIt is clear that this proof relies in the fact that the two topologies can't coincide in the unit ball because the non-strong continuity of the involution operator, and it is clearly an extendable method to the $$*$$-subalgebras setting. But, if we have a subalgebra $$A$$, not necessarily closed by the adjoint map, are there any 'invariants' of the relative topologies that we can analyze to decide if the two topologies cannot coincide in the ball of $$A$$?\n\n## 1 Answer\n\nI'm not sure there is a definitive answer here, as it's asked if there are \"invariants\", not a complete characterisation. Further, it's not clear to me if the question is asking about compactness or just whether the WOT and SOT are different.\n\nIndeed, if all we are interested in is to show that the SOT is not the same as the WOT when restricted to the unit ball of $$B(H)$$, then we don't need to consider compactness at all. As you say, Murphy observes (this is Example 4.1.1 in his book) that the adjoint is WOT continuous but not SOT continuous.\n\nActually, there is another example which doesn't use the adjoint. We can just consider $$S$$ the unilateral shift acting on $$\\ell^2$$, and let $$S_n = S^n$$. Then $$S_n\\rightarrow 0$$ weakly, but of course as each $$S_n$$ is an isometry, not subnet of $$(S_n)$$ can converge to $$0$$ in the SOT.\n\nI don't think it is obvious that the same argument will work for self-adjoint subalgebras $$A\\subseteq B(H)$$. For example, if $$A$$ is commutative, then the adjoint is continuous for the SOT restricted to the unit ball of $$A$$.\n\n1. Let $$A=L^\\infty(\\mathbb T)$$ acting on $$H=L^2(\\mathbb T)$$ by multiplication, and let $$a_n(e^{i\\theta}) = e^{in\\theta}$$. The Riemann-Lebesgue Lemma shows that $$(a_n)$$ is WOT convergent to $$0$$, but clearly $$(a_n)$$ is not SOT convergent to $$0$$.\n\n2. However, let $$A=\\ell^\\infty$$ acting on $$H=\\ell^2$$. Then a bounded sequence $$(x_n)$$ in $$A$$ is WOT convergent to $$0$$ exactly when $$x_n(k)\\rightarrow 0$$ for each $$k$$ (thinking of $$x_n$$ as a bounded function $$\\mathbb N\\rightarrow\\mathbb C$$). This implies that $$x_n\\rightarrow 0$$ in the SOT.\n\nLet $$X\\subseteq B(H)$$ be just a subspace. Identify $$B(H\\oplus H)$$ with $$2\\times 2$$-matrices with entries in $$B(H)$$. Define $$Y(X) = \\left\\{ \\begin{pmatrix} \\lambda & x \\\\ 0 & \\lambda \\end{pmatrix} : x\\in X, \\lambda\\in\\mathbb C \\right\\}.$$ The $$Y(X)$$ is a subalgebra, not self-adjoint, and $$Y(X)$$ is closed in all the various operator topologies if and only if $$X$$ is closed in that topology. While the unit ball of $$Y(X)$$ and $$X$$ do not biject, bounded subsets to correspond to bounded subsets. In this way, we can transport (counter-)examples from subspaces to subalgebras.\n\nLet $$X$$ be a \"row space\" of $$H$$. That is, fixed a unit vector $$\\xi_0\\in H$$ and let $$X$$ be the collection of rank-one operators $$\\{ \\theta_\\xi : \\xi\\in H \\}$$ where $$\\theta_\\xi:H\\rightarrow H; \\eta\\mapsto (\\eta\|\\xi)\\xi_0$$. Then $$X$$ is anti-linearly isomorphic to $$H$$. A bounded net $$(\\theta_{\\xi_n})$$ converges to $$x\\in B(H)$$ in the WOT if and only if $$(\\eta\|\\xi_n) (\\xi_0\|\\eta') \\rightarrow (x(\\eta)\|\\eta') \\qquad (\\eta, \\eta'\\in H).$$ This is equivalent to $$(\\xi_n)$$ being weakly-convergent in $$H$$, say to $$\\xi$$, in which case $$\\theta_{\\xi_n}\\rightarrow\\theta_\\xi$$. Similarly, $$(\\theta_{\\xi_n})$$ converges strongly to $$x\\in B(H)$$ when $$\\\|(\\eta\|\\xi_n) \\xi_0 - x(\\eta)\\\|\\rightarrow 0 \\qquad (\\eta\\in H).$$ This gives the same condition on $$(\\xi_n)$$. So, the WOT and SOT agree on the bounded subsets of $$X$$, and hence on the unit ball of $$Y(X)$$." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8486312,"math_prob":0.9999478,"size":3012,"snap":"2021-21-2021-25","text_gpt3_token_len":952,"char_repetition_ratio":0.10638298,"word_repetition_ratio":0.008213553,"special_character_ratio":0.31407702,"punctuation_ratio":0.0995106,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000076,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-18T07:01:37Z\",\"WARC-Record-ID\":\"<urn:uuid:24193e05-5f5a-43a6-8618-70eee676e183>\",\"Content-Length\":\"135395\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4b0f166d-4f06-4692-badc-c3a5ea7d40a0>\",\"WARC-Concurrent-To\":\"<urn:uuid:3f9ac61e-e04a-4179-8ad7-e0d34f015707>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/342926/coinciding-weak-operator-and-strong-operator-topologies\",\"WARC-Payload-Digest\":\"sha1:PTXULODI4P4JSFJ37DEVEAZIWSMKUNQU\",\"WARC-Block-Digest\":\"sha1:O7W7XI3YCHKQPMOKHZSPF2QMTKKLO3R7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623487635724.52_warc_CC-MAIN-20210618043356-20210618073356-00263.warc.gz\"}"}
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equations and inequalities she loves math.\n\nanalytical geometry examples.\n\narea under a curve mathematics a level revision.\n\n2x2 free shipping available.\n\nhow to solve quadratic equation by completing the square method.\n\nchi square test of independence spss tutorials libguides at kent.\n\nact form 74f math answer explanations act helper.\n\nex square a 2x2 matrix youtube.\n\nintegration algebraic fractions integration from a level maths tutor.\n\nceiling tiles the home depot canada.\n\nvariance wikipedia." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7054857,"math_prob":0.9937634,"size":3759,"snap":"2019-26-2019-30","text_gpt3_token_len":862,"char_repetition_ratio":0.12037284,"word_repetition_ratio":0.20672269,"special_character_ratio":0.2037776,"punctuation_ratio":0.12554744,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9991636,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T07:05:09Z\",\"WARC-Record-ID\":\"<urn:uuid:d1dc8dc3-f67d-49f8-83e1-902df0b4ad73>\",\"Content-Length\":\"76647\",\"Content-Type\":\"application/http; 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https://socratic.org/questions/what-does-the-first-law-of-reflection-state	[ "# What does the first law of reflection state?\n\nJan 1, 2015\n\nThe First Law of reflection States that the angle made by the incident light ray with the normal to the surface at the point of incidence is equal to angle made by the reflected light ray with the normal.\n\nThe following figures are examples of this law under different circumstances:\n1) A flat Mirror", null, "2)Curved Mirrors", null, "One note of caution though always take the Normal at the point of incidence, telling this is trivial for plane mirrors as normal is always the same but in curved mirrors the normal changes from point to point so always remember to take the normal at the point of incidence.\n\nNice applet on this website: http://www.physicsclassroom.com/mmedia/optics/lr.cfm" ]	[ null, "https://d2jmvrsizmvf4x.cloudfront.net/ixWP16kYQxGbxtK3Jy5b_Screenshot_2015-01-02-01-50-29.png", null, "https://d2jmvrsizmvf4x.cloudfront.net/4aSy8aZeRBuCAdhj9UDz_Screenshot_2015-01-02-01-49-50.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89606005,"math_prob":0.9187474,"size":837,"snap":"2019-51-2020-05","text_gpt3_token_len":171,"char_repetition_ratio":0.13565426,"word_repetition_ratio":0.0,"special_character_ratio":0.19235365,"punctuation_ratio":0.06369427,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95686394,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-16T12:50:42Z\",\"WARC-Record-ID\":\"<urn:uuid:5de0fff6-c799-45e2-8ada-3b93b9a693e4>\",\"Content-Length\":\"35162\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3f2b53fc-943b-431a-94f3-87ce3f93573e>\",\"WARC-Concurrent-To\":\"<urn:uuid:47eea308-2ca3-4c9d-995a-9edd2ac6a368>\",\"WARC-IP-Address\":\"54.221.217.175\",\"WARC-Target-URI\":\"https://socratic.org/questions/what-does-the-first-law-of-reflection-state\",\"WARC-Payload-Digest\":\"sha1:KVQEFUFZ2KZ6247WDAMUMJZAOTRAWYXC\",\"WARC-Block-Digest\":\"sha1:EREWH33W5V6PBJAIRBY6B6UD3M7HV5PQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540565544.86_warc_CC-MAIN-20191216121204-20191216145204-00067.warc.gz\"}"}
https://smartwritings.org/let-true-population-model-be-y-01xu-0-not-0-consider-following-empirical-models-correctly-sp/	[ "# let true population model be y 01xu 0 not 0 consider following empirical models correctly sp\n\nLet the true population model be: y = ?0+?1x+u with ?0 not= 0. Consider the following empirical models: ! The correctly specied model: y = ?^0 + ?^1x ! The misspecied model: y = ?^x Answer the following questions. 1. Derive the expression for the OLS estimators ?^ and ?^1. 2. Is the OLS estimator ?^ an unbiased estimator of ?1? If your answer is no, determine the bias. 3. Derive an expression for the variances of ?^ and ?^1. 4. Compare the expressions for the variances of ?^ with the variance of ?^1 from the correctly specified regression. Is Var(?^) > V ar( ?^1)? 5. Comment on the tradeoff between bias and variance when choosing between these estimators." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.65191597,"math_prob":0.99052066,"size":1040,"snap":"2022-40-2023-06","text_gpt3_token_len":304,"char_repetition_ratio":0.11003861,"word_repetition_ratio":0.15584415,"special_character_ratio":0.31346154,"punctuation_ratio":0.20982143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987271,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-03T18:38:32Z\",\"WARC-Record-ID\":\"<urn:uuid:0cca9b2a-87b0-4eb7-af45-cbb8abe2ec03>\",\"Content-Length\":\"40306\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3c8c2b50-bf5d-457b-aff8-34c1e0c985a4>\",\"WARC-Concurrent-To\":\"<urn:uuid:54493901-c66a-4c54-9be4-ead7ccffcd42>\",\"WARC-IP-Address\":\"162.0.232.59\",\"WARC-Target-URI\":\"https://smartwritings.org/let-true-population-model-be-y-01xu-0-not-0-consider-following-empirical-models-correctly-sp/\",\"WARC-Payload-Digest\":\"sha1:XQHW6FUYOKKSPBHGP2W3VIMG26C27XMR\",\"WARC-Block-Digest\":\"sha1:HFTDK2STZ6OFPHUEMRGHOJSHC5OQYMR2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030337428.0_warc_CC-MAIN-20221003164901-20221003194901-00603.warc.gz\"}"}
https://www.metafunction.co.uk/post/all-about-digital-oscillators-part-1-aliasing-foldover	[ "Search\n\n# All About Digital Oscillators Part 1 – Aliasing & Foldover", null, "At the core of any synth is it’s oscillator – the device that produces the raw waveforms that form the building blocks of synthesis. In the analogue world, these are typically voltage controlled and built via electronic components such as op-amps and diodes. In the digital world, such as in a plug-in, there are many ways to build an oscillator and one of the most popular methods is via an algorithmic technique (i.e. a mathematical equation). The most important and deployed approach in digital synthesizer design within the modern landscape is the use of quasi band-limited oscillators to reduce aliasing. Quasi what? To answer that question we first need to consider aliasing in terms of synthesis….\n\n## Folding Over\n\nYou are most likely familiar with aliasing from your experience of the world of digital audio theory. We typically tend to set our DAW to operate with a sampling frequency (i.e. 44.1kHz or 48kHz) that conforms to the Nyquist-Shannon sampling theorem. This theorem states that we need to sample a signal at twice its highest frequency to be able to uniquely reconstruct it. This frequency is called the Nyquist frequency. When recording an acoustic signal into our DAW via an audio interface, the A/D converter in the interface has a brick-wall lowpass filter which is set to our Nyquist frequency. The filters job is to only allow content lower than the Nyquist frequency in, therefore providing us with a signal that has been appropriately band-limited (i.e. at 22,050 Hz or 24,000 Hz). So, in the digital audio world, aliasing happens if you do not properly band-limit the input signal. Alias signals ‘foldover’ at the Nyquist frequency and are reflected back down along the frequency spectrum (see figure 1). They are most noticeable as distortion in our audio signal and not a lot can be done to remove them.", null, "Figure 1: Frequency domain representation of aliasing\n\nHere the dotted red line shows foldover, the mirror-like reflection of aliased signal components, at a Nyquist frequency of 24kHz\n\nWithin our digital synthesizer, we simply replace the input signal that exists when we are recording an acoustic signal with the output of our oscillator algorithm. The output of the oscillator algorithm IS our input signal – but it has not been sent through the brick-wall lowpass filter in our A/D converter. So technically, when using algorithmic techniques to create a digital oscillator within a computer, it is possible to create a signal that extends well beyond the Nyquist frequency and contains aliased components. We perceive this aliasing as subtle distortion.\n\n## Time & Frequency\n\nAs we can see in figure 1, aliased components foldover around the Nyquist frequency and are visible in the frequency domain on a spectrogram. It is easy to see and hear this via a simple Max patch as shown in figure 2. On the left hand side is the Max Saw~ object (a band-limited oscillator), whilst on the right hand side we have the Max Phasor~ object which provides an ‘ ideal mathematical ramp without filtering out frequency information above Nyquist’ (i.e. a non band-limited saw oscillator). On the left side we can see a clean signal in the spectrogram, whilst on the right side we can see aliased components folding over around the Nyquist frequency and being reflected back down the frequency spectrum.", null, "Figure 2: Using Max to show a frequency domain representation of aliasing\n\nOn the left we can see that the band-limited Saw~ object produces a nice clean signal on the spectrogram whilst on the right we can see that the non band-limited Phasor~ produces alias components that foldover around the Nyquist frequency and are reflected back down the frequency spectrum\n\nThe easiest wave to see aliasing in the frequency domain is via 3D spectrogram plots. These give us a very clear indication of the amount of aliasing and how it is represented as distortion spread across the frequency spectrum, down from the Nyquist frequency. In fact, using these type of plots the amount of foldover and distortion in the signal is really surprising. To show this off, the authour developed a band-limited and non band-limited oscillator algorithm in C++ and generated plots. Figure 3 shows a 3D spectrogram plot of a non band-limited square wave and figure 4 shows a 3D spectrogram plot of a band-limited square wave. As aliasing is often much more noticeable at higher frequencies, these plots are of the oscillator running at C6 (MIDI note number 96), which has a fundamental frequency of approx~ 2093Hz.", null, "Figure 3: 3D spectrogram representation of aliased oscillator Non band-limited square wave plot at approx~ 2kHz. Notice the high levels of aliasing appearing as ripples down the frequency spectrum", null, "Figure 4: 3D spectrogram representation of non aliased oscillator Band-limited square wave plot at approx~ 2kHz. Notice the reduced levels of aliasing\n\n‘OK, that all makes sense, but what about the time domain?’ I hear you ask. As it happens, in the time domain, aliased components are visualized a bit differently. In the time domain aliased components are seen as discontinuities in the edges of the synthesized waveforms. That is, at the fast changing edges of waveforms. For example, during the cycle of a waveform there are moments when the waveform changes amplitude over time. These moments give the characteristic shape to our waveform (i.e. square). In the digital world, if we were to plot out a series of instantaneous values for frequency and amplitude for many samples over time we would see our waveshape. The higher the frequency or pitch of the oscillator the more rapid the edges of waveforms change, and hence the more discontinuities in the edges of the synthesized waveforms. Figure 5 shows a time domain oscilloscope plot of a non band-limited square wave. Notice the amount of discontinuity at the edges of the waveshape. This is the time domain representation of aliasing – visible at the fast changing edges of waveforms. A non band-limited oscillator is often known as a ‘naive’ oscillator. Figure 6 shows a time domain oscilloscope plot of a band-limited square wave. Notice the reduced amount of discontinuity at the edges of the waveshape.", null, "Figure 5: Time domain oscilloscope plot of a naive, non band-limited square wave Notice the amount of discontinuity at the edges of the waveshape", null, "Figure 6: Time domain oscilloscope plot of a band-limited square wave Notice the reduced amount of discontinuity at the edges of the waveshape\n\n## Quasi Solution?\n\nThere are a number of ways around this problem. The most obvious may be to use different techniques to generate our oscillator in the first place that bypass the problem. This is a common solution and several options exist. A frequency limited Fourier series representation can be used to generate an aliasing free signal, or additive synthesis techniques could be used. However, these may both be computationally expensive. Another technique is that a wave is precalculated using a mathematical equation, and stored in memory (i.e. a table). We then use a look-up table technique to read the values from the table to generate our oscillator. This is a technique known as a wavetable (or look-up table) oscillator. However, clever tricks are needed here to ensure the wavetable oscillator is band-limited and free from other types of distortion (i.e. quantization & interpolation distortion). So what’s the solution? Well, one of the most popular techniques is really a compromise – we allow a little bit of aliasing as long as the aliased components are far away from the fundamental frequency and low in amplitude. For example, if we trigger our band-limited oscillator with a fundamental frequency of approx~ 2kHz (which is C6 on a MIDI keyboard), we may expect to see aliased components stretching down to somewhere in the region of -60dB at approx~ 10kHz. Remember, the dynamic range of the human ear is considered to be 130dB and arguably aliased components at -60dB would be difficult, if not impossible, to hear. Don’t believe me? Take this test to find out.\n\nAs these modern techniques don’t really remove aliasing, they suppress it enough so we cant hear it, they are known as quasi band-limited techniques. So this links nicely back to our opening paragraph – the most important and deployed approach in digital synthesizer design within the modern landscape is the use of quasi band-limited oscillators to reduce aliasing. ‘Ok – got it, but how does it work?’ I hear you say. In our next blog post, entitled ‘All About Digital Oscillators Part 2 – BLITS & BLEPS’ we will explore how these techniques work. So for now, happy wiggling." ]	[ null, "https://static.wixstatic.com/media/bbc0f5_fe0df30f2fb94b76abe0b75a102523d3~mv2.png/v1/fill/w_967,h_670,al_c/bbc0f5_fe0df30f2fb94b76abe0b75a102523d3~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_2247d64e5ba24c9eb16c1ba2650af53a~mv2.png/v1/fill/w_843,h_296,al_c/bbc0f5_2247d64e5ba24c9eb16c1ba2650af53a~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_4d0e2d1a80ce4d958cda3a0ccbccbffb~mv2.png/v1/fill/w_771,h_375,al_c/bbc0f5_4d0e2d1a80ce4d958cda3a0ccbccbffb~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_b2130d839c384775b614137f63abd64e~mv2.png/v1/fill/w_967,h_669,al_c/bbc0f5_b2130d839c384775b614137f63abd64e~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_fe0df30f2fb94b76abe0b75a102523d3~mv2.png/v1/fill/w_967,h_670,al_c/bbc0f5_fe0df30f2fb94b76abe0b75a102523d3~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_3c4a426bd63142bc92b1342f5e14381f~mv2.png/v1/fill/w_968,h_670,al_c/bbc0f5_3c4a426bd63142bc92b1342f5e14381f~mv2.png", null, "https://static.wixstatic.com/media/bbc0f5_5cf265c891d14f91a7ac9e354c4fc43e~mv2.png/v1/fill/w_967,h_668,al_c/bbc0f5_5cf265c891d14f91a7ac9e354c4fc43e~mv2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9161733,"math_prob":0.9251915,"size":8681,"snap":"2022-05-2022-21","text_gpt3_token_len":1858,"char_repetition_ratio":0.15247205,"word_repetition_ratio":0.10111576,"special_character_ratio":0.19894022,"punctuation_ratio":0.084059775,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9531759,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,4,null,2,null,2,null,4,null,4,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-27T19:27:11Z\",\"WARC-Record-ID\":\"<urn:uuid:c499acc5-ba88-4176-88b0-121ffa269dd8>\",\"Content-Length\":\"779995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ce5ef4b5-b824-4815-b11e-c9cb4eabb114>\",\"WARC-Concurrent-To\":\"<urn:uuid:239d91a6-24d4-431f-92f0-5b39a0801c62>\",\"WARC-IP-Address\":\"34.117.168.233\",\"WARC-Target-URI\":\"https://www.metafunction.co.uk/post/all-about-digital-oscillators-part-1-aliasing-foldover\",\"WARC-Payload-Digest\":\"sha1:UFAG3OADUCB72RRAAIPLHZAPV5Q3RMM7\",\"WARC-Block-Digest\":\"sha1:L43D6N3YKK3S2VDJAODIKMNY4MLYSDDN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662675072.99_warc_CC-MAIN-20220527174336-20220527204336-00451.warc.gz\"}"}
https://projecteuclid.org/proceedings/geometry-integrability-and-quantization/Proceedings-of-the-Seventeenth-International-Conference-on-Geometry-Integrability-and/Chapter/Functionals-on-Toroidal-Surfaces/10.7546/giq-17-2016-270-283	[ "Translator Disclaimer\nVOL. 17 \| 2016 Functionals on Toroidal Surfaces\nMetin Gürses\n\nEditor(s) Ivaïlo M. Mladenov, Guowu Meng, Akira Yoshioka\n\nAbstract\n\nWe show that the torus in ${\\mathbb R}^3$ is a critical point of a sequence of functionals ${\\mathcal F}_{n}$ ($n=1,2,3, \\ldots$) defined over compact surfaces (closed membranes) in ${\\mathbb R}^3$. When the Lagrange function ${\\mathcal E}$ is a polynomial of degree $n$ of the mean curvature $H$ of the torus, the radii ($a,r$) of the torus are constrained to satisfy $\\frac{a^2}{r^2}=\\frac{n^2-n}{n^2-n-1},~~ n \\ge 2$. A simple generalization of torus in ${\\mathbb R}^3$ is a tube of radius $r$ along a curve ${\\bf \\alpha}$ which we call it toroidal surface (TS). We show that toroidal surfaces with non-circular curve ${\\bf \\alpha}$ do not provide minimal energy surfaces of the functionals ${\\mathcal F}_{n}$ ($n=2,3$) on closed surfaces. We discuss possible applications of the functionals discussed in this work on cell membranes.\n\nInformation\n\nPublished: 1 January 2016\nFirst available in Project Euclid: 15 December 2015\n\nzbMATH: 1346.53012\nMathSciNet: MR3445435\n\nDigital Object Identifier: 10.7546/giq-17-2016-270-283", null, "", null, "" ]	[ null, "https://projecteuclid.org/Content/themes/SPIEImages/Share_black_icon.png", null, "https://projecteuclid.org/images/proceedings/cover_pgiq.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7152463,"math_prob":0.99529225,"size":912,"snap":"2022-05-2022-21","text_gpt3_token_len":276,"char_repetition_ratio":0.13105726,"word_repetition_ratio":0.030534351,"special_character_ratio":0.3125,"punctuation_ratio":0.07734807,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9987033,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-26T18:41:18Z\",\"WARC-Record-ID\":\"<urn:uuid:851eb695-eb81-4ba0-b644-73f2de2f0f98>\",\"Content-Length\":\"119552\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:68d6c644-c60d-4f1e-b8f8-0738d7d3011b>\",\"WARC-Concurrent-To\":\"<urn:uuid:826e1e73-304e-4d4f-b5df-43d2ef709cba>\",\"WARC-IP-Address\":\"107.154.79.145\",\"WARC-Target-URI\":\"https://projecteuclid.org/proceedings/geometry-integrability-and-quantization/Proceedings-of-the-Seventeenth-International-Conference-on-Geometry-Integrability-and/Chapter/Functionals-on-Toroidal-Surfaces/10.7546/giq-17-2016-270-283\",\"WARC-Payload-Digest\":\"sha1:TO6O2WFSMGJ7HY6WBY5LYWBTOC4LF5HY\",\"WARC-Block-Digest\":\"sha1:3ZDTUAFYWUFQC67ATZ4KC6DR4DRJT2CK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304959.80_warc_CC-MAIN-20220126162115-20220126192115-00347.warc.gz\"}"}
https://www.thelearningpoint.net/computer-science/python-for-data-5-data-analysis-with-pandas-basic	[ "Computer Science‎ > ‎\n\n### Why pandas ?\n\nThere are many data analysis tools available, so why pandas ?\nIf you've spent time in a spreadsheet software like Microsoft Excel, Apple Numbers, or Google Sheets and are eager to take your data analysis skills to the next level, this tutorial is for you! and you are at the right place.\n\nPandas is a powerful tool that allows you to do anything and everything with colossal data sets. These are the major task that we performs, analyzing, organizing, sorting, filtering, pivoting, aggregating, munging, cleaning, calculating, and more! We can say it's \"Excel on steroids\"!\n\nHere one can find data set (also available in the reference) on which we are gonna analyse some important statistics. Here we are gonna use 3 python libraries. pandas for data analysis, numpy for calculation and mathematics and lastly matplotlib for visualization of graphs.\nYou can see that we have 128 rows and 8 columns here. we are going to perform following task with this 'house.csv' data set.\n\n1. Load the data using pandas\n\n2. Summarize each field in the data, i.e. mean, average etc.\n\n3. Group data by the field [nbhd].\n\n(a) Give average sqft, average price and average bedroom of each group.\n\n(b) Plot for each field ([sqft], [bedroom], [price] etc). Use a boxplot that visualizes the statistical information about them.\n\n(c) For each group of [nbhd], draw a prediction line for [price] vs [sqft]\n\nHere we are gonna see the data description i.e. Mean, max, standard deviation etc. of each column. 'Transpose' here is working similar to matrix transpose where it rotate our columns. see below - Now let's summarize our data using pandas 'groupby'Here is the output of this line of our code. we want to plot the dependency of prices on the size of the house and number of bathrooms ans so on. Therefore, we used 'groupby' here. Boxplots of (sqft, bedroom, price) with respect to \"nbhd\" groupby field.Prediction line plot \"price vs sqft\" for \"nbhd\"= nbhd01Here one can see the plot and nice visualization by 'matplotlib' libraryhere you see, how price is increasing with respect to square fit/size of the house for group 01 i.e. [nbhd01]. Now let's see for second group. here is the plot. Similarly for last group [nbhd03Here we are done with basic pandas operations. We have seen how we can visualize data and it's analysis. We saw how price is dependent on size of the house and other factor among 3 groups of house data.\n\n### What's Next\n\nLinear Regression via Normal Equations using same 'house.csv' data.\nwhere we will learn below matrix decomposition methods to perform Linear regression.\n(a) Gaussian elimination\n(b) Cholesky decomposition\n(c) QR decomposition" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.849575,"math_prob":0.9420206,"size":1785,"snap":"2020-45-2020-50","text_gpt3_token_len":408,"char_repetition_ratio":0.101628296,"word_repetition_ratio":0.0,"special_character_ratio":0.21904762,"punctuation_ratio":0.14497042,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99397606,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-26T10:26:20Z\",\"WARC-Record-ID\":\"<urn:uuid:51dcf2d6-de3e-4037-8384-950c23e9c4f9>\",\"Content-Length\":\"121795\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ceb9c208-d7e9-4302-81df-88c160a659c7>\",\"WARC-Concurrent-To\":\"<urn:uuid:6313d9d5-ac30-4f12-b27a-8a31b1a7aece>\",\"WARC-IP-Address\":\"104.27.187.64\",\"WARC-Target-URI\":\"https://www.thelearningpoint.net/computer-science/python-for-data-5-data-analysis-with-pandas-basic\",\"WARC-Payload-Digest\":\"sha1:B5J4CP7BFVBUDJ5PUGZW42RLPETZ332B\",\"WARC-Block-Digest\":\"sha1:XNRIC7AMY37FBPBRZF5WEVJROPMYXCTM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107891203.69_warc_CC-MAIN-20201026090458-20201026120458-00520.warc.gz\"}"}
https://www.findfilo.com/maths-question-answers/if-lim-xvec0-a-e-x-b-x-2-then-find-the-value-of-aa3m2	[ "", null, "If (\"lim\")_(xvec0)(a e^x-b)/x = 2 , then find the value of aa n d \| Filo", null, "", null, "Class 11\n\nMath\n\nCalculus\n\nLimits And Derivatives", null, "563\n\nIf then find the value of", null, "563", null, "Connecting you to a tutor in 60 seconds." ]	[ null, "https://www.facebook.com/tr", null, "https://www.findfilo.com/images/logo.svg", null, "https://www.findfilo.com/images/icons/navbar.png", null, "https://www.findfilo.com/images/icons/view.svg", null, "https://www.findfilo.com/images/icons/view.svg", null, "https://www.findfilo.com/images/common/mobile-widget.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.71354574,"math_prob":0.9958686,"size":671,"snap":"2021-31-2021-39","text_gpt3_token_len":241,"char_repetition_ratio":0.11694153,"word_repetition_ratio":0.0,"special_character_ratio":0.29955292,"punctuation_ratio":0.11612903,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97772956,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T17:51:32Z\",\"WARC-Record-ID\":\"<urn:uuid:0c0c1a64-8d40-4eeb-9b69-8f5998d0701f>\",\"Content-Length\":\"71374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:99aea219-a432-411c-9777-c7bbe8c9254e>\",\"WARC-Concurrent-To\":\"<urn:uuid:500234a2-fdb6-4453-9ad9-ad86d35fda5d>\",\"WARC-IP-Address\":\"34.117.94.82\",\"WARC-Target-URI\":\"https://www.findfilo.com/maths-question-answers/if-lim-xvec0-a-e-x-b-x-2-then-find-the-value-of-aa3m2\",\"WARC-Payload-Digest\":\"sha1:64B252GUV4BOMWCFPI6QQMBGTOV2E5RS\",\"WARC-Block-Digest\":\"sha1:BS7WAXZYYXKUKN2VIFJL3CSISFXCLEQO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153971.20_warc_CC-MAIN-20210730154005-20210730184005-00319.warc.gz\"}"}
http://irerahihu3.megafile.ru.net/fe_sample-stoichiometry-problems-and-answers-gy/qezer.htm	[ "##### Sample stoichiometry problems and answers", null, "Stoichiometry sat ii chemistry.", null, "Stoichiometry example problem 1 (video) \| khan academy.", null, "Stoichiometry 7: limiting reagents and percentage yield calculations.", null, "## Icechart.", null, "Solution stoichiometry practice problems & examples finding.", null, "##### Stoichiometry with solutions problems.", null, "### Stoichiometry and balancing reactions chemistry libretexts.", null, "Ideal stoichiometry (practice) \| khan academy.", null, "Chapter 13 – stoichiometry.", null, "###### Practice problems: stoichiometry (answer key).", null, "##### Stoichiometry problem \| wyzant ask an expert.", null, "How do you solve a stoichiometry problem? + example.", null, "#### Worksheet stoichiometry ap level.", null, "Sparknotes: stoichiometric calculations: problems.", null, "# Solving stoichiometry problems.", null, "#### Stoichiometry (solutions, examples, videos).", null, "# Gas stoichiometry sample problems, assignment, and answers.", null, "Chemistry and more practice problems with answers.", null, "", null, "Chemteam: stoichiometry: mass-mass examples." ]	[ null, "https://i.ytimg.com/vi/TdpOZTPAKbg/maxresdefault.jpg", null, "https://i.ytimg.com/vi/Ab3wfKjaWWQ/maxresdefault.jpg", null, "http://intro.chem.okstate.edu/1314F97/Chapter4/Molarity7.1.GIF", null, "https://images.sampletemplates.com/wp-content/uploads/2016/11/17162322/Mass-to-Mass-Stoichiometry-Worksheet.jpg", null, "https://i.ytimg.com/vi/5RRVfI2fItk/maxresdefault.jpg", null, "http://www.chem.tamu.edu/class/fyp/stone/tutorialnotefiles/stoich9.gif", null, "http://alistairtheoptimist.org/wp-content/uploads/2018/02/ap-chemistry-page-practice-test-ch-14-wanswers-word-acrobat-and-worked-out-answers-.gif", null, "https://images.sampletemplates.com/wp-content/uploads/2016/11/17162238/Molarity-Stoichiometry-Worksheet.jpg", null, "https://images.sampletemplates.com/wp-content/uploads/2016/11/17162546/Stoichiometry-Practice-Worksheet.jpg", null, "http://intro.chem.okstate.edu/1314F97/Chapter4/Molarity8.GIF", null, "https://i.ytimg.com/vi/aB_S-C02bgM/maxresdefault.jpg", null, "http://colgurchemistry.com/Chem12/SCANNED%20KEYS/Worksheet3-2/Assignment3-2_KEYp3.jpg", null, "https://images.sampletemplates.com/wp-content/uploads/2016/11/17162152/Gas-Stoichiometry-Worksheet.jpg", null, "https://s3.studylib.net/store/data/008351429_1-690533bd35cd021cad902f8855cbaa92-260x520.png", null, "https://s3.studylib.net/store/data/008711874_1-d12f29b6b62688faeae95fc3bd7c9543-260x520.png", null, "http://www.colgurchemistry.com/Chem11/SCANNED%20KEYS/Un6ReviewKEY/Un6ReviewKEYp4.jpg", null, "https://i.ytimg.com/vi/F_TiekZIFgU/maxresdefault.jpg", null, "https://images.sampletemplates.com/wp-content/uploads/2016/11/17162413/Mole-to-Mole-Stoichiometry-Worksheet.jpg", null, "https://i.ytimg.com/vi/iTwQ9jX3xwM/hqdefault.jpg", null, "https://img.yumpu.com/52027820/1/358x462/stoichiometry-worksheet-answers.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7771953,"math_prob":0.57873577,"size":908,"snap":"2020-34-2020-40","text_gpt3_token_len":204,"char_repetition_ratio":0.29756638,"word_repetition_ratio":0.0,"special_character_ratio":0.17070484,"punctuation_ratio":0.22727273,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97113854,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40],"im_url_duplicate_count":[null,2,null,3,null,2,null,4,null,3,null,3,null,2,null,4,null,5,null,2,null,6,null,2,null,5,null,2,null,2,null,5,null,2,null,null,null,3,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-13T12:31:02Z\",\"WARC-Record-ID\":\"<urn:uuid:a9ff913a-25f9-4ab2-b6d4-a0127415274b>\",\"Content-Length\":\"15876\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7dc6530-9c82-4f2f-ab1a-d2fc0f18f453>\",\"WARC-Concurrent-To\":\"<urn:uuid:a59718a5-484f-4221-afa3-1be9094e20a7>\",\"WARC-IP-Address\":\"172.67.212.238\",\"WARC-Target-URI\":\"http://irerahihu3.megafile.ru.net/fe_sample-stoichiometry-problems-and-answers-gy/qezer.htm\",\"WARC-Payload-Digest\":\"sha1:635YJFYWON4NJM2PU66GUTVVDLYY5Q4L\",\"WARC-Block-Digest\":\"sha1:664E4SPOXPBJYSDJLLGLCTUAFX4O5ACE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738982.70_warc_CC-MAIN-20200813103121-20200813133121-00261.warc.gz\"}"}
https://www.ques10.com/p/16293/programming-in-c-data-structures-question-paper--1/	[ "0\nProgramming in C & Data Structures : Question Paper Dec 2014 - First Year Engineering (C Cycle) (Semester 2) \| Visveswaraya Technological University (VTU)\n\n## Programming in C and Data Structures - Dec 2014\n\n### First Year Engineering (C Cycle) (Semester 2)\n\nTOTAL MARKS: 100\nTOTAL TIME: 3 HOURS\n(1) Question 1 is compulsory.\n(2) Attempt any four from the remaining questions.\n(3) Assume data wherever required.\n(4) Figures to the right indicate full marks.\n1(a) What is pseudo code? How it is used as a prob lem-solving tool. (6 marks) 1(b) What is an operator? Explain the arithmetic, relational, logical, and assignment operators in C language. (10 marks) 1(c) Write a program in C to print the numbers from 4 to 9 and their squares. (4 marks) 10(a) Explain the array of pointers with example. (4 marks) 10(b) Write and explain any two preprocessor directives in C.(4 marks) 10(c) What is a stack? Explain it with its applications. (4 marks) 10(d) Write a C program to read n unsorted numbers to a n array of size n and pass the address of this array to a function to sort the numbers in ascending order using bubble sort technique. (8 marks) 2(a) Write and explain the basic concepts of a C program. (8 marks) 2(b) Write the guidelines to use printf() function in C language. (8 marks) 2(c) Write a program in C to find the area and perimeter of a circle. (4 marks) 3(a) Explain the two way selection (if, if-else, nested if- lse, cascaded if-else) in C language with syntax . (8 marks) 3(b) Explain the switch statement with syntax and example. (8 marks) 3(c) Design and develop a C program to read a year as an input and find whether it is leap year or not. Also consider end of the centuries. (4 marks) 4(a) Explain the different types of loops in C with syntax a nd example. (8 marks) 4(b) Explain the use of break and continue statement in loops with example. (6 marks) 4(c) Design and develop a C program to reverse of an integer number NUM and check whether it is PALINDROME or NOT. (6 marks) 5(a) What is an array? Explain the declaration and initialization of one and two dimensional arrays with example. (6 marks) 5(b) Explain void and parameter less functions in C with examples. (6 marks) 5(c) Write a C program that:\ni. Implements string copy operation STRCOPY (str1,str2) that copies a string str1 to another string str2 without using library function.\nii. Reads a sentence and prints frequency of each of the vowels and total count of consonants.\n(8 marks)\n6(a) Explain any five string manipulation library functions with examples. (10 marks) 6(b) What is function parameter? Explain different types of parameters in C functions. (4 marks) 6(c) Write a C function isprime (num) that accepts an integer argument and returns 1 if the argument is prime, a 0 otherwise. Write a C program that invokes this function to generate prime numbers between the given ranges. (6 marks) 7(a) What is a structure? Explain the syntax of structure declaration with example. (4 marks) 7(b) What is a file? Explain how the file open and file close functions handled in C. (6 marks) 7(c) Write a C program to maintain a record of n\"student details using an array of structures with four fields (Roll number(10 marks) 8(a) Explain array of structures and structure within a structure with examples. (8 marks) 8(b) Explain how the structure variable passed as a param eter to a function with example. (6 marks) 8(c) Write a C program to read and display a text from the file.(6 marks) 9(a) What is a pointer? Explain how the pointer variable de clared and initialized. (4 marks) 9(b) What is dynamic memory allocation? Write and explain the different dynamic memory allocation functions in C. (6 marks) 9(c) What are primitive and non-primitive data types? (4 marks) 9(d) Write a C program to swap two numbers using call by pointers method. (6 marks)\n\n0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8019006,"math_prob":0.8828657,"size":3682,"snap":"2021-04-2021-17","text_gpt3_token_len":908,"char_repetition_ratio":0.17808592,"word_repetition_ratio":0.02643857,"special_character_ratio":0.2593699,"punctuation_ratio":0.08042896,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96494776,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-15T20:38:44Z\",\"WARC-Record-ID\":\"<urn:uuid:23024d93-7230-415b-933f-34c17669430b>\",\"Content-Length\":\"25257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4057e4d3-9419-47d3-908a-8ba32c3f7b23>\",\"WARC-Concurrent-To\":\"<urn:uuid:a0b86e14-76e3-4d1e-bd34-e2fef35be52d>\",\"WARC-IP-Address\":\"172.67.155.223\",\"WARC-Target-URI\":\"https://www.ques10.com/p/16293/programming-in-c-data-structures-question-paper--1/\",\"WARC-Payload-Digest\":\"sha1:HOGYYBOQ3FEE5OK4QX5KTUWPBBC54ONB\",\"WARC-Block-Digest\":\"sha1:QWWEYLQRP56LC3YBPIZW2CGLKFH6PYWY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703496947.2_warc_CC-MAIN-20210115194851-20210115224851-00064.warc.gz\"}"}
https://xianblog.wordpress.com/tag/ila/	[ "## Hyper-g priors\n\nPosted in Books, R, Statistics with tags , , , , , , on August 31, 2010 by xi'an\n\nEarlier this month, Daniel Sabanés Bové and Leo Held posted a paper about g-priors on arXiv. While I glanced at it for a few minutes, I did not have the chance to get a proper look at it till last Sunday. The g-prior was first introduced by the late Arnold Zellner for (standard) linear models, but they can be extended to generalised linear models (formalised by the late John Nelder) at little cost. In Bayesian Core, Jean-Michel Marin and I do centre the prior modelling in both linear and generalised linear models around g-priors, using the naïve extension for generalised linear models,", null, "$\\beta \\sim \\mathcal{N}(0,g \\sigma^2 (\\mathbf{X}^\\text{T}\\mathbf{X})^{-1})$\n\nas in the linear case. Indeed, the reasonable alternative would be to include the true information matrix but since it depends on the parameter", null, "$\\beta$ outside the normal case this is not truly an alternative. Bové and Held propose a slightly different version", null, "$\\beta \\sim \\mathcal{N}(0,g \\sigma^2 c (\\mathbf{X}^\\text{T}\\mathbf{W}\\mathbf{X})^{-1})$\n\nwhere W is a diagonal weight matrix and c is a family dependent scale factor evaluated at the mode 0. As in Liang et al. (2008, JASA) and most of the current literature, they also separate the intercept", null, "$\\beta_0$ from the other regression coefficients. They also burn their “improperness joker” by choosing a flat prior on", null, "$\\beta_0$, which means they need to use a proper prior on g, again as Liang et al. (2008, JASA), for the corresponding Bayesian model comparison to be valid. In Bayesian Core, we do not separate", null, "$\\beta_0$ from the other regression coefficients and hence are left with one degree of freedom that we spend in choosing an improper prior on g instead. (Hence I do not get the remark of Bové and Held that our choice “prohibits Bayes factor comparisons with the null model“. As argued in Bayesian Core, the factor g being an hyperparameter shared by all models, we can use the same improper prior on g in all models and hence use standard Bayes factors.) In order to achieve closed form expressions, the authors use Cui and George ‘s (2008) prior", null, "$\\pi(g) \\propto (1+g)^{1+a}\\exp\\{-b/(1+g)\\}$\n\nwhich requires the two hyper-hyper-parameters a and b to be specified.\n\nThe second part of the paper considers computational issues. It compares the ILA solution of Rue, Martino and Chopin (2009, Series B) with an MCMC solution based on an independent proposal on g resulting from linear interpolations (?). The marginal likelihoods are approximated by Chib and Jeliazkov (2001, JASA) for the MCMC part. Unsurprisingly, ILA does much better, even with a 97% acceptance rate in the MCMC algorithm.\n\nThe paper is very well-written and quite informative about the existing literature. It also uses the Pima Indian dataset (The authors even dug out a 1991 paper of mine I had completely forgotten!) I am actually thinking of using the review in our revision of Bayesian Core, even though I think we should stick to our choice of including", null, "$\\beta_0$ within the set of parameters…" ]	[ null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null, "https://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91915125,"math_prob":0.9290526,"size":2744,"snap":"2021-04-2021-17","text_gpt3_token_len":614,"char_repetition_ratio":0.09854015,"word_repetition_ratio":0.0,"special_character_ratio":0.21027696,"punctuation_ratio":0.08429119,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99097747,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-15T05:11:48Z\",\"WARC-Record-ID\":\"<urn:uuid:1fa7eb40-1800-4c6d-b45d-7510228fbccf>\",\"Content-Length\":\"56856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:971489ef-9c2c-4f00-bec2-3bd46bbdcd83>\",\"WARC-Concurrent-To\":\"<urn:uuid:da44ef2e-2615-4e9e-ab09-4357a0981b92>\",\"WARC-IP-Address\":\"192.0.78.13\",\"WARC-Target-URI\":\"https://xianblog.wordpress.com/tag/ila/\",\"WARC-Payload-Digest\":\"sha1:33XIHRORMNGLJ4VEBCVOX5NMLHLN2A6K\",\"WARC-Block-Digest\":\"sha1:732PPRBR27R4SCYMSJBXYIKEX3RAMGUS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038083007.51_warc_CC-MAIN-20210415035637-20210415065637-00295.warc.gz\"}"}
https://mathhelpboards.com/threads/some-very-basic-questions-on-topology.2796/	[ "# Some very basic questions on topology!\n\n#### Mathelogician\n\n##### Member\nhello everybody!\nMy knowledge in the field is so basic. So please answer as clear and simple as possible:\nThe following 4 questions are the ones i couldn't solve after trying!\n++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\n\n1-Prove [0,1] is not isometric to [0,2].\n2-For which intervals [a,b] in R is the intersection of [a,b] and Q a clopen (closed and open) subset of the metric space Q?\n3-Find a metric space in which the boundary of Mr(p)={xEM :d(x,p)< r} is not equal to the sphere of radius r at p ,{xEM : d(x,p)=r}.\n4- For a subspace N of a metric space M,prove that if the openness of S c N is equivalent to openness of S in M, then N is open in M.\n============================================================\n\nThe next is about my proof of the following problem:\n\nLet (An) be a nested decreasing sequence of non-empty closed sets in the metric space M.\nIf M is complete(every Cauchy is convergent) and diam(An) -> 0 as n -> infinity, show that S=Intersection of (An) is exactly one point( note that diam X= sup{d(x,y): x,y E X} ).\nMy proof : (I use Euclidean metric on R for simplicity of the proof- it can be done for any other metric on R)\nLets x,y E S. we must show that d(x,y)=0.\nLets d(x,y)= t > 0. By the red part: there exists a N1 such that for all n>N1 , we have diam(An)<t\nLet k > n. So diam(Ak)<t.\nIn other hand, d(x,y) =< diam(Ak)< t. So d(x,y)< t which contradicts the blue.\n\nNow my question is : What was the usage of supposing Ai's as Closed subspaces of M and M's being complete, in this proof?\n\n#### Opalg\n\n##### MHB Oldtimer\nStaff member\nRe: Some very basic quastions on topology!\n\nhello everybody!\nMy knowledge in the field is so basic. So please answer as clear and simple as possible:\nThe following 4 questions are the ones i couldn't solve after trying!\n++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++\n\n1-Prove [0,1] is not isometric to [0,2]. The interval [0,2] contains two points whose distance apart is 2, but the interval [0,1] does not.\n2-For which intervals [a,b] in R is the intersection of [a,b] and Q a clopen (closed and open) subset of the metric space Q? Think about what difference it makes whether a and b are rational or irrational.\n3-Find a metric space in which the boundary of Mr(p)={xEM :d(x,p)< r} is not equal to the sphere of radius r at p ,{xEM : d(x,p)=r}. Think about a space with a discrete metric.\n4- For a subspace N of a metric space M, prove that if the openness of S c N is equivalent to openness of S in M, then N is open in M. What happens if you take S = N?\n============================================================\n(See the hints in red.)\n\nThe next is about my proof of the following problem:\nLet (An) be a nested decreasing sequence of non-empty closed sets in the metric space M.\nIf M is complete(every Cauchy is convergent) and diam(An) -> 0 as n -> infinity, show that S=Intersection of (An) is exactly one point( note that diam X= sup{d(x,y): x,y E X} ).\nMy proof : (I use Euclidean metric on R for simplicity of the proof- it can be done for any other metric on R)\nLets x,y E S. we must show that d(x,y)=0.\nLets d(x,y)= t > 0. By the red part: there exists a N1 such that for all n>N1 , we have diam(An)<t\nLet k > n. So diam(Ak)<t.\nIn other hand, d(x,y) =< diam(Ak)< t. So d(x,y)< t which contradicts the blue.\n\nNow my question is : What was the usage of supposing Ai's as Closed subspaces of M and M's being complete, in this proof?\nYour proof shows that S cannot contain more than one point. But how do you know that it contains any points at all? Answer: You need those extra conditions to ensure that S is not empty." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9178247,"math_prob":0.99695724,"size":2841,"snap":"2021-31-2021-39","text_gpt3_token_len":789,"char_repetition_ratio":0.17377512,"word_repetition_ratio":0.8023256,"special_character_ratio":0.36149243,"punctuation_ratio":0.13364056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9979302,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-20T23:07:22Z\",\"WARC-Record-ID\":\"<urn:uuid:fe8c3530-1a2e-4edc-9912-e80e2dad2f21>\",\"Content-Length\":\"60301\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d4e9c8c4-0c35-4afc-8085-c800a7c0066b>\",\"WARC-Concurrent-To\":\"<urn:uuid:574cd4be-034f-462b-874a-67e40e2be1f9>\",\"WARC-IP-Address\":\"50.31.99.218\",\"WARC-Target-URI\":\"https://mathhelpboards.com/threads/some-very-basic-questions-on-topology.2796/\",\"WARC-Payload-Digest\":\"sha1:AQ3WQX6AKUVWPG7IO6KT7NCVD5AGQAKP\",\"WARC-Block-Digest\":\"sha1:GC64WXQYLGXQEA2WWE56I7A6IHFHKYLZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057119.85_warc_CC-MAIN-20210920221430-20210921011430-00099.warc.gz\"}"}
https://neon-lang.dev/html/files/binary-neon.html	[ "# binary\n\nBitwise binary logical operations on 32-bit words.\n\nSummary\n binary Bitwise binary logical operations on 32-bit words. Functions and32 Bitwise logical “and” of two 32-bit words. and64 Bitwise logical “and” of two 64-bit words. andBytes Bitwise logical “and” of two Bytes buffers. extract32 Extract a range of bits from a 32-bit word. extract64 Extract a range of bits from a 64-bit word. get32 Get a specific bit from a 32-bit word. get64 Get a specific bit from a 64-bit word. not32 Invert all bits in a 32-bit word. not64 Invert all bits in a 64-bit word. notBytes Invert all bits in a Bytes buffer. or32 Bitwise logical “or” of two 32-bit words. or64 Bitwise logical “or” of two 64-bit words. orBytes Bitwise logical “or” of two Bytes buffers. replace32 Replace a range of bits in a 32-bit word. replace64 Replace a range of bits in a 64-bit word. set32 Set a specific bit in a 32-bit word. set64 Set a specific bit in a 64-bit word. shiftLeft32 Shift a 32-bit word left by a given number of bits. shiftLeft64 Shift a 64-bit word left by a given number of bits. shiftRight32 Shift a 32-bit word right by a given number of bits. shiftRight64 Shift a 64-bit word right by a given number of bits. shiftRightSigned32 Shift a 32-bit word right by a given number of bits, preserving the sign bit. shiftRightSigned64 Shift a 64-bit word right by a given number of bits, preserving the sign bit. xor32 Bitwise logical “exclusive-or” of two 32-bit words. xor64 Bitwise logical “exclusive-or” of two 64-bit words. xorBytes Bitwise logical “exclusive-or” of two Bytes buffers.\n\n### and32\n\n DECLARE NATIVE FUNCTION and32( x, y: Number ): Number\n\nBitwise logical “and” of two 32-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 0 \|\n---+---+---+\n1 \| 0 \| 1 \|\n---+---+---+\n^\ny```\n\n### and64\n\n DECLARE NATIVE FUNCTION and64( x, y: Number ): Number\n\nBitwise logical “and” of two 64-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 0 \|\n---+---+---+\n1 \| 0 \| 1 \|\n---+---+---+\n^\ny```\n\n### andBytes\n\n DECLARE NATIVE FUNCTION andBytes( x, y: Bytes ): Bytes\n\nBitwise logical “and” of two Bytes buffers. The size of both input parameters must be the same.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 0 \|\n---+---+---+\n1 \| 0 \| 1 \|\n---+---+---+\n^\ny```\n\n### extract32\n\n DECLARE NATIVE FUNCTION extract32( x, n, w: Number ): Number\n\nExtract a range of bits from a 32-bit word.\n\n#### Parameters\n\n x number to extract bits from n bit position of the lowest order bit to extract w width of bit range to extract\n\n#### Description\n\nThis function extracts a given range of bits. The bits in an word are numbered starting at 0 for the least significant bit.\n\nThe function call\n\n`binary.extract(x, 5, 3)`\n\nwill extract the following bits from the word\n\n```31 28 24 20 16 12 8 4 0\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \|\n+-----+```\n\n### extract64\n\n DECLARE NATIVE FUNCTION extract64( x, n, w: Number ): Number\n\nExtract a range of bits from a 64-bit word.\n\n#### Parameters\n\n x number to extract bits from n bit position of the lowest order bit to extract w width of bit range to extract\n\n#### Description\n\nThis function extracts a given range of bits. The bits in an word are numbered starting at 0 for the least significant bit.\n\nThe function call\n\n`binary.extract(x, 5, 3)`\n\nwill extract the following bits from the word\n\n```31 28 24 20 16 12 8 4 0\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \|\n+-----+```\n\n### get32\n\n DECLARE NATIVE FUNCTION get32( x, n: Number ): Boolean\n\nGet a specific bit from a 32-bit word.\n\n#### Parameters\n\n x number to extract the bit from n bit position of the bit to extract\n\n#### Description\n\nThis function returns a Boolean value indicting whether the indicated bit is set (TRUE) or not (FALSE).\n\n### get64\n\n DECLARE NATIVE FUNCTION get64( x, n: Number ): Boolean\n\nGet a specific bit from a 64-bit word.\n\n#### Parameters\n\n x number to extract the bit from n bit position of the bit to extract\n\n#### Description\n\nThis function returns a Boolean value indicting whether the indicated bit is set (TRUE) or not (FALSE).\n\n### not32\n\n DECLARE NATIVE FUNCTION not32( x: Number ): Number\n\nInvert all bits in a 32-bit word.\n\n### not64\n\n DECLARE NATIVE FUNCTION not64( x: Number ): Number\n\nInvert all bits in a 64-bit word.\n\n### notBytes\n\n DECLARE NATIVE FUNCTION notBytes( x: Bytes ): Bytes\n\nInvert all bits in a Bytes buffer.\n\n### or32\n\n DECLARE NATIVE FUNCTION or32( x, y: Number ): Number\n\nBitwise logical “or” of two 32-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 1 \|\n---+---+---+\n^\ny```\n\n### or64\n\n DECLARE NATIVE FUNCTION or64( x, y: Number ): Number\n\nBitwise logical “or” of two 64-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 1 \|\n---+---+---+\n^\ny```\n\n### orBytes\n\n DECLARE NATIVE FUNCTION orBytes( x, y: Bytes ): Bytes\n\nBitwise logical “or” of two Bytes buffers. The size of both input parameters must be the same.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 1 \|\n---+---+---+\n^\ny```\n\n### replace32\n\n DECLARE NATIVE FUNCTION replace32( x, n, w, y: Number ): Number\n\nReplace a range of bits in a 32-bit word.\n\n#### Parameters\n\n x number to replace bits into n bit position of the lowest order bit to replace w width of bit range to replace y new bits to replace\n\n#### Description\n\nThis function replaces a given range of bits with new bits. The bits in an word are numbered starting at 0 for the least significant bit.\n\nThe function call\n\n`binary.replace(x, 5, 3, 7)`\n\nwill replace the following bits in the word\n\n```31 28 24 20 16 12 8 4 0\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \|\n+-----+```\n\nEach of the bits numbered 5, 6, and 7 would be set to 1 (from the input value 7).\n\n### replace64\n\n DECLARE NATIVE FUNCTION replace64( x, n, w, y: Number ): Number\n\nReplace a range of bits in a 64-bit word.\n\n#### Parameters\n\n x number to replace bits into n bit position of the lowest order bit to replace w width of bit range to replace y new bits to replace\n\n#### Description\n\nThis function replaces a given range of bits with new bits. The bits in an word are numbered starting at 0 for the least significant bit.\n\nThe function call\n\n`binary.replace(x, 5, 3, 7)`\n\nwill replace the following bits in the word\n\n```31 28 24 20 16 12 8 4 0\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \|\n+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+\n\| \|\n+-----+```\n\nEach of the bits numbered 5, 6, and 7 would be set to 1 (from the input value 7).\n\n### set32\n\n DECLARE NATIVE FUNCTION set32( x, n: Number, v: Boolean ): Number\n\nSet a specific bit in a 32-bit word.\n\n#### Parameters\n\n x number to set the bit n bit position of the bit to set v bit value to set, 0 (FALSE) or 1 (TRUE)\n\n#### Description\n\nThis function returns a new value with the given bit set according to the parameter v.\n\n### set64\n\n DECLARE NATIVE FUNCTION set64( x, n: Number, v: Boolean ): Number\n\nSet a specific bit in a 64-bit word.\n\n#### Parameters\n\n x number to set the bit n bit position of the bit to set v bit value to set, 0 (FALSE) or 1 (TRUE)\n\n#### Description\n\nThis function returns a new value with the given bit set according to the parameter v.\n\n### shiftLeft32\n\n DECLARE NATIVE FUNCTION shiftLeft32( x, n: Number ): Number\n\nShift a 32-bit word left by a given number of bits.\n\n#### Parameters\n\n x word to shift n number of bits to shift left\n\n#### Description\n\nNew bits shifted into the right hand side of the word are 0.\n\n### shiftLeft64\n\n DECLARE NATIVE FUNCTION shiftLeft64( x, n: Number ): Number\n\nShift a 64-bit word left by a given number of bits.\n\n#### Parameters\n\n x word to shift n number of bits to shift left\n\n#### Description\n\nNew bits shifted into the right hand side of the word are 0.\n\n### shiftRight32\n\n DECLARE NATIVE FUNCTION shiftRight32( x, n: Number ): Number\n\nShift a 32-bit word right by a given number of bits.\n\n#### Parameters\n\n x word to shift n number of bits to shift right\n\n#### Description\n\nNew bits shifted into the left hand side of the word are 0.\n\n### shiftRight64\n\n DECLARE NATIVE FUNCTION shiftRight64( x, n: Number ): Number\n\nShift a 64-bit word right by a given number of bits.\n\n#### Parameters\n\n x word to shift n number of bits to shift right\n\n#### Description\n\nNew bits shifted into the left hand side of the word are 0.\n\n### shiftRightSigned32\n\n DECLARE NATIVE FUNCTION shiftRightSigned32( x, n: Number ): Number\n\nShift a 32-bit word right by a given number of bits, preserving the sign bit.\n\n#### Parameters\n\n x word to shift n number of bits to shift right\n\n#### Description\n\nNew bits shifted into the left hand side of the word are the same as the original leftmost bit.\n\n### shiftRightSigned64\n\n DECLARE NATIVE FUNCTION shiftRightSigned64( x, n: Number ): Number\n\nShift a 64-bit word right by a given number of bits, preserving the sign bit.\n\n#### Parameters\n\n x word to shift n number of bits to shift right\n\n#### Description\n\nNew bits shifted into the left hand side of the word are the same as the original leftmost bit.\n\n### xor32\n\n DECLARE NATIVE FUNCTION xor32( x, y: Number ): Number\n\nBitwise logical “exclusive-or” of two 32-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 0 \|\n---+---+---+\n^\ny```\n\n### xor64\n\n DECLARE NATIVE FUNCTION xor64( x, y: Number ): Number\n\nBitwise logical “exclusive-or” of two 64-bit words.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 0 \|\n---+---+---+\n^\ny```\n\n### xorBytes\n\n DECLARE NATIVE FUNCTION xorBytes( x, y: Bytes ): Bytes\n\nBitwise logical “exclusive-or” of two Bytes buffers. The size of both input parameters must be the same.\n\nFor each bit in the inputs, the following truth table determines the output bit:\n\n``` \| 0 \| 1 \| <- x\n---+---+---+\n0 \| 0 \| 1 \|\n---+---+---+\n1 \| 1 \| 0 \|\n---+---+---+\n^\ny```\n DECLARE NATIVE FUNCTION and32( x, y: Number ): Number\nBitwise logical “and” of two 32-bit words.\n DECLARE NATIVE FUNCTION and64( x, y: Number ): Number\nBitwise logical “and” of two 64-bit words.\n DECLARE NATIVE FUNCTION andBytes( x, y: Bytes ): Bytes\nBitwise logical “and” of two Bytes buffers.\n DECLARE NATIVE FUNCTION extract32( x, n, w: Number ): Number\nExtract a range of bits from a 32-bit word.\n DECLARE NATIVE FUNCTION extract64( x, n, w: Number ): Number\nExtract a range of bits from a 64-bit word.\n DECLARE NATIVE FUNCTION get32( x, n: Number ): Boolean\nGet a specific bit from a 32-bit word.\n DECLARE NATIVE FUNCTION get64( x, n: Number ): Boolean\nGet a specific bit from a 64-bit word.\n DECLARE NATIVE FUNCTION not32( x: Number ): Number\nInvert all bits in a 32-bit word.\n DECLARE NATIVE FUNCTION not64( x: Number ): Number\nInvert all bits in a 64-bit word.\n DECLARE NATIVE FUNCTION notBytes( x: Bytes ): Bytes\nInvert all bits in a Bytes buffer.\n DECLARE NATIVE FUNCTION or32( x, y: Number ): Number\nBitwise logical “or” of two 32-bit words.\n DECLARE NATIVE FUNCTION or64( x, y: Number ): Number\nBitwise logical “or” of two 64-bit words.\n DECLARE NATIVE FUNCTION orBytes( x, y: Bytes ): Bytes\nBitwise logical “or” of two Bytes buffers.\n DECLARE NATIVE FUNCTION replace32( x, n, w, y: Number ): Number\nReplace a range of bits in a 32-bit word.\n DECLARE NATIVE FUNCTION replace64( x, n, w, y: Number ): Number\nReplace a range of bits in a 64-bit word.\n DECLARE NATIVE FUNCTION set32( x, n: Number, v: Boolean ): Number\nSet a specific bit in a 32-bit word.\n DECLARE NATIVE FUNCTION set64( x, n: Number, v: Boolean ): Number\nSet a specific bit in a 64-bit word.\n DECLARE NATIVE FUNCTION shiftLeft32( x, n: Number ): Number\nShift a 32-bit word left by a given number of bits.\n DECLARE NATIVE FUNCTION shiftLeft64( x, n: Number ): Number\nShift a 64-bit word left by a given number of bits.\n DECLARE NATIVE FUNCTION shiftRight32( x, n: Number ): Number\nShift a 32-bit word right by a given number of bits.\n DECLARE NATIVE FUNCTION shiftRight64( x, n: Number ): Number\nShift a 64-bit word right by a given number of bits.\n DECLARE NATIVE FUNCTION shiftRightSigned32( x, n: Number ): Number\nShift a 32-bit word right by a given number of bits, preserving the sign bit.\n DECLARE NATIVE FUNCTION shiftRightSigned64( x, n: Number ): Number\nShift a 64-bit word right by a given number of bits, preserving the sign bit.\n DECLARE NATIVE FUNCTION xor32( x, y: Number ): Number\nBitwise logical “exclusive-or” of two 32-bit words.\n DECLARE NATIVE FUNCTION xor64( x, y: Number ): Number\nBitwise logical “exclusive-or” of two 64-bit words.\n DECLARE NATIVE FUNCTION xorBytes( x, y: Bytes ): Bytes\nBitwise logical “exclusive-or” of two Bytes buffers." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5753112,"math_prob":0.9860071,"size":9373,"snap":"2020-24-2020-29","text_gpt3_token_len":2615,"char_repetition_ratio":0.23844594,"word_repetition_ratio":0.8153127,"special_character_ratio":0.3640243,"punctuation_ratio":0.13481894,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9946891,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-07-09T04:28:21Z\",\"WARC-Record-ID\":\"<urn:uuid:edf344dc-b84d-4496-821d-db2d8bb9affd>\",\"Content-Length\":\"62725\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:94a81dfd-be35-452c-aa79-19f3459a0cce>\",\"WARC-Concurrent-To\":\"<urn:uuid:954f029c-5cf9-4407-b1da-e94864bf8de1>\",\"WARC-IP-Address\":\"185.199.111.153\",\"WARC-Target-URI\":\"https://neon-lang.dev/html/files/binary-neon.html\",\"WARC-Payload-Digest\":\"sha1:DW4TI6RFXMBUD4ZAZIS4PE7KPPNQ57TP\",\"WARC-Block-Digest\":\"sha1:42GRX27VSTDD524IEEOGXYZOMJP3JCLX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-29/CC-MAIN-2020-29_segments_1593655898347.42_warc_CC-MAIN-20200709034306-20200709064306-00200.warc.gz\"}"}
https://www.economicsdiscussion.net/demand-curve/single-firms-demand-curve-for-capital/20670	[ "Keynes defined the marginal efficiency of capital (MEC) as the rate of discount, e, that will just equate the present value of the flow of receipts a capital good (such as a machine) generates equal to its purchase price.\n\nIf the future cash flow or monetary return that can be obtained (expected) from a piece of capital is assumed to remain uniform throughout we can calculate the MEC by applying the following simple formula:\n\nC = R/e\n\nwhere C is the purchase price of the piece of capital, R is its constant flow of gross returns and e is the unknown value of MEC.\n\nNow we can easily solve for e by using this formula:\n\ne = R/C\n\nIf we assume that a firm is having a large number of capital equipment and is making a decision, at the margin, to acquire one extra unit of capital, we may call the e associated with this marginal addition to the firm’s stock of capital the MEC.\n\nHere e is the efficiency of any unit of capital, and MEC is the efficiency of the marginal unit. What is true of a firm is equally true of society. So we can refer to MEC for the whole economy.\n\nWe have noted earlier that the value to the firm now of a flow of future cash flows is its present value while the purchase price is the cost to the firm now of the capital that produces that flow of receipts. While the former is the addition to revenue made by an extra cost of acquiring the piece of capital the latter is the extra cost of acquiring the piece of capital.\n\nAccording to the famous marginal productivity theory of income distribution a profit-maximizing firm will go on adding units of capital so long as its marginal revenue product (MRP) exceeds it cost. Thus a firm will go on adding to its existing stock of capital as long as the present value of the flow of receipts from the marginal units exceeds the purchase price.\n\nThe value of e is the rate of discount which makes the two (R) and (C) equal. But we calculate the present value on the basis of the market rate of interest (r). (This is exactly what we have done earlier in this section).\n\nThus, when MEC is exactly equal to r, the present value of future cash flows from the marginal unit of capital equals its current purchase price (or acquisition cost). When this happens the firm’s capital stock is said to be in equilibrium and firm would no longer consider it worthwhile to add another unit of capital.\n\nAn interesting question that arises at this stage is the following: How does the firm take decision regarding whether or not to acquire (buy) one extra unit of capital?\n\nThe usual method used by modern business firms is to use the market rate of interest to arrive at the present value of the gross return. Then, in the next step, it has to be compared with the purchase price of the piece of equipment. A simple example, may make the point clear.\n\nLet us suppose that a firm can buy a machine for Rs. 8,000. It is expected to yield Rs. 1,000 per annum, net of all non-capital costs, for an indefinite period. Also suppose that the market rate of interest at which it can borrow (and lend) money is 10 per cent. Thus, the present value of the future cash flow generated by the machine (i.e., the capitalized value of the machine) is R/r = Rs. 1000/0.10= Rs. 10,000.\n\nOn the other hand the present value of Rs. 8,000 today is (obviously) Rs. 8,000. The firm can make a gain if it purchases the machine for Rs. 8,000 to derive an income of Rs. 10,000. We may alternatively suppose that the firm is left with two options—either to buy the machine or lend the money (Rs. 8,000) at 10 per cent interest.\n\nThe firm would prefer to buy the machine because, by doing do, it can get an income of Rs. 1000. On the other hand if it lends the money at 10 per cent interest, it gets, at the end of a year, a sum of Rs. 800 as interest.\n\nThus, it is quite clear that the first option is better than the second one.\n\nSo as a general rule a profit- maximizing firm will purchase a capital good like a machine if the following condition holds:\n\nR/r > C\n\nwhere R is the infinite stream of future (gross) cash flows. The term R/r is the present value of the stream of gross cash flows generated by the capital good, i.e., the capitalized value of the machine.\n\nIt will, therefore, be worthwhile, on the part of a profit-maximizing firm, to purchase a capital good as and when the present value of its future stream of gross returns exceeds the purchase price of the good.\n\nThe above inequality may also be expressed in the an alternative way:\n\nR/C > r\n\nBut R/C is the MEC. So we obtain the following rule of investment decision at the micro level, i.e., at the level of an individual firm: it is always profitable for a firm to acquire extra units of capital whenever its MEC exceeds the market rate of interest.\n\nLooked at from this angle, MEC measures the return on an extra (marginal unit of capital to the firm, while r is a measure of the opportunity cost of capital (always assuming that the firm can borrow and lend at the existing market rate of interest, which is 10 per cent in our example).\n\nA firm’s MEC curve is shown is Fig. 32.5. It is a schedule that relates the efficiency of each additional rupee’s worth of capital to the size of the capital stock. On the horizontal axis we measure the value of the firm’s capital stock. On the vertical axis we measure the money value of the product of capital, expressed in terms of percentage yields on that capital.\n\nThe MEC curve is downward sloping, as stated earlier, due to the operation of the law of diminishing returns. .Thus, in Fig. 32.5 when the stock of capital is k1 the MEC is r1. When the stock of capital increases to k2 the MEC falls to r2 and so on.\n\nIn fact, the MEC curve is known as the firm’s demand curve for capital. It shows the amount of capital that a firm would choose to employ at different costs of employing it (i.e., at different rates of interest). The MEC curve is indeed the demand curve for capital of a firm because it shows how much capital the firm will demand at different prices, or yields, of capital.\n\nWe may note here that the cost of capital is the market rate of interest, i.e., the interest that the firm has to pay to borrow funds to acquire capital (or, for a firm that has generated funds internally, the cost of capital is the rate of interest that could be earned by lending them elsewhere).\n\nKeynes pointed out that so long as the MEC exceeds the rate of interest, net return from a project will be positive. So the firm will employ more capital. But as MEC gradually falls, it will ultimately be equal to r and the firm’s stock of capital will be optimum. This point may also be illustrated diagrammatically.\n\nSuppose in Fig. 32.5 the rate of interest (the cost of capital to the firm) is r1. Now the firm will certainly employ all capital upto k1 because the MEC is greater than the rate of interest. At the same time, it will not be profitable for the firm to employ, say, units of capital because the MEC of the k2-th unit of capital is less than the rate of interest.\n\nIn a like manner, it will not pay the firm to employ any capital in excess of k1. Hence k1 represents the optimum stock of capital of the firm, when the rate of interest is r1. Thus the equilibrium amount of capital of a profit-making firm is that at which the MEC is exactly equal to the rate of interest.\n\nThere is another way of establishing the fundamental principle stated above. In our previous example, a machine producing children’s toys was supposed to generate a yield of 10 per cent. If the market rate of interest is 10 per cent or less, it will be profitable to buy the machine: If the rate of interest is more than 10 per cent, it will not be profitable to buy or install it." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.944666,"math_prob":0.97365594,"size":8043,"snap":"2023-40-2023-50","text_gpt3_token_len":1826,"char_repetition_ratio":0.18173902,"word_repetition_ratio":0.032150034,"special_character_ratio":0.22553773,"punctuation_ratio":0.09882353,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9952561,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-29T08:43:52Z\",\"WARC-Record-ID\":\"<urn:uuid:9c76bb31-40fb-4e3e-96b8-9a0aeb4908f1>\",\"Content-Length\":\"100691\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6cb3394a-529f-49aa-8eb2-52504d7e6e04>\",\"WARC-Concurrent-To\":\"<urn:uuid:4b6db450-2e16-4a99-8790-8b99f8dbf4e5>\",\"WARC-IP-Address\":\"104.26.2.203\",\"WARC-Target-URI\":\"https://www.economicsdiscussion.net/demand-curve/single-firms-demand-curve-for-capital/20670\",\"WARC-Payload-Digest\":\"sha1:QK5LH4UOEIRJBDGC6ESXMJ5QA3OOVXF2\",\"WARC-Block-Digest\":\"sha1:SCUAE3TCHQRZIQ6FLJCLYIOK4UX5Q2QO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100057.69_warc_CC-MAIN-20231129073519-20231129103519-00642.warc.gz\"}"}
https://www.colorhexa.com/59f1dc	[ "# #59f1dc Color Information\n\nIn a RGB color space, hex #59f1dc is composed of 34.9% red, 94.5% green and 86.3% blue. Whereas in a CMYK color space, it is composed of 63.1% cyan, 0% magenta, 8.7% yellow and 5.5% black. It has a hue angle of 171.7 degrees, a saturation of 84.4% and a lightness of 64.7%. #59f1dc color hex could be obtained by blending #b2ffff with #00e3b9. Closest websafe color is: #66ffcc.\n\n• R 35\n• G 95\n• B 86\nRGB color chart\n• C 63\n• M 0\n• Y 9\n• K 5\nCMYK color chart\n\n#59f1dc color description : Soft cyan.\n\n# #59f1dc Color Conversion\n\nThe hexadecimal color #59f1dc has RGB values of R:89, G:241, B:220 and CMYK values of C:0.63, M:0, Y:0.09, K:0.05. Its decimal value is 5894620.\n\nHex triplet RGB Decimal 59f1dc `#59f1dc` 89, 241, 220 `rgb(89,241,220)` 34.9, 94.5, 86.3 `rgb(34.9%,94.5%,86.3%)` 63, 0, 9, 5 171.7°, 84.4, 64.7 `hsl(171.7,84.4%,64.7%)` 171.7°, 63.1, 94.5 66ffcc `#66ffcc`\nCIE-LAB 87.094, -44.849, -1.74 48.489, 70.198, 78.7 0.246, 0.356, 70.198 87.094, 44.883, 182.222 87.094, -59.816, 4.531 83.784, -43.317, 2.956 01011001, 11110001, 11011100\n\n# Color Schemes with #59f1dc\n\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #f1596e\n``#f1596e` `rgb(241,89,110)``\nComplementary Color\n• #59f190\n``#59f190` `rgb(89,241,144)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #59baf1\n``#59baf1` `rgb(89,186,241)``\nAnalogous Color\n• #f19059\n``#f19059` `rgb(241,144,89)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #f159ba\n``#f159ba` `rgb(241,89,186)``\nSplit Complementary Color\n• #f1dc59\n``#f1dc59` `rgb(241,220,89)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #dc59f1\n``#dc59f1` `rgb(220,89,241)``\n• #6ef159\n``#6ef159` `rgb(110,241,89)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #dc59f1\n``#dc59f1` `rgb(220,89,241)``\n• #f1596e\n``#f1596e` `rgb(241,89,110)``\n• #14eacc\n``#14eacc` `rgb(20,234,204)``\n• #2aedd2\n``#2aedd2` `rgb(42,237,210)``\n• #41efd7\n``#41efd7` `rgb(65,239,215)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #71f3e1\n``#71f3e1` `rgb(113,243,225)``\n• #88f5e6\n``#88f5e6` `rgb(136,245,230)``\n• #a0f7eb\n``#a0f7eb` `rgb(160,247,235)``\nMonochromatic Color\n\n# Alternatives to #59f1dc\n\nBelow, you can see some colors close to #59f1dc. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #59f1b6\n``#59f1b6` `rgb(89,241,182)``\n• #59f1c3\n``#59f1c3` `rgb(89,241,195)``\n• #59f1cf\n``#59f1cf` `rgb(89,241,207)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #59f1e9\n``#59f1e9` `rgb(89,241,233)``\n• #59edf1\n``#59edf1` `rgb(89,237,241)``\n• #59e0f1\n``#59e0f1` `rgb(89,224,241)``\nSimilar Colors\n\n# #59f1dc Preview\n\nThis text has a font color of #59f1dc.\n\n``<span style=\"color:#59f1dc;\">Text here</span>``\n#59f1dc background color\n\nThis paragraph has a background color of #59f1dc.\n\n``<p style=\"background-color:#59f1dc;\">Content here</p>``\n#59f1dc border color\n\nThis element has a border color of #59f1dc.\n\n``<div style=\"border:1px solid #59f1dc;\">Content here</div>``\nCSS codes\n``.text {color:#59f1dc;}``\n``.background {background-color:#59f1dc;}``\n``.border {border:1px solid #59f1dc;}``\n\n# Shades and Tints of #59f1dc\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010f0d is the darkest color, while #fcfffe is the lightest one.\n\n• #010f0d\n``#010f0d` `rgb(1,15,13)``\n• #03211d\n``#03211d` `rgb(3,33,29)``\n• #04332d\n``#04332d` `rgb(4,51,45)``\n• #06453c\n``#06453c` `rgb(6,69,60)``\n• #07574c\n``#07574c` `rgb(7,87,76)``\n• #09695c\n``#09695c` `rgb(9,105,92)``\n• #0a7b6c\n``#0a7b6c` `rgb(10,123,108)``\n• #0c8e7c\n``#0c8e7c` `rgb(12,142,124)``\n• #0da08b\n``#0da08b` `rgb(13,160,139)``\n• #0fb29b\n``#0fb29b` `rgb(15,178,155)``\n• #11c4ab\n``#11c4ab` `rgb(17,196,171)``\n• #12d6bb\n``#12d6bb` `rgb(18,214,187)``\n• #14e8cb\n``#14e8cb` `rgb(20,232,203)``\n• #23ecd1\n``#23ecd1` `rgb(35,236,209)``\n• #35eed4\n``#35eed4` `rgb(53,238,212)``\n• #47efd8\n``#47efd8` `rgb(71,239,216)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #6bf3e0\n``#6bf3e0` `rgb(107,243,224)``\n• #7df4e4\n``#7df4e4` `rgb(125,244,228)``\n• #8ff6e7\n``#8ff6e7` `rgb(143,246,231)``\n• #a1f7eb\n``#a1f7eb` `rgb(161,247,235)``\n• #b3f9ef\n``#b3f9ef` `rgb(179,249,239)``\n• #c6faf3\n``#c6faf3` `rgb(198,250,243)``\n• #d8fcf7\n``#d8fcf7` `rgb(216,252,247)``\n• #eafdfb\n``#eafdfb` `rgb(234,253,251)``\n• #fcfffe\n``#fcfffe` `rgb(252,255,254)``\nTint Color Variation\n\n# Tones of #59f1dc\n\nA tone is produced by adding gray to any pure hue. In this case, #9eacaa is the less saturated color, while #4bffe6 is the most saturated one.\n\n• #9eacaa\n``#9eacaa` `rgb(158,172,170)``\n• #97b3af\n``#97b3af` `rgb(151,179,175)``\n• #90bab4\n``#90bab4` `rgb(144,186,180)``\n• #89c1b9\n``#89c1b9` `rgb(137,193,185)``\n• #83c7be\n``#83c7be` `rgb(131,199,190)``\n• #7ccec3\n``#7ccec3` `rgb(124,206,195)``\n• #75d5c8\n``#75d5c8` `rgb(117,213,200)``\n• #6edccd\n``#6edccd` `rgb(110,220,205)``\n• #67e3d2\n``#67e3d2` `rgb(103,227,210)``\n``#60ead7` `rgb(96,234,215)``\n• #59f1dc\n``#59f1dc` `rgb(89,241,220)``\n• #52f8e1\n``#52f8e1` `rgb(82,248,225)``\n• #4bffe6\n``#4bffe6` `rgb(75,255,230)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #59f1dc is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5296015,"math_prob":0.6727387,"size":3711,"snap":"2021-43-2021-49","text_gpt3_token_len":1703,"char_repetition_ratio":0.12489884,"word_repetition_ratio":0.011111111,"special_character_ratio":0.53058475,"punctuation_ratio":0.23809524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9812372,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T15:03:07Z\",\"WARC-Record-ID\":\"<urn:uuid:a696cd22-76ed-4f8e-897d-6dc0070be04b>\",\"Content-Length\":\"36203\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5764220f-28bd-4f99-891f-45da2cdb5c09>\",\"WARC-Concurrent-To\":\"<urn:uuid:a662d2c0-2e12-4f2b-b561-07b114b7b0cf>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/59f1dc\",\"WARC-Payload-Digest\":\"sha1:VP6O263H2BG2YIU3XJXCKNXIO7MMRTCW\",\"WARC-Block-Digest\":\"sha1:XM6Y7UWIKSC2FTDOMSAGDRJDO5KD3UBQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363515.28_warc_CC-MAIN-20211208144647-20211208174647-00264.warc.gz\"}"}
https://www.jiskha.com/questions/1674867/1-how-fast-does-the-radius-of-a-spherical-soap-bubble-change-when-you-blow-air-into-it-at	[ "# 4 Calculus Related-Rates Problems\n\n1. How fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dV/dt, the change in volume with respect to time, which is 15 cubic centimeters per second. The rate we want to find is dr/dt, the change in the radius with respect to time. Remember that the volume of a sphere is V=4/3 pi r^3.\n\n2. A baseball diamond is a square 90 feet on a side. A player runs from first base to second base at a rate of 15 feet per second. At what rate is the player's distance from third base decreasing when the player is halfway between first and second base? We've already set part of this problem up. If we let x be the distance between the player and second base,and y be the distance between the player and third base, then dx/dt=-15 feet per second, and dy/dt will tell us what we want to know. Use the picture to find a relationship that will help you answer the question.\n\n3. A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate is the tip of his shadow moving? We've already set this up part of the way. We know that dx/dt=-2 meters per second, and we're looking for dv/dt . Use the picture to help you find the relationship between x and y, and use it to answer the question asked here.\n\n4. Here's one we haven't worked with before: A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = pi(r^2)h, where r is the radius and h is the height of this cylinder.)\n\n1. 👍\n2. 👎\n3. 👁\n1. These are all just applications of the chain rule. If y is a function of u, and u is a function of t, then\ndy/dt = dy/du * du/dt\n\nSo, now we see what do do:\n#1\nv=4/3 π r^3\ndv/dt = 4πr^2 dr/dt\ndr/dt = 15/(4πr^2)\nAs expected, if the volume changes at a steady rate, the bigger the balloon gets, the slower the radius increases.\n\n#2\nUsing the Pythagorean Theorem, we know that\n\nx^2 + 90^2 = y^2\nx dx/dt = y dy/dt\nSo, find y when x=45, and then just plug in dx/dt = -15\n\n#3\nYou don't say what y is, but suppose y is the distance of the tip of the shadow from the man. In that case, using similar triangles,\n\ny/2 = (x+y)/5\ny = 2/3 x\nSo,\ndy/dt = 2/3 dx/dt\n\n#4\n1st way: use the product rule\nv = πr^2h\ndv/dt = 2πrh dr/dt + πr^2 dh/dt\nYou know that dv/dt=0, since the volumem of oil is constant.\nNow, what is h? πr^2h = 1, so h = 1/(π.01^2)\nNow you can plug in the numbers.\n\n2nd way: use the fact that v is constant\nπr^2h = 1, so\nπh = 1/r^2\nπ dh/dt = -2/r^3 dr/dt\nnow just plug in your numbers to find dh/dt\n\ndon't forget to watch the units.\n1m^3 = 10^6 cm^3\n\n1. 👍\n2. 👎\n2. Steve is tha goat\n\n1. 👍\n2. 👎\n3. Thanks bestie Steve! Hope ur thriving bc im not! <3\n\n1. 👍\n2. 👎\n\n## Similar Questions\n\n1. ### Calculus\n\na spherical balloon is inflated with gas at the rate of 500 cubic centimeters per minute. how fast is the radius of the balloon increasing at the instant the radius is 30 centimeters?\n\n2. ### Calculus\n\nAir is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm? V= 4/3pi*r^3 S= 4 pi r^2\n\n3. ### calculus\n\nA spherical balloon is losing air at the rate of 2 cubic inches per minute. How fast is the radius of the ballon shrinking when the radius is 8 inches.\n\n4. ### math-calculus\n\nA spherical party balloon is being inflated with helium pumped in at a rate of 12 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 1 ft?\n\n1. ### physics\n\nA 1.6 air bubble is released from the sandy bottom of a warm, shallow sea, where the gauge pressure is 1.6 . The bubble rises slowly enough that the air inside remains at the same constant temperature as the water. What is the\n\n2. ### Calculus: need clarification to where the #'s go\n\nAir is being pumped into a spherical balloon so that its volume increases at a rate of 80 \\mbox{cm}^3\\mbox{/s}. How fast is the surface area of the balloon increasing when its radius is 14 \\mbox{cm}? Recall that a ball of radius r\n\n3. ### Math( Can you help me with starting this problem)\n\nHow fast does the radius of a spherical soap bubble change when you blow air into it at the rate of 15 cubic centimeters per second? Our known rate is dv/dt , the change in volume with respect to time, which is 15 cubic\n\n4. ### calculus\n\nThe radius of a spherical balloon is increasing at a rate of 2 centimeters per minute. How fast is the volume changing when the radius is 8 centimeters? Note: The volume of a sphere is given by 4(pi)r^3\n\n1. ### calc\n\nAir is being pumped into a spherical balloon so that its volume increases at a rate of 100 {cm}^3}. How fast is the surface area of the balloon increasing when its radius is 16{cm}?\n\n2. ### physics\n\nCalculate the gauge pressure inside a soap bubble 0.0002m in radius using the surface tension for soapy water\n\n3. ### algebra\n\nSara goes on a slingshot ride in an amusement park. She is strapped into a spherical ball that has a radius of centimeters. What is the volume of air in the spherical ball?\n\n4. ### science\n\n1) In a syringe, what happens to the air paricles when you push the plunger? 2) In a syring, what happens the the air particles in the bubble from the bubble wrap when you pull up the plunger? 3) Are there more air particles in" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9098016,"math_prob":0.9906278,"size":2711,"snap":"2021-31-2021-39","text_gpt3_token_len":704,"char_repetition_ratio":0.15256742,"word_repetition_ratio":0.17567568,"special_character_ratio":0.24640355,"punctuation_ratio":0.06284658,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99894434,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-30T16:26:02Z\",\"WARC-Record-ID\":\"<urn:uuid:79cdff40-078e-4e36-a5ad-0747dab03b62>\",\"Content-Length\":\"27390\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7c71fad-0318-43c0-9af0-333142717158>\",\"WARC-Concurrent-To\":\"<urn:uuid:7cfb1a94-0fe2-49f4-acaa-ff99c1d9d52a>\",\"WARC-IP-Address\":\"66.228.55.50\",\"WARC-Target-URI\":\"https://www.jiskha.com/questions/1674867/1-how-fast-does-the-radius-of-a-spherical-soap-bubble-change-when-you-blow-air-into-it-at\",\"WARC-Payload-Digest\":\"sha1:EKG4ECNXAKLRDDD47IHH6ZRSKEQDCFL4\",\"WARC-Block-Digest\":\"sha1:TWVPQ74FP22NWY5CF7R6S27KEP5BCJYU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153971.20_warc_CC-MAIN-20210730154005-20210730184005-00309.warc.gz\"}"}
http://discountgiftcards.club/draw-log-math/	[ "# Draw Log Math", null, "Draw log math dot plot of data of figure 2 shown on a log scale with base of math games online.", null, "Draw log math ordinary axis versus semi log axis mathematics definition.", null, "Draw log math learning logs mathies skyrim.", null, "Draw log math above is the graph of log x base is e all function graphs are made with the help of it now y log a y log x log e a y log math curriculum kindergarten.", null, "Draw log math to draw math graphs and more mathletics london ontario.", null, "Draw log math video thumbnail mathiesen.", null, "Draw log math linear graph mathies skyrim.", null, "Draw log math logarithm loop trick notes my interactive notebook pages math algebra and algebra 2 math prodigy toys.", null, "Draw log math students keep a calendar and weather log that uses reading math science and technology to collect and analyze data make comparisons observations and mathies money.", null, "Draw log math sign in to download full size image mathematics of data management.", null, "Draw log math log graphs math playground duck life.", null, "Draw log math dot plot of revenues of the top fortune companies data source magazines fortune fortune full list mathway app.", null, "Draw log math math prodigy play.", null, "Draw log math semi log graph math playground time games.", null, "Draw log math draw math equations and get the latex code for it thanks mathies tools." ]	[ null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-dot-plot-of-data-of-figure-2-shown-on-a-log-scale-with-base-of-math-games-online.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-ordinary-axis-versus-semi-log-axis-mathematics-definition.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-learning-logs-mathies-skyrim.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-above-is-the-graph-of-log-x-base-is-e-all-function-graphs-are-made-with-the-help-of-it-now-y-log-a-y-log-x-log-e-a-y-log-math-curriculum-kindergarten.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-to-draw-math-graphs-and-more-mathletics-london-ontario.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-video-thumbnail-mathiesen.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-linear-graph-mathies-skyrim.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-logarithm-loop-trick-notes-my-interactive-notebook-pages-math-algebra-and-algebra-2-math-prodigy-toys.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-students-keep-a-calendar-and-weather-log-that-uses-reading-math-science-and-technology-to-collect-and-analyze-data-make-comparisons-observations-and-mathies-money.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-sign-in-to-download-full-size-image-mathematics-of-data-management.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-log-graphs-math-playground-duck-life.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-dot-plot-of-revenues-of-the-top-fortune-companies-data-source-magazines-fortune-fortune-full-list-mathway-app.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-math-prodigy-play.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-semi-log-graph-math-playground-time-games.jpg", null, "http://discountgiftcards.club/wp-content/uploads//2019/03/draw-log-math-draw-math-equations-and-get-the-latex-code-for-it-thanks-mathies-tools.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7662531,"math_prob":0.9146902,"size":1275,"snap":"2019-35-2019-39","text_gpt3_token_len":259,"char_repetition_ratio":0.23524784,"word_repetition_ratio":0.026785715,"special_character_ratio":0.19137254,"punctuation_ratio":0.061728396,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99946064,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T11:41:43Z\",\"WARC-Record-ID\":\"<urn:uuid:55226a5b-1a62-4e77-b797-611fac5969b3>\",\"Content-Length\":\"24368\",\"Content-Type\":\"application/http; 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https://socratic.org/questions/simplify-this-index-law-question-1#455314	[ "Simplify this index law question?\n\n${\\left(2 x {y}^{3}\\right)}^{2} / \\left(7 {x}^{3}\\right) \\times \\frac{3 {x}^{5} {y}^{2}}{4 y}$\n\nJul 23, 2017\n\n$= \\frac{3 {x}^{4} {y}^{7}}{7}$\n\nExplanation:\n\n${\\left(2 x {y}^{3}\\right)}^{2} / \\left(7 {x}^{3}\\right) \\cdot \\frac{3 {x}^{5} {y}^{2}}{4 y}$\n\nexpand\n\n$= \\frac{{2}^{2} {x}^{2} {\\left({y}^{3}\\right)}^{2} \\cdot \\left(3 {x}^{5} {y}^{2}\\right)}{7 {x}^{3} \\cdot 4 y}$\n\n$= \\frac{4 \\cdot {x}^{2} \\cdot {y}^{6} \\cdot 3 \\cdot {x}^{5} \\cdot {y}^{2}}{7 \\cdot {x}^{3} \\cdot 4 \\cdot y}$\n\n$= \\frac{12 \\cdot {x}^{2 + 5} \\cdot {y}^{6 + 2}}{28 \\cdot {x}^{3} \\cdot y}$\n\n$= \\frac{12 \\cdot {x}^{7 - 3} \\cdot {y}^{8 - 1}}{28}$\n\n$= \\frac{3 \\cdot {x}^{4} \\cdot {y}^{7}}{7}$\n\n$= \\frac{3 {x}^{4} {y}^{7}}{7}$" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57259524,"math_prob":1.0000097,"size":242,"snap":"2022-05-2022-21","text_gpt3_token_len":75,"char_repetition_ratio":0.12605043,"word_repetition_ratio":0.0,"special_character_ratio":0.30165288,"punctuation_ratio":0.06666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99998665,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T22:22:57Z\",\"WARC-Record-ID\":\"<urn:uuid:f2d91e43-8026-456e-add0-ac2655f2cf73>\",\"Content-Length\":\"31983\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e9ab2d7d-f3a2-4db1-9e9d-205c91911aa2>\",\"WARC-Concurrent-To\":\"<urn:uuid:6faefcbf-c5eb-416f-a5f1-1a1dc8190e50>\",\"WARC-IP-Address\":\"216.239.38.21\",\"WARC-Target-URI\":\"https://socratic.org/questions/simplify-this-index-law-question-1#455314\",\"WARC-Payload-Digest\":\"sha1:KWXKOOT3YX3R2SAJZR6QWXCFABMWDCTD\",\"WARC-Block-Digest\":\"sha1:IZ2VYRXDZKZ6QAIZIZZZXGB2VHS4QKSP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304686.15_warc_CC-MAIN-20220124220008-20220125010008-00391.warc.gz\"}"}
http://topcoder.bgcoder.com/print.php?id=183	[ "### Problem Statement\n\nThis problem involves an elaborate marble maze game. The maze is composed of choosers that look like this:\n``` Right Oriented Left Oriented\n\| \| \| \|\n\| \| \| \|\n\| \| \| \|\n/ \\ \\ / / \\\n/ \\ \\ / / \\\n/ \\ \\ / / \\\n/ O \\ / O \\\n/ / \\ \\ / / \\ \\\n/ / \\ \\ / / \\ \\ ```\nA marble is put in the top of the chooser. The bar in the middle of the chooser determines whether the marble will leave to the left or the right. After a marble passes through the chooser the bar moves. For example, if a marble enters the right oriented chooser pictured above, it would leave toward the right but, the next marble to pass through would leave toward the left. The exact opposite would hold true for the left oriented chooser pictured above. We can make a game by networking a bunch of choosers together using tubes. For example:\n\ngame = {\"L 1 2\",\"R 2 0\",\"L X X\"}\n\nThis means that chooser 0 begins oriented to the left. Its left path leads to chooser 1, and its right path leads to chooser 2. Chooser 1 begins oriented to the right. Its left path leads to chooser 2, and its right path leads to chooser 0. Chooser 2 begins oriented to the left. Its left and right path both leave the game. If I place a marble in chooser 0 it will pass through 4 choosers before leaving the game (namely 0 then 1 then 0 then 2). Given a network of choosers, and a chooser that we drop the marble into, determine how many choosers the marble will pass through before leaving the game. If it will never leave return -1.\n\nCreate a class Choosers that contains the method length, which takes a String[] game, and an int chooser, and returns an int representing how many choosers the marble will pass through before leaving the game. Return -1 if it will never leave.\n\n### Definition\n\n Class: Choosers Method: length Parameters: String[], int Returns: int Method signature: int length(String[] game, int start) (be sure your method is public)\n\n### Constraints\n\n-game will contain between 1 and 15 elements, inclusive.\n-Each element of game will be formatted as \"<dir> <left> <right>\" with no extra, leading or trailing spaces, or extra leading 0's.\n-<dir> is either 'L' or 'R'\n-<left> and <right> are each either the character 'X' or integers between 0 and the length of game - 1, inclusive.\n\n### Examples\n\n0)\n\n `{\"L 1 2\",\"R 2 0\",\"L X X\"}` `0`\n`Returns: 4`\n The marble goes from 0 to 1 to 0 to 2 and then leaves.\n1)\n\n `{\"L 1 2\",\"R 2 0\",\"L X X\"}` `2`\n`Returns: 1`\n The marble leaves immediately.\n2)\n\n `{\"L 0 0\"}` `0`\n`Returns: -1`\n Clearly, the marble never leaves.\n\n#### Problem url:\n\nhttp://www.topcoder.com/stat?c=problem_statement&pm=1125\n\n#### Problem stats url:\n\nhttp://www.topcoder.com/tc?module=ProblemDetail&rd=4375&pm=1125\n\nbrett1479\n\n#### Testers:\n\nalexcchan , lbackstrom\n\nSimulation" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9176009,"math_prob":0.80833364,"size":2103,"snap":"2019-51-2020-05","text_gpt3_token_len":542,"char_repetition_ratio":0.1676989,"word_repetition_ratio":0.16747573,"special_character_ratio":0.27960056,"punctuation_ratio":0.09701493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95061094,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T02:15:53Z\",\"WARC-Record-ID\":\"<urn:uuid:01ff2fd2-d75c-4894-9c27-7f3e31ce8128>\",\"Content-Length\":\"7631\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2e1c8199-4198-481a-abcd-2f90aed3d14f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0fdb1001-9713-4bb9-ada9-245efb09c2f9>\",\"WARC-IP-Address\":\"78.128.6.10\",\"WARC-Target-URI\":\"http://topcoder.bgcoder.com/print.php?id=183\",\"WARC-Payload-Digest\":\"sha1:RPDV3RTAUZV6ULEBBXPJKEBYOWUJOC2P\",\"WARC-Block-Digest\":\"sha1:S7L5CSRKXL6WUM6R53CTKXKT3DXBR7XY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250608062.57_warc_CC-MAIN-20200123011418-20200123040418-00555.warc.gz\"}"}
https://brainmass.com/math/basic-algebra/multiples-quadratic-equations-167033	[ "Explore BrainMass\nShare\n\n### Explore BrainMass\n\nThis content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!\n\nTwo quadratic equations that are multiples of each other are actually the same equation, and that they don't count as being \"more than one QE that have the same solution set\". However, if you plot the two equations on a graph, will they be different from each other, but have the same x-axis intercepts, which are the answer set?\nPlease explain, If they are different graphs, does that mean that they are really two different equations?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.96904886,"math_prob":0.9923105,"size":1164,"snap":"2019-51-2020-05","text_gpt3_token_len":245,"char_repetition_ratio":0.15517241,"word_repetition_ratio":0.5363129,"special_character_ratio":0.20360824,"punctuation_ratio":0.11818182,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99181545,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T04:43:56Z\",\"WARC-Record-ID\":\"<urn:uuid:c9cecdaa-dfac-45a5-98b9-11d6000623fe>\",\"Content-Length\":\"52126\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e7adb1cc-440e-4910-9010-34b1ef1284fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad311966-e8cf-47f9-a353-39a748f70393>\",\"WARC-IP-Address\":\"65.39.198.123\",\"WARC-Target-URI\":\"https://brainmass.com/math/basic-algebra/multiples-quadratic-equations-167033\",\"WARC-Payload-Digest\":\"sha1:EMRPKM6A44PDGHICKGCUCM2SJPWEKETR\",\"WARC-Block-Digest\":\"sha1:IYRJT36VRFTMXNPN345ZFRR2YZVYIQFU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251773463.72_warc_CC-MAIN-20200128030221-20200128060221-00366.warc.gz\"}"}
https://wohin-woher.com/jp/contents/uk/mo1-5/F5_monograph_1-5.htm	[ "< = back \| = > volume 2 (overview)\n\nBlue links lead to the fully translated html versions of the page, purple links lead to pages whose start pages (as well as introductions and tables of contents at least) are already set up, green links lead to extern sites, grey means that no file is available yet).\n\n/Notes in this color and between two / are from the operator of the German mirror site and translator/.\n\nChapter F: The Oscillatory Chamber\n\nF5. Mathematical model of the Oscillatory Chamber\n\nOur present knowledge of magnetic and electric phenomena enables us to deduce the equations expressing the values of the resistance, inductance and capacitance of the Oscillatory Chamber. Further combination of these equations will lead to the prediction of the behaviour of this device.\n\nThis subsection is just intended to describe the Oscillatory Chamber in such language of mathematics. Therefore it supplies the vital interpretative foundations for all the researchers experimenting with this device. Unfortunately for the readers less oriented towards mathematics, it may spoil the pleasure of familiarizing themselves with the content of this monograph. For this reason, those readers who experience a revival of sleepiness each time they encounter a mathematical equation are recommended to shift from this point directly into the beginning of subsection F6.\n\nF5.1. Resistance of the Oscillatory Chamber\n\nThe general form of the equation for the resistance of any resistor of cross-section \"A\" and length \"l\" is as follows:\n\nR = Ω(l/A)\n\nIn this equation the \"Ω\" represents the resistivity of a material from which the resistor is made. In our case it will be the maximal resistivity of the dielectric gas that fills the Oscillatory Chamber, determined for the conditions of the initial moment of electric breakdown. In turn operators \"\" and \"/\" adopted from computer programming, mean \"multiply\" and \"divide\".\n\nIf in the above general equation, we replace the variables by the specific parameters determined for the Oscillatory Chamber, i.e. l=a and A=a2 (compare with Img.012 (F1b), this gives:\n\nR = Ω/a (F1)\n\nThe equation (F1) received here represents the resistance of the Oscillatory Chamber, which is a function of the chamber's side wall dimension \"a\".\n\nF5.2. Inductance of the Oscillatory Chamber\n\nThe determination of the chamber's inductance is an extremely difficult and complex task. Completing it with total accuracy is beyond the author's knowledge of the subject. Also a number of experts consulted in this matter were unable to help. (Perhaps some of the readers know how to resolve this problem - in such a case the author would warmly welcome a review of their deductions and the final equation they derived.) Being unable to find the exact solution, the author decided to apply temporarily a simplified one. To justify this simplification it should be stated that the deducted equation (F2) for the value of inductance will be used only once in the entire monograph, when the meaning of factor \"s\" (see (F5)) is interpreted. Therefore all the vital equations in this work remain unaffected.\n\nIn the simplified deductions of the chamber's inductance an assumption is made that a unitary inductance of a stream of sparks (i.e. the inductance related to the unit of a spark's length) will be equal to the inductance of the equivalent strand of wires. This assumption allows for the application of a well-known equation for the inductance of a solenoid (see the book [1F5.2] by David Halliday et al, \"Fundamentals of Physics\", John Willey & Sons, 1966):\n\nL = μn2lA\n\nWhen in this equation we substitute: n=p/a, l=a, and A=a2 (where \"p\" is the number of segments in each of the chamber's plates, whereas \"a\" is the dimension of the chamber's walls), the simplified equation for the inductance of the Oscillatory Chamber is derived:\n\nL = μp2a (F2)\n\nIt can be theoretically asserted that the unitary electrical inertia of a stream of sparks should be greater than such an inertia in the equivalent strand of wires. The justification for this assertion can be obtained from the analysis of the inertia mechanism. The inertia reveals itself only when the motion involves the reversible phenomena or media which absorb energy in the initial stage of the motion's development, and which release this energy when the motion declines. The greater the number of such phenomena and media involved, and the higher their energy absorption, the larger is the resultant inertia. The stream of sparks jumping through gas in every aspect manifests better potentials for causing an inertia higher than the one of a current flowing through wires. The first reason for this lies in the more efficient energy absorption and releasing by sparks, occurring because:\n\na) The speed of electrons in a spark can be higher than in a wire,\n\nb) The contiguous sparks can pass closer to each other because they do not require thick insulation layers in between them (as is the case for wires).\nThe second reason for the higher inertia of sparks in gas results from them involving a variety of reversible phenomena - not appearing at all during flows of currents through wires. These are:\n\nc) The ionization of surrounding gases. This, due to the returning of the absorbed energy, supports the inertia of the process at the moment of the sparks' decline.\n\nd) The motion of heavy ions, whose mass absorbs and then releases the kinetic energy.\n\ne) The initiation of hydrodynamic phenomena (e.g. dynamic pressure, rotation of the gas) which also will be the cause of the charges' dislocation and energy return at the moment of the sparks' decline.\n\nThe above theoretical premises should not be difficult to verify by experiments described in subsection F8.2. (e.g. stage 1c).\n\nF5.3. Capacitance of the Oscillatory Chamber\n\nWhen we use the well-known equation for the capacitance of a parallel-plate capacitor, of the form:\n\nC = εA/l\n\n(where \"ε\" is the dielectric constant of this capacitor, \"A\" is the surface area of electrodes, while \"l\" is the distance between these electrodes), and when we apply the substitutions: A=a5, l=a, this yields the final equation for the capacitance \"C\" of a cubical Oscillatory Chamber:\n\nC = εa (F3)\n\n(i.e. the capacitance \"C\" of a cubical Oscillatory Chamber is equal to the value of the dielectric constant \"ε\" for the dielectric gas that fills this chamber, multiplied by the side dimension \"a\" of this chamber).\n\nF5.4. The \"sparks' motivity factor\" and its interpretation\n\nEach of the equations (F1), (F2) and (F3) describes only one selected parameter of the Oscillatory Chamber. On the other hand, it would be very useful to obtain a single complex factor which would express simultaneously all electromagnetic and design characteristics of this device. Such a factor is now introduced, and will be called a \"sparks' motivity factor\". Its defining equation is the following:\n\ns = p(R/2)√(C/L) (F4)\n\nNotice, that after expressing this in the notation of computer languages, in which the symbol \"\" means multiplication, the symbol \"/\" means division, the symbol \"+\" means addition, the symbol \"-\" means subtraction, while the symbol \"sqrt()\" means the square roof from the parameter provided in brackets \"()\", the above equation (F4) takes the following form:\n\ns = p(R/2)sqrt(C/L).\n\nNotice that, according to the definition, this \"s\" factor is dimensionless.\nIndependently from the above defining equation (F4), the \"s\" factor has also an interpretative description. This is obtained when in (F4) the variables R, L and C are substituted by the values expressed by equations (F1), (F2) and (F3). When this is done, the following interpretative equation for \"s\" is received:\n\ns = (1/(2a))Ω√(ε/μ). (F5)\n\nNotice, that after expressing this in the notation of computer languages, in which the symbol \"\" means multiplication, the symbol \"/\" means division, the symbol \"+\" means addition, the symbol \"-\" means subtraction, while the symbol \"sqrt()\" means the square roof from the parameter provided in brackets \"()\", the above equation (F5) takes the following form:\n\ns = (1/(2a))Ωsqrt(ε/μ).\n\nEquation (F5) reveals that the \"s\" factor perfectly represents the current state of all environmental conditions in which the sparks occur, and which determine their course and effectiveness. It describes the type and consistency of the gas used as a dielectric, and the actual conditions under which this gas is stored. It also describes the size of the chamber. Therefore the \"s\" factor constitutes a perfect parameter which is able to inform exactly about the working situation existing within the chamber at any particular instant in time.\n\nThe value of the \"s\" factor can be controlled at the design stage and at the exploitation stage. At the design stage it is achieved by changing the size \"a\" of a cubical chamber. At the exploitation stage it requires the change of the pressure of a gas within the chamber or altering its composition. In both cases this influences the constants Ω, μ and ε, describing the properties of this gas. (Note that constants \"Ω\", \"μ\", and \"ε\", are: Ω = resistivity of a dielectric gas within the chamber determined at the moment of electric breakdown in [Ohmmeter], μ = magnetic permeability of a dielectric in [Henry/metre], ε = dielectric constant for a gas filling the chamber in [Farad/metre].)\n\nF5.5. Condition for the oscillatory response\n\nFrom the electric point of view the Oscillatory Chamber represents a typical RLC circuit. The research on Electric Networks has determined for such circuits the condition under which, once they are charged, they will maintain the oscillatory response. This condition, presented in the book [1F5.5] by Hugh H. Skilling, \"Electric Network\" (John Willey & Sons, 1974), takes the form:\n\nR2 < 4L/C\n\nIf the above relation is transformed and then its variables are substituted by the equation (F4), it takes the final form:\n\np > s (F6)\n\nThe above condition describes the design requirement for the number \"p\" of segments separated within the plates of the Oscillatory Chamber, in relation to the environmental conditions \"s\" existing in the area where the sparks appear. If this condition is fulfilled, the sparks produced within the Oscillatory Chamber will acquire an oscillatory character.\n\nTo interpret the condition (F6), a possible range of values taken by the factor \"s\" should be considered (compare with the equation (F5)).\n\nF5.6. The period of pulsation of the chamber's field\n\nFrom the RLC circuits we know that the period of their oscillations is described by the equation:", null, "Notice, that after expressing this in the notation of computer languages, in which the symbol \"\" means multiplication, the symbol \"/\" means division, the symbol \"+\" means addition, the symbol \"-\" means subtraction, the symbol \"x*2\" means \"x\" to the power of \"2\", while the symbol \"sqrt()\" means the square roof from the parameter provided in brackets \"()\", the above equation takes the following form:\n\nT = (2π)/(sqrt(1/(LC) - (R/(2*2) = 2πsqrt(LC/(1 - ((R*2)C)/(4L))).\n\nIf the defining equation (F4) on the factor \"s\" replaces in the above a combination of R, L, and C parameters, whereas equation (F1) and equation (F3) provide the values for R and C, then this period is described as:", null, "(F7)\n\nNotice, that after expressing this in the notation of computer languages, in which the symbol \"\" means multiplication, the symbol \"/\" means division, the symbol \"+\" means addition, the symbol \"-\" means subtraction, division, while the symbol \"sqrt()\" means the square root from the parameter provided in brackets \"()\", the equation (F7) takes the following form:\n\nT = (π(p/s)Ωε)/sqrt(1 - (s/p)*2).\n\nThe final equation (F7) not only illustrates which parameters determine the value of the period of pulsations \"T\" in the Oscillatory Chamber, but also shows how the value of \"T\" can practically be controlled. Thus this equation will be highly useful for the understanding of the amplifying control of the period \"T\" of field pulsation described in subsection F6.5.\n\nIf we know the period \"T\" of chamber's field pulsations, then we can easily determine the frequency \"f\" of pulsations of this field. The well known equation linking these two parameters is as follows:\n\nf = 1/T (F8)\n\nOf course, according to the above equation (F8), the control over the frequency \"f\" of the field's pulsations will be achieved via influencing the value of the period \"T\" of this field pulsations.\n\nF6.\n\nVisitors since 15.12.22: (english sites)" ]	[ null, "https://wohin-woher.com/jp/contents/uk/mo1-5/formelF7a.jpg", null, "https://wohin-woher.com/jp/contents/uk/mo1-5/formelF7b.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9085277,"math_prob":0.97799104,"size":12192,"snap":"2023-40-2023-50","text_gpt3_token_len":2725,"char_repetition_ratio":0.16401379,"word_repetition_ratio":0.10465116,"special_character_ratio":0.2245735,"punctuation_ratio":0.09606792,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99743485,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T13:26:39Z\",\"WARC-Record-ID\":\"<urn:uuid:0727f321-8ec5-43f4-b7fa-cc19b39b8f74>\",\"Content-Length\":\"16309\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:41abc999-1b0d-4fa8-9808-de3189c2ea4e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e39fdecc-f103-4b3a-b560-5bc2f127cd19>\",\"WARC-IP-Address\":\"178.77.80.187\",\"WARC-Target-URI\":\"https://wohin-woher.com/jp/contents/uk/mo1-5/F5_monograph_1-5.htm\",\"WARC-Payload-Digest\":\"sha1:RO2QS6WEIGNIQJTCMU2LS3RQ4GI5JK4Y\",\"WARC-Block-Digest\":\"sha1:FJGGCNZDWR32FPAK633UXQTW5MYB2BOE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506027.39_warc_CC-MAIN-20230921105806-20230921135806-00746.warc.gz\"}"}
https://gradeup.co/undecidability-i-32c748fa-bea5-11e5-8fa3-b0415042f6e8	[ "# Study Notes on Undecidability\n\nUpdated : Oct 13, 2019, 9:00\nBy : Richa Tiwari\n\n## There are two types of TMs (based on halting):\n\n1. Halting TM : (Accepts Recursive languages) : TMs that always halt, no matter accepting or non no matter accepting or non-accepting (called as decidable problems)\n2. TM : (Accepts Recursively enumerable): TMs that are guaranteed to halt are guaranteed to halt only on acceptance only on acceptance. If non-accepting, it may or may not halt (i.e., could loop forever). (Either decidable or partially decidable)\n\n## Decidable Problem\n\n• If there is a Turing machine that decides the problem, called as Decidable problem.\n• A decision problem that can be solved by an algorithm that halts on all inputs in a finite number of steps.\n• A problem is decidable, if there is an algorithm that can answer either yes or no.\n• A language for which membership can be decided by an algorithm that halts on all inputs in a finite number of steps.\n• Decidable problem is also called as totally decidable problem, algorithmically solvable, recursively solvable.\n\n## Undecidable Problem (Semi-dedidable or Totally not decidable)\n\n• A problem that cannot be solved for all cases by any algorithm whatsoever.\n• Equivalent Language cannot be recognized by a Turing machine that halts for all inputs.\n\nThe following problems are undecidable problems:\n\n• Halting Problem: A halting problem is undecidable problem. There is no general method or algorithm which can solve the halting problem for all possible inputs.\n• Emptyness Problem: Whether a given TM accepts Empty?\n• Finiteness Problem: Whether a given TM accepts Finite?\n• Equivalence Problem: Whether Given two TM’s produce same language?. Is L(TM1) = L(TM2) ?\n• Is L(TM1) ⊆ L(TM2) ? (Subset Problem)\n• Is L(TM1) Ո L(TM2) = CFL?\n• Is L(TM1) = Σ* ? (Totality Problem)\n• Is the complement of L(G1) context-free ?\nUndecidable problems are two types: Partially decidable (Semi-decidable) and Totally not decidable.\n• Semi decidable: A problem is semi-decidable if there is an algorithm that says yes. if the answer is yes, however it may loop infinitely if the answer is no.\n• Totally not decidable (Not partially decidable): A problem is not decidable if we can prove that there is no algorithm that will deliver an answer.\n\n### Decidability table for Formal Languages:\n\n Problems RL DC CFL Rec RE Membership Y Y Y Y N Finiteness Y Y Y N N Emptiness Y Y Y N N Equivalence Y Y N N N Is L1 ⊆ L2 ?(SUBSET) Y N N N N Is L = REGULAR? Y Y N N N Is L Ambiguous? Y N N N N L=∑* ?(UNIVERSAL) Y Y N N N L1 ∩ L2= Ф ?(DISJOINT) Y N N N N Is L= Regular? Y Y N N N L1 ∩ L2= L Y N N Y Y Is L' also same type? Y Y N Y N\n\nNotes : RL= Regular Language , DC = Deterministic context-free languages (DCFL), CFL= Context Free Languages (CFL), Rec = Recursive language, RE= Recusively Enumerable Language.\n\nAll the Best\n\nThanks,\n\nPrep Smart, Score Better. Go Gradeup.\n\nOct 13GATE & PSU CS\n\nPosted by:", null, "Want to create my own success story✌️\nMember since Feb 2019", null, "GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor,Sector 125, Noida,Uttar Pradesh [email protected]" ]	[ null, "https://gs-post-images.grdp.co/2018/12/screenshot-2018-12-21-at-3-img1545387500435-75.png-rs-high-webp.png", null, "https://gs-post-images.grdp.co/2018/12/screenshot-2018-12-21-at-3-img1545387500435-75.png-rs-high-webp.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84935325,"math_prob":0.8997585,"size":2753,"snap":"2020-34-2020-40","text_gpt3_token_len":786,"char_repetition_ratio":0.16769734,"word_repetition_ratio":0.102,"special_character_ratio":0.24300763,"punctuation_ratio":0.11721612,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9909439,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-01T20:19:01Z\",\"WARC-Record-ID\":\"<urn:uuid:4898f01f-593e-4b8c-90a9-a6077a16fea2>\",\"Content-Length\":\"145740\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0894b8ed-3647-4d10-b59a-ed3e1a77b928>\",\"WARC-Concurrent-To\":\"<urn:uuid:f4fce1f8-5222-455f-b5a8-69a994fb3390>\",\"WARC-IP-Address\":\"104.17.13.38\",\"WARC-Target-URI\":\"https://gradeup.co/undecidability-i-32c748fa-bea5-11e5-8fa3-b0415042f6e8\",\"WARC-Payload-Digest\":\"sha1:VMETIELCSVHHIJSN6O65CLHMLCH22TJA\",\"WARC-Block-Digest\":\"sha1:6O55LY7NOAMAVVVWIQOV2B6TXLBDTKFR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600402131986.91_warc_CC-MAIN-20201001174918-20201001204918-00006.warc.gz\"}"}
https://cstheory.stackexchange.com/questions/44273/efficient-graph-isomorphism-for-similar-graph-queries	[ "# Efficient graph isomorphism for similar graph queries\n\nGiven the graph G1, G2 and G3, we want to perform isomorphism test F between G1 and G2 as well as G1 and G3. If G2 and G3 are very similar such that G3 is formed by deleting one node and inserting one node from G2, and we have the result of F(G1,G2), can we compute F(G1,G3) without computing it from scratch by extending any existing state-of-the-art methods?\n\nFor example, if G2 is formed by nodes 2,3,4,5 and G3 is formed by nodes 3,4,5,6, can we make use of the result of F(G1,G2) to compute F(G1,G3) more efficiently?\n\nThis is a simple polynomial time reduction to show that the problem is GI complete: even if you know that $$G_1, G_2$$ are isomorphic, checking if $$G_3$$, built from $$G_2$$ deleting and adding a node, is isomorphic to $$G_1$$ is as hard as graph isomorphism itself (in the worst case).\n\nGiven two graphs $$G = (V, E), G'= (V', E')$$ build\n\n$$G_1 = ( V \\cup V' \\cup \\{u\\},\\; E \\cup E' \\cup \\{ (v_i,u) \\mid v_i \\in V \\})$$\n\ni.e. the union of the two graphs plus an extra node $$u$$ connected to all the vertices of $$V$$\n\npick $$G_2 = G_1$$; and clearly they are isomorphic.\n\nNow build $$G_3$$ deleting $$u$$ and adding $$u'$$ connected to all the vertices of $$V'$$:\n\n$$G_3 = ( V \\cup V' \\cup \\{u'\\},\\; E \\cup E' \\cup \\{ (v_i',u') \\mid v_i' \\in V'\\} )$$\n\n$$G_1, G_3$$ are isomorphic iff $$G, G'$$ are isomorphic.\n\n• This is a nice reduction! However, I would add that GI-completeness alone doesn't necessarily mean there is no advantage, only that in the worst case their complexities are polynomially related. As another example, note that vertex-colored GI is also GI-complete, but most algorithms I know of can still take advantage of vertex colors in a useful way. – Joshua Grochow Jul 16 at 17:40\n• @JoshuaGrochow: thanks, I clarified that point. – Marzio De Biasi Jul 16 at 19:02\n• @MarzioDeBiasi: thank you for the explanations. Based on my understanding about your explanations, we can no take any advantage to compute F(G1,G3) knowing F(G1,G2) if the vertices connected to u and u' are different (not necessarily connected to all vertices of V or V') even if we know G and G' are isomorphic. Is that correct? In this case, is this problem as hard as the graph isomorphism itself? – Eric Huang Jul 17 at 3:06\n• @EricHuang: the reduction says that given two isomorphic graphs $G_1, G_2$ and an explicit way to build $G_3$ by deleting/adding one node (and some edges) to $G_2$ the problem of checking whether $G_1, G_3$ are isomorphic is as hard as graph isomorphism. BTW, this result also extends to the associated promise problem in which you're not given the explicit way to build $G_3$, but only the promise that $G_1,G_3$ are isomorphic up to a node delete/add operation. – Marzio De Biasi Jul 17 at 6:48\n• You may try the Weisfeiler-Lehman method or its variations, especially if your original graphs have structures like planar, tree, interval graph or bounded treewidth graph, their Weisfeiler-Lehman dimension is a small constant, in the refinement step, I guess you can take advantage of the relationship between the two graphs. – Rupei Xu Jul 29 at 0:44" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9646398,"math_prob":0.99462855,"size":521,"snap":"2019-51-2020-05","text_gpt3_token_len":158,"char_repetition_ratio":0.13926499,"word_repetition_ratio":0.0,"special_character_ratio":0.29942417,"punctuation_ratio":0.14179105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993518,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-08T07:31:58Z\",\"WARC-Record-ID\":\"<urn:uuid:c5bd8e00-ceb5-453b-87c6-8e6718fe72f9>\",\"Content-Length\":\"139032\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e6f2698f-4c96-472f-89f2-f8de26e3947e>\",\"WARC-Concurrent-To\":\"<urn:uuid:7b8c224b-ef9a-4358-ba66-ce974aa41c04>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://cstheory.stackexchange.com/questions/44273/efficient-graph-isomorphism-for-similar-graph-queries\",\"WARC-Payload-Digest\":\"sha1:FEHNZOSKSHHBJ7JUE4FFGEMWRU2CBMJG\",\"WARC-Block-Digest\":\"sha1:REWFCLHS6LD2S7AJ4CBX7BW5XMJ67KO3\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540507109.28_warc_CC-MAIN-20191208072107-20191208100107-00348.warc.gz\"}"}
https://mran.revolutionanalytics.com/snapshot/2020-12-31/web/packages/RestRserve/vignettes/ContentHandlers.html	[ "# Encoding\n\nLet’s consider an example. We develop an application which calculates factorial of a number:\n\nlibrary(RestRserve)\nbackend = BackendRserve$new() application = Application$new()\napplication$add_get(path = \"/factorial\", function(.req, .res) { x = .req$get_param_query(\"x\")\nx = as.integer(x)\n.res$set_body(factorial(x)) }) Here is how request will be processed: request = Request$new(\npath = \"/factorial\",\nmethod = \"GET\",\nparameters_query = list(x = 10)\n)\nresponse = application$process_request(request) response #> <RestRserve Response> #> status code: 200 OK #> content-type: text/plain #> <Headers> #> Server: RestRserve/0.4.0 Let’s take a closer look to the response object and its body property: str(response$body)\n#> chr \"3628800\"\n\nAs we can see it is a numeric value. HTTP response body however can’t be an arbitrary R object. It should be something that external systems can understand - either character vector or raw vector. Fortunately application helps to avoid writing boilerplate code to encode the body. Based on the content_type property it can find encode function which will be used to transform body into a http body.\n\nresponse$content_type #> \"text/plain\" response$encode\n#> NULL\n\nTwo immediate questions can arise:\n\n1. Why content_type is equal to text/plain?\n• This is because we can specify default content_type in Application constructor. It is text/plain by default, which means all the responses by default will have text/plain content type.\n2. How does application know how to encode text/plain? Can it encode any arbitrary content type?\n• Application by default is initialized with pre-defined ?EncodeDecodeMiddleware middleware. The logic on how to encode and decode request and response body is controlled by its ContentHandlers property. Out of the box it supports two content types - text/plain and application/json.\n\nFor instance app1 and app2 are equal:\n\nencode_decode_middleware = EncodeDecodeMiddleware$new() app1 = Application$new(middleware = list())\napp1$append_middleware(encode_decode_middleware) app2 = Application$new()\n\nHere is example on how you can get the actual function used for application/json encoding:\n\n\nFUN = encode_decode_middleware$ContentHandlers$get_encode('application/json')\nFUN\n#> function(x, unbox = TRUE) {\n#> res = jsonlite::toJSON(x, dataframe = 'columns', auto_unbox = unbox, null = 'null', na = 'null')\n#> unclass(res)\n#> }\n#> <bytecode: 0x7fdb7f31db98>\n#> <environment: namespace:RestRserve>\n\nWe can manually override application default content-type:\n\napplication$add_get(path = \"/factorial-json\", function(.req, .res) { x = as.integer(.req$get_param_query(\"x\"))\nresult = factorial(x)\n.res$set_body(list(result = result)) .res$set_content_type(\"application/json\")\n})\nrequest = Request$new( path = \"/factorial-json\", method = \"GET\", parameters_query = list(x = 10) ) response = application$process_request(request)\nresponse$body #> \"{\\\"result\\\":3628800}\" And here is a little bit more complex example where we store a binary object in the body. We will use R’s native serialization, but one can use protobuf, messagepack, etc. application$add_get(path = \"/factorial-rds\", function(.req, .res) {\nx = as.integer(.req$get_param_query(\"x\")) result = factorial(x) body_rds = serialize(list(result = result), connection = NULL) .res$set_body(body_rds)\n.res$set_content_type(\"application/x-rds\") }) However function above won’t work correctly. Out of the box ContentHndlers doesn’t know anything about application/x-rds: request = Request$new(\npath = \"/factorial-rds\",\nmethod = \"GET\",\nparameters_query = list(x = 10)\n)\nresponse = application$process_request(request) response$body\n#> \"500 Internal Server Error: can't encode body with content_type = 'application/x-rds'\"\n\nIn order to resolve problem above we would need to either register application/x-rds content handler with ContentHandlers$set_encode() or manually specify encode function (identity in our case): application$add_get(path = \"/factorial-rds2\", function(.req, .res) {\nx = as.integer(.req$get_param_query(\"x\")) result = factorial(x) body_rds = serialize(list(result = result), connection = NULL) .res$set_body(body_rds)\n.res$set_content_type(\"application/x-rds\") .res$encode = identity\n})\n\nrequest = Request$new( path = \"/factorial-rds2\", method = \"GET\", parameters_query = list(x = 10) ) response = application$process_request(request)\nunserialize(response$body) #>$result\n#> 3628800\n\n# Decoding\n\nRestRserve facilitates with parsing incoming request body as well. Consider a service which expects JSON POST requests:\n\napplication = Application$new(content_type = \"application/json\") application$add_post(\"/echo\", function(.req, .res) {\n.res$set_body(.req$body)\n})\n\nrequest = Request$new(path = \"/echo\", method = \"POST\", body = '{\"hello\":\"world\"}', content_type = \"application/json\") response = application$process_request(request)\nresponse$body #> \"{\\\"hello\\\":\\\"world\\\"}\" The logic behind decoding is also controlled by ?EncodeDecodeMiddleware and its ContentHandlers property. # Extending encoding and decoding Here is an example which demonstrates on how to extend ?EncodeDecodeMiddleware to handle additional content types: encode_decode_middleware = EncodeDecodeMiddleware$new()\n\nencode_decode_middleware$ContentHandlers$set_encode(\n\"text/csv\",\nfunction(x) {\ncon = rawConnection(raw(0), \"w\")\non.exit(close(con))\nwrite.csv(x, con, row.names = FALSE)\nrawConnectionValue(con)\n}\n)\n\nencode_decode_middleware$ContentHandlers$set_decode(\n\"text/csv\",\nfunction(x) {\nres = try({\ncon = textConnection(rawToChar(x), open = \"r\")\non.exit(close(con))\n}, silent = TRUE)\n\nif (inherits(res, \"try-error\")) {\nraise(HTTPError$bad_request(body = attributes(res)$condition$message)) } return(res) } ) Extended middleware needs to be provided to the application constructor: data(iris) app = Application$new(middleware = list(encode_decode_middleware))\n\nNow let’s test it:\n\napp$add_get(\"/iris\", FUN = function(.req, .res) { .res$set_content_type(\"text/csv\")\n.res$set_body(iris) }) req = Request$new(path = \"/iris\", method = \"GET\")\nres = app$process_request(req) iris_out = read.csv(textConnection(rawToChar(res$body)))\n#> Sepal.Length Sepal.Width Petal.Length Petal.Width Species\n#> 1 5.1 3.5 1.4 0.2 setosa\n#> 2 4.9 3.0 1.4 0.2 setosa\n#> 3 4.7 3.2 1.3 0.2 setosa\n#> 4 4.6 3.1 1.5 0.2 setosa\n#> 5 5.0 3.6 1.4 0.2 setosa\n#> 6 5.4 3.9 1.7 0.4 setosa\napp$add_post(\"/in\", FUN = function(.req, .res) { str(.req$body)\n})\nreq = Request$new(path = \"/in\", method = \"POST\", body = res$body, content_type = \"text/csv\")\napp$process_request(req) #> 'data.frame': 150 obs. of 5 variables: #>$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...\n#> $Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ... #>$ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...\n#> $Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ... #>$ Species : chr \"setosa\" \"setosa\" \"setosa\" \"setosa\" ...\n#> <RestRserve Response>\n#> status code: 200 OK\n#> content-type: text/plain\n#> Server: RestRserve/0.4.0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.52659184,"math_prob":0.920139,"size":6747,"snap":"2023-14-2023-23","text_gpt3_token_len":1920,"char_repetition_ratio":0.14696723,"word_repetition_ratio":0.1120283,"special_character_ratio":0.30324587,"punctuation_ratio":0.19262294,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9685267,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-27T13:17:23Z\",\"WARC-Record-ID\":\"<urn:uuid:273d772e-8422-4216-97f3-e9a03a773bf6>\",\"Content-Length\":\"41850\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09e4d222-2231-459c-8de9-a9b29fd60510>\",\"WARC-Concurrent-To\":\"<urn:uuid:36271904-02d8-4e9c-b4c4-4e7f0f1cda5e>\",\"WARC-IP-Address\":\"40.118.246.51\",\"WARC-Target-URI\":\"https://mran.revolutionanalytics.com/snapshot/2020-12-31/web/packages/RestRserve/vignettes/ContentHandlers.html\",\"WARC-Payload-Digest\":\"sha1:WRTXV4ZDIRBG7XWVVGRECULMIZG2RJK3\",\"WARC-Block-Digest\":\"sha1:5AVDR4Z6ZUJZINS44ULP3SS7AQMBKPRW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296948632.20_warc_CC-MAIN-20230327123514-20230327153514-00666.warc.gz\"}"}
https://numbermatics.com/n/460/	[ "# 460\n\n## 460 is an even composite number composed of three prime numbers multiplied together.\n\nWhat does the number 460 look like?\n\nThis visualization shows the relationship between its 3 prime factors (large circles) and 12 divisors.\n\n460 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twelve divisors.\n\n## Prime factorization of 460:\n\n### 22 × 5 × 23\n\n(2 × 2 × 5 × 23)\n\nSee below for interesting mathematical facts about the number 460 from the Numbermatics database.\n\n### Names of 460\n\n• Cardinal: 460 can be written as Four hundred sixty.\n\n### Scientific notation\n\n• Scientific notation: 4.6 × 102\n\n### Factors of 460\n\n• Number of distinct prime factors ω(n): 3\n• Total number of prime factors Ω(n): 4\n• Sum of prime factors: 30\n\n### Divisors of 460\n\n• Number of divisors d(n): 12\n• Complete list of divisors:\n• Sum of all divisors σ(n): 1008\n• Sum of proper divisors (its aliquot sum) s(n): 548\n• 460 is an abundant number, because the sum of its proper divisors (548) is greater than itself. Its abundance is 88\n\n### Bases of 460\n\n• Binary: 1110011002\n• Base-36: CS\n\n### Squares and roots of 460\n\n• 460 squared (4602) is 211600\n• 460 cubed (4603) is 97336000\n• The square root of 460 is 21.4476105895\n• The cube root of 460 is 7.7194426293\n\n### Scales and comparisons\n\nHow big is 460?\n• 460 seconds is equal to 7 minutes, 40 seconds.\n• To count from 1 to 460 would take you about three minutes.\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 460 cubic inches would be around 0.6 feet tall.\n\n### Recreational maths with 460\n\n• 460 backwards is 064\n• 460 is a Harshad number.\n• The number of decimal digits it has is: 3\n• The sum of 460's digits is 10\n• More coming soon!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86606205,"math_prob":0.9825132,"size":3048,"snap":"2019-51-2020-05","text_gpt3_token_len":826,"char_repetition_ratio":0.1130092,"word_repetition_ratio":0.05027933,"special_character_ratio":0.31167978,"punctuation_ratio":0.14681892,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99452126,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-11T14:04:38Z\",\"WARC-Record-ID\":\"<urn:uuid:c1f33756-1888-4a7d-adec-31a04f7c7aa6>\",\"Content-Length\":\"16006\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:09c542ef-4b52-4cc9-b69c-e32267391224>\",\"WARC-Concurrent-To\":\"<urn:uuid:003d713f-340f-426e-9c97-090c6c5d622d>\",\"WARC-IP-Address\":\"72.44.94.106\",\"WARC-Target-URI\":\"https://numbermatics.com/n/460/\",\"WARC-Payload-Digest\":\"sha1:OQE7BSALF7EYFSSYMLQD3ZFSPSG3KWZO\",\"WARC-Block-Digest\":\"sha1:LQ6BDVDYNCETKFBFU43KN5K6GGTO6SXY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540531917.10_warc_CC-MAIN-20191211131640-20191211155640-00268.warc.gz\"}"}
http://api.fmrchallenge.org/classintegrator__chains_1_1Polytope.html	[ "fmrbenchmark 0.0.4\nintegrator_chains::Polytope Class Reference\n\nBasic half-space representation of polytopes. More...\n\n`#include <polytope.hpp>`\n\nInheritance diagram for integrator_chains::Polytope:", null, "[legend]\n\n## Public Member Functions\n\nvoid dumpJSONcore (std::ostream &out) const\nOutput this polytope in JSON excluding opening { and closing }. More...\n\nbool is_consistent () const\nCheck that defining matrices have consistent dimensions. More...\n\nbool is_in (Eigen::VectorXd X) const\nIs X contained in this polytope? More...\n\nPolytope operator& (const Polytope &P2)\nIntersect two polytopes that are defined in the same dimension space. More...\n\nPolytope (Eigen::MatrixXd incoming_H, Eigen::VectorXd incoming_K)\n\nPolytope (const Polytope &other)\n\n## Static Public Member Functions\n\nstatic Polytopebox (const Eigen::VectorXd &bounds)\nConstruct Polytope that is an axis-aligned rectangle. More...\n\nstatic PolytoperandomH (int n)\nCreate random Polytope in R^n using Eigen::MatrixXd::Random(). More...\n\n## Friends\n\nstd::ostream & operator<< (std::ostream &out, const Polytope &P)\nOutput this polytope in JSON to given stream. More...\n\n## Detailed Description\n\nBasic half-space representation of polytopes.\n\n## Member Function Documentation\n\n Polytope * integrator_chains::Polytope::box ( const Eigen::VectorXd & bounds )\nstatic\n\nConstruct Polytope that is an axis-aligned rectangle.\n\nParameters\n bounds an array of the form [x1_min, x1_max, x2_min, x2_max, ..., xn_min, xn_max], which defines a polytope in terms of intervals along each axis in R^n. The interval along the first axis is [x1_min, x1_max], the interval along the second axis is [x2_min, x2_max], etc. Thus the length of the given array is 2n.\n\nE.g., a unit square in R^2 can be created using\n\nEigen::Vector4d bounds;\nbounds << 0, 1,\n0, 1;\nPolytope square = Polytope::box( bounds );\n void integrator_chains::Polytope::dumpJSONcore ( std::ostream & out ) const\n\nOutput this polytope in JSON excluding opening { and closing }.\n\nThis method facilitates inheritance from Polytope without crowding the JSON representations of other classes.\n\n bool integrator_chains::Polytope::is_consistent ( ) const\n\nCheck that defining matrices have consistent dimensions.\n\nThe \"defining matrices\" are those involved in the inequality providing the half-space representation, i.e., H and K in {x \| Hx <= K}. N.B., this check does not occur during instantiation of Polytope.\n\n bool integrator_chains::Polytope::is_in ( Eigen::VectorXd X ) const\n\nIs X contained in this polytope?\n\nThe polytope is a closed set. No numerical tolerance is used.\n\n Polytope integrator_chains::Polytope::operator& ( const Polytope & P2 )\n\nIntersect two polytopes that are defined in the same dimension space.\n\n Polytope integrator_chains::Polytope::randomH ( int n )\nstatic\n\nCreate random Polytope in R^n using Eigen::MatrixXd::Random().\n\nParameters\n n dimension of the containing space.\n\nRandom matrices as provided by Eigen::MatrixXd::Random() are used to instantiate Polytope in half-space representation. The matrices have `n+1` rows, corresponding to `n+1` inequalities.\n\n## Friends And Related Function Documentation\n\n std::ostream& operator<< ( std::ostream & out, const Polytope & P )\nfriend\n\nOutput this polytope in JSON to given stream.\n\nThe documentation for this class was generated from the following file:\n• polytope.hpp" ]	[ null, "http://api.fmrchallenge.org/classintegrator__chains_1_1Polytope__inherit__graph.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74885833,"math_prob":0.8558041,"size":2359,"snap":"2023-14-2023-23","text_gpt3_token_len":625,"char_repetition_ratio":0.14479831,"word_repetition_ratio":0.1402439,"special_character_ratio":0.24713862,"punctuation_ratio":0.24107143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97926277,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-04-01T08:44:33Z\",\"WARC-Record-ID\":\"<urn:uuid:bf457017-ef65-4186-b698-dfef1ad033cb>\",\"Content-Length\":\"21223\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:416cc007-563d-45f9-a9c5-6d05e5f29fcb>\",\"WARC-Concurrent-To\":\"<urn:uuid:2ecd2426-cf98-4050-86e2-d36d83295fec>\",\"WARC-IP-Address\":\"172.67.210.94\",\"WARC-Target-URI\":\"http://api.fmrchallenge.org/classintegrator__chains_1_1Polytope.html\",\"WARC-Payload-Digest\":\"sha1:GTKCXD3MK7RA7NKNLD2E2NQSVOYVYOVP\",\"WARC-Block-Digest\":\"sha1:7TOXFUXIANV4XNJXOJAI4RUVWHD3TFXO\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949701.56_warc_CC-MAIN-20230401063607-20230401093607-00180.warc.gz\"}"}
https://www.codespeedy.com/difference-of-days-between-two-dates-in-python/	[ "# How to find the difference or gap of days between two dates in python\n\nIn this tutorial, We will see How to find the difference or gap of days between two dates in python. So Guy’s, No need to fear about it that how we will solve this problem. Here, I am going to provide a simple solution to this problem in python by using datetime module. In order to solve this problem in python language, we have to know datetime module in python language.\n\nYou may also like to read:\n\nWhat is datetime module in python?\n\nPython provide inbuilt module datetime which allow us to solve various problem related to date and time.\n\n## To find the difference or gap of days between two dates in Python\n\nSo, let’s start to understand this problem in easy ways. We will follow these steps in python language:\n\n1. First of all, we will include the date in our python program from datetime module by using import function.\n\n`from datetime import date`\n\n2. We will take the starting date and the ending date of calendar between which you have to find the difference or gap of days.\n\n```Start=date(2019,7,1)\nEnd=date(2019,8,15)```\n\n3. Now, we will find the difference or gap of days between two dates as we have learned to Subtract two number by using minus sign (-) and print gap of days on output page.\n\n```Gap=(End-Start).days\nprint(\"Days gap:\")\nprint(Gap)```\n\n#### Python program to find difference or gap of days\n\nPython program:\n\n```from datetime import date\nStart=date(2019,7,1)\nEnd=date(2019,8,15)\nGap=(End-Start).days\nprint(\"Days gap:\")\nprint(Gap)```\n\nOutput of code:\n\n```Days gap:\n45```\n\nSo Guy’s, I hope you really enjoy this tutorial and feel free to leave a comment if you have any doubt.\n\n### One response to “How to find the difference or gap of days between two dates in python”\n\n1. Vishal says:\n\nHello,\nWhat if I want to create a code where the user can put in 2 dates?\nEverywhere I see, I see the dates are already a part of the code, is there a way to let a user “input” 2 dates?\nThanks!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88840854,"math_prob":0.8010707,"size":1820,"snap":"2020-45-2020-50","text_gpt3_token_len":435,"char_repetition_ratio":0.13381058,"word_repetition_ratio":0.10869565,"special_character_ratio":0.2434066,"punctuation_ratio":0.11278196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9932301,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T11:38:59Z\",\"WARC-Record-ID\":\"<urn:uuid:974d5f50-621d-4394-a6d3-17f653d2be63>\",\"Content-Length\":\"25731\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1bc7dba-85f8-41cf-afb6-91dd5ebb2012>\",\"WARC-Concurrent-To\":\"<urn:uuid:41021375-ca18-4226-9c96-d0b572747671>\",\"WARC-IP-Address\":\"74.207.248.30\",\"WARC-Target-URI\":\"https://www.codespeedy.com/difference-of-days-between-two-dates-in-python/\",\"WARC-Payload-Digest\":\"sha1:4LJHA2AVZWO7A2MP4E6QWIVNFEJZ4YYW\",\"WARC-Block-Digest\":\"sha1:QHYN7YTIW5LRWGN2AACAMQXACQCDBIUD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107894175.55_warc_CC-MAIN-20201027111346-20201027141346-00195.warc.gz\"}"}
https://www.exyb.cn/news/show-4555798.html	[ "## 分支语句（选择结构）——“C”\n\nnews/2023/6/6 4:26:01\n\n• 如果你在家，我去拜访你\n• 如果考试不及格，要补考\n• 如果遇到红灯，要停车等待\n\nC语言中有两种选择语句：\n\n• if语句，比较适合实现两个分支的选择结构（比如上课或者不上课，向左还是向右......）\n• switch语句，比较适合实现多分支的选择结构（比如从多个选修课中选择一门，选择一个最喜欢的科目）\n\nif(表达式)\n\n语句;\n\nif(表达式)\n\n语句1;\n\nelse\n\n语句2;\n\n//多分支\n\nif(表达式1)\n\n语句1;\n\nelse if(表达式2)\n\n语句2;\n\nelse\n\n语句3;\n\n例1：输入两个实数，按代数值由小到大的顺序输出这两个数\n\n解题思路：\n\n只需要做一次比较，然后进行一次交换即可\n\n用if语句实现条件决断\n\n想要实现两个变量值的互换，则要定义第三个变量\n\n这是这个题目的一段小代码，我们一起来看看\n\n``````#include<stdio.h>\nint main()\n{float a=0.0;float b=0.0;float tmp=0.0;scanf(\"%f %f\",&a,&b);if(a>b){tmp=a;a=b;b=tmp;}printf(\"%.2f,%.2f\\n\",a,b);return 0;\n}``````\n\n解题思路：可以先用伪代码写出算法\n\nif a>b,a和b对换（a是a、b中的小者）\n\nif a>c,a和c对换 (a是三者中的最小者）\n\nif b>c,b和c对换 (b是三者中次小者）\n\n顺序输出a,b,c\n\n``````#include<stdio.h>\nint main()\n{int a=0;int b=0;int c=0;int tmp=0;scanf(\"%d %d %d\",&a,&b,&c);if(a>b){tmp=a; a=b;b=tmp;}if(a>c){tmp=a;a=c;c=tmp;}if(b>c){tmp=b;b=c;c=tmp;}printf(\"%d %d %d\",a,b,c);return 0;\n}``````\n\n当你写了这样一个具有迷惑性的代码：\n\n``````#include<stdio.h>\nint main()\n{int a=0;int b=2;if(a==1)if(b==2)printf(\"hehe\\n\");else printf(\"haha\\n\");return 0;\n}``````\n\n``````//适当地使用{}可以使代码的逻辑更加清楚\n//代码风格很重要\n#include<stdio.h>\nint main()\n{int a=0;int b=2;if(a==1){if(b==2){printf(\"hehe\\n\");}}else{printf(\"haha\\n\");}return 0;\n}``````\n\n哈哈，这个代码的输出结果就确确实实为haha了\n\nswitch是“开关”的意思，它也是一种选择语句，常常用于多分支的情况\n\nswitch和if\n\n从功能上说，switch语句和if语句完全可以相互替代。但从编程的角度，它们又各有各的特点，所以至今为止，也不能说谁可以完全取代谁。\n\n当嵌套的if比较少时（三个以内）,用if编程程序会比较简洁。但是当选择的分支比较多时，嵌套的if语句层数就会很多，导致程序冗长，可读性下降。因此C语言提供switch语句来处理多分支选择。在很多大型的项目中，多分支选择的情况经常会遇到，所以switch语句用得还是比较多的。\n\nswitch语句说明\n\nswitch语句的作用是根据表达式的值，使流程跳转到不同的语句\n\n下面，来给大家举几个例子来深入说明一下switch语句\n\n解题思路：\n\n判断出这是一个多分支选择问题\n\n根据百分制分数将学生成绩分为4个等级\n\n如果用if语句，至少要用三层嵌套的if，进行3次检查判断\n\n用switch语句进行一次检查即可得到结果\n\n``````#include<stdio.h>\nint main()\n}``````\n\nbreak语句的实际效果是把语句列表划分为不同的分支部分。\n\n在最后一个 case 语句的后面加上一条 break语句。\n\n（之所以这么写是可以避免出现在以前的最后一个 case 语句后面忘了添加 break语句）\n\ndefault子句\n\n s<250 没有折扣 250==3000 15%折扣\n\n解题思路：\n\n设每吨每千米货物的基本运费为p，货物重量为w，距离为s，折扣为d\n\n总运费的计算公式为f=pws(1-d)\n\n计算等式中p、w、s为用户输入，所以关键是要找到折扣d的变化规律，由题可知折扣d与输入距离s有关，而且“变化点”都是250的倍数，故引入一折扣变量c，令c的值为s/250，即c代表250的倍数。\n\n即当c<1时，表示s<250,无折扣\n\n1=<c<2时，表示250=<s<500,折扣d=2%\n\n2=<c<4时，表示500=<s<1000，折扣d=5%\n\n4=<c<8时，表示1000=<s<2000，折扣d=8%\n\n8=<c<12时，表示2000=<s<3000，折扣d=10%\n\nc>=12时，表示s>=3000，折扣d=15%\n\n``````#include<stdio.h>\nint main()\n{int c=0;//倍数int s=0;//距离float p=0.0;//每吨每千米货物的基本运费float w=0.0;//货物的重量float d=0.0;//折扣float f=0.0;//总运费printf(\"please enter price,weight,distance:\");scanf(\"%f %f %d\",&p,&w,&s);if(s>=3000){c=12;}else {c=s/250;}switch(c){case 0:d=0;break;case 1:d=2;break;case 2:case 3:d=5;break;case 4:case 5:case 6:case 7:d=8;break;case 8:case 9:case 10:case 11:d=10;break;case 12:d=15;break;}f=pws(1-d/100);printf(\"freight=%10.2f\\n\",f);return 0;\n}``````\n\n### 运算放大器：运放的平衡电阻\n\n1．主要运放参数集成运放的基本参数很多，包括静态技术指标（直流参数）和动态技术指标（交流参数）。。例如，输入失调电压、输入失调电流、输入偏置电流、输入失调电压温漂、最大差模输入电压、最大…" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.8239567,"math_prob":0.9927211,"size":3900,"snap":"2023-14-2023-23","text_gpt3_token_len":2877,"char_repetition_ratio":0.11524641,"word_repetition_ratio":0.013953488,"special_character_ratio":0.30564103,"punctuation_ratio":0.1948718,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95071197,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T20:26:01Z\",\"WARC-Record-ID\":\"<urn:uuid:3feaeb6e-a9e7-4d08-982b-b3093774d4b3>\",\"Content-Length\":\"30540\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5f8cfcec-a6ba-40da-82c0-66e37a50d3cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:4ae82759-21f3-4540-9c29-20f222d730b2>\",\"WARC-IP-Address\":\"43.154.132.222\",\"WARC-Target-URI\":\"https://www.exyb.cn/news/show-4555798.html\",\"WARC-Payload-Digest\":\"sha1:OQ65VXFUTYLKDLOOBJPKZ3WNTS75VOTI\",\"WARC-Block-Digest\":\"sha1:24MFO355XRGJHT3XLJIOAD3IFWXVV6HC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652161.52_warc_CC-MAIN-20230605185809-20230605215809-00625.warc.gz\"}"}
https://alb.codehs.com/info/standards/explore/OH-CTE	[ "# Ohio CTE Framework\n\n## Standards\n\nStandard Description\n2.3.1. Identify and explain coding information and representation of characters (e.g., American Standard Code for Information Interchange [ASCII], Extended Binary Coded Decimal Interchange Code [EBCDIC], Unicode). Lessons\n2.3.2. Convert between numbering systems (e.g., binary, hexadecimal, decimal). Lessons\n2.9.1. Identify and incorporate branding strategies. Lessons\n2.9.2. Determine the scope and purpose of the project. Lessons\n2.9.3. Determine the target audience, client needs, expected outcomes, objectives, and budget. Lessons\n2.9.4. Develop a conceptual model and design brief for the project. Lessons\n2.9.5. Develop a timeline, a communication plan, a task breakdown, costs (e.g., equipment, labor), deliverables, and responsibilities for completion. Lessons\n2.9.6. Develop and present a comprehensive proposal to stakeholders. Lessons\n2.11.1. Identify the problem. Lessons\n2.11.2. Select troubleshooting methodology (e.g., top down, bottom up, follow the path, spot the differences). Lessons\n2.11.3. Investigate symptoms based on the selected methodology. Lessons\n2.11.4. Gather and analyze data about the problem. Lessons\n2.11.5. Design a solution. Lessons\n2.11.6. Test a solution. Lessons\n2.11.7. Implement a solution. Lessons\n2.11.8. Document the problem and the verified solution. Lessons\n2.12.1. Create a written procedure agreed by the stakeholders and project team for determining the acceptability of the project deliverables. Lessons\n2.12.2. Develop a test system that accurately mimics external interfaces. Lessons\n2.12.3. Develop test cases that are realistic, compare with expected performance, and include targeted platforms and device types. Lessons\n2.12.4. Develop, perform, and document usability and testing integration. Lessons\n2.12.5. Make corrections indicated by test results. Lessons\n2.12.6. Seek stakeholder acceptance upon successful completion of the test plan. Lessons\n2.13.1. Include overall project goals and timelines in the rollout plan. Lessons\n2.13.2. Communicate rollout plans to key stakeholders in a timely manner. Lessons\n2.13.3. Conduct final review and approvals according to company standards. Lessons\n2.13.4. Identify support staff, training needs, and contingency plans in the rollout plan. Lessons\n2.13.5. Test delivered application to assure that it is fully functional for the customer or user and meets all requirements. Lessons\n2.13.6. Deliver support and training materials. Lessons\n5.1.1. Describe how computer programs and scripts can be used to solve problems (e.g., desktop, mobile, enterprise). Lessons\n5.1.2. Explain how algorithms and data structures are used in information processing. Lessons\n5.1.3. Model the solution using both graphic tools (e.g., flowcharts) and pseudocode techniques. Lessons\n5.1.4. Describe, compare, and contrast the basics of procedural, structured, object-oriented (OO), and event driven programming. Lessons\n5.1.5. Describe the concepts of data management through programming languages. Lessons\n5.1.6. Analyze the strengths and weaknesses of different languages for solving a specific problem. Lessons\n5.1.7. Compare the functions and operations of compilers and interpreters. Lessons\n5.1.8. Describe version control and the relevance of documentation. Lessons\n5.2.1. Compare primitive types of numeric and nonnumeric data (e.g., integers, floats, Boolean, strings). Lessons\n5.2.2. Identify the scope of data (e.g., global versus local, variables, constants, arrays). Lessons\n5.2.3. Write code that uses arithmetic operations. Lessons\n5.2.4. Write code that uses subtotals and final totals. Lessons\n5.2.5. Write code that applies string operations (e.g., concatenation, pattern matching, substring). Lessons\n5.3.1. Explain Boolean logic. Lessons\n5.3.2. Solve a truth table. Lessons\n5.3.3. Write code that uses logical operators (e.g., and, or, not). Lessons\n5.3.4. Write code that uses relational operators and compound conditions. Lessons\n5.3.5. Write code that uses conditional control structures (e.g., if, if-then-else). Lessons\n5.3.6. Write code that uses repetition control structures (e.g., while, for). Lessons\n5.3.7. Write code that uses selection control structures (e.g., case, switch). Lessons\n5.3.8. Write code that uses nested structures and recursion. Lessons\n5.3.9. Write code that creates and calls functions. Lessons\n5.3.10. Code error handling techniques. Lessons\n5.3.11. Write code to access data repositories. Lessons\n5.3.12. Write code to create classes, objects, and methods. Lessons\n5.4.1. Configure options, preferences, and tools. Lessons\n5.4.2. Write and edit code in the integrated development environment (IDE). Lessons\n5.4.3. Compile or interpret a working program. Lessons\n5.4.4. Define test cases. Lessons\n5.4.5. Test the program using defined test cases. Lessons\n5.4.6. Correct syntax and runtime errors. Lessons\n5.4.7. Debug logic errors. Lessons\n5.5.1. Develop programs using data validation techniques. Lessons\n5.5.2. Develop programs that use reuse libraries. Lessons\n5.5.3. Develop programs using operating system calls. Lessons\n5.5.4. Develop programs that call other programs. Lessons\n5.5.5. Use appropriate naming conventions and apply comments. Lessons\n5.5.6. Format output (e.g., desktop, mobile, enterprise, reports, data files). Lessons\n5.6.1. Determine requirements specification documentation. Lessons\n5.6.2. Identify constraints and system processing requirements. Lessons\n5.6.3. Develop and adhere to timelines. Lessons\n5.6.4. Identify a programming language, framework, and an integrated development environment (IDE). Lessons\n5.6.5. Identify input and output (I/O) requirements. Lessons\n5.6.6. Design system inputs, outputs, and processes. Lessons\n5.6.7. Document a design using the appropriate tools (e.g., program flowchart, dataflow diagrams, Unified Modeling Language [UML]). Lessons\n5.6.8. Create documentation (e.g., implementation plan, contingency plan, data dictionary, user help). Lessons\n5.6.9. Review the design (e.g., peer walkthrough). Lessons\n5.6.10. Present the system design to stakeholders. Lessons\n5.6.11. Develop the application. Lessons\n5.6.12. Compare software methodologies (e.g., agile, waterfall). Lessons\n5.6.13. Perform code reviews (e.g., peer walkthrough, static analysis). Lessons\n5.6.14. Ensure code quality by testing and debugging the application (e.g., system testing, user acceptance testing). Lessons\n5.6.15. Train stakeholders. Lessons\n5.6.16. Deploy the application. Lessons\n5.6.17. Collect application feedback and maintain the application. Lessons\n5.7.1. Explain version management and interface control. Lessons\n5.7.2. Explain baseline and software lifecycle phases. Lessons\n5.7.3. Analyze the impact of changes. Lessons\n6.3.1. Select and apply scripting languages used in web development. Lessons" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6287717,"math_prob":0.7769306,"size":7132,"snap":"2022-27-2022-33","text_gpt3_token_len":2042,"char_repetition_ratio":0.23835579,"word_repetition_ratio":0.008088979,"special_character_ratio":0.29080203,"punctuation_ratio":0.29904762,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9695059,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-07T23:01:22Z\",\"WARC-Record-ID\":\"<urn:uuid:0eac5f73-6a8c-441f-989b-541720272a24>\",\"Content-Length\":\"191317\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:649e4a12-09aa-4f56-bf82-cbc9f51ba43c>\",\"WARC-Concurrent-To\":\"<urn:uuid:78a57bc5-8423-44f5-9b0b-519deb430a0a>\",\"WARC-IP-Address\":\"52.85.132.81\",\"WARC-Target-URI\":\"https://alb.codehs.com/info/standards/explore/OH-CTE\",\"WARC-Payload-Digest\":\"sha1:X7UIRPIXXXT6HKJVXV5Y3SIMYU2U2MXA\",\"WARC-Block-Digest\":\"sha1:KSRQ3RZZR3F2HTF6IONKPZ6NLXCBR6EC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570730.59_warc_CC-MAIN-20220807211157-20220808001157-00441.warc.gz\"}"}
https://gis.stackexchange.com/questions/115822/calculate-fields-with-null-values-using-arcpy-da-updatecursor-of-arcgis-for-desk	[ "Calculate Fields with Null Values using arcpy.da.UpdateCursor of ArcGIS for Desktop?\n\nHow do you calculate fields with null values? I have multiple fields that may have null values, that aren't being calculated in the third field.", null, "import arcpy\nwith arcpy.da.UpdateCursor(r\"...gdb\\test.gdb\\test\",[\"A\", \"B\", \"C\"]) as cursor:\nfor row in cursor:\nrow = row + row\ncursor.updateRow(row)\n\nThe result of that calculation is this error:\n\nTraceback (most recent call last):\nFile \"C:\\Users\\Desktop\\updatecursor2.py\", line 4, in <module>\nrow = row + row\nTypeError: unsupported operand type(s) for +: 'NoneType' and 'float'\n\nThe result of the script are the same results when I try to use the field calculator:", null, "How do you use the data update cursor, or field calculator to calculate the sum of the values in a new field if you have null values?\n\nI'm using ArcGIS 10.1 SP1 for Desktop.\n\n• What are the data types of these fields? It looks like you are trying to set the value of field \"A\" (2) with the sum of the values from ObjectID (0), and Shape (1). I also don't think you are setting the values correctly. – jbchurchill Oct 1 '14 at 13:37\n• @jbchurchill The data type is Double. – user3767931 Oct 1 '14 at 13:38\n• @jbchurchill FYI, when using data access cursors, the index numbers correspond to the user-supplied attribute list (in this case [\"A\",\"B\",\"C\"]) and not the list of all fields in the table, so they were being referenced correctly. – nmpeterson Oct 1 '14 at 14:31\n• Ah I see that now. So those values are being set correctly after all. – jbchurchill Oct 1 '14 at 14:34\n• What do you want to do if A and/or B is Null? Give C the value of the non-Null field, if there is one, or skip it, leaving C Null? – recurvata Oct 1 '14 at 15:05\n\nHere's a way to replace all the nulls with zeroes:\n\nimport arcpy\nwith arcpy.da.UpdateCursor(r\"...gdb\\test.gdb\\test\",[\"A\", \"B\", \"C\"]) as cursor:\nfor row in cursor:\na_value = row if row else 0 # Use 0 when \"A\" is falsy\nb_value = row if row else 0 # Use 0 when \"B\" is falsy\nrow = a_value + b_value\ncursor.updateRow(row)\n\nNote: this doesn't actually update the null values in the table; it merely replaces them with zeroes during calculations. If you want to replace the nulls with zeroes permanently, you could change the penultimate line to row = [a_value, b_value, a_value + b_value].\n\n• You could take advantage of the way Python does boolean expressions and simply do a_value = row or 0 – Jason Scheirer Oct 3 '14 at 22:00\n\nLets start with the field calculator. it should look like these 2 screenshots.", null, "", null, "When I calculate fields inside of an Update cursor (update_cursor) it looks more like this. Here I have a SearchCursor (row is set to searchCursor.next) and an update Cursor (row2 is set to update_cursor.next). This is actually slightly different since I am using values in a different layer for the updates and a selection.\n\nwhile row:\nwhile row2:\ngrpnumValue = row2.getValue(\"A\")\nif (grpnumValue > 0):\nrow.setValue(\"MAILING_GROUP\", grpnumValue)\nrow2 = searchCursor.next\nrow = update_cursor.next()" ]	[ null, "https://i.stack.imgur.com/bWNCw.jpg", null, "https://i.stack.imgur.com/hZwOU.jpg", null, "https://i.stack.imgur.com/PXRUL.jpg", null, "https://i.stack.imgur.com/9H7sK.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.816297,"math_prob":0.8318572,"size":932,"snap":"2019-43-2019-47","text_gpt3_token_len":255,"char_repetition_ratio":0.13038793,"word_repetition_ratio":0.013513514,"special_character_ratio":0.2832618,"punctuation_ratio":0.16256158,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97314256,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-24T02:10:17Z\",\"WARC-Record-ID\":\"<urn:uuid:66a087ba-a10e-4595-a7c4-58b66f422934>\",\"Content-Length\":\"153831\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9554e577-7ba9-43bd-9a85-50f540c783a3>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e566a43-a811-4b52-80db-788a4b7d21b6>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://gis.stackexchange.com/questions/115822/calculate-fields-with-null-values-using-arcpy-da-updatecursor-of-arcgis-for-desk\",\"WARC-Payload-Digest\":\"sha1:B3CGWBY7GHMIRJD3PVTF3BRUUZQB2EQF\",\"WARC-Block-Digest\":\"sha1:RL6KURMILABFXXSGXZB72IHYJFYQUSCY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987838289.72_warc_CC-MAIN-20191024012613-20191024040113-00521.warc.gz\"}"}
https://www.thebalancesmb.com/liquidity-position-analysis-with-ratios-393233	[ "# Analysis of Liquidity Position Using Financial Ratios\n\nIn this article, we will consider some commonly used liquidity ratios used in the financial analysis of a company. A balance sheet is provided as an example for calculating a company's financial position by measuring its liquidity, which is the ability to pay its current debt with its current assets. The information reflects two years of data for a hypothetical company.\n\nThe balance sheet data will also be used to calculate the current ratio, quick ratio, and net working capital, as well as provide an explanation of each as well as the meaning of changes from year to year. The results can be replicated for your own firm or one that you are interested in investing in.\n\n## Calculate the Company's Current Ratio\n\nThe first step in liquidity analysis is to calculate the company's current ratio. The current ratio shows how many times over the firm can pay its current debt obligations based on its assets. \"Current\" usually means a short time period of less than twelve months. The formula is:\n\nCurrent Ratio = Current Assets/Current Liabilities\n\nIn the balance sheet, you can see the highlighted numbers. Those are the ones you use for the calculation. For 2018, the calculation would be:\n\nCurrent Ratio = \\$708/\\$540 = 1.311 X\n\nThis means that the firm can meet its current short-term debt obligations 1.311 times over. In order to stay solvent, the firm must have a current ratio of at least 1.0 X, which means it can exactly meet its current debt obligations. So, this firm is solvent.\n\nHowever, in this case, the firm is a little more liquid than that. It can meet its current debt obligations and have a little left over. If you calculate the current ratio for 2017, you will see that the current ratio was 1.182 X. So, the firm improved its liquidity in 2018 which, in this case, is good as it is operating with relatively low liquidity.\n\n## Calculate the Company's Quick Ratio or Acid Test\n\nThe second step in liquidity analysis is to calculate the company's quick ratio or acid test. The quick ratio is a more stringent test of liquidity than the current ratio. It looks at how well the company can meet its short-term debt obligations without having to sell any of its inventory to do so.\n\nInventory is the least liquid of all the current assets because you have to find a buyer for your inventory. Finding a buyer, especially in a slow economy, is not always possible. Therefore, firms want to be able to meet their short-term debt obligations without having to rely on selling inventory. The formula is:\n\nQuick Ratio = Current Assets-Inventory/Current Liabilities.\n\nIn the balance sheet, you can see the highlighted numbers. Those are the ones you use for the calculation. For 2018, the calculation would be:\n\nQuick Ratio = \\$708-\\$422/\\$540 = 0.529 X.\n\nThis means that the firm cannot meet its current short-term debt obligations without selling inventory because the quick ratio is 0.529 X, which is less than 1.0 X. In order to stay solvent and pay its short-term debt without selling inventory, the quick ratio must be at least 1.0 X, which it is not.\n\nHowever, in this case, the firm will have to sell inventory to pay its short-term debt. If you calculate the quick ratio for 2017, you will see that it was 0.458 X. So, the firm improved its liquidity by 2018 which, in this case, is good, as it is operating with relatively low liquidity. It needs to improve its quick ratio to above 1.0 X so it won't have to sell inventory to meet its short-term debt obligations.\n\n## Calculate the Company's Net Working Capital\n\nA company's net working capital is the difference between its current assets and current liabilities:\n\nNet Working Capital = Current Assets - Current Liabilities\n\nFor 2018, this company's net working capital would be:\n\n\\$708 - 540 = \\$168\n\nFrom this calculation, you know you have positive net working capital with which to pay short-term debt obligations before you even calculate the current ratio. You should be able to see the relationship between the company's net working capital and its current ratio.\n\nFor 2017, the company's net working capital was \\$99, so its net working capital position, and, thus, its liquidity position, has improved from 2017 to 2018." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9627623,"math_prob":0.9341934,"size":5181,"snap":"2021-31-2021-39","text_gpt3_token_len":1100,"char_repetition_ratio":0.15549546,"word_repetition_ratio":0.12429378,"special_character_ratio":0.22254391,"punctuation_ratio":0.10858835,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98586506,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-22T13:29:46Z\",\"WARC-Record-ID\":\"<urn:uuid:c5b4890a-1bee-4aee-8040-f806dd1b84e3>\",\"Content-Length\":\"145193\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02d46a1e-52dc-42cc-a4a6-f301acaf974b>\",\"WARC-Concurrent-To\":\"<urn:uuid:726f56c7-7fba-418a-b32c-4f101ceeb109>\",\"WARC-IP-Address\":\"199.232.66.137\",\"WARC-Target-URI\":\"https://www.thebalancesmb.com/liquidity-position-analysis-with-ratios-393233\",\"WARC-Payload-Digest\":\"sha1:VB7NSKMMIRI7GKRQZ4Y4AON4S4L47V2D\",\"WARC-Block-Digest\":\"sha1:KQCMWTIMRG7CWDEHB72PIVPIRKDQK3AN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057366.40_warc_CC-MAIN-20210922132653-20210922162653-00641.warc.gz\"}"}
https://www.intmath.com/blog/videos/friday-math-movie-how-to-use-the-quadratic-formula-by-hector-the-battle-droid-1248	[ "Search IntMath\nClose\n\n# How To Use the Quadratic Formula by Hector the Battle Droid\n\nBy Murray Bourne, 20 Jun 2008\n\nThis week's math movie is fun.\n\nThe title says it all...\n\nIt's quite a clever compilation of social commentary with an important mathematical concept. The quadratic formula arises a lot when solving math problems, because the curve of a quadratic (a parabola) is an elegant simple shape, and can be used to model many real-life situations.\n\nWell anyway, the video is not a bad way to revise the quadratic formula.\n\nBe the first to comment below.\n\n### Comment Preview\n\nHTML: You can use simple tags like <b>, <a href=\"...\">, etc.\n\nTo enter math, you can can either:\n\n1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):\na^2 = sqrt(b^2 + c^2)\n(See more on ASCIIMath syntax); or\n2. Use simple LaTeX in the following format. Surround your math with $$ and $$.\n$$\\int g dx = \\sqrt{\\frac{a}{b}}$$\n(This is standard simple LaTeX.)\n\nNOTE: You can mix both types of math entry in your comment." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87548465,"math_prob":0.97498626,"size":531,"snap":"2023-40-2023-50","text_gpt3_token_len":118,"char_repetition_ratio":0.121442124,"word_repetition_ratio":0.0,"special_character_ratio":0.21657251,"punctuation_ratio":0.11320755,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995673,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T06:00:16Z\",\"WARC-Record-ID\":\"<urn:uuid:55a2166b-8bae-421f-af7c-c142ba50016c>\",\"Content-Length\":\"123023\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4fb9ccb4-0d11-49f5-90d0-503d0a5aca0d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c592aedc-c518-40b1-b883-236460ad0345>\",\"WARC-IP-Address\":\"104.26.13.103\",\"WARC-Target-URI\":\"https://www.intmath.com/blog/videos/friday-math-movie-how-to-use-the-quadratic-formula-by-hector-the-battle-droid-1248\",\"WARC-Payload-Digest\":\"sha1:QJDBCT3YTUPDUIHZIBQDS4HGCQ3P772Z\",\"WARC-Block-Digest\":\"sha1:43A6NE6Y7AV26ZFJHHLSTQVY4FQMUBBZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100650.21_warc_CC-MAIN-20231207054219-20231207084219-00572.warc.gz\"}"}
https://theshippingforecastmusic.com/qa/is-ranking-categorical-or-quantitative.html	[ "", null, "# Is Ranking Categorical Or Quantitative?\n\n## Is rating categorical or quantitative?\n\nSatisfaction ratings on a scale from 1 to 5.\n\nAlthough these are represented by numbers, they do not represent a count or true measurement.\n\nSuch ratings are categorical..\n\n## Is height categorical or quantitative?\n\nIn our medical example, age is an example of a quantitative variable because it can take on multiple numerical values. It also makes sense to think about it in numerical form; that is, a person can be 18 years old or 80 years old. Weight and height are also examples of quantitative variables.\n\n## What are the 3 types of variables?\n\nA variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled.\n\n## What are the two types of categorical data?\n\nThere are two types of categorical data, namely; the nominal and ordinal data. Nominal Data: This is a type of data used to name variables without providing any numerical value.\n\n## What is categorical data used for?\n\nCategorical data is a good way of collecting information that doesn’t present itself in the form of numbers or if the numbers needed to be grouped to analyze. Categorical data is data that is collected in groups or topics; the number of events in each group is counted numerically.\n\n## What is categorical data with examples?\n\nCategorical variables represent types of data which may be divided into groups. Examples of categorical variables are race, sex, age group, and educational level. There are 8 different event categories, with weight given as numeric data. …\n\n## Can time be a categorical variable?\n\nCategorical data might not have a logical order. For example, categorical predictors include gender, material type, and payment method. Discrete variables are numeric variables that have a countable number of values between any two values. … A continuous variable can be numeric or date/time.\n\n## What are the 5 types of variables?\n\nThere are six common variable types:DEPENDENT VARIABLES.INDEPENDENT VARIABLES.INTERVENING VARIABLES.MODERATOR VARIABLES.CONTROL VARIABLES.EXTRANEOUS VARIABLES.\n\n## Are yes or no questions categorical or quantitative?\n\nThe Categorical Variable Another instance of categorical variables is answers to yes and no questions.\n\n## What are the 4 types of data?\n\nIn statistics, there are four data measurement scales: nominal, ordinal, interval and ratio. These are simply ways to sub-categorize different types of data (here’s an overview of statistical data types) .\n\n## Is age categorical or continuous?\n\nMondal suggests that age can be viewed as a discrete variable because it is commonly expressed as an integer in units of years with no decimal to indicate days and presumably, hours, minutes, and seconds." ]	[ null, "https://mc.yandex.ru/watch/66677008", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8343634,"math_prob":0.9604684,"size":3179,"snap":"2020-45-2020-50","text_gpt3_token_len":638,"char_repetition_ratio":0.20409448,"word_repetition_ratio":0.11787819,"special_character_ratio":0.19723183,"punctuation_ratio":0.13559322,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9835442,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-26T23:44:41Z\",\"WARC-Record-ID\":\"<urn:uuid:e1585146-6b4b-4759-8eec-722b7264bb9d>\",\"Content-Length\":\"34255\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:77dbde19-0db5-4ac3-9a03-c9733904971b>\",\"WARC-Concurrent-To\":\"<urn:uuid:f1b8a5bf-9d88-4dbf-ac76-b317cbccad73>\",\"WARC-IP-Address\":\"87.236.16.235\",\"WARC-Target-URI\":\"https://theshippingforecastmusic.com/qa/is-ranking-categorical-or-quantitative.html\",\"WARC-Payload-Digest\":\"sha1:R75JV7ZRRYLMDJ6VAGU2DD3VDQJSO7QG\",\"WARC-Block-Digest\":\"sha1:YEN3CYUZZS4OM5G33C7362AJ3UOD3SMS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141189030.27_warc_CC-MAIN-20201126230216-20201127020216-00593.warc.gz\"}"}
https://www.analyticsvidhya.com/blog/2017/05/questions-python-for-data-science/	[ "Faizan Shaikh — May 4, 2017\n\nPython is increasingly becoming popular among data science enthusiasts, and for right reasons. It brings the entire ecosystem of a general programming language. So you can not only transform and manipulate data, but you can also create strong pipelines and machine learning workflows in a single ecosystem.\n\nAt Analytics Vidhya, we love Python. Most of us use Python as our preferred tool for machine learning. Not only this, if you want to learn Deep Learning, Python clearly has the most mature ecosystem among all other languages.\n\nLearning Python is the first step in your Data Science Journey. Want to know what are the milestones in Data Science Journey and how to achieve them? Check out the complete Data Science Roadmap! Click here to Download.\n\nIf you are learning Python for Data Science, this test was created to help you assess your skill in Python. This test was conducted as part of DataFest 2017. Close to 1,300 people participated in the test with more than 300 people taking this test.\n\nBelow are the distribution scores of the people who took the test:", null, "You can access the final scores here. Here are a few statistics about the distribution.\n\nMean Score: 14.16\n\nMedian Score: 15\n\nMode Score: 0", null, "Question Context 1\n\nYou must have seen the show “How I met your mother”. Do you remember the game where they played, in which each person drinks a shot whenever someone says “but, um”. I thought of adding a twist to the game. What if you could use your technical skills to play this game?\n\nTo identify how many shots a person is having in the entire game, you are supposed to write a code.\n\nBelow is the subtitle sample script.\n\nNote: Python regular expression library has been imported as re.\n\n```txt = '''450\n00:17:53,457 --> 00:17:56,175\nOkay, but, um,\nthanks for being with us.\n\n451\n00:17:56,175 --> 00:17:58,616\nBut, um, if there's any\ncollege kids watching,\n\n452\n00:17:58,616 --> 00:18:01,610\nBut, um, but, um, but, um,\nbut, um, but, um,\n\n453\n00:18:01,610 --> 00:18:03,656\nWe have to drink, professor.```\n```454\n00:18:03,656 --> 00:18:07,507\nIt's the rules.\nShe said \"But, um\"\n\n455\n00:18:09,788 --> 00:18:12,515\nBut, um, but, um, but, um...\ngod help us all.\n'''```\n\n1) Which of the following codes would be appropriate for this task?\n\nA) len(re.findall(‘But, um’, txt))\n\nB) re.search(‘But, um’, txt).count()\n\nC) len(re.findall(‘[B,b]ut, um’, txt))\n\nD) re.search(‘[B,b]ut, um’, txt)).count()\n\nQuestion Context 2\n\nSuppose you are given the below string\n\n[email protected],aa,Owner,2014\n[email protected],bb,Member,2015\n[email protected],cc,Member,2017\n[email protected],dd,Member,2016\n[email protected],ee,Member,2020\n“””\n\nIn order to extract only the domain names from the email addresses from the above string (for eg. “aaa”, “bbb”..) you write the following code:\n\n```for i in re.finditer('([a-zA-Z]+)@([a-zA-Z]+).(com)', str):\nprint i.group(__)```\n\n2) What number should be mentioned instead of “__” to index only the domains?\n\nNote: Python regular expression library has been imported as re.\n\nA) 0\n\nB) 1\n\nC) 2\n\nD) 3\n\nQuestion Context 3\n\nYour friend has a hypothesis – “All those people who have names ending with the sound of “y” (Eg: Hollie) are intelligent people.” Please note: The name should end with the sound of ‘y’ but not end with alphabet ‘y’.\n\nNow you being a data freak, challenge the hypothesis by scraping data from your college’s website. Here’s data you have collected.\n\n Name Marks Andy 0 Mandi 10 Sandy 20 Hollie 18 Molly 19 Dollie 15\n\nYou want to make a list of all people who fall in this category. You write following code do to the same:\n\n```temp = []\nfor i in re.finditer(pattern, str):\ntemp.append(i.group(1))```\n\n3) What should be the value of “pattern” in regular expression?\n\nNote: Python regular expression library has been imported as re.\n\nA) pattern = ‘(i\|ie)(,)’\n\nB) pattern = ‘(i\\$\|ie\\$)(,)’\n\nC) pattern = ‘([a-zA-Z]+i\|[a-zA-Z]+ie)(,)’\n\nD) None of these\n\nQuestion Context 4\n\nAssume, you are given two lists:\n\na = [1,2,3,4,5]\n\nb = [6,7,8,9]\n\nThe task is to create a list which has all the elements of a and b in one dimension.\n\nOutput:\n\na = [1,2,3,4,5,6,7,8,9]\n\n4) Which of the following option would you choose?\n\nA) a.append(b)\n\nB) a.extend(b)\n\nC) Any of the above\n\nD) None of these\n\n5) You have built a machine learning model which you wish to freeze now and use later. Which of the following command can perform this task for you?\n\nNote: Pickle library has been imported as pkl.\n\nA) push(model, “file”)\n\nB) save(model, “file”)\n\nC) dump(model, “file”)\n\nD) freeze(model, “file”)\n\nQuestion Context 6\n\nWe want to convert the below string in date-time value:\n\n```import time\nstr = '21/01/2017'\ndatetime_value = time.strptime(str,date_format)```\n\n6) To convert the above string, what should be written in place of date_format?\n\nA) “%d/%m/%y”\n\nB) “%D/%M/%Y”\n\nC) “%d/%M/%y”\n\nD) “%d/%m/%Y”\n\nSolution: (D)\n\nOption D is correct\n\nQuestion Context 7\n\nI have built a simple neural network for an image recognition problem. Now, I want to test if I have assigned the weights & biases for the hidden layer correctly. To perform this action, I am giving an identity matrix as input. Below is my identity matrix:\n\nA = [ 1, 0, 0\n0, 1, 0\n0, 0, 1]\n\n7) How would you create this identity matrix in python?\n\nNote: Library numpy has been imported as np.\n\nA) np.eye(3)\n\nB) identity(3)\n\nC) np.array([1, 0, 0], [0, 1, 0], [0, 0, 1])\n\nD) All of these\n\nSolution: (A)\n\nOption B does not exist (it should be np.identity()). And option C is wrong, because the syntax is incorrect. So the answer is option A\n\n8) To check whether the two arrays occupy same space, what would you do?\n\nI have two numpy arrays “e” and “f”.\n\nYou get the following output when you print “e” & “f”\n\n```print e\n[1, 2, 3, 2, 3, 4, 4, 5, 6]\nprint f\n[[1, 2, 3], [2, 3, 4], [4, 5, 6]]```\n\nWhen you change the values of the first array, the values for the second array also changes. This creates a problem while processing the data.\n\nFor example, if you set the first 5 values of e as 0; i.e.\n\n```print e[:5]\n\n0\n```\n\nthe final values of e and f are\n\n```print e\n\n[0, 0, 0, 0, 0, 4, 4, 5, 6]\n\nprint f\n\n[[0, 0, 0], [0, 0, 4], [4, 5, 6]]```\n\nYou surmise that the two arrays must have the same space allocated.\n\nA) Check memory of both arrays, if they match that means the arrays are same.\n\nB) Do “np.array_equal(e, f)” and if the output is “True” then they both are same\n\nC) Print flags of both arrays by e.flags and f.flags; check the flag “OWNDATA”. If one of them is False, then both the arrays have same space allocated.\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct\n\nQuestion Context 9\n\nSuppose you want to join train and test dataset (both are two numpy arrays train_set and test_set) into a resulting array (resulting_set) to do data processing on it simultaneously. This is as follows:\n\n```train_set = np.array([1, 2, 3])\ntest_set = np.array([[0, 1, 2], [1, 2, 3]])\nresulting_set --> [[1, 2, 3], [0, 1, 2], [1, 2, 3]]```\n\n9) How would you join the two arrays?\n\nNote: Numpy library has been imported as np\n\nA) resulting_set = train_set.append(test_set)\n\nB) resulting_set = np.concatenate([train_set, test_set])\n\nC) resulting_set = np.vstack([train_set, test_set])\n\nD) None of these\n\nSolution: (C)\n\nBoth option A and B would do horizontal stacking, but we would like to have vertical stacking. So option C is correct\n\nQuestion Context 10\n\nSuppose you are tuning hyperparameters of a random forest classifier for the Iris dataset.\n\n Sepal_length Sepal_width Petal_length Petal_width Species 4.6 3.2 1.4 0.2 Iris-setosa 5.3 3.7 1.5 0.2 Iris-setosa 5.0 3.3 1.4 0.2 Iris-setosa 7.0 3.2 4.7 1.4 Iris-versicolor 6.4 3.2 4.5 1.5 Iris-versicolor\n\n10) What would be the best value for “random_state (Seed value)”?\n\nA) np.random.seed(1)\n\nB) np.random.seed(40)\n\nC) np.random.seed(32)\n\nD) Can’t say\n\nSolution: (D)\n\nThere is no best value for seed. It depends on the data.\n\nQuestion 11\n\nWhile reading a csv file with numpy, you want to automatically fill missing values of column “Date_Of_Joining” with date “01/01/2010”.\n\n Name Age Date_Of_Joining Total_Experience Andy 20 01/02/2013 0 Mandy 30 01/05/2014 10 Sandy 10 0 Bandy 40 01/10/2009 20\n\n11) Which command will be appropriate to fill missing value while reading the file with numpy?\n\nNote: numpy has been imported as np\n\nA) filling_values = (“-“, 0, 01/01/2010, 0)\ntemp = np.genfromtxt(filename, filling_values=filling_values)\n\nB) filling_values = (“-“, 0, 01/01/2010, 0)\n\nC) filling_values = (“-“, 0, 01/01/2010, 0)\ntemp = np.gentxt(filename, filling_values=filling_values)\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\n12) How would you import a decision tree classifier in sklearn?\n\nA) from sklearn.decision_tree import DecisionTreeClassifier\n\nB) from sklearn.ensemble import DecisionTreeClassifier\n\nC) from sklearn.tree import DecisionTreeClassifier\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct\n\n13) You have uploaded the dataset in csv format on google spreadsheet and shared it publicly. You want to access it in python, how can you do this?\n\nNote: Library StringIO has been imported as StringIO.\n\n```A)\n```B)\n\n```\n\nC)\n\n```link = https://docs.google.com/spreadsheets/d/...source = StringIO(requests.get(link).content))\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\nQuestion Context 14\n\nImagine, you have a dataframe train file with 2 columns & 3 rows, which is loaded in pandas.\n\nimport pandas as pd\n\n`train = pd.DataFrame({'id':[1,2,4],'features':[[\"A\",\"B\",\"C\"],[\"A\",\"D\",\"E\"],[\"C\",\"D\",\"F\"]]})`\n\nNow you want to apply a lambda function on “features” column:\n\n`train['features_t'] = train[\"features\"].apply(lambda x: \" \".join([\"_\".join(i.split(\" \")) for i in x]))`\n\n14) What will be the output of following print command?\n\n`print train['features_t']`\n\nA)\n\n0 A B C\n1 A D E\n2 C D F\n\nB)\n\n0 AB\n\n2 CDF\n\nC) Error\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\nQuestion Context 15\n\nWe have a multi-class classification problem for predicting quality of wine on the basis of its attributes. The data is loaded in a dataframe “df”\n\n fixed acidity volatile acidity citric acid residual sugar chlorides free sulfur dioxide total sulfur dioxide density pH sulphates Alcohol quality 0 7.4 0.70 0.00 1.9 0.076 11 34 0.9978 3.51 0.56 9.4 5 1 7.8 0.88 0.00 2.6 0.098 25 67 0.9968 3.20 0.68 9.8 5 2 7.8 0.76 0.04 2.3 0.092 15 54 0.9970 3.26 0.65 9.8 5 3 11.2 0.28 0.56 1.9 0.075 17 60 0.9980 3.16 0.58 9.8 6 4 7.4 0.70 0.00 1.9 0.076 11 34 0.9978 3.51 0.56 9.4 5\n\nThe quality column currently has values 1 to 10, but we want to substitute this by a binary classification problem. You want to keep the threshold for classification to 5, such that if the class is greater than 5, the output should be 1, else output should be 0.\n\nNote: Numpy has been imported as np and dataframe is set as df.\n\nA)\n\n```Y = df[quality].values\nY = np.array([1 if y >= 6 else 0 for y in Y])```\n\nB)\n\n```Y = df[quality].values()\nY = np.array([0 if y >= 6 else 1 for y in Y])```\n\nC)\n\n```Y = df[quality]\nY = np.array([0 if y >= 6 else 1 for y in Y])```\n\nD)None of these\n\nSolution: (A)\n\nOption A is correct\n\nQuestion Context 16\n\nSuppose we make a dataframe as\n\n```df = pd.DataFrame(['ff', 'gg', 'hh', 'yy'],\n[24, 12, 48, 30],\ncolumns = ['Name', 'Age'])```\n\n16) What is the difference between the two data series given below?\n\n1. df[‘Name’] and\n2. df.loc[:, ‘Name’]\n\nNote: Pandas has been imported as pd\n\nA) 1 is view of original dataframe and 2 is a copy of original dataframe.\n\nB) 2 is view of original dataframe and 1 is a copy of original dataframe.\n\nC) Both are copies of original dataframe.\n\nD) Both are views of original dataframe\n\nSolution: (B)\n\nOption B is correct. Refer the official docs of pandas library.\n\nQuestion Context 17\n\nConsider a function “fun” which is defined below:\n\n```def fun(x):\nx = 5\nreturn x```\n\nNow you define a list which has three numbers in it.\n\ng = [10,11,12]\n\n17) Which of the following will be the output of the given print statement:\n\n`print fun(g), g`\n\nA) [5, 11, 12] [5, 11, 12]\n\nB) [5, 11, 12] [10, 11, 12]\n\nC) [10, 11, 12] [10, 11, 12]\n\nD) [10, 11, 12] [5, 11, 12]\n\nSolution: (A)\n\nOption A is correct\n\nQuestion Context 18\n\nSigmoid function is usually used for creating a neural network activation function. A sigmoid function is denoted as\n\n```def sigmoid(x):\nreturn (1 / (1 + math.exp(-x)))```\n\n18) It is necessary to know how to find the derivatives of sigmoid, as it would be essential for backpropagation. Select the option for finding derivative?\n\nA)\n\n```import scipy\n\nDv = scipy.misc.derive(sigmoid)```\n\nB)\n\n```from sympy import \n\nx = symbol(x)\n\ny = sigmoid(x)\n\nDv = y.differentiate(x)```\n\nC)\n\n`Dv = sigmoid(x) (1 - sigmoid(x))`\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct\n\nQuestion Context 19\n\nSuppose you are given a monthly data and you have to convert it to daily data.\n\nFor example,", null, "For this, first you have to expand the data for every month (considering that every month has 30 days)\n\n19) Which of the following code would do this?\n\nNote: Numpy has been imported as np and dataframe is set as df.\n\nA) new_df = pd.concat([df]30, index = False)\n\nB) new_df = pd.concat([df]30, ignore_index=True)\n\nC) new_df = pd.concat([df]30, ignore_index=False)\n\nD) None of these\n\nSolution: (B)\n\nOption B is correct\n\nContext: 20-22\n\nSuppose you are given a dataframe df.\n\n`df = pd.DataFrame({'Click_Id':['A','B','C','D','E'],'Count':[100,200,300,400,250]})`\n\n20) Now you want to change the name of the column ‘Count’ in df to ‘Click_Count’. So, for performing that action you have written the following code.\n\n`df.rename(columns = {'Count':'Click_Count'})`\n\nWhat will be the output of print statement below?\n\n`print df.columns`\n\nNote: Pandas library has been imported as pd.\n\nA) [‘Click_Id’, ‘Click_Count’]\n\nB) [‘Click_Id’, ‘Count’]\n\nC) Error\n\nD) None of these\n\nSolution: (B)\n\nOption B is correct\n\nContext: 20-22\n\nSuppose you are given a data frame df.\n\n`df = pd.DataFrame({'Click_Id':['A','B','C','D','E'],'Count':[100,200,300,400,250]})`\n\n21) In many data science projects, you are required to convert a dataframe into a dictionary. Suppose you want to convert “df” into a dictionary such that ‘Click_Id’ will be the key and ‘Count’ will be the value for each key. Which of the following options will give you the desired result?\n\nNote: Pandas library has been imported as pd\n\nA) set_index(‘Click_Id’)[‘Count’].to_dict()\n\nB) set_index(‘Count’)[‘Click_Id’].to_dict()\n\nC) We cannot perform this task since dataframe and dictionary are different data structures\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\n22) In above dataframe df. Suppose you want to assign a df to df1, so that you can recover original content of df in future using df1 as below.\n\n`df1 = df`\n\nNow you want to change some values of “Count” column in df.\n\n`df.loc[df.Click_Id == 'A', 'Count'] += 100`\n\nWhich of the following will be the right output for the below print statement?\n\n`print df.Count.values,df1.Count.values`\n\nNote: Pandas library has been imported as pd.\n\nA) [200 200 300 400 250] [200 200 300 400 250]\n\nB) [100 200 300 400 250] [100 200 300 400 250]\n\nC) [200 200 300 400 250] [100 200 300 400 250]\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\n23) You write a code for preprocessing data, and you notice it is taking a lot of time. To amend this, you put a bookmark in the code so that you come to know how much time is spent on each code line. To perform this task, which of the following actions you would take?\n\n1. You put bookmark as time.sleep() so that you would know how much the code has “slept” literally\n2. You put bookmark as time.time() and check how much time elapses in each code line\n3. You put bookmark as datetime.timedelta(), so that you would find out differences of execution times\n4. You copy whole code in an Ipython / Jupyter notebook, with each code line as a separate block and write magic function %%timeit in each block\n\nA) 1 & 2\n\nB) 1,2 & 3\n\nC) 1,2 & 4\n\nD) All of the above\n\nSolution: (C)\n\nOption C is correct\n\n24) How would you read data from the file using pandas by skipping the first three lines?\n\nNote: pandas library has been imported as pd In the given file (email.csv), the first three records are empty.\n\n```,,,\n\n,,,\n\n,,,\n\n[email protected],aa,Owner,2014\n\n[email protected],bb,Member,2015\n\n[email protected],cc,Member,2017\n\n[email protected],dd,Member,2016```\n\nD) None of these\n\nSolution: (B)\n\nOption B is correct\n\n25) What should be written in-place of “method” to produce the desired outcome?\n\nGiven below is dataframe “df”:", null, "Now, you want to know whether BMI and Gender would influence the sales.\n\nFor this, you want to plot a bar graph as shown below:", null, "The code for this is:\n\n```var = df.groupby(['BMI','Gender']).Sales.sum()\n\nvar.unstack().plot(kind='bar', method, color=['red','blue'], grid=False)```\n\nA) stacked=True\n\nB) stacked=False\n\nC) stack=False\n\nD) None of these\n\nSolution: (A)\n\nIt’s a stacked bar chart.\n\n26) Suppose, you are given 2 list – City_A and City_B.\n\nCity_A = [‘1′,’2′,’3′,’4’]\n\nCity_B = [‘2′,’3′,’4′,’5’]\n\nIn both cities, some values are common. Which of the following code will find the name of all cities which are present in “City_A” but not in “City_B”.\n\nA) [i for i in City_A if i not in City_B]\n\nB) [i for i in City_B if i not in City_A]\n\nC) [i for i in City_A if i in City_B]\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\nQuestion Context 27\n\nSuppose you are trying to read a file “temp.csv” using pandas and you get the following error.\n\n```Traceback (most recent call last):\nFile \"<input>\", line 1, in<module>\nUnicodeEncodeError: 'ascii' codec can't encode character.```\n\n27) Which of the following would likely correct this error?\n\nNote: pandas has been imported as pd\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct, because encoding should be ‘utf-8’\n\n28) Suppose you are defining a tuple given below:\n\ntup = (1, 2, 3, 4, 5 )\n\nNow, you want to update the value of this tuple at 2nd index to 10. Which of the following option will you choose?\n\nA) tup(2) = 10\n\nB) tup = 10\n\nC) tup{2} = 10\n\nD) None of these\n\nSolution: (D)\n\nA tuple cannot be updated.\n\n29) You want to read a website which has url as “www.abcd.org”. Which of the following options will perform this task?\n\nA) urllib2.urlopen(www.abcd.org)\n\nB) requests.get(www.abcd.org)\n\nC) Both A and B\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct\n\nQuestion Context 30\n\nSuppose you are given the below web page\n\n```html_doc = “””\n<!DOCTYPE html>\n<htmllang=\"en\">\n<metacharset=\"utf-8\">\n<metaname=\"viewport\" content=\"width=device-width\">\n<title>udacity/deep-learning: Repo for the Deep Learning Nanodegree Foundations program.</title>\n<metaproperty=\"fb:app_id\" content=\"1401488693436528\">\n...\n\n“””```\n\n30) To read the title of the webpage you are using BeautifulSoup. What is the code for this?\n\nHint: You have to extract text in title tag\n\n1. from bs4 import BeautifulSoup\nsoup =BeautifulSoup(html_doc,’html.parser’)\nprint soup.title.name\n2. from bs4 import BeautifulSoup\nsoup =BeautifulSoup(html_doc,’html.parser’)\nprint soup.title.string\n3. from bs4 import BeautifulSoup\nsoup=BeautifulSoup(html_doc,’html.parser’)\nprint soup.title.get_text\n4. None of these\n\nSolution: (B)\n\nOption B is correct\nQuestion Context 31\n\nImagine, you are given a list of items in a DataFrame as below.\n\nD = [‘A’,’B’,’C’,’D’,’E’,’AA’,’AB’]\n\nNow, you want to apply label encoding on this list for importing and transforming, using LabelEncoder.\n\n```from sklearn.preprocessing import LabelEncoder\n\nle = LabelEncoder()```\n\n31) What will be the output of the print statement below ?\n\n`print le.fit_transform(D)`\n1. array([0, 2, 3, 4, 5, 6, 1])\n2. array([0, 3, 4, 5, 6, 1, 2])\n3. array([0, 2, 3, 4, 5, 1, 6])\n4. Any of the above\n\nSolution: (D)\n\nOption D is correct\n\n32) Which of the following will be the output of the below print statement?\n\n`print df.val == np.nan`\n\nAssume, you have defined a data frame which has 2 columns.\n\n```import numpy as np\n\ndf = pd.DataFrame({'Id':[1,2,3,4],'val':[2,5,np.nan,6]})```\n\nA) 0 False\n1 False\n2 False\n3 False\n\nB) 0 False\n1 False\n2 True\n3 False\n\nC) 0 True\n1 True\n2 True\n3 True\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\n33) Suppose the data is stored in HDFS format and you want to find how the data is structured. For this, which of the following command would help you find out the names of HDFS keys?\n\nNote: HDFS file has been loaded by h5py as hf.\n\nA) hf.key()\n\nB) hf.key\n\nC) hf.keys()\n\nD) None of these\n\nSolution: (C)\n\nOption C is correct\n\nQuestion Context 34\n\nYou are given reviews for movies below:\n\nreviews = [‘movie is unwatchable no matter how decent the first half is . ‘, ‘somewhat funny and well paced action thriller that has jamie foxx as a hapless fast talking hoodlum who is chosen by an overly demanding’, ‘morse is okay as the agent who comes up with the ingenious plan to get whoever did it at all cost .’]\n\nYour task is to find sentiments from the review above. For this, you first write a code to find count of individual words in all the sentences.\n\n```counts = Counter()\n\nfor i in range(len(reviews)):\nfor word in reviews[i].split(value):\ncounts[word] += 1```\n\n34)What value should we split on to get individual words?\n\n1. ‘ ‘\n2. ‘,’\n3. ‘.’\n4. None of these\n\nSolution: (A)\n\nOption A is correct\n\n35) How to set a line width in the plot given below?", null, "For the above graph, the code for producing the plot was\n\n```import matplotlib.pyplot as plt\nplt.plot([1,2,3,4])\nplt.show()```\n1. In line two, write plt.plot([1,2,3,4], width=3)\n2. In line two, write plt.plot([1,2,3,4], line_width=3\n3. In line two, write plt.plot([1,2,3,4], lw=3)\n4. None of these\n\nSolution: (C)\n\nOption C is correct\n\n36) How would you reset the index of a dataframe to a given list? The new index is given as:\n\nnew_index=[‘Safari’,’Iceweasel’,’Comodo Dragon’,’IE10′,’Chrome’]\n\nNote: df is a pandas dataframe\n\n http_status response_time Firefox 200 0.04 Chrome 200 0.02 Safari 404 0.07 IE10 404 0.08 Konqueror 301 1.00\n\nA) df.reset_index(new_index,)\n\nB) df.reindex(new_index,)\n\nC) df.reindex_like(new_index,)\n\nD) None of these\n\nSolution: (A)\n\nOption A is correct\n\n37) Determine the proportion of passengers survived based on their passenger class.\n\n PassengerId Survived Pclass Name Sex Age SibSp Parch Ticket Fare Cabin Embarked 0 1 0 3 Braund, Mr. Owen Harris male 22.0 1 0 A/5 21171 7.2500 NaN S 1 2 1 1 Cumings, Mrs. John Bradley (Florence Briggs Th… female 38.0 1 0 PC 17599 71.2833 C85 C 2 3 1 3 Heikkinen, Miss. Laina female 26.0 0 0 STON/O2. 3101282 7.9250 NaN S 3 4 1 1 Futrelle, Mrs. Jacques Heath (Lily May Peel) female 35.0 1 0 113803 53.1000 C123 S 4 5 0 3 Allen, Mr. William Henry male 35.0 0 0 373450 8.0500 NaN S\n\n1. crosstab(df_train[‘Pclass’], df_train[‘Survived’])\n2. proportion(df_train[‘Pclass’], df_train[‘Survived’])\n3. crosstab(df_train[‘Survived’], df_train[‘Pclass’])\n4. None of these\n\nSolution: (A)\n\nOption A is correct\n\n38) You want to write a generic code to calculate n-gram of the text. The 2-gram of this sentence would be [[“this, “is”], [“is”, “a”], [“a, “sample”], [“sample”, “text”]]\n\nWhich of the following code would be correct?\n\nFor a given a sentence:\n‘this is a sample text’.\n\n1. def generate_ngrams(text, n):\nwords = text.split(‘\\n’)\noutput = [] for i in range(len(words)-n+1):\nappend(words[i+1:i+n])\nreturn output\n2. def generate_ngrams(text, n):\nwords = text.split()\noutput = [] for i in range(len(words)-n+1):\nappend(words[i:i+n])\nreturn output\n3. def generate_ngrams(text, n):\nwords = text.split()\noutput = [] for i in range(len(words)-n+1):\nappend(words[i+1:i+n])\nreturn output\n4. None of these\n\nSolution: (B)\n\nOption B is correct\n\n39) Which of the following code will export dataframe (df) in CSV file, encoded in UTF-8 after hiding index & header labels.\n\n4. None of these\n\nSolution: (C)\n\nOption C is correct\n\n40) Which of the following is a correct implementation of mean squared error (MSE) metric?\n\nNote: numpy library has been imported as np.\n\n1. def MSE(real_target, predicted_target):\nreturn np.mean((np.square(real_target) – np.square(predicted_target)))\n2. def MSE(real_target, predicted_target):\nreturn np.mean((real_target – predicted_target)*2)\n3. def MSE(real_target, predicted_target):\nreturn np.sqrt(np.mean((np.square(real_target) – np.square(predicted_target))))\n4. None of the above\n\nSolution: (B)\n\nOption B is correct\n\n### End Notes\n\nIf you are learning Python, make sure you go through the test above. It will not only help you assess your skill. You can also see where you stand among other people in the community. If you have any questions or doubts, feel free to post them below.\n\n### Learn, compete, hack and get hired!", null, "###### Faizan Shaikh\n\nFaizan is a Data Science enthusiast and a Deep learning rookie. A recent Comp. Sc. undergrad, he aims to utilize his skills to push the boundaries of AI research.\n\n## 4 thoughts on \"40 Questions to test your skill in Python for Data Science\"", null, "###### Lia says:May 07, 2017 at 10:39 am\nFor Question Context 7: np.identity(3), which is option (B) does exist: https://docs.scipy.org/doc/numpy/reference/generated/numpy.identity.html But option (A) seems to be incorrect as we've got to write np.eye(3), don't we? Reply", null, "###### Faizan Shaikh says:May 08, 2017 at 5:14 am\nThank you for letting us know the typo. Actually option B does not exist because it should have been np.identity(), whereas it should have been np.eye() in option A. Reply", null, "###### issac says:June 06, 2017 at 10:48 am\nFor Question Context 16： I found there is instruction in python documnet about this issue http://pandas.pydata.org/pandas-docs/stable/indexing.html#returning-a-view-versus-a-copy and I think it's a little different from your answer. Could you explain why option B is corret ? Thanks Reply", null, "###### Faizan Shaikh says:July 08, 2017 at 6:21 pm\nThanks for the feedback. I have updated the same Reply" ]	[ null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/05/03143531/Python_graph.png", null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/04/29130118/29-mar-2017_720.png", null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/05/03141613/Image4.png", null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/05/03140933/image2.png", null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/05/03140941/image3.png", null, "https://cdn.analyticsvidhya.com/wp-content/uploads/2017/05/03135922/Image1.jpg", null, "https://secure.gravatar.com/avatar/fd90df6aafd127cc243609d0e1a97621", null, "https://secure.gravatar.com/avatar/8d05eee5fd922007b14b4e07676f12c6", null, "https://secure.gravatar.com/avatar/fd90df6aafd127cc243609d0e1a97621", null, "https://secure.gravatar.com/avatar/31eba86a1ff39432cacdd8026d9e521b", null, "https://secure.gravatar.com/avatar/fd90df6aafd127cc243609d0e1a97621", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78266305,"math_prob":0.86941147,"size":25783,"snap":"2021-43-2021-49","text_gpt3_token_len":7711,"char_repetition_ratio":0.12665348,"word_repetition_ratio":0.118395776,"special_character_ratio":0.3270372,"punctuation_ratio":0.17626588,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9855077,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,4,null,10,null,4,null,4,null,4,null,4,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-26T21:52:02Z\",\"WARC-Record-ID\":\"<urn:uuid:8cdb2dee-4c2c-47f0-871a-dbab03722b04>\",\"Content-Length\":\"156155\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7eb43d2d-d282-4433-88b3-8f057c780c55>\",\"WARC-Concurrent-To\":\"<urn:uuid:3e2d75c7-e7cd-4447-92c4-d67857e7ec2f>\",\"WARC-IP-Address\":\"104.22.54.101\",\"WARC-Target-URI\":\"https://www.analyticsvidhya.com/blog/2017/05/questions-python-for-data-science/\",\"WARC-Payload-Digest\":\"sha1:Z7IXCQFF2CHQFR4XFVP73MVEUYDAJOJD\",\"WARC-Block-Digest\":\"sha1:DA3VDSHZTJD2VQ5WHON4UKFI4XSO2LXM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323587926.9_warc_CC-MAIN-20211026200738-20211026230738-00663.warc.gz\"}"}
https://www.fossilhunters.xyz/thermodynamics/info-fza.html	[ "## Info\n\nFor T > 423 K: use ideal-gas law for v h = 2783 + 1.62 x (T-423) + 3.33 x KM CP -4231) s = 8.69 + 1.62 x bi(T/423) + 6.65 x 104 (T-423)\n\nFor T > 573 K: use ideal gas law for v h = 3065 + 1.71 x (T - 573) + 2.94 x 10\" CP - 573») s = 7.71 + 1.71 x 111(7/573) + 5.87 x IQ-« (T-S13)\n\nFor T > 423 K: use ideal-gas law for v h = 2783 + 1.62 x (T-423) + 3.33 x KM CP -4231) s = 8.69 + 1.62 x bi(T/423) + 6.65 x 104 (T-423)\n\nFor T > 573 K: use ideal gas law for v h = 3065 + 1.71 x (T - 573) + 2.94 x 10\" CP - 573») s = 7.71 + 1.71 x 111(7/573) + 5.87 x IQ-« (T-S13)\n\nT, K\n\nv, m3/kg\n\nh, kj/kg\n\ns, kJ/kg-K\n\nT, K\n\nv, m3/kg\n\nh, kj/kg\n\ns, kJ/kg-K", null, "" ]	[ null, "https://www.fossilhunters.xyz/images/downloads/eJw9yksKgCAQANDbuPQHagTSUWJSyaF0pAyvH23avNXLvbdZiIEHjrRJ6XghIh6oiEijngRxLVBhT5e4U4285bZA6EjVNwz9uRL7I0ZvrLITBKeM0Zplr41k4_MF7K0j-w.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73997325,"math_prob":0.99996173,"size":1060,"snap":"2019-35-2019-39","text_gpt3_token_len":432,"char_repetition_ratio":0.10227273,"word_repetition_ratio":0.65,"special_character_ratio":0.504717,"punctuation_ratio":0.12676056,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9997347,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T09:03:14Z\",\"WARC-Record-ID\":\"<urn:uuid:a485351e-6c41-42c4-8822-d139cc4733a9>\",\"Content-Length\":\"20481\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:414714eb-4269-42a1-bebb-88002bcd03f2>\",\"WARC-Concurrent-To\":\"<urn:uuid:4a6f4385-5323-477d-82c9-6335220cac82>\",\"WARC-IP-Address\":\"104.24.124.40\",\"WARC-Target-URI\":\"https://www.fossilhunters.xyz/thermodynamics/info-fza.html\",\"WARC-Payload-Digest\":\"sha1:DOCNXJPFLJ2UVBZAAK4P2QYFGHSPI4U4\",\"WARC-Block-Digest\":\"sha1:4WPU656ZJDR32FHTPRHOTZZJ6QFUIV3T\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573264.27_warc_CC-MAIN-20190918085827-20190918111827-00375.warc.gz\"}"}
https://math.stackexchange.com/questions/3192319/what-is-the-characteristic-of-a-local-ring/3192449	[ "What is the Characteristic of a local ring? [closed]\n\nWhat is the Characteristic of a local ring ?\n\nWe define Characteristic of a Commutative ring with $$1$$ say, $$A$$ in the following way: Define a ring homomorphism $$\\phi: \\mathbb{Z} \\to A$$ by $$\\phi(n)=n \\cdot 1.$$ Since $$\\mathbb{Z}$$ is a PID, $$\\text{ker}(\\phi)$$ is a principal ideal. If $$\\text{ker}(\\phi)=m\\mathbb{Z},$$ we define the Characteristic of the ring $$A$$ to be $$m.$$ We know that Characteristic of a domain is either $$0$$ or a prime $$p.$$\n\nHow do I classify the Characteristic of a local ring $$(A,m)$$\n\nclosed as unclear what you're asking by Arnaud D., max_zorn, Javi, José Carlos Santos, RamiroApr 19 at 16:25\n\nPlease clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.\n\n• The characteristic of a ring need not be prime: for instance, the characteristic of $\\mathbb Z/(6)$ is $6$. It only needs to be a prime if the ring is an integral domain. – Claudius Apr 18 at 12:23\n• I don't really understand the question. You have the definition of the characteristic of a ring, so theoretically you now how to find it, right? – Arnaud D. Apr 18 at 12:28\n• It is not clear to me what you mean by “find the characteristic“. Are you asking about the definition of the characteristic of a local ring? – Claudius Apr 18 at 12:29\n• Like integral domain you can classify the characteristic of a local ring. – user371231 Apr 18 at 12:45\n• I much as I guess $6$ cannot be characteristic of a local ring – user371231 Apr 18 at 12:49\n\nThe characteristic of a local ring is a power of a prime or $$0$$, and any of these happens in some local rings.\n\nThat they all happen is easy : you may look at fields for characteristic $$0$$, and $$\\mathbb{Z}/p^n\\mathbb{Z}$$ for powers of primes.\n\nNow let $$(R,m)$$ be a local ring, and $$n$$ its characteristic, which we assume to be $$>0$$. Suppose $$n=ab, a\\land b = 1$$. Then the ideals $$I=\\{x\\in R, ax = 0\\}$$ and $$J=\\{x\\in R, bx=0\\}$$ are comaximal : indeed $$a\\in J, b\\in I$$ and there are $$u,v$$ with $$au+bv=1$$ so $$1\\in I+J$$.\n\nTherefore by locality, one of them is $$R$$ (otherwise they would both be $$\\subset m$$). If it is $$I$$, then $$a = 0$$ in $$R$$ and so $$R$$ has characteristic $$\\mid a$$ so $$b=1$$. If it's $$J$$, then $$a=1$$. In any case, $$a=1 \\lor b=1$$, so that $$n$$ is a power of a prime.\n\nIf you already know a local ring has only trivial idempotents, then you can reason this way:\n\nSuppose the characteristic of a ring $$R$$ is finite, say $$n$$, and is divisible by more than one prime. The ring contains a copy of $$\\mathbb Z/n\\mathbb Z$$. So to show $$R$$ isn't local, it suffices to show that $$\\mathbb Z/n\\mathbb Z$$ contains a nontrivial idempotent, so that $$R$$ will also contain a nontrivial idempotent.\n\nBy reasoning with the Chinese remainder theorem, you can quickly see that $$\\mathbb Z/n\\mathbb Z$$ is isomorphic to $$\\mathbb Z/p^k\\mathbb Z$$ for primes $$p$$ dividing $$n$$ and powers $$k$$ depending on $$p$$. Since there is more than one prime dividing $$n$$ (our assumption) there is at least one nontrivial idempotent splitting $$\\mathbb Z/n\\mathbb Z$$ into two pieces. This is clearly a nontrivial idempotent of $$R$$ too.\n\nSo by contrapositive, we have shown that a local ring $$R$$ must have either characteristic $$0$$, or else it has finite characteristic that is a power of a prime.\n\nSince we already have examples of such rings for every such finite characteristic, we can see these are precisely the characteristics that are possible." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73721415,"math_prob":0.9998975,"size":499,"snap":"2019-26-2019-30","text_gpt3_token_len":156,"char_repetition_ratio":0.18181819,"word_repetition_ratio":0.05,"special_character_ratio":0.3046092,"punctuation_ratio":0.10377359,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000027,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-17T19:21:28Z\",\"WARC-Record-ID\":\"<urn:uuid:60b0fa4e-8592-4c20-929f-ed7970222ac4>\",\"Content-Length\":\"133671\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a74ef8f2-b99e-49be-802d-cd62079b87b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:913fb40a-a8e1-41f8-97f7-66db343e057f>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/3192319/what-is-the-characteristic-of-a-local-ring/3192449\",\"WARC-Payload-Digest\":\"sha1:RQPP7RQDNDMIZB2XA2X7BSGBS27EY3TS\",\"WARC-Block-Digest\":\"sha1:WAPBJHNDIOMN2FL5WHYKJU2XXTTWQPOZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998558.51_warc_CC-MAIN-20190617183209-20190617205209-00447.warc.gz\"}"}