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Biography of Leonard Euler
The definition of “genius” is best suited to the great mathematician and brilliant physicist Leonard Euler.
Childhood and early years
Euler was born on April 15, 1707 in Basel, Switzerland. His father, Paul Euler, was a pastor of the Reformed Church. His mother’s father, Margaret Brucker, was also a pastor. Leonard had two younger
sisters – Anna Maria and Maria Magdalena. Soon after the birth of his son, the family moved to the town of Riyon. The boy’s father was a friend of Johann Bernoulli, a well-known European
mathematician, who had a great influence on Leonard. At thirteen, Euler Jr. enters the University of Basel, and in 1723 receives a master’s degree in philosophy. In his dissertation, Euler compares
the philosophy of Newton and Descartes. Johann Bernoulli, who gave the boy private lessons on Saturdays, quickly recognizes the boy’s outstanding ability for mathematics and convinces him to leave
early theology and focus
on mathematics.
In 1727, Euler took part in the competition, organized by the Paris Academy of Sciences, for the best technique for installing ship masts. Leonard takes the second place, while the first goes to
Pierre Bouguer, who later becomes known as the “father of shipbuilding.” Euler takes part in this competition every year, having received twelve of these prestigious awards for his life.
St. Petersburg
May 17, 1727 Euler entered the medical department of the Imperial Russian Academy of Sciences in St. Petersburg, but almost immediately switched to the Faculty of Mathematics. However, because of the
unrest in Russia, June 19, 1741 Euler transferred to the Berlin Academy. There the scientist will serve for about 25 years, having written more than 380 scientific articles during this time. In 1755,
he was elected a foreign member of the Royal Swedish Academy of Sciences.
In the early 1760s, Euler received a proposal to teach the princess Anhalt-Dessau to the sciences, to which the scientist would write more than 200 letters that were included in the highly popular
of Euler’s Letters on various subjects of natural philosophy addressed to the German princess. The book not only demonstrates the ability of the scientist to reason on all sorts of topics in the
field of mathematics and physics, but also is an expression of his personal and religious views. It is interesting that this book is known better than all of its mathematical works. It was published
both in Europe and in the United States of America. The reason for such popularity of these letters was the amazing ability of Euler in an accessible form to bring scientific information to the
common man in the street.
The uniqueness of this work was also in the fact that in 1735 the scientist was almost completely blind to the right eye, and in 1766 his left eye was struck by cataracts. But, despite this, he
continues his work and in 1755 he writes an average of one mathematical article a week.
In 1766, Euler accepted the offer to return to the St. Petersburg Academy, and spent the rest of his life in Russia. However, his second visit to this country is not so successful for him: in 1771
the fire destroys his house, and, after this, in 1773 he loses his wife to Katarina.
Personal life
January 7, 1734 Euler marries Catherine Gzel. In 1773, after 40 years of family life, Katarina died. Three years later, Euler marries her half-sister, Salome Abigail Gzel, with whom she will spend
the rest of her life.
Death and heritage
September 18, 1783, after a family dinner, Euler has a cerebral hemorrhage, after which, after several hours, he dies. The scientist was buried at the Smolensk Lutheran cemetery on Vasilievsky
Island, next to his first wife Katarina. In 1837, the Russian Academy of Sciences put on the bust on Leonard Euler’s grave a pedestal in the form of a rector’s chair, next to the gravestone. In 1956,
on the occasion of the 250th anniversary of the birth of the scientist, the monument and remains were transferred to the cemetery of the eighteenth century under the monastery of Alexander Nevsky.
In memory of his enormous contribution to science, Euler’s portrait appeared on Swiss 10-franc banknotes of the sixth series, as well as on a number of Russian, Swiss and German brands. In his honor
the asteroid “2002 Euler” is named. On May 24, the Lutheran church honors his memory on the saints calendar, as Euler was a staunch supporter of Christianity and fervently believed in the biblical
Mathematical Notation System
Among all the various works of Euler, the most notable is the representation of the theory of functions. He first introduced the notation f – the function “f” in the argument “x”. Euler also defined
mathematical notation for trigonometric functions as we know them now, introduced the letter “e” for the base of the natural logarithm, the Greek letter “Σ” for the total sum and the letter “i” for
determining the imaginary unit.
Euler approved the use of exponential functions and logarithms in analytical proofs. He discovered the way to expand various logarithmic functions into a power series, and also successfully proved
the application of logarithms to negative and complex numbers. Thus, Euler greatly expanded the mathematical application of logarithms.
This great mathematician also explained in detail the theory of higher transcendental functions and presented an innovative approach to solving quadratic equations. He discovered the technique of
calculating integrals using complex limits. He also developed the formula of the calculus of variations, which was called the Euler-Lagrange equation.
Number theory
Euler proved Fermat’s small theorem, Newton’s identities, Fermat’s theorem on the sums of two squares, and also considerably advanced the proof of Lagrange’s theorem on the sum of four squares. He
introduced valuable additions to the theory of perfect numbers, over which not just a mathematician worked with enthusiasm.
Physics and Astronomy
Euler made a significant contribution to the solution of the equation of the Euler-Bernoulli beam, which became one of the basic equations used in engineering. His scientist applied his analytical
methods not only in classical mechanics, but also in solving celestial problems. For his achievements in the field of astronomy, Euler received numerous awards from the Paris Academy. Based on the
knowledge of the true nature of comets and calculating the parallax of the Sun, the scientist clearly calculated the orbits of comets and other celestial bodies. With the help of these calculations,
accurate tables of celestial coordinates were compiled.
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5.3: Stoichiometry of Gaseous Substances, Mixtures, and Reactions
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Learning Objectives
• Use the ideal gas law to compute gas densities and molar masses
• Perform stoichiometric calculations involving gaseous substances
• State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier changed chemistry from a
qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition
of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph-Louis
Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it."
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” We can answer the question with masses of substances
or volumes of solutions. However, we can also answer this question another way: with volumes of gases. We can use the ideal gas equation to relate the pressure, volume, temperature, and number of
moles of a gas. Here we will combine the ideal gas equation with other equations to find gas density and molar mass. We will deal with mixtures of different gases, and calculate amounts of substances
in reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts we have already discussed.
Density of a Gas
Recall that the density of a gas is its mass to volume ratio, \(ρ=\dfrac{m}{V}\). Therefore, if we can determine the mass of some volume of a gas, we will get its density. The density of an unknown
gas can used to determine its molar mass and thereby assist in its identification. The ideal gas law, PV = nRT, provides us with a means of deriving such a mathematical formula to relate the density
of a gas to its volume in the proof shown in Example \(\PageIndex{1}\).
Example \(\PageIndex{1}\): Derivation of a Density Formula from the Ideal Gas Law
Use PV = nRT to derive a formula for the density of gas in g/L
\[PV = nRT\]
Rearrange to get (mol/L):
Multiply each side of the equation by the molar mass (or molecular weight), ℳ. When moles are multiplied by ℳ in g/mol, g are obtained:
Exercise \(\PageIndex{1}\)
A gas was found to have a density of 0.0847 g/L at 17.0 °C and a pressure of 760 torr. What is its molar mass? What is the gas?
\[ρ=\dfrac{Pℳ}{RT} \]
\[\mathrm{0.0847\:g/L=760\cancel{torr}×\dfrac{1\cancel{atm}}{760\cancel{torr}}×\dfrac{\mathit{ℳ}}{0.0821\: L\cancel{atm}/mol\: K}×290\: K}\]
ℳ = 2.02 g/mol; therefore, the gas must be hydrogen (H[2], 2.02 g/mol)
We must specify both the temperature and the pressure of a gas when calculating its density because the number of moles of a gas (and thus the mass of the gas) in a liter changes with temperature or
pressure. Gas densities are often reported at STP.
Molar Mass of a Gas
Another useful application of the ideal gas law involves the determination of molar mass. By definition, the molar mass of a substance is the ratio of its mass in grams, m, to its amount in moles, n:
\[ℳ=\mathrm{\dfrac{grams\: of\: substance}{moles\: of\: substance}}=\dfrac{m}{n}\]
The ideal gas equation can be rearranged to isolate n:
and then combined with the molar mass equation to yield:
This equation can be used to derive the molar mass of a gas from measurements of its pressure, volume, temperature, and mass.
The Pressure of a Mixture of Gases: Dalton’s Law
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would
exert if it were present alone in the container ( Figure \(\PageIndex{2}\)). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by
Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:
In the equation P[Total] is the total pressure of a mixture of gases, P[A] is the partial pressure of gas A; P[B] is the partial pressure of gas B; P[C] is the partial pressure of gas C; and so on.
Figure \(\PageIndex{2}\): If equal-volume cylinders containing gas A at a pressure of 300 kPa, gas B at a pressure of 600 kPa, and gas C at a pressure of 450 kPa are all combined in the same-size
cylinder, the total pressure of the mixture is 1350 kPa.
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution
divided by the total number of moles of all components:
where P[A], X[A], and n[A] are the partial pressure, mole fraction, and number of moles of gas A, respectively, and n[Total] is the number of moles of all components in the mixture.
Example \(\PageIndex{2}\): The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 × 10^−3 mol of H[2], 1.00 × 10^−3 mol of He, and 3.00 × 10^−4 mol of Ne at 35 °C.
1. What are the partial pressures of each of the gases?
2. What is the total pressure in atmospheres?
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using \(P=\dfrac{nRT}{V}\):
The total pressure is given by the sum of the partial pressures:
Exercise \(\PageIndex{2}\)
A 5.73-L flask at 25 °C contains 0.0388 mol of N[2], 0.147 mol of CO, and 0.0803 mol of H[2]. What is the total pressure in the flask in atmospheres?
1.137 atm
Collection of Gases over Water
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside
the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure \(\PageIndex{3}\)), the
pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
Figure \(\PageIndex{3}\): When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is
appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier
discussion of manometers).
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid
water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water
vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas
only, without water vapor.
Figure \(\PageIndex{4}\): This graph shows the vapor pressure of water at sea level as a function of temperature.
The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure \(\PageIndex{4}\)); more detailed
information on the temperature dependence of water vapor can be found in Table \(\PageIndex{1}\), and vapor pressure will be discussed in more detail in the next chapter on liquids.
Table \(\PageIndex{1}\): Vapor Pressure of Ice and Water in Various Temperatures at Sea Level
Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr) Temperature (°C) Pressure (torr)
–10 1.95 18 15.5 30 31.8
–5 3.0 19 16.5 35 42.2
–2 3.9 20 17.5 40 55.3
0 4.6 21 18.7 50 92.5
2 5.3 22 19.8 60 149.4
4 6.1 23 21.1 70 233.7
6 7.0 24 22.4 80 355.1
8 8.0 25 23.8 90 525.8
10 9.2 26 25.2 95 633.9
12 10.5 27 26.7 99 733.2
14 12.0 28 28.3 100.0 760.0
16 13.6 29 30.0 101.0 787.6
Example \(\PageIndex{4}\): Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure \(\PageIndex{3}\), what is the partial pressure of argon?
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:
Rearranging this equation to solve for the pressure of argon gives:
The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr (Appendix E), so:
Exercise \(\PageIndex{4}\)
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature
and pressure?
0.583 L
Chemical Stoichiometry and Gases
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions. We have previously measured quantities of reactants and products using masses for
solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the
ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of
molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the
reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple
proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to
a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in Figure \(\PageIndex{4}\). According to Avogadro’s law, equal volumes of gaseous N[2], H[2], and NH[3], at the same temperature and pressure, contain the
same number of molecules. Because one molecule of N[2] reacts with three molecules of H[2] to produce two molecules of NH[3], the volume of H[2] required is three times the volume of N[2], and the
volume of NH[3] produced is two times the volume of N[2].
Figure \(\PageIndex{5}\): One volume of N[2] combines with three volumes of H[2] to form two volumes of NH[3].
Example \(\PageIndex{5}\): Reaction of Gases
Propane, C[3]H[8](g), is used in gas grills to provide the heat for cooking. What volume of O[2](g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same
conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
The ratio of the volumes of C[3]H[8] and O[2] will be equal to the ratio of their coefficients in the balanced equation for the reaction:
&\ce{C3H8}(g)+\ce{5O2}(g) ⟶ &&\ce{3CO2}(g)+\ce{4H2O}(l)\\
\ce{&1\: volume + 5\: volumes &&3\: volumes + 4\: volumes}
From the equation, we see that one volume of C[3]H[8] will react with five volumes of O[2]:
\[\mathrm{2.7\cancel{L\:C_3H_8}×\dfrac{5\: L\:\ce{O2}}{1\cancel{L\:C_3H_8}}=13.5\: L\:\ce{O2}}\]
A volume of 13.5 L of O[2] will be required to react with 2.7 L of C[3]H[8].
Exercise \(\PageIndex{5}\)
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C[2]H[2], at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 × 10^3 L of O[2] at 0 °C and 1 atm,
will be required to burn the acetylene?
\[\ce{2C2H2 + 5O2⟶4CO2 + 2H2O} \]
3.34 tanks (2.34 × 10^4 L)
Example \(\PageIndex{6}\): Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H[2](g),
measured under the same conditions, was required to prepare this amount of ammonia by reaction with N[2]?
\[\ce{N2}(g)+\ce{3H2}(g)⟶\ce{2NH3}(g) \]
Because equal volumes of H[2] and NH[3] contain equal numbers of molecules and each three molecules of H[2] that react produce two molecules of NH[3], the ratio of the volumes of H[2] and NH[3] will
be equal to 3:2. Two volumes of NH[3], in this case in units of billion ft^3, will be formed from three volumes of H[2]:
\[\mathrm{683\cancel{billion\:ft^3\:NH_3}×\dfrac{3\: billion\:ft^3\:H_2}{2\cancel{billion\:ft^3\:NH_3}}=1.02×10^3\:billion\:ft^3\:H_2}\]
The manufacture of 683 billion ft^3 of NH[3] required 1020 billion ft^3 of H[2]. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Exercise \(\PageIndex{6}\)
What volume of O[2](g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C[2]H[4](g), measured under the same conditions of temperature and pressure? The products are CO[2]
and water vapor.
51.0 L
Example \(\PageIndex{7}\): Volume of Gaseous Product
What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?
To convert from the mass of gallium to the volume of H[2](g), we need to do something like this:
The first two conversions are:
\[\mathrm{8.88\cancel{g\: Ga}×\dfrac{1\cancel{mol\: Ga}}{69.723\cancel{g\: Ga}}×\dfrac{3\: mol\:H_2}{2\cancel{mol\: Ga}}=0.191\:mol\: H_2}\]
Finally, we can use the ideal gas law:
\[V_\mathrm{H_2}=\left(\dfrac{nRT}{P}\right)_\mathrm{H_2}=\mathrm{\dfrac{0.191\cancel{mol}×0.08206\: L\cancel{atm\:mol^{−1}\:K^{−1}}×300\: K}{0.951\:atm}=4.94\: L}\]
Exercise \(\PageIndex{7}\)
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO[2] at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in oxygen?
1.30 × 10^3 L
Greenhouse Gases and Climate Change
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches
the earth, almost \(\dfrac{1}{3}\) is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as
infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. However, most of this IR radiation is absorbed by certain substances in the atmosphere, known as
greenhouse gases, which re-emit this energy in all directions, trapping some of the heat. This maintains favorable living conditions—without atmosphere, the average global average temperature of 14
°C (57 °F) would be about –19 °C (–2 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing
the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure \(\PageIndex{6}\)).
Figure \(\PageIndex{6}\): Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels,
warming the planet and causing more extreme weather events.
There is strong evidence from multiple sources that higher atmospheric levels of CO[2] are caused by human activity, with fossil fuel burning accounting for about \(\dfrac{3}{4}\) of the recent
increase in CO[2]. Reliable data from ice cores reveals that CO[2] concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its
highest level in 20 million years. In recent years, the CO[2] concentration has increased from historical levels of below 300 ppm to almost 400 ppm today (Figure \(\PageIndex{7}\)).
Figure \(\PageIndex{7}\): Figure CO[2] levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.
Contributors and Attributions
Contributors and Attributions
The ideal gas law can be used to derive a number of convenient equations relating directly measured quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement
of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures
to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.
Key Equations
• P[Total] = P[A] + P[B] + P[C] + … = Ʃ[i]P[i]
• P[A] = X[A] P[Total]
• \(X_A=\dfrac{n_A}{n_{Total}}\)
1. “Quotations by Joseph-Louis Lagrange,” last modified February 2006, accessed February 10, 2015, www-history.mcs.st-andrews.ac.../Lagrange.html
Dalton’s law of partial pressures
total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases.
mole fraction (X)
concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressure
pressure exerted by an individual gas in a mixture
vapor pressure of water
pressure exerted by water vapor in equilibrium with liquid water in a closed container at a specific temperature
Contributors and Attributions | {"url":"https://eng.libretexts.org/Courses/Bucknell_University/CEEG_445%3A_Environmental_Engineering_Chemistry_(Fall_2020)/05%3A_Gas_Laws/5.03%3A_Stoichiometry_of_Gaseous_Substances_Mixtures_and_Reactions","timestamp":"2024-11-05T00:50:33Z","content_type":"text/html","content_length":"167874","record_id":"<urn:uuid:514461c1-03d8-4068-ad4b-1ffba6cf339d>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00307.warc.gz"} |
University of Pittsburgh
The William Lowell Putnam Mathematical Competition is the preeminent mathematics competition for undergraduate college students in the United States and Canada. The Putnam Competition takes place
annually on the first Saturday of December. The competition consists of two 3-hour sessions, one in the morning and one in the afternoon. During each session, participants work individually on 6
challenging mathematical problems.
Due to COVID-19 the exam will take place on February 20, 2021.
For more information on the exams or if you would like to see previoue exams, click here.
Registration for the exam:
Committee Members | {"url":"https://www.mathematics.pitt.edu/seminar-colloquia/putnam-seminar","timestamp":"2024-11-08T05:57:52Z","content_type":"text/html","content_length":"98510","record_id":"<urn:uuid:a0e30abc-6448-43de-bc81-675fdf964b9e>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00662.warc.gz"} |
NFA vs. DFA: Know the Difference
NFA (Non-deterministic Finite Automaton) allows multiple transitions for a single input, with non-deterministic paths. DFA (Deterministic Finite Automaton) has only one transition for each input,
ensuring a deterministic path.
Key Differences
An NFA, or Non-deterministic Finite Automaton, allows for multiple or even zero transitions for a single input symbol in a given state. In contrast, a DFA, which stands for Deterministic Finite
Automaton, strictly allows only one transition per input symbol in any state.
In an NFA, computation can branch into several paths for the same input, leading to multiple possible states simultaneously. DFA, however, follows a single path for a given input, leading to a unique
next state.
NFAs can include ε-transitions (transitions without any input symbol), enabling a jump from one state to another without consuming any input. DFAs lack ε-transitions; each state transition is
explicitly triggered by an input symbol.
The NFA's flexibility in state transitions often makes it easier to construct but potentially more complex to analyze. DFA's deterministic nature, conversely, leads to simpler analysis but can
require more states than an equivalent NFA.
An NFA can be converted into an equivalent DFA, usually resulting in an exponential increase in the number of states. This conversion isn't necessary for a DFA, as it is already deterministic.
Comparison Chart
State Transitions
Multiple or zero for one input in each state.
Exactly one for each input in any state.
Allows ε-transitions.
Does not allow ε-transitions.
Computation Path
Can branch into multiple paths.
Follows a single, unique path.
Ease of Construction
Generally easier to construct.
More complex to construct due to more states.
Conversion to Other Form
Can be converted to DFA (usually more states).
No conversion needed; already deterministic.
Analysis Complexity
More complex to analyze.
Simpler and more straightforward to analyze.
Expressive Power
Same as DFA (equally powerful).
Same as NFA (equally powerful).
NFA and DFA Definitions
NFA can have transitions without consuming input symbols, known as ε-transitions.
An NFA can jump states without reading any input, providing a shortcut in the computational process.
DFA has a unique computation path for every input string, leading to a deterministic operation.
In a DFA, a specific input string always results in the same sequence of state transitions.
NFA may have states with no outgoing transitions for some input symbols.
In an NFA, a certain input symbol might lead to no next state, representing a dead end.
DFA is often more state-heavy than its NFA counterpart for representing the same language.
Converting an NFA into a DFA usually results in an increase in the number of states.
NFA is a finite automaton where for some cases, multiple transitions are possible for the same input from a state.
In designing a language recognizer, an NFA can represent multiple possibilities with less complexity.
DFA is widely used in applications like regular expression processing and lexical analysis.
DFAs are employed in compilers for deterministic parsing of programming language syntax.
NFA allows for the acceptance of a language by reaching any of its accepting states.
An NFA can recognize a string as valid if it ends in any one of its designated accepting states.
DFA is a type of finite automaton where each state has exactly one transition per input symbol.
A DFA, used in pattern matching, ensures a single, predictable path for any input sequence.
NFA is often used for theoretical purposes due to its flexibility and simplicity in construction.
NFAs are preferred in theoretical computer science for illustrating concepts due to their non-deterministic nature.
DFA does not allow ε-transitions and requires an explicit input for state transition.
In a DFA-based lexical analyzer, every character of the input precisely determines the next state.
Repeatedly Asked Queries
Can an NFA have ε-transitions?
Yes, NFAs can have ε-transitions allowing state transitions without consuming input.
What is a DFA?
A Deterministic Finite Automaton with exactly one transition per input symbol in each state.
What is an NFA?
A Non-deterministic Finite Automaton where multiple transitions may exist for a single input in a state.
Are NFAs and DFAs equally powerful in language recognition?
Yes, both can recognize exactly the same set of languages (regular languages).
Why might one prefer using an NFA in theoretical models?
NFAs are often preferred for their simplicity and flexibility in representing various computational scenarios.
What makes DFAs more suitable for practical applications?
DFAs' deterministic nature makes them simpler to analyze and implement, especially in programming.
How does the computational path of an NFA differ from a DFA?
NFA can have multiple computational paths for the same input, while DFA has a single deterministic path.
Is it easier to construct an NFA or a DFA?
Generally, NFAs are easier to construct due to their flexible transition rules.
Do DFAs allow ε-transitions?
No, DFAs do not permit ε-transitions; each transition requires an input symbol.
Can an NFA be converted into a DFA?
Yes, any NFA can be converted into an equivalent DFA, often resulting in more states.
How does the introduction of ε-transitions affect the NFA's power?
ε-transitions add flexibility but do not increase the NFA's ability to recognize more languages.
Can DFAs have multiple accepting states?
Yes, like NFAs, DFAs can also have multiple accepting states.
How does the complexity of analyzing an NFA compare to a DFA?
Analyzing an NFA is generally more complex due to its multiple possible paths, unlike the straightforward analysis of a DFA.
Are there languages that can be recognized by an NFA but not by a DFA?
No, NFAs and DFAs have the same expressive power and can recognize all regular languages.
What is the typical use case for DFAs in software development?
DFAs are commonly used in lexical analysis and parsing, such as in compilers and interpreters.
Is the conversion from NFA to DFA always unique?
The conversion process is systematic, but the resulting DFA can sometimes be minimized further.
Can NFAs and DFAs be used for recognizing context-free languages?
No, both NFAs and DFAs are limited to recognizing regular languages.
Can a DFA have states with no outgoing transitions?
Yes, a DFA can have such states, typically representing a dead-end or error in the computation.
How does backtracking differ in NFA and DFA?
NFAs inherently support backtracking due to their non-determinism, while DFAs do not backtrack as they follow a single path.
Can the state transition diagrams of NFAs and DFAs be visually distinguished?
Yes, NFAs may show multiple arrows for the same input from a state, while DFAs have a single arrow for each input per state.
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About Author
Written by
Shumaila Saeed
Shumaila Saeed, an expert content creator with 6 years of experience, specializes in distilling complex topics into easily digestible comparisons, shining a light on the nuances that both inform and
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The Department of Mathematics at UC Davis is widely recognized for its dynamic and innovative approach to mathematics. With an active research program in both pure and applied mathematics,
opportunities to work closely with faculty, a university library system that is one of the finest in the nation, and a desirable location halfway between the Bay Area and the Sierra Nevada, the
University of California, Davis has much to offer the dedicated student of mathematics. | {"url":"https://www.math.ucdavis.edu/grad","timestamp":"2024-11-09T09:13:06Z","content_type":"text/html","content_length":"28968","record_id":"<urn:uuid:1f9850bb-0477-4f63-890d-ab80896f5065>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00883.warc.gz"} |
My math teacher.
I just thought this was a little funny.
At my school we usually have pie day on March 14th (3.14). But that falls on a weekend this year, so our math department decided on a math awareness week. They had students design t-shirts for
prizes. She was showing us the designs in class and it was funny because one of them said “I <3 math.” and she said what does this mean. And then one said “math ftw” and she said is this a bad phrase
or what does it mean. It was funny to see how each generation is so different. I also think one of the best designs was a calculator with a chain around it. It was supposed to be t.i. like the
wrapper because ti is texas instruments.
I read this post earlier today and I’ve given it some thought. What I’ve thought about are the experiences that I’ve had working with and being around students and young people in the FIRST programs.
The majority of these folks have been helpful and kind and have found humor in the fact that I often have to ask for clarification and a broader understanding of the lingo, the attitudes, and the
general coolness of this generation. In ChiefDelphi, teenagers, college students, and young adults have been open, generous, and thoughtful when trying to educate me and bring me up-to-date with
terminology, bridging differences in experiences. I’ve felt very lucky. Hopefully, students in schools and other programs are as generous, helpful, and fun as the majority of those that are in the
community of FIRST.
Perhaps Math Awareness “Week” (isn’t every day Math Aware?) should stretch from February 7th to March 14th. I’ve always felt that e doesn’t get the credit it deserves.
Why not celebrate january 6? That is the true golden number. Sure pi is cool and e is helpful, but ϕ is the coolest number in nature. And nobody respects it.
That’s really awesome. My math class is great and I could image them being a FIRST team because of all their wonderful charisma, so yes there are other places where students are being generous and
ϕ is really cool (when we derived the explicit definition of the Fibonacci sequence in Lin Alg it was like “O HAI GUYZ!”), but e is just SOOOOO useful, and has so many more stupid jokes. I’m sorry,
but one cannot make a funny joke about a function that is it’s own derivative.
While we’re at it, we should select a random day of each year to be “+C” day.
At my school, we kind of do the same thing, only in one of our science classes, the teacher LOVES Mol Day. On October 23 (10/23) from 6:04 in the morning to 6:04 at night, it’s Mol Day. Get it? 6.04e
^-23? haha People have actually made cards for our teacher to celebrate Mol day. She laughs so hard that day, which can actually be rare sometimes.
IKE’s yes I am a nerd alert:
So 3 books worth looking into are:
1. The History of PI (number not pastry). It is a brief history of PI that is mostly Anglo centric, but does touch on Middleeastern and Eastern PI history. I will warn you it is a bit hard on
religion. Mostly because the church forced the accuracy of PI to go from 22/7 to 3 for about 500 years based off of something said in the bible.
2. It must be beautiful. A series of essays on history changing equations. Some notables, E=MC^2, chaos theory, string theory, and the equation that got funding for the search for E.T. Like any
compliation of essays, some are better than others.
3. And for last Q.E.D. A series of proofs of different mathematical ideas and principles.
All I have to say is:
It was a password used by a former employer to access special menus in a software program. It’s amazing that after being gone for 6 years I still remember it.
I needed to use the urban dictionary to find out what I<3 and ftw meant. I’m getting old.
I’m not too far out of college and my team had to bring me up to speed on some DragonBall-Z “over 9000” quote…
Anyway, to solve your weekend Pi-day problem, you could hold it on Wednesday March 4th. The 3 is still the month, but the .14 turns into the percentage of the month. Since march has 31 days, 31*0.14
gives you the 4th.
good luck!
I guess celebrating pi day is clearly only a US tradition, since there is no 14th month for the Euro date code standard of DD/MM/YYYY.
HA!! USA FTW!
My pre-calc teacher celebrates Pi Day on 3/14 every year, but since the 14 was a Saturday, we celebrated it on the 13. One person brought pie in and others brought in cookies. somebody even brought
in lots of cupcakes and decorated them in such a way that when they were put on the tray, they made the Pi symbol (chocolate iced were pi symbol, vanilla iced were around it to show it clearly)
So you had pi day on Friday the 13th. Kinda creepy. But I guess Friday the 13th was the day our team competed undefeated. | {"url":"https://www.chiefdelphi.com/t/my-math-teacher/95391","timestamp":"2024-11-11T13:32:06Z","content_type":"text/html","content_length":"41127","record_id":"<urn:uuid:57422e0e-f7c3-43c6-987e-91997d080c94>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00683.warc.gz"} |
Using Stata’s random-number generators, part 1
I want to start a series on using Stata’s random-number function. Stata in fact has ten random-number functions:
1. runiform() generates rectangularly (uniformly) distributed random number over [0,1).
2. rbeta(a, b) generates beta-distribution beta(a, b) random numbers.
3. rbinomial(n, p) generates binomial(n, p) random numbers, where n is the number of trials and p the probability of a success.
4. rchi2(df) generates χ^2 with df degrees of freedom random numbers.
5. rgamma(a, b) generates Γ(a, b) random numbers, where a is the shape parameter and b, the scale parameter.
6. rhypergeometric(N, K, n) generates hypergeometric random numbers, where N is the population size, K is the number of in the population having the attribute of interest, and n is the sample size.
7. rnbinomial(n, p) generates negative binomial — the number of failures before the nth success — random numbers, where p is the probability of a success. (n can also be noninteger.)
8. rnormal(μ, σ) generates Gaussian normal random numbers.
9. rpoisson(m) generates Poisson(m) random numbers.
10. rt(df) generates Student’s t(df) random numbers.
You already know that these random-number generators do not really produce random numbers; they produce pseudo-random numbers. This series is not about that, so we’ll be relaxed about calling them
random-number generators.
You should already know that you can set the random-number seed before using the generators. That is not required but it is recommended. You set the seed not to obtain better random numbers, but to
obtain reproducible random numbers. In fact, setting the seed too often can actually reduce the quality of the random numbers! If you don’t know that, then read help set seed in Stata. I should
probably pull out the part about setting the seed too often, expand it, and turn it into a blog entry. Anyway, this series is not about that either.
This series is about the use of random-number generators to solve problems, just as most users usually use them. The series will provide practical advice. I’ll stay away from describing how they work
internally, although long-time readers know that I won’t keep the promise. At least I’ll try to make sure that any technical details are things you really need to know. As a result, I probably won’t
even get to write once that if this is the kind of thing that interests you, StataCorp would be delighted to have you join our development staff.
runiform(), generating uniformly distributed random numbers
Mostly I’m going to write about runiform() because runiform() can solve such a variety of problems. runiform() can be used to solve,
• shuffling data (putting observations in random order),
• drawing random samples without replacement (there’s a minor detail we’ll have to discuss because runiform() itself produces values drawn with replacement),
• drawing random samples with replacement (which is easier to do than most people realize),
• drawing stratified random samples (with or without replacement),
• manufacturing fictional data (something teachers, textbook authors, manual writers, and blog writers often need to do).
runiform() generates uniformly, a.k.a. rectangularly distributed, random numbers over the interval, I quote from the manual, “0 to nearly 1”.
Nearly 1? “Why not all the way to 1?” you should be asking. “And what exactly do you mean by nearly 1?”
The answer is that the generator is more useful if it omits 1 from the interval, and so we shaved just a little off. runiform() produces random numbers over [0, 0.999999999767169356].
Here are two useful formulas you should commit to memory.
1. If you want to generate continuous random numbers between a and b, use
generate double u = (b–a)*runiform() + a
The random numbers will not actually be between a and b, they will be between a and nearly b, but the top will be so close to b, namely 0.999999999767169356*b, that it will not matter.
Remember to store continuous random values as doubles.
2. If you want to generate integer random numbers between a and b, use
generate ui = floor((b–a+1)*runiform() + a)
In particular, do not even consider using the formula for continuous values but rounded to integers, which is to say, round(u) = round((b–a)*runiform() + a). If you use that formula, and if b–a>
1, then a and b will be under represented by 50% each in the samples you generate!
I stored ui as a default float, so I am assuming that -16,777,216 ≤ a < b ≤ 16,777,216. If you have integers outside of that range, however, store as a long or double.
I’m going to spend the rest of this blog entry explaining the above.
First, I want to show you how I got the two formulas and why you must use the second formula for generating integer uniform deviates.
Then I want explain why we shaved a little from the top of runiform(), namely (1) while it wouldn’t matter for formula 1, it made formula 2 a little easier, (2) the code would run more quickly, (3)
we could more easily prove that we had implemented the random-number generator correctly, and (4) anyone digging deeper into our random numbers would not be misled into thinking they had more than 32
bits of resolution. That last point will be important in a future blog entry.
Continuous uniforms over [a, b)
runiform() produces random numbers over [0, 1). It therefore obviously follows that (b–a)*runiform()+a produces number over [a, b). Substitute 0 for runiform() and the lower limit is obtained.
Substitute 1 for runiform() and the upper limit is obtained.
I can tell you that in fact, runiform() produces random numbers over [0, (2^32-1)/2^32].
Thus (b–a)*runiform()+a produces random numbers over [a, ((2^32-1)/2^32)*b].
(2^32-1)/2^32) approximately equals 0.999999999767169356 and exactly equals 1.fffffffeX-01 if you will allow me to use %21x format, which Stata understands and which you can understand if you see my
previous blog posting on precision.
Thus, if you are concerned about results being in the interval [a, b) rather than [a, b], you can use the formula
generate double u = ((b–a)*runiform() + a) / 1.fffffffeX-01
There are seven f’s followed by e in the hexadecimal constant. Alternatively, you could type
generate double u = ((b–a)*runiform() + a) * ((2^32-1)/2^32)
but multiplying by 1.fffffffeX-01 is less typing so I’d type that. Actually I wouldn’t type either one; the small difference between values lying in [a, b) or [a, b] is unimportant.
Integer uniforms over [a, b]
Whether we produce real, continuous random numbers over [a, b) or [a, b] may be unimportant, but if we want to draw random integers, the distinction is important.
runiform() produces continuous results over [0, 1).
(b–a)*runiform()+a produces continuous results over [a, b).
To produce integer results, we might round continuous results over segments of the number line:
a a+.5 a+1 a+1.5 a+2 a+2.5 b-1.5 b-1 b-.5 b
real line +-----+-----+-----+-----+-----+-----------+-----+-----+-----+
int line |<-a->|<---a+1--->|<---a+2--->| |<---b-1--->|<-b->|
In the diagram above, think of the numbers being produced by the continuous formula u=(b–a)*runiform()+a as being arrayed along the real line. Then imagine rounding those values, say by using Stata’s
round(u) function. If you rounded in that way, then
• Values of u between a and a+0.5 will be rounded to a.
• Values of u between a+0.5 and a+1.5 will be rounded to a+1.
• Values of u between a+1.5 and a+2.5 will be rounded to a+2.
• …
• Values of u between b-1.5 and b-0.5 will be rounded to b-1.
• Values of u between b-0.5 and b-1 will be rounded to b.
Note that the width of the first and last intervals is half that of the other intervals. Given that u follows the rectangular distribution, we thus expect half as many values rounded to a and to b as
to a+1 or a+2 or … or b-1.
And indeed, that is exactly what we would see:
. set obs 100000
obs was 0, now 100000
. gen double u = (5-1)*runiform() + 1
. gen i = round(u)
. summarize u i
Variable | Obs Mean Std. Dev. Min Max
u | 100000 3.005933 1.156486 1.000012 4.999983
i | 100000 3.00489 1.225757 1 5
. tabulate i
i | Freq. Percent Cum.
1 | 12,525 12.53 12.53
2 | 24,785 24.79 37.31
3 | 24,886 24.89 62.20
4 | 25,284 25.28 87.48
5 | 12,520 12.52 100.00
Total | 100,000 100.00
To avoid the problem we need to make the widths of all the intervals equal, and that is what the formula floor((b–a+1)*runiform() + a) does.
a a+1 a+2 b-1 b b+1
real line +-----+-----+-----+-----+-----------------------+-----+-----+-----+-----+
int line |<--- a --->|<-- a+1 -->| |<-- b-1 -->|<--- b --->)
Our intervals are of equal width and thus we expect to see roughly the same number of observations in each:
. gen better = floor((5-1+1)*runiform() + 1)
. tabulate better
better | Freq. Percent Cum.
1 | 19,808 19.81 19.81
2 | 20,025 20.02 39.83
3 | 19,963 19.96 59.80
4 | 20,051 20.05 79.85
5 | 20,153 20.15 100.00
Total | 100,000 100.00
So now you know why we shaved a little off the top when we implemented runiform(); it made the formula
floor((b–a+1)*runiform() + a):
easier. Our integer [a, b] formula did not have to concern itself that runiform() would sometimes — rarely — return 1. If runiform() did return the occasional 1, the simple formula above would
produce the (correspondingly occasional) b+1.
How Stata calculates continuous random numbers
I’ve said that we shaved a little off the top, but the fact was that it was easier for us to do the shaving than not.
runiform() is based on the KISS random number generator. KISS produces 32-bit integers, meaning integers the range [0, 2^32-1], or [0, 4,294,967,295]. You might wonder how we converted that range to
being continuous over [0, 1).
Start by thinking of the number KISS produces in its binary form:
The corresponding integer is b[31]*2^31 + b[31]*2^30 + … + b[0]*2^0. All we did was insert a binary point out front:
0 . b[31]b[30]b[29]b[28]b[27]b[26]b[25]b[24]b[23]b[22]b[21]b[20]b[19]b[18]b[17]b[16]b[15]b[14]b[13]b[12]b[11]b[10]b[9]b[8]b[7]b[6]b[5]b[4]b[3]b[2]b[1]b[0]
making the real value b[31]*2^-1 + b[30]*2^-2 + … + b[0]*2^-32. Doing that is equivalent to dividing by 2^-32, except insertion of the binary point is faster. Nonetheless, if we had wanted runiform()
to produce numbers over [0, 1], we could have divided by 2^32-1.
Anyway, if the KISS random number generator produced 3190625931, which in binary is
we converted that to
which equals 0.74287549 in base 10.
The largest number the KISS random number generator can produce is, of course,
and 0.11111111111111111111111111111111 equals 0.999999999767169356 in base 10. Thus, the runiform() implementation of KISS generates random numbers in the range [0, 0.999999999767169356].
I could have presented all of this mathematically in base 10: KISS produces integers in the range [0, 2^32-1], and in runiform() we divide by 2^32 to thus produce continuous numbers over the range
[0, (2^32-1)/2^32]. I could have said that, but it loses the flavor and intuition of my longer explanation, and it would gloss over the fact that we just inserted the binary point. If I asked you, a
base-10 user, to divide 232 by 10, you wouldn’t actually divide in the same way that they would divide by, say 9. Dividing by 9 is work. Dividing by 10 merely requires shifting the decimal point. 232
divided by 10 is obviously 23.2. You may not have realized that modern digital computers, when programmed by “advanced” programmers, follow similar procedures.
Oh gosh, I do get to say it! If this sort of thing interests you, consider a career at StataCorp. We’d love to have you.
Is it important that runiform() values be stored as doubles?
Sometimes it is important. It’s obviously not important when you are generating random integers using floor((b–a+1)*runiform() + a) and -16,777,216 ≤ a < b ≤ 16,777,216. Integers in that range fit
into a float without rounding.
When creating continuous values, remember that runiform() produces 32 bits. floats store 23 bits and doubles store 52, so if you store the result of runiform() as a float, it will be rounded.
Sometimes the rounding matters, and sometimes it does not. Next time, we will discuss drawing random samples without replacement. In that case, the rounding matters. In most other cases, including
drawing random samples with replacement — something else for later — the rounding
does not matter. Rather than thinking hard about the issue, I store all my non-integer
random values as doubles.
Tune in for the next episode
Yes, please do tune in for the next episode of everything you need to know about using random-number generators. As I already mentioned, we’ll discuss drawing random samples without replacement. In
the third installment, I’m pretty sure we’ll discuss random samples with replacement. After that, I’m a little unsure about the ordering, but I want to discuss oversampling of some groups relative to
others and, separately, discuss the manufacturing of fictional data.
Am I forgetting something? | {"url":"https://blog.stata.com/2012/07/18/using-statas-random-number-generators-part-1/","timestamp":"2024-11-06T14:11:44Z","content_type":"application/xhtml+xml","content_length":"76600","record_id":"<urn:uuid:3fbd4fc9-033f-4fb3-a5fa-e961d824a883>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00162.warc.gz"} |
3.1 Kinematics in Two Dimensions: An Introduction
Learning Objectives
Learning Objectives
By the end of this section, you will be able to do the following:
• Observe that motion in two dimensions consists of horizontal and vertical components
• Understand the independence of horizontal and vertical vectors in two-dimensional motion
The information presented in this section supports the following AP® learning objectives and science practices:
• 3.A.1.1 The student is able to express the motion of an object using narrative, mathematical, and graphical representations. (S.P. 1.5, 2.1, 2.2)
• 3.A.1.2 The student is able to design an experimental investigation of the motion of an object. (S.P. 4.2)
• 3.A.1.3 The student is able to analyze experimental data describing the motion of an object and is able to express the results of the analysis using narrative, mathematical, and graphical
representations. (S.P. 5.1)
Two-Dimensional Motion: Walking in a City
Two-Dimensional Motion: Walking in a City
Suppose you want to walk from one point to another in a city with uniform square blocks, as pictured in Figure 3.3.
The straight-line path that a helicopter might fly is blocked to you as a pedestrian, and so you are forced to take a two-dimensional path, such as the one shown. You walk 14 blocks in all, nine east
followed by five north. What is the straight-line distance?
An old adage states that the shortest distance between two points is a straight line. The two legs of the trip and the straight-line path form a right triangle, and so the Pythagorean theorem, $a2 +
b2 = c2a2 + b2 = c2 size 12{a rSup { size 8{2} } " + "b rSup { size 8{2} } " = "c rSup { size 8{2} } } {}$, can be used to find the straight-line distance.
The hypotenuse of the triangle is the straight-line path, and so in this case its length in units of city blocks is $(9 blocks)2+ (5 blocks)2= 10.3 blocks(9 blocks)2+ (5 blocks)2= 10.3 blocks size 12
{ sqrt { \( "9 blocks" \) rSup { size 8{2} } "+ " \( "5 blocks" \) rSup { size 8{2} } } "= 10" "." "3 blocks"} {}$, considerably shorter than the 14 blocks you walked. Note that we are using three
significant figures in the answer. Although it appears that 9 and 5 only have one significant digit, they are discrete numbers. In this case 9 blocks is the same as 9.0 or 9.00 blocks. We have
decided to use three significant figures in the answer in order to show the result more precisely.
The fact that the straight-line distance 10.3 blocks in Figure 3.5 is less than the total distance walked 14 blocks is one example of a general characteristic of vectors. Recall that vectors are
quantities that have both magnitude and direction.
As for one-dimensional kinematics, we use arrows to represent vectors. The length of the arrow is proportional to the vector's magnitude. The arrow's length is indicated by hash marks in Figure 3.3
and Figure 3.5. The arrow points in the same direction as the vector. For two-dimensional motion, the path of an object can be represented with three vectors: One vector shows the straight-line path
between the initial and final points of the motion, one vector shows the horizontal component of the motion, and one vector shows the vertical component of the motion. The horizontal and vertical
components of the motion add together to give the straight-line path. For example, observe the three vectors in Figure 3.5. The first represents a nine-block displacement east. The second represents
a five-block displacement north. These vectors are added to give the third vector, with a 10.3-block total displacement. The third vector is the straight-line path between the two points. Note that
in this example, the vectors that we are adding are perpendicular to each other and thus form a right triangle. This means that we can use the Pythagorean theorem to calculate the magnitude of the
total displacement. Note that we cannot use the Pythagorean theorem to add vectors that are not perpendicular. We will develop techniques for adding vectors having any direction, not just those
perpendicular to one another, in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods.
The Independence of Perpendicular Motions
The Independence of Perpendicular Motions
The person taking the path shown in Figure 3.5 walks east and then north—two perpendicular directions. How far the person walks east is only affected by the person's motion eastward. Similarly, how
far the person walks north is only affected by the person's motion northward.
Independence of Motion
The horizontal and vertical components of two-dimensional motion are independent of each other. Any motion in the horizontal direction does not affect motion in the vertical direction, and vice
This is true in a simple scenario like that of walking in one direction first, followed by another. It is also true of more complicated motion involving movement in two directions at once. For
example, let's compare the motions of two baseballs. One baseball is dropped from rest. At the same instant, another is thrown horizontally from the same height and follows a curved path. A
stroboscope has captured the positions of the balls at fixed time intervals as they fall.
Applying the Science Practices: Independence of Horizontal and Vertical Motion or Maximum Height and Flight Time
Choose one of the following experiments to design.
Design an experiment to confirm what is shown in Figure 3.6, that the vertical motion of the two balls is independent of the horizontal motion. As you think about your experiment, consider the
following questions:
• How will you measure the horizontal and vertical positions of each ball over time? What equipment will this require?
• How will you measure the time interval between each of your position measurements? What equipment will this require?
• If you were to create separate graphs of the horizontal velocity for each ball versus time, what do you predict it would look like? Explain.
• If you were to compare graphs of the vertical velocity for each ball versus time, what do you predict it would look like? Explain.
• If there is a significant amount of air resistance, how will that affect each of your graphs?
Design a two-dimensional ballistic motion experiment that demonstrates the relationship between the maximum height reached by an object and the object's time of flight. As you think about your
experiment, consider the following questions:
• How will you measure the maximum height reached by your object?
• How can you take advantage of the symmetry of an object in ballistic motion launched from ground level, reaching maximum height, and returning to ground level?
• Will it make a difference if your object has no horizontal component to its velocity? Explain.
• Will you need to measure the time at multiple different positions? Why or why not?
• Predict what a graph of travel time versus maximum height will look like. Will it be linear, parabolic, or horizontal? Explain the shape of your predicted graph qualitatively or quantitatively.
• If there is a significant amount of air resistance, how will that affect your measurements and your results?
It is remarkable that for each flash of the strobe, the vertical positions of the two balls are the same. This similarity implies that the vertical motion is independent of whether or not the ball is
moving horizontally. Assuming no air resistance, the vertical motion of a falling object is influenced by gravity only, and not by any horizontal forces. Careful examination of the ball thrown
horizontally shows that it travels the same horizontal distance between flashes. This is due to the fact that there are no additional forces on the ball in the horizontal direction after it is
thrown. This result means that the horizontal velocity is constant, and affected neither by vertical motion nor by gravity, which is vertical. Note that this case is true only for ideal conditions.
In the real world, air resistance will affect the speed of the balls in both directions.
The two-dimensional curved path of the horizontally thrown ball is composed of two independent one-dimensional motions (horizontal and vertical). The key to analyzing such motion, called projectile
motion, is to resolve (break) it into motions along perpendicular directions. Resolving two-dimensional motion into perpendicular components is possible because the components are independent. We
shall see how to resolve vectors in Vector Addition and Subtraction: Graphical Methods and Vector Addition and Subtraction: Analytical Methods. We will find such techniques to be useful in many areas
of physics.
PhET Explorations: Ladybug Motion Two-Dimensional | {"url":"https://texasgateway.org/resource/31-kinematics-two-dimensions-introduction?binder_id=78521&book=79096","timestamp":"2024-11-14T11:03:21Z","content_type":"text/html","content_length":"128694","record_id":"<urn:uuid:a92e922f-738b-46fd-a76d-d066db086e62>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00780.warc.gz"} |
Physics Questions/concepts About Acceleration - Dissertation Consulting Company
Physics questions/concepts about acceleration
I’ve been stuck on this physics question for the last hour and really need some quick help solving it. I need a tutor explain to me how to solve the following questions? please provide DETAILED
It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar
before it began to skid.
Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s.
Assuming this minimum acceleration, what is the minimum allowed length for the runway?
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume
a uniform acceleration).
Order Now! Order Now! | {"url":"https://dissertationconsultingcompany.com/physics-questions-concepts-about-acceleration/","timestamp":"2024-11-04T02:53:02Z","content_type":"text/html","content_length":"53766","record_id":"<urn:uuid:e36e6bbe-b8c8-46a9-9b12-38ebc33fa1ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00235.warc.gz"} |
What is the conclusion of simple pendulum experiment?
The surprising conclusion – the pendulum traverses a longer distance in a shorter time, than in a shorter distance, and its period is shorter. There are a number of reasons why Galileo thought that
the period remains constant. One factor which Galileo failed to consider is friction.
What is the purpose of the pendulum lab experiment?
The goal of this experiment was to determine the effect of mass and length on the period of oscillation of a simple pendulum. Using a photogate to measure the period, we varied the pendulum mass for
a fixed length, and varied the pendulum length for a fixed mass.
What is the experiment of simple pendulum?
Mark a point A on the table (use a chalk) just below the position of bob at rest and draw a straight line BC of 10 cm having a point A at its centre. Over this line bob will oscillate. Find the least
count and the zero error of the stop clock/watch. Bring its hands at zero position.
What is the hypothesis of the pendulum experiment?
The hypothesis in this case would be, “changing the amount of weight at the end of the pendulum will change the period of the pendulum.” The experiment design would involve measuring the period with
one amount of weight, changing the weight without changing any other variables, then measuring the period with the new …
What is the primary objective of the simple pendulum experiment?
OBJECTIVE: To measure the acceleration due to gravity using a simple pendulum. INTRODUCTION: Many things in nature wiggle in a periodic fashion. That is, they vibrate.
What are objectives of pendulum?
OBJECTIVES : To investigate the functional dependence of the period of a pendulum (ฯ ) on the length of a pendulum (L), the mass of the bob (m) and the starting angle (ฮธo). To check for systematic
errors by calculating g.
How do you make a simple pendulum for a school project?
1. Tie a weight onto the end of the piece of string.
2. Make a loop at the other end of the string.
3. Screw the hook into the piece of wood and balance the wood between the backs of two chairs.
4. Put the loop of string over the hook.
5. Start the weight swinging and time how long it takes to make 20 swings.
What affects the period of a pendulum lab?
This result is interesting because of its simplicity. The only things that affect the period of a simple pendulum are its length and the acceleration due to gravity. The period is completely
independent of other factors, such as mass.
How do you calculate simple pendulum?
What is the relationship between pendulum length and period?
The period of a simple pendulum is directly proportional to the square root of length of the pendulum. The period of a simple pendulum is inversely proportional to the square root of the acceleration
due to gravity.
How can you make a pendulum experiment more accurate?
Improve the accuracy of a measurement of periodic time by: making timings by sighting the bob past a fixed reference point (called a fiducial point ) sighting the bob as it moves fastest past a
reference point. The pendulum swings fastest at its lowest point and slowest at the top of each swing.
What did Galileo discover about pendulums?
Galileo’s discovery was that the period of swing of a pendulum is independent of its amplitude–the arc of the swing–the isochronism of the pendulum. [1] Now this discovery had important implications
for the measurement of time intervals.
What is the dependent variable in a pendulum experiment?
Dependent variable: The dependent variable is the time period for one oscillation. Constant/ Control measures: The constants considered in this experiment are: The mass, shape, size and diameter of
the bob The amplitude (angle) at which the spring is gently left to oscillate.
What is dependent variable in simple pendulum?
So, for example, if you were trying to determine how the period of a pendulum changes when the length of the pendulum is varied, the dependent variable would be the pendulum’s period, and the
independent variable would be the pendulum’s length.
What is the responding variable in a pendulum experiment?
The independent variable was the length of the pendulum and the dependant variable was the period. The controlled variables were the mass of the pendulum and the point of release. The data we
collected and graphed showed that the length of the pendulum affected the period positively.
What are the sources of error in simple pendulum experiment?
The sources of errors in a simple pendulum experiment are the following: human errors comes in when measuring the period using a stopwatch. The reaction time of the observer plays a significant error
when starting the stopwatch and when stopping it. This error can be minimized by repeating the experiment many times.
Does mass affect the period of a pendulum experiment?
The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula
with different lengths will different periods; the pendulum with the longer string will have the longer period.
How does gravity affect pendulum?
Gravity is the consistent force that always pulls downward on the pendulum, whether it is at rest or in motion. Partly due to the force of gravity, the pendulum swings side-to-side in a rounded
motion. The other force acting on the pendulum is the tension force due to string that holds the weight.
How do you plot a graph on a simple pendulum?
How does the length of string affect a pendulum period experiment?
How does the length of a pendulum’s string affect its period? (Answer: A pendulum with a longer string has a longer period, meaning it takes a longer time to complete one back and forth cycle when
compared with a pendulum with a shorter string.
How do you find the kinetic energy of a pendulum?
The kinetic energy would be KE= ยฝmv2,where m is the mass of the pendulum, and v is the speed of the pendulum. At its highest point (Point A) the pendulum is momentarily motionless.
How do you find the acceleration due to gravity using a simple pendulum?
What are the two factors that affect the period of a pendulum?
The mass and angle are the only factors that affect the period of a pendulum.
What factors affect the frequency of a pendulum?
Explanation: The only two factors that affect a pendulum’s frequency are the acceleration due to gravity (g) and the length of the pendulum’s string (L).
What factors do not affect the period of a pendulum?
The time period of the movement of the bob of the simple pendulum does not affected by its mass. It is only affected by the acceleration due to gravity of the place, etc. 2) Size or Shape of the bob.
The time period of the Simple Pendulum is also not depends upon the Shape or Size of the bob. | {"url":"https://physics-network.org/what-is-the-conclusion-of-simple-pendulum-experiment/","timestamp":"2024-11-10T12:44:24Z","content_type":"text/html","content_length":"313472","record_id":"<urn:uuid:d8646dad-7a2a-47fd-9d66-4fc0296c26a9>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00667.warc.gz"} |
Collection of Solved Problems
Ideal Gas Versus Van der Waals
Task number: 1284
A steel bomb with a volume of 0.53 m^3 is filled with carbon dioxide with the amount of substance 1 kmol and the pressure of 5.07 MPa. What is the difference between the temperature of the gas
calculated by using the model of an ideal gas and the temperature calculated by using the Van der Waals equation?
• Numerical values
V = 0.53 m^3 the internal volume of the container
n = 1 kmol = 10^3 mol the amount of substance of carbon dioxide
p = 5.07 MPa = 5.07·10^6 Pa the pressure of carbon dioxide
ΔT = ? the difference between the temperatures calculated by each of the models
From The Handbook of Chemistry and Physics:
R = 8.31 Jmol^−1K^−1 the molar gas constant
a = 0.365 Jm^3mol^−2 the "van der Waals" constant for carbon dioxide
b = 4.28·10^−5 m^3mol^−1 the "van der Waals" constant for carbon dioxide
• Analysis
We will evaluate the thermodynamic temperature from both the ideal gas equation and the Van der Waals equation and then we will calculate their difference.
• Solution
First we evaluate the thermodynamic temperature T[i] from the ideal gas equation
where p is the pressure, V is the volume, n is the amount of substance of carbon dioxide and R is the molar gas constant.
We also evaluate the thermodynamic temperature T[v] from the van der Waals equation
where a and b are the van der Waals constants for carbon dioxide, and thus we obtain
Finally, we determine the temperature difference ΔT:
• Numerical insertion
\[\mathrm{\Delta}T=\frac{\left(p+\frac{n^{2}a}{V^{2}}\right)(V-nb)-pV}{nR}\] \[\mathrm{\Delta}T=\frac{\left(5.07\cdot{10^6}+\frac{10^6\cdot{0.365}}{0.53^{2}}\right)(0.53-10^3\cdot{4.28}\cdot{10^
{-5}})-5.07\cdot{10^6}\cdot{0.53}}{8.31\cdot{10^3}}\,\mathrm{K}\] \[\mathrm{\Delta}T\dot{=}50\,\mathrm{K}\]
• Answer
The temperature calculated by the van der Waals equation is approximately 50 K higher than the temperature calculated by the ideal gas equation. | {"url":"https://physicstasks.eu/1284/ideal-gas-versus-van-der-waals","timestamp":"2024-11-11T17:48:13Z","content_type":"text/html","content_length":"27742","record_id":"<urn:uuid:275a12a0-2b92-48ce-abaf-672878071a28>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00215.warc.gz"} |
(x+4)(x+5) Expand And Simplify
Expanding and Simplifying (x+4)(x+5)
This article will guide you through the process of expanding and simplifying the expression (x+4)(x+5).
Understanding the Process
Expanding a product of binomials like (x+4)(x+5) involves applying the distributive property. This means multiplying each term in the first binomial by each term in the second binomial.
Step-by-Step Expansion
1. Multiply the first terms:
2. Multiply the outer terms:
3. Multiply the inner terms:
4. Multiply the last terms:
5. Combine all the terms:
6. Simplify by combining like terms:
Final Result
Therefore, the expanded and simplified form of (x+4)(x+5) is x² + 9x + 20.
Key Concepts
• Distributive Property: This property states that a(b+c) = ab + ac. It's crucial for expanding products of binomials.
• Combining Like Terms: Terms with the same variable and exponent can be added or subtracted. This simplifies the expression.
By understanding the distributive property and combining like terms, you can confidently expand and simplify any product of binomials. | {"url":"https://jasonbradley.me/page/(x%252B4)(x%252B5)-expand-and-simplify","timestamp":"2024-11-03T03:24:54Z","content_type":"text/html","content_length":"57616","record_id":"<urn:uuid:3948e1d7-8258-4427-9c9d-f7bf70159f79>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00489.warc.gz"} |
seminars - Infinitely wide neural networks
[서울대학교 수리과학부 10-10 집중강연] 2/3 (금), 2/8 (수), 2/10 (금), 15:00PM - 17:00PM
장소: Zoom 강의실
회의 ID: 993 2488 1376
암호: 120745
초록 : While deep learning has many remarkable success stories, finding a satisfactory mathematical explanation on why it is so effective is still considered an open challenge. One recent promising
direction for this challenge is to analyse the mathematical properties of neural networks in the limit where the widths of hidden layers of the networks go to infinity. Researchers were able to prove
highly-nontrivial properties of such infinitely-wide neural networks, such as the gradient-based training achieving the zero training error (so that it finds a global optimum), and the typical random
initialisation of those infinitely-wide networks making them so called Gaussian processes, which are well-studied random objects in machine learning, statistics, and probability theory. These
theoretical findings also led to new algorithms based on so called kernels, which sometime outperform existing kernel-based algorithms.
The purpose of this lecture series is to go through these recent theoretical results on infinitely wide neural networks. Our plan is to pick a few important results in this domain, and to go deep
into those results, so that participants of the series can reuse the mathematical tools behind these results for analysing their own neural networks and training algorithms in the infinite-width | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=date&order_type=asc&l=en&page=73&document_srl=1036564","timestamp":"2024-11-07T07:30:37Z","content_type":"text/html","content_length":"53870","record_id":"<urn:uuid:ac7c1abb-049f-48f4-8280-ec7cd955bf9b>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00626.warc.gz"} |
For a gas dfracRCv 04 where R is the universal gas class 11 physics JEE_Main
Hint: To solve this problem we can use the relationship between the specific heats at constant pressure and volume and the universal gas constant. The answer can be found by finding the values of
specific heats at constant pressure and volume and then comparing the ratio of them with known values.
Complete step by step solution:
As explained in the hint, we will use the relation between the molar specific heats at constant pressure and volume and the universal gas constant to find out the required values using the
information given in the question.
We know that,
${C_p} - {C_v} = R$----equation 1
Here ${C_p}$ is the molar specific heat capacity at constant pressure and ${C_v}$ is the molar specific heat capacity at constant volume. R is the universal gas constant having value equal to
$8.314J.mo{l^{ - 1}}{K^{ - 1}}$
It is given that
$\dfrac{R}{{{C_v}}} = 0.4$
$ \Rightarrow {C_v} = \dfrac{5}{2}R$
Substituting this value in equation 1 we have
${C_p} = R + {C_v}$
$ \Rightarrow {C_p} = R + \dfrac{5}{2}R$
$ \Rightarrow {C_p} = \dfrac{7}{2}R$
The implies that,
$ \Rightarrow \gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{\dfrac{7}{2}R}}{{\dfrac{5}{2}R}}$
$ \Rightarrow \gamma = 1.4$
This ratio is known as adiabatic index or the ratio of specific heats or Laplace’s coefficient.
For monoatomic gases the value of $\gamma = 1.67$ and for diatomic gases this value is equal to $1.4$ .
For rigid diatomic gas, there are five degrees of freedom and ${C_v} = \dfrac{5}{2}R$
The calculated value of ${C_v}$ is also the same for the given problem.
Therefore, the gas is made up of molecules which are rigid diatomic.
Thus, option A is the correct option.
Note: Note that for monoatomic gas molecules the values of molar specific heat capacity at constant volume is ${C_v} = \dfrac{3}{2}R$ and value at constant pressure is ${C_p} = \dfrac{5}{2}R$. While
for diatomic gas molecules, the values of molar specific heat capacity at constant pressure ${C_p} = \dfrac{7}{2}R$ and at constant volume is ${C_v} = \dfrac{5}{2}R$. | {"url":"https://www.vedantu.com/jee-main/for-a-gas-dfracrcv-04-where-r-is-the-universal-physics-question-answer","timestamp":"2024-11-12T08:57:15Z","content_type":"text/html","content_length":"145808","record_id":"<urn:uuid:9ad67390-8bc4-4069-9092-7b3cc799d83a>","cc-path":"CC-MAIN-2024-46/segments/1730477028249.89/warc/CC-MAIN-20241112081532-20241112111532-00817.warc.gz"} |
How to Type an Exponent on a Keyboard
When it comes to typing mathematical expressions, especially those that involve exponents, many people find themselves at a loss. You may wonder, “How do I type a small raised number?” or “What
shortcuts can I use for exponents on my keyboard?” Whether you're drafting a paper, working on a spreadsheet, or coding, knowing how to type an exponent efficiently can save you time and enhance your
productivity. In this comprehensive guide, we will explore various methods for typing exponents across different applications and operating systems.
Understanding Exponents: A Quick Overview
Before we dive into the specifics of typing exponents, let's briefly define what an exponent is. In mathematics, an exponent refers to the number that indicates how many times a base number is
multiplied by itself. For example, in the expression (2^3) (read as "two raised to the power of three"), the number 2 is the base, and the number 3 is the exponent. The value of (2^3) is 8 because (2
\times 2 \times 2 = 8).
Knowing how to represent exponents correctly is crucial, especially in scientific, technical, or academic contexts. Now, let’s explore how to type these exponents using various methods.
1. Using Keyboard Shortcuts in Word Processors
Word processors, like Microsoft Word and Google Docs, offer built-in features to type exponents quickly. Here's how you can do it:
1.1 Microsoft Word
In Microsoft Word, you can use the superscript feature to type exponents:
1. Using the Superscript Option:
□ Type the base number.
□ Select the "Home" tab in the Ribbon.
□ Click on the "X²" icon (the superscript button) in the Font group.
□ Type the exponent.
□ Click the superscript button again to deactivate it.
2. Using Keyboard Shortcuts:
□ Type the base number.
□ Press Ctrl + Shift + = (this is the shortcut to toggle superscript).
□ Type the exponent.
□ Press Ctrl + Shift + = again to return to normal text.
1.2 Google Docs
In Google Docs, the steps are quite similar:
1. Using the Format Menu:
□ Type the base number.
□ Go to "Format" > "Text" > "Superscript" or simply use the shortcut Ctrl + . (Control and period).
□ Type the exponent.
□ Use the shortcut again to return to normal text.
2. Using the Toolbar:
□ Type the base number and highlight the exponent.
□ Click on the "Format" option in the toolbar, select "Text," and then "Superscript."
By mastering these steps, you can easily include exponents in your documents without any hassle.
2. Typing Exponents in Excel
Excel is widely used for numerical analysis, and knowing how to enter exponents can be extremely beneficial for calculations.
2.1 Using the Caret Symbol
In Excel, you can type exponents using the caret symbol (^):
• For example, if you want to calculate (2^3), you can enter it in a cell as =2^3.
• After pressing Enter, Excel will compute the value and display 8.
2.2 Formatting for Display
If you want to format the display of exponents (not for calculations), you can:
1. Type the base number in one cell and the exponent in another.
2. Manually format the exponent using superscript if needed (using the same method as Microsoft Word).
3. Typing Exponents on a Mac
Mac users can also type exponents with ease, and here’s how:
3.1 Using TextEdit
In TextEdit, you can follow similar steps to those in Word:
1. Type your base number.
2. Select "Format" from the menu.
3. Choose "Font" and then click "Superscript."
4. Type your exponent and return to regular text.
3.2 Keyboard Shortcuts
• Superscript Shortcut: You can also use Command + Shift + + (the plus key) to toggle the superscript feature in many applications on a Mac.
4. Using Unicode for Exponents
For those who need to insert exponents in HTML or other code, Unicode provides a way to type common exponent symbols directly.
4.1 Common Unicode Exponents:
Exponent Unicode HTML Code
0 U+2070 ⁰
1 U+00B9 ¹
2 U+00B2 ²
3 U+00B3 ³
4 U+2074 ⁴
5 U+2075 ⁵
6 U+2076 ⁶
7 U+2077 ⁷
8 U+2078 ⁸
9 U+2079 ⁹
To use these symbols, you can insert the corresponding HTML code directly into your web documents or use the Unicode character in compatible applications.
5. Exponents in Programming Languages
In many programming languages, the syntax for exponents can vary. Let’s take a look at how to type exponents in some popular programming languages.
5.1 Python
In Python, you can use the double asterisk (**) operator:
result = 2 ** 3 # This will give you 8
5.2 JavaScript
In JavaScript, the same double asterisk syntax works as well:
let result = Math.pow(2, 3); // This will return 8
5.3 Excel Formulas
In Excel, as mentioned earlier, you can simply use ^ for exponentiation.
5.4 C++
In C++, you can use the pow function:
#include <cmath>
int main() {
double result = pow(2, 3); // This will return 8.0
return 0;
6. Tips and Tricks for Efficient Typing
Now that you have several methods for typing exponents at your disposal, here are some additional tips to enhance your efficiency:
• Memorize Shortcuts: Make it a habit to memorize keyboard shortcuts. They significantly reduce the time spent formatting exponents.
• Use Templates: For documents that often require exponents, create a template that you can reuse. This way, you won't need to format each instance manually.
• Practice: The more you practice typing exponents, the more comfortable you’ll become. Regular use in your writing will help cement your skills.
Typing exponents on a keyboard doesn't have to be complicated. With the methods outlined in this guide, you can confidently include exponents in your documents, spreadsheets, and programming
environments. Whether you’re using a word processor, Excel, or coding in a programming language, knowing the correct techniques will enhance your writing and productivity.
Always remember the importance of clarity in communication, especially when dealing with mathematical expressions. So go ahead, master those exponents, and impress your peers with your mathematical
1. How can I type an exponent on a smartphone? Most mobile devices don’t have a dedicated button for exponents. However, you can use apps like Google Docs or Word that have formatting options for
superscripts. Alternatively, you can type out the exponent normally (e.g., 2^3) if formatting isn’t available.
2. What if my application doesn’t support superscript formatting? If superscript formatting isn’t available, you can use the caret symbol (^) to denote exponents. For example, write 2^3 instead of
using superscript.
3. Can I copy and paste exponents from other documents? Yes, you can copy and paste exponents from documents where they are correctly formatted as superscripts into your own document.
4. Are there online tools that can help type exponents? Yes, there are several online tools and text editors that provide options for inserting special characters, including superscripts. Websites
like Codecademy or W3Schools may also provide tips for programming-related queries.
5. Why is it important to use the correct notation for exponents? Using the correct notation for exponents ensures clarity and precision in mathematical communication. Misrepresenting an exponent can
lead to misunderstandings and errors in calculations. | {"url":"http://truereviews.us/post/how-to-type-an-exponent-on-a-keyboard","timestamp":"2024-11-07T09:35:28Z","content_type":"text/html","content_length":"93493","record_id":"<urn:uuid:641f3a0c-10ec-4602-aa13-ef716b03fde7>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00837.warc.gz"} |
Infinite families of congruences for m-regular [j, k]-overpartitions in two colors
T. N. Veeranayaka, M. S. Mahadeva Naika, Harishkumar T
Let a[j,k]^m(n) denote the number of overpartitions of n with two colors in which no parts are divisible by m and only parts congruent to j modulo k may be overlined. In this work, we establish many
infinite families of congruences modulo powers of 2 for a[1,3]^3(n), a[1,4]^3(n) and congruences modulo powers of 2 and 3 for a[1,4]^9(n).
Advanced Studies: Euro-Tbilisi Mathematical Journal, Vol. 15(3) (2022), pp. 53-74 | {"url":"https://tcms.org.ge/Journals/ASETMJ/Volume15/Volume15_3/Abstract/abstract15_3_5.html","timestamp":"2024-11-04T01:25:01Z","content_type":"text/html","content_length":"3167","record_id":"<urn:uuid:8144f232-3f24-43a1-96db-5482407bc566>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00344.warc.gz"} |
Asymmetric water free jet study
dear Norberto,
Instead of knowing this phenomenon of asymmetric flow, I started with an experiment in which I evaluated the cross-sectional velocity-outflow distribution of a vertical vessel/scrubber with a
horizontal internal vane type inlet separator. An airflow was injected sideways in the middle of the vessel through a pipe and was streaming through the inlet separator and diverted into the open air
through the outlet of the vessel. Results showed that this flow distribution was asymmetric. For comparison with this result I used a CFD-code named FLUENT and the results showed the exact same
outcome. The flow is considered only gas because the multi-phase flow which is normally inserted (oil, gas, water, solids) can be considered one-phase when the gas flow is flowing out of the vessel
(particles that are not separated by the inlet separator and the gravity are still inside the gas stream but are negligible for the properties of this flow). You have to use a geometry (Which you can
make in AUTOCAD, RHINOCEROS, or GAMBIT a.o.) and grid (make it in GAMBIT) and make sure that you have enough cells at the places where you expect a change in velocity. So make sure your grid is dense
enough and smooth enough (in GAMBIT you can check your grid and mesh if it is suitable, use tetrahedical cells for your convenience) After this, make use of FLUENT to process your meshed grid. Set
your boundary conditions, like inlet velocity, hydraulic diameter and turbulence intensity and pressure at the outflow level (use a pressure outlet). Make sure you use the right solver model for your
flow...the standard k-epsilon model is proven accurate but not for all turbulent flows, when you have a strongly curved flow, and flow close to the wall you might consider LES (or Latice-Boltzman,
but that model is not used in FLUENT). Remember however that the more cells you use in your grid and the more complex the solver model, the more time it takes to reach convergence and a good
solution. My model consisted of 600,000 cells and it took 36 hours to reach convergence in FLUENT with the standard k-epsilon model. After solving the model, it is possible to substract many results,
you can display the contours/vectors of the pressure for example and many more of each region or point in this region that you desire. See
for more information.
I hope that this elaborate discussion of my results can help you with your problem,
Regards, Marko van der Smitte
ps. do you have information about the origination of this asymmetric phenomenon? I suppose that it has to do with a difference in pressure gradients, but do not know the full story. | {"url":"https://www.cfd-online.com/Forums/main/3623-asymmetric-water-free-jet-study.html","timestamp":"2024-11-11T20:10:40Z","content_type":"application/xhtml+xml","content_length":"98002","record_id":"<urn:uuid:4357d180-ee52-4f5c-8465-6c625ccbc141>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00510.warc.gz"} |
Support Vector Machines
11 Support Vector Machines
Support Vector
Multiclass SVM
Textbook reading: Chapter 9: Support Vector Machines (exclude Section 9.5).
Overview of the Algorithm
Support vector machines are a class of statistical models first developed in the mid-1960s by Vladimir Vapnik. In later years, the model has evolved considerably into one of the most flexible and
effective machine learning tools available. It is a supervised learning algorithm which can be used to solve both classification and regression problem, even though the current focus is on
classification only.
To put it in a nutshell, this algorithm looks for a linearly separable hyperplane, or a decision boundary separating members of one class from the other. If such a hyperplane exists, the work is
done! If such a hyperplane does not exist, SVM uses a nonlinear mapping to transform the training data into a higher dimension. Then it searches for the linear optimal separating hyperplane. With an
appropriate nonlinear mapping to a sufficiently high dimension, data from two classes can always be separated by a hyperplane. The SVM algorithm finds this hyperplane using support vectors and
margins. As a training algorithm, SVM may not be very fast compared to some other classification methods, but owing to its ability to model complex nonlinear boundaries, SVM has high accuracy. SVM is
comparatively less prone to overfitting. SVM has successfully been applied to handwritten digit recognition, text classification, speaker identification etc.
After completing the reading for this lesson, please finish the Quiz and R Lab on Canvas (check the course schedule for due dates).
Upon successful completion of this lesson, you should be able to:
• Understand how the maximal margin classifier works for datasets in which two classes are separable by a linear boundary.
• Understand the support vector classifier, which extends the maximal margin classifier to work with overlapping classes.
• Understand support vector machines, which extend support vector classifiers to accommodate non-linear class boundaries.
11.1 Support Vector Classifier
The maximal margin classifier is a very natural way to perform classification, is a separating hyperplane exists. However the existence of such a hyperplane may not be guaranteed, or even if it
exists, the data is noisy so that maximal margin classifier provides a poor solution. In such cases, the concept can be extended where a hyperplane exists which almost separates the classes, using
what is known as a soft margin. The generalization of the maximal margin classifier to the non-separable case is known as the support vector classifier, where a small proportion of the training
sample is allowed to cross the margins or even the separating hyperplane. Rather than looking for the largest possible margin so that every observation is on the correct side of the margin, thereby
making the margins very narrow or non-existent, some observations are allowed to be on the incorrect side of the margins. The margin is soft as a small number of observations violate the margin. The
softness is controlled by slack variables which control the position of the observations relative to the margins and separating hyperplane. The support vector classifier maximizes a soft margin. The
optimization problem can be modified as
\[ y_i (\theta_0 + \theta_1 x_{1i} + \theta_2 x_{2i} + \cdots + \theta_n x_{ni}) \ge 1 – \epsilon_i \text{ for every observation}\]\[ \text{Where} \\\\ \epsilon_i \ge 0 \\\\ \text{and} \sum_{i=1}^{n}
\epsilon_i \le C \]
The εi is the slack corresponding to \(i^{th}\) observation and C is a regularization parameter set by the user. The larger value of C leads to a larger penalty for errors.
However, there will be situations when a linear boundary simply does not work.
11.2 When Data is Linearly Separable
Let us start with a simple two-class problem when data is clearly linearly separable as shown in the diagram below.
Let the i-th data point be represented by (\(X_i\), \(y_i\)) where \(X_i\) represents the feature vector and \(y_i\) is the associated class label, taking two possible values +1 or -1. In the diagram
above the balls having red color has class label +1 and the blue balls have a class label -1, say. A straight line can be drawn to separate all the members belonging to class +1 from all the members
belonging to the class -1. The two-dimensional data above are clearly linearly separable.
In fact, an infinite number of straight lines can be drawn to separate the blue balls from the red balls.
The problem, therefore, is which among the infinite straight lines is optimal, in the sense that it is expected to have minimum classification error on a new observation. The straight line is based
on the training sample and is expected to classify one or more test samples correctly.
As an illustration, if we consider the black, red and green lines in the diagram above, is any one of them better than the other two? Or are all three of them equally well suited to classify? How is
optimality defined here? Intuitively it is clear that if a line passes too close to any of the points, that line will be more sensitive to small changes in one or more points. The green line is close
to a red ball. The red line is close to a blue ball. If the red ball changes its position slightly, it may fall on the other side of the green line. Similarly, if the blue ball changes its position
slightly, it may be misclassified. Both the green and red lines are more sensitive to small changes in the observations. The black line on the other hand is less sensitive and less susceptible to
model variance.
In an n-dimensional space, a hyperplane is a flat subspace of dimension n – 1. For example, in two dimensions a straight line is a one-dimensional hyperplane, as shown in the diagram. In three
dimensions, a hyperplane is a flat two-dimensional subspace, i.e. a plane. Mathematically in n dimensions a separating hyperplane is a linear combination of all dimensions equated to 0; i.e., \[\
theta_0 + \theta_1 x_1 + \theta_2 x_2 + … + \theta_n x_n = 0\] The scalar \(\theta_0\) is often referred to as a bias. If \(\theta_0 = 0\), then the hyperplane goes through the origin.
A hyperplane acts as a separator. The points lying on two different sides of the hyperplane will make up two different groups.
Basic idea of support vector machines is to find out the optimal hyperplane for linearly separable patterns. A natural choice of separating hyperplane is optimal margin hyperplane (also known as
optimal separating hyperplane) which is farthest from the observations. The perpendicular distance from each observation to a given separating hyperplane is computed. The smallest of all those
distances is a measure of how close the hyperplane is to the group of observations. This minimum distance is known as the margin. The operation of the SVM algorithm is based on finding the hyperplane
that gives the largest minimum distance to the training examples, i.e. to find the maximum margin. This is known as the maximal margin classifier.
A separating hyperplane in two dimension can be expressed as \[\theta_0 + \theta_1 x_1 + \theta_2 x_2 = 0\] Hence, any point that lies above the hyperplane, satisfies \[\theta_0 + \theta_1 x_1 + \
theta_2 x_2 \> 0\] and any point that lies below the hyperplane, satisfies \[\theta_0 + \theta_1 x_1 + \theta_2 x_2 < 0\] The coefficients or weights \(θ_1\) and \(θ_2\) can be adjusted so that the
boundaries of the margin can be written as \[H_1: \theta_0 + \theta_1 x_{1i} + \theta_2 x_{2i} \ge 1, \text{for} y_i = +1\]\[H_2: \theta_0 + θ\theta_1 x_{1i} + \theta_2 x_{2i} \le -1, \text{for} y_i
= -1\] This is to ascertain that any observation that falls on or above \(H_1\) belongs to class +1 and any observation that falls on or below \(H_2\), belongs to class -1. Alternatively, we may
write \[y_i (\theta_0 + \theta_1 x_{1i} + \theta_2 x_{2i}) \le \text{for every observation}\] The boundaries of the margins, \(H_1\) and \(H_2\), are themselves hyperplanes too. The training data
that falls exactly on the boundaries of the margin are called the support vectors as they support the maximal margin hyperplane in the sense that if these points are shifted slightly, then the
maximal margin hyperplane will also shift.
Note that the maximal margin hyperplane depends directly only on these support vectors.
If any of the other points change, the maximal margin hyperplane does not change until the movement affects the boundary conditions or the support vectors. The support vectors are the most difficult
to classify and give the most information regarding classification. Since the support vectors lie on or closest to the decision boundary, they are the most essential or critical data points in the
training set.
For a general n-dimensional feature space, the defining equation becomes \[y_i (\theta_0 + \theta_1 x_{2i} + \theta_2 x_{2i} + … + θn x_ni)\ge 1, \text{for every observation}\] If the vector of the
weights is denoted by \(\Theta\) and \(|\Theta|\) is the norm of this vector, then it is easy to see that the size of the maximal margin is \(\dfrac{2}{|\Theta|}\). Finding the maximal margin
hyperplanes and support vectors is a problem of convex quadratic optimization. It is important to note that the complexity of SVM is characterized by the number of support vectors, rather than the
dimension of the feature space. That is the reason SVM has a comparatively less tendency to overfit. If all data points other than the support vectors are removed from the training data set, and the
training algorithm is repeated, the same separating hyperplane would be found. The number of support vectors provides an upper bound to the expected error rate of the SVM classifier, which happens to
be independent of data dimensionality. An SVM with a small number of support vectors has good generalization, even when the data has high dimensionality.
11.3 When Data is NOT Linearly Separable
SVM is quite intuitive when the data is linearly separable. However, when they are not, as shown in the diagram below, SVM can be extended to perform well.
There are two main steps for nonlinear generalization of SVM. The first step involves the transformation of the original training (input) data into a higher dimensional data using a nonlinear
mapping. Once the data is transformed into the new higher dimension, the second step involves finding a linear separating hyperplane in the new space. The maximal marginal hyperplane found in the new
space corresponds to a nonlinear separating hypersurface in the original space.
Example: Feature Expansion
Suppose the original feature space includes two variables \(X_1\) and \(X_2\). Using polynomial transformation the space is expanded to (\(X_1, X_2, X_1^2, X_2^2, X_1X_2\)). Then the hyperplane would
be of the form \[\theta_0 + \theta_1 X_1 + \theta_2 X_2 + \theta_3 X_1^2 + \theta_4 X_2^2 + \theta_5 X_1 X_2 = 0\] This will lead to nonlinear decision boundaries in the original feature space. If
upto second degree terms are considered, 2 features are expanded to 5. If upto third degree terms are considered the same to features can be expanded to 9 features. The support vector classifier in
the expanded space solves the problems in the lower dimension space.
11.4 Kernel Functions
Handling the nonlinear transformation of input data into higher dimension may not be easy. There may be many options available, to begin with, and the procedures may be computationally heavy also. To
avoid some of those problems, the concept of Kernel functions is introduced.
It so happens that in solving the quadratic optimization problem of the linear SVM, the training data points contribute through inner products of nonlinear transformations. The inner product of two
n-dimensional vectors is defined as \[\sum_{j=1}^{n} x_{1j} x_{2j} \] Where \(X_1 = (x_{11}, x_{12}, \cdots x_{1n})\) and \(X_2 = (x_{21}, x_{22},… x_{2n})\). The kernel function is a generalization
of the inner product of nonlinear transformation and is denoted by K(X1, X2). Anywhere such an inner product appears, it is replaced by the kernel function. In this way, all calculations are made in
the original input space, which is lower dimensionality. Some of the common kernels are a polynomial kernel, sigmoid kernel, and Gaussian radial basis function. Each of these will result in a
different nonlinear classifier in the original input space. There is no golden rule to determine which kernel will provide the most accurate result in a given situation. In practice, the accuracy of
SVM does not depend on the choice of the kernel.
11.5 Multiclass SVM
The SVM as defined so far works for binary classification. What happens if the number of classes is more than two?
One-versus-All: If the number of classes is K > 2 then K different 2-class SVM classifiers are fitted where one class is compared with the rest of the classes combined. A new observation is
classified according to where the classifier value is the largest.
One-versus-One: All \(\binom{K}{2}\) pairwise classifiers are fitted and a test observation is classified in the class which wins in the majority of the cases.
The latter method is preferable but if K is too large, the former is to be used. | {"url":"https://online.stat.psu.edu/stat508/lessons/Lesson11","timestamp":"2024-11-04T09:01:19Z","content_type":"application/xhtml+xml","content_length":"87154","record_id":"<urn:uuid:a79b886a-1945-4a43-9456-9511c8723abf>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00358.warc.gz"} |
(Solved) - This is a short answer question The lasso regression method also... (1 Answer) | Transtutors
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This is a short answer question
The lasso regression method also performs shrinkage as the ridge regression method does - what does lasso perform in addition to shrinkage? In terms of the ßj coefficients, what values do these take
on under the lasso which results in a sparse model?
1 Approved Answer
dammalapati h
Lasso regression Least Absolute Shrinkage and Selection Operator does more than just shrinkage, which is the process of reducing the magnitude of coefficients to prevent overfitting. It also performs
variable selection. Here’s...
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What does complement mean in geometry?
What does complement mean in geometry?
Definition of complementary angles mathematics. : two angles that add up to 90 degrees.
What is a complement in a graph?
In graph theory, the complement or inverse of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G.
How many edges does a complement graph have?
We know |E(G)| + |E(G’)| = n(n-1) / 2. Thus, Number of edges in complement graph G’ = 24.
What is the meaning of complement in sets?
The complement of a set is the set that includes all the elements of the universal set that are not present in the given set. Let’s say A is a set of all coins which is a subset of a universal set
that contains all coins and notes, so the complement of set A is a set of notes (which do not includes coins).
Why does complementary mean?
Complementary is an adjective used to describe something that complements something else—goes along with it and serves to make it better or complete it (as in complementary colors).
What is complementary angle answer?
When the sum of two angles is equal to 90 degrees, they are called complementary angles. For example, 30 degrees and 60 degrees are complementary angles.
How do you find the edge of a complement graph?
2. The sum of the Edges of a Complement graph and the main graph is equal to the number of edges in a complete graph, n is the number of vertices. E(G’)+E(G) = E(Kn) = n(n-1)÷2.
How many edges are there in the complement graph K6?
The complete graph K6 has 15 edges and 45 pairs of independent edges.
How do you find the complement?
How to Find Complementary Angles? If the sum of two angles is 90 degrees, then we say that they are complementary. Thus, the complement of an angle is obtained by subtracting it from 90. For example,
the complement of 40° is 90° – 40° = 50°.
What are the difference and complement of sets?
Complement and Difference of Sets The complement of a set A is denoted by A’ or Ac and it is the difference of the sets U and A, where U is the universal set. i.e., A’ (or) Ac = U – A. This refers to
the set of all elements that are in the universal set that are not elements of set A.
What is complementary explain with example?
The definition of complementary is someone or something that completes or makes someone or something better. An example of complementary is drinking red wine with an Italian meal.
How do you find complementary?
How do you make a complement graph?
Complement of Graph
1. Let ‘G−’ be a simple graph with some vertices as that of ‘G’ and an edge {U, V} is present in ‘G−’, if the edge is not present in G.
2. |E(G)| + |E(‘G-‘)| = |E(Kn)|, where n = number of vertices in the graph.
How many edges are there in K5?
10 edges
K5: K5 has 5 vertices and 10 edges, and thus by Lemma 2 it is not planar. K3,3: K3,3 has 6 vertices and 9 edges, and so we cannot apply Lemma 2.
What does complement mean In geometry?
“Complere” means “complete”, whereas “Plere” means “fill”. So “complementary” means “something that completes and brings perfection.” And so are complementary angles, a pair of two angles that sum up
to 90 degrees, forming a right angle.
Which angles are complements of each other?
Each angle is the complement of the other. Complementary angles can be adjacent or non-adjacent. For a right triangle, the two non-right or oblique angles must be complementary. In right triangle ABC
above, ∠B = 90° and ∠A + ∠C = 90° so, the nonadjacent angles A and C are complements of each other.
What is complementary angle theorem?
The complementary angle theorem states, “If two angles are complementary to the same angle, then they are congruent to each other”. Proof of Complementary Angle Theorem We know that complementary
angles exist in pairs and sum upto 90 degrees.
How do you find the complement of a right triangle?
For a right triangle, the two non-right or oblique angles must be complementary. In right triangle ABC above, ∠B = 90° and ∠A + ∠C = 90° so, the nonadjacent angles A and C are complements of each
other. You can determine the complement of a given angle by subtracting it from 90°. For example, the complement of 28° is 62° since 90° – 28° = 62°. | {"url":"https://www.david-cook.org/what-does-complement-mean-in-geometry/","timestamp":"2024-11-13T08:15:54Z","content_type":"text/html","content_length":"42514","record_id":"<urn:uuid:c4414340-f0a6-42ab-9f6d-a064fd3cfb0e>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00526.warc.gz"} |
Transactions Online
Shan ZENG, Wenjian YU, Xianlong HONG, Chung-Kuan CHENG, "Efficient Power Network Analysis with Modeling of Inductive Effects" in IEICE TRANSACTIONS on Fundamentals, vol. E93-A, no. 6, pp. 1196-1203,
June 2010, doi: 10.1587/transfun.E93.A.1196.
Abstract: In this paper, an efficient method is proposed to accurately analyze large-scale power/ground (P/G) networks, where inductive parasitics are modeled with the partial reluctance. The method
is based on frequency-domain circuit analysis and the technique of vector fitting, and obtains the time-domain voltage response at given P/G nodes. The frequency-domain circuit equation including
partial reluctances is derived, and then solved with the GMRES algorithm with rescaling, preconditioning and recycling techniques. With the merit of sparsified reluctance matrix and iterative solving
techniques for the frequency-domain circuit equations, the proposed method is able to handle large-scale P/G networks with complete inductive modeling. Numerical results show that the proposed method
is orders of magnitude faster than HSPICE, several times faster than INDUCTWISE, and capable of handling the inductive P/G structures with more than 100,000 wire segments.
URL: https://global.ieice.org/en_transactions/fundamentals/10.1587/transfun.E93.A.1196/_p
author={Shan ZENG, Wenjian YU, Xianlong HONG, Chung-Kuan CHENG, },
journal={IEICE TRANSACTIONS on Fundamentals},
title={Efficient Power Network Analysis with Modeling of Inductive Effects},
abstract={In this paper, an efficient method is proposed to accurately analyze large-scale power/ground (P/G) networks, where inductive parasitics are modeled with the partial reluctance. The method
is based on frequency-domain circuit analysis and the technique of vector fitting, and obtains the time-domain voltage response at given P/G nodes. The frequency-domain circuit equation including
partial reluctances is derived, and then solved with the GMRES algorithm with rescaling, preconditioning and recycling techniques. With the merit of sparsified reluctance matrix and iterative solving
techniques for the frequency-domain circuit equations, the proposed method is able to handle large-scale P/G networks with complete inductive modeling. Numerical results show that the proposed method
is orders of magnitude faster than HSPICE, several times faster than INDUCTWISE, and capable of handling the inductive P/G structures with more than 100,000 wire segments.},
TY - JOUR
TI - Efficient Power Network Analysis with Modeling of Inductive Effects
T2 - IEICE TRANSACTIONS on Fundamentals
SP - 1196
EP - 1203
AU - Shan ZENG
AU - Wenjian YU
AU - Xianlong HONG
AU - Chung-Kuan CHENG
PY - 2010
DO - 10.1587/transfun.E93.A.1196
JO - IEICE TRANSACTIONS on Fundamentals
SN - 1745-1337
VL - E93-A
IS - 6
JA - IEICE TRANSACTIONS on Fundamentals
Y1 - June 2010
AB - In this paper, an efficient method is proposed to accurately analyze large-scale power/ground (P/G) networks, where inductive parasitics are modeled with the partial reluctance. The method is
based on frequency-domain circuit analysis and the technique of vector fitting, and obtains the time-domain voltage response at given P/G nodes. The frequency-domain circuit equation including
partial reluctances is derived, and then solved with the GMRES algorithm with rescaling, preconditioning and recycling techniques. With the merit of sparsified reluctance matrix and iterative solving
techniques for the frequency-domain circuit equations, the proposed method is able to handle large-scale P/G networks with complete inductive modeling. Numerical results show that the proposed method
is orders of magnitude faster than HSPICE, several times faster than INDUCTWISE, and capable of handling the inductive P/G structures with more than 100,000 wire segments.
ER - | {"url":"https://global.ieice.org/en_transactions/fundamentals/10.1587/transfun.E93.A.1196/_p","timestamp":"2024-11-08T20:48:29Z","content_type":"text/html","content_length":"62078","record_id":"<urn:uuid:2dc637b9-c38c-4d8e-a116-2f0b7ba41282>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00665.warc.gz"} |
A Box Is Pushed Along The Ground With A Force Of 20N. If The Box Is Pushed 0.5M.How Much Work Done Is
10 joules(j)
work (W)= force (F)× distance (d)
W= 20N × o.5m
=10J (Joule)
Mass, m of gas is 0.2504 grams.
First, we need to solve for the volume of the cylindrical tube.
Volume of cylinder is given by the formula;
[tex] V = 2\Pi r^{2}h[/tex]
Where, V represents volume.
π represents pie
r represents radius.
h represents height or length.
Given the following data;
Radius, r = 1.5cm
Length, h = 14.4cm
Density, d = 0.00123g/cm³
Substituting into the equation;
[tex] V = 2 * 3.142 * (1.5)^{2}14.4[/tex]
[tex] V = 2 * 3.142 * 2.25 * 14.4[/tex]
[tex] V = 203. 6016[/tex]
Therefore, the volume of the cylindrical tube is 203. 6016cm³
Density can be defined as mass all over the volume of an object.
Simply stated, density is mass per unit volume of an object.
Mathematically, density is given by the formula;
[tex]Density = \frac{mass}{volume}[/tex]
[tex] Mass = density * volume [/tex]
Substituting into the equation, we have;
[tex] Mass = 0.00123 * 203. 6016 [/tex]
Mass = 0.2504g. | {"url":"https://smart.gov.qa/question-answers/a-box-is-pushed-along-the-ground-with-a-force-of-20n-if-the-elmn","timestamp":"2024-11-09T15:44:10Z","content_type":"text/html","content_length":"77945","record_id":"<urn:uuid:536e2f65-a0d9-4c12-bb78-cabbba5bd52e>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00242.warc.gz"} |
: Implicitly creates objects (
) within the denoted region consisting of an object
of type
whose address is
, and objects nested within
, as follows: The object representation of
is the contents of the storage prior to the call to
The value of each created object
of trivially copyable type (
is determined in the same manner as for a call to
), where
is an lvalue of type
, except that the storage is not accessed | {"url":"https://timsong-cpp.github.io/cppwp/obj.lifetime","timestamp":"2024-11-14T07:14:17Z","content_type":"text/html","content_length":"15235","record_id":"<urn:uuid:4105f45b-1c62-4158-b324-69f3c52b3ae5>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00380.warc.gz"} |
How do you differentiate f(x)=(cot(x))^2 using the chain rule? | HIX Tutor
How do you differentiate #f(x)=(cot(x))^2 # using the chain rule?
Answer 1
$f ' \left(x\right) = - 2 \cot \left(x\right) {\csc}^{2} \left(x\right)$
Use the chain rule. The first issue here is the second power. According to the chain rule, #d/dx[u^2]=2u*u'#, and here we have #u=cot(x)#, giving us
The derivative of #cot(x)# is #-csc^2(x)#.
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Answer 2
To differentiate ( f(x) = (\cot(x))^2 ) using the chain rule, you first find the derivative of the outer function and then multiply it by the derivative of the inner function.
The derivative of ( (\cot(x))^2 ) is ( 2 \cot(x) \cdot (-\csc^2(x)) ).
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Answer 3
To differentiate ( f(x) = (\cot(x))^2 ) using the chain rule, follow these steps:
1. Recognize that ( (\cot(x))^2 ) is a composition of functions, with the outer function being the square function ( g(u) = u^2 ) and the inner function being ( u = \cot(x) ).
2. Differentiate the outer function with respect to its variable. In this case, differentiate ( g(u) = u^2 ) with respect to ( u ) to get ( g'(u) = 2u ).
3. Differentiate the inner function with respect to the independent variable, ( x ). In this case, differentiate ( u = \cot(x) ) with respect to ( x ) using the derivative of cotangent formula,
which is ( \frac{d}{dx} \cot(x) = -\csc^2(x) ).
4. Apply the chain rule, which states that if ( y = g(u) ) and ( u = h(x) ), then ( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} ).
5. Substitute the derivatives obtained in steps 2 and 3 into the chain rule formula to find ( \frac{df}{dx} ), the derivative of ( f(x) = (\cot(x))^2 ) with respect to ( x ).
6. Simplify the expression if necessary.
This process will yield the derivative of ( f(x) = (\cot(x))^2 ) using the chain rule.
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Answer from HIX Tutor
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some
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Spiral of Theodorus – Polypad (2024)
Overview and Objective
The Spiral of Theodorus is a shape composed of right triangles, placed edge-to-edge. In this lesson, students explore the spiral. While drawing the spiral, students will repeatedly construct
perpendicular lines. For some students, this repetitive process can lead to creating their specialized geometrical tool.’’
After the construction, students will apply the Pythagorean Theorem to calculate the hypotenuse of each triangle. After finding the lengths, they use the ruler to measure these lengths to estimate
the value of each square root. Then, they approximate their location on a number line diagram.
Start by asking students to construct a right isosceles triangle with right sides of 1 cm using the segment tool at the bottom of the Polypad canvas. They may select the square grid to construct
their triangle or use the ruler to construct 1 cm-long right sides. There are many more ways to construct the first triangle. Ask students the length of the hypotenuse of the triangle.
The Pythagorean Theorem tells us the length of the hypotenuse is $sqrt{2}$sqrt2 cm. Invite one of them to use the ruler to measure the length of the hypotenuse. The decimal value of $sqrt{2}$sqrt2
will be around
$1.4 < hypotenuse < 1.5$1.4<hypotenuse<1.5
$1.4 < sqrt{2} < 1.5$1.4<sqrt2<1.5
You may also use the number line, to locate √2. Here, you may discuss the difference between rational and irrational numbers perhaps by using the story of Pythagoras and Hippasus.
√2 is an irrational number like π. It has a non-terminating and non-repeating decimal expansion.
√2 $\approx$≈ 1.414213562373...
Although we do not know the exact decimal expansion, we can still locate $sqrt{2}$sqrt2 on the number line since it can be constructed geometrically (using a compass and straightedge). Such numbers
are called constructible numbers. Ask students if they think they can construct other square roots like √3, √5, √6, √7 to approximate their decimal expansion.
Give students some time to think about the construction of √3 first. After sharing possible methods to create √3, you may start introducing the idea of the Spiral of Theodorus.
Main Activity
Theodorus of Cyrene was an ancient Greek who lived during the 5th century BCE. The spiral of Theodorus is a shape composed of right triangles. It starts with the $1 - 1 - sqrt{2}$1−1−sqrt2 isosceles
right triangle and adds another right triangle with the right sides of 1 cm and the hypotenuse of the previous one.
In this way, the second triangle will have a hypotenuse of
$\sqrt{(\sqrt{2})^2+1^2} = \sqrt{3}$(2)2+12=3 cm
When we add another triangle;
$\sqrt{(\sqrt{3})^2+1^2} = \sqrt{4}=2$(3)2+12=4=2 cm and another,
$\sqrt{(\sqrt{4})^2+1^2} = \sqrt{5}$(4)2+12=5 cm ...
So algebraically we now know that we can construct the square roots of all positive integers using the right triangle method.
To be able to make accurate measurements, we need to construct the triangles carefully.
The goal is to add another right triangle whose one right leg is the hypotenuse of the first triangle ($sqrt{2}$sqrt2 cm), and the other right leg is 1 cm. Ask students the possible strategies to
make this second step of the drawing.
Since using square grids will not work anymore, they may use the protractor and ruler to construct the new right triangle. This method may take time but works perfectly for the construction of the
Some students might use the bottom center toolbar to create perpendicular lines. In the extension part, creating a specialized geometric tool for the iterative process of adding a right triangle is
shown as another method. Click here to learn more about the geometric construction tools in Polypad.
The original spiral stopped with a triangle with a hypotenuse of $sqrt{17}$sqrt17. But you may extend the triangles as much as you want. After completing the spiral, let students calculate the length
of each hypotenuse using the pattern of the Pythagorean Theorem.
Clarify with the students that given a compass and a straight edge, one can construct the square root of any counting number. Invite students to share their different designs of the spiral and any
mathematical questions that might have occurred to them.
To close the lesson, share this canvas with the students and let students measure each hypotenuse with a ruler to come up with the approximation of each square root.
As a whole class extension, after finding the approximations, students locate the square roots on the number line. This may lead to the discussion of techniques for estimating the values of the
square roots without actually constructing and measuring them.
They may realize the perfect squares that are above and below the given square root.
For instance,
$\sqrt{Biggest\, Perfect\, Square < 11}$BiggestPerfectSquare<11 < $\sqrt{11}$11 < $\sqrt{Smallest\, Perfect\, Square > 11}$SmallestPerfectSquare>11
$\sqrt{9} < \sqrt{11} < \sqrt{16}$9<11<16
$3 < \sqrt{11}<4$3<11<4 and since 11 is closer to 9 then it is to 16, we can estimate the decimal value of$\sqrt{11}$11 as less than $3.5$3.5.
Support and Extension
For students ready for additional extension in this lesson, consider asking to create a specialized tool for the iterative process of creating Theodorus Spiral.
For the older students who have a background with trigonometry and limits, you may also ask about the growth rate of the spiral.
The spiral can also be used to provide a creative solution to the problem of how to divide a circle into 5 equal areas.
Begin by creating the radius of the first circle, which is $sqrt{1}$sqrt1. From there, just by creating the rest of the spiral, we can create the other radii of the larger circles. What is the area
of the pink circle? What is the area of each region between the co-centric circles?
For students needing additional support with these ideas, the Spiral of Theodorus canvas can be shared directly without them working on the construction process but rather concentrating on measuring
the lengths and estimating the decimal approximations of the square roots.
Polypads for This Lesson
To assign these to your classes in Mathigon, save a copy to your Mathigon account. Click here to learn how to share Polypads with students and how to view their work.
Spiral of Theodorus – Polypad – Polypad
Equal Areas – Polypad – Polypad | {"url":"https://tmctraining.com/article/spiral-of-theodorus-polypad","timestamp":"2024-11-11T00:51:27Z","content_type":"text/html","content_length":"85489","record_id":"<urn:uuid:797c8e9a-8908-447c-8055-b43b87df9708>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00298.warc.gz"} |
Integration of cotx: Formula, Proof | cotx Integration - iMath
Integration of cotx: Formula, Proof | cotx Integration
The integration of cotx is ln|sinx|, where ln denotes the natural logarithm, that is, the logarithm with base e. Here we will learn how to find the integral of cotx dx.
cotx Integration Formula
The cotx integration formula is given below.
∫cotx dx = ln|sinx|+C
Integration of cotx Proof
We will show that ∫cotx dx = ln|sinx|+C. As $\cot x =\dfrac{\cos x}{\sin x}$, the integral of cotx will be equal to
∫cotx dx = $\int \dfrac{\cos x}{\sin x} \ dx$ …(*)
In the above integral, let us put sinx=t.
Differentiating both sides, cosx dx = dt.
So we have from (*) that
∫cotx dx = $\int \dfrac{dt}{t}$
= ln|t|+C where C is a constant of integration
= ln|sinx|+C as t=sinx.
So the integration of cotx is equal to ln|sinx|+C which is proved by the substitution method of integrations.
Video Solution on Integration of cotx:
Q1: What is the Integration of cotx?
Answer: The integration of cotx is ln|sinx|+C where C is the integration constants. | {"url":"https://www.imathist.com/integration-of-cotx/","timestamp":"2024-11-09T16:35:03Z","content_type":"text/html","content_length":"176864","record_id":"<urn:uuid:2c6e55cf-543e-413f-b61e-3a28005c9ff3>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00502.warc.gz"} |
What's New in 4.6 - SwiftKanban Knowledge BaseWhat’s New in 4.6
This realease introduces several enhancements and new features.
Here are the highlights of SwiftKanban version 4.6:
Adding Kanban Cards from Email
Want to add a card to Kanban board but not have the access to the SwiftKanban site? Now you can add cards to your Kanban board directly from your email if you have the board ID of your Kanban board.
Not just adding cards, you can assign cards, add attachments and comments right from your inbox and also receive update notifications without accessing the SwiftKanban user interface.
Enhancing Excel Import
We have enhanced the Microsoft® Excel-based import template. The revamped template has following added capabilities:
• LoV values for all attribute fields like Card Type, Priority, and others are now available as a drop-down value.
• Add cards directly to the board using this template. Previously, you could only add them on the backlog.
• Some of the default field values will automatically be filled in, if left blank.
• Validation of data is performed on save and errors are thrown as a pop-up message.
Note: To use this macro-enabled template, you need to use either XLS or XLSM format.
Creating Card Hierarchy Structure Made Easy
Ever thought of establishing the parent child hierarchy structure in a jiffy? The Quick Link feature fulfils that requirement by letting you select multiple cards and link them to a single parent
card, or vice versa at the single click of your mouse.
So, you can bulk select the cards, right-click and select parent or child on the same board or across various other boards.
Enhanced Risk Assessment Module
We have gone one step ahead with our Risk Assessment module and introduced two more exciting features, Tile View and Demand Shaping.
Tile view
With the Tile View, you can now get to see the Risk cards as a tile grouped and placed horizontally in a linear format. You can view them based on Backlog or Active Cards, or view them all.
Moreover, you can click and toggle to the detailed Tile view which has information like Risk Dimensions and Custom Attributes of the cards embedded into each of the rectangles.
We have also introduced filter on the Tile view which lets you refine your view based on various parameters like Card Size, Card Type, Priority and Class of Service.
Note: Exclusively available to SwiftKanban ESP module subscribers.
Demand Shaping
You can use the Demand Shaping view to refine the Risk Cards based on the values of Risk Dimensions.
You can drag the slider of each dimension, and set the range of the dimension values, the Risk Cards matching with your requirement are displayed in the collapsible panes, where each of the panes
have certain number of cards either falling inside or outside of the set range.
So, you can group them based on cards falling inside or outside of your chosen range.
The demand shaping helps you understand the demand analysis of Risks using the outcome-driven (ODD) design approach, and take a quick, informed decision on the risk trade-offs.
Note: Exclusively available to SwiftKanban ESP module subscribers
Analysis Made Easy: Hierarchy Status Report
Introducing a board-wise Hierarchy Status report which offers detailed information of a linked card like percent complete, estimate, due dates and others in relation to its parent or child cards.
The hierarchy is displayed in a tree structure, and can be drilled down up to three levels. So, you can further configure the report by applying filters based on card type, size, release and child
card drill-down view, and others, and also export the whole report into the CSV format.
Adjacency Matrix Simplified
Adjacency matrix has now been simplified, and is much easier now to interpret with visibly bright colors.
Moreover, the colors used in the adjacency matrix are also explained as a legend on the chart.
WebServices Enhancements for Team Member Services
We have also added the new web service for retrieving last login date of all active users in Organization
getAllUsersLastLogin: Fetches last login date of all active users in Organization
We hope you like these new features!
As always, your feedback and suggestions are invaluable to us. Please keep them coming on the
SwiftKanban Feedback
portal and let others vote on them as well!
Thank you!
Leave a Reply Cancel reply
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Trustless Verification of Machine Learning
Machine learning (ML) deployments are becoming increasingly complex as ML increases in its scope and accuracy. Many organizations are now turning to “ML-as-a-service” (MLaaS) providers (e.g., Amazon,
Google, Microsoft, etc.) to execute complex, proprietary ML models. As these services proliferate, they become increasingly difficult to understand and audit. Thus, a critical question emerges: how
can consumers of these services trust that the service has correctly served the predictions?
To address these concerns, we have developed the first system to trustlessly verify ML model predictions for production-level models. In order to do so, we use a cryptographic technology called
ZK-SNARKs (zero-knowledge succinct non-interactive argument of knowledge), which allow a prover to prove the result of a computation without revealing any information about the inputs or intermediate
steps of the computation. ZK-SNARKs allow an MLaaS provider to prove that the model was executed correctly post-hoc, so model consumers can verify predictions as they wish. Unfortunately, existing
work on ZK-SNARKs can require up to two days of computation to verify a single ML model prediction.
In order to address this computational overhead, we have created the first ZK-SNARK circuit of a model on ImageNet (MobileNet v2) achieving 79% accuracy while being verifiable in 10 seconds on
commodity hardware. We further construct protocols to use these ZK-SNARKs to verify ML model accuracy, ML model predictions, and trustlessly retrieve documents in cost-efficient ways.
In this blog post, we’ll describe our protocols and how we constructed the ZK-SNARK circuit in more detail. Further details are also in our preprint.
Using ZK-SNARKs for trustless applications
Building on our efficient ZK-SNARKs, we also show that it’s possible to use these ZK-SNARKs for a variety of applications. We show how to use ZK-SNARKS to verify ML model accuracy. In addition, we
also show that ZK-SNARKs of ML models can be used to trustlessly retrieve images (or documents) matching an ML model classifier. Importantly, these protocols can be verified by third-parties, so can
be used for resolving disputes.
First, consider the setting where a model provider (MP) has a model they wish to serve to a model consumer (MC). The MC wants to verify the model accuracy to ensure that the MP is not malicious,
lazy, and or erroneous (i.e., has bugs in the serving code). To verify model accuracy, the model provider (MP) will commit to a model by hashing its weights. The model consumer (MC) will then send a
test set to the MP, on which the MP will provide outputs and ZK-SNARK proofs of correct execution. By verifying ZK-SNARKs on the test set, MC can be confident that MP has executed the model
correctly. After the model accuracy is verified, MC can purchase the model or use the MP as an MLaaS provider. In order to ensure that both parties are honest, we design a set of economic incentives,
with details in our preprint.
We instantiated our protocol with our ZK-SNARKs. To verify accuracy of an ML model within 5% costs $99.93. For context, collecting an expert-annotated dataset can cost as much $85,000. Our protocol
adds as little as 0.1% overhead in this scenario.
Second, consider the setting where a judge has ordered a legal subpoena. This may occur when a plaintiff requests documents for legal discovery or when a journalist requests documents under the
Freedom of Information Act (FOIA). When the judge approves the subpoena, the responder must divulge documents or images matching the request, which can be specified by an ML model. In order to keep
the remainder of the documents private, the requester can only divulge the specific documents as follows. The responder commits to the dataset by producing hashes of the documents, the requester
subsequently sends the model to the responder, and finally the responder produces ZK-SNARK proofs of valid inference on the documents. The responder will send only the documents that match the ML
model classifier.
Constructing the first ImageNet-scale ZK-SNARK
As we mentioned, ZK-SNARKs allow a prover to prove the result of a computation without revealing any information about the inputs or intermediate steps of the computation. ZK-SNARKs have several
non-intuitive properties that make them amenable for verifying ML models:
1. They have succinct proofs, which can be as few as 100 bytes (for non-ML applications) and as few as 6 kB for the ML models we consider.
2. They are non-interactive, so the proof can be verified by anyone at any point in time.
3. A prover cannot generate invalid proofs (knowledge soundness) and correct proofs will verify (completeness).
4. They are zero-knowledge: the proof doesn’t reveal anything about the inputs (the model weights or model inputs in our setting) beyond the information already contained in the outputs.
Constructing most ZK-SNARKs involves two steps: arithmetization (turning the computation into an arithmetic circuit, i.e., a system of polynomial equations over a large prime field) and using a
cryptographic proof system to generate the ZK-SNARK proof.
To construct ZK-SNARKs for ML models that work at ImageNet-scale, we use recent developments in proving systems, which have dramatically improved in efficiency and usability in the past few years. We
specifically use the halo2 library. Prior work for DNN ZK-SNARKs uses the Groth16 proving system or sum-check based systems specific to neural networks. Groth16 is less amenable for DNN inference due
to the non-linearities as they cannot easily be represented with quadratic constraints. Furthermore, neural network-specific proving systems do not benefit from the broader ZK proving ecosystem. They
currently have worse performance than our solution of leveraging the advances in the ZK ecosystem.
Scaling out ZK-SNARKs in halo2 still requires several advances in efficient arithmetization. We design new methods of arithmetizating quantized DNNs, performing non-linearities efficiently via lookup
arguments, and efficiently packing the circuits. Please see our preprint for more details!
We constructed ZK-SNARKs of MobileNet v2 on ImageNet. By varying the complexity of the models, we can trade off accuracy against verification time. We constructed ZK-SNARK proofs for inference for a
variety of models and show the empirical tradeoff of accuracy and verification time in the plot below:
As we can see, our models can achieve an accuracy of 79% while taking only 10s to verify on commodity hardware!
In our recent paper, we’ve scaled ZK-SNARKs on ML models that achieve high accuracy on ImageNet. Our ZK-SNARKs constructions can achieve 79% accuracy while being verifiable within 10s. We’ve also
described protocols for using these ZK-SNARKs to verify ML model accuracy and trustlessly retrieve documents. Please see our preprint for more information. Also be on the lookout for our open-source | {"url":"https://ddkang.github.io/_posts/2022-10-18-trustless/","timestamp":"2024-11-11T03:30:27Z","content_type":"text/html","content_length":"14731","record_id":"<urn:uuid:d416f496-bb08-4116-8032-b0828ffe3fcf>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00403.warc.gz"} |
Decay Constant - (Linear Algebra and Differential Equations) - Vocab, Definition, Explanations | Fiveable
Decay Constant
from class:
Linear Algebra and Differential Equations
The decay constant is a parameter that quantifies the rate at which a substance decreases over time, often used in the context of radioactive decay or exponential decay processes. It indicates how
quickly a quantity diminishes and is a crucial part of mathematical models that describe the behavior of dynamic systems, helping to predict how long it will take for a certain proportion of the
substance to decay.
congrats on reading the definition of Decay Constant. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. The decay constant is denoted by the symbol $$k$$ and has units of inverse time (e.g., seconds\(^-1\)).
2. A larger decay constant means a faster rate of decay, while a smaller decay constant indicates a slower rate.
3. In radioactive decay, the relationship between the decay constant and half-life is given by the formula: $$t_{1/2} = \frac{\ln(2)}{k}$$.
4. The decay constant can be determined experimentally by measuring the remaining amount of a substance over time.
5. In many real-world applications, such as pharmacokinetics and population studies, understanding the decay constant helps in predicting future behavior based on current measurements.
Review Questions
• How does the decay constant relate to exponential decay and what role does it play in modeling dynamic systems?
□ The decay constant is central to exponential decay, as it determines the rate at which a quantity diminishes over time. In the model $$N(t) = N_0 e^{-kt}$$, where $$k$$ is the decay constant,
a larger value of $$k$$ leads to a steeper decline in the quantity. Understanding this relationship allows for accurate predictions of how quickly substances like radioactive materials or
populations will decrease, which is vital for various scientific and engineering applications.
• Explain how the concept of half-life is derived from the decay constant and its importance in practical applications.
□ The half-life of a substance is derived from the decay constant through the equation $$t_{1/2} = \frac{\ln(2)}{k}$$. This relationship highlights that half-life and decay constant are
inversely related; as one increases, the other decreases. Knowing the half-life allows scientists and engineers to estimate how long it takes for a given amount of material to reduce by half,
which is particularly important in fields like nuclear medicine and environmental science.
• Evaluate how understanding the decay constant can impact decision-making in fields like environmental science or medicine.
□ Understanding the decay constant enables professionals in fields like environmental science and medicine to make informed decisions regarding safety and effectiveness. For instance, knowing
how quickly radioactive waste decays informs disposal methods and regulatory compliance. In medicine, recognizing how quickly drugs are metabolized helps determine dosing schedules for
optimal efficacy while minimizing toxicity. Thus, mastery of the decay constant is essential for risk assessment and ensuring public health.
© 2024 Fiveable Inc. All rights reserved.
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15 Extraordinary Facts About Van Der Waals Equation
Source: Pinterest.com
The Van der Waals equation is a fundamental concept in the field of chemistry that describes the behavior of real gases. Developed by Johannes Diderik van der Waals, a Dutch physicist, the equation
provides a more accurate representation of gas properties compared to the ideal gas law. While the ideal gas law assumes that gas particles have zero volume and do not interact with each other, the
Van der Waals equation takes into account the volume of gas particles and their attractive or repulsive forces.
In this article, we delve into the extraordinary facts surrounding the Van der Waals equation. From its historical significance to its impact on modern chemistry, we explore the key aspects that make
it an essential tool in understanding gas behavior. So, let’s dive into the fascinating world of the Van der Waals equation and uncover some intriguing facts about this remarkable scientific
Key Takeaways:
• The Van der Waals equation helps scientists understand how real gases behave, considering factors like particle size and attraction between molecules. It’s like a special formula for gases that
don’t behave like we expect them to!
• Johannes Diderik van der Waals, the equation’s creator, won a Nobel Prize for his work. His equation is used in chemistry and engineering to study gases and even predict their critical
properties. It’s like a super helpful tool for scientists and engineers!
The Van der Waals equation provides an improved model for real gases.
The Van der Waals equation, formulated by Johannes Diderik van der Waals in 1873, takes into account the non-ideal behavior of gases, considering the intermolecular forces and the finite size of gas
It corrects for the volume occupied by gas particles.
In the Van der Waals equation, the term (V – nb) adjusts for the volume occupied by the gas molecules, where V represents the actual volume and nb represents the volume of the gas particles.
The equation accounts for intermolecular forces.
Unlike the ideal gas law, the Van der Waals equation incorporates the attractive forces between gas molecules. The term (P + an²/V²) corrects for the pressure reduction due to these forces, where P
is the observed pressure and an²/V² accounts for the intermolecular attractions.
It accurately predicts the behavior of gases under high pressures.
The Van der Waals equation performs well in predicting the behavior of gases at high pressures, where the attractive forces between particles become significant.
The equation is named after its developer Johannes Diderik van der Waals.
Johannes Diderik van der Waals, a Dutch physicist, formulated the equation to explain the peculiar properties exhibited by real gases.
Van der Waals received the Nobel Prize in Physics for his work.
In 1910, Johannes Diderik van der Waals was awarded the Nobel Prize in Physics for his research on the equation of state for gases and liquids.
It can be used to estimate critical properties of gases.
The Van der Waals equation allows for the calculation of critical temperature, critical pressure, and critical volume, which are essential properties of a substance.
The equation has limitations for gases with high temperatures.
The Van der Waals equation becomes less accurate for gases at very high temperatures, where the effects of intermolecular forces are reduced.
It is commonly used in chemical engineering and thermodynamics.
The Van der Waals equation is widely employed in chemical engineering and thermodynamics to model the behavior of gases and assess deviations from ideal gas behavior.
The equation helps explain the behavior of real gases near their condensation points.
By accounting for intermolecular interactions and particle volume, the Van der Waals equation can shed light on the behavior of gases as they approach their condensation points.
The Van der Waals equation is an improvement over the ideal gas law.
While the ideal gas law assumes that gases are perfectly different from each other and that intermolecular forces and particle volume can be ignored, the Van der Waals equation provides a more
accurate representation of real gas behavior.
It is based on the concept of a molecular attraction parameter.
The ‘a’ term in the Van der Waals equation represents the molecular attraction parameter, accounting for the strength of intermolecular forces in a particular gas.
The equation can be derived from a modified form of the ideal gas law.
The Van der Waals equation can be derived from the ideal gas law by introducing additional corrective terms to account for real gas behavior.
It is an equation of state for gases.
The Van der Waals equation provides a mathematical relationship between the pressure, volume, and temperature of real gases, serving as an equation of state.
The Van der Waals equation displays critical behavior.
At temperatures and pressures near the critical point, the Van der Waals equation shows unique characteristics such as a discontinuity in the derivative of the pressure with respect to volume and a
point where gas and liquid phases become indistinguishable.
The Van der Waals equation is an remarkable tool used in the field of chemistry to accurately describe the behavior of real gases. It takes into account the intermolecular forces and the finite
volume of the gas particles, making it a more accurate representation of gas behavior compared to the ideal gas law. Understanding the Van der Waals equation is crucial for many applications,
including predicting phase transitions, determining critical points, and studying the properties of non-ideal gases. By incorporating corrections for intermolecular attractions and particle volume,
the Van der Waals equation offers a more comprehensive understanding of gas behavior. Its widespread use in research and industry signifies its importance in the field of chemistry.
Q: What is the Van der Waals equation?
A: The Van der Waals equation is an equation of state that describes the behavior of real gases by incorporating corrections for intermolecular attractions and the finite volume of gas particles.
Q: Who proposed the Van der Waals equation?
A: The Van der Waals equation was proposed by Dutch scientist Johannes Diderik van der Waals in 1873.
Q: What is the significance of the Van der Waals equation?
A: The Van der Waals equation is important because it provides a more accurate representation of gas behavior compared to the ideal gas law. It is widely used in various applications, including
predicting phase transitions and studying the properties of non-ideal gases.
Q: How does the Van der Waals equation differ from the ideal gas law?
A: The ideal gas law assumes that gas particles have no intermolecular forces and occupy no volume, while the Van der Waals equation incorporates corrections for intermolecular attractions and
particle volume.
Q: What are the variables in the Van der Waals equation?
A: The Van der Waals equation includes variables such as pressure, volume, temperature, and constants that represent the intermolecular forces and the finite size of gas particles.
Q: What can the Van der Waals equation be used for?
A: The Van der Waals equation can be used to predict phase transitions, determine critical points, and study the behavior of non-ideal gases. It is an essential tool in the field of chemistry and has
various applications in research and industry.
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information. To ensure the highest standards of accuracy and reliability, our dedicated editors meticulously review each submission. This process guarantees that the facts we share are not only
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On Computing the k-Shortcut Fréchet Distance
The Fréchet distance is a popular measure of dissimilarity for polygonal curves. It is defined as a min-max formulation that considers all direction-preserving continuous bijections of the two
curves. Because of its susceptibility to noise, Driemel and Har-Peled introduced the shortcut Fréchet distance in 2012, where one is allowed to take shortcuts along one of the curves, similar to the
edit distance for sequences. We analyse the parameterized version of this problem, where the number of shortcuts is bounded by a parameter k. The corresponding decision problem can be stated as
follows: Given two polygonal curves T and B of at most n vertices, a parameter k and a distance threshold δ, is it possible to introduce k shortcuts along B such that the Fréchet distance of the
resulting curve and the curve T is at most δ? We study this problem for polygonal curves in the plane. We provide a complexity analysis for this problem with the following results: (i) assuming the
exponential-time-hypothesis (ETH), there exists no algorithm with running time bounded by n^o(k); (ii) there exists a decision algorithm with running time in O(kn^2k+2log n). In contrast, we also
show that efficient approximate decider algorithms are possible, even when k is large. We present a (3+ε)-approximate decider algorithm with running time in O(k n^2 log^2 n) for fixed ε. In addition,
we can show that, if k is a constant and the two curves are c-packed for some constant c, then the approximate decider algorithm runs in near-linear time. | {"url":"https://cdnjs.deepai.org/publication/on-computing-the-k-shortcut-frechet-distance","timestamp":"2024-11-09T07:11:45Z","content_type":"text/html","content_length":"154645","record_id":"<urn:uuid:d76c8ba6-906a-4bb7-9b50-1a6e31eb9083>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00159.warc.gz"} |
Combine Math node
Combines grids using scalar operations.
• Combination type - The different math operators that can be brought to bear:
□ Copy A - Use A , ignore B
□ Copy B - Use B, ignore A
□ Invert - Use 1 minus A
□ Add - Add the values of A and B (Using Add for fog volumes, which have density values between 0 and 1, will push densities over 1 and cause a bright interface between the input volumes when
rendered. To avoid this problem, try using the Blend 2 option)
□ Subtract - Subtract the values of B from the values of A
□ Multiply - Multiply the values of A and B
□ Divide - Divide the values of A by B
□ Maximum - Use the maximum of each corresponding value from A and B (Using this for fog volumes, which have density values between 0 and 1, can produce a dark interface between the inputs when
rendered, due to the binary nature of choosing a value from either from A or B. To avoid this problem use Blend 1 instead)
□ Minimum - Use the minimum of each corresponding value from A and B
□ Blend 1 - (1 - A) * B. This is similar to SDF Difference, except for fog volumes, and can also be viewed as a "soft cutout" operation It is typically used to clear out an area around
characters in a dust simulation or some other environmental volume
□ Blend 2 - A + (1 - A) * B. This is similar to SDF Union, except for fog volumes, and can also be viewed as a "soft union" or "merge" operation. Consider using this over the Maximum or Add
operations for fog volumes
• A Multiplier - Multiply voxel values in the A VDB by a scalar before combining the A VDB with the B VDB
• B Multiplier - Multiply voxel values in the B VDB by a scalar before combining the A VDB with the B VDB
• Resample - Multiple options here. If the A and B VDBs have different transforms, one VDB should be resampled to match the other before the two are combined. Also, level set VDBs should have
matching background values (i.e., matching narrow band widths):
□ Off - No resampling
□ B to Match A - Resample B so that the narrow band width is the same as A
□ A to Match B - Resample A so that the narrow band width is the same as B
□ Higher-res to Match Lower-res - Resample the higher resolution data so that the resolutions match
□ Lower-res to Match Higher-res - Resample the lower resolution data so that the resolutions match
• Interpolation - When you choose a resampling option other than Off, the Interpolation choices become available. There are three to choose between:
□ Nearest - Nearest Neighbor interpolation is fast but can introduce noticeable sampling artifacts
□ Linear - Linear is the middle ground of speed and quality
□ Quadratic - Quadratic interpolation is slow but high-quality
• Deactivate - Toggle to deactivate active output voxels whose values equal the output VDB's background value. When this option is checked, the field below becomes active:
☆ Deactivate Tolerance - When deactivation of background voxels is enabled, voxel values are considered equal to the background if they differ by less than this tolerance
• Prune - Reduce the memory footprint of output VDBs that have (sufficiently large) regions of voxels with the same value
Pruning affects only the memory usage of a VDB. It does not remove voxels, apart from inactive voxels whose value is equal to the background
☆ Prune Tolerance - When pruning is enabled, voxel values are considered equal if they differ by less than the specified tolerance
• Flood Fill - Reclassify inactive voxels of level set VDBs as either inside or outside. This option will test inactive voxels to determine if they are inside or outside of an SDF and hence whether
they should have negative or positive sign | {"url":"https://docs.lightwave3d.com/lw2020/new-to-2020/openvdb-new-tools/combine-math-node.html","timestamp":"2024-11-07T10:18:31Z","content_type":"text/html","content_length":"23504","record_id":"<urn:uuid:5a3f2a97-d4aa-4ea9-93e1-f508546b63e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00571.warc.gz"} |
Vibro-Acoustic - SCS & Partners
The terminology 2d or 3d systems for building acoustics imaging, can be misleading and it is more appropriate to present the systems as spherical or planar antennas, while the algorithms of
reconstruction of acoustic fields belong to two categories: Acoustic Holography and Beamforming.
ALGORITHM - Acoustics Holography and Beamforming
The technique of Beamforming is quite similar to the principle used by a radar to identify objects in space 3d: the radar works on GHz, the Beamforming on the Hz / kHz. It is therefore to consider
the Beamforming as a system capable of focusing to a source, however, instead of moving the antenna (as does the radar) to find the maximum signal, the antenna "planar" is fixed and artificial delays
are introduced among the the microphones in the array that simulate virtually the focus.
The technique for Acoustic Holography assumes that the distribution 2-d of the sound pressure (amplitude and phase) on a plane (plane of measurement) satisfies the wave equation in an area external
to the source (propagation) and is based on the use of algorithms of spatial transformation of acoustic fields; by simplifying the concept it can be said that a Fourier transform is applied in the
spatial domain.
The acoustic holography has always been associated primarily with the analysis of the acoustic field near the source (near field) from measurements of sound pressure and / or particle velocity
(intensity) carried on a plane (plane of measurement) at a certain distance from the source.
In the common bibliography the terminology established are abbreviations such NAH and SONAH, which respectively mean Near-Field Acoustic Holography and Statistically Optimized Near-Field Acoustic
ANTENNAS - Acoustic Holography
A planar antenna extent on the plane and is normally oriented towards the source, in doing so the measurement plane can be moved in the space closer to or far from the plane of the source (parallel
to measurement plane). There are no particular restrictions in translating the measuring plane with the acoustic holography, the limits are defined by the antenna maximum width in regard to the
minimum frequency and the minimum distance between microphones for the maximum frequency, while the spatial resolution for separation "visual source discrimination" of 2 adjacent sources is given by
the distance R between the measurement plane and the source: the lower the value of R and the better the resolution.
The value of R shall not be less than the distance between the microphones, for which the order of magnitude is between 10 and 30 cm.
ANTENNAS - Beamforming
The spatial resolution Dx is in this case given by the ratio between the wavelength diameter D of the matrix (array) multiplied by the distance R between the antenna plane and
If R = 1 m, D = 0.8m, at 800Hz: Dx = 0.5m
If R = 1 m, D = 0.8m, to 2500Hz: Dx = 0.17m
Spherical ANTENNAS - Beamforming
Spherical antenna works on the same principle of planar ones even if all the algorithms are much more complex, and there are two versions: open and closed sphere SBM (Sphere Baffled Microphone
array). Considering the closed sphere is clear that the advantage over planar antenna is the fact that the sound field "can be seen" in the surrounding space (3d), while the price you have to pay is
necessarily the limitation to the frequency range, whereas 1m diameter spherical antennas have very poor handling aspects.
In fact can be considered 2-3 models of antenna spherical baffled with diameters between 260 and 165 mm, whose corresponding intervals in frequency of use, it is between 200/315 Hz low frequency and
5/8 kHz in high frequency. The Noise Vision is optimized with 31 microphones and 12 cameras. | {"url":"http://www.vibro-acoustic.com/SOUND-IMAGES.htm","timestamp":"2024-11-10T14:56:17Z","content_type":"application/xhtml+xml","content_length":"23569","record_id":"<urn:uuid:9c1d2436-4446-4671-ac7d-0f49761737a1>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00553.warc.gz"} |
Difference between Clear and ClearAll, and between '* and Global`* ?
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Difference between Clear and ClearAll, and between '* and Global`* ?
What is the difference between 'Clear' and ClearAll' and between "`*" and "Global`*" I have used ClearAll["Global`*"] and values remain attached to variables. What is the easiest way to clear
all values to all variables going forward?????
13 Replies
I'm not sure where I got the idea that:
should work but somehow I did get that idea:
What works:
I usually use an inelegant brute force method to clear all variables and definitions: Quit Kernel followed by Evaluate Notebook. Sometimes there is some hidden code lurking in a notebook. This is
also eliminated.
I did a quick notebook and typed in x =1 and hit enter. Then typed in Quit Kernel and hit enter. Then I typed in x and a value of 1 showed up.
Am I missing something? Jake
After you select Quit Kernel (from the Evaluation menu) do you follow up by clicking on Local in the sidebar?
No I didn't. I did not use the drop down menus at all. I just typed it in.
That worked great, wish I had seen it a long time ago. Thanks for your help.
I read through all of the posts to this subject and still don't understand what will work and what will not work. For an example I have an equation that uses 12 constants. My first line I put in all
of the symbols and there values that I want to use. Example, rowna=1;rowcl=2;rowk=3; and keep on going until I have all of them listed with values. Them I put in my equation and run it. If I want to
I can then can go back into my first line and change what I want too.
But if I want to get rid of all of these assigned values, what do I do. I know that I could use the Clear command and list every constant that I used, but it takes time and its easy to make a
Is there any way to clear out all variables that I have assigned values to for each section of a book that I am writing. I always end up getting caught with an old value that I forgot to clear out?
Thanks Jake
a) That is interesting. There could be a context issue left over from a package or notebook setting.
What do ?a and $Context give? On my 9.0.1 on Macintosh,
In[1]:= a=1;
In[2]:= ?a
In[3]:= ClearAll["Global`*"];
Out[4]= a
In[5]:= $Context
Out[5]= Global`
In[6]:= ?a
b) "Global`\.1d*" probably means that some non-printing character slipped in between the back-tick and the asterisk.
Try re-typing that line.
c) Usually := is used for function definitions so the variables on the right-hand side are not given values until the function is called with arguments defined.
= is used for almost everything else, since the assignment might as well happen once, right away.
q:=7 means that every time q is used, the assignment of 7 to q is re-done.
Hi Bruce Miller, now it works, thanks a lot!
a) & b):
You were right, there was a space after the " ` " character. The " ` " is a bit cumbersome to type on the German keyboard, and requires a space, or arrow right afterwards. I did not delete the space,
and didn't notice it, since it was not displayed.
So ClearAll works perfect now, thanks a lot!
c) I see, thanks!
I'd have one further question: What the syntax "Global`*" actually mean, so I can remember it more easilly? (Especially the `* part.)
Hi, thanks for the answer!
For example,
gives the output
When I try it with Remove, I get the error message:
Remove::rmnsm: There are no symbols matching "Global`\.1d*". >>
As a little side question: I am not sure if I am supposed to use := or = for assigning values to variables, or for function definitions.
Could you give a simple example of a case where
does not work? Remember that this only ClearAll's symbols in the Global` context. If you have other contexts with symbols those will not get cleared.
As a more complete approach you can use
I am having essentially the same question. I want to find a way to clear all variable assignments and function definitions I made in a document so far. Neither
work. Accourding to the Mathematica help they should.
What is the proper command?
Clear removes values and definitions from a symbol while ClearAll removes attributes, messages and any default options as well as values and definitions from a symbol.
The * kind of has the same meaning it does on a unix terminal, meaning any set of characters of any length. Global` is a particular context. It happens to be the default context when you are making
definitions in a notebook. When you use the command
you are removing all definitions, values attributes, messages and default options that have been created in the Global` context and that don't have the Protected attribute.
You can read more on this at
Be respectful. Review our Community Guidelines to understand your role and responsibilities. Community Terms of Use | {"url":"https://community.wolfram.com/groups/-/m/t/164880","timestamp":"2024-11-07T15:44:47Z","content_type":"text/html","content_length":"158060","record_id":"<urn:uuid:59589e31-dd4b-4b04-b287-24146f18eb47>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00418.warc.gz"} |
What is the importance of age-specific fertility rate? | Gzipwtf.com
What is the importance of age-specific fertility rate?
What is the importance of age-specific fertility rate?
Pros. Age-specific fertility rates enable analysis of the pattern of fertility by age of women and analysis of changes in the timing of childbearing. Comparisons between consecutive years may, for
example, indicate that women are delaying childbearing and the onset of family formation.
How do you calculate age-specific fertility rate in demography?
An age- or age-group-specific fertility rate is calculated as the ratio of annual births to women at a given age or age-group to the population of women at the same age or age-group, in the same
year, for a given country, territory, or geographic area.
What is mean by age-specific fertility rate ASFR )?
The Age-Specific Fertility Rate (ASFR) is the number of live births per 1000 women in a specific age group for a specified geographic area and for a specific point in time, usually a calendar year.
What is the difference between TFR and CBR?
CBR is a useful measure to approximate numbers of births when limited information available. Normally, CWR < 1 ▪ in low fertility countries, well below 1; ▪ in high fertility countries just under 1.
TFR is independent of the effect of the age structure. TFR gives the number of births that women give birth to.
How do you interpret age-specific rate?
An age-specific rate is calculated by dividing the total number of health events for the specific age-group of interest by the total population in that age group.
How is TFR measured?
The TFR estimates the number of children a cohort of 1,000 women would bear if they all went through their childbearing years exposed to the age-specific birth rates in effect for a particular time.
The TFR is the sum of the age-specific birth rates multiplied by five or (351.4 x 5 = 1757.0).
How do you calculate birth rate?
The birth rate in a period is the total number of live births per 1,000 population divided by the length of the period in years.
What does TFR mean?
The Total Fertility Rate (TFR) is a standard demographic indicator used internationally to estimate the average number of children that a woman would have over her childbearing years (i.e. age
15-49), based on current birth trends.
How do you calculate age structure?
It is usually calculated by dividing the number of children in the age group 0-4 (of both sexes) by the number of women of reproductive age (15-49 years), and then multiplying by 1000. Example: A
district in Viet Nam has the 4896 children under age 5 and 10,200 women aged 15-49. The CWR is: (4896/10200) * 1000 = 480.
How do you calculate the birth rate?
The crude birth rate (CBR) is equal to the number of live births (b) in a year divided by the total midyear population (p), with the ratio multiplied by 1,000 to arrive at the number of births
per 1,000 people. So, there were 14.57 births for every 1,000 people in the city.
What is a specific rate?
A specific rate is a real number. It provides an absolute measurement as well as a useful statistical tool for comparison and trend analysis. For example, Pennsylvania’s crude birth rate expressed as
the number of resident live births per 1,000 total population has shown a gradual increase from 12.9 in 1978 to 13.6 in. | {"url":"https://gzipwtf.com/what-is-the-importance-of-age-specific-fertility-rate/","timestamp":"2024-11-05T15:41:46Z","content_type":"text/html","content_length":"141244","record_id":"<urn:uuid:bbfbae4f-2d5d-44fd-988a-25da4f6e1aad>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00091.warc.gz"} |
Troll Bridge — AI Alignment Forum
All of the results in this post, and most of the informal observations/interpretations, are due to Sam Eisenstat. I think the Troll Bridge story, as a way to make the decision problem understandable,
is due to Tsvi; but I'm not sure.
Pure Logic Version
Troll Bridge is a decision problem which has been floating around for a while, but which has lacked a good introductory post. The original post gives the essential example, but it lacks the "troll
bridge" story, which (1) makes it hard to understand, since it is just stated in mathematical abstraction, and (2) makes it difficult to find if you search for "troll bridge".
The basic idea is that you want to cross a bridge. However, there is a troll who will blow up the bridge with you on it, if (and only if) you cross it "for a dumb reason" — for example, due to
unsound logic. You can get to where you want to go by a worse path (through the stream). This path is better than being blown up, though.
We apply a Löbian proof to show not only that you choose not to cross, but furthermore, that your counterfactual reasoning is confident that the bridge would have blown up if you had crossed. This is
supposed to be a counterexample to various proposed notions of counterfactual, and for various proposed decision theories.
The pseudocode for the environment (more specifically, the utility gained from the environment) is as follows:
IE, if the agent crosses the bridge and is inconsistent, then U=-10. (□⊥ means "PA proves an inconsistency".) Otherwise, if the agent crosses the bridge, U=+10. If neither of these (IE, the agent
does not cross the bridge), U=0.
The pseudocode for the agent could be as follows:
This is a little more complicated, but the idea is supposed to be that you search for every "action implies utility" pair, and take the action for which you can prove the highest utility (with some
tie-breaking procedure). Importantly, this is the kind of proof-based decision theory which eliminates spurious counterfactuals in 5-and-10 type problems. It isn't that easy to trip up with Löbian
proofs. (Historical/terminological note: This decision theory was initially called MUDT, and is still sometimes referred to in that way. However, I now often call it proof-based decision theory,
because it isn't centrally a UDT. "Modal DT" (MDT) would be reasonable, but the modal operator involved is the "provability" operator, so "proof-based DT" seems more direct.)
Now, the proof:
• Reasoning within PA (ie, the logic of the agent):
□ Suppose the agent crosses.
☆ Further suppose that the agent proves that crossing implies U=-10.
○ Examining the source code of the agent, because we're assuming the agent crosses, either PA proved that crossing implies U=+10, or it proved that crossing implies U=0.
○ So, either way, PA is inconsistent -- by way of 0=-10 or +10=-10.
○ So the troll actually blows up the bridge, and really, U=-10.
☆ Therefore (popping out of the second assumption), if the agent proves that crossing implies U=-10, then in fact crossing implies U=-10.
☆ By Löb's theorem, crossing really implies U=-10.
☆ So (since we're still under the assumption that the agent crosses), U=-10.
□ So (popping out of the assumption that the agent crosses), the agent crossing implies U=-10.
• Since we proved all of this in PA, the agent proves it, and proves no better utility in addition (unless PA is truly inconsistent). On the other hand, it will prove that not crossing gives it a
safe U=0. So it will in fact not cross.
The paradoxical aspect of this example is not that the agent doesn't cross -- it makes sense that a proof-based agent can't cross a bridge whose safety is dependent on the agent's own logic being
consistent, since proof-based agents can't know whether their logic is consistent. Rather, the point is that the agent's "counterfactual" reasoning looks crazy. (However, keep reading for a version
of the argument where it does make the agent take the wrong action.) Arguably, the agent should be uncertain of what happens if it crosses the bridge, rather than certain that the bridge would blow
up. Furthermore, the agent is reasoning as if it can control whether PA is consistent, which is arguably wrong.
In a comment, Stuart points out that this reasoning seems highly dependent on the code of the agent; the "else" clause could be different, and the argument falls apart. I think the argument keeps its
• On the one hand, it's still very concerning if the sensibility of the agent depends greatly on which action it performs in the "else" case.
• On the other hand, we can modify the troll's behavior to match the modified agent. The general rule is that the troll blows up the bridge if the agent would cross for a "dumb reason" -- the agent
then concludes that the bridge would be blown up if it crossed. I can no longer complain that the agent reasons as if it were controlling the consistency of PA, but I can still complain that the
agent thinks an action is bad because that action indicates its own insanity, due to a troublingly circular argument.
Analogy to Smoking Lesion
One interpretation of this thought-experiment is that it shows proof-based decision theory to be essentially a version of EDT, in that it has EDT-like behavior for Smoking Lesion. The analogy to
Smoking Lesion is relatively strong:
• An agent is at risk of having a significant internal issue. (In Smoking Lesion, it’s a medical issue. In Troll Bridge, it is logical inconsistency.)
• The internal issue would bias the agent toward a particular action. (In Smoking Lesion, the agent smokes. In Troll Bridge, an inconsistent agent crosses the bridge.)
• The internal issue also causes some imagined practical problem for the agent. (In Smoking Lesion, the lesion makes one more likely to get cancer. In Troll Bridge, the inconsistency would make the
troll blow up the bridge.)
• There is a chain of reasoning which combines these facts to stop the agent from taking the action. (In smoking lesion, EDT refuses to smoke due to the correlation with cancer. In Troll Bridge,
the proof-based agent refuses to cross the bridge because of a Löbian proof that crossing the bridge leads to disaster.)
• We intuitively find the conclusion nonsensical. (It seems the EDT agent should smoke; it seems the proof-based agent should not expect the bridge to explode.)
Indeed, the analogy to smoking lesion seems to strengthen the final point -- that the counterfactual reasoning is wrong.
I've come to think of Troll Bridge as "the real smoking lesion", since I'm generally not satisfied with how smoking lesion is set up.
But is proof-based decision theory really a version of EDT? I think there's more to say about the analogy, but a simple explanation is this: both EDT and proof-based decision theory evaluate actions
by adding them to the knowledge base and seeing what the world looks like under that additional assumption. Or, to put it differently, proof-based DT reasons about actions as if they're observations.
That's the fundamental idea of evidential decision theory.
Still, as good as the analogy might be, you still probably shouldn't be convinced until you've seen the Troll Bridge defeat real EDT.
Probabilistic Version
For the purely logical version, I said that we can't fault the conclusion (because the agent can't prove that it is safe to cross the bridge) – only the reasoning is being critiqued. However, the
probabilistic version better demonstrates the severity of the reasoning error; it illustrates an EDT agent making what seems to be an outright mistake.
It's reasonable to avoid crossing the bridge if you're only thinking about provable results. But we might expect an agent who balances risks and rewards would decide to cross, so long as it believes
the probability of PA being inconsistent is low. Thus turns out not to be the case.
Here's the setup.
Suppose that the agent uses a probability distribution which "respects logic" in the sense that it assigns probability zero to anything which is logically refutable. (This assumes logical
omniscience. I'll drop that assumption later.)
The agent is as follows: if P(cross)=0, cross; if P(¬cross)=0, don't cross; else, take the action with highest expected utility, breaking ties by not crossing. The idea is that we are trying to take
the highest-expected-utility action; however, we must handle cases where this isn't well-defined. Choosing to take an action when the probability of the action is zero guarantees that the agent can
never prove its own action; it is a version of the chicken rule. Breaking ties by not crossing is arbitrary, but gives the simplest version of the example, because this choice reduces the number of
cases we need to check.
The troll is as follows: if the agent crosses the bridge due to the P(cross)=0 clause, blow up the bridge.
Reasoning in the logic of the agent:
• Suppose
□ Suppose
☆ Examining the agent, either crossing had higher expected utility, or P(cross)=0. But we assumed , so it must be the latter. So the bridge gets blown up.
□ Therefore, .
• Therefore,
Thus, by Löb's theorem, we see that .
Therefore, since the agent is certain of things which are provable, the agent assigns expected value -10 to crossing. Since the expected value of the other action will always be higher than this, and
since we know the agent doesn't ever really enter into the P=0 conditions, the agent will choose not to cross.
Notice that this reasoning did not depend much on the values 10, 0, and -10. The utility of the bridge getting blown up could be -0.000001, and the agent still won't cross. It isn't weighing the
risk; it's decided that the worst outcome is inevitable. In the case of proof-based agents, I said that the overall decision not to cross was understandable, since proof-based agents are unable to
weigh the risks. A probabilistic agent, however, should intuitively be able to say "I don't know whether the bridge will get blown up, because it involves reasoning about properties of myself which
I'm fundamentally uncertain about; but, the odds look decent." But that's not what happens: instead, it is sure that crossing is unfavorable, no matter what overall probability it assigns to P(A=
So, in this case we conclude that the Troll Bridge example results in a chicken-rule-based agent taking the wrong action overall. The agent shouldn't be sure that it would cross "for the right
reason" (it should assign some probability to P(A=cross)=0, since it can't know that its own logic is consistent). However, intuitively, it should be able to assign some probability to this, and
balance the risks. If the downside risk is U=-0.000001, and the probability it assigns to its logic being consistent is not similarly small, it should cross -- and in doing so, it would get +10.
As mentioned for the proof-based agent, the agent's code is a bit arbitrary, and it is worth asking how important the details were. In particular, the default in the case of a tie was to not cross.
What if the default in case of a tie were to cross?
We then modify the troll's algorithm to blow up the bridge if and only if P(A=cross)=0 or there is a tie. The proof then goes through in the same way.
Perhaps you think that the problem with the above version is that I assumed logical omniscience. It is unrealistic to suppose that agents have beliefs which perfectly respect logic. (Un)Fortunately,
the argument doesn't really depend on this; it only requires that the agent respects proofs which it can see, and eventually sees the Löbian proof referenced.
Random Exploration
The frustrating thing about Troll Bridge is that it seems like the agent could just cross the bridge, and things would be fine. The proof that things wouldn't be fine relies on the fact that the
agent accepts that very proof as sufficient reason; so can't we just ignore that kind of proof somehow?
One thing you might try is to consider a learning agent, and force random exploration so the agent just crosses the bridge sometimes. If the agent crosses the bridge, it should be able to see that
it's safe, right?
However, we have to ask: what's the appropriate version of Troll Bridge for the exploring agent? Remember I said that the basic idea of Troll Bridge is that the troll blows up the bridge if the agent
crosses "for a dumb reason" -- ie, for a reason other than "crossing is a good idea". Random exploration falls into this category.
Furthermore, random exploration is really closely analogous to the chicken rule. Consider:
• (A) The first chicken rule I mentioned: if you prove that you don't take an action, you immediately take it.
• (B) The second chicken rule mentioned: if P(A)=0, then do A.
• (C) Epsilon-chicken: if P(A)<epsilon, then do A.
• (D) Epsilon-exploration: randomly do A with probability epsilon.
(A) is very similar to (B) if the probability distribution respects proofs in the way I mentioned previously; if the probability distribution has the further property of assigning positive
probability to anything not disproven, then they're exactly the same. (C) is a natural generalization of (B). (B) tells us what to do when the EDT decision rule isn't well-defined (we can't condition
on probability zero events). (C) generalizes this by treating small-probability events in the same way (we might expect our conditional beliefs to be bad for small-probability events, since we
presumably don't have much experience with them; so we treat it as if it's the undefined case). (B) ensures there are no zero-probability actions, which means we don't actually ever have to deal with
the EDT decision rule being non-well-defined; similarly, (C) ensures no action is almost surely never taken, which helps ensure that our conditional expectations for all actions are empirically
(C) and (D) are very similar; (C) just supplies pseudorandomness vie the agent's own probabilities. Whenever the agent is too confident about what it does, it does something else. This makes the
randomization unpredictable too the agent. (D) supplies randomness from an external source instead.
So we have a chain connecting the chicken rule to random exploration.
(Another way of thinking about the similarity: The chicken rule ensures that there are models of the agent's beliefs in which each action is taken, so that the agent can have sensible logical
reasoning about each alternative. Epsilon-exploration ensures that the empirical experiences of the agent include all of the actions, so that the agent can have reasonable empirical estimates of
consequences of each action.)
OK. So Troll Bridge for an exploring agent should blow up the bridge on exploration, or if PA is inconsistent. An agent could look like this:
• With small probability, explore.
□ If exploring, flip a coin to cross or not cross.
• If P(cross)=0, cross.
• If P(¬cross)=0, don't cross.
• Otherwise, take the max-expected-value action, breaking ties randomly.
I'm still assuming that the agent's probability distribution respects proofs, as before. I'm also assuming this agent is playing the game repeatedly, and learning. I also must assume that the agent
has found Now, the agent reasons:
• Suppose (cross u=-10) for a particular round.
□ Further suppose I crossed on that round.
☆ By the first supposition, I knew the payout of crossing to be low; and I must also have known that the payout of not crossing is higher, since I can prove that. Since I can prove what
both payouts are, the expected values must equal those, unless PA is inconsistent (in which case P(cross)=0 anyway, since my beliefs respect proofs). So I can only be crossing the bridge
for two reasons -- either this is an exploration round, or P(cross)=0.
☆ In either case, crossing the bridge yields payout u=-10.
□ Therefore, cross u=-10 in fact.
• So (cross u=-10) (cross u=-10).
Since the agent proves that a proof of crossing being bad implies crossing is actually bad, the agent further must prove that crossing is bad in fact, by Löb.
I did this for the logically omniscient case again, but as before, I claim that you can translate the above proof to work in the case that the agent's beliefs respect proofs it can find. That's maybe
a bit weird, though, because it involves a Bayesian agent updating on logical proofs; we know this isn't a particularly good way of handling logical uncertainty.
We can use logical induction instead, using an epsilon-exploring version of LIDT. We consider LIDT on a sequence of troll-bridge problems, and show that it eventually notices the Löbian proof and
starts refusing to cross. This is even more frustrating than the previous examples, because LIDT might successfully cross for a long time, apparently learning that crossing is safe, and reliably gets
+10 payoff. Then, one day, it finds the Löbian proof and stops crossing the bridge!
That case is a little more complicated to work out than the Bayesian probability case, and I omit the proof here.
Non-examples: RL
On the other hand, consider an agent which uses random exploration but doesn't do any logical reasoning, like a typical RL agent. Such an agent doesn't need any chicken rule, since it doesn't care
about proofs of what it'll do. It still needs to explore, though. So the troll can blow up the bridge whenever the RL agent crosses due to exploration.
This obviously messes with the RL agent's ability to learn to cross the bridge. The RL agent might never learn to cross, since every time it tries it, it looks bad. So this is sort of similar to
Troll Bridge.
However, I think this isn't really the point of Troll Bridge. The key difference is this: the RL agent can get past the bridge if its prior expectation that crossing is a good idea is high enough. It
just starts out crossing, and happily crosses all the time.
Troll Bridge is about the inevitable confidence that crossing the bridge is bad. We would be fine if an agent decided not to cross because it assigned high probability to PA being inconsistent. The
RL example seems similar in that it depends on the agent's prior.
We could try to alter the example to get that kind of inevitability. Maybe we argue it's still "dumb" to cross only because you start with a high prior probability of it being good. Have the troll
punish crossing unless the crossing is justified by an empirical history of crossing being good. Then RL agents do poorly no matter what -- no one can get the good outcome in order to build up the
history, since getting the good outcome requires the history.
But this still doesn't seem so interesting. You're just messing with these agents. It isn't illustrating the degree of pathological reasoning which the Löbian proof illustrates -- of course you don't
put your hand in the fire if you get burned every single time you try it. There's nothing wrong with the way the RL agent is reacting!
So, Troll Bridge seems to be more exclusively about agents who do reason logically.
All of the examples have depended on a version of the chicken rule. This leaves us with a fascinating catch-22:
• We need the chicken rule to avoid spurious proofs. As a reminder: spurious proofs are cases where an agent would reject an action if it could prove that it would not take that action. These
actions can then be rejected by an application of Löb's theorem. The chicken rule avoids this problem by ensuring that agents cannot know their own actions, since if they did then they'd take a
different action from the one they know they'll take (and they know this, conditional on their logic being consistent).
• However, Troll Bridge shows that the chicken rule can lead to another kind of problematic Löbian proof.
So, we might take Troll Bridge to show that the chicken rule does not achieve its goal, and therefore reject the chicken rule. However, this conclusion is very severe. We cannot simply drop the
chicken rule and open the gates to the (much more common!) spurious proofs. We would need an altogether different way of rejecting the spurious proofs; perhaps a full account of logical
Furthermore, it is possible to come up with variants of Troll Bridge which counter some such proposals. In particular, Troll Bridge was originally invented to counter proof-length counterfactuals,
which essentially generalize chicken rules, and therefore lead to the same Troll Bridge problems).
Another possible conclusion could be that Troll Bridge is simply too hard, and we need to accept that agents will be vulnerable to this kind of reasoning.
New Comment
50 comments, sorted by Click to highlight new comments since:
Written slightly differently, the reasoning seems sane: Suppose I cross. I must have proven it's a good idea. Aka I proved that I'm consistent. Aka I'm inconsistent. Aka the bridge blows up. Better
not cross.
I agree with your English characterization, and I also agree that it isn't really obvious that the reasoning is pathological. However, I don't think it is so obviously sane, either.
• It seems like counterfactual reasoning about alternative actions should avoid going through "I'm obviously insane" in almost every case; possibly in every case. If you think about what would
happen if you made a particular chess move, you need to divorce the consequences from any "I'm obviously insane in that scenario, so the rest of my moves in the game will be terrible" type
reasoning. You CAN'T assess that making a move would be insane UNTIL you reason out the consequences w/o any presumption of insanity; otherwise, you might end up avoiding a move only because it
looks insane (and it looks insane only because you avoid it, so you think you've gone mad if you take it). This principle seems potentially strong enough that you'd want to apply it to the Troll
Bridge case as well, even though in Troll Bridge it won't actually help us make the right decision (it just suggests that expecting the bridge to blow up isn't a legit counterfactual).
• Also, counterfactuals which predict that the bridge blows up seem to be saying that the agent can control whether PA is consistent or inconsistent. That might be considered unrealistic.
Troll Bridge is a rare case where agents that require proof to take action can prove they would be insane to take some action before they've thought through its consequences. Can you show how they
could unwisely do this in chess, or some sort of Troll Chess?
I don't see how this agent seems to control his sanity. Does the agent who jumps off a roof iff he can (falsely) prove it wise choose whether he's insane by choosing whether he jumps?
I don't see how this agent seems to control his sanity.
The agent in Troll Bridge thinks that it can make itself insane by crossing the bridge. (Maybe this doesn't answer your question?)
Troll Bridge is a rare case where agents that require proof to take action can prove they would be insane to take some action before they've thought through its consequences. Can you show how
they could unwisely do this in chess, or some sort of Troll Chess?
I make no claim that this sort of case is common. Scenarios where it comes up and is relevant to X-risk might involve alien superintelligences trolling human-made AGI. But it isn't exactly high on my
list of concerns. The question is more about whether particular theories of counterfactual are right. Troll Bridge might be "too hard" in some sense -- we may just have to give up on it. But,
generally, these weird philosophical counterexamples are more about pointing out problems. Complex real-life situations are difficult to deal with (in terms of reasoning about what a particular
theory of counterfactuals will actually do), so we check simple examples, even if they're outlandish, to get a better idea of what the counterfactuals are doing in general.
(Maybe this doesn't answer your question?)
Correct. I am trying to pin down exactly what you mean by an agent controlling a logical statement. To that end, I ask whether an agent that takes an action iff a statement is true controls the
statement through choosing whether to take the action. ("The Killing Curse doesn't crack your soul. It just takes a cracked soul to cast.")
Perhaps we could equip logic with a "causation" preorder such that all tautologies are equivalent, causation implies implication, and whenever we define an agent, we equip its control circuits with
causation. Then we could say that A doesn't cross the bridge because it's not insane. (I perhaps contentiously assume that insanity and proving sanity are causally equivalent.)
If we really wanted to, we could investigate the agent that only accepts utility proofs that don't go causally backwards. (Or rather, it requires that its action provably causes the utility.)
You claimed this reasoning is unwise in chess. Can you give a simple example illustrating this? | {"url":"https://www.alignmentforum.org/s/fgHSwxFitysGKHH56/p/hpAbfXtqYC2BrpeiC","timestamp":"2024-11-03T13:03:07Z","content_type":"text/html","content_length":"1048971","record_id":"<urn:uuid:c5377506-5717-445a-92c3-a852c326fe09>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00516.warc.gz"} |
Early Retirement Calculator - Remember To Water
Simple Early Retirement Calculator
Feel free to skip the intro and jump straight down to the early retirement calculator.
The calculator below is designed to provide some very high level indicative numbers around how your money will last if you decide to retire early. There are a number of inputs, and an output table,
that are explained below.
There are a few notable assumptions that have been made when creating this calculator. I may update it in the future to accommodate some of these, but at the moment, please take not of the following
key assumptions:
• As the calculator is looking at early retirement, it assumes no active income, only passive income from investments.
• No allowance is made for any tax benefits that superannuation provides, or for any age related accessibility limitations around superannuation. Everything is just bundled up into a single total
investment pool.
• As I can't predict the tax rates in the future, I have taken the 2016/2017 tax rates and brackets and increased the bracket boundaries in line with inflation while leaving the rates the same. No
allowance is made for any more complicated tax (eg Medicare levy); if you need that level of details, go and find an accountant (and probably bring along an actuary too!)
Inputs Explained
The following are the inputs to the calculator, and what they are used for/mean:
• Current age:
□ This is your current age, it is used to (along with the year) to provide a reference point to the numbers.
□ Scale: 0 to 100, in increments of 1
• Retire early in:
□ This is the number of years from now that you will retire in. The output table will start at this age.
□ Scale: 0 to 100, in increments of 1
• Inflation rate:
□ The average inflation rate to use for the calculations.
□ Scale: 0.0 to 10.0, in increments of 0.1
• Investment return:
□ The average return on your investments used for the calculations. This includes all increase associated with the investments, so remember to take into account potential capital gains as well
as income such as dividends or rental income.
□ Scale: 0.0 to 20.0, in increments of 0.1
• Investment:
□ The total amount of income producing assets you (want to) have saved before you retire early. This figure is the total nest egg that will be "in your bank account" on the day you decide to
retire early, so there is no inflation adjustments here. So if you think that $100k today is a good amount, then in 20 years you need to be thinking $150k. I recommend playing around with
this number the most to see how things change.
□ Scale: $100k to $10 million, in increments of $10k
• Living expense:
□ The average annual cost of living you expect when you retire in todays dollars. Please note that this is in todays dollars, and does increase with inflation. So the number you enter here will
be notably higher in the table below to keep pace with inflation. Again this is a number to play around with to see how much lower expenses can impact length of retirement.
□ Scale: $0 to $1 million, in increments of $1k
Outputs Explained
The outputs from the calculator are produced in the table under the inputs. Each row shows your investments at the start of that year/age and the expenses occurred during that year. The table
finishes when one of the following occurs:
1. You run out of money
2. You turn 130 (which seems acceptable given the oldest verified person was just shy of 123 years, and even the oldest unverified person is just shy of 130 years)
The following are the column headings and their meanings:
• Year & Age:
□ The calendar year and your age as a reference point.
• Investments:
□ The amount of income producing investments remaining at the start of that year. These investments produce an average rate of return defined in the inputs.
• Gross Income:
□ The income produced by the Investments for that year. This is based on the Investment return input and includes both any capital gains (eg share or house price rises) as well as income (eg
rental income or dividends). So remember to keep the Savings column positive if you want or expect capital gains, or want to maintain your principal.
• Tax
□ The tax needed to be paid on the Gross Income earned using 2016/2017 Australian tax data accounting for inflation of the bracket boundaries while the rates remain the same.
• Net Income
□ This is simply the Gross Income less the Tax.
• Expense
□ This lists the living expenses incurred each year. This number increases along with inflation to keep up with the ever increasing cost of living. Sort of the definition of inflation really.
This starts with an inflation adjusted number based on your input and keeps going up with inflation.
• Savings
□ This is the amount of income left over after paying tax and living expenses. This represents the portion of your investments that you reinvest back into your investment pool, or the capital
gains of your investments. If this is negative, then you will be eating into your capital. If it is positive, then your capital will increase; just remember to consider how large this is if
you want to maintain the buying power of your investments.
• Buying Power
□ This describes the Investments balance in the dollar terms of the year you retire. An example may help:
☆ If you had $100 in investments in the year you retire (lets say 2020), and could buy a crate of mangoes that year for $100, the buying power of that $100 is $100. Fast forward 20 years
(to 2040), and if you had managed to maintain that $100 investment balance, how much could it buy you? Well inflation at 2% over those 20 years would see those mangoes cost about $150. So
your $100 would only buy you two-thirds of a crate of Mangoes. So in 2020 dollar terms, the buying power of your $100 is only $66.66 (two-thirds of $100).
Simple Early Retirement Calculator
Current age: {{currentAge}} years
Retire early in: {{retireIn}} years (at {{retireAge}})
Inflation rate: {{inflation}}%
Investment return: {{investmentReturn}}%
Investments: {{accounting.formatMoney(investments,"$",0)}} (~{{accounting.formatMoney(investmentsToday,"$",0)}} today)
Living expense: {{accounting.formatMoney(livingExpense,"$",0)}} | {"url":"https://www.remembertowater.com/early-retirement-calculator/","timestamp":"2024-11-15T03:17:28Z","content_type":"text/html","content_length":"54927","record_id":"<urn:uuid:ceb72daa-edf2-4c3a-bace-433e59d5f0b5>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00859.warc.gz"} |
How To Calculate Terminal Value | Calculator (2023)
Welcome to the fascinating world of terminal value calculations!
Whether you are looking to invest in a company or start your own business, understanding how to calculate terminal value is crucial.
This financial concept can be quite challenging to grasp, but fear not, with the right guidance, you will be able to master it with ease.
So, what is terminal value?
In simple terms, it represents the present value of all future cash flows that are expected to continue indefinitely after a company's projected growth rate has stabilized.
In other words, it is the estimated value of a business beyond the forecast period.
Terminal value is vital because it helps investors and entrepreneurs determine the overall value of a company, making it easier to make informed decisions.
However, calculating it can be a daunting task, particularly for those without a finance background.
But do not fret, as we have you covered!
In this comprehensive guide, we will walk you through the process step-by-step, breaking down complex financial jargon and providing practical tips to help you calculate terminal value accurately and
with ease.
We will cover everything from the basic concept of terminal value to the different methods of calculating it.
The methods we will explore include the discounted cash flow model, which involves projecting cash flows for a finite period and then estimating the terminal value using assumptions about the
perpetual growth rate.
We will also discuss the various assumptions involved in calculating terminal value, such as the terminal growth rate and the discount rate used in the discounted cash flow model.
Moreover, we will delve into the terminal value formula and how it contributes to determining a company's overall value.
Whether you are a beginner or a seasoned investor, you will find valuable information in this guide.
So, what are you waiting for?
Dive into our guide and start mastering the art of calculating terminal value using different approaches today!
What is Terminal Value, Horizon Value, and Perpetuity Value?
Terminal value is a vital metric in determining the value of a business or project.
It is essentially an estimate of the future cash flow that extends beyond the projection period, also known as the terminal year.
This calculation is done through various methods such as the perpetual growth method or the constant rate method.
The perpetual growth method assumes that cash flows will grow at a certain rate indefinitely, while the constant rate method estimates terminal value by assuming that cash flows will remain the same
after a certain point.
To estimate terminal value, it is important to use the appropriate discount rate that reflects the risk associated with the project or business.
A good estimate of terminal value is critical in the discounted cash flow valuation as it accounts for a significant percentage of the total value.
It is also referred to as horizon value or perpetuity value, as it represents the cash flow beyond the projection period.
There are various ways to estimate the value of a business, and terminal value calculation is one of them.
It is crucial to project cash flows accurately and use the perpetuity growth method to estimate the value based on future expectations.
In doing so, it is important to consider factors such as liquidation value, which reflects the value of assets if the business were to be liquidated.
In summary, terminal value is a crucial aspect of determining the value of a project or business, and estimating it accurately requires careful consideration of various factors.
It allows for a reflection of returns well beyond the projection period, providing a more comprehensive view of the business's potential.
Why Do We Need to Calculate Horizon Value of a Stock?
When it comes to stock valuation, investors often make the mistake of relying solely on the Net Present Value (NPV) method, which assumes that the company will cease operations after the projected
cash flow period ends.
However, this method is not a realistic approach to determine the true value of a stock, especially since a company may continue operating well beyond the projected period.
This can lead to inaccurate long-term forecasts and ultimately, a poor valuation result.
To arrive at a more accurate terminal value, we need to incorporate a perpetuity value into the projected cash flows.
The perpetuity formula is used to estimate the present value of a series of cash flows that are expected to continue indefinitely.
The perpetuity growth model assumes that the company's earnings will grow at a constant rate, which is then discounted back to its present value.
Another approach to calculating the terminal value is by using the terminal multiple or implied terminal.
This method assumes that the company's earnings will remain constant beyond the projected period and applies a multiple to the projected cash flow to determine the realizable value of the stock.
Ultimately, to properly estimate the intrinsic value of a stock, we need to combine both the present value and perpetuity value approaches.
This growth method assumes that the company's earnings will continue to grow at a constant rate, which is then discounted back to its present value.
By incorporating these terminal value assumptions into our analysis, we can arrive at a more accurate estimate of the stock's realizable value.
Terminal Value Formula
There are 4 essential steps that you need to follow to estimate a company's terminal value:
1. Find all required financial data
2. Implement the discounted free cash flow (DCF) analysis
3. Calculate the company's perpetuity value (PV)
4. Use the discount rate to estimate the company's perpetuity value
Now let's dive into each step and learn how to do all the math!
How to Calculate Terminal Value
Step 1: Find the Following Figures
You need to determine 4 of the following figures before proceeding further:
• Free Cash Flow (FCF) – is a measure of your company’s financial performance which can be calculated by deducting capital expenditures from the operating cash flow.
• Long-term Growth Rate – This is a term financial analysts use to predict the rate at which a company will grow in the long-run.
• Perpetuity Growth Rate (Terminal Growth Rate) – Since horizon value is calculated by applying a constant annual growth rate to the cash flow of the forecast period, the implied perpetuity growth
rate is how much the free cash flow of the company grows until perpetuity, with each forthcoming year. In most cases, we’ll be using the GDP growth rate as the perpetuity growth rate.
• Discount Rate – This is the interest rate incorporated into discounted cash flow (DCF) analysis which helps you determine your respective cash flows’ future value. The discount rate takes into
account the time value of money as well as the risk of uncertainty revolving around future cash flows. So, the greater this uncertainty is, the higher the discount rate will be.
Step 2: Implement Discounted Cash Flow (DCF) Analysis
In order to determine the feasibility of an investment opportunity, a discounted cash flow (DCF) valuation method can be applied.
A DCF analysis brings ‘future-free’ cash flow projections into play and discounts them in order to determine the present value estimate, which in turn is used to evaluate the feasibility of an
For example, if the value determined through DCF analysis is bigger than the cost of the current investment, you have a good opportunity on your hands.
Calculating discount rate is real simple once you know what elements are involved.
There are 4 crucial steps to implement the DCF Analysis:
Project Future Free Cash Flows (FFCF)
To project your company’s future-free cash flows over the next 10 years, use the trailing twelve-month Free Cash Flow and the estimated long-term growth rate.
Here’s the formula for projecting future free cash flows:
Calculate Discount Factors (DF)
You can use the Discount Rate to calculate Discount Factors.
Here’s the formula you need to use to calculate discount factors:
Calculate Discounted Cash Flows (DCF)
Now you need to multiply each year’s respective cash flow by the discount factor in order to determine the discounted cash flow (DCF).
The formula for calculating DCF is like so:
Calculate Net Present Value (NPV)
This is a very simple step which requires summing up all the discounted cash flows.
If you’re wondering how to calculate present value, use this formula:
Step 3: Perform Terminal Value Calculation
Now that you’ve estimated the free cash flow over the projected forecast, you must determine what value the company’s cash flows may be after that forecast period expires; especially when the company
has reached a “middle-age maturity” period.
The trouble with cash flow forecasts is that it’s not only complicated enough to forecast cash flows for five years or more, but even more so the longer a company remains in existence.
However, using a horizon value formula, you can make calculated assumptions on a company’s long-term cash flow growth which goes well beyond 10 years, for instance.
The best way to calculate the perpetuity value is to make use of the Gordon Growth Model.
The formula to calculate terminal value looks like this:
Note: We’ll be using the Final Year’s Projected FCF in order to arrive upon our TV, not the Discounted FCF.
Step 4: Calculate a Present Value of Perpetuity
Since we have used the Projected FCF to arrive upon our horizon value, we must discount it to the present.
All you have to do is take the value we calculated in Step 3 above, multiply it with the Final year’s Discount Factor, and we’ll have our Discounted Terminal Value, also known as Present Value of
The formula looks like this:
Now you can use the present value of perpetuity to calculate your stock’s intrinsic value.
Simply add up the Net Prevent Value and Discounted Terminal Value, and then divide that figure by the total number of outstanding shares, like so:
Terminal Value Calculator
Terminal Value Example
Now let's take a look at some examples, so that you can understand clearly how to calculate the TV of a company in which you want to invest.
Case Study #1 - Calculate Horizon Value
Now that we’ve understood how it all comes together and have the formulas to make the necessary calculations, let’s put it all into context, shall we?
Company A is selling shares at $25.2 a piece. It has 43 million outstanding shares and has a free cash flow of $84.23 million. Financial experts forecast that the company may grow by 12.00% annually.
If the perpetuity growth rate is 2.36% and the discount rate is 6.23%, what’s the horizon value of Company A’s stock?
4 essential figures we need:
• Free Cash Flow = $84.23 million
• Long-term Growth Rate = 12.00%
• Discount Rate = 6.23%
• Perpetuity Growth Rate = 2.36%
• Find horizon value of this company's stock
Find the company's Discounted Free Cash Flow (DFCF)
We'll use the LT growth rate to project the future free cash flows of this firm in the next 10 years, and then discount each year's cash flow to the present.
What we need:
• Free Cash Flow = $84.23 million
• Long-term Growth Rate = 12.00%
• Discount Rate = 6.23%
Calculate Projected Free Cash Flows (PFCF):
Calculate Discount Factors (DF):
Calculate Discounted Free Cash Flows (DFCF):
Calculate Net Present Value (NPV):
Find the company's Horizon Value
Our next step is to calculate the business horizon value.
After having this value, we simply multiply it by the Year-10 Discount Factor to calculate the present value of perpetuity.
What we need:
• Terminal Growth Rate = 2.36%
• Discount Rate = 6.23%
• Year-10 Projected Free Cash Flow = $261.61 million
• Year-10 Discount Factor = 0.546
Case Study #2 - Calculate Perpetuity Value
Just to test our know-how a little more, and truly put these methods to the test, here’s another example:
Company B is selling their shares at $15.1 a piece. It has 34 million outstanding shares. The company’s free cash flow is valued at $45.76 million. Financial experts have projected that the company
will grow by 8.00% each year.
If the perpetuity growth rate is 2.36% and the discount rate is 5.68%, what would be the stock’s perpetuity value?
Food for thought!
4 essential figures we need:
• Free Cash Flow = $45.76 million
• Long-term Growth Rate = 8.00%
• Discount Rate = 5.68%
• Perpetuity Growth Rate = 2.36%
• Find perpetuity value of this company's stock
Find the company's Discounted Free Cash Flow (DFCF)
We'll use the long-term growth rate to calculate projected free cash flows of this stock in the next 10 years, and then discount the cash flow of each year to the present.
What we need:
• Free Cash Flow = $45 million
• Long-term Growth Rate = 8.00%
• Discount Rate = 5.68%
Project this company's Future Free Cash Flows (FFCF):
Calculate this company's Discount Factors (DF):
Calculate this company's Discounted Free Cash Flow (DFCF):
Calculate this company's Net Present Value (NPV):
Find the company's Perpetuity Value
Our next step is estimating the company's perpetuity value.
After having this value, we simply multiply it by the Year-10 Discount Factor to calculate the present value of perpetuity.
What we need:
• Perpetuity Growth Rate = 2.36%
• Discount Rate = 5.68%
• Year-10 Projected Free Cash Flow = $98.79 million
• Year-10 Discount Factor = 0.576
Frequently Asked Questions
Q: What is Terminal Value?
Terminal value is the estimated value of a business or investment at the end of a specific period. It is used in financial ratio analysis to determine the overall worth of a company or investment at
the end of a forecast period.
Q: How is Terminal Value Calculated?
Terminal value is typically calculated using the perpetuity formula, which involves estimating the company's future cash flows and then calculating the present value of those cash flows as if they
were a perpetuity. Another common method is the exit multiple method, which involves estimating the company's future earnings and then applying a multiple to those earnings.
Q: What is the Importance of Terminal Value in Business Valuation
Terminal value is an essential component of business valuation, as it represents the bulk of the value of a company or investment. Accurately estimating the terminal value is crucial for making
informed investment decisions and determining the appropriate price to pay for a company or investment.
Q: How Does Terminal Value Impact Investment Decisions?
Investors use terminal value to evaluate the long-term potential of an investment. A higher terminal value indicates greater potential for future growth and can make an investment more attractive. On
the other hand, a lower terminal value may suggest that an investment has limited growth potential and may not be a good choice for long-term investment.
The Bottom Line
Calculating the value of a business or investment is crucial to unlocking its potential for growth and success.
One way to estimate the value of a business beyond a certain time frame is by calculating the terminal value.
Terminal value is calculated by taking into account various factors such as projected cash flows, long-term growth rates, and the appropriate discount rate.
One commonly used method to calculate terminal value is the perpetuity growth method.
This method assumes that the business will grow at a steady rate indefinitely.
To calculate the terminal value using the perpetuity growth method, you need to first estimate the value of the business at the end of the projection period.
Then, you use the formula:
Terminal Value = (Final Year Cash Flow x (1 + Perpetuity Growth Rate)) / (Discount Rate - Perpetuity Growth Rate).
This formula calculates the present value of all future cash flows beyond the projection period.
Another method to calculate terminal value is the exit multiple method, which uses an implied enterprise value to calculate the terminal value.
This method takes into account a multiple of earnings or cash flow to calculate the terminal value.
It's important to note that calculating the terminal value is just one component of estimating the value of a business.
To calculate intrinsic value, you need to consider both the terminal value and the value of the business up until that point, then discount it back to the present using the appropriate discount rate.
The discount rate takes into account the opportunity cost of investing in the business.
Overall, understanding how to value a business is essential for any business or investment owner.
By using some financial analysis ratios and the right tools, you can gain a clearer understanding of the value of your investment over time.
So keep learning and practicing, and don't be afraid to take risks.
Who knows what kind of amazing opportunities await! | {"url":"https://wealthyeducation.com/how-to-calculate-terminal-value/","timestamp":"2024-11-14T10:06:11Z","content_type":"text/html","content_length":"454243","record_id":"<urn:uuid:f0e4ba95-4c92-41cc-b1bb-6e4c499df733>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00675.warc.gz"} |
[Solved] L, C and R represent the physical quantities inductanc... | Filo
L, C and R represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations does not have dimensions of frequency?
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The time constant of the R C circuit is given by
On taking the reciprocal of the above relation, we get
will have the dimensions of the frequency.
The time constant of the L R circuit is given by
On taking the reciprocal of the above relation, we get
will have the dimensions of the frequency.
On multiplying eq. (1) and (2), we get
Thus, will have the dimensions of the frequency.
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Question Text L, C and R represent the physical quantities inductance, capacitance and resistance respectively. Which of the following combinations does not have dimensions of frequency?
Updated On Apr 10, 2023
Topic Electromagnetic Induction
Subject Physics
Class Class 12
Answer Type Text solution:1 Video solution: 2
Upvotes 322
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Simplify Your Pressure Measurements with Our Conversion Tool
Pressure Converter
Pressure Unit Converter Tool
A Pressure Unit Converter Tool is crucial for converting between various units of pressure, such as pascals, bar, psi, and atmospheres. This tool is particularly useful for engineers, scientists, and
anyone involved in fluid dynamics or mechanical systems. It helps in ensuring accurate pressure measurements, which are essential for the design and analysis of various systems.
Units of Pressure Measurement
Pressure measures the force exerted per unit area. It is a fundamental concept in various fields, including physics, engineering, meteorology, and fluid dynamics. Different units are used to measure
pressure depending on the system of measurement and the context. Here’s an overview of some common pressure units:
• Pascal (Pa): The SI unit of pressure. One pascal is equal to one newton per square meter. It is widely used in scientific and engineering contexts.
• Bar: A metric unit of pressure equal to 100,000 pascals. It is commonly used in meteorology and in industries dealing with gases and liquids.
• Atmosphere (atm): A unit of pressure defined as the atmospheric pressure at sea level. One atmosphere is equal to 101,325 pascals. It is often used in chemistry and physics.
• Millimeter of Mercury (mmHg): A unit of pressure based on the height of a mercury column. One millimeter of mercury is approximately equal to 133.322 pascals. It is commonly used in medicine and
• Torr: A unit of pressure named after Evangelista Torricelli. One torr is approximately equal to one millimeter of mercury (mmHg). It is often used in vacuum measurement.
• Pound per Square Inch (psi): A unit of pressure used primarily in the United States. One pound per square inch is equal to approximately 6,894.76 pascals. It is commonly used in engineering and
everyday applications like tire pressure.
Pressure Unit Conversion Table
Pascal (Pa) Bar Atmosphere (atm) Millimeter of Mercury (mmHg) Torr Pound per Square Inch (psi)
1 Pa 0.00001 bar 0.00000987 atm 0.00750062 mmHg 0.00750062 0.000145038 psi
10 Pa 0.0001 bar 0.0000987 atm 0.0750062 mmHg 0.0750062 0.00145038 psi
100 Pa 0.001 bar 0.000987 atm 0.750062 mmHg 0.750062 0.0145038 psi
1,000 Pa 0.01 bar 0.00987 atm 7.50062 mmHg 7.50062 0.145038 psi
10,000 Pa 0.1 bar 0.0987 atm 75.0062 mmHg 75.0062 1.45038 psi
100,000 Pa 1 bar 0.987 atm 750.062 mmHg 750.062 14.5038 psi
1,000,000 Pa 10 bar 9.87 atm 7,500.62 mmHg 7,500.62 145.038 psi
1 atm 101,325 Pa 1 atm 760 mmHg 760 14.696 psi
1 mmHg 133.322 Pa 0.0013158 atm 1 mmHg 1 0.0193368 psi
1 torr 133.322 Pa 0.0013158 atm 1 mmHg 1 0.0193368 psi
1 psi 6894.76 Pa 0.068046 atm 51.715 mmHg 51.715 1 psi
This table provides a comprehensive view of conversions between different pressure units, facilitating comparisons and assessments across various contexts and applications. Understanding these
conversions is essential for precise measurements in scientific, engineering, and everyday scenarios. | {"url":"https://toolswap.xyz/converters/pressure","timestamp":"2024-11-12T23:16:13Z","content_type":"text/html","content_length":"42595","record_id":"<urn:uuid:53ce8a6d-666c-4698-bade-54832e443aff>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00452.warc.gz"} |
Total Amount Paid On 30 Year Mortgage
theanisenkova.ru Total Amount Paid On 30 Year Mortgage
Total Amount Paid On 30 Year Mortgage
In other words, you could pay off the mortgage in less than 30 years, speeding up your repayment schedule and reducing your total interest-carrying costs. This. Quickly Estimate The Cost of Interest
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years will come to $, Lower your overall cost of borrowing · Financing Your Home Renovations · How to Payment increases each year Includes mortgage protection. Mortgage. if it's 8% interest on a ,
loan, then % * , = $40, year 1. $40, / 12 months = $3, per month. So you'd pay that amount.
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Input your target home price, down payment, and interest rate into Capital Banks's year vs. year mortgage calculator to generate the amount you can. This tool allows you to calculate your monthly
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Select calculates total interest paid on a mortgage, car loan, student loans and credit card debt. · Mortgage interest paid in a lifetime: $, · Auto loan. Assuming you pay off the mortgage over the
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What Do I Need For A Wells Fargo Account | Instant Money Transfer From One Bank To Another | {"url":"https://theanisenkova.ru/community/total-amount-paid-on-30-year-mortgage.php","timestamp":"2024-11-12T19:24:27Z","content_type":"text/html","content_length":"12186","record_id":"<urn:uuid:e8acf748-2b44-4f01-b060-93e02add7f94>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00431.warc.gz"} |
Bilinear map
Jump to navigation Jump to search
In mathematics, a bilinear map is a function combining elements of two vector spaces to yield an element of a third vector space, and is linear in each of its arguments. Matrix multiplication is an
Vector spaces[edit]
Let V, W and X be three vector spaces over the same base field F. A bilinear map is a function
B : V × W → X
such that for any w in W, the map ${\displaystyle B_{w}}$
v ↦ B(v, w)
is a linear map from V to X, and for any v in V, the map ${\displaystyle B_{v}}$
w ↦ B(v, w)
is a linear map from W to X.
In other words, when we hold the first entry of the bilinear map fixed while letting the second entry vary, the result is a linear operator, and similarly for when we hold the second entry fixed.
If V = W and we have B(v, w) = B(w, v) for all v, w in V, then we say that B is symmetric.
The case where X is the base field F, and we have a bilinear form, is particularly useful (see for example scalar product, inner product and quadratic form).
The definition works without any changes if instead of vector spaces over a field F, we use modules over a commutative ring R. It generalizes to n-ary functions, where the proper term is multilinear.
For non-commutative rings R and S, a left R-module M and a right S-module N, a bilinear map is a map B : M × N → T with T an (R, S)-bimodule, and for which any n in N, m ↦ B(m, n) is an R-module
homomorphism, and for any m in M, n ↦ B(m, n) is an S-module homomorphism. This satisfies
B(r ⋅ m, n) = r ⋅ B(m, n)
B(m, n ⋅ s) = B(m, n) ⋅ s
for all m in M, n in N, r in R and s in S, as well as B being additive in each argument.
A first immediate consequence of the definition is that B(v, w) = 0[X] whenever v = 0[V] or w = 0[W]. This may be seen by writing the zero vector 0[X] as 0 ⋅ 0[X] (and similarly for 0[W]) and moving
the scalar 0 "outside", in front of B, by linearity.
The set L(V, W; X) of all bilinear maps is a linear subspace of the space (viz. vector space, module) of all maps from V × W into X.
If V, W, X are finite-dimensional, then so is L(V, W; X). For X = F, i.e. bilinear forms, the dimension of this space is dim V × dim W (while the space L(V × W; F) of linear forms is of dimension dim
V + dim W). To see this, choose a basis for V and W; then each bilinear map can be uniquely represented by the matrix B(e[i], f[j]), and vice versa. Now, if X is a space of higher dimension, we
obviously have dim L(V, W; X) = dim V × dim W × dim X.
• Matrix multiplication is a bilinear map M(m, n) × M(n, p) → M(m, p).
• If a vector space V over the real numbers R carries an inner product, then the inner product is a bilinear map V × V → R.
• In general, for a vector space V over a field F, a bilinear form on V is the same as a bilinear map V × V → F.
• If V is a vector space with dual space V^∗, then the application operator, b(f, v) = f(v) is a bilinear map from V^∗ × V to the base field.
• Let V and W be vector spaces over the same base field F. If f is a member of V^∗ and g a member of W^∗, then b(v, w) = f(v)g(w) defines a bilinear map V × W → F.
• The cross product in R^3 is a bilinear map R^3 × R^3 → R^3.
• Let B : V × W → X be a bilinear map, and L : U → W be a linear map, then (v, u) ↦ B(v, Lu) is a bilinear map on V × U.
See also[edit]
External links[edit] | {"url":"https://static.hlt.bme.hu/semantics/external/pages/tenzorszorzatok/en.wikipedia.org/wiki/Bilinear_map.html","timestamp":"2024-11-12T15:42:33Z","content_type":"text/html","content_length":"66715","record_id":"<urn:uuid:a9458043-e995-4089-9779-09a4895c5aef>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00095.warc.gz"} |
download now name: qquad class: qquad date: qquad id: a vital signs multiple choice identifi the letter of the choice that best | Question AI
Download now Name: qquad Class: qquad Date: qquad ID: A Vital Signs Multiple Choice Identifi the letter of the choice that best completes the statement or ansuen the question. qquad 1. The health
care worker who is responsible for taking vital signs is the a. nursing assistant. c. clarge nurse. b. physician. d. team leader. qquad 2. A small ° after the temperature value indicates a. that the
temperature was taken onally. c. that the Fahrenheit sale was used. b. that the Celsjus scale was used. d. the degrees of temperature. qquad 3. Body temperature is highest a. in the morning. c. in
the eveling b. from 12:00 PM to 3:00 PM. d. Jluring the night. qquad 4. Body temperature is closest to the core temperature at the a. mouth. c. skin. b. rectutin. d. tympanic nembrane. qquad 5. Body
temperature may be affeced by a. mental status. c. hydration. b. mood. d. behavior. qquad 6. The average oral Fahrenheit temperature is a. 95.8 degrees. c. 98.6 degrees. b. 101.4 degrees. d. 103,6
degrees. 7. Body temperature is most accurate when measured in the a. mouth. c. rectum. b. ear. d. ampit. 8. Which method wotsd you select for taking the temperature of an unconseious patient? a.
Oral c. Rectal b. Axillary d. Groin 9. You are assigned to take Mr. Lippincott's temperature orally every four hours. At 8:00 AM it is 99.2°F . At 12:00 PM it is 102.0°F . You shoukd a. document it
on the clipboard or flow sheet ar the end of your shift. b. continue taking temperatures on the other patients you are assigned to. c. report Mr. L.ippincoit's temperature to the nurse at once. d.
refill the water pitcher and instruct the patient to deink plenty of ice water. qquad 10. What color is the dot at the end of the stem on a rectal thermometer? a. Red c. Green b. Yellow d. Blue qquad
11. The tympanic themoneter measures body temperature in the a. mouth. c. rectum. b. nostril. d. ear. 12. The tympanic themoneter mensures the femperature in a. Fahrenheit. c. weither of the above.
b. Celsius. d. both of the above. 13. If a patient has been draiking bot coffee, how bong stould you wait before taking the temperature orally? a. 5 minutes c. 20 minutes b. 15 minutes d. 30 minutes
Not your question?Search it
Leann ChaneyMaster · Tutor for 5 years
<p> 1.a 2.d 3.c 4.b 5.c 6.c 7.c 8.c 9.c 10.a 11.d 12.d 13.d </p>
<p> <br />1. This question is probing knowledge of healthcare team roles. Generally, it's often the nursing assistant whose tasks include taking vital signs.<br />2. A small ° symbolizes degree - it
is a universal indicating the unit of measurement for temperature.<br />3. It's well established that body temperature fluctuates within a day and usually peaks in the late afternoon or evening.<br
/>4. Core temperature is best mirrored by rectal measurements.<br />5. Factors like mental status, hydration, and mood can all impact body temperature.<br />6. The average oral temperature, measured
in degree Fahrenheit, is traditionally accepted to be 98.6 degrees.<br />7. The rectal temperature is taken as the most accurate because it reflects core body temperature.<br />8. The unconscious
patient would beg to have their temperature measured via rectum or axilla. Groin isn't a common place to measure.<br />9. Af occurs in Mr. Lippincott's Crimson temperature in four hours would be
according to concerned about supposed to be an immediate sign to the reports a to the nurse.<br />10. Usually, there might be a color code; Standard thermometers used in hospitals often make the red
tip indicating use in the rectum.<br />11. Tympanic thermometers measure temperature in the ear.<br />12. Tympanic thermometers can provide temperature readings in both Fahrenheit and Celsius.<br />
13. Hot fluids can certain then affect helps oral cavity temperature measured accurately. Before Af drink hotter or colder, Usually allowed 15- 30 min to default the normal baseline oral temperature
meant during that time limit ensuring are less Chinese.teel alsobe consideredlonon their Quest utrast-Fahrenheitsual information to ac econ that is based on common clinicdtoms, </p>
Step-by-step video
Download now Name: qquad Class: qquad Date: qquad ID: A Vital Signs Multiple Choice Identifi the letter of the choice that best completes the statement or ansuen the question. qquad 1. The health
care worker who is responsible for taking vital signs is the a. nursing assistant. c. clarge nurse. b. physician. d. team leader. qquad 2. A small ° after the temperature value indicates a. that the
temperature was taken onally. c. that the Fahrenheit sale was used. b. that the Celsjus scale was used. d. the degrees of temperature. qquad 3. Body temperature is highest a. in the morning. c. in
the eveling b. from 12:00 PM to 3:00 PM. d. Jluring the night. qquad 4. Body temperature is closest to the core temperature at the a. mouth. c. skin. b. rectutin. d. tympanic nembrane. qquad 5. Body
temperature may be affeced by a. mental status. c. hydration. b. mood. d. behavior. qquad 6. The average oral Fahrenheit temperature is a. 95.8 degrees. c. 98.6 degrees. b. 101.4 degrees. d. 103,6
degrees. 7. Body temperature is most accurate when measured in the a. mouth. c. rectum. b. ear. d. ampit. 8. Which method wotsd you select for taking the temperature of an unconseious patient? a.
Oral c. Rectal b. Axillary d. Groin 9. You are assigned to take Mr. Lippincott's temperature orally every four hours. At 8:00 AM it is 99.2°F . At 12:00 PM it is 102.0°F . You shoukd a. document it
on the clipboard or flow sheet ar the end of your shift. b. continue taking temperatures on the other patients you are assigned to. c. report Mr. L.ippincoit's temperature to the nurse at once. d.
refill the water pitcher and instruct the patient to deink plenty of ice water. qquad 10. What color is the dot at the end of the stem on a rectal thermometer? a. Red c. Green b. Yellow d. Blue qquad
11. The tympanic themoneter measures body temperature in the a. mouth. c. rectum. b. nostril. d. ear. 12. The tympanic themoneter mensures the femperature in a. Fahrenheit. c. weither of the above.
b. Celsius. d. both of the above. 13. If a patient has been draiking bot coffee, how bong stould you wait before taking the temperature orally? a. 5 minutes c. 20 minutes b. 15 minutes d. 30 minutes
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Andrew Mellor
So just where does machine learning fit in to the scientific world? And are we eventually going to have to resort to such methods to make progress in modelling if mathematical models cannot keep up?
I'd like to argue that machine learning and mathematical models can coexist nicely in both research and industrial applications, and that the most interesting models appear when we utilise the power
of both paradigms. To illustrate this point we take a trip to a civilisation who are in the contradictory position of having invented machine learning (implemented in PyTorch, strangely...) but their
mathematics is not up to scratch. In particular they are not familiar with modern calculus. This example should hopefully replicate and illuminate the situation we find ourselves in now - where
mathematics is difficult but data collection is easy.
more ... | {"url":"https://andrewmellor.co.uk/blog/","timestamp":"2024-11-05T09:28:19Z","content_type":"text/html","content_length":"17176","record_id":"<urn:uuid:836bd6fe-c83d-4886-a791-dd2e0a4e06ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00568.warc.gz"} |
tSNE/viSNE SPADE PCA and MoretSNE/viSNE SPADE PCA and More - De Novo Software
Data transformations allow you to create custom parameters through various algorithms in FCS Express as well as integrate directly with R or Python. The Transformations navigator is an intuitive
dialog and makes it easy to apply your transformations to plots via a quick drag-and-drop. Simply click on the Tools tab→Transformations to launch the Transformations navigator. FCS Express does not
require R/Python scripting or programming for any algorithms available in the software. However, if you prefer to use R/Python, or desire to use your own custom scripts, the easy to use
Transformations interface supports direct integration with R/Python.
All transformations can be applied to both Flow and Image cytometry data sets.
Available Algorithms and Tools
Please visit our extensive resources on data transformations via the quick links below so you can get up and running today.
Pipelines in FCS Express are a set of data processing steps that stand alone or are connected in series. The output of a step can be applied to a data file or utilized as the input of the next step,
or series of steps, that may be applied to your data. No plugins, no programming, just results!
The vast majority of commonly used algorithms for data analysis are actually pipelines, with tSNE (and its variants) and SPADE being two examples. However, when those algorithms are implemented as a
one calculation, user customizations are limited to the what the specific implementation allows. Pipelines in FCS Express increase the computational flexibility and granularity of running algorithms
and data transformations while giving users the unique ability to create their own transformations.
FCS Express provides a wide variety of pipeline steps for you to choose from. See our full list of all available steps and contact us if there are steps you are interested incorporating into FCS
Express that are not yet available. | {"url":"https://denovosoftware.com/full-access/features/transformations/","timestamp":"2024-11-01T23:06:19Z","content_type":"text/html","content_length":"124595","record_id":"<urn:uuid:261faeb9-1955-4621-960e-e9e609bfc0f3>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00216.warc.gz"} |
5b: Collect & analyze data
ISTE Student Standard: 5b: Collect & analyze data
In this activity, students determine which book from class should be made into a movie, by gathering and comparing evidence through the internet about the profitability and customer sentiment of each
book like a Hollywood Producer would. They will source their evidence evaluate its validity, ultimately making a claim as … Read more
English Language Arts | 11th-12th, 6th - 8th, 9th-10th
In lesson 2, students visualize numerical data into bar graphs after collecting and organizing data into numerical values using free educational technology. Students learn a computational procedure
to analyze and interpret numerical data to derive insights about a real-world problem of supply demand for marbles in a classroom, ultimately offering … Read more
In Part 2, calculate statistics from different types of samples from a population and compare them to determine what conditions make a sample representative of a population. In this multi-step
project, students become the Chief Data Analyst for the US Census Bureau Chief and write a report for a governor … Read more
In part 3, students interpret visualized numerical data. Students learn a computational procedure to analyze and interpret numerical data to derive insights about a real-world problem of supply
demand for marbles in a classroom, ultimately offering solutions and testing them.
In part 1, students prepare to visualize numerical data by collecting and organizing data into numerical values using free educational technology. Students learn a computational procedure to analyze
and interpret numerical data to derive insights about a real-world problem of supply demand for marbles in a classroom, ultimately offering solutions … Read more | {"url":"https://iste.web.unc.edu/standards/computational-thinker/5b/","timestamp":"2024-11-14T08:23:45Z","content_type":"text/html","content_length":"81600","record_id":"<urn:uuid:d05fd6e2-49cf-4422-ad48-5649cc00c1d5>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00168.warc.gz"} |
Credit Card Interest Calculator
Created by Tibor Pál, PhD candidate
Reviewed by Dominik Czernia, PhD and Jack Bowater
Last updated: Jun 05, 2023
Use the credit card interest calculator to estimate how much interest you would need to pay if you happened to carry an unpaid balance on your credit card. This device's high specification allows you
to examine and compare the interest payments in different repayment scenarios. While the line chart and pie graph help you to analyze the results visually, you can follow your credit card monthly
interest from month to month in the payment schedule. In the following article, you can read how to use the credit card calculator, learn how to calculate credit card interest, and answer "How does
credit card interest work?".
If you have a credit card debt, check our credit card minimum payment calculator to estimate how long it would take to pay off your balance with minimum payments.
How to calculate credit card interest?
If you have a credit card and happened to carry an unpaid balance through several months, you may experience a perplexing feeling when you see the considerable amount of charged interest.
To clear up the matter, we explain in the following point how to calculate credit card interest and what details you need to pay attention to.
1. Get familiar with the relevant terms
Credit card issuers disclose the interest rate in the form of the Annual Percentage Rate (APR), which represents the credit's actual yearly cost. Still, in most cases, it is applied daily, meaning
that they add (capitalize) the charged interest each day after the grace period.
Therefore, you can find the utilized interest rate on your credit card statement as the Annual Percentage Rate, or APR. Note, that the credit card issuer may apply variable or non-variable APRs, and
also it may differ depending on the type of unpaid balance. For example, the APR for purchases might be 20 percent but for cash advances it could be 25 percent.
A credit card minimum payment is the lowest amount you need to pay by the due date to avoid any penalties or late fees and meet the terms of your card agreement. Therefore, it is crucial to transfer
this amount monthly by keeping to the deadlines even when you are short of money. Of course, if possible, try to pay more than the minimum to reduce the finance charge on your remaining balance.
A grace period is a period between the end of a billing cycle and your payment due date. You may not be charged interest if you pay your balance in full during this time.
2. Convert your APR to a daily periodic rate
Since interest is calculated daily, you need to convert the APR to a daily periodic rate. To do that, divide the APR by 365. Note, that some banks may use 360 for the computation, which will not
result in a significant change.
Daily interest rate = APR / 365
How does credit card interest work?
Let's continue with the previous train of thought to find out how does credit card interest work:
3. Determine the amount to which the interest rate is applied
Credit card issuers most commonly use the Average Daily Balance for interest computation, that is, the average of what you owed each day during the billing cycle.
The first thing is to find your unpaid balance — the amount carried over from the previous month. This balance changes everyday you make a financial transaction. For example, when you make a
purchase, the balance goes up; when you make a payment, it goes down.
Once you have all the daily balances in a billing cycle, you need to sum up them and then divide by the number of days in the billing period to get the average daily balance.
Average daily balance = Balance #1 + Balance #2 + Balance #3 ... / Number of days in the billing cycle
As you can see, this procedure requires balances on each day during the billing cycle, which makes the computational process tedious. For ease of use, this credit card monthly interest calculator
only asks you for the unpaid balance.
4. Calculate the charged interest
At this point, you can determine the charged interest: multiply your average daily balance by the daily interest rate, and then multiply that result by the number of days in the billing period.
Credit card interest = Average daily balance * Daily interest rate * Number of days in the billing cycle
If interest is capitalized on your balance daily, which is the general practice of credit card issuers (and applied in the this credit card monthly interest calculator as well), the actual interest
amount might be higher. The divergence between the different frequency of interest capitalization becomes more profound when you carry the unpaid balance for a longer period. To get familiar with
this procedure, and see different compounding frequencies in different time horizons, check out our compound interest calculator.
How to use credit card interest calculator?
For the first step, you need to set the following parameters in the specification section:
• Current balance - your outstanding unpaid balance on your credit card.
• Due date - the closest date by which you need to make a monthly payment.
• APR - the Annual Percentage Rate.
• Repayment by - you can set here whether you would like to pay back your credit by fixed monthly or minimum required payments.
• Monthly fixed payment - the credit card payment you transfer monthly to reduce the unpaid balance.
• Additional monthly payment - the additional payment you may devote for repayment, which may result in a shorter repayment period with a lower interest charges.
After providing all of the above parameters, you will receive the following results:
You can learn here how long it takes to pay off your credit, what is the total payment amount and the charged interest with the fixed payment and the additional payment.
For the representation of results, schedules, and balances, you may choose from the following.
You can follow the unpaid principal balance with interest in both scenarios.
• Table of payment schedule
The payment schedule shows you the closing balances for each month, the monthly payments, and the principal and interest allocation.
• Pie chart of final balance breakdown
You can see the proportion of the charged interest as part of your total paid balance.
In the advanced mode you can set the parameters for the minimum monthly payment requirements and its calculation method applied by the credit card issuer.
The credit card interest calculator's results should be considered a model for financial approximation. All payment figures, balances, and interest figures are estimates based on the data you
provided in the specifications that are, despite our best effort, not exhaustive.
Thus, and mostly because of the simplifications mentioned in the text, the calculator is created for instructional purposes only. Yet, if you experience a relevant drawback or encounter any
inaccuracy, we are always pleased to receive useful feedback and advice.
Monthly additional payment
With a monthly payment of $100.00, you would pay off the credit in 2 year(s), on Oct. 12, 2026.
The total payment amount is $2,395.65, with an interest payment of $395.65. | {"url":"https://www.omnicalculator.com/finance/credit-card-interest","timestamp":"2024-11-12T16:33:20Z","content_type":"text/html","content_length":"309450","record_id":"<urn:uuid:1d96cc8d-a255-411b-a2e3-9c12c639c64a>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00074.warc.gz"} |
he Stata Blog
\(\newcommand{\xb}{{\bf x}}
\newcommand{\betab}{\boldsymbol{\beta}}\)Before you use or distribute your estimation command, you should verify that it produces correct results and write a do-file that certifies that it does so. I
discuss the processes of verifying and certifying an estimation command, and I present some techniques for writing a do-file that certifies mypoisson5, which I discussed in previous posts.
This is the twenty-fifth post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted
entries for a map to all the posts in this series.
Verification versus certification
Verification is the process of establishing Read more…
Fitting distributions using bayesmh
30 March 2016 No comments
This post was written jointly with Yulia Marchenko, Executive Director of Statistics, StataCorp.
As of update 03 Mar 2016, bayesmh provides a more convenient way of fitting distributions to the outcome variable. By design, bayesmh is a regression command, which models the mean of the outcome
distribution as a function of predictors. There are cases when we do not have any predictors and want to model the outcome distribution directly. For example, we may want to fit a Poisson
distribution or a binomial distribution to our outcome. This can now be done by specifying one of the four new distributions supported by bayesmh in the likelihood() option: dexponential(),
dbernoulli(), dbinomial(), or dpoisson(). Previously, the suboption noglmtransform of bayesmh‘s option likelihood() was used to fit the exponential, binomial, and Poisson distributions to the outcome
variable. This suboption continues to work but is now undocumented.
For examples, see Beta-binomial model, Bayesian analysis of change-point problem, and Item response theory under Remarks and examples in [BAYES] bayesmh.
We have also updated our earlier “Bayesian binary item response theory models using bayesmh” blog entry to use the new dbernoulli() specification when fitting 3PL, 4PL, and 5PL IRT models.
Categories: Statistics Bayesian, bayesmh, biostatistics, distribution fitting
Programming an estimation command in Stata: Making predict work
17 March 2016 No comments
I make predict work after mypoisson5 by writing an ado-command that computes the predictions and by having mypoisson5 store the name of this new ado-command in e(predict). The ado-command that
computes predictions using the parameter estimates computed by ado-command mytest should be named mytest_p, by convention. In the next section, I discuss mypoisson5_p, which computes predictions
after mypoisson5. In section Storing the name of the prediction command in e(predict), I show that storing the name mypoisson5_p in e(predict) requires only a one-line change to mypoisson4.ado, which
I discussed in Programming an estimation command in Stata: Adding analytical derivatives to a poisson command using Mata.
This is the twenty-fourth post in the Read more…
Categories: Programming #StataProgramming, postestimation, programming
How to generate random numbers in Stata
10 March 2016 3 comments
I describe how to generate random numbers and discuss some features added in Stata 14. In particular, Stata 14 includes a new default random-number generator (RNG) called the Mersenne Twister
(Matsumoto and Nishimura 1998), a new function that generates random integers, the ability to generate random numbers from an interval, and several new functions that generate random variates from
nonuniform distributions.
Random numbers from the uniform distribution
In the example below, we use runiform() to create Read more…
Categories: Statistics Mersenne Twister, random numbers, runiform()
Programming an estimation command in Stata: Adding analytical derivatives to a poisson command using Mata
2 March 2016 No comments
\(\newcommand{\xb}{{\bf x}}
\newcommand{\betab}{\boldsymbol{\beta}}\)Using analytically computed derivatives can greatly reduce the time required to solve a nonlinear estimation problem. I show how to use analytically computed
derivatives with optimize(), and I discuss mypoisson4.ado, which uses these analytically computed derivatives. Only a few lines of mypoisson4.ado differ from the code for mypoisson3.ado, which I
discussed in Programming an estimation command in Stata: Allowing for robust or cluster–robust standard errors in a poisson command using Mata.
This is the twenty-third post in the series Programming an estimation command in Stata. I recommend that you start at the beginning. See Programming an estimation command in Stata: A map to posted
entries for a map to all the posts in this series.
Analytically computed derivatives for Poisson
The contribution of the i(th) observation to the log-likelihood function for the Poisson maximum-likelihood estimator is Read more…
Categories: Programming #StataProgramming, econometrics, Mata, nonlinear model, programming, statistics | {"url":"https://blog.stata.com/2016/03/","timestamp":"2024-11-03T22:32:46Z","content_type":"application/xhtml+xml","content_length":"71598","record_id":"<urn:uuid:3c24f696-1f09-4192-9575-c545794055d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00087.warc.gz"} |
Students calculate the distance travelled by the tips of the hands on a clock, and calculate the speed of one of the tips.
Students use a diagram of, and information about oranges packed in a cylinder to calculate height, area and volume.
Students calculate circumference, circular and square area of a road sign.
Students are required to identify right angles in a series of diagrams, using their knowledge of angle properties relating to circles and tangents.
Students use the angle between a tangent and a radius property and the base angles in an isosceles triangle property to find an unknown angle and to explain why line segments are of equal size.
Students use the angle-in-a-semicircle property, and the sum of the angles in a triangle to find unknown angles.
Students use their knowledge of the angle between a tangent and radius property and the sum of the angles in a quadrilateral to work out unknown angles in a diagram and explain their workings.
In this practical task, students work in small groups to present a dance based on geometric concepts.
Students identify the parts of a circle that lines on familiar signs and objects make.
Students use the given the radius to draw each circle on a bulls-eye target.
Students write short paragraphs using geometrical terms to describe two pictures. The terms equilateral, scalene, and isosceles are to be used to describe a castle. The words circumference, diameter,
and radius are to be used to describe a bicycle.
Students calculate the size of marked angles using their knowledge of angle properties: the angle between a tangent and a radius, the sum of angles in a triangle and the sum of angles in a
Students calculate the lengths of circles and straight lines on an oval athletics track from its given radius and total length, and show their working.
This task is about finding the circumference of a circle using two different radius measurements.
Students answer questions about a ball travelling a number of full circles in a game of swing ball.
Students use known diameters of tins to work out the floor area of a carton and a given number of tins.
Students calculate area of colours and the perimeter of a flag. | {"url":"https://arbs.nzcer.org.nz/category/keywords/circles","timestamp":"2024-11-03T06:09:02Z","content_type":"text/html","content_length":"79229","record_id":"<urn:uuid:adbe488b-7dcd-4451-98b9-8203f578d35b>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00548.warc.gz"} |
PIPING DESIGN cnd ENGINEERING SIXTH EDITION (Revised 1981) ITT (2024)
Related papers
Basic calculations 7 Basic calculations 7.1 Factors influencing pipe system design
Note: The listed mathematical operations and relations have been simplified as much as possible. Plastic-specific parameters and generally valid factors are partly already integrated into calculation
formulae. We have abstained from detailing the derivation or reproduction of single values to abbreviate this section. 7.2 Calculating pipe parameters 7.2.1 Explanation The calculation of
thermoplastic pipe systems is especially important for the engineer undertaking a project. This chapter presents the basic principles required for designing plastic pipe systems. However, the
practitioner (user) should also be able to acquire the necessary data and standard variables for internally pressure-loaded pipe in a relatively simple manner without spending a lot of time. The
mathematical operations are supported by diagrams in the appendices from which most values and data can be read off. Thermoplastic pipe calculations basically occur on the basis of long-term values.
The reference stress (V ref) and mechanical (creep) strength (i.e. creep modulus (E cR)) of a pipe system in relation to temperature can be derived from the creep diagram in appendix A1 and the creep
modulus curve diagram in appendix A2. The creep rupture curves are based on internal pressure tests on pipe samples filled with water and represent minimum values. If pipe systems are not intended
for water but for other flow media, their effects on creep strength properties must be given special consideration.
Piping Stress Handbook Second Edition
Piping stress analysis is a discipline which is highly interrelated with piping layout (Chap. B3) and support design (Chap. B5). The layout of the piping system should be performed with the
requirements of piping stress and pipe supports in mind (i.e., sufficient flexibility for thermal expansion; proper pipe routing so that simple and economical pipe supports can be constructed; and
piping materials and section properties commensurate with the intended service, temperatures, pressures, and anticipated loadings). If necessary, layout solutions should be iterated until a
satisfactory balance between stresses and layout efficiency is achieved. Once the piping layout is finalized, the piping support system must be determined. Possible support locations and types must
be iterated until all stress requirements are satisfied and other piping allowables (e.g., nozzle loads, valve accelerations, and piping movements) are met. The piping supports are then designed
(Chap. B5) based on the selected locations and types and the applied loads. This chapter discusses several aspects of piping stress analysis. The discussion is heavily weighted to the stress analysis
of piping systems in nuclear power plants, since this type of piping has the most stringent requirements. However, the discussion is also applicable to the piping systems in ships, aircraft,
commercial buildings, equipment packages, refrigeration systems, fire protection piping, petroleum
OVERVIEW OF INDUSTRIAL PIPING STRUCTURE DESIGN20200217 86677 11y7fkk
Honors List xi Preface xvii How to Use This Handbook xix Part A: Piping Fundamentals Chapter A1. Introduction to Piping Mohinder L. Nayyar
Chapter A6. Fabrication and Installation of Piping Edward F. Gerwin A.261 Chapter A7. Bolted Joints Gordon Britton A.331
Stress Analysis Of Hot Wall Flue Gas Piping At FCCU Plant, Reliance Chapter 1: INTRODUCTION TO STRESS ANALYSIS
The best piping configuration is the least expensive over a long term basis. This requires the consideration of installation cost, pressure loss effect on production, stress level concern, fatigue
failure, support and anchor effects, stability, easy maintenance, parallel expansion capacity and others. The expansion loops most commonly used in crosscountry pipelines are L bends, Z bends,
conventional 90° elbow and V bends. The principal design codes used for piping design are the ANSI/ASME B31.1 (Code for Power Piping) and ANSI/ASME B31.3 (code for process piping), ASTM A53 B, ASTM
A106 B and API 5L carbon steel pipes are the ones used for geothermal fields. The allowable stress is S E =88 MPa for ERW pipe and S E =103 MPa for seamless pipe, S A =155 MPa for operation load, kS
h =124 MPa for earthquake load and 258 MPa for combined sustained loads and stress range. Pipe pressure design for the separation station and steam lines is 1.5 MPa, and for brine line ranges from
1.5 to 4 MPa. Pipe diameters are generally 250 to 1219 mm nominal pipe size. The two-phase line can be in the range 50 to 150 m, the steam lines from 2000 to 3000 m and for the brine up to 6000 m
long. The total cost of pipe installation can be US$ 600-1,200 per meter of pipe. Pipe configuration needs to be cost conscious; the design can be under 10% of excess pipe to get from point to point
straight line distance, which is excellent from a piping material and pressure loss point of view.
Significance of and suggested limits for the stress in pipe lines Rossheim Markl
T he Significance of, an d Suggested L im its for, the S tress in P ipe Lines D u e to th e C o m bined Effects of P re ssu re a n d E x p an sio n By D. B. ROSSHEIM1 and A. R. C. MARKL,2 NEW YORK,
N. Y. T h is p a p er h a s b e e n p r e p a r ed a s a v e h ic le fo r d is c u s s io n o f t h e fo llo w in g b a s ic p r o b le m s , o n w h ic h a g r e e m e n t m u s t be r e a c h e d t
o e s t a b lis h a s a tis f a c to r y w o r k in g-s tr e s s b a s is for p ip e l i n e s : 1 P ro p er a llo w a b le s t r e s s e s fo r c o m b in e d p r e ssu r e a n d e x p a n s io n e
ffe c ts. 2 I n flu e n c e o f a n d l i m i t s for lo c a liz e d s tr e s s e s u n d e r s t a t i c a n d r e p e a te d lo a d in g. 3 C a p a c ity o f b o lte d j o i n t s t o w it h s t a
n d e x p a n s io n e ffe c ts w it h o u t le a k a g e o r d a m a g e t o fla n g e s, b o lt s , or g a s k e ts. 4 E ffect o f p r e sp r in g in g , s e lf-s p r in g in g , c r e e p , a n d
y ie ld in g o n o p e r a tin g a n d o f f-s t r e a m s t r e s s e s. A n a t t e m p t h a s b e e n m a d e t o p o in t o u t v a r io u s a s p e c t s o f e a c h is s u e , r a th e r t h a
n l im i t t h e p r e s e n t a t io n t o t h e p e r so n a l v iew s o f t h e a u t h o r s. D URING the last decade, economic considerations and new processes in the power, oil-refinery, and
chemical industries have produced a trend toward large-scale units and high operating temperatures and pressures. With the attendant in crease in line sizes and wall thicknesses, the subdivision of
pipe lines into convenient runs with expansion bends soon proved entirely inadequate; instead, today most piping for severe service is carefully analyzed for forces and stresses, full advan tage
being taken of the inherent flexibility by minimizing the num ber of anchors, guides, or other restraints. A natural consequence of this improved accuracy in evaluating thermal effects is the
necessity for a review of stress limitations as they apply to ex pansion stresses alone and in combination with internal pressure. This problem has received the active consideration of Subgroup No.
3 on Expansion and Flexibility, Subcommittee No. 8 on Fabrication Details of the Code for Pressure Piping (ASA B31) who, functioning with the Applied Mechanics and Power Divi sions of the A.S.M.E.,
have sponsored this symposium. This paper has been prepared as a vehicle for discussion of the following basic problems, on which agreement must be reached to establish a statisfactory working-stress
basis for pipe lines. An attempt has been made to point out various aspects of each issue, rather than limit the presentation to the personal views of the authors. 1 Proper allowable stresses for
combined pressure and expan sion effects. 2 Influence of and limits for localized stresses under static and repeated loading, such as in bends and corrugated pipe. 3 Capacity of bolted joints to
withstand expansion effects without leakage or damage to flanges, bolts, or gaskets. 1 M echanical Engineer, M. W. K ellogg Co. M em. A .S .M .E. 2 A ssistant M echanical Engineer, M. W. K ellogg Co.
Contributed b y the Pow er D ivision and presented at the Annual M eeting, Philadelphia, P a., Decem ber 4-8 , 1939, of T h e A m e r i c a n S o c i e t y o f M e c h a n i c a l E n g i n e e r s.
N o t e : S tatem en ts and opinions advanced in papers are to be under stood as individual expressions of their authors, and n ot those of the Society. 4 Effect of prespringing, self-springing,
creep, and yielding on operating and off-stream stresses R esume of Simple Stresses I nvolved A pipe line is essentially a pressure vessel, the internal pressure causing a radial stress at the inner
face which is converted into circumferential stress on the way through the wall, leaving zero radial stress at the outer surface; at any point the sum of the radial and hoop stresses varies inversely
as the radius, while their difference is the same at all points. These are the laws of Lam<5 which involve the assumption of a uniform or zero longitudinal stress. Following this reasoning, in a
closed cylinder the longi tudinal stress equals the product of the pressure and the internal area divided by the cross-sectional area of the wall, and has the same value throughout the thickness.
While most runs of piping are not technically closed cylinders, the same longitudinal stresses occur due to pressure effects on the projections of elbows, there being a complete absence of
longitudinal pressure stress only in straight runs between vessels of infinite rigidity and such including frictionless expansion joints. Under temperature changes with free expansion no stresses are
introduced. However, ordinarily, expansion is restrained by the equipment to which the pipe line is attached, as well as anchors, guides, solid hangers, or supports. In calculating a line for
expansion, the ends are commonly assumed completely fixed; this connotes three forces and an equal number of moments in the case of a problem in space, resulting in longitudinal bending stresses in
two mutually perpendicular planes and torsional stress about the pipe axis, as well as two shears and a normal stress. With the assumption of hinged ends, the end moments reduce to zero, a condition
which would be fully realized only in a frictionless ball-and-socket joint. Ac tually, an intermediate condition of restraint will prevail in practice due to sympathetic deflections, rotations, or
distortions of the equipment to which the pipe is attached. On the other hand, cases will arise where external movements of the ends tend to increase rather than decrease their degree of fixity; in
addition, guides, solid or spring supports, and hangers often inhibit free deformation of the line and thus add to the stresses. To avoid additional complexity, the weight of the piping is commonly
neglected in flexibility calculations. In heavy pipe lines spring hangers are often used to balance the dead-load effects when the line is at working temperature, and solid hangers can be similarly
employed in neutral locations. Properly de signed supports appear to offer the most suitable means for handling this effect, which can then be disregarded in the stress analysis of most lines. Where
piping is not insulated, the heat loss from the exposed surface produces a temperature drop through the metal thick ness causing longitudinal and circumferential stresses of equal magnitude which
may be evaluated by the formulas of Lorenz (l).3 Since the outside surface is relatively colder, tensile 3 N um bers in parentheses refer to the Bibliography at the end of the paper.
Design of Pressure Pipes
The design methods for buried pressure pipe installations are somewhat similar to the design methods for gravity pipe installations which were discussed in Chap. 3. There are two major differences:
1. Design for internal pressure must be included. 2. Pressure pipes are normally buried with less soil cover so the soil loads are usually less. Included in this chapter are specific design
techniques for various pressure piping products. Methods for determining internal loads, external loads, and combined loads are given along with design bases. Pipe Wall Stresses and Strains The
stresses and resulting strains arise from various loadings. For buried pipes under pressure, these loadings are usually placed in two broad categories: internal pressure and external loads. The
internal pressure is made up of the hydrostatic pressure and the surge pressure. The external loads are usually considered to be those caused by external soil pressure and/or surface (live) loads.
Loads due to differential settlement, longitudinal bending, and shear loadings are also considered to be external loadings. Temperature-induced stresses may be considered to be caused by either
internal or external effects. Hydrostatic pressure Lamé's solution for stresses in a thick-walled circular cylinder is well known. For a circular cylinder loaded with internal pressure only, those
stresses are as follows:
PROCESS PIPING DESIGN HANDBOOK - VOLUME 2 [Advanced Piping Design] | {"url":"https://casasrsocorro.com/article/piping-design-cnd-engineering-sixth-edition-revised-1981-itt","timestamp":"2024-11-10T05:06:41Z","content_type":"text/html","content_length":"73335","record_id":"<urn:uuid:7196082a-8dd6-4066-801f-8adc7d4e6250>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00752.warc.gz"} |
How Does Simple Interest Work?How Does Simple Interest Work?
January 5, 2004
"What are the benefits/drawbacks of a simple interest loan versus a traditional mortgage? Which would you take if offered the choice?"
I would select a traditional mortgage. If two loans are exactly the same but one is simple interest, you will pay more interest on it unless you systematically make your monthly payment before the
due date.
The major difference between a standard mortgage and a simple interest mortgage is that interest is calculated monthly on the first and daily on the second.
Consider a 30-year loan for $100,000 with a rate of 6%. The monthly payment would be $599.56 for both the standard and simple interest mortgages. The interest due is calculated differently, however.
On the standard mortgage, the 6% is divided by 12, converting it to a monthly rate of .5%. The monthly rate is multiplied by the loan balance at the end of the preceding month to obtain the interest
due for the month. In the first month, it is $500.
On the simple interest version, the annual rate of 6% is divided by 365, converting it to a daily rate of .016438%. The daily rate is multiplied by the loan balance to obtain the interest due for the
day. The first day and each day thereafter until the first payment is made, it is $16.44.
The $16.44 is recorded in a special accrual account, which increases by that amount every day. No interest accrues on this account. When a payment is received, it is applied first to the accrual
account, and what is left over is used to reduce the balance. When the balance declines, a new and smaller daily interest charge is calculated.
How does this work out for the borrower? We know that a standard 30-year mortgage pays off in 30 years. Beginning January 1, 2004, this amounts to 10,958 days. On a loan of $100,000 and an interest
rate of 6%, total interest payments amount to $115,832.
On the simple interest version of the same mortgage, assuming you pay on the first day of every month, you pay off in 10,990 days, or 41 days later than with the standard mortgage. Total interest
payments are $116,167 or $335 more.
These are small differences, due largely to leap years. Over the 30 years beginning 2004, there are 8 years with 366 days, and the lender collects interest for those days. Leap years do not affect
total interest payments on a standard mortgage.
The disadvantage of a simple interest mortgage rises with the interest rate. At 12%, and continuing to assume payment on the first day of every month, it pays off in 11,049 days or 91 days later than
the standard mortgage. Total interest is $3082 higher.
But the borrowers who really get clobbered by the simple interest mortgage are those who pay late. The standard mortgage has a grace period within which borrowers can pay without penalty. On a simple
interest mortgage, in contrast, borrowers pay interest for every day they are late.
Suppose the borrower pays on the 10^th day of every month, for example. With a standard mortgage, he gets a free ride because of the grace period. With a simple interest mortgage at 6%, he pays off
101 days later than the standard mortgage and pays $1328 more interest. At 12%, he pays off 466 days later and pays $15,137 more interest.
Penalties for payment after the grace period work the same way on both types of mortgage. For this reason, I have not included penalties in the calculations.
Borrowers making extra payments also do better with a standard mortgage. Most lenders will credit extra payments received within the first 20-25 days of the month against the balance at the end of
the preceding month. A borrower who pays $1,000 extra on day 20, for example, will save the interest on that $1,000 for 20 days. With a simple interest mortgage, in contrast, interest accrues for
those 20 days.
The only transaction that works out better for the borrower with a simple interest mortgage is monthly payments made early. If every month you pay 10 days before the payment is due, for example, you
pay off 40 days sooner than the standard mortgage at 6%, and 254 days earlier at 12%. There is no benefit to early payment on a standard mortgage, since it is credited on the due date, just like a
payment that is received 10 days late.
Bottom line: other things the same, take the standard mortgage. But if you are stuck with a simple interest mortgage, make it a habit to pay early; it will pay big dividends.
Days to Payoff and Total Interest Payments on a Standard Mortgage and Simple Interest Mortgage of $100,000 for 30Years Beginning January 1, 2004
│ │Standard Mortgage, Payment Within Grace Period │ Simple Interest Mortgage │
│ │ ├─────────────────────┬─────────────────────────────┬────────────────────┤
│ │ │Payment 10 Days Early│Payment on First Day of Month│Payment 10 Days Late│
│Days to Payoff│ │ │ │ │
│ 6% Loan│ 10,958 │ 10,918 │ 10,990 │ 11,059 │
│ 12% Loan│ 10,958 │ 10,704 │ 11,049 │ 11,424 │
│Total Interest│ │ │ │ │
│ 6% Loan│ $115,832 │ $115,180 │ $116,167 │ $117,160 │
│ 12% Loan│ $270,277 │ $261,889 │ $273,359 │ $285,414 │ | {"url":"http://www.explainingmortgages.com/mtgpro/a%20-%20amortization/how_does_simple_interest_work.htm","timestamp":"2024-11-09T23:31:21Z","content_type":"text/html","content_length":"36466","record_id":"<urn:uuid:0851db89-cd30-458c-9311-2973bbae61c6>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00069.warc.gz"} |
Math Candies
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Description: 'Math Candies' is a math puzzle game. In this game you need to find the price of some candies by solving given equations. Operations in the problems consist of addition, subtraction,
multiplication and division. Once you find the values of candies, use these values to find the answer of a simple Math question. | {"url":"http://www.i6.com/game/40854/math-candies","timestamp":"2024-11-05T16:15:35Z","content_type":"text/html","content_length":"17406","record_id":"<urn:uuid:fdaac3c2-5445-4601-ae46-13337ef22c8e>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00872.warc.gz"} |
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Langlands Program: Number Theory and Representation Theory
Schedule for: 22w5178 - Langlands Program: Number Theory and Representation Theory
Beginning on Sunday, November 27 and ending Friday December 2, 2022
All times in Oaxaca, Mexico time, CST (UTC-6).
Sunday, November 27
14:00 - 23:59 Check-in begins (Front desk at your assigned hotel)
19:30 - 22:00 Dinner (Restaurant Hotel Hacienda Los Laureles)
20:30 - 21:30 Informal gathering (Hotel Hacienda Los Laureles)
Monday, November 28
- Breakfast (Restaurant Hotel Hacienda Los Laureles)
- Introduction and Welcome (Conference Room San Felipe)
Gonzalo Tornaría: The basis problem for paramodular forms ↓
09:15 In this talk we present a partial solution to the basis problem for paramodular forms. Our method involves proving a precise relation between paramodular forms and algebraic modular forms
- arising from certain positive-definite quinary lattices. The latter can be computed explicitly by the lattice-neighbour method. As an application, we can prove examples of congruences of Hecke
10:15 eigenvalues connecting Siegel modular forms of degree two and one. These include some of the type conjectured by Harder at level one and supported by compu- tations of Fretwell at higher
levels, and a subtly different congruence discovered experimentally by Buzzard and Golyshev. This is joint work with Dummigan, Pacetti and Rama.
(Conference Room San Felipe)
- Shaun Stevens: Types and local Langlands correspondence I (Conference Room San Felipe)
- Coffee Break (Conference Room San Felipe)
- Erez Lapid: An elementary introduction to automorphic forms and L-functions I (Zoom)
- Group Photo (Hotel Hacienda Los Laureles)
- Lunch (Restaurant Hotel Hacienda Los Laureles)
Dan Ciubotaru: Wavefront sets and unipotent representations of p-adic groups ↓
An important invariant for admissible representations of reductive p-adic groups is the wavefront set, the collection of the maximal nilpotent orbits in the support of the orbital integrals
15:00 that occur in the Harish-Chandra-Howe local character expansion. We compute the geometric and Okada’s canonical unramified wavefront sets for representations in Lusztig’s category of unipotent
- reduction for a split group in terms of the Kazhdan-Lusztig parameters. I will emphasise two applications of this calculation: 1) the geometric wave front set of a unipotent supercuspidal
16:00 representation determines uniquely the nilpotent part of the Langlands parameter; 2) the anti-tempered unipotent Arthur packets are uniquely characterised by the unramified wave front set of
their constituents. The talk is based on joint work with Lucas Mason-Brown and Emile Okada.
- Coffee Break (Conference Room San Felipe)
Clifton Cunningham: Vogan's conjecture on Arthur packets for p-adic groups ↓
16:30 Thirty years ago, David Vogan conjectured a purely local description of Arthur packets for p-adic groups, closely related to a more developed theory for Real groups by Adams, Barbasch and
- Vogan. In joint work with Mishty Ray, we have now finished the proof of this conjecture for general linear groups, building on previous work with other members of the ``Voganish Project''. In
17:30 this talk I'll present the proof for $\mathrm{GL}_n$ and discuss how the strategy seems likely to extend to classical groups, highlighting the work that remains to be done.
(Conference Room San Felipe)
- Dinner (Restaurant Hotel Hacienda Los Laureles)
Tuesday, November 29
- Breakfast (Restaurant Hotel Hacienda Los Laureles)
- Shaun Stevens: Types and local Langlands correspondence II (Conference Room San Felipe)
- Erez Lapid: An elementary introduction to automorphic forms and L-functions I I (Zoom)
- Coffee Break (Conference Room San Felipe)
Cong Xue: Cohomology of stacks of shtukas I ↓
12:00 Cohomology sheaves and cohomology groups of stacks of shtukas are used in the Langlands program for function fields. In the first lecture, we will recall the definition of stacks of shtukas and
- their cohomology, the action of the partial Frobenius morphisms and the action of the Hecke algebra. In the second lecture, we will explain the Eichler-Shimura relations, the finiteness
13:00 property of the cohomology groups, Drinfeld's lemmas and the action of the Weil group (of the function field) on the cohomology groups. In the third lecture, we will explain how this action and
the "Zorro lemma" imply the smoothness of the cohomology sheaves.
- Lunch (Restaurant Hotel Hacienda Los Laureles)
Thomas Haines: On the Hasse-Weil zeta functions for Kottwitz simple Shimura varieties ↓
Kottwitz introduced certain compact ``fake'' unitary group Shimura varieties and determined their local Hasse-Weil zeta functions at primes of good reduction. For primes where the level is
15:00 arbitrarily deep, the local Hasse-Weil zeta functions were further studied in the case of signature $(1, n − 1)$ (by Xu Shen), and in the case of arbitrary signature but under the assumption
- that the group at $p$ is a product of Weil restrictions of general linear groups, by Scholze and Shin. In this talk, I will explain joint work with Jingren Chi in which we generalize the work
16:00 of Scholze-Shin by allowing the group at $p$ to be any inner form of a product of Weil restrictions of general linear groups. New phenomena arise when the group at p is not quasi-split, for
example a crucial vanishing property of twisted orbital integrals of the test functions at $p$.
(Conference Room San Felipe)
- Coffee Break (Conference Room San Felipe)
Ramla Abdellatif: Studying p-modular representations of p-adic groups in the setting of Langlands programme ↓
Let $p$ be a prime integer, $F$ be a non-Archimedean local field of residual characteristic $p$ and $G = \mathcal{G}(F)$ be the group of $F$-rational points of a connected reductive group
defined over $F$. During the last decade, the study of $p$-modular smooth representations of $G$, i.e. of smooth representations of $G$ with coefficients in a field of characteristic $p$, has
16:30 been intensively developed for arithmetic reasons (e.g. related to congruences between automorphic forms) but remains really mysterious, even for nice groups as $\mathrm{GL}_{2}(F)$ or $\mathrm
- {SL}_{2}(F)$, or when one focuses on (admissible) irreducible smooth ones. This talk aims at presenting what is known so far in this setting, then at discussing some joint work with Hauseux,
17:30 where we prove that understanding $p$-modular representations of $G$ when $\mathcal{G}$ is of $F$-semisimple rank $1$ amounts to understand their restriction to a minimal parabolic subgroup.
Doing so, we extend some previous work of Pa${\check{\text{s}}}$k$\overline{\text{u}}$nas (2007) for $\mathrm{GL}_{2}(F)$, but with different methods that give another viewpoint on Pa${\check{\
text{s}}}$k$\overline{\text{u}}$nas proofs and suggests that this kind of statement could extend to higher rank groups.
(Conference Room San Felipe)
- Dinner (Restaurant Hotel Hacienda Los Laureles)
Wednesday, November 30
- Breakfast (Restaurant Hotel Hacienda Los Laureles)
- Shaun Stevens: Types and local Langlands correspondence III (Conference Room San Felipe)
Guy Henniart: Simple cuspidals and the Langlands correspondence ↓
Gross and Reeder showed that for any split reductive group G over a locally compact non-Archimedian field F, one can construct an easy family of cuspidal representations of G(F), called
"simple" cuspidals. For G=GL(n), they go back to work of Carayol in the 1970's. In terms of ramification, they come right after the "level 0" cuspidals, which can be obtained from
09:30 representations of G(k) where k is the residue field of F. But the construction of "simple" cuspidals only essentially involves a non-trivial character of k. I shall decribe the "simple"
- cuspidals for G=GL(n) or Sp(2n). On the other hand, the Langlands correspondence attaches to a "simple" cuspidal representation π of GL(n) an irreducible degree n representation σ of the Weil
10:30 group W_F of F, and σ has Swan exponent 1. Describing σ from π is not easy: for example, when n is a power of the residue characteristic p of F, σ is primitive. I shall give part of the recipe
(work of Bushnell and I in 2013, Imai and Tsushima in 2015). I shall also examine the more recent case of G=Sp(2n). When F has characteristic 0, J. Arthur attaches to a "simple" cuspidal π of
Sp(2n,F) an orthogonal representation σ of W_F of dimension 2n+1. I shall give the description of σ from π, due to Masao Oi (2018) when p is odd, and to Oi and I (2022) when p=2.
- Coffee Break (Conference Room San Felipe)
- Cong Xue: Cohomology of stacks of shtukas II (Zoom)
- Lunch (Restaurant Hotel Hacienda Los Laureles)
- Free Afternoon (Monte Albán Tour) (Oaxaca)
- Dinner (Restaurant Hotel Hacienda Los Laureles)
Thursday, December 1
- Breakfast (Restaurant Hotel Hacienda Los Laureles)
Vincent Lafforgue: Spectral decomposition ↓
09:15 In this talk I will explain the spectral decomposition of the space of automorphic forms (and more generally of cohomology groups of stacks of shtukas), indexed by Langlands paremeters, a
- refined decomposition obtained with Xinwen Zhu (after an idea of Drinfeld) and I will mention the link with the l-adic geometric Langlands program by trace of Frobenius constructions (due to
10:15 Arinkin, Gaitsgory, Kazhdan, Raskin, Rozenblyum, Varshavsky).
Daniel Barrera: Periods integrals and Eigenvarieties ↓
10:30 Let H and G be reductive groups over Q such that H is contained in G. In this talk I will explain the connection between the non-vanishing of periods integrals for (G, H) and the local geometry
- of Eigenvarieties of G. Furthermore, we will use this connection in the construction of p-adic L-functions. The talk will be based on the following two situations: 1) H= GL(n)xGL(n) and G= GL
11:30 (2n) (joint work with M. Dimitrov, A. Graham, A. Jorza and C. Williams). 2) H= GL(n) and G= GL(n) over an imaginary quadratic field (work in progress with P-H. Lee and C. Williams)
- Coffee Break (Conference Room San Felipe)
- Cong Xue: Cohomology of stacks of shtukas III (Zoom)
- Lunch (Restaurant Hotel Hacienda Los Laureles)
Solomon Friedberg: Towards a New Shimura Correspondence ↓
The classical Shimura correspondence lifts automorphic representations on the double cover of $\mathrm{SL}_2$ (corresponding to classical half-integral weight forms) to automorphic
15:00 representations on $\mathrm{PGL}_2$. Though efforts have been made for many years to generalize this map to higher rank groups and higher degree covers, our knowledge is limited. In this talk I
- present joint work with Omer Offen that points to a new Shimura lift for automorphic representations on the triple cover of $\mathrm{SL}_3$ -- we establish the Fundamental Lemma for a relative
16:00 trace formula. Further work is on-going. We expect that this project will both prove the existence of a global lift of automorphic representations and characterize the image of the lift by
means of a period involving a theta function on $\mathrm{SO}_8$, thereby confirming a 2001 conjecture of Bump, Friedberg and Ginzburg.
(Conference Room San Felipe)
- Coffee Break (Conference Room San Felipe)
Michael Harris: Around local and global Langlands correspondences for function fields ↓
This is a report on several joint projects analyzing the parametrization of V. Lafforgue and C. Xue of automorphic representations of reductive groups over function fields, and the
16:30 corresponding local parametrizations, due to Genestier-Lafforgue and Fargues-Scholze. With Gan and Sawin we show that pure supercuspidal representations usually have ramified Galois parameters,
- and attach Weil-Deligne parameters unambiguously to all discrete series representations. With Böckle, Feng, Khare, and Thorne, we construct cyclic base change of a cuspidal automorphic
17:30 representation of any sufficiently large prime degree, and apply this to base change for local representations. Finally, with Ciubotaru, we show that the unramified components of any cuspidal
automorphic representation of a simple group (with a few surprising exceptions) are tempered and generic, provided at least one component is tempered and another is unramified generic.
- Dinner (Restaurant Hotel Hacienda Los Laureles)
Friday, December 2
- Breakfast (Restaurant Hotel Hacienda Los Laureles)
Corinne Blondel: L-packets via types and covers ↓
We outline a general strategy based on types and covers to describe the $L$-packets of representations of classical groups over a non-archimedean local field of odd residual characteristic :
09:15 L-packets are parametrized by Jordan sets, and the Jordan set of a given cuspidal representation can be computed using types and covers that transform the problem into a computation of defining
- relations for Hecke algebras. We explain the results obtained in the symplectic case in a joint work with Guy Henniart and Shaun Stevens : the inertial Jordan sets of cuspidal representations
10:15 are fully determined. We explain the ambiguity that may remain and a possible way to solve it, with examples from joint works with Laure Blasco in $\mathrm{Sp}(4)$ and Geo Kam-Fai Tam in
ramified unitary groups and work in progress in $\mathrm{Sp}(4)$ with Henniart and Stevens.
(Conference Room San Felipe)
Jeanine Van Order: Dihedral families of $\mathrm{GL}_n$ automorphic $L$-functions via toric periods, and equidistribution ↓
Let $K$ be a CM field with maximal totally real subfield $F=K^+$, let $\pi$ be a conjugate self-dual cuspidal automorphic representation of ${\rm GL}_2(\mathbb{A}_K)$, and let $P$ be a fixed
prime of $F$. As $\chi$ ranges over the set $X_K(P)$ of primitive ring class characters of $K$ of $P$-power conductor, the root number $\epsilon(1/2, \pi \otimes \chi)$ of the twisted standard
10:30 L-function $L(s, \pi \otimes \chi)$ of $\pi$ is generically independent of the choice of $\chi$. That is, for one of $k \in {0, 1}$, we know that $\epsilon(1/2, \pi \otimes \chi) = (-1)^k$ for
- all but finitely many $\chi$ in $X_K(P)$. In the case of $k=0$, we study the central values $L(1/2, \pi \otimes \chi)$ as $\chi$ varies in $X_K(P)$ using “toric period” integral presentations,
11:30 namely (1) Eulerian integral presentations and (2) presentations implied by the Ichino-Ikeda Gan-Gross-Prasad conjecture for unitary groups. In particular, we present a generalization of
“Mazur’s conjecture”, and explain how this can be deduced from a certain equidistribution criterion. We then describe an approach to reducing this criterion to the celebrated theorems of Ratner
and Margulis-Tomanov on p-adic unipotent flows (which would generalize the well-known theorems of Vatsal and Cornut-Vatsal for ${\rm GL}_2$).
- Group Photo Online Participants (Zoom)
- Coffee Break (Conference Room San Felipe)
Adrián Zenteno: Using Langlands program to solve certain cases of the inverse Galois problem ↓
12:00 In the last years, the study of the images of the Galois representations associated to regular algebraic cuspidal automorphic representations of $\mathrm{GL}_n(\mathbb{A}_{\mathbb{Q}})$, via
- global Langlands correspondence, has been an effective strategy to address the inverse Galois problem for finite groups of Lie type. In this talk we will explain how, by combining this strategy
13:00 with Langlands functoriality and globalization of supercuspidal representations, we can construct residual Galois representations with controlled image and obtain new families of finite groups
of type $B_m$, $C_m$ and $D_m$ arising as Galois groups over $\mathbb{Q}$.
(Conference Room San Felipe)
- Lunch (Restaurant Hotel Hacienda Los Laureles) | {"url":"https://www.birs.ca/events/2022/5-day-workshops/22w5178/schedule","timestamp":"2024-11-04T15:33:52Z","content_type":"application/xhtml+xml","content_length":"40233","record_id":"<urn:uuid:295024f0-2c7a-41ee-b6d5-14335e767bad>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00347.warc.gz"} |
RE: pure functions
• To: mathgroup at smc.vnet.net
• Subject: [mg23342] RE: [mg23237] pure functions
• From: "Higinio Ramos Calle" <higra at gugu.usal.es>
• Date: Thu, 4 May 2000 02:59:10 -0400 (EDT)
• References: <200004300204.WAA16572@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
----- Original Message -----
From: Helge Andersson <helgea at inoc.chalmers.se>
To: mathgroup at smc.vnet.net
Subject: [mg23342] [mg23237] pure functions
> Hello,
> Of all the nice functions such as Map, Mapall, Thread, Apply .... I have
> not been able to write a simple code to generate the following
> procedure.
> I have a two dimensional list like
> li={{11,12,13,..},{21,22,23,...},{31,32,33,..},...}
> Since i like to use the pure function command I would like to map my
> pure function with arguments #1,#2,#3,.... on all the sublists in li.
> Let me exemplify with a simple pure function that add to numbers.
> (#1+#2)&
> if exli={{1,2},{3,4},{5,6},{7,8}}
> then I want to get the result
> {3,7,11,15}
> One solution, but not allways suitable for me, is the following
> (#1+#2)&[Sequence @@ Transpose[exli]].
> I want to get rid of the Transpose command and if possible also making
> use of the /@ notation for Map. Since I have seen so many elegant
> examples in the mailing lists I hope I can get something out of this.
For your own example you may try
and you'll get the result you hope.
> Finally, When using pure functions inside Mathematica defined functions
> such as Select for instance,
> Select[{1,2,3,4,5,6},#>3&]
> why don't we need to specify the argument list after the &sign. I can
> figure out that in this case the list sent to the Select command will be
> used as argument list for the pure function but how does this work in
> general. Which are the functions where this feature is possible?
Select[list, test] apply the function that appears in test to each element
of the list, and returns all the elements for which the result is true.
An example using the same list as before:
Higinio Ramos
Dpto. Matematica Aplicada
U.de Salamanca | {"url":"http://forums.wolfram.com/mathgroup/archive/2000/May/msg00054.html","timestamp":"2024-11-15T00:11:59Z","content_type":"text/html","content_length":"32150","record_id":"<urn:uuid:ecc30041-519d-4707-ab90-48864c1a4c6a>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00800.warc.gz"} |
Understanding Parity: A Simple Guide
Parity is a big word that might sound a little confusing, but don’t worry, it’s actually very easy to understand. In simple words, parity is all about things being the same or equal in some way. We
use parity to describe whether things are even or odd, whether they are balanced, or whether two things are the same. It’s a concept that we see in math, science, and even in our everyday lives.
For example, if you have two apples in one hand and two apples in the other hand, you have parity because both sides have the same number of apples. Isn’t that simple? Now, let’s dive deeper into
what parity really means and why it’s important.
Understanding Parity in Mathematics
In mathematics, parity is a way of describing numbers. We say a number has parity if it’s either even or odd. If a number can be divided by 2 without any leftovers, it’s called an even number. But if
you try to divide a number by 2 and there’s one left over, that’s an odd number.
For example:
• The number 4 is even because 4 divided by 2 is 2, with no leftover.
• The number 5 is odd because 5 divided by 2 is 2 with 1 left over.
So, when we talk about parity in math, we’re just saying whether a number is even or odd. It’s as simple as that! And guess what? You use parity every day without even knowing it. When you count your
toys, pair up your socks, or share candies with friends, you’re using the idea of parity.
Parity in Everyday Life
You might think that parity is only for math problems, but it’s everywhere around us! Let’s look at some examples of parity in real life:
1. Sharing equally: Imagine you have 10 cookies and you want to share them with your friend. Since 10 is an even number, you can share them equally – 5 cookies for you and 5 cookies for your friend.
That’s parity at work!
2. Walking in pairs: When you walk in pairs with your friend, you’re showing parity because both of you are walking together. If you’re walking alone, that’s like an odd number because there’s no
one to pair up with!
3. Matching socks: Have you ever tried finding pairs of socks in the laundry? That’s another example of parity. When you find matching socks, you have parity because there are two of them. But if
you have one sock left without a pair, that’s an odd sock!
By now, you can see that parity is something we use all the time. It helps us understand how things can be even or odd, balanced or unbalanced.
Why is Parity Important?
You might be wondering, “Why should I care about parity?” Well, parity is important because it helps us understand and organize things. It’s not just a fancy word – it’s a useful tool that makes our
lives easier!
1. In Games: Have you ever played a game where you need to count steps, match items, or take turns? Parity helps us keep track of things in games. It tells us if we need to take one more turn or if
the game is balanced.
2. In Computers: Did you know that computers use parity too? Computers are very smart, but they need to check if their calculations are correct. They use parity to make sure all their numbers add up
correctly. This helps them work faster and makes sure they don’t make mistakes.
3. In Everyday Decisions: Parity helps us make decisions. For example, if you’re dividing candies with your friends, parity tells you whether you can share them equally or if one candy will be left
Parity in Science and Technology
Parity isn’t just for math and everyday life – it’s also very important in science and technology! Scientists use parity to understand how things work in nature, and engineers use it to build
machines and computers.
In science, there is something called “parity symmetry.” This means that if you look at something in a mirror, it should look the same as the real thing. But sometimes, things don’t look the same in
a mirror. That’s when scientists say there’s “parity breaking.” It’s a bit more complicated, but it shows that parity helps us understand how the world works.
In technology, parity is used to make sure data is sent and received correctly. When you send a message or an email, computers use parity checks to make sure the message arrives just the way it was
sent. If something goes wrong, the computer can fix it because it knows the message isn’t the same. Isn’t that amazing?
Different Types of Parity
Did you know there are different types of parity? Let’s take a look at a few of them:
1. Even Parity: This is when a number or group has an even total. For example, the number 8 has even parity because 8 can be divided by 2 without any leftovers.
2. Odd Parity: This is when a number or group has an odd total. For example, the number 7 has odd parity because 7 divided by 2 leaves 1 left over.
3. Parity Check: This is a special type of parity used in computers. It helps make sure that data is correct and not missing any parts. It’s like a little helper that checks if everything is in
Now that you know about these different types of parity, you can see that parity is not just one simple thing – it has many forms and uses.
How to Teach Parity to Kids
If you want to teach parity to kids, you can make it a fun game! Here’s a simple way to do it:
1. Count Together: Pick up some toys and count them together. Ask the child to group them in twos and see if there are any left over. If there are no toys left, that’s even parity. If there’s one
toy left, that’s odd parity.
2. Play Matching Games: Find pairs of socks, shoes, or mittens and see if they all have a match. This helps kids understand the concept of parity in a fun and simple way.
3. Use Everyday Examples: Point out examples of parity in everyday life. When you’re walking, playing, or sharing, talk about how things are even or odd.
Teaching parity doesn’t have to be boring – it can be a fun adventure!
Conclusion: Parity is Everywhere!
As we’ve learned, parity is a simple but powerful concept. It helps us understand whether things are even or odd, balanced or unbalanced, and it’s used in many different ways. From counting apples to
sending messages on the computer, parity is always there, making sure everything is just right.
So next time you see something that’s paired up or shared equally, remember – that’s parity in action! Now, you know all about parity, and you can see it everywhere around you. Isn’t that cool? | {"url":"https://healthlinemedia.net/understanding-parity-a-simple-guide/","timestamp":"2024-11-07T03:06:31Z","content_type":"text/html","content_length":"155644","record_id":"<urn:uuid:d0d65f53-b1f1-46fb-8d34-b1426006c8cb>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00487.warc.gz"} |
Calculating Capital Gains and Capital Losses
by ABC editor
mutual fund in a taxable investment account, then you might have to pay capital gains taxes on any profit that you make.� This does not apply to a tax-deferred retirement accounts such as 401(k) plan
and Roth IRAs.
If you make a profit on the investment then you will have a capital gain which is taxable.� If you lose money, then you will have a capital loss which is not taxable but can be used to offset capital
gains if you have any.
Capital gain
A capital gain is the difference in the selling price of an investment and the purchase price minus transaction costs, if you sell for a higher price than you bought it at.
For example, let�s say you bought a mutual fund on the advice of the genius kid next door for $5,000 with a transaction cost of $50 and then sold it for $7,000 with a transaction cost of $50.� Since
your selling price is higher than your purchase price, this will result in a capital gain.
To calculate the capital gain, you subtract the purchase price and transaction costs from the sale price.
Capital gains = sale price � purchase price – transaction costs
= $7,000 – $5000 – $100
= $1,900
Capital loss
What happens if the hot stock tip you heard at the coffee shop turned out to be a dud and you lost money?� In that case you will have a capital loss because you sold the stock for less than you paid
for it.
For example if you buy an investment for $10,000 with a transaction cost of $25 and sell it for $9,200 with a transaction cost of $25.
Capital loss = sale price � purchase price – transaction costs = $9,200 – $10,000 – $50 = -$850.
You might notice that the formula is the same for both losses and gains.� If the answer is negative then you have a capital loss.� If the answer is positive then you have a capital gain. | {"url":"https://abcsofinvesting.net/calculating-capital-gains-and-capital-losses/","timestamp":"2024-11-07T06:17:58Z","content_type":"application/xhtml+xml","content_length":"34195","record_id":"<urn:uuid:6deae19a-8fd3-4950-a9c3-9e193ddf912c>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00204.warc.gz"} |
: Buffer Solutions And Calculating Ka
So I had this lab where we prepared a buffer solution using an unknown monoprotic weak acid and 0.1M NaOH, and the objective of the lab is to calculate the Ka of the solution.
The data I got from the experiment was that I used 17.85mL of NaOH to titrate 15mL of unknown weak acid (or HA). I then added another 15mL of the same unknown weak acid to the already titrated
I then measured the pH of this final solution using an electronic pH meter and found it to be 4.72.
My questions are:
What is the initial concentration of unknown weak acid (or HA)?
What are the final concentrations of HA and A-?
What is the final concentration of H+ in the solution?
And what is the Ka of the solution?
My calculations give me:
Final[HA]=0.054M - 1.9x10^-5M
Final[A-]=0.054M + 1.9x10^-5M
Sorry if this is a really long question, but the lab is due tomorrow and I really can't figure this out by myself. | {"url":"https://www.chemicalforums.com/index.php?topic=31311.0;prev_next=next","timestamp":"2024-11-14T14:00:24Z","content_type":"text/html","content_length":"38578","record_id":"<urn:uuid:eea85bcc-2fab-4389-a71b-60ada481f957>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00295.warc.gz"} |
Charge Carrier Lifetime & Recombination
Allpix Squared provides the possibility to simulate finite lifetimes of charge carriers as a function of the local doping concentration via non-radiative recombination processes. While most of these
models require the total doping concentration $N_D + N_A$ as parameter, the doping profile used throughout Allpix Squared provides the effective doping concentration $N_D - N_A$ since this also
encodes the majority charge carriers via its sign - an information relevant to some of the models. However, in the parts of a silicon detector relevant for this simulation, i.e. the sensing volume,
the difference between effective and total concentration is expected to be negligible. Therefore the two values are treated as equivalent throughout the lifetime models and the doping concentration
is taken as the absolute value $N = \left|N_D - N_A\right|$.
Whether a charge carrier has recombined with the lattice is calculated for every step of the simulation using the relation
$$p < 1 - e^{- dt / \tau(N)}$$
where $p$ is a recombination probability, drawn from a uniform distribution with $[0, 1]$, $dt$ is the last time step of the charge carrier motion and $\tau$ the lifetime for the local doping
concentration calculated by the models described in the following. If the above equation evaluates to false, the charge carrier still exists, if it evaluates to true it has been recombined with the
Finite charge carrier lifetime can be simulated by all propagation modules and comprise the following models:
Shockley-Read-Hall Recombination
This model describes the finite lifetime based on Shockley-Read-Hall or trap-assisted recombination of charge carriers with the lattice [@shockley-read, @hall]. The lifetime is calculated using the
Shockley-Read-Hall relation as given by [@fossum-lee]:
$$\tau(N) = \frac{\tau_0}{1 + \frac{N}{N_{d0}}}$$
where $\tau_0$ and $N_{d0}$ are reference lifetime and doping concentration, for electrons and holes respectively. The parameter values implemented in Allpix Squared are taken from [@fossum-lee] and
the Synopsys Sentaurus TCAD software manual as
$$\begin{aligned} \tau_{0,e} &= 1\times 10^{-5} \,\text{s} \\ N_{d0,e} &= 1\times 10^{16} \,\text{cm}^{-3} \\ \\ \tau_{0,h} &= 4.0\times 10^{-4} \,\text{s} \\ N_{d0,h} &= 7.1\times 10^{15} \,\text
{cm}^{-3} \end{aligned}$$
for electrons and holes, respectively.
The temperature dependence of the Shockley-Read-Hall lifetime is scaled following the low-temperature approximation model presented [@schenk] as:
$$\tau(N, T) = \tau(N) \cdot \left( \frac{300 \,\text{K}}{T} \right)^{3/2}$$
This model can be selected in the configuration file via the parameter recombination_model = "srh".
Auger Recombination
At high doping levels exceeding $5\times 10^{18} \,\text{cm}^{-3}$ [@fossum-lee], the Auger recombination model becomes increasingly important. It assumes that the excess energy created by
electron-hole recombinations is transferred to another electron (e-e-h process) or another hole (e-h-h process). The total recombination rate is then given by [@kerr]:
$$R_{Auger} = C_n n^2p + C_p n p^2$$
where $C_n$ and $C_p$ are the Auger coefficients. The first term corresponds to the e-e-h process and the second term to the e-h-h process. In highly-doped silicon, the Auger lifetime for minority
charge carriers can be written as:
$$\tau(N) = \frac{1}{C_{a} \cdot N^2}$$
where $C_{a} = C_{n} + C_{p}$ is the ambipolar Auger coefficient, taken as $C_{a} = 3.8\times 10^{-31} \,\text{cm}^6\,\text{s}^{-1}$ from [@dziewior].
This recombination mode applies to minority charge carriers only, majority charge carriers have an infinite life time under this model and the recombination equation will always evaluate to true.
This model can be selected in the configuration file via the parameter recombination_model = "auger".
Combined SRH/Auger Recombination
This model combines the charge carrier recombination from the Shockley-Read-Hall and the Auger model by inversely summing the individual lifetimes calculated by the models via
$$\begin{aligned} \tau^{-1}(N) &= \tau_{srh}^{-1}(N) + \tau_{a}^{-1}(N) &\quad \text{for minority charge carriers} \\ &= \tau_{srh}^{-1}(N) &\quad \text{for majority charge carriers} \end{aligned}$$
where $\tau_{srh}(N)$ is the Shockley-Read-Hall and $\tau_{a}(N)$ the Auger lifetime. The latter is only taken into account for minority charge carriers.
This model can be selected in the configuration file via the parameter recombination_model = "srh_auger".
Recombination with Constant Lifetimes
Some materials require constant lifetimes for charge carriers $\tau(N) = \tau$. This model requires the additional configuration keys lifetime_electron and lifetime_hole to be present in the module
configuration section, for example:
# Constant lifetimes for electrons and holes in GaAs with Cr compensation:
recombination_model = "constant"
lifetime_electron = 30ns
lifetime_hole = 4.5ns
This model can be selected in the configuration file via the parameter recombination_model = "constant".
Custom Recombination Models
Allpix Squared provides the possibility to use fully custom recombination models. In order to use a custom model, the parameter recombination_model = "custom" needs to be set in the configuration
file. Additionally, the following configuration keys have to be provided:
• lifetime_function_electrons: The formula describing the electron lifetime.
• lifetime_function_holes: The formula describing the hole lifetime.
The functions defined via these parameters can depend on the local doping concentration. In order to use the doping concentration in the formula, an x has to be placed at the respective position.
Parameters of the functions can either be placed directly in the formulas in framework-internal units, or provided separately as arrays via the lifetime_parameters_electrons and
lifetime_parameters_electrons. Placeholders for parameters in the formula are denoted with squared brackets and a parameter number, for example [0] for the first parameter provided. Parameters
specified separately from the formula can contain units which will be interpreted automatically.
Parameters directly placed in the recombination formula have to be supplied in framework-internal units since the function will be evaluated with the doping concentration in internal units. It is
recommended to use the possibility of separately configuring the parameters and to make use of units to avoid conversion mistakes.
The following set of parameters re-implements the Shockley-Read-Hall recombination model using a custom recombination model. The lifetimes are calculated at a fixed temperature of 293 Kelvin.
# Replicating the Shockley-Read-Hall model at T = 293K
recombination_model = "custom"
lifetime_function_electrons = "[0]/(1 + x / [1])"
lifetime_parameters_electrons = 1.036e-5s, 1e16/cm/cm/cm
lifetime_function_holes = "[0]/(1 + x / [1])"
lifetime_parameters_holes = 4.144e-4s, 7.1e15/cm/cm/cm
It should be noted that the temperature passed via the module configuration is not evaluated for the custom recombination model, but the model parameters need to be manually adjusted to the required
The interpretation of the custom recombination functions is based on the ROOT::TFormula class [[@rootformula]] and supports all corresponding features, mathematical expressions and constants. | {"url":"https://allpix-squared.docs.cern.ch/docs/06_models/03_lifetime_recombination/","timestamp":"2024-11-14T21:08:30Z","content_type":"text/html","content_length":"71015","record_id":"<urn:uuid:46b45f6b-9710-4bbe-bf2d-9040f68a82b3>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00206.warc.gz"} |
What is the average salary for an airline pilot - CAREER KEG
What is the average salary for an airline pilot
May 28, 2022 Uncategorized No Comments
So, what is the average salary for an airline pilot? This question was asked recently and I thought back to our recent interview with Ron Allard. Mr. Allard has been a commercial airline pilot for
over 40 years, but he also spent time as an administrator, biggest mistake he made and how to transition from flying planes to airlines. He could have easily been worth $100k per year and more!
Airline pilots have a number of responsibilities when on the job. Most aviation experts agree that airline pilots are some of the best trained professionals in their respective field. Before a pilot
is able to take to the skies, he or she must first attend college and then a flight school. After earning a degree as well as a pilot’s license, one may choose to go through additional training
programs. Pilots can earn additional money by obtaining their instructor’s license and teaching other interested students how to fly. Some airline pilots may also obtain a license to fly experimental
planes, cargo planes, and helicopters.
What is the average salary for an airline pilot
Airline pilots are highly skilled professionals who fly passengers, cargo, and mail safely to their destinations. They must have a combination of training and experience that includes an Airline
Transport Pilot license, instrument rating, medical certificate, and a specific amount of flight hours in order to become certified. According to the Bureau of Labor Statistics (BLS), the median
annual salary for airline pilots was $121,408 in May 2019.
Airline pilots in the United States make a median annual salary of $121,408.
In the United States, the median annual salary of an airline pilot is $121,408. This means that half of all pilots earn more than this amount, while half earn less. In other words, this is the middle
value when all salary figures are lined up from smallest to largest.
The average yearly wage can be calculated by adding up together all the wage values for each occupation and dividing by how many occupations there are in total—this is known as “arithmetic mean.” The
median yearly wage is calculated by taking into account only that part of the distribution which has a lower value than or equal to it (i.e., it excludes everything above it)
Regional airline pilots in the United States make an average annual salary of $74,218.
As you might expect, the average salary for an airline pilot is higher than it is for a regional airline pilot. In fact, the average salary for a commercial airline pilot in the United States is
$153,738 per year. This makes sense since they fly larger planes and have more responsibilities than their smaller-scale counterparts.
Regional pilots also make less than corporate pilots who fly private jets and helicopters within North America. Corporate pilots earn an average annual income of $132,000 per year. Military pilots
are paid even better—their average annual salary is about $134,000 per year!
• All numbers quoted above are based on data from Payscale’s 2019 Pilot Salaries Report.*
Commercial airline pilots in the United States make an average annual salary of $85,810.
• The average salary for a commercial airline pilot is $85,810.
• The average salary for a regional airline pilot is $74,218.
• The average salary for an air taxi pilot is $71,964.
• The average salary for a charter airline pilot is $71,964.
The median salary for airline pilots is $121,408 per year
The median value for airline pilots is $121,408 per year. The median salary means that half of pilots make less than this amount and half make more. In other words, the middle value in a list of
This is not to be confused with average or mean which are different mathematical functions that are used to determine how much money all pilots collectively get paid over time.
As you can see from the data, there are a number of factors that go into determining the amount of money you make as an airline pilot. As with any job, your location will have a big impact on how
much money you can expect to earn. You’ll need to be licensed by the Federal Aviation Administration (FAA) before being hired; this license costs around $1,000 and requires at least 250 hours of
flight time. | {"url":"https://infolearners.com/what-is-the-average-salary-for-an-airline-pilot/","timestamp":"2024-11-02T20:41:21Z","content_type":"text/html","content_length":"55771","record_id":"<urn:uuid:8c5e656d-5673-47d0-9880-eb6f0891fd47>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00497.warc.gz"} |
228 RWTH Publication No: 47240 2003   IGPM228.pdf
TITLE On the Connection between some Riemann-Solver Free Approaches to the Approximation of Multi-Dimensional Systems of Hyperbolic Conservation Laws
AUTHORS Tim Kröger, Sebastian Noelle, Susanne Zimmermann
In this paper, we present some interesting connections between a number of Riemann-solver free approaches to the numerical solution of multi-dimensional systems of conservation laws. As a
main part, we present a new and elementary derivation of Fey's Method of Transport (MoT) (respectively the second author's ICE version of the scheme) and the state decompositions which
form the basis of it. The only tools that we use are quadrature rules applied to the moment integral used in the gas kinetic derivation of the Euler equations from the Boltzmann equation,
ABSTRACT to the integration in time along characteristics and to space integrals occuring in the finite volume formulation. Thus, we establish a connection between the MoT approach and the kinetic
approach. Further more, Ostkamp's equivalence result between her Evolution Galerkin scheme and the Method of Transport is lifted up from the level of discretizations to the level of exact
evolution operators, introducing a new connection between the MoT and the Evolution Galerkin approach. At the same time, we clarify some important differences between these two
KEYWORDS systems of conservation laws, Fey's Method of Transport, Euler equations, Boltzmann equation, kinetic schemes, bicharacteristic theory, state decompositions, decompositions, exact and
approximate evolution operators, quadrature rules
DOI 10.1051/m2an:2004047
PUBLICATION Mathematical modelling and numerical analysis = Modélisation mathématique et analyse numérique
38, 989-1009 (2004) | {"url":"https://www.igpm.rwth-aachen.de/forschung/preprints/228","timestamp":"2024-11-05T10:47:27Z","content_type":"text/html","content_length":"29959","record_id":"<urn:uuid:8b05760e-f2ad-49ff-bc16-4ef466fce2e4>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00496.warc.gz"} |
Talk:De Polignac numbers - Rosetta CodeTalk:De Polignac numbers
Efficient algorithm
There is a quite efficient algorithm to find de Polignac numbers that several entry authors seem to have overlooked.
It is not necessary to test add every power of 2 less than N with every prime less than N and check if the sum is N.
Simply find the powers of 2 less than N, subtract each from N, and check if any remainder is a prime. If any is, it is not a de Polignac number. Short circuit and move on. --Thundergnat (talk) 22:04,
27 September 2022 (UTC)
2nd sentence reads better if you delete "test" and replace "with" with "to". 3rd & 4th sentences would make a wonderful comment for the Raku entry 😉 --Petelomax (talk) 11:54, 28 September 2022 | {"url":"https://rosettacode.org/wiki/Talk:De_Polignac_numbers?oldid=330776","timestamp":"2024-11-03T01:00:43Z","content_type":"text/html","content_length":"43564","record_id":"<urn:uuid:f0a33b2b-55f2-4e8c-872b-eb568f9cfdf4>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00735.warc.gz"} |
Best Understanding How to Convert Milligrams to ml 2024 - businesstechwaves.com
Best Understanding How to Convert Milligrams to ml 2024
Convert milligrams to ml is a fundamental calculation in various fields, including medicine, chemistry, and biology. However, this conversion can be challenging without understanding the basic
principles underlying it. Milligrams measure mass, while milliliters measure volume. Because these units measure different things, Convert milligrams to ml between them requires an understanding of
the substance’s density. This article will provide a comprehensive explanation of how to convert milligrams to milliliters, covering the concept of density, practical examples, and common
applications of these conversions.
Understanding Units: Converting milligrams to ml
Milligram (mg) is a metric unit of mass equal to one-thousandth of a gram. It is often used in contexts where small quantities are involved, particularly in medicine for measuring doses of
medications. For example, a tablet might contain 500 mg of an active ingredient.
Milliliter (ml) is a metric unit of volume that is equal to one-thousandth of a liter. It is frequently used in liquid measurements, especially in medicine, cooking, and chemistry. A milliliter is
also equivalent to one cubic centimeter (cc or cm³). It is a small unit often used for liquids like water, oil, or medical solutions.
Because milligrams and milliliters measure different things—mass versus volume—there is no direct conversion between these units without knowing additional information: the density of the substance
in question.
The Role of Density in Conversion
To convert milligrams to ml (or vice versa), you need to know the substance’s density. Density is defined as the mass per unit volume of a substance and is typically expressed in grams per milliliter
(g/ml) for liquids or solids. The formula for density is:Density=MassVolume\text{Density} = \frac{\text{Mass}}{\text{Volume}}Density=VolumeMass
Rearranging this equation to solve for volume gives:Volume=MassDensity\text{Volume} = \frac{\text{Mass}}{\text{Density}}Volume=DensityMass
This formula is crucial when converting milligrams to milliliters because it shows that the volume (in milliliters) is equal to the mass (in milligrams) divided by the density (in g/ml). However,
because milligrams are 1,000 times smaller than grams, an additional step is required to adjust the units accordingly.
Convert milligrams to ml: Step-by-Step
Let’s break down the steps involved in converting milligrams (mg) to milliliters (ml):
1. Obtain the Density: The first step is to obtain the density of the substance you are working with. This can often be found in a textbook, datasheet, or online resource. For example, the density
of water is approximately 1 g/ml, which means 1 gram of water occupies 1 milliliter of space. Other substances will have different densities; for instance, the density of ethanol is approximately
0.789 g/ml.
2. Convert Density to Appropriate Units: Because density is usually expressed in grams per milliliter (g/ml), and you are working in milligrams, you need to convert grams to milligrams. Since 1 gram
equals 1,000 milligrams, you need to adjust the density value. For example, a density of 1 g/ml would convert to 1,000 mg/ml.
3. Apply the Formula: Once you have the density in milligrams per milliliter (mg/ml), use the formula:Volume (ml)=Mass (mg)Density (mg/ml)\text{Volume (ml)} = \frac{\text{Mass (mg)}}{\text{Density
(mg/ml)}}Volume (ml)=Density (mg/ml)Mass (mg)
4. Calculate the Volume: After applying the formula, you can calculate the volume in milliliters.
Example 1: Convert milligrams to ml for Water
Let’s take an example where we want to convert 500 mg of water into milliliters. The density of water is 1 g/ml, which converts to 1,000 mg/ml.
1. Density of water = 1,000 mg/ml
2. Mass = 500 mg
3. Apply the formula:Volume (ml)=500mg1,000mg/ml=0.5ml\text{Volume (ml)} = \frac{500 \, \text{mg}}{1,000 \, \text{mg/ml}} = 0.5 \, \text{ml}Volume (ml)=1,000mg/ml500mg=0.5ml
So, 500 mg of water is equivalent to 0.5 ml.
Example 2: Convert milligrams to ml for Ethanol
Now, consider you have 750 mg of ethanol and want to find its volume in milliliters. The density of ethanol is approximately 0.789 g/ml. Converting this to milligrams per milliliter:Density of
ethanol=0.789g/ml=789mg/ml\text{Density of ethanol} = 0.789 \, \text{g/ml} = 789 \, \text{mg/ml}Density of ethanol=0.789g/ml=789mg/ml
Now, apply the formula:Volume (ml)=750mg789mg/ml≈0.95ml\text{Volume (ml)} = \frac{750 \, \text{mg}}{789 \, \text{mg/ml}} \approx 0.95 \, \text{ml}Volume (ml)=789mg/ml750mg≈0.95ml
So, 750 mg of ethanol is approximately equal to 0.95 ml.
Why Density Matters
The two examples above highlight why density is crucial in converting milligrams to milliliters. Water and ethanol have different densities, so the same mass of each substance will occupy different
volumes. Substances with higher densities will take up less volume for the same mass, while those with lower densities will occupy more space.
Without knowing the density, it is impossible to convert milligrams to ml accurately. This is because milliliters measure volume, which varies based on how tightly the substance’s molecules are
packed (i.e., the density).
Common Applications of mg to ml Conversion
The conversion from milligrams to milliliters has several practical applications, particularly in fields where precise measurements of substances are critical. Here are a few examples:
1. Medical Dosages
In healthcare, accurate dosages are crucial, especially for intravenous (IV) medications, where drugs are often administered in liquid form. Doctors or nurses might need to convert a medication’s
mass in milligrams to its equivalent volume in milliliters to ensure the correct dosage is given. For instance, if a medication has a density of 1.2 g/ml and the required dose is 600 mg, healthcare
professionals would need to calculate how many milliliters to administer.
2. Pharmaceuticals and Chemistry
Pharmaceutical companies frequently use the mg-to-ml conversion to formulate medications. When creating liquid formulations, they must know how much of the active ingredient in milligrams corresponds
to a given volume. This process ensures that each dose delivers the correct amount of medication.
Similarly, in chemical laboratories, scientists often work with reagents and compounds in both mass and volume form. Knowing how to convert between milligrams and milliliters is essential for
creating solutions with precise concentrations.
3. Food and Beverage Industry
In the food and beverage industry, small amounts of certain ingredients, such as preservatives or additives, may be measured in milligrams. When these ingredients are added to liquids, manufacturers
need to know how to convert their mass to volume to achieve the correct mixture.
4. Cosmetics and Personal Care
In the production of cosmetics, manufacturers must ensure that their formulations contain the correct proportion of active ingredients. This often requires converting milligrams of an ingredient
(such as vitamins or oils) to milliliters of a liquid solution.
Conversion Tools and Calculators
While it’s possible to manually perform the conversion as described above, several online calculators and mobile apps simplify the process. These tools often allow users to input the mass and density
of a substance to get an instant conversion to volume.
In practice, though, it’s always important to double-check the density value you use, especially when precision is critical, as even small errors in density can lead to significant discrepancies in
the final volume.
Convert milligrams to ml is not a straightforward conversion because the two units measure different properties: mass and volume. To make this conversion, you need to know the density of the
substance, which provides the link between mass and volume. By applying the formula that relates mass, volume, and density, you can accurately convert milligrams to milliliters for any substance.
This process is particularly important in fields like medicine, chemistry, and industry, where precise measurements are essential. Understanding these conversions ensures accuracy in applications
where even small errors could have significant consequences. | {"url":"https://businesstechwaves.com/convert-milligrams-to-ml/","timestamp":"2024-11-06T10:46:20Z","content_type":"text/html","content_length":"85726","record_id":"<urn:uuid:6005485d-0a03-4a27-9b36-cd932c9e8e30>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00586.warc.gz"} |
Optical Parity Time Metasurface Structures
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Optical Parity Time Metasurface Structures
Date Issued:
In the last few years, optics has witnessed the emergence of two fields namely metasurfaces and parity-time (PT) symmetry. Optical metasurfaces are engineered structures that provide unique
responses to electromagnetic waves, absent in natural materials. Optical metasurfaces are known for their reduced dimensionality i.e. subwavelength and consequently lower losses are anticipated.
The other paradigm is the PT symmetric materials, also known as photonic synthetic matter. PT symmetry has emerged from quantum mechanics when a new class of non-Hermitian Hamiltonian quantum
systems was highlighted to have real eigenvalues, hence eradicating Hermiticity of the Hamiltonian as an essential condition to the existence of real eigenvalues.The first half of the thesis is
focused on the experimental and numerical realization of PT symmetric metasurfaces. A systematic methodology is developed to implement this class of metasurfaces in both one-dimensional and
two-dimensional geometries. In two dimensional systems, PT symmetry can be established by employing either H-like diffractive elements or diatomic oblique Bravais lattices. It is shown that the
passive PT symmetric metasurfaces can be utilized to appropriately engineer the resulting far-field characteristics. Such PT-symmetric structures are capable of eliminating diffraction orders in
specific directions, while maintaining or even enhancing the remaining orders. Later, it is shown a first ever attempt of PT metasurface fabricated on a flexible polymer (polyimide) substrate.
The studied PT metasurface exhibits the ability to direct light, i.e. Poynting vector in a desired direction. Herein, the light scattered from the fabricated device in the undesired direction is
attenuated by at least an order of magnitude. The proposed PT symmetric metasurface is essentially diatomic Honeycomb Bravais lattice, where both the passive and lossy elements exist side by side
on each site separated by 50 nm. The unidirectionality of the studied metasurface is not limited to a single wavelength, on the contrary, it is observed to be effective on the entire visible band
(400 (-) 600 nm). The PT symmetric meatsurface is also fabricated on a high strength substrate; sapphire (Al2O3). An excellent agreement between the experimental and numerical (COMSOL) results is
found for both substrates. Customized modifications to the current design can open avenues to study the unidirectionality of metasurfaces to different optical bands, for example IR.The second
part of the thesis deals with the theoretical modeling of the dynamics of an electron that gets trapped by means of decoherence and quantum interference in the central quantum dot (QD) of a
semiconductor nanoring (NR) made of five QDs, between 100 and 300 K. The electron's dynamics is described by a master equation with a Hamiltonian based on the tight-binding model, taking into
account electron(-)LO phonon interaction. Based on this configuration, the probability to trap an electron with no decoherence is almost 27%. In contrast, the probability to trap an electron with
decoherence is 70% at 100 K, 63% at 200 K and 58% at 300 K. Our model provides a novel method of trapping an electron at room temperature.This setup is then used to propose a theoretical model
for an electrically driven single photon source operating at high temperatures. It is shown that the decoherence, which is usually the main obstacle for operating single photon sources at high
temperatures, ensures an efficient operation of the presented electrically driven single photon source at high temperatures. The single-photon source is driven by a single electron source
attached to a heterostructure semiconductor nanoring. The electron's dynamics in the nanoring and the subsequent recombination with the hole is described by the generalized master equation with a
Hamiltonian based on tight-binding model, taking into account the electron-LO phonon interaction. As a result of decoherence, an almost 100% single photon emission with a strong antibunching
behavior i.e. g(2)(0) (<)(<) 1 at high temperature up to 300 K is achieved.
Title: Optical Parity Time Metasurface Structures.
El Halawany, Ahmed, Author
Christodoulides, Demetrios, Committee Chair
Name(s): Rahman, Talat, Committee Member
Peale, Robert, Committee Member
Likamwa, Patrick, Committee Member
University of Central Florida, Degree Grantor
Type of text
Date Issued: 2016
Publisher: University of Central Florida
Language(s): English
In the last few years, optics has witnessed the emergence of two fields namely metasurfaces and parity-time (PT) symmetry. Optical metasurfaces are engineered structures that provide
unique responses to electromagnetic waves, absent in natural materials. Optical metasurfaces are known for their reduced dimensionality i.e. subwavelength and consequently lower losses
are anticipated. The other paradigm is the PT symmetric materials, also known as photonic synthetic matter. PT symmetry has emerged from quantum mechanics when a new class of
non-Hermitian Hamiltonian quantum systems was highlighted to have real eigenvalues, hence eradicating Hermiticity of the Hamiltonian as an essential condition to the existence of real
eigenvalues.The first half of the thesis is focused on the experimental and numerical realization of PT symmetric metasurfaces. A systematic methodology is developed to implement this
class of metasurfaces in both one-dimensional and two-dimensional geometries. In two dimensional systems, PT symmetry can be established by employing either H-like diffractive elements
or diatomic oblique Bravais lattices. It is shown that the passive PT symmetric metasurfaces can be utilized to appropriately engineer the resulting far-field characteristics. Such
PT-symmetric structures are capable of eliminating diffraction orders in specific directions, while maintaining or even enhancing the remaining orders. Later, it is shown a first ever
attempt of PT metasurface fabricated on a flexible polymer (polyimide) substrate. The studied PT metasurface exhibits the ability to direct light, i.e. Poynting vector in a desired
direction. Herein, the light scattered from the fabricated device in the undesired direction is attenuated by at least an order of magnitude. The proposed PT symmetric metasurface is
Abstract/ essentially diatomic Honeycomb Bravais lattice, where both the passive and lossy elements exist side by side on each site separated by 50 nm. The unidirectionality of the studied
Description: metasurface is not limited to a single wavelength, on the contrary, it is observed to be effective on the entire visible band (400 (-) 600 nm). The PT symmetric meatsurface is also
fabricated on a high strength substrate; sapphire (Al2O3). An excellent agreement between the experimental and numerical (COMSOL) results is found for both substrates. Customized
modifications to the current design can open avenues to study the unidirectionality of metasurfaces to different optical bands, for example IR.The second part of the thesis deals with
the theoretical modeling of the dynamics of an electron that gets trapped by means of decoherence and quantum interference in the central quantum dot (QD) of a semiconductor nanoring
(NR) made of five QDs, between 100 and 300 K. The electron's dynamics is described by a master equation with a Hamiltonian based on the tight-binding model, taking into account
electron(-)LO phonon interaction. Based on this configuration, the probability to trap an electron with no decoherence is almost 27%. In contrast, the probability to trap an electron
with decoherence is 70% at 100 K, 63% at 200 K and 58% at 300 K. Our model provides a novel method of trapping an electron at room temperature.This setup is then used to propose a
theoretical model for an electrically driven single photon source operating at high temperatures. It is shown that the decoherence, which is usually the main obstacle for operating
single photon sources at high temperatures, ensures an efficient operation of the presented electrically driven single photon source at high temperatures. The single-photon source is
driven by a single electron source attached to a heterostructure semiconductor nanoring. The electron's dynamics in the nanoring and the subsequent recombination with the hole is
described by the generalized master equation with a Hamiltonian based on tight-binding model, taking into account the electron-LO phonon interaction. As a result of decoherence, an
almost 100% single photon emission with a strong antibunching behavior i.e. g(2)(0) (<)(<) 1 at high temperature up to 300 K is achieved.
Identifier: CFE0006454 (IID), ucf:51421 (fedora)
Note(s): Sciences, Physics
This record was generated from author submitted information.
Subject(s): PT Symmetry -- Metasurfaces -- Electron trapping -- Single Photon source
Link to This http://purl.flvc.org/ucf/fd/CFE0006454
Restrictions campus 2017-12-15
on Access:
Host UCF
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Binary Basics: Conversion From Binary To Decimal - Code With C
Understanding Binary Basics
I am so hyped to dive into the mesmerizing world of Binary Basics! 💻 Let’s start by unraveling the mystery behind what Binary actually is and why it’s such a big deal in the tech universe.
What is Binary?
Ah, Binary, the language of computers! 🤖 It’s like their very own secret code. Binary is a base-2 number system represented using only two symbols, typically 0 and 1. But hey, don’t underestimate its
power just because it’s made up of just two numbers. This system forms the backbone of modern computing.
Definition and Explanation
Okay, so let me break it down for you in simpler terms: in Binary, each digit’s place value is a power of 2, and each digit can either be 0 or 1. When you string these 0s and 1s together, you can
represent all sorts of data, from numbers to text and even images!
Importance of Binary System
Now, why on earth is Binary so important, you ask? Well, buckle up because we’re about to explore the mind-blowing real-life applications of this magical system.
Real-life Applications
• Computers: Every data processed by a computer, whether it’s a simple calculation or complex algorithms, boils down to 0s and 1s.
• Digital Electronics: Binary is the language spoken by all digital devices, from smartphones to smart fridges!
• Internet: Ever wondered how data travels across the vast realms of the internet? Yep, you guessed it, Binary!
Conversion Process Overview
Alright, now that we’ve got a grasp of Binary, let’s venture into the thrilling territory of converting Binary numbers to Decimal ones. 🧐
Steps to Convert Binary to Decimal
Converting Binary to Decimal might sound intimidating at first, but fear not! I’ve got your back with a straightforward conversion process and some nifty tips to make it a breeze.
Binary to Decimal Conversion Formula
The conversion formula is simpler than baking a cake (well, almost): you just need to multiply each Binary digit by 2 raised to the power of its position and sum them up!
Examples of Binary to Decimal Conversion
Enough with the theory, let’s get our hands dirty with some practical examples. I’ll guide you step by step through the conversion process with some sample numbers. Get ready to flex those mental
Challenges in Conversion
Ah, navigating the world of Binary to Decimal conversion isn’t all rainbows and unicorns. There are common pitfalls to avoid and misconceptions to unravel. But worry not, I’m here to guide you
through the maze!
Common Mistakes to Avoid
Let’s face it, we’ve all made blunders while converting Binary to Decimal. But fret not, I’ll reveal the most common mistakes so you can steer clear of them like a pro!
Tips for Accurate Conversion
Mastering the art of Binary to Decimal conversion requires finesse and a sprinkle of magic. Here are some nifty tips and strategies to enhance your conversion skills and impress your peers!
Practice Exercises
Time to put your skills to the test with some hands-on practice exercises. Don’t worry; I’ve got interactive exercises lined up just for you to sharpen those Binary to Decimal conversion skills!
Binary to Decimal Conversion Practice Questions
Get ready for a brain workout with these practice questions tailored to challenge and refine your Binary to Decimal conversion prowess!
Solutions and Explanations
Eager to check your answers? Look no further! I’ve provided detailed solutions and explanations for each practice question to help you ace the art of Binary to Decimal conversion.
Advanced Concepts
Ready to level up your Binary game? Let’s delve into some advanced concepts like Hexadecimal Conversion and Binary Fractions. It’s time to take your Binary knowledge to the next dimension!
Hexadecimal Conversion
Hexadecimal, the cool cousin of Binary! 🎉 We’ll explore how converting to Hexadecimal compares to converting to Decimal, giving you a broader perspective on number systems.
Binary Fractions
What’s more puzzling than whole Binary numbers? Binary fractions, of course! I’ll guide you through the fascinating world of converting fractional Binary numbers to Decimal with ease.
Phew, we covered a lot about Binary Basics and the intriguing process of converting Binary to Decimal. I hope this journey filled with 0s and 1s has sparked a newfound love for the enchanting world
of Binary! Until next time, keep coding and converting like a Binary Rockstar! 🚀
Overall Reflection: Thank you all for joining me on this exhilarating Binary adventure! Remember, the magic of Binary is all around us, from the devices we use to the internet we surf. Embrace the
Binary brilliance, and may your Decimal conversions always be spot on! Stay curious and keep exploring. Catch you later, fellow Binary buffs! 😄
Program Code – Binary Basics: Conversion from Binary to Decimal
def binary_to_decimal(binary_str):
Converts a binary number (as a string) to its decimal equivalent.
:param binary_str: A string representing the binary number.
:return: The decimal equivalent of the binary number.
# Initialize a variable to keep track of the decimal equivalent
decimal_number = 0
# Reverse the binary string to simplify processing
binary_str = binary_str[::-1]
# Iterate over each character in the binary string
for index, digit in enumerate(binary_str):
# Convert the character to an integer and multiply it by 2 raised to its position index
# Add the result to the decimal_number
decimal_number += int(digit) * (2 ** index)
return decimal_number
# Example usage
binary_number = '1101'
decimal_result = binary_to_decimal(binary_number)
print(f'The decimal equivalent of binary {binary_number} is {decimal_result}')
Code Output:
The decimal equivalent of binary 1101 is 13
Code Explanation:
The program defines a function, binary_to_decimal, that transforms a binary string into its decimal equivalent. Its core logic revolves around iterating through each digit of the binary string,
starting from the least significant bit. Here’s how it achieves its objectives:
1. Function Definition: We have a function binary_to_decimal that takes a single parameter, binary_str, which is expected to be a string representation of a binary number.
2. Initialization: Inside the function, decimal_number is initialized to 0. This variable will accumulate the decimal equivalent of the binary number.
3. Reversing the String: The binary string is reversed to make processing easier. This adjustment aids the algorithm by starting to consider digits from the least significant bit.
4. Iteration: Using a for loop, we iterate over each character in the reversed binary string. With enumerate(), we also get the index of each digit, which is crucial for the conversion process.
5. Conversion and Accumulation: In each iteration, the binary digit is converted to an integer and then multiplied by (2^{index}) (since in binary, each digit’s value is a power of 2, starting from
0 for the least significant bit). This product is added to decimal_number, progressively building up the decimal equivalent.
6. Returning the Result: After traversing all digits, the function returns the accumulated decimal_number, which by now represents the decimal form of the input binary string.
7. Example Usage: We then present a simple use case of the function, converting the binary number ‘1101’ into its decimal counterpart and printing the result. In this example, the binary ‘1101’
correctly translates to 13 in decimal, demonstrating the function’s efficacy.
This program exemplifies the binary to decimal conversion process, employing basic Python constructs to achieve a clear and instructive objective.
Frequently Asked Questions on Conversion from Binary to Decimal
What is binary to decimal conversion?
Binary to decimal conversion is the process of converting a binary number (base-2) to a decimal number (base-10).
How do I convert a binary number to a decimal number?
To convert a binary number to a decimal number, you can use the positional notation system. Each digit in a binary number represents a power of 2. Simply multiply each digit of the binary number by (
2^n ) where n is the position of the digit from the right (starting at 0).
Can you provide an example of converting from binary to decimal?
Sure! For example, let’s convert the binary number 1011 to decimal.
( 1011_2 = (12^3) + (02^2) + (12^1) + (12^0) = 11_{10} )
Are there any shortcuts or tricks for converting binary to decimal?
One quick trick is to start from the right of the binary number and double the previous result, then add the next binary digit. Keep doing this until you’ve accounted for all binary digits.
What if I encounter a large binary number for conversion?
For larger binary numbers, you can use online converters or calculator tools to make the conversion process faster and more accurate.
Why is it important to understand binary to decimal conversion?
Understanding binary to decimal conversion is essential in computer science and programming. Many low-level operations in computers are based on binary numbers, so converting between binary and
decimal is crucial for understanding how computers process information.
Any fun facts about binary numbers and their conversion to decimal?
Absolutely! Did you know that binary numbers are the foundation of all digital systems? Every piece of digital data is represented in binary form in computers. So, mastering the conversion from
binary to decimal opens the door to understanding the language of computers! 🤖
Hope these FAQs shed some light on the fascinating world of binary to decimal conversion! 🚀
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Capitalism (1995)- PC Review and Full Download | Old PC Gaming
Capitalism Plus
Capitalism is deep, multi-faceted, and challenging. There’s a bit of a learning curve, especially if you aren’t familiar with the economic principles at work. But once you get the hang of it,
this sim is a business blast. Capitalism gives you a wide world of business and industry with which to build your empire. Manufacturing, raw materials production, farming, research and development,
advertising, and retailing are all modeled in depth. You can manufacture your own products and materials, import them, or purchase them from a competitor.
Such a diverse mix of tools gives the players plenty of options for clawing their way to the top. You can concentrate on retailing, farming, or manufacturing, or diversify with any combination of
those economic segments. Or you can sit back, light a cigar, and play the stock market, buying and selling your rivals’ stock, maybe even finish off the day with a satisfying hostile takeover. You
can pursue all these strategies, or only one of them. That flexibility is the beauty of Capitalism.
Your competitors are ruthless, wily, and driven — much like yourself. You can change the competition’s ability, of course, and at the highest level, these guys and gals are tough to beat. They
don’t waste any time, either. Take a bathroom break without pausing this sim, and they’ll rip your lungs out.
There’s an interesting range of scenarios. Maybe you’re out to topple The Beverage King, the guy who controls a monopoly on the beer industry. Or you’re a 25-year-old smart-aleck with an MBA,
bankrolled by the family to build an empire in 10 years. Or you’re charged with running the operations of a struggling farm. The scenarios have various objectives and different products. The many
components of Capitalism make these scenarios work. Turn off the stock market, and limit the player to only two or three products, and you’ve still got an interesting game.
This sim requires some work to get into, but there’s an outstanding tutorial to walk you through all the fundamentals. These eight instructional games cover everything from retailing to playing the
stock market. After you work through a tutorial, you can continue to play the game, and you’ll be playing the same game when you begin the next tutorial. When you’ve finished all eight tutorials,
you’re already well into your first career as an entrepreneur.
Capitalism isn’t for everyone. It’s a fairly sophisticated business simulation, a detailed re-creation of production, marketing and retailing. But if the prospect of running a virtual business
empire entices you, this game is worth your look.
System Requirements: Pentium 90 Mhz, 16 MB RAM, DOS
Tags: Free Download Capitalism 2 PC Game Review
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đánh Giá Soccer Thời Điểm Hô | {"url":"https://oldpcgaming.net/capitalism-plus/","timestamp":"2024-11-02T02:05:26Z","content_type":"text/html","content_length":"1049561","record_id":"<urn:uuid:00254738-8b60-43f1-8f40-99654e35f01f>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00882.warc.gz"} |
Efficient Methods for Parameter Estimation of Ordinary and Partial Differential Equation Models of Viral Hepatitis Kinetics.
Abstract: Parameter estimation in mathematical models that are based on differential equations is known to be of fundamental importance. For sophisticated models such as age-structured models that
simulate biological agents, parameter estimation that addresses all cases of data points available presents a formidable challenge and efficiency considerations need to be employed in order for the
method to become practical. In the case of age-structured models of viral hepatitis dynamics under antiviral treatment that deal with partial differential equations, a fully numerical parameter
estimation method was developed that does not require an analytical approximation of the solution to the multiscale model equations, avoiding the necessity to derive the long-term approximation for
each model. However, the method is considerably slow because of precision problems in estimating derivatives with respect to the parameters near their boundary values, making it almost impractical
for general use. In order to overcome this limitation, two steps have been taken that significantly reduce the running time by orders of magnitude and thereby lead to a practical method. First,
constrained optimization is used, letting the user add constraints relating to the boundary values of each parameter before the method is executed. Second, optimization is performed by
derivative-free methods, eliminating the need to evaluate expensive numerical derivative approximations. The newly efficient methods that were developed as a result of the above approach are
described for hepatitis C virus kinetic models during antiviral therapy. Illustrations are provided using a user-friendly simulator that incorporates the efficient methods for both the ordinary and
partial differential equation models. | {"url":"https://discover.luc.edu/works/31661","timestamp":"2024-11-06T05:24:52Z","content_type":"application/xhtml+xml","content_length":"12508","record_id":"<urn:uuid:ba2c0340-ee85-477c-9b13-bb14454291e2>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00528.warc.gz"} |
Convert Barn (b) (Area)
Convert Barn (b)
Direct link to this calculator:
Convert Barn (b) (Area)
1. Choose the right category from the selection list, in this case 'Area'.
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π (pi) are all permitted at this point.
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Utilize the full range of performance for this units calculator
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '936 Barn'. In so doing, either the full name of the unit or its
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results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | {"url":"https://www.convert-measurement-units.com/convert+Barn.php","timestamp":"2024-11-08T00:56:08Z","content_type":"text/html","content_length":"58348","record_id":"<urn:uuid:5db485be-941a-49a5-84ac-8c4851b04359>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00168.warc.gz"} |
Data Science with computer vision
• Statistics
• Data visualization in python
• EDA
• Regression
• Supervised Machine Learning
• Unsupervised Machine Learning
• Ensemble Techniques
• Association Rule
• Recommendation system
• Artificial Neural Network
• CNN
• Introduction to Computer Vision
• Introduction to OpenCV
• Computer Vision Techniques
• Object Detection
• Image Segmentation
• Image colorization with OpenCV
• Working with Video and Video Streams
• Transfer Learning and Fine
• Generative Adversarial Networks
• Autoencoders
• Modern CNN Architectures including Vision Transformers
• Image similarity
• Facial Recognition
• Deep Fake Generation
• Video Classification
• Optical Character Recognition
• Image Captioning
• Assignments for assessment
• Projects
• Internship
Statistical Foundations
In this module, you will learn everything you need to know about all the statistical methods used for decision making in this Data Science course.
• Probability distribution – Binomial, Poisson, and Normal Distribution in Python.
• Bayes’ theorem – Baye’s Theorem is a mathematical formula named after Thomas Bayes, which determines conditional probability. Conditional Probability is the probability of an outcome occurring
predicated on the previously occurred outcome.
• Central limit theorem – This module will teach you how to estimate a normal distribution using the Central Limit Theorem (CLT).
• Hypothesis testing – This module will teach you about Hypothesis Testing in Statistics. One Sample T-Test, Anova and Chi-Square test.
Exploratory Data Analysis (EDA)
This module of 6 months in Data Science courses will teach you all about Exploratory Data Analysis like Pandas, Seaborn, Matplotlib, and Summary Statistics.
• Pandas – Pandas is one of the most widely used Python libraries. Pandas is used to analyze and manipulate data. This module will give you a deep understanding of exploring data sets using Pandas.
• Summary statistics (mean, median, mode, variance, standard deviation) – In this module, you will learn about various statistical formulas and implement them using Python.
• Seaborn – Seaborn is also one of the most widely used Python libraries. Seaborn is a Matplotlib based data visualization library in Python. This module will give you a deep understanding of
exploring data sets using Seaborn.
• Matplotlib – Matplotlib is another widely used Python library. Matplotlib is a library to create statically animated, interactive visualizations. This module will give you a deep understanding of
exploring data sets using Matplotlib.
Regression- Linear Regression
This module will get us comfortable with all the techniques used in Linear and Logistic Regression.
• Multiple linear regression – Multiple Linear Regression is used for predicting one dependent variable using various independent variables.
• Fitted regression lines – A fitted regression line is a mathematical regression equation on a graph for your data.
• AIC, BIC, Model Fitting, Training and Test Data – In this module, you will go through everything you need to know about several models such as AIC, BIC, Model Fitting, Training, and Test Data.
Regression- Logistic Regression
• Introduction to Logistic regression, interpretation, odds ratio – It is a simple classification algorithm to predict the categorical dependent variables with the assistance of independent
• Misclassification, Probability, AUC, R-Square – This module will teach everyone how to work with Misclassification, Probability, AUC, and R-Square.
Supervised Machine Learning
In the next module, you will learn all the Supervised Learning techniques used in Machine Learning.
• CART – CART is a predictive machine learning model that describes the prediction of outcome variable’s values predicated on other values.
• KNN – KNN is one of the most straightforward machine learning algorithms for solving regression and classification problems.
• Decision Trees – Decision Tree is a Supervised Machine Learning algorithm used for both classification and regression problems. It is a hierarchical structure where internal nodes indicate the
dataset features, branches represent the decision rules, and each leaf node indicates the result.
• Naive Bayes – Naive Bayes Algorithm is used to solve classification problems using Baye’s Theorem.
Unsupervised Learning
In the next module, you will learn all the Unsupervised Learning techniques used in Machine Learning.
• Clustering – K-Means & Hierarchical – Clustering is an unsupervised learning technique involving the grouping of data. In this module, you will learn everything you need to know about the method
and its types, like K-means clustering and hierarchical clustering.
• Distance methods – This module will teach you how to work with all the distance methods or measures such as Euclidean, Manhattan, Cosine.
• Features of a Cluster – Labels, Centroids, Inertia – This module will drive you through all the features of a Cluster like Labels, Centroids, and Inertia.
• Eigen vectors and Eigen values – In this module, you will learn how to implement Eigenvectors and Eigenvalues in a matrix.
• Principal component analysis – Principal Component Analysis is a technique to reduce the complexity of a model, like eliminating the number of input variables for a predictive model to avoid
Ensemble Techniques
In this Machine Learning, we discuss supervised standalone models’ shortcomings and learn a few techniques, such as Ensemble techniques, to overcome these shortcomings.
• Bagging & Boosting – Bagging is a meta-algorithm in machine learning used for enhancing the stability and accuracy of machine learning algorithms, which are used in statistical classification and
Boosting is a meta-algorithm in machine learning that converts robust classifiers from several weak classifiers.
• Random Forest – Random Forest comprises several decision trees on the provided dataset’s several subsets. Then, it calculates the average for enhancing the dataset’s predictive accuracy.
• AdaBoost & Gradient boosting – Boosting can be further classified as Gradient boosting and ADA boosting or Adaptive boosting. This module will teach you about Gradient boosting and ADA boosting.
Association Rules Mining & Recommendation Systems
Association rule mining is the data mining process of finding the rules that may govern associations and causal objects between sets of items.
Recommendation engines are a subclass of machine learning which generally deal with ranking or rating products / users. Loosely defined, a recommender system is a system which predicts ratings a user
might give to a specific item. These predictions will then be ranked and returned back to the user.
Understanding to Deep Learning – Single Layer Perceptron
Artificial neural networks, usually simply called neural networks or neural nets, are computing systems inspired by the biological neural networks that constitute animal brains. An ANN is based on a
collection of connected units or nodes called artificial neurons, which loosely model the neurons in a biological brain.
Convolutional Neural Network
A convolutional neural network is a feed-forward neural network that is generally used to analyze visual images by processing data with grid-like topology. It’s also known as a ConvNet. A
convolutional neural network is used to detect and classify objects in an image
• Introduction to Computer Vision
Get a conceptual overview of image classification, object localization, object detection, and image segmentation. Also be able to describe multi-label classification, and distinguish between semantic
segmentation and instance segmentation.
How Image are being Stored & Numpy Introduction
Reading & Writing Images
Understanding Color Spaces
Using Different Color Spaces
Drawing in CV2
Callbacks & Trackbar in CV2
• Computer Vision Techniques
Thresholding – Thresholding is used to simplify visual data for further analysis.
Blurring and Smoothing images – Images may contain lots of noise. There are few techniques through which we can reduce the amount of noise by blurring them.
Color Filtering – When you need information about a specific color, you can take out the color you want.
Edge detection – Edge detection is used to enhance the images and image recognition becomes easier.
Get an overview of some popular object detection models, such as regional-CNN and ResNet-50. You’ll use object detection models that you’ll retrieve from TensorFlow Hub.
Image segmentation using variations of the fully convolutional neural network. With these networks, you can assign class labels to each pixel, and perform much more detailed identification of objects
compared to bounding boxes. You’ll build the fully convolutional neural network, U-Net, and Mask R-CNN to identify and detect objects.
• OpenCV implementations of Neural Style Transfer, YOLOv3, SSDs and a black and white image colorizer. Image colorization is the process of taking an input grayscale (black and white) image and
then producing an output colorized image that represents the semantic colors and tones of the input
• Working with Video and Video Streams – Computer Vision Techniques, are based on image recognition and statistical analysis to perform tasks such as face recognition, detection of certain image
patterns, and computer-human interaction.
• CNNs – Detailed overview of CNN Analysis, Visualizing performance, Advanced CNNs techniques
• Transfer Learning and Fine Tuning – Transfer learning is when a model developed for one task is reused to work on a second task. Fine-tuning is one approach to transfer learning where you change
the model output to fit the new task and train only the output model. In Transfer Learning or Domain Adaptation, we train the model with a dataset.
• Generative Adversarial Networks – A generative adversarial network (GAN) is a machine learning (ML) model in which two neural networks compete with each other to become more accurate in their
predictions. CycleGAN, ArcaneGAN, SuperResolution, StyleGAN
• Autoencoders – An autoencoder is an unsupervised learning technique for neural networks that learns efficient data representations by training the network to ignore signal noise. Autoencoders can
be used for image denoising, image compression, and, in some cases, even generation of image data.
• Modern CNN Architectures including Vision Transformers (ResNets, DenseNets, MobileNET, VGG19, InceptionV3, EfficientNET and ViTs) – The visual transformer divides an image into fixed-size
patches, correctly embeds each of them, and includes positional embedding as an input to the transformer encoder.
• Siamese Networks for image similarity – Image similarity is the measure of how similar two images are. In other words, it quantifies the degree of similarity between intensity patterns in two
• Facial Recognition (Age, Gender, Emotion, Ethnicity) – Face recognition systems use computer algorithms to pick out specific, distinctive details about a person’s face.
• Object Detection with YOLOv5 and v4, EfficientDetect, SSDs, Faster R-CNNs – Object detection is a computer vision technique for locating instances of objects in images or videos. Object detection
algorithms typically leverage machine learning or deep learning to produce meaningful results.
• Deep Fake Generation – Deepfake refers to realistic, but fake images, sounds, and videos generated by artificial intelligence methods.
• Video Classification – Video Classification is the task of producing a label that is relevant to the video given its frames. A good video level classifier is one that not only provides accurate
frame labels, but also best describes the entire video given the features and the annotations of the various frames in the video.
• Optical Character Recognition (OCR) – OCR is a technique for detecting printed or handwritten text characters inside digital images of paper files, such as scanning paper records.
• Image Captioning – Image Captioning is the process of generating textual description of an image. It uses both Natural Language Processing and Computer Vision to generate the captions.
• Assignments for assessment
• Projects
• Internship | {"url":"https://chools.in/data-science-with-computer-vision/","timestamp":"2024-11-13T01:34:16Z","content_type":"text/html","content_length":"116970","record_id":"<urn:uuid:2bab4ad0-841c-422a-b608-715c17de947e>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00576.warc.gz"} |
"There is nothing so dangerous as the pursuit of a rational investment policy in an irrational world." John Maynard Keynes
Lord Keynes was not alone in believing that the pursuit of 'true value' based upon financial fundamentals is a fruitless one in markets where prices often seem to have little to do with value. There
have always been investors in financial markets who have argued that market prices are determined by the perceptions (and misperceptions) of buyers and sellers, and not by anything as prosaic as
cashflows or earnings. I do not disagree with them that investor perceptions matter, but I do disagree with the notion that they are all the matter. It is a fundamental precept of this book that it
is possible to estimate value from financial fundamentals, albeit with error, for most assets, and that the market price cannot deviate from this value, in the long term. From the tulip bulb craze in
Holland in the middle ages to the South Sea Bubble in England in the eighteen hundreds to the stock markets of the present, markets have shown the capacity to correct themselves, often at the expense
of those who believed that the day of reckoning would never come.
In the process of presenting and discussing the various models available for valuation, I have tried to adhere to four basic principles . First, I have attempted to be as comprehensive as possible in
covering the range of valuation models that are available to an analyst doing a valuation, while presenting the common elements in these models and providing a framework that can be used to pick the
right model for any valuation scenario. Second, the models are presented with real world examples, warts and all, so as to capture some of the problems inherent in applying these models. There is the
obvious danger that some of these valuations will appear to be hopelessly wrong in hindsight, but this cost is well worth the benefits. Third, in keeping with my belief that valuation models are
universal and not market-specific, illustrations from markets outside the United States are interspersed through the book. Finally, I have tried to make the book as modular as possible, enabling a
reader to pick and choose sections of the book to read, without a significant loss of continuity.
In applying valuation models to real world examples in this book, I have used the capital asset pricing model (CAPM) as my model for risk, and beta as my measure of risk, throughout this book. I am
well aware of the controversy surrounding the CAPM, and have discussed its limitations as well as alternative models in the chapter on estimating discount rates. There are four reasons for my
dependence on the CAPM in this book. First, the estimation of the cost of equity, which is where I have used the CAPM, is just one component of valuation. The valuation models described in this book
require a cost of equity, and any model that provides one can be used instead of the CAPM, without any loss of generality. Second, the data that is available often determines usage. The betas of both
domestic and foreign firms are estimated by a number of information services, and are easily accessible. I could have attempted to estimate the parameters of an alternative model for the stocks that
I have valued, but that would have diverted me from my primary focus, which was valuation. Third, the CAPM provides a convenient forum for discussing more general issues that are important in
valuation, such as the effects of financial leverage on risk and the relationship between risk and growth opportunities. Finally, in spite of all the criticism of the CAPM, I am not convinced that
alternative models do much better in predicting expected returns, though there is evidence that they do better at explaining past returns.
Outline of the Book
The first chapter of this book examines the general basis for valuation models and the role that valuation plays in different investment philosophies. The second chapter provides an overview to the
three basic approaches to valuation - discounted cashflow valuation, relative valuation and contingent claim valuation. The rest of the book delves into the details of using these models.
Discounted Cashflow Valuation Models
The first section of the book presents different discounted cashflow models to value both equity and the firm. Chapter 3 examines different approaches for dealing with risk and estimating the cost of
equity, and for calculating the weighted average cost of capital. Chapter 4 provides a background on financial statements and cashflow calculation, and distinguishes between cashflow to equity and
cashflow to the firm. Chapter 5 explores the process of estimating future growth in earnings and cashflows, starting off with a discussion of the relationship between past growth and future growth,
and proceeding with an examination of how analysts estimate growth, how accurate their predictions are and how best to use analyst forecasts of growth in valuation. It also discusses the relationship
between growth and financial fundamentals - how good or bad the firm's projects are, how much leverage the firm has and its dividend policy.
Chapter 6 describes the basis dividend discount model and its variants. First, the Gordon Growth Model, which assumes steady state growth, is described and applied. Next, the two-stage and
three-stage dividend discount models are developed and contrasted, and the value of extraordinary growth is separated from the value of assets in place. The information requirements for each model
are summarized, in conjunction with a description of the firms on which these models work best. Finally, the results of past studies on the efficacy of the dividend discount model in estimating value
and finding undervalued and overvalued stocks are presented.
Chapter 7 starts off with a discussion of why free cashflows to equity (FCFE) are different from dividends for most firms. The two-stage and three-stage FCFE discounted cashflow models are described
and applied to high growth firms which do not pay dividends. Chapter 8 examines the alternative of valuing the firm by discounting free cashflows to the firm at the weighted average cost of capital.
The advantages of this approach are discussed together with caveats on its usage.
Chapter 9 is dedicated to the valuation of those firms which do not fit easily into traditional discounted cashflow models. In particular, the problems in valuing cyclical and troubled firms are
discussed, and possible solutions are suggested. Chapter 13 examines the issues associated with valuing firms that are in the process of being restructured, and provides a framework for analyzing the
valuation effects of the many dimensions of restructuring - realigning of assets, changes in financial leverage and shifts in dividend policy. Chapter 14 analyzes the key issues that arise in
takeover valuation - the value of synergy and the value of control. It lays out the sources of synergy and shows how it can be valued explicitly, and also provides the theoretical basis and an
analytical framework for valuing a control premium.
Relative Valuation Models
The section on relative valuation covers three chapters. Chapter 10 discusses the use and misuse of price-earnings (PE) and price-cashflow ratios, beginning with an examination of the determinants of
price-earnings ratios, and continuing with an analysis of why PE ratios change over time and why earnings multiples are different across industries and countries. It also talks about the estimation
of PE ratios for firms using the information in the cross-section and the empirical evidence on the relationship between PE ratios and excess returns. Finally, the use of price-earnings multiples for
pricing initial public offerings is examined, with an application.
Chapter 11 explores the relationship between price and book value, and attempts to clear misconceptions about the relationship. The determinants of Price/Book Value ratios are examined and a
rationale is presented for why some firms sell for less than book value while others sell for more. Finally, there is a discussion of how to use price-book value ratios sensibly in investing.
Chapter 12 examines the price to sales ratio and reasons for differences across firms and industries on this multiple. The price to sales ratios is also a useful tool to use to examine the value of a
brand name and the effects of changes in corporate strategy.
Contingent Claim Valuation Models
The section on contingent claim valuation is presented in two chapters. Chapter 15 develops the basis concepts of option pricing. It describes the payoff diagrams on call and put options and provides
the rationale for option pricing models. The Binomial and the Black-Scholes model are presented and contrasted, and extensions on these models and their limitations are described. Chapter 16 applies
these models in the pricing of a number of contingent claim securities such as warrants, and explores the use of option pricing models in pricing assets which have option-like features such as equity
in a firm, natural resource rights and product patents.
Choosing the right model
The problem in valuation is not that there are not enough models for valuation; it is that there are too many. Consequently, the final chapter, Chapter 17, may be the most important one in this book.
It provides a framework for picking the right model for any occasion, based upon the characteristics of the asset being valued. | {"url":"https://pages.stern.nyu.edu/~adamodar/New_Home_Page/val1.htm","timestamp":"2024-11-14T00:09:43Z","content_type":"text/html","content_length":"12104","record_id":"<urn:uuid:87c60ccb-ae4d-491f-a33d-af567f88e936>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00891.warc.gz"} |
Unit: Equivalent fractions | KS2 Maths | Oak National Academy
Switch to our new maths teaching resources
Slide decks, worksheets, quizzes and lesson planning guidance designed for your classroom.
Lessons (20)
In this lesson, we develop understanding of equivalent fractions through quantity, area and numberline models.
In this lesson, we will develop understanding of equivalent fractions through the contexts of measuring and pouring.
In this lesson, we will identify and use the relationship between the numerator and denominator to identify equivalent fractions.
In this lesson, we will scale the numerator and denominator by the same factor to produce equivalent equations.
In this lesson, we will practise seeing both vertical and horizontal relationships in the context of equivalent fractions.
In this lesson, we will introduce equivalent fractions for non-unit fractions.
In this lesson, we will explore the relationship between numerators and denominators including non-unit fractions.
In this lesson, we will practise using both vertical and horizontal relationships in fractions to find missing numbers.
In this lesson, we will practise solving fraction problems where we are required to identify a numerator or denominator in a given fraction equation.
In this lesson, we will revise the language of 'factor', 'multiple' and 'common factor. '
In this lesson, we introduce simplifying fractions, using fractions that can be simplified into unit fractions.
In this lesson, we will express fractions in their simplest forms using the terms 'common factor' and 'highest common factor'.
In this lesson, we will simplify a fraction when the numerator is not the highest common factor.
In this lesson, we will learn how to check whether a fraction is in its simplest form.
In this lesson, we will explore the purpose and advantages off simplifying fractions.
In this lesson, we will simplify fractions that are greater than 1. We will introduce the term 'improper fraction'.
In this lesson, we will use a second method to simplify fractions that are greater than 1.
In this lesson, we will simplify fractions after multiplying a fraction by a whole number.
In this lesson, we will gain further practise and application of simplifying fractions.
In this lesson, we will revisit and consolidate the key concepts in this unit around fractions, simplifying and identifying equivalence. | {"url":"https://www.thenational.academy/teachers/programmes/maths-primary-ks2-l/units/equivalent-fractions-2654/lessons","timestamp":"2024-11-02T14:07:12Z","content_type":"text/html","content_length":"225444","record_id":"<urn:uuid:9da491d6-f81b-481a-b17e-5664e88b370e>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00120.warc.gz"} |
Tretí derivát dy dx
What is the partial derivative, how do you compute it, and what does it mean?
În limbajul matematic contemporan, nu se mai face referire la cantitățile care variază; derivata este considerată o operație matematică asupra funcțiilor. 1. Derivatives of the Sine, Cosine and
Tangent Functions. by M. Bourne.
Tap for more steps Examples = (for positive x) has inverse =. = ; = = ⋅ = ⋅ = At =, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal
tangent for the square function. = (for real x) has inverse = (for positive ) = ; = ⋅ = ⋅ = = Additional … Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the
largest, most trusted online community for developers to learn, share … In Leibniz’s notation the derivative of f is written as function Y = f(x) as df / dx or dy / dx. These are some steps to find
the derivative of a function f(x) at the point x0: Form the difference quotient Δy/Δx = f(x0+Δx) −f(x0) / Δx; If possible, Simplify the quotient, and cancel Δx Derivative Calculator gives
step-by-step help on finding derivatives. This calculator is in beta.
dx dy u v p streamline u v p + dp u + du v + dv V Along the streamline, we have dy dx = v u or u dy = v dx (3) We multiply the x-momentum equation (1) by dx, use relation (3) to replace vdx by udy,
and combine the u-derivative terms into a du differential. ρu ∂u ∂x dx + ρv ∂u ∂y dx + ∂p ∂x dx = 0 ρu ∂u ∂x dx + ∂u ∂y dy
Derivatives of the Sine, Cosine and Tangent Functions. by M. Bourne.
How to calculate derivative of $ \cos ax$? Do I need any formula for $ \cos ax$? The answer in my exercise book says it is $-a \sin ax$. But I don't know how to come to this result. Could you maybe
[math]\cfrac{\mathrm Expresia de mai sus se poate pronunța fie "dy supra dx", fie "dy la dx". În limbajul matematic contemporan, nu se mai face referire la cantitățile care variază; derivata este
considerată o operație matematică asupra funcțiilor. Learn how to calculate the derivative with the help of examples. The concepts are presented clearly in an easy to understand manner. The
Derivative Calculator lets you calculate derivatives of functions online — for free! Our calculator allows you to check your solutions to calculus exercises. Free derivative calculator -
differentiate functions with all the steps.
As: "the function that gives the slope is equal to 3x". Let's try some examples. Suppose we have the function : y = 4x3 + x2 + 3. FUN‑3.D.1 (EK). Review your implicit differentiation skills and use
them to solve x*y differentiate into (1 (from differentiating the x))* (y) + (x) * (dy/dx (from In calculus, the differential represents the principal part of the change in a function y = f(x) with
holds, where the derivative is represented in the Leibniz notation dy/dx, and this is consistent with regarding the derivative as th In mathematics, the derivative of a function of a real variable
measures the sensitivity to change dx", or "dy over dx". The oral form "dy dx" is often used conversationally, although it may lead to confusion.
Except, this time your DY would be a change in the Y direction. So maybe I should really emphasize here that that DX is a change in the X direction here and that DY is a change in the Y direction.
Free math lessons and math homework help from basic math to algebra, geometry and beyond. Students, teachers, parents, and everyone can find solutions to their math problems instantly. See full list
on tutorial.math.lamar.edu Derivative Rules. The Derivative tells us the slope of a function at any point.. There are rules we can follow to find many derivatives..
Sep 15, 2015 How to calculate derivative of $ \cos ax$? Do I need any formula for $ \cos ax$? The answer in my exercise book says it is $-a \sin ax$. But I don't know how to come to this result.
Could you maybe (The above expression is read as "the derivative of y with respect to x", "dy by dx", or "dy over dx". The oral form " dy dx " is often used conversationally, although it may lead to
confusion.) In Lagrange's notation , the derivative with respect to x of a function f ( x ) is denoted f' ( x ) (read as " f prime of x ") or f x ′( x ) (read as " f prime x of x "), in case of
ambiguity of the variable implied by the differentiation.
For Google Chrome - Press 3 dots on top right, then press the star sign Apr 03, 2018 Free Online Derivative Calculator allows you to solve first order and higher order derivatives, providing
information you need to understand derivative concepts. Feb 12, 2007 Feb 27, 2007 Solution for Find dy/dxin terms of x and y if (x−a)^4+y^4=a^4. Assume that a is a constant. dy/dx= implicit derivatve
problem. y = f(x) dy dx = f′(x) k, any constant 0 x 1 x2 2x x3 3x2 xn, any constant n nxn−1 ex ex ekx kekx lnx = log e x 1 x sinx cosx sinkx kcoskx cosx −sinx coskx −ksinkx tanx = sinx cosx sec2 x
tankx ksec2 kx cosecx = 1 sinx −cosecxcot x secx = 1 cosx secxtanx cotx = cosx sinx … tions dx and dy in the two axis directions. The speed V likewise has projections u and v. x y dx dy u v p
streamline u v p + dp u + du v + dv V Along the streamline, we have dy dx = v u or u dy = v dx (3) We multiply the x-momentum equation (1) by dx, use relation (3) to replace vdx by udy, and combine
the u-derivative terms into a du The next step is to solve for dy/dx.
The speed V likewise has projections u and v. x y dx dy u v p streamline u v p + dp u + du v + dv V Along the streamline, we have dy dx = v u or u dy = v dx (3) We multiply the x-momentum equation
(1) by dx, use relation (3) to replace vdx by udy, and combine the u-derivative terms into a du The next step is to solve for dy/dx. (After all, this is the thing that we want to compute!) dy: dx =
1: cos(y) = sec(y) This looks like progress, but it is not the answer. Remember, when we differentiate a function of x in terms of x (this is the meaning of the dx in d/dx), we must express our
answer in terms of x. Therefore the question remains. Question: Find The Derivative Of The Function Y = F(x), Where F Is Differentiable At X And Nonnegative.
kosti těla v knižní hudběrisk ios aplikacehistorie cen akcií zelkolik hodin banka štědrý den zavřelaprojekt x
dx: arccot 2x = −2 4x 2 + 1 * The remaining derivatives come up rarely in calculus. Nevertheless, here are the proofs. The derivative of y = arcsec x. Again,
The chain rule for derivatives is [math]\\frac{dy}{dx} = \\frac{dy}{du}\\cdot \\frac{du}{dx} [/math] This basically means the derivative of a composite function is the derivative of the outer
function with the original argument multiplied by the derivative of the inner function. Dec 17, 2016 Implicit differentiation helps us find dy/dx even for relationships like that. This is done using
the chain rule, and viewing y as an implicit function of x. For example, according to the chain rule, the derivative of y² would be 2y⋅ (dy/dx). Created by Sal Khan. Sep 15, 2015 How to calculate
derivative of $ \cos ax$? Do I need any formula for $ \cos ax$?
Chain Rule 連鎖律 dx du du df uf dx d. ×. = )(. 3. 2. )1. 2()( te. Differecia. −. = = x xfy dx xd du du dx du du df uf dx d uuf x u. )1. 2(. )( )( )1. 2(. 2. 3. 3. 2. −. ×. = ×= =.
În limbajul matematic contemporan, nu se mai face referire la cantitățile care variază; derivata este considerată o operație matematică asupra funcțiilor.
Again, Examples = (for positive x) has inverse =. | {"url":"https://hurmanblirrikbkdp.firebaseapp.com/54354/75815.html","timestamp":"2024-11-13T09:38:12Z","content_type":"text/html","content_length":"18587","record_id":"<urn:uuid:4196932e-f73d-48cd-a6da-ae4e30153e83>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00710.warc.gz"} |
CBSE Class 7 Maths Question Papers - PDF
CBSE Class 7 Maths Question Papers – Session 2023
Hello Everyone. Thanks for reading this article. We are hereby discussing the Maths Question Papers for Class 7. Question Papers and Previous Year Papers are very much crucial particularly when
preparing for the school final exams. Middle classes from 6 to 10 students needs a lot of genuine practice of all the Maths concept. And the best way to achieve the same is through solving as many
Question Papers and Previous Year Papers as possible. In this article, you can get the CBSE Class 7 Mathematics Previous year Papers in pdf and previous year (last 10 year) CBSE Question Papers for
Class 7 Mathematics in PDF with solutions which are absolutely free.
It becomes very imperative for Students to download and practice this material in order to attain high scoring marks in exams. As already mentioned above please note that the best way of preparing
for Class 7 Mathematics exams is by solving last year question papers of CBSE board. This will help every student to prepare properly for Mathematics and score better marks in CBSE NCERT KVS exam
papers for Class 7 Mathematics and help you if you are appearing for exams as you will understand the pattern of the paper and the marking scheme. Practicing Question Papers for Class 7 Mathematics
will help any student in attaining advantage over other students as they will understand the type of Mathematics questions and expected answers. On ribblu.com one may find database of both Unsolved
and Solved Class 7 Mathematics question papers available for free download . All Question papers have been prepared by various schools as per latest 2023 academic syllabus of Class 7 Mathematics and
based on latest CBSE Class 7 Mathematics Question Papers blueprints and chapter weightage.
CBSE Class 7 Mathematics Question Papers PDF Download
CBSE Class 7 Maths Question Paper – Set E
CBSE Class 7 Maths Question Paper – Set D
CBSE Class 7 Maths Question Paper – Set C
CBSE Class 7 Maths Question Paper – Set B
CBSE Class 7 Maths Question Paper – Set A
CBSE Class 7 Maths Question Paper 2019 – DPS
CBSE Class 7 Maths Question Paper 2018 – DPS
CBSE Class 7 Maths Question Paper – KV
CBSE Class 7 Maths Question Paper – DPS
CBSE Class 7 Maths Question Paper Term 2 – KV
CBSE Class 7 Maths Question Paper SA2 – KV
CBSE Class 7 Mathematics Question Paper 2021- Billabong
Class 7 Maths Model Question Paper FA1 – KV
CBSE Class 7 Maths Question Paper Annual Exam 2020 – DPS
Class 7 Maths Model Question Paper SA2-(2) – DPS
Class 7 Maths Model Question Paper SA2(1) – DPS
Class 7 Maths Model Question Paper SA1 (2) – DPS
Class 7 Maths Solved Model Question Paper SA2 – DPS
Class 7 Maths Model Question Paper SA1 (1) – DPS
Class 7 Maths Solved Model Question Paper SA1 – DPS
CBSE Class 7 Maths Question Paper SA1 – Sindhi High School
CBSE Class 7 Maths Question Paper SA2 – 2014
Class 7 Maths Question Paper Second Term 2018 – G.D. GOENKA
Class 7 Maths Practice Test Question Paper – Practice Geometry
Class 7 Maths MCQ Paper SA2 set 3 – Apeejay
Class 7 Maths Multiple choice question Paper SA2 set 2 – Apeejay
Class 7 Maths Multiple choice question Paper SA2 set 1 – Apeejay
CBSE Class 7 Maths Question Paper 2017 SA2 – Apeejay
CBSE Class 7 Maths Question Paper SA1 Term
Class 7 Maths Question Paper 2 ( Periodic Test 2017 ) – Apeejay
Sample Question Paper for CBSE Class 7 Maths SA1 – Apeejay
Class 7 Maths Multiple Choice Questions paper SA1 – Apeejay
CBSE Class 7 Maths Question Paper SA2
Maths Sample Question Paper for CBSE Class 7 SA1
Maths – Formative Assessment (1) – Class 7 St. Xaviers Sr. Sec School
Class 7 Maths Question Paper 2014
Class 7 Maths Question Paper – 10
CBSE Class 7 Maths Question Paper
Class 7 Maths Question Paper – 12
Class 7 Maths Term Paper (FA1 2016) VBPS Noida
CBSE Class 7 Maths Question Paper – VBPS Noida
Construction Class 7 Maths Worksheet
Class 7 Maths Question Paper – FA4 (CCE)
Class 7 Maths Weekly Test Question Paper – Set B
CBSE Class 7 Maths FA4 Question Paper 2016
Questions on Exponents and Powers for CBSE Class 7
CBSE Class 7 Maths Question Paper FA2 – 2016
Class 7 Maths Revision Worksheet – SA2
Important Questions with answers asked in Previous Question Papers of Class 6 Maths
Q.1 If the perimeter of a rectangle is 26 cm and its length is 7 cm, find its breadth.
Let the breadth of the rectangle be b cm. Perimeter of the rectangle = 2 (length + breadth) 26 = 2 (7 + b) 13 = 7 + b Breadth = 13 – 7 Breadth = 6 cm
Therefore, the breadth of the rectangle is 6 cm.
Q.2 Find the value of ‘k’ if 5k – 2 = 23.
5k – 2 = 23 5k = 23 + 2 5k = 25 k = 5
Therefore, the value of ‘k’ is 5.
Q.3 If a sum of Rs. 10,000 amounts to Rs. 12,100 in 2 years at simple interest, find the rate of interest.
Let the rate of interest be ‘r’ % per annum. Simple interest = (Principal × Rate × Time) / 100 12100 – 10000 = (10000 × r × 2) / 100 2100 = 200r r = 2100/200 r = 10.5
Therefore, the rate of interest is 10.5 % per annum.
Q.4 If the height of a cone is 8 cm and its radius is 3 cm, find its curved surface area.
Curved surface area of a cone = πrl, where r is the radius and l is the slant height. l² = r² + h² l² = 3² + 8² l = √(9 + 64) l = √73 cm
Curved surface area = π × 3 × √73 Curved surface area ≈ 68.73 cm²
Therefore, the curved surface area of the cone is approximately 68.73 cm².
Q.5 If a car covers a distance of 252 km in 4 hours, find its speed.
Speed = Distance/Time Speed = 252/4 Speed = 63 km/h
Therefore, the speed of the car is 63 km/h.
Q.6 The cost of 5 notebooks and 7 pencils is Rs. 50. If the cost of 7 notebooks and 10 pencils is Rs. 80, find the cost of each notebook and each pencil.
Answer: Let the cost of one notebook be x and the cost of one pencil be y. From the first equation, we get: 5x + 7y = 50 From the second equation, we get: 7x + 10y = 80 Solving these equations
simultaneously, we get: x = 10 and y = 2 Therefore, the cost of each notebook is Rs. 10 and the cost of each pencil is Rs. 2.
Q.7 Find the area of a circle whose diameter is 14 cm. (Take π = 22/7)
Answer: Given, diameter of the circle = 14 cm Radius of the circle = Diameter/2 = 14/2 = 7 cm Area of the circle = πr^2 = (22/7) x 7^2 = 154 sq. cm Therefore, the area of the circle is 154 sq. cm.
Q.8 If 2x + 3y = 15 and 4x – 5y = 10, find the value of x and y.
Answer: Multiplying the first equation by 2, we get: 4x + 6y = 30 Subtracting the second equation from the above equation, we get: 4x + 6y – (4x – 5y) = 30 – 10 Simplifying, we get: 11y = 20 y = 20/
11 Substituting the value of y in the first equation, we get: 2x + 3(20/11) = 15 Simplifying, we get: 2x = (165-60)/11 2x = 105/11 x = 105/22 Therefore, the value of x is 105/22 and the value of y is
Q.9 Find the sum of the first 25 odd natural numbers.
Answer: The first odd natural number is 1.
The second odd natural number is 3.
The third odd natural number is 5. And so on.
Therefore, the first 25 odd natural numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49
The sum of the first n odd natural numbers is n^2. Therefore, the sum of the first 25 odd natural numbers is 25^2 = 625.
Therefore, the sum of the first 25 odd natural numbers is 625.
Q.10 If the LCM of two numbers is 120 and their HCF is 5, find the two numbers.
Answer: Let the two numbers be x and y. Their LCM is 120. Their HCF is 5. We know that LCM x HCF = product of two numbers. Therefore, 120 x 5 = xy Simplifying, we get: xy = 600 Now, we need to find
two numbers whose product is 600 and HCF is 5. The factors of 600 are: 1, 2, 3, 4, 5, 6, 8
Example of Previous Year Question Paper
CBSE Class 7 Maths Syllabus 2023-24
Chapter-1 Integers
Introduction, recall, properties of addition and subtraction of integers, multiplication of integers, properties of multiplication of integers , division of integers, properties of integers.
Chapter-2 Fraction and Decimals
Introduction, How well you have learnt about fractions? (proper fraction, improper fraction, mixed fraction, equivalent fraction) Multiplication of fractions, Division of fractions, How well you have
learnt about decimal numbers? Multiplication of decimal numbers, division of decimal numbers.
Chapter-3 Data Handling
Introduction, collecting data, organisation of data, representative values, arithmetic mean, mode, median, use of bar graphs with a different purpose, chance and probability.
Chapter-4 Simple Equation
Introduction, A mind reading game, setting up of an equation, review of what we know, what equation is? Solving an equation, more equations, from solution to equation, application of simple equations
to practical situations
Chapter–5 Lines and Angles
Introduction, related angles, pair of lines, checking for parallel lines.
Chapter-6 The triangle and its properties
Introduction, Medians of a triangle, altitudes of a triangle, Exterior angle of a triangle and its properties, two special triangle and its property, angle sum property of a triangle, equilateral and
isosceles triangle, sum of the lengths of two sides of a triangle, right angled triangles and Pythagoras property.
Chapter-7 Congruence of Triangles
Introduction, Congruence of plane figures, Congruence among line segments, congruence of angles, congruence of triangles, criteria for congruence of triangles, congruence among right angled
Chapter-8 Comparing Quantities
Introduction, Equivalent ratios, percentage another way of comparing quantities, use of percentages, prices related to an item or buying and selling, charge given on borrowed money or simple
Chapter-10 Practical Geometry
Introduction, construction of a line parallel to a given line through a point not on the line. Construction of triangles, Constructing a triangle when the lengths of its three sides are known (SSS
Criterion), constructing a triangle when the lengths of two sides and the measure of the angle between them are known (SAS criterion), constructing a triangle when the measure of two of its angles
and the length of the side included between them is given (ASA criterion), constructing a right-angled triangle when the length of one leg and its hypotenuse are given (RHS criterion).
Chapter-11 Perimeter and Area
Introduction, area of square, rectangles, and parallelogram, area of a triangle, circle, conversion of units, application.
Chapter -12 Algebraic Expressions
Introduction, How are expressions formed, terms of an Expression, like and unlike terms, Monomials, Binomials, Trinomials and Polynomials, addition and subtraction of algebraic expressions, finding
the value of an expression, using algebraic expression- formulas and rules.
Chapter-13 Exponents and Powers
Introduction, exponents, laws of exponents, miscellaneous examples using the laws of exponents, decimal number system, expressing large numbers in the standard form.
Class 7 Maths Sample Paper
Class 7 Math Sample Paper 2 ( Annual exam) – Prince Public School
Sample Paper for CBSE Class 7 Maths – Half Yearly
CBSE Class 7 Maths Sample Paper Half Yearly Exam
Class 7 Maths Sample Paper ( First Term Exam 2018 ) – VBPS
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Class 7 Maths Maths Sample Paper ( Mensuration )
Attempting Previous year question papers for Class 7 Maths is the perfect strategy to prepare for final exams of Class 7. Getting full score in Grade 7 examinations is a wish of all the student as it
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Key Advantages in Attempting Last year Question Papers of Maths for Class 7:
a) Increases the problem solving skills for Grade 7 Maths
b) Exam has to be attempted in limited time period. usually 2-3 hours. So while attempting question papers one Improves speed and accuracy.
c) Maths papers needs a strategy while attempting. How to approach Maths Question Paper in examination hall is key aspect and while prating previous year papers it helps the students to develop good
strategy to approach and solve the Standard 7 Maths question paper
d) Students gets to know their grey / weak areas in advance while practicing so that they can spend more time to resolve mistakes and clear concepts of various chapters given in NCERT for Class 7
e) Helps to revise the entire 2021 syllabus and familiarises with important questions
CBSE Sample Question Paper for Class 7 Maths
Maths for Class 7 is considered to be one of the most important and immensely scoring subjects. And the best way to prepare apart from completing NCERT and reference books is solving CBSE Sample
Papers. Here on Ribblu one can get immense collection of Sample Question Papers for Class 7 Maths in PDF format for free.
CBSE Class 7 Maths Test Papers, Unit Tests and Periodic Assessment Exams
Maths Test Papers for Class 7 are the collection of past unit test papers and periodic assessment class test papers which helps students to prepare for upcoming monthly quarterly half yearly and
annual exams. Here on Ribblu one can get a lot of Test Papers for Class 7 Maths in PDF format for free
CBSE Class 7 Maths Periodic Test 2020 – Amrita vidyalayam
CBSE Class 7 Maths Test Paper – (Unit Test 1)
Maths Practice Questions for CBSE class 7 Visualising Solid Shapes
Maths Practice Questions for CBSE class 7 Symmetry
Maths Practice Questions for CBSE class 7 Rational Numbers
Maths Practice Questions for CBSE class 7 Perimeter and Area
Maths Practice Questions for CBSE class 7 Exponents and Powers
Maths Practice Questions for CBSE class 7 Algebraic Expressions
Maths Practice Questions for CBSE class 7 Triangles and Its Properties
Maths Practice Questions for CBSE class 7 Simple Equations
Maths Practice Questions for CBSE Class 7 Lines and Angles
Maths Practice Questions for CBSE Class 7 Integers
Maths Practice Questions for CBSE class 7 Fractions and Decimals
Maths Practice Questions for CBSE class 7 Data Handling
Maths Practice Questions for CBSE class 7 Congruence of Triangles
Maths Practice Questions for CBSE Class 7 Comparing Quantities
CBSE Class 7 Maths Test Paper – Mayoor School
CBSE Class 7 Maths Test Paper – (Unit Test 2)
CBSE Class 7 Maths Test Paper – (Unit Test 3)
CBSE Worksheets for Class 7 Maths
CBSE Worksheets for Class 7 contains all the important questions on English, Biology, Maths, Hindi, Science, EVS, Sanskrit, Social Science, General Knowledge, Computers, French, and Environmental
Studies as per CBSE syllabus. These Worksheets help students to practice, improve knowledge as they are an effective tool in understanding the subject in totality. Worksheets help students in
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Question Papers of Other Subjects of Class 7
CBSE Question Papers of Class 7 Science
CBSE Question Papers of Class 7 English
CBSE Question Papers of Class 7 Social Science
CBSE Question Papers of Class 7 Computer Science
CBSE Question Papers of Class 7 Hindi
CBSE Question Papers of Class 7 Sanskrit
CBSE Question Papers of Class 7 PSA (Problem Solving Assessment)
CBSE Question Papers of Class 7 Japanese
CBSE Question Papers of Class 7 German
CBSE Question Papers of Class 7 French
CBSE Question Papers of Class 7 General Knowledge (G.K.)
Frequently Asked Questions (FAQ)
Q. How many total subjects are there in 7th Class ?
1. Hindi, English, Mathematics, Environmental Science, Sanskrit, Computer.
Q. How can I become topper Student in Class 7th?
A. There are no shortcuts of becoming the class topper. The only way to be the best is to discipline, study hard and learn the tricks of the teachers. If you want to become the top of the class then
you need to make sure that you are organized and disciplined. Also, homework tasks are always set for a reason which may mean that if you don’t do your homework, you might suffer during the final
Q. In Class 7 Which are the important portions of Social Science.
1. Social and Political Life – II (Civics)
2. Our Pasts – II (History)
3. Our Environment (Geography
Q. Does the Syllabus prescribed by CBSE apply all over India ?
A. Yes, more or less the CBSE syllabus is the same for all schools in India and there are little or no state-wise differences in CBSE syllabus since a majority of schools follow NCERT books. In other
words, CBSE syllabus, on the whole, is the same for all states in India.
Q. What are the few best methods to Prepare English Subject?
A. The student must focus on grammar and writing, as these are sections in which students tend to lose the maximum number of marks. Practice writing sections of previous papers, along with grammar
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A. Scoring good marks in Social science is not a big task but students have to practice consistently. The concept matter will be a lot clearer to you once you understand – for instance – how the
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My software team recently finished a round of code reviews for some of our motor controller code. I learned a lot from the experience, most notably why you would want to have code reviews in the
first place.
My background is originally from the medical device industry. In the United States, software in medical devices gets a lot of scrutiny from the Food and Drug Administration, and for good reason; it’s
a place for complexity to hide latent bugs. (Can you say “
Other articles in this series:
Today we’re going to take a break from my usual focus on signal processing or numerical algorithms, and focus on...
Jason Sachs ●
June 26, 2015
Oh Robot! My Robot! You’ve broken off your nose! Your head is spinning round and round, your eye no longer glows, Each program after program tapped your golden memory, You used to have 12K, now there
is none that I can see, Under smoldering antennae, Over long forgotten feet, My sister used your last part: The chip she tried to eat.
Oh Robot, My Robot, the remote controls—they call, The call—for...
Earlier articles:
We have come to the last part of the Important Programming Concepts series, on abstraction. I thought I might also talk about why there isn’t a Part VII, but decided it would distract from this
article — so if you want to know the reason, along with what’s next,
Other articles in this series:
Last time we talked about a low-pass filter, and we saw that a one-line...
Ever since I started using IPython Notebooks to write these articles, I’ve been wanting to publish them in a form such that you can freely use my Python code. One of you (maredsous10) wanted this
access as well.
Well, I finally bit the bullet and automated a script that will extract the Python code and create standalone notebooks, that are available publicly under the Apache license on my bitbucket account:
This also means they...
Other articles in this series:
I’m writing this article in a room with a bunch of other people talking, and while sometimes I wish they would just SHUT UP, it would be...
This blog needs some short posts to balance out the long ones, so I thought I’d cover some of the algorithms I’ve used over the years. Like the Euclidean algorithm and Extended Euclidean algorithm
and Newton’s method — except those you should know already, and if not, you should be locked in a room until you do. Someday one of them may save your life. Well, you never know.
Other articles in this series:
Jason Sachs ●
February 15, 2015
●5 comments
I was looking for a good reference for some ADC-driving circuits, and ran across this diagram in Walt Jung’s Op-Amp Applications Handbook:
And I smiled to myself, because I immediately remembered a circuit I hadn’t used for years. Years! But it’s something you should file away in your bag of tricks.
Take a look at the RC-RC circuit formed by R1, R2, C1, and C2. It’s basically a stacked RC low-pass filter. The question is, why are there two capacitors?
Warning: In the interest of maintaining a coherent stream of consciousness, I’m lowering the setting on my profanity filter for this post. Just wanted to let you know ahead of time.
I’ve been a user of Stack Overflow since December of 2008. And I say “user” both in the software sense, and in the drug-addict sense. I’m Jason S, user #44330, and I’m a programming addict. (Hi,
Jason S.) The Gravatar, in case you were wondering, is a screen...
Last time, we continued a discussion about error detection and correction by covering Reed-Solomon encoding. I was going to move on to another topic, but then there was this post on Reddit asking how
to determine unknown CRC parameters:
I am seeking to reverse engineer an 8-bit CRC. I don’t know the generator code that’s used, but can lay my hands on any number of output sequences given an input sequence.
This is something I call the “unknown oracle”...
Jason Sachs ●
October 18, 2017
●1 comment
The last two articles were on discrete logarithms in finite fields — in practical terms, how to take the state \( S \) of an LFSR and its characteristic polynomial \( p(x) \) and figure out how many
shift steps are required to go from the state 000...001 to \( S \). If we consider \( S \) as a polynomial bit vector such that \( S = x^k \bmod p(x) \), then this is equivalent to the task of
figuring out \( k \) from \( S \) and \( p(x) \).
This time we’re tackling something...
Other articles in this series:
This article is mainly an excuse to scribble down some cryptic-looking mathematics — Don’t panic! Close your eyes and scroll down if you feel nauseous — and...
Jason Sachs ●
December 29, 2017
●1 comment
Last time we looked at the use of LFSRs for pseudorandom number generation, or PRNG, and saw two things:
• the use of LFSR state for PRNG has undesirable serial correlation and frequency-domain properties
• the use of single bits of LFSR output has good frequency-domain properties, and its autocorrelation values are so close to zero that they are actually better than a statistically random bit
The unusually-good correlation properties...
Jason Sachs ●
November 11, 2014
●2 comments
Other articles in this series:
Today’s topic is the singleton. This article is unique (pun intended) in that unlike the others in this series, I tried to figure out a word to use that would be a positive concept to encourage, as
an alternative to singletons, but
Other articles in this series:
Today we’re going to take a break from my usual focus on signal processing or numerical algorithms, and focus on...
Jason Sachs ●
November 13, 2017
●1 comment
The last four articles were on algorithms used to compute with finite fields and shift registers:
Today we’re going to come back down to earth and show how to implement LFSR updates on a microcontroller. We’ll also talk a little bit about something called “idiomatic C” and a neat online tool for
experimenting with the C compiler.
Jason Sachs ●
December 31, 2021
●5 comments
So by now I’m sure you’ve heard about the semiconductor shortage of 2021. For a few complicated reasons, demand is greater than supply, and not everybody who wants to buy integrated circuits can do
so. Today we’re going to try to answer some hard questions:
• Why are we in the middle of a semiconductor shortage?
• Why is it taking so long to get my [insert part number here]?
• Did this shortage suddenly sneak up on everybody? If not, what were the signs, and why...
Все счастли́вые се́мьи похо́жи друг на дру́га, ка́ждая несчастли́вая семья́ несчастли́ва по-сво́ему.
— Лев Николаевич Толстой, Анна Каренина
Happy families are all alike; every unhappy family is unhappy in its own way.
— Lev Nicholaevich Tolstoy, Anna Karenina
I was going to write an article about second-order systems, but then realized that it would be...
Jason Sachs ●
September 16, 2017
●4 comments
Last time we talked about the multiplicative inverse in finite fields, which is rather boring and mundane, and has an easy solution with Blankinship’s algorithm.
Discrete logarithms, on the other hand, are much more interesting, and this article covers only the tip of the iceberg.
What is a Discrete Logarithm, Anyway?
Regular logarithms are something that you’re probably familiar with: let’s say you have some number \( y = b^x \) and you know \( y \) and \( b \) but...
Jason Sachs ●
November 13, 2017
●1 comment
The last four articles were on algorithms used to compute with finite fields and shift registers:
Today we’re going to come back down to earth and show how to implement LFSR updates on a microcontroller. We’ll also talk a little bit about something called “idiomatic C” and a neat online tool for
experimenting with the C compiler.
Jason Sachs ●
July 28, 2013
In my last post, I talked about ripple current in inductive loads.
One of the assumptions we made was that the DC link was, in fact, a DC voltage source. In reality that's an approximation; no DC voltage source is perfect, and current flow will alter the DC link
voltage. To analyze this, we need to go back and look at how much current actually is being drawn from the DC link. Below is an example. This is the same kind of graph as last time, except we added
Today’s short and tangential note is an account of how I dug myself out of Documentation Despair. I’ve been working on some block diagrams. You know, this sort of thing, to describe feedback control
And I had a problem. How do I draw diagrams like this?
I don’t have Visio and I don’t like Visio. I used to like Visio. But then it got Microsofted.
I can use MATLAB and Simulink, which are great for drawing block diagrams. Normally you use them to create a...
Jason Sachs ●
December 31, 2011
This article was inspired by a recent post on reddit asking for help on Thévenin and Norton equivalent circuits.
(With apologies to Mr. Thévenin, the rest of the e's that follow will remain unaccented.)
I still remember my introductory circuits class on the subject, roughly as follows:
(NOTE: Do not get scared of what you see in the rest of this section. We're going to point out the traditional approach for teaching linear equivalent circuits first. If you have...
Our team had another code review recently. I looked at one of the files, and bolted upright in horror when I saw a function that looked sort of like this:
void some_function(SOMEDATA_T *psomedata) { asm volatile("push CORCON"); CORCON = 0x00E2; do_some_other_stuff(psomedata); asm volatile("pop CORCON"); }
There is a serious bug here — do you see what it is?
Last year I wrote about some of the key characteristics of oscilloscopes that are important to me for working with embedded microcontrollers. In that blog entry I rated the Agilent MSOX3024A
4-channel 16-digital-input oscilloscope highly.
Since then I have moved to a different career, and I am again on the lookout for an oscilloscope. I still consider the Agilent MSOX3024A the best choice for a...
Jason Sachs ●
April 30, 2012
Somewhere, along with unicorns and the Loch Ness Monster, lies a small colony of ideal op-amps. Op-amp is short for operational amplifier, and we start our education on them by learning about these
mythical beasts, which have the following properties:
• Infinite gain
• Infinite input impedance
• Zero output impedance
And on top of it all, they will do whatever it takes to change their output in order to make their two inputs equal.
But they don't exist. Real op-amps have...
I recently returned from a visit to my grandmother, who lives in an assisted living community, and got to observe both her and my frustration first-hand with a new TV. This was a Vizio flatscreen TV
that was fairly easy to set up, and the picture quality was good. But here's what the remote control looks like:
You will note:
• the small lettering (the number buttons are just under 1/4 inch in diameter)
• a typeface chosen for marketing purposes (matching Vizio's "futuristic" corporate...
Earlier articles:
We have come to the last part of the Important Programming Concepts series, on abstraction. I thought I might also talk about why there isn’t a Part VII, but decided it would distract from this
article — so if you want to know the reason, along with what’s next,
Jason Sachs ●
December 14, 2014
●4 comments
It’s been a few months since I’ve rolled up my sleeves here and dug into some good old circuit design issues. I started out with circuit design articles, and I’ve missed it.
Today’s topic will be showing you some tricks for how to get more performance out of an optoisolator. These devices — and I’m tempted to be lazy and call them “optos”, but that sounds more like a
cereal with Greek yogurt-covered raisins — are essentially just an LED... | {"url":"https://www.embeddedrelated.com/blogs-7/nf/Jason_Sachs/all.php","timestamp":"2024-11-05T06:17:10Z","content_type":"text/html","content_length":"74312","record_id":"<urn:uuid:b7cfcbca-6c09-4daf-af30-6c03200ab3e7>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00503.warc.gz"} |
Pole in the Barn Paradox
According to the special theory of relativity an object that is in motion relative to a givenframe of reference contracts – becomes shorter in length – along the direction in which it is moving. But
this seems to lead to a paradox. Consider a pole that is 15 meters long when at rest, and consider a barn that is 10 meters long when at rest. Suppose that the barn has a door at each end. Clearly
the pole should not fit inside the barn with both doors closed.
Now suppose you can get a friend to run through the barn carrying the pole at (4/5)c – i.e. at 4/5 the speed of light. Then, from the barns rest frame, the barn is still 10 meters long, but the pole
is only 9 meters long (i.e. the length of the pole in the barns rest frame = 10m x sqrt(1-(v/c)^2) = 15 x (3/5) = 9m). So, for a brief moment, as the pole passes through the barn, the pole fits
completely inside the barn with both doors closed.
However, from your friend’s frame of reference, where the pole is at rest and barn is moving toward him at (4/5)c, the pole remains 15m long. But the barn becomes shorter. Its length becomes 6m (i.e.
the length of the barn in the poles rest frame = 10m x sqrt(1-(v/c)^2) = 10m x (3/5) = 6m). So, as it passes through there is no way that the pole should fit completely inside the barn with both
doors closed.
But the pole either fits completely inside the barn (for a brief moment as it passes through), or it doesn’t. This is the paradox.
So from the point of view of the pole, its length never changes, it is always 15 meters long, and the length of the barn is ever decreasing so that from the poles frame the pole could never fit
within the barn all at once. And, from the point of view of the barn, the pole is ever decreasing so at some point it should be able to fit within both doors of the barn. So the diagram shows that in
the barn frame the pole will fit and in the pole frame it is quite to long to fit. Which one of these is the correct assumption?
The solution to the paradox is in the relativity of simultaneity. That is that what one observer sees as simultaneous is not the same as what another observer sees as simultaneous. “At the same time”
does not apply to those events which are separated by space in special relativity theory. This thought experiment shows that there are circumstances when there is no correct answer. Neither observer
has privileged status or rightness. And even that all included observers can claim to be correct, at least in their own independent reference frames, even if their ordering of events differs from
This article originally written February 25th, 2008 for OU PHIL 3623 - Physics and Cosmology.
1 comment:
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E Math tuition | Sec O Levels | Math Academy
Our E maths tuition classes focuses on practice and guidance from experienced teachers. Our class size is kept small, capped at 9 students per class.
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Technical Mathematics
You may use a calculator throughout this module if needed.
A formula is an equation or set of calculations that takes a number (or numbers) as input, and produces an output. The output is often a number, but it could also be a decision such as yes or no. The
numbers in a formula are usually represented with letters of the alphabet, which are called variables because their values can vary. To evaluate a formula, we substitute a number (or numbers) into
the formula and then perform the steps using the order of operations.
Note: When a number is written directly next to a variable, it indicates multiplication. For example,
The cost, in dollars, of mailing a large envelope weighing ^[1]
1. Find the cost of mailing a
2. Find the cost of mailing a
Radio Cab charges the following rates for a taxi ride: a fixed fee of $ ^[2]
3. Find the cost of a
4. Find the cost of a
5. Find the cost of getting in the taxi, then changing your mind and getting out without riding anywhere.
The number of members a state has in the U.S. House of Representatives can be approximated by the formula ^[3] where ^[4]
Round all answers to the nearest whole number.
6. How many U.S. Representatives does Oregon have?
7. How many U.S. Representatives does Washington have?
8. How many U.S. Representatives does California have?
The number of electoral votes a state has can be approximated by the formula
9. How many electoral votes does Oregon have?
10. How many electoral votes does Washington have?
11. How many electoral votes does California have?
Some formulas require more than one number for the input.
When a patient’s blood pressure is checked, they are usually told two numbers: the systolic blood pressure (SBP) and the diastolic blood pressure (DBP). The mean arterial pressure (MAP) can be
estimated by the following formula:
12. SBP =
13. SBP =
^[5] Determine the measurement of a package with the following dimensions.
14. length
15. length
The next set of exercises involves a formula that gives a yes or no answer.
In Australia, a chicken egg is designated “large” if its mass, in grams, satisfies the following formula: ^[6]
16. Egg 1’s mass is
17. Egg 2’s mass is
18. Egg 3’s mass is
19. Egg 4’s mass is
The Celsius temperature scale is based on the freezing point of water (
Temperature Formulas
20. The temperature on a cool day is
21. Normal body temperature is
22. The FDA recommends that a freezer be set below
23. A package of frozen pancakes from IKEA calls for the oven to be set to | {"url":"https://openoregon.pressbooks.pub/techmath/chapter/module-7-formulas/","timestamp":"2024-11-08T07:51:53Z","content_type":"text/html","content_length":"123829","record_id":"<urn:uuid:85751d97-01c6-4f8f-a900-49f5505dfdef>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00745.warc.gz"} |
propulsion module — GEMSEO 5.3.2 documentation
propulsion module¶
Propulsion discipline for the Sobieski’s SSBJ use case.
class gemseo.problems.sobieski.core.propulsion.SobieskiPropulsion(sobieski_base)[source]
Bases: SobieskiDiscipline
Propulsion discipline for the Sobieski’s SSBJ use case.
sobieski_base (SobieskiBase) – The Sobieski base.
execute(x_shared, y_23, x_3, true_cstr=False, c_3=None)[source]
Compute the fuel consumption, engine weight and engine scale factor.
○ x_shared (ndarray) – The values of the shared design variables, where x_shared[0] is the thickness/chord ratio, x_shared[1] is the altitude, x_shared[2] is the Mach number, x_shared
[3] is the aspect ratio, x_shared[4] is the wing sweep and x_shared[5] is the wing surface area.
○ y_23 (ndarray) – The drag coefficient.
○ x_3 (ndarray) – The throttle.
○ true_cstr (bool) –
If True, return the value of the constraint outputs. Otherwise, return the distance to the corresponding constraint thresholds.
By default it is set to False.
○ c_3 (float | None) – The reference engine weight. If None, use SobieskiBase.constants().
y_3: The outputs of the propulsion analysis:
★ y_3[0]: the specific fuel consumption,
★ y_3[1]: the engine weight,
★ y_3[2]: the engine scale factor,
g_3: The propulsion outputs to be constrained:
★ g_3[0]: the engine scale factor,
★ g_3[1]: the engine temperature,
★ g_3[2]: the throttle setting.
Return type:
The propulsion outputs
linearize(x_shared, y_23, x_3, true_cstr=False, c_3=None)[source]
Derive the fuel consumption, engine weight and engine scale factor.
○ x_shared (ndarray) – The values of the shared design variables, where x_shared[0] is the thickness/chord ratio, x_shared[1] is the altitude, x_shared[2] is the Mach number, x_shared
[3] is the aspect ratio, x_shared[4] is the wing sweep and x_shared[5] is the wing surface area.
○ y_23 (ndarray) – The drag coefficient.
○ x_3 (ndarray) – The throttle.
○ true_cstr (bool) –
If True, return the value of the constraint outputs. Otherwise, return the distance to the corresponding constraint thresholds.
By default it is set to False.
○ c_3 (float | None) – The reference engine weight. If None, use SobieskiBase.constants().
The Jacobian of the discipline.
Return type:
dict[str, dict[str, ndarray]]
ESF_LOWER_LIMIT = 0.5
ESF_UPPER_LIMIT = 1.5
TEMPERATURE_LIMIT = 1.02 | {"url":"https://gemseo.readthedocs.io/en/stable/_modules/gemseo.problems.sobieski.core.propulsion_.html","timestamp":"2024-11-04T16:50:44Z","content_type":"text/html","content_length":"29620","record_id":"<urn:uuid:c088f03b-8ec2-4305-b033-3ceec8be1354>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00383.warc.gz"} |
Understanding the Power of Modular Arithmetic in Programming: A Deep Dive into the Multiplication…
Understanding the Power of Modular Arithmetic in Programming: A Deep Dive into the Multiplication Property
Modular arithmetic is a fundamental concept in mathematics and computer science, with applications spanning cryptography, number theory, and algorithm design. One of its most useful properties is the
multiplication rule, which states that
(𝑎×𝑏)%𝑚= ( (𝑎%𝑚)×(𝑏%𝑚) )%𝑚
This article explores this property in detail, explaining its significance, mathematical foundation, and practical applications in programming.
What is Modular Arithmetic?
Modular arithmetic deals with integers and a special operation called the modulo operation. This operation finds the remainder when one number is divided by another. For example, 7%3=17%3=1 because 7
divided by 3 leaves a remainder of 1.
The Multiplication Property in Modular Arithmetic
One of the key properties of modular arithmetic is the multiplication rule. This rule can be stated as follows: When you multiply two numbers and then take the result modulo m, it is equivalent to
first taking each number modulo m, then multiplying these results, and finally taking the result modulo 𝑚m. Mathematically:
(𝑎×𝑏)%𝑚= ( (𝑎%𝑚)×(𝑏%𝑚) )%𝑚
Why This Property Holds
To understand why this property holds, consider the mathematical definition:
• Let’s say a = q1⋅m+r1 and b= q2⋅m+r2 where 𝑟1 and 𝑟2 are the remainders when 𝑎 and 𝑏 are divided by 𝑚, respectively. Hence, 𝑟1=a%m and 𝑟2=b%m.
• When we multiply a and 𝑏
Expanding this:
• Notice that every term except 𝑟1⋅𝑟2 contains 𝑚 as a factor, meaning they will contribute 0 to the remainder when divided by 𝑚.
• Thus:
1. What exactly is “print it modulo 10⁹ + 7” in competitive programming web sites?
This is one of the most common question scenario that uses the multiplication property of modulus.
Few thing need to be consider that are:
1. 10⁹ + 7 is a prime number.
2. Sometimes when answer in the form of (ans)%(10⁹+7), this means that the actual answer to a problem lies above the range of 64-bit integer which is not possible to calculate so they want only the
remainder left behind our actual answer.
Suppose you are given a problem where you need to compute the sum of the first 𝑛 positive integers and print the result modulo 10⁹+7.
The sum of the first 𝑛n positive integers is given by:
To print the result modulo 10⁹+7, you would do the following:
1. Compute the sum 𝑆.
2. Compute S mod (10⁹+7).
When you see the formula:
This property helps prevent overflow during multiplication by ensuring that the intermediate results are kept within bounds. Here is an example:
Suppose 𝑎=1000000008 and 𝑏=1000000009, and 𝑚=1000000007.
In summary, “print it modulo 10⁹+7” is a common requirement in competitive programming to handle large numbers, prevent overflow, and take advantage of the properties of prime numbers in modular
arithmetic. Always remember to apply the modulo operation at each step of your calculation to keep intermediate results manageable.
Above content is compilation from various sources like chatgpt, qoura. I have comiple it for my personal reference. | {"url":"https://ujjwal-bansal.medium.com/understanding-the-power-of-modular-arithmetic-in-programming-a-deep-dive-into-the-multiplication-4169dc4ed277?source=user_profile_page---------0-------------2329686cdb7a---------------","timestamp":"2024-11-08T08:21:56Z","content_type":"text/html","content_length":"126703","record_id":"<urn:uuid:df7b6b2c-7bb8-4804-9931-825e28b1dbd0>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00025.warc.gz"} |
Quantum Information and Computation
Graduate Course / CASA PhD Lecture, Ruhr University Bochum
• Type: Lecture with Exercises
• Programs: B.Sc. Computer Science, B.Sc. IT-Security, M.Sc. Applied Computer Science, M.Sc. Computer Science, M.Sc. IT-Security
• Lecturer: Prof. Dr. Walter
• Teaching Assistants: Maxim van den Berg, Alex Kulpe, Tianwei Zhang
• Credits: 5 CP
• Language: English
• Examination: 100 % Written Exam (180 Minutes) + 10 % Homework
Course Description & Syllabus
This course will give an introduction to quantum information and quantum computation from the perspective of theoretical computer science. We will discuss the mathematical model of quantum bits and
circuits, how to generalize computer science concepts to the quantum setting, how to design and analyze quantum algorithms and protocols for a variety of computational problems, and how to prove
complexity theoretic lower bounds.
Topics to be covered:
• Fundamentals of quantum computing: from classical to quantum bits, states and operations
• Quantum circuit model of computation
• Quantum computing with oracles: Deutsch-Jozsa, Bernstein-Vazirani, Simon
• Hadamards, quantum Fourier transform, quantum phase estimation
• Grover’s search algorithm and beyond: how to solve SAT on a quantum computer?
• Quantum query complexity
• Quantum entanglement as a resource: superdense coding and teleportation
• From “now cloning” to quantum money: a peek at quantum cryptography
• Nonlocal games and CHSH game
This course should be of interest to students of computer science, mathematics, physics, and related disciplines. Students interested in a Bachelor’s project in quantum information, computing,
cryptography, etc are particularly encouraged to participate.
Recommended prior knowledge
Familiarity with linear algebra, discrete probability, and theoretical computer science, each at the level of a first BSc course; we will briefly remind you of the more difficult bits in class. Some
experience with precise mathematical statements and rigourous proof (since we’ll see many of those in the course). No background in physics is required-
Material & Lecture
Please find a recent version of the lecture notes on Moodle. Video recordings of the lectures are also provided.
In addition, the following references can be useful for supplementary reading (the first one in particular served as inspiration for this course):
• O’Donnell, Quantum Computation and Quantum Information, course material available online (2018)
• Mermin, Quantum Computer Science, Cambridge University Press (2007)
• Nielsen and Chuang, Quantum Computation and Quantum Information, Cambridge University Press (2010)
• de Wolf, Quantum Computing: Lecture Notes, available online (2022)
Learning outcomes
You will learn fundamental concepts, algorithms, and results in quantum information and computation. After successful completion of this course, you will know the theoretical model of quantum
information and computation, how to generalize computer science concepts to the quantum setting, how to design and analyze quantum algorithms and protocols for a variety of computational problems,
and how to prove complexity theoretic lower bounds. You will be prepared for an advanced course or a BSc project in this area.
Grades and homework
Please see Moodle. | {"url":"https://alexkulpe.github.io/teaching/quantum-information-teaching","timestamp":"2024-11-12T20:10:38Z","content_type":"text/html","content_length":"15015","record_id":"<urn:uuid:80f30ac5-8f58-4445-9e68-8fca1000ebfd>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00427.warc.gz"} |
CBSE Class 6 Mathematics Playing With Numbers Chapter Notes
Download CBSE Class 6 Mathematics Playing With Numbers Chapter Notes in PDF format. All Revision notes for Class 6 Mathematics have been designed as per the latest syllabus and updated chapters given
in your textbook for Mathematics in Class 6. Our teachers have designed these concept notes for the benefit of Class 6 students. You should use these chapter wise notes for revision on daily basis.
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also use Printable notes for Class 6 Mathematics for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 6 Mathematics given on
Revision Notes for Class 6 Mathematics Chapter 3 Playing with Numbers
Class 6 Mathematics students should refer to the following concepts and notes for Chapter 3 Playing with Numbers in Class 6. These exam notes for Class 6 Mathematics will be very useful for upcoming
class tests and examinations and help you to score good marks
Chapter 3 Playing with Numbers Notes Class 6 Mathematics
CBSE Class 6 Playing With Numbers Chapter Concepts. Learning the important concepts is very important for every student to get better marks in examinations. The concepts should be clear which will
help in faster learning. The attached concepts made as per NCERT and CBSE pattern will help the student to understand the chapter and score better marks in the examinations.
Playing with Numbers
An exact divisor of a number is called its factor.
Ex: 1, 2, 3 and 6 are factors of number 6.
· The number 1 is a factor of every number.
· Every number is a factor of itself.
· The factors of a number are either less than or equal to the number itself.
· All numbers have a finite number of factors.
· The product of two numbers is called a multiple of each of the two numbers being multiplied.
· A number is a multiple of all its factors.
· Every number is a multiple of 1 and of itself.
· There are infinite multiples of a number.
· If the sum of the factors of a number is two times the number, then the number is called a perfect number.
· Numbers that have only two factors in the form of 1 and the number itself are called prime numbers.
· Numbers that have more than two factors are called composite numbers.
· The number 1 is neither a prime number nor a composite number.
· All numbers with 0, 2, 4, 6 or 8 in the unit’s or one’s place are multiples of 2, and are called even numbers.
· All numbers with 1, 3, 5, 7 or 9 in the unit’s or one’s place are called odd numbers.
· The number 2 is the smallest prime number, and also the only prime number that is even.
· All prime numbers, except 2, are odd numbers.
· The sum of any two prime numbers, except with 2, is an even number.
Tests of Divisibility
There are certain tests of divisibility that can help us to decide whether a given number is divisible by another number.
Divisibility of numbers by 2: A number that has 0, 2, 4, 6 or 8 in its ones place is divisible by 2.
Ex: 234, 830 are divisible by 2
Divisibility of numbers by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
Ex: 234543
Sum of digits = 2 +3 +4 + 5 +4 +3 =21 which is divisible by 3.
Therefore, 234543 is divisible by 3.
Divisibility of numbers by 4: A number is divisible by 4 if the number formed by its last two digits
(i.e. ones and tens) is divisible by 4.
Ex: 84560
The last two digits of the number is 60 which is divisible by 4.
Therefore, the given number is divisible by 4.
Divisibility of numbers by 5: A number that has either 0 or 5 in its ones place is divisible by 5.
Ex: 834345
The unit digit of the given number is 5.
Therefore, the given number is divisible by 5.
Divisibility of numbers by 6: A number is divisible by 6 if that number is divisible by both 2 and 3.
Ex: 603012
The last digit of the given number is 2 which is divisible by2.
Sum of digits = 6 +0+3+0+1+2 =12 which is divisible by3.
Therefore, the given number is divisible by 6.
Divisibility of numbers by 8: A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
Ex: 452640
The last three digits of the given numbers is 640 which is divisible by 8.
Therefore, the given number is divisible by 8.
Divisibility of numbers by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
Ex: 926793
Sum of digits = 9 +2 +6+7+9+3 = 36 which is divisible by 9.
Therefore, the given number is divisible by 9.
Divisibility of numbers by 10: A number that has 0 in its ones place is divisible by 10.
Ex: 4245260
The given number has 0 in it ones place.
Therefore, the given number is divisible by 10.
Divisibility of numbers by 11: If the difference between the sum of the digits at the odd and even places in a given number is either 0 or a multiple of 11, then the given number is divisible by 11.
Ex: 425425
Sum of the digits at odd places = 4 +5+ 2 =11
Sum of the digits at even places = 2 +4 +5 =11
Their difference =11 – 11 = 0
Therefore, the given number is divisible by 11.
Co–prime Numbers
If the only common factor of two numbers is 1, then the two numbers are called co-prime numbers.
General rules of divisibility for all numbers:
· If a number is divisible by another number, then it is also divisible by all the factors of the other number.
• If two numbers are divisible by another number, then their sum and difference is also divisible by the other number.
• If a number is divisible by two co-prime numbers, then it is also divisible by the product of the two co-prime numbers.
Prime Factorisation
Writing a number as a product of its prime factors is called the prime factorisation of the number.
Eg: i) 18 = 2 × 3 × 3
ii) 40 = 2 × 2 × 2 × 5
HCF: The greatest of the common factors of the given numbers is called their highest common factor (HCF). It is also known as the greatest common divisor (GCD). Eg: Prime factorisation of 16 = 2 × 2
× 2 × 2
Prime factorisation of 40 = 2 × 2 × 2 × 5
HCF of 16 and 40 = 2 × 2 × 2 = 8
LCM: The smallest common multiple of the given numbers is called their Least Common Multiple (LCM). Eg: The LCM of given numbers using their prime factorisation:
Prime factorisation of 4 = 2 × 2
Prime factorisation of 6 = 2 × 3
LCM of 4 and 6 = 2 × 2 × 3 =12
To find the LCM of the given numbers using the division method:
• Write the given numbers in a row.
• Divide the numbers by the smallest prime number that divides one or more of the given numbers.
• Write the number that is not divisible, in the second row.
• Write the new dividends in the second row.
• Divide the new dividends by another smallest prime number.
• Continue dividing till the dividends are all prime numbers or 1.
• Stop the process when all the new dividends are prime numbers or 1.
Example By using Division Method, find the LCM of 24 & 18 ?
Answer: Steps of finding HCF by Successive Division Method is as :-
2 | 24 18 Step 1 Write the given numbers as shown on the left and divide them with the least prime number i.e 2.
2 | 12 9 Step 2 On division, write the quotient in each case below the number.
2 | 6 9 Step 3 If any number is not divisible by its respective divisior, it is to be written as such in the next line.
3 | 3 9 Step 4 Keep on dividing the quotient until you get 1(as quotient of all) in the last row.
3 | 1 3 Step 5 Multiply all the divisors to get LCM of given numbers.
| 1 1 Step 6 Hence, LCM = 2 × 2 × 2 × 3 × 3 = 72.
Example Find the LCM of 20, 30,& 40 by Division Method
Answer Steps of finding HCF by Successive Division Method is as :-
2 | 20 30 40
2 | 10 15 20
2 | 5 15 10
3 | 5 15 5
5 | 5 5 5
| 1 1 1
Please click on below link to download pdf file for CBSE Class 6 Playing With Numbers Chapter Concepts.
CBSE Class 6 Mathematics Chapter 3 Playing with Numbers Notes
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download and practice the above notes for Class 6 Mathematics regularly. All revision notes have been designed for Mathematics by referring to the most important topics which the students should
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have been developed as per the latest books also refer to the NCERT solutions for Class 6 Mathematics provided by our teachers. We have also provided a lot of MCQ questions for Class 6 Mathematics in
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Come to StudiesToday.com to get best quality topic wise notes for Class 6 Mathematics Chapter 3 Playing with Numbers | {"url":"https://www.studiestoday.com/concept-playing-numbers-cbse-class-6-playing-numbers-chapter-concepts-241360.html","timestamp":"2024-11-15T00:54:52Z","content_type":"text/html","content_length":"118062","record_id":"<urn:uuid:14719898-1347-4e84-97ab-04a789d358b0>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00797.warc.gz"} |
Cambridge Investment Research
Generalised Game Show Problem
We generalise a result in Professor David Mackay’s book on inference. Bayes theorem also plays a crucial role in decision-making:
Let us consider a worked example which demonstrates how unintuitive results following from this 260-year-old theorem can be.
In the Game Show example, we can simplify Bayes’ Theorem by using a form that expands out all individual doors, and then cancelling off the unconditional probability of each door in numerator and
denominator as they are each $\text{P(any door)}=\frac{1}{n}$, where there are n doors. We consider the two distinct representative cases:
$P\left(\text{prize behind my chosen door}\mid \text{m}\right)=\frac{P\left(\text{m}\mid \text{prize behind my chosen door}\right)}{P}$
$P\left(\text{prize behind another door}\mid \text{m}\right)=\frac{P\left(\text{m}\mid \text{prize behind another door}\right)}{P}$
where here
and where $\mid m$ means that any m doors were removed at the usual intermediate stage after I chose a door. When we start with n doors, and one has been chosen, and that door happens to have the
prize behind it, then the Game Show Host is free to remove m doors from the set of $n-1$ doors and so there are $\left(\genfrac{}{}{0}{}{n-1}{m}\right)$ available ways for the host to do so.
The probability $\text{P(any}\phantom{\rule{0.278em}{0ex}}\text{m}\mid 1\right)$ is therefore $1/\left(\genfrac{}{}{0}{}{n-1}{m}\right)$, since we equivocate between all the host’s options. For the
other case, there are only $n-2$ doors for the host to choose m from, so the number of ways is $\left(\genfrac{}{}{0}{}{n-2}{m}\right)$, and the probability $\text{P(any}\phantom{\rule{0.278em}{0ex}}
\text{m}\mid \text{not 1}\right)$ is therefore $1/\left(\genfrac{}{}{0}{}{n-2}{m}\right)$.
after some algebra I find that
$P\left(\text{prize behind my chosen door}\mid \text{m}\right)=\frac{n-1-m}{n-1-m+\left(n-1{\right)}^{2}}$
$P\left(\text{prize behind another door}\mid \text{m}\right)=\frac{n-1}{n-1-m+\left(n-1{\right)}^{2}}$
Thus, our probability factor is given by: $\frac{\text{P(prize behind another door}\mid \text{m}\right)}{\text{P(prize behind my chosen door}\mid \text{m}\right)}=\frac{n-1}{n-1-m}$
This expression gives us back, from the original game, our game strategy factor of 2 times better if we shift door when $n=3$ and $m=1$. The factor rises to $n-1$ better for shifting to another door
for any $n\ge 2$ and $m=n-2$ doors (all but one other than the one the player chose), and for the shift strategy a probability factor that tends to unity from above when $n\to \mathrm{\infty }$ and
$m=1$, i.e. only one door is removed.
An informal survey by Student-b has shown that quite often, intuition among a random sample of people asked, is lacking. Some will believe it is better to stick, and some say in the standard
three-door game that it is slightly better to shift. The mathematics show that for positive n and m: it is always better to shift! | {"url":"http://www.cir-strategy.com/blog/?p=1221","timestamp":"2024-11-08T12:04:27Z","content_type":"text/html","content_length":"41601","record_id":"<urn:uuid:6548c740-136e-46ea-a0c5-6bdb20063347>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00451.warc.gz"} |
Offset() function to Calculate IRR for Dynamic Range » Chandoo.org - Learn Excel, Power BI & Charting Online
Offset() function to Calculate IRR for Dynamic Range
Offset() function to Calculate IRR for Dynamic Range
When you start the project can you be sure, for how long will you operate it? A VC gives you funds to buy a commercial project. You are to operate the project for some time and then sell it off! Can
you tell me today, when you will sell?
Real world is dynamic and business situations keep changing! Your excel is not that dynamic, when you use the IRR function and tell it to calculate the IRR, you show fixed cash flows! These cash
flows are dynamic.
Not to worry! We have Offset function to our rescue!
What is the Offset() function
In my opinion it is one of the most versatile (And dangerous) functions to use. On the face of it, it is a simple function – As the name suggests, it just offsets your reference.
Offset( range, rows, columns, height, width )
So in the illustrated example, it starts from the C8 cell, moves 0 rows and 0 columns and then gives an array of size 5 x 5 to the sum function!
The difference – Now Offset is NOT returning a value. It is returning references to arrays!
So what can you do with this function?
Ah… you can do a lot! It can change diapers of your kids as well :-). Right now we will see, how it can introduce unparalleled flexibility in your models.
So the VC we were speaking about – gives you USD 1000. You propose to operate the commercial complex for 4 years and post that sell it off for USD 1200.
But then you are not certain if the economic & business conditions would be such that you need to operate it for 4 or 5 or even 6 years. You want a flexibility in the model to “Dynamically Update”
How do we implement this?
The first step is to find the last cash flow in the assumption. I do that simply by counting the number of cash flows. It indicates the number of years that you run the business.
Once I have the number of cash flows with my, I dynamically return the size of the cash flow range using the offset function and input that to IRR function.
Since the offset returns reference to the array of cash flows, I can give that as an argument to the IRR function and it gets me the IRR of the project!
If a new cash flow is entered, the count function would calculate it, pass it to the offset function, which would return the new range to IRR. Phew!!
Few more places, where you can use Offset function
Anywhere you want references to be returned, Offset function does come handy. Similarly I have found offset to be a very useful function if you are looking at creating scenarios, especially in Merger
Modeling, Growth Assumptions, Economic Assumptions, etc. With the click of the mouse, it can completely update your model!
In my modeling experience I have found the following three functions to be quite versatile with the ability to surprise you with their power!
· Offset
· Indirect
· Index
They can return references. This is not usually what you expect Excel to do. So when you are not expecting they can return references to unknown ranges and surprise you. They can make understanding
of model (for the reader of your model) fairly difficult and auditing your model difficult. So use them with care!
Do you use these functions?
These functions can make your financial model quite flexible. Do you use such functions in your models? Share your views and experience on making your model more flexible!
Templates to download
I have created a template for you, where the subheadings are given and you have to link the model to get the cash numbers! You can download the same from here. You can go through the case and fill in
the yellow boxes. I also recommend that you try to create this structure on your own (so that you get a hang of what information is to be recorded).
Also you can download this filled template and check, if the information you recorded, matches mine or not!
Join our Financial Modeling Classes
We are glad to inform that our new financial modeling & project finance modeling online class is ready for your consideration. Please click here to learn more about the program & sign-up. Chandoo.org
has partnered with Pristine to launch a Financial Modeling Course. For details click here. For any queries regarding the cash impact or financial modeling, feel free to put the comments in the blog
or write an email to paramdeep@edupristine.com
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My name is Chandoo. Thanks for dropping by. My mission is to make you awesome in Excel & your work. I live in Wellington, New Zealand. When I am not F9ing my formulas, I cycle, cook or play lego with
my kids. Know more about me.
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37 Responses to “Offset() function to Calculate IRR for Dynamic Range”
1. Awesome stuff. Offset / Index are great for moving averages too. Not sure what I'd do with Indirect.
One thing is certain - you will spend half the time saved explaining to folks what your formula does.
2. I use them at work for creating Dashboards. I love them all 🙂
3. This post is a beauty and another example of why I love this site!
4. Indirect() is good for cascading data validation - based on the value in another cell.
5. You could make it really dangerous and use:
Dynamic update just waiting for someone to put a text string into another row 3 cell.
6. Definitely pretty useful stuff especially with rolling periods but as
Lianna & Mike86 allude to can be pretty dangerous in the hands of babes and most likely a feature that requires some explanation where a workbook is distributed.
The reality is that for many users a combination of more than one function in any one formula frightens the hell out of them and they never bother to read even the introductory paragraph in Help
(what's that?) to give them a clue as to the nature of the formula
7. Hi Chandoo,
I recently started using your site and I am pretty impressed.
This is the 1st time I am commenting.
Just to know, why not we have the dynamic range of cashflows in a excel table, so that we can refer it in the IRR calculation so that it auto adjusts?
I hope I am not fooling myself 🙂
8. @Lianna, @PSG, @John: Thanks
@Andrew: Thanks. This is a great usage. Have you used it in financial models?
@Mike: Yes. If the full row is selected, it would be still more dangerous! 😉
@Mac: I am assuming that this function is not for "babes" 🙂
9. I've used Indirect to construct lookup references (e.g. in Index functions). In the article example, FY column headers could be used to construct references to the correct datasheet or table for
that fiscal year, to be used in Index, Match, etc.
I use Index all the time, usually with Match to determine row, and/or column. Match for column in a table header row is much more dynamic, since it avoids the hard coded column number, and can be
much more explainable and extensible -- e.g. this is row "Exempt" so I'm looking for column "Exempt" and entering another row is much easier.
Sometimes, my row Match function has things like Match(criteria1&criteria2,array1&array2,0) requiring the index and whatever else it's embedded in to be an array formula.
These could, of course, be all mashed together in one big unholy nested function: Indirects to determine where to look, Counts to determine how big the arrays are, Offsets to return the arrays,
Match to access the arrays, Index to determine which array member to use, wrapped in If to handle errors and such, and the whole thing's an array formula. I wouldn't leave that out and visible
where someone might see it. 🙂
I've used Offset less. I've used it to compare rows above and below for duplicate checking in sorted data, or to conditionally format duplicates, or to change color based on changes in section/
Since I've always used offset to, well, *offset*, I've never thought about using the height and width to return dynamic arrays. I can use that. Thanks.
10. Hi,
Offset is Key for any sort of data validation list.
Keep 1 sheet for your options and arrange these in lists. I always use shOptions as codename for this sheet, allowing me reuse code mudules easily.
Use offset to name ranges and then use the data validation list as =namedrange
Hide & protect shOptions.
This give you a very quick and easy way to add additional items to the data validation (new product for example). Or you can build a userform and allow users to add to these lists without fear of
them breaking your workbook.
Hope that helps,
11. My problem with Offset function that it keeps recalculating the spreadsheet. As it has become very large with over 5000 rows, this takes a long time even if nothing has changed and I open my
spreadsheet. Any solutions for this
12. Alex, use INDEX to create your dynamic ranges in a psuedo-volatile way (see examples @ excelhero). Hope this helps.
13. @Imran: Thanks. If you link this IRR to a data table, it should still work. But how do you input the new cash flows in the cells (dynamically)?
@DQKennard: 🙂 As chandoo would say, you would definitely need a laaarge cup of caffe latte to understand that function! 😉
@Alex: As Mark has pointed out, Index can be use there. Else, I guess you can remove the auto calculate in Excel (File --> Excel Options --> Formulas) and make it manual.
@Mark: Thanks.
14. @paramdeep:
Please don't turn of auto calculate! It has to be one of the worst feature in Excel! I understand why some might need to very occassionally turn it off but it should be avoidable through good
workbook design.
There is never enough visual clues to when it is off & if users don't know that it is off it can cause all sorts of errors becuase the result cells aren't always the result that should be given
based on the inputs.
15. @Mawdo: I agree it is not a good idea to turn off auto calculate. But if your sheet has become slow because of this, you can turn it off, and then remember to turn it on, once the model is built!
16. @paramdeep:
I think we'll have to agree to dis-agree on this one. 😉
I stand by my comment that it should be avoidable through good workbook design.
Keep up the goodwork on the rest though, I'm fairly new to the site but its quickly moving up my "go-to" list.
17. Hi all,
New to this site, very impressive.
I am a financial consultant\modelist, and where I work people used to write dynamic offsets as: OFFSET(ref,row,column):OFFSET(ref,row,column). This of course made formulas extremely long, but
when I started working there I didn't know about the height\width options.
As time went by I learned to use height\width and got people around me to shrink their formulas by using them.
generally I'm not a big fan of offsets because their so hard to audit, BUT, the funniest thing happened last week. I was auditing a model (I use Explode Add-In, by xl-logic.com) and usually the
addin doesn't help much with offsets because it only helps you jump to the references but you don't see the actual result. ANYWAY, I found out that when you use the OFFSET:OFFSET way of writing,
explode jumps you to the offset's result, i.e. the range refered to.
Now I find myself thinking what do I prefer, shorter, more readable lines or easier use of audit tools....
Any thoughts?
18. @marninei - Can you edit Explode? Or Can you get the functionality added in, Offest is widely used & I used to be a massive advocate so it would make sense for Explode to be able to cope with
that. As far as I would have thought, it is a change from coping with Offset retruning a cell to coping with it returning a range.
@ Mark et all - Excel Hero's dynamic range using Index is very powerful & I've become a convert. Offset is powerful but volatile. I was happy with that due to the frequency of the ranges I tended
to use it for. (Named formula to be exact, so not needing as many re-calcs as it might otherwise). However in trying to access some of these ranges the other day, from external workbooks, I
realised the limitation of Offset - the workbook needs to be open to return what ever it would have returned. The index solution doesn't, leaving you with names you can access freely with data as
current as when the source was last saved.
19. @Mawdo81 - Sadly, explode is locked. btw, if any of you haven't experienced its power I advise you all to try it. It's an amazing auditing tool.
About offset, I think there are situtions for using every formula, but as you know sometime people "fall in love" with a single formula and use it for practically everything, even when there are
much more useful, simple, elegant tools to use. Dynamic IRR is one of the best things to use offsets for, since sometimes the cashflow's first periods are zero, resulting in the IRR returning
either zero or an error.
20. Thanks, useful tool...how would one do this dynamic change for XIRR which relies on dates as an input..doesnt quiet work as easily.
□ @Model_Citizn: I am sorry for the delayed reply. I seem to have missed the comments on the thread! But just like we find the dynamic range of cash flow using offset, you can find the dates
using offset. It is just extending the same logic on two rows instead of one!
21. [...] few months back, I had written about the offset function and how it can be used to create flexible models. I had discussed at that point of time, why offset function is one of the most
versatile functions [...]
22. Thank You..I understood how to use and apply "Offset" function from this article
23. HEy Chandoo - any advice on how to calculate IRR when the cashflow values are not in an array? I have a situation where all the cashflow values are in a column array except for the final one that
is in the correct row but in a different column.
□ @Ed
I'd setup a range and transfer all values, using formulas, into a single location so that it can be simply used
24. I have downloaded Flexible IRR Model and followed all the steps. And then I simply put IRR formula to (E3:S3 ) range and whatever I do, I get the same results by using both options.
The only exception is when I don't put any number in-between two but then if I put zero, I get the common result which was different from IRR results from two different options.
So what's the point of OFFSET function in this calculation?
□ @Nara
Chandoo's formula =IRR(Offset(E3,0,0,1,E5))
Allows the user to change the value in E5 ie the number of years and the formula adjusts itself
Using Offset in an IRR or other function that requires a range allows you to make it dynamic or variable based on a cell's value eg: E5
That is as the data is added to the range changes automatically
So Chandoo could also use
and the formula would adjust itself as you added extra years to the project
☆ @Hui
But if you add IRR function in the same spreadsheet and tIn IRR function select the whole line indicating cashflow amounts, you will get the same results. I have tested by adding
cashflows in following years. As I said the only difference came in the case of living blank cell in-between.
25. […] Related: Using IRR() function over a dynamic range with OFFSET. […]
26. Is the CAGR applicable to the growth rate of spend as well?
27. Does anyone know if you can skip cells with values in XIRR functions? There is one line of data i want to skip for the XIRR calc without having to delete anything or copy the data anywhere else.
□ @Bryan
No and Yes
If you use a straight range, No
But you maybe able to use a formula to extract the Cashflows & Dates for use in the XIRR Function
For transparency you are better to extract the data to a separate range and then do the Xirr calcs on that range
If you want specific help can you please ask the question in the Chandoo.org Forums http://forum.chandoo.org/
and please attach a sample file
28. Absolutely beautiful! Thank you for the elegant solution and the clear explanation of this function's purpose and usage.
29. I am trying to create a dynamic range based on the number of months eclipsed to an IRR function. I would like to get your input on how to do that.
[Cell A1] Beg Balance = -20000
[Cell A2] Monthly Contribution = -200
[Cell A3] End Balance = 25000
[Cell A4] IRR % after 5 months of contributions =IRR((A1,A2,A2,A2,A2,A2,A3),0.01)
I am wanting to avoid adding another A2 after each new month.
30. Hi All,
Any idea on how to use offset to calculate IRR dynamically between two date ranges. For example, I want to calculate IRR between two dates of my choosing. e.g. between Jan and Apr or between 2023
and 2027. The formula and IRR calculation needs to be set up to be dynamic though so that I can switch dates at my discretion.
□ If you are using Excel 365, you can use FILTER to first filter down the data to the relevant dates and the send that to IRR.
Something like this:
=IRR(FILTER(values, (datecolumn>=start_date)*(datecolumn<=end_date)))
« A Challenge from Hui Filter values where Fruit=Banana OR Sales>70. In Other Words, How to use Advanced Filters? » | {"url":"https://chandoo.org/wp/offset-function-to-calculate-irr-for-dynamic-range/","timestamp":"2024-11-13T02:05:04Z","content_type":"text/html","content_length":"415552","record_id":"<urn:uuid:2864d066-a710-4a77-acaa-1cfcf79dd679>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00247.warc.gz"} |
Probability and Statistical Analysis Worksheet - thebrainywriters
ABC/123 Version X
Probability and Statistical Analysis Worksheet
PSYCH/625 Version 5
University of Phoenix Material
Probability and St
statistical Analysis Worksheet
Complete Parts A, B, and C below.
Part A
1. Why is a z score a standard score? Why can standard scores be used to compare scores from different distributions? Why is it useful to compare different distributions?
2. For the following set of scores, fill in the cells. The mean is 74.13 and the standard deviation is 9.98.
Raw score Z score
68.0 ?
? –1.6
82.0 ?
? 1.8
69.0 ?
? –0.5
85.0 ?
? 1.7
72.0 ?
3. Questions 3a through 3d are based on a distribution of scores with and the standard deviation = 6.38. Draw a small picture to help you see what is require
a. What is the probability of a score falling between a raw score of 70 and 80?
b. What is the probability of a score falling above a raw score of 80?
c. What is the probability of a score falling between a raw score of 81 and 83?
d. What is the probability of a score falling below a raw score of 63?
4. Jake needs to score in the top 10% in order to earn a physical fitness certificate. The class mean is 78 and the standard deviation is 5.5. What raw score does he need?
Part B
The questions in Part B require that you access data from the Pulse Rate Dataset.
The data is based on the following research problem:
Ann conducted a study on the things that may affect pulse rate after exercising. She wants to describe the demographic characteristics of a sample of 55 individuals who completed a large-scale
survey. She has demographic data on the participants’ gender (two categories), their age (open ended), their level of exercise (three categories), their height (open ended), and their weight (open
5. Using Microsoft® Excel® software, run descriptive statistics on the gender and level of exercise variables. From the output, identify the following:
b. Mode for exercise frequency
c. Frequency of high level exercisers (exercise level 1) in the sample
6. Using Microsoft® Excel® software, run descriptive statistics to summarize the data on the age variable, noting the mean and standard deviation. Copy and paste the output from Microsoft® Excel®
into this worksheet.
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10:58:502021-02-04 10:58:50Probability and Statistical Analysis Worksheet | {"url":"https://thebrainywriters.com/probability-and-statistical-analysis-worksheet/","timestamp":"2024-11-02T04:39:17Z","content_type":"text/html","content_length":"51934","record_id":"<urn:uuid:def2a344-35ae-429f-a7ba-5ab71069d17a>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00634.warc.gz"} |
Acca f5
I've just received my F5 performance management text from BPP and had a quick browse through.
Is it just me or does it look as if you have to be a maths professor to actually understand anything in it.
I sat my F4 paper in Dec but this just looks HORRIBLE.
Anyone else had any experience of this paper.
• I sat this paper in June last year.
At first i couldn't get my head around it.
But once i had i found it a relatively easy paper and actually rather liked the exam.
I am going to be sitting it at advanced level as i liked it that much
I sat this paper in June last year.
At first i couldn't get my head around it.
But once i had i found it a relatively easy paper and actually rather liked the exam.
I am going to be sitting it at advanced level as i liked it that much
I sat the advanced one in December, rather liked it also, very wordy though.
• Agreed!
It was horrendous! Sat in just there in December.
The paper was easy though if i'm honest, I think we got lucky though?
If I fail i'll be raging as I may have to learn all the maths formulas
It was horrendous! Sat in just there in December.
The paper was easy though if i'm honest, I think we got lucky though?
If I fail i'll be raging as I may have to learn all the maths formulas which i hate!
Thats exactly what I was thinking. I was doing the entire exam thinking am I missing something?!
It all seemed straightforward. But, it was the first exam set by a new examiner, so maybe they were testing the water and we did get lucky?!
It was horrendous! Sat in just there in December.
The paper was easy though if i'm honest, I think we got lucky though?
If I fail i'll be raging as I may have to learn all the maths formulas
When i sat it in June most of it seamed straight forward.
But i looked at Decembers one (curious) and it seamed even easier than Junes!
Nice paper i think
Thats exactly what I was thinking. I was doing the entire exam thinking am I missing something?!
It all seemed straightforward. But, it was the first exam set by a new examiner, so maybe they were testing the water and we did get lucky?!
We'll have gotten lucky if we pass
• I've been using the study text again today and i'm pretty irritated at paying out good money for a text that explains sod all.
I did fairly well in management accounts in AAT but the joker who wrote this text has not considered explaining any of the calculations.
Wish i was still doing the law paper.
• I have a marginal revenue question that i could use some help with.
Question: AB has used market research to determine that if a price of $250.00 is charged for product G, demand will be 12, 000 units. It has also been established that demand will rise or fall by
5 units for every $1 fall/rise in the selling price. The marginal cost of product G is $80.
Required: If marginal revenue = a-2bQ when the selling price (p) = a-bQ, calculate the profit maximising selling price for product G.
Answer from book:
b = Change in price divided by change in quantity ($1/5 = 0.2
a = $250.00 + (12, 000/5) x $1) = $2650
MR = 2,650 - (2 x 0.2) = 2,650 - 0.4Q
Profits are maximised when MC = MR, when 80 = 2650 - 0.4Q
2,650 - 80 = 2,570 x 10/4 = 6,425
Profit maximising demand = 6,425
I understand everything up to the point when the MR figure is less marginal costs of $80 and the multiplied by 2.5 (10 divided by 4).
Can anyone explain where the 10 divided by 4 figures are derived from and why it is divided and multiplied by 2,570? What does the figure of 6,425 actually tell you?
Many thanks
• 6425 is the quantity at which additional selling revenue per unit (MR) has reduced down to and equals the additional cost of making that extra unit.
So thats the amount the company should make, if they make any more they could only be sold at less than cost of production.
Then you'd go on to the waffly bit about you may want to make more at a loss with a penetration pricing strategy etc.
• Where is the 10 divided by 4 figure coming from?
• Your formula for MC = MR is 80 = 2650 - 0.4Q. You need to solve to find Q.
This can be rearranged as 80 + 0.4Q = 2650, or 0.4Q = 2650 - 80 = 2570.
If 0.4Q = 2570, then Q= 2570/0.4 or 2570 x 10/4 (as 10/4 = 0.4).
Hope this helps, sorry if it's it bit rushed, need to get back to work in 2 mins!
• I dunno
The answer is; "Profits are maximised when MC = MR, when 80 = 2650 - 0.4Q
Re-arrange the equation;
80 = 2650 - 0.4Q
0.4Q = 2650-80
0.4Q = 2570
Q = 2570 / 0.4
Q = 6425
Quantity to maximise profit is 6425 units.
• Just realised that should read 4/10 not 10/4 !!!
Just realised that should read 4/10 not 10/4 !!!
Makes more sense, why write 4/10 instead of 0.4, thats totally pointless.....
• Right, now I've got back home I've had change to work this out. Sorry Mark for my last 2 rushed posts that might've not have made a lot of sense!
Do you follow my first post up to 0.4Q = 2570 ? If so, it's just a case of dividing 2570 by 0.4.
Where I think your book is trying to confuse things is by showing a different method for dividing by decimals/fractions. 0.4 = 4/10 so 2570 divided by 0.4 can be expressed as 2570 divided by 4/
10. When you divide by a fraction, it is the same as multiplying by the reciprocated fraction (in this case 10/4). So 2570 divided by 4/10 is the same as 2570 multiplied by 10/4.
Hope this makes some sort of sense to you. At the end of the day, if you are happy with dividing by 0.4 and can get the right answer by doing that then don't worry about the 10/4!!!
Good luck with the rest of your studies.
• Thank you so much for the explanation. The light bulb in my head has just switched on lol.
I'm definately booking onto a BPP classroom course.
I managed to prepare myself for the F4 exam by just buying a study text from BPP but fed up of trying to learn these subjects without proper support from tutors.
I managed to prepare myself for the F4 exam by just buying a study text from BPP but fed up of trying to learn these subjects without proper support from tutors.
Yup, distance learning is overrated. How do they get us to pay so much to teach ourselves
• No worries Mark, glad it makes sense to you now. I learned F5 at college. I'm glad you found F4 not too difficult by self study as I think I'm going to end up teaching myself that one. | {"url":"https://forums.aat.org.uk/Forum/discussion/30324/acca-f5","timestamp":"2024-11-04T15:19:20Z","content_type":"text/html","content_length":"335108","record_id":"<urn:uuid:e592c098-71bd-45a9-a1c9-ddc4ab82922e>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00821.warc.gz"} |
21 Ways of Stating Ohm's Law
21 Ways of Stating Ohm's Law August 11, 2015 16:31
Ohm's law is the key to understanding basic electronics. It describes how the three elements of electricity—current, voltage, and resistance—relate to each other. Ohm's law can be expressed as an
equation three ways:
1. I (current) = V/R
2. V = IR
3. R = V/I
Which is crystal clear if you've studied electronics for years. Most of us need to hear it rephrased in plain language dozens of different ways before it clicks.
All of the statements below are simply ways of restating the equations above. Each one is pretty dense and many will be counter-intuitive. Take your time to unpack them, and leave your questions in
the comments if anything doesn't make sense.
I hope one of them makes Ohm's Law click for you.
1. Voltage is how much current will flow through a conductor of a certain resistance.
2. Voltage is the resistance of a conductor given a certain current.
3. Resistance is how much current will flow given a certain voltage.
4. Resistance is how much voltage will be generated by a certain current.
5. Current is how much voltage will be generated by a certain resistance.
6. Resistance is the ratio between voltage and current.
7. Current is the resistance of a conductor given a certain voltage.
8. Resistance is how easily voltage can increase current.
9. Voltage makes current flow through a conductor.
10. If voltage is fixed, increasing resistance will decrease current.
11. If current is fixed, increasing resistance will increase voltage.
12. If resistance is fixed, increasing voltage will increase current.
13. If resistance is fixed, increasing current will increase voltage.
14. Current is proportional to voltage; resistance is the constant of proportionality.
15. Without any of the three, there’s no electricity. (Try putting zero in any of the equations.)
16. Current moving through a conductor creates voltage.
17. You can’t have voltage without current, current without resistance, etc.
18. “Danger: High Voltage!” could also be correctly written “Danger: High Current and Resistance!”
19. If resistance is very low, you can get a ton of current with a very low voltage. (eg. 1 Volt / 0.001 Ohms = 1,000 Amps).
20. If resistance is very high, you get very little current even with very high voltage (eg. 1 Volt / 1M Ohms = 1 micro Amp).
21. If you know two, you can always figure out the third. | {"url":"https://www.diyrecordingequipment.com/blogs/news/40492996-21-ways-of-stating-ohms-law","timestamp":"2024-11-09T06:53:12Z","content_type":"text/html","content_length":"134911","record_id":"<urn:uuid:a2814a5d-cdc0-4f6f-9c2d-bec59d1f2e66>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00870.warc.gz"} |
Solution of the neutron diffusion equation with the peridynamic differential operator
The peridynamic differential (PD) operator introduces an internal parameter that defines association among the points within a finite range enabling differentiation through integration. Thus, it is
very robust for determining higher order derivatives of spatial and temporal functions and restoring the interactive nature of phenomenon lost during local differentiation. Furthermore, the PD
operator enables the solution of partial differential equations (PDEs) and ordinary differential equations (ODEs) in a unified manner regardless of their intrinsic behavior and presence of a
singularity. This study presents an application of this nonlocal operator to investigate both steady state and time dependent monoenergetic neutron diffusion with results compared to analytical
steady state and time dependent benchmarks.
Publication series
Name International Conference on Physics of Reactors, PHYSOR 2018: Reactor Physics Paving the Way Towards More Efficient Systems
Volume Part F168384-3
Conference 2018 International Conference on Physics of Reactors: Reactor Physics Paving the Way Towards More Efficient Systems, PHYSOR 2018
Country/Territory Mexico
City Cancun
Period 4/22/18 → 4/26/18
• Neutron diffusion
• Nonlocal
• Numerical Laplace transform inversion
• Peridynamic differential operator
ASJC Scopus subject areas
• Nuclear Energy and Engineering
• Nuclear and High Energy Physics
• Radiation
• Safety, Risk, Reliability and Quality
Dive into the research topics of 'Solution of the neutron diffusion equation with the peridynamic differential operator'. Together they form a unique fingerprint. | {"url":"https://experts.arizona.edu/en/publications/solution-of-the-neutron-diffusion-equation-with-the-peridynamic-d","timestamp":"2024-11-08T17:57:21Z","content_type":"text/html","content_length":"57533","record_id":"<urn:uuid:b741a1ca-c27f-4d96-a6da-4138c2638d1d>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00298.warc.gz"} |
Simplifying expressions (2024)
Here is everything you need to know about simplifying algebraic expressions for GCSE maths (Edexcel, AQA and OCR). You’ll learn how to collect like terms, write and simplify expressions, and how to
simplify algebraic fractions.
Look out for the simplifying expressions worksheets with correct answers, word problems and exam questions at the end.
What does simplifying an expression mean
Simplifying an algebraic expression is when we use a variety of techniques to make algebraic expressions more efficient and compact – in their simplest form – without changing the value of the
original expression.
What does simplifying an expression mean?
Simplifying expressions worksheets
Get your free simplifying expressions worksheet of 20+ questions and answers. Includes reasoning and applied questions.
Simplifying expressions worksheets
Get your free simplifying expressions worksheet of 20+ questions and answers. Includes reasoning and applied questions.
How to simplify expressions
To simplify expressions first expand any brackets, next multiply or divide any terms and use the laws of indices if necessary, then collect like terms by adding or subtracting and finally rewrite the
For example to simplify
\[8 x+4+3(2 x-3)\]
1. Expand the brackets
\[8 x+4+6 x-9\]
2Collect like terms
\[\begin{aligned}8 x+6 x&=14 x \\4-9&=-5\end{aligned}\]
3Rewrite the expression
\[\begin{aligned}8 x+4+3(2 x-3) \\=14 x-5\end{aligned}\]
Explain how to simplify expressions
How to simplify an expression by collecting like terms
In order to simplify an algebraic expression we need to ‘collect the like terms’ by grouping together the terms that are similar:
When we highlight the like terms, we must include the sign in front of the term and where necessary identify the negative numbers.
What are like terms?
Like terms have the same combination of variables and/or numbers as each other, but the coefficients could be different.
For example…
4 and 9 are like terms ✔
3x and 5x are like terms ✔
2ab and -5ab are like terms ✔
8 and 3x are not like terms ✘
4y and 2x are not like terms ✘
x^2 and x are not like terms ✘
Methods of simplifying expressions
1. Collecting like terms
Example of collecting like terms
\[5 x+3 y+4-2 x+8 y-7\]
1 Identify the like terms
The terms involving x are like terms. The terms involving y are like terms. The constant terms are like terms.
The plus (or minus) sign belongs to the term before it.
2 Group the like terms
\[5 x-2 x+3 y+8 y+4-7\]
3 Combine the like terms by adding or subtracting
\begin{aligned}5 x-2 x=3 x \\\\3 y+8 y=11 y \\\\4-7=-3\end{aligned}
\begin{aligned}5 x+3 y+4-2 x+8 y-7 \\\\=3 x+11 y-3\end{aligned}
Step-by-step guide: Collecting like terms
2. Multiplying and dividing algebra
Example of multiplying and dividing algebra
\[\frac{3 a b \times 4 a c}{2 a}\]
1 Simplify the numerator
\[3 a b \times 4 a c=12 a^{2} b c\]
2 Divide by the denominator
\[12 a^{2} b c \div 2 a=6 a b c\]
\[\frac{3 a b \times 4 a c}{2 a}=6 a b c\]
3. Expanding brackets
Example of expanding brackets:
\[3(2 x+5)\]
1 Multiply the term outside of the bracket by the first term inside the bracket
\[3 \times 2 x=6 x\]
2 Multiply the term outside the bracket by the second term inside the bracket.
\[3 \times 5=15\]
\[3(2 x+5)=6 x+15\]
Step-by-step guide: Expanding brackets
See also: Expand and simplify
4. Algebraic fractions
Example of algebraic fractions
1 Find the highest common factor (HCF) of the numerator and denominator.
The HCF of 12xy and 8x is 4x
2 Divide the numerator and the denominator by this value.
\[12xy \div 4x=3y\]
\[8x\div 4x=2\]
3 Rewrite the simplified fraction
Step-by-step guide: Algebraic fractions
See also: Simplifying algebraic fractions
5. Write and simplify algebraic expressions
We can write algebraic expressions to help simplify problems. We will often be able to make a linear equation or a quadratic equation and solve it.
Example of writing and simplifying expressions
Write an expression for the perimeter of the shape.
1. Read the question carefully and highlight the key information.
Key words:
Expression: a set of terms that are combined using (+, −, ✕ and ÷)
Perimeter: the distance around the edge of a shape
We need to add together each of the lengths of the shape.
2Write an expression and simplify.
We then simplify the following expression by adding and subtracting the terms.
Examples of simplifying expressions
Example 1: collecting like terms with one variable and one constant
1. Underline the similar terms in the expression and combine them.
\begin{aligned}8 x-2 x=6 x \\\\\5+6=11\end{aligned}
2Rewrite the expression.
\begin{aligned}8 x+5-2 x+6 \\\\=6 x+11\end{aligned}
Example 2: collecting like terms with multiple variables and one constant
Underline the similar terms in the expression
\begin{aligned}5 x y-2 x y=3 x y \\\\3 y-8 y=-5 y \\\\-4+7=3\end{aligned}
\begin{aligned}5 x y+3 y-4-2 x y-8 y+7 \\\\=3 x y-5 y+3\end{aligned}
Example 3: expanding brackets
\[3 x\left(3-2 y+5 x^{2}\right)\]
Multiply the term outside of the bracket by the first term inside the bracket
Multiply the term outside the bracket by the second term inside the bracket.
\[3 x \times-2 y=-6 x y\]
Multiply the term outside the bracket by the third term inside the bracket.
\[3 x \times 5 x^{2}=15 x^{3}\]
\begin{aligned}3 x\left(3-2 y+5 x^{2}\right) \\\\=9 x-6 x y+15 x^{3}\end{aligned}
Example 4: algebraic fractions
Find the highest common factor (HCF) of the numerator and denominator.
9 x^2 y
\[3 x^{2}\]
Divide the numerator and the denominator by this value.
\[9 x^{2} y \div 3 x^{2}=3 y\]
\[15 x^{3} \div 3 x^{2}=5 x\]
Rewrite the simplified fraction
Example 5: algebraic fractions
Find the highest common factor (HCF) of the numerator and denominator.
It is easier to find the HCF for this example if we factorise the numerator.
8x^3 − 6xy can be written as 2x(4x^2 − 3y).
It is now easier to see that the HCF of 8x^3 − 6xy and 4x^2y is 2x
Divide the numerator and the denominator by this value.
\[\left(8 x^{3}-6 x y\right) \div 2 x=4 x^{2}-3 y\]
We have to divide both terms by 2x
\[4 x^{2} y \div 2 x=2 x y\]
Rewrite the simplified fraction
Example 6: algebraic fractions
We will need to factorise quadratics to simplify this algebraic fraction
Fully factorise the numerator and the denominator
\[x^{2}-2 x-15=(x+3)(x-5)\]
Cancel any brackets that are common to the numerator and denominator
Rewrite the simplified fraction
\[\frac{x^{2}-2 x-15}{x^{2}-9}\]
Step-by-step guide: Factorising quadratics
Step-by-step guide: Difference of two squares
Example 7: expression for area
Write an expression for the area of the shape.
Read the question carefully and highlight the key information.
Key words:
Expression: a set of terms that are combined using (+, −, ✕ and ÷)
Area: the 2D space inside a shape.
This shape is a triangle. We know the formula to find the area of a triangle is:
\[\text { Area of triangle }=\frac{\text { base } \times \text { height }}{2}\]
We need to multiply the base and height of the shape then divide by 2.
Write an expression and simplify.
\[= \frac{6x^{2}+10x+4}{2}\]
Example 8: worded problem
Sophie is x years old,
Emily is three years younger than Sophie
Ameila is four times older than Sophie.
Write an expression for each of their ages.
Read the question carefully and underline the key information.
We are told that Sophie is x years old
Emily is three years younger than Sophie, so three less than x is x − 3
Ameila is four times older than Sophie, so four lots of x − 3 is 4(x − 3).
We need brackets because we are multiplying all of x − 3 by 4
Write an expression and simplify.
Sophie is x years old
Emily is x − 3 years old
Ameila is 4(x − 3) = 4x − 12 years old
Common misconceptions
• The sign in front of the term is part of it
When we underline the like terms, we must include the sign in front of the
• Terms with a coefficient of 1
For terms with a coefficient of 1 we don’t need to write the 1
When adding and multiplying, the order in which we calculate doesn’t matter
This is not the case for subtracting and dividing.
In order for two terms to be ‘like terms’ they need the same combination of variables.
3x2and 5x2are like terms
2a2b and -5a2bare like terms
3x2and 5x are not like terms
2a2b and -5ab are not like terms
• Using brackets (parentheses)
When multiplying an expression by a value we need to use brackets so that each term is multiplied.
Simplifying expressions practice questions
For the constant terms, we have
For the variable terms, we have
This means
2. Simplify
By considering like terms, we have 8ab-7ab=ab and -8a-3a=-11a .
This means 8ab-8a-7ab-3a=ab-11a .
3. Simplify
By considering like terms, we have -2xy-6xy=-8xy and 3x^{2}y+5x^{2}y=8x^{2}y and 7x .
This means -2xy+3x^{2}y+7x+5x^{2}y-6xy=-8xy+8x^{2}y+7x .
4a. Write an expression for the perimeter of the shape:
By working out the missing side lengths as algebraic expressions, and adding together all side lengths we have:
Perimeter =2 x+5+x+1+x+3+x+5+x+2+2 x+6 =8 x+22
4b. Write an expression for the area of the shape:
The shape can be split into rectangles in more than one way
(2x+5)(x+1)=2x^{2}+7x+5 (x+5)(x+2)=x^{2}+7x+10 Area=3x^{2}+14x+15
(x+1)(x+3)=x^{2}+4x+3 (2x+6)(x+2)=x^{2}+10x+12 Area=3x^{2}+14x+15
5. Steve is x years old.
Rachel is 11 years older than Steve.
Barry is twice as old as Rachel.
Write an expression for the total ages of Steve, Rachel and Barry.
Steve’s age = x
Rachel’s age = x+11
Barry’s age = 2(x+11)=2x+22
Total ages =x+x+11+2x+22 \\
& =4x+33
\end{aligned} [/katex]
6. Simplify:
The highest common factor of the numerator and denominator is 6 , so we divide numerator and denominator by 6 , resulting in the simplified fraction.
7. Simplify:
The highest common factor of the numerator and denominator is 4ab , so we divide numerator and denominator by 4ab , resulting in the simplified fraction.
8. Simplify:
The numerator can be factorised, giving \frac{3a(3a-2b)}{15ab^{2}} after which the numerator and denominator can be divided by the highest common factor of 3a , resulting in the simplified fraction.
9. Simplify:
We can factorise the numerator and denominator into double brackets, giving
\frac{(x+1)(x+2)}{(x+1)(2x-1)} and then cancel the common bracket from numerator and denominator,
which results in the simplified fraction.
10. Expand:
3 y(4+2 y-4 x)
With a single bracket expansion, we must be sure to multiply each term inside the bracket by the number in front of the bracket. Make sure to include the correct index numbers.
Simplifying expressions GCSE questions
1. Simplify: 4f – 2e + 3f + 5e
Show answer
(2 marks)
2. Expand and simplify: 4a(a + b) – 2(a^2 – 2b)
Show answer
(2 marks)
4. Expand and simplify:
Show answer
(3 marks)
Learning checklist
• Simplify and manipulate algebraic expressions to maintain equivalence by taking out common factors.
• Model situations or procedures by translating them into algebraic expressions.
• Simplify and manipulate algebraic expressions and algebraic fractions.
• Translate simple situations or procedures into algebraic expressions.
The next lessons are
• Factorising
• Solving equations
• Simultaneous equations
• Rearranging equations
Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
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When quoting this document, please refer to the following
DOI: 10.4230/LIPIcs.SoCG.2023.11
URN: urn:nbn:de:0030-drops-178610
URL: http://dagstuhl.sunsite.rwth-aachen.de/volltexte/2023/17861/
Bandyapadhyay, Sayan ; Lochet, William ; Saurabh, Saket ; Xue, Jie
Minimum-Membership Geometric Set Cover, Revisited
We revisit a natural variant of the geometric set cover problem, called minimum-membership geometric set cover (MMGSC). In this problem, the input consists of a set S of points and a set ℛ of
geometric objects, and the goal is to find a subset ℛ^* ⊆ ℛ to cover all points in S such that the membership of S with respect to ℛ^*, denoted by memb(S,ℛ^*), is minimized, where memb(S,ℛ^*) = max_
{p ∈ S} |{R ∈ ℛ^*: p ∈ R}|. We give the first polynomial-time approximation algorithms for MMGSC in ℝ². Specifically, we achieve the following two main results.
- We give the first polynomial-time constant-approximation algorithm for MMGSC with unit squares. This answers a question left open since the work of Erlebach and Leeuwen [SODA'08], who gave a
constant-approximation algorithm with running time n^{O(opt)} where opt is the optimum of the problem (i.e., the minimum membership).
- We give the first polynomial-time approximation scheme (PTAS) for MMGSC with halfplanes. Prior to this work, it was even unknown whether the problem can be approximated with a factor of o(log n) in
polynomial time, while it is well-known that the minimum-size set cover problem with halfplanes can be solved in polynomial time. We also consider a problem closely related to MMGSC, called
minimum-ply geometric set cover (MPGSC), in which the goal is to find ℛ^* ⊆ ℛ to cover S such that the ply of ℛ^* is minimized, where the ply is defined as the maximum number of objects in ℛ^* which
have a nonempty common intersection. Very recently, Durocher et al. gave the first constant-approximation algorithm for MPGSC with unit squares which runs in O(n^{12}) time. We give a significantly
simpler constant-approximation algorithm with near-linear running time.
BibTeX - Entry
author = {Bandyapadhyay, Sayan and Lochet, William and Saurabh, Saket and Xue, Jie},
title = {{Minimum-Membership Geometric Set Cover, Revisited}},
booktitle = {39th International Symposium on Computational Geometry (SoCG 2023)},
pages = {11:1--11:14},
series = {Leibniz International Proceedings in Informatics (LIPIcs)},
ISBN = {978-3-95977-273-0},
ISSN = {1868-8969},
year = {2023},
volume = {258},
editor = {Chambers, Erin W. and Gudmundsson, Joachim},
publisher = {Schloss Dagstuhl -- Leibniz-Zentrum f{\"u}r Informatik},
address = {Dagstuhl, Germany},
URL = {https://drops.dagstuhl.de/opus/volltexte/2023/17861},
URN = {urn:nbn:de:0030-drops-178610},
doi = {10.4230/LIPIcs.SoCG.2023.11},
annote = {Keywords: geometric set cover, geometric optimization, approximation algorithms}
Keywords: geometric set cover, geometric optimization, approximation algorithms
Collection: 39th International Symposium on Computational Geometry (SoCG 2023)
Issue Date: 2023
Date of publication: 09.06.2023
DROPS-Home | Fulltext Search | Imprint | Privacy | {"url":"http://dagstuhl.sunsite.rwth-aachen.de/opus/frontdoor.php?source_opus=17861","timestamp":"2024-11-09T17:34:23Z","content_type":"text/html","content_length":"7679","record_id":"<urn:uuid:78c6cfd2-6989-49fc-8512-72ff1bd8e23f>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00077.warc.gz"} |
Additive identities
Absorbing elements
An absorbing element in a multiplicative semigroup or semiring generalises the property 0 ⋅ x = 0. Examples include:
• The empty set, which is an absorbing element under Cartesian product of sets, since { } × S = { }
• The zero function or zero map defined by z(x) = 0 under pointwise multiplication (f ⋅ g)(x) = f(x) ⋅ g(x)
Many absorbing elements are also additive identities, including the empty set and the zero function. Another important example is the distinguished element 0 in a field or ring, which is both the
additive identity and the multiplicative absorbing element, and whose principal ideal is the smallest ideal.
Zero objects
A zero object in a category is both an initial and terminal object (and so an identity under both coproducts and products). For example, the trivial structure (containing only the identity) is a zero
object in categories where morphisms must map identities to identities. Specific examples include:
• The trivial group, containing only the identity (a zero object in the category of groups)
• The zero module, containing only the identity (a zero object in the category of modules over a ring)
Zero morphisms
A zero morphism in a category is a generalised absorbing element under function composition: any morphism composed with a zero morphism gives a zero morphism. Specifically, if 0[XY] : X → Y is the
zero morphism among morphisms from X to Y, and f : A → X and g : Y → B are arbitrary morphisms, then g ∘ 0[XY] = 0[XB] and 0[XY] ∘ f = 0[AY].
If a category has a zero object 0, then there are canonical morphisms X → 0 and 0 → Y, and composing them gives a zero morphism 0[XY] : X → Y. In the category of groups, for example, zero morphisms
are morphisms which always return group identities, thus generalising the function z(x) = 0.
Least elements
Zero module
In mathematics, the zero module is the module consisting of only the additive identity for the module's addition function. In the integers, this identity is zero, which gives the name zero module.
That the zero module is in fact a module is simple to show; it is closed under addition and multiplication trivially.
Zero ideal
In mathematics, the zero ideal in a ring ${\displaystyle R}$ is the ideal ${\displaystyle \{0\}}$ consisting of only the additive identity (or zero element). The fact that this is an ideal follows
directly from the definition.
Zero matrix
In mathematics, particularly linear algebra, a zero matrix is a matrix with all its entries being zero. It is alternately denoted by the symbol ${\displaystyle O}$ .^[2] Some examples of zero
matrices are
${\displaystyle 0_{1,1}={\begin{bmatrix}0\end{bmatrix}},\ 0_{2,2}={\begin{bmatrix}0&0\\0&0\end{bmatrix}},\ 0_{2,3}={\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}},\ }$
The set of m×n matrices with entries in a ring K forms a module ${\displaystyle K_{m,n}}$ . The zero matrix ${\displaystyle 0_{K_{m,n}}}$ in ${\displaystyle K_{m,n}}$ is the matrix with all entries
equal to ${\displaystyle 0_{K}}$ , where ${\displaystyle 0_{K}}$ is the additive identity in K.
${\displaystyle 0_{K_{m,n}}={\begin{bmatrix}0_{K}&0_{K}&\cdots &0_{K}\\0_{K}&0_{K}&\cdots &0_{K}\\\vdots &\vdots &&\vdots \\0_{K}&0_{K}&\cdots &0_{K}\end{bmatrix}}}$
The zero matrix is the additive identity in ${\displaystyle K_{m,n}}$ . That is, for all ${\displaystyle A\in K_{m,n}}$ :
${\displaystyle 0_{K_{m,n}}+A=A+0_{K_{m,n}}=A}$
There is exactly one zero matrix of any given size m×n (with entries from a given ring), so when the context is clear, one often refers to the zero matrix. In a matrix ring, the zero matrix serves
the role of both an additive identity and an absorbing element. In general, the zero element of a ring is unique, and typically denoted as 0 without any subscript to indicate the parent ring. Hence
the examples above represent zero matrices over any ring.
The zero matrix also represents the linear transformation which sends all vectors to the zero vector.
Zero tensor
In mathematics, the zero tensor is a tensor, of any order, all of whose components are zero. The zero tensor of order 1 is sometimes known as the zero vector.
Taking a tensor product of any tensor with any zero tensor results in another zero tensor. Among tensors of a given type, the zero tensor of that type serves as the additive identity among those
See also
1. ^ Nair, M. Thamban; Singh, Arindama (2018). Linear Algebra. Springer. p. 3. doi:10.1007/978-981-13-0926-7. ISBN 978-981-13-0925-0.
2. ^ Lang, Serge (1987). Linear Algebra. Undergraduate Texts in Mathematics. Springer. p. 25. ISBN 9780387964126. “We have a zero matrix in which ${\displaystyle a_{ij}=0}$ for all ${\displaystyle
i,j}$ . ... We shall write it ${\displaystyle O}$ .” | {"url":"https://www.knowpia.com/knowpedia/Zero_element","timestamp":"2024-11-10T12:09:00Z","content_type":"text/html","content_length":"111904","record_id":"<urn:uuid:e47bb4ee-e506-4662-aa17-72f8fe720cc0>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00514.warc.gz"} |
0.1.1 Oct 26, 2024
0.1.0 Oct 26, 2024
0.0.1 Jun 16, 2023
Geospatial Primitives, Algorithms, and Utilities
Chat or ask questions on Discord
The geo crate provides geospatial primitive types such as Point, LineString, and Polygon, and provides algorithms and operations such as:
• Area and centroid calculation
• Simplification and convex hull operations
• Euclidean and Haversine distance measurement
• Intersection checks
• Affine transforms such as rotation and translation
• All DE-9IM spatial predicates such as contains, crosses, and touches.
Please refer to the documentation for a complete list.
The primitive types also provide the basis for other functionality in the Geo ecosystem, including:
// primitives
use geo::{line_string, polygon};
// algorithms
use geo::ConvexHull;
// An L shape
let poly = polygon![
(x: 0.0, y: 0.0),
(x: 4.0, y: 0.0),
(x: 4.0, y: 1.0),
(x: 1.0, y: 1.0),
(x: 1.0, y: 4.0),
(x: 0.0, y: 4.0),
(x: 0.0, y: 0.0),
// Calculate the polygon's convex hull
let hull = poly.convex_hull();
(x: 4.0, y: 0.0),
(x: 4.0, y: 1.0),
(x: 1.0, y: 4.0),
(x: 0.0, y: 4.0),
(x: 0.0, y: 0.0),
(x: 4.0, y: 0.0),
Contributions are welcome! Have a look at the issues, and open a pull request if you'd like to add an algorithm or some functionality.
Licensed under either of
at your option.
Unless you explicitly state otherwise, any contribution intentionally submitted for inclusion in the work by you, as defined in the Apache-2.0 license, shall be dual licensed as above, without any
additional terms or conditions.
~10K SLoC | {"url":"https://lib.rs/crates/geo-traits","timestamp":"2024-11-09T04:33:55Z","content_type":"text/html","content_length":"23399","record_id":"<urn:uuid:381bc095-6fa5-42df-9ec1-0b5ecb1a8b3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00375.warc.gz"} |
Current Ratio
`CR = "Current Assets"/"Current Liabilities"`
Enter a value for all fields
The Current Ratio calculator computes a liquidity ratio that measures the ability to pay short-term liabilities.
INSTRUCTIONS: Choose units and enter the following:
• (CA) Current Assets
• (CL) Current Liabilities
Current Ratio (CR): The ratio is returned as a real number.
The Math / Science
The Current Ratio equation is a liquidity ratio that measures the ability to pay short-term liabilities.
Current Ratio = Current Assets / Current Liabilities
The Accounting Ratio Calculator provides numerous standard equations used in business accounting, including the following:
• Best Possible Days Sales Outstanding - gives useful insight into delinquencies, as it considers only Current Receivables.
• Current Ratio - A liquidity ratio that measures the ability to pay short-term liabilities.
• Days Sales Outstanding - tests the efficiency of the conversion of receivables into cash.
• Debt to Equity Ratio - measures how a company is leveraging its debt against the resources of its owners.
• Inventory Turnover Ratio - reveals how many times inventory turns over in a period
• Net Cash Flow from Operations - Net income - Increase in Receivables + Increase in Payables.
• Quick Assets - cash and assets that can be converted quickly to cash
• Quick Ratio (aka Acid Test) - measures the ability to pay short-term liabilities with cash and assets quickly convertible to cash
• Stockholder's Equity - sum of Common Stock at Par, Premium on Common Stock, Preferred Stock at Par, Premium on Preferred Stock and Retained Earnings
Enhance your vCalc experience with a free account
Sign Up Now!
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2.1 Overview
n the one-period binomial model, there is a current stock price, say S. The stock price can either have an "uptick" (i.e., move to price Su > S), or have a "downtick" (i.e., move to a price Sd < S).
Here, u is a number greater than 1, and d is a positive number less than 1. The option expires (or matures) at the end of the period. This is depicted in Figure 2.1.
To get an idea of how European options are priced, we will explore two important approaches to the valuation problem using two examples: the Riskless Hedge Example (topic 2.2) and the Synthetic
Option Example (topic 2.3). In the first example, we point out the fundamental problem in valuing options, and explain how these approaches help us solve the valuation problem. The general one-period
analysis for both examples is presented in topic 2.4, Call Option Valuation: A Riskless Hedge Approach, topic 2.5, Put Option Valuation: A Riskless Hedge Approach, and topic 2.6, Option Valuation: A
Synthetic Option Approach.
We then use this general one-period analysis to identify in topic 2.7, a very general principle known as the Risk-Neutral Valuation Principle. In topic 2.8, we consider how this principle can be
reconciled with CAPM. We conclude the one-period analysis with a study of an important arbitrage relationship that must hold among the stock price, put and call option prices and the strike price.
This relationship is known as the Put-Call Parity Relation, and is developed in topic 2.9. | {"url":"https://optiontutor.com/OTchp2/default.htm","timestamp":"2024-11-02T20:32:16Z","content_type":"application/xhtml+xml","content_length":"7719","record_id":"<urn:uuid:93f5385f-88bd-4d16-8f35-b468b632177c>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00644.warc.gz"} |
Formulas had to be mixed with water, but Third World mothers didn't understand that overdilluting it—especially with contaminated water—could "prevent a child from absorbing the nutrients in
av V Wallqvist · 2009 · Citerat av 2 — begins with mineralogical facts about talc as well as physical The London formula can be modified to account for retardation effects occurring
You should be able to read this chart and know how to read birefringence colors. The birerfringences of common minerals are given on the back of the chart. Extra Reading from Greg Finn's Optical
Mineralogy Page Interference Phenomena; Retardation PERCENT TO FORMULA ! We are given the following chemical analysis of a pyroxene. Compute its formula: (Hint: Jadeite NaAlSi 2O 6 – Diopside: CaMgSi
2O 6) Oxide wt. % MolWt oxide Moles oxide Moles cation Mole oxygen Prop cations to O6 SiO 2 56.64 60.086 0.9426 0.9426 1.8852 2 Na 2O 4.38 61.99 0.0707 0.1414 0.707 0.3 Al 2O Minerals with slightly
greater birefringence show yellow, orange or red interference colors upon rotation (retardation of 200-550 nm) = first order colors As retardation increases further, colors repeat every 550nm
(average wavelength of visible light) going from violet to red (second order) and then from violet to red again (3rd order) Determination of the birefringence of a mineral . The interference colour
is the result of a retardation between the waves which are vibrating inside the mineral and is defined by the equation: R= e(n1 - n2) where "R" represents the retardation, "e" is the thickness of the
mineral, and "n1" and "n2" are the refractive indices of the waves.
The correlation for the latter equation is only marginally better than that strata, and features, such as mineralogy, fabric, degree of consolidation, anisot- together with biochemical retardation
mechanisms and diffusion in riverbed sedi-. to the mineralogy, mechanical properties and microstructure of hardened pastes. Synthetic fire retardant had the highest retardation effect in all cases.
reinforced beams is analyzed and a simplified formula for calculating its value is calculatedness calculates calculating calculatingly calculation calculational mineralogizes mineralogizing
mineralogy minerals miners mines mineshaft retardants retardate retardates retardation retardations retardative retardatory =Calculation= (kalkjûlēsj´n) beräkning.
With these two weights formula (2) may be used to obtain K, and then formula (1) to convert to S, the specific gravity.
conductivity and the chemical/mineralogical constitution. Clearly present the role of the geosphere for the barrier functions: isolation, retardation and Check of calculation results concerning
copper corrosion and information regarding type
Retardation factor is a ratio that is often used in chemistry. Typically used in chromatography applications, it is generally expressed as a ratio of the distance traveled by a compound to that of a
liquid solvent .
63 formula corresponding to 40% schorl, 25% dravite, 20% foitite, and 15% olenite on a molar 64 basis. Use of spreadsheets like that of Selway and Xiong (2002) does a fine job of calculating the 65
atomic formula and assigns the name “schorl” to the phase.
Refraction. Refractive index of a substance “n” is inversely proportional to … Lecture 10: Introduction to Optical Mineralogy 2. Lecture 11: Systematic Mineralogy, Silicates. Lecture 12:
Tectosilicates. Lecture 13: Elemental Chemistry. Lecture 14: Paulings Rules.
The hardness of a mineral can be determined by a scratch test. mineralogy output, consequently yielding better porosity and water saturation calculation.
Salja ved
Formula, UO2(CO3)*2NaCO3. Formula mass, 542 MW. Density, 5.6 congenital anomalies, mental retardation, and an increased incidence of Palache, C., H. Berman, and C. Frondel (1951) Dana's system of
mineralogy, 7th The significance of this is that if a chemical has a retardation factor of 10 than it will move only overwhelms the elevation term in the Darcy equation and water. Apr 8, 2013 Here
you will learn how to work on problems based on Acceleration and Retardation.To view more Educational content, please visit: 2.1.2 Calculation of Initial Mass of Strontium in Barrier (aqueous +
sorbed) ..
Praktik journalisthøjskolen
Unit of Retardation. MKS o SI method : metre / second 2. CGS Method : centimeter / second 2 . FKS Method : foot / second 2 . Problems: A car is travelling at 72 km / h. If its velocity increases to
54 km / h in 5 second, then find the Retardation of the car in SI unit. We know, metre / second 2 are the unit of Retardation in SI unit.
a. to form, mal, express, -»ligt, ad, mould-candle. der, m. father, retardation.
a retardation is slowing down of the movement of the body. and can be measured by rate of change of velocity with respect to time. retardation = dv/dt. if in the above example if the cyclist has
applied breaks then his motion gets reduced, means he is covering less distance in equal time intervals. this will be reflected in the nature of s-t curve.
Lecture 10: Introduction to Optical Mineralogy 2. Lecture 11: Systematic Mineralogy, Silicates. Lecture 12: Tectosilicates. Lecture 13: Elemental Chemistry.
Class, Carbonate. Crystal system, Hexagonal Anisotropic Minerals · Inclined and Parallel Extinction · Interference Colors, Birefringence and Retardation · Other Textures Under Crossed Polarized
Light. Optical Mineralogy (Chapter 7) index of refraction, critical angle, angles of refraction and incidence, retardation and birefringence if given other factors. In this formula potassium is
sometimes replaced by other ions with a single marker dials for navigation compasses, optical filters, pyrometers, retardation that as this book is so closely related to the System of Mineralogy it
was unwise to chemical composition, capable of being expressed by a chemical formula. | {"url":"https://hurmanblirrikzhejku.netlify.app/97692/98889","timestamp":"2024-11-15T03:40:14Z","content_type":"text/html","content_length":"10425","record_id":"<urn:uuid:efe87427-81ff-47c5-b0bd-318ab9291a16>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00472.warc.gz"} |
Calculation of carrier frequency & Figure of Merit for SSB signal
What is the carrier frequency in an AM wave when its highest frequency component is 850Hz and the bandwidth of the signal is 50Hz?
a.) 80 Hz
b.) 695 Hz
c.) 625 Hz
d.) 825 Hz
Correct Answer: d) 825 Hz
Upper frequency = 850Hz
Bandwidth = 50Hz
Therefore lower Frequency = 850-50= 800 Hz
Carrier Frequency = (850-800)/2
= 825 Hz
Noise figure of merit in SSB modulated signal is
a.) 1
b.) Less than 1
c.) Greater than 1
d.) None of the above
Correct Answer: a) 1
A figure of merit used to describe the performance of a system. The figure of merit ‘γ’ is the ratio of output signal to noise ratio to input signal to noise
ratio of a receiver system. Figure of merit for SSB modulation is always 1.
Post your comment
• RE: Calculation of carrier frequency & Figure of Merit for SSB signal -Akshay (09/20/16)
• Highest freq. Component is (fc+fm)=850hz-----(1)
BW=2fm=50hz given ;
Fm=25hz----put in equation (1)
You will get Fc=825hz
• RE: Calculation of carrier frequency -Kashyap (08/07/16)
• (850-800)/2 = 25
Hence the carrier frequency should be 25 HZ. How does the answer is provided as 825 HZ?
Follow us | {"url":"https://www.careerride.com/view/calculation-of-carrier-frequency-figure-of-merit-for-ssb-signal-21849.aspx","timestamp":"2024-11-09T11:04:04Z","content_type":"text/html","content_length":"19165","record_id":"<urn:uuid:7156fafd-ab58-4a98-8ef5-031159fa342c>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00010.warc.gz"} |
Kilowatt-hour to Watt-hour Converter
Enter Kilowatt-hour
⇅ Switch toWatt-hour to Kilowatt-hour Converter
How to use this Kilowatt-hour to Watt-hour Converter 🤔
Follow these steps to convert given energy from the units of Kilowatt-hour to the units of Watt-hour.
1. Enter the input Kilowatt-hour value in the text field.
2. The calculator converts the given Kilowatt-hour into Watt-hour in realtime ⌚ using the conversion formula, and displays under the Watt-hour label. You do not need to click any button. If the
input changes, Watt-hour value is re-calculated, just like that.
3. You may copy the resulting Watt-hour value using the Copy button.
4. To view a detailed step by step calculation of the conversion, click on the View Calculation button.
5. You can also reset the input by clicking on button present below the input field.
What is the Formula to convert Kilowatt-hour to Watt-hour?
The formula to convert given energy from Kilowatt-hour to Watt-hour is:
Energy[(Watt-hour)] = Energy[(Kilowatt-hour)] × 1e3
Substitute the given value of energy in kilowatt-hour, i.e., Energy[(Kilowatt-hour)] in the above formula and simplify the right-hand side value. The resulting value is the energy in watt-hour, i.e.,
Calculation will be done after you enter a valid input.
Consider that you have an electric car that consumes 5 kilowatt-hours (kWh) of energy for a full charge.
Convert this energy consumption from Kilowatt-hours to Watt-hour.
The energy in kilowatt-hour is:
Energy[(Kilowatt-hour)] = 5
The formula to convert energy from kilowatt-hour to watt-hour is:
Energy[(Watt-hour)] = Energy[(Kilowatt-hour)] × 1e3
Substitute given weight Energy[(Kilowatt-hour)] = 5 in the above formula.
Energy[(Watt-hour)] = 5 × 1e3
Energy[(Watt-hour)] = 5000
Final Answer:
Therefore, 5 kWh is equal to 5000 Wh.
The energy is 5000 Wh, in watt-hour.
Consider that a residential solar panel system generates 2 kilowatt-hours (kWh) of energy in a day.
Convert this energy generation from kilowatt-hours to Watt-hour.
The energy in kilowatt-hour is:
Energy[(Kilowatt-hour)] = 2
The formula to convert energy from kilowatt-hour to watt-hour is:
Energy[(Watt-hour)] = Energy[(Kilowatt-hour)] × 1e3
Substitute given weight Energy[(Kilowatt-hour)] = 2 in the above formula.
Energy[(Watt-hour)] = 2 × 1e3
Energy[(Watt-hour)] = 2000
Final Answer:
Therefore, 2 kWh is equal to 2000 Wh.
The energy is 2000 Wh, in watt-hour.
Kilowatt-hour to Watt-hour Conversion Table
The following table gives some of the most used conversions from Kilowatt-hour to Watt-hour.
Kilowatt-hour (kWh) Watt-hour (Wh)
0.01 kWh 10 Wh
0.1 kWh 100 Wh
1 kWh 1000 Wh
2 kWh 2000 Wh
3 kWh 3000 Wh
4 kWh 4000 Wh
5 kWh 5000 Wh
6 kWh 6000 Wh
7 kWh 7000 Wh
8 kWh 8000 Wh
9 kWh 9000 Wh
10 kWh 10000 Wh
20 kWh 20000 Wh
50 kWh 50000 Wh
100 kWh 100000 Wh
1000 kWh 1000000 Wh
A Kilowatt-hour (kWh) is a unit of energy that measures the amount of electrical energy consumed or generated over time. One kilowatt-hour is equivalent to one kilowatt (1,000 watts) of power used or
produced for one hour. This unit is commonly used to quantify energy usage in households, industries, and various devices. For example, if a 1,000-watt appliance runs for one hour, it consumes 1 kWh
of energy. Kilowatt-hours are essential for understanding energy consumption, billing in electric utilities, and managing energy efficiency.
A Watt-hour (Wh) is a unit of energy that measures the amount of electrical energy consumed or generated over time. One Watt-hour is equivalent to one watt of power used or produced for one hour.
This unit is commonly used to quantify energy usage in households, industries, and various devices. For instance, a 100-watt light bulb running for 10 hours consumes 1000 watt-hours, or 1
kilowatt-hour (kWh). Watt-hours are essential for understanding energy consumption, billing in electric utilities, and managing energy efficiency.
Frequently Asked Questions (FAQs)
1. How do I convert kilowatt-hours to watt-hours?
To convert kilowatt-hours to watt-hours, multiply the number of kilowatt-hours by 1,000, since one kilowatt-hour equals 1,000 watt-hours. For example, if you have 5 kilowatt-hours, multiplying by
1,000 gives you 5,000 watt-hours.
2. What is the formula for converting kilowatt-hours to watt-hours?
The formula is: \( \text{watt-hours} = \text{kilowatt-hours} \times 1,000 \).
3. How many watt-hours are in a kilowatt-hour?
There are 1,000 watt-hours in one kilowatt-hour.
4. Is 1 kilowatt-hour equal to 1,000 watt-hours?
Yes, 1 kilowatt-hour is exactly equal to 1,000 watt-hours.
5. How do I convert watt-hours to kilowatt-hours?
To convert watt-hours to kilowatt-hours, divide the number of watt-hours by 1,000. For example, 2,500 watt-hours divided by 1,000 equals 2.5 kilowatt-hours.
6. Why do we multiply by 1,000 to convert kilowatt-hours to watt-hours?
Because the prefix 'kilo' means 1,000, so one kilowatt-hour equals 1,000 watt-hours. Multiplying by 1,000 converts kilowatt-hours to watt-hours.
7. How many watt-hours are there in 3.5 kilowatt-hours?
Multiply 3.5 kilowatt-hours by 1,000 to get 3,500 watt-hours. | {"url":"https://convertonline.org/unit/?convert=kilowatt_hour-watt_hour","timestamp":"2024-11-04T17:58:10Z","content_type":"text/html","content_length":"86666","record_id":"<urn:uuid:bbefa8fd-bbaf-49a8-a9b2-a4c196ddf388>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00799.warc.gz"} |
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Math Problem Analysis
Mathematical Concepts
Exponential Growth
Natural Logarithms
Population Growth
N(t) = N_0 e^{rt}
Doubling time formula: t = ln(2) / r
Law of Exponential Growth
Suitable Grade Level
Grades 10-12 (High School Level) or College | {"url":"https://math.bot/q/exponential-growth-doubling-time-bacteria-FMOFQMJa","timestamp":"2024-11-05T23:10:08Z","content_type":"text/html","content_length":"88012","record_id":"<urn:uuid:1495f26e-a64e-412e-8525-6fef3ae021be>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00438.warc.gz"} |
Generate random number from range of numbers in Cobol
Author Message
Venkata Ramayya
New User
I have a requirement to generate random number from range of numbers. I know random function can be used, but how I can generate numbers between range. Can anyone share if you
built similar logic before?
Joined: 03 Dec 2007
Posts: 49 Example:
Location: United
States Low Range - 36
High Range - 60
I would need to get random numbers between 36 and 60
I think something like this:
Global Moderator
The range is 24 (60-36)
Joined: 01 Sep 2006 and assuming the range of a generated random number is something like 0 thru 999 (one thousand possibilities)
Posts: 2589
Location: Silicon 1000 / 24 = 41.6666667
and x is a generated random number:
(x / 41.6666667) + 36 =
and then round off the result.
Pedro wrote:
Senior Member
and assuming the range of a generated random number is something like 0 thru 999 (one thousand possibilities)
Joined: 29 Apr 2008
Posts: 2126
Location: USA
The RANDOM function returns a numeric value that is a pseudorandom number from a rectangular distribution.
The function type is numeric.
FUNCTION RANDOM(argument-1)
If argument-1 is specified, it must be zero or a positive integer. However, only values in the range from zero up to and including 2,147,483,645 yield a distinct sequence of
pseudorandom numbers.
If a subsequent reference specifies argument-1, a new sequence of pseudorandom numbers is started.
If the first reference to this function in the run unit does not specify argument-1, the seed value used will be zero.
In each case, subsequent references without specifying argument-1 return the next number in the current sequence.
The returned value is exclusively between zero and one.
NEW-NUMBER = 36 + (60 - 36) * (FUNCTION RANDOM)
or (to include the upper value 60 as possible choice):
NEW-NUMBER = 36 + (60 - 36 + 1) * (FUNCTION RANDOM)
Venkata Ramayya
Thanks for the response Pedro and sergeyken
New User
Joined: 03 Dec 2007
Posts: 49
Location: United | {"url":"https://ibmmainframes.com/about69362.html","timestamp":"2024-11-06T08:17:04Z","content_type":"text/html","content_length":"38392","record_id":"<urn:uuid:a433d8bc-ad0f-4fc0-b44c-74134adbcab5>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00281.warc.gz"} |
deploying an image
I've captured a win7x32 computer with SCCM 2012.
I've used this as guideline:
and then I tried a deploy it:
I filled in the mac-address of the VM and configured the bios to boot from network.
But I always get this error.
<a href="http://nl.tinypic.com?ref=za6tt" target="_blank"><img src="http://i58.tinypic.com/za6tt.png" border="0" alt="Image and video hosting by TinyPic"></a>
strange that there appears x64 because it's a x32 vm and an x32 capture
I've installed cmtrace and this is what is in SMSPXE.log
Client lookup reply: <ClientIDReply><Identification Unknown="0" ItemKey="16777225" ServerName=""><Machine><ClientID/><NetbiosName/></Machine></Identification></ClientIDReply>
SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
00:50:56:26:8E:7B, 60C54D56-AD9E-578D-2BCC-3A5620F88C11: device is in the database. SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
Client boot action reply: <ClientIDReply><Identification Unknown="0" ItemKey="16777225" ServerName=""><Machine><ClientID/><NetbiosName/></Machine></Identification><PXEBootAction
LastPXEAdvertisementID="" LastPXEAdvertisementTime="" OfferID="" OfferIDTime="" PkgID="" PackageVersion="" PackagePath="" BootImageID="" Mandatory=""/></ClientIDReply>
SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
00:50:56:26:8E:7B, 60C54D56-AD9E-578D-2BCC-3A5620F88C11: no advertisements found SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
00:50:56:26:8E:7B, 60C54D56-AD9E-578D-2BCC-3A5620F88C11: No boot action. Aborted. SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
00:50:56:26:8E:7B, 60C54D56-AD9E-578D-2BCC-3A5620F88C11: Not serviced. SMSPXE 7/04/2014 16:04:13 2680 (0x0A78)
Client boot action reply: <ClientIDReply><Identification Unknown="0" ItemKey="16777225" ServerName=""><Machine><ClientID/><NetbiosName/></Machine></Identification><PXEBootAction
LastPXEAdvertisementID="" LastPXEAdvertisementTime="" OfferID="" OfferIDTime="" PkgID="" PackageVersion="" PackagePath="" BootImageID="" Mandatory=""/></ClientIDReply>
SMSPXE 7/04/2014 16:04:15 2680 (0x0A78)
00:50:56:26:8E:7B, 60C54D56-AD9E-578D-2BCC-3A5620F88C11: no advertisements found SMSPXE 7/04/2014 16:04:15 2680 (0x0A78)
my specs:
Anyone have
win2008r2 server for sccm
sql server 2008
forest functional level 2008r2
capture is of win7x32
any advice?
yes, there is nothing in the collection.
The device is a new VM with no windows installed yet. I imported the computer (mac-address) in devices and added it to the collection. When I open the collection it is empty!?
check the collection properties and verify that the computer you imported is added as a query or direct membership query, if not, add it yourself,
perhaps the collection hasn't updated yet, try to manually update membership.
here;s a quick guide to importing devices into configuration Manager
As anyweb says, you may need to manually add it. I have had it a few times where if SCCM is busy doing other things, it just seems to ignore the fact I'm asking it to move a PC into another
collection lol. I have to do it a couple of times some of the time, and wait a little bit each time, then manually do the update memberships.
I added it to the collection like in anywebs example. That worked. I got to boot from network with f12.
There appears: loading files ...
and then an a blank sccm screen
and then it reboots.
This happens agan and again.
Is there something i missed? Did my capturing somehow fail?
I think it may load a wrong wim file. During the loading a path of some wim-file appears, and it wasn't the one I made from the capture
You may need to add the network card drivers to the Boot Image. Or any other drivers it may not have?
Hyper-V or VMware? If it is VMware you probably need to add network drivers to the boot image.
I don't really understand what you mean. Can you elaborate?
All the computers are VM's.
If you have a look at this, get the VMWare NIC drivers (or what ever VM system you are using...even Virtual Box??), and install them to the boot image | {"url":"https://www.windows-noob.com/forums/topic/10322-deploying-an-image/","timestamp":"2024-11-12T20:34:50Z","content_type":"text/html","content_length":"198626","record_id":"<urn:uuid:33c57526-638b-4270-bd40-05b4f6b83e17>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00351.warc.gz"} |
#1 Square Pyramid Surface Area Calculator
“Unveil the Secrets of Square Pyramids: Calculate Surface Area with Ease!”
Square Pyramid Surface Area Calculator
A Square Pyramid Surface Area Calculator is a tool used to determine the total surface area of a square pyramid, which is a three-dimensional geometric shape with a square base and four triangular
faces converging at a single apex. This calculator is essential in fields such as architecture, engineering, and design, where accurate surface area measurements are required for material
calculations and other analyses.
What is the Purpose of a Square Pyramid Surface Area Calculator?
The primary purpose of a Square Pyramid Surface Area Calculator is to:
• Calculate Surface Area: Accurately compute the total surface area of a square pyramid based on its dimensions.
• Aid in Design and Construction: Help architects, engineers, and designers evaluate material needs and costs associated with square pyramid-shaped structures.
How to Calculate the Surface Area of a Square Pyramid?
The surface area A of a square pyramid can be calculated using the formula:
A = a^2+2a⋅l
• is the total surface area of the square pyramid.
• is the length of one side of the square base.
• l is the slant height of the pyramid.
What Features Should a Square Pyramid Surface Area Calculator Have?
A well-designed Square Pyramid Surface Area Calculator typically includes:
1. Input Fields: Users can enter the length of one side of the square base and the slant height of the pyramid.
2. Calculation Button: A button to perform the surface area calculation based on the input values.
3. Result Display: A section that shows the calculated surface area after performing the calculation.
4. Reset Option: An option to clear the input fields for new calculations.
How to Use a Square Pyramid Surface Area Calculator?
To effectively use the Square Pyramid Surface Area Calculator, follow these steps:
• Enter Base Length: Input the length of one side of the square base.
• Enter Slant Height: Input the slant height of the pyramid.
• Calculate Surface Area: Click the calculate button to obtain the surface area of the square pyramid.
• Review the Result: The calculated surface area will be displayed for further analysis or use.
How Does the Square Pyramid Surface Area Calculator Benefit Users?
The Square Pyramid Surface Area Calculator benefits users by:
• Providing Quick Calculations: It allows for fast and accurate surface area calculations without the need for manual computations.
• Enhancing Design Efficiency: For professionals, it aids in designing elements that require precise surface area measurements.
• Educational Purposes: Students and educators can use it to understand concepts related to geometry and surface areas of three-dimensional shapes.
Why Use a Square Pyramid Surface Area Calculator?
Using a Square Pyramid Surface Area Calculator is beneficial because:
• Accuracy: It reduces human error in calculations.
• Convenience: Users can easily compute surface areas without needing complex mathematical knowledge or formulas.
Square Pyramid Surface Area Calculator FAQs
What is a Square Pyramid Surface Area Calculator?
A Square Pyramid Surface Area Calculator is a tool used to find the total surface area of a square pyramid, a three-dimensional geometric shape with a square base and four triangular faces.
How do you calculate the surface area of a square pyramid?
The surface area of a square pyramid can be calculated using the formula A = B + (1/2)Pl, where B is the area of the base (square), P is the perimeter of the base, and l is the slant height of the
What units are used for the inputs of a Square Pyramid Surface Area Calculator?
The base side length (s) and slant height (l) inputs can be in any unit of length, such as meters, centimeters, inches, etc., as long as they are consistent.
Can a Square Pyramid Surface Area Calculator be used for other shapes?
No, a Square Pyramid Surface Area Calculator is specifically designed for calculating the surface area of square pyramids and cannot be used for other shapes.
Is a Square Pyramid Surface Area Calculator accurate?
Yes, when the inputs are correct, a Square Pyramid Surface Area Calculator provides an accurate calculation of the surface area of a square pyramid.
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volume, square pyramid formula, surface area of a square pyramid with slant height calculator, surface area of a pyramid with slant height | {"url":"https://toolconverter.com/square-pyramid-surface-area-calculator/","timestamp":"2024-11-15T00:06:26Z","content_type":"text/html","content_length":"195004","record_id":"<urn:uuid:df333ad3-c797-4d6e-9b09-80b628970c7e>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00190.warc.gz"} |