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https://boardgames.stackexchange.com/questions/9446/free-city-with-babylon-b-how-to-play-the-special-ability-2nd-stage?noredirect=1 | 1,718,992,626,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862132.50/warc/CC-MAIN-20240621160500-20240621190500-00779.warc.gz | 121,846,252 | 36,683 | # 'Free City' with Babylon (B): How to play the special ability (2nd stage)?
You can use the special ability of Babylon (side B, 2nd stage) to get a 7th turn (EDIT: well, apparently not) where you are allowed to play your last (7th) card (which would otherwise be discarded). You can use this ability in each age (if the 2nd step is built).
How does this work in a 2-player game where the 'Free City' (NPC) uses Babylon?
Let's say the 2nd wonder stage is already built. It's the 6th turn. Player Bob manages the 'Free City'. After playing his and the NPC's card, he has left 1 card on his hand. The 'Free City' pile has 1 hidden card left. Player Alice has also 1 card left on her hand. Now the NPC gets a 7th turn.
• Bob plays his last card?
• Bob draws the last pile card and chooses which of his two cards to play?
• Bob discards his card, draws the last pile card and plays it?
• Bob transfers his card to Alice, and Alice …
• discards her card and plays the one handed over by Bob?
• discards her card, draws the last pile card and chooses which of her two cards to play?
• Alice plays her last card?
• Alice draws the last pile card and chooses which of her two cards to play?
• Alice discards her card, draws the last pile card and plays it?
(did I miss a possibility?)
Note that even if there would be no choice which card to play, it's still important which player has to play it for the 'Free City' (because this player has to decide whether to build the card or the 3rd wonder stage).
Also, if the last card of one of the players has to be played, it's important which player's hand is used, because one of the players might have the wonder Halicarnassus (which allows to build discarded cards).
• Note: The same question is asked on boardgamegeek.com.
– unor
Commented Dec 2, 2012 at 2:35
• Started formulating a read-as-written answer based on the idea that the Free City can't use special abilities (play from discard, double-play, copy guild) since you only play its action, but then there's no clear definition of whether these abilities are used as part of an action. In the end, I really don't know, but I would think it most sensible to have the player ending the round look at the last card in the draw pile and decide whether or not Babylon should play it. Commented Dec 10, 2012 at 16:44
## 1 Answer
The player controlling the Free City during the 6th turn plays the last remaining card in the Free City's draw pile.
The 7-wonders rules define each Wonder boards special abilities. (page 9)
The Hanging Gardens of Babylon the second stage gives the player the option of playing their seventh Age card instead of discarding it. This card can be played by paying its costs, discarded to earn 3 coins or used in the building of the third phase of the Wonder.
Clarification : during the sixth turn, the player can therefore play both cards they have in hand. If the second stage of the Wonder is not built, the Babylon player can then build it on their sixth turn and then play the seventh card instead of discarding it.
There isn't really a 7th turn in an Age, but it helps to think of it as an extra turn, since the resources used to build the 6th card as well as the 6th card itself are available to build the 7th and last card. This 7th card is normally the card that you would discard during the 6th turn of an Age, so that is the card that the player controlling the Free City can play to construct a structure, build their wonder, or if both if those are impossible to discard for 3 coins. (Page 7)
Clarification : The Free City cannot discard a card to get 3 coins unless it cannot play a card.
Note : during the sixth turn of the age, there is only one card remaining in the Free city’s draw pile. This last card is discarded at the same time as the seventh and last card of the players. | 923 | 3,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.97899 |
https://oeis.org/A078486 | 1,632,041,856,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00345.warc.gz | 490,960,758 | 4,192 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A078486 Expansion of (x-7*x^2+19*x^3-21*x^4+10*x^5-6*x^6) / (1-9*x+31*x^2-53*x^3+44*x^4-16*x^5+6*x^6). 1
0, 1, 2, 6, 24, 102, 414, 1598, 5982, 22102, 81442, 300562, 1111638, 4117382, 15259738, 56561346, 209629750, 776850166, 2878660394, 10666717442, 39524757670, 146456879830, 542691221946, 2010931777154, 7451478924278, 27611353095414, 102313463160906 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Number of irreducible indecomposable pop-stack permutations of a certain type. LINKS Colin Barker, Table of n, a(n) for n = 0..1000 M. D. Atkinson and T. Stitt, Restricted permutations and the wreath product, Preprint, 2002. M. D. Atkinson and T. Stitt, Restricted permutations and the wreath product, Discrete Math., 259 (2002), 19-36. Index entries for linear recurrences with constant coefficients, signature (9,-31,53,-44,16,-6). MATHEMATICA CoefficientList[Series[(x-7x^2+19x^3-21x^4+10x^5-6x^6)/(1-9x+31x^2- 53x^3+ 44x^4- 16x^5+6x^6), {x, 0, 40}], x] (* Harvey P. Dale, Aug 26 2019 *) PROG (PARI) concat(0, Vec((x-7*x^2+19*x^3-21*x^4+10*x^5-6*x^6)/(1-9*x+31*x^2-53*x^3+44*x^4-16*x^5+6*x^6) + O(x^50))) \\ Colin Barker, May 27 2016 CROSSREFS Sequence in context: A141253 A306672 A324063 * A129817 A230797 A128652 Adjacent sequences: A078483 A078484 A078485 * A078487 A078488 A078489 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Jan 04 2003 EXTENSIONS Replaced definition with g.f. given by Atkinson and Stitt (2002). - N. J. A. Sloane, May 24 2016 STATUS approved
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Last modified September 19 03:31 EDT 2021. Contains 347550 sequences. (Running on oeis4.) | 757 | 2,021 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | latest | en | 0.588108 |
https://socratic.org/questions/544a902c581e2a79716d266f | 1,660,538,106,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572127.33/warc/CC-MAIN-20220815024523-20220815054523-00363.warc.gz | 483,264,840 | 6,015 | # Question #d266f
Oct 25, 2014
$| z - \left(2 + i\right) | < | z + 1 |$
by replacing $z$ by $x + i y$,
$R i g h t a r r o w | x + i y - \left(2 + i\right) | < | x + i y + 1 |$
by separating real parts and imaginary parts,
$R i g h t a r r o w | \left(x - 2\right) + i \left(y - 1\right) | < | \left(x + 1\right) + i y |$
by find the norms,
$R i g h t a r r o w \sqrt{{\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2}} < \sqrt{{\left(x + 1\right)}^{2} + {y}^{2}}$
by squaring,
$R i g h t a r r o w {\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} < {\left(x + 1\right)}^{2} + {y}^{2}$
by multiplying out,
$R i g h t a r r o w {x}^{2} - 4 x + 4 + {y}^{2} - 2 y + 1 < {x}^{2} + 2 x + 1 + {y}^{2}$
by cleaning up a bit,
$R i g h t a r r o w - 2 y < 6 x - 4$
by dividing by -2,
$R i g h t a r r o w y > - 3 x + 2$
Hence, this is a region above the line $y = - 3 x + 2$.
I hope that this was helpful. | 424 | 917 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-33 | latest | en | 0.679738 |
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When trend and seasonality is present in a time series, instead of decomposing it manually to fit an ARMA model using the Box Jenkins method, another very popular method is to use the seasonal autoregressive integrated moving average (SARIMA) model which is a generalization of an ARMA model. SARIMA models are denoted SARIMA(p,d,q)(P,D,Q)[S], where S refers to the number of periods in each season, d is the degree of differencing (the number of times the…
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6 | 847 | 3,534 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-30 | longest | en | 0.888995 |
https://handwiki.org/wiki/Kilowatt | 1,723,735,578,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00756.warc.gz | 228,621,176 | 27,208 | # Watt
(Redirected from Kilowatt)
Short description: SI derived unit of power
watt
Unit systemSI
Unit ofpower
SymbolW
Named afterJames Watt
Conversions
1 W in ...... is equal to ...
SI base units 1 kgm2s−3
CGS units 107 ergs−1
English Engineering Units 0.7375621 ft⋅lbf/s = 0.001341022 hp
The watt (symbol: W) is the unit of power or radiant flux in the International System of Units (SI), equal to 1 joule per second or 1 kg⋅m2⋅s−3.[1][2][3] It is used to quantify the rate of energy transfer. The watt is named in honor of James Watt (1736–1819), an 18th-century Scottish inventor, mechanical engineer, and chemist who improved the Newcomen engine with his own steam engine in 1776. Watt's invention was fundamental for the Industrial Revolution.
## Overview
When an object's velocity is held constant at one meter per second against a constant opposing force of one newton, the rate at which work is done is one watt. $\displaystyle{ \mathrm{1 ~ W = 1 ~ J {/} s = 1 ~ N {\cdot} m {/} s = 1 ~ kg {\cdot} m^2 {\cdot} s^{-3}}. }$
In terms of electromagnetism, one watt is the rate at which electrical work is performed when a current of one ampere (A) flows across an electrical potential difference of one volt (V), meaning the watt is equivalent to the volt-ampere (the latter unit, however, is used for a different quantity from the real power of an electrical circuit). $\displaystyle{ \mathrm{1 ~ W = 1 ~ V \times 1 ~ A}. }$
Two additional unit conversions for watt can be found using the above equation and Ohm's law. $\displaystyle{ \mathrm{1 ~ W = 1 ~ V^2 / \Omega = 1 ~ A^2 {\cdot} \Omega}, }$ where ohm ($\displaystyle{ \Omega }$) is the SI derived unit of electrical resistance.
### Examples
• A person having a mass of 100 kg who climbs a 3-meter-high ladder in 5 seconds is doing work at a rate of about 600 watts. Mass times acceleration due to gravity times height divided by the time it takes to lift the object to the given height gives the rate of doing work or power.[lower-roman 1]
• A laborer over the course of an eight-hour day can sustain an average output of about 75 watts; higher power levels can be achieved for short intervals and by athletes.[4]
## Origin and adoption as an SI unit
The watt is named after the Scottish inventor James Watt.[5] The unit name was proposed by C. William Siemens in August 1882 in his President's Address to the Fifty-Second Congress of the British Association for the Advancement of Science.[6] Noting that units in the practical system of units were named after leading physicists, Siemens proposed that watt might be an appropriate name for a unit of power.[7] Siemens defined the unit within the existing system of practical units as "the power conveyed by a current of an Ampère through the difference of potential of a Volt".[8]
In October 1908, at the International Conference on Electric Units and Standards in London,[9] so-called international definitions were established for practical electrical units.[10] Siemens' definition was adopted as the international watt. (Also used: 1 A2 × 1 Ω.)[5] The watt was defined as equal to 107 units of power in the practical system of units.[10] The "international units" were dominant from 1909 until 1948. After the 9th General Conference on Weights and Measures in 1948, the international watt was redefined from practical units to absolute units (i.e., using only length, mass, and time). Concretely, this meant that 1 watt was defined as the quantity of energy transferred in a unit of time, namely 1 J/s. In this new definition, 1 absolute watt = 1.00019 international watts. Texts written before 1948 are likely to be using the international watt, which implies caution when comparing numerical values from this period with the post-1948 watt.[5] In 1960, the 11th General Conference on Weights and Measures adopted the absolute watt into the International System of Units (SI) as the unit of power.[11]
## Multiples
Submultiples Multiples Value SI symbol Name Value 10−1 W dW deciwatt 101 W daW decawatt 10−2 W cW centiwatt 102 W hW hectowatt 10−3 W mW milliwatt 103 W kW kilowatt 10−6 W µW microwatt 106 W MW megawatt 10−9 W nW nanowatt 109 W GW gigawatt 10−12 W pW picowatt 1012 W TW terawatt 10−15 W fW femtowatt 1015 W PW petawatt 10−18 W aW attowatt 1018 W EW exawatt 10−21 W zW zeptowatt 1021 W ZW zettawatt 10−24 W yW yoctowatt 1024 W YW yottawatt Common multiples are in bold face
Attowatt
The sound intensity in water corresponding to the international standard reference sound pressure of 1 μPa is approximately 0.65 aW/m2.[12]
Femtowatt
Powers measured in femtowatts are typically found in references to radio and radar receivers. For example, meaningful FM tuner performance figures for sensitivity, quieting and signal-to-noise require that the RF energy applied to the antenna input be specified. These input levels are often stated in dBf (decibels referenced to 1 femtowatt). This is 0.2739 microvolts across a 75-ohm load or 0.5477 microvolt across a 300-ohm load; the specification takes into account the RF input impedance of the tuner.
Picowatt
Powers measured in picowatts are typically used in reference to radio and radar receivers, acoustics and in the science of radio astronomy. One picowatt is the international standard reference value of sound power when this quantity is expressed in decibels.[13]
Nanowatt
Microwatt
Powers measured in microwatts are typically stated in medical instrumentation systems such as the electroencephalograph (EEG) and the electrocardiograph (ECG), in a wide variety of scientific and engineering instruments and also in reference to radio and radar receivers. Compact solar cells for devices such as calculators and watches are typically measured in microwatts.[14]
Milliwatt
A typical laser pointer outputs about five milliwatts of light power, whereas a typical hearing aid uses less than one milliwatt.[15] Audio signals and other electronic signal levels are often measured in dBm, referenced to one milliwatt.
Kilowatt
The kilowatt is typically used to express the output power of engines and the power of electric motors, tools, machines, and heaters. It is also a common unit used to express the electromagnetic power output of broadcast radio and television transmitters.
One kilowatt is approximately equal to 1.34 horsepower. A small electric heater with one heating element can use 1 kilowatt. The average electric power consumption of a household in the United States is about 1 kilowatt.[lower-roman 2]
A surface area of 1 square meter on Earth receives typically about one kilowatt of sunlight from the Sun (the solar irradiance) (on a clear day at midday, close to the equator).[17]
Megawatt
Many events or machines produce or sustain the conversion of energy on this scale, including large electric motors; large warships such as aircraft carriers, cruisers, and submarines; large server farms or data centers; and some scientific research equipment, such as supercolliders, and the output pulses of very large lasers. A large residential or commercial building may use several megawatts in electric power and heat. On railways, modern high-powered electric locomotives typically have a peak power output of 5 or 6 MW, while some produce much more. The Eurostar e300, for example, uses more than 12 MW, while heavy diesel-electric locomotives typically produce and use 3 and 5 MW. U.S. nuclear power plants have net summer capacities between about 500 and 1300 MW.[18]:{{{1}}}
The earliest citing of the megawatt in the Oxford English Dictionary (OED) is a reference in the 1900 Webster's International Dictionary of the English Language. The OED also states that megawatt appeared in a 28 November 1947 article in the journal Science (506:2).
Gigawatt
A gigawatt is typical average power for an industrial city of one million habitants and also the output of a large power station. The GW unit is thus used for large power plants and power grids. For example, by the end of 2010, power shortages in China's Shanxi province were expected to increase to 5–6 GW[19] and the installation capacity of wind power in Germany was 25.8 GW.[20] The largest unit (out of four) of the Belgian Doel Nuclear Power Station has a peak output of 1.04 GW.[21] HVDC converters have been built with power ratings of up to 2 GW.[22]
Terawatt
The primary energy used by humans worldwide was about 160,000 terawatt-hours in 2019, corresponding to an average continuous power consumption of 18 TW that year.[23] The most powerful lasers from the mid-1960s to the mid-1990s produced power in terawatts, but only for nanosecond intervals. The average lightning strike peaks at 1 TW, but these strikes only last for 30 microseconds.
Petawatt
A petawatt can be produced by the current generation of lasers for time scales on the order of picoseconds. One such laser is the Lawrence Livermore's Nova laser, which achieved a power output of 1.25 PW by a process called chirped pulse amplification. The duration of the pulse was roughly 0.5 ps, giving a total energy of 600 J.[24] Another example is the Laser for Fast Ignition Experiments (LFEX) at the Institute of Laser Engineering (ILE), Osaka University, which achieved a power output of 2 PW for a duration of approximately 1 ps.[25][26]
Based on the average total solar irradiance of 1.361 kW/m2,[27] the total power of sunlight striking Earth's atmosphere is estimated at 174 PW. The planet's average rate of global warming, measured as Earth's energy imbalance, reached about 0.5 PW (0.3% of incident solar power) by 2019.[28]
Yottawatt
The power output of the Sun is 382.8 YW, about 2 billion times the power estimated to reach Earth's atmosphere.[29]
## Conventions in the electric power industry
In the electric power industry, megawatt electrical (MWe[30] or MWe)[31] refers by convention to the electric power produced by a generator, while megawatt thermal or thermal megawatt[32] (MWt, MWt, or MWth, MWth) refers to thermal power produced by the plant. For example, the Embalse nuclear power plant in Argentina uses a fission reactor to generate 2109 MWt (i.e. heat), which creates steam to drive a turbine, which generates 648 MWe (i.e. electricity). Other SI prefixes are sometimes used, for example gigawatt electrical (GWe). The International Bureau of Weights and Measures, which maintains the SI-standard, states that further information about a quantity should not be attached to the unit symbol but instead to the quantity symbol (e.g., Pth = 270 W rather than P = 270 Wth) and so these unit symbols are non-SI.[33] In compliance with SI, the energy company Ørsted A/S uses the unit megawatt for produced electrical power and the equivalent unit megajoule per second for delivered heating power in a combined heat and power station such as Avedøre Power Station.[34]
When describing alternating current (AC) electricity, another distinction is made between the watt and the volt-ampere. While these units are equivalent for simple resistive circuits, they differ when loads exhibit electrical reactance.
Radio stations usually report the power of their transmitters in units of watts, referring to the effective radiated power. This refers to the power that a half-wave dipole antenna would need to radiate to match the intensity of the transmitter's main lobe.
## Distinction between watts and watt-hours
The terms power and energy are closely related but distinct physical quantities. Power is the rate at which energy is generated or consumed and hence is measured in units (e.g. watts) that represent energy per unit time.
For example, when a light bulb with a power rating of 100W is turned on for one hour, the energy used is 100 watt hours (W·h), 0.1 kilowatt hour, or 360 kJ. This same amount of energy would light a 40-watt bulb for 2.5 hours, or a 50-watt bulb for 2 hours.
Power stations are rated using units of power, typically megawatts or gigawatts (for example, the Three Gorges Dam in China, is rated at approximately 22 gigawatts). This reflects the maximum power output it can achieve at any point in time. A power station's annual energy output, however, would be recorded using units of energy (not power), typically gigawatt hours. Major energy production or consumption is often expressed as terawatt hours for a given period; often a calendar year or financial year. One terawatt hour of energy is equal to a sustained power delivery of one terawatt for one hour, or approximately 114 megawatts for a period of one year:
Power output = energy / time
1 terawatt hour per year = 1×1012 W·h / (365 days × 24 hours per day) ≈ 114 million watts,
equivalent to approximately 114 megawatts of constant power output.
The watt-second is a unit of energy, equal to the joule. One kilowatt hour is 3,600,000 watt seconds.
While a watt per hour is a unit of rate of change of power with time,[lower-roman 3] it is not correct to refer to a watt (or watt-hour) as a watt per hour.[35]
## Explanatory notes
1. The energy in climbing the stairs is given by mgh. Setting m = 100 kg, g = 9.8 m/s2 and h = 3 m gives 2940 J. Dividing this by the time taken (5 s) gives a power of 588 W.
2. Average household electric power consumption is 1.19 kW in the US, 0.53 kW in the UK. In India it is 0.13 kW (urban) and 0.03 kW (rural) – computed from GJ figures quoted by Nakagami, Murakoshi and Iwafune.[16]
3. Watts per hour refers to the rate of change of power being used (or generated). For example, a power plant that changes its power output from 100 MW to 200 MW in 15 minutes would have a ramp-up rate of 400 MW/h. Gigawatts per hour are used to characterize the ramp-up required of the power plants on an electric grid to compensate for loss of output from other sources, such as when solar power generation drops to zero as the sun sets. See duck curve.
## References
1. Newell, David B; Tiesinga, Eite (2019). The international system of units (SI) (Report). Gaithersburg, MD: National Institute of Standards and Technology. doi:10.6028/nist.sp.330-2019. §2.3.4, Table 4.
2. Yildiz, I.; Liu, Y. (2018). "Energy units, conversions, and dimensional analysis". in Dincer, I.. Comprehensive energy systems. Vol 1: Energy fundamentals. Elsevier. pp. 12–13. ISBN 9780128149256.
3. Avallone, Eugene A, ed. (2007), Marks' Standard Handbook for Mechanical Engineers (11th ed.), New York: Mc-Graw Hill, pp. 9–4, ISBN 978-0-07-142867-5 .
4. Klein, Herbert Arthur (1988). The Science of measurement: A historical survey. New York: Dover. p. 239. ISBN 9780486144979.
5. "Address by C. William Siemens". Report of the Fifty-Second meeting of the British Association for the Advancement of Science. 52. London: John Murray. 1883. pp. 1–33.
6. Siemens supported his proposal by asserting that Watt was the first who "had a clear physical conception of power, and gave a rational method for measuring it." "Siemens, 1883, p. 6"
7. "Siemens", 1883, p. 5"
8. Tunbridge, P. (1992). Lord Kelvin: His Influence on Electrical Measurements and Units. Peter Peregrinus: London. p. 51. ISBN 0-86341-237-8.
9. Fleming, John Ambrose (1911). "Units, Physical". in Chisholm, Hugh. Encyclopædia Britannica. 27 (11th ed.). Cambridge University Press. pp. 738–745; see page 742.
10. "Resolution 12 of the 11th CGPM (1960)". Bureau International des Poids et Mesures (BIPM).
11. Ainslie, M. A. (2015). A century of sonar: Planetary oceanography, underwater noise monitoring, and the terminology of underwater sound. Acoustics Today.
12. Morfey, C.L. (2001). Dictionary of Acoustics.
13. "Bye-Bye Batteries: Radio Waves as a Low-Power Source", The New York Times, Jul 18, 2010 .
14. Stetzler, Trudy; Magotra, Neeraj; Gelabert, Pedro; Kasthuri, Preethi; Bangalore, Sridevi. "Low-Power Real-Time Programmable DSP Development Platform for Digital Hearing Aids". Datasheet Archive.
15. Nakagami, Hidetoshi; Murakoshi, Chiharu; Iwafune, Yumiko (2008). "International Comparison of Household Energy Consumption and Its Indicator". ACEEE Summer Study on Energy Efficiency in Buildings. Pacific Grove, California: American Council for an Energy-Efficient Economy. Figure 3. Energy Consumption per Household by Fuel Type. 8:214–8:224. Retrieved 14 February 2013.
16. Elena Papadopoulou, Photovoltaic Industrial Systems: An Environmental Approach, Springer 2011 ISBN:3642163017, p.153
17. (in en-us) 2007–2008 Information Digest (Report). 19. United States Nuclear Regulatory Commission. 2007-08-01. pp. 84-101.
18. Bai, Jim; Chen, Aizhu (11 November 2010). "China's Shanxi to face 5–6 GW power shortage by yr-end – paper". in Lewis, Chris. Reuters.
19. "Not on my beach, please". The Economist. 19 August 2010.
20. "Chiffres clés" (in fr). Electrabel. 2011.
21. Davidson, CC; Preedy, RM; Cao, J; Zhou, C; Fu, J (October 2010), "Ultra-High-Power Thyristor Valves for HVDC in Developing Countries", 9th International Conference on AC/DC Power Transmission, London: IET .
22. Hannah Ritchie; Max Roser (2020). "Global Direct Primary Energy Consumption". Our World in Data (Published online at OurWorldInData.org.). Retrieved 2020-02-09.
23. "Crossing the Petawatt threshold". Livermore, California: Lawrence Livermore National Laboratory.
24. Loeb, Norman G.; Johnson, Gregory C.; Thorsen, Tyler J.; Lyman, John M. et al. (15 June 2021). "Satellite and Ocean Data Reveal Marked Increase in Earth's Heating Rate". Geophysical Research Letters 48 (13). doi:10.1029/2021GL093047. Bibcode2021GeoRL..4893047L.
25. Williams, David R.. "Sun Fact Sheet". NASA.
26. Cleveland, CJ (2007). "Watt". Encyclopedia of Earth.
27. "Solar Energy Grew at a Record Pace in 2008 (excerpt from EERE Network News". Department of Energy). 25 March 2009.
28. "Inverter Selection". Northern Arizona Wind and Sun. | 4,643 | 17,725 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.850935 |
https://mathematica.stackexchange.com/questions/172275/checking-rationality-using-findinstance | 1,722,962,028,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640497907.29/warc/CC-MAIN-20240806161854-20240806191854-00197.warc.gz | 302,477,560 | 42,284 | # Checking Rationality Using FindInstance
I need to find example if $x,\sqrt{x^2+6},\sqrt{x^2+12}$ all can be rational at once or not. But the following command FindInstance[Element[Sqrt[x + 6], Rationals], {x}]returns an errors saying
The methods available to FindInstance are insufficient to find the requested instances or prove they do not exist.
Anyway to bypass this or modify the input to get result?
Edit FindInstance[Element[y - 8.4, Rationals], {y}] also returns the same error. Is this a bug or what?
• Please, elaborate what means "all can be rational at once or not" Commented Apr 30, 2018 at 17:53
• @JoséAntonioDíazNavas for example both √y and √(y+9) are rational for y =16. I need to x such that all three of those terms are rational Commented Apr 30, 2018 at 18:35
Maybe this?:
FindInstance[6 + x^2 == y^2 && 12 + x^2 == z^2, {x, y, z}, Rationals]
{{x -> -(1/2), y -> -(5/2), z -> 7/2}}
Given that by definition $|c|=\sqrt{c^2}$ for $c\in\mathbb{R}$, then the solutions should be:
{{x -> -(1/2), y -> 5/2, z -> 7/2}} || {{x -> 1/2, y -> 5/2, z -> 7/2}}
You might try brute-force search, enumerating all (positive) rationals starting from 0:
x = 0; While[! (Element[Sqrt[x^2 + 6], Rationals] &&Element[Sqrt[x^2 + 12], Rationals]), x = 1/(Floor[x] + 1 - FractionalPart[x])]; x
with simple answer: (* x = 1/2 *).
• relying on a bit of luck that the answer occurs early in the sequence. This could take approximately forever. Commented May 2, 2018 at 19:29
• Above one-liner check 10^6 distinct rationals/second. You can easily get 10^8/sec rewriting in C. Commented May 3, 2018 at 18:41
• Ah, but the space of "all" rationals is infinite. It is a fair point that such a puzzle likely has a "nice" answer though. Commented May 3, 2018 at 19:07
• Space of all rationals representable in computer is finite, e.g. limited by memory. FindInstance probably uses algorithm like this for Integers, so why not for Rationals? Commented May 4, 2018 at 5:27
A systematic look-up for x^2 == y && 9 + y == z^2
sol = Union[{x1/x2, x1^2/x2^2, z1/z2} /.
Solve[(9 + (x1/x2)^2 == (z1/z2)^2 && 0 <= x1 <= a && 0 < x2 <= a &&
0 <= z1 <= a && 0 < z2 <= a) /. a -> 100, {x1, x2, z1, z2},
Integers, Method -> Reduce]]
(* {{0, 0, 3}, {13/28, 169/784, 85/28}, {11/20, 121/400, 61/20},
{16/21, 256/441, 65/21}, {7/8, 49/64, 25/8}, {5/4, 25/16, 13/4},
{8/5, 64/25, 17/5}, {28/15, 784/225, 53/15}, {9/4, 81/16, 15/4},
{65/24, 4225/576, 97/24}, {20/7, 400/49, 29/7}, {63/20, 3969/400, 87/20},
{55/16, 3025/256, 73/16}, {4, 16, 5}, {56/11, 3136/121, 65/11},
{45/8, 2025/64, 51/8}, {80/13, 6400/169, 89/13}, {77/12, 5929/144, 85/12},
{36/5, 1296/25, 39/5}, {35/4, 1225/16, 37/4}, {72/7, 5184/49, 75/7},
{40/3, 1600/9, 41/3}} *)
and for 6 + x^2 == y^2 && 12 + x^2 == z^2
sol2 = Union[{x1/x2, y1/y2, z1/z2} /.
Solve[(6 + (x1/x2)^2 == (y1/y2)^2 && 12 + (x1/x2)^2 == (z1/z2)^2 &&
0 <= x1 <= a && 0 < x2 <= a && 0 <= y1 <= a && 0 < y2 <= a &&
0 <= z1 <= a && 0 < z2 <= a) /. a -> 100, {x1, x2, y1, y2, z1,
z2}, Integers, Method -> Reduce]]
(* {{1/2, 5/2, 7/2}} *)
Seems to the the only solution. | 1,284 | 3,105 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.864373 |
http://mathcrush.com/math_art_worksheets.html | 1,506,417,105,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818695375.98/warc/CC-MAIN-20170926085159-20170926105159-00107.warc.gz | 206,808,144 | 17,917 | # Math Art - Worksheets
Scroll down to see all choices.
## Ordinal Art - Level 1
These one page worksheets introduce ordinals. Example: 1st, 2nd, 3rd, etc. Students color certain objects to recognize their place in a set. There is also a spelling section on each handout.
## Word and Number Form - Art Level 1
This one page, art worksheet reviews word and number forms. Students read the word form of a number and then find and color the flower that matches each answer.
Key concept: Students need to learn how to spell numbers and see both the number and word form of numbers.
## Answer, Find, and ColorPlace ValueAll Levels
This CHRISTMAS art worksheet reviews place value. Students write the correct number for each place value and then color the object with the corresponding letter.
## Greater Than and Less ThanLevel 1
This one page worksheet introduces greater than and less than. It uses alligators to help explain it. Students cut out alligator heads and glue the faces to show which number is bigger.
Key concept: Understanding inequalities and their symbols (<, >).
This art worksheet has students write numbers in standard form and includes a few place value questions.
Key concept: Students should be able to count, read, and write whole numbers up to 100,000, and identify the place value of specific digits in numbers.
## Basic Subtraction - Level 1
This two page worksheet introduces subtraction. The first page is basic practice and the second page has two Answer, Find, and Color activities.
Important note: These handouts were designed to go with a short basic subtraction video which is currently located on YouTube, but can also be used alone.
This art worksheet reviews USA Currency. Students count dollars and change to calculate an amount.
Key concept: Students should be able to solve problems using different combinations of coins and bills.
## Stained GlassMoney Review - Level 1
This art worksheet reviews money (USA Currency). It includes basic change, rounding to the nearest dollar, and adding and subtracting money. Each student chooses their own colors to create a unique art project.
Key concept: Understand money amounts in decimal notation.
This art worksheet reviews time. Students read clocks, count minutes, and estimate hours.
Key concept: Students should be able to tell time, and determine the duration of intervals of time in minutes and hours.
## Time Art - Level 2
This one page, art worksheet helps students practice time. Students answer the word problems, find the answers in the grid, and then shade the squares that match the answers.
These one page art worksheets have students practice reading a schedule. The schedule and picture become more difficult as the levels get higher. For example: Information is left out of the schedule and students must calculate what is missing.
These one page art worksheets review multi-digit addition. Students answer the problems, find the answers at the bottom of the page, and shade in the corresponding number in the picture. The picture is symmetric.
## Multiple Digit Subtraction - Art Level 3
This one page, art worksheet reviews multiple digit subtraction with borrowing. Students answer the problems, and then find and shade in the square that matches each answer.
Helpful idea: Students can use different colors to create more interesting balloons.
These one page art worksheets review multi-digit subtraction. Students answer the problems, and color in the corresponding number for each answer. Odd answers will be one color and even answers another. Questions are different but the picture is the same.
## Add and Subtract ArtLevel 2
This one page worksheet covers addition and subtraction. Students answer the problems, and then find and color the cat that matches each answer. It includes borrowing and carrying.
This art worksheet has students use their prior knowledge of addition to get a better understanding of multiplication. It includes multiple choice questions, grouping, and number lines.
Key concept: Use repeated addition, grouping, and counting by multiples to do multiplication.
This art worksheet reviews basic multiplication. It includes solving multiplication with grouping, basic algebra, money, and a few word problems.
Key concept: Practice makes perfect, especially with multiplication.
## The Basics Art - Level 1
This one page worksheet covers addition, subtraction, multiplication, and division. Students choose two colors and answer the questions. After they answer the problems they color the corresponding shapes to create a picture. This worksheet can be used as a HALLOWEEN handout.
## Fill in the Blanks ArtLevel 2
This one page HALLOWEEN worksheet is an addition, subtraction, multiplication, and division practice page. It uses fill in the blanks to help students review. Students answer the problems, and then find and color the object that matches each answer.
These one page art worksheets review multi-digit multiplication. The problems are multiple choice to help with test taking skills. The problems are different but the picture is the same for each level.
These one page CHRISTMAS worksheets review long division. The drawing is the same but the questions are different for each level.
These one page art worksheets review multi-digit division. Students should already know long division without remainders or decimals. They answer the problems, find the answers at the bottom of the page, and shade in the corresponding number in the picture. Questions different but the picture the same.
## Multiplying and Dividing by Multiples of 10Level 2
This one page worksheet reviews multiplying and dividing by multiples of 10. It uses simple multiples (10, 100, and 1,000), and also includes a few problems with slightly more difficult ones (300 and 400).
Key concept: Students should know basic multiplication and division using multiples of ten which will help them develop better strategies for multiplication and division with larger numbers, and estimating.
## Help Video
Vimeo: https://vimeo.com/171025484
Links for you to copy and paste for students
This math art page uses color shading to create a picture. Students divide by multiples of ten. This is great practice for estimating two digit division.
Helpful idea: Use worksheet for special occasion like Mother's Day, birthday, or thank you note. Students can fill in the empty banner.
This math art page uses shading to create a picture. Students use long division to solve the problems.
Student misunderstanding: Most students get lost with two digit division. Estimate the divisor (number on the outside) to a multiple of ten, and use that number to help find the quotient (answer). Use Division Book - Level 2 pages 38 to 47 for review.
These one page, art worksheets review the four basic operations of arithmetic (addition, subtraction, multiplication, and division). The drawing is the same but the questions and answers are different for each level.
## Stained GlassRounding Review - Level 1
This art worksheet reviews rounding. Students round to the nearest ten, hundred, and thousand. Each student chooses their own colors to create a unique art project.
Helpful idea: Students should be able to color or shade in shapes neatly, but if students are struggling with the art use it as extra credit to motivate practice.
This one-page art worksheet has two pictures hidden inside. Students will need to round each number to a specific value. They must answer all the questions but choose between shading in the odd or even numbered problems to generate one of the two different pictures hidden in the worksheet.
## Estimating - Level 1
This one page HALLOWEEN worksheet is an estimation practice page. Students estimate the problems to the nearest whole number. After they answer the problems they shade in the corresponding shapes to create a scary picture.
Key concept: Students need to understand the difference between whole numbers and decimals (pieces).
This free art worksheet reviews basic arithmetic (add, subtract, multiply, and divide) and estimation.
Student misunderstanding: Students should estimate before doing any operations. Example: 13+29 should be estimated to 10+30=40, NOT 13+29=42 and then estimated to 40.
These one page art worksheets review adding with decimals. Students will need to line up the decimals and digits and add.
Helpful idea: Use picture as extra credit.
These one page art worksheets review subtracting with decimals. Students will need to line up the decimals and digits and subtract.
Key concept: Subtract with decimals.
These one page worksheets cover adding and subtracting decimals. Each worksheet includes basic practice and an Answer, Find, and Shade Activity.
Important note: These handouts are designed to go with a short Adding and Subtracting Decimals Video which is currently located on YouTube, but can also be used alone.
These one page, art worksheets review multiplying decimals. The drawing is the same but the questions are different for each level.
Student misunderstanding: Count how many digits are to the right of the decimal in all the numbers before multiplying. After the final answer is calculated put the decimal back so the same number of digits are to the right.
This art worksheet reviews the basic operations (add, subtract, multiply, and divide) with decimals.
Student misunderstanding: Multiplying: the total number of digits to the right of the decimal should be the same in the final answer. Dividing: cannot have a decimal in the divisor (outside number). Whatever you do to the outside you have to do to the inside.
This one page art worksheet reviews basic decimals. It includes changing simple fractions to decimals, understanding the word form, plotting decimals on a number line, and two word problems.
Student misunderstanding: Students should understand the difference between tenths and hundredths (Example: 0.4 and 0.04.) and know that 0.4 and 0.40 are equal.
## Answer, Find, and ColorDecimal Review - Level 2
This art worksheet reviews decimals. It covers word form of decimals, greatest decimal value, rounding, and problem solving. Students choose two different colors and color the odd numbered problems one color and the even ones the other color.
Key concept: Students understand decimals and solve problems involving decimals.
This math art page uses shading to create a picture. Students find the common denominator to add fractions.
Student misunderstanding: Students need to know their basic multiples to find the common denominator. It's a good idea to practice basic multiplication facts on a regular basis.
## Answer, Find, and ShadeSubtracting Fractions with Unlike Denominators - Level 2
This math art page uses shading to create a picture. Students find the common denominator to subtract fractions.
Student misunderstanding: Students need to know their basic multiples to find the common denominator. It's a good idea to practice basic multiplication facts on a regular basis.
This math art page uses shading to create a picture. Students divide fractions using the reciprocal method and cross canceling to simplify.
Student misunderstanding: When dividing fractions change the division to multiplication and flip the fraction on the right.
## Mixed Problems with FractionsLevel 1
This one-page art worksheet covers arithmetic with fractions. Students use their knowledge of basic operations to solve fraction computations.
Key concept: Students perform calculations and solve problems involving addition, subtraction, and simple multiplication and division of fractions.
## Stained GlassEstimating Fractions - Level 2
This art worksheet covers estimating fractions. Students need to estimate the mixed numbers to the nearest whole number and then solve. Each student chooses their own colors to create a unique art project.
Student misunderstanding: Estimating is something to make your life easier, so students need to estimate before trying to solve the problem.
## Stained Glass - Level 2Comparing Fractions and Decimals
This one page art worksheet covers comparing fractions and decimals. Students circle the greatest amount in each group and then color the related shapes in the picture. Each student chooses their own colors to create a unique art project.
Key concept: Students need to be able to compare fractions and decimals and be able to visualize the size of a number.
This math art page uses shading to create a picture. Students use their factoring skills to find the GCF.
Key concept: Factoring is the key to simplifying fractions. If a student can factor / divide then they usually understand simplifying.
## Mean, Median, Mode, RangeLevel 2
This one page HOLIDAY art worksheet reviews mean, median, mode, and range. Students answer the problems, and then find and color the turkey that matches each answer.
Key concept: Students should know what each term means and how to find the solution.
## Answer, Find, and Shade (2 in 1)Mean, Median, Mode, and RangeLevel 2
This one-page art worksheet has two pictures hidden inside. Students will need to find the mean, median, mode, and range for each set. They must answer all the questions but get to choose to only shade in the odd numbered or even numbered problems to get one of the two different pictures.
## Stained GlassUnit Rate - Level 2
We call this art worksheet Stained Glass. The picture is symmetrical and students choose their own colors to create a unique art project. The worksheet covers unit rate and students answer the problems, find the numbers in the picture, and color the shapes based on the colors they chose.
Key concept: Understand that rate is a measure of one quantity per unit value of another quantity.
These one page art worksheets review adding integers. Students are given two or three numbers to add. Each level has different questions, but the picture and answers are the same.
These one page art worksheets review subtracting integers. Students are given two or three numbers to subtract. Each level has different questions, but the picture and answers are the same.
These one page abstract art worksheets have two pictures hidden inside. Students are given two or three numbers to multiply or divide. They must answer all the questions but get to choose to only shade in the odd numbered or even numbered problems to get one of the two different pictures. They will be more motivated to finish because they will want to compare their work.
## IntegersMultiplying and Dividing Art
These one page, art worksheets cover multiplying and dividing integers. Students pick their own colors and answer each problem. Then they find and color the character that matches each answer.
Helpful idea: Even if students have not learned about integers or order of operations you can use it as a quick introduction and then review their basics.
This one page art worksheet reviews arithmetic with rational numbers, positive and negative integers, fractions, and decimals.
Key concept: Students need to know and review their basics to progress through higher math.
An alternative version of Dividing by 10's (above). Students use the Order of Operations to solve expressions.
Teacher misunderstanding: Many teachers like to use the phrase, "Please Excuse My Dear Aunt Sally," to help students remember the Order of Operations. This is helpful, but many students then think that multiplication comes before division and addition comes before subtraction.
This one page art HALLOWEEN worksheet reviews integers and order of operations.
Key concept: Understanding positive and negative numbers and the order of operations.
These one page Halloween worksheets review verbal sentences and variables. Students read the sentences (equations) and find the missing number. The drawing is the same but the questions are different for each level.
These one page, art worksheets review the distributive property. Level one has students fill in the blanks, and level 2 and 3 has them solve equations. The drawing is the same but the questions are different for each level.
These one page worksheets have students use the distributive, associative, and commutative properties to simplify expressions. The drawing is the same but the shading and questions are different for each worksheet.
Version 1
Version 2
This one page art worksheet reviews squares and square roots. The square roots are perfect squares so students will not have to estimate and no irrational numbers will be left over. Watch out for the decimals.
## Answer, Find, and Color - Level 1Fill in the Blank
This free winter holiday, one page art worksheet reviews basic order of operations by having students fill in the blanks to complete various equations.
Help idea: Allow students to choose their own colors so the art work becomes more personal and unique. Just shade in lightly.
These one, page art worksheets use fill in the blank questions to review order of operations and solving equations. The drawing is the same but the questions are different for each level.
Helpful idea: Use the picture to help students see patterns. If they do the worksheet incorrectly than the shape will not look correct.
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding:
1. Variables next to each other or a number next to
a variable is the same as multiplication.
For Example: abc = a x b x c or 2h = 2 x h
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding:
1. Variables next to each other mean multiplication.
For Example: abc = a x b x c
2. Be careful replacing the decimal after multiplying.
For Example: 0.6 x 0.6 = 0.36, but 0.6 x 6 = 3.6
This one page art worksheet reviews substitution, order of operations, and basic algebra. Students substitute the values into the expressions and solve.
Student misunderstanding: Variables next to each other mean multiplication.
For Example: abc = a x b x c.
These one page art worksheets review solving equations, substitution, charts, and learning real world equations. Students are given a chart and must fill in the empty spaces. Level one and two include equations to help students, but the level 3 worksheet makes students research or use prior knowledge.
## Solving One Step EquationsLevel 2
This one page HALLOWEEN worksheet is a basic algebra practice page. Students use addition, subtraction, multiplication, and division to solve one step equations. After they answer the problems they shade in the corresponding shapes to create a scary picture.
Key concept: Solving variables and basic equations.
## One-Step EquationsLevel 2
This one page WINTER HOLIDAY worksheet covers solving one-step equations. It includes addition, subtraction, multiplication, and division. Students answer the problems and then shade in the corresponding shapes to create a picture.
## Combining Like Terms ArtLevel 1
This one page, art worksheet has students practice combining like terms. It includes using the Distributive Property to simplify the expressions.
Student misunderstanding: Even though a part of an expression or equation is surrounded by parenthesis, it can still be pulled apart using the Distributive Property. For example: 3(y + 4) = 3y + 12.
## Two-Step Fill in the Blanks
These one page art worksheets cover solving two-step equations. They help students visualize the process which should lead to solving equations faster.
Helpful idea: If students are confused change the blanks to a variable and solve the equation.
## Answer, Find, and ColorMulti-Step Equations
These one page art worksheets review multi-step equations. Level one includes combining like terms, Level two includes The Distributive Property, and Level three includes fractions.
This math art worksheet reviews solving equations. Students answer the problems, and then find and shade in the area that matches each answer.
Key concept: Understanding that we can change equations or "move numbers and variables" as long as we do the same to both sides.
These one page Scary Halloween art worksheets review solving functions. Students need to substitute the given number into the function and solve. The picture and answers are the same but the problems are different for each level.
This one page, art worksheet reviews solving inequalities. Students solve two inequalities and find the answer they have in common.
Student misunderstanding: Multiplying or dividing each side of an inequality by a negative number reverses the inequality sign. To prove this have students check their answers.
This art worksheet covers decimals and percents. Students look at squares and write the decimal and percent of the shaded part. In addition, there are a few find the percent of a number problems.
Key concepts: Interpret percents as a part of a hundred, find decimal and percent equivalents for common fractions, and compute a given percent of a number.
These one page art worksheets have two pictures hidden inside. Students need to use their knowledge of percent of numbers to solve various equations. Examples: 42.5% of 59 = ?, ? % of 85 = 10, or 40% of ? = 81. They must answer all the questions but for the shading students have the option to only shade in the odd numbered or even numbered problems to get one of the two different pictures.
These one page, art worksheets review percent of a number. Students need to find the percent of a number, what percent one number is of another, and find the original number when a percent is given.
Helpful idea: The two versions have the same questions, but in slightly different order and the pictures have been altered. Give half the class one version and the rest the other.
These one page art worksheets review percent of change. Students are given an original and new number and must figure out if the change is an increase or decrease and then calculate the percent of change. After solving, students find the answers in the picture and shade in the corresponding shapes.
These one page art worksheets review fractions, decimals, and percent. Problems include finding the greatest number, percent and fraction of a number, arithmetic with fractions and decimals, and a word problem.
This FREE one page art worksheet reviews Probability. It is a tribute to the 2014 FIFA World Cup. Students read word problems and determine the probability of a certain event occurring.
Key concept: Students make predictions for simple probability situations.
These one page, art worksheets review proportions. Students find the missing value in each proportion. The drawing is the same but the questions are different for each level.
Helpful idea: The pattern is seen best from a distance.
This art worksheet covers linear measurement. Students find the distance of line segments using number lines and equations.
Key concept: Students should be able to calculate the distance between two points and read, understand, and visualize numbers on a number line.
This art worksheet includes word problems on distance. Students use a distant chart in kilometers to answer various questions, and convert kilometers to miles or miles to kilometers.
Important note: To shade the picture correctly students must use 1 kilometer = 0.62 miles when converting.
This one page, art worksheet reviews finding the slope from two points. Students calculate the slope using the change in x and y.
Helpful idea: When subtracting the coordinates, change the second point to its opposite.
Example: Given two points: (3 , 4);(2 , -2) - subtract
( 3 , 4)
(-2 , +2) opposite of the numbers
( 1 , 6) , answer: slope is 6.
## Word Problems Art - All Levels
These one page WINTER HOLIDAY worksheets are simple words problems designed for all levels. Note: The picture and answers are the same for each level, but the questions are different.
Level 1 Print
Level 2 Print
Level 3 Print
These one page, art worksheets review word problems. The drawing is the same but the questions are different for each level.
Can you see it?
This one-page art worksheet covers vertical and adjacent angles. Students use their knowledge of lines and vertical angles to calculate missing angles.
Key concept: Use facts about supplementary, complementary, vertical, and adjacent angles to calculate unknown angles in a figure.
## Graphing / Coordinate Plane - A
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - B
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Where is the origin? The origin is where the x-axis and y-axis cross. It is also where the zero goes for both axes.
## Graphing / Coordinate Plane - C
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - D
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Where is the origin? The origin is where the x-axis and y-axis cross. It is also where the zero goes for both axes.
## Graphing / Coordinate Plane - E
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Graphing / Coordinate Plane - FLevel 3
This one page worksheet is on plotting ordered pairs. Students graph and connect the points to create a picture.
Student misunderstanding: Which number is the x-axis and which one is the y-axis? Have to go into the elevator before you can go up or down, or have to move the ladder before you climb it. So the first number is the x-axis (number going across).
## Line of Symmetry Art- All Levels
These one page Easter worksheets use line of symmetry and art to review student knowledge. Students re-draw the second half of the shape to create the full rabbit.
Key concept: Line of symmetry is a line such that if a figure is folded about the line, then one half of the figure matches the other half.
## Angle Art - Level 2
Students need to find the missing angle for each polygon. This one page worksheet covers triangles, quadrilaterals, pentagons, and hexagons. Students choose two colors and answer the questions. After they answer the problems they color the corresponding shapes to create a picture. This worksheet can be used as a HALLOWEEN handout.
Alternative version to The Basics Art - Level 1.
## Answer, Find, and Shade - Level 3Distance of a Line Segment
This one page art worksheet reviews finding the distance of a line segment. Students will need to know the Distance Formula, but also can use the Pythagorean Theorem on most of the problems. The problems include line segments on graphs and two coordinates with no graph.
This one page art worksheet reviews the distance formula. Students find the distance between two points. All answers are rational numbers. This worksheet can also be used during Halloween.
Helpful idea: Students should be allowed to use a calculator.
These one page worksheets cover converting capacity. There is one worksheet for the USA Customary System of Measurement and one for the Metric System. The drawing is the same but the questions and answers are different.
USA Customary System of Measurement
Metric System | 5,631 | 28,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-39 | latest | en | 0.912684 |
http://docs.itascacg.com/3dec700/common/models/cwipp/doc/modelcwipp.html | 1,670,591,177,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00265.warc.gz | 11,378,138 | 7,396 | WIPP-Salt Model
A crushed-salt constitutive model is implemented in FLAC3D/3DEC to simulate volumetric and deviatoric creep compaction behaviors. The model is a variation of the WIPP model, and is based on the model described by Sjaardema and Krieg (1987), with an added deviatoric component as proposed by Callahan and DeVries (1991).
Formulations
In the crushed-salt constitutive model, the material density, $$\rho$$, is a variable that evolves as a function of compressive volumetric strain, $$\epsilon_v$$, from the initial crushed-salt emplacement value, $${\rho}_i$$, to the ultimate intact salt density, $${\rho}_f$$. The relation between the rate of change of volumetric strain and density for use in the FLAC3D/3DEC incremental Lagrangian formulation may be outlined as follows. (Remember that, as a convention, stresses and strains are negative in compression.)
Consider a given material domain of mass, $$m$$ (which at time, $$t$$, has volume, $$V_{\circ}$$, and density, $${\rho}_{\circ}$$), and let the volumetric strain increment, $$\Delta \epsilon_v$$, correspond to a change in volume, $$\Delta V$$, and in density from $$\rho_{\circ}$$ to $$\rho$$ during the time interval, $$\Delta t$$. By virtue of mass conservation, we have
(1)$\rho_{\circ} V_{\circ} = \rho (V_{\circ} + \Delta V)$
and, by definition of volumetric strain, we obtain
(2)$\rho = {{\rho_{\circ}}\over{1 + \Delta \epsilon_v}}$
Also, from a continuum approach, we may write
(3)$\rho = {{m}\over{V}}$
and the rate-of-change of density of the given mass is
(4)$\dot{\rho} = -{{m}\over{V^2}} \dot{V}$
Using $$\dot{\epsilon_v} = \dot{V} / V$$ together with Equation (3), after some manipulation we obtain
(5)$\dot{\epsilon_v} = -{{\dot{\rho}}\over{\rho}}$
A measure of the crushed-salt compaction is given by the fractional density, $$F_d$$, defined as the ratio between actual and ultimate salt densities:
(6)$F_d = {{\rho}\over{\rho_f}}$
In the model implementation, it is assumed that the creep-compaction mechanism is irreversible (the density can only increase and cannot decrease) and bounded (no further compaction occurs after the intact salt value has been reached).
Constitutive Equations
In the crushed-salt model, elastic stress and strain rates are related by means of the incremental expression of Hooke’s law:
(7)$\dot{\sigma}_{ij} = 2G \left[\dot{\epsilon}_{ij}^e -{{\dot{\epsilon}_{kk}^e}\over{3}} \delta_{ij} \right] + K \dot{\epsilon}_{kk}^e \delta_{ij}$
where $${\delta}_{ij}$$ is the Kronecker delta.
In this expression, the bulk modulus, $$K$$, and shear modulus, $$G$$, are related to the density by a nonlinear empirical law of the form
(8)$K = K_f e^{K_1(\rho - \rho_f)}$
(9)$G = G_f e^{G_1(\rho - \rho_f)}$
where $$\rho_f$$, $$K_f$$, and $$G_f$$ are properties of the intact salt, and $$K_1$$, $$G_1$$ are two constants determined from the condition that bulk and shear must take their initial values at the initial value of the density.
It is assumed that, for density values below that of the intact salt, the total strain-rate $$\dot\epsilon_{ij}$$ can be expressed as the sum of three contributions: nonlinear elastic, $$\dot\epsilon_{ij}^e$$; viscous compaction, $$\dot\epsilon_{ij}^c$$; and viscous shear, $$\dot\epsilon_{ij}^v$$. The elastic strain-rate takes the form
(10)$\dot\epsilon_{ij}^e = \dot\epsilon_{ij} - \dot\epsilon_{ij}^c - \dot\epsilon_{ij}^v$
The viscous compaction term is based on an experimental compaction-rate law of the form
(11)$\dot{\rho}^c = -B_0 \left[ 1 - e^{-B_1 \sigma} \right] e^{B_2 \rho}$
where $$\sigma = \sigma_{kk} / 3$$ is the mean stress, and $$B_0$$, $$B_1$$, $$B_2$$ are constants determined experimentally from results of isotropic compaction tests.
The volumetric compaction strain-rate $$\dot\epsilon_v^c$$ may be derived after substitution of Equation (11) for $$\dot{\rho}$$ in Equation (5):
(12)$\dot{\epsilon_v^c} = {{1}\over{\rho}} B_0 \left[ 1 - e^{-B_1 \sigma} \right] e^{B_2 \rho}$
In the this implementation, it is assumed that volumetric compaction can only take place if the mean stress is compressive. Furthermore, a cap is assumed for the preceding expression so that no further compaction arises once the intact salt density has been reached.
Viscous Compaction
The total compaction strain rate has the expression
(13)$\dot\epsilon_{ij}^c = \dot\epsilon_v^c \left[ {{\delta_{ij}}\over{3}} - \beta {{{\sigma_{ik}^d}{\delta_{kj}}}\over{\bar{\sigma}}} \right]$
where $$\sigma_{ij}^d$$ is the deviatoric stress tensor, $$\bar\sigma = \sqrt{3J_2}$$ is the von Mises stress, and $$J_2 = \sigma_{ij}^d\sigma_{ij}^d / 2$$. In this formula, the parameter $$\beta$$ is a constant set equal to one, so that in a uniaxial compression test, the lateral compaction strain-rate components vanish.
Viscous Shear
The viscous shear strain rate corresponds to that of the WIPP model. The primary creep strain rate is the same as that given in the WIPP model, but the secondary creep strain-rate has the deviatoric stress magnitude, $$\bar\sigma$$, divided by the fractional density (Equation (6)). It has the form
(14)$\dot\epsilon_s=D\,{\left( {{\bar\sigma}\over{F_d}} \right)}^n e^{(-Q/RT)}$
where the parameters are as previously defined.
As the material approaches full compaction, the fractional density approaches one. Because a cap is introduced to eliminate further creep compaction when the intact salt density is reached, the viscous shear behavior evolves toward that of the intact salt. Note that in the framework of the WIPP model, the intact salt creep behavior is triggered by deviatoric stresses, while the volumetric behavior is elastic.
Implementation
In the this implementation of the crushed-salt model, the total stresses and strain rates are decomposed into volumetric and deviatoric components. The incremental equations governing the volumetric behavior are linearized and solved explicitly for the mean stress increment. The creep compaction strain-rate is then derived and used in the expression for the deviatoric behavior whose implementation otherwise closely follows that adopted for the WIPP model. Finally, total stresses for the step are evaluated from the updated volumetric and deviatoric components.
References
Callahan, G.D., and K.L. DeVries. Analysis of Backfilled Transuranic Waste Storage Rooms, RE/SPEC, Inc., report to Sandia National Laboratories SAND91-7052 (1991).
Sjaardema, G.D., and R.D. Krieg. A Constitutive Model for the Consolidation of WIPP Crushed Salt and Its Use in Analyses of Backfilled Shaft and Drift Configurations, Sandia National Laboratories, SAND87-1977 (1987).
wipp-salt Model Properties
Use the following keywords with the zone property (FLAC3D) or zone property (3DEC) command to set these properties of the WIPP-salt model.
wipp-salt
activation-energy f
activation energy, $$Q$$
bulk f
bulk modulus, $$K$$
bulk-final f
final, intact salt, bulk modulus, $$K_f$$
constant-a f
WIPP model constant, $$A$$
constant-b f
WIPP model constant, $$B$$
constant-d f
WIPP model constant, $$D$$
constant-gas f
gas constant, $$R$$
compaction-0 f
creep compaction parameter, $$B_0$$
compaction-1 f
creep compaction parameter, $$B_1$$
compaction-2 f
creep compaction parameter, $$B_2$$
creep-rate-critical f
critical steady-state creep rate, $$ε̇^*_{ss}$$
density-final f
final, intact salt, density, $$ρ_f$$
density-salt f
density, $$ρ$$
exponent f
WIPP model exponent, $$n$$
poisson f
Poisson’s ratio, $$v$$
shear f
elastic shear modulus, $$G$$
shear-final f
final, intact salt, shear modulus, $$G_f$$
temperature f
zone temperature, $$T$$
young f
Young’s modulus, $$E$$
compaction-bulk f (r)
creep compaction parameter, $$K$$
compaction-shear f (r)
creep compaction parameter, $$G$$
density-fractional f (r)
current fractional density, $$F_d$$
Key
• Only one of the two options is required to define the elasticity: bulk modulus $$K$$ and shear modulus $$G$$, or Young’s modulus $$E$$ and Poisson’s ratio $$v$$.
• The two densities: density-salt $$ρ$$, and density-final $$ρ_f$$ , should be considered as internal variables for the constitutive level calculation only, which should not be confused with the real material density. | 2,275 | 8,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2022-49 | latest | en | 0.869354 |
http://www.slideserve.com/bernad/zuhone-flash | 1,493,306,353,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122174.32/warc/CC-MAIN-20170423031202-00631-ip-10-145-167-34.ec2.internal.warc.gz | 679,986,073 | 19,963 | # Novae and Mixing - PowerPoint PPT Presentation
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Novae and Mixing John ZuHone ASCI/Alliances Center for Thermonuclear Flashes University of Chicago Overview Purpose What is FLASH? Mixing in Novae Setting Up the Problem Doing the Problem Conclusions Purpose
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## Novae and Mixing
John ZuHone
ASCI/Alliances Center for Thermonuclear Flashes
University of Chicago
### Overview
• Purpose
• What is FLASH?
• Mixing in Novae
• Setting Up the Problem
• Doing the Problem
• Conclusions
### Purpose
• To develop a numerical simulation using the FLASH code to simulate mixing of flulds at the surface of a white dwarf star
• Understanding this mixing will contribute to our understanding of novae explosions in binary systems containing a white dwarf star
### What is FLASH?
• FLASH is a three dimensional hydrodynamics code that solves the Euler equations of hydrodynamics
• FLASH uses an adaptive mesh of points that can adjust to areas of the grid that need more refinement for increased accuracy
• FLASH also can account for other physics, such as nuclear reactions and gravity
### What is FLASH?
• Euler equations of hydrodynamics
¶r/¶t + Ñ • rv = 0
¶rv/¶t + Ñ • rvv + ÑP = rg
¶rE/¶t + Ñ • (rE + P) v = rv • g
where
E = e + ½v2
### What is FLASH?
• Pressure obtained using equation of state
• ideal gas
P = (g- 1)re
• other equations of state (i.e. for degenerate Fermi gases, radiation, etc.)
• For reactive flows track each species
¶rXl/¶t + Ñ • rXlv = 0
### Mixing in Novae
• What is a nova?
• novae occur in binary star systems consisting of a white dwarf star and a companion star
• the white dwarf accretes material into an accretion disk around it from the companion
• some of this material ends up in a H-He envelope on the surface of the white dwarf
### Mixing in Novae
• this material gets heated and compressed by the action of gravity
• at the base of this layer, turbulent mixing mixes the stellar composition with the white dwarf composition (C, N, O, etc.)
• temperatures and pressures are driven high enough for thermonuclear runaway to occur (via the CNO cycle) and the radiation causes the brightness increase and blows the layer off
### Setting Up the Problem
• Initial Conditions
• what we want is a stable model of a white dwarf star and an accretion envelope in hydrostatic equilibrium
• we get close enough to the surface where Cartesian coordinates (x, y, z) and a constant gravitational field are valid approximations
### Setting Up the Problem
• Hydrostatic Equilibrium
• to ensure a stable solution we must set up the initial model to be in hydrostatic equilibrium, meaning v = 0 everywhere
• momentum equation reduces to
ÑP = rg
• set this up using finite difference method, taking an average of densities
### Setting Up the Problem
• Procedure for initial model
• set a density at the interface
• set temperature, elemental abundances
• call equation of state to get pressure
• iterate hydrostatic equilibrium condition and equation of state to get pressure, density, etc. in rest of domain
### Setting Up the Problem
• Region I: 50% C, 50% O, T1 = 107 K
• Region II: 75% H, 25% He, T2 = 108 K
• Density at interface:
ro = 3.4 × 103 g cm-3
### Doing the Problem
• load the model in and see if the simulation is in hydrostatic equilibrium
• it ISN’T!
• high velocities at interface and boundary
• begin to examine the model for possible flaws
### Doing the Problem
• Question: Is the model itself really in hydrostatic equilibrium?
• test the condition, discover that the model is in hydrostatic equilibrium to about one part in 1012
• Question: Is the resolution high enough?
• try increasing number of points read in, increase refinement, still no change
### Doing the Problem
• Question: Is the density jump across the interface hurting accuracy?
• smooth out density jump by linearly changing temperature and abundances
• velocities slightly lower, but still present
• try this for a number of different sizes of smoothing regions, still no change
### Doing the Problem
• Check the equation of state
• the Helmholtz equation of state we were using was complex
• accounts for gas, degenerate electrons, and radiation
• switch to gamma equation of state to see if anything improves
• NO IMPROVEMENT!
## MovieTime!
(maybe)
### Doing the Problem
• Two important resolutions
• there was an error in temperature calculation which was caused by a mismatch in precision of numerical constants
• we found that if we used the same number of points in FLASH as we did the initial model some of the inconsistency was resolved
### Doing the Problem
• Which brings us to where we currently are…
• we believe that by our linear interpolation for the density is too imprecise
• we are currently implementing a quadratic interpolation for the density
### Conclusions
• What have we learned?
• stability is important
• the need for there to be a check within FLASH itself for hydrostatic equilibrium
• the need to carefully examine all parts of a code to look for possible mistakes
• consistency!
### Conclusions
• Thanks to:
• Mike Zingale and Jonathan Dursi
• Prof. Don Lamb
• the ASCI FLASH Center
• the University of Chicago | 1,377 | 5,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-17 | longest | en | 0.862321 |
http://ask.cvxr.com/t/concave-function/4813 | 1,544,526,569,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823618.14/warc/CC-MAIN-20181211104429-20181211125929-00603.warc.gz | 27,068,533 | 2,749 | # Concave function?
(Chethan Kumar Anjinappa) #1
Hi All,
Is the function `f(x,y,z) = y*log(1+(c*x/(y*z)))` is concave ? If so, how can I implement it in cvx (using rel_entr)?
with x,y,z>=0
(Mark L. Stone) #2
f(x,y,z) is neither convex nor concave. For instance, at x = y = z = 1, its Hessian has 2 negative eigenvalues and 1 positive eigemvalue, so is indefinite there.
If z were a constant, not a (CVX) variable, f could be expressed as `-rel_entr(y,y+c*x/z)` and would be concave as a function of x and y. | 169 | 514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-51 | latest | en | 0.897466 |
http://www.ck12.org/arithmetic/Problem-Solving-Plan-Diagrams/lesson/Problem-Solving-Plan-Diagrams/r11/ | 1,455,358,745,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701166261.11/warc/CC-MAIN-20160205193926-00253-ip-10-236-182-209.ec2.internal.warc.gz | 341,016,973 | 39,363 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Problem Solving Plan, Diagrams
## Develop selected strategies for problem solving.
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Problem Solving Plan, Diagrams
On his last day with Uncle Larry, Travis worked with Mr. Wilson on laying tile on the kitchen floor. Travis worked hard all morning and he was a bit discouraged when he reached his first break and realized that he had only finished about one-third of the floor. It had taken Travis two hours to tile one-third of the floor. He thought about this as he drank from his water bottle and ate an apple. “If it took me this long to tile one-third, how long will it take me to finish?” Travis wondered.
The floor is divided into 12 sections. If he has finished one-third of them, how many sections has he completed? This is the number that he completed in the two hours.
How many sections does he have left to complete? About how long will it take him to finish the rest?
There are many different strategies you could use to help Travis solve this problem, but drawing a diagram is probably the most useful. This Concept will show you how to effectively use a diagram to solve a problem.
### Guidance
You have been learning about fractions and mixed numbers and about how to add and subtract them. Many of the examples in these Concepts have used pictures to help you learn to solve them. Drawing a diagram or a picture is a strategy to help you solve many different problems. The first thing that you have to do when approaching a problem is to read and understand the problem and how to solve it.
John ate 15\begin{align*}\frac{1}{5}\end{align*} of the cake. What fraction is left?
First, you can see that we have the amount of cake the John ate and we need to know how much he has left. We are going to be subtracting. Let’s draw a diagram to show what we know about John and his cake.
Now that we have looked at what we know and what we need to know, we can draw the diagram. This is a diagram of fraction bars to represent John’s cake. The blue section shows how much of the cake John has eaten. The white bars represent the amount of cake that is left.
Here is the one-fifth that John ate. You can see that there are four-fifths left.
The answer to the problem is four-fifths.
Sometimes, we can set up a problem as addition and sometimes we can set it up as subtraction. Often times both ways will work but one will make more sense than the other.
Shannon jogged 1320\begin{align*}1 \frac{3}{20}\end{align*} miles yesterday. Today, she jogged 12\begin{align*}\frac{1}{2}\end{align*} mile. How many total miles did Shannon jog?
Method one –– Draw a diagram:
One way to solve this problem is to draw a diagram. Let’s start by looking at the first distance that Shannon jogged. Draw two same-sized rectangles. Divide one rectangle into 20 equal-sized sections. Then shade 1320\begin{align*}1 \frac{3}{20}\end{align*} of the diagram.
This represents the 1320\begin{align*}1 \frac{3}{20}\end{align*} miles that Shannon jogged yesterday.
Shannon also jogged 12\begin{align*}\frac{1}{2}\end{align*} mile today.
So, shade 12\begin{align*}\frac{1}{2}\end{align*} of the partially filled rectangle to represent the distance she jogged today.
The diagram is 11320\begin{align*}1 \frac{13}{20}\end{align*} shaded. So, Shannon jogged a total of 11320\begin{align*}1 \frac{13}{20}\end{align*} miles on those two days.
Method two –– Set up an addition problem:
To find out how many miles she jogged all together, add 1320+12\begin{align*}1 \frac{3}{20} + \frac{1}{2}\end{align*}. The fractional part of the mixed number has a different denominator than 12\begin{align*}\frac{1}{2}\end{align*}. Find the least common multiple (LCM) of both denominators. The least common multiple of 20 and 2 is 20. Next, we rename the fractions.
12=1020
Now we can add the two together.
1320+1020=11320
Notice that our answer is the same. Both methods will produce the same result. You can choose the method that you find easiest when working on problems like this.
Now it's time for you to try a few. Draw a diagram and solve each problem.
#### Example A
213+4\begin{align*}2 \frac{1}{3} + 4\end{align*}
Solution:613\begin{align*}6 \frac{1}{3}\end{align*}
#### Example B
4515\begin{align*}\frac{4}{5} - \frac{1}{5}\end{align*}
Solution:35\begin{align*}\frac{3}{5}\end{align*}
#### Example C
34+24\begin{align*}\frac{3}{4} + \frac{2}{4}\end{align*}
Solution:114\begin{align*}1 \frac{1}{4}\end{align*}
Let’s use a diagram to help Travis with his tiling project. Here is the original problem once again.
On his last day with Uncle Larry, Travis worked with Mr. Wilson on laying tile on the kitchen floor. Travis worked hard all morning and he was a bit discouraged when he reached his first break and realized that he had only finished about one-third of the floor. It had taken Travis two hours to tile one-third of the floor. He thought about this as he drank from his water bottle and ate an apple. “If it took me this long to tile one-third, how long will it take me to finish?” Travis wondered. The floor is divided into 12 sections. If he has finished one-third of them, how many sections has he completed? This is the number that he completed in the two hours. How many sections does he have left to complete? About how long will it take him to finish the rest?
First, let’s underline all of the important information to help us read and understand the problem. Let’s figure out how much of the floor Travis has finished. First, let’s find an equivalent fraction for one-third with a denominator of 12.
13=412
Next, we can draw a diagram of the finished part of the floor.
Here is a picture of what Travis has finished.
How much does he have left?
We can count the units and see that he has 812\begin{align*}\frac{8}{12}\end{align*} of the floor left to tile. This is double what he did in two hours.
Travis has about four hours of work left.
Travis finishes his break and gets back to work. If he continues working at the same pace, he will finish working around 2 pm just in time for some pizza for lunch.
### Vocabulary
Problem Solving
using key words and operations to solve mathematical dilemmas written in verbal language
Diagram
a drawing used to represent a mathematical problem.
### Guided Practice
Here is one for you to try on your own.
Teri ran 112\begin{align*}1 \frac{1}{2}\end{align*} miles yesterday, and she ran 212\begin{align*}2 \frac{1}{2}\end{align*} miles today. How many miles did she run in all?
If John ran 7 miles, what is the difference between his total miles and Teri’s total miles? How many miles have they run altogether?
We can solve this problem a couple of different ways. First, we could draw a diagram of the path of both runners.
112+212=4\begin{align*}1 \frac{1}{2} + 2 \frac{1}{2} = 4\end{align*} miles
Terri ran 4 miles.
John ran 7 miles.
There is a difference of 3 miles.
Together, they ran 11 miles.
### Practice
Directions: Solve each of the following problems by using a problem solving strategy.
1. Tyler has eaten one-fifth of the pizza. If he eats another two-fifths of the pizza, what part of the pizza does he have left?
2. What part has he eaten in all?
3. How many parts of this pizza make a whole?
4. Maria decides to join Tyler in eating pizza. She orders a vegetarian pizza with six slices. If she eats two slices of pizza, what fraction has she eaten?
5. What fraction does she have left?
6. If Tyler was to eat half of Maria’s pizza, how many pieces would that be?
7. If Maria eats one-third, and Tyler eats half, what fraction of the pizza is left?
8. How much of the pizza have they eaten altogether?
9. John and Terri each ran 18 miles. If Kyle ran half the distance that both John and Teri ran, how many miles did he run?
10. If Jeff ran 312\begin{align*}3 \frac{1}{2}\end{align*} miles, how much did he and Kyle run altogether?
11. What is the distance between Jeff and Kyle’s combined mileage and John and Teri’s combined mileage?
12. Sarah gave Joey one-third of the pie. Kara gave him one-fourth of another pie. How much pie did Joey receive altogether?
13. Is this less than or more than one-half of a pie?
14. Who gave Joey a larger part of the pie, Kara or Sarah?
15. What is the difference between the two fractions of pie?
### Vocabulary Language: English
Diagram
Diagram
A diagram is a drawing used to represent a mathematical problem.
Problem Solving
Problem Solving
Problem solving is using key words and operations to solve mathematical dilemmas written in verbal language. | 2,253 | 8,843 | {"found_math": true, "script_math_tex": 22, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2016-07 | latest | en | 0.966313 |
https://codeproject.freetls.fastly.net/Articles/706136/Csharp-Bin-Packing-Cutting-Stock-Solver?msg=5722453 | 1,631,969,236,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00304.warc.gz | 226,116,508 | 32,300 | 15,030,744 members
Articles / General Programming / Algorithms
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Posted 5 Jan 2014
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# C# Bin Packing - Cutting Stock Solver
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Application for solving Bin Packing and Cutting Stock problem
## Introduction
The original idea behind this article is a proposal for the solution of two well known classes of Operation Research problems: the Bin Packing and the Cutting Stock with many sizes of Bin/Stock. Altough this work meets the traditional one dimension problem, the exposed approach could be employed to solve the two and three dimensions problem. This work also include some useful features, such as the possibility of considering cost and quantities constraints in searching solutions. Finally, remaining in the very spirit of CodeProject, two more non- design time features has been added: constraints on minimum reject, and a method for improving solutions by making "local moves" - see the History section below and the related Comments and Discussion section.
## Background
In the Bin Packing Problem a set of items must be grouped into bins of a certain size (capacity of the bin): the aim is to minimize the total number of bins.
In the Cutting Stock Problem a set of items must be cutted from Stocks of a certain size (length of the stock): the aim is to minimize the total number of stocks.
In the past several different methods were developed by researcher in order to solve this couple of very interesting and practical problems, including linear programming, heuristics, and evolutionary algorithms such as genetic algorithms and ant colony optimization systems [1, 2, 3, 4, 5]. Being known that in Operation Research only small size problems can be exhaustively investigated in order to find the "Absolute optimum" solution, it's obvious that any solver attempt aims to find a more general "Optimum solution" which only sometimes can be also an "Absolute optimum".
To complicate things it should be noticed that in practice it's fairly common to tackle problems with bins of multiple sizes, or stocks of multiple length. This leads to an exponential growth of the problem size.
It should be clear by definitions given that the Bin Packing Problem and the Cutting Stock Problem share a common nature. The algorithm proposed in this work is able to handle both of them.
## Using the application
To understand how this application works we follow an example that I took from
http://www.codeproject.com/Questions/553711/optimizedpluscuttingplussolutionplus-knapsack-2fbi
For simplicity, I write directly here the question:
Quote:
Hi,
I need the following solution and have been battling to get it right. I have also searched high and low but could not find anything that I could use. Perhaps I am not understanding my own problem well enough.
I need to be able to programmatically find the best way to cut lengths of Rope/Pipe.
Here is the scenario:
I have the following lenthgs in stock: 3 X 4.7m 5 X 7.4m 3 X 11.2m 4 X 9.8m (these are lengths of rope that I currently have in stock).
I need to cut the following lengths for an order: 2 X 3m 4 X 1.2m 1 X 6m 2 X 5.4m
The application exploits a Windows Form which is suitable for a comfortable data input. In this example, at the original problem has been added dummies constraints on minimum rejects:
The Form exposes two `DataGridView` with their respective `BindingNavigators` and some `Buttons`
• SEARCH, which starts the solving process for the current problem
• RESET, which clears both the DataGridView for the set up of a new problem
• OPEN, which load a file saved problem
• SAVE, which save the current input data into a file
There is also a `CheckBox `with a red text that will appear at some point of the researching process, and in the lowest part of the Form there is a `ToolStrip` which brings a `ProgressBar` and a `Label` that give information about what is happening.
In the top-right part of the Form a message in the downright spirit of CodeProject can be red.
The user should insert the size and the number of pieces in the `DataGridViewItems`.
By default, if there isn’t explicit input by the user, the program will automatically consider a minimum of 1 piece for the just inserted item. A `TextBox` embedded in the `BindingNavigatorItems` will show the real-time total sum of the items.
The `DataGridViewStocks` should be filled with the size, the cost and the maximum number of available pieces for each stock (if it’s limited). To effectively consider this upper limit in the calculation a confirm by `CheckBox` is required. This is why it’s possible to simulate different scenarios simply checking or unchecking checkboxes.
Be aware that only the stock size is always taken into consideration during the algorithm, as will be clear after seen the flow chart.
One more thing is about the cost. Obviously this is the cost of the stocks, but for further analysis this value could include other parameters, such as cost for shipment and so on. In this way, a same size of stocks can be evaluated with two different costs and maximum number of pieces.
In the driver example written above, and downloadable at the top of the article, there isn’t information regarding the costs, so I fixed them in order to show how many result can be found.
However when the input is accomplished the “SEARCH” button must be pressed, and the program runs according to the following flow chart:
The First step is to build a quick solution, regardless the cost or the limited number of available pieces.
This solution has name “Bound Solution” and will be an upper bound for any further attempt of improvement.
The second step consists in a deeper look for a better solution, and it’s now that come into play the ‘optional parameters’ cost and max number of pieces.
As one can see running the application with the given input data, at this point our driver example finds two solution: one is a “Best Cost Solution”, while the other is a “Best Size Solution”.
Now if the problem is enough small and if the “Absolute Best Size Solution” (mathematically speaking regarding the total size of the solutions found) hasn’t been met, a further search is performed. The aim of this final step is to fit one of the many bin/stock combinations which have minimum sum. There is no guarantee that such a solution exists, and this process can be time consuming. At this point appears the above mentioned `CheckBox` with red text “End Search”, as the picture below shows. When checked, the search algorithm stops and results are immediately shown.
This snapshot gives an idea of the result:
:
By the way, for the driver example doesn’t exist any “Absolute Best Size Solution”. Howewer, if you are curious to see this event happen, you can slightly modify the item of size 5,4: change its size into 5,8 and ask for only 1 piece of it, then press “SEARCH” and wait a few seconds.
## A look to the algorithms
When the “SEARCH” `Button` is pressed, a CuttingStock object is constructed and a solving process begins according to the previous flow chart .
The Bound Solution is obtained exploiting a “Greedy First Fit” algorithm. In simple word, this algorithm processes all the items and assigns them to the biggest bin/stock available. When the bin/stock can’t hold the item, it tries to assign this item at the first bin/stock that can hold it. If no bin/stock has enough space, than a new bin/stock is used. After the assignement procedure the algorithm look for an improvement following these ordered steps:
1. Trying to reduce the size of some bin/stock if it's employ is less or equal the size of another bin/stock available. This is performed calling the DownSize(List<Bin> mySolution) method.
2. Trying to qualify the solution: the QualifySolution(List<Bin> mySolution) method is called, and it explores the possibility to free some space in the bin/stock making a series of local moves. The following, short paragraph will give more details about this idea.
3. Trying again to DownSize the solution .
Once got a Bound Solution, the application build a set of potential improving solution exploiting the BranchAndBound class. This class is inspired to the Branch & Bound strategy, which is fairly common in Operation Research and gives back a `List` of potential better solutions which is both sorted on Cost and on Size and is passed to a “Greedy Best Fit” algorithm until new solutions will be found. This greedy algorithm works in a similar way to the “Greedy First Fit”, the main differences being that now it tries to fit the items with the set derived by the Branch & Bound and that when a bin/stock can’t hold the item, it tries to assign this item at the bin/stock which can hold it giving the lowest reject. Hower, if it runs succesfully, it is able to return a so called "Best Size Solution" and "Best Cost Solution", both of them being precessed by the already told QualifySolution(List<Bin> mySolution) method.
At this point a new search will be performed only to achieve the Absolute Best Size with a "Greedy Next Fit" algorithm applied to the branches of the Branch & Bound space with size equal to the "Absolute optimum" value. This step will not be executed if "Best Size Solution" is also "Absolute Best", or if there are too much items to process (the design choice is 12). This greedy algorithm is the simplest and the quickest of all: the items are placed in a bin/stock until it’s enough empty, than a new bin/stock is added. However, to be sure every possible permutation of the items in ListOfItems must be tried to fit a given set of bins/stocks. This is the main reason for which I made the choice to introduce a constraint on the number of items to process. This seeming simple idea took several lines of code, and I leave the comments in it to better clarify the concept.
## What is a Qualified Solution?
According to the statements given in the Background section, the aim of the Bin Packing or Cutting Stock problem is to minimize the amount of row material - bins or stocks. Provided that we are able to get this purpose, we can consider a more advanced target: to get a solution which allow us a good reusability of its reject. Given a problem and its solutions, provided that they have the same total size of bins/stocks, we can consider as "more qualified" a solution which gives a reject still reusable in a next job process. At the opposite, we can consider as "less qualified" a solution whose reject is not reusable and so it's completely useless. Altough this aspect seems to be interesting, I haven’t found references to it in the work coming from the field of research. At the same way, I didn’t find any attention in optimizing according to the “Cost” or giving a limitation at the number of some stock - the two features I decided to set in the application at design time. To fit the need to get a "more qualified" solution the dedicated method QualifySolution(List<Bin> mySolution) has been added to the code. The basic plan behind this action in "qualifying" is to free space in the bins/stocks trying to move items from one bin/stock to another: this one is chosen with a "Best Fit" criteria, whilst the items to move are selected from the greatest to the lowest running back from the last bin/stock to the first bin/stock in the examined solution.
## Is it a good idea to ask for a minimum rejects?
As I said in the introduction, the main purpose of this article is to find a good, and possibly quick, solution at the traditional Bin Packing/Cutting Stock Problem, whose solving time may be unacceptable even for small set of bin/stocks and small number of items. This concept implies the objective function is the minimization of the row materials employed. Thus we can consider as good choices all that ideas which doesn't increase the row material consumption - regardless to what could happen in the next job, or in the next of the next... Inspired to this principle is the "local move" that I called "QualifySolution". The constraint of "Minimum Reject" could lead to a not worst solution. In this case it would be another good "local moves" for our job. On the other hand, more often the result could be worst. In this case, I suggestat least to compare this results with those obtained excluding the minimum rejects. Here is a point where readers and users can make their experiences, possibly giving reasonable feedback at the CodeProject community through the Comments and Discussion section.
## Points of Interest
This is my first C# application, and I must say that although I’m not at all a software developer I met several interesting feature in C# and .NET framework which I hope could be useful for other beginners:
• The use of `DataGridView `control with in-memory data structure instead of a physical database;
• The implementation of several sorting method for the `List` collections;
• The useful `Linq` query and the method to put the query result in a separated collection;
• The powerful couple of classes `BinaryFormatter` and` FileStream `which make simple storing in one file many collections of objects.
## References
[1] P.C. Gilmore, R.E. Gomory. A linear programming approach to the cutting stock problem. Operations Research, 9:848-859, 1961.
[2] Silvano Martello, Paolo Toth. Knapsack Problems, Algorithms and Computer Implementations. John Wiley and Sons Ltd., England, 1990.
[3] Emanuel Falkenauer, Alain Delchambre. A genetic algorithm for bin packing and line balancing. Proceedings of the IEEE 1992 International Conference on Robotics and Automation, Nice, France, May 1992.
[4] Emanuel Falkenauer. A hybrid grouping genetic algorithm for bin packing. Journal of Heuristics, 2:5-30, 1996.
[5] Rhyd Lewis. A General-Purpose Hill-Climbing Method for Order Independent Minimum grouping Problems: A Case Study in Graph Colouring and Bin Packing. Computers and Operations Research, vol. 36(7), pp. 2295-2310.
## History
• 2014-01-04
• Original article
• 2014-01-06
• 2014-08-02
• Modified both the article and the code due to the post dated 25-May-2014:
• Updated article
## Share
Engineer Italy
I am an Engineer experienced in designing instruments and controls for industrial polluted air and V.O.C treatment plants. Currently I work in designing radiant heating/cooling systems and related controls, solar systems and other solutions for buildings energy efficiency.
My first personal computer was a ZX Spectrum Sinclair that in the early ‘80s I used so much in playing games and learning the bases of Basic. Then in my career I met other languages. Other than programming my hobbies are graphology, acoustic guitar, good books, my orchard, astronomy and play chess.
I live in the North West of Italy, very close to mountains and lakes and I like simple things. In summer you can find me somewhere walking on the mountains or going around with my sport bike. In winter you can probably meet me in my chessclub.
If you are planning to visit Italy I will be glad to show you some of this beautiful places and landscapes.
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source code Matin Pirasteh26-Jul-21 2:31 Matin Pirasteh 26-Jul-21 2:31
source code Member 1278414925-Jul-21 22:45 Member 12784149 25-Jul-21 22:45
source code AlexPentagon1-Jun-21 6:15 AlexPentagon 1-Jun-21 6:15
source code Member 31451707-Apr-21 23:58 Member 3145170 7-Apr-21 23:58
Source Code Member 1479207927-Dec-20 0:44 Member 14792079 27-Dec-20 0:44
I would like the code in c # andreire23-Nov-20 5:54 andreire 23-Nov-20 5:54
ciao Alberto mi manderesti il codice ? andreire22-Nov-20 22:10 andreire 22-Nov-20 22:10
Re: ciao Alberto mi manderesti il codice ? OriginalGriff22-Nov-20 22:13 OriginalGriff 22-Nov-20 22:13
Source Code please shanojscp20-Oct-20 4:34 shanojscp 20-Oct-20 4:34
delete it! db_developer14-May-20 21:01 db_developer 14-May-20 21:01
My vote of 1 db_developer14-May-20 20:58 db_developer 14-May-20 20:58
This resource supposes to post a SOURCE CODE! I m looking for solution not something else. Possibly it is even not his app at all.
Source code Member 1420334617-Nov-19 3:50 Member 14203346 17-Nov-19 3:50
Source code Member 1156117628-Oct-19 4:52 Member 11561176 28-Oct-19 4:52
Source code Member 1106619316-Oct-19 14:47 Member 11066193 16-Oct-19 14:47
My vote of 5 Member 1453822722-Jul-19 10:59 Member 14538227 22-Jul-19 10:59
source Member 1453822722-Jul-19 10:36 Member 14538227 22-Jul-19 10:36
Hi Alberto allviewsystem from Bayamon15-May-19 9:26 allviewsystem from Bayamon 15-May-19 9:26
Re: Hi Alberto Member 145177741-Jul-19 9:22 Member 14517774 1-Jul-19 9:22
Source code aselan3-May-19 21:52 aselan 3-May-19 21:52
Source Code? Jeff.Bk3-May-19 0:04 Jeff.Bk 3-May-19 0:04
Source code? tampicoscott5-Mar-19 20:17 tampicoscott 5-Mar-19 20:17
Re: Source code? Member 145177741-Jul-19 9:23 Member 14517774 1-Jul-19 9:23
where is source code? Member 1358481814-Feb-19 17:35 Member 13584818 14-Feb-19 17:35
Source Code Member 1231561128-Jan-19 5:34 Member 12315611 28-Jan-19 5:34
No sources available wwwings17-Jan-18 5:35 wwwings 17-Jan-18 5:35
Last Visit: 31-Dec-99 18:00 Last Update: 18-Sep-21 2:47 Refresh 123 Next ᐅ | 4,096 | 17,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-39 | latest | en | 0.942619 |
https://matcomgrader.com/problem/9576/chinese-curves/ | 1,582,724,382,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146342.41/warc/CC-MAIN-20200226115522-20200226145522-00455.warc.gz | 448,520,309 | 52,917 | C - Chinese curves
Languages: C, C++, Java, Python, ... (details)
You are in a world with curves of the form $f(x) = \arctan(e ^ x + a) \sqrt{b \cdot x^2 + c}$, where $a$, $b$ and $c$ are integers. In this world, there are two types of queries:
• Given three integers $a$, $b$ and $c$, create a new curve with parameters $a$, $b$ and $c$.
• Given an integer $p$, determine the minimum value of $f(p)$ among all curves.
The following image corresponds to the sample test case, where the blue curve is the last curve added.
Here is an explanation of the sample input:
1. From the input, mask $m$ is intially set to $0$.
2. Query 1 creates a curve with parameter $a = 1000$, $b = 1$, $c = 1$ (since mask $m = 0$).
3. Query 2 asks for the minimum value of $f(1)$ among all curves. Since there is only one curve, answer is simply $\arctan(e ^ 1 + 1000) \sqrt{1^2 + 1}$, which is about $2.22$.
4. After answering query 2, mask $m$ becomes $\lfloor 2.22 \rfloor \oplus 0 = 2$.
5. Query 3 creates a curve with parameters $a$, $b$, and $c$ equal to $3 \oplus 2 = 1$.
6. As $p = 3 \oplus 2 = 1$, query 4 asks for the minimum value of $f(1)$ among the two existing curves. From the image, we can see that the last added curve yields a minimum value of about $1.85$.
7. After answering query 4, mask $m$ is updated to $\lfloor 1.85 \rfloor \oplus 2 = 3$.
8. The last query asks to find the minimum value of $f(41 \oplus 3) = f(42)$ among the two curves.
Input
In the first line, there are two integers $q$ and $m$, where $q$ is the number of queries to follow $(1 \le q \le 10 ^ 5)$ and $m$ is a mask $(0 \le m \le 10 ^ 5)$. Each of the following $q$ lines contains a query using one of these two formats:
• $1$ $a$ $b$ $c$: if the query is of the first type, where $0 \le b \le 10 ^ 6$, $0 \le a, c \le 10 ^{18}$.
• $2$ $p$: if the query is of the second type, where $0 \le p \le 10 ^ 6$.
In the input, the values of $a$, $b$, $c$ and $p$ are not given directly. Instead, you should restore them from the inputs $a'$, $b'$, $c'$ and $p'$ by computing $a = a' \oplus m$, $b = b' \oplus m$, $c = c' \oplus m$, and $p = p' \oplus m$, where $m$ is the current value of mask $m$ and $\oplus$ is the bitwise xor operator (available as the ^ symbol in many programming languages).
After each query of type 2, the mask $m$ changes to $\lfloor s \rfloor \oplus m$, where $s$ is the answer to the query, $\oplus$ is the bitwise xor operator, and $\lfloor s \rfloor$ is the floor function that yields the integer part of number $s$.
It is guaranteed that the first query is of type 1.
Output
For each query of type 2, output a line with the required answer, rounding to two digits of precision after the decimal point.
Sample test(s)
Input
5 0 1 1000 1 1 2 1 1 3 3 3 2 3 2 41
Output
2.22 1.85 65.99 | 937 | 2,792 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-10 | longest | en | 0.783241 |
https://aic-components.com/useful-articles/can-460-volts-be-single-phase.html | 1,638,318,100,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00047.warc.gz | 162,770,555 | 18,079 | # Can 460 volts be single phase?
Contents
## Is 460 volts the same as 480?
RE: Difference between 460V and 480V regarding…
480 volts is the system voltage. 460 volts is the rated voltage of your equipment. 460 volt rated equipment is always used on a 480 volt system to account for voltage drop due to starting or line losses.
## Is 440 volt the same as 460 volt?
460V was frequently used as the nameplate voltage for motors used on a 480V system. So a motor rated at 460V (3-phase) will work at 440V, indefinitely, without overheating or not delivering rated output.
## How many volts is a single-phase motor?
Common 60hz voltages for single-phase motors are 115 volt, 230 volt, and 115/230 volt. Common 60hz voltages for three-phase motors are 230 volt, 460 volt and 230/460 volt. Two hundred volt and 575 volt motors are sometimes encountered. In prior NEMA standards these voltages were listed as 208 or 220/440 or 550 volts.
## Is 480V always 3 phase?
480V can be classified as single and 3 phase circuits. 480V 3 phase circuits are the most common power systems used in US industrial plants and are considered to be low voltage power systems.
## Is 480 single phase?
Single phase 480 is 277 volt. To get 480 you need two phase at any given time to give you 480 volts.
THIS IS IMPORTANT: Which of the following code is used to give cutter offset compensation right of a part surface?
## Is 480 volts considered High Voltage?
Generac states that generators less than and equal to 600 volts are medium-voltage and generators greater than 600 volts are considered high voltage.
## Can a 460V motor run on 440V?
The current will adjust accordingly and as voltage drops the motor will become less efficient, but the motor will still deliver rated output. So a motor rated at 460V (3-phase) will work at 440V, indefinitely, without overheating or not delivering rated output.
## Can a 440V motor run on 480V?
A standard NEMA motor should be good for continuous operation at 110% of nameplate. Even if the voltage at the MCC is 480V, the voltage at the motor will be less. If these are old 440 V motors, they will probably be OK.
## Why 3 phase is called 440 volts not 660 volts?
In 3 phase supply, there are 3 supply lines phase shifted at 120 degrees from each other. So the net voltage difference between the two phases in accordance with the phase angle of 120 degrees is 440V. | 582 | 2,401 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2021-49 | latest | en | 0.886062 |
http://www.pharmtech.com/pharmtech/Analytics/Multifactor-Non-linear-Modeling-for-Accelerated-St/ArticleStandard/Article/detail/848456?contextCategoryId=50207 | 1,416,777,089,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400380037.17/warc/CC-MAIN-20141119123300-00033-ip-10-235-23-156.ec2.internal.warc.gz | 809,292,106 | 40,836 | Multifactor Non-linear Modeling for Accelerated Stability Analysis and Prediction - Pharmaceutical Technology
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Multifactor Non-linear Modeling for Accelerated Stability Analysis and Prediction
The right approach can provide a clear, statistically defendable method for determining dissolution and accelerated stability.
Jul 2, 2014 Pharmaceutical Technology Volume 38, Issue 7, pp. 46-49
Accelerated stability analysis is a strategy used to quickly evaluate alternative formulations, packaging, and processes. Accelerated linear studies are commonly performed and modeled; however, accelerated multiple-factor non-linear modeling has been a gap, and statistical software tools such as SAS/JMP do not directly have any provision to model multiple factor nonlinear responses. This paper outlines an approach to model and predict non-linear multiple factor stability/tablet dissolution data under accelerated and nominal storage conditions.
RELATED CONTENT Data Integrity in the Analytical Laboratory More Rapid Extractable & Leachable Analyses with Advanced Mass Spectrometers More Analytics
There are many non-linear stability cases such as dissolution, leachables, and moisture. Being able to model these non-linear processes is crucial to proper drug development. In addition to general non-linear modeling, there are multiple factors that may influence the non-linear curve. The following is a list of factors that may impact the asymptotes, growth rate, and inflection point of a curve:
• Stability storage temperature and humidity
• Particle size
• pH
• Amount of an excipient
• Processing conditions and/or set points
• Packaging materials/method
Study design
Proper design of experiments for data collection and curve isolation is crucial for building non-linear models. Figure 1 illustrates the factors that should all be square relative to all other factors and have zero correlation relative to each other. For this tablet dissolution example, multiple time points (minutes), multiple storage conditions (temperature), multiple drug substance particle sizes, and multiple weeks were measured. Percent dissolution was the response of interest.
Analysis method
The following is a step-by-step procedure for non-linear stability modeling and expiry determination.
Step one. Measure the data at multiple time periods, using multiple particle sizes and at multiple temperatures. Generate a plot of the data to visualize the relationship of the curves over time (25-10-0: 25=Temperature, 10=Particle Size, and 0 = days) (see
Figure 2).
Step two. Fit each curve individually. In this example, each dissolution curve was fit using a four-parameter logistics (4PL) curve. The four parameters are: upper asymptote, lower asymptote, inflection point, and slope of the dissolution curve. R-Square should be high (typically above 0.95) and RMSE error should be low for each curve. Outliers should be checked using the residuals. Save the parameters of the curve. In this example, there are four parameters, upper asymptote, lower asymptote, growth rate (slope), and inflection point (see Figure 3).
Step three. Save the parameters from the curve and the factors that influence them into a table. For this example, the growth rate and inflection point are the coefficients that may change the most based on the factors under consideration. The upper and lower asymptotes are not of concern for this problem as all of the curves have similar lower asymptotes and similar upper asymptotes, but upper and lower asymptotes could be important for other problems, so generally it is best to model all of the curve parameters (see Figure 4).
Step four. Fit the curve parameters with a least-squares multivariate regression. Growth rate, the slope of the 4PL fit, and inflection point are the most important as the dissolution starting point and the upper asymptote are essentially the same for all curves. Main effects and interaction models generally work best and p-values and F tests can be used to evaluate each model term (see Figure 5).
Step five. Save the equation from the multivariate parameter model. An example of the inflection point model is in Equation 1.
[Eq. 1]
Step six. Substitute the growth rate (slope) and inflection point coefficients from the multivariate model into the nonlinear prediction. The red box in Figure 6 shows the substitution for the inflection point.
Step seven. Check the model to make sure it matches the actual data. Correct any modeling errors by adding or modifying a multivariate model term. A simple YX regression plot of the model versus the measurement will indicate model quality and any systematic errors (see Figure 7).
Step eight. Create a profiler from the equation to predict future dissolution rates. This can be done using a modern statistical package such as SAS/JMP (see Figure 8).
Step nine. Predict dissolution at any time, temperature, or particle size combination using the profiler. For this example, particle size was set to 5 μm, temp to 25 and time in weeks to zero. The dissolution time was fixed at 100 min with a specification of not less than 90% (see Figure 9).
Step ten. From the profiler at each time point (weeks), make sure the time (min) and particle size (5 um) are fixed, determine dissolution at the nominal temperature (non-accelerated condition). Fit the rate of degradation using either a linear or nonlinear model from the profiler predicted data. In this case, the rate of dissolution is not linear so a non-linear curve is fit to the data and the expiry is determined based on the extrapolated curve (see Figure 10).
The same method is used for predicting both the nominal expiry and the 95% CI expiry.
Step eleven. Finally, long-term stability evaluation at nominal storage conditions will be used to confirm the early model prediction and will provide an independent secondary determination of stability and changes in dissolution. Understanding rates of change should factor into shelf life and release specification limits (1).
Summary
Non-linear multiple factor analysis has long been a problem in a variety of process and product modeling and prediction. The novel procedure discussed in this paper for the characterization of multiple factor non-linear product performance provides a clear, statistically defendable method for determining dissolution and accelerated stability. Long-term verification of accelerated conditions should always follow early determinations of expiry, acceleration rates, and rates of degradation.
Reference
1. ICH, Q6A Specifications: Test Procedures and Acceptance Criteria for New Drug Substances and New Drug Products: Chemical Substances (Oct. 6, 1999).
Mark Alasandro, PhD, is director, Allergan Irvine. Thomas A. Little PhD, is president, Thomas A. Little Consulting, drlittle@dr-tom.com.
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https://math.stackexchange.com/questions/727662/a-planar-brownian-motion-has-area-zero/732272 | 1,558,557,894,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256958.53/warc/CC-MAIN-20190522203319-20190522225319-00434.warc.gz | 556,828,021 | 42,541 | # A planar Brownian motion has area zero
I'm looking for proofs of Paul Lévy's theorem that a planar Brownian motion has Lebesgue measure $0$. I know of only two proofs: one is in Lévy's original paper (Théorème 12, p. 532) and the other is in Mörters & Peres's "Brownian Motion" (Theorem 2.24, p. 46).
Unfortunately, Mörters & Peres's proof leaves out many technical details, without which I find the proof hopelessly difficult to understand.
I have not attempted to read Lévy's original proof, partly because it's in French (though I can read French, with effort) and partly because it's in the middle of a long article that was not written for students, but rather for professional mathematicians, and hence I suspect that it would be a pain for me to read it.
I have not been able to find any other sources containing a proof of this theorem. If someone knows of any such sources, in English, French or German, ideally textbooks (but any other source will be helpful too), please let me know.
Thank you.
Unfortunately, I'm not aware of an alternative proof (or a more detailed version of the proof by Mörters/Peres). So, instead of providing you with a source, I'll follow the lines of Paul Lévy; hopefully filling the gaps in his proof.
Let $(B(t))_{t \geq 0}=((B^1(t),B^2(t)))_{t \geq 0}$ be a two-dimensional Brownian motion and denote by $\lambda$ the (two-dimensional) Lebesgue measure.
First, we show that $L := \lambda(B([0,1]))$ satisfies $\mathbb{E}L<\infty$. To this end, note that
\begin{align*} \{L>4r^2\} &\subseteq \left\{ \max_{t \in [0,1]}(|B^1(t)|,|B^2(t)|) > r \right\} \\ &= \left\{ \max_{t \in [0,1]} B^1(t) > r \right\} \cup \left\{ \max_{t \in [0,1]} B^2(t) > r \right\} \\ &\quad \cup \left\{ \min_{t \in [0,1]} B^1(t) < -r \right\}\cup \left\{ \min_{t \in [0,1]} B^2(t) < -r \right\}.\end{align*}
From the reflection principle we know that $\max_{t \in [0,1]} B^j(t) \sim -\min_{t \in [0,1]} B^j(t) \sim |B^j(1)|$ for $j=1,2$. Hence,
$$\mathbb{P}(L>4r^2) \leq 4 \sqrt{\frac{2}{\pi}} \int_r^{\infty} \exp \left( - \frac{y^2}{2} \right) \, dy \leq \frac{4 \sqrt{2}}{\pi} \frac{1}{r} \exp \left(- \frac{r^2}{2} \right).$$
Combining this estimate with the fact that $\mathbb{E}L = \int_0^{\infty} \mathbb{P}(L >r) \, dr$, we get $\mathbb{E}L<\infty$.
Next, we note that for the restarted Brownian motion $W_t := B_{t+1}-B_1$, we have $\lambda(W([0,1])) = \lambda(B([1,2]))$ and, since $W \sim B$, we conclude that $$\mathbb{E}(\lambda(B([0,1]))) = \mathbb{E}(\lambda(B([1,2]))).$$ Similarly, as $\tilde{W}_t := \frac{1}{\sqrt{2}} B_{2t}$ is a Brownian motion, we have
$$\lambda(\tilde{W}([0,1])) = \lambda \left( \frac{1}{\sqrt{2}} B([0,2]) \right) = \frac{1}{2} \lambda(B([0,2])$$
i.e. $\lambda(B([0,2])) \sim 2 \lambda(B([0,1]))=2L$. Therefore,
\begin{align*} 2\mathbb{E}L = \mathbb{E}(\lambda(B([0,2]) &= \mathbb{E}(\lambda(B([0,1])) + \mathbb{E}(\lambda(B([1,2])) - \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) \\ &= 2\mathbb{E}L + \mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))). \end{align*}
Hence, $$\mathbb{E}\big(\lambda(B([0,1]) \cap B([1,2]))\big)=0. \tag{1}$$
By Fubini's theorem, we have
$$\mathbb{E}L = \int \int 1_{B([0,1])}(x,y) d\lambda(x,y) \, d\mathbb{P} = \int \underbrace{\mathbb{P}((x,y) \in B([0,1]))}_{=:p(x,y)} \, d\lambda(x,y). \tag{2}$$
If we set $W_t := B_{1-t}-B_1$ and $\tilde{W}_t := B_{t+1}-B_1$, then both processes are Brownian motions and
$$W([0,1]) = B([0,1])-B_1 \qquad \qquad \tilde{W}([0,1]) = B([1,2])-B_1.$$
Using that $(W_t)_{t \geq 0}$ and $(\tilde{W}_t)_{t \geq 0}$ are independent, we get
\begin{align*}\mathbb{E}(\lambda(B([0,1]) \cap B([1,2]))) &= \mathbb{E}(\lambda(W([0,1]) \cap \tilde{W}([0,1]))) \\ &= \int \mathbb{P}((x,y) \in W([0,1]) \cap \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int \mathbb{P}((x,y) \in W([0,1])) \mathbb{P}((x,y) \in \tilde{W}([0,1])) \, d\lambda(x,y) \\ &= \int p^2(x,y) \, d\lambda(x,y). \end{align*}
Now $(1)$ implies $p(x,y)=0$ $\lambda$-almost everywhere and therefore, by $(2)$, $\mathbb{E}L=0$. Hence, as $L \geq 0$, we finally conclude $L=0$ almost surely.
Remark It is not obvious that $\omega \mapsto L(\omega)$ and $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y))$ are random variables; see this question and Evan Aad's answer for the measurability of $L$ and this question for the measurability of the second mapping. (Note that measurability of $((x,y),\omega) \mapsto 1_{B([0,1],\omega)}((x,y)$ entails the measurability of $L$, by Fubini's theorem.)
• Thank you very much. Unfortunately, I won't be able to read your answer until tomorrow. – Evan Aad Mar 30 '14 at 12:19
• @EvanAad No problem, I'm not in a hurry. – saz Mar 30 '14 at 14:57
• Before I read your answer, I'd like to understand why $L$ is a random variable. I have read Lévy's proof on p. 533, but there are a couple points I didn't get. I have written then as another question here. – Evan Aad Apr 1 '14 at 6:55
• Do you understand why $L$ is a random variable? Cause I sure don't. – Evan Aad Apr 3 '14 at 19:05
• @EvanAad Sorry, I don't have that much time to think about it right now; I'll take a look at it this weekend. – saz Apr 3 '14 at 19:16
I will complement saz's answerby proving that $L$ (to use saz's terminology) is a random variable.
EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end).
Introduction & proof outline 0
Following is a proof that the measure of the path of an $n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable function. The proof is based on ideas presented by Paul Lévy in [L] (p. 533), but mine is not exactly the same as the proof given there, simply because I didn't quite understand that proof.
I first establish some notations (Notation 1), which are used in the sequel. I then introduce three constants (Constants 2) that will be in effect throughout. A few definitions and lemmas lead to the main result (theorem 9). The functions introduced in Definition 5 and Definition 7 correspond, more or less, to objects used in Lévy's proof. Lemma 6 and Lemma 8 study the properties and inter-relationships between these objects.
The idea of the proof is to cover the Brownian path segment between the times $0$ and $t$ ($g_t(\omega)$) by two sets: $f_{m, r, t}(\omega)$ and $h_{r, t}(\omega)$. $f_{m, r, t}(\omega)$'s measure $F_{m, r, t}(\omega)$ can readily be shown to be a measurable function of $\omega$, whereas $h_{r, t}(\omega)$'s measure $H_{r, t}(\omega)$ can readily be shown to converge to $g_t(\omega)$'s measure $G_t(\omega)$, as $r$ diminishes to $0$. These two desireable properties ($F_{m, r, t}$'s measurability and $H_{r, t}$'s convergence to $G_t$) are combined (Theorem 9.1) using a result that I proved in another post (theorem 1 here).
The reason why $F_{m, r, t}$'s measurability is relatively easy to demonstrate, is that $f_{m, r, t}(\omega)$ has a simple structure: it is the union of a finite number ($m$) of balls of the same radius ($r$). Such unions are denoted below by $e_r(x_1, \dots, x_m)$ with $x_1, \dots, x_m$ marking the centers of the balls. To assist in the study of the properties of $e_r(x_1, \dots, x_m)$'s measure $E_r(x_1, \dots, x_m)$, we use bounds derived in Lemma 4.
All the functions listed in the previous two paragraphs are introduced in Definition 5 and studied in Lemma 6. The functions introduced in Definition 7, whose properties are studied in Lemma 8, are used in the main result (Theorem 9.1) to construct a sequence of covers $f_{m_k, 1 / k, t}$ converging to $g_t$ and to quantify the rate of convergence of $F_{m_k, 1/k, t}$ to $G_t$. To quantify this rate of convergence, a uniform upper bound ($u$) is calculated (Lemmas 8.2(c)) on the maximal distance ($d_{a, b}(\omega)$) between two points that lie on the path segment $g_t(\omega)$ between times $a$ and $b$.
The entire proof is carried out under the assumption (Constants 2.1) that $n \geq 2$. In the end (Remark 10) I indicate how to extend the proof to the case $n = 1$.
The published works referenced in the post are listed at the end of the post in the section "Works cited".
Notation 1
For every $n \in \mathbb{N}_1 := \{1, 2, \dots\}$,
1. Denote the Borel field on $\mathbb{R}^n$ by $\mathcal{B}_n$,
2. Denote the Lebesgue measure on $\mathcal{B}_n$ by $\lambda_n$ and the outer measure on $\mathbb{R}^n$ by $\lambda_n^*$,
3. Denote by $|\cdot|_n$ the standard Euclidean norm on $\mathbb{R}^n$,
4. For every $x \in \mathbb{R}^n$ and every $r \in (0,\infty)$, define \begin{align} \mathbf{B}_x(r) & := \{y \in \mathbb{R}^n \mid: |y - x|_n < r\} \\ \overline{\mathbf{B}_x(r)} & := \{y \in \mathbb{R}^n \mid: |y - x|_n \leq r\} \end{align}
5. Denote the topology induced by $|\cdot|_n$ on $\mathbb{R}^n$ by $\mathcal{O}_n$ and for every $m \in \mathbb{N}_1$, denote the product topology induced by $|\cdot|_n$ on $(\mathbb{R}^n)^m$ by $\mathcal{O}_{n;m}$.
6. If $S$ is a set, we denote its power set by $2^S$.
7. Denote by $\mathbf{C}_{(0; n)}$ the set of continuous functions $f : [0, \infty) \rightarrow \mathbb{R}^n$, such that $f(0) = 0$. Denote by $\mathcal{B}(\mathbf{C}_{(0; n)})$ the Borel $\sigma$-algebra on $\mathbf{C}_{(0; n)}$ (cf. [S], p. 41).
Constants 2
Fix the following for the remaining of this post.
1. $n \in \mathbb{N}_1$, such that $n \geq 2$,
2. $S := (\Omega, \mathcal{F}, P)$ - A probability space,
3. $W : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ - a standard, $n$-dimensional Brownian motion over $S$. Denote $W$'s components thus: $W = (W_1, \dots, W_n)$.
Definition 3
Set $$c := \frac{\pi^{n / 2}}{\Gamma(\frac{n}{2} + 1)}$$ (see the next lemma for the rationale.)
Lemma 4
1. Let $x, y \in \mathbb{R}^n$, let $\varepsilon \in (0,1]$ and let $r \in (0,\infty)$. If $|x - y|_n < \varepsilon$, $$\lambda_n(\mathbf{B}_x(r) \Delta \mathbf{B}_y(r)) < 2c(r + 1)^n\varepsilon$$
2. Let $m \in \mathbb{N}_1$ and consider some pair of $m$-length sequences in $\mathcal{B}_n$: $(B_1, \dots, B_m)$, $(C_1, \dots, C_m)$ with $B_i, C_i \in \mathcal{B}_n$ for every $i \in \{1, \dots, m\}$. Let $\varepsilon \in (0,\infty)$. If, for each $i \in \{1, \dots, m\}$, we have $\lambda_n(B_i \Delta C_i) < \varepsilon$, then $$\lambda_n\left(\left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right)\right) < m\varepsilon$$
3. Let $m\in \mathbb{N}_1$, let $\varepsilon \in (0,\infty)$, let $x := (x_1, \dots, x_m), y := (y_1, \dots, y_m) \in (\mathbb{R}^n)^m$ and let $r \in (0,\infty)$. If, for every $i \in \{1, \dots, m\}$, $|x_i - y_i|_n < \varepsilon$, we have $$\lambda_n\left(\left(\bigcup_{i = 1}^m \mathbf{B}_{x_i}(r)\right) \Delta \left(\bigcup_{j = 1}^m \mathbf{B}_{y_j}(r)\right)\right) < 2mc(r + 1)^n\varepsilon$$
Proof
1. By the triangle inequality, $\mathbf{B}_y(r) \subseteq \mathbf{B}_x(r + |y - x|_n)$. Hence $\mathbf{B}_y(r) \setminus \mathbf{B}_x(r) \subseteq \mathbf{B}_x(r + |y - x|_n) \setminus \mathbf{B}_x(r)$. Hence $$\lambda_n(\mathbf{B}_y(r) \setminus \mathbf{B}_x(r)) \leq \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r))$$
By the formula for the volume of an $n$-dimensional ball, \begin{aligned} \lambda_n(\mathbf{B}_x(r + |y - x|_n)) - \lambda_n(\mathbf{B}_x(r)) & = c(r + |y - x|_n)^n - cr^n \\ & = c \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^k \\ & = c |y - x|_n \sum_{k = 1}^n \binom{n}{k}r^{n - k}|y - x|_n^{k - 1} \\ & \leq c \varepsilon \sum_{k = 1}^n\binom{n}{k}r^{n - k} \\ & < c \varepsilon (r + 1)^n \end{aligned}
By an analogous argument, $\lambda_n(\mathbf{B}_x(r) \setminus \mathbf{B}_y(r)) < c \varepsilon (r + 1)^n$.
2. By properties of the symmetric difference, $$\left(\bigcup_{i = 1}^m B_i\right) \Delta \left(\bigcup_{i = 1}^m C_i\right) \subseteq \bigcup_{i = 1}^m \left(B_i \Delta C_i\right)$$
3. The result follows from the first two parts.
Q.E.D.
Definition 5
For every $m \in \mathbb{N}_1$, every $t \in (0,\infty)$ and every $r \in (0,\infty)$,
1. Define $e_{m,r} : (\mathbb{R}^n)^m \rightarrow 2^{\mathbb{R}^n}$ and $E_{m,r} : (\mathbb{R}^n)^m \rightarrow [0,\infty)$ as follows \begin{align} e_{m,r}(x_1, \dots, x_m) & := \bigcup_{i = 1}^m\overline{\mathbf{B}_{x_i}(r)} \\ E_{m,r}(x_1, \dots, x_m) & := \lambda_n(e_m(x_1, \dots, x_m)) \end{align}
2. Define $f_{m, r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $F_{m, r, t} : \Omega \rightarrow [0,\infty)$ as follows \begin{align} f_{m, r, t}(\omega) & := e_{m,r}(W(\omega, 0), W(\omega, \frac{t}{m}), W(\omega, 2\frac{t}{m}), \dots, W(\omega, t)) \\ F_{m, r, t}(\omega) & := \lambda_n(f_{m, r, t}(\omega)) \end{align}
3. Define $g_t : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G_t : \Omega \rightarrow [0,\infty]$ as follows \begin{align} g_t(\omega) & := \{W(\omega, s) :\mid s \in [0,t]\} \\ G_t(\omega) & := \lambda_n^*(g_t(\omega)) \end{align}
4. Define the functions $g^* : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $G^* : \Omega \rightarrow [0,\infty]$ as follows \begin{aligned} g^*(\omega) & := \{W(\omega, t) :\mid t \in [0,\infty)\} \\ G^*(\omega) & := \lambda_n^*(g^*(\omega)) \end{aligned}
5. Define $h_{r, t} : \Omega \rightarrow 2^{\mathbb{R}^n}$ and $H_{r, t} : \Omega \rightarrow [0,\infty]$ as follows \begin{align} h_{r, t}(\omega) & := \bigcup_{s \in [0, t]} \overline{\mathbf{B}_{W(\omega, s)}(r)} \\ H_{r, t}(\omega) & := \lambda_n^*(h_{r, t}(\omega)) \end{align}
Lemma 6
For every $m \in \mathbb{N}_1$, every $r \in (0,\infty)$, every $t \in (0,\infty)$, every sequence $r_1, r_2, \dots \in (0,\infty)$ such that $\lim_{k \rightarrow \infty} r_k = 0$ and every $\omega \in \Omega$,
1. $h_{r, t}(\omega)$ is compact (and therefore $\in \mathcal{B}_n$ and $\lambda_n^*(h_{r, t}(\omega)) = \lambda_n(h_{r, t}(\omega))$).
2. For every $r' \in [r,\infty)$, we have \begin{aligned} h_{r, t}(\omega) & \subseteq h_{r', t}(\omega) \\ H_{r, t}(\omega) & \leq H_{r', t}(\omega) \end{aligned}
3. We have $$g_t(\omega) = \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$$ (and therefore $g_t(\omega) \in \mathcal{B}_n$ and $\lambda_n^*(g_t(\omega)) = \lambda_n(g_t(\omega)$).
4. We have $$G_t(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega)$$
5. We have \begin{align} f_{m, r, t}(\omega) & \subseteq h_{r, t}(\omega) \\ F_{m, r, t}(\omega) & \leq H_{r, t}(\omega) \end{align}
6. $E_{m, r}$ is $\mathcal{O}_{n; m}/\mathcal{O}_1$-continuous (and hence $(\otimes_{i = 1}^m\mathcal{B}_n)/\mathcal{B}_1$-measurable).
7. $F_{m, r, t}$ is $\mathcal{F}/\mathcal{B}_1$-measurable.
8. $g^*(\omega) \in \mathcal{B}_n$ (and therefore $\lambda_n^*(g^*(\omega)) = \lambda_n(g^*(\omega))$.
Proof
1. Define $d : \mathbb{R}^n \rightarrow [0,\infty)$ as follows $$d(x) := \inf \{|x - y|_n :\mid y \in g_t(\omega)\}$$ Then $h_{r, t}(\omega) = d^{-1}([0,r])$. Hence, by the fact that $d$ is $\mathcal{O}_n / \mathcal{O}_1$-continuous ([M2] p. 175), $h_{r, t}(\omega)$ is closed. Since $[0,t]$ is compact, so is $g_t(\omega)$. In particular, $g_t(\omega)$ is bounded. Let $R \in (0,\infty)$ be such that $g_t(\omega) \subseteq \overline{\mathbf{B}_0(R)}$. Then $h_{r, t}(\omega) \subseteq \overline{\mathbf{B}_0(R + r)}$. In particular, $h_{r, t}(\omega)$ is bounded. $h_{r, t}(\omega)$ is closed and bounded, hence compact.
2. Immediate.
3. It is evident that $$g_t(\omega) \subseteq \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$$ I will prove the converse containment. Let $x \in \bigcap_{k = 1}^\infty h_{r_k, t}(\omega)$. For every $k \in \mathbb{N}_1$, let $y_k \in g_t(\omega)$ be such that $|y_k - x|_n \leq r_k$. Since $g_t(\omega)$ is compact, there is, by the Bolzano-Weierstrass theorem, a convergent subsequence $(y_{k_1}, y_{k_2}, \dots)$. Set $y := \lim_{i \rightarrow \infty} y_{k_i}$. Since $g_t(\omega)$ is closed, $y \in g_t(\omega)$. I will show that $x = y$.
Let $\varepsilon \in (0,\infty)$ and let $i \in \mathbb{N}_1$ be such that $r_{k_i} < \varepsilon / 2$ and $|y_{k_i} - y|_n < \varepsilon / 2$. We have, by the triangle inequality, $$|y - x|_n \leq |y - y_{k_i}|_n + |y_{k_i} - x|_n < \varepsilon$$ Since $\varepsilon$ was chosen arbitrarily in $(0,\infty)$, $|y - x|_n = 0$.
4. Choose a strictly decreasing subsequence $(r_{k_1}, r_{k_2}, \dots)$. Then by 6.2 and 6.3, $$g_t(\omega) = \bigcap_{i = 1}^\infty h_{r_{k_i}, t}(\omega)$$ Hence, by the continuity of measures, and by the fact (6.1) that $h_{r_{k_1}, t}$ is bounded and therefore of finite Lebesgue measure, $$G_t(\omega) = \lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega)$$ But by 6.2, $\lim_{i \rightarrow \infty} H_{r_{k_i}, t}(\omega) = \lim_{k \rightarrow \infty} H_{r_k, t}(\omega)$.
5. Immediate.
6. Fix some $x_1, \dots, x_m \in \mathbb{R}^n$. For every $y_1, \dots, y_m \in \mathbb{R}^n$, we have $$|E_{m, r}(y_1, \dots, y_m) - E_{m, r}(x_1, \dots, x_m)| \leq \lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m))$$ Let $\varepsilon \in (0,\infty)$. Define $$\delta := \frac{\varepsilon}{2 m c (r + 1)^n}$$ If $|y_i - x_i|_n < \delta$ for all $i \in \{1, \dots, m\}$, then, by lemma 4.3, $$\lambda_n(e_{m, r}(y_1, \dots, y_m) \Delta e_{m, r}(y_1, \dots, y_m)) < \varepsilon$$
7. Observe that, for all $\omega \in \Omega$, $$F_{m, r, t}(\omega) = E_{m, r}(W(\omega, 0), W(\omega, \frac{t}{m}), \dots, W(\omega, t))$$ The result now follows by 6.6.
8. The result follows from 6.3 by observing that $$g^*(\omega) = \bigcup_{k = 1}^\infty g_k(\omega)$$
Q.E.D.
Definition 7
1. Define $u : (n - 2, \infty) \rightarrow (0,\infty)$ as follows $$u(s) := \frac{1}{\sqrt{\pi}} \frac{s}{s - (n - 2)} \exp\left(-\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right)$$ (for the rationale - see the next lemma.)
2. For every $a, b \in [0, \infty)$ with $a < b$, define $d_{a, b}: \Omega \rightarrow [0,\infty]$ as follows $$d_{a, b}(\omega) := \max \{\left|W(\omega, s) - W(\omega, a)\right|_n :\mid s \in [a, b]\}$$
3. For every $m \in \mathbb{N}_1$ and every $t \in (0,\infty)$, define $D_{m, t} : \Omega \rightarrow [0,\infty)$ as follows $$D_{m, t}(\omega) := \max(d_{0, \frac{1}{m}t}(\omega), d_{\frac{1}{m}t, \frac{2}{m}t}(\omega), \dots, d_{\frac{m - 1}{m}t, t}(\omega))$$
Lemma 8
1. $\lim_{s \rightarrow \infty} s u(s) = 0$
2. Let $a, b \in [0, \infty)$ with $a < b$.
a) $d_{a,b}$ is $\mathcal{F}/\mathcal{B}_1$ - measurable (and, therefore, so is $D_{m, t}$ for all $m \in \mathbb{N}_1$, $t \in (0, \infty)$).
b) $d_{a, b} \sim d_{0, b - a}$,
c) For every $r \in (0,\infty)$, $$\frac{r^2}{b} > n - 2 \implies P(d_{0, b} > r) \leq u\left(\frac{r^2}{b}\right)$$
3. Let $r \in (0,\infty)$ and $t \in (0,\infty)$.
a) For every $m \in \mathbb{N}_1$, $$\{D_{m , t} \leq r \} \subseteq \{g_t \subseteq f_{m, r, t}\} = \{G_t \leq F_{m, r, t}\}$$ b) There exists some $M \in \mathbb{N}_1$, such that, for all $m \in \{M, M + 1, \dots\}$, $$P(D_{m, t} > r) \leq m u\left(\frac{r^2}{t} m\right)$$
Proof
1. Since, for every $s \in (0,\infty)$, $$\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right) = \frac{1}{2} \frac{s}{n} \left(n - (n - 2) \frac{\ln(s / n)}{s / n} + \frac{\ln(n) - n}{s / n} \right)$$
we have
$$\lim_{s \rightarrow \infty} \frac{\frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)}{\frac{s}{2\sqrt{\pi}}} = \sqrt{\pi} > 1$$
Hence $$\limsup_{s \rightarrow \infty} \frac{\exp\left(- \frac{1}{2} \left(s - n - (n - 2) \ln\left(\frac{s}{n}\right) + \ln(n)\right)\right)}{\exp(- \frac{s}{2\sqrt{\pi}})} \leq 1$$
Hence $$\limsup_{s \rightarrow \infty} \frac{s u(s)}{\frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}})} \leq 1$$
But $$\lim_{s \rightarrow \infty} \frac{s}{\sqrt{\pi}} \exp(- \frac{s}{2\sqrt{\pi}}) = 0$$
Therefore, $\limsup_{s \rightarrow \infty} s u(s) = 0$.
2. a) Let $\omega \in \Omega$. Since $W(\omega)$ is continuous and since the rational numbers are dense in $\mathbb{R}$, we have $$d_{a, b}(\omega) = \sup \{\left|W(\omega, q) - W(\omega, a)\right|_n :\mid q \in [a, b] \cap \mathbb{Q}\}$$ The result now follows from the fact ([K2] theorem 1.92, p. 40) that the supremum of a denumerable family of random variables is a random variable.
b) Define $\varphi: \mathbf{C}_{(0; n)} \rightarrow [0,\infty)$ as follows $$\varphi(f) := \max f([0,t])$$ Firstly, I will show that $\varphi$ is $\mathcal{B}(\mathbf{C}_{(0; n)})/\mathcal{B}_1$-measurable. It suffices to show that for all $M \in \mathbb{R}$, $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$. Let $M \in \mathbb{R}$. We have $$\{\varphi \leq M\} = \bigcap_{s \in [0,t]} \{f \in \mathbf{C}_{(0; n)} \mid: f(s) \leq M\} = \bigcap_{q \in [0,t] \cap \mathbb{Q}} \{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\}$$ For every $q \in [0,t] \cap \mathbb{Q}$, $\{f \in \mathbf{C}_{(0; n)} \mid: f(q) \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$, hence $\{\varphi \leq M\} \in \mathcal{B}(\mathbf{C}_{(0; n)})$.
Now, define the process $V : \Omega \times [0,\infty) \rightarrow \mathbb{R}^n$ as follows: $$V(\omega, t) := W(\omega, a + t) - W(\omega, a)$$ $V$ is a standard, $n$-dimensional Brownian motion ([S] 2.9, p. 12), hence $V \sim W$. Therefore, $$d_{a, b} = \varphi \circ V \sim \varphi \circ W = d_{0, b - a}$$ c) For every $k \in \{1, \dots, n\}$, define $$d_{0, b; k}(\omega) := \max \{\left|W_k(\omega, s) - W_k(\omega, a)\right|_1 :\mid s \in [0, b]\}$$ We have $$d_{0, b} \leq \sqrt{d_{0, b; 1}^2 + \cdots + d_{0, b; n}^2}$$ Hence, \begin{aligned} P(d_{0, b} > r) & = P(d_{0, b}^2 > r^2) & \\ & \leq P(\sum_{k = 1}^n d_{0, b; k}^2 > r^2) & \\ & = P(\sum_{k = 1}^n W_k^2(b) > r^2) & (1) \\ & = \chi^2(n)\left(\left(r^2/b, \infty\right)\right) & (2) \\ & \leq u\left(\frac{r^2}{b}\right) & (3) \end{aligned} where (1) is by Bachelier's maximum process theorem ([K1] Proposition 13.13, p. 256) together with the fact that the components of an $n$-dimensional Brownian motion are independent; $\chi^2(n)$ in (2) denotes the (central) chi-squared distribution with $n$ degrees of freedom; and (3) is from [I2] equation (3.1), p. 341 (see also [I1] Lemma 1, p. 586; a proof can be found there in the appendix).
3. a) Let $\omega \in \{D_{m, r} \leq r\}$ and let $x \in g_t(\omega)$. Let $s \in [0,t]$ be such that $x = W(\omega, s)$. Define $$K := \max \{k \in \{0, 2, \dots, m - 1\} \mid: s \geq \frac{k}{m}\}$$ Since $d_{K, K + 1}(\omega) \leq r$, $g_s(\omega) \in \mathbf{B}_{g_{K / m}(\omega)}(r) \subseteq f_{m, r, t}(\omega)$.
b) Let $m \in \mathbb{N}_1$ be such that $$\frac{r^2}{t} m > n - 2$$ Then \begin{aligned} P(D_{m,t} > r) & \leq \sum_{k = 0}^{m - 1} P(d_{k\frac{t}{m}, (k + 1) \frac{t}{m}} > r) & \\ & = m P(d_{0, \frac{t}{m}} > r) & (1)\\ & \leq m u\left(\frac{r^2}{t} m\right) & (2) \end{aligned} where (1) is by 8.2 (b) and (2) is by 8.2 (c). Now set $$M := \lceil\frac{t}{r^2} (n - 2)\rceil + 1$$
Q.E.D.
Theorem 9
For every $t \in (0, \infty)$,
1. $G_t$ is $\mathcal{F}/\mathcal{B}_1$-measurable in the sense that there is an $\mathcal{F}/\mathcal{B}_1$-measurable function $G'_t$ and an event $B \in \mathcal{F}$, with $P(B) = 0$, such that $$\{G'_t \neq G_t\} \subseteq B$$
2. $G^*$ is $\mathcal{F}/\mathcal{B}_1$-measurable (in the same sense as in 9.1).
Proof
1. For every $k \in \mathbb{N}_1$, use 8.1 to choose $I_k \in \mathbb{N}_1$ such that, for all $i \in \{I_k, I_k + 1, \dots\}$, $$\frac{k^{-2}}{t} i \cdot u(\frac{k^{-2}}{t} i) < \frac{k^{-2}}{t} \cdot \frac{1}{k}$$ and use 8.3 (b) to choose $J_k \in \mathbb{N}_1$, such that, for all $j \in \{J_k, J_k + 1, \dots\}$, $$P(D_{j, t} > \frac{1}{k}) \leq j u\left(\frac{k^{-2}}{t} j\right)$$ Set $m_k := \max(I_k, J_k)$, and define $$A_k := \left\{D_{m_k, t} \leq \frac{1}{k}\right\}$$ Then $P(A_k) \geq 1 - \frac{1}{k}$.
For every $k \in \mathbb{N}_1$ we have, by 6.5, \begin{aligned} &F_{m_k, \frac{1}{k}, t} \leq H_{\frac{1}{k}, t} & (1) \end{aligned}
and by 6.2 and 6.4 \begin{aligned} &G_t = \lim_{j \downarrow \infty} H_{\frac{1}{j}, t} & (2) \end{aligned}
and by 8.3 (a), \begin{aligned} &\forall \omega \in A_k, G_t(\omega) \leq F_{m_k, \frac{1}{k}, t}(\omega) & (3) \end{aligned}
The result now follows from (1), (2) and (3) by theorem 1 here.
2. Observe that for every $\omega \in \Omega$, $$g^*(\omega) = \lim_{k \uparrow \infty} g_k(\omega)$$ Therefore, by continuity of measures, $G^* = \lim_{k \rightarrow \infty} G_k$. The result now follows from 9.1.
Q.E.D.
Remark 10
The only snag that prevents this proof from extending to the case $n = 1$, is the requirement that $n \geq 2$ in equation (3.1) of [I2], which is used to justify the upper bound used in 8.2 (c).
To extend to $n = 1$, the following bound on the expression that is tagged with "(1)" in the proof of 8.2 (c) can be used (cf. [M1] lemma 12.9, p. 349): $$1 - \Phi(r) \leq \frac{1}{r} \frac{1}{\sqrt{2\pi}} e^{- r^2 / 2}$$ where $\Phi$ is the c.d.f. of the standard normal distribution.
WORKS CITED
• [I1] Inglot, Tadeusz and Ledwina, Teresa. Asymptotic optimality of a new adaptive test in regression model, Annales de l'Institut Henri Poincaré 42 (2006), pp. 579–590.
• [I2] Inglot, Tadeusz. Inequalities for Quantiles of the Chi-Square Distribution. Probability and Mathematical Statistics, Vol. 30, Fasc. 2 (2010), pp. 339-351
• [K1] Kallenberg, Olav. Foundations of modern probability. 2nd edition. Springer, 2001
• [K2] Klenke, Achim. Probability Theory, A Comprehensive Course. Springer, 2008
• [L] Lévy, Paul M. Le mouvement Brownien plan. American Journal of Mathematics, Vol 62, No. 1 (1940), pp. 487-550
• [M1] Mörters, Peter and Peres, Yuval. Brownian motion. Published by Cambridge University Press, but I used the online version from Yuval Peres's homepage, retrieved 2014-02-18
• [M2] Munkres, James R. Topology. 2nd edition. Prentice Hall, 2000
• [S] Schilling, René L. and Partzsch, Lothar. Brownian Motion, An Introduction to Stochastic Processes. De Gruyter, 2012
• Wow, what an answer. Just for future reference, you may find it useful to draft your response using our sandbox for long answers. It helps the community in that by lowering the number of edits, you free up some real-estate on the front page for other questions. Thanks! – Willie Wong Apr 10 '14 at 7:19
• @WillieWong: Thanks, will do. I did actually use a separate sandbox (stackedit.io) to prepare this answer, which I've been working on for several days, but there are always bugs, omissions and pitfalls that I find after I post my answers. It can't be helped! – Evan Aad Apr 10 '14 at 7:30 | 10,558 | 26,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-22 | latest | en | 0.88312 |
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GRE vs. GMAT
Author Message
VP
Joined: 08 Apr 2009
Posts: 1212
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
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01 Mar 2010, 07:57
4
KUDOS
With top schools now accepting both GRE and GMAT. I thought this might be a common question.
GRE and GMAT Score Comparison Tool
Attachment:
GRE_Comparison_Tool.xls [124.5 KiB]
GRE vs GMAT, Which Test to Take?
Intern
Joined: 17 Mar 2010
Posts: 4
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17 Mar 2010, 05:24
GMAT is still more popular between MBA programs as well as Students
Founder
Joined: 04 Dec 2002
Posts: 16886
Location: United States (WA)
GMAT 1: 750 Q49 V42
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17 Mar 2010, 17:47
Sure by far: gre-gets-little-traction-in-bschools-91370.html
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05 May 2010, 07:44
Yes, both have their advantages and disadvantages. On one hand, if you ever decide that you want to go for a master's in an academic field, the GRE is vital. On the other hand, not every business school out there takes a GRE score. However, one part of the article I disagree with is that the GMAT provides more security. The GRE administrators take just as many measures and those of the GMAT to make sure that the person actually taking the test is the person whose name is on the form. Personally, since I don't know exactly where I want to go yet, I plan on taking both.
Intern
Joined: 08 Mar 2010
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06 May 2010, 13:16
1
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Actually, that's a good thing to mention - the fact that not all business schools accept the GRE score. It might not be time well-spent to study and take a gre practice test when working toward a good GMAT score is time better spent.
ahillis wrote:
Yes, both have their advantages and disadvantages. On one hand, if you ever decide that you want to go for a master's in an academic field, the GRE is vital. On the other hand, not every business school out there takes a GRE score.
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Joined: 05 Jan 2010
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06 May 2010, 13:57
joeleitz wrote:
Actually, that's a good thing to mention - the fact that not all business schools accept the GRE score. It might not be time well-spent to study and take a gre practice test when working toward a good GMAT score is time better spent.
This. Plus its unclear how GRE performance is evaluated relative to GMAT performance in the eyes of each admissions committee. Do they take them both at face value? Do they map your GRE Q/V/Total percentiles onto the equivalent GMAT scores? Do they weight the splits differently? Is not taking the GMAT or not sending the GMAT score you received a negative, in their eyes? Is sending both scores a negative? Is it a positive?
It adds another variable to the decision equation and creates more uncertainty for future applicants -- something you might not want during a process that many see as already being a bit of a crapshoot for certain profiles.
That said, I took both because I was a dual degree applicant and some of my non-MBA applications required the GRE.
Manager
Joined: 22 Oct 2009
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GMAT 1: 760 Q49 V44
GPA: 3.88
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03 Jul 2010, 07:33
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I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
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03 Jul 2010, 17:34
YourDreamTheater wrote:
I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
Honestly, I didn't find this to be the case, but perhaps that's just me. The GMAT may offer problems that are somewhat more complex, but it also allows test takers 25% more time to solve them. I mean, the material is exactly the same. I suppose if someone prefers the GRE's "Which value is bigger?" to the GMAT's "Which of these things do you need to solve the problem?" he/she might find the GRE more intuitive. But I wouldn't say either is inherently more difficult.
IMO, most MBA applicants should still prepare for the GMAT, unless they have a concrete reason otherwise, such as a plan to apply to a degree program that won't accept it.
Manager
Joined: 22 Oct 2009
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06 Jul 2010, 15:02
coaks wrote:
YourDreamTheater wrote:
I think that the GRE should be taken advantage of by people who struggle with the GMAT quant section. The GRE math section is a cakewalk in comparison
Honestly, I didn't find this to be the case, but perhaps that's just me. The GMAT may offer problems that are somewhat more complex, but it also allows test takers 25% more time to solve them. I mean, the material is exactly the same. I suppose if someone prefers the GRE's "Which value is bigger?" to the GMAT's "Which of these things do you need to solve the problem?" he/she might find the GRE more intuitive. But I wouldn't say either is inherently more difficult.
IMO, most MBA applicants should still prepare for the GMAT, unless they have a concrete reason otherwise, such as a plan to apply to a degree program that won't accept it.
To each their own!
Intern
Joined: 03 Jul 2010
Posts: 8
Schools: Stanford, MIT
WE 1: 1600 GRE
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06 Jul 2010, 15:50
My \$0.02, since this is near and dear to me as a soon-to-be GRE score submitter... The "official" reason that some bschools are taking the GRE is that they want to encourage non-standard applicants (attracted by the lower price of the GRE or the fact that they've already taken it) to apply.
Also, at a Stanford info session I asked about the GMAT vs. GRE and they said "either's fine; we actually think the GRE is harder."
AJ
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09 Jul 2010, 13:26
The GMAT is still preferred, but the GRE is becoming more widely accepted.
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Current Student
Joined: 02 Jul 2010
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14 Jul 2010, 16:02
i took gre
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Intern
Joined: 08 May 2010
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GMAT 1: 750 Q47 V47
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14 Jul 2010, 18:20
I appeared for GRE 5 years ago and got 1460/1600 with 2 weeks prep approx... i'm finding GMAT much much harder.. sometimes I'm wondering whether I should take the GRE again instead of GMAT but i guess its too late
Intern
Joined: 30 Apr 2010
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17 Jul 2010, 04:43
Thanks for the comparision kit ...
Moderator
Status: battlecruiser, operational...
Joined: 25 Apr 2010
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03 Aug 2010, 22:15
GMAT > GRE
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14 Aug 2010, 06:54
Could anyone recommend a similar forum like gmatclub for the GRE. My brother is interested in completing his GRE and is on the lookout for good forums that he can learn from.
Mhasan
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Joined: 06 Nov 2009
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14 Aug 2010, 07:38
Just search in google for gre forum. There are plenty of good gre forums available!
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30 Aug 2010, 10:58
occamsrazor wrote:
I appeared for GRE 5 years ago and got 1460/1600 with 2 weeks prep approx... i'm finding GMAT much much harder.. sometimes I'm wondering whether I should take the GRE again instead of GMAT but i guess its too late
Wow!! 1460 with 2 weeks is amazing
I guess you have an awesome vocabulory. Most candidates take upto 1.5-2 months to study the Barrons wordlists alone.
Intern
Joined: 16 Sep 2011
Posts: 4
Concentration: Entrepreneurship, Strategy
GRE 1: 1450 Q790 V660
WE: Project Management (Consulting)
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16 Sep 2011, 15:59
Is there any way to compare the GRE verbal score to the GMAT verbal score?
Senior Manager
Joined: 25 Jul 2011
Posts: 298
Location: India
Concentration: Strategy, General Management
GMAT 1: 730 Q50 V40
GPA: 3
WE: Operations (Non-Profit and Government)
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22 Sep 2011, 10:05
I'm not sure if there was a tool for comparing the verbal scores, but there definitely isn't one at the moment for the revised GRE.
I'm not sure how relevant it would be, but I've taken a crack at the approximate coversions:
Verbal - GRE Verbal - GMAT 600 37 650 39 700 43 750 47 800 51
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Re: GRE vs. GMAT [#permalink] 22 Sep 2011, 10:05
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Started Apr 12, 2013 | Discussions thread
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Re: Equivalent focal length for MFT lenses In reply to KenBalbari, Apr 15, 2013
KenBalbari wrote:
Detail Man wrote:
KenBalbari wrote:
... there was a recent thread where a less experienced user seemed to be confused about whether getting closer to a subject would change the exposure, with the idea that the inverse square law should apply to the reflected light the same way it does to the lighting source. Now everyone could tell him that this was wrong, and that changing the distance of the camera doesn't change the exposure.
What a piece of work that bloke was. The worst combination ever is dense as well as vitriolic.
Yes, he was trying to BS hs way through, with a "never admit you were wrong" mentality. Basically the world's least successful con man.
When one refers to oneself as a "seasoned professional" after having been run out of town on a rail, it probably helps to have a portfolio of more than wretchedly exposed "bootay shots" to show.
But some of the explanations of why this is so were quite confusing. I read through a couple and was still baffled as to why.
But the simple explanation that clicked for me was that if you move the camera position and keep the focal length the same, the size of the area captured changes so that the total light captured is the same, and if you change the focal length to capture the same scene, the aperture diameter changes to keep the total light (and the exposure) the same. But if you change the distance of your strobe light from the scene, nothing automatically adjusts in the camera, so you have to change your exposure to account for the difference in total light being captured.
Puzzled a bit myself. Found this web-page (linked in the last post on the thread) to be helpful:
http://www.scantips.com/lights/flashbasics1b.html
Regards, DM
I kept wanting to think there was a difference between reflected light and a source light, in that a source light is dispersing, whereas reflected light reaching the camera all tends to be coming off a surface in one direction (and even tends to be polarized for this reason, that only waves hitting the object in a certain way bounce in that direction), which I think is all basically true, but apparently wasn't really that critical here.
In my humble understanding (recently gained from researching that subject a bit), such (reflected) sources represent the integral (in space) of a large number of point-sources (the paths that happen to be reflected in a uniform direction) - similar to the case of a non-point-source radiator.
It's interesting that when the dimension of such a (non-point) source is around 1/5 of the distance from that source, the inverse-square law holds to with around 1% accuracy (or so some state). As ever, though, the area (from a single point-source) decreases with distance, so the intensity (from each individual point-source comprising the whole) remains constant.
It seems that the integral (in space) of a large number of point-sources (might) have different characteristics ? Seems that are still "superpositioned", though. Interesting stuff to try to envision.
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Color scheme? Blue / Yellow | 741 | 3,384 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2014-52 | longest | en | 0.9645 |
https://classroom.thenational.academy/lessons/reading-analogue-time-to-the-nearest-minute-cdgkjd | 1,701,858,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100593.71/warc/CC-MAIN-20231206095331-20231206125331-00688.warc.gz | 202,726,938 | 25,349 | Reading analogue time to the nearest minute
In this lesson, we will learn to read an analogue clock to the nearest minute. We will use vocabulary around 'to' and 'past', and utilise linear time lines to compare times on clocks to number lines.
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.All scales on a clock are red and blue.
1/5
Q2.Which artist made the melting clock sculpture that is in Shanghai, China?
2/5
Q3.How many scales does the clock use?
3/5
Q4.Gemma said we can estimate the time using just the minute hand. Toby said we can estimate the time using just the hour hand. Who is right?
4/5
Q5.Calendars, diaries and birthdays do not have a connection with time.
5/5
Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.All scales on a clock are red and blue.
1/5
Q2.Which artist made the melting clock sculpture that is in Shanghai, China?
2/5
Q3.How many scales does the clock use?
3/5
Q4.Gemma said we can estimate the time using just the minute hand. Toby said we can estimate the time using just the hour hand. Who is right?
4/5
Q5.Calendars, diaries and birthdays do not have a connection with time.
5/5
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Lesson summary: Reading analogue time to the nearest minute
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Dance | 556 | 2,322 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-50 | latest | en | 0.908948 |
https://www.weibull.com/hotwire/issue181/article181.htm | 1,537,930,090,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267163146.90/warc/CC-MAIN-20180926022052-20180926042452-00243.warc.gz | 914,202,344 | 5,292 | # Derivations of Failure Rate Equations for Series and Parallel Systems
This article shows the derivations of the system failure rates for series and parallel configurations of constant failure rate components in Lambda Predict.
## Series System Failure Rate Equations
Consider a system consisting of n components in series. For this configuration, the system reliability, Rs, is given by:
where R1, R2, ..., Rn are the values of reliability for the n components. If the failure rates of the components are λ1, λ2,..., λn, then the system reliability is:
Therefore, the system reliability can be expressed in terms of the system failure rate, λS, as:
where and λS is constant. Note that since the component failure rates are constant, the system failure rate is constant as well. In other words, the system failure rate at any mission time is equal to the steady-state failure rate when constant failure rate components are arranged in a series configuration. If the components have identical failure rates, λC, then:
It should be pointed out that if n blocks with non-constant (i.e., time-dependent) failure rates are arranged in a series configuration, then the system failure rate has a similar equation to the one for constant failure rate blocks arranged in series and is given by:
where λS(t) and λi(t) are functions of time. Please see the HotWire article "Failure Rate of a Series System Using Weibull++" for more details about this equation.
## Parallel System Failure Rate Equations
Consider a system with n identical constant failure rate components arranged in a simple parallel configuration. For this case, the system reliability equation is given by:
where RC is the reliability of each component. Substituting the expression for component reliability in terms of the constant component failure rate, λC, yields:
Notice that this equation does not reduce to the form of a simple exponential distribution like for the case of a system of components arranged in series. In other words, the reliability of a system of constant failure rate components arranged in parallel cannot be modeled using a constant system failure rate model.
To find the failure rate of a system of n components in parallel, the relationship between the reliability function, the probability density function and the failure rate is employed. The failure rate is defined as the ratio between the probability density and reliability functions, or:
Because the probability density function can be written in terms of the time derivative of the reliability function, the previous equation becomes:
The reliability of a system of n components in parallel is:
and its time derivative is:
Substituting into the expression for the system failure rate yields:
For constant failure rate components, the system failure rate becomes:
Thus, the failure rate for identical constant failure rate components arranged in parallel is time-dependent. Taking the limit of the system failure rate as t approaches infinity leads to the following expression for the steady-state system failure rate:
Applying L'Hopital's rule one obtains:
So the steady-state failure rate for a system of constant failure rate components in a simple parallel arrangement is the failure rate of a single component. It can be shown that for a k-out-of-n parallel configuration with identical components: | 644 | 3,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2018-39 | latest | en | 0.92174 |
https://www.cuemath.com/ncert-solutions/exercise-14-4-factorization-class-8-maths/ | 1,623,515,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00300.warc.gz | 671,612,116 | 12,994 | # Exercise 14.4 Factorization- NCERT Solutions Class 8
## Chapter 14 Ex.14.4 Question 1
Find and correct the errors in the statement:
$4(x - 5) = 4x - 5$
### Solution
What is known?
Incorrect mathematical statement.
What is unknown?
Correct mathematical statement.
Reasoning:
Solve L.H.S.
Steps:
${\rm{S}} = 4(x - 5) \ne \rm R.H.S.$
The correct statement is $$4(x - 5) = 4x - 20$$
## Chapter 14 Ex.14.4 Question 2
Find and correct the errors in the statement:
$x(3x + 2) = 3{x^2} + 2$
### Solution
What is known?
Incorrect mathematical statement.
What is unknown?
Correct mathematical statement.
Reasoning:
Solve L.H.S.
Steps:
\begin{align} {\rm{L.H.S}} &= x(3x + 2)\\&= 3{x^2} + 2x\\ {\rm{L.H.S}} &\ne {\rm{R.H.S}} \end{align}
The correct statement is $$x(3x + 2) = 3{x^2} + 2x$$
Instant doubt clearing with Cuemath Advanced Math Program | 293 | 865 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-25 | latest | en | 0.679413 |
http://forum.woodenboat.com/archive/index.php/t-5822.html?s=25b7443f0cd382c89f8af375f9dc5ee8 | 1,563,670,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526799.4/warc/CC-MAIN-20190720235054-20190721021054-00342.warc.gz | 58,627,126 | 3,903 | PDA
View Full Version : How lumber is priced?
03-17-2002, 02:27 AM
Scratching my head after looking a little more closely at the reciept I just recieved for some wood I recently ordered. 8/4" x 6"UP x 12'UP = 39 board feet?!? Am I missing something here? Isn't a "board foot" usually 4/4" x 12" x 1'? So my order should have been more like 12 board feet, no? (2 x .5 x 12?) What do the letters "UP" signify? Uncut product perhaps? I had this order re-sawn for me, and it didn't dawn on me when I picked it up to do the math to figure out just how big a hunk of wood it must have originally come from. I WAS a bit taken aback by the total however! Is it possible they might have made such a basic mistake, or am I just not reading this right? Even waste from resawing couldn't acount for that big a discrepency could it? :confused:
[ 03-17-2002, 02:41 AM: Message edited by: Art Read ]
Roger Stouff
03-17-2002, 07:40 AM
Sounds like 12 bf to me. Dunno what they did.
Here's a neat tool I found:
Board foot calculator (http://www.woodbin.com/calcs/tabulator.htm)
htom
03-17-2002, 09:30 AM
Looks like 12 bf to me, too.
If they'd cut down a 10/4 x 8" x 12', I could understand a bill for 20 bf, but I wouldn't like it.
Maybe they price resawing by multiplying the actual board feet x some factor? Maybe they just made a mistake, billed you for a plank on someone else's order?
Don Maurer
03-17-2002, 11:04 AM
Was the invoice for a single piece or several? It sounds like the price per board foot depends on thickness, width and length. I would expect from the bill that you got 39 bd ft from rough stock of 8/4 x 6"+ x 12'+. Roughly 3 pieces 12' long or equivalent.
NormMessinger
03-17-2002, 11:49 AM
Sounds like 12 bd.ft. to me but only the guy that made up the invoice can tell you for sure. Please let us know what he says.
--Norm
Bill Perkins
03-17-2002, 12:05 PM
I think UP means UnPlaned after the resaw .The bill makes no sence , unless they took a slice off a 6 by 6 and charged you for the whole thing .
RGM
03-17-2002, 09:47 PM
So Art, did you get charged for the fall-off from the re-saw (that you didn't receive) in addition to a handling and re-saw fee that was not itemized separately on your invoice? If I could make one guess as to who you purchased this stock from I bet it would be .... , I better not say here. Reach me offline, inquiring minds want to know.
03-18-2002, 12:12 AM
RGM... Based on our phone conversation about plywood a while back, I'd say you've guessed correctly about the supplier. I'm gonna call 'em tommorow and see what's up, so I won't mention any names either until I hear what they have to say. I was, in fact, charged a seperate re-saw fee, and it's on the invoice. I expected that. But I think the board foot amount must be just a ""clerical" error. I had called for priceing on a "ballparked" amount before actually measuring for it on the boat. A couple hours later, after shopping around a bit and checking my measurements, I called back and placed the order, but for about half the amount I'd had quoted to me earlier. The guy who took the order was still refering to his old notes when he wrote it up. I think he must have just plugged the earlier "board foot" amount in while entering the correct information in the "cut" order. It's the only thing that makes sense to me. The numbers would be about right if it had been the original amount. Still, it's hard to imagine them not catching it... (Harder still to imagine ME not "catching" it!) The amount I actually received seems just about right. (plus the re-saw fall off, which they did include...) The 39 board feet would have been WAY too much. And I just can't see that the waste could have possibly amounted to DOUBLE the yeild?
[ 03-18-2002, 12:51 AM: Message edited by: Art Read ]
03-21-2002, 12:57 PM
Just got a call back from the guy that put my order together. "Somebody" goofed. Got a nice apology and a credit for 14 board feet's worth on my card. This will teach me to look at the invoices more carefully from now on!
Paul Scheuer
03-21-2002, 06:55 PM
Art: 39 - 14 = 12 ? What am I missing ? | 1,151 | 4,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-30 | latest | en | 0.945348 |
https://www.physicsforums.com/threads/symmetry-hints-need-please.98827/ | 1,519,079,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812841.74/warc/CC-MAIN-20180219211247-20180219231247-00480.warc.gz | 925,149,795 | 14,652 | # Symmetry - hints need please
1. Nov 7, 2005
### Natasha1
Sorry I am not sure if I should put this thread here.. looks like I am going to be told off by the boss
Anyway, here is my question in a triangle ABC, AB = AC and D is a point on AC such that AD = DB = BC. Find the size of the angle BAD? Find the angles of triangle ABC?
Just need a few hints to get this problem started please?
2. Nov 7, 2005
### Fermat
You triangle, ABC, involves three related/interdepandent similar triangles.
Draw out the triangle(s). Try doing it to scale (i.e. try making the triangles look similar).
Label the angles. It should fall out pretty easily.
3. Nov 7, 2005
### Natasha1
I have ADB being isoceles where AD=DB but from there how do you mesure angle A? I know that A and B will be the same in that triangle but how do u get a value?
4. Nov 8, 2005
### Fermat
There are three isosceles triangles. ABC and ADB are two of them. You have to find the third one before you can solve for the angles.
One property of isosceles triangles is that, not only are two sides of equal length, but the angles oppposite them are also equal.
Label all the (equal) angles and it will fall out. | 314 | 1,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-09 | longest | en | 0.937078 |
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