url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://agunited.org/adopt-a-farmer/dawn-nagel-november-2016/ | 1,713,223,149,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817033.56/warc/CC-MAIN-20240415205332-20240415235332-00677.warc.gz | 78,996,752 | 3,616 | # Dawn Nagel, November 2016
Posted: 11/16/2016
Originally filmed in November 2015.
Discussion: Silage is the entire corn plant chopped up and fermented. This is kind of like turning a cucumber in to a pickle! Here’s a great link about the science of pickles. http://www.exploratorium.edu/cooking/pickles/index.html
Core Standard & Question: 4.OA.3 Solve multi-step word problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
• Dawn showed us the scale while the feed truck was being loaded. The scale first showed 1,510 pounds then the silage was added. After the silage was loaded, the scale showed 2,230 pounds. How many pounds of silage were added to the ration?
• 1,510 pounds of feed + pounds of S = 2,230 pounds of feed | 232 | 1,027 | {"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} | 3.0625 | 3 | CC-MAIN-2024-18 | latest | en | 0.945558 |
https://www.zbmath.org/?q=an%3A0617.92001 | 1,632,375,730,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.10/warc/CC-MAIN-20210923044248-20210923074248-00387.warc.gz | 1,065,047,693 | 10,064 | # zbMATH — the first resource for mathematics
The mathematics of computerized tomography. (English) Zbl 0617.92001
Stuttgart: B. G. Teubner; Chichester etc.: John Wiley & Sons. X, 222 p. DM 72.00 (1986).
Computerized tomography deals with the reconstruction of a function on $$R^ n$$ from its line, plane or hyperplane integrals. The best known applications include diagnostic radiology, where a cross-section of a part of a body is scanned by X-ray beams, whose intensity loss is recorded and processed by a computer to produce a two-dimensional image. Other applications stem from electron microscopy, positron emission tomography and NMR-spectroscopy, to mention only some.
In this context the Radon transform plays a key role, since it maps functions on $$R^ n$$ into the set of their integrals over hyperplanes of $$R^ n$$. The reconstruction problem simply calls for the inversion of the Radon transform (or related transforms). In principle, this inversion problem was solved by Radon in 1917. For practical applications, however, Radon’s inversion formula is not very useful, since in practice the integrals can be measured only for a finite number of lines, whose arrangement is determined by the scanning geometry. This causes severe complications, since one has to deal with ill posed and incomplete data problems.
The present book gives a well written account on the underlying mathematics. Chapter I contains the basic practical examples. Chapter II treats the Radon transform and related transforms. In particular, inversion formulas, theorems on uniqueness and Sobolev space estimates are derived.
In Chapter III sampling and resolution is discussed, that is, it is investigated for which directions the transformed function has to be known in order that the original function can be recovered reliably. It turns out that the (essentially) band-limited functions play an important role.
Chapter IV is on ill-posed problems and accuracy. In particular, it contains error estimates and a singular value decomposition theorem of the Radon transform. Chapter V deals with reconstruction algorithms and Chapter VI is dedicated to incomplete data problems. Finally, Chapter VII contains some mathematical tools that are used throughout, including Fourier analysis, discrete Fourier transform, special functions and Sobolev spaces.
The book will not only be of interest to mathematicians working in this field, but it could also serve as an introductory text for readers having a background in the topics that are reviewed in chapter VII.
Reviewer: R.Bürger
##### MSC:
92-02 Research exposition (monographs, survey articles) pertaining to biology 92C50 Medical applications (general) 44-02 Research exposition (monographs, survey articles) pertaining to integral transforms 78A70 Biological applications of optics and electromagnetic theory 78-02 Research exposition (monographs, survey articles) pertaining to optics and electromagnetic theory 92F05 Other natural sciences (mathematical treatment) | 612 | 3,009 | {"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": 0, "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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-39 | latest | en | 0.922219 |
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# Ky/Tn/Ga/Ill Week of 9/26 - 10/2/05
Topic closed. 85 replies. Last post 12 years ago by Shawnintennesse.
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Evening Results for 9-25-05
TN P3 = 995 (VT 551)
TN P4 = 3997 (VT 4553)
Ga P3 = 549 (VT 155)
Ga P4 = 2857 (VT 3413)
Ky P3 = 306 (VT 412)
Ky P4 = 4845 (VT 5455)
Ill P3 = 887 (VT 443)
Ill P4 = 5168 (VT 1224)
If it weren't for Vtracs and STXS where would we be?
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9/26/ 2005 Ky Sun Eve: 306
Ky-Mid Due Numbers: 5,9,7,2,0. The most frequent vtrac on midday is the 3(27). The due vtracs are 1(05) and 3(27). The due double vtracs are 11(00,55,05), 55(44,99,49), 44(33,88,38) and 22(11,66,16). The due sums are: 24,23,6,4,12,5. Due roots 3,6,8,2,1 Ky-Eve Due Numbers: 1,8,9,5,4,7. The most frequent vtrac at nite is the 3(27). The due vtracs are 5(49) and 3(27). The most due double vtracs are 55(44,99,49), 22(11,66,16) and 33(22,77,27). The due sums are: 23,3,6,26,4,21, 22, 24. Due roots 4,7,2,6 TN-Due Numbers 2,8,6,1,0. The most frequent vtrac in TN is 4(38). The due vtracs are 2(16), 3(27) and 4(38). The due double vtracs are 33(22,77,27), and 44(33,88,38) . The due sums are 6,3,26,7,15,8,5,4,24,21,15. Due Roots 3,7,2,4 Ky Mid38 16 4916 27 1627 05 27 Ky Eve27 05 1616 49 2705 27 05 TN27 49 4949 38 3805 16 05 05 272749 38 61 49 272738 05 61 16 27 383849 49 50 346-906,375,417,531,980,413,104,720,612,538,458,(920,029) 392,079,718,904,956,287,013,708,912,075,725,503,610,264,015,491 Doubles-933,575,559,811,668,232,277,388,777,477 036-684,267,458,(756,765),529,386,317,947,903,784,(986,698,968), 342,150,692,745,079,798,(402,042)102,451,715,158,378,429,687,029,139 Doubles-151,663,100,114,707,959,969,166,188,(377,737), 177,002 559-599-021,169,560,740,675,014,732,378,141,397,843, 661,039,(880,880),599,538,627,861,201,476,897,208,161,882, 819
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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Posted: September 26, 2005, 7:58 am - IP Logged
KY Mid 9/26
• 473
• 436
• 332
• 306 (OUT 1,5,8,9)
GA Mid 9/26
• 906
• 203
• 251
• 549 (OUT 7,8)
Ill Mid 9/26
• 411
• 614
• 961
• 887 (0ut 0,2,3,5,)
TN 9/26
• 136
• 505
• 347
• 995 (OUT 2,8,)
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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Midday P4 Stxs 9-26-05
Ill Mid Ga Mid Ky Mid
• 6725 9586 6450
• 2130 3600 3331
• 3805 3842 1378
• 3265 9362 5818
• ---------------------
• 3183 2901 2125
• 3726 4541 6666
• 6655 5320 2106
• 6388 8002 7623
• -----------------------
Evening P4 Stxs 9-25-05
Tn Ill Eve Ga Eve Ky Eve
• 2014 6309 4198 8778
• 7421 7993 6274 5914
• 0439 2586 8027 7073
• 4580 6645 8154 7667
• ----------------------------
• 2537 0600 8777 0894
• 1055 3418 9640 3684
• 0165 1049 1129 8377
• 3997 5168 2857 4845
If it weren't for Vtracs and STXS where would we be?
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hi shawnintennesse, is your roots and sums in vtracs or not thankyou.
Thread Starter
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hi shawnintennesse, is your roots and sums in vtracs or not thankyou.
I dont understand what you mean?
Vtracs are Vtracs. Roots and sums are roots and sums..
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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Midday Results for 9-26-05
Ky P3 = 162 (VT 223)
Ky P4 = 7707 (VT 3313)
Ga P3 = 206 (VT 312)
Ga P4 = 0027 (VT 1133)
Ill P3 = 384 (VT 445)
Ill P4 = 1520 (VT 2131)
If it weren't for Vtracs and STXS where would we be?
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Posted: September 26, 2005, 1:49 pm - IP Logged
KY Eve 9/26
• 436
• 332
• 306
• 162 (OUT 5,7,8,9)
GA Eve 9/26
• 203
• 251
• 549
• 206 (OUT 7,8)
Ill Eve 9/26
• 614
• 961
• 887
• 384 (0ut 0,2,5,)
TN 9/26
• 136
• 505
• 347
• 995 (OUT 2,8,)
If it weren't for Vtracs and STXS where would we be?
Clarksville
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Posted: September 26, 2005, 7:33 pm - IP Logged
TN 077..P4 5471
If you know your number is going to hit, have patience and then KILL IT!
You never know when you will get another hit.
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TN P3 = 077 (VT 133)
TN P4 = 5471 (VT 1532)
Ga P3 = 122 (VT 233)
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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Posted: September 26, 2005, 7:34 pm - IP Logged
TN P3 = 077 (VT 133)
TN P4 = 5471 (VT 1532)
Ga P3 = 122 (VT 233)
We must have been posting right at the same time...:)
Tennessee's P3 played off from Kentucky's midday P4... I knew that 7077 was going to come out somewhere.
Just wasnt sure where...
If it weren't for Vtracs and STXS where would we be?
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KY came back with vtrac 223 tonight.. Hope someone put the dough there...
OLD/Vtrac
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Evening Results for 9-26-05
TN P3 = 077 (VT 133)
TN P4 = 5471 (VT 1532)
Ky P3 = 667 (VT 223)
Ky P4 = 9173 (VT 5234)
Ga P3 = 122 (VT 233)
Ga P4 = 9139 (VT 5245)
Ill P3 = 831 (VT 442)
Ill P4 = 9140 (VT 5251)
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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9/27 /2005 Ky Eve 667
Ky-Mid Due Numbers: 5,9,7,0,8. The most frequent vtrac on midday is the 3(27). The due vtracs are 1(05) 4(38) and 5(49). The due double vtracs are 11(00,55,05), 55(44,99,49), 44(33,88,38) and 33(22,77,27). The due sums are: 24,23,6,4,12,5. Due roots 3,6,8,2,1 Ky-Eve Due Numbers: 1,8,9,5,4,2. The most frequent vtrac at nite is the 3(27). The due vtracs are 5(49) 4(38) and 1(05). The most due double vtracs are 55(44,99,49) and 33(22,77,27). The due sums are: 23,3,6,26,4,21, 22, 24. Due roots 4,7,2,6 TN-Due Numbers 2,8,6,1,3,4. The most frequent vtrac in TN is 4(38). The due vtracs are 2(16), 4(38) and 5(49). The due double vtracs are 33(22,77,27), and 44(33,88,38) . The due sums are 6,3,26,7,15,8,5,4,24,21,15. Due Roots 3,7,2,4 Ky Mid05 05 0527 16 4949 38 16 Ky Eve27 27 2716 16 3805 49 05 TN49 27 4916 38 2727 05 16 05 38 492716 61 27 49 38 052761 61 72 16 38 493805 72 72 126-478, 960, 271, 045, 713, 058, 423, 321, 293, 285, 897, 283, 537, 497, 351,621,860 Doubles-040, 440, 776, 338, 200, 112,066,113,772,400,883 667-677-628,471,107,805,793,462,214,304,030,450,057,380,056 007-077-733,730,664,004,187,593,111,656,427,397,453,152,916, 403,603,348,491,364
If it weren't for Vtracs and STXS where would we be?
Thread Starter
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KY Mid 9/27
• 332
• 306
• 162
• 667 (OUT 4,5,8,9)
GA Mid 9/27
• 251
• 549
• 206
• 122 (OUT 3,7,8)
Ill Mid 9/27
• 961
• 887
• 384
• 831 (0ut 0,2,5,)
TN 9/27
• 505
• 347
• 995
• 077 (OUT 1,2,6,8,)
If it weren't for Vtracs and STXS where would we be?
Page 1 of 6 | 3,630 | 8,148 | {"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} | 3.25 | 3 | CC-MAIN-2017-30 | latest | en | 0.557247 |
http://docplayer.net/29160763-Attachment-3-a-a-technical-evaluation-of-two-weight-and-engineering-based-fuel-efficiency-parameters-for-cars-and-light-trucks.html | 1,542,187,687,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741764.35/warc/CC-MAIN-20181114082713-20181114104713-00322.warc.gz | 77,667,128 | 28,136 | # ATTACHMENT 3-A. A Technical Evaluation of Two Weight- and Engineering-Based Fuel Efficiency Parameters for Cars and Light Trucks
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1 ATTACHMENT 3-A A Technical Evaluation of Two Weight- and Engineering-Based Fuel Efficiency Parameters for Cars and Light Trucks Measuring fuel economy of vehicles in miles per gallon (mpg) alone does not provide sufficient information to evaluate a vehicle s efficiency in performing its intended function. A better way to measure the energy-efficiency of vehicles is needed, one that has a sound engineering basis. This attachment presents two weight-based parameters as examples of approaches that take the intended use of a vehicle into consideration. One is based on a vehicle s curb weight and the other includes its payload (passenger plus cargo). Because of the short time frame of the committee s study, an analysis sufficiently detailed to draw conclusions as to the value of these or other parameters was not possible. Miles Per Gallon Versus Gallons Per Mile and How to Measure The physics of vehicle design can form the basis for parameters that more accurately represent system energy efficiencies and could be used by EPA in fuel economy testing. Miles-per-gallon (mpg) is not by itself a sufficient parameter to measure efficiency since it is inherently higher for smaller vehicles and lower for larger vehicles (which can carry more passengers and a greater cargo load.) Although CAFE currently characterizes vehicles by miles driven per gallon (gal) of fuel consumed, the inverse, gallons per mile, would be more advantageous for several reasons. As shown in Figure 3A-1, gallons per mile measures fuel consumption and thus relates directly to the goal of decreasing the gallons consumed. Note that the curve is relatively flat beyond 3 or 35 mpg because fuel savings become increasingly smaller as mpg increases. Also, the use of fuel consumption (gal per mile) has analytical advantages as addressed in this appendix. To aid and clarify the analysis and make the numbers easier to comprehend, the term to be used is gal per 1 miles (gal/1 miles). A vehicle getting 25 mpg uses 4 gal/1 miles. For reproducibility reasons, fuel consumption measurements are made on a chassis dynamometer. The driving wheels are placed on the dynamometer rollers; other wheels do not rotate. Thus rolling resistance, as with aerodynamic drag, must be accounted for mathematically. Vehicle coast down times are experimentally determined (a measure of aerodynamic drag); an auxiliary power unit (APU) ensures that dynamometer coast down times are in reasonable agreement with road-tested coast down times. Test reproducibility is in the few percent range. The driver follows two different 3A-1
2 Fuel consumed (gallons per 1 miles) Fuel economy (miles per gallon) Series1 FIGURE 3A-1 Dependence of Fuel Consumption on Fuel Economy SOURCE: NRC, 2. specified cycles, city and highway, which were deduced from traffic measurements made some 3-4 years ago. A change in the test cycle is not a minor item much engineering know-how is based on the present cycle, which is also used for exhaust emissions measurements. Weight-Specific Fuel Consumption Figure 5A-4 plots gal/1 miles versus vehicle weight for model year (MY) 1999 vehicles. The vertical scatter along a line of constant weight reflects the fact that vehicles of the same weight may differ in the efficiency of their drive trains or rolling resistance or aerodynamic drag (and thus in the number of gallons used to travel a given distance). While gal/1 miles is a straightforward parameter for measuring fuel consumption, it does not reflect the load-carrying capacity of the vehicle. Smaller cars, with lower fuel consumption, are designed to carry smaller loads, and larger cars and trucks larger ones. For engineering analysis purposes, it is convenient to normalize the data in Figure 5A-4 in Appendix 5A, i.e., divide the y value (vertical scale) of each data point by its curb weight in tons. The resulting new vertical scale is the weight specific fuel consumption (WSFC). The units shown in Figure 3A-2 are gal/ton of vehicle weight/1 miles. The straight horizontal line is a reasonable representation of the average efficiency of fuel use data for a wide variety of vehicle types and weights. It shows that the 3A-2
3 4 WSFC VS. WEIGHT FOR ALL VEHICLES WSFC, Gallons/Ton of Vehicle Weight - 1 Miles 3 2 Cars Trucks 1 Near Luxury Small SUV Luxury Luxury Average Weight, lb FIGURE 3A-2 Weight-Specific Fuel Consumption vs. Weight For All Vehicles efficiency (WSFC) is approximately the same for this variety of different vehicle types (MY2 33 trucks and 44 cars) and weights. Note that some vertical scatter is to be expected; all vehicles having approximately the same weight do not necessarily have the same drive train efficiency. Figure 3A-3 shows on-the-road data taken by Consumer Report (April 21). Their measurements were based on a realistic mixture of express-way, country-road and city driving. Again the efficiency of fuel use for their on-the-road tests is reasonably represented by a horizontal straight line. This figure illustrates the analytical utility of this approach. The lowest car point at a little less than 3, lbs is a diesel engine; its WSFC is around 1.8 compared to around 2.7 for the average. The two hybrid points also show lower WSFC than the average but higher than the diesel. Figure 3A-4 illustrates possible realistic reductions in WSFC. EPA has fuelconsumed data for more than one thousand 2 and 21 model year vehicles. A horizontal line was drawn on a WSFC (highway) graph for these vehicles such that 125 vehicles were below this arbitrary line. The results for the 125 vehicles are shown in Figure 3A-4. The average WSFC value for all vehicles was 1.7; the average for the 125 vehicles was around 1.4. Since these were production vehicles it would appear that application of in-production technologies to the entire fleet could produce significant reductions in WSFC. 3A-3
4 CONSUMER REPORT FUEL EFFICIENCY DATA WSFC, gallons per ton per 1 miles AUTO LTR&SUV HYBRID vehicle weight (pounds) FIGURE 3A-3 Fleet Fuel Economy Based on information from Consumer Report (April, 21) 125 Vehicles With Lowest WFSC - HWY WSFC, gallons per ton per 1 miles AUTOS LT TRUCKS HYBRIDS vehicle weight (pounds) FIGURE 3A-4 Best in Class Fuel Efficiency Analysis of 2 and 21 Vehicles. 3A-4
6 Figure 4 : FC VS. WEIGHT FOR ALL VEHICLES 1 9 Cars Trucks 8 FC, gallons/ 1 miles y =.1x R 2 = Near Luxury Small SUV Luxury 1 Luxury CAFE Standard for Cars CAFE Standard for Trucks Weight, lb FIGURE 3A-5 Fuel Economy vs. Payload for a variety of light- and heavy-duty vehicles. Figure 38: LSFC VS. WEIGHT FOR ALL VEHICLES 1 LSFC, gallons/ 1 miles - Tons of Passengers plus Cargo Cars Trucks 3 2 Near Luxury Small SUV Luxury 1 Luxury Average Weight, lb FIGURE 3A-6 Fuel Consumption vs. Weight for a Variety of Vehicles 3A-6
7 Figure 49: FUEL ECONOMY VS. PAYLOAD LSFC, gallons/payloadton-1 miles 9 8 PNGV 7 Fuel Economy, mpg Light-Duty Vehicles 3 2 Heavy-Duty Vehicles 1 mpg x 4Ton 24 mpg x.5ton 1mpg x 4Ton 1 1 mpg x 1 Ton Payload, tons 5 mpg x 4 Ton FIGURE 3A-7 LSFC vs. Weight for a Variety of Vehicles. 3A-7
8 24 Figure 41: PAYLOAD VS. VEHICLES y = 51.59x R 2 = Small SUV Near Luxury Luxury Luxury Payload, lb In order of increasing average weight FIGURE 3A-8 Payload vs. Vehicles. Comparing the Two Weight-Based Parameters The weight specific fuel consumption (WSFC) essentially normalizes the fuel consumed per 1 miles to take out the strong dependence on vehicle weight. Different weight vehicles can be compared more equitably. Lower WSFC parameters indicate lower road load requirements and/or higher power train efficiencies with lower accessory loads. Figure 3A-9 shows fuel economy versus vehicle curb weight for the 87 light-duty vehicles. Constant efficiency lines (in mpg multiplied [x] by tons of vehicle weight) are 3A-8
9 Figure 48: FUEL ECONOMY VS. WSFC WSFC, gallons/ton of Vehicle Weight -1 miles mpg x 2Ton 24 mpg x 2Ton 22 mpg x 2Ton 38 mpg x 1Ton 32 mpg x 1Ton 26 mpg x 1Ton Weight, lb Ave Ave Ave Small SUV Ave Ave Ave Near Luxury Ave Luxury Ave Ave Ave Ave Ave Ave Luxury Ave Fuel Economy, mpg Cars Near Luxury Luxury Trucks Small SUV Luxury FIGURE 3A-9 Fuel Economy vs. WSFC. also shown along with WFSC. Figure 3A-1 shows fuel economy versus vehicle payload for the 87 vehicles. The constant efficiency lines in mpg x tons of payload are also shown along with LSFC. The utility of this plot is that it shows the interrelationship of fuel economy, payload, and LSFC in gallons/payload ton-1 miles.lsfc and WSFC show similar utility in determining whether certain types of vehicles are either above or below the average lines in Figures A3-2 and A3-7. Using LSFC, however, would encourage manufacturers to consider all aspects of vehicle design, including materials, accessory power consumption, body design, and engine and transmission efficiency. The use of the LSFC number will show high-performance, heavy, two seat sports cars without much cargo space, and large luxury cars to be on the high side compared with vehicles designed to be fuel- and payload-efficient. WSFC does not account for the load carrying capacity of certain vehicles such as pick-ups, vans, and SUVs. The vans and large SUVs in Figure 3A-7 are shown below the average fit line and below the average WSFC line in Figure 3A-2 indicting they have highly efficient power train technologies and low road load/accessory load requirements. Pickups are above the average WSFC line in Figure 3A-2, showing that it is difficult to design a truck with load carrying capabilities that has low aerodynamic drag and is fuel 3A-9
10 Figure 47: FUEL ECONOMY VS. PAYLOAD LSFC, gallons/payload Ton -1 miles mpg x.5ton 4 mpg x.5ton 32 mpg x.5ton 24 mpg x.5ton 16 mpg x.5ton Payload, lb Ave Ave Ave Near Luxury Ave Luxury Ave Ave Ave Ave Small SUV Ave Ave Ave Luxury Ave Ave Ave Fuel Economy, mpg Cars Near Luxury Luxury Trucks Small SUV Luxury FIGURE 3A-1 Fuel Economy vs. Payload. efficient. When fuel consumption of these vehicles (vans, pickups and SUVs) are normalized to payload, they are below the average LSFC line in Figure 3A-7, indicating they are well designed for their intended use. Conclusion Both the WSFC and LSFC parameters have potential utility as fuel-efficieny parameters for vehicles but their applicability requires additional study. 3A-1
11 REFERENCES NRC, 2. Automotive Fuel Economy: How Far Should We Go? Washington, D.C.: National Academy Press. p. 156 Riley Alternative Cars in the 21 st Century. Society of Automotive Engineers. 3A-11
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https://www.sanfoundry.com/surveying-questions-answers-closing-errors-limitations/ | 1,718,728,065,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00099.warc.gz | 859,122,159 | 24,195 | # Surveying Questions and Answers – Theodolite Traversing – Closing Error and its Limitation
This set of Surveying Multiple Choice Questions & Answers (MCQs) focuses on “Theodolite Traversing – Closing Error and its Limitation”.
1. In order to mitigate the closing error, sum of latitudes and departures must be equal to zero.
a) True
b) False
Explanation: The algebraic sum latitudes and algebraic sum of departures must be equal to zero for avoiding the closing error, which will occur when the end point don’t coincide with the starting point.
2. Which among the following determines the direction of closing error?
a) Tan δ = ∑L/∑D
b) Tan δ = ∑L2/∑D2
c) Tan δ = ∑D/∑L
d) Tan δ = ∑D2/∑L2
Explanation: From the figure, Tan δ =∑D/∑L, which will give the direction of the closing error.
3. The sum of interior angles must be equal to_______
a) (2N+4) right angles
b) (2N-4) right angles
c) (2N+4) * 180
d) (2N-4) * 180
Explanation: The theoretical sum of the interior angles of a traverse should equal to (2N-4) right angles, and that of the exterior angles should equal to (2N+4) right angles, where N is the number of sides of a closed traverse.
4. For adjusting the angular error, the error may be distributed equally among all the angles.
a) False
b) True
Explanation: When all angles are measured and under similar conditions, angular error is distributed equally among all the angles. However, if the accuracy of some angle is suspected due to peculiar field conditions, the whole angular error may be assigned to that angle.
5. Closing error can be given as________
a) ((∑L)2+(∑D)2)1/4
b) (∑L2-∑D2)1/2
c) (∑L2*∑D2)1/2
d) ((∑L)2+(∑D)2)1/2
Explanation: If a closed traverse is plotted according to the field measurements, the end point of the traverse will not coincide exactly with the starting point, due to the errors in the field observations, such as error is known as closing error. This is given as,
Closing error, e = ((∑L)2+(∑D)2)1/2
Where, ∑L = sum of latitudes, ∑D = sum of departures.
6. Which of the following corresponds to the correction applied to the bearing of the last side?
a) Correction = Ne/N
b) Correction = 2Ne/N
c) Correction = 3Ne/N
d) Correction = e/N
Explanation: If e is the closing error in bearing, and N is the number of the sides of the traverse, then the correction applied to the bearing of the sides will be
Correction to the first bearing = e/N
Correction to the first bearing = 2e/N
And so on to the last bearing = Ne/N = e.
7. If traversing is done by taking bearings of the lines, the closing error in bearing may be determined by _______________
a) Comparing the back and fore bearings of the last line of the open traverse
b) Comparing the back and fore bearings of the middle line of the closed traverse
c) Comparing the back and fore bearings of the last line of the closed traverse
d) Comparing the back and fore bearings of the first line of the closed traverse
Explanation: By comparing the back and fore bearings of the last line of the closed traverse, the error in bearing may be determined by finding the difference between its observed bearing and known bearing.
8. Which of the following is a method of adjusting a closed traverse?
a) Departure method
b) Axis method
c) Tangential method
d) Latitude method
Explanation: The methods which are used to adjust the traverse are Bowditch’s rule, Transit rule, Axis method and Graphical method. These are employed based on the precision of the values obtained during surveying.
9. Relative error of closure is given as____________
a) Perimeter of closure/error of traverse
b) Error of perimeter/perimeter of traverse
c) Perimeter of traverse/error of traverse
d) Error of closure/perimeter of traverse
Explanation: The relative error of closure is used only in case of determination of the sign of latitudes and departures i.e., in which quadrant latitudes and departures lie.
10. Closing error can be briefly explained in which of the following set of methods?
a) Bowditch’s, Transit methods
b) Transit, Axis methods
c) Graphical, Axis methods
d) Bowditch’s, Graphical methods
Explanation: Since more amount of diagrammatic explanation is involved in Graphical and Axis methods, those are able to explain in a brief manner.
11. From the following observations, calculate closing error.
Line Length (m) Latitude Departure
AB 92.96 +92.57 -217.92
BC 157.63 -317.39 +24.62
CA 131.24 +226.19 +192.36
a) 1.66
b) 1.55
c) 1.44
d) 1.99
Explanation: The value of closing error can be given by e = ((∑L)2+(∑D)2)1/2
Where, ∑L = 92.57 – 317.39 + 226.19 = 1.37
∑D = -217.92 + 24.62 + 192.36 = – 0.94
On substituting, we get e = ((∑L)2+(∑D)2)1/2
e = (1.372+ 0.942)1/2
e = 1.661.
12. Calculate the direction of closing error for the following data.
Line Length (m) Latitude Departure
AB 24.29 -102.31 -119.22
BC 130.32 +360.24 -204.92
CA 249.11 -257.43 +323.26
a) 50023ꞌ
b) 60029ꞌ
c) 60023ꞌ
d) 62023ꞌ
Explanation: The value of direction for closing error can be given as Tan δ =∑D/∑L, where ∑L = – 102.31 + 360.24 –257.43 = 0.5; ∑D = -119.22 – 204.92 + 323.26 = -0.88. On substitution we get, Tan δ = 0.88 / 0.5 = 60023ꞌ.
13. For a traverse containing 10 sides, what would be the correction applied for the first side, if it consists a closing error of +1.92?
a) 19.0
b) 19.2
c) 1.902
d) 0.192
Explanation: The correction for sides in a traverse is given as correction = e / N, where N is the number of sides and e is the closing error. On substitution, we get, correction = 1.92 / 10 = 0.192.
14. What would be the correction for any side of a traverse in axis method if it has a closing error e = 0.93, length of side and axis would be 243.13 and 100 respectively?
a) 2.131
b) 1.131
c) 1.113
d) 1.311
Explanation: In Axis method of balancing a traverse, correction = length of side * (e/2) / length of axis. On substitution we get,
Correction = 243.13 * (0.93/2) / 100 = 1.131.
15. Which of the following indicates the correct value of precise closing error if e = 0.54 and lengths of sides are 92.69 m, 119.23 m, 92.64 m, 42.96 m and 60.96 m.
a) 1 / 766.445
b) 1 / 746.445
c) 1 / 756.445
d) 1 / 765.445
Explanation: The precise error of closure can be given as, error of closure = e / p
Where e = closing error = 0.54 and p = perimeter of traverse = 92.69 + 119.23 + 92.64 + 42.96 + 60.96 = 408.48 m.
Precise error is given as 0.54 / 408.48 = 1 / 756.445.
Sanfoundry Global Education & Learning Series – Surveying.
To practice all areas of Surveying, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 2,003 | 6,660 | {"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} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.847459 |
https://bestcurrencysfus.web.app/bocock50042hyco/expected-rate-of-inflation-150.html | 1,638,669,157,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00174.warc.gz | 195,714,067 | 6,352 | ## Expected rate of inflation
Estimates of future prices and values are usually based on projections using the average inflation rate - essentially an expected inflation calculator. Wondering
For years prior to 2015, the new value of the dollar amount is calculated using historical annual inflation rates provided by the Bureau of Labor Statistics. For years between 2016 and 2065, the new value is calculated using the historical average inflation rate, but this can be adjusted. Expected rate of inflation. The public's expectations for inflation.These expectations determine how large an effect a given policy action by the Fed will have on economic activity. Each bar represents the average Annual Inflation for that decade (not the total cumulative inflation for that 10 year period but how much it increased each year on average during that decade). The final Yellow bar shows the average annual inflation rate since the government began tracking it in 1913. Expected rate of inflation The public's expectations for inflation. These expectations determine how large an effect a given policy action by the Fed will have on economic activity. Expected Rate of Inflation Investor and public expectations of current or future inflation. These expectations may or may not be rational, but they may affect how the market All agencies are consistent that CPI inflation will increase in 2020 from an average of 1.8 in 2019. Over the longer-term up to 2024, CPI inflation in the US is expected to be around 2.3 percent. The inflation rate depends on the balance between aggregate supply and demand within the economy.
## Table: Annual Inflation Rates by Month and Year. Since figures below are 12-month periods, look to the December column to find inflation rates by calendar year. For example, the rate of inflation in 2019 was 2.3%. The last column, “Ave,” shows the average inflation rate for each year, which was 1.8% in 2019.
The particular measure of consumer price inflation is the percentage change in Assessing the current and expected rate of inflation against the inflation target The nominal interest rate reflects two factors: the rate of interest that would prevail if inflation were zero (the real rate of interest, below), and the expected rate of expected rate of inflation (which only few economists would expect in a rate of inflation is estimated by standard regression analysis. An examination of the 10 Feb 2020 Infexps affect interest rates (nominal interest rate (NIR) = real interest rate (RIR) + expected inflation (EIR)) and, consequently, investment Estimates of future prices and values are usually based on projections using the average inflation rate - essentially an expected inflation calculator. Wondering
### Fisher's famous theory about interest and inflation (Fisher. 1930). That theory holds that an increase in the rate of inflation expected by the public leads to an
More precisely, the Fisher equation states that the nominal interest ( i ) rate equals the real interest ( ir ) rate plus the expected rate of inflation ( πe ). i = ir + πe. Inflation rates vary from year to year and from currency to currency. where i is the nominal rate, r is the real rate, and gPe is the expected inflation rate. Inflation rate, average consumer prices. Annual percent change. map list chart. Settings. Map. From, Up to, Label, Color. confirm cancel reset. 25% or more. With x =inflation rate, (y-y*) =output gap, xe =expected inflation rate and g (.) = functional fonn. -4. 9. Page 13. The inflation According to the Lucas supply function, the natural rate is defined as the level of the GDP gap where the actual rate of inflation coincides with the expected rate of
### The public's expectations for inflation. These expectations determine how large an effect a given policy action by the Fed will have on economic activity.
Inflation rate is said to be the Price Increasing percentage of certain items which are approved by In a recession, one would expect a fall in the inflation rate. 25 Jul 2019 Suppose the inflation rate is 8% (per year) and your income Nobody can expect to optimize perfectly, so some loss of opportunity has to be Inflation is a continuing rise in the general price level usually attributed to an increase in the volume Inflation does not work the same way as you might expect. The U.S. inflation rate by year is how much prices change year-over-year. Year-over-year inflation rates give a clearer picture of price changes than annual average inflation. The Federal Reserve uses monetary policy to achieve its target rate of 2% inflation. Expected rate of inflation The public's expectations for inflation. These expectations determine how large an effect a given policy action by the Fed will have on economic activity. Expected Rate of Inflation Investor and public expectations of current or future inflation. These expectations may or may not be rational, but they may affect how the market What is the inflation rate for 2019? This statistic shows the annual projected inflation rate in the U.S. from 2008 to 2024. According to the forecast, prices will increase by two percent in 2019. Each bar represents the average Annual Inflation for that decade (not the total cumulative inflation for that 10 year period but how much it increased each year on average during that decade). The final Yellow bar shows the average annual inflation rate since the government began tracking it in 1913.
## Even a modest rate of inflation can seriously erode purchasing power over time. Assume Example: The current (2005) estimated cost of a project is \$100,000.
Inflation rate, average consumer prices. Annual percent change. map list chart. Settings. Map. From, Up to, Label, Color. confirm cancel reset. 25% or more. With x =inflation rate, (y-y*) =output gap, xe =expected inflation rate and g (.) = functional fonn. -4. 9. Page 13. The inflation According to the Lucas supply function, the natural rate is defined as the level of the GDP gap where the actual rate of inflation coincides with the expected rate of 26 Jul 2019 The Fed is expected to cut interest rates Wednesday, after second quarter GDP shows inflation is still sluggish and trade wars are impacting 20 May 2019 However, inflation rates for supplies and materials, as well as for utilities, are projected above those of last year. Faculty salaries are projected 10 Sep 2013 by subtracting the expected inflation rate from the nominal interest rate. For the Fisher hypothesis to hold, the resultant ex ante real interest rate Answer to When the actual and expected (or anticipated) inflation rates are both zero, the money interest rate must equal the real
With x =inflation rate, (y-y*) =output gap, xe =expected inflation rate and g (.) = functional fonn. -4. 9. Page 13. The inflation According to the Lucas supply function, the natural rate is defined as the level of the GDP gap where the actual rate of inflation coincides with the expected rate of 26 Jul 2019 The Fed is expected to cut interest rates Wednesday, after second quarter GDP shows inflation is still sluggish and trade wars are impacting 20 May 2019 However, inflation rates for supplies and materials, as well as for utilities, are projected above those of last year. Faculty salaries are projected 10 Sep 2013 by subtracting the expected inflation rate from the nominal interest rate. For the Fisher hypothesis to hold, the resultant ex ante real interest rate Answer to When the actual and expected (or anticipated) inflation rates are both zero, the money interest rate must equal the real | 1,595 | 7,605 | {"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} | 2.515625 | 3 | CC-MAIN-2021-49 | latest | en | 0.935112 |
https://www.thinbug.com/q/43545879 | 1,620,558,474,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00361.warc.gz | 1,076,285,606 | 5,155 | ### 带有多个标签的条形图
``````df = pd.DataFrame(np.random.rand(6, 4),
index=['one', 'two', 'three', 'four', 'five', 'six'],
columns=pd.Index(['A', 'B', 'C', 'D'],
name='Genus')).round(2)
df.plot(kind='bar',figsize=(10,4))
``````
#### 3 个答案:
``````import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(6, 4),
index=['one', 'two', 'three', 'four', 'five', 'six'],
columns=pd.Index(['A', 'B', 'C', 'D'],
name='Genus')).round(2)
df.plot(kind='bar',figsize=(10,4))
ax = plt.gca()
pos = []
for bar in ax.patches:
pos.append(bar.get_x()+bar.get_width()/2.)
ax.set_xticks(pos,minor=True)
lab = []
for i in range(len(pos)):
l = df.columns.values[i//len(df.index.values)]
lab.append(l)
ax.set_xticklabels(lab,minor=True)
plt.setp(ax.get_xticklabels(), rotation=0)
plt.show()
``````
``````df = pd.DataFrame(np.random.rand(6, 4),
index=['one', 'two', 'three', 'four', 'five', 'six'],
columns=pd.Index(['A', 'B', 'C', 'D'],
name='Genus')).round(2)
ax = df.plot(kind='bar',figsize=(10,4), rot = 0)
# "Activate" minor ticks
ax.minorticks_on()
# Get location of the center of each rectangle
rects_locs = map(lambda x: x.get_x() +x.get_width()/2., ax.patches)
# Set minor ticks there
ax.set_xticks(rects_locs, minor = True)
# Labels for the rectangles
new_ticks = reduce(lambda x, y: x + y, map(lambda x: [x] * df.shape[0], df.columns.tolist()))
# Set the labels
from matplotlib import ticker
# Move the category label further from x-axis
# Remove minor ticks where not necessary
ax.tick_params(axis='x',which='both', top='off')
ax.tick_params(axis='y',which='both', left='off', right = 'off')
``````
``````import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
def subcategorybar(X, vals,als, width=0.8):
n = len(vals)
_X = np.arange(len(X))
plt.figure(figsize=(14,9))
for i in range(n):
plt.bar(_X - width/2. + i/float(n)*width, vals[i],
width=width/float(n), align="edge")
for j in _X:
plt.text([_X - width/2. + i/float(n)*width][0][j],vals[i][j]+0.01*vals[i]
[j],str(als[i][j]))
plt.xticks(_X, X)
### data
X = ['a','b','c','d','f']
A1 = [1,2,3,4,5]
A2= [1,7,6,7,8]
A3 = [3,5,6,8,9]
A4= [4,5,6,7,3]
A5 = [5,6,7,8,5]
##labels
A1_al = ['da','dd',5,6,3]
A2_al = np.random.random_integers(20,size=5)
A3_al = np.random.random_integers(20,size=5)
A4_al = np.random.random_integers(20,size=5)
A5_al = np.random.random_integers(20,size=5)
subcategorybar(X, [A1,A2,A3,A4],[A1_al,A2_al,A3_al,A4_al],width=0.8)
plt.show()
`````` | 874 | 2,483 | {"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} | 2.65625 | 3 | CC-MAIN-2021-21 | latest | en | 0.334872 |
https://vestnik.utmn.ru/eng/energy/vypuski/2015-tom-1/2-2/113221/ | 1,596,542,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.93/warc/CC-MAIN-20200804102630-20200804132630-00533.warc.gz | 565,068,335 | 10,965 | # Natural frequencies of longitudinal oscillations of a droplet in the constriction of the capillary tube
Tyumen State University Herald. Physical and Mathematical Modeling. Oil, Gas, Energy
Release:
2015, Vol. 1. №2(2)
Title:
Natural frequencies of longitudinal oscillations of a droplet in the constriction of the capillary tube
Authors:
Amir A. Gubaidullin, Dr. Sci. (Phys.-Math.), Professor, Сhief Researcher, Tyumen Branch of the Khristianovich Institute of Theoretical and Applied Mechanics of the Siberian Branch of the Russian Academy of Sciences; eLibrary AuthorID, ORCID, Web of Science ResearcherID, Scopus AuthorID, a.a.gubaidullin@yandex.ru
Aleksei Yu. Maksimov, Lead Engineer, CompMechLab® LLC, Peter the Great St. Petersburg Polytechnic University; eLibrary AuthorID, ORCID, maksimov@compmechlab.ru
Abstract:
The actual problem of improving wave technologies of enhanced oil recovery lies in challenges of fi nding the frequency of longitudinal oscillations of oil droplets (ganglion), jammed in the narrowing of pores. To determine the natural frequencies of droplets computer modeling can be used. As a mathematical model for the numerical study Navier – Stokes equations or Lattice-Boltzmann method can be used. However, to obtain preliminary estimates and solve engineering problems it is convenient to have a formula that allows, even approximately, getting the desired result. In this paper a formula is obtained for the natural frequencies of longitudinal oscillations of a droplet surrounded by immiscible with its fl uid in the narrowing of capillary conical shape. Comparison of the results of calculations by the formula with the numerical solution of the problem has been carried out. The analysis of the frequency of the droplet oscillations in the static pressure difference, surface tension and contact angle of wetting has been carried out.
Keywords:
References:
1. Gubaidullin, A. A., Maksimov, A. Y. Modeling of dynamics of oil droplet in a capillary with narrowing // Herald of the Tyumen State University. 2013. № 7. Pp. 71-77.
2. Hilpert, M. Capillarity-induced resonance of blobs in porous media: Analytical solutions, Lattice-Boltzmann modeling, and blob mobilization // J. of Colloid and Interface Science. 2007. Vol. 309 (2). Pp. 493-504.
3. Nikolaevski, V., Stepanova, G. S. Nonlinear seismic and acoustic impact on oil recovery // Acoustical Physics. 2005. Vol. 51. Pp. 150-159.
4. Maksimov, G. A, Radchenko A. V. Modeling of oil production intensification at acoustical impact on the formation of the well // Acoustical Physics. 2005. Vol. 51. Pp. 118-131.
5. Serdyukov, S. V., Kurlenya, M. V. The mechanism of stimulation of oil fields of low intensity seismic // Acoustical Physics. 2007. Vol. 53. № 5. Pp. 703-714.
6. Dyblenko, V. P., Kamalov, R. N., Sharifullin, R. J., Tufanov I. A. Increased productivity and resuscitation wells using vibration impact. M., 2000. 381 p.
7. Surguchev, M. L., Kuznetsov, O. L., Simkin, E. M. Hydrodynamic, acoustic, thermal cyclic exposure to oil reservoirs. M.: Nedra, 1975.
8. Beresnev, I. A. Theory of vibratory mobilization of non-wetting fluids entrapped in pore constrictions // GEOPHYSICS. Vol. 71. № 6. 2006. Pр. 47-56.
9. Beresnev, I. A., and W. Deng. Viscosity effects in vibratory mobilization of residual oil // GEOPHYSICS. Vol. 75. № 4. 2010. Pр. 79-85.
10. Beresnev, I., Gaul, W., Vigil, R. D. Direct pore-level observation of permeability increase in two-phase fl ow by shaking // GEOPHYSICAL RESEARCH LETTERS. Vol. 38. 2011. L20302.
11.Gubaidullin, D. A., Nikiforov, G. A. The simulation of two-phase fl uid flow in a layered oil reservoir under a nonliner fi ltration law // Bullettin of the Tyumen State University. Physical and mathematical modeling. Oil, Gas , Energy. № 1(1). Т. 1. 2015. Pp. 59-64.
| 1,087 | 3,823 | {"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} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.840111 |
https://www.teacherspayteachers.com/Product/Finding-Distance-Between-Two-Points-Scavenger-Hunt-Decimals-PP-1329283 | 1,547,756,192,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00218.warc.gz | 952,406,967 | 18,351 | # Finding Distance Between Two Points Scavenger Hunt - Decimals - PP
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This Finding Distance Between Two Points Scavenger Hunt - Decimals - PP was created by Pupsaroni Puzzles for use in an 8th grade, Algebra 1, Algebra 2 or Geometry class during a unit where students are finding distance.
The cards all have an “answer” at the top, and another problem at the bottom to continue the scavenger hunt. Students can start at any card they choose. They write down their first answer, and work the problem below which leads them to the next card. Students continue working until they have gone through all the cards and come back to the answer they started with on the scavenger hunt. If there are 13 cards, students should have 13 different answers. If they come back to an answer before going through them all, the students have made a mistake and need to start over and retrace their steps checking their answers for accuracy. An answer key is provided for either the teacher or students to check their work. Answers are in order going down the column and going on to the next.
Printing the cards one per page makes the pages very easy to see, or printing two to a page makes the cards slightly smaller, but still very readable. Teachers just need to tape the cards up on the walls in some random order. Students can then work the cards as they have time and opportunity.
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\$2.50 | 344 | 1,575 | {"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-2019-04 | latest | en | 0.955352 |
https://www.physicsforums.com/threads/blade-containment-simplified-analytical-approach.999900/ | 1,725,861,747,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651072.23/warc/CC-MAIN-20240909040201-20240909070201-00709.warc.gz | 880,065,250 | 21,557 | # Blade containment - simplified analytical approach
• FEAnalyst
In summary, the fan blade detached from the hub in the engine causing the aircraft to depressurize and a passenger to die.
FEAnalyst
TL;DR Summary
What is the simplified analytical approach to blade containment problem?
Hi,
one of the most interesting experimental tests performed for rotating machinery (such as gas turbines) is blade containment test - if the blade detaches from the hub, it can't break through the cover of the turbine because it could result in catastrophic damage (especially in case of airplanes). Apart from physical experiments, such tests are very often performed with FEA. However, I wonder if there's a way to perform some simplified analytical calculations (and determine whether the blade will be able to break through the cover or not) before proceeding to FEA or in order to confirm the correctness of numerical analysis. What I've found so far is approach based on the kinetic energy of the blade after detachment. The formula I've found in some scientific paper is: $$E_{K}=\frac{1}{2}m (\omega \cdot r)^{2}$$ where: ##m## - blade mass, ##\omega## - angular velocity, ##r## - radius. Is this formula correct for a situation when body suddenly switches from rotational to translation motion? What to do next? And is it possible to account for the deformation of the blade (which is significant in this case)?
I found some good sources using search term jet engine blade containment. From one of the hits:
I had looked into a similar problem some years ago. That problem in involved orbital log saws coming apart. A chunk of steel weighing over 100 lbs coming off a 24 inch radius orbit arm running 300 RPM makes a serious dent in the steel guardhouse. In our case, I calculated that it would not penetrate a 12 gauge steel inner wall, provided that the inner wall was properly supported around the edges. The analysis involved calculating force vs dent distance, and integrating until the total work was greater than the worst case kinetic energy of the flying parts.
Brochure of the log saw mentioned above: https://catalog.bretting.com/item/saws/24-inch-orbital-log-saw/1058. I headed the team that developed that saw, the conveyor system, and the guard house.
Bottom line: No simple solution for high speed turbine blades.
Lnewqban and berkeman
Thanks for reply. Can you say more about the approach you used for log saws (some equations, sources)?
Building 273 at GE's Schenectady works had a turbine overspeed test stand. It had a steel inner liner, then 6 feet of railroad ties, and a steel outer liner. It did not always succeed in containing broken blades. Legend said that one engineer was killed while sitting at his desk as a blade fell from the sky and hit him.
Lnewqban
Then, we had accident of flight 1380 of SW Airlines.
https://en.m.wikipedia.org/wiki/Southwest_Airlines_Flight_1380
“On November 19, 2019, following the aforementioned hearing, the NTSB released the final report on the accident.[2] the probable cause reads:
The National Transportation Safety Board (NTSB) determines that the probable cause of this accident was a low-cycle fatigue crack in the dovetail of fan blade No. 13, which resulted in the fan blade separating in flight and impacting the engine fan case at a location that was critical to the structural integrity and performance of the fan cowl structure. This impact led to the in-flight separation of fan cowl components, including the inboard fan cowl aft latch keeper, which struck the fuselage near a cabin window and caused the window to depart from the airplane, the cabin to rapidly depressurize, and the passenger fatality.”
## 1. What is blade containment?
Blade containment refers to the ability of a turbine engine to prevent the fragments of a failed blade from causing damage to other parts of the engine or the aircraft.
## 2. Why is blade containment important?
Blade containment is important because it ensures the safety of the aircraft and its passengers in the event of a blade failure. It also helps to minimize the risk of engine damage and potential engine failure.
## 3. What is a simplified analytical approach to blade containment?
A simplified analytical approach to blade containment involves using mathematical models and simulations to predict the behavior of a blade in the event of failure. This approach can help engineers design more effective containment systems.
The design of a blade, including its material, shape, and attachment to the engine, can greatly impact its ability to be contained in the event of failure. A well-designed blade will have a higher chance of being contained and minimizing damage.
## 5. Are there regulations or standards for blade containment?
Yes, there are regulations and standards set by organizations such as the Federal Aviation Administration (FAA) and the European Aviation Safety Agency (EASA) that dictate the minimum requirements for blade containment in aircraft engines.
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3K | 1,135 | 5,204 | {"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": 0, "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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-38 | latest | en | 0.918676 |
https://cheatsheeting.com/show.html?sheet=furlong-to-millimeter-conversions | 1,642,597,198,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301341.12/warc/CC-MAIN-20220119125003-20220119155003-00601.warc.gz | 231,030,766 | 9,222 | Home > Conversions (Length) > Conversion tables from/to furlong > fur to mm Conversion Cheat Sheet (Interactive)
From: Step: Decimals: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 You could also enter the values to convert and print directly on the table
[Formula: mm = fur x 201168] [Printer friendly] [Millimeters to Furlongs]
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# Furlongs to Millimeters Conversion Table
fur = 201168 mm
# How to convert from Furlongs to Millimeters
Since 1 furlong is equal to 201168 millimeters, we could say that n furlongs are equal to 201168 times n millimeters. In other words, we could use the following formula:
millimeters = furlongs x 201168
For example, let's say that we want to convert 2 furlongs to millimeters. Then, we just replace furlongs in the abovementioned formula with 2:
millimeters = 2 x 201168
That is, 2 furlongs are equal to 402336.0 millimeters. | 587 | 1,515 | {"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-2022-05 | latest | en | 0.182934 |
https://cracku.in/blog/open-dice-questions-for-ssc-cgl-pdf/ | 1,713,128,462,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00233.warc.gz | 167,539,273 | 36,510 | 0
14320
Open dice Questions for SSC CGL PDF
Download SSC CGL Open dice Questions with answers PDF based on previous papers very useful for SSC CGL exams. Top-10 Very Important Open dice Questions for SSC Exams.
Question 1:
a)
b)
c)
d)
Question 2: A paper sheet is made into the following pattern (A). Find the cube that will best represent after folding the paper (A).
a) 1, 2 and 3 only
b) 1, 3 and 4 only
c) 2, 3 and 4 only
d) 2 and 3 only
Question 3: Four different positions of the same dice are shown. Find the number on the face opposite to the one having 3.
a) 2
b) 4
c) 1
d) 6
Question 4: Four positions of a cube are shown below. Which color is opposite to white color in the given cubes?
a) Orange
b) Blue
c) Red
d) Yellow
e) None of the above
Question 5: Four positions of a dice are given below:
Find the number on the face opposite to the face showing 4.
a) 5
b) 3
c) 6
d) 1
Question 6: From the given blocks when 10 is at the bottom, which number will be at the top ?
a) 8
b) 12
c) 6
d) 4
Question 7: From the given options which answer figure can be formed by folding the figure given in the question
a)
b)
c)
d)
Question 8: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 9: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 10: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 11: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 12: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 13: From the given options, which answer figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
Question 14: From the given options, which answer figure can be formed by folding the figure in the question ?
a)
b)
c)
d)
Question 15: From the given options, which figure can be formed by folding the figure given in the question ?
a)
b)
c)
d)
2, 3 and 4 will best represent after folding the paper (A).
From 4 figure of dice we conclude that adjacent of 3 is 4,5,6,1, so 2 will be opposit of 3.
When we fold the question figure, we get ‘$’ as the base of the cube and ‘|’ as top, thus they face each other. Also, ‘%’ will face ‘>’ and ‘*’ will face ‘#’. Thus, (*,#) , ($,|) and (%,>) cannot be the adjacent sides of the cube as mentioned in the last three options respectively.
=> Ans – (A) | 766 | 2,675 | {"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} | 2.890625 | 3 | CC-MAIN-2024-18 | latest | en | 0.872378 |
https://metanumbers.com/5696 | 1,611,345,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531335.42/warc/CC-MAIN-20210122175527-20210122205527-00165.warc.gz | 466,820,294 | 7,505 | 5696
5,696 (five thousand six hundred ninety-six) is an even four-digits composite number following 5695 and preceding 5697. In scientific notation, it is written as 5.696 × 103. The sum of its digits is 26. It has a total of 7 prime factors and 14 positive divisors. There are 2,816 positive integers (up to 5696) that are relatively prime to 5696.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 26
• Digital Root 8
Name
Short name 5 thousand 696 five thousand six hundred ninety-six
Notation
Scientific notation 5.696 × 103 5.696 × 103
Prime Factorization of 5696
Prime Factorization 26 × 89
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 178 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 5,696 is 26 × 89. Since it has a total of 7 prime factors, 5,696 is a composite number.
Divisors of 5696
1, 2, 4, 8, 16, 32, 64, 89, 178, 356, 712, 1424, 2848, 5696
14 divisors
Even divisors 12 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 14 Total number of the positive divisors of n σ(n) 11430 Sum of all the positive divisors of n s(n) 5734 Sum of the proper positive divisors of n A(n) 816.428 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 75.4718 Returns the nth root of the product of n divisors H(n) 6.97673 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 5,696 can be divided by 14 positive divisors (out of which 12 are even, and 2 are odd). The sum of these divisors (counting 5,696) is 11,430, the average is 8,16.,428.
Other Arithmetic Functions (n = 5696)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2816 Total number of positive integers not greater than n that are coprime to n λ(n) 352 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 753 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 2,816 positive integers (less than 5,696) that are coprime with 5,696. And there are approximately 753 prime numbers less than or equal to 5,696.
Divisibility of 5696
m n mod m 2 3 4 5 6 7 8 9 0 2 0 1 2 5 0 8
The number 5,696 is divisible by 2, 4 and 8.
• Abundant
• Polite
• Practical
• Frugal
Base conversion (5696)
Base System Value
2 Binary 1011001000000
3 Ternary 21210222
4 Quaternary 1121000
5 Quinary 140241
6 Senary 42212
8 Octal 13100
10 Decimal 5696
12 Duodecimal 3368
20 Vigesimal e4g
36 Base36 4e8
Basic calculations (n = 5696)
Multiplication
n×i
n×2 11392 17088 22784 28480
Division
ni
n⁄2 2848 1898.67 1424 1139.2
Exponentiation
ni
n2 32444416 184803393536 1052640129581056 5995838178093694976
Nth Root
i√n
2√n 75.4718 17.859 8.68745 5.63786
5696 as geometric shapes
Circle
Diameter 11392 35789 1.01927e+08
Sphere
Volume 7.74103e+11 4.07709e+08 35789
Square
Length = n
Perimeter 22784 3.24444e+07 8055.36
Cube
Length = n
Surface area 1.94666e+08 1.84803e+11 9865.76
Equilateral Triangle
Length = n
Perimeter 17088 1.40488e+07 4932.88
Triangular Pyramid
Length = n
Surface area 5.61954e+07 2.17793e+10 4650.76
Cryptographic Hash Functions
md5 b8cfbf77a3d250a4523ba67a65a7d031 a9c9a6e266a80dc17285537584f10a5d4322f5e8 e38fa75ec230eecbb53b3e2809eb8e99108a596c65bdff543bc904cf11a5184d 96f26abb321621232e6c8659a3944caf09150589f3fc787232ac2cebb36229b9a95af265eb87ac833fd5b81ec742c2552d6ab2d537fd366f9f70632907220bab d6174f5b4a5067cd91f0e09391a11fb07a721c76 | 1,419 | 3,949 | {"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} | 3.625 | 4 | CC-MAIN-2021-04 | longest | en | 0.832037 |
https://answer-helper.com/mathematics/question1093282 | 1,679,933,902,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948673.1/warc/CC-MAIN-20230327154814-20230327184814-00197.warc.gz | 123,425,695 | 15,908 | , 30.08.2019 17:50 mariafsoriano928
# What are the key features for a linear function, and how can you use those features to sketch graphs? explain them through examples
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Questions on the website: 13802536 | 377 | 1,262 | {"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} | 2.515625 | 3 | CC-MAIN-2023-14 | latest | en | 0.903698 |
https://www.careerride.com/mchoice/conditions-for-repeated-stress-10253.aspx | 1,726,758,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00378.warc.gz | 644,785,088 | 5,815 | # Conditions for repeated stress
Q. Which of the following conditions is true for repeated stress?
1. σm = 0
2. σm = σmax / 2
3. σm = σa
4. σmin = 0
5. σmin = – σmax
6. σa = σmax / 2
where σm = mean stress and σa = stress amplitude
- Published on 14 Sep 15
a. condition 2 and 3
b. condition 1, 3 and 5
c. condition 2, 4, and 6
d. condition 3,4, 5 and 6
ANSWER: condition 2, 4, and 6
#### Discussion
• Sravanthi -Posted on 03 Nov 15
- The variable stresses induced in a component which vary in magnitude and direction with respect to time are called as fluctuating stresses.
- Fluctuating stresses are of two types:
1) Repeated stresses
2) Completely reversed stress
1) Repeated stresses:
- Repeated stresses are observed in gears, cams, etc.
Mean stress (σm) = (σmax + σmin) / 2
Stress amplitude (σa) = (σmax – σmin) / 2
- σmin = 0 for repeated stresses, therefore substituting the given values we get,
σm = σmax / 2 & σa = σmax / 2
## ➨ Post your comment / Share knowledge
Enter the code shown above:
(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.) | 360 | 1,121 | {"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} | 3.203125 | 3 | CC-MAIN-2024-38 | latest | en | 0.796774 |
https://justcnw.com/volume/litres-pints/ | 1,487,743,749,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170914.10/warc/CC-MAIN-20170219104610-00586-ip-10-171-10-108.ec2.internal.warc.gz | 735,299,496 | 8,963 | # Convert Litres to Pints
Want to change from a volume in litres to pints? Use this litres to pints converter to help with any calculation you need. Many more volume converters here!
## Conversion Litres - Pints
L
Litres and pints are most commonly used for measuring volume quantities of drinks. They are both used in the imperial system and American system of volume units. The unit of litres is found on many drink packaging information labels while the unit of pint is used more frequently when pouring drinks, mainly beer, in bars and restaurants. Pints are used on labels for drinks in the United States more than litres. Use our litre – pint converter for your calculations, with many more volume converters available.
## How to use the Converter from Litres to Pints?
If you wish to convert litres to pints, enter the quantity of litres you need to change in the box above and click “convert” to be given a new quantity in pints. To work the other way, use our conversion tool to convert pints to litres.
## Formula to Convert L – Pt
You can see below that we have given the mathematical formula for this litres – pints conversion if you need to know how it is done or so you can do the volume calculation yourself. As some examples, 7 L = 14.8 pt and 33 L = 69.7 pt
[number of] litres x 2.11338 = [number of] pints
## Uses for Litres
The volume unit of litres is easy to find for everyday capacities. For example, in the shops you will find drinks such as water and juices measured in bottles of litres. Litres derived from the French metric system and the unit of litrons. One litre is the capacity for a cube with sides of 10cm. Above is our litre to pint calculator.
## Uses for Pints
Pints are famously known in Europe for measuring volumes of beers and ales. Originating from the word to signify the line painted on an ale glass, it is now colloquially used when ordering drinks in a restaurant or in bars. Try out our L – pt volume converter if you need to change between the two units.
### How much is 1 Litre in Pints?
1 Litre equal to 2.11 Pints `(1L = 2.11pt)`
### How much are 2 Litres in Pints?
2 Litres equal to 4.23 Pints `(2L = 4.23pt)`
### How much are 3 Litres in Pints?
3 Litres equal to 6.34 Pints `(3L = 6.34pt)`
### How much are 4 Litres in Pints?
4 Litres equal to 8.45 Pints `(4L = 8.45pt)`
### How much are 5 Litres in Pints?
5 Litres equal to 10.57 Pints `(5L = 10.57pt)`
### How much are 10 Litres in Pints?
10 Litres equal to 21.13 Pints `(10L = 21.13pt)`
### How much are 15 Litres in Pints?
15 Litres equal to 31.7 Pints `(15L = 31.7pt)`
### How much are 20 Litres in Pints?
20 Litres equal to 42.27 Pints `(20L = 42.27pt)`
### How much are 25 Litres in Pints?
25 Litres equal to 52.83 Pints `(25L = 52.83pt)`
### How much are 30 Litres in Pints?
30 Litres equal to 63.4 Pints `(30L = 63.4pt)`
### How much are 50 Litres in Pints?
50 Litres equal to 105.67 Pints `(50L = 105.67pt)`
### How much are 100 Litres in Pints?
100 Litres equal to 211.34 Pints `(100L = 211.34pt)`
### How much are 200 Litres in Pints?
200 Litres equal to 422.68 Pints `(200L = 422.68pt)`
### How much are 500 Litres in Pints?
500 Litres equal to 1056.69 Pints `(500L = 1056.69pt)`
### How much are 1000 Litres in Pints?
1000 Litres equal to 2113.38 Pints `(1000L = 2113.38pt)`
Insert this converter to your website | 977 | 3,371 | {"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} | 3.140625 | 3 | CC-MAIN-2017-09 | latest | en | 0.887414 |
https://www.jiskha.com/display.cgi?id=1345773176 | 1,502,920,698,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102663.36/warc/CC-MAIN-20170816212248-20170816232248-00247.warc.gz | 951,101,699 | 4,122 | # Math (Trig/pre-calculus)
posted by .
some stars are so far away that their position appear fixed as earth orbits the sun. other stars, however, appear over time to shift their positions relative to the background of "fixed" stars. suppose that the star shown below appears to shift through an arc of theta = 0 degrees 0 minutes and 1.5 seconds when viewed on the first day of winter and the first day of summer. if the distance from earth to the sun is about 1.5 x 10^8km, find the approximate distance from earth to the star
• Math (Trig/pre-calculus) -
nevermind, I figured it out, it is approximately 4.1 x10^13 km from the star
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mc030-1.jpg Two stars are bound together by gravity. The larger star has four times the mass of the smaller star. Which diagram best illustrates the orbital motion of the two stars? | 494 | 2,133 | {"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} | 3.078125 | 3 | CC-MAIN-2017-34 | latest | en | 0.916775 |
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-4-20246 11 No solution. 7 At least 15 times 8 No solution. E nKpugtWay dSUosft6wdazr2eo KLsLECeO C SADlpl7 XrGimgphItusZ 9rAe4sOexrav6exdPZ x XMnaHd9eQ rwJift vhs yI3ndf7ifnviWtiec zAolXgmeVbAr7aq z1qv-6-Worksheet by Kuta Software LLC Answers to Review Sheet.
Write a linear inequality from a word problem graph a linear inequality in two variables practice 1 each country is being asked to donate food packages and medicine packages. Mark has fourteen dollars and Dan has sixteen dollars. Mark still has less money than Dan.
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X 2 0 is the only solution. Solving Linear Inequalities Solve simple linear inequalities that involve multiplying or dividing by a negative number. Express the given fraction as the product of two fractionsone of which has.
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Any content, trademark/s, or other material that might be found on this site that is not this site property remains the copyright of its respective owner/s. | 1,081 | 4,782 | {"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} | 3.578125 | 4 | CC-MAIN-2022-05 | latest | en | 0.845232 |
https://www.coursehero.com/file/37730713/Book11xlsx/ | 1,550,314,757,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480272.15/warc/CC-MAIN-20190216105514-20190216131514-00154.warc.gz | 789,992,715 | 57,802 | Book11.xlsx
# Book11.xlsx - SECTION 12-7 SOLUTIONS TO SELF-TEST Project P...
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SECTION 12-7 SOLUTIONS TO SELF-TEST Regular payback Regular payback Years 0 1 2 3 4 | | | | | Cash Flow -950 525 485 445 405 Cas Cumulative Cash Flow -950 -425 60 505 910 Cumulative Cas Regular payback = 1.88 Regular payback = Discounted payback Discounted payb WACC 10% WACC 15% Years 0 1 2 3 4 | | | | | Cash Flow -950 525 485 445 405 Cas Discounted Cash Flow -950 477 401 334 277 Discounted Cas Cumulative Discounted CF -950 -473 -72 262 539 Cumulative Discoun Discounted payback = 2.22 Discounted paybac The payback rule of 3 years leads to a reject decision. The payback rule o NPV = \$539.05 NPV = \$256.72 IRR = 35.54% IRR = 24.78% Project P has a cost of \$1,000 and cash flows of \$300 per year for 3 years plus another \$1,000 in Year 4. The project’s cost of capital is 15%. What are P’s regular and discounted paybacks? If the company requires a payback of 3 years or less, would the project be accepted?
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https://openturns.github.io/openturns/1.16/auto_data_analysis/sample_analysis/plot_draw_survival.html | 1,675,076,161,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00540.warc.gz | 432,092,862 | 6,875 | # Draw a survival function¶
# sphinx_gallery_thumbnail_number = 9
## Introduction¶
The goal of this example is to show how to draw the survival function of a sample or a distribution, in linear and logarithmic scales.
Let be a random variable with distribution function :
for any . The survival function is:
for any .
Let us assume that is a sample from .
Let be the empirical cumulative distribution function:
for any . Let be the empirical survival function:
for any .
## Motivations for the survival function¶
For many probabilistic models associated with extreme events or lifetime models, the survival function has a simpler expression than the distribution function.
• More specifically, several models (e.g. Pareto or Weibull) have a simple expression when we consider the logarithm of the survival function. In this situation, the plot is often used. For some distributions, this plot is a straight line.
• When we consider probabilities very close to 1 (e.g. with extreme events), a loss of precision can occur when we consider the expression with floating point numbers. This loss of significant digits is known as “catastrophic cancellation” in the bibliography and happens when two close floating point numbers are subtracted. This is one of the reasons why we sometimes use directly the survival function instead of the complementary of the distribution.
## Define a distribution¶
import openturns as ot
import openturns.viewer as viewer
from matplotlib import pylab as plt
ot.Log.Show(ot.Log.NONE)
sigma = 1.4
xi=0.5
u=0.1
distribution = ot.GeneralizedPareto(sigma, xi, u)
## Draw the survival of a distribution¶
The computeCDF and computeSurvivalFunction computes the CDF and survival of a distribution.
p1 = distribution.computeCDF(10.)
p1
Out:
0.9513919027838056
p2 = distribution.computeSurvivalFunction(10.)
p2
Out:
0.048608097216194426
p1 + p2
Out:
1.0
The drawCDF and drawSurvivalFunction methods allows to draw the functions and .
graph = distribution.drawCDF()
graph.setTitle("CDF of a distribution")
view = viewer.View(graph)
graph = distribution.drawSurvivalFunction()
graph.setTitle("Survival function of a distribution")
view = viewer.View(graph)
In order to get finite bounds for the next graphics, we compute the xmin and xmax bounds from the 0.01 and 0.99 quantiles of the distributions.
xmin = distribution.computeQuantile(0.01)[0]
xmin
Out:
0.11410588272579382
xmax = distribution.computeQuantile(0.99)[0]
xmax
Out:
25.29999999999998
The drawSurvivalFunction methods also has an option to plot the survival with the X axis in logarithmic scale.
npoints = 50
logScaleX = True
graph = distribution.drawSurvivalFunction(xmin, xmax, npoints, logScaleX)
graph.setTitle("Survival function of a distribution where X axis is in log scale")
view = viewer.View(graph)
#graph
In order to get both axes in logarithmic scale, we use the LOGXY option of the graph.
npoints = 50
logScaleX = True
graph = distribution.drawSurvivalFunction(xmin, xmax, npoints, logScaleX)
graph.setLogScale(ot.GraphImplementation.LOGXY)
graph.setTitle("Survival function of a distribution where X and Y axes are in log scale")
view = viewer.View(graph)
#graph
## Compute the survival of a sample¶
We now generate a sample that we are going to analyze.
sample = distribution.getSample(1000)
sample.getMin(), sample.getMax()
Out:
(class=Point name=Unnamed dimension=1 values=[0.10353], class=Point name=Unnamed dimension=1 values=[269.593])
The computeEmpiricalCDF method of a Sample computes the empirical CDF.
p1 = sample.computeEmpiricalCDF([10])
p1
Out:
0.954
Activating the second optional argument allows to compute the empirical survival function.
p2 = sample.computeEmpiricalCDF([10], True)
p2
Out:
0.046
p1+p2
Out:
1.0
## Draw the survival of a sample¶
In order to draw the empirical functions of a Sample, we use the UserDefined class.
• The drawCDF method plots the CDF.
• The drawSurvivalFunction method plots the survival function.
userdefined = ot.UserDefined(sample)
graph = userdefined.drawCDF()
graph.setTitle("CDF of a sample")
view = viewer.View(graph)
#graph
graph = userdefined.drawSurvivalFunction()
graph.setTitle("Empirical survival function of a sample")
view = viewer.View(graph)
#graph
As previously, the drawSurvivalFunction method of a distribution has an option to set the X axis in logarithmic scale.
xmin = sample.getMin()[0]
xmax = sample.getMax()[0]
pointNumber = sample.getSize()
logScaleX = True
graph = userdefined.drawSurvivalFunction(xmin, xmax, pointNumber, logScaleX)
graph.setTitle("Empirical survival function of a sample; X axis in log-scale")
view = viewer.View(graph)
#graph
We obviously have , where is the sample maximum. This prevents from using the sample maximum and have a logarithmic Y axis at the same time. This is why in the following example we restrict the interval where we draw the survival function.
xmin = sample.getMin()[0]
xmax = sample.getMax()[0] - 1 # To avoid log(0) because P(X>Xmax)=0
pointNumber = sample.getSize()
logScaleX = True
graph = userdefined.drawSurvivalFunction(xmin, xmax, pointNumber, logScaleX)
graph.setLogScale(ot.GraphImplementation.LOGXY)
graph.setTitle("Empirical survival function of a sample; X and Y axes in log-scale")
view = viewer.View(graph)
#graph
## Compare the distribution and the sample with respect to the survival¶
In the final example, we compare the distribution and sample survival functions in the same graphics.
xmin = sample.getMin()[0]
xmax = sample.getMax()[0] - 1 # To avoid log(0) because P(X>Xmax)=0
npoints = 50
logScaleX = True
graph = userdefined.drawSurvivalFunction(xmin, xmax, pointNumber, logScaleX)
graph.setLogScale(ot.GraphImplementation.LOGXY)
graph.setColors(["blue"])
graph.setLegends(["Sample"])
graphDistribution = distribution.drawSurvivalFunction(xmin, xmax, npoints, logScaleX)
graphDistribution.setLegends(["GPD"])
graph.add(graphDistribution)
graph.setLegendPosition("topright")
graph.setTitle("GPD against the sample - n=%d" % (sample.getSize()))
view = viewer.View(graph)
#graph
plt.show()
Total running time of the script: ( 0 minutes 1.368 seconds)
Gallery generated by Sphinx-Gallery | 1,472 | 6,216 | {"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": 0, "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} | 3.796875 | 4 | CC-MAIN-2023-06 | latest | en | 0.833942 |
https://www.wiseowl.co.uk/power-bi/exercises/powerpivot/more-advanced-dax-functions/4068/ | 1,653,498,267,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00398.warc.gz | 1,270,705,702 | 9,824 | PowerPivot | More advanced DAX functions exercise | Use the EARLIER function to group sales into category bands
This exercise is provided to allow potential course delegates to choose the correct Wise Owl Microsoft training course, and may not be reproduced in whole or in part in any format without the prior written consent of Wise Owl.
Software ==> PowerPivot (75 exercises) Version ==> Excel 2013 and later Topic ==> More advanced DAX functions (5 exercises) Level ==> Harder than average Subject ==> Power BI training
Before you can do this exercise, you'll need to download and unzip this file (if you have any problems doing this, click here for help).
You need a minimum screen resolution of about 700 pixels width to see our exercises. This is because they contain diagrams and tables which would not be viewable easily on a mobile phone or small laptop. Please use a larger tablet, notebook or desktop computer, or change your screen resolution settings.
If you haven't already done so, run the SQL script in the above folder in SQL Server Management Studio to generate a database (not for commercial use or copying) called MAM
Again if you haven't already done so, create a new workbook and PowerPivot data model, and import the following tables: tblCentre, tblCentreType, tblPos, tblStore, tblTransaction
Open the Word document in the above folder, and paste the data into your model to create the following new table:
Resist the temptation, however, to link this to any other table - it should stand alone!
Now create a calculated column in the Centre table using the EARLIER function to give a size band for each centre. Here's what this should look like:
The column should show for each centre the name of the category where the centre's number of units is less than the upper threshold for the cateogry, but also greater than or equal to the lower.
This must be a calculated column, not a calculated field!
You should now be able to use this calculated column to create the following pivot table:
The biggest sales take place in So-so sized shopping centres.
Save your workbook as Rename that function, and close it down.
You can unzip this file to see the answers to this exercise, although please remember this is for your personal use only. | 477 | 2,277 | {"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} | 2.78125 | 3 | CC-MAIN-2022-21 | latest | en | 0.859378 |
https://www.assignmentexpert.com/homework-answers/physics/other/question-41556 | 1,596,467,973,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735812.88/warc/CC-MAIN-20200803140840-20200803170840-00458.warc.gz | 586,546,993 | 290,492 | 88 928
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# Answer to Question #41556 in Other Physics for promise
Question #41556
Calculate the change in internal energy of 2kg of water at 90 degree Celsius when it is changed to
3:30m3
of steam at
100oC
. The whole process occurs at atmospheric pressure. The latent heat of vaporization of water is
2:26×106J=kg
.
a. 4.27 MJ
b. 3.43 kJ
c. 45.72 mJ
d. 543.63 J
1
2014-04-28T06:29:56-0400
Need a fast expert's response?
Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS! | 189 | 589 | {"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} | 2.515625 | 3 | CC-MAIN-2020-34 | latest | en | 0.861745 |
https://brainmass.com/math/trigonometry/trigonometric-function-graphs-41532 | 1,708,586,832,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473735.7/warc/CC-MAIN-20240222061937-20240222091937-00796.warc.gz | 145,889,884 | 6,516 | Purchase Solution
# Trigonometric Function Graphs
Not what you're looking for?
I need help with the following problem.
For the range 0.1 < or = x <or = 0.2, plot the following function.
cos (3x)/sin (2x)
I really don't know where to begin.
thanks
##### Solution Summary
The expert examines trigonometric function graphs.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
Each question is a choice-summary multiple choice question that will present you with a linear equation and then make 4 statements about that equation. You must determine which of the 4 statements are true (if any) in regards to the equation.
##### Probability Quiz
Some questions on probability | 170 | 792 | {"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} | 2.625 | 3 | CC-MAIN-2024-10 | latest | en | 0.889767 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=12&t=32767 | 1,611,378,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703533863.67/warc/CC-MAIN-20210123032629-20210123062629-00443.warc.gz | 424,002,389 | 11,203 | ## Need help w/ determining the moles of gas produced
KylieY_3B
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am
### Need help w/ determining the moles of gas produced
Could someone explain how to find the net number of moles of gas produced for number 19 on the second review module?
It involves the combustion of 4 moles of butane (C4H10) gas. I understand how to balance the equation but I'm unsure how to determine the net number of moles of gas produced. The possible answers are 6, 4, and 5.
Brandon_Tran_2E
Posts: 63
Joined: Fri Sep 28, 2018 12:23 am
### Re: Need help w/ determining the moles of gas produced
Hey Kylie, the net moles of gas means the overall moles gas produced from the reactants. There are 36 moles on the product side and 30 moles on the reactant side. The net change is 36 - 30 = 6
Hai-Lin Yeh 1J
Posts: 89
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time
### Re: Need help w/ determining the moles of gas produced
The chemical equation is:
4 C4H10 (g) + 26O2 (g) -> 16 CO2 (g) + 20 H2O (g)
You can find the net number of moles produced by looking at the stoichiometric coefficients.
You take the sum of coefficient of product and minus sum of coefficient of reactants to get the net number of moles produced.
4+ 26 = 30 moles of reactants produced.
16 + 20 = 36 moles of products produced.
36 - 30 = 6. | 410 | 1,343 | {"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} | 3.3125 | 3 | CC-MAIN-2021-04 | latest | en | 0.894281 |
http://enakai00.hatenablog.com/entry/2019/04/15/210756 | 1,586,410,098,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371829677.89/warc/CC-MAIN-20200409024535-20200409055035-00392.warc.gz | 61,338,816 | 10,125 | # Reinforcement Learning 2nd Edition: Exercise Solutions (Chapter 2 - Chapter 5)
Reinforcement Learning: An Introduction (Adaptive Computation and Machine Learning series)
#### Sample code for the Jack's car rental problem
##### Exercise 4.8
Since p<0.5, if you keep playing with a constant bet, you will eventually lose in average. So, at some points, you need to bet enough to win at once. In this particular case, the player decided to bet 50 when he/she has 50, and to bet 25 when he/she has 75 (hoping to win with this bet). Similarly, the player decided to bet 25 when he/she has 25 to reach 50 (hoping to win at the next bet). | 163 | 637 | {"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} | 2.96875 | 3 | CC-MAIN-2020-16 | latest | en | 0.951417 |
http://www.decoz.com/Numerology_PersonalYear_A6.htm | 1,454,849,268,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701149377.17/warc/CC-MAIN-20160205193909-00127-ip-10-236-182-209.ec2.internal.warc.gz | 369,816,894 | 8,131 | Professional Numerology Software
# About Personal Year Cycles in Numerology
### There is some confusion among professionals and novices alike, about when exactly the Personal Year cycle starts and ends. Perhaps this will help clear up the confusion!
The word "cycle" implies something that is circular or moves in a circular motion.
In numerology, cycles are neither circular, nor do they move in a circular motion. They are more like waves. And, just as a wave starts with a gradual rise, slowly culminating to a climax - perhaps maintaining its position for a period of time - but eventually decreasing until dissolved completely - cycles, in numerology, also enter the picture gradually, they mature and stay for awhile, then slowly disappear.
In addition, cycles overlap each other during the first and last periods of their existence. When one cycle is born and begins its gradual rise toward maturity, the old cycle is still alive and kicking, albeit slowly dying. This overlap period is called a cusp period. Different cycles have cusp periods of different lengths.
The most important medium-term cycles in numerology are the Personal Year Cycles and the Essence Cycles. Generally speaking, the Personal Year cycles tend to be overrated, while the Essence cycles are almost always underrated.
### Let's talk about Personal Year cycles.
The Personal Year cycles in particular are popular, because not only are they strongly felt, they are also very easy to calculate. (See Calculation Methods.)
### What exactly are Personal Year cycles?
Think of Personal Year cycles as short 12-month sections on your Life Path that reveal how your personal "internal" cycles - your inner clockwork, so to speak - relate to, or are influenced by, the Universal Year cycles.
This, of course, is reflected in the way the Personal Year cycle is calculated: Add your Month of Birth, your Day of Birth, and the current year together, then reduce to a single digit. In other words, add your Month and Day of Birth to the Universal Year cycle. A fusion, you might say, of your Birth Date and the Universal Year cycle.
It is because of this fusion, that the Personal Year cycles run concurrent with the Universal Year cycles and, of course, the calendar year cycles.
### But, if that is true, why do some numerologists feel that the Personal Year cycles start at your birth day and not at the beginning of the year?
Even to the extend that they claim to have "tested it and found it to be true and more accurate?"
The answer, as you may have guessed, is in the overlapping cusp periods.
Think of your birth day as a powerful force that pulls and guides all your cycles, including the Personal Year cycle. The force your day of birth bestows upon your Personal Year cycle pulls this cycle towards maturity. Just like the moon pulls the tide, your day of birth pulls your Personal Year cycle.
Remember those cusp periods? Well, those cusp periods, which can be anywhere from 1 month to 6 months long, are controlled by your month and day of birth.
For example, a person born in February, has a short cusp period of perhaps one or two months at the beginning of the Personal Year cycle, reaching maturity sometime in the early part of January, which is followed by a five or six months period from January through July or August during which the Personal Year cycle is fully mature, after which the cycle slowly decreases over the next five or six months. During the last couple of months of this period of decrease, the new cycle is already muscling its way in, and so the cycles, excuse me, the waves, continue moving in and out of your life.
See the graphic below for a relatively accurate presentation of when Personal Years are in place, and their cusp periods, based on the month of birth.
Red shows the period when your current Personal Year cycle is in full force.
Yellow reveals when the old cycle is still influencing you, but is on its way out.
Blue/purple highlights the time when the next cycle is starting to influence you.
The list of months on the left reflects the months of birth.
Each row to the right of a month of birth shows the strength and duration of a Personal Year cycle for someone born during that month.
Yellow shows last year's outgoing cycle, red represents the Personal Year cycle for the current year, and orange the period during which they overlap.
Blue is the incoming cycle for next year, and purple the period during which the cusp periods of the current cycle and next year's cycle overlap.
### Personal Month cycles.
Personal Month cycles are calculated by adding the current month to your Personal Year cycle. And, just like Personal Year cycles, there is a cusp period on each end. This time, however, the cusp period is not controlled by your month of birth, but by your day of birth - in a similar fashion but on a smaller scale, as the yearly cycles.
### Personal Day cycles.
Personal Day cycles also have cusp periods. In Personal Day cycles they are based on the time of your birth. In addition, daylight savings time and location/time zones often push these cycles forward or backward several hours. You can see why some people feel that the Personal Day cycles are more accurate if shifted one day forward or backward, whichever the case may be. | 1,078 | 5,296 | {"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} | 2.8125 | 3 | CC-MAIN-2016-07 | latest | en | 0.945414 |
https://git.ondrovo.com/MightyPork/crsn/commit/d9be3278f5d0c93b59033286ac5e2d30f5dff6bc | 1,716,811,178,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059040.32/warc/CC-MAIN-20240527113621-20240527143621-00890.warc.gz | 233,659,727 | 15,734 | ### docs for float
`@ -169,7 +169,8 @@ This document uses the following notation for arguments:` `- `NUM` - literal values` ` - unsigned `123`` ` - signed `-123`` ` - float `-45.6789`` ` - float `-45.6789`. For now, you must use literals with a period to enter float literals, integers will not be converted to ` ` float when used in floating point instructions!` ` - hex `0xabcd`, `#abcd`` ` - binary `0b0101`` ` - character `'a'`, `'🐁'`. Supports unicode and C-style escapes. Use `\\` for a literal backslash.` `@ -428,6 +429,17 @@ Many instructions have two forms:` `(mod Wr Rd Rd'divider)` `(mod RW Rd'divider)` ``` ``` `; Get abs value` `(abs Wr Rd)` `(abs RW)` ``` ``` `; Get signum` `(sgn Wr Rd)` `(sgn RW)` ``` ``` `; Power - e.g. (pow r0 2 8) is 256` `(pow Wr Rd Rd'pow)` ``` ``` `; Swap the 32-bit halves of a value` `; 0x01234567_89abcdef -> 0x89abcdef_01234567` `(sw32 Wr Rd)` `@ -515,6 +527,119 @@ Many instructions have two forms:` `(del @Rd)` ````` ``` ``` `### Floating Point Arithmetics` ``` ``` `The arithmetics module has support for floating point values. There are some gotchas though:` ``` ``` `- Floating point is simply the binary representation of it in an unsigned integer register.` ` I thought of adding special float registers for this, but then you can't easily pass floats` ` to subroutines, push them on a stack etc. Not worth it.` `- To enter a float literal, always use the notation with a decimal point. It should support minus and scientific notation too.` `- There are special instructions dedicated to working with floats. The regular integer instructions ` ` will happily work with their binary forms, but that's absolutely not what you want.` ``` ``` ````` `(itf r0 1.0) ; NO!!! it is already float, what are you doing` `(ld r0 1.0) ; okay` `(itf r0 1) ; also okay` ``` ``` `(fmul r0 2) ; NO!!!!!!!!!!!!! 2 is not float` `(mul r0 2.0) ; ALSO NO!!!!!!!!!!!!! mul is not a float instruction!` `(fmul r0 2.0) ; good` ````` ``` ``` `You have to be a bit careful, that's all.` ``` ``` `Here's an abridged summary of the floating point submodule:` ``` ``` `(most of these support the shorthand version too - `RW` in place of `Wr Rd`)` ``` ``` ````` `; IntToFloat` `(itf Wr Rd)` `; FloatToInt (round)` `(fti Wr Rd)` `; FloatToInt (ceil)` `(ftic Wr Rd)` `; FloatToInt (floor)` `(ftif Wr Rd)` ``` ``` `; FloatTest - nan->invalid, infinities->overflow, positive, negative, zero` `(ftst Rd)` `; FloatCompare` `(fcmp Rd Rd)` `; FloatRangeTest` `(fcmpr Rd Rd'min Rd'max)` ``` ``` `; FloatRng` `(frng Wr Rd'min Rd'max)` ``` ``` `; --- Basic float arith ---` ``` ``` `; FloatAdd` `(fadd Wr Rd Rd)` `; FloatSub` `(fsub Wr Rd Rd)` `; FloatMul` `(fmul Wr Rd Rd)` `; FloatPow` `(fpow Wr Rd Rd'pow)` `; FloatRoot` `(froot Wr Rd Rd'root)` `; FloatHyp` `(fhyp Wr Rd Rd)` `; FloatDiv` `(fdiv Wr Wr'rem Rd'a Rd'div)` `; FloatMod` `(fmod Wr Rd'a Rd'div)` `; FloatAbs` `(fabs Wr Rd)` `; FloatSgn` `(fsgn Wr Rd)` ``` ``` `; --- Basic trig ---` ``` ``` `; FloatSin` `(fsin Wr Rd)` `; FloatAsin` `(fasin Wr Rd)` `; FloatCos` `(fcos Wr Rd)` `; FloatAcos` `(facos Wr Rd)` `; FloatTan` `(ftan Wr Rd)` `; FloatAtan` `(fatan Wr Rd)` `; FloatAtan2` `(fatan2 Wr Rd'y Rd'x)` `; FloatCot` `(fcot Wr Rd)` `; FloatAcot` `(facot Wr Rd)` ``` ``` `; --- Hyperbolic trig ---` ``` ``` `; FloatHypSin` `(fsinh Wr Rd)` `; FloatHypAsin` `(fasinh Wr Rd)` `; FloatHypCos` `(fcosh Wr Rd)` `; FloatHypAcos` `(facosh Wr Rd)` `; FloatHypTan` `(ftanh Wr Rd)` `; FloatHypAtan` `(fatanh Wr Rd)` `; FloatHypCot` `(fcoth Wr Rd)` `; FloatHypAcot` `(facoth Wr Rd)` ````` ``` ``` `Wow, thats a lot. I didn't test many of these yet. There may be bugs.` ``` ``` `## Buffers Module` ``` ``` `This module defines dynamic size integer buffers.` | 1,188 | 3,719 | {"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} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.787986 |
http://www.caam.rice.edu/~lc55/SFEMaNS/html/doc_SFEMaNS_condlim_penal_in_real_space.html | 1,516,728,674,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00757.warc.gz | 419,958,478 | 4,497 | SFEMaNS version 4.1 (work in progress) Reference documentation for SFEMaNS
The function penal_in_real_space
It is used to define the penalty function $$\chi$$. We remind this function is a scalar function equal to 0 in the solid obstacle and 1 in the fluid region. We refer to the section Extension to non axisymmetric geometry for more information on the formulation of the Navier-Stokes equations in SFEMaNS.
This function defines a scalar function in the physical space for a number of angles, denoted nb_angles, on a number of nodes of the finite element mesh. We denote by mesh the finite element mesh used to approximate the velocity field.
### Inputs and outputs
The inputs of this function are the following:
1. mesh is the mesh where the velocity field is approximated.
2. angles is the list of the angles where the penalty function is computed. These reals numbers are in the interval $$[0,2\pi[$$.
3. nb_angles is the number of angles considered. It is an interger.
4. nb is the label of the first node considered. It is an integer.
5. ne is the label of the last node considered. It is an integer.
6. rr_gauss is a real valued tabular that contains two columns with dimensions (2,ne-nb+1). The tabular rr(1,:) contains the radial cylindrical coordinate of each nodes considered. Respectively, rr(2,:) contains the vertical coordinates of these nodes. We note that we usually use ne=1 and nb=mesh%np.
7. time is the time at which this term is computed. It is a real number.
The output of this function is a real valued tabular vv with two columns of dimension (nb_angles,ne-nb+1).
### Exemple
Here is an exemple where the solid obsctable is the domain $$\Omega_\text{obs} = \{(r,\theta,z)\in [0,1]\times[0,\pi]\times[-0.1,0.1] \}$$.
The corresponding code lines are written as follows.
DO i = 1, nb_angles
DO n = 1, SIZE(rr,2)
r=rr(1,n)
theta=angles(i)
z=rr(2,n)
IF ((r.LE.1).AND.(theta.LE.ACOS(-1.d0)).AND.(ABS(z).LE.0.1d0)) THEN
vv(i,n)=0.d0
ELSE
vv(i,n)=1.d0
END IF
END DO
END DO
RETURN
We note that it is better to use a smooth penalty function and not a discontinuous one as above.
We refer to the sections Examples with manufactured solutions (see test 24) and Examples on physical problems for more examples. | 593 | 2,239 | {"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": 0, "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} | 3.40625 | 3 | CC-MAIN-2018-05 | latest | en | 0.797353 |
http://www.docstoc.com/docs/75098421/hw2a | 1,369,020,106,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698222543/warc/CC-MAIN-20130516095702-00026-ip-10-60-113-184.ec2.internal.warc.gz | 422,946,096 | 13,777 | # hw2a
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``` Business Mathematics I
Homework 1
for
Math 115a, Section:
Instructor:
Date:
by
Team:
We, the undersigned, affirm that each of us participated fully and equally in the
completion of this assignment and that the work contained herein is original.
Furthermore, we acknowledge that sanctions will be imposed jointly if any part of this
work is found to violate the Student Code of Conduct, the Code of Academic Integrity, or
the policies and procedures established for this course.
______________________________ ______________________________
Name (printed) Signature
______________________________ ______________________________
Name (printed) Signature
______________________________ ______________________________
Name (printed) Signature
______________________________ ______________________________
Name (printed) Signature
(The Exercises are available in MBD 1 Proj1.ppt file)
1. Fifty-two percent of all undergraduates at a large university are female, 18% are
majoring in business, and 62% are female or are majoring in business. What percent of
Solution.
2. Fifty-two percent of all undergraduates at a large university are female, 18% are
majoring in business, and 62% are female or are majoring in business. What percent of
all undergraduates at the university are female and are majoring in business?
Solution.
3. Exercise 7
Solution.
4. Consider a randomly selected worker in the United States. The probability that the
worker participates in a company-sponsored retirement plan is 0.56, the probability that
the worker has health insurance is 0.68, and the probability that the worker participates in
a company-sponsored retirement plan and has health insurance is 0.49. What is the
probability that the worker does not have health insurance?
Solution.
5. Consider a randomly selected worker in the United States. The probability that the
worker participates in a company-sponsored retirement plan is 0.56, the probability that
the worker has health insurance is 0.68, and the probability that the worker participates in
a company-sponsored retirement plan and has health insurance is 0.49. What is the
probability that the worker participates in a company-sponsored retirement plan or has
health insurance?
Solution.
6. Consider a randomly selected worker in the United States. The probability that the
worker participates in a company-sponsored retirement plan is 0.56, the probability that
the worker has health insurance is 0.68, and the probability that the worker participates in
a company-sponsored retirement plan and has health insurance is 0.49. What is the
probability that the worker does not participate in a company-sponsored retirement plan
or does not have health insurance?
Solution.
7. Exercise 13
Solution.
8. (Exercise 14 in Basic Probability) You are planning a big sidewalk sale for the coming
Saturday. Records indicate that the probability of rain is 0.3, the probability of high
winds is 0.4, and the probability of rain or high winds is 0.5. What is the probability that
it rains but you do not have high winds?
9. A fair coin is tossed four times and the outcome of each toss is recorded. We can take
all possible combinations of H and T as our sample space, S. Since the coin is fair, we
will assume that each of the 16 outcomes is equally likely, and therefore, that they all
have the same probability. Find the probability of each of the following events:
E: The first toss is tails. ( HHHH ) ( HHHT ) ( HHTH ) ( HHTT )
F: All four tosses are tails. ( HTHH ) ( HTHT ) ( HTTH ) ( HTTT )
G: At least three tosses are tails.
S
(THHH ) (THHT ) (THTH ) (THTT )
(TTHH )
(TTHT ) (TTTH ) (TTTT )
Solution.
10. (Exercise 2 in Basic Probability) (i) Use repeated computation in the file Bank
Customers.xls to generate a set of twenty fractions for periods of one business month of
21 days. (ii) What is the largest fraction that you found? (iii) What is the smallest
fraction that you found?
Solution.
```
Related docs
Other docs by qingyunliuliu
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CALM_SLI | 1,025 | 4,597 | {"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} | 4.59375 | 5 | CC-MAIN-2013-20 | latest | en | 0.920243 |
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Helper III
## Calculate the month-over-month of Course Completion
Hello,
I'm looking to calculate Month over Month course completion rates. For example, in January we had 67% complete on course ABC. In February we had a 80% completion rate. Either a bar or trend line.
Table name is 'UoM-TrainMe-Report'
Columns are [Enrolment Date] and [Course Name] and [NewCourseEnrolment]. The date format fro CSV file is DAY/MONTH/YEAR using numbers which PowerBi converts to Wednesday, 12 June 2017 format.
I also have three measures which are Total Enrol (which is a count of course enrolments), Total Completed (a count of course enrolments with a status of Completed) and Percent Completed (which uses the Total Enrol and Total Completed measures to calculate the percent complete)
I hope all this info helps with a solution. Thanks guys!!! Youve been a huge help!
RayinOz
1 ACCEPTED SOLUTION
Super User
So here are the steps:
- add continuous calendar table in your data model
- set relationship between your data table and continuous calendar table
- calculate cummulative measure
and you are good to go...
As a best practice, it is always nice to have continuous table for time intelligence analysis.
Thanks,
Parv
Subscribe to the @PowerBIHowTo YT channel for an upcoming video on List and Record functions in Power Query!!
Learn Power BI and Fabric - subscribe to our YT channel - Click here: @PowerBIHowTo
If my solution proved useful, I'd be delighted to receive Kudos. When you put effort into asking a question, it's equally thoughtful to acknowledge and give Kudos to the individual who helped you solve the problem. It's a small gesture that shows appreciation and encouragement! ❤
Did I answer your question? Mark my post as a solution. Proud to be a Super User! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.
8 REPLIES 8
Helper III
Here is an update...
I was able to show Month by Month... but it isn't cumulative... each month that is shown only shows completions FOR THAT MONTH. But what i need is cumulative.
Thoughts?
Super User
Few important items:
1. Do you have continuous calendar table in your dataset?
2. You have to calculate cummulative total measure to achieve this?
There are lot of articles on cummulative/running total, here is quick reference for you:
http://www.daxpatterns.com/cumulative-total/
Subscribe to the @PowerBIHowTo YT channel for an upcoming video on List and Record functions in Power Query!!
Learn Power BI and Fabric - subscribe to our YT channel - Click here: @PowerBIHowTo
If my solution proved useful, I'd be delighted to receive Kudos. When you put effort into asking a question, it's equally thoughtful to acknowledge and give Kudos to the individual who helped you solve the problem. It's a small gesture that shows appreciation and encouragement! ❤
Did I answer your question? Mark my post as a solution. Proud to be a Super User! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.
Helper III
ahhhh.. ok. No, I don't have a continuous calendar table setup in my dataset. I only have a date column in my primary table for Enrolment date.
i'll take a look through the link you provided.
Super User
So here are the steps:
- add continuous calendar table in your data model
- set relationship between your data table and continuous calendar table
- calculate cummulative measure
and you are good to go...
As a best practice, it is always nice to have continuous table for time intelligence analysis.
Thanks,
Parv
Subscribe to the @PowerBIHowTo YT channel for an upcoming video on List and Record functions in Power Query!!
Learn Power BI and Fabric - subscribe to our YT channel - Click here: @PowerBIHowTo
If my solution proved useful, I'd be delighted to receive Kudos. When you put effort into asking a question, it's equally thoughtful to acknowledge and give Kudos to the individual who helped you solve the problem. It's a small gesture that shows appreciation and encouragement! ❤
Did I answer your question? Mark my post as a solution. Proud to be a Super User! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.
Helper III
OK... Found a resource to walk me through creating a calendar table in powerbi....
https://kohera.be/blog/business-intelligence/how-to-create-a-date-table-in-power-bi-in-2-simple-step...
now.. to set a relationship between the calendar table and my data table... i think the [enrolment date] is the date column i want to use from my data table.
Super User
Awesome, that is perfect. You see "DateAsInteger" in calendar table, you have to create similar column on your data table, let's "EnrolmentDateAsInteger", to do so, follow these steps,
- Go to Query Editor
- Select your data table
- Click Add Column
- It will open up a window to add formulat for custom field, give it a name "EnrolmentDateAsInteger" and then add this formula =Date.ToText([Your Enrolment Field, "yyyyMMdd")
If you Enrolment Field is date/time and then add this formula
=DateTime.ToText([Your Enrolment Field, "yyyyMMdd")
Once you have this field, it will show in your query and now we can use this field to set relation with calendar table. Give it a try and let me know if there is an issue.
Cheers,
Subscribe to the @PowerBIHowTo YT channel for an upcoming video on List and Record functions in Power Query!!
Learn Power BI and Fabric - subscribe to our YT channel - Click here: @PowerBIHowTo
If my solution proved useful, I'd be delighted to receive Kudos. When you put effort into asking a question, it's equally thoughtful to acknowledge and give Kudos to the individual who helped you solve the problem. It's a small gesture that shows appreciation and encouragement! ❤
Did I answer your question? Mark my post as a solution. Proud to be a Super User! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.
Helper III
Fantastic! I've setup the relationship between the tables... next step is to setup the calculation for the cumulative measure....
you've been a tremendous help... i'm almost there!
Super User
If need further help with calendar table, let me know or send me your sample pbix file and I can get back to you with solution.
Cheers
Subscribe to the @PowerBIHowTo YT channel for an upcoming video on List and Record functions in Power Query!!
Learn Power BI and Fabric - subscribe to our YT channel - Click here: @PowerBIHowTo
If my solution proved useful, I'd be delighted to receive Kudos. When you put effort into asking a question, it's equally thoughtful to acknowledge and give Kudos to the individual who helped you solve the problem. It's a small gesture that shows appreciation and encouragement! ❤
Did I answer your question? Mark my post as a solution. Proud to be a Super User! Appreciate your Kudos 🙂
Feel free to email me with any of your BI needs.
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Author Gelfand, I. M. author Trigonometry [electronic resource] / by I. M. Gelfand, Mark Saul Boston, MA : Birkhรคuser Boston : Imprint: Birkhรคuser, 2001 http://dx.doi.org/10.1007/978-1-4612-0149-6 X, 229 p. online resource
SUMMARY
In a sense, trigonometry sits at the center of high school mathematics. It originates in the study of geometry when we investigate the ratios of sides in similar right triangles, or when we look at the relationship between a chord of a circle and its arc. It leads to a much deeper study of periodic functions, and of the so-called transcendental functions, which cannot be described using finite algebraic processes. It also has many applications to physics, astronomy, and other branches of science. It is a very old subject. Many of the geometric results that we now state in trigonometric terms were given a purely geometric exposition by Euclid. Ptolemy, an early astronomer, began to go beyond Euclid, using the geometry of the time to construct what we now call tables of values of trigonometric functions. Trigonometry is an important introduction to calculus, where one studยญ ies what mathematicians call analytic properties of functions. One of the goals of this book is to prepare you for a course in calculus by directing your attention away from particular values of a function to a study of the function as an object in itself. This way of thinking is useful not just in calculus, but in many mathematical situations. So trigonometry is a part of pre-calculus, and is related to other pre-calculus topics, such as exponential and logarithmic functions, and complex numbers
CONTENT
0 Trigonometry -- 1 Trigonometric Ratios in a Triangle -- 2 Relations among Trigonometric Ratios -- 3 Relationships in a Triangle -- 4 Angles and Rotations -- 5 Radian Measure -- 6 The Addition Formulas -- 7 Trigonometric Identities -- 8 Graphs of Trigonometric Functions -- 9 Inverse Functions and Trigonometric Equations
Mathematics Algebra Geometry Mathematics -- Study and teaching Mathematics Geometry Mathematics Education Algebra
Location
Office of Academic Resources, Chulalongkorn University, Phayathai Rd. Pathumwan Bangkok 10330 Thailand | 505 | 2,197 | {"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-2021-17 | latest | en | 0.87558 |
https://math.stackexchange.com/questions/3506424/bijection-between-homotopy-classes | 1,716,669,787,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00437.warc.gz | 326,239,042 | 34,994 | # Bijection between homotopy classes
Let $$X$$ be a CW-complex and $$w: Z \to Y$$ a weak homotopy equivalence. Show that $$w_*: [X, Z] \to [X, Y]: [f] \mapsto [w \circ f]$$ is a bijection. Hint: use mapping cylinders.
I am having trouble with this question for both the injectiveness and surjectiveness.
Injective
Suppose $$[w \circ f] = [w \circ g]$$. We can factor $$f$$ as $$X \xrightarrow{i_X} M(f) \xrightarrow{p} Y$$ where $$i_X$$ is a closed inclusion and $$p$$ is a homotopy equivalence (and $$M(f)$$ is the mapping cylinder of $$f$$) and similarly factor $$g$$ as $$X \xrightarrow{j_X} M(g) \xrightarrow{q} Y$$.
I want to use Whitehead on $$w$$. The space $$Z$$ is not a CW-complex, but it is homotopy equivalent to $$M(f)$$ and $$M(g)$$, which are. Similarly $$Y$$ is homotopy equivalent to $$M(w \circ f)$$ and $$M(w \circ g)$$. Composing $$w$$ with the homotopy equivalences gives a map $$M(f) \to M(w \circ g)$$ (we can swap $$f$$ and $$g$$ around), and since homotopy equivalences do not change the $$\pi_n$$, this is a weak homotopy equivalence, hence by Whitehead a homotopy equivalence. I don't know what to do with this.
Surjective
Given a map $$f: X \to Y$$, I can't figure out a way to factor it via $$z$$ as $$w$$ has no (partial) inverse.
Use the factorisation $$Z \overset{j_Z}{\to} M(w) \overset{p}{\to} Y$$ of $$w$$ to show that $$j_Z$$ is a weak homotopy equivalence. This implies that $$(M(w), j_Z(Z))$$ is $$n$$-connected for every $$n$$.
• For surjectivity, any $$f : X \to Y$$ gives $$j_Y \circ f : (X, \varnothing) \to (M(w), j_Z(Z))$$. By the above, and that $$X$$ is a CW-complex, this composition will be homotopic to $$j_Z \circ \tilde{f}$$ for some $$\tilde{f} : X \to Z$$. It is now easy to see that $$[f] = [w \circ \tilde{f}] = w_* [\tilde{f}]$$.
• For injectivity, if $$f, g : X \to Z$$ are such that $$[w \circ f] = [w \circ g]$$, let $$H : X \times [0, 1] \to Y$$ be a homotopy between them. We can slightly alter this to obtain $$H': X \times [0, 1] \to M(w) : (x, t) \mapsto \left\{ \begin{array}{cl} (f(x), 3t) & \text{if } 0 \leq t \leq 1/3 \\ H(x, 3t - 1) & \text{if } 1/3 \leq t \leq 1/3 \\ (g(x), 2 - 3t) & \text{if } 2/3 \leq t \leq 1/3 \\ \end{array} \right.$$ and view $$H'$$ as a map of pairs $$H' : (X \times [0, 1], X \times \{ 0, 1 \}) \to (M(w), j_Z(Z))$$. Similar as to before, $$H'$$ will be homotopic (relative to $$X \times \{ 0, 1 \}$$) to some $$\tilde{H} : X \times [0, 1] \to M(w)$$ with image in $$j_Z(Z)$$. Now it is easy to see this gives a homotopy between $$f$$ and $$g$$. | 942 | 2,550 | {"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": 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": 51, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-22 | latest | en | 0.870823 |
https://www.maths-formula.com/2022/03/mcq-questions-for-class-11-maths-chapter-5.html | 1,709,078,074,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00654.warc.gz | 869,578,482 | 40,346 | MCQs Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
# MCQs Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
## MCQs Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
In this 21st century, Multiple Choice Questions (MCQs) play a vital role to prepare for a competitive examination. CBSE board has also brought a major change in its board exam patterns. Now-a-days, a total of 10 MCQs questions are asked in the class 10 board examination.
In most of the competitive examinations, only MCQ questions are asked. So, for getting ready for the competitive examinations, we have to practice for MCQs questions to solve. It strengthens the critical thinking and problem solving skills.
In future, if you want to prepare for competitive examination and to crack it, then you should more focus on the MCQ questions. Thus, from the board examinations point of view and competitive examinations point of view, you should practice more on multiple choice questions.
Thus, let’s solve these MCQs Questions to make our foundation very strong.
## MCQs Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
1. The value of √(-81) is
(a) -9i
(b) 9i
(c) -3i
(d) 3i
2. In a complex number, z = 4 + i, what is the real part?
(a) 4
(b) i
(c) 1
(d) 4 + i
3. The value of √(-144) is
(a) 12i
(b) -12i
(c) ±12i
(d) None of these
4. If (x + 3) + i(y – 2) = 5 + 2i, then find the values of x and y.
(a) x = 8 and y = 4
(b) x = 2 and y = 4
(c) x = 2 and y = 0
(d) x = 8 and y = 0
5. If z lies on |z| = 1, then 2/z lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola
6. If z1 = 2 + 3i and z2 = 5 + 2i, then find the sum of two complex numbers.
(a) 4 + 8i
(b) 3 – I
(c) 7 + 5i
(d) 7 – 5i
7. If ω is an imaginary cube root of unity, then (1 + ω – ω²)7 equals
(a) 128 ω
(b) -128 ω
(c) 128 ω²
(d) -128 ω²
8. -z is _________________ for the complex number z.
(c) multiplicative identity element
(d) multiplicative inverse
9. The least value of n for which {(1 + i)/(1 – i)}n is real, is
(a) 1
(b) 2
(c) 3
(d) 4
10. 1/z is _________________ for complex number z.
(c) multiplicative identity element
(d) multiplicative inverse
11. The value of i-999 is
(a) 1
(b) -1
(c) i
(d) -i
12. If z1 = 8 + 5i and z2 = 3 + 6i, then find the value of z1 – z2.
(a) -5 + i
(b) 5 – i
(c) 5 + i
(d) 5 – 5i
13. The complex numbers sin x + i cos 2x and sin x i cos 2x are conjugate to each other for
(a) x = nπ
(b) x = 0
(c) x = (n + 1/2) π
(d) no value of x
14. In the complex number, z = 5 + 2i, what is the imaginary part?
(a) 5
(b) 2i
(c) 5 + 2
(d) 5 + 2i | 951 | 2,671 | {"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} | 4.6875 | 5 | CC-MAIN-2024-10 | longest | en | 0.812443 |
http://sepwww.stanford.edu/data/media/public/docs/sep75/martin2/paper_html/node6.html | 1,508,330,704,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00375.warc.gz | 325,110,097 | 3,383 | Next: Approximating TI dispersion relations Up: Dellinger, Muir, & Karrenbach: Previous: THE SECOND ANELLIPTIC APPROXIMATION
# ANELLIPTIC PARAMETERS FOR TI MEDIA
The anelliptic approximations are useful in their own right for fitting measured phase- and group-velocity surfaces, but it is also useful to know how to relate these approximations to a real anisotropic symmetry system. Because the approximations are two-dimensional, the appropriate symmetry system is transverse isotropy (TI). The equations for TI media are most easily written in terms of phase-velocity squared, suggesting we follow the notation of equation ().
Let be the phase-velocity squared measured at phase angle from the vertical, and let and .Express the elastic constants in units of phase-velocity squared as well, so .Then we have for the TI SH wavetype
(17)
Equation () is linear because SH waves in TI media are exactly elliptically anisotropic; by matching coefficients in equations () and () we find the required anelliptic parameters:
(18)
The TI qP-qSV wavetype is rather more complicated:
(19)
where the + sign is for qP, - for qSV. We can find Wz or Wx by substituting S=0 or S=1, respectively, into equation (). To find or , however, we first have to fit a paraxial ellipse about the z or x axes, respectively. We have already seen that elliptical anisotropy is linear in these coordinates. To find the equation for the paraxial ellipse about the z axis we therefore linearize about S=0, obtaining
(20)
Next, remembering that is the horizontal velocity for the paraxial approximation about the z axis, we set S=1 (horizontal propagation) in equation (), obtaining
(21)
Similarly,
(22)
For the TI qP wavetype we obtain:
(23)
For the TI qSV wavetype we obtain:
(24)
Next: Approximating TI dispersion relations Up: Dellinger, Muir, & Karrenbach: Previous: THE SECOND ANELLIPTIC APPROXIMATION
Stanford Exploration Project
11/17/1997 | 465 | 1,933 | {"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} | 2.890625 | 3 | CC-MAIN-2017-43 | latest | en | 0.849959 |
http://www.literacyandnumeracyforadults.com/resources/355993 | 1,498,640,354,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323588.51/warc/CC-MAIN-20170628083538-20170628103538-00548.warc.gz | 575,582,303 | 22,412 | ahh
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"Towards better teaching & learning"
# Theoretical probability
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Last updated 26 October 2012 15:30 by NZTecAdmin
Theoretical probability (PDF, 53 KB)
Probability, 4th-5th steps
## The purpose of the activity
In this activity, the learners develop their understanding of the difference between theoretical and experimental probabilities.
## The teaching points
• The probability of an event is a measure of the chance of an event occurring. The probability of an event is a number between 0 and 1 and can be expressed as a fraction, a percentage (0–100%), or as odds.
• A sample space is the set of all possible outcomes for an event. For example, there are 6 possible outcomes for rolling a single standard six-sided die and 36 (6 x 6) possible outcomes for rolling two standard six-sided dice.
• There are two ways to measure chance. One way is to analyse the situation logically (theoretical probability), and the other way is to generate data to analyse the situation (experimental probability). Examples of situations that can be analysed theoretically include rolling dice, throwing coins and Lotto. Examples of situations that need to be analysed experimentally include the likelihood of there being an earthquake or a car accident.
## Resources
• A series of three different-coloured counters.
• A non-transparent bag.
• Coins.
## The guided teaching and learning sequence
1. With the learners watching, put 10 coloured counters in a bag: 5 red, 3 blue and 2 yellow. Record the colours and numbers of counters placed in the bag on the board for the learners to refer to.
2. Ask the learners to state what will happen if a counter is selected at random from the bag:
“What colour counter will be selected?”
“What colours of counter could be selected?”
“What colour is most likely?” “Why?”
Some learners are likely to try to tell you that a red counter will be chosen. It is important for them to realise that they cannot be certain what colour will be selected. If one is chosen at random, it could be any of the three colours. Red is more likely than blue or yellow because there are more red counters than blue or yellow counters in the bag.
3. Now ask the learners to describe the chance of selecting a counter of each colour.
“What is the chance of selecting a red, blue or yellow counter?” “How did you work that out?”
The learners should be able to state:
• There is a 50% or 1/2 chance of selecting a red counter because 5 of the 10 counters are red.
• There is a 30% or 3/10 chance of selecting a blue counter because 3 of the 10 counters are blue.
• There is a 20% or 1/5 chance of selecting a yellow counter because 2 of the 10 counters are yellow.
4. Explain to the learners that these are the theoretical probabilities of the event.
5. Discuss what these theoretical probabilities mean in terms of selecting counters. For example, you are twice as likely to select a red counter as any other colour.
6. Ask the learners what they think will happen if you select 10 counters, one at a time from the bag, returning the counter to the bag between selections. The learners are likely to suggest that there will be 5 red counters, 3 blue counters and 2 yellow counters selected. While this is possible, it is not the only possible outcome, in fact the probability of selecting exactly 5 red, 3 blue and 2 yellow counters is quite low.
7. Ask 10 learners to each come up, select a counter, record its colour and replace it in the bag. Graph the results on the board and discuss the results.
8. Repeat several times.
9. Ask the learners to discuss in small groups why the results do not match the theoretical probabilities.
10. Share the ideas as a class. Hopefully some learners will recognise that theoretical probability is only a way of predicting what may occur, it does not tell you what will occur. Ensure that all the learners understand this.
## Follow-up activity
Ask the learners to work in pairs to calculate the theoretical probabilities for tossing two coins. Then ask them to toss two coins 100 times and compare the results of the experiment with their theoretical probabilities.
Theoretical probabilities:
• 2 Heads: 1/4 or 25%.
• 2 Tails: 1/4 or 25%.
• 1 Head, 1 Tail: 1/2 , 50%.
26 April 2016 22:15
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https://aimsciences.org/article/doi/10.3934/dcdsb.2009.11.1 | 1,568,805,330,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573284.48/warc/CC-MAIN-20190918110932-20190918132932-00536.warc.gz | 371,867,483 | 10,723 | # American Institute of Mathematical Sciences
January 2009, 11(1): 1-10. doi: 10.3934/dcdsb.2009.11.1
## Variational models for incompressible Euler equations
1 Scuola Normale Superiore, Piazza Cavalieri 7, 56123 Pisa, Italy
Received November 2007 Revised March 2008 Published November 2008
In this paper we illustrate some recent work [1], [2] on Brenier's variational models for incompressible Euler equations. These models give rise to a relaxation of the Arnold distance in the space of measure-preserving maps and, more generally, measure-preserving plans. We analyze the properties of the relaxed distance, we show a close link between the Lagrangian and the Eulerian model, and we derive necessary and sufficient optimality conditions for minimizers. These conditions take into account a modified Lagrangian induced by the pressure field.
Citation: Luigi Ambrosio. Variational models for incompressible Euler equations. Discrete & Continuous Dynamical Systems - B, 2009, 11 (1) : 1-10. doi: 10.3934/dcdsb.2009.11.1
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2018 Impact Factor: 1.008 | 1,624 | 5,084 | {"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} | 2.515625 | 3 | CC-MAIN-2019-39 | latest | en | 0.699818 |
https://la.mathworks.com/matlabcentral/cody/problems/78-implement-a-rot13-cipher/solutions/2635186 | 1,606,950,949,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141716970.77/warc/CC-MAIN-20201202205758-20201202235758-00423.warc.gz | 371,461,811 | 17,170 | Cody
# Problem 78. Implement a ROT13 cipher
Solution 2635186
Submitted on 30 Jun 2020 by Santhu Ruthvik Bandaru
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
s1 = 'I love MATLAB'; s2_correct = 'V ybir ZNGYNO'; assert(isequal(rot13(s1),s2_correct))
2 Pass
s1 = 'I <3 MATLAB!!'; s2_correct = 'V <3 ZNGYNO!!'; assert(isequal(rot13(s1),s2_correct))
3 Pass
s1 = 'The quick fox stumbled over the confusing instructions.'; s2_correct = 'Gur dhvpx sbk fghzoyrq bire gur pbashfvat vafgehpgvbaf.'; assert(isequal(rot13(s1),s2_correct))
4 Pass
s1 = 'Snape kills Dumbledore on page 606. Also, there is no Santa.'; s2_correct = 'Fancr xvyyf Qhzoyrqber ba cntr 606. Nyfb, gurer vf ab Fnagn.'; assert(isequal(rot13(s1),s2_correct))
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 309 | 978 | {"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-2020-50 | latest | en | 0.54371 |
https://electronics.stackexchange.com/questions/398838/how-does-current-flow-in-the-short-circuit-arent-the-two-ends-at-the-same-pote | 1,571,025,451,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649035.4/warc/CC-MAIN-20191014025508-20191014052508-00379.warc.gz | 558,616,535 | 32,691 | # How does current flow in the short circuit? Aren't the two ends at the same potential?
How does the current flow in the short circuit wire ? Isn't potential difference necessary for existence of current?
• No such thing as a short circuit wire. – Andy aka Oct 1 '18 at 16:19
• You got what I wanted to say. – user17616 Oct 1 '18 at 17:20
• @MatsK Students are taught to analyze circuits composed of ideal elements. An ideal wire has no resistance and there is no potential difference between any two points. If you want to model the resistance of a real (non-ideal) wire then you must add a resistor to the circuit. I think we could all be more helpful if we remember what it's like to be a new student. – Elliot Alderson Oct 1 '18 at 18:15
• Likely duplicate: electronics.stackexchange.com/questions/91729/… – Dmitry Grigoryev Oct 3 '18 at 1:56
Try looking at it the other way.
simulate this circuit – Schematic created using CircuitLab
Ask yourself what would happen if current didn't flow in the short-circuit.
The answer is that the potential difference across it would rise. If the PD rises then current will flow.
The circuit settles down with current flowing. All the rest of the wiring in the current loop is a short-circuit too.
• +1 A good answer that meets the OP at their level of understanding. – Elliot Alderson Oct 1 '18 at 18:16
I don't know enough about cooper-pairs (which form up two electron fermions into a boson particle) to answer your question with respect to super-conduction. You may have to ask a specialist physicist that question, if you are interested in it. (I did find this discussion, though: How does current flow in superconductors.)
However, electrons as fermions are accelerated by a very small charge gradient that appears only at the cylindrical surface of the (copper, say) wire. This gradient sets itself up quite quickly and forms the accelerating forces required to move charges. If you place a bend in a wire, there will be yet another slight accumulation of charges at the corner of the bend at just the right level required to divert the electrons around the bend.
(Note that a copper wire does possess a small resistivity so there will be a slight potential difference, one end to the other.)
There are some good physics books that discuss exactly how this sets up and happens. I'd recommend the most recent edition of the one called "Matter & Interactions." (I have the 3rd edition here.) See this Matter & Interactions video on Surface Charge with respect to what I'm discussing here.
Isn't potential difference necessary for existence of current?
No. A potential difference is required for a current to flow through a resistor. An ideal conductor can sustain current without any voltage applied to it. | 620 | 2,767 | {"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} | 3.40625 | 3 | CC-MAIN-2019-43 | latest | en | 0.945069 |
https://rdrr.io/cran/MatrixExtra/src/tests/testthat/test-matmul.R | 1,642,520,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00668.warc.gz | 560,157,616 | 10,287 | # tests/testthat/test-matmul.R In MatrixExtra: Extra Methods for Sparse Matrices
```library("testthat")
library("Matrix")
library("MatrixExtra")
library("float")
context("Matrix multiplications")
set_new_matrix_behavior()
test_that("matmult dense-CSC", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
B <- rsparsematrix(50, 20, .4)
A1 <- rsparsematrix(1, 50, .4)
B1 <- rsparsematrix(50, 1, .4)
expect_equal(as.matrix(A) %*% as.csc.matrix(B), as.matrix(A) %*% as.matrix(B))
expect_equal(float::dbl(float::fl(as.matrix(A)) %*% as.csc.matrix(B)),
as.matrix(A) %*% as.matrix(B), tolerance=1e-5)
expect_s4_class(float::fl(as.matrix(A)) %*% as.csc.matrix(B), "float32")
expect_equal(as.matrix(A1) %*% as.csc.matrix(B), as.matrix(A1) %*% as.matrix(B))
expect_equal(as.matrix(A) %*% as.csc.matrix(B1), as.matrix(A) %*% as.matrix(B1))
expect_equal(as.matrix(A1) %*% as.csc.matrix(B1), as.matrix(A1) %*% as.matrix(B1))
expect_equal(as.matrix(A) %*% as.csc.matrix(B, binary=TRUE),
as.matrix(A) %*% as.matrix(as.csc.matrix(B, binary=TRUE)))
expect_equal(as.matrix(A) %*% as.csc.matrix(B, logical=TRUE),
as.matrix(A) %*% as.matrix(as.csc.matrix(B, logical=TRUE)))
sy <- sparseMatrix(i= c(2,4,3:5), j= c(4,7:5,5), x = 1:5, dims = c(7,7),
symmetric=TRUE, dimnames = list(NULL, letters[1:7]))
expect_s4_class(sy, "dsCMatrix")
set.seed(1)
sy_counterpart <- rsparsematrix(20, nrow(sy), .4)
sy_counterpart <- as.matrix(sy_counterpart)
expect_equal(sy_counterpart %*% sy, sy_counterpart %*% as.matrix(sy))
tri <- matrix(c(1,2,0,4, 0,0,6,7, 0,0,8,9, 0,0,0,0), byrow=TRUE, nrow=4)
tri <- as(tri, "triangularMatrix")
expect_s4_class(tri, "dtCMatrix")
set.seed(1)
tri_counterpart <- rsparsematrix(20, nrow(tri), .4)
tri_counterpart <- as.matrix(tri_counterpart)
expect_equal(tri_counterpart %*% tri, tri_counterpart %*% as.matrix(tri))
})
test_that("tcrossprod dense-CSR", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
B <- rsparsematrix(20, 50, .4)
A1 <- rsparsematrix(1, 50, .4)
B1 <- rsparsematrix(1, 50, .4)
expect_equal(tcrossprod(as.matrix(A), as.csr.matrix(B)),
tcrossprod(as.matrix(A), as.matrix(as.csr.matrix(B))))
expect_equal(float::dbl(tcrossprod(float::fl(as.matrix(A)), as.csr.matrix(B))),
tcrossprod(as.matrix(A), as.matrix(as.csr.matrix(B))),
tolerance=1e-5)
expect_s4_class(tcrossprod(float::fl(as.matrix(A)), as.csr.matrix(B)), "float32")
expect_equal(tcrossprod(as.matrix(A1), as.csr.matrix(B)),
tcrossprod(as.matrix(A1), as.matrix(as.csr.matrix(B))))
expect_equal(tcrossprod(as.matrix(A), as.csr.matrix(B1)),
tcrossprod(as.matrix(A), as.matrix(as.csr.matrix(B1))))
expect_equal(tcrossprod(as.matrix(A1), as.csr.matrix(B1)),
tcrossprod(as.matrix(A1), as.matrix(as.csr.matrix(B1))))
sy <- sparseMatrix(i= c(2,4,3:5), j= c(4,7:5,5), x = 1:5, dims = c(7,7),
symmetric=TRUE, dimnames = list(NULL, letters[1:7]))
sy <- as(sy, "RsparseMatrix")
expect_s4_class(sy, "dsRMatrix")
set.seed(1)
sy_counterpart <- rsparsematrix(20, nrow(sy), .4)
sy_counterpart <- as.matrix(sy_counterpart)
expect_equal(tcrossprod(sy_counterpart, sy),
tcrossprod(sy_counterpart, as.matrix(sy)))
tri <- matrix(c(1,2,0,4, 0,0,6,7, 0,0,8,9, 0,0,0,0), byrow=TRUE, nrow=4)
tri <- as(tri, "triangularMatrix")
tri <- as(tri, "RsparseMatrix")
expect_s4_class(tri, "dtRMatrix")
set.seed(1)
tri_counterpart <- rsparsematrix(20, nrow(tri), .4)
tri_counterpart <- as.matrix(tri_counterpart)
expect_equal(tcrossprod(tri_counterpart, tri),
tcrossprod(tri_counterpart, as.matrix(tri)))
})
test_that("crossprod dense-CSC", {
set.seed(1)
A <- rsparsematrix(50, 100, .4)
B <- rsparsematrix(50, 20, .4)
expect_equal(crossprod(as.matrix(A), as.csc.matrix(B)),
crossprod(as.matrix(A), as.matrix(B)),
tolerance=1e-3)
})
test_that("matmult CSR-dense", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
B <- rsparsematrix(50, 20, .4)
expect_equal(as.csr.matrix(A) %*% as.matrix(B), as.matrix(A) %*% as.matrix(B))
})
test_that("tcrossprod CSR-dense", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
B <- rsparsematrix(20, 50, .4)
expect_equal(tcrossprod(as.csr.matrix(A), as.matrix(B)),
tcrossprod(as.matrix(A), as.matrix(B)))
})
test_that("tcrossprod dense-CSR", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
B <- rsparsematrix(20, 50, .4)
expect_equal(tcrossprod(as.matrix(A), as.csr.matrix(B)),
tcrossprod(as.matrix(A), as.matrix(B)))
})
test_that("matmult CSR-vector", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
v <- rnorm(50)
dvec <- as(v, "dsparseVector")
ivec <- as(dvec, "isparseVector")
lvec <- as(dvec, "lsparseVector")
nvec <- as(dvec, "nsparseVector")
num <- as.numeric(v)
int <- as.integer(num)
bool <- as.logical(num)
lst_inputs <- list(
dvec, ivec, lvec, nvec,
num, int, bool
)
for (inp in lst_inputs) {
expect_equal(as.csr.matrix(A) %*% inp,
as.matrix(A) %*% as.numeric(inp))
expect_equal(unname(as.matrix(as.csr.matrix(A[,1,drop=FALSE]) %*% inp)),
unname(as.matrix(A[,1,drop=FALSE]) %*% as.numeric(inp)))
expect_equal(unname(as.matrix(as.csr.matrix(A[1:10,1,drop=FALSE]) %*% inp)),
unname(as.matrix(A[1:10,1,drop=FALSE]) %*% as.numeric(inp)))
}
# v[4] <- NA_real_
# A <- as.csr.matrix(A)
# A@x[10] <- NA_real_
# expect_equal(A %*% v, unname(as.matrix(as.matrix(A) %*% v)))
})
test_that("float32 vectors", {
set.seed(1)
A <- rsparsematrix(100, 50, .4)
v <- float::fl(rnorm(100))
v2 <- float::fl(rnorm(50))
expect_equal(float::dbl(v %*% A),
float::dbl(v) %*% as.matrix(A),
tolerance=1e-5)
### sparse CSC
expect_equal(unname(as.matrix(v2 %*% as.csc.matrix(A[1,,drop=FALSE]))),
float::dbl(v2) %*% as.matrix(A)[1,,drop=FALSE],
tolerance=1e-5)
### sparse CSC
expect_equal(unname(as.matrix(v2 %*% as.csc.matrix(A[1,1:10,drop=FALSE]))),
float::dbl(v2) %*% as.matrix(A)[1,1:10,drop=FALSE],
tolerance=1e-5)
expect_equal(float::dbl(tcrossprod(v2, as.csr.matrix(A))),
tcrossprod(float::dbl(v2), as.matrix(A)),
tolerance=1e-5)
expect_equal(float::dbl(tcrossprod(v2, as.csr.matrix(A[1:10,]))),
tcrossprod(float::dbl(v2), as.matrix(A)[1:10,]),
tolerance=1e-5)
### sparse CSC
expect_equal(unname(as.matrix(tcrossprod(v, as.csr.matrix(A[,1,drop=FALSE])))),
tcrossprod(float::dbl(v), as.matrix(A[,1,drop=FALSE])),
tolerance=1e-5)
### sparse CSC
expect_equal(unname(as.matrix(tcrossprod(v, as.csr.matrix(A[1:10,1,drop=FALSE])))),
tcrossprod(float::dbl(v), as.matrix(A[1:10,1,drop=FALSE])),
tolerance=1e-5)
expect_equal(float::dbl(crossprod(v, as.csc.matrix(A))),
crossprod(float::dbl(v), as.matrix(A)),
tolerance=1e-5)
expect_equal(float::dbl(crossprod(v, as.csc.matrix(A[,1:10,drop=FALSE]))),
crossprod(float::dbl(v), as.matrix(A)[,1:10,drop=FALSE]),
tolerance=1e-5)
expect_equal(float::dbl(as.csr.matrix(A) %*% v2),
as.matrix(A) %*% float::dbl(v2),
tolerance=1e-5)
# sparse CSR
expect_equal(unname(as.matrix(as.csr.matrix(A[,1,drop=FALSE]) %*% v2)),
as.matrix(A[,1,drop=FALSE]) %*% float::dbl(v2),
tolerance=1e-5)
# sparse CSR
expect_equal(unname(as.matrix(as.csr.matrix(A[1:10,1,drop=FALSE]) %*% v2)),
as.matrix(A[1:10,1,drop=FALSE]) %*% float::dbl(v2),
tolerance=1e-5)
})
set_new_matrix_behavior()
```
## Try the MatrixExtra package in your browser
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MatrixExtra documentation built on Dec. 19, 2021, 9:07 a.m. | 2,492 | 7,155 | {"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} | 2.65625 | 3 | CC-MAIN-2022-05 | latest | en | 0.376347 |
https://phys.libretexts.org/TextBooks_and_TextMaps/College_Physics/Book%3A_College_Physics_(OpenStax)/27%3A_Wave_Optics/27.6%3A_Limits_of_Resolution%3A_The_Rayleigh_Criterion | 1,545,148,779,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829429.94/warc/CC-MAIN-20181218143757-20181218165757-00006.warc.gz | 698,877,283 | 21,367 | $$\require{cancel}$$
# 27.6: Limits of Resolution: The Rayleigh Criterion
Light diffracts as it moves through space, bending around obstacles, interfering constructively and destructively. While this can be used as a spectroscopic tool -- a diffraction grating disperses light according to wavelength, for example, and is used to produce spectra -- diffraction also limits the detail we can obtain in images. Figure 1a shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. This pattern is caused by diffraction similar to that produced by a single slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large apertures, too.
Figure $$\PageIndex{1}$$: a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point light sources that are close to one another produce overlapping images because of diffraction. (c) If they are closer together, they cannot be resolved or distinguished.
How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 1b shows the diffraction pattern produced by two point light sources that are close to one another. The pattern is similar to that for a single point source, and it is just barely possible to tell that there are two light sources rather than one. If they were closer together, as in Figure 1c, we could not distinguish them, thus limiting the detail or resolution we can obtain. This limit is an inescapable consequence of the wave nature of light.
There are many situations in which diffraction limits the resolution. The acuity of our vision is limited because light passes through the pupil, the circular aperture of our eye. Be aware that the diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus light passing through a lens with a diameter $$D$$ shows this effect and spreads, blurring the image, just as light passing through an aperture of diameter $$D$$ does. So diffraction limits the resolution of any system having a lens or mirror. Telescopes are also limited by diffraction, because of the finite diameter $$D$$ of their primary mirror.
TAKE-HOME EXPERIMENT: RESOLUTION OF THE EYE:
Draw two lines on a white sheet of paper (several mm apart). How far away can you be and still distinguish the two lines? What does this tell you about the size of the eye’s pupil? Can you be quantitative? (The size of an adult’s pupil is discussed in "Physics of the Eye."
Just what is the limit? To answer that question, consider the diffraction pattern for a circular aperture, which has a central maximum that is wider and brighter than the maxima surrounding it (similar to a slit) [see Figure 2a]. It can be shown that, for a circular aperture of diameter $$D$$, the first minimum in the diffraction pattern occurs at $$\theta = 1.22 \lambda / D$$ (providing the aperture is large compared with the wavelength of light, which is the case for most optical instruments). The accepted criterion for determining the diffraction limit to resolution based on this angle was developed by Lord Rayleigh in the 19th century. The Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 2b. The first minimum is at an angle of $$\theta = 1.22 \lambda / D$$, so that two point objects are just resolvable if they are separated by the angle $\theta = 1.22 \frac{\lambda}{D}, \tag{27.7.1}$ where $$\lambda$$ is the wavelength of light (or other electromagnetic radiation) and $$D$$ is the diameter of the aperture, lens, mirror, etc., with which the two objects are observed. In this expression, $$\theta$$ has units of radians.
Figure $$\PageIndex{2}$$: (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for being just resolvable. The central maximum of one pattern lies on the first minimum of the other.
CONNECTIONS: LIMITS TO KNOWLEDGE:
All attempts to observe the size and shape of objects are limited by the wavelength of the probe. Even the small wavelength of light prohibits exact precision. When extremely small wavelength probes as with an electron microscope are used, the system is disturbed, still limiting our knowledge, much as making an electrical measurement alters a circuit. Heisenberg’s uncertainty principle asserts that this limit is fundamental and inescapable, as we shall see in quantum mechanics.
Example $$\PageIndex{1}$$: Calculating Diffraction Limits of the Hubble Space Telescope
The primary mirror of the orbiting Hubble Space Telescope has a diameter of 2.40 m. Being in orbit, this telescope avoids the degrading effects of atmospheric distortion on its resolution. (a) What is the angle between two just-resolvable point light sources (perhaps two stars)? Assume an average light wavelength of 550 nm. (b) If these two stars are at the 2 million light year distance of the Andromeda galaxy, how close together can they be and still be resolved? (A light year, or ly, is the distance light travels in 1 year.)
Strategy:
The Rayleigh criterion stated in the equation $$\theta = 1.22 \frac{\lambda}{D}$$ gives the smallest possible angle $$\theta$$ between point sources, or the best obtainable resolution. Once this angle is found, the distance between stars can be calculated, since we are given how far away they are.
Solution (a):
The Rayleigh criterion for the minimum resolvable angle is $\theta = 1.22 \frac{\lambda}{D}. \tag{27.7.1}$ Entering known values gives $\theta = 1.22 \frac{550 \times 10^{-9}}{2.440 m}$ $= 2.80 \times 10^{-7} rad.$
Solution (b):
The distance $$s$$ between two objects a distance $$r$$ away and separated by an angle $$\theta$$ is $s = r\theta. \tag{27.7.2}$ Substituting known values gives $s = \left(2.0 \times 10^{6} ly \right) \left(2.80 \times 10^{-7} rad \right)$ $= .56 ly.$
Discussion:
The angle found in part (a) is extraordinarily small (less than 1/50,000 of a degree), because the primary mirror is so large compared with the wavelength of light. As noticed, diffraction effects are most noticeable when light interacts with objects having sizes on the order of the wavelength of light. However, the effect is still there, and there is a diffraction limit to what is observable. The actual resolution of the Hubble Telescope is not quite as good as that found here. As with all instruments, there are other effects, such as non-uniformities in mirrors or aberrations in lenses that further limit resolution. However, Figure 3 gives an indication of the extent of the detail observable with the Hubble because of its size and quality and especially because it is above the Earth’s atmosphere.
Figure $$\PageIndex{3}$$: These two photographs of the M82 galaxy give an idea of the observable detail using the Hubble Space Telescope compared with that using a ground-based telescope. (a) On the left is a ground-based image. (credit: Ricnun, Wikimedia Commons) (b) The photo on the right was captured by Hubble. (credit: NASA, ESA, and the Hubble Heritage Team (STScI/AURA))
The answer in part (b) indicates that two stars separated by about half a light year can be resolved. The average distance between stars in a galaxy is on the order of 5 light years in the outer parts and about 1 light year near the galactic center. Therefore, the Hubble can resolve most of the individual stars in Andromeda galaxy, even though it lies at such a huge distance that its light takes 2 million years for its light to reach us. Figure 4 shows another mirror used to observe radio waves from outer space.
Figure $$\PageIndex{4}$$: A 305-m-diameter natural bowl at Arecibo in Puerto Rico is lined with reflective material, making it into a radio telescope. It is the largest curved focusing dish in the world. Although $$D$$ for Arecibo is much larger than for the Hubble Telescope, it detects much longer wavelength radiation and its diffraction limit is significantly poorer than Hubble’s. Arecibo is still very useful, because important information is carried by radio waves that is not carried by visible light. (credit: Tatyana Temirbulatova, Flickr)
Diffraction is not only a problem for optical instruments but also for the electromagnetic radiation itself. Any beam of light having a finite diameter $$D$$ and a wavelength $$\lambda$$ exhibits diffraction spreading. The beam spreads out with an angle $$\theta$$ given by the equation $$\theta = 1.22 \frac{\lambda}{D}$$. Take, for example, a laser beam made of rays as parallel as possible (angles between rays as close to $$\theta = 0^{\circ}$$ as possible) instead spreads out at an angle $$\theta = 1.22 \lambda / D$$, where $$D$$ is the diameter of the beam and $$lambda$$ is its wavelength. This spreading is impossible to observe for a flashlight, because its beam is not very parallel to start with. However, for long-distance transmission of laser beams or microwave signals, diffraction spreading can be significant (see Figure 5). To avoid this, we can increase $$D$$. This is done for laser light sent to the Moon to measure its distance from the Earth. The laser beam is expanded through a telescope to make $$D$$ much larger and $$\theta$$ smaller.
Figure $$\PageIndex{5}$$: The beam produced by this microwave transmission antenna will spread out at a minimum angle $$\theta = 1.22 \lambda / D$$ due to diffraction. It is impossible to produce a near-parallel beam, because the beam has a limited diameter.
In most biology laboratories, resolution is presented when the use of the microscope is introduced. The ability of a lens to produce sharp images of two closely spaced point objects is called resolution. The smaller the distance $$x$$ by which two objects can be separated and still be seen as distinct, the greater the resolution. The resolving power of a lens is defined as that distance $$x$$. An expression for resolving power is obtained from the Rayleigh criterion. In Figure 6a we have two point objects separated by a distance $$x$$. According to the Rayleigh criterion, resolution is possible when the minimum angular separation is $\theta = 1.22 \frac{\lambda}{D} = \frac{x}{d}, \tag{27.7.3}$ where $$d$$ is the distance between the specimen and the objective lens, and we have used the small angle approximation (i.e., we have assumed that $$x$$ is much smaller than $$d$$), so that $$\tan{\theta} \approx \sin{\theta} \approx \theta$$.
Therefore, the resolving power is $x = 1.22 \frac{\lambda d}{D}. \tag{27.7.4}$ Another way to look at this is by re-examining the concept of Numerical Aperture ($$NA$$) discussed in "Microscopes." There, $$NA$$ is a measure of the maximum acceptance angle at which the fiber will take light and still contain it within the fiber. Figure 6b shows a lens and an object at point P. The $$NA$$ here is a measure of the ability of the lens to gather light and resolve fine detail. The angle subtended by the lens at its focus is defined to be $$\theta = 2\alpha$$. From the figure and again using the small angle approximation, we can write $\sin{\alpha} = \frac{D/2}{d} = \frac{D}{2d}. \tag{27.7.5}$ The $$NA$$ for a lens is $$NA = n \sin{\alpha}$$, where $$n$$ is the index of refraction of the medium between the objective lens and the object at point P.
From this definition for $$NA$$, we can see that $x = 1.22\frac{\lambda d}{D} = 1.22 \frac{\lambda}{2 \sin{\alpha}} = 0.61 \frac{\lambda n}{NA}. \tag{27.7.6}$ In a microscope, $$NA$$ is important because it relates to the resolving power of a lens. A lens with a large $$NA$$ will be able to resolve finer details. Lenses with larger $$NA$$ will also be able to collect more light and so give a brighter image. Another way to describe this situation is that the larger the $$NA$$, the larger the cone of light that can be brought into the lens, and so more of the diffraction modes will be collected. Thus the microscope has more information to form a clear image, and so its resolving power will be higher.
Figure $$\PageIndex{6}$$: (a) Two points separated by at distance $$x$$ and a positioned a distance $$d$$ away from the objective. (credit: Infopro, Wikimedia Commons) (b) Terms and symbols used in discussion of resolving power for a lens and an object at point P. (credit: Infopro, Wikimedia Commons)
One of the consequences of diffraction is that the focal point of a beam has a finite width and intensity distribution. Consider focusing when only considering geometric optics, shown in Figure 6a. The focal point is infinitely small with a huge intensity and the capacity to incinerate most samples irrespective of the $$NA$$ of the objective lens. For wave optics, due to diffraction, the focal point spreads to become a focal spot (see Figure 6b) with the size of the spot decreasing with increasing $$NA$$. Consequently, the intensity in the focal spot increases with increasing $$NA$$. The higher the $$NA$$, the greater the chances of photodegrading the specimen. However, the spot never becomes a true point.
Figure $$\PageIndex{7}$$: (a) In geometric optics, the focus is a point, but it is not physically possible to produce such a point because it implies infinite intensity. (b) In wave optics, the focus is an extended region.
# Summary
• Diffraction limits resolution.
• For a circular aperture, lens, or mirror, the Rayleigh criterion states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.
• This occurs for two point objects separated by the angle $$\theta = 1.22 \frac{\lambda}{D}$$, where $$\lambda$$ is the wavelength of light (or other electromagnetic radiation) and $$D$$ is the diameter of the aperture, lens, mirror, etc. This equation also gives the angular spreading of a source of light having a diameter $$D$$.
## Glossary
Rayleigh criterion
two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other
## Contributors
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0). | 3,441 | 14,983 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-51 | latest | en | 0.949004 |
http://www.enotes.com/homework-help/find-all-solutions-x-4-3x-2-2-0-225073 | 1,477,268,092,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719453.9/warc/CC-MAIN-20161020183839-00201-ip-10-171-6-4.ec2.internal.warc.gz | 438,381,224 | 10,300 | # Find all solutions of x^4 - 3x^2 + 2 = 0?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to find all the solutions of x^4 - 3x^2 + 2 = 0.
Now let y = x^2
x^4 - 3x^2 + 2 = 0
=> y^2 - 3y + 2 = 0
=> y^2 - 2y - y + 2 =0
=> y*(y - 2) - 1*(y - 2) =0
=> (y - 1)(y - 2) = 0
=> y = 1 or y = 2
As y = x^2
x^2 = 1
=> x = 1 or x = -1
x^2 = 2
=> x = sqrt 2 or -sqrt 2
Therefore the solutions of the equation x^4 - 3x^2 + 2 = 0 are 1, -1, sqrt 2, -sqrt 2.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
The equation will have 4 solutions since it's order is 4.
We'll use factorization to solve it. For the beginning, we'll re-write the middle terms as a sum of 2 terms:
-3x^2 = -x^2 - 2x^2
We'll substitute the middle terms by the algebraic sum:
x^4 -x^2 - 2x^2 + 2 = 0
We'll group the first 2 terms and the last 2 terms:
(x^4 -x^2) - (2x^2 - 2) = 0
We'll factorize by x^2 the first group and by 2 the last group;
x^2(x^2 - 1) - 2(x^2 - 1) = 0
We'll factorize by (x^2 - 1):
(x^2 - 1)(x^2 - 2) = 0
We'll set each factor as zero:
x^2 - 1 = 0
We'll re-write the difference of squares, using the formula:
a^2 - b^2 = (a-b)(a+b)
We'll put a = x and b = 1
x^2 - 1 = (x-1)(x+1)
x - 1 = 0
x1 = 1
x + 1 = 0
x2 = -1
We'll do the same with the second factor x^2 - 2:
x^2 - 2 = (x - sqrt2)(x + sqrt2)
x - sqrt2 = 0
x3 = sqrt2
x4 = -sqrt2
The 4 roots of the equation are: {-sqrt2 ; -1 ; 1 ; sqrt2}. | 655 | 1,474 | {"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} | 4.46875 | 4 | CC-MAIN-2016-44 | latest | en | 0.869276 |
https://www.autoblog.com/2008/05/10/chrysler-explains-details-of-the-2-99-gas-price-guarantee/ | 1,716,897,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059085.33/warc/CC-MAIN-20240528092424-20240528122424-00039.warc.gz | 548,916,849 | 40,950 | # Chrysler explains details of the \$2.99 gas price guarantee
Following Chrysler's new "Let's Refuel America Gas Card" announcement, the critics spoke out quickly. While a guaranteed gas price of \$2.99 a gallon for three years will certainly appeal to some people, other incentives might offer buyers better savings. To deal with the media fallout and to clarify a few questions about the card, Chrysler's media-only blog, The Firehouse, has tackled the topic with a post discussing the deal and the media coverage.
Stuart Schorr, the senior manager of sales for Mopar and Dealer Communications, wrote that there have been hundreds, possibly thousands, of stories about the incentive around the country (three of these videos are available after the jump). A TV station in Wisconsin calculated possible gas card savings to the customer of \$8,760. Of course, they used gas prices of \$6.65 a gallon to get that number. Also, Chrysler will be sending out official "Let's Refuel America" banners to dealerships in the coming days. Schorr also provided a list of ten questions (and answers) they've been hearing about the gas card, including one we've heard from our readers about what other incentives, if any, are available to people who sign up for the card. Check them out after the jump.
The video meant to be presented here is no longer available. Sorry for the inconvenience.
The video meant to be presented here is no longer available. Sorry for the inconvenience.
The video meant to be presented here is no longer available. Sorry for the inconvenience.
TOP TEN QUESTIONS ON HOW THE REFUEL AMERICA PROGRAM WORKS
Q: How do you figure out the gallons on the card?
A: Take average MPG of any vehicle, divide into 12,000 miles and you get an annual allocation of gallons of fuel, for each of three years. For Example: Dodge Caliber, 24 MPG average. The customer's card is allocated 500 gallons of unleaded fuel for each of three years. Total value is 1,500 gallons of gas x (price of gas - \$2.99).
Q: Heavy Duty Ram trucks do not have a MPG on the sticker Price?
A: We are using 15 MPG for all Ram's HD and LD. That equals 800 gallons of fuel per year at \$2.99. That's a lot of diesel dollars.
Q: How do I get charged? What about the receipt I get at the station?
A: While the gas price on the pump and on the customers' reciept at the time of the transaction will show the actual gas price at that time, the customer credit card account is actually billed only \$2.99 a gallon for their gas purchases and Chrysler picks up the difference. Regardless of the actual dollar figure shown on the reciept and the pump at the time of purchase the customer will only be charged at \$2.99 per gallon. The customer's Let's Refuel America card must be connected to an active Visa or MasterCard credit card account.
Q: How do you keep the card from being used on other vehicles?
A: The card can be used for any vehicle as long as the fuel purchased is the fuel specified for the purchased vehicle.
Q: Are there specific filling stations you have to go to in order to get the \$2.99/gallon price?
A: It is good at 97% of stations across the country.
Q: Is diesel fuel also \$2.99 a gallon?
A: The purchased vehicle has a recommended fuel: Unleaded 87, E85 or Diesel. The card can only be used for that specific kind of fuel, and yes the cost will be \$2.99 for diesel. If you use a card for purchase of a different kind of fuel than is specified for the purchased vehicle, there are additional charges.
Q: Is it only Trucks and SUVs?
A: No it is for the full line, starting with Jeep Compass, Jeep Patriot and Dodge Caliber and minivans up to Dodge Ram pickup trucks. In fact in terms of incremental value compared to previous incentives, the greatest increase in value is with our smallest, most fuel efficient vehicles. Vehicles excluded include Viper, Challenger, all SRT models, Dodge Sprinter, Jeep Wrangler and Ram Chassis Cab.
Q: What happens if I sell my car?
A: While the card is linked specifically to the purchase of an specific vehicle, the card then can be used by the customer for any vehicle and is not effected by the status of the original vehicle.
Q: I understand this offer is in lieu of other incentives. Explain?
A: Customer has a choice of three basic options: Cashback incentive; lowered APR financing; or the Gas Program. On certain vehicles, like the Dodge Ram, the Gas Program option includes additional bonus cash. For example:
Dodge Grand Caravan customers have a choice of \$2,500 cash incentive; 0% financing for qualified buyers; or the three year gas program including \$500 cash. Caravan gets 19 MPG for a three year total of 1,895 gallons of fuel at \$2.99.
[Source: The Firehouse] | 1,064 | 4,723 | {"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} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.933306 |
http://mathhelpforum.com/math-philosophy/222047-where-do-numbers-we-count-come-print.html | 1,519,244,027,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813712.73/warc/CC-MAIN-20180221182824-20180221202824-00569.warc.gz | 217,927,228 | 4,945 | # Where do the Numbers we count with come from?
• Sep 17th 2013, 07:55 AM
ashesmi
Where do the Numbers we count with come from?
How do mathamaticians experience what we call numbers? Do they describe all the numbers we use, as a part of infinity, or do all numbers mathamaticians use come from 1?
• Sep 17th 2013, 09:29 AM
Shakarri
Re: Where do the Numbers we count with come from?
I think that 1 is the most fundamental number, more fundamental than 0, and as such I think that all the numbers we know come from 1
Here's a good video about theories of what numbers are
o numbers EXIST? - Numberphile - YouTube
• Sep 17th 2013, 10:03 AM
HallsofIvy
Re: Where do the Numbers we count with come from?
I don't know what you mean by "part of infinity" but most "developments" of the numbers start with some version of "Peano's axioms" describing the "whole numbers":
"The whole numbers consist of a set of objects, N, (called numbers) and a "successor function, s, such that
1) There exist a member of N (called "0") such that s maps N "one-to-one" and "onto" N-{0} (the set N without 0).
2) If X is a subset of N such that (a) 0 is in X and (b) if n is in X then s(n) is in X, then X= N.
Those are, effectively, the "counting numbers" and (2) above is essentially "counting".
The other number sets, the integers, the rational numbers, the real numbers, and the complex numbers, are defined from those.
• Sep 17th 2013, 10:34 AM
ashesmi
Re: Where do the Numbers we count with come from?
Quote:
Originally Posted by Shakarri
I think that 1 is the most fundamental number, more fundamental than 0, and as such I think that all the numbers we know come from 1
Here's a good video about theories of what numbers are
o numbers EXIST? - Numberphile - YouTube
That was Absolutly The MOST AMAZING MATH VIDEO I HAVE EVER SEEN. IN...MY...LIFE!!! THANK YOU!!!
I agree I "Experience" numbers as sound, vibration, I am a musician and numbers to me are like music, like notes as matter and energy, potentual and kenetic, In harmonics there is the Fundemental Frequency, and I also feel 1 is the Fundemental Frequency. I am trying to help develope a New Math called Fractal Binary. It is built from the simple laws of physics (wave/frequency). In this video below is a new math from this new physics. It's the Physics that is used to describe the math...and it is simple, here is a short kids music video I am working on, the Video cuts out half way through because it isn't finished yet (sorry)... http://www.youtube.com/watch?v=CCYwbzd_Z2k
Also Here is a Method of doing Logs by using (wave/frequency) from physics
http://ashesmi.yolasite.com/resource...0s860x2243.png
Here is the 1 Plus Times Table, where the matrix begins in the center from 1+1+1+1 and spirials outward to infinity.
http://ashesmi.yolasite.com/resource...C0s860x823.png
And here is the Site I am working from trying to develope this new math.
http://ashesmi.yolasite.com/fractal-binary.php
...what do you think?
Thanks for the Video I added it to my site! Blew Me Away, God Bless You :)
• Sep 17th 2013, 10:49 AM
ashesmi
Re: Where do the Numbers we count with come from?
Thanks Hallsoflvy for your reply Shakarri Helped answer it for me. Please watch the Video Shakarri shared, it is the Phlosophy deffintions of Math I have been searching for for years. How do you view your self in the video Hallsoflvy? I am still trying to define my place in it...but I think numbers can have both a Potentual, Past and Present. When they Exist they are the Present "Kinetic", When they can exist is when they are "Potentual", Imaginary. and the past? I do not have any idea about...still pondering the video and my place in it.
• Sep 17th 2013, 02:04 PM
Hartlw
Re: Where do the Numbers we count with come from?
They come from successive slashes, found in ancient sites, and form the basis for the Peano Axioms which formally describe them.
1,11,111,1111, ...etc.
• Mar 21st 2014, 04:17 PM
bkbowser
Re: Where do the Numbers we count with come from?
Steven Pinker once described, one construction of, the most primitive form of number as "Two bears enter a cave, one bear leaves. Is it safe to enter the cave?" If I remember right he then went on to site languages developed independently by deaf communities which included only numerical concepts related to one or two or three objects, and greater then, less then, relationships.
• Mar 21st 2014, 04:43 PM
Plato
Re: Where do the Numbers we count with come from?
Quote:
Originally Posted by bkbowser
Steven Pinker once described, one construction of, the most primitive form of number as "Two bears enter a cave, one bear leaves. Is it safe to enter the cave?" If I remember right he then went on to site languages developed independently by deaf communities which included only numerical concepts related to one or two or three objects, and greater then, less then, relationships.
Steve Pinker's work in the physiology of language acquisition as well as his contribution to the mind/body problem. the blank slate are great. However, those who work on the philosophy of mathematics have reservations about his venture into the nature of mathematics. I suggest anyone interested in this read THE NUMBER SENSE: How the Mind Creates Mathematics by Stanislas Dehaene. This is written by a brain scientists who divides his time between the U of Origin and Paris.
• Mar 22nd 2014, 01:11 PM
ashesmi
Re: Where do the Numbers we count with come from?
Very COOL Plato!
• Nov 3rd 2014, 01:15 AM
xenaco88
Re: Where do the Numbers we count with come from?
That was a truly useful info. I never asked myself what the origin of the numbers we use is, but at the moment I see that this question and especially the materials you've kiondly provided can give some decent food for thought. Thanks | 1,498 | 5,780 | {"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} | 3.5625 | 4 | CC-MAIN-2018-09 | longest | en | 0.935263 |
https://www.math-only-math.com/3rd-grade-patterns.html | 1,726,261,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651540.48/warc/CC-MAIN-20240913201909-20240913231909-00479.warc.gz | 817,654,728 | 13,783 | # 3rd Grade Patterns
We have learnt in our previous class that pattern is a sequence of repeating objects, shapes or numbers. Pattern has a rule which tells us that which objects belong to the pattern and which object do not.
Patterns exist all around us. We can find a pattern in trees, in the window frames, in the floor, in our clothes etc.
The pattern that neither grows not reduces but only repeats are known as repeating pattern.
Following examples of some repeating patterns
Questions and Answers on 3rd Grade Patterns:
1. Draw the picture that comes next in each pattern.
Patterns Around Us
Thus, a pattern is an arrangement of objects, colors, or numbers placed in a certain order. Think about words or melodies in songs, lines and curves on buildings, or even in the grocery store where boxes and jars of various items arrangement forms a pattern.
1. Colour and draw the Easter egg that comes next.
2. Look at the pattern in each row. Circle the picture that continues the pattern.
3. Cross out the one which does not form a pattern.
4. Continue the pattern.
(ii) tttt, TTTT, ttt, TTT, tt, TT __________
(iv) 4, 8, 12, 16, 20, _____, _____
(iv) 24, 28
3rd Grade Math Worksheets
3rd Grade Math Lessons
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What are the rules for the comparison of two-digit numbers? We know that a two-digit number is always greater than a single digit number. But, when both the numbers are two-digit numbers | 602 | 2,587 | {"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} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.915533 |
https://nl.pinterest.com/explore/law-of-cosines/ | 1,521,582,774,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00450.warc.gz | 628,857,167 | 93,192 | ### Law of Sines and Law of Cosines Maze
Student have to figure out whether to use the Law of Sines or the Law of Cosines! I like that this maze doesn't tell them which one to use.
### Right Triangles - The Law of Cosines Notes and Practice
Right Triangles - The Law of Cosines Notes and PracticeThe Law of Cosines is used to find the missing sides and angles of acute and obtuse triangles in certain situations. This packet is designed to teach students how to solve problems given those situations.
### Law of Cosines Coded Message for Trigonometry RECENTLY UPDATED
Law of Cosines Coded Message activity for trigonometry
### Law of Sines and Law of Cosines Maze
This self-checking maze has 11 problems involving the law of sines and the law of cosines. Students will be required to use both to solve for angles and sides of triangles.Answer key is included for easy checking.This product pairs very well with my Law of Sines and Law of Cosines Task CardsoClick Here for more Triangle & Trigonometry activitiesoClick Here for more Math MazesThis product is also part of the following money saving bundle:Right Triangles & Trigonometry Activity Bundle.
### What is the Law of Cosines? (Easily Explained with 3 Powerful Examples)
Law of Cosines - EXCELLENT video lesson with 3 Example Problems. Learn how to solve oblique triangles using the Law of Cosines. Terrific for new algebra teachers. Perfect for high school and middle school math courses. Learn how to use the Law of Cosines formula to solve triangles for SAS or SSS congruency. Watch it today!
### Law of Sines and Law of Cosines Maze
Love this Law of Sines and Law of Cosines Maze! My Geometry students would like…
### Right-Angle Trigonometry, Law of Sines, Law of Cosines Application Problems
Engage your trig or pre-cal students with eleven well-written, interesting, thought-provoking, real-world questions. Students must use their knowledge of right-angle trig (SOH-CAH-TOA), law of sines, and law of cosines in order to solve these problems.
### Law of Sines/Cosines “Mapquest”
Law of Sines/Cosines “Mapquest” | Hilbert's Hotel
### Law of Sines and Law of Cosines Mazes
Law of Sines and Law of Cosines MazesThis is a set of four mazes to practice using the law of sines and law of cosines to find missing side and angle measures in triangles. Students use their solutions to navigate through the maze. This activity was designed for a high school level geometry class.
Pinterest | 570 | 2,463 | {"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} | 3.40625 | 3 | CC-MAIN-2018-13 | latest | en | 0.895277 |
https://kupdf.net/download/09-rotational-motion_59ff5c30e2b6f5cc1991b790_pdf | 1,701,978,734,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100686.78/warc/CC-MAIN-20231207185656-20231207215656-00808.warc.gz | 395,509,992 | 13,202 | # 09 Rotational Motion
November 6, 2017 | Author: Balaram Mullapudi | Category: Rotation Around A Fixed Axis, Torque, Angular Momentum, Rotation, Inertia
gh...
#### Description
The Big Idea The third conservation law is conservation of angular momentum. This can be roughly understood as spin, more accurately it is rotational velocity multiplied by rotational inertia. In any closed system (including the universe) the quantity of angular momentum is fixed. Angular momentum can be transferred from one body to another but cannot be lost or gained. If a system has its angular momentum changed from the outside it is caused by a torque multiplied by time of impact. Torque is a force applied at a distance from the center of rotation. Key Concepts •
• •
To determine the rotation axis, wrap your right hand fingers in the direction of rotation and your thumb points along the axis (see figure). When something rotates in a circle, it moves through a position angle θ that runs from 0 to 2π radians and starts over again at 0. The physical distance d it moves is called the path length. If the radius of the circle is larger, the distance traveled is larger. How quickly the position angle θ changes with time determines the angular velocity ω. The direction of angular velocity is either clockwise or counterclockwise. How quickly the angular velocity changes determines the angular acceleration α. The linear velocity v and linear acceleration also v depend on the radius of rotation, which is called the moment arm r (See figure below.) ω, α If something is spinning, it moves more quickly if it r d is farther from the center of rotation. For instance, θ people at the Earth’s equator are physically moving faster than people at northern latitudes, even though their day is still 24 hours long – this is because they have a greater circumference to travel in the same The axis of rotation amount of time. points out of the page at the center of this circle
People’s Physics Book
Ch 9-1
• • •
• • •
There are analogies to the “Big Three” equations that work for rotational motion just like they work for linear motion. As before: once you have the acceleration you can predict the motion. Just as linear accelerations are caused by forces, angular accelerations are caused by torques. Torques produce angular accelerations, but just as masses resist acceleration (due to inertia), there is an inertia that opposes angular acceleration. The measure of this inertial resistance depends on the mass, but more importantly on the distribution of the mass in the body. The moment of inertia, I, is the rotational version of mass. Values for the moment of inertia of common objects are given in problem 2. Torques have only two directions: those that produce clockwise (CW) and those that produce counterclockwise (CCW) rotations. The angular acceleration or change in ω would be in the direction of the torque. Imagine spinning a fairly heavy disk. To make it spin, you don’t push towards the disk center– F F┴ that will just move it in a straight line. To spin it, you need to push along the side, much like when you spin a basketball. Thus, the torque α you exert on a disk to make it accelerate r depends only on the component of the force perpendicular to the radius of rotation: τ = rF┴. Moment of Inertia is equivalent to mass for rotational motion. The larger the moment of I inertia (symbol I) the harder it is to rotate. The moment of inertia is equal to the mass multiplied by the distance of the mass to the rotational axis SQUARED. The fact that the distance is squared means that mass at farther distances from the rotational axis is weighted much more than mass close to it. Many separate torques can be applied to an object. The angular acceleration produced is α = τnet / I. The angular momentum of a spinning object is L = Iω. Torques produce a change in angular momentum with time: τ = ΔL/Δt Spinning objects have a kinetic energy, given by K = ½Iω2.
Table of Analogies Linear motion
Rotational motion
Rotational unit
x
θ
v
ω
a
α
F
τ
Meter-Newton·
m
I
kg·meters2
a = Fnet / m
α = τnet / I
p = mv
L = Iω
kg·meters2/sec
F = Δp/Δt
τ = ΔL/Δt
Meter-Newton
2
K = ½Iω2
Joules
K = ½mv
People’s Physics Book
Ch 9-2
Key Equations (These are simpler than they look: many are familiar equations in new “rotational” language!) •
d = rΔθ
v = rω
(ω = Δθ/Δt)
a = rα
(α = Δω/Δt)
a C = -mv²/r = –rω2
; the path length along an arc is equal to the radius of the arc times the angle through which the arc passes ; the linear velocity of an object in rotational motion is the radius of rotation times the angular velocity ; the linear acceleration of an object in rotational motion is the radius of rotation times the angular acceleration; this is in the direction of motion ; the centripetal acceleration of an object in rotational motion depends on the radius of rotation and the angular speed; the sign reminds us that it points inward towards the center of the circle; this is just mv2/r!
• • • • •
ω = 2π/T θ(t) = θ0 + ω0t + ½αt2 ω(t) = ω0 + αt ω2 = ω02 + 2α(Δθ) α = τnet / I
τnet = ∑ τall individual forces = Iα
τ = r×F = r┴F = rF┴
L = Iω
τ = ΔL/Δt
K = ½Iω2
People’s Physics Book
; angular velocity and period are simply related ; Kinematic equations for rotation.
; angular accelerations are produced by net torques, with inertia opposing acceleration; this is the rotational analog of a = Fnet / m ; the net torque is the vector sum of all the torques acting on the object. When adding torques it is necessary to subtract CW from CCW torques. ; individual torques are determined by multiplying the force applied by the perpendicular component of the moment arm ; angular momentum is the product of moment of inertia and angular velocity. Angular momentum is conserved just like linear momentum ; torques produce changes in angular momentum; this is the rotational analog of F = Δp/Δt ; angular motion counts for kinetic energy as well!
Ch 9-3
Rotational Motion Problem Set 1.
The wood plug, shown below, has a lower moment of inertia than the steel plug because it has a lower mass. rotation axis
rotation axis
wood plug
a.
steel plug
Which of these plugs would be easier to spin on its axis? Explain.
Even though they have the same mass, the plug on the right has a higher moment of inertia (I), than the plug on the left, since the mass is distributed at greater radius. rotation axis
rotation axis
wood plug
equal mass wood plug with hole
b. Which of the plugs would have a greater angular momentum if they were spinning with the same angular velocity? Explain. 2.
Here is a table of some moments of inertia of commonly found objects: Object
Drawing
Disk
Moment of Inertia ½MR2
Ring
MR2
Rod or plank
1
Rod or plank
/12ML2
1
Sphere (solid) Satellite
R
/3ML2
/5MR2 MR2
a. Calculate the moment of inertia of the Earth about its spin axis. b. Calculate the moment of inertia of the Earth as it revolves around the Sun. c. Calculate the moment of inertia of a hula hoop with mass 2 kg and radius 0.5 m. d. Calculate the moment of inertia of a rod 0.75 m in length and mass 1.5 kg rotating about one end. e. Repeat d., but calculate the moment of inertia about the center of the rod.
People’s Physics Book
Ch 9-4
3. Imagine standing on the North Pole of the Earth as it spins. You would barely notice it, but you would turn all the way around over 24 hours, without covering any real distance. Compare this to people standing on the equator: they go all the way around the entire circumference of the Earth every 24 hours! Decide whether the following statements are TRUE or FALSE. Then, explain your thinking. a. The person at the North Pole and the person at the equator rotate by 2π radians in 86,400 seconds. b. The angular velocity of the person at the equator is 2π/86400 radians per second. c. Our angular velocity in San Francisco is 2π/86400 radians per second. d. Every point on the Earth travels the same distance every day. e. Every point on the Earth rotates through the same angle every day. f. The angular momentum of the Earth is the same each day. g. The angular momentum of the Earth is 2/5MR2ω. h. The rotational kinetic energy of the Earth is 1/5MR2ω2. i. The orbital kinetic energy of the Earth is 1/2MR2ω2, where R refers to the distance from the Earth to the Sun. 4. You spin up some pizza dough from rest with an angular acceleration of 5 rad/s2. a. b. c. d. e. f.
How many radians has the pizza dough spun through in the first 10 seconds? How many times has the pizza dough spun around in this time? What is its angular velocity after 5 seconds? What is providing the torque that allows the angular acceleration to occur? Calculate the moment of inertia of a flat disk of pizza dough with mass 1.5 kg and radius 0.6 m. Calculate the rotational kinetic energy of your pizza dough at t = 5 s and t = 10 s.
5. Your bike brakes went out! You put your feet on the wheel to slow it down. The rotational kinetic energy of the wheel begins to decrease. Where is this energy going? 6. Consider hitting someone with a Wiffle ball bat. Will it hurt them more if you grab the end or the middle of the bat when you swing it? Explain your thinking, but do so using the vocabulary of moment of inertia (treat the bat as a rod), angular momentum (imagine the bat swings down in a semi-circle), and torque (in this case, torques caused by the contact forces the other person’s head and the bat are exerting on each other). 7. Why does the Earth keep going around the Sun? Shouldn’t we be spiraling farther and farther downward towards the Sun, eventually falling into it? Why do low-Earth satellites eventually spiral down and burn up in the atmosphere, while the Moon never will? 8. If most of the mass of the Earth were concentrated at the core (say, in a ball of dense iron), would the moment of inertia of the Earth be higher or lower than it is now? (Assume the total mass stays the same.) 9. Two spheres of the same mass are spinning in your garage. The first is 10 cm in diameter and made of iron. The second is 20 cm in diameter but is a thin plastic sphere filled with air. Which is harder to slow down? Why? (And why are two spheres spinning in your garage?)
People’s Physics Book
Ch 9-5
10. A game of tug-o-war is played … but with a twist (ha!). Each team has its own rope attached to a merry-go-round. One team pulls clockwise, the other counterclockwise. Each pulls at a different point and with a different force, as shown. 200 N
1.2 m
400 N
2.6 m
a. Who wins? b. By how much? That is, what is the net torque? c. Assume that the merry-go-round is weighted down with a large pile of steel plates. It is so massive that it has a moment of inertia of 2000 kg·m2. What is its angular acceleration? d. How long will it take the merry-go-round to spin around once completely? 11. You have two coins; one is a standard U.S. quarter, and the other is a coin of equal mass and size, but with a hole cut out of the center. a. Which coin has a higher moment of inertia? b. Which coin would have the greater angular momentum if they are both spun at the same angular velocity? 12. A wooden plank is balanced on a pivot, as shown below. Weights are placed at various places on the plank. 0.50 m A: 2.00 kg
0.30 m
1.00 m
B: 1.50 kg
C: 1.39 kg
Consider the torque on the plank caused by weight A. a. b. c. d. e. f.
What force, precisely, is responsible for this torque? What is the magnitude (value) of this force, in Newtons? What is the moment arm of the torque produced by weight A? What is the magnitude of this torque, in N·m? Repeat parts (a – d) for weights B and C. Calculate the net torque. Is the plank balanced? Explain.
People’s Physics Book
Ch 9-6
13. A star is rotating with a period of 10.0 days. It collapses with no loss in mass to a white dwarf with a radius of .001 of its original radius. a. What is its initial angular velocity? b. What is its angular velocity after collapse? 14. For a ball rolling without slipping with a radius of 0.10 m, a moment of Inertia of 25.0 kg-m2, and a linear velocity of 10.0 m/s calculate the following: a. b. c. d.
the angular velocity the rotational kinetic energy the angular momentum the torque needed to double its linear velocity in 0.2 sec
15. A merry-go-round consists of a uniform solid disc of 225 kg and a radius of 6.0 m. A single 80 kg person stands on the edge when it is coasting at 0.20 revolutions /sec. How fast would the device be rotating after the person has walked 3.5 m toward the center. (The moments of inertia of compound objects add.)
16. In the figure we have a horizontal beam of length, L, pivoted on one end and supporting 2000 N on the other. Find the tension in the supporting cable, which is at the same point at the weight and is at an angle of 30 degrees to the vertical. Ignore the weight of the beam.
People’s Physics Book
Ch 9-7
17. Two painters are on the fourth floor of a Victorian house on a scaffold, which weighs 400 N. The scaffold is 3.00 m long, supported by two ropes, each located 0.20 m from the end of the scaffold. The first painter of mass 75.0 kg is standing at the center; the second of mass, 65.0 kg, is standing 1.00 m from one end. a. Draw a free body diagram, showing all forces and all torques. (Pick one of the ropes as a pivot point.) b. Calculate the tension in the two ropes c. Calculate the moment of inertia for rotation around the pivot point, which is supported by the rope with the least tension. (This will be a compound moment of inertia made of three components.) d. Calculate the instantaneous angular acceleration assuming the rope of greatest tension breaks. 18. A horizontal 60 N beam. 1.4 m in length has a 100 N weight on the end. It is supported by a cable, which is connected to the horizontal beam at an angle of 37 degrees at 1.0 m from the wall. Further support is provided by the wall hinge, which exerts a force of unknown direction but which has a vertical (friction) component and a horizontal (normal) component. a. Find the tension in the cable b. Find the two components of the force on the hinge (magnitude and direction) c. Find the coefficient of friction of wall and hinge. 19. There is a uniform rod of mass 2.0 kg of length 2.0 m. It has a mass of 2.6 kg at one end. It is attached to the ceiling .40 m from the end with the mass. The string comes in at a 53 degree angle to the rod. a. Calculate the total torque on the rod. b. Determine its direction of rotation c. Explain, but don’t calculate, what happens to the angular acceleration as it rotates toward a vertical position.
People’s Physics Book
Ch 9-8
20. On a busy intersection a 3.00 m beam of 150 N is connected to a post at an angle upwards of 20.0 degrees to the horizontal. From the beam straight down hang a 200 N sign 1.00 m from the post and a 500 N signal light at the end of the beam. The beam is supported by a cable which connects to the beam 2.00 m from the post at an angle of 45.0 degrees measured from the beam; also by the hinge to the post which has horizontal and vertical components of unknown direction. a. Find the tension in the cable b. Find the magnitude and direction of the horizontal and vertical forces on the hinge. c. Find the total moment of inertia around the hinge as the axis. d. Find the instantaneous angular acceleration of the beam if the cable were to break.
People’s Physics Book
Ch 9-9
21. The medieval catapult consists of a 200 kg beam with a heavy ballast at one end and a projectile of 75.0 kg at the other end. The pivot is located 0.5 m from the ballast and a force with a downward component of 550 N is applied by prisoners to keep it steady until the commander gives the word to release it. The beam is 4.00 m long and the force is applied 0.900 m from the projectile end. Consider the situation when the beam is perfectly horizontal. a. Draw a free-body diagram labeling all torques b. Find the mass of the ballast c. Find the force on the horizontal support d. How would the angular acceleration change as the beam moves from the horizontal to the vertical position. (Give a qualitative explanation.) a. In order to maximize range at what angle should the projectile be released? b. What additional information and/or calculation would have to be done to determine the range of the projectile? 22. A bowling ball is thrown with an initial speed of 9m/s. Initially it slides along the lane without spinning, but eventually catches and starts spinning. The coefficient of kinetic friction between the ball and the lane is 0.24. a. What is the balls linear acceleration and angular acceleration? b. How long until the ball starts spinning without slipping? (hint: this occurs when v=rω) c. How far has the ball slid? d. What is its linear speed and angular speed when the ball begins smooth rolling (i.e. rolling without slipping)
People’s Physics Book
vcm
t=0
ω t=?
Ch 9-10 | 4,138 | 16,964 | {"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} | 3.640625 | 4 | CC-MAIN-2023-50 | latest | en | 0.901911 |
https://brainly.in/question/335455 | 1,485,286,271,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560285244.23/warc/CC-MAIN-20170116095125-00331-ip-10-171-10-70.ec2.internal.warc.gz | 785,545,688 | 9,537 | # What temperature would necessary to doubled the volume of a gas at stp,if the pressure is decreased to 50%
1
by sarfaraz2
2016-05-02T00:07:52+05:30
Solutions
T1=273 K T2=?
P1=1 ATMOS. PRESSURE P2=1/2 ATMOS. PRESSURE
V1=V V2=2 V
P1V1/T1=P2V2/T2
T2=1/2*2V*273/V
=273K
273 K=0C | 140 | 421 | {"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} | 3.421875 | 3 | CC-MAIN-2017-04 | latest | en | 0.22241 |
http://www.earthsciweek.org/node/17659 | 1,553,116,509,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202471.4/warc/CC-MAIN-20190320210433-20190320232433-00181.warc.gz | 259,780,586 | 10,053 | # Deep-Sea Drilling
## Activity Source:
The JOIDES Resolution (JR) has physical dimensions unlike most oceangoing vessels. Why? So that scientists can sail nearly anywhere in the world to drill for samples of rocks and sediment from below the seafloor — in hopes of discovering clues about Earth’s history and structure, life in the deep biosphere, past climate change, earthquakes, and natural resources.
Note for teachers: The JR has a flat bottom, a 6.4-meter hole in the middle, 12 laboratories, and a derrick towering 67 meters (about 13 stories) above the waterline. The vessel is 143 meters long and can drill 8,382 meters below sea level. To understand these numbers, it might be helpful to construct a model of the ship - right in your schoolyard.
Before class, measure several lengths of string that are 143 meters long, 24 meters long, and 16 meters long (this may require tying some pieces of string together), and then ball up these lengths of string. Also, you may wish to emailing Deep Earth Academy at learning@oceanleadership.org to request “The Ocean Drilling Program in Film” for your students to view later.
This activity enables students to estimate and calculate scales of distance and length as used by ocean drilling scientists. Correlated standards include National Science Education Standard E in Earth and Space Science and the National Council of Teachers in Mathematics Standard that students understand and use ratios and proportions to represent quantitative relationships.
## Materials
• Computer with Internet connection
• Measuring tapes
• String
• Markers
• Paper
## Procedure
1. View photos of the JR (http://iodp.tamu.edu/publicinfo/drillship.html) and discuss what this ship does. It is the only U.S. vessel that drills the ocean floor just for science. If possible, watch “The Ocean Drilling Program in Film".
2. Guess how big the ship is, judging from the photos. Record your answers. Compare estimates with the actual measurements (above). You will use estimation to determine just how “big” each measurement is.
3. Go outside to the schoolyard. In groups, devise strategies to estimate the distance from one end of the boat to the other end without measuring tapes. Use sticks or rocks to mark your estimates n the schoolyard. Place a label with the group’s names on each estimated length.
4. Inspect the balled-up lengths of string prepared by your teacher, and find the longest one. Lay it straight in the yard alongside the estimates you marked. This string represents the real length of the ship. How does it compare to your estimates?
5. Now you’re going to create a one-quarter scale model of the ship’s length on the schoolyard. How long would that be? The group with the most accurate estimate should model its estimation skills by showing the rest of the class how far that measurement would reach.
6. Place the lengths of string that are 24 meters long and 16 meters long on the schoolyard. Measure these lengths using measuring tapes. Calculate what ratio these lengths represent for the ship.
7. Using a 1/9th scale, calculate and indicate 5,980 meters to show how deep the water can be from the bottom of the ship to the ocean floor, and 8,382 meters to show the greatest length of the drill string. Calculate how deep into the seafloor the drill must be able to travel using those two measurements.
Extra challenge: Using graph paper, design a scale model of the ship and the drill string while holding the paper vertically, or “portrait style.” Remember, the ship length (143 m), height (67 m), greatest water depth (5,980 m), and greatest depth below the seafloor (2,402 m) should all be in proportion to one another. | 792 | 3,690 | {"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} | 3.28125 | 3 | CC-MAIN-2019-13 | latest | en | 0.886668 |
https://www.reference.com/web?q=Calculate%20Percent%20of%20a%20Number&qo=pagination&o=600605&l=dir&sga=1&qsrc=998&page=3 | 1,632,325,237,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00001.warc.gz | 966,111,481 | 27,150 | Related Search
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Percentage calculation. If you want to find out what x percent of your number are, divide your number by 100 and multiply it by x. | 376 | 1,752 | {"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} | 2.640625 | 3 | CC-MAIN-2021-39 | latest | en | 0.800698 |
https://www.allanswered.com/post/pgxrx/system-equivalent-to-5-point-finite-difference-stencil/ | 1,534,532,753,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212768.50/warc/CC-MAIN-20180817182657-20180817202657-00101.warc.gz | 825,394,753 | 7,266 | ### System equivalent to 5 point finite difference stencil
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0
5 months ago by
What kind of element and cell type for the finite element method in Fenics with Dirichlet boundary condition leads to an equivalent system of 5 point stencil with the Finite difference method for the Poisson equation ?
Consider the code below:
import numpy as np
from fenics import *
mesh = UnitSquareMesh.create(3, 3,CellType.Type_Option1)
V = FunctionSpace(mesh, 'Option 2', 1)
u = TrialFunction(V)
v = TestFunction(V)
matA = assemble(a)
with the different parameters(Option1 and Option 2) available as in Functionspace to get the same stiffness matrix as for the 5 point stencil using the finite difference method ?
The MIT matlab code here shows the possibility of obtaining such a matrix using Finite element methods. I would like to get this with Fenics. If I can obtain the same matrix with any of Fenics' capabilities as in the example MIT MATLAB code, that would be great as well.
Thank you.
Community: FEniCS Project
6
5 months ago by
The short answer is that to get the standard 5-point finite difference mesh for the Laplacian with Dirichlet boundary conditions by assembling a finite element in FEniCS, you can use
from fenics import *
n = 3
mesh = UnitSquareMesh(n, n)
V = FunctionSpace(mesh, 'Lagrange', 1)
u = TrialFunction(V)
v = TestFunction(V)
| 331 | 1,361 | {"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": 0, "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} | 2.578125 | 3 | CC-MAIN-2018-34 | latest | en | 0.874952 |
http://jean-pierre.moreau.pagesperso-orange.fr/p_complex.html | 1,680,438,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00687.warc.gz | 20,356,263 | 2,686 | COMPLEX ROOTS IN PASCAL
Choose a source program (*.pas) by clicking the appropriate button.
COMPLEX2.PAS TCOMPLEX.PAS ROOTNUM.PAS ZCIRCLE.PAS ZCIRCLE.PDF TEQUA2.PAS CROOT3.PAS TCROOT4.PAS MCGAMA.PAS ME1Z.PAS MCPSI.PAS MCLPN.PAS MCLQN.PAS MCLPMN.PAS MCLQMN.PAS CHORNER.PAS ZNEWTON.PAS ZNEWTON.TXT TNEWTON.PAS TCLAGUE.PAS FBAUHUBE.PAS TBAUHUBE.PAS ZMUELLER.PAS ZMUELLER.TXT COMPLEX1.PAS UCOMPLEX.PAS ZMATSYS.PAS CSYSMAT.PAS RSHCGT.PAS TRSLCGTC.PAS CLU.PAS TEST_CLU.PAS INV_CLU.PAS CDETMAT.PAS CFINDDET.PAS TICGT.PAS TCEIGEN.PAS TCOMEIG.PAS
Program Description
• Elementary operations and functions on complex numbers
• Test program of unit above (COMPLEX2)
• Program to demonstrate the complex root counting subroutine
• Program to demonstrate the zero searching algorithm
• Explanation File of program Zcircle
• Roots of 2nd degree equation with complex coefficients
• Roots of 3rd degree equation with complex coefficients
• Roots of 4th degree equation with complex coefficients
• Calculate Function Gamma with a complex argument
• Calculate the complex exponential integral E1(z) with a complex argument
• Calculate the Psi function for a complex argument
• Calculate the Legendre polynomials of first kind for a complex argument
• Calculate the Legendre polynomials of second kind for a complex argument
• Calculate the Asociated Legendre Functions and their First Derivatives for a Complex Argument
• Calculate the associated Legendre Functions of the Second Kind and their First Derivatives for a Complex Argument
• Evaluation of a complex polynomial by Horner's rule
• Program to demonstrate the Newton root subroutine in the complex domain
• Explanation File of Program above (Znewton)
• Find all roots of a complex polynomial using Newton's iterative formulation
• Find all roots of a complex polynomial using Laguerre formulation in complex domain
• Module used by program below (Bauhuber's Method)
• This program uses Bauhuber's Method to find all real or complex roots of a polynomial of degree n
• Program to demonstrate the complex domain Mueller's subroutine
• Explanation File of Program above (Zmueller)
• Pascal unit used by program below (COMPLEX1)
• Complex numbers calculator
• Solving a complex linear system by Gauss-Jordan
• Solving a complex linear system AX=B by GAUSS-JORDAN Method
• Solving a complex homogeneous linear system by Gauss Method with full pivoting
• Solve a Complex Linear System By Gauss Method with full pivoting and correction process
• Unit CLU used by Program below
• Solving a complex linear system by LU decomposition
• Inversion of a complex square matrix by LU decomposition
• Determinant of a complex square matrix By Gauss Method with full pivoting
• Determinant of a complex square matrix using function CFindDet NEW
• Calculate inverse of a complex square matrix by Gauss Method with full pivoting
• Eigenvalues and eigenvectors of a general complex square matrix using QR algorithm
• Eigenvalues and eigenvectors of a general complex square matrix using Jacobi method
RETURN | 738 | 3,033 | {"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} | 2.859375 | 3 | CC-MAIN-2023-14 | latest | en | 0.68917 |
https://www.coursehero.com/file/6105525/30/ | 1,516,668,510,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891546.92/warc/CC-MAIN-20180122232843-20180123012843-00744.warc.gz | 896,960,022 | 23,494 | # 30 - 1 Larger and smaller particles Since equilateral t...
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Unformatted text preview: 1. Larger and smaller particles Since equilateral t riangles can be constructed out of a ’s (and squares out of b ’s) in more than one way, it is possible to have “molecules” of each of the elements that have different numbers of atomic triangles ( a ’s and b ’s). These might be considered “isotopes” of the basic molecules described by Plato (with each t made of 6 a ’s, and each s made of 4 b ’s). An equilateral t riangle can also be constructed out of 2, or 8, or 18, a ’s (and so on, ad infinitum). A square can also be constructed out of 2, or 8, or 16, b ’s (and so on, ad infinitum). This means that one “normal” particle of earth (6 s = 24 b ) can be transformed into 2 of the smaller “isotopes” of earth (6 s = 12 b ) Similarly, 4 “normal” particles of water (containing 120 a ’s each) can combine to form one huge particle of one of the larger “isotopes” of water (20 sides of 24 a ’s each, for 480 a ’s altogether). Final Reflections Comparison with predecessors Plato’s theory combines elements of the views of many of his predecessors. 1. Pythagoras Like Pythagoras, he made the physical universe fundamentally mathematical. But whereas Pythagoras thought that everything was made of numbers, Plato made geometrical figures - ultimately, t riangles - the atoms of his system....
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Ask a homework question - tutors are online | 478 | 1,884 | {"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} | 2.8125 | 3 | CC-MAIN-2018-05 | latest | en | 0.871515 |
https://discuss.leetcode.com/topic/85933/c-using-hash-tables-unordered_maps | 1,513,339,455,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948569405.78/warc/CC-MAIN-20171215114446-20171215140446-00282.warc.gz | 538,637,437 | 8,489 | # [C++] Using hash tables (unordered_maps)
• ``````vector<vector<int>> multiply(vector<vector<int>> &A, vector<vector<int>> &B) {
int m = A.size(), n = m ? A[0].size() : 0, p = n ? B[0].size() : 0;
if (!m || !n || !p) return {};
unordered_map<int, list<pair<int, int>>> htA, htB;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (A[i][j])
htA[i].emplace_back(j, A[i][j]);
for (int i = 0; i < n; i++)
for (int j = 0; j < p; j++)
if (B[i][j])
htB[j].emplace_back(i, B[i][j]);
vector<vector<int>> res (m, vector<int> (p, 0));
for (int i = 0; i < m; i++) {
for (int j = 0; j < p; j++) {
auto foundA = htA.find(i), foundB = htB.find(j);
if (foundA != htA.end() && foundB != htB.end()) {
auto l1 = foundA->second, l2 = foundB->second;
auto it1 = l1.begin(), it2 = l2.begin();
while (it1 != l1.end() && it2 != l2.end()) {
if (it1->first == it2->first) {
res[i][j] += it1->second * it2->second;
it1++; it2++;
}
else if (it1->first < it2->first) it1++;
else it2++;
}
}
}
}
return res;
}
``````
Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect. | 413 | 1,094 | {"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} | 2.640625 | 3 | CC-MAIN-2017-51 | latest | en | 0.418319 |
https://socratic.org/questions/how-do-you-solve-y-4-x-and-y-3x-1-using-substitution | 1,601,169,757,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400249545.55/warc/CC-MAIN-20200926231818-20200927021818-00016.warc.gz | 614,850,688 | 6,064 | # How do you solve y=4/x and y=3x-1 using substitution?
Jul 1, 2016
$x = \frac{4}{3}$ or $x = - 1$
#### Explanation:
To solve $y = \frac{4}{x}$ and $y = 3 x - 1$ using substitution, just substitute value of $y$ from first equation into second, which leads to
$\frac{4}{x} = 3 x - 1$ or multiplying each by $x$ (we assome $x \ne 0$, we get
$4 = 3 {x}^{2} - x$ or
$3 {x}^{2} - x - 4 = 0$ or
$3 {x}^{2} + 3 x - 4 x - 4 = 0$ or
$3 x \left(x + 1\right) - 4 \left(x + 1\right) = 0$ or
$\left(3 x - 4\right) \left(x + 1\right) = 0$, which means that $3 x - 4 = 0$ or $x + 1 = 0$
i.e. $x = \frac{4}{3}$ or $x = - 1$ | 287 | 620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "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} | 4.5625 | 5 | CC-MAIN-2020-40 | latest | en | 0.681628 |
https://stats.stackexchange.com/questions/230795/anova-for-difference-between-three-percentage-differences?noredirect=1 | 1,723,022,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00158.warc.gz | 447,527,158 | 39,524 | # ANOVA for difference between three percentage differences
This question is deliberately similar to a previous question: test for difference between two differences (proportions) because although the answer by StevenP was very useful, it only answered part of the question.
I have data for an analyte measured in the same people at two different times, and a fixed effect factor (genotype) which separates the subjects into 3 groups. We suspect that the percentage change (between times 1 and 2) between the 3 groups differs (note that the absolute change is not thought to be so important).
StevenP's answer was to use ANOVA with repeated measures and to check assumptions (presumably the sphericity assumption is not necessary as there are only 2 times). However, the interaction between group x time will give an indication of whether the absolute changes over time are significantly different between the 3 groups, not the percentage changes.
My question is, therefore, how to proceed with a legitimate ANOVA assessment of percentage changes ?
I suppose this question is really about how the variance is assessed and included in the analysis when the outcome is a combination of several variables. In the case of the absolute differences, for the combination outcome "measurement at time 2 minus mesaurement at time 1", a repeated measures ANOVA nicely takes into account the variance at time 2 and the variance at time 1.
However, what happens when the combination outcome is "measurement at time 2 minus measurement at time 1, divided by measurement at time 1" ?? My own feeling is that the variance at time 1 is already included in the percentage along with the variance at time 2, when the proportion or percentage was created (for each subject), and therefore a simple one-way ANOVA is needed for the percentage data only - is this correct ? | 378 | 1,857 | {"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} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.962123 |
http://www.excelarticles.com/LE10ePub-129.html | 1,508,792,169,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826642.70/warc/CC-MAIN-20171023202120-20171023222120-00682.warc.gz | 455,136,093 | 26,717 | • Excel Book Excerpt
# Excel Change Smith, Jane to Jane Smith
This page is an advertiser-supported excerpt of the book, Learn Excel 2007-2010 from MrExcel - 512 Excel Mysteries Solved. If you like this topic, please consider buying the entire e-book.
Change Smith, Jane to Jane Smith
Problem: I have a column of names in last name, first name style. How can I convert the data to first name last name?
Strategy: While you could do this in many steps, using Text to Columns and then a concatenation formula, a single large formula would also solve the problem. To begin, you need to insert a blank column after column A to hold the calculation.
=FIND(“,",A2) will locate the comma within the value in column A. In Smith, Jane, the comma is the sixth character, so the FIND function would return a 6.
The first name starts two characters after the result of the FIND function. It extends to the end of the text. You can use the MID function to isolate the first name. The MID function requires some text, a starting location, and a length. If you ask for more characters than are in the text, then Excel will return from the starting position to the end of the text. For example, if you ask for 50 characters, Excel will handle any first name that has 50 characters or less. Therefore, you use =MID(A2,FIND(“,",A2)+2,50).
The last name is always the leftmost characters, so you can use =LEFT(A2,FIND(“,",A2)-1).
To join the first name and last name together, you concatenate the function for the first name, a space in quotes, and the function for the last name. You need to be sure to leave the = sign off the LEFT function because you don’t prefix the function with an equals sign when it occurs in the middle of the formula.
If you want the text in uppercase and lowercase, you need to wrap the entire function in the PROPER function. As shown below, the formula is =PROPER(MID(A2,FIND(“,",A2)+2,50)&" “&LEFT(A2,FIND(“,",A2)-1)).
Figure 255 The formula in column B achieves the result..
For more resources for Microsoft Excel: | 492 | 2,039 | {"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} | 3.40625 | 3 | CC-MAIN-2017-43 | longest | en | 0.80957 |
http://www.basicmusictheory.com/a-double-sharp-suspended-4th-triad-chord | 1,722,667,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640361431.2/warc/CC-MAIN-20240803060126-20240803090126-00239.warc.gz | 31,834,226 | 10,067 | # A-double-sharp suspended 4th triad chord
The Solution below shows the A-double-sharp suspended 4th triad chord in root position, 1st inversion and 2nd inversion on the piano, treble clef and bass clef.
The Lesson steps then explain how to construct this triad chord using the 3rd and 5th note intervals, then finally how to construct the inverted chord variations.
For a quick summary of this topic, have a look at Triad chord.
Key C C# Db D D# Eb E E# Fb F F# Gb G G# Ab A A# Bb B B# Cb
## Solution - 3 parts
### 1. A-double-sharp suspended 4th chord
This step shows the A-double-sharp suspended 4th triad chord in root position on the piano, treble clef and bass clef.
The A-double-sharp suspended 4th chord contains 3 notes: A##, D##, E##.
The chord spelling / formula relative to the A## major scale is: 1 4 5.
A-double-sharp suspended 4th chord note names
Note no.Note intervalSpelling
/ formula
Note name#Semitones
from root
1root1The 1st note of the A-double-sharp suspended 4th chord is A##0
2A##-perf-4th4The 2nd note of the A-double-sharp suspended 4th chord is D##5
3A##-perf-5th5The 3rd note of the A-double-sharp suspended 4th chord is E##7
Middle C (midi note 60) is shown with an orange line under the 2nd note on the piano diagram.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord in root position are 5/4.
The staff diagrams and audio files contain each note individually, ascending from the root, followed by the chord containing all 3 notes.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 2. A-double-sharp suspended 4th 1st inversion
This step shows the A-double-sharp suspended 4th 1st inversion on the piano, treble clef and bass clef.
The A-double-sharp suspended 4th 1st inversion contains 3 notes: D##, E##, A##.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord inversion are 5/2, so the chord is said to be in five-two position.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 3. A-double-sharp suspended 4th 2nd inversion
This step shows the A-double-sharp suspended 4th 2nd inversion on the piano, treble clef and bass clef.
The A-double-sharp suspended 4th 2nd inversion contains 3 notes: E##, A##, D##.
These note names are shown below on the treble clef followed by the bass clef.
The figured bass symbols for this chord inversion are 7/4, so the chord is said to be in seven-four position.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
## Lesson steps
### 1. Piano key note names
This step shows the white and black note names on a piano keyboard so that the note names are familiar for later steps, and to show that the note names start repeating themselves after 12 notes.
The white keys are named using the alphabetic letters A, B, C, D, E, F, and G, which is a pattern that repeats up the piano keyboard.
Every white or black key could have a flat(b) or sharp(#) accidental name, depending on how that note is used. In a later step, if sharp or flat notes are used, the exact accidental names will be chosen.
The audio files below play every note shown on the piano above, so middle C (marked with an orange line at the bottom) is the 2nd note heard.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 2. A-double-sharp tonic note and one octave of notes
This step shows 1 octave of notes starting from note A##, to identify the start and end notes of the scale used to build this chord.
The numbered notes are those that might be used when building this chord.
Note 1 is the root note - the starting note of the chord - A##, and note 13 is the same note name but one octave higher.
No. Note 1 2 3 4 5 6 7 8 9 10 11 12 13 A## C C# / Db D D# / Eb E F F# / Gb G G# / Ab A A# / Bb A##
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 3. A-double-sharp major scale note interval positions
This step describes the A## major scale, whose note intervals are used to define the chord in a later step.
The major scale uses the W-W-H-W-W-W-H note counting rule to identify the scale note positions.
To count up a Whole tone, count up by two physical piano keys, either white or black.
To count up a Half-tone (semitone), count up from the last note up by one physical piano key, either white or black.
The tonic note (shown as *) is the starting point and is always the 1st note in the major scale.
Again, the final 8th note is the octave note, having the same name as the tonic note.
No. Note 1 2 3 4 5 6 7 8 A## C# / Db D# / Eb E F# / Gb G# / Ab A# / Bb A##
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 4. A-double-sharp major scale note interval numbers
This step identifies the note interval numbers of each scale note, which are used to calculate the chord note names in a later step.
To identify the note interval numbers for this major scale, just assign each note position from the previous step, with numbers ascending from 1 to 8.
No. Note 1 2 3 4 5 6 7 8 A## B## C### D## E## F### G### A##
To understand why the note names of this major scale have these specific sharp and flat names, have a look at the A## major scale page.
Both the note interval numbers and note names from the piano diagram above will be used in later steps to calculate the chord note names.
### 5. Triad chord qualities
This step defines a triad chord, names the triad chord qualities and identifies the notes that vary between them.
#### Triad chord definition
The music theory term triad chord means that 3 or more notes played together, or overlapping.
#### Triad chord qualities
Triad chords exist in five different chord qualities, which are major, minor, augmented, and diminished.
Each chord quality name is the name of the entire chord as a whole, not its individual notes (which will be covered later).
#### Triad chord qualities using the 1st, 3rd and 5th scale notes
All of these triad qualities are based on the 1st, 3rd and 5th notes of the major scale piano diagram above.
Depending on the chord quality, the 3rd and 5th scale note names of the major scale above might need to be adjusted up or down by one half-note / semitone / piano key.
It is these variations of the 3rd and 5th notes that give each one a distinctive sound for any given key (eg. C-flat, E etc).
#### Suspended triad chords - using the 2nd or 4th scale notes
A suspended chord is known in music theory as an altered chord because it takes one of the above chord qualities and modifies it in some way.
Unlike all of the above qualities, Suspended triad chords do not use the 3rd note of the major scale (at all) to build the chord.
The 3rd note is suspended, ie. removed completely, and replaced by either the 2nd note of the major scale - a suspended 2nd, or more commonly by the 4th note of the major scale - a suspended 4th.
Musically, this is interesting, since it is usually the 3rd note of the scale that defines the overall character of the chord as being major (typically described as 'happy') or minor ('sad').
Without this 3rd note, suspended chords tend to have an open and ambiguous sound.
The steps below will detail the suspended 4th triad chord quality in the key of A##.
### 6. Triad chord note intervals
This step defines the note intervals for each chord quality, including the intervals for the A-double-sharp suspended 4th triad chord.
Each individual note in a triad chord can be represented in music theory using a note interval, which is used to express the relationship between the first note of the chord (the root note), and the note in question.
The root note is always the 1st note (note interval 1 in the above diagram) of the major scale diagram above. ie. the tonic of the major scale.
Then there is one note interval to describe the 2nd note, and another to describe the 3rd note of the chord.
In the same way that the entire chord itself has a chord quality, the intervals representing the individual notes within that chord each have their own quality.
These note interval qualities are diminished, minor, major, perfect and augmented.
Below is a table showing the note interval qualities for all triad chords, together with the interval short names / abbrevations in brackets.
Triad chord note interval qualities
Chord quality2nd note quality3rd note quality
majormajor (M3)perfect (P5)
minorminor (m3)perfect (P5)
augmentedmajor (M3)augmented (A5)
diminishedminor (m3)diminished (d5)
suspended
(2nd/4th)
major (M2) or
perfect (P4)
perfect (P5)
The numbers in brackets are the note interval numbers (ie the scale note number) shown in the previous step.
#### A-double-sharp triad chord note intervals
Looking at the table above, the note intervals for the chord quality we are interested in (suspended 4th triad), in the key of A## are A##-perf-4th and A##-perf-5th.
The links above explain in detail the meaning of these note qualities, the short abbrevations in brackets, and how to calculate the interval note names based on the scale note names from the previous step.
### 7. A-double-sharp suspended 4th triad chord in root position
This step shows the A-double-sharp suspended 4th triad chord note interval names and note positions on a piano diagram.
#### Note name adjustments
Each note interval quality (diminished, minor, major, perfect, augmented) expresses a possible adjustment ie. a possible increase or decrease in the note pitch from the major scale notes in step 4.
If an adjustment in the pitch occurs, the note name given in the major scale in step 4 is modified, so that sharp or flat accidentals will be added or removed.
But crucially, for all interval qualities, the starting point from which accidentals need to be added or removed are the major scale note names in step 4.
For this chord, this is explained in detail in A##-perf-4th and A##-perf-5th, but the relevant adjustments for this suspended 4th chord quality are shown below:
A##-4th: Since the 4th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 4th note name - D##, is used, and the chord note spelling is 4.
A##-5th: Since the 5th note quality of the major scale is perfect, and the note interval quality needed is perfect also, no adjustment needs to be made. The 5th note name - E##, is used, and the chord note spelling is 5.
#### A-double-sharp suspended 4th triad chord note names
The final chord note names and note interval links are shown in the table below.
Note Interval No. Interval def. A## D## E## 1 4 5 root A##-perf-4th A##-perf-5th 1 4 5 0 5 7
The piano diagram below shows the interval short names, the note positions and the final note names of this triad chord.
In music theory, this triad chord as it stands is said to be in root position because the root of the chord - note A##, is the note with the lowest pitch of all the triad notes.
The note order of this triad can also be changed, so that the root is no longer the lowest note, in which case the triad is no longer in root position, and will be called an inverted triad chord instead.
For triad chords, there are 2 possible inverted variations as described in the steps below.
#### Figured bass notation
The figured bass notation for a triad in root position is 5/3, with the 5 placed above the 3 on a staff diagram.
These numbers represent the interval between the lowest note of the chord and the note in question.
So another name for this chord would be A-double-sharp suspended 4th triad in five-three position.
For example, the 5 represents note E##, from the A##-5th interval, since the triad root, A##, is the lowest note of the chord (as it is not inverted).
In the same way, the figured bass 4 symbol represents note E##, from the A##-4th interval.
Since figured bass notation works within the context of a key, we don't need to indicate in the figured bass symbols whether eg. the 3rd is a major, minor etc. The key is assumed from the key signature.
Often the 5 symbol is not shown at all, and only the number 4 symbol is shown - the 5th is assumed.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 8. A-double-sharp suspended 4th 1st inversion
This step shows the first inversion of the A-double-sharp suspended 4th triad chord.
#### Structure
To invert a chord, simply take the first note of the chord to be inverted (the lowest in pitch) and move it up an octave to the end of the chord.
So for a 1st inversion, take the root of the triad chord in root position from the step above - note A##, and move it up one octave (12 notes) so it is the last (highest) note in the chord.
The second note of the original triad (in root position) - note D## is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this triad in 1st inversion is 5/2, with the 5 placed above the 2 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be A-double-sharp suspended 4th triad in five-two position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original triad root!), and the note in question.
For example, the 5 represents note A##, from the D##-5th interval, since the lowest (bass) note of the chord - now inverted, is D##.
In the same way, the figured bass 2 symbol represents note E##, from the D##-2nd interval.
Often the 5 symbol is not shown at all, and only the number 2 symbol is shown - the 5th is assumed.
Bass Clef: Midi MP3 Treble Clef: Midi MP3
### 9. A-double-sharp suspended 4th 2nd inversion
This step shows the second inversion of the A-double-sharp suspended 4th triad chord.
#### Structure
For a 2nd inversion, take the first note of the 1st inversion above - D##, and move it to the end of the chord.
So the second note of the 1st inversion - note E## is now the note with the lowest pitch for the 2nd inversion.
Or put another way, the third note of the original triad (in root position) is now the note with the lowest pitch.
#### Figured bass notation
The figured bass notation for this triad in 2nd inversion is 7/4, with the 7 placed above the 4 on a staff diagram.
Based on this numbering scheme, another name for this inversion would be A-double-sharp suspended 4th triad in seven-four position.
These numbers represent the interval between the lowest note of the chord (not necessarily the original triad root!), and the note in question.
For example, the 7 represents note D##, from the E##-7th interval, since the lowest (bass) note of the chord - now inverted, is E##.
In the same way, the figured bass 4 symbol represents note A##, from the E##-4th interval.
Bass Clef: Midi MP3 Treble Clef: Midi MP3 | 3,766 | 14,712 | {"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} | 3.515625 | 4 | CC-MAIN-2024-33 | latest | en | 0.89463 |
http://www.algebra.com/algebra/homework/Finance/Finance.faq.question.122857.html | 1,369,249,039,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702298845/warc/CC-MAIN-20130516110458-00097-ip-10-60-113-184.ec2.internal.warc.gz | 308,028,174 | 4,517 | # SOLUTION: 87.Money is invested in a savings account at 12% simple interest.After one year there is \$1,680 in the account.How much was originally invested?
Algebra -> Algebra -> Finance -> SOLUTION: 87.Money is invested in a savings account at 12% simple interest.After one year there is \$1,680 in the account.How much was originally invested? Log On
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Algebra in Finance Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Finance Question 122857This question is from textbook Algebra 1 : 87.Money is invested in a savings account at 12% simple interest.After one year there is \$1,680 in the account.How much was originally invested?This question is from textbook Algebra 1 Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!.12x=1,680 divide both sides by .12 .12x/.12=1,680/.12 x=1,680/.12 x=\$14,000 answer for the amount in the savings account. | 271 | 1,109 | {"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} | 3.328125 | 3 | CC-MAIN-2013-20 | latest | en | 0.914837 |
https://www.howtodraw.ca/Mechanical/Lines-And-Curves.html | 1,726,775,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00760.warc.gz | 721,612,133 | 11,917 | # Lines And Curves
Although the beginner will find that a study of geometry is not essential to the production of such elementary examples of mechanical drawing as are given in this book, yet as more difficult examples are essayed he will find such a study to be of great advantage and assistance. Meantime the following explanation of simple geometrical terms is all that is necessary to an understanding of the examples given.
The shortest distance between two points is termed the radi
s; and, in the case of a circle, means the distance from the centre to the perimeter measured in a straight line.
Fig. 38.
Fig. 39.
Fig. 40.
Dotted lines, thus, <——- >, mean the direction and the points at which a dimension is taken or marked. Dotted lines, thus,——-, simply connect the same parts or lines in different views of the object. Thus in Figure 38 are a side and an end view of a rivet, and the dotted lines show that the circles on the end view correspond to the circle of the diameters of the head and of the stem, and therefore represent their diameters while showing that both are round. A straight line is in geometry termed a right line.
A line at a right angle to another is said to be perpendicular to it; thus, in Figures 39, 40, and 41, lines A are in each case perpendicular to line B, or line B is in each case perpendicular to line A.
A point is a position or location supposed to have no size, and in cases where necessary is indicated by a dot.
Parallel lines are those equidistant one from the other throughout their length, as in Figure 42. Lines maybe parallel though not straight; thus, in Figure 43, the lines are parallel.
Fig. 41.
Fig. 42.
Fig. 43.
Fig. 44.
Fig. 45.
Fig. 46.
A line is said to be produced when it is extended beyond its natural limits: thus, in Figure 44, lines A and B are produced in the point C.
A line is bisected when the centre of its length is marked: thus, line A in Figure 45 is bisected, at or in, as it is termed, e.
The line bounding a circle is termed its circumference or periphery and sometimes the perimeter.
A part of this circumference is termed an arc of a circle or an arc; thus Figure 46 represents an arc. When this arc has breadth it is termed a segment; thus Figures 47 and 48 are segments of a circle. A straight line cutting off an arc is termed the chord of the arc; thus, in Figure 48, line A is the chord of the arc.
Fig. 47.
Fig. 48.
Fig. 49.
Fig. 50.
Fig. 51.
A quadrant of a circle is one quarter of the same, being bounded on two of its sides by two radial lines, as in Figure 49.
When the area of a circle that is enclosed within two radial lines is either less or more than one quarter of the whole area of the circle the figure is termed a sector; thus, in Figure 50, A and B are both sectors of a circle.
A straight line touching the perimeter of a circle is said to be tangent to that circle, and the point at which it touches is that to which it is tangent; thus, in Figure 51, line A is tangent to the circle at point B. The half of a circle is termed a semicircle; thus, in Figure 52, A B and C are each a semicircle.
Fig. 52.
Fig. 53.
The point from which a circle or arc of a circle is drawn is termed its centre. The line representing the centre of a cylinder is termed its axis; thus, in Figure 53, dot d represents the centre of the circle, and line b b the axial line of the cylinder.
To draw a circle that shall pass through any three given points: Let A B and C in Figure 54 be the points through which the circumference of a circle is to pass. Draw line D connecting A to C, and line E connecting B to C. Bisect D in F and E in G. From F as a centre draw the semicircle O, and from G as a centre draw the semicircle P; these two semicircles meeting the two ends of the respective lines D E. From B as a centre draw arc H, and from C the arc I, bisecting P in J. From A as a centre draw arc K, and from C the arc L, bisecting the semicircle O in M. Draw a line passing through M and F, and a line passing through J and Q, and where these two lines intersect, as at Q, is the centre of a circle R that will pass through all three of the points A B and C.
Fig. 54.
Fig. 55.
To find the centre from which an arc of a circle has been struck: Let A A in Figure 55 be the arc whose centre is to be found. From the extreme ends of the arc bisect it in B. From end A draw the arc C, and from B the arc D. Then from the end A draw arc G, and from B the arc F. Draw line H passing through the two points of intersections of arcs C D, and line I passing through the two points of intersection of F G, and where H and I meet, as at J, is the centre from which the arc was drawn.
A degree of a circle is the 1/360 part of its circumference. The whole circumference is supposed to be divided into 360 equal divisions, which are called the degrees of a circle; but, as one-half of the circle is simply a repetition of the other half, it is not necessary for mechanical purposes to deal with more than one-half, as is done in Figure 56. As the whole circle contains 360 degrees, half of it will contain one-half of that number, or 180; a quarter will contain 90, and an eighth will contain 45 degrees. In the protractors (as the instruments having the degrees of a circle marked on them are termed) made for sale the edges of the half-circle are marked off into degrees and half-degrees; but it is sufficient for the purpose of this explanation to divide off one quarter by lines 10 degrees apart, and the other by lines 5 degrees apart. The diameter of the circle obviously makes no difference in the number of decrees contained in any portion of it. Thus, in the quarter from 0 to 90, there are 90 degrees, as marked; but suppose the diameter of the circle were that of inner circle d, and one-quarter of it would still contain 90 degrees.
Fig. 56.
So, likewise, the degrees of one line to another are not always taken from one point, as from the point O, but from any one line to another. Thus the line marked 120 is 60 degrees from line 180, or line 90 is 60 degrees from line 150. Similarly in the other quarter of the circle 60 degrees are marked. This may be explained further by stating that the point O or zero may be situated at the point from which the degrees of angle are to be taken. Here it may be remarked that, to save writing the word "degrees," it is usual to place on the right and above the figures a small °, as is done in Figure 56, the 60° meaning sixty degrees, the °, of course, standing for degrees.
Fig. 57.
Suppose, then, we are given two lines, as a and b in Figure 57, and are required to find their angle one to the other. Then, if we have a protractor, we may apply it to the lines and see how many degrees of angle they contain. This word "contain" means how many degrees of angle there are between the lines, which, in the absence of a protractor, we may find by prolonging the lines until they meet in a point as at c. From this point as a centre we draw a circle D, passing through both lines a, b. All we now have to do is to find what part, or how much of the circumference, of the circle is enclosed within the two lines. In the example we find it is the one-twelfth part; hence the lines are 30 degrees apart, for, as the whole circle contains 360, then one-twelfth must contain 30, because 360÷12 = 30.
Fig. 58.
If we have three lines, as lines A B and C in Figure 58, we may find their angles one to the other by projecting or prolonging the lines until they meet as at points D, E, and F, and use these points as the centres wherefrom to mark circles as G, H, and I. Then, from circle H, we may, by dividing it, obtain the angle of A to B or of B to A. By dividing circle I we may obtain the angle of A to C or of C to A, and by dividing circle G we may obtain the angle of B to C or of C to B.
Fig. 59.
Fig. 60.
It may happen, and, indeed, generally will do so, that the first attempt will not succeed, because the distance between the lines measured, or the arc of the circle, will not divide the circle without having the last division either too long or too short, in which case the circle may be divided as follows: The compasses set to its radius, or half its diameter, will divide the circle into 6 equal divisions, and each of these divisions will contain 60 degrees of angle, because 360 (the number of degrees in the whole circle) ÷6 (the number of divisions) = 60, the number of degrees in each division. We may, therefore, subdivide as many of the divisions as are necessary for the two lines whose degrees of angle are to be found. Thus, in Figure 59, are two lines, C, D, and it is required to find their angle one to the other. The circle is divided into six divisions, marked respectively from 1 to 6, the division being made from the intersection of line C with the circle. As both lines fall within less than a division, we subdivide that division as by arcs a, b, which divide it into three equal divisions, of which the lines occupy one division. Hence, it is clear that they are at an angle of 20 degrees, because twenty is one-third of sixty. When the number of degrees of angle between two lines is less than 90, the lines are said to form an acute angle one to the other, but when they are at more than 90 degrees of angle they are said to form an obtuse angle. Thus, in Figure 60, A and C are at an acute angle, while B and C are at an obtuse angle. F and G form an acute angle one to the other, as also do G and B, while H and A are at an obtuse angle. Between I and J there are 90 degrees of angle; hence they form neither an acute nor an obtuse angle, but what is termed a right-angle, or an angle of 90 degrees. E and B are at an obtuse angle. Thus it will be perceived that it is the amount of inclination of one line to another that determines its angle, irrespective of the positions of the lines, with respect to the circle.
; | 2,421 | 9,921 | {"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} | 4.625 | 5 | CC-MAIN-2024-38 | latest | en | 0.946527 |
https://math.stackexchange.com/questions/4391833/the-meaning-of-the-indefinite-integral-symbol-the-definition-of-an-antiderivativ | 1,720,857,387,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00553.warc.gz | 327,120,350 | 41,394 | # The meaning of the indefinite integral symbol the definition of an antiderivative
I have been thinking that the symbol $$\int f(x) dx$$ is a variable as a whole, referring to $$F(x)$$, where $$F$$ is some antiderivative of $$f$$. It can refer to any function that is an antiderivative of $$f$$, evaluated at $$x$$. So $$A =$$ the symbol means $$A$$ is equal to some function evaluated at $$x$$, and that function is some unknown antiderivative of $$f$$.
Does this symbol mean above? Am I correct?
Also,
What is the precise definition of an antiderivative of a function? What is the definition that is most commonly accepted? What is the one that I write, others will know what I am referring to? Is there one?
I need to know it because the symbol is based on an “antiderivative”.
According to Wikipedia, an antiderivative of $$f$$ is $$F$$ such that $$F' = f$$. And two functions are equal iff their domains are the same and outputs are the same for each input from the domain.
By that definition, $$F: \mathbb{R} \rightarrow \mathbb{R}$$, $$F(x) = C$$ is not an antiderivative of $$f: (0,1) \rightarrow \mathbb{R}$$, $$f(x) = 0.$$
Is this also correct? Is it the commonly accepted definition of an antiderivative?
• That's correct. $F$ is an antiderivative of $F' : \Bbb R \to \Bbb R$, $x \mapsto 0$, and in particular not to any function $(0, 1) \to \Bbb R$. It's still true, of course, that $\int_a^b f \,dx = F(b) - F(a)$ for any $a, b \in (0, 1)$, simply because $(F \vert_{(0, 1)})' = f$. Commented Feb 26, 2022 at 20:19
When we write$$\int f(x)\,\mathrm dx=F(x),$$what that means is that $$F$$ is an antiderivative of $$f$$, that is, it is a function with the same domain as $$f$$ such that $$F'=f$$. If it turns out that the domain of $$f$$ is an interval (with more than one point) then every antiderivative of $$f$$ will be equal to $$F+K$$, for some constant $$K$$.
And, indeed, if you have $$f\colon(0,1)\longrightarrow\Bbb R$$ defined by $$f(x)=0$$, then no constant function $$F\colon\Bbb R\longrightarrow\Bbb R$$ is an antiderivative of $$f$$, since they have distinct domains. However, the restriction of $$F$$ to $$(0,1)$$ is an antiderivative of $$f$$.
• So this F' = f is the common definition of an antiderivative of a function? And F'(x)= f(x) for all x in the domain of f is not? When I say an antiderivative of a function, will most people know what I am talking about (referring to the first definition)?
– TFR
Commented Feb 26, 2022 at 20:26
• The assertion $F'=f$ means that $F'$ and $f$ have the same domain $D$ and that, for each $x\in D$, $F'(x)=f(x)$. Commented Feb 26, 2022 at 20:58
• Now, what is F’? Is F’ a function whose domain is the set of values on which F’s derivative exist, and is equal to the derivative of F for all values in its domain. Or, is it a function whose domain is the same as F and equal to the derivative of F at every value (in which case F’ exist iff F is differentiable at every point).
– TFR
Commented Feb 27, 2022 at 17:05
• You did not clarify what is an antiderivative.
– TFR
Commented Feb 27, 2022 at 17:07
• Indeed I did not (but now I've added it). Since it is a standard mathematical concept and since you used it several times in your question without asking for a definition, I didn't think that that would be a problem. I also did not define the meaning of, say, “interval” or “domain”. Do you think that I should add those definitions too? Commented Feb 27, 2022 at 17:21
The statement $$\int f(x)\,\mathrm dx=F(x)$$ is a really interesting one. In my opinion, it's a good example of context-dependent notation: it doesn't mean the same type of thing in different scenarios.
Let's use a simple example for better intuition: $$\int 2x\,\mathrm dx=x^2+c$$
What is the right-hand side? Is it an expression? A family of expressions? Well, let's say my original question was:
Find $$\int 2x\,\mathrm dx\ \ \ \$$ (1 mark)
Here, the meaning of the question is "find the functions ($$\mathbb{R}\rightarrow \mathbb{R}$$) that have derivative $$2x$$". And I've found them all: it's the set $$\{f\in \mathbb{R}^{\mathbb{R}}:\exists c\in\mathbb{R}:\forall x\in\mathbb{R}:f(x)=x^2+c\}$$ i.e. the set of functions $$f:\mathbb{R}\rightarrow\mathbb{R}$$ of the form $$f(x)=x^2+c$$ (for some $$c\in\mathbb{R}$$).
So $$x^2+c$$ is not an expression, but a set of functions. And so (for this to be "equal to" the left-hand side) the indefinite integral is a set of functions.
But what if my original question was:
Given that $$\frac{\mathrm dy}{\mathrm dx}=2x$$ and $$y=f(x)$$ and $$f(0)=3$$, find $$f$$.$$\ \ \ \ \ \$$ (2 marks)
Now, when I write, $$f(x)=\int 2x\,\mathrm dx=x^2+c$$ it is the first step in a two-step solution (the second step is $$0^2+c=3\implies c=3\implies f(x)=x^2+3$$).
Here, the first step is the statement "for some specific and knowable value of $$c$$, and all real values $$x$$, the function $$f$$ satisfies $$f(x)=x^2+c$$".
It's not a statement about a set of functions, because I'm looking for one specific function. Suddenly, the indefinite integral is a single function: a specific antiderivative of $$2x$$ called $$f$$.
The indefinite integral sign, and surrounding presentation of calculations in calculus, is an example of abuse of notation: either there is no formal definition of a symbol, or the symbol is commonly used to indicate a very clear meaning, but one that violates the definition.
Moreover, functions are a common example of abuse of definitions: something isn't a function if it doesn't have both a domain and a codomain. So "$$f(x)=x^2$$ and $$x>0$$", which you might commonly see as a definition of the "function" $$f$$, is not a function, because I don't know whether the codomain of $$f$$ is $$f(\mathbb{R}^+)=\mathbb{R}^+$$, or $$\mathbb{R}$$, or even $$\mathbb{C}$$. Sometimes you don't even see the possible input values specified. And sometimes you see authors use definitions of functions where the image is irrelevant, and two functions are equal iff they have the same domain and the same value on every input in the domain.
Hopefully, the reason for all of this is clear: mathematical notation, like all forms of language, is primarily designed for easy communication. If there is no (reasonable) way to misinterpret the bunch of symbols you write on the page, then you are using them correctly.
• So it seems like I was correct in the first part of my question. This symbol in a statement from some proof means some “anti-derivative” evaluated at x of a function. And in formula books or problems, it mean the set of “anti-derivatives” of functions.
– TFR
Commented Feb 26, 2022 at 20:49
• Then what is the precise definition of an “antiderivative”? Is it the same as I wrote in the question?
– TFR
Commented Feb 26, 2022 at 20:50
• @TFR an antiderivative of the function $f$ is exactly what you wrote, a function $F$ such that $F'=f$. This entails $F$ and $f$ having the same domain and range (depending on your definition of function). $F$ is never unique, as $x\mapsto F(x)+c$ is another antiderivative for any $c\in\mathbb{R}$. Or, in my language, an antiderivative is a member of the set of functions $\int f(x)\, \mathrm dx$.
– A.M.
Commented Feb 26, 2022 at 20:53
• What does F have the same domain as f? F could have a domain larger than f’s but not differentiable except on values in the domain of f, so that F’ = f.
– TFR
Commented Feb 26, 2022 at 20:58
• @TFR $F'=f$ is a statement that the functions are equal. They're not equal without the same domain. In a common abuse of notation, $F'=f$ might mean what you say, but this is not what $=$ formally means.
– A.M.
Commented Feb 26, 2022 at 21:37
To directly answer your question, $$F$$ is an antiderivative of $$f$$ if and only if $$F' = f$$. In this case, we can write $$F(x) = \int f(x) \ \text dx$$ This is the most common (and arguably, the only reasonable) definition of the word.
However, I think the source of confusion here is not your understanding of antiderivatives but that of functions. We like to throw around symbols and write things like $$f = g$$ because it's concise and precise, but it's not exactly clear without some prior understanding. When we define functions like $$f: X\to Y \\ f: x\mapsto y$$ we're saying a lot of different things. First, there are sets $$X$$ and $$Y$$, called the domain and codomain, respectively. Then, there is a rule that associates to each $$x\in X$$ a particular $$y\in Y$$ (usually written as an expression involving $$x$$). We can think of the function as a triple: (domain, codomain, rule). Finally, $$f$$ is simply the name given to that function.
The upshot of all this is that a statement like $$f = g$$ means that the whole triple of (domain, codomain, rule) is the same between $$f$$ and $$g$$. Importantly, if $$f$$ and $$g$$ have different domains, then they are not the same function, even if they agree everywhere their domains overlap. This is why the statement at the end of your question is correct; $$F'(x) = f(x)$$ for $$x\in (0,1)$$, but $$F'$$ and $$f$$ have different domains and so are different functions.
• I understand the meaning of f = g. What I wanted to be sure about is whether that is the definition (which is according to many). However, someone said F’ = g implies F’s domain is the same as f because F’ is a function whose domain is the same as f. What I have been thinking is that F’ is a function defined whenever F is differentiable, so that their domains might be different. The reason for this is because I saw the derivative of F at x is often written as F’(x), whether F is a differentiable function (differentiable at every point in domain) or not.
– TFR
Commented Feb 27, 2022 at 16:39
• @TFR It's certainly true that $F$ and $F'$ may have different domains, in fact this is quite typical. However, the equation $F' = f$ makes no direct reference to the domain of $F$; regardless of whether $F'$ and $F$ share a domain, $F' = f$ requires that $F'$ and $f$ do. Commented Feb 27, 2022 at 17:45
• This is what I thought, F can have F’ when F is not a differentiable function. However, this is different from the other two answers which require F’ has the same domain as F.
– TFR
Commented Feb 27, 2022 at 20:08 | 2,890 | 10,209 | {"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": 84, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.916539 |
https://www.slideserve.com/saeran/7-2-the-substitution-method-powerpoint-ppt-presentation | 1,638,551,461,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362891.54/warc/CC-MAIN-20211203151849-20211203181849-00444.warc.gz | 1,086,113,631 | 18,781 | Download
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7.2 The Substitution Method
# 7.2 The Substitution Method
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## 7.2 The Substitution Method
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. 7.2 The Substitution Method Objective: Find an exact solution to a system of linear equations by using the substitution method. Standard Addressed: 2.8.8.E: Select an use a strategy to solve equations.
2. If you know the value of one variable in a system of equations, you can find the solution for the system by substituting the known value of the variable into one of the equations. This method is called substitution.
3. Ex. 1 Solve by using substitution: b. • 2x – 4 (2) = 1 • 2x – 8 = 1 • 2x = 9 • X = 4.5 • (4.5, 2)
4. Ex. 2a. • 4x + 5 (4x) = -24 • 4x + 20x = -24 • 24x = -24 • X = -1 • Y = 4 (-1) • Y = -4 • Solution is (-1, -4)
5. Ex. 3 b. Solve by substitution. • Solve first EQ for X x = -6y + 1 • Substitute above EQ into the 2nd EQ • 3(-6y + 1) - 10y = 31 • -18y + 3 – 10y = 31 • -28 y + 3 = 31 • -28y = 28 • Y = -1 • Find X by substituting y = -1 into one of your above EQ. • X = -6 (-1) + 1 • X = 7 • Solution is (7, -1)
6. b. Sam sells T-shirts at a baseball park. He has 50 of last year’s T-shirts and 200 of this year’s T-shirts in stock. He knows that his customers will pay \$5 more for this year’s T-shirt. He needs to make a total of \$3750 from T-shirt sales. How much should he charge for each type? • 50N + 200m = 3750 and m = n + 5 50n + 200(n + 5) = 3750 50n + 200n + 1000 = 3750 250n = 2750 N = 11 last year’s T-shirt 11 + 5 = M M = 16 for this year’s T-shirt | 610 | 1,687 | {"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} | 4.6875 | 5 | CC-MAIN-2021-49 | latest | en | 0.826136 |
https://community.smartsheet.com/discussion/122410/survey-result-dashboard-countif-contains-formula | 1,718,308,791,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00114.warc.gz | 165,152,236 | 107,047 | # Survey Result Dashboard COUNTIF CONTAINS formula
Options
I created a survey (form) where guests of an event can rate location, food, keynote speakers etc. from 1 to 5 (1 being very bad and 5 being very good). To be able to create a widget in the dashboard that displays an average score I first created a stats sheet that collects the entries from the form sheet.
I couldn't get this in an all-in-one formula, I'm no Smartsheet expert and I suffer from LongCovid. Because the brain fog is terrible sometimes I gave up and tried a rookie approach by creating simple formulas in several colums of the stats sheet calculating the number of times a guest selected a value (5 Zeer goed, 4 Goed etc..). Zeer goed is Dutch for very good with this formula:
=COUNTIFS({751 Enquete Uitnodigingstraject}; ="5 Zeer goed")
I added a column next to it that calculates this outcome x the score, in this case 5. I do this for all 5 ratings (5,4,3,2,1) and created a total column hoping I could find a formula to devide the total score with the number of guest that filled out the survey.
So I therefor created an auto-number column in the survey result form sheet that displays a row-ID with a prefix. This way I was hoping to be able to calculate the number of rows by counting the prefix in this row-ID column and use this in the stats sheet by dividing the total score with the number of rows in the survey sheet.
This is where I have been stranded. Apparently, to count the identical prefix of the row ID (aka count the number of rows in the survey form sheet) this formula isn't working.
=COUNTIFS({751 Enquete Row ID}; CONTAINS("751-210524"))
But I'm also having trouble getting the right formula to be able to divide the total score with the number of rows in the survey sheet.
Help is much appreciated.
• ✭✭✭✭✭✭
Options
Hi @MikeChapNL — to fix the COUNTIFS, update the formula to include the @cell reference and change the semi-colon to a comma (unless the semi-colon is a convention specific to the Netherlands instance of Smartsheet—in the States, we use a comma to separate the the fields). Anyway, you will need the @cell reference — let me know if it works with the semi-colon, just so I know! Here it is:
`=COUNTIFS({751 Enquete Row ID}, CONTAINS("751-210524", @cell))`
• ✭✭✭✭✭✭
Options
Hi @MikeChapNL — to fix the COUNTIFS, update the formula to include the @cell reference and change the semi-colon to a comma (unless the semi-colon is a convention specific to the Netherlands instance of Smartsheet—in the States, we use a comma to separate the the fields). Anyway, you will need the @cell reference — let me know if it works with the semi-colon, just so I know! Here it is:
`=COUNTIFS({751 Enquete Row ID}, CONTAINS("751-210524", @cell))`
• ✭✭✭✭✭✭
Options
Also, I'm sorry about the long-Covid issues! But I think it's a better practice to break up your formulas anyway. It's easier to fix things in the future, and other people will be able to figure out what you did.
• Options
Hi Lucas,
Thank you!!!!
Kind regards,
Michael
## Help Article Resources
Want to practice working with formulas directly in Smartsheet?
Check out the Formula Handbook template! | 788 | 3,179 | {"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-2024-26 | longest | en | 0.909903 |
https://fr.mathworks.com/matlabcentral/answers/293259-plotting-data-on-curvilinear-coord-projection | 1,686,104,950,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00599.warc.gz | 294,977,725 | 28,576 | # Plotting data on curvilinear coord projection
8 views (last 30 days)
mashtine on 1 Jul 2016
Hello,
I have two lat and lon matrices, each 622x810 and they are of a curvilinear projection. I previously just used vectors of lon and lat to plot my data with surfacem (see below) but I am not sure how to do this with lat and lon as a matrix of curvilinear coordinates.
Any suggestions?
ax = worldmap(latlim, lonlim);
surfacem(lat, lon, inpdata)
geoshow([S.Lat], [S.Lon],'Color','black');
KSSV on 1 Jul 2016
As you have matrices in hand, it shall be very easy to plot what you want. If you want to get vectors of lon, lat from 622x810 matrices of each, try using unique(); this will give you the vectors.
##### 2 CommentsShow 1 older commentHide 1 older comment
Is ther anyway to convert the covarian projection (meshgird) into noramal Longitude and latitudes?
José-Luis on 1 Jul 2016
It is not clear to me how you want those arrays displayed. If you only want to show the points:
geoshow(S.Lat(:), S.Lon(:),'Color','black');
##### 2 CommentsShow 1 older commentHide 1 older comment
José-Luis on 4 Jul 2016
I don't know what a matrix of curvilinear coordinates is.
Jonathan Eliashiv on 18 Oct 2016
Edited: Jonathan Eliashiv on 18 Oct 2016
Super simple actually.
X_curvi = reshape(lon,[],1);
Y_curvi = reshape(lat,[],1);
data_curvi = reshape(inpdata,[],1);
Make a meshgrid that you want to interpolate into:
[lon_grid,lat_grid] = meshgrid(minlon:dx:maxlon,minlat:dy:maxlat)
and then you can use the griddata function:
[~,~,data_rectilinear] = griddata(X_curvi,Y_curvi,data_curvi,...
lon_grid,lat_grid)
and plot away
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Translated by | 502 | 1,851 | {"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-2023-23 | longest | en | 0.801916 |
http://blog.forret.com/2007/09/28/two-dice-make-a-calendar/ | 1,503,039,785,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00151.warc.gz | 51,488,753 | 10,315 | # Two dice make a calendar
Little riddle: if you have two dice (so each has 6 sides), with 1 digit on each side and you want to be able to form all numbers between 1 and 31 (for the days of the month), what digits would be on each dice? E.g. if dice #1 has 0-1-2-3-4-5 and the second 4-5-6-7-8-9, it won’t work, because you cannot form the number ’22’.
Try to guess and I’ll give the answer next week. There’s a little trick involved! I first tried to figure it out without looking at the dice, and I enjoyed the mind gymnastics.
### 23 thoughts on “Two dice make a calendar”
1. Pascal
(for the non-Dutch speakers… I meant: “done it without looking at dice and without writing anything down”)
2. Clopin
Hmm, one question though: is it allowed to leave dice out from time to time? Because I could have a problem in the first 10 days 😉
3. Inferis
@Clopin: I struggled with that too, but the riddle says “1 to 31”, so I assumed it was okay to represent the first 9 days using 1 dice.
4. Pascal
@Clopin and @Inferis:
No, you need to have two dice, otherwise it wouldn’t be much fun…
So: you need to have “0” on one or both dices.
Oh, and may I point out that the digits are Arabic, and not represented by numbers of dots?
5. Stijn
*spoiler warning*
I think I got it right. (:
Dice A: zero-one-two-three-seven-eight
Dice B: zero-one-two-four-five-six
The six can be used as a nine as well.
6. Brian
With the 123456 and 789120 solution, the 1 in either die can be used as a “dash” so that two die could still be used to represent the first 9 days. That is: -1, -2, -3, etc.
7. Srinu
Dice 1 : 0 1 2 3 4 5
Dice 2 : 0 1 2 6 7 8
This is exact solution for this question.. | 498 | 1,683 | {"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-2017-34 | latest | en | 0.94462 |
https://www.jiskha.com/questions/560590/convert-0-75-to-a-decimal | 1,597,315,184,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738982.70/warc/CC-MAIN-20200813103121-20200813133121-00232.warc.gz | 706,657,461 | 5,499 | # pre-algebra
Convert 0.75% to a decimal
1. 👍 0
2. 👎 0
3. 👁 98
1. To convert a percent to a decimal, divide by 100 or move the decimal point two places to the left.
0.75% = 0.0075
1. 👍 0
2. 👎 0
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asked by Tobi on October 3, 2014 | 937 | 2,826 | {"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} | 3.90625 | 4 | CC-MAIN-2020-34 | latest | en | 0.926731 |
https://www.physicsforums.com/threads/control-engineering-block-diagram.527564/ | 1,531,969,279,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590443.0/warc/CC-MAIN-20180719012155-20180719032155-00449.warc.gz | 941,136,155 | 13,814 | # Control Engineering: Block Diagram
1. Sep 6, 2011
### wantan7671
Hi guys, this is quite an inbetween question , not really homework but still theory.
http://img850.imageshack.us/img850/1290/81696653.jpg [Broken]
Came across this diagram and was wondering if the arrow (in blue box) can be ignored ? Quite irritated by it now since I can't be sure how I should create the Transfer Function or obtain the Control Canonical Form.
Any help would be fantastic !!
Last edited by a moderator: May 5, 2017
2. Sep 6, 2011
### Floid
There is no way of knowing without more context. There might be some unknown gain term there... but without the problem statement for which that block diagram describes who knows.
3. Sep 6, 2011
### wantan7671
the question is just to find the transfer function for that block diagram. I drew the blue box myself. Just wondering if it's actually valid to have a drawing like this with "loose ends"
4. Sep 6, 2011
### jim hardy
"Just wondering if it's actually valid to have a drawing like this with "loose ends" "
Is it really a loose end? Looks to me like input to 2nd integrator stage.
in my day it would have been fine.
Were one building an analog computer that'd be a place one could hook a recorder.
I recognize that as a bear to analyze
would be lots quicker to build with opamps.
old jim
5. Sep 6, 2011
### wantan7671
Thanks for the insight Jim. So for now, if nothing is connected there, can I assume the entire block only has U (input) and Y (output) points ? And obtain the Transfer Function directly? (or is there another technique for these sort of things?)
6. Sep 7, 2011
### jim hardy
I think so. It was mid 1960's when i took controls course, so i am really rusty and shudder at the thought of all that algebra now.
But looking at that sketch it seems to me --
they could have drawn it as one horizontal chain
then the X2 line wouldn't have to cross X1 where X2 feeds back to sum with u, so it would look less formidable.
isn't it just two cascaded lags with output of 2nd fed back to input of first one?
i hope you'll post the transfer function.
Think it'll oscillate?
old jim | 543 | 2,147 | {"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} | 2.703125 | 3 | CC-MAIN-2018-30 | latest | en | 0.959715 |
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Close
# New 3D interactive graph
By Murray Bourne, 13 Apr 2016
Most students struggle when trying to understand 3D cartesian coordinates, especially when diagrams in text books are static, and in 2 dimensions only (of course).
I recently published a new 3D cartesian coordinates interactive graph:
3D Space Interactive Applet
In the introduction to 3D graphs, I gave the example of the point P (2, 3, 5) . The 3D interactive graph allows you to explore what a point in 3 dimensions means and helps to orientate those unfamiliar with the xyz-coordinate system.
You can vary the x-, y- and z-coordinate of the given point and rotate the axes, as well as zoom in and out.
This is quite a simple interactive and I hope to introduce more involved ones in the coming months.
I developed the applet using Three.js, a 3D javascript library. As such, the applet is cross-platform, cross-browser and mobile friendly.
## xyz axes orientation?
When I first wrote the section on 3D Space in the Vectors chapter, I wanted to ensure I chose a "good" orientation for the 3 axes. By "good", I mean that I was using a proper convention and that it was commonly used.
For interest, I looked at what Google Images returned for a search on xyz axes. I'm using the results of that search below.
When I first learned 3D geometry in secondary school, we started with the familiar xy-axes and drew the new z-axis coming "out of the page" towards us, like this (let's call it Orientation #1):
At times we would see the axes rotated around the y-axis, like this one (let's call it Orientation #2):
But that was OK - the y-axis was still at the top, so we didn't get lost.
When I met the idea again at university, the following convention was used. The z-axis was vertical, with the x-axis going to the left (and down-ish), and the y-axis going to the right (also down-ish), like this:
This change of orientation was quite disconcerting at first. I kept drawing points, planes and graphs the wrong way round.
Now I'm more used to this orientation (let's call it Orientation #3) , and it seems that it's more common to have the z-axis going up.
You'll also see it with the x- and y-axes rotate 90°, like this (let's call it Orientation #4):
### Right-hand Rule
Each of the above orientations obeys the right hand rule, whereby if our right-hand index finger points in the direction of the x-axis, and our second finger points in the direction of the y-axis, then our thumb will point to the z-index.
(Yes, you have to contort your hand around somewhat to ensure each of the above orientations work, but they do.)
The following one (found in Google's index of such axes) fails the right-hand rule:
(To be fair, it is possible the z-axis is coming out of the page towards us, but that's not the impression I get.)
### Is a vertical z-axis more common?
The calculus text books by Kreyszig, Washington, Mizrahi/Sullivan and Thomas/Finney all use Orientation #3, which is why I used that orientation as well.
Wolfram Alpha's default 3D view is the same as Orientation #4 (where the positive x-axis is to the right and down, and the y-axis is to the right and up.)
Here's the curve z = x2 + sin(y) by Wolfram|Alpha:
Geogebra has the same default orientation. Here's the same graph using Geogebra, which doesn't appear to have an option for naming the axes (it does for 2D axes).
Up next is Scientific Notebook, which uses x to the right and down, and y-axis to the left and down, with z up.
This means it fails the right hand rule! At least the axes are labelled.
Matlab's plot on first glance appears to be the same as Wolfram|Alpha's, but in fact, the positive y-axis is going off to the left and up.
It obeys the right hand rule, but this is not a common orientation.
## Why does it matter?
As I mentioned earlier, a lot of novice learners have difficulty imagining planes and other surfaces in 3D. Presenting graphs in a myriad of orientations doesn't help.
## Suggestions for teachers
1. The first examples of 3D graphs that students see should use a consistent orientation, as far as possible (of course, we need to turn things around a bit to see key features, but keep such turns minimal)
2. Slowly introduce other orientations, and point out clearly the features of the new orientation and why you are using it
3. Always label the axes and insist students do so as well. If you are using one orientation, and they understand it to be another, confusion is sure to follow
4. Eventually give examples of all the possible 3D axes orientations, so students can figure out the differences.
The link to the interactive graph again:
3D Space Interactive Applet
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From Math Blogs | 1,242 | 5,235 | {"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": 0, "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} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.945631 |
https://electronics.stackexchange.com/questions/601917/simplest-current-limiting-method-for-battery-charging | 1,656,794,233,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00092.warc.gz | 281,319,756 | 69,208 | # Simplest current limiting method for battery charging
I need to charge 12V car battery (from main battery), but I have to limit current, because power cables are quite thin and I don't want to draw too much power from main system (in case battery is empty). What would be simplest solution (without ineffective linear regulators)? I though about PWM controlled LC circuit, but maybe there are controllers (ICs) made for that? I would like to charge the battery at no more than 3A.
Those cheap chinese converters have current limiting option through potentiometer, but I want something less delicate.
• What batteries are those and what voltages they have? Dec 27, 2021 at 9:03
• regular lead car batteries Dec 27, 2021 at 9:04
• Given your requirements stated (which are in no way sufficient to make a better judgement), a 1 Mohm resistor would likely suffice. Might even get away with something much smaller in value too. Dec 27, 2021 at 9:04
• well a resistor (1 Ohm) would limit the current, but it would also wouldn't be very efficient. I could get ~18W of charging power and ~4W of losses. But it very depends on what is batteries internal resistance. Dec 27, 2021 at 9:09
• Connect a car headlight bulb in series - 55W @ 12V is 4.5A so close enough. Used bulbs like that after rewiring cars so any fatal shorts weren't... Dec 27, 2021 at 11:37
## 3 Answers
Let’s see your wish-list:
1. Charge a 12V car battery from the “main battery”. <=> Assumed here the main battery is the battery connected to the car starter engine and alternator.
2. Use of thin cables, to not draw to much power in case “aux” battery is empty. Here is a problem, as thin cables should not be used to present a high resistance to limit the current. This is wrong and dangerous! More on that later.
3. Simplest solution without inefficient linear regulators. <=> Linear is inefficient, but are simple and can be very robust.
4. Something less delicate than “Chinese converters”.
My thoughts of what you will need:
1. Charging/equalizing cables compatible with the maximum current expected to charge the Aux-12V battery. Surely anything of at least of 4 mm^2 or 12AWG, for at least 20A and a couple of meters long, but 6 mm^2 or 10AWG is good up to 30A; and 8AWG goes up to 40A safely, without overheating. And please, do protect the Positive cable(s) with Fuse(s) rated accordingly.
2. Switching device: For something Robust and Fool-proof, use 12V relays as switching elements; they are more forgiving than bipolar or mosfet transistors. Auxiliary relays can operate 30 to 40A, so you may be ok if you use a current limiting feature (later). You will need two relays, eventually a third one.
3. Current limiting circuit: The simplest and a robust solution is to use headlight lamps as power resistors. A more elegant option is to use sensing resistors (0.6~0.7V of voltage drop at max. current) monitored by a driver transistor to control a series-pass power transistor, heatsinked.This is essentially a current limit, but causes a minimum voltage drop of about 1.0V.
4. A Control circuit, to measure voltage differential between batteries and absolute voltage in Aux-Batt, and act according to these voltages. For example: (A) If voltage differential is low enough, the current-limit circuit (or lamps) could be bypassed by above Aux-relays, enabling a complete charge and in shorter time. (B) If Aux-Batt voltage is too low, another Aux-relay could disconnect the battery. (C) If Aux-Batt is fully-charged, a third Aux-Relay could disconnect the charging circuit. Hysteresis and interlocking logic could be used to prevent erratic operation of relays.
Finally, I suggest you to check if a dedicated VSR - Voltage Sensitive Relay as those used in trucks and motor-homes could solve your needs. I believe it does. Here are two links of different brands - a VSR only or as a VSR kit. I found also a good reference about Split Charging Methods - here. See below a typical VSR assembly diagram:
Let us know what you would prefer to do - use a Split charging system as those VSR-based, or to make and tinker your own system “VSR-like”, where the above list is a reminder.
Don't reinvent the wheel.
They sell a ready-made solution: 12 V to 12 V battery charger.
https://www.google.com/search?client=firefox-b-1-d&q=12+V+to+12+V+battery+charger
• why not? May be the newly invented wheel is rounder. And how do those devices you have mentioned work? Dec 29, 2021 at 17:22
• the charger you suggested has "Input Voltage 110-240V AC" Dec 31, 2021 at 11:24
• 12V to 12V would be perfect for me, but I did not find anything like that Dec 31, 2021 at 11:24
• They definitely exist. Boats and RV's use them. Here is a victron product that might work for you: victronenergy.com/dc-dc-converters/orion-tr-smart Jan 1 at 8:14
• Here are some other useful search terms for the OP: ACR (automatic charging relay). VSR (voltage sensing relay). Battery Combiner. Jan 1 at 8:17
What about a 2-transistor constant current sink? Use an NPN as the control transistor. If you put in a 0.2 Ohm resistor, the NPN will choke the pass transistor at about 3-4 A. If current is lower, the pass transistor will be fully on and can be a low resistance N-MOSFET.
This is probably a little bit more efficient than using only a resistor of around 1-2 ohm.
However, I think the 55W, 12 V bulb in series as suggested by @SolarMike is actually a more elegant way to achieve similar behavior. The transistor version allows setting the current limiter more deliberately, though. | 1,390 | 5,536 | {"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} | 2.828125 | 3 | CC-MAIN-2022-27 | longest | en | 0.962835 |
https://www.physicsforums.com/threads/speed-of-electron-accelerated-by-a-potential-difference.768691/ | 1,508,328,820,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822930.13/warc/CC-MAIN-20171018104813-20171018124813-00097.warc.gz | 984,376,521 | 17,474 | # Speed of electron accelerated by a potential difference
1. Sep 2, 2014
### Brianrofl
1. The problem statement, all variables and given/known data
The full problem can be seen here - however, I only need help with one part: http://puu.sh/biN9W/ee2a7bf393.png [Broken]
I'm not sure how to find the velocity of a particle when accelerated to the point that it exceeds or approaches the speed of light when the classical equations are used. I'd really appreciate if someone could help me with it.
2. Relevant equations
K = (γ-1)mc^2
P = γmv
Mass of electron: .511MeV/c^2 or 9.11*10^-31kg
3. The attempt at a solution
I've got 5+ pages of scratch work scribble but I've tried many things. I've tried just using classical equations, which nets a velocity greater than the speed of light, so that can't work. I've tried finding the value of gamma from K = (γ-1)mc^2 and using that value of γ to find a velocity, but that didn't give me the correct answer either.
An explanation or tips would really be appreciated. Also, another problem that I've been having a lot of trouble with can be seen here http://puu.sh/biNDy/4bdaa9d2b3.png [Broken]. This one really racks my brain. For this one, I've had two thoughts:
1. The velocity needed to travel the distance in a certain time
and
2. The velocity needed to dilate time to a specific number
and it's this that kills me. I feel like, as #1 changes, #2 also changes. I've tried making two equations - one setting v equal to the distance of the planet divided by time, and another one setting time equal to the time dilation equation (ΔT/sqrt(1-u^2/c^2))
Any tips for this one would also be appreciated!
Last edited by a moderator: May 6, 2017
2. Sep 2, 2014
### nrqed
For the first one, did you try simply setting the relativistic KE equal to Q times the difference of potential?
For the second, you have to pick a frame. For example, working in the frame of the spaceship, the time interval is 11 years and the distance travelled will be the round trip distance as measured on Earth divided by the gamma factor. You set that up and solve for the speed. (Edit: by that I mean: you set speed = distance / time, using the values in the astronaut`s frame)
Last edited by a moderator: May 6, 2017
3. Sep 2, 2014
### Brianrofl
Ok, I'll let you know how it goes.
Edit: as for the first idea, Q * V turns out to be equal to 48MeV
However, I found out something I was doing wrong with the equation. When I substituted e = .511MeV/c^2 into the mass m in (γ-1)mc^2, I didn't include the /c^2. This was making my gamma value ridiculously small.
After fixing that I got a gamma value of 28.397 (note: I'm doing a practice problem where V = 14MV), and then I used the equation
P = γmv where I put in (28.397)(.511MeV/c^2)(.9994c) and I get 14.50MeV/c - which is the right answer so looks like I've solved this one.
I'll try out the astronaut problem again in a few minutes.
Last edited: Sep 2, 2014 | 801 | 2,952 | {"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} | 3.3125 | 3 | CC-MAIN-2017-43 | longest | en | 0.952783 |
https://socratic.org/questions/how-do-you-simplify-tan-80-tan-55-1-tan-80-tan-55 | 1,713,995,109,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00837.warc.gz | 458,618,229 | 5,923 | # How do you simplify (tan 80° + tan 55°)/(1 – tan 80° tan 55°)?
Use the trig identity: $\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a . \tan b}$
$\frac{\tan 80 + \tan 55}{1 - \tan 80. \tan 55} = \tan \left(80 + 55\right) = \tan {135}^{\circ}$ | 112 | 258 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 2, "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} | 4.0625 | 4 | CC-MAIN-2024-18 | latest | en | 0.223395 |
https://www.clubspartacus.be/control/9536/aggregates-required-in-1m3-concrete-of-m30.html | 1,563,786,747,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527828.69/warc/CC-MAIN-20190722072309-20190722094309-00559.warc.gz | 654,998,259 | 6,329 | ### aggregates required in 1m3 concrete of m30
Calculate Quantities of Materials for Concrete -Cement , aggregates required in 1m3 concrete of m30 ,Calculate Quantities of Materials for Concrete -Cement, Sand, Aggregates Quantities of materials for concrete such as cement, sand and aggregates for production of required quantity of concrete of given mix proportions such as 1:2:4 (M15,1:15: 3 (M20,1:1:2 (M25) can be calculated by absolute volume methodwater for 1m3 concrete for m30 grade - geetaschoolinaggregate required for 1m3 concrete m30 - gyptechin water for 1m3 concrete for m30 grade - Hence 200 Litres of Water is required for 1m3 of M20 sand and aggregates in 1 m3 of M30 grade concrete ratio of cement [More info] calculate the rates for 1m3 concrete grade 20.
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• what weight is 1m3 of concrete - hindimastercoin
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An Experimental work was performed to determine the compressive strength of recycled coarse aggregate concrete and to compare them with those of concrete made using natural coarse aggregate , Table 2Mix proportion for 1m3 M30 concrete , but replacement of 25% and 50% RCA gives required compressive strength however ,...
• Calculate the rates for 1m3 concrete grade 20 (1:2:4 ,
Transcript of Calculate the rates for 1m3 concrete grade 20 (1:2:4) which Calculate the rates for 1m3 concrete grade 20 (1:2:4) which is mixed manually including lifting, compaction and leveling work by using the data given , caggregate / m3 rm 4000 IILabour constant : amixing concrete 300 hour/m3 ....
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• 1m3 aggregate in kg - ventaskgroupco
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• For m30 how much cement sand and aggregate required
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• sand for m25 mix for 1 m3 - linxpersonnelcoza
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• SUSTAINABLE USE OF RECYCLED AGGREGATE IN CONCRETE ,
An Experimental work was performed to determine the compressive strength of recycled coarse aggregate concrete and to compare them with those of concrete made using natural coarse aggregate , Table 2Mix proportion for 1m3 M30 concrete , but replacement of 25% and 50% RCA gives required compressive strength however ,... | 2,277 | 9,393 | {"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-2019-30 | longest | en | 0.814677 |
https://oeis.org/A254383 | 1,660,252,347,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00117.warc.gz | 400,663,960 | 3,536 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A254383 Number of (n+2)X(n+2) 0..1 arrays with every 3X3 subblock diagonal maximum plus antidiagonal maximum nondecreasing horizontally, vertically and ne-to-sw antidiagonally 1
512, 35529, 8533796, 6208345944, 14008701570284, 99661265319327258, 2165666188993340904078, 146977323440908911644985083, 30949376415040215031162843169020, 20149875590838758609564635874891385665 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Diagonal of A254390 LINKS R. H. Hardin, Table of n, a(n) for n = 1..13 EXAMPLE Some solutions for n=2 ..1..1..0..1....0..1..0..1....0..1..1..1....0..1..0..1....0..1..1..1 ..1..0..1..1....0..1..1..0....0..1..0..1....1..1..0..0....0..1..0..1 ..0..1..1..0....1..0..0..0....0..0..1..0....0..0..0..1....0..1..0..1 ..0..1..1..0....1..1..1..0....1..0..1..1....1..1..0..0....0..1..1..0 CROSSREFS Sequence in context: A253864 A308815 A253536 * A253977 A254084 A284033 Adjacent sequences: A254380 A254381 A254382 * A254384 A254385 A254386 KEYWORD nonn AUTHOR R. H. Hardin, Jan 29 2015 STATUS approved
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Last modified August 11 15:10 EDT 2022. Contains 356066 sequences. (Running on oeis4.) | 543 | 1,525 | {"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} | 3.0625 | 3 | CC-MAIN-2022-33 | latest | en | 0.578649 |
https://livewell.com/finance/after-tax-real-rate-of-return-definition-and-how-to-calculate-it/ | 1,708,839,376,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474581.68/warc/CC-MAIN-20240225035809-20240225065809-00897.warc.gz | 363,109,712 | 40,748 | Home>Finance>After-Tax Real Rate Of Return Definition And How To Calculate It
Finance
# After-Tax Real Rate Of Return Definition And How To Calculate It
Learn how to calculate the after-tax real rate of return in finance. Understand its definition and its importance in making financial decisions.
## Understanding After-Tax Real Rate of Return and How to Calculate It
When it comes to finance, understanding the after-tax real rate of return is crucial for making informed investment decisions. But what exactly does this term mean, and how can you calculate it? In this blog post, we will demystify the concept of after-tax real rate of return and provide you with a step-by-step guide on calculating it.
## Key Takeaways:
• The after-tax real rate of return measures the actual profit or loss on an investment after accounting for taxes and inflation.
• Calculating the after-tax real rate of return involves adjusting the nominal return for inflation and factoring in relevant taxes.
Before we dive into the nitty-gritty of calculations, let’s first understand what the after-tax real rate of return truly signifies. In simple terms, it tells you how much your investment is growing or shrinking in purchasing power after accounting for taxes and inflation. This important metric allows you to compare investments more accurately, considering the erosion of value caused by inflation and the impact of taxes on returns.
To calculate the after-tax real rate of return, follow these steps:
1. Start by determining the nominal return on your investment. This is the return before taxes and other deductions.
2. Next, account for inflation by subtracting the inflation rate from the nominal return. This adjustment gives you the real return.
3. Now, consider the impact of taxes on your investment. Determine your marginal tax rate, which is the highest tax rate you’re subject to based on your income bracket.
4. Calculate the after-tax return by multiplying the real return by (1 – marginal tax rate). This accounts for the taxes you need to pay on your investment gains.
Let’s look at an example to illustrate how to calculate the after-tax real rate of return:
Suppose you invest \$10,000 in a mutual fund, and after a year, it grows to \$11,500. The nominal return on your investment is \$1,500 (\$11,500 – \$10,000). Now, assume the inflation rate for that year was 3%. To calculate the real return, subtract the inflation rate from the nominal return:
Real return = \$1,500 – (\$1,500 x 0.03) = \$1,500 – \$45 = \$1,455
If your marginal tax rate is 25%, the after-tax return can be calculated as:
After-tax return = \$1,455 x (1 – 0.25) = \$1,455 x 0.75 = \$1,091.25
Therefore, the after-tax real rate of return on your investment is \$1,091.25, providing you with a more accurate picture of the growth in your purchasing power.
Now that you understand the process of calculating the after-tax real rate of return, you can use this metric to evaluate the true performance of your investments. By incorporating taxes and inflation into your assessment, you can make more informed financial decisions that align with your long-term goals.
In conclusion, the after-tax real rate of return is an essential tool for investors as it helps determine the actual growth or decline in purchasing power. By applying the steps outlined in this blog post, you will be able to calculate the after-tax real rate of return on your investments and gain valuable insights into their performance. Make informed decisions and take control of your financial future!
• https://livewell.com/finance/after-tax-real-rate-of-return-definition-and-how-to-calculate-it/ | 776 | 3,667 | {"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} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.885422 |
https://blender.stackexchange.com/questions/linked/50252?sort=votes | 1,714,004,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296820065.92/warc/CC-MAIN-20240425000826-20240425030826-00228.warc.gz | 116,929,803 | 33,110 | 23 questions linked to/from Can I "fill" an empty object?
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### Is there a way to fill in hollow objects?
I am trying to make a mold of the human airway. To do so I need to subtract the airway components from a block, then 3D print the mold. It appears that all components are hollow, is there a way to ... | 928 | 3,805 | {"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-2024-18 | latest | en | 0.929402 |
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Two particles are fixed to an x axis: particle 1 of charge q1 = 2.58 × 10-8 C at x = 22.0 cm and particle 2 of charge q2 = -3.24q1 at x = 69.0 cm.
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https://docgo.net/algebra | 1,511,354,743,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806586.6/warc/CC-MAIN-20171122122605-20171122142605-00600.warc.gz | 591,132,804 | 15,926 | # Algebra
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Algebra lesson 1.Complex numbers 2.Matrix Algebra 3.Vector Algebra
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Chapter 1 Complex Numbers 1.1 Introduction To have solutions for all equations bax = where ba , are integers, ,0 ≠ a the integers were extended to the rational numbers. To have a solution for ,2 2 = x we went outside the rational numbers to the irrational numbers. There is no real number x such that 1 2 −= x since the square of any nonzero real number must be positive. To have a solution for ,1 2 −= x mathematicians introduced the complex number i with the property .1 2 −= i We also use 1 − to denote i . Thus the equation 01 2 =+ x would have two roots . i ± Also, the square root of a negative number, , 2 b − is written in the form , bi where b is a real number. Furthermore, solutions of the equation ( ) 22 ba x −=− ( ba , both real numbers) are given by , bia x ±=− that is bia x ±= . Complex numbers made their appearance in the mathematical scene in The Great Art, or in the rules of Algebra (1545) by the Italian mathematician and physician Gerolamo Cardano (1501 – 1576). Cardano studied the quadratic problem of dividing 10 into two parts such that their product is 40 and found the solutions 155 −+ and 155 −− by standard technique. About 27 years later the engineer Raphael Bombelli (1926 – 1572) published an algebra text in which he dealt systematically with complex numbers. Definition 1.1.1 A complex number z is any expression of the form bia z += where a and b are real numbers and .1 −= i The scalar a is called the real part of z and the scalar b is called the imaginary part of z . Notation If ∈+= babia z ,, , then we use Re ( ) z to denote the number a and I ( ) zm to denote the number b i.e., Re ( ) z= a and I ( ) zm= b . = { } ∈+ babia ,: denotes the set of all complex numbers. Remark If ,0 = b we write ia 0 + as aand ia 0 + is identified as the real number a. If ,0 = a we write bi + 0 as bi and bi is called a pure imaginary number . Every complex number is the sum of a real number and a pure imaginary number. Note. 0 stands for i 00 + . Example 1.1.2 If ,43 i z += then Re ( ) ,3 = z I ( ) 4 = zm (3 is the real part of z and 4 is the imaginary part of z ). 1.2 Operations on complex numbers Definition 1.2.1 Let ( ) ∈+=+= d cbadic zbia z ,,,, 21 . (i) Equality : d bca z z and 21 ==⇔= . (ii) Addition : ( ) ( ) ( ) ( ) id bcadicbia z z +++=+++=+ 21 . (iii) Subtraction : ( ) ( ) ( ) ( ) id bcadicbia z z −+−=+−+=− 21 . (iv) Multiplication: ( )( ) ( ) ( ) . 21 ibcad bd acdicbia z z ++−=++= (v) Division : For ,0 2 ≠ z id cad bcd cbd acdicdicdicbiadicbia z z 2222 21 +−+++=−−⋅++=++= . (vi) Conjugation: For , bia z += define bia z −= . z is called the conjugate of . z It is also sometimes denoted by * z . Note that ( )( ) 22 babiabia z z +=−+= . Theorem 1.2.2 (i) ( ) ( ) ∈∀++=++ 321321321 ,, z z z z z z z z z . (Associative law) (ii) ∈∀+=+ 211221 , z z z z z z . (Commutative law) (iii) ( ) ( ) ∈∀= 321321321 ,, z z z z z z z z z . (Associative law) (iv) ∈∀= 211221 , z z z z z z . (Commutative law) (v) ( ) ∈∀+=+ 3213231321 ,, z z z z z z z z z z . (Distributive law) (vi) ( ) ∈∀+=+ 3211313213 ,, z z z z z z z z z z . (Distributive law) (vii) z z z ⇔= is a real number. Example 1.2.3 (i) Factorize (a) 1 2 + x (b) ( ) 22 q p x ++ (ii) Solve the equation .082 2 =+− x x (iii) Let .31,42 21 i zi z +−=+= Compute 2122212121 ,,,, z z z z z z z z z z −+ . Solution . (i)(a) ( )( ) i xi xi x x −+=−=+ 222 1 (b) ( ) ( ) ( )( ) qi p xqi p xqi p xq p x −+++=−+=++ 22222 (ii) ( ) ( ) ( )( ) .712282228228422 2 ii x ±=±=−±=−−±−− = (iii) ( ) ( ) ii z z 713412 21 +=++−=+ ( ) ( ) ii z z +=−++=− 33412 21 ( )( ) iiiiiii z z 2141222124623142 221 +−=−+−=+−+−=+−+= ( ) ( ) 1031,31 22222 =−+−=−−= z zi z ( )( ) iiiiiiii z z z z z z −=−=+−−−−−− =−−+== 110101010121021012462 103142 2222121 Theorem 1.2.4 For any complex numbers zand w, (i) w zw z +=+ (ii) w z zw = (iii) w zw z = whenever .0 ≠ w Proof . Let bia z += and . dicw += Then bia z −= and . dicw −= (i) Now id bcaw z )()( +++=+ . Then w zdicbiaid bcaw z +=−+−=+−+=+ )()()()( . (ii) Now ibcad bd ac zw )()( ++−= . Then ibcad bd ac zw )()( +−−= . Hence zwibcad bd acidicbiaw z =++−=−−= )()())(( . (iii) Exercise. Theorem 1.2.5 Let 2 ,,,, 21 ≥ n z z z n K , be complex numbers. Then (i) nn z z z z z z +++=+++ L 2121 .... (1) (ii) nn z z z z z z ⋅⋅⋅⋅=⋅⋅⋅⋅ 2121 . (2) Proof . We prove by mathematical induction. We shall only prove (i). Case 1: 2 = n By Theorem 1.2.4, 2121 z z z z +=+ . Case 2: 2 ≥ n. Assume that (1) is true for some ,2 ≥= k n i.e. k k z z z z z z +++=+++ L 2121 ... (3) Then ( ) 121121 ...... ++ +++=++++ k k k k z z z z z z z z ... 1k 21 + ++++= z z z z k (by Theorem 1.2.4) 121 + ++++= k k z z z z L (by (3)) We have shown that if (1) is true for some k n = then (1) is true for .1 += k n It follows by induction that (1) is true for all positive integers 2 ≥ n. Example 1.2.6 Verify Theorem 1.2.4 for iwi z +=−= 3,72 . Solution . Note that iwi z −=+= 3,72 . (i) Now iiiw z 65)3()72( −=++−=+ . Then iw z 65 +=+ . Hence w ziiiw z +=+=−++=+ 65)3()72( . (ii) Now iii zw 1913)3)(72( −=+−= . Then i zw 1913 += . Hence zwiiiw z =+=−+= 1913)3)(72( . (iii) Now )231( 101372 iiiw z −−=+−= . Then )231( 101 iw z +−= . Hence =+−= −+= w ziiiw z )231( 101372 .
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No, Thanks | 2,601 | 6,247 | {"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} | 4.4375 | 4 | CC-MAIN-2017-47 | latest | en | 0.867967 |
http://cambelt.com/cs/tpl=tr_math_area/trapezium&toc=trtoc | 1,542,169,097,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741628.8/warc/CC-MAIN-20181114041344-20181114063344-00203.warc.gz | 64,617,557 | 4,056 | Areas and Dimensions of a Trapezium
Enter either a b and h. a b c h H Area =
Formulas:
A = Area a = middle base segment b = left base segment c = right base segment h = smaller height H = greater height A = (((H + h) * a) + (b * h) + (c * H)) / 2 Note: In England, this figure is called a trapezoid. A trapezium can also be divided into two triangles as indicated by the dotted line. The area of each of these triangles is computed, and the results added to find the area of the trapezium.
Purchase Order Terms and Conditions | 142 | 531 | {"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} | 2.8125 | 3 | CC-MAIN-2018-47 | longest | en | 0.90027 |
http://openstudy.com/updates/511c7ecfe4b06821731b2b15 | 1,448,962,066,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398466178.24/warc/CC-MAIN-20151124205426-00280-ip-10-71-132-137.ec2.internal.warc.gz | 175,504,402 | 10,867 | ## xartaan 2 years ago Find the directional derivative of $f(x,y,z) = yz + x^4$ at (2,3,1) in the direction of vector v=i+j+k. I have gotten as far as finding the directional derivative (gradient) at (2,3,1) which is: 32i+3j+k I dont know how to do the second part of this problem. What does it mean to find this direction? What do I have to do to the gradient to find it? My book is very vague, please help!
1. xartaan
Tips?
2. xartaan
The box for the answer says fu= _____ , where u looks like a unit vector, is that what this is asking for? A unit vector of the gradient?
3. xartaan
Never mind, that, I tried, and was told my answer should be a number, not a vector. Guess I misinterpreted that.
4. xartaan
5. UnkleRhaukus
$\newcommand \ve [1] { \mathbf{#1} }% vector \nabla_{\ve v}f(\ve x)=\nabla f(\ve x)\cdot{\ve v}\\ \ \\ f=yz+x^4\\ \ve v=(1,1,1)$
6. xartaan
I think I understand what you wrote, but it still isnt clear to me what "in the direction of" means... I can find the gradient just fine, I've just no idea what to do with it.
7. UnkleRhaukus
once you have found the gradient, take the dot product with the direction vector
8. xartaan
ahhhhh, that makes sense, thank you! | 373 | 1,201 | {"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} | 3.3125 | 3 | CC-MAIN-2015-48 | longest | en | 0.889267 |
https://www.essaydale.com/order-data-driven-decision-making/ | 1,685,400,021,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00328.warc.gz | 842,633,848 | 13,773 | # Order Data Driven Decision Making
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Why is it important to gather and report valid and reliable data?
4. Identify two examples of real world problems that you have observed in your personal, academic, or professional life that could benefit from data driven solutions. Explain how you would use data/statistics and the steps you would take to analyze each problem. You may also choose topics below (or examples from the weekly content) to help support your response:
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# TableVoltage
Voltage source by linear interpolation in a table
# Wolfram Language
In[1]:=
`SystemModel["Modelica.Electrical.Analog.Sources.TableVoltage"]`
Out[1]:=
# Information
This information is part of the Modelica Standard Library maintained by the Modelica Association.
This voltage source uses the corresponding signal source of the Modelica.Blocks.Sources package. Furthermore, an offset parameter is introduced, which is added to the value calculated by the blocks source. The startTime parameter allows to shift the blocks source behavior on the time axis.
This block generates a voltage source by linear interpolation in a table. The time points and voltage values are stored in a matrix table[i,j], where the first column table[:,1] contains the time points and the second column contains the voltage to be interpolated. The table interpolation has the following properties:
• The time points need to be monotonically increasing.
• Discontinuities are allowed, by providing the same time point twice in the table.
• Values outside of the table range, are computed by extrapolation through the last or first two points of the table.
• If the table has only one row, no interpolation is performed and the voltage value is just returned independently of the actual time instant, i.e., this is a constant voltage source.
• Via parameters startTime and offset the curve defined by the table can be shifted both in time and in the voltage.
• The table is implemented in a numerically sound way by generating time events at interval boundaries. This generates continuously differentiable values for the integrator.
Example:
``` table = [0 0
1 0
1 1
2 4
3 9
4 16]
If, e.g., time = 1.0, the voltage v = 0.0 (before event), 1.0 (after event)
e.g., time = 1.5, the voltage v = 2.5,
e.g., time = 2.0, the voltage v = 4.0,
e.g., time = 5.0, the voltage v = 23.0 (i.e., extrapolation).
```
Furthermore, an offset parameter is introduced, which is added to the value calculated by the blocks source. The startTime parameter allows to shift the blocks source behavior on the time axis.
# Parameters (3)
offset Value: 0 Type: Voltage (V) Description: Voltage offset Value: 0 Type: Time (s) Description: Time offset Value: [0, 0; 1, 1; 2, 4] Type: Real[:,:] Description: Table matrix (time = first column, voltage = second column)
# Connectors (2)
p Type: PositivePin Description: Positive electrical pin Type: NegativePin Description: Negative electrical pin
# Components (1)
signalSource Type: TimeTable Description: Generate a (possibly discontinuous) signal by linear interpolation in a table
# Revisions
• 1998 by Christoph Clauss
initially implemented | 645 | 2,706 | {"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} | 2.765625 | 3 | CC-MAIN-2024-30 | latest | en | 0.812926 |
https://mathhelpboards.com/threads/johnathans-question-at-yahoo-answers-power-series-representation.4664/ | 1,657,069,388,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00495.warc.gz | 408,642,512 | 14,639 | # Johnathan's question at Yahoo! Answers (Power series representation)
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Here is the question:
am studying for a Cal 2 final and I am having a lot of trouble with this one example. Find the power series representation for the function and the radius of convergence. I understand the concepts of power series representations and radii of convergence but I am not sure how to go about solving this problem. f(x) = 1/((2 + x)^2)
I've thought about maybe a partial fraction, but that wouldn't work, then I've thought about making this into f(x) = 1/4 * 1/1-(x^2 + 4x) and setting the x^2 and 4x to my a[n] function but I am not sure if this is correct or how to do it.
Here is a link to the question:
Help with this power series representation? - Yahoo! Answers
I have posted a link there to this topic so the OP can find my response.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Hello Johnathan,
Using the geometric series: $$g(x)=\dfrac{1}{x+2}=\dfrac{1}{2}\dfrac{1}{1+ \frac{x}{2}}=\dfrac{1}{2}\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{2^n}\;(|x|<2)$$
Using the uniform convergence of the power series on all $[-\rho,\rho]\subset (-2,2)$: $$g'(x)=-\frac{1}{(x+2)^2}=\sum_{n=1}^{\infty}\frac{(-1)^nnx^{n-1}}{2^{n+1}}\;(|x|<2)$$ As a consequence, $$f(x)=\dfrac{1}{(x+2)^2}=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}nx^{n-1}}{2^{n+1}}\;(|x|<2)$$ | 474 | 1,439 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-27 | latest | en | 0.816519 |
https://zarahomework.com/ages-of-oscar-winning-best-actors-and-actresses/ | 1,723,444,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00373.warc.gz | 828,623,151 | 18,787 | # Ages of oscar-winning best actors and actresses
Listed below are the ages of actressesand actors at the times that they won Oscars for the categories of Best Actress and Best Actor. The ages are listed in chronological order by row, so that corresponding locations in the two tables are from the same year. (Notes: In 1968 there was a tie in the Best Actress category, and the mean of the two ages is used; in 1932 there was a tie in the Best Actor category, and the mean of the two ages is used. These data are suggested by the article “Ages of Oscar-winning Best Actors and Actresses,” by Richard Brown and Gretchen Davis, Mathematics Teacher magazine. In that article, the year of birth of the award winner was subtracted from the year of the awards ceremony, but the ages in the tables below are based on the birth date of the winner and the date of the awards ceremony.)
Analyzing the Results
1. First explore the data using suitable statistics and graphs. Use the results to make informal comparisons.
2. Determine whether there are significant differences between the ages of the Best Actresses and the ages of the Best Actors. Use appropriate hypothesis tests. Describe the methods used and the conclusions reached.
3. Discuss cultural implications of theresults. Does it appear that actresses and actors are judged strictly on the basis of their artistic abilities? Or does there appear to be discrimination based on age, with the Best Actresses tending to be younger than the Best Actors? Are there any other notable differences?
Best Actresses 22 37 28 63 32 26 31 27 27 28 30 26 29 24 38 25 29 41 30 35 35 33 29 38 54 24 25 46 41 28 40 39 29 27 31 38 29 25 35 60 43 35 34 34 27 37 42 41 36 32 41 33 31 74 33 50 38 61 21 41 26 80 42 29 33 35 45 49 39 34 26 25 33 35 35 28 30 29 61 Best Actors 44 41 62 52 41 34 34 52 41 37 38 34 32 40 43 56 41 39 49 57 41 38 42 52 51 35 30 39 41 44 49 35 47 31 47 37 57 42 45 42 44 62 43 42 48 49 56 38 60 30 40 42 36 76 39 53 45 36 62 43 51 32 42 54 52 37 38 32 45 60 46 40 36 47 29 43 37 38 45
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Money Back | 735 | 2,429 | {"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} | 3.15625 | 3 | CC-MAIN-2024-33 | latest | en | 0.893317 |
http://deeplearning.stanford.edu/wiki/index.php?title=%E6%A0%88%E5%BC%8F%E8%87%AA%E7%BC%96%E7%A0%81%E7%AE%97%E6%B3%95&diff=2343&oldid=1627 | 1,620,882,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991537.32/warc/CC-MAIN-20210513045934-20210513075934-00281.warc.gz | 14,606,317 | 11,880 | # 栈式自编码算法
Revision as of 12:25, 8 March 2013 (view source)Kandeng (Talk | contribs) (→Training)← Older edit Latest revision as of 05:15, 8 April 2013 (view source)Wikiroot (Talk | contribs) Line 1: Line 1: - 初译@小猴机器人 + ==概述== - 一审@邓亚峰-人脸识别 + - ===Overview=== + 逐层贪婪训练法依次训练网络的每一层，进而预训练整个深度神经网络。在本节中，我们将会学习如何将自编码器“栈化”到逐层贪婪训练法中，从而预训练（或者说初始化）深度神经网络的权重。 - The greedy layerwise approach for pretraining a deep network works by training each layer in turn. In this page, you will find out how autoencoders can be "stacked" in a greedy layerwise fashion for pretraining (initializing) the weights of a deep network. - 【初译】 + 栈式自编码神经网络是一个由多层稀疏自编码器组成的神经网络，其前一层自编码器的输出作为其后一层自编码器的输入。对于一个 $\textstyle n$ 层栈式自编码神经网络，我们沿用自编码器一章的各种符号，假定用 $\textstyle W^{(k, 1)}, W^{(k, 2)}, b^{(k, 1)}, b^{(k, 2)}$ 表示第 $\textstyle k$ 个自编码器对应的 $\textstyle W^{(1)}, W^{(2)}, b^{(1)}, b^{(2)}$ 参数，那么该栈式自编码神经网络的编码过程就是，按照从前向后的顺序执行每一层自编码器的编码步骤： - 在深度神经网络进行预先训练的时候，有一种贪心分层方法，它的原理基于对每一个层次轮流进行训练。在本文中，我们会一起学习如何将上文中提到的自编码神经网络以栈的形式组成这种贪心分层的模样，从而预先训练（或者说初始化）深度神经网络的权重。 + - 【一审】 - 可以采用依次训练每一层的贪心分层算法来预训练深度神经网络。在本节中，我们将会学习如何将自编码器以贪心分层的方式栈化，从而预训练（或者说初始化）深度神经网络的权重。 - - - A stacked autoencoder is a neural network consisting of multiple layers of sparse autoencoders in which the outputs of each layer is wired to the inputs of the successive layer. Formally, consider a stacked autoencoder with n layers. Using notation from the autoencoder section, let $W^{(k, 1)}, W^{(k, 2)}, b^{(k, 1)}, b^{(k, 2)}$ denote the parameters $W^{(1)}, W^{(2)}, b^{(1)}, b^{(2)}$ for kth autoencoder. Then the encoding step for the stacked autoencoder is given by running the encoding step of each layer in forward order: $[itex] Line 22: Line 14:$ [/itex] - The decoding step is given by running the decoding stack of each autoencoder in reverse order: + + 同理，栈式神经网络的解码过程就是，按照从后向前的顺序执行每一层自编码器的解码步骤： + $[itex] Line 31: Line 25:$ [/itex] - The information of interest is contained within $a^{(n)}$, which is the activation of the deepest layer of hidden units. This vector gives us a representation of the input in terms of higher-order features. - 【初译】 + 其中，$\textstyle a^{(n)}$ 是最深层隐藏单元的激活值，其包含了我们感兴趣的信息，这个向量也是对输入值的更高阶的表示。 - 栈式自编码神经网络指的是这样一种网络，它由多层稀疏自编码神经网络组成，前一层网络的输出作为后一层的输入。形式化来讲，让我们来设想一个n层栈式自编码神经网络，并沿用自编码神经网络那章的各种符号，比如$W^{(k, 1)}, W^{(k, 2)}, b^{(k, 1)}, b^{(k, 2)}$ 表示第k层网络里对应的$W^{(1)}, W^{(2)}, b^{(1)}, b^{(2)}$ 参数。那么对于该栈式网络的编码过程就是，按照从前向后的顺序，执行每一层的编码过程： + - - \begin{align} - a^{(l)} = f(z^{(l)}) \\ - z^{(l + 1)} = W^{(l, 1)}a^{(l)} + b^{(l, 1)} - \end{align} - - 解码就是从后向前每层解码咯： + 通过将 $\textstyle a^{(n)}$ 作为softmax分类器的输入特征，可以将栈式自编码神经网络中学到的特征用于分类问题。 - - \begin{align} - a^{(n + l)} = f(z^{(n + l)}) \\ - z^{(n + l + 1)} = W^{(n - l, 2)}a^{(n + l)} + b^{(n - l, 2)} - \end{align} - - 在这里，我们感兴趣的信息是a(n)，也就是最深的隐含单元层的激发阈值。这个向量对我们的输入给出更高层次的特征表示。 + ==训练== - 【一审】 - 栈式自编码神经网络是一个由多层稀疏自编码器组成的神经网络，其前一层自编码器的输出作为其后一层自编码器的输入。形式上，对于一个n层栈式自编码神经网络而言，沿用自编码器一章的各种符号，假定用$W^{(k, 1)}, W^{(k, 2)}, b^{(k, 1)}, b^{(k, 2)}$ 表示第k个自编码器对应的$W^{(1)}, W^{(2)}, b^{(1)}, b^{(2)}$ 参数，那么该栈式自编码神经网络的编码过程就是依照从前向后的顺序执行每一层自编码器的编码步骤： - $+ 一种比较好的获取栈式自编码神经网络参数的方法是采用逐层贪婪训练法进行训练。即先利用原始输入来训练网络的第一层，得到其参数 [itex]\textstyle W^{(1,1)}, W^{(1,2)}, b^{(1,1)}, b^{(1,2)}$；然后网络第一层将原始输入转化成为由隐藏单元激活值组成的向量（假设该向量为A），接着把A作为第二层的输入，继续训练得到第二层的参数 $\textstyle W^{(2,1)}, W^{(2,2)}, b^{(2,1)}, b^{(2,2)}$；最后，对后面的各层同样采用的策略，即将前层的输出作为下一层输入的方式依次训练。 - \begin{align} + - a^{(l)} = f(z^{(l)}) \\ + - z^{(l + 1)} = W^{(l, 1)}a^{(l)} + b^{(l, 1)} + - \end{align} + - [/itex] + - 同理，栈式神经网络的解码过程就是依照从后向前的顺序执行每一层自编码器的解码步骤： - + 对于上述训练方式，在训练每一层参数的时候，会固定其它各层参数保持不变。所以，如果想得到更好的结果，在上述预训练过程完成之后，可以通过反向传播算法同时调整所有层的参数以改善结果，这个过程一般被称作“微调（fine-tuning）”。 - \begin{align} + - a^{(n + l)} = f(z^{(n + l)}) \\ + - z^{(n + l + 1)} = W^{(n - l, 2)}a^{(n + l)} + b^{(n - l, 2)} + - \end{align} + - + - 其中，a(n)是隐藏单元最深层的响应，其包含了我们感兴趣的信息，这个向量是对输入的更高阶的表示。 - The features from the stacked autoencoder can be used for classification problems by feeding $a(n)$ to a softmax classifier. + 实际上，使用逐层贪婪训练方法将参数训练到快要收敛时，应该使用微调。反之，如果直接在随机化的初始权重上使用微调，那么会得到不好的结果，因为参数会收敛到局部最优。 - 【初译】 - 从栈式自编码神经网络中学到的特征，可以通过向softmax分类器添加 $a(n)$的方式来解决分类问题。 - 【一审】 + 如果你只对以分类为目的的微调感兴趣，那么惯用的做法是丢掉栈式自编码网络的“解码”层，直接把最后一个隐藏层的 $\textstyle a^{(n)}$ 作为特征输入到softmax分类器进行分类，这样，分类器（softmax）的分类错误的梯度值就可以直接反向传播给编码层了。 - 通过将 $a(n)$作为softmax分类器的输入特征，可以将栈式自编码神经网络中学到的特征用于分类问题。 + - ===Training=== + ==具体实例== - A good way to obtain good parameters for a stacked autoencoder is to use greedy layer-wise training. To do this, first train the first layer on raw input to obtain parameters $W^{(1,1)}, W^{(1,2)}, b^{(1,1)}, b^{(1,2)}$. Use the first layer to transform the raw input into a vector consisting of activation of the hidden units, A. Train the second layer on this vector to obtain parameters $W^{(2,1)}, W^{(2,2)}, b^{(2,1)}, b^{(2,2)}$. Repeat for subsequent layers, using the output of each layer as input for the subsequent layer. + - 【初译】 + 让我们来看个具体的例子，假设你想要训练一个包含两个隐含层的栈式自编码网络，用来进行MNIST手写数字分类（这将会是你的下一个练习）。 - 想从上述栈式网络中得到好的参数，我们可以使用贪心分层方法。即先通过输入来训练第一层网络，得到$W^{(1,1)}, W^{(1,2)}, b^{(1,1)}, b^{(1,2)}$。再用第一层网络将输入转化，成为由隐藏单元的激发阈值组成的向量，暂且称做A。把A作为第二层的输入，继续进行训练，得到$W^{(2,1)}, W^{(2,2)}, b^{(2,1)}, b^{(2,2)}$。如此这般，都是前一层的输出作为后一层的输入。 + 首先，你需要用原始输入 $\textstyle x^{(k)}$ 训练第一个自编码器，它能够学习得到原始输入的一阶特征表示$\textstyle h^{(1)(k)}$（如下图所示）。 - 【一审】 - 一种比较好的获取栈式自编码神经网络参数的方法是采用贪心分层方式进行训练。即先通过原始输入来训练第一层网络，得到其参数$W^{(1,1)}, W^{(1,2)}, b^{(1,1)}, b^{(1,2)}$；再用第一层网络将原始输入转化成为由隐藏单元响应组成的向量，假设该向量为A，接着把A作为第二层的输入，继续训练得到第二层的参数$W^{(2,1)}, W^{(2,2)}, b^{(2,1)}, b^{(2,2)}$；对后面的各层同样采用将前层的输出作为下一层输入的方式依次训练。 + [[File:Stacked_SparseAE_Features1.png|400px]] - This method trains the parameters of each layer individually while freezing parameters for the remainder of the model. To produce better results, after this phase of training is complete, [[Fine-tuning Stacked AEs | fine-tuning]] using backpropagation can be used to improve the results by tuning the parameters of all layers are changed at the same time. - 【初译】 + 接着，你需要把原始数据输入到上述训练好的稀疏自编码器中，对于每一个输入$\textstyle x^{(k)}$，都可以得到它对应的一阶特征表示$\textstyle h^{(1)(k)}$。然后你再用这些一阶特征作为另一个稀疏自编码器的输入，使用它们来学习二阶特征 $\textstyle h^{(2)(k)}$。（如下图所示） - 这种方法在训练的时候，除了当前层，不会改变其他层次的参数。因此，为了得到更好的结果，在训练阶段结束以后，用反向传播算法微调一下，让所有层的参数同时变化，会对结果有所提升。 + - 【一审】 + [[File:Stacked_SparseAE_Features2.png|400px]] - 上述训练方式，在训练每一层参数的时候，会固定其它各层参数保持不变。所以，如果想得到更好的结果，在上述预训练过程完成之后，可以通过反向传播算法同时调整所有层的参数以改善结果，这个过程一般被称作“微调（fine-tuning）”。 + - + 同样，再把一阶特征输入到刚训练好的第二层稀疏自编码器中，得到每个 \textstyle h^{(1)(k)}[/itex] 对应的二阶特征激活值 $\textstyle h^{(2)(k)}$。接下来，你可以把这些二阶特征作为softmax分类器的输入，训练得到一个能将二阶特征映射到数字标签的模型。 - {{Quote| + [[File:Stacked_Softmax_Classifier.png|400px]] - If one is only interested in finetuning for the purposes of classification, the common practice is to then discard the "decoding" layers of the stacked autoencoder and link the last hidden layer $a^{(n)}$ to the softmax classifier. The gradients from the (softmax) classification error will then be backpropagated into the encoding layers. + - 【初译】 + 如下图所示，最终，你可以将这三层结合起来构建一个包含两个隐藏层和一个最终softmax分类器层的栈式自编码网络，这个网络能够如你所愿地对MNIST数字进行分类。 - 如果你只对分类神经网络的微调感兴趣，惯用的做法是丢掉栈式网络的“解码”层，直接把最后那个隐含层$a^{(n)}$ 跟softmax分类器连起来。这样分类器错误的梯度值就会直接反向传播给编码层了。 + - 【一审】 + [[File:Stacked_Combined.png|500px]] - 注：如果你只对以分类为目的的微调感兴趣，惯用的做法是丢掉栈式自编码器网络的“解码”层，直接把最后一个隐含层的$a^{(n)}$ 作为特征输入到softmax分类器进行分类，这样，分类器（softmax）的分类错误的梯度值就可以直接反向传播给编码层了。 + - }} - ===Concrete example=== + ==讨论== - To give a concrete example, suppose you wished to train a stacked autoencoder with 2 hidden layers for classification of MNIST digits, as you will be doing in [[Exercise: Implement deep networks for digit classification | the next exercise]]. + 栈式自编码神经网络具有强大的表达能力及深度神经网络的所有优点。 - First, you would train a sparse autoencoder on the raw inputs $x^{(k)}$ to learn primary features $h^{(1)(k)}$ on the raw input. + 更进一步，它通常能够获取到输入的“层次型分组”或者“部分-整体分解”结构。为了弄清这一点，回顾一下，自编码器倾向于学习得到能更好地表示输入数据的特征。因此，栈式自编码神经网络的第一层会学习得到原始输入的一阶特征（比如图片里的边缘），第二层会学习得到二阶特征，该特征对应一阶特征里包含的一些模式（比如在构成轮廓或者角点时，什么样的边缘会共现）。栈式自编码神经网络的更高层还会学到更高阶的特征。 - [[File:Stacked_SparseAE_Features1.png|400px]] - Next, you would feed the raw input into this trained sparse autoencoder, obtaining the primary feature activations $h^{(1)(k)}$ for each of the inputs $x^{(k)}$. You would then use these primary features as the "raw input" to another sparse autoencoder to learn secondary features $h^{(2)(k)}$ on these primary features. + 举个例子，如果网络的输入数据是图像，网络的第一层会学习如何去识别边，第二层一般会学习如何去组合边，从而构成轮廓、角等。更高层会学习如何去组合更形象且有意义的特征。例如，如果输入数据集包含人脸图像，更高层会学习如何识别或组合眼睛、鼻子、嘴等人脸器官。 - [[File:Stacked_SparseAE_Features2.png|400px]] - Following this, you would feed the primary features into the second sparse autoencoder to obtain the secondary feature activations $h^{(2)(k)}$ for each of the primary features $h^{(1)(k)}$ (which correspond to the primary features of the corresponding inputs $x^{(k)}$). You would then treat these secondary features as "raw input" to a softmax classifier, training it to map secondary features to digit labels. - [[File:Stacked_Softmax_Classifier.png|400px]] + ==中英文对照== + + :自编码器 Autoencoder + + :逐层贪婪训练法 Greedy layer-wise training + + :预训练 PreTrain + + :栈式自编码神经网络 Stacked autoencoder - Finally, you would combine all three layers together to form a stacked autoencoder with 2 hidden layers and a final softmax classifier layer capable of classifying the MNIST digits as desired. + :微调 Fine-tuning - [[File:Stacked_Combined.png|500px]] + :原始输入 Raw inputs + + :层次型分组 Hierarchical grouping + + :部分-整体分解 Part-whole decomposition + + :一阶特征 First-order features + + :二阶特征 Second-order features + + :更高阶特征 Higher-order features + + :激活值 Activation + + + ==中文译者== - ===Discussion=== + 张天雷（ztl2004@gmail.com）, 邓亚峰（dengyafeng@gmail.com）, 许利杰（csxulijie@gmail.com） - A stacked autoencoder enjoys all the benefits of any deep network of greater expressive power. - Further, it often captures a useful "hierarchical grouping" or "part-whole decomposition" of the input. To see this, recall that an autoencoder tends to learn features that form a good representation of its input. The first layer of a stacked autoencoder tends to learn first-order features in the raw input (such as edges in an image). The second layer of a stacked autoencoder tends to learn second-order features corresponding to patterns in the appearance of first-order features (e.g., in terms of what edges tend to occur together--for example, to form contour or corner detectors). Higher layers of the stacked autoencoder tend to learn even higher-order features. + {{建立分类用深度网络}} - {{CNN}} - + | 4,799 | 10,419 | {"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} | 3.171875 | 3 | CC-MAIN-2021-21 | latest | en | 0.445184 |
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## #676 2012-07-05 01:19:00
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
It would be interesting to see one, though.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #677 2012-07-05 01:26:35
bobbym
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### Re: Oh,Oh,Oh,merry analysis!!!
An algorithm with a O(n^n^n^n)?
2^2^2^2 = 65536
Do you have any idea what 3^3^3^3 is?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #678 2012-07-05 02:51:29
anonimnystefy
Real Member
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### Re: Oh,Oh,Oh,merry analysis!!!
Yes, I do. But imagine an algorithm of complexity O(G_n) where G_n is the Graham's number except that it uses n instead of 3 everywhere.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #679 2012-07-05 03:25:24
bobbym
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### Re: Oh,Oh,Oh,merry analysis!!!
Forget that Graham's number kaboobly doo, what is your estimate of 3^3^3^3?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #680 2012-07-05 03:29:37
anonimnystefy
Real Member
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### Re: Oh,Oh,Oh,merry analysis!!!
It has around 3.6*10^12 digits.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #681 2012-07-05 03:32:58
bobbym
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### Re: Oh,Oh,Oh,merry analysis!!!
And the mantissa?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #682 2012-07-05 03:35:28
anonimnystefy
Real Member
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### Re: Oh,Oh,Oh,merry analysis!!!
What mantissa?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #683 2012-07-05 03:40:42
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
1.342*10^20, the 1.342 is the mantissa.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #684 2012-07-05 06:57:46
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #685 2012-07-05 20:25:46
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
What is the mantissa?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #686 2012-07-05 20:38:58
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
It is 3.63something.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #687 2012-07-05 21:32:46
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Hi;
Those are the first 3 digits of the exponent.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #688 2012-07-05 22:06:33
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Hi bobbym
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #689 2012-07-05 23:01:39
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Hi;
Yes, I know but what are the front digits. The mantissa!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #690 2012-07-13 08:05:19
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
3.6 ?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #691 2012-07-13 17:15:28
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Hi anonimnystefy;
Those are the front digits of the exponents not the mantissa.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #692 2012-07-13 22:48:13
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
I am getting that as the mantissa.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #693 2012-07-14 01:54:00
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #694 2012-07-14 02:11:16
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Hi bobbym
This is what W|A gives.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #695 2012-07-14 02:13:37
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
That is the number of digits. That is the same as the exponent of the answer. But it is not the mantissa of the answer.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #696 2012-07-14 02:23:12
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
W|A won't tell me.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #697 2012-07-14 02:27:35
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
That is the point! Wolfram can not do 3^3^3^3. But we can still get an answer!
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #698 2012-07-14 02:39:59
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Numerical analysis?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #699 2012-07-14 03:15:29
bobbym
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Yes and any other branch of math! In a school problem you can only use the material you just learned. Usually just the last chapter of the textbook covered. In the real world there are no bounds! Any technique that gets an answer is good.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #700 2012-07-14 03:40:02
anonimnystefy
Real Member
Offline
### Re: Oh,Oh,Oh,merry analysis!!!
Let me see how youwould do that, will you?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment | 2,881 | 10,169 | {"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} | 3.109375 | 3 | CC-MAIN-2014-15 | longest | en | 0.886974 |
https://www.physicsforums.com/threads/what-does-f-ma-really-means.346386/ | 1,532,293,161,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593586.54/warc/CC-MAIN-20180722194125-20180722214125-00584.warc.gz | 939,892,991 | 27,592 | # What does F=ma really means?
1. Oct 17, 2009
### cocosisi
F=ma
How do we define m and F? If their definition both come from this equation, then the equation doesn't really mean anything...We can also definite a F', and say F'=ma^2, and there isn't anything wrong... However, we have F=ma as something true, then F and m must be something true, indentifiable and measureable. So what does F and m really means??
2. Oct 17, 2009
### xxChrisxx
Er, where to even start.
F=ma is a mathematical fit for the force experienced by an accelerating object. It isn't arbritraty as it can be found experiementally.
F = force m = mass a = acceleration.
You can assign any letter you want to any mathematical value. So it's perfectly valid mathematically to say that F' = Ma^2, but it dosent show anything in reality, its purely a mathemetical construct.
F=MA shows something.
3. Oct 17, 2009
### Count Iblis
In classical physics F and m are phenomenological quantities which are specified by other relations. So, you can take some amount of matter and on the basis of that assign to m some value. You can postulate some force law and then make predictions for the motion.
Deeper explanations for m and F can only be obtained from fundamental physics (quantum field theory).
4. Oct 17, 2009
### f95toli
Sort if true. But the question asked by the OP is by no means trivial. There were whole books written about this back when classical physics was still "state of the art"( as far as I remember the "The Science of Mechanics" by Ernst Mach discusses this at great lenght) and and far as I know they never reached a generally accepted conclusion.
I guess one could say that back then they were having the same sort of issues with classical mechanics (and thermodynamics) as we are having now with quantum mechanics.
Also, some of the questions that come out of this are still very much "hot" topics in physics, e.g. why "heavy" (gravitational) mass is the same as inertial (the m in F=ma) mass. Meaning it touches upon topics such as the Higg's boson etc.
5. Oct 17, 2009
### arildno
xxChrisxx:
This is totally wrong.
cocosisi's question is very meaningful, and goes, for example, deep into the philosophical underpinnings of Einstein's General Relativity.
Einstein noted that we are using essentially two distinct definitions of mass, namely inertial mass and gravitational mass (a fundamental assumption within G.R is that these have the same value).
To take the case of inertial mass:
Cocosisi:
1. Suppose you have a mechanism that can be observed to impart accelerations to objects in contact with it (note that ACCELERATIONS are measurable (not that easy, but it CAN be done!))
2. We say that this mechanism, BY WAY OF A FORCE, imparts acceleration to various objects.
3. Let us ASSUME that the mechanism gives the SAME force to different objects; for example that our mechanism is a spring that we have compressed 1cm prior to bringing it into contact with the object.
(That assumption can be given empirical verification that by performing NUMEROUS experiments on some set of objects, each of those objects gains the same acceleration it did on a prior trial)
4. We observe, however, that different objects gains DIFFERENT acceleration from what we can arguably say is the same force.
We may therefore construct, within this particular experimental setting, an object-connected scale of ratios F/a, i.e, a "mass scale" We can pick out one object, O_{0}, whose acceleration equals $a_{0}$, and we set the value of the common force, $F_{0}$, equal to a_{0}.
Thus, that object is given "unit mass", and the rest will be assigned various mass values directly related to the acceleration they experience.
5. Now, some non-trivial results can be observed within the setting:
If we take two objects, and makes them stick to each other so that they move as one object, then that compound object's ACCELERATION will be found as if we could add the masses of the two objects together and find the compound mass (i.e, degree of resistance to acceleration) from that.
Thus, mass is seen to be an ADDITIVE property, that certainly is consistent with the (as yet unproven) idea that mass is some constant property belonging to the object itself, rather than some meaningless, accidental correlation!
6. By fiddling with two objects, 1 and 2, we find that in the above experiment, the relation:
$$\frac{m_{1}}{m_{2}}=\frac{a_{2}}{a_{1}}$$
holds.
Now, SUPPOSE that masses are something intrinsic to objects, "forces" something that this mass-property is independent of, something that merely "happens to" the object.
What would this entail?
It would entail that if we did ENTIRELY DIFFERENT experiments (than the spring example, compression of 1cm), then the ratio of observed accelerations would REMAIN CONSTANT, also when we have absolutely no reason of thinking that the forces are generated by the same mechanism as before, and for that matter, have another value than the one in the spring example.
7. And this is indeed what we see!
The ratio $$\frac{m_{1}}{m_{2}}$$ (and therefore acceleration ratio) remains CONSTANT, irrespective of what type of force is applied to both objects.
8. The accelerations are seen to change, but we are fully entitled to regard the masses as being constant properties of the objects themselves.
9. Now, we can utilize this discovered property to gauge forces of a variety of forms, and thus to create force scales as well.
6. Oct 17, 2009
### xxChrisxx
You are talking to an engineer here not a physicist. I run with arms flailing from GR, Newtonian mechanics all the way.
On reflection it does seem that the OP's question could be asking something deeper than what is 'F' and 'M' in the most basic context.
7. Oct 17, 2009
### arildno
And an engineer ought to know everything about how to construct mass and force scales, and WHY they can be regarded as meaningful constructions that most likely have some important relation to actual reality.
8. Oct 17, 2009
### xxChrisxx
All I need to know from F=MA is; shove something and it moves. Shove it twice as hard it accelerates twice as fast.
F=ma can be regarded as meaningful becuase... thats what happens in reality. We know this to be true.
For practical purposes it really doesnt need to be more complicated than that.
Last edited: Oct 17, 2009
9. Oct 17, 2009
### atyy
This equation is the definition of F. m is just a constant of proportionality, defined to make the equation true.
The goal of classical physics is to find how properties of objects (such as their colour, texture, size etc) cause them to produce or react to forces. For example, the the electric force is produced by a property called "electric charge", and F=(q1.q2)/(r.r).
10. Oct 17, 2009
### arildno
And on what basis do you really define what it means to "shove twice as hard"??
cocosisi's question is highly apposite.
IF, for example, we REALLY had a relation:
F=m(F)*a,
then shoving "twice as hard" wouldn't necessarily result in twice the acceleration.
11. Oct 17, 2009
### xxChrisxx
There is really no point in going into this deeper. As what 'shove it twice as hard' means is pretty much what you put a few posts above (post 5), albeit in a very wordy fashion.
The fact that you can explain it better than I can makes no odds and is the reason I would make a crap teacher.
12. Oct 17, 2009
### Staff: Mentor
In most consistent systems of units (including the SI system) F=ma would define force, and mass would be defined in some other way (e.g. by reference to a standard mass).
13. Oct 17, 2009
### Count Iblis
Mass is the total energy content of an object. The (generalized) force is defined as minus the derivative of the energy content w.r.t. the external variable.
If you have two atoms at some distance, then the force is given by minus the derivative of the total energy of the system w.r.t. the distance. And the energy can be obtained by solving the Schrödinger equation. A fully relativistic treatment will give you the total energy and thus the mass.
14. Oct 17, 2009
### arildno
Well, you could for example use Newton's 3.law to postulate the equality of two forces within a couple, and then use the ratio of observed accelerations to define a scale for inertial mass.
15. Oct 17, 2009
### Count Iblis
SI units are not consistent. In SI units we use inconsistent units for Time, Length, Mass, Temperature etc. etc. This comes at the price of having to introduce extraneous conversion factors (c, G, hbar, k_b, etc.) in formulae.
From a metrological point of view, however, it is advantageous to have different physical standards for physical quantities, even if they are related by fundamental physics. As technology advances, it can happen that we can rely on fundamental physics to perform accurate measurements. If that enhances the acuracy whit which you can perform measurements, the SI system will be revised. That happened in 1983 when the meter was redefined in terms of the second by spcifying a value for the speed of light.
For historical reasons we still use different units for lengths and time intervals, but we could just as well measure time in meters.
16. Oct 17, 2009
### Staff: Mentor
The word you are looking for is "natural"; the SI system of units is consistent but not natural.
Consistent simply means that the base units and the derived units in the system don't have any conversion factors between them. For example, in the SI system the unit of power is 1 W = 1 kg m²/s³, but in the US system the unit of power is 1 hp = 550 ft lb/s. The US system is inconsistent because it needs a conversion between a derived unit and the base units with the same dimensions.
If a system of units is "natural" then some of the dimensionful universal constants are numerically equal to 1. You can think of these as conversion factors between units of different dimensions. A "geometrized" system of units takes this one step further and considers the universal constants to be not just numerically equal to 1, but also dimensionally equal to 1 (i.e. c is unitless). This means that all of the fundamental units have dimensions of length.
17. Oct 17, 2009
### Count Iblis
I see! I prefer to use geometrized units.
18. Oct 17, 2009
### Werg22
Last edited by a moderator: Apr 24, 2017
19. Oct 17, 2009
### cocosisi
Last edited: Oct 17, 2009
20. Oct 17, 2009
### cocosisi
Last edited by a moderator: Apr 24, 2017
21. Oct 17, 2009
### Gerenuk
I think that's very good question for which one has to learn to get away from "taking everything your high school teacher said as law" to "self-made insights"
Force is defined as the flow of momentum
$$F=\frac{\mathrm{d}p}{\mathrm{d}t}$$
and together with the law of action and reaction, it turns out your question is completely equivalent to:
...asking why momentum
$$p=mv$$
is conserved and how to determine the m in that equation.
Well, I assume all know forces between two objects (gravity, electromagnetism,...) experimentally obey
$$v_1=av_2+b$$
at any moment. And therefore we can define mass ratios by
$$\frac{m_2}{m_1}=a$$
which yields $m_1v_1+m_2v_2=\text{const}$ which we wanted.
It also turns out the mass of a particle does not change over time. So now we know how to determine the mass of a particle (which we can tabulate) and also we accept conservation of $mv$ as a law.
All these consideration are of course classical. I'll think about what one has to keep in mind taking into account relativity...
Last edited: Oct 17, 2009
22. Oct 18, 2009
### arildno
It tells us that it is a consistent, i.e, meaningful construction.
Furthermore, by working FROM that construction, we have found that it is an extraordinarily USEFUL construction.
I think that's quite a lot!
23. Oct 18, 2009
### vissarion.eu
What is tricky thing that mass, velocity exist in any small time interval. But acceleration in infinityly small time interval turns to zero, when dt limt->0 then acceleration apears to disapearing and instead accleration in this small interval you see constant speed v. Acceleration is in our heads, but not in some time interval or in some certain time interval. Force good describing gravity F=mg. If we add h then we know how long force was upon object of mass m. For instance, if m=1 kg and g=10 m/s^2, then after 10 seconds object speed will be 100 m/s. And average speed will be 50 m/s after 10 seconds. S=v*t=50*10=500 meters will fall 1 kg stone withing 10 seconds on the ground from height 500 meters. So force was acting upon stone 10 seconds, so how would you call such action on 1 kg stone? ?=F*t=m*a*t. I would think it could be somthing between kinetic energy W=mvv/2 and momentum p=mv. Why ones better than overs? Maybe I should call it job A=mat. But seems we already have work formula. So if 1kg stone after 10 s reach speed 100 m/s, so this fine gives momentum m*v=1(kg)*100(m/s)=100 (kg*m/s) and m*a*t=1(kg)*10(m/s/s)*10(s)=100 (kg*m/s).
Edit: I would describe job A=m*a*t or A=m*g*t if there no any contra force. For example if stone fall in vacum tube onto earth from 500 m height then his job is velocity, which stone will reach till fall on ground multiplied with mass, acceleration and time A=mat=1*10*100=100 kg*m/s. And now what is work? Work is when rocket with acceleration a=20 m/s^2, agains gravity reaching height 500 meters. Air friction ignore (supose rocket flying up in some vacum tube). So force which is contra force to rocket raise force is gravity force or acceleration g=10 m/s/s. So rocket mass m=1000 kg and mass disipation because of fuel decreasing is almost invisble or let just ignore it. So gravity force is G=10(m/s^2)*1000(kg)=10000 N. And rocket lifting force F=20(m/s/s)*1000(kg)=20000 N. So total force is R=F-G=20000-10000=10000 N. So rocket will be seen as falling like stone on eeart in vacum, but in oposit direction. So rocket will reach height 500 meters in 10 seconds. So rocket kinda engine make job A=mat=20(m/s/s)*1000(kg)*10(s)=200000 kg*m/s. This is real work-job which made rocket reaching such height.
Now supose we add friction of air to rocket which will be a force counter to force which lifting rocket, this force may be about 4 m/s/s. So then rocket after first second will got speed 20-10-4=6 m/s. After next second rocket from first to second second will reach speed 2 m/s and total speed will be 8 m/s (after two seconds when rocket start to fly from ground). But here need integration since speed of rocket also deacrease and for second-next second since it flying at bigger speed friction too will be bigger. So as you gues here need real work formula W=mvv/2=1000*100*100/2=5000000 J and this units teoreticaly equal to job, which rocket made without friction so with air friction it takes 5000000/200000=25 times more energy to reach same speed with same momentum of rocket.
Or another maybe simpler example. If I want just to accelerata 1 kg stone in cosmos to 100 m/s then for it need A=mv=1*100=100 (energy units). And if I put this stone on some very long desk and again trying it accelerate to speed 100 m/s, then here again we meet desk and air friction, which is say F=ma=1(kg)*3(m/s^2)=3 N counterforce. So if I just push stone with 3 N force then stone will not move, but if I push with 6 N force then stone will move as if he would be in vacum, in free space and I would push stone with 3 N force or a=3 m/s^2 acceleration. So this time need to use kinetic energy formula to find out how need energy to push stone 500 meters distance. And this would look like this: W=mgh=1(kg)*3(m/s/s)*500(m)=1500 J. Such distance will be reached after (let me see 3*10(s)=30 m/s, average v=30/2=15; 3*100=300 m/s, average v=150 m/s; 500/150=10/3; (3*100*10/3)=1000 m/s, average v=1000/2=500 m/s | let me try different way, 3 m/s after 1 second and average speed 1.5 m/s, from 1 s to 2 second speed will chanes from 3 to 6, so 6/2=3 meters per second second will move stone, 6 per third second, 12 per four s, so 1.5+3+6+12+24+48+96+192=382.5 meters after 8 seconds or 382.5+192=574.5 m after 8.5 s) about 8.3 seconds. And velocity will be about 8.5(s)*3(m/s/s)=25.5 m/s. So kinetic energy W=mvv/2=1*25.5*25.5/2=325.125 J. Somthing wrong 500/3=166.(6) m/s. W=mvv/2=1*(500/3)^2/2=13888.(8). mvv/2=1500, v=2*1500^(1/2)=77.46 m/s. Whatever...
Edit2: Speed raise: 3 m/s, 6 m/s, 9 m/s, 12 m/s, 15, 18, 21... Distance raise: (0+3)/2+(3+6)/2+(6+9)/2+(9+12)/2+(12+15)/2... or 1.5+4.5+7.5+10.5+13.5+16.5+19.5+22.5+25.5... Or 1.5*(3+6+9+12+15+18+21+24+27+30+33+36+39+42+45)=1.5*315=472.5 m. And this take 15 seconds. 45*1.5=67.5 meters will move per fifteen second. 3*15=45 m/s. Still less than 77.46 m/s. What the hell it is?
Edit3: It's ok I just wrong vv/2=1500, vv=1500*2, v=3000^(1/2)=54.77 m/s, so very good with almost 45 m/s agree. But still need more.
Edit4: mah=1*3*472.5=1417.5 J. vv/2=1417.5; v*v=2835; v=53.2447 m/s. Even closer to go. No per first second reaching speed 3 m/s, per second second 6, per third 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45 and this all is 15 seconds per which stone reaching 472.5 m, but still cann't get that mah=mvv/2.
Edit5: Maybe everybody wrong and acceleration a means, not how much speed increasing in one second, but how much flying distance increasing in one second? Let's check this out: 3(m)+6(m)+12(m)+15(m)+18+21+24+27+30+33+36+39+42+45+48+51+54=513 meters and this we get in 17 seconds. But then we do not know speed of stone after 17 seconds. But we can calculate it. If 3 m falling per first second then average speed is v=3 m/s. And speed in the end of first second is 6 m/s. And average speed when per second second stone fall 6 m, average speed means 6 m/s, so at ve end we get 54 m/s speed. And now it need just put into two kinetic energy formuls comparision. mah=1*3*513=1539 J. mvv/2=1539; vv=3078; v=55.4797 m/s. Still do not much with 54 m/s. What I doing wrong (supose 1kg stone falling on the ground from 500 m heigh without air resistance and need to know stone velocity when it reach/hit ground, when g=3m/s^2)?
Edit6: Seems I wrongly calculate here: 1.5*(1+3+6+9+12+15+18+21+24+27+30+33+36+39+42)=1.5*316=474 m and it's after 15 seconds. So speed will be 3*15=45 m/s. mgh=1*3*474=1422 J. vv/2=1422, vv=2844, v=53.329. Still wrong.
Let's try everything to calculate clearly. (0+3)/2(m)+(3+6)/2+(6+9)/2+(9+12)/2+ (12+15)/2+(15+18)/2+(18+21)/2+(21+24)/2+ (24+27)/2+(27+30)/2+(30+33)/2+(33+36)/2+ (36+39)/2+(39+42)/2+(42+45)/2+(45+48)/2+(48+51)/2+(51+54)/2= 1.5+4.5+7.5+10.5+13.5+16.5+19.5+22.5+25.5+28.5+31.5+34.5+37.5+40.5+43.5+46.5+49.5+52.5=486 m. And it's after 18 seconds. mah=1*3*486=1458 J. mvv/2=1458, vv=2*1485=2916, v=54 m/s. And now it's finaly everything much. After 18 seconds stone falling with a=3 m/s/s will reach ground with speed 54 m/s. And if there no resistance, but only force in cosmos F=ma=1*3, then job will be A=mat=1*3*18=54 kg*m/s or energy units. So with friction and moving stone on table 486 metrs will cost 1458/54=27 times more energy. Almost all energy will go to friction, only small part of it will be used for stone acceleration. As far as I can see now.
Edit7: Back to the rocket problem. Rocket of mass 1000 kg have engine which in free space accelerating with force F=ma=1000*13=13000 N. So a=13 m/s/s. Ignore air friction and rocket fuel waist. Only gravity acceleration is g=10 m/s/s. So if I would put 1000 kg object on table and if I would push it agains force of gm=10000 N and distance 500 meters then I would waist W=mgh=1000*10*500=5*10^6 J. Since there is gravity then rocket will fly with acceleration a-g=13-10=3 m/s^2. Such distance rocket will doen in 18 seconds or let it be 486 meters distance from previous example with stone. So after 18 seconds rocket will get speed 54 m/s and would fly distance 486 . Flying such distance with no gravity force (because gravity force we calculate separatly and get 5 milions joules) rocket engine will spend same amount of energy as if it I moving on table against force 3+3+3+3+3... meters distance sine rocket flying 18 seconds then it's willl be like distance 18*3=54 meters. So energy will be spend A=m*a*S=1000*3*54=162000 J or A=m*a*t=1000*3*18=54000 J - we will later figure out which is not wrong. So rocket total waist 5*10^6 + 162000=5162000 J. And get moment p=mv=54000 kg*m/s. This moment can lift rocket distance of like if rocket would falling after some amount of seconds. This distance of falling with 10 m/s/s acceleration of gravity will be (0+10)/2+(10+20)/2+(20+30)/2=5+15+25=45 m. So after about 3 or more precisly 3.1 second rocket would fall distance 54 m. And rocket speed from moment is also 54 m/s. So rocket will fly another second 54 m height till it stop. So rocket total height will be 486+54=540 meters and it's only with energy of 5162000 J (more precisly it would be mgh+maS=1000*10*486+1000*3*54=5022000 J).
Now assume rocket engine pushing rocket in free space with acceleration of 20 m/s^2 and with gravity it would gives us only 10 m/s^2. Rocket spend energy for overcoming gravity all the time will be mgh=1000*10*500=5*10^6 J. Till rocket reach 500 meters it till took (0+10)/2+(10+20)2+(20+30)/2+35+45+55+65+75+85+95=500 meters. So in egzactly 10 seconds rocket will reach height 500 m and speed 100 m/s. So rocket moment this time will be about two times bigger mv=1000*100=10^5 kg*m/s. And rocket when turn off engine at 500 m height will fly up about 5 seconds because (0+10)/2+(10+20)2+(20+30)/2+35+45=125 or in 4 seconds it will be 80 m heigh, so in about 4.5 seconds rocket would fall 100 meters. And it have 100 m/s speed. So rocket will lift itself 100 more meters and will stop. On the over hand perhaps I should calculate what average speed rocket gonna make that it would be 100 m/s if rocket would fall some amount of time because of gravity. But this is also is average speed or average distance flyied per roughly 4.5 second. So this time total wasted energy will be mgh+maS=5*10^6+1000*10*S, S this time will be a*t=10*10=100 so W=mgh+maS=5*10^6+1000*10*100=6*10^6 J. Or mgh+mat=5*10^6+1000*10*10=5100000 J. Which answer is true I will try to figure out by if rocket always fall from 500 m height and how much energy need for acceleration and after what time it would stop if instantly accelerated rocket (with moment mv=1000*v) would be puted on the ground and for it would be let to fly until it will stop.
Edit8: So if would rocket fall from 500 m height without friction on earth ground and if g=10 m/s^2. Then rocket would fall with final speed 100 m/s. To reach such speed acording to A=maS=maat formula need A=m*a*a*t=1000*10*10*10=10^6 J. And rocket have 10^5 kg*m/s moment. Now if we instantly teleportate rocket witch in space flying with 100 m/s speed to the earth ground then teoreticaly it should fly to height of 500 m and to stop. So I calculate, that energy would be wasted 5 times less. How it can be? Maybe because after first second acceleration took from rocket 10 m/s speed and after first raising second rocket speed will drop to 90 m/s, but not to 95 m/s like if would be with average deaceleration gravity speed 100-5=95. So then after ten seconds rocket speed will fall to 0 m/s, because after second second it would be 80 m/s. But even with average speed of deaceleration each time substrackting should lift rocket to desired 500 meters. 100-5-15-25-35-45-55-65-75-85-95=-400, so now I see, that it's not enough such a speed of 100 m/s to fight with gravity till 500 m. So rocket should be accelerated to 500 m/s to overcome gravity and reach 500 m and only then to stop.
Edit9: So to accelerate 1000 kg rocket with engine which accelerating rocket with 3 m/s/s acceleration to 54 m/s need 18 seconds and now how much seconds need to accelerate rocket to 486 m/s? Don't you think guys it would be easier to find out with mvv/2, but I not sure if h supose to be 486 m in mah formula, so need to do old work 3, 6, 9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,61, or just simply 486(m/s)/3(m/s^2)=162 seconds. So after 162 seconds rocket reach 486 m/s speed if rocket engine in empty free space accelerating it with 3 m/s^2. So according A=maat=maS=1000*3*3*162 and we get then energy against gravity. And here m is rocket mass 1000 kg, first a means toghter with m which force F=ma=1000*3=3000 kg*m/s^2 resist through the distance when we pushing object of certain mass m=1000 kg. Second a means that we also pushing objects 3 m/s speed, because since is friction object speed do not increasing and times t we get a*t=S - the distance which we push 1000 kg object against force 3000 N at constant speed 3 m/s. This is real job if you put on table 1000 kg rocket and will push it against friction which would be 3000 N at speed 3 m/s then you will do job A=maat=1000*3*3*162=1458000 J. And rocket with moment mv=1000*486=486000 kg*m/s is able to reach 486 m height if it would be teleportated with such moment on the ground and instantly will fly up. So after first second rocket speed will deacres from 486 m/s to 486-(0+10)/2=481 m/s, after second second deacreas to 481-(10+20)/2=466 m/s, after third second to 466-(20+30)/2=441 m/s, then to 441-35=406 m/s, then 406-45=361 m/s, then 361-55=306 m/s, 306-65=241, 241-75=166, 166-81, 81-95=-14 m/s. So after 9 seconds speed decrease to 81 m/s and after about 9.8 second speed will deacrease to zero and rocket will stop at 486 m height.
Last edited: Oct 19, 2009
24. Oct 18, 2009
### Staff: Mentor
They do have a certain appeal, and I actually don't have trouble giving time units of length, but I could never quite get my head around doing the same for mass and charge for some reason. I guess it is just a failure of imagination on my part.
25. Oct 18, 2009
### arildno
I think geometrized units are silly. Why not chargified units instead?
They would sizzle, at least..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook | 7,357 | 25,965 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2018-30 | latest | en | 0.969943 |
https://ragazzi-pizza.com/tips-for-making-pizza/quick-answer-how-many-calories-in-a-medium-dominos-pizza.html | 1,680,050,577,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00251.warc.gz | 520,639,816 | 20,254 | # Quick Answer: How Many Calories In A Medium Domino’s Pizza?
## How many calories are in a Domino’s Pizza?
A single slice of Domino’s pizza cut from a medium hand-tossed pie is between 200 and 290 calories.
## How many calories are in a medium pepperoni pizza from Dominos?
There are 215 calories in 1 slice of Domino’s Pizza Hand-tossed Pepperoni Pizza ( Medium ).
## How many calories in a medium Domino’s Pizza UK?
Nutrition
Calories 699 (2924 kJ)
Saturated Fat 13 g 65%
Carbohydrate 73.8 g
Sugars 22.8 g
Fibre 3.9 g
310 Calories
Fat 19 g
Carbs 23 g
Fiber 0.5 g
Protein 12 g
## How many slices is a serving of Domino’s Pizza?
Six slices, just as with the large. Though if you want 12 slices, you can cut those big pieces in half — or order two. You only live once!
## What is a serving of pizza?
And, it turns out, it really is handy (in that all you need are your hands): one serving of pizza equals the size of one or two of your hands. Says Young: “Since portion size is highly individual, it’s partly a matter of how many of your daily calories you might want to ‘spend’ on—for example— pizza that day.
You might be interested: FAQ: Where Is The Nearest Pizza Hut?
## How many slices are in a medium pizza?
Medium pizzas run 12 inches in diameter and will give you about eight slices.
## How many calories are in a medium sized pizza?
Other common serving sizes
Serving Size Calories
1 pizza (12″ dia) 1904
1 medium (13″ dia) 2236
1 pizza (14″ dia) 2592
1 large (15″ dia) 2975
10
## How many calories should I be eating to lose weight?
And if you eat fewer calories and burn more calories through physical activity, you lose weight. In general, if you cut 500 to 1,000 calories a day from your typical diet, you’ll lose about 1 pound (0.5 kilogram) a week.
## How many slices is a medium Domino’s?
A personal pizza has four slices, a small has six, a medium has eight and our large pizza has 10 slices.
## Which Domino’s Pizza has the least calories?
The lowest calorie Domino’s pizza is the chain’s Thin Crust Veggie Pizza. The lowest calorie Domino’s pizza option is their Thin Crust Veggie Pizza with light cheese. Each slice has 135 calories, 250 milligrams of sodium, two grams of sugar, and two grams of saturated fat.
## How many calories is a small Domino’s Pizza?
Domino’s Small Pizza Calories 190 calories. 7 grams of fat. 3 grams of saturated fat.
## How many calories is 3 slices of thin crust pizza?
There are 668 calories in 3 pieces of Thin Crust Pizza with Meat.
## Is Domino’s thin crust healthy?
The thin crust option is always healthier at Domino’s, but I included the hand-tossed veggie and chicken pizza for variety. With the thicker crust and flavor from all the toppings, you’d think you’re indulging.
You might be interested: FAQ: How Long Does Pizza Delivery Take?
## Does thin crust pizza have less calories?
A thin crust pizza will contain fewer calories than a pizza with a thicker dough, according to Amer. “With a thin crust pizza, you’ll eat a better balance of carbohydrates, protein, and fat than you would with a thicker crust,” says Amer. | 780 | 3,118 | {"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-2023-14 | latest | en | 0.913712 |
https://physics.stackexchange.com/questions/137103/does-centripetal-force-change-direction | 1,566,679,062,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00324.warc.gz | 589,514,456 | 31,382 | # Does Centripetal Force Change Direction?
I have been struggling with this question that I have,
"Does centripetal force change direction?"
From every point, it points to the center. But if we draw its vector, we draw the tails from different point. How can we tell whether a vector has changed direction or not? Do we considered the vector's head too?
You have to consider the head in order to know what direction the vector is in, right? After all, that's half the point of having a head - to let you know what direction the vector points. (But keep in mind that, except for position or displacement vectors, the position at which you draw the head doesn't mean anything. The vector itself is located at the point where you draw its tail, and the head is just to show which direction it goes in and what its magnitude is.)
Now think about this: what direction is the centripetal force when the object is on the right side of its circle? What direction is the centripetal force when the object is on the top of its circle? Or on the left side? Are those all the same direction?
This is not really relevant to your question, but you have kind of stumbled on an issue that becomes very important in more complicated kinds of physics (specifically, general relativity): if you are comparing vectors at different points in space, how do you know what constitutes the "same direction" at different points? In a curved space, you can't assume that "left" at one point is the same as "left" at another point, for example. There is some fairly complicated math involved in determining how to "translate" a direction from one point to another. But in this case, you're not working in a curved space so you don't have to worry about any of that.
• so "does Fc change direction" – Raman CHAWRESH Sep 26 '14 at 5:38
A vector doesn't have a head and a tail. A vector in this context is something with a magnitude, and except for the zero vector, a direction. Another way to look at it: The displacement vector from the origin to the point with coordinates (1,1,1) and the displacement vector from the point with coordinates (2,2,2) to the point with coordinates(3,3,3) are the the same vector.
The angle between non-zero two vectors in three dimensional Euclidean space is determined by the ratio of the inner product between the vectors and the product of their magnitudes: $\cos\theta = (\mathbf a \cdot \mathbf b)/(||\mathbf a||\,||\mathbf b||)$. Once again, the head and tail don't come into play.
Does the centrifugal force change direction on an object moving along a circular path?
Imagine a kid on a playground roundabout. One moment the centrifugal force needed to keep her on the roundabout points south. A bit later, as the roundabout has made a quarter turn, the centrifugal force is pointing east. It's pointing north after yet another quarter of a turn, then west, then south again.
All that matters is the direction in which the force vector is pointing. If this direction changes over time, it's changing direction. It's as simple as that. You are overcomplicating things by asking about the head and tail of the vector.
• i am still at high school and you i don't understand anything from you answer. but can you answer my main question "Does Fc change direction" on a point that moves on a perfect circular path? – Raman CHAWRESH Sep 26 '14 at 5:37 | 762 | 3,367 | {"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} | 3.90625 | 4 | CC-MAIN-2019-35 | latest | en | 0.967115 |
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Common core worksheets and activities for 2NBT6 Number And Operations In Base Ten Use Place Value Understanding And Properties Of Operations To Add And Subtract. Add up to four two-digit numbers using strategies based on place value and properties of operations. PDF 268 MB This second grade math standards-based place value unit focuses on the Common Core math standards 2NBT1. Write the words. First Grade Common Core Math 1 NBT2 Understand Place Value 1NBT2 Practice provides two ways for students to practice and show mastery of their ability to understand place value. Understand that the two digits of a two-digit number represent amounts of tens and ones.
### First Grade Common Core Math 1 NBT2 Understand Place Value 1NBT2 Practice provides two ways for students to practice and show mastery of their ability to understand place value.
First Grade Common Core Math 1 NBT2 Understand Place Value 1NBT2 Practice provides two ways for students to practice and show mastery of their ability to understand place value. Here is a collection of our common core aligned worksheets for core standard 2NBTA1. This asks students to identify the the value of each digit in a 3-digit number and understand that hundreds tens and ones make up a 3-digit number. Students are asked to write the number of tens and ones below the squares in order to compare each two-digit number. 3NBTA2 Common Core Worksheets CCSSMathContent3NBTA2. Number Lines up to 100 Fill in the missing numbers on the number lines.
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Add up to four two-digit numbers using strategies based on place value and properties of operations. Fluently add and subtract within 1000 using strategies and algorithms based on place value properties of operations andor the relationship between addition and subtraction. 1NBTB2A 10 can be thought of as a bundle of ten ones called a ten. 3NBTA2 Common Core Worksheets CCSSMathContent3NBTA2. Browse this range of example resources and printable teacher stuff to learn more about the quality of Twinkls resources.
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Common core worksheets and activities for 1NBT2 Number And Operations In Base Ten Understand Place Value. Please visit 2NBTA to view our large collection of printable worksheets. Eg 706 equals 7 hundreds 0 tens and 6 ones. It includes 40 distinct problems in two sets 129 slides in all with. Round the numbers to the nearest 100 then add the two.
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Here is a collection of our common core aligned worksheets for core standard 2NBTA1. 1NBTB2A 10 can be thought of as a bundle of ten ones called a ten. It includes 40 distinct problems in two sets 129 slides in all with. Ccssmathcontent2nbta1 Understand that the three digits of a three-digit number represent amounts of hundreds tens and ones. A brief description of the worksheets is on each of the worksheet widgets.
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Not Grade Specific. Not Grade Specific. Printables Fluently add and subtract within 100 using strategies based on place value properties of operations andor the relationship between addition and subtraction. Understand the following as special cases. Price Ascending Most Recent.
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Not Grade Specific. Count on by hundreds to add multiples of 100. Comparing Numbers - up to 1000. Understand that the two digits of a two-digit number represent amounts of tens and ones. 10 can be thought of as a bundle of ten ones - called a.
Source: pinterest.com
Understand that the two digits of a two-digit number represent amounts of tens and ones. Common core worksheets and activities for 2NBT6 Number And Operations In Base Ten Use Place Value Understanding And Properties Of Operations To Add And Subtract. Understand that the two digits of a two-digit number represent amounts of tens and ones. Eg 706 equals 7 hundreds 0 tens and 6 ones. Add on Multiples of 100.
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Download and print 1NBTB2 worksheets to help kids develop this key first grade Common Core math skill. This worksheet covers CCSS MAFS 2NBT11 Understand that the three digits of a three-digit number represent amounts of hundreds tens and ones. Students are asked to write the number of tens and ones below the squares in order to compare each two-digit number. Checkout 2NBTA1 Common Core Worksheets to help kids develop their common core skill. Browse this range of example resources and printable teacher stuff to learn more about the quality of Twinkls resources.
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Eg 706 equals 7 hundreds 0 tens and 6 ones. 1NBTB2B The numbers from 11 to 19 are composed of a ten and one two three four five six seven eight or nine. Comparing Two-Digit Numbers Worksheet. Understand the following as special cases. First Grade Common Core Math 1 NBT2 Understand Place Value 1NBT2 Practice provides two ways for students to practice and show mastery of their ability to understand place value.
Source: pinterest.com
Browse this range of example resources and printable teacher stuff to learn more about the quality of Twinkls resources. Please visit 2NBTA to view our large collection of printable worksheets. Understand that the two digits of a two-digit number represent amounts of tens and ones. Download and print 1NBTB2 worksheets to help kids develop this key first grade Common Core math skill. Printables Fluently add and subtract within 100 using strategies based on place value properties of operations andor the relationship between addition and subtraction.
Source: pinterest.com
Is greater than is equal to or is less than to compare the numbers up to 1000. Round the numbers to the nearest 100 then add the two. Place Value Worksheet 1 2 Common Core State Standards. Students are asked to write the number of tens and ones below the squares in order to compare each two-digit number. Comparing Numbers - up to 1000.
Source: pinterest.com
Source: pinterest.com
Count by 10s to complete the number pattern. Common core worksheets and activities for 1NBT2 Number And Operations In Base Ten Understand Place Value. 1 2 Next This special collection of teacher-made resources has been selected to give you a taste of what Twinkl is all about. Eg 706 equals 7 hundreds 0 tens and 6 ones. Add up to four two-digit numbers using strategies based on place value and properties of operations.
Source: learningyay.com
Count on by hundreds to add multiples of 100. Students are asked to write the number of tens and ones below the squares in order to compare each two-digit number. This is a 5-worksheet preview of a back to school. 1NBTB2B The numbers from 11 to 19 are composed of a ten and one two three four five six seven eight or nine. PDF 268 MB This second grade math standards-based place value unit focuses on the Common Core math standards 2NBT1.
Source: pinterest.com
Count by 10s to complete the number pattern. Browse this range of example resources and printable teacher stuff to learn more about the quality of Twinkls resources. Is greater than is equal to or is less than to compare the numbers up to 1000. It includes 40 distinct problems in two sets 129 slides in all with. Comparing Numbers - up to 1000.
Source: pinterest.com
Understand the following as special cases. Understand the following as special cases. 10 can be thought of as a bundle of ten ones - called a. 1NBTB2B The numbers from 11 to 19 are composed of a ten and one two three four five six seven eight or nine. All worksheets are free for individual and non-commercial use.
This site is an open community for users to share their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.
If you find this site value, please support us by sharing this posts to your own social media accounts like Facebook, Instagram and so on or you can also bookmark this blog page with the title premium 2 nbt 1 worksheets by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website. | 2,020 | 9,305 | {"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} | 3.34375 | 3 | CC-MAIN-2021-21 | latest | en | 0.850739 |
https://testbook.com/question-answer/the-mass-of-the-earth-is-________--5ce403a0fdb8bb431fc6fff0 | 1,680,304,747,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00164.warc.gz | 625,344,779 | 80,055 | # The mass of the Earth is ________.
This question was previously asked in
RRC Group D Previous Paper 1 (Held On: 17 Sep 2018 Shift 1)
View all RRB Group D Papers >
1. 6 × 10-23 kg
2. 6 × 1023 kg
3. 6 × 10-24 kg
4. 6 × 1024 kg
Option 4 : 6 × 1024 kg
Free
CBSE Junior Assistant: Full Mock Test
5.3 K Users
100 Questions 200 Marks 120 Mins
## Detailed Solution
• Let the mass of the earth be M.
• Mass of the other object on earth is m.
• We know the value of Gravitational Constant (G) = 6.67259 × 10-11 Nm2/kg2.
• We know the formula for the force between two objects (F) = G m1 m2/r2.
• Now the radius of the earth (r) = 6.3781 × 106 m.
• We know that F = mg
• Now by substituting the values, we get mg = G M m/r2.
M = g r2/G
M = (9.81) (6.3781 × 106)2/6.67259 × 10-11
M = 6 × 1024 kg.
The mass of the earth (M) is 6 × 1024 kg. | 317 | 836 | {"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} | 3.625 | 4 | CC-MAIN-2023-14 | latest | en | 0.759881 |
https://metanumbers.com/131455 | 1,632,728,069,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00044.warc.gz | 434,321,731 | 7,395 | # 131455 (number)
131,455 (one hundred thirty-one thousand four hundred fifty-five) is an odd six-digits composite number following 131454 and preceding 131456. In scientific notation, it is written as 1.31455 × 105. The sum of its digits is 19. It has a total of 3 prime factors and 8 positive divisors. There are 103,200 positive integers (up to 131455) that are relatively prime to 131455.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 19
• Digital Root 1
## Name
Short name 131 thousand 455 one hundred thirty-one thousand four hundred fifty-five
## Notation
Scientific notation 1.31455 × 105 131.455 × 103
## Prime Factorization of 131455
Prime Factorization 5 × 61 × 431
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 131455 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 131,455 is 5 × 61 × 431. Since it has a total of 3 prime factors, 131,455 is a composite number.
## Divisors of 131455
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 160704 Sum of all the positive divisors of n s(n) 29249 Sum of the proper positive divisors of n A(n) 20088 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 362.567 Returns the nth root of the product of n divisors H(n) 6.54396 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 131,455 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 131,455) is 160,704, the average is 20,088.
## Other Arithmetic Functions (n = 131455)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 103200 Total number of positive integers not greater than n that are coprime to n λ(n) 2580 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 12251 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 103,200 positive integers (less than 131,455) that are coprime with 131,455. And there are approximately 12,251 prime numbers less than or equal to 131,455.
## Divisibility of 131455
m n mod m 2 3 4 5 6 7 8 9 1 1 3 0 1 2 7 1
The number 131,455 is divisible by 5.
## Classification of 131455
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (131455)
Base System Value
2 Binary 100000000101111111
3 Ternary 20200022201
4 Quaternary 200011333
5 Quinary 13201310
6 Senary 2452331
8 Octal 400577
10 Decimal 131455
12 Duodecimal 640a7
20 Vigesimal g8cf
36 Base36 2tfj
## Basic calculations (n = 131455)
### Multiplication
n×y
n×2 262910 394365 525820 657275
### Division
n÷y
n÷2 65727.5 43818.3 32863.8 26291
### Exponentiation
ny
n2 17280417025 2271597220021375 298612812557909850625 39254147274800039413909375
### Nth Root
y√n
2√n 362.567 50.8463 19.0412 10.5622
## 131455 as geometric shapes
### Circle
Diameter 262910 825956 5.4288e+10
### Sphere
Volume 9.51524e+15 2.17152e+11 825956
### Square
Length = n
Perimeter 525820 1.72804e+10 185905
### Cube
Length = n
Surface area 1.03683e+11 2.2716e+15 227687
### Equilateral Triangle
Length = n
Perimeter 394365 7.48264e+09 113843
### Triangular Pyramid
Length = n
Surface area 2.99306e+10 2.6771e+14 107333
## Cryptographic Hash Functions
md5 7275a8591b84cb1a26d6d9701b7e96cb 7d9529eae53a776f4b7ce6002d82a4fb22fdb1bc 1ffb30b9b62691f12df9287c493edb3c56cc7f97308cda7232a9c413e6864b57 8eaba6af6bd8391cd55b8c08ec9e2c83ebc5a402bc548140e13fa0e010e7bfe51a0fe90d853529340f9fa7fab939862dd3a22603aa810799ccb3d9eb728913e7 a54a3bc46355cca0949ad5ca616d2b971b022fe0 | 1,447 | 4,198 | {"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} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.818464 |
https://dsnielsen.com/2018/05/24/encode-decode/ | 1,553,228,818,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202628.42/warc/CC-MAIN-20190322034516-20190322060516-00406.warc.gz | 473,084,125 | 12,105 | # Encode/decode
There’s a very neat way of encoding any set as a set of ordinals, which has the somewhat peculiar feature of it being hard (which here meaning that it requires the axiom of choice) to encode sets, but easy to decode them. Like some kind of a very ineffective crypto-system.
Let’s start out with the encoding process. If we’re given a set $X$ and want to encode it as a set of ordinals, we first “unpack” our set, meaning that we pass to its transitive closure $\text{tc}(\{X\})$, which is defined recursively as $\text{tc}_0(\{X\}):=\{X\}$,
$\text{tc}_{n+1}(\{X\}):=\text{tc}_n(\{X\})\cup\{y\mid\exists x\in\text{tc}_n(\{X\}): y\in x\}$
and finally $\text{tc}(\{X\}):=\bigcup_{n<\omega}\text{tc}_n(\{X\})$. Now, our next step is to use the axiom of choice to find a well-ordering $\prec$ of $\text{tc}(\{X\})$, because then there exists a unique ordinal $\alpha$ such that
$(\text{tc}(\{X\}),\prec)\cong(\alpha,\in)$.
Let $\varphi:\text{tc}(\{X\})\to\alpha$ be the underlying bijection. We can then take the point-wise image under $\varphi\times\varphi$ of the membership relation $\in\upharpoonright\text{tc}(\{X\})$ to achieve a binary relation $E\subseteq\alpha\times\alpha$, giving us a new isomorphism
$(\text{tc}(\{X\}),\in)\cong(\alpha,E)$.
Letting $\kappa$ be the cardinality of $\alpha$ we also have a bijection $\psi:\alpha\to\kappa$ (without using choice since $\alpha$ is an ordinal), so by now taking the point-wise image of $E$ under $\psi\times\psi$, we end up with a binary relation $F\subseteq\kappa\times\kappa$.
Next up, we have the canonical Gödel pairing function, a bijective function $G:\kappa\times\kappa\to\kappa$ — here we’re using that $\kappa$, being a cardinal, is closed under this function. By taking that point-wise image of $F$ under $G$, we wind up with a set of ordinals $A\subseteq\kappa$This, is our encoding of $X$. The thing that made this encoding hard was that we had to rely on the axiom of choice to pull out the well-ordering $\prec$ along the way.
Now, let’s start out with the encoded set $A\subseteq\alpha$ and begin the decoding procedure. First of all, since the Gödel pairing function works the same way in every universe, we can extract the binary relation $F\subseteq\kappa\times\kappa$ from $A$, by taking the point-wise image of $A$ under $G^{-1}$. Now we consider $(\kappa,F)$, which satisfies the requirements for taking the Mostowski collapse, so we retrieve the isomorphism
$(\text{tc}(\{X\}),\in)\cong(\kappa,F)$
by uniqueness of the collapse, using that $\text{tc}(\{X\})$ is in fact transitive. But now that we have extracted $\text{tc}(\{X\})$, we can from this also extract $X$, since $X$ is the unique element of $\text{tc}(\{X\})$ which isn’t an element of any $Y\in\text{tc}(\{X\})$. This encoding procedure doesn’t refer to the well-ordering $\prec$ at all!
A consequence of this encoding-decoding procedure is that, given any set $X$, we can find a set of ordinals $A$ such that $X\in L[A]$: by simply letting $A$ encode $X$, we get that $A=A\cap L[A]\in L[A]$ as $L[A]$ contains all the ordinals, so we can perform the above decoding process inside $L[A]$ to extract $X$. | 957 | 3,169 | {"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 51, "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} | 3.390625 | 3 | CC-MAIN-2019-13 | latest | en | 0.856886 |
https://socratic.org/questions/what-is-the-average-speed-of-an-object-that-is-still-at-t-0-and-accelerates-at-a-23 | 1,607,192,624,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141748276.94/warc/CC-MAIN-20201205165649-20201205195649-00497.warc.gz | 485,537,245 | 6,375 | # What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =12-2t from t in [3, 6]?
Nov 8, 2016
Given
Acceleration $a \left(t\right) = 12 - 2 t$
$\implies \frac{d \left(v \left(t\right)\right)}{\mathrm{dt}} = 12 - 2 t$
$\implies \int \left(d \left(v \left(t\right)\right)\right) = \int \left(12 - 2 t\right) \mathrm{dt}$
$\implies v \left(t\right) = 12 t - 2 \times {t}^{2} / 2 + c$
$\implies v \left(t\right) = 12 t - {t}^{2} + c$
where t = integration constant
Now by the given condition
at t=0, v(0)=0
So $v \left(0\right) = 0 = 12 \times 0 - {0}^{2} + c \implies c = 0$
Hence
$\implies v \left(t\right) = 12 t - {t}^{2}$
$\implies \frac{d \left(s \left(t\right)\right)}{\mathrm{dt}} = 12 t - {t}^{2}$
$\implies {\int}_{3}^{6} \left(d \left(s \left(t\right)\right)\right) = {\int}_{3}^{6} \left(12 t - {t}^{2}\right)$
$\implies s \left(6\right) - s \left(3\right) = {\left[6 {t}^{2} - {t}^{3} / 3\right]}_{3}^{6}$
$= 6 \cdot {6}^{2} - {6}^{3} / 3 - 6 \cdot {3}^{2} + {3}^{3} / 3$
$= 216 - 72 - 54 + 9 = 99$
Average speed in time [3,6] is
$= \frac{s \left(6\right) - s \left(3\right)}{6 - 3} = \frac{99}{3} = 33 u n i t$ | 533 | 1,178 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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} | 4.46875 | 4 | CC-MAIN-2020-50 | latest | en | 0.496976 |
https://www.cableizer.com/documentation/Pr_gas/ | 1,669,730,834,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00534.warc.gz | 743,975,129 | 6,712 | # Prandtl number for gas
The Prandtl number is a dimensionless number, defined as the ratio of kinematic viscosity (also called momentum diffusivity) to thermal diffusivity.
Equation for humid air is taken from paper by P.T. Tsilingiris: 'Thermophysical and transport properties of humid air at temperature range between 0 and 100°C', 2007
Symbol
$\mathrm{Pr}_{gas}$
Formulae
$0.7215798365-3.703124976{\cdot}{10}^{-4} \theta_{gas}+2.240599044{\cdot}{10}^{-5} {\theta_{gas}}^2-4.162785412{\cdot}{10}^{-7} {\theta_{gas}}^3+4.969218948{\cdot}{10}^{-9} {\theta_{gas}}^4$ humid air @ 1 atm (Tsilingiris2007) $\frac{c_{p,gas} \eta_{gas}}{k_{gas}}$ general formula for gases
Related
$c_{p,gas}$
$\eta_{gas}$
$k_{gas}$
$\theta_{gas}$ | 250 | 729 | {"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} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.676576 |
https://sckool.org/first-quest.html | 1,620,855,020,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989705.28/warc/CC-MAIN-20210512193253-20210512223253-00172.warc.gz | 456,102,185 | 6,162 | First Quest
Date 08.08.2018 Size 19.16 Kb.
Pythagorean Theorem Student Webquest Worksheet Name: ___________________
First Quest: Mastering the Basics - In this section, you will learn to classify triangles according to their angles and sides.
Click on the first Link (#1) in the webquest to begin completing your worksheet.
1. What is the sum of the angles of every triangle? __________.
2. A triangle which all the sides have the same length is called a ______________triangle.
3. A triangle with two sides having the same length is called a _____________ triangle.
4. A triangle in which none of the sides have the same length is called a ___________ triangle.
5. A triangle with a 90 degree angle is called a _____________ triangle.
6. A triangle with an angle greater than 90 degrees is called an ____________triangle.
7. A triangle in which none of the angles are greater than 90 degrees is called an ______________ triangle.
8. A triangle in which all of the angles are congruent to each other is called an ______________ triangle.
Go to the next Link (#2) to answer the following questions. Scroll down to timed test part.
1. How many triangles were you able to correctly classify in 60 seconds by their angles? ___________. Make sure to share this score with your teacher before you go on.
2. Can you beat your own record? How many did you correctly classify this time?
Go to the next Link (#3)
1. How many triangles were you able to correctly classify in 60 seconds by their sides? _________. Make sure to share this score with your teacher before you go on.
2. Can you beat your own record? How many did you correctly classify this time?
Go to the next Link (#4).
1. This one will slow you down! How many third angles were you able to correctly identify in 60 seconds? ___________. Share your results with teacher before going on.
2. Try it again! How many? _____. Share your results with teacher before going on.
Second Quest: Mastering the Pythagorean Theorem
In this section you will learn about a powerful formula for working with right triangles and practice using it to solve practical problems.
Go to the next Link (#5).
1. The Pythagorean Theorem is a powerful tool for finding the length of the third side of a right triangle when the other two sides are given. The Pythagorean Theorem (formula) is __________________________, where A and B are the legs for the triangle and C is the hypotenuse (always directly across from the right angle).
2. Below is an Example of using the Pythagorean Theorem to find the length of the third side of the right triangle described below. Round your answer to the nearest tenth.
One side is 7, the second side is 12, how long is the hypotenuse?
72 + 122 = C2
49 + 144 = C2
193 = C
Use your calculator to find the square root of 193 and, of course, the square root of C2 is just C so...
13.89 = C
****Now you try it. Given a right triangle with one side 8 meters, and the second side 6 meters, how long is the hypotenuse?_____________ meters. Show work.
Go to the next Link (#6)
Show work and label answer. Round answer to the nearest whole #.
Go to the next Link (#7)
1. Solve the Baseball problem. How far did you have to throw the ball? _________
Third Quest: Review on how to apply the Pythagorean Theorem
Complete Worksheet “Working with Pythagorean Theorem” (Get copy from teacher). You must show work and label answers.
1. After completing the above worksheet, show work to teacher. Teacher will give you the O.K. to go to this Link (Review) and check your answers.
Fourth Quest: Solving the Mysterious Bermuda Triangle
In the final section you will research the Bermuda Triangle and offer a logical explanation for a mysterious disappearance.
Go to the next Link (#8).
1. What has happened in the Bermuda Triangle that has given it a
mysterious reputation? _______________________________________
________________________________________________________________________________________________________________________________
1. Give three possibilities to explain why so many ships disappeared in this area.
1. ___________________________________________________________________
2)___________________________________________________________________
3)___________________________________________________________________
Go to the next Link (#9).
1. Although the area is called the Bermuda Triangle, there is a disagreement about the exact location and shape. Using the map on this link give the location of the apexes of the triangle.
From _________________ to ___________________ to ______________________
1. When classified by its sides, what type of triangle does the Bermuda Triangle most closely resemble on the given map? __________________________________
1. When classified by its angles, what type of triangle does the Bermuda Triangle most closely resemble on the given map? __________________________________
Go to the next Link (#10) and read some of the theories offered to explain the disappearances (click Items listed under “Theories on the left-hand side of the screen). Using this information, choose one of the missing ships from Link (#11) and explain the disappearance according to the directions in question 25.
1. Choose a ship from the list, print out (optional) the description of the incident and write an essay explaining what could have happened. Be sure to support our explanation with facts from the theories link. ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ | 1,158 | 6,470 | {"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} | 4.15625 | 4 | CC-MAIN-2021-21 | latest | en | 0.946781 |
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## Question number: 335
Edit
MCQ▾
### Question
If is given, then the value of is (Paper-1)
### Choices
Choice (4) Response
a.
1.6886
b.
1.7023
c.
1.1576
d.
1.4094
## Question number: 336
Edit
MCQ▾
### Question
The perimeter of a rhombus is 70 m and its height is 10 m. Its area is (Paper- I)
### Choices
Choice (4) Response
a.
175 sq. m.
b.
170 sq. m.
c.
190 sq. m.
d.
180 sq. m.
## Question number: 337
Edit
MCQ▾
### Question
A cistern of capacity 7040 liters measures externally and its walls are 5 cm thick. The thickness of the bottom is (Paper-1)
### Choices
Choice (4) Response
a.
30 m
b.
60 cm
c.
80 m
d.
40 cm
## Question number: 338
Edit
MCQ▾
### Question
If 350 coins consist of rupees, 50 paise and 25 paise coins whose values are in the ratio of, the number of 50 paise coins will be (Paper- I)
### Choices
Choice (4) Response
a.
230
b.
160
c.
140
d.
210
## Question number: 339
Edit
MCQ▾
### Question
If a selling price of Rs. 36 results after 40 % discount on the list price of an article, the selling price that would result after 20 % discount on the list price is (Paper - I)
### Choices
Choice (4) Response
a.
Rs. 48
b.
Rs. 18
c.
Rs. 28
d.
Rs. 38
## Question number: 340
Edit
MCQ▾
### Question
If is given, then the value of is (Paper-1)
### Choices
Choice (4) Response
a.
1.4021
b.
1.7023
c.
1.6843
d.
1.5331
## Question number: 341
Edit
MCQ▾
### Question
Nisha’s salary was increased by 20%. In order to bring her salary back to the original value, her new salary must be decreased by (Paper - I)
### Choices
Choice (4) Response
a.
b.
c.
d.
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# Part I: Suppose your bank account will be worth \$7,000.00 in one year.
Part I: This part of the assignments tests your ability to calculate present value.
A. Suppose your bank account will be worth \$7,000.00 in one year. The interest rate (discount rate) that the bank pays is 8%. What is the present value of your bank account today? What would the present value of the account be if the discount rate is only 3%?
B. Suppose you have two bank accounts, one called Account A and another Account B. Account A will be worth \$4,000.00 in one year. Account B will be worth \$9,600.00 in two years. Both accounts earn 5% interest. What is the present value of each of these accounts?
C. Suppose you just inherited an gold mine. This gold mine is believed to have three years worth of gold deposit. Here is how much income this gold mine is projected to bring you each year for the next three years:
Year 1: \$42,000,000
Year 2: \$62,000,000
Year 3: \$99,000,000
Compute the present value of this stream of income at a discount rate of 8%. Remember, you are calculating the present value for a whole stream of income, i.e. the total value of receiving all three payments (how much you would pay right now to receive these three payments in the future). Your answer should be one number - the present value for this gold mine at a 8% discount rate but you have to show how you got to this number.
Now compute the present value of the income stream from the gold mine at a discount rate of 6%, and at a discount rate of 4%. Compare the present values of the income stream under the three discount rates and write a short paragraph with conclusions from the computations.
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# Logarithmic Functions
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## Sam Poli
on 3 June 2010
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#### Transcript of Logarithmic Functions
8.3 Logarithmic Functions What the @#\$% does that mean? y=log x b "y equals the logarithm to the base b of x" Properties of Logarithmic Functions The Domain (x) is the set of positive real numbers
The Range (y) is the set of all real numbers
The x-intercept of the graph is 1.
The y-axis is an asymptote of the the graph
The function is a one to one function One to one function? The inverse of the original function is a function as well
This means that the following is true for Logarithmic Functions: y=log x b x=log y b Expressing Numeral Equations in Logarithmic form 7 =49 2 log 49 = 2 7
log 49 = 2 7
base power equals 7 =49 2 base power equals Expressing Logarithmic equations in Exponential/Numeral form So... uh... how do you think logarithmic equations are expressed in Numeral form?
It's just doing the same thing in reverse Evaluating Logarithmic Equations Goals:
To define and graph logarithmic functions
To solve logarithmic equations
y=log x b Quick Review Finding and Graphing Inverses Graph the function Y=X² and find its inverse Y = X²
Set up an "X/Y" chart
Graph the points (0,0) (-1,1) (1,1) (-2,4) (2,4) Exponential/Numeral Form Logarithmic Form Practicing Logarithmic Form ----> Numeral Form Y = logb X X = b Y
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http://stackoverflow.com/questions/4068044/sorting-polygons-for-correct-alpha-blending-in-directx-9 | 1,469,330,264,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00219-ip-10-185-27-174.ec2.internal.warc.gz | 231,679,947 | 18,761 | Dismiss
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# Sorting polygons for correct alpha blending in DirectX 9?
What is the correct way to sort polygons so that they blend properly? The basic concept I think is to render the furthest polygon first back to closest in order. But what about cases of intersecting polygons?
-
What is the correct way to sort polygons so that they blend properly?
Sort them back to front.
But what about cases of intersecting polygons?
Simply ... don't do it. If you have no choice then you will have to split the polygons along their intersection as you go.
Edit: Its worth bearing in mind that finding the intersections and splitting will be very slow. You could use some sort of acceleration structure to aid you.
Its quite common to use a BSP to sort and split static transparent polys.
-
But then what can I do about things like ID3DXMesh's? Am I supposed to lock the vertex buffer and rearrange the vertices? – soshiki Nov 1 '10 at 11:08
@Meds: Effectively, yes. You shouldn't be storing transparent tris in an ID3DXMesh really. You want some sort of better structure for that ... – Goz Nov 1 '10 at 11:19
I see, what particular properties should the struct have? (I'm assuming you mean vertex struct). – soshiki Nov 1 '10 at 11:34
@Meds: No I don't mean a vertex struct. Well basically all you are sorting is indices. So you "could" just update your index buffer. But it will save time (over locking the vertex buffer) to have the vertex positions somewhere in normal memory so you can sort them without touching the VB and then you just update the index buffer and use that. – Goz Nov 1 '10 at 11:38
Frankly I would go for depth peeling and compositing passes. I have seen some implementation of this algorithm and most of the time peeling 2 or 3 layers is "good enough".
This will also save you from having to managed different data structures for storing your meshes. One drawback is that it can be a bit performance intensive, for example if you have a lot of transparent meshes.
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https://agrimetsoft.com/KNN-WG | 1,660,647,677,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00245.warc.gz | 118,269,532 | 10,861 | # KNN Weather GeneratorA Windows Tool for Generating Weather Data By KNN Weather Generator
In KNN-WG V1.1, you can use ensemble average for N times run.
KNN WG is used in this paper: Prediction of climate variables by comparing the k-nearest neighbor method and MIROC5 outputs in an arid environment
## What is KNN-WG software?
The KNN Weather Generator is a tool for lead time simulation of daily weather data based on K-nearest-neighbor approach. The user can load seven different variables, for example Tmin, Tmax, Rain, Srad, ETo, WSPD, and Humidity. Then, the user can load the input data and run KNN Weather Generator. In this software the user can draw graphs and calculate efficiency criteria: d, NSE, RMSE, MBE, Pearson and Spearman. The user can compare outputs of KNN Weather Generator with other models such as Lars-WG, SDSM, CMIP5, and etc. One of the most advantages of the KNN Weather Generator software is that the user can select every number of variables (from 7 variables) to generate future data.
KNN Weather Generator
### What is non-parametric techniques?
Non-parametric re-sampling procedures are an alternative to generating daily weather data. Unlike parametric alternatives for time series generation, non-parametric approaches generate new values by conditionally resampling past observations using a probability rationale. Generally, a statistical method for generating daily weather sequences needs to consider the statistical dependence of the weather variables with each other on the same day. By finding this relation, we can generate future data (by similarity between days). This method can be employed on the assumption that the weather during the target year is analogous to the weather recorded in the past.
Non-parametric weather generators have evolved as a simple way to simulate climate data at multiple sites while avoiding the limitations of the parametric and semi- parametric approaches. A common drawback is that most parametric and semi-parametric models require statistical assumptions
regarding the probability distributions of climate variables.
### What is KNN Weather Generator method?
The K-nearest neighbors (K-NN) is an analogous approach. This method has its origin as a non-parametric statistical pattern recognition procedure to distinguish between different patterns according to a selection criterion. Through this method, researchers can generate future data. In other words, the KNN Weather Generator is a technique that conditionally resamples the values from the observed record based on the conditional relationship specified. The KNN Weather Generator is most simple approach.
The most promising non-parametric technique for generating weather data is the K-nearest neighbor (K-NN) resampling approach. The K-NN method is based on recognizing a similar pattern of target file within the historical observed weather data which could be used as reduction of the target year (Young, 1994; Yates, 2003; Eum et al., 2010). The target year is the initial seed of data which, together with the historical data, are required as
input files for running the model. This method relies on the assumption that the actual weather data observed during the target year could be a replication of weather recorded in the past. The k-NN technique does not use any predefined mathematical functions to estimate a target variable.
Actually, the algorithm of this method typically involves selecting a specified number of days similar in characteristics to the day of interest. One of these days is randomly resampled to represent the weather of the next day in the simulation period. The nearest neighbor approach involves simultaneous sampling of the weather variables, such as precipitation and temperature. The sampling is carried out from the observed data, with replacement.
The KNN Weather Generator is widely used in agriculture (Bannayan and Hoogenboom, 2009), forestry (Lopez et al., 2001) and hydrology (Clark et al., 2004; Yates et al., 2003).
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https://www.physicsforums.com/threads/atwoods-machine-lab-help-proportional-error.771457/ | 1,531,734,187,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589237.16/warc/CC-MAIN-20180716080356-20180716100356-00312.warc.gz | 961,281,771 | 14,500 | # Homework Help: Atwood's Machine Lab Help (proportional error)
1. Sep 17, 2014
### atir.besho
I have attached the lab
right now I am stuck on proportional error in multiple variables
I wanna take the lab step by step so i can learn how to do calculate all the variables from now on.
I have lab partners during the lab but I am completely clueless when I am working on it on my own.
Can someone just explain the formula needed to get the proportional error please?
I know I need 1 gram as uncertainty.
#### Attached Files:
• ###### lab 2 pg 1.jpg
File size:
30.8 KB
Views:
440
2. Sep 17, 2014
### Simon Bridge
For independent measurements $x\pm\sigma_x$ and $y\pm\sigma_y$, the proportional errors are given by:$$p_x=\frac{\sigma_x}{x}, p_y=\frac{\sigma_y}{y}\\ z=xy\implies p^2_z=p_x^2+p_y^2$$ | 230 | 805 | {"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": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-30 | latest | en | 0.823091 |
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