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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.scribd.com/document/79166430/Assignment-a-1 | 1,566,642,804,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027320156.86/warc/CC-MAIN-20190824084149-20190824110149-00234.warc.gz | 959,964,680 | 57,319 | You are on page 1of 9
# Engineering Economics
Content:
Introduction 3 Overview .3 Methodology ...4 Results and recommendations 5 Case Study 1 ..5 Case Study 2 ..6 References ..6 Appendix A ..7 Appendix B ..8 Appendix C ..9
Introduction:
This is an engineering economics analysis of a co-generation project. With hydro costs spiraling out of control, companys decision is to build their own facilities to produce enough power, to satisfy their own needs and some surplus to sell to other users. The facility would be a steam generator to drive the turbines with natural gas fuel, considering that consumption of the company is around 1 MW (megawatt) per week. The power plant was estimated to have a life for 20 years. In additional, toward the purchase of hydrogenation equipment the company would qualify for tax credits. This analyst will estimate the most that company should pay for the installation for such power plant to yield capital project return of 35%. IRR feature in excel should be used.
Overview:
The decision is made to install and use their own facility to satisfy their own needs for electrical energy for 20 years with following reference data of the assignment (See Appendix C) 1. \$100Million paid last year in power.
2. Natural gas cost estimated \$59Million per year. 3. Maintenance approximate cost \$1 million per year (six new employees hired). 4. Sales to other consumers, cash inflow for The Co expected \$5Million per year. 5. Ontario governments tax credit toward equipment purchase \$10Million. 6. Capital project return 35%(MARR=35%)
7. Power plant life 20 years and must be dismantled at a time \$20 Million.; 8. Overhaul in year 11must be performed \$20 Million. I decided to show the case if company continue to pay to Hydro for the energy consumed, if they do not install their own producing energy facility. It was interesting to me to see what the Present Value is in this case: (Series Present Worth factor is used), (P/A, 0.35, 20) =100000000*(1+0.35)^20/0.35(1+0.35)^20=285007841.4 or approximately 285 million Present Worth. There is cash flow diagram to visualize this case: See Appendix A
Methodology:
1. First is defined what is the cash flow within the project, producing electrical
energy for 20 years. Bellow all in and out cash flows are brought to one pattern of cash flow. Every year the steel company consumes energy for \$100Million .If the produce their energy this amount is inflow for them, plus \$5Million from sales of energy. Accordingly the costs (outflows) for their energy are, \$59Million for naturalgas, \$1Million for maintenance, totally \$60Million per year. Two times outflow considered in year 11and year 20, \$20Million each. Total inflow per year: \$100Million+\$5Million=\$105Million Total outflow per year (All except year 11 and 20): \$59Millin+ \$1Million=\$60Million Outflow per year (year 11 and 20 only): \$59Millin+\$1Million+ \$20Million=\$80Million In an Outflow loaded on Excel. See Appendix B (Cash Flow Diagram See Appendix A)
2. The theory of the method used in the project is as follows:
Present worth of disbursements=present worth of receipts*(P/A, i*, N) or (P/A, i*, N) = Present worth of disbursements/ Present worth of receipts
3. As the assignment reference suggests methods used in this work are doing the problem
using the IRR feature in Excel. The principle of IRR method is analogous. This is when the project just breaks even. We will invest in any project that has IRR equal or exceeding the MARR. (The MARR of my project is accepted as35% according to reference) Next is shown excel calculation of NPV (Net Present Value) .Calculations include only positive cash flow without first payment or negative cash flow(formulae above could be used). For the purpose calculations 35% interest is used.
NPV(without first cost or how much we are willing to pay) \$127,467,111.24 4. Tax credits consideration: It results from the fact that in taxed company the first cost
immediately gives rise to future tax savings as defined by the host countrys tax rules. In this particular case since we already know amount of 10 million tax deduction. The future cost(FV) translated as present worth( Present Worth Factor ) is used, because taxes are paid after one year ( end of the tax year).Therefore: P=\$10000000(P/F, 0.35, 1) =\$7407407.40
## Results and recommendations:
The case studies presented in diagrams and text show a classical project. This is a typical example repeated two times with different diagrams and charts, and provides us with a good sense what is happening within the project. The installation cost is negligible and is considered to take no significant time for the project. Case 1 and 2 have IRR higher than the MARR. Both cases could be used to answer what is the most that Dofasco should pay max for the equipment. The number that is in this analysis is \$120000000, (\$120Million) Cash flow Diagram (See Appendix A)
Case Study 1:
The interest rate at which curve cross the horizontal axis where Present Worth is zero (PW=0) is by definition the IRR. In this case I visualize with the included graph Fig.1 situation shown in Methodology part as Excel calculation. Theoretically could be explained as receipts are my annuities for 20 years equal to disbursements (which are my onetime first payment). It declares my effort to find out how much to pay for the equipment, tax credits subtracted, solely in order to solve the problem. P=\$120059703.8 first cost used (See Appendix B)
IRR is 37%>MARR35%
Fig.1
Case Study 2:
This is the same case only difference is amount of the first payment(price of the equipment).The case is with less first payment than case study 1 which give as higher IRR=45% and bigger Present Worth .This case gives better and more stable project in long term regarding eventual future concussion. This case is another version to compare and see more clearly the situation in my view. Fig 2 visualizes the case. P=\$120059703.8 first cost used (See Appendix B).
IRR is 45%>MARR35%
Fig.2
References:
Global Engineering Economics, Fourth Edition Fraser, Jewkes, Bernhardt, Tajima Engineering Economics EIT Review Cash Flow Evaluation Miller
Appendix A
## PW (Do Nothing) =\$100000000*(P/A, 0.35, 20) = \$285007841.4
Cash Flow Diagram (Do nothing) \$ 100M \$ 100M \$100M \$ 100M \$100M
## \$100M \$100M \$100M \$100M
\$100M
1 2 3 4 5 6 Cash7Flow Diagram 8 9 10 15 16 17 18 19 20 \$45M \$45M \$45M \$45M \$45M \$45M \$45M \$25M \$45M \$45M \$25M
11
12
13
14
1 15
2 16
3 4 17 18 19
5 20
10
11
12
13
14
## First cost (Equipment price)
\$120Million
Table Shows how Fig 1 and 2 are made NPV(\$) as function of Interest Rate (interest rate)i \$152,511,699.43 15% \$91,861,708.20 20% \$52,081,253.37 25% \$24,198,179.42 30% \$3,559,928.94 35% -\$12,470,796.32 40% -\$25,542,142.42 45% -\$36,833,437.86 50% -\$47,386,869.10 55% -\$58,423,723.77 60%
Appendix B
Case Study 1: Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 In 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 Out -120059703.8 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -80000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -80000000 Cash Flow -120059703.8 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 25000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 25000000 Year 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Case Study 2 In 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 10500000 0 Out -107407407.4 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -80000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -60000000 -80000000 Cash Flow -107407407.4 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 25000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 45000000 25000000
IRR
37%
IRR
45%
Appendix C
CO-GENERATION CATCHING ON media, In a move mirroring recent operational initiatives at Medical Centre, Countryss largest steelmaker announced intention to study the introduction of hydro co-generation equipment at its sity operations. The spokesman H,s reported that Company, Countrys largest user of hydro, consumed just over \$100 million in power last year. We have to consider our options with hydro costs spiralling out of control, he said. It estimates that it would be able to produce enough power to satisfy its own needs as well sell some surplus power to energy hungry users in the Sity area including rival Telco. While sketchy on specific costs, Smith indicated that operating the plant would involve hiring six new employees and natural gas would be the fuel of choice for generating steam to drive the turbines. Natural gas costs were estimated at \$59 million a year while operating and maintenance manpower along with general power plant maintenance would cost approximately \$1 million a year. In addition to satisfying The companys voracious appetite for power (it consumes some 1 megawatts per week), Smith indicated that energy sales could generate some \$5 million dollars per year for The Company. In a related story, the Government today announced that tax credits would be available to companies considering the introduction of hydro generation equipment. The officials were confident that it might qualify for up to \$10M in credits towards the purchase of equipment. The Company, as a matter of policy, demands that all capital projects return 35%. The power plant was estimated to have a life of 20 years and must be dismantled at a cost of \$20M at that time. In addition, an overhaul must be performed sometime in year 11 to promote efficient operations at a cost of \$20M. As The company starts down the road of inviting vendors in to provide project cost estimates, what is the most that The Co should pay for the installation of such a power plant to yield 35%? Make the decision based on:
1. Doing the problem long hand including a time diagram 2. Doing the problem using the IRR feature in Excel | 2,932 | 10,523 | {"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-2019-35 | latest | en | 0.9264 |
http://www.cfd-online.com/Forums/fluent/50793-swirl-tumble-ratios-ic-chamber-turbulence-print.html | 1,477,444,703,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988720471.17/warc/CC-MAIN-20161020183840-00365-ip-10-171-6-4.ec2.internal.warc.gz | 361,518,219 | 2,170 | CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- FLUENT (http://www.cfd-online.com/Forums/fluent/)
- - Swirl and Tumble Ratios in IC Chamber - Turbulence (http://www.cfd-online.com/Forums/fluent/50793-swirl-tumble-ratios-ic-chamber-turbulence.html)
Hussam March 7, 2009 00:41
Swirl and Tumble Ratios in IC Chamber - Turbulence
I am a novice who is contrasting the turbulence characteristics of the flow field inside a combustion chamber of a 4 stroke direction injection IC engine during the intake and compression stroke resulting from crankshaft RPM variation. (i.e. 2000 RPM vs 3000 RPM, for example) In order to analyse the turbulence intensity, I need to find out the swirl and tumble ratios in the control volumes. The experiment is performed in Fluent 6.3.26 2DDP on a single cylinder four stroke engine under unsteady turbulent conditions using the k-ε turbulence model.
Coming to the crux of my dilemma, I have been unable to find an option for calculating either tumble or swirl ratios in the Solve --> Monitors ---> Volume option boxes. Mathematically, the swirl and tumble ratios (in a non-dimensional form) are calculated by the effective angular speed of in-cylinder air motion divided by the engine speed and the effect angular speed is the ratio of the angular momentum to the angular inertia of moment. Is it possible to get Fluent to give me either the swirl/tumble ratios directly or the angular momentum/angular inertia in the x, y and z directions? I am at a complete loss as to which parameters to use.
Thank you for your help in advance!
All times are GMT -4. The time now is 21:18. | 389 | 1,645 | {"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} | 2.796875 | 3 | CC-MAIN-2016-44 | latest | en | 0.833918 |
https://discuss.codecademy.com/t/confusion-with-perimeterbox/76438 | 1,537,506,016,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00520.warc.gz | 485,361,706 | 4,880 | # Confusion with perimeterBox
#1
So I'm on the exercise that tackles creating a perimeter box, and I've played around with it and eventually got it to work but I have no idea why and I'm certainly sure this is not what I should've done.
Can anyone explain why this works and what exactly I did here because I have no idea:
``````var perimeterBox = function(length, width) {
return length (+ 3); + width (+ 9);;
};``````
the root of my confusion is that the exercise tells me to create a parameter with length and width, however I'm not sure if it's asking me to write length, width or if it's asking me to actually put a number for length and width.
I've tried another code more in-line with what it's asking, but it's come back with an error and honestly I'm just not getting it or understanding this part at all:
``````var perimeterBox = function(length, width) {
return length + length + width + width;
};``````
"Oops, try again. Make sure to call your perimeterBox function with any value for length and width. "
This doesn't work, and I'm not sure why - in the info section it mentions adding length + length + width + width but that's just confusing.
at the end it asks me to call the function perimeterBox but I've not managed to figure out how to do that, I feel I'm missing something here.
#2
Woah that code raises all sorts of red flags I would say you just confused the SCT until it let you pass,
This exercise is just based on the formula for perimeter which is `P = 2l + 2w`,
programmatically that would be written assuming you have thearguents length and width,
`perimeter = (length *2) + (width * 2)`
#3
Hah, figured as much -
What you've said makes some sense to me, I'll read over it a few more times and give it another bash.
Thanks!
#4
Ah! I get it now, okay. I'm slow - Thanks bandit.
#5
Programming takes time and patience, so in a way we're all slow
#6
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 491 | 1,998 | {"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-2018-39 | latest | en | 0.965198 |
http://www.convertaz.com/convert-28-c-to-pt/ | 1,560,985,583,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999066.12/warc/CC-MAIN-20190619224436-20190620010436-00136.warc.gz | 227,225,485 | 16,969 | # Convert 28 c to pt
## Conversion details
To convert c to pt use the following formula:
1 c equals 0.4399385 pt
So, to convert 28 c to pt, multiply 0.4399385 by 28 i.e.,
28 c = 0.4399385 * 28 pt = 12.3182779 pt
For conversion tables, definitions and more information on the c and pt units scroll down or use the related c and pt quick access menus located at the top left side of the page.
c is the symbol for cup (metric)
Pt is a common alias of the unit 'pint (Imperial)'
### From 0.10 to 4.00 c, 40 entries
0.1 c = 0.0439938497 pt
0.2 c = 0.0879876993 pt
0.3 c = 0.131981549 pt
0.4 c = 0.1759753986 pt
0.5 c = 0.2199692483 pt
0.6 c = 0.263963098 pt
0.7 c = 0.3079569476 pt
0.8 c = 0.3519507973 pt
0.9 c = 0.3959446469 pt
1 c = 0.4399384966 pt
1.1 c = 0.4839323463 pt
1.2 c = 0.5279261959 pt
1.3 c = 0.5719200456 pt
1.4 c = 0.6159138952 pt
1.5 c = 0.6599077449 pt
1.6 c = 0.7039015946 pt
1.7 c = 0.7478954442 pt
1.8 c = 0.7918892939 pt
1.9 c = 0.8358831435 pt
2 c = 0.8798769932 pt
2.1 c = 0.9238708429 pt
2.2 c = 0.9678646925 pt
2.3 c = 1.0118585422 pts
2.4 c = 1.0558523918 pts
2.5 c = 1.0998462415 pts
2.6 c = 1.1438400912 pts
2.7 c = 1.1878339408 pts
2.8 c = 1.2318277905 pts
2.9 c = 1.2758216401 pts
3 c = 1.3198154898 pts
3.1 c = 1.3638093395 pts
3.2 c = 1.4078031891 pts
3.3 c = 1.4517970388 pts
3.4 c = 1.4957908884 pts
3.5 c = 1.5397847381 pts
3.6 c = 1.5837785878 pts
3.7 c = 1.6277724374 pts
3.8 c = 1.6717662871 pts
3.9 c = 1.7157601367 pts
4 c = 1.7597539864 pts
Click here for a list of all conversion tables of c to other compatible units.
## cup (metric)
Cup (metric) is a unit of measurement of volume. The definition for cup (metric) is the following:
A metric cup is equal to 250 × 10-6 cubic meters.
The symbol for cup (metric) is c
## pint (Imperial)
Pint (imperial) is a unit of measurement of volume. The definition for pint (imperial) is the following:
An Imperial pint is equal to 1/8 Imperial gallon.
The symbol for pint (Imperial) is pt (Imp)
## Other people are also searching for information on c conversions.
Following are the most recent questions containing c. Click on a link to see the corresponding answer.
1 Dm =how many cm
120 ml to ounces
cm to liters
yards to inches
hectares
256 cm to MM
525 mL is how many cL
convert 20 kilograms to pounds
convert 55 centimeters to inches
centimeter
Home | Base units | Units | Conversion tables | Unit conversion calculator | 936 | 2,414 | {"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-2019-26 | longest | en | 0.651469 |
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1021 | 1,519,374,793,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814538.66/warc/CC-MAIN-20180223075134-20180223095134-00445.warc.gz | 7,874,021 | 4,413 | Welcome to ZOJ
Problem Sets Information Select Problem Runs Ranklist
ZOJ Problem Set - 1021
The Willy Memorial Program
Time Limit: 2 Seconds Memory Limit: 65536 KB
Willy the spider used to live in the chemistry laboratory of Dr. Petro. He used to wander about the lab pipes and sometimes inside empty ones. One night while he was in a pipe, he fell asleep. The next morning, Dr. Petro came to the lab. He didn't notice Willy while opening the valve to fill the pipes with hot water. Meanwhile, Stanley the gray mouse got what was going to happen. No time to lose! Stan ran hard to reach the valve before Willy gets drawn, but... Alas! He couldn't make it!
Poor Willy was boiled in hot water, but his memory is still in our hearts. Though Stan tried his best, we want to write a program, in the memory of Willy, to compute the time Stan had, to rescue Willy, assuming he started to run just when the doctor opened the valve.
To simplify the problem, assume the pipes are all vertical cylinders with diameter 1 cm. Every pipe is open from the top and closed at the bottom. Some of the pipes are connected through special horizontal pipes named links. The links have very high flow capacity, but are so tiny that at any given time, the volume of water inside them is negligible. The water enters from top of one of the pipes with a constant rate of 0.25�� cm3/sec and begins to fill the pipe from the bottom until the water reaches a link through which it flows horizontally and begins to fill the connected pipe. From elementary physics we know if two pipes are connected and the surface of the water is above the connecting link, the level of water in both pipes remains the same when we try to fill one of them. In this case the water fills each pipe with a rate equal to half of the rate of incoming water. As an example, consider the following configuration:
First, the lower 2 centimeters of the left pipe is filled with water at full rate, then, the lower 3 centimeters of the right pipe is filled, and after that, the upper part of the two pipes are filled in parallel at half rate. The input to your program is a configuration of pipes and links, and a target level in one of the pipes (the heavy dotted line in the above figure). The program should report how long it takes for the level of water to reach the target level. For the above configuration, the output is 9 seconds.
It is assumed that the water falls very rapidly, such that the time required for the water to fall can be neglected. The target level is always assumed to be a bit higher than the specified level for it. As an example, if we set the target point to level 4 in the left pipe in the figure above, the elapsed time for water to reach that target is assumed to be 5 (not 2), Also note that if the water reaches to the top of a pipe (say in level x), it won't pour out outside the pipe until empty spaces in connected pipes below level x are filled (if can be filled, i.e. the level of water reaches the connecting links). (Note that there may be some links at level x, to which water is entered). After all such spaces are filled; the water level would not go up further.
Input
To describe positions, we assume the coordinates are expressed as (x, y) and the origin lies in the top-left of all pipes and links. (Note that y coordinates are increased downwards). All coordinates are integer numbers between 0 and 100, inclusive.
The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is p (1 <= p <= 20), the number of pipes, followed by p lines, each describing a pipe. Each pipe description line consists of three numbers. The first two are (x, y) coordinates of the upper-left corner of the pipe and the third number is the height of the pipe (at least 1 cm and at most 20 cm). Note that diameter of each pipe is 1 cm.
After input data describing the pipes, there is a line containing a single integer l, which is the number of links (0 <= l <= 50). After it, there are l lines describing links. Each link description contains 3 integers. The first two are (x, y) coordinates of the left end-point of the link and the third is the length of the link (at least 1 cm and at most 20 cm). It is assumed that the width of the link is zero.
The last line for each test case contains two numbers. The first is the number of target pipe (starting from one, with the order appeared in test data). The second line is the desired y for the level of water in the target pipe (note that the specified level may be out of the pipe at all).
You can assume the following about the input:
The water enters into the first pipe.
No two links have the same y coordinates.
No two pipes have the same upper-left x coordinates.
Both endpoints of each link are connected to pipes.
Output
The output should contain exactly t lines with no blank lines in between, each corresponding to one test case. Each output line should contain the time required for the water to reach the target level in the target pipe (an integer number). If in a specific test case, the water never reaches the target level, the line should contain No Solution string in it.
Sample Input
1
2
2 0 6
5 1 6
1
3 4 2
2 2
Sample Output
9
Source: Asia 2001, Tehran (Iran)
Submit Status | 1,227 | 5,368 | {"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} | 3.609375 | 4 | CC-MAIN-2018-09 | longest | en | 0.942961 |
https://physics.stackexchange.com/tags/linear-algebra/hot | 1,579,874,280,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250620381.59/warc/CC-MAIN-20200124130719-20200124155719-00332.warc.gz | 600,562,330 | 28,656 | # Tag Info
51
A second-order tensor can be represented by a matrix, just as a first-order tensor can be represented by an array. But there is more to the tensor than just its arrangement of components; we also need to include how the array transforms upon a change of basis. So tensor is an n-dimensional array satisfying a particular transformation law. So, yes, a third-...
39
Matrices are often first introduced to students to represent linear transformations taking vectors from $\mathbb{R}^n$ and mapping them to vectors in $\mathbb{R}^m$. A given linear transformation may be represented by infinitely many different matrices depending on the basis vectors chosen for $\mathbb{R}^n$ and $\mathbb{R}^m$, and a well-defined ...
36
Coming from a math perspective, I would define a dimension as "any property which is orthogonal to all other properties." "Orthogonal" here means you cannot get to one property by applying scalar operations on another. For example, the x-axis dimension can never become a y-axis value, and similarly for time vs. spatial dimensions. For that matter, it'...
32
The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the ...
28
In this context, I usually explain it (non-mathematically) by saying that the number of dimensions is the number of values you need to specify where an event occurs. For most people this involves space and time (but for particle physicists it might involve more values ;). Anyway, certainly even people before Einstein would need to specify the time as well ...
26
There are at least three notions of basis depending on the mathematical structure you are considering. I will quickly discuss three cases relevant in physics (topological vector spaces are relevant too, but I will not consider them for the shake of brevity). (1) Pure algebraic structure (i.e. vector space structure over the field $\mathbb K=$ $\mathbb R$ ...
24
The dual of a tensor you refer to is the Hodge dual, and has nothing to do with the dual of a vector. The word "dual" is used in too many different contexts, and in this case it is even used the same $*$ symbol. One usually specifies "Hodge dual", or "Hodge star operator", to avoid confusion. Both these "duals" are isomorphisms between vector spaces endowed ...
24
Quantum mechanics "lives" in a Hilbert space, and Hilbert space is "just" an infinite-dimensional vector space, so that the vectors are actually functions. Then the mathematics of quantum mechanics is pretty much "just" linear operators in the Hilbert space. Quantum mechanics Linear algebra ----------------- -------------- wave function vector ...
22
To take the partial trace you need to build the sum over the matrix elements w.r.t. the same input and output basis, as you probably already used to calculate the partial traces you gave. In Dirac notation this is often written as: tr_A(L_{AB}) =\sum_i \langle i|_A L_{AB} |i\rangle_A=\langle0|0\rangle\langle 0|0\rangle (|1\rangle\langle0|)_B+\langle1|0\... 21 The wording used in your textbook was sloppy. A acts as A^* on a bra, as \langle u\rvert A\lvert v\rangle:=\langle u\lvert Av\rangle~ is the same as \langle u\rvert A\lvert v\rangle=\langle A^*u\lvert v\rangle~, by definition of the adjoint. The latter formula also shows that \langle A^*u\rvert=\langle u\rvert A. Everything becomes very simple ... 17 As previous answers have correctly noted gamma matrices do not forma a basis of M(4,\mathbb{C}). Nevertheless you can construct one from them in the following way 1 the identity matrix \mathbb{1} 4 matrices \gamma^\mu 6 matrices \sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]} 4 matrices \sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]} 1 ... 17 Let \{|i\rangle\} be an orthonormal basis for the Hilbert space of the system. Then the trace of an operator O is given by (See the Addendum below) \begin{align} \mathrm {tr}(O) = \sum_i \langle i|O|i\rangle \end{align} For a given state |\psi\rangle, we define an operator P_\psi by \begin{align} P_\psi|\phi\rangle = \langle\psi|\phi\rangle|\psi\... 16 You are starting from the incorrect point. The argument follows by linearity of the equation. Suppose \Psi_k(x,t) is solution of the time dependent Schr\ddot{\hbox{o}}dinger equation: i\hbar \frac{\partial }{\partial t}\Psi_k(x,t)=-\frac{\hbar^2}{2m}\frac{\partial^2\Psi_k(x,t)}{\partial x^2}+U(x)\Psi_k(x,t)\, . $$Then:$$ \Phi(x,t)=a_1\Psi_1(x,t)+...
16
Your top-line question can be answered at many levels. Setting aside issues of forms and covariant/contravariant, the answer is: The dot product is the product of the magnitudes of the two vectors, times the cosine of the angle between them. No matter what basis you compute that in, you have to get the same answer because it's a physical quantity. The ...
15
Dot products, or inner products are defined axiomatically, or abstractly. An inner product on a vector space $V$ over $R$ is a pairing $V\times V\to R$, denoted by $\langle u,v\rangle$, with properties $\langle u,v\rangle=\langle v,u\rangle$, $\langle u+cw,v\rangle= \langle u,v\rangle+c\langle w,v\rangle$, and $\langle u,u\rangle\gt0$ if $u\ne0$. In ...
14
For several years I have been teaching Clifford (geometric) algebra as part of the Vector Analysis Course for undergraduate physics majors in Ateneo de Manila University. I strictly use Cl_{n,0}, even for Special Relativity. 18-year old students do not complain how difficult geometric algebra is. They just learn the math and the geometric interpretations: ...
14
If the Hilbert space of the system in question is finite-dimensional, then in a given basis for the Hilbert space, the Hamiltonian (and every other observable for that matter), will be represented by a matrix. If the Hilbert space is infinite-dimensional, the situation is a bit different. In Quantum Mechanics, we typically assume that the Hilbert spaces we ...
14
I) More generally, Let $V$ be a (say, finite dimensional) vector space over a field $\mathbb{F}$. Let $(e_i)_{i\in I}$ be a basis for $V$. Let $A\in {\rm End}(V)$ be an endomorphism in $V$, i.e. a $\mathbb{F}$-linear map $A:V\to V$. Let the matrix $(M^i{}_j)_{i,j\in I}$ be the unique $\mathbb{F}$-valued matrix that represents the linear map $A$ in the ...
14
“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural ...
13
Ron Maimon is entirely correct when he says that GA is precisely Clifford algebra from a mathematical perspective, as any book or paper using the phrase "Geometric Algebra" is sure to say. But I think he misses both the point of the question and the point of "GA" — which is different from Clifford algebra from a pedagogical perspective. The question I'll ...
13
Your doubt is not ridiculous, it is probably simply due to the confused way often mathematics is taught in physics. (I am a physicist too and, during my career, I had to bear ridiculous misconceptions, wasting lot of time in tackling non-existent pseudo-mathematical problems instead of focusing on genuine physical issues). There are sensible mathematical ...
13
There are two concepts of duality for vector spaces. One is the algebraic dual that is the set of all linear maps. Precisely, given a vector space $V$ over a field $\mathbb{K}$, the algebraic dual $V_{alg}^*$ is the set of all linear functions $\phi:V\to \mathbb{K}$. This is a subset of $\mathbb{K}^V$, the set of all functions from $V$ to $\mathbb{K}$. The ...
13
Does this mean that the operator $\hat O$ (an observable) is special in some way? I believe it means there is no such $\hat O$. If $\hat O$ corresponds to an observable, we require the eigenvalues to be real. Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$: $$\hat O |o\rangle = o |o\rangle$$ Now consider the following $$\hat O \... 13 Let H_A \otimes H_B be your Hilbert space, and O be an operator acting on this composite space. Then O can be written has$$ O = \sum_{i,j} c_{ij} M_i \otimes N_j$$where the M_i's and N_j's act on H_A and H_B respectively. Then the partial trace over H_A defined as$$tr_{H_A}(O) = \sum_{i,j} c_{ij} tr(M_i) N_j ,$$and similarly for H_B. 12 To complement V. Moretti's excellent answer, it's worth emphasizing that the dimension of the four-by-four complex matrices \mathbb C^{4\times 4}, when seen as a vector space over \mathbb C, is 4\!\times\!4=16. As such, a set of four matrices (i.e. vectors in \mathbb C^{4\times 4}) can never be a basis for it. It's also worth saying that the general ... 12 Call u_1, u_2, u_3, u_4 the eigenvectors described by you, respectively. Your claims are all right, but realize that both u_1 and u_2 share the same eigenvalue, that is 1, i.e., Pu_1=u_1 and Pu_2=u_2. Hence, any linear combination of u_1 and u_2 will also be eigenvectors with the same eigenvalue 1. Try to find eigenvectors of H of the ... 12 In a vector space over the field of complex numbers the notion of complex conjugation is basis dependent. You might say a vector is "real" if its components in some basis are real numbers, but if you change to another basis and the matrix expressing the new basis in terms of the old has complex entries then the "real" vector will have complex components in ... 11 1) Let us replace the momentum indices {\bf k} and -{\bf k} with abstract indices 1 and 2, and ignore the momentum summation. The Hamiltonian then reads$$\tag{1}H ~=~ \begin{pmatrix}a_1^{\dagger} & a_2\end{pmatrix} M\begin{pmatrix}a_1 \\ a_2^{\dagger}\end{pmatrix}, $$where$$\tag{2} M~:=\begin{pmatrix}A & B\\B^{*} & A \end{pmatrix}~...
11
The ground state of the harmonic oscillator $|0\rangle$ obeys $$a|0\rangle = 0$$ which means that the action of $a$ can't be undone: once you act with it on a state, you set to zero the coefficient in front of $|0\rangle$ in the decomposition into occupation eigenstates. Any candidate inverse operator $a^{-1}$ acting on zero will give you zero again; you ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2,746 | 10,537 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-05 | latest | en | 0.932061 |
https://www.theproblemsite.com/pro-problems/math/statistics-probabilities/counting-principles/how-many-arrangements | 1,719,331,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866143.18/warc/CC-MAIN-20240625135622-20240625165622-00854.warc.gz | 904,641,922 | 7,459 | Games
Problems
Go Pro!
# How Many Arrangements?
Pro Problems > Math > Statisics and Probabilities > Counting Principles
## How Many Arrangements?
The number 5435 is made up of the digits 3, 4, and 5, with the 5 used two times. The numbers 3455 and 5534 are two examples of other ways to arrange these digits.
In how many ways can you arrange the digits?
Presentation mode
Problem by Mr. Twitchell
## Solution
In order to make it feasible for teachers to use these problems in their classwork, no solutions are publicly visible, so students cannot simply look up the answers. If you would like to view the solutions to these problems, you must have a Virtual Classroom subscription.
Assign this problem
Click here to assign this problem to your students.
## Similar Problems
### 161 - Digits and Letters
In a basket I have slips of paper with several different digits on them. In another basket I have slips of paper with several different letters on them. If I reach into the first basket and pull out a number, and then reach into the second basket and pull out a letter, there are 161 possible results. How many letters of the alphabet are not in the second basket?
### Making License Plates
In the State of Confusion, all car license plates are made with the following stipulations:
• The license plate will have six characters.
• The first character will be a digit between one and nine inclusive.
• The next two characters will be digits between zero and nine inclusive.
• The fourth character will be a vowel
• The fifth character will be a consonant
• The sixth character could be any letter.
How many different license plates are possible in the State of Confusion?
### Twelve Days of Christmas
We all know the song "The Twelve Days of Christmas," and how the gift giving works:
Day One: 1 partridge in a pear tree
Day Two: 2 turtle doves and 1 partridge in a pear tree
Day Three: 3 calling birds, 2 turtle doves, and 1 partridge in a pear tree
Thus, after three days, the singer has 1 + 3 + 6 = 10 gifts. [Note that some people interpret the song as 1 gift the first day, 2 the second day, three the third day, etc., but that is not what the song says, so we will go with the literal interpretation of the song!]
After Mrs. Claus heard the elves singing this song one Christmas, she decided that the next year she would take this song to the extreme, and extend it to 20 days of Christmas, giving gifts (in the same pattern as the song) for 20 days.
Santa Claus caught wind of what she was doing and, since he is considered to be the epitome of the spirit of giving at Christmas time, decided he couldn't be out-given, and did the 25 days of Christmas instead.
By how many gifts did Santa "outgive" Mrs. Claus?
### Andy and Roger
Andy picks three letters from his name and arrange them on the table. Roger picks two letters from his name and arrange them. If Roger puts his two letters after Andy’s three letters, how many possible arrangements of five letters are there?
### STOPS!
In how many different ways can the letters STOPS be arranged?
### Guitar Pickin'
Owen is picking out random tunes on his guitar. He can play either half notes, quarter notes, or eighth notes. Each pluck can be one of the tones of the scale (C, C#, D, D#, E, F, F#, G, G#, A, A#, B). If he wanted to play every possible three-note tune, and he could do one tune every 20 seconds, how many days would it take him to play all of them?
[No calculators, please]
### Borders
On an island with seven towns, one town borders on five of the other towns, three towns border on four other towns, two towns border on three other towns, and one town only borders one other town.
Every day postmen from each town walk to each of their borders and exchange a sack of mail with a postman from the neighboring town. How many such exchanges happen each day?
### Tic Tac Toe Grid
Begin with a 3 x 3 grid, like the one used to play Tic Tac Toe. You have a blue square, a green square, and a red square, that you can place within the grid, according to the following rules:
• The green square must be touching the blue square on a side.
• The red square must be touching the green square on either a side or a vertex.
• No square can be placed on top of another square.
For example, the following arrangement is valid, because the green is horizontally connected to the blue square, and the red square is diagonally connected to the green square.
In how many different ways could the three squares be placed on the grid?
### Buttons and Levers
There are three buttons on a wall, each of which can be pressed or unpressed. There are four levers on the wall, and each of them can be either flipped up or down, or in a central position. How many possible arrangements of levers and buttons are there?
### Building a Burger
I can have either mustard, ketchup, or relish for my burger. I can choose either swiss, Monterey, or cheddar cheese. I can choose either pickles, tomatoes, or lettuce. I can choose wheat bread or white (but I must have bread). I dislike the combination of mustard and swiss cheese. I like all other combinations, including combinations which include no cheese, condiments or veggies. How many combinations are there which I will like?
Thanksgiving Dinner, Burgers and Condiments
# Blogs on This Site
Reviews and book lists - books we love!
The site administrator fields questions from visitors.
Like us on Facebook to get updates about new resources | 1,256 | 5,488 | {"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-2024-26 | latest | en | 0.920879 |
http://www.pinkmonkey.com/studyguides/subjects/eco/chap9/e0909403.asp | 1,511,192,703,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806070.53/warc/CC-MAIN-20171120145722-20171120165722-00358.warc.gz | 475,888,229 | 5,373 | Support the Monkey! Tell All your Friends and Teachers
Home MonkeyNotes Printable Notes Digital Library Study Guides Study Smart Parents Tips College Planning Test Prep Fun Zone Help / FAQ How to Cite New Title Request
ii) Average costs: Average cost is a ratio of Total Cost to Total Output units produced. Since total cost has fixed and variable costs as two components, we have three types of average costs. These are average fixed cost, average variable cost and average total cost.
I II III IV V Production Total (TFC ¸TO) (TVC ¸ TO) (TC¸ TO) = (III+IV) units output = AFC = AVC = ATC 1 - - - - 2 4 10 2.5 12.5 3 11 3.63 1.82 5.45 4 18 2.22 1.67 3.89 5 23 1.73 1.73 3.46 6 25 1.6 2 3.6 7 26 1.54 2.3 3.84
In the table, column I shows Production Units 1 to 7. Total Output is shown in column II as 4,11…26. In column III we find AFC which is a ratio of Total Fixed Cost to Total Output (TFC¸ TO). Therefore it is determined as 40 ¸ 4 = 10, 40 ¸ 11 = 3.63…etc. Column IV reads Average Variable Cost. It is a ratio of Total Variable Cost to Total Output (TVC ¸ TO). It is obtained as 10 ¸ 4 = 2.5, 20 ¸ 11 = 1.82...etc. Finally, in column V we notice Average Total Cost for different units. It is a ratio of Total Cost to Total Output. (TC ¸ TO). Therefore, it is determined as 50 ¸ 4, 60 ¸ 11…etc. Note that ATC is exactly equal to AFC + AVC. The alternate method of computing ATC is to add the values in columns III and IV.
The behavior of the three average cost varieties is an interesting and important part of Cost Analysis. We are mainly interested in the behavior of ATC (column V) which has two components, AFC and AVC. The behavior of ATC is jointly determined by AFC and AVC. AFC continuously falls. Therefore AFC tends to pull AFC in its own direction and causes its fall. AFC falls sharply in the initial stage from 10 to 3.63 but it slows down in its rate of fall towards the end such as from 1.60 to 1.54. The effect of the fall in AFC is the progressive reduction in the value of ATC. AVC initially decreases from 2.5 to 1.82 to 1.7 but it subsequently rises from 1.67 to 1.73 to 2 to 2.30. The effect of AVC on ATC is initially reductive but when it starts rising, it attempts to pull ATC in its own direction. ATC initially falls sharply from 12.5 to 5.45 to 3.89 when both AFC and AVC reduce in value. Then it becomes moderate in its fall us present the cost curves with the help of diagrams to gain further familiarity with them.
Index
9.1 - Concept of a Firm
9.2 - Factors of Production and Product Output
9.3 - Costs and Profits
9.4 - Costs Analysis
Chapter 10 | 749 | 2,592 | {"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} | 4.25 | 4 | CC-MAIN-2017-47 | latest | en | 0.917556 |
https://math.answers.com/Q/If_3x_plus_9_equals_5_what_is_x | 1,643,295,038,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00300.warc.gz | 425,185,819 | 63,042 | 0
# If 3x plus 9 equals 5 what is x?
Wiki User
2011-10-19 20:59:53
Well if 3x+9=5 x=?
Well you minus 9 from the side and then do 5-9 and that equals -4. Then you have to do -4 divided by 3. Which equals -1 so I believe x will equal -1.
Wiki User
2011-10-19 20:59:53
🙏
0
🤨
0
😮
0
Study guides
20 cards
➡️
See all cards
3.73
389 Reviews | 145 | 342 | {"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} | 3.390625 | 3 | CC-MAIN-2022-05 | latest | en | 0.734309 |
https://www.jobilize.com/course/section/color-interpolation-textures-in-processing-by-openstax?qcr=www.quizover.com | 1,606,424,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141188947.19/warc/CC-MAIN-20201126200910-20201126230910-00686.warc.gz | 720,735,635 | 20,054 | # 0.7 Textures in processing
Page 1 / 2
A brief guide to texture mapping in Processing
## Color interpolation
As seen in Graphic Composition in Processing , one can obtain surfaces as collections of polygons by means of the definition of a vertexwithin the couple beginShape() - endShape() . It is possible to assign a color to one or more vertices, in order to make the color variations continuous ( gradient ). For example, you can try to run the code size(200,200,P3D);beginShape(TRIANGLE_STRIP); fill(240, 0, 0); vertex(20,31, 33);fill(240, 150, 0); vertex(80, 40, 38); fill(250, 250, 0); vertex(75, 88, 50);vertex(49, 85, 74); endShape(); in order to obtain a continuous nuance from red to yellow in the strip of two triangles.
## Bilinear interpolation
The graphical system performs an interpolation of color values assigned to the vertices. This type of bilinear interpolation is defined in the following way:
• For each polygon of the collection
• For each side of the polygon one assigns to each point on the segment the color obtained by means of linear interpolation of the colors of the vertices $i$ e $j$ that define the polygon: ${C}_{ij}(\alpha )=\left(1-\mathrm{\alpha \right)}{C}_{i}+\alpha {C}_{j}$
• A scan line scans the polygon (or, better, its projection on the image window) intersecting at each step two sides in two points $l$ ed $r$ whose colors have already been identified as ${C}_{l}$ e ${C}_{r}$ . In each point of the scan line the color is determined by linear interpolation ${C}_{lr}(\beta )=\left(1-\mathrm{\beta \right)}{C}_{l}+\beta {C}_{r}$
A significative example of interpolation of colors associated to the vertices of a cube can be found in examples of Processing , in the code RGB Cube .
## Texture
When modeling a complex scene by means of a composition of simple graphical elements one cannot go beyond a certainthreshold of complexity. Let us think about the example of a modelization of a natural scene, where one has to representeach single vegetal element, including the grass of a meadow. It is unconceivable to do this manually. It would bepossible to set and control the grass elements by means of some algorithms. This is an approach taken, for example, inrendering the hair and skin of characters of the most sophisticated animation movies (see for example, the Incredibles ). Otherwise, especially in case of interactive graphics, one has to resort to using textures . In other words, one employs images that represent the visual texture of the surfaces and map them onthe polygons that model the objects of the scene. In order to have a qualitative rendering of the surfaces it is necessaryto limit the detail level to fragments not smaller than one pixel and, thus, the texture mapping is inserted in the rendering chain at the rastering level of the graphic primitives, i.e. where one passes from a 3Dgeometric description to the illumination of the pixels on the display. It is at this level that the removal of thehidden surfaces takes place, since we are interested only in the visible fragments.
In Processing, a texture is defined within a block beginShape() - endShape() by means of the function texture() that has as unique parameter a variable of type PImage . The following calls to vertex() can contain, as last couple of parameters, the point of the texture correspondingto the vertex. In fact, each texture image is parameterized by means of two variables $u$ and $v$ , that can be referred directly to the line and column of a texel (pixel of a texture) or, alternatively, normalized between $0$ and $1$ , in such a way that one can ignore the dimension as well as the width and height of the texture itself. Themeaning of the parameters $u$ and $v$ is established by the command textureMode() with parameter IMAGE or NORMALIZED .
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers! | 1,633 | 7,133 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 14, "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} | 3.609375 | 4 | CC-MAIN-2020-50 | latest | en | 0.846728 |
http://www.convertit.com/Go/BusinessConductor/Measurement/Converter.ASP?From=DEM | 1,670,387,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00340.warc.gz | 71,856,456 | 3,790 | New Online Book! Handbook of Mathematical Functions (AMS55)
Conversion & Calculation Home >> Measurement Conversion
Measurement Converter
Convert From: (required) Click here to Convert To: (optional) Examples: 5 kilometers, 12 feet/sec^2, 1/5 gallon, 9.5 Joules, or 0 dF. Help, Frequently Asked Questions, Use Currencies in Conversions, Measurements & Currencies Recognized Examples: miles, meters/s^2, liters, kilowatt*hours, or dC.
Conversion Result: ```German mark on 4-28-2017 = 0.5570144336164 United States dollar on 4-28-2017 (currency)``` Related Measurements: Try converting from "DEM" to AUD (Australian dollar on 4-28-2017), BEF (Belgian franc on 4-28-2017), CHF (Swiss franc on 4-28-2017), EUR (European Union euro on 4-28-2017), FKP (Falkland pound on 4-28-2017), GIP (Gibraltar pound on 4-28-2017), HUF (Hungarian forint on 4-28-2017), IDR (Indonesian rupiah on 4-28-2017), IEP (Irish pound on 4-28-2017), ILS (Israeli new shekel on 4-28-2017), JMD (Jamaican dollar on 4-28-2017), LKR (Sri Lankan rupee on 4-28-2017), LSL (Lesotho loti on 4-28-2017), OMR (Omani rial on 4-28-2017), SHP (Saint Helenian pound on 4-28-2017), SZL (Swazi lilangeni on 4-28-2017), THB (Thai baht on 4-28-2017), tick, TWD (Taiwanese new dollar on 4-28-2017), ZAR (South African rand on 4-28-2017), or any combination of units which equate to "currency" and represent currency. Sample Conversions: DEM = .55701443 ADP (Andorran peseta on 4-28-2017), .74513972 AUD (Australian dollar on 4-28-2017), .99148134 AWG (Aruban guilder on 4-28-2017), 20.63 BEF (Belgian franc on 4-28-2017), .77820287 BND (Bruneian dollar on 4-28-2017), 1.11 BZD (Belizean dollar on 4-28-2017), 3.8 DKK (Danish krone on 4-28-2017), .55701443 EEK (Estonian kroon on 4-28-2017), 1.16 FJD (Fijian dollar on 4-28-2017), .4305692 GIP (Gibraltar pound on 4-28-2017), 59.19 ISK (Icelandic krona on 4-28-2017), 62.08 JPY (Japanese yen on 4-28-2017), 2.23 LTL (Lithuanian litas on 4-28-2017), 1.13 NLG (Netherlands guilder on 4-28-2017), .81199085 NZD (New Zealand dollar on 4-28-2017), 1.81 PEN (Peruvian nuevo sol on 4-28-2017), 7.43 SZL (Swazi lilangeni on 4-28-2017), 16.82 TWD (Taiwanese new dollar on 4-28-2017), 12,686 VND (Vietnamese dong on 4-28-2017), 1.5 XCD (East Caribbean dollar on 4-28-2017).
Feedback, suggestions, or additional measurement definitions?
Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 908 | 2,414 | {"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-2022-49 | latest | en | 0.589141 |
https://web2.0calc.com/questions/help-ratio_4 | 1,685,830,916,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649343.34/warc/CC-MAIN-20230603201228-20230603231228-00493.warc.gz | 643,218,042 | 5,487 | +0
# help ratio
0
199
1
Our company parking lot has cars, trucks, and motorcycles. The ratio of cars, trucks, and motorcycles is \(1:6:4\) If there are a total of 42 cars and trucks, then how many motorcycles are there?
Jun 21, 2022
#1
+124718
+1
Cars and trucks are ( 1 + 6) / (1 + 6 + 4) = 7/11 of the total number, N, of vevicles
So...
(7/11) N = 42
N = 42 ( 11/7) = 66 vehicles in total
So....there are 66 - 42 = 24 motorcycles
Jun 21, 2022 | 187 | 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} | 3.859375 | 4 | CC-MAIN-2023-23 | latest | en | 0.821167 |
https://forum.solidworks.com/message/298698 | 1,547,829,128,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00372.warc.gz | 529,871,621 | 24,464 | 5 Replies Latest reply on Jun 20, 2012 1:31 PM by Corey Takehara
# Solidworks smart dimensions wrong?
Hello everybody,
I have to say I'm a little stumped here about smart dimensions. I'm trying to build some trim around my cabinets at 1.5" thick with the trim having a 45 degree anlge. Well when I go to build the one side with a 45 deg at 1.5" thick everything works fine. But when I go to make the other side it tells me that the new side can only be at 1.18" thick without over defining. I guess I don't see how that's possible. Is this a glich with solidworks dimensioning or is it me?
Thanks,
~Corey
• ###### Re: Solidworks smart dimensions wrong?
It is possible there is another dimension or relation that is fully defining this sketch. Would it be possible for you to upload the file? If not, perhaps taking another screen shot showing the entire sketch could help diagnose the issue. Ensure the relations and dimensions are all being shown.
Regards,
Jesse B.
• ###### Re: Solidworks smart dimensions wrong?
Hi Jesse,
Thanks for the response . Here are some more pics of whats going on. On the last image is when I deleted the line to get it to set straight at 1.5" but the only way that it shows that its straight is at 1.18". To me this is really simple thing to accomplish since one side is done already but yet it's giving trouble lol.
~ Corey
• ###### Re: Solidworks smart dimensions wrong?
Ok after going back and forth to each one for some reason I went to set it at 45 deg angle and it didn't end up saving it so it had the angle at like 52 deg. Uggg I must have misclicked to get rid of the change oh well.
Sorry for posting
~Corey
• ###### Re: Solidworks smart dimensions wrong?
Try removing one of these relations. It looks like your coincident relation is tying your horizontal line to a previously created feature. So the thickness of this sketch is being defined by this feature. I'd recommend removing the coincident relation. But then you will have to define the angle of that line. But you should be able to define the thickness to 1.5".
I hope this helps!
Regards,
Jesse B.
• ###### Re: Solidworks smart dimensions wrong?
Thanks Jesse for all the help! Unforunately it was a user error lol. When I went to set at 45deg angle I must not have hit the accept button so it left it at some weird angle which is why I couldnt get the other part of the trim to be 1.5" thick. Thanks for the information though I really appreciate it! Have an awesome day!!
~Corey | 600 | 2,505 | {"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-2019-04 | longest | en | 0.955512 |
https://studiegids.universiteitleiden.nl/en/courses/123530/introductie-moderne-natuurkunde | 1,726,768,293,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652055.62/warc/CC-MAIN-20240919162032-20240919192032-00594.warc.gz | 492,000,066 | 6,557 | # Introductie Moderne Natuurkunde
Course
2024-2025
Mathematics and Physics on Dutch high school VWO level
## Description
The remarkable revolution in physics at the beginning of the twentieth century has influenced our worldview at all levels. In this first course on modern physics, we will introduce the counter-intuitive principles, relativity and quantum mechanics and relativity, that are the foundations of the laws of physics. Starting with Einstein’s insight that the speed of light is the same for all observers, even when they move with respect to each other, we deduce the theory of special relativity, with E = mc2 as it most well-known consequence. Next we show how the photo-electric effect indicates the existence of a smallest quantum of light. This makes clear that waves are particles as well, but the principles of quantum mechanics demand that the converse is also true. Particles are also waves, and this explains the stability of atoms. We explain the important role of quanta in the explanation of blackbody radiation, en the structure of the atomic nucleus. We conclude by describing how these new insights have shown that forces and particles are in essence the same thing, en how daring experiments such as the Large Hadron Collider at CERN test these amazing principles with astonishing precision. The outlook towards modern physics given in this course is the first step to the research frontier.
## Course Objectives
-Special Relativity (postulates of Einstein, time dilation, length Contraction, Lorentz transformations, Doppler shift, relativistic energy and momentum)
-Basis of Quantummechanics (photoelectric effect, De Broglie waves, wavefunction, uncertainty principle)
-Atomic structure (electron orbits and the Bohr atom)
-From single to many body physics (statistical distributions/Maxwell-Boltzmann, the ideal gas, blackbody radiation and Planck)
-Nuclear structure (binding energy, liquid-drop model, radioactive decay, nuclear reactions)
-Elementary Particles (forces and particles, the standard model, relativistic collisions)
## Timetable
In MyTimetable, you can find all course and programme schedules, allowing you to create your personal timetable. Activities for which you have enrolled via MyStudyMap will automatically appear in your timetable.
Questions? Watch the video, read the instructions, or contact the ISSC helpdesk.
Note: Joint Degree students from Leiden/Delft need to combine information from both the Leiden and Delft MyTimetables to see a complete schedule. This video explains how to do it.
See Brightspace
## Assessment method
Midterm test: 30% of the final grade
Final exam: 70% of the final grade
11 Weekly Problem assignents: max 1.0 on top of the final grade (if final grade > 5.5)
Only the final exam can be retaken and will then count for 100% of the final grade.
H.D. Young, R.A. Freedman, University Physics; Pearson; 15th ed. (2019)
## Registration
As a student, you are responsible for enrolling on time through MyStudyMap.
In this short video, you can see step-by-step how to enrol for courses in MyStudyMap.
Extensive information about the operation of MyStudyMap can be found here.
There are two enrolment periods per year:
• Enrolment for the fall opens in July
• Enrolment for the spring opens in December
Note:
• It is mandatory to enrol for all activities of a course that you are going to follow.
• Your enrolment is only complete when you submit your course planning in the ‘Ready for enrolment’ tab by clicking ‘Send’.
• Not being enrolled for an exam/resit means that you are not allowed to participate in the exam/resit.
## Contact
Contact: Prof.dr. M.P. van Exter (Martin)
## Remarks
Software
Starting from the 2024/2025 academic year, the Faculty of Science will use the software distribution platform Academic Software. Through this platform, you can access the software needed for specific courses in your studies. For some software, your laptop must meet certain system requirements, which will be specified with the software. It is important to install the software before the start of the course. More information about the laptop requirements can be found on the student website. | 897 | 4,203 | {"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-2024-38 | latest | en | 0.862025 |
https://forum.allaboutcircuits.com/threads/simple-question.1693/#post-11076 | 1,670,118,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710953.78/warc/CC-MAIN-20221204004054-20221204034054-00161.warc.gz | 302,263,790 | 23,012 | Simple Question
Tekker
Joined Apr 22, 2005
33
I'm currently taking math 111 right now (algebra) so this should be an easy question to answer for you guys. I haven't had math in over a couple years, and I definitely wasn't very into math at the time so I've long forgotten all this "review" stuff. LOL
Basically, what I want to know is for this equation:
Rich (BB code):
1 1 3
--- + --- = ----
t 6 2t
Why can't I just invert both sides of the equation and solve for t? When I do that I get:
Rich (BB code):
2t
3 * (t + 6) = ---- * 3
3
3t + 18 = 2t
t = -18..... ???? (the answer is supposed to be 3)
I know what the proper procedure is, I just want to know why this doesn't work, since it apparently doesn't. I know that you can do anything to one side of the equation as long as you do it to the other side as well.... So is there some step or rule that I'm missing here, or am I messing something up when solving for t? I need to know why this doesn't work, so I know when not to use this method.
Thanks.
-tkr
hgmjr
Joined Jan 28, 2005
9,027
Hi tekker,
Rich (BB code):
To invert the left side of your equation requires that you invert the left side expression as a whole.
Since:
1
---------- does not equal (t + 6)
1 1
--- + ---
t 6
Then the techinique you attempted to employ did not yield the correct answer.
Rich (BB code):
Instead :
1 6t
---------- = ----------
1 1 (t + 6)
--- + ---
t 6
Inverting the individual terms in the left side expression is not the same as inverting the expression as a whole.
Math can be a bit tricky but the real key is to solve many example problems so that the rules govening the algebraic manipulation of expressions becomes clear.
At least you were curious enough to try something different to arrive at a solution. That shows you are keen to explore new ways to do things. You should do fine in your Math course with that approach.
Good Luck,
hgmjr
Tekker
Joined Apr 22, 2005
33
Originally posted by hgmjr@Oct 17 2005, 04:56 AM
Inverting the individual terms in the left side expression is not the same as inverting the expression as a whole.
I did invert the whole expression (on the right side it used to be 3/2t, then I inverted to be 2t/3 with the rest of the equation on the left side)..... That's why I don't understand why it doesn't come out the same.
This math stuff is tricksy.
-tkr
hgmjr
Joined Jan 28, 2005
9,027
Hi tekker,
You did correctly invert the expression to the right of the equal sign.
Inverting the individual terms in the left side expression is not the same as inverting the expression as a whole.
By the term "expression as a whole", I was referring to the expression to the left of the equal sign. I can see how my statement could have been interpreted as referring to the entire expression.
I plead guilty to not being clear with my statement.
Rich (BB code):
1 1
---- + ----
t 6
By adding these two terms you get:
6 + t
--------
6t
In this quotient form you can them invert this expression along with the term on the right side of the expression and you will arrive at the correct answer the variable t.
hgmjr
Tekker
Joined Apr 22, 2005
33
Ahhhh..... *Light comes on* Now I get it!
In that form it would probably be easier to just solve it now without inverting, whereas before it appeared easier to just invert to get rid of the fractions. But now it makes sense why that wouldnt work.
Rich (BB code):
6 + t 3
------- = ----
6t 2t
18t
6 + t = -----
2t
6 + t = 9
t = 3
Fantastic! The way the book mentioned it was to multiply by the lowest common factor (6t) to cancel out the denominator. But in general, this is the way I think. Just solve the problem, so I had to know why my previous solving method didnt work.... and now I know. Thanks a lot! B)
-tkr
hgmjr
Joined Jan 28, 2005
9,027
Hi tekker,
Those "light bulb" moments are always exciting. The neat part about math is that it is full of them.
Good luck with the remainder of your math studies.
hgmjr
Tekker
Joined Apr 22, 2005
33
Originally posted by hgmjr@Oct 18 2005, 04:37 AM
Hi tekker,
Those "light bulb" moments are always exciting. The neat part about math is that it is full of them.
Yep the light bulb moments are great, but the bulb doesn't seem to last for very long. I'm also taking chemistry 201 at the same time so I've got quite a work load this term. Lots of light bulb moments in chemistry also that's for sure.
Good luck with the remainder of your math studies.
Thanks. It's funny because before I got really interested in electronics, I probably would have just looked the answer up in the book, wrote the answer down and not given it another thought.... Which is exactly what I did with all of my previous math classes. But now I'm actually looking forward to learning as much as I can about this stuff (and even chemistry) because I know it'll be worth it when I can apply it all to electronics. It's amazing how much more enjoyable it is when you can actually apply it to something you enjoy. I see how knowlegable you guys are here on the forum and it just blows my mind! :wacko: That's the level that I want to get to eventually. B)
-tkr
hgmjr
Joined Jan 28, 2005
9,027
Hi tekker,
It has been my personal experience that one always excels in any endeavour for which one has a genuine passion.
It sounds like your enthusiasm is high toward the courses you are taking. That will fuel your dedication to the task of absorbing as much knowledge as you can.
One of the most important things you can learn is how to continue to expand your knowledge once you complete your formal studies.
Continue to ask questions along the way.
Good Luck,
hgmjr | 1,490 | 5,748 | {"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} | 3.953125 | 4 | CC-MAIN-2022-49 | latest | en | 0.945443 |
https://gmatclub.com/forum/which-of-the-following-most-logically-completes-the-143684.html | 1,513,568,282,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948604248.93/warc/CC-MAIN-20171218025050-20171218051050-00762.warc.gz | 564,043,345 | 51,012 | It is currently 17 Dec 2017, 19:38
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Which of the following most logically completes the
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Manager
Joined: 27 Jul 2011
Posts: 183
Kudos [?]: 294 [3], given: 103
Which of the following most logically completes the [#permalink]
### Show Tags
06 Dec 2012, 04:08
3
KUDOS
2
This post was
BOOKMARKED
00:00
Difficulty:
45% (medium)
Question Stats:
66% (01:33) correct 34% (01:41) wrong based on 248 sessions
### HideShow timer Statistics
Which of the following most logically completes the argument?
Some teenagers, citing Freedom of Speech, say they oppose parents and teachers who chastise them for speaking their mind, provided what they say is their opinion, is honest, and is not derogatory to any of their peers or guardians. The authors of the Amendments to the Constitution made no such provisions. Freedom of speech carries with it no responsibility to meet standards of honesty or respect or truth, especially those established by outside arbiters: students, teachers, and parents.
(A) One's speech is required only to present the facts according to common sense standards of opinion, honesty, and respect for peers and family.
(B) One's speech must consider first and foremost the wishes and interests of one's peers, the audience that will listen closely to every word
(C) One's speech is free to be as "dishonest" and "derogatory" as one wishes.
(D) One's speech must simply provide space for one's opponents to present their point of view in order to guarantee the watching public that they are fully and honestly informed.
(E) One's speech must take public responsibility to be truthful, since one's word has a profound impact on the public's understanding of a situation.
OA after discussion
[Reveal] Spoiler: OA
_________________
If u can't jump the 700 wall , drill a big hole and cross it .. I can and I WILL DO IT ...need some encouragement and inspirations from U ALL
Kudos [?]: 294 [3], given: 103
Manager
Joined: 17 Jul 2008
Posts: 245
Kudos [?]: 495 [2], given: 29
Re: Which of the following most logically completes the [#permalink]
### Show Tags
07 Dec 2012, 16:05
2
KUDOS
It is a tough question . Thanks for your post
+1 Kudos
My answer is "C" . The argument say that . Parents and teachers punish the teenagers when they speak in mind, although they are hones and not derogatory to anyone. Teenagers oppose to parents and teachers' action (chastising) . am.of constitution have no rule for that and also the freedom of speech has no responsibility for students ,teachers...etc. Therefore taking the views from am.of constitution and freedom of speech(no support to students) it can be said One's speech is free to be as "dishonest" and "derogatory" as one wishes. What is the OA
sujit2k7 wrote:
Which of the following most logically completes the argument?
Some teenagers, citing Freedom of Speech, say they oppose parents and teachers who chastise them for speaking their mind, provided what they say is their opinion, is honest, and is not derogatory to any of their peers or guardians. The authors of the Amendments to the Constitution made no such provisions. Freedom of speech carries with it no responsibility to meet standards of honesty or respect or truth, especially those established by outside arbiters: students, teachers, and parents.
(A) One's speech is required only to present the facts according to common sense standards of opinion, honesty, and respect for peers and family.
(B) One's speech must consider first and foremost the wishes and interests of one's peers, the audience that will listen closely to every word
(C) One's speech is free to be as "dishonest" and "derogatory" as one wishes.
(D) One's speech must simply provide space for one's opponents to present their point of view in order to guarantee the watching public that they are fully and honestly informed.
(E) One's speech must take public responsibility to be truthful, since one's word has a profound impact on the public's understanding of a situation.
OA after discussion
_________________
Please give kudos if you enjoy the explanations that I have given. Thanks
Kudos [?]: 495 [2], given: 29
VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1074
Kudos [?]: 672 [0], given: 70
Location: India
GMAT 1: 410 Q35 V11
GMAT 2: 530 Q44 V20
GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)
Re: Which of the following most logically completes the [#permalink]
### Show Tags
08 Dec 2012, 14:00
Hi Sujit
IMO answer to the above question is C.
But i used strengthening technique to get to the answer. I could not find the line for completing the argument. Believing that it is towards the end i have answered the question.
Kudos [?]: 672 [0], given: 70
Manager
Joined: 27 Jul 2011
Posts: 183
Kudos [?]: 294 [0], given: 103
Re: Which of the following most logically completes the [#permalink]
### Show Tags
08 Dec 2012, 20:08
OA: C,
Kudos to perfectstranger for good explanation
_________________
If u can't jump the 700 wall , drill a big hole and cross it .. I can and I WILL DO IT ...need some encouragement and inspirations from U ALL
Kudos [?]: 294 [0], given: 103
Manager
Joined: 21 Sep 2012
Posts: 235
Kudos [?]: 417 [1], given: 63
Re: Which of the following most logically completes the [#permalink]
### Show Tags
08 Dec 2012, 23:05
1
KUDOS
Archit143 wrote:
Hi Sujit
IMO answer to the above question is C.
But i used strengthening technique to get to the answer. I could not find the line for completing the argument. Believing that it is towards the end i have answered the question.
Some teenagers, citing Freedom of Speech, say they oppose parents and teachers who chastise them for speaking their mind, provided what they say is their opinion, is honest, and is not derogatory to any of their peers or guardians. The authors of the Amendments to the Constitution made no such provisions. Freedom of speech carries with it no responsibility to meet standards of honesty or respect or truth, especially those established by outside arbiters: students, teachers, and parents.
Doesn't this line help answer the question? if you ask yourself why after that statement, only C answers the question.
Kudos [?]: 417 [1], given: 63
Manager
Joined: 31 May 2012
Posts: 157
Kudos [?]: 203 [0], given: 69
Re: Which of the following most logically completes the [#permalink]
### Show Tags
09 Dec 2012, 00:14
+1 Kudo. Its difficult question, I went wrong on first read. Convinced with OA.
Teachers and parents chastise the teenagers for speaking their opinion, which is honest and not derogatory. Teenagers oppose this fact.
-> Open minded speech should not be dishonest and derogatory. BUT, constitution made no such provisions.
=> Speech may be dishonest or derogatory. (implied meaning)<<- Answer.
constitution tells that "The speech should carry no responsibility to meet standards of honesty,respect or truth, when it is spoken by outside arbiters. ".. Idea can be extended by saying Speech is open to be dishonest and derogatory.. only Option (C) says this.
Kudos [?]: 203 [0], given: 69
Non-Human User
Joined: 01 Oct 2013
Posts: 10226
Kudos [?]: 277 [0], given: 0
Re: Which of the following most logically completes the [#permalink]
### Show Tags
22 Feb 2015, 21:08
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Kudos [?]: 277 [0], given: 0
Intern
Joined: 23 Dec 2014
Posts: 23
Kudos [?]: 2 [0], given: 5
Re: Which of the following most logically completes the [#permalink]
### Show Tags
24 Feb 2015, 09:23
In my opinion C, because in the passage it is mentioned that the Person beats no responsibility, hence the Person does not have to think if it Alriht to say certain things
Kudos [?]: 2 [0], given: 5
Intern
Joined: 08 Feb 2015
Posts: 28
Kudos [?]: 1 [0], given: 244
Re: Which of the following most logically completes the [#permalink]
### Show Tags
01 Mar 2015, 05:37
Some teenagers, citing Freedom of Speech, say they oppose parents and teachers who chastise them for speaking their mind, provided what they say is their opinion, is honest, and is not derogatory to any of their peers or guardians. The authors of the Amendments to the Constitution made no such provisions. Freedom of speech carries with it no responsibility to meet standards of honesty or respect or truth, especially those established by outside arbiters: students, teachers, and parents.
(C) One's speech is free to be as "dishonest" and "derogatory" as one wishes.
only C is logically matched... IMO
Kudos [?]: 1 [0], given: 244
Non-Human User
Joined: 01 Oct 2013
Posts: 10226
Kudos [?]: 277 [0], given: 0
Re: Which of the following most logically completes the [#permalink]
### Show Tags
12 Sep 2016, 05:08
Hello from the GMAT Club VerbalBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Kudos [?]: 277 [0], given: 0
Re: Which of the following most logically completes the [#permalink] 12 Sep 2016, 05:08
Display posts from previous: Sort by
# Which of the following most logically completes the
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,666 | 10,782 | {"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} | 2.71875 | 3 | CC-MAIN-2017-51 | latest | en | 0.930438 |
https://www.newworldencyclopedia.org/entry/Infinitesimal | 1,586,468,757,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371876625.96/warc/CC-MAIN-20200409185507-20200409220007-00401.warc.gz | 1,031,320,379 | 18,477 | # Infinitesimal
An infinitesimal is a quantity that is so small that it cannot be seen or measured. In mathematics, it is a non-zero quantity that approaches zero as a limit. When used as an adjective in the vernacular, infinitesimal means extremely small. In everyday life, an infinitesimal object is one that is smaller than any possible measure, whether we measure size, time, chemical concentration, or other property.
Before the nineteenth century, none of the mathematical concepts as we know them today were formally defined, but many of these concepts were already there. The founders of calculus—Leibniz, Newton, Euler, Lagrange, the Bernoullis and many others—used infinitesimals in the way shown below and achieved essentially correct results, although no formal definition was available. (Likewise, there was no formal definition of real numbers at the time).
## History of the infinitesimal
The first mathematician to make use of infinitesimals was Archimedes (around 250 B.C.E.).[1] The Archimedean property is the property of an ordered algebraic structure having no nonzero infinitesimals.
In India, from the twelfth to the sixteenth century, infinitesimals were discovered for use with differential calculus by Indian mathematician Bhaskara and various Keralese mathematicians.
When Newton and Leibniz developed calculus, they made use of infinitesimals. A typical argument might go as follows:
To find the derivative f′(x) of the function f(x) = x2, let dx be an infinitesimal. Then,
$f'(x)\,$ $=\frac{f(x + \mathrm dx) - f(x)}{\mathrm dx}\,$ $=\frac{x^2 + 2x \cdot \mathrm dx + \mathrm dx^2 -x^2}{\mathrm dx}\,$ $=2x + \mathrm dx\,$ $=2x\,$
since dx is infinitely small.
This argument, while intuitively appealing, and producing the correct result, is not mathematically rigorous. The use of infinitesimals was attacked as incorrect by Bishop Berkeley in his work The Analyst.[2] The fundamental problem is that dx is first treated as non-zero (because we divide by it), but later discarded as if it were zero.
When we consider numbers, the naive definition is clearly flawed: an infinitesimal is a number whose modulus is less than any non-zero positive number. Considering positive numbers, the only way for a number to be less than all numbers would be to be the least positive number. If h is such a number, then what is h/2? Or, if h is indivisible, is it still a number? Also, intuitively, one would require the reciprocal of an infinitesimal to be infinitely large (in modulus) or unlimited. That step should yield the "largest" number, but clearly there is no "last" biggest number.
It was not until the second half of the nineteenth century that the calculus was given a formal mathematical foundation by Karl Weierstrass and others using the notion of a limit. In the twentieth century, it was found that infinitesimals could, after all, be treated rigorously. Neither formulation is wrong, and both give the same results if used correctly.
## Modern uses of infinitesimals
Infinitesimal is necessarily a relative concept. If epsilon is infinitesimal with respect to a class of numbers, it means that epsilon cannot belong to that class. This is the crucial point: infinitesimal must necessarily mean infinitesimal with respect to some other type of numbers.
### The path to formalization
Proving or disproving the existence of infinitesimals of the kind used in nonstandard analysis depends on the model and which collection of axioms are used. We consider here systems where infinitesimals can be shown to exist.
In 1936 Maltsev proved the compactness theorem. This theorem is fundamental for the existence of infinitesimals as it proves that it is possible to formalize them. A consequence of this theorem is that if there is a number system in which it is true that for any positive integer n there is a positive number x such that 0< x < 1/n, then there exists an extension of that number system in which it is true that there exists a positive number x such that for any positive integer n we have 0 < x < 1/n. The possibility to switch “for any" and "there exists" is crucial. The first statement is true in the real numbers as given in ZFC set theory: for any positive integer n it is possible to find a real number between 1/n and zero, only this real number will depend on n. Here, one chooses n first, then one finds the corresponding x. In the second expression, the statement says that there is an ‘x’’ (at least one), chosen first, which is between 0 and 1/n for any n. In this case x is infinitesimal. This is not true in the real numbers (R) given by ZFC. Nonetheless, the theorem proves that there is a model (a number system) in which this will be true. The question is: what is this model? What are its properties? Is there only one such model?
There are in fact many ways to construct such a one-dimensional linearly ordered set of numbers, but fundamentally, there are two different approaches:
1) Extend the number system so that it contains more numbers than the real numbers.
2) Extend the axioms (or extend the language) so that the distinction between the infinitesimals and non-infinitesimals can be made in the real numbers.
In 1960, Abraham Robinson provided an answer following the first approach. The extended set is called the hyperreals and contains numbers less in absolute value than any positive real number. The method may be considered relatively complex but it does prove that infinitesimals exist in the universe of ZFC set theory. The real numbers are called standard numbers and the new non-real hyperreals are called nonstandard.
In 1977 Edward Nelson provided an answer following the second approach. The extended axioms are IST, which stands either for Internal Set Theory or for the initials of the three extra axioms: Idealization, Standardization, Transfer. In this system we consider that the language is extended in such a way that we can express facts about infinitesimals. The real numbers are either standard or nonstandard. An infinitesimal is a nonstandard real number which is less, in absolute value, than any positive standard real number.
In 2006 Karel Hrbacek developed an extension of Nelson's approach in which the real numbers are stratified in (infinitely) many levels, i.e, in the coarsest level there are no infinitesimals nor unlimited numbers. Infinitesimals are in a finer level and there are also infinitesimals with respect to this new level and so on.
All of these approaches are mathematically rigorous.
This allows for a definition of infinitesimals which refers to these approaches:
### A definition
An infinitesimal number is a nonstandard number whose modulus is less than any nonzero positive standard number.
What standard and nonstandard refer to depends on the chosen context.
Alternatively, we can have synthetic differential geometry or smooth infinitesimal analysis with its roots in category theory. This approach departs dramatically from the classical logic used in conventional mathematics by denying the law of excluded middle—i.e., not (ab) does not have to mean a = b. A nilsquare or nilpotent infinitesimal can then be defined. This is a number x where x2 = 0 is true, but x = 0 need not be true at the same time. With an infinitesimal such as this, algebraic proofs using infinitesimals are quite rigorous, including the one given above.
## Notes
1. Archimedes, The Method of Mechanical Theorems. see the Archimedes palimpsest.
2. George Berkeley, The Analyst; or a discourse addressed to an infidel mathematician.
## Credits
New World Encyclopedia writers and editors rewrote and completed the Wikipedia article in accordance with New World Encyclopedia standards. This article abides by terms of the Creative Commons CC-by-sa 3.0 License (CC-by-sa), which may be used and disseminated with proper attribution. Credit is due under the terms of this license that can reference both the New World Encyclopedia contributors and the selfless volunteer contributors of the Wikimedia Foundation. To cite this article click here for a list of acceptable citing formats.The history of earlier contributions by wikipedians is accessible to researchers here:
The history of this article since it was imported to New World Encyclopedia:
Note: Some restrictions may apply to use of individual images which are separately licensed. | 1,813 | 8,374 | {"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": 5, "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} | 4.03125 | 4 | CC-MAIN-2020-16 | longest | en | 0.933503 |
https://community.powerbi.com/t5/Desktop/Loop-Through-Columns-To-Create-a-Set/m-p/837856 | 1,585,532,439,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496330.1/warc/CC-MAIN-20200329232328-20200330022328-00265.warc.gz | 380,292,579 | 128,529 | cancel
Showing results for
Did you mean:
Frequent Visitor
## Loop Through Columns To Create a Set
Hello there!
I have five columns with the following data:
Baltimore Cincinnati Cleveland Pittsburgh AFCNorth 74 83 62 71 2
My requirement is: if each column has a number > 0, then make a set that equals '1'. So, my output would look like this:
Baltimore Cincinnati Cleveland Pittsburgh AFCNorth # of Sets 74 83 62 71 2 2
How do I do this in Power BI using either a measure or column?
4 REPLIES 4
Highlighted
Frequent Visitor
## Re: Loop Through Columns To Create a Set
Oh, I should mention this would be based on a unique id. So, I've updated the columns:
What it looks like now:
ID Baltimore Cincinnati Cleveland Pittsburgh AFCNorth 1 74 83 62 71 2
What I would like for it to look like:
ID Baltimore Cincinnati Cleveland Pittsburgh AFCNorth # of Sets 1 74 83 62 71 2 2
Microsoft
## Re: Loop Through Columns To Create a Set
hi @kreneec ,
If I understand you correctly, # of sets is the sum of set, so I’m confused about why # of sets is 2 in your table.
1. Unpivot table by columns except ID.
2. Create measure:
# of sets = CALCULATE(COUNT('Table'[Value]),FILTER('Table','Table'[Value]>0))
3. Create matrix visual and result would be shown as below:
BTW, Pbix as attached, hopefully works for you.
Best Regards,
Jay
Community Support Team _ Jay Wang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Frequent Visitor
## Re: Loop Through Columns To Create a Set
Hi Jay.
Thank you for responding. The number of sets is defined as "each column has to have greater than or equal to a 1 to define a set." So, as example:
Baltimore = 74; true;
Cincinnati = 83; true;
Cleveland = 62; true;
Pittsburgh = 71; true;
AFCNorth = 2; true
Since every column has more than 1, a set has been completed, i.e. add '1' to the # of sets column. Now, I need to reiterate again to check to see if another set has been completed. It has, which means add another '1' to the # of sets column, which means there are only 2 completed sets. Hope that makes sense.
Microsoft
## Re: Loop Through Columns To Create a Set
Hi @kreneec ,
Thank you for your patient explanation. Please check following steps as blew and see if the result meet your expectations:
1. Create calculated column:
set = IF('Table'[Pittsburgh]*'Table'[Cincinnati]*'Table'[Cleveland]*'Table'[Baltimore]*'Table'[AFCNorth]>0,1,0)
2. Create Measure:
# of sets = CALCULATE(SUM('Table'[set]),FILTER(ALLSELECTED('Table'),'Table'[ID]<=MAX('Table'[ID])))
3. Result would be shown as below:
Pbix as attached.
Best Regards,
Jay
Community Support Team _ Jay Wang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Announcements
#### New Ranks Launched March 24th!
The time has come: We are finally able to share more details on the brand-new ranks coming to the Power BI Community!
#### ‘Better Together’ Contest Finalists Announced!
Congrats to the finalists of our ‘Better Together’-themed T-shirt design contest! Click for the top entries.
#### Arun 'Triple A' Event Video, Q&A, and Slides
Missed the Arun 'Triple A' event or want to revisit it? We've got you covered! Check out the video, Q&A, and slides now.
#### Join THE global Power Platform event series.
Attend for two days of expert-led learning and innovation on topics like AI and Analytics, powered by Dynamic Communities.
#### Community Summit North America
Innovate, Collaborate, Grow. The top training and networking event across the globe for Microsoft Business Applications
Top Solution Authors
Top Kudoed Authors | 935 | 3,714 | {"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} | 2.734375 | 3 | CC-MAIN-2020-16 | latest | en | 0.848525 |
http://www.listserv.uga.edu/cgi-bin/wa?A2=ind0409&L=spssx-l&D=0&P=34353 | 1,386,774,420,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164037762/warc/CC-MAIN-20131204133357-00036-ip-10-33-133-15.ec2.internal.warc.gz | 413,292,663 | 4,327 | ```Date: Thu, 23 Sep 2004 12:38:32 +1000 Reply-To: paulandpen@optusnet.com.au Sender: "SPSSX(r) Discussion" From: Paul Dickson Subject: Re: Odd SPSS Results - Zero Coefficient Comments: To: Hector Maletta Content-Type: text/plain Hector You stated the following: "You reject the null hypothesis that the true value is zero whenever the confidence interval does not include the zero value". That is correct as far as I am concerned. But, I am wondering based on the thread of this discussion whether you meant in cases where the t-test (p-value) of the beta weight is significant, but your confidence intervals overlap with zero. If they are different (CI contains a zero value within the range) but (the P-value is significant) I myself would defer to the t-test, because while both CI and P-values assume probability sampling because they are inferential statistics (most data evaluated is not drawn from individuals with an equal likelihood of being selected) I would think in a case like this that CI's are less robust to violations of this assumption. Correct me if I am wrong here. Would this also not make the p-value a more robust criterion for decision making. What do others do if the confidence interval around the beta weight includes 0, but the p-value for the same beta-weight is significant. Is this possible, since the t-tests in spss are based on a similar decision criteria ideal to the CI where it is assumed that each coefficient = 0 (null hypothesis) when the t-test is run on the beta weight. Regards Paul > Hector Maletta wrote: > > Richard, > In fact I have reread the thread and think I misunderstood the question > and > consequently confused the matter more than clarifying it, and apologize > for > it. > The 95% confidence interval of an estimate of a regression coefficient > is > based on 1.96 times its standard error. If that confidence interval > includes > the zero value, you cannot reject the null hypothesis that the true > value is > zero. Thus, if your estimate is (nearly) zero and nonetheless you > obtain a > very low p value, it means the probability of the true value being zero > is > very low. You should rescale the results and will probably see that the > estimate is not exactly zero. > > However, I have my suspicions about part of Richard Ristow's comments, > namely: > > > At 02:41 PM 9/22/2004, Hector Maletta wrote: > > > > >If there is a population value for the coefficient, and you > > estimate it > > >obtaining a value of zero or any other, the significance > > tells you that > > >the true value is 95% likely to be within two standard > > errors from your > > >estimate. > > > > Is this really true? I thought that the population value > > would fall within two standard-errors-of-estimate of the > > estimated value, in any case. > > "In any case" seems too much for me. You have a probability that it > lies > within a certain interval (usually 2SE corresponding to a 95% > probability) > and a probability (usually 5%) that it lies outside that interval. If > the > probability is larger than 95% the interval is wider, say 3SE or > whatever, > and it is narrower if the probability is less than 95%. You reject the > null > hypothesis that the true value is zero whenever the confidence interval > does > not include the zero value. Suppose the estimate for the coefficient is > 0.25, and the SE is 0.1. A 95% interval means an interval of 2 SE, i.e. > approximately 0.2 to each side of your estimate, ranging from 0.05 to > 0.45. > Since this interval does not include the zero, you have 95% confidence > that > the null hypothesis (which says the true value is zero) is false > (though > there is a 5% probability that it is true). > > Am I so terribly wrong in this? > > Hector ```
Back to: Top of message | Previous page | Main SPSSX-L page | 909 | 3,810 | {"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-2013-48 | latest | en | 0.923961 |
https://forum.ansys.com/discussion/26498/pulsatile-flow-over-an-hexagon-residuals-convergence-problem | 1,619,164,926,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00553.warc.gz | 350,981,846 | 14,212 | # Pulsatile Flow over an Hexagon - Residuals Convergence Problem
Member Posts: 3
Hi,
I am new in simulating pulsatile flows and I am having some troubles with convergence.
My problem constist of an hexagon (2D problem) which is inmersed in a pulsating flow determined at the inlet by the formula -> vx = Vamp·sin(w·t) ; vy = 0.0.
At first, the solver uses the initial time step which is set to 1e-6 seconds. When some time steps has ocurred the solver cannot hold the specified residuals and it reachs the maximum iterations per time step which are set to 50. In that moment the residuals (specially of continuity) starts to rise up to something in the range of 1e-1 - 1. I set the iterations limit to 50 because I see that residuals do not drop down more from the value reached in 40 iterations, it gets stuck there. Below, I attach a example snapshot of what I get.
I have done some researching specially removing the hexagon from the domain, and I have seen a pressure wave propagating over the domain due to the change in the velocity at the inlet (the flow is accelerated in positive X direction when the flow comes in and in the other way around when it is going out), so it logic this pressure wave appears. In this case (without hexagon in the domain) the solver converges to the required residuals accuracy.
Although the pressure wave is logic to appear, I suspect it is doing the velocity-pressure coupling to be more difficult to achieve, so I reach the maximum iterations limit. Obviusly, when I put the hexagon inside it is even more dificult to achive, so I have these poor convergence problems.
Furthermore, I have noticed that lowering the CFL, it gets more stable but at the end it goes to same way.
The general setup of Fluent is described below:
• Solver type: pressure-based
• Pressure Coupling: PISO
• Pressure: Second Order
• Momentum: Second Order Upwind
• Viscosity Model: Laminar
• Fluid: Water (Incompressilbe)
• Time Step -> Adaptive CFL Based -> CFL <= 1.0 (I test with various values: 1.0 - 0.5 - 0.25)
• Residuals set to -> Continuity = 1e-3 - Velocities = 1e-6
This is a general view of my mesh
And it is a closer look near the hexagon
I know that the problem should be more rough than a constant uniform flow but, I think I am missing something, because the computational time for this "simple" object is quite high...
Any ideas??
Best Regards,
Sergio
• Posts: 200Forum Coordinator
Hello,
Since your inlet boundary condition is periodic with time, you will see some residual fluctuations as well.
Did you try to make use of Coupled Solver for this case?
Do you see reversed flow from the outlet? If so, then you might consider extending the outlet further downstream.
Regards,
SD | 651 | 2,734 | {"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-2021-17 | latest | en | 0.915769 |
https://www.teacherspayteachers.com/Product/Measurement-and-Data-2nd-Grade-Math-Short-Answer-Word-Problems-2839877 | 1,542,384,974,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743105.25/warc/CC-MAIN-20181116152954-20181116174954-00465.warc.gz | 1,036,190,628 | 23,653 | Subject
Resource Type
Common Core Standards
Product Rating
File Type
PDF (Acrobat) Document File
4 MB|60+
Share
Also included in:
1. Math Short Answer- 2nd Grade This math short answer pack is a constructed response resource to use when teaching common core math! This is not a curriculum, but a resource to help supplement your lessons! There are between 2 and 5 short answers PER standard to practice towards the end of a unit or
\$18.00
\$15.00
Save \$3.00
2. 2nd Grade Math Common Core Bundle This is a common core aligned math bundle for 2nd grade. There are four bundles in this mega bundle- one bundle with all PRINTABLES, one with INTERACTIVE NOTEBOOKS, one with SHORT ANSWERS and another with BLACK AND WHITE CENTERS. Each packet included standard spec
\$90.50
\$63.50
Save \$27.00
Product Description
This Measurement and Data MD pack is a Short Answer resource to use when teaching measurement, telling time, counting money, and graphing. This is perfect to use when teaching students how to answer constructed response questions (which is great for test prep or simply teaching students how to write about their math and understanding).
In this pack, you are able to choose the format that you would like to use for your students. You can either print the pages of as traditional worksheets to put into a folder or a binder. Or you could print the strips off for an interactive notebook.
There is also a problem solving strategy page at the beginning. One is a poster to print and keep in your classroom. The other are small bookmarks for students to hold onto while they are learning.
This item aligns to the Common Core standards for the Measurement and Data MD domain, but you don't have to be a Common Core classroom to use this pack!
Topics Covered:
-Selecting Appropriate Tools
-Measuring Lengths
-Comparing Two Measurements
-Estimating Lengths
-Comparing Two Objects’ Lengths
-Measurement Word Problems
-Number Lines
-Telling Time
-AM vs. PM
-Counting Money
-Money Word Problems
-Creating Measurement Graphs
You can save money by buying this short answer pack in a GROWING bundle of four other CCSS math short answers!
You can also save even MORE money by buying this pack within the Second Grade Common Core Math Mega Bundle!
Not interested in the bundles? But want more 2nd Grade MD products?
This purchase is for one single classroom only.
If you're interested in sharing with other classrooms, make sure to buy the extra licenses for 50% off through the TeachersPayTeachers tool. If you are interested in a site license, please contact me for a quote at jessica.L.tobin@gmail.com.
**There is a 1st grade version of this. The standards that are similar may have similar problems, but they use different numbers. There are different short answer problems for the standards that are totally different.**
CCSS.Math.Content.2.MD.A.1 (2.MD.1)
Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes.
CCSS.Math.Content.2.MD.A.2 (2.MD.2)
Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen.
CCSS.Math.Content.2.MD.A.3 (2.MD.3)
Estimate lengths using units of inches, feet, centimeters, and meters.
CCSS.Math.Content.2.MD.A.4 (2.MD.4)
Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. Relate addition and subtraction to length.
CCSS.Math.Content.2.MD.B.5 (2.MD.5)
Use addition and subtraction within 100 to solve word problems involving lengths that are given in the same units, e.g., by using drawings (such as drawings of rulers) and equations with a symbol for the unknown number to represent the problem.
CCSS.Math.Content.2.MD.B.6 (2.MD.6)
Represent whole numbers as lengths from 0 on a number line diagram with equally spaced points corresponding to the numbers 0, 1, 2, ..., and represent whole-number sums and differences within 100 on a number line diagram. Work with time and money.
CCSS.Math.Content.2.MD.C.7 (2.MD.7)
Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m.
CCSS.Math.Content.2.MD.C.8 (2.MD.8)
Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have?
Represent and interpret data.
CCSS.Math.Content.2.MD.D.9 (2.MD.9)
Generate measurement data by measuring lengths of several objects to the nearest whole unit, or by making repeated measurements of the same object. Show the measurements by making a line plot, where the horizontal scale is marked off in whole-number units.
CCSS.Math.Content.2.MD.D.10 (2.MD10)
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems using information presented in a bar graph.
Total Pages
60+
N/A
Teaching Duration
N/A
Report this Resource
Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 1,223 | 5,220 | {"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-2018-47 | latest | en | 0.914043 |
http://mathhelpforum.com/calculus/19886-various-derivative-problems.html | 1,500,593,441,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423512.93/warc/CC-MAIN-20170720222017-20170721002017-00656.warc.gz | 188,425,951 | 10,794 | 1. ## Various Derivative Problems
I'm having trouble with these problems not sure where to start. Any help is greatly appreciated.
1.) Find all values of where the tangent lines to and are parallel.
2.) Find an equation for the line tangent to the graph of at the point for .
3.) Calculate for , where (with a,b,c,d the constants)
Thanks again!
2. Hello,jwebb19!
1) Find all values of $x$ where the tangent lines to $y = x^4$ and $y = x^5$ are parallel.
We have: . $\begin{array}{ccc}y' & = & 4x^3 \\ y' & = & 5x^4\end{array}$
Since the slopes are equal: . $5x^4 \:=\:4x^3\quad\Rightarrow\quad 5x^4 - 4x^3\:=\:0\quad\Rightarrow\quad x^3(5x - 4)\:=\:0$
Therefore: . $\boxed{x \:=\:0,\:\frac{4}{5}}$
2) Find an equation for the line tangent to the graph of $f(x) = -4xe^x$at the point $(1,\,f(1)).$
$f(1)\:=\:-4(1)(e^1) \:=\:-4e$ . . . The point is: . $\left(1,\:-4e\right)$
The slope is: . $f'(x)\;=\;-4xe^x - 4x^x\:=\:-4e^x(x+1)$
When $x = 1$, we have: . $f'(1) \:=\:-4e^1(1 + 1) \:=\;-8e$
The equation is: . $y - (-4e) \:=\:-8e(x - 1) \quad\Rightarrow\quad y + 4e \:=\:-8ex + 8e$
Therefore: . $\boxed{y \;=\;-8ex + 4e}$
3) Calculate $y^{(k)}$ .for $0 \leq k \leq 5$
where $y \:=\:9x^4 + ax^3 + bx^2 + cx + d$ (with a,b,c,d constant)
$\begin{array}{ccc}y^{(0)} & = & 9x^4 + ax^3 + bx^2 + cx + d \\
y' & = & 36x^3 + 3ax^2 + 2bx + c \\
y'' & = & 108x^2 + 6x + 2b \\
y''' & = & 216x + 6 \\
y^{(4)} & = & 216 \\
y^{(5)} & = & 0 \end{array}$
3. Thanks a million! I have been trying the following problem to no avail.
If then
I believe f'(x) = [IMG]file:///C:/Users/Jthan/AppData/Local/Temp/moz-screenshot-8.jpg[/IMG] [sec(x)*tan(x)*(x^3)+3*(x^2)*sec(x)]/[(x^3)^2].
Thanks again! | 704 | 1,693 | {"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": 19, "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-2017-30 | longest | en | 0.558958 |
http://appliedresearch.cancer.gov/surveys/chis/dietscreener/uses.html | 1,386,930,812,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164929439/warc/CC-MAIN-20131204134849-00041-ip-10-33-133-15.ec2.internal.warc.gz | 7,851,220 | 7,424 | # Diet Screener in CHIS 2005: Uses of Screener Estimates in CHIS
## Introduction
Dietary intake estimates from the California Health Interview Survey (CHIS) Diet Screener are rough estimates of usual intake of fruits and vegetables and added sugar. They are not as accurate as more detailed methods (e.g. 24-hour recalls). However, validation research suggests that the estimates may be useful to characterize a population's median intakes, to discriminate among individuals or populations with regard to higher vs. lower intakes, to track dietary changes in individuals or populations over time, and to allow examination of interrelationships between diet and other variables. In addition, diet estimates from the CHIS could be used to augment national data using similar methods.
### What is the variance adjustment estimate and why is it needed?
Data from the CHIS Diet Screener are individuals' reports about their intake and, like all self-reports, contain some error. The algorithms we use to estimate servings of fruits and vegetables and added sugar calibrate the data to 24-hour recalls. The screener estimate of intake represents what we expect the person would have reported on his 24-hour recall, given what he reported on the individual items in the screener. As a result, the mean of the screener estimate of intake should equal the mean of the 24-hour recall estimate of intake in the population. (It would also equal the mean of true intake in the population if the 24-hour recalls were unbiased. However, there are many studies suggesting that recalls underestimate individuals' true intakes).
When describing a population's distribution of dietary intakes, the parameters needed are an estimate of central tendency (i.e., mean or median) and an estimate of spread (i.e., variance). The variance of the screener, however, is expected to be smaller than the variance of true intake because the screener prediction formula estimates the conditional expectation of true intake given the screener responses, and in general, the variance of a conditional expectation of a variable X is smaller than the variance of X itself.
As a result, the screener estimates of intake cannot be used to estimate quantiles (other than median) or prevalence estimates of true intake unless it is first adjusted so that it has approximately the same variance as true intake.
### When is it appropriate to use variance adjustment estimates?
The appropriate use of the screener information depends on the analytical objective. Following is a characterization of suggested procedures for various analytical objectives.
Analytical Objective Procedure
Estimate mean or median intake in the population or within subpopulations. Use the unadjusted screener estimate of intake.
Estimate quantiles (other than median) of the distribution of intake in the population; estimate prevalence of attaining certain levels of dietary intake. Use the variance-adjusted screener estimate.
Classify individuals into exposure categories (e.g., meeting recommended intake vs. not meeting recommended intake) for later use in a regression model. Use the variance-adjusted screener estimates to determine appropriate classification into categories.
Use the screener estimate as a continuous covariate in a multivariate regression model. Use the unadjusted screener estimate.
### How were the variance adjustment factors estimated?
We developed procedures to estimate the variance of true intake using data from 24-hour recalls, by taking into consideration within-person variability1,2. We extended these procedures to allow estimation of the variance of true intake using data from the screener. The resulting variance adjustment factors adjust the screener variance to approximate the variance of true intake in the population.
We used two external validation datasets available to us to estimate the adjustment factors: the Eating at America's Table Study (EATS) and the Observing Protein and Energy Nutrition Study (OPEN). The results indicate that the adjustment factors differ by gender for each dietary variable. Under the assumption that the variance adjustment factors appropriate to the California Health Interview Survey are similar to those in these external studies, the variance-adjusted screener estimates of intake should have variances closer to the estimated variance of true intake than would have been obtained from repeat 24-hour recalls.
Dietary Variable Variance Adjustment Factors
Men Women
Total fruits and vegetables (includes French fries and dried beans) Pyramid servings 1.3 1.1
cup equivalents 1.2 1.1
Fruits and vegetables without French fries Pyramid servings 1.3 1.1
cup equivalents 1.2 1.1
Fruits and vegetables without dried beans Pyramid servings 1.2 1.1
cup equivalents 1.2 1.1
Fruits and vegetables without French fries and dried beans Pyramid servings 1.3 1.1
cup equivalents 1.2 1.1
Added sugar (tsp) 1.5 1.3
### How are the variance adjustment factors applied?
Adjust the screener estimate of intake by:
• multiplying intake by an adjustment factor (an estimate of the ratio of the standard deviation of true intake to the standard deviation of screener intake); and
• adding a constant so that the overall mean is unchanged.
The formula for the variance-adjusted screener is:
This procedure is performed on the normally distributed version of the variable (i.e., Pyramid servings of fruits and vegetables is square-rooted; teaspoons of added sugar is cube-rooted). The results can then be back-transformed to obtain estimates in the original units.
A similar variance adjustment procedure is used to estimate prevalence of intakes for the 2000 NHIS in:
Thompson FE, Midthune D, Subar AF, McNeel T, Berrigan D, Kipnis V. Dietary intake estimates in the National Health Interview Survey, 2000: Methodology, results, and interpretation. J Am Dietet Assoc 2005;105:352-63.
## Attenuation of Regression Parameters Using Screener Estimates
When the screener estimate of dietary intake is used as a continuous covariate in a multivariate regression, the estimated regression coefficient will typically be attenuated (biased toward zero) due to measurement error in the screener. This "attenuation factor"3 can be estimated in a calibration study and used to deattenuate the estimated regression coefficient (by dividing the estimated regression coefficient by the attenuation factor).
We estimated attenuation factors in the EATS and OPEN data (see the following table).
Dietary Variable Attenuation factors for screener-predicted intake
Men Women
(Square-root) Total fruits and vegetables Pyramid servings 0.81 0.66
cup equivalents 0.79 0.63
(Square-root) Fruits and vegetables without French fries Pyramid servings 0.87 0.69
cup equivalents 0.84 0.65
(Square-root) Fruits and vegetables without dried beans Pyramid servings 0.78 0.65
cup equivalents 0.78 0.63
(Square-root) Fruits and vegetables without French fries and dried beans Pyramid servings 0.85 0.69
cup equivalents 0.83 0.65
(Cube-root) Added sugar 0.95 0.89
If the screener values are categorized into quantiles and the resulting categorical variable is used in a linear or logistic regression, the bias (due to misclassification) is more complicated because the categorization can lead to differential misclassification in the screener4. Although methods may be available to correct for this5,6, it is not simple, nor are we comfortable suggesting how to do it at this time.
Even though the estimated regression coefficients are biased (due to measurement error in the screener or misclassification in the categorized screener), tests of whether the regression coefficient is different from zero are still valid. For example, if one used the SUDAAN REGRESS procedure with fruit and vegetable intake (estimated by the screener) as a covariate in the model, one could use the Wald F statistic provided by SUDAAN to test whether the regression coefficient were statistically significantly different from zero. This assumes that only one covariate in the model is measured with error; when multiple covariates are measured with error, the Wald F test that a single regression coefficient is zero may not be valid, although the test that the regression coefficients for all covariates measured with error are zero is still valid. | 1,710 | 8,321 | {"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-2013-48 | longest | en | 0.923639 |
https://www.essaytown.com/subjects/paper/attic-fan-essential/84657 | 1,601,221,681,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00600.warc.gz | 808,160,262 | 15,240 | # Attic FanTerm Paper
Pages: 10 (2857 words) · Bibliography Sources: 5 · File: .docx · Level: College Senior · Topic: Physics
SAMPLE EXCERPT . . .
A high quality thermostat is also necessary since it helps in energy conservation and is also time saving since it eliminates the need of the home owner climbing up to the attic in order to reset the fan. It is also necessary to include a firestat that helps in shutting off the fan incase of a fire outbreak. The fan's motor should be thoroughly lubricated at all times to ensure smooth operation.
There exist two types of attic fans. The roof top type and a gable-end type. The diagram below shows the effect of the attic fan on the overall ventilation of the house.
Determination of the attic ventilation requirement for the attic fan
1.Determination of the attic area's square footage to be ventilated
In order to determine the area of the attic to be ventilated by the attic fan, one need to multiply the length of the attic by the attic's width.
For example: Let's assume we have an attic whose length is 50 feet by 25 feet. Then it follows that the attic area to be ventilate d would be 50' x X 25'= 1250 square feet.
2.Determination of the required total net free area..
After determining the attic square footage, divide it by 150 in order to get the net free area needed to effectively ventilate the attic
Calculation:
1250 sq. ft + 150 = 8.333 square feet of net free area.
3.Determination of the required amount of intake and exhaust net free area.
for \$19.77
In order to achieve an optimum performance in the attic ventilation, a balance must be achieved between the intake and the exhaust vents. This is achieved in calculation by dividing the result of step 2 by two.
Calculation
8.333 + 2= 4.16
4. Conversion to square inches
Ventilation product's specifications are usually listed in square inches. Therefore we must convert our calculations to square inches. To do this we multiply the results obtained by 1.
Calculation:
4.16 sq. ft x 144 = 599 sq. in. Of intake net free area and The attic fan design
## Term Paper on Attic Fan Is an Essential Assignment
The fan should be highly efficient in the exhaustion of air from the attic. The fan includes highly optimized airflow blades with a twisted configuration. The blades should be capable of achieving a rotational speed of up to 600 RPM.The choice of material for the blades should be plastic. That means that the blades can be remolded on to a hub in order to form one piece (unit).Metal too can be used in the design of the blades.
The assembly can then be mounted on an exhaust outlet possessing a conical shaped diffuser on the roof. The other alternative is to have the fan assembly in a portable form. This arrangement allows for the fan to be moved anywhere where need for ventilation arises. The blades are put in motion by the action of a solar powered motor. The blade and motor assembly should be capable of achieving approximately 1050 cfm while at the same time utilizing no more than 17 Watts.
The arrangement below portrays the basic configuration of the attic fan design.
The following factors are the most important in the design of the attic fan:
a) Uniform flow
In order for the fun wheel to impart a uniform stream of air of required velocity and pressure over the entire net attic area, the wheel must possess two very important characteristics.
i. The blades should be narrow at the tip as that is the area where the velocity is highest. The blade should then widen towards the hub. The widening of the blade towards the hub should is done in order to achieve more blade area that is required because of lower blade velocity.
ii. The blade's angle of rotation to the plane of rotation should be minimum at the tip and should increase towards the hub. It is necessary to calculate the width and angle for each point of the blade in this manner in order to achieve uniform velocity and pressure.
It is paramount to maintain uniform velocity and pressure over the blades because of the following reasons:
In case certain areas of the blade do not have same pressure as other portions of the blade, then there will be back flow of air in those points. Such points are located near the hub in blades such as the ones for an airplane propeller.
Inequality in the discharge velocity also results in power wastage in excess velocity pressure. Because velocity pressure increases as the square of velocity, inequality in discharge velocity, if translated converted into velocity pressure, would result to a higher velocity pressure. A good example being that fans which develop most of their velocity and pressure near the tip of the blade usually have very low efficiency.
b) Hub size.
The hub's aerodynamic purpose is to prevent back flow of air through the center of the fun and hence it increases the fan's efficiency. The hub should never be too large or too small. A large hub would result in an increase in velocity pressure. That would result in loss of efficiency due to the small net opening. A very small hub also results in the deterioration of air flow near the hub.
The selection of the aerofoil shape of the blade should be done to ensure maximum efficiency. That is achieved by ensuring that the blade exhibits high lift and low drag. It is also important to consider the abruptness at which stalling occurs. It is important to consider the following factors while selecting a blade for the fan.Lift, the drag and the angle of attack.
d) Pitch and the blade angle
Pitch is the advance of air stream per revolution of the fan wheel. It may be expressed in terms of feet per revolution. For convenience, it is expressed as a dimensionless quantity. It is often expressed in terms of diameter of the hub.
It is worth noting that blade setting and blade angle is not the same thing as pitch. Blade angle shows the angle of the blade with the plane of rotation at a particular point on the blade
The number of blades that a fan has is not important. What is of paramount importance is the effective surface area of the blades. The total blade surface area for each radius is what really matters. The width of the blade is always limited by physical considerations in an effort to prevent the fan wheel from being too deep. Because the importance attached to the fan's efficiency, it is important to avoid excess or insufficient blade surface area. It is advantageous to select any particular number of blades up to the point when there is a limitation posed by the physical conditions of the design.
It is important to correlate the design of the wheel with the speed of the blades. This is because the average pressure imparted by any wheel is directly proportional to the square of the speed of rotation in RPM (Rotation per minute)
Conclusion
A solar powered attic fan is an essential part of every household due to its applicability and affordability. The current global financial crisis coupled with the eminent environmental and energy crisis therefore call for more affordable, efficient and environmentally friendly alternatives.
The solar powered attic fan is therefore an appropriate use of renewable solar energy and its use should be advocated for and even be adopted by the governments and corporations in order to aid in the use of more efficient technologies.Further research and development in the use of solar power is necessary to match equal improvements in the design of the fan blades and hub assemblies.
A solar powered attic fan is therefore a work of true ingenuity.
'Bibliography
ANSI/AMCA Standard 210-99, "Laboratory Methods Of Testing Fans for Aerodynamic Performance Rating"
Eckert, Michael, "The Dawn of Fluid Dynamics: A Discipline Between Science and Technology"
Frank Bleier, "Fan Handbook: Selection, Application, and Design"
Vermass, Wim. "An Introduction to Photosynthesis and Its Applications."… [END OF PREVIEW] . . . READ MORE
Two Ordering Options:
1. Buy full paper (10 pages)
- or -
2. Write a NEW paper for me!✍🏻
### How to Cite "Attic Fan" Term Paper in a Bibliography:
APA Style
Attic Fan. (2009, December 4). Retrieved September 27, 2020, from https://www.essaytown.com/subjects/paper/attic-fan-essential/84657
MLA Format
"Attic Fan." 4 December 2009. Web. 27 September 2020. <https://www.essaytown.com/subjects/paper/attic-fan-essential/84657>.
Chicago Style
"Attic Fan." Essaytown.com. December 4, 2009. Accessed September 27, 2020.
https://www.essaytown.com/subjects/paper/attic-fan-essential/84657. | 1,852 | 8,567 | {"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} | 4.34375 | 4 | CC-MAIN-2020-40 | latest | en | 0.900278 |
https://espanol.libretexts.org/Biologia/Ecolog%C3%ADa/Biolog%C3%ADa_de_la_Conservaci%C3%B3n_en_%C3%81frica_Subsahariana_(Wilson_y_Pripack)/05%3A_La_retazos_por_el_espacio | 1,723,472,704,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641039579.74/warc/CC-MAIN-20240812124217-20240812154217-00813.warc.gz | 194,607,339 | 30,622 | Saltar al contenido principal
5: La retazos por el espacio
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
This page titled 5: La retazos por el espacio is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by John W. Wilson & Richard B. Primack (Open Book Publishers) via source content that was edited to the style and standards of the LibreTexts platform. | 1,838 | 4,689 | {"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} | 2.78125 | 3 | CC-MAIN-2024-33 | latest | en | 0.181071 |
https://www.mathconcentration.com/profile/AnthonyJohnHawken | 1,560,993,269,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999130.50/warc/CC-MAIN-20190620004625-20190620030625-00070.warc.gz | 839,694,514 | 20,250 | One Stop Shop for Math Teacher Resources
Anthony John Hawken
• teacher
• Blog Posts
• Discussions
• Events
• Groups
• Photos
• Photo Albums
• Videos
# Anthony John Hawken's Page
Give Anthony John Hawken a Gift
## Profile Information
How did you hear about Math Concentration?
Are you a parent, teacher, student, other?
teacher
I am a retired former teacher of mathematics and computing. I still do a bit of tuition and help my 16 year old son with his studies.
## Comment Wall
Join Math Concentration
## Make a Difference
Please support our community of students, parents, and teachers or caregivers who all play vital roles in the homework process by contributing whatever you can to keep our site alive :)
## Notes
### Figure This Challenge #56
• Complete Solution will be given on May 17, 2015
Complete Solution:
…
Continue
Created by Wanda Collins May 10, 2015 at 1:56pm. Last updated by Wanda Collins May 10, 2015.
## Math Homework Help Online
Professional math homework help get your math solved today.
Do you need help with math homework? Our reliable company provides only the best math homework help.
# Math Limerick
Question: Why is this a mathematical limerick?
( (12 + 144 + 20 + 3 Sqrt[4]) / 7 ) + 5*11 = 92 + 0 .
A dozen, a gross, and a score,
plus three times the square root of four, divided by seven, plus five times eleven, is nine squared and not a bit more.
---Jon Saxton (math textbook author)
Presentation Suggestions:
Challenge students to invent their own math limerick!
The Math Behind the Fact:
It is fun to mix mathematics with poetry.
Resources:
Su, Francis E., et al. "Math Limerick." Math Fun Facts.
funfacts
• View All | 414 | 1,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-26 | latest | en | 0.873137 |
https://brainly.in/question/85904 | 1,484,785,119,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00302-ip-10-171-10-70.ec2.internal.warc.gz | 801,742,862 | 11,227 | # How to balance chemical equation?
2
by herorock
• Brainly User
2015-03-12T20:09:03+05:30
. Write the unbalanced chemical equation.
Left would be the reactant side, while right would be the product side.
C3H8 + O2 ---> H2O + CO2
2. Start by balancing elements which is NOT Hydrogen or Oxygen. Hydrogen should be second to the last and Oxygen last.
As we can see, Carbon is not balanced. There are 3 carbon atoms while in the product side, only one. So to balance it, we need to write 3 as a stoichiometric coefficient. It is the number in the front of a compound.
This will now be:
C3H8 + O2 ---> H2O + 3CO2
3. Next, we're now only left with H and O, we must first balance Hydrogen. In the reactant side, there are 8 H atoms, while, there are only 2 H atoms at the right. So in order to make 2, eight, we need to multiply it by 4. So you have to put a coefficient in front of H in the product side.
C3H8 + O2 ---> 4H2O + 3CO2
4. Now, we finally balance O. In the left, there are 2 atoms while at the right, Oxygen has 4 atoms in H2O plus 6 atoms in CO2 (3*2) so at the product side, we have 10 atoms at the product side.
To balance it, we'll need to multiply 2 by 5 in the reactant side.
C3H8 + 5O2 ---> 4H2O + 3CO2
hope it helps a bit!!!
edited the ans
http://www.ncert.nic.in/ncerts/textbook/textbook.htm?jesc1=1-16
herez the url
She mean in 10th book.
yes
2015-03-12T20:09:47+05:30
A chemical equation is a written symbolic representation of a chemical reaction. The reactant chemical(s) are given on the left-hand side and the product chemical(s) on the right-hand side. The law of conservation of mass states that no atoms can be created or destroyed in a chemical reaction, so the number of atoms that are present in the reactants has to balance the number of atoms that are present in the products. Follow this guide to learn how to balance chemical equations differently.
1.
Write the unbalanced equation.
2.
Balance the equation.
3.
Indicate the states of matter of the reactants and products.
Example Problem:
1.Write the unbalanced equation.SnO2 + H2 → Sn + H2O
2.
Balance the equation.Look at the equation and see which elements are not balanced. In this case, there are two oxygen atoms on the lefthand side of the equation and only one on the righthand side. Correct this by putting a coefficient of 2 in front of water:SnO2 + H2 → Sn + 2 H2OThis puts the hydrogen atoms out of balance. Now there are two hydrogen atoms on the left and four hydrogen atoms on the right. To get four hydrogen atoms on the right, add a coefficient of 2 for the hydrogen gas. Remember, coefficients are multipliers, so if we write 2 H2O it denotes 2x2=4 hydrogen atoms and 2x1=2 oxygen atoms.SnO2 + 2 H2 → Sn + 2 H2O
3.
Indicate the physical states of the reactants and products.To do this, you need to be familiar with the properties of various compounds or you need to be told what the phases are for the chemicals in the reaction. Oxides are solids, hydrogen forms a diatomic gas, tin is a solid, and the term 'water vapor' indicates that water is in the gas phase:SnO2(s) + 2 H2(g) → Sn(s) + 2 H2O(g) | 882 | 3,107 | {"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} | 3.296875 | 3 | CC-MAIN-2017-04 | latest | en | 0.910395 |
http://en.wikipedia.org/wiki/Bump_mapping | 1,432,695,093,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928865.24/warc/CC-MAIN-20150521113208-00143-ip-10-180-206-219.ec2.internal.warc.gz | 79,435,396 | 12,839 | # Bump mapping
A sphere without bump mapping (left). A bump map to be applied to the sphere (middle). The sphere with the bump map applied (right) appears to have a mottled surface resembling an orange. Bump maps achieve this effect by changing how an illuminated surface reacts to light without actually modifying the size or shape of the surface.
Bump mapping[1] is a technique in computer graphics for simulating bumps and wrinkles on the surface of an object. This is achieved by perturbing the surface normals of the object and using the perturbed normal during lighting calculations. The result is an apparently bumpy surface rather than a smooth surface although the surface of the underlying object is not actually changed. Bump mapping was introduced by James Blinn in 1978.[2]
Normal mapping is the most common variation of bump mapping used.[3]
## Bump mapping basics
Bump mapping is limited in that it does not actually modify the shape of the underlying object. On the left, a mathematical function defining a bump map simulates a crumbling surface on a sphere, but the object's outline and shadow remain those of a perfect sphere. On the right, the same function is used to modify the surface of a sphere by generating an isosurface. This actually models a sphere with a bumpy surface with the result that both its outline and its shadow are rendered realistically.
Bump mapping is a technique in computer graphics to make a rendered surface look more realistic by simulating small displacements of the surface. However, unlike displacement mapping, the surface geometry is not modified. Instead only the surface normal is modified as if the surface had been displaced. The modified surface normal is then used for lighting calculations (using, for example, the Phong reflection model) giving the appearance of detail instead of a smooth surface.
Bump mapping is much faster and consumes less resources for the same level of detail compared to displacement mapping because the geometry remains unchanged.
There are also extensions which modify other surface features in addition to increasing the sense of depth.[clarification needed] Parallax mapping is one such extension.
The primary limitation with bump mapping is that it perturbs only the surface normals without changing the underlying surface itself.[4] Silhouettes and shadows therefore remain unaffected, which is especially noticeable for larger simulated displacements. This limitation can be overcome by techniques including the displacement mapping where bumps are actually applied to the surface or using an isosurface.
### Methods
There are two primary methods to perform bump mapping. The first uses a height map for simulating the surface displacement yielding the modified normal. This is the method invented by Blinn[2] and is usually what is referred to as bump mapping unless specified. The steps of this method are summarized as follows.
Before a lighting calculation is performed for each visible point (or pixel) on the object's surface:
1. Look up the height in the heightmap that corresponds to the position on the surface.
2. Calculate the surface normal of the heightmap, typically using the finite difference method.
3. Combine the surface normal from step two with the true ("geometric") surface normal so that the combined normal points in a new direction.
4. Calculate the interaction of the new "bumpy" surface with lights in the scene using, for example, the Phong reflection model.
The result is a surface that appears to have real depth. The algorithm also ensures that the surface appearance changes as lights in the scene are moved around.
The other method is to specify a normal map which contains the modified normal for each point on the surface directly. Since the normal is specified directly instead of derived from a height map this method usually leads to more predictable results. This makes it easier for artists to work with, making it the most common method of bump mapping today.[3]
## Realtime bump mapping techniques
Realtime 3D graphics programmers often use variations of the technique in order to simulate bump mapping at a lower computational cost.
One typical way was to use a fixed geometry, which allows one to use the heightmap surface normal almost directly. Combined with a precomputed lookup table for the lighting calculations the method could be implemented with a very simple and fast loop, allowing for a full-screen effect. This method was a common visual effect when bump mapping was first introduced. | 871 | 4,554 | {"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-2015-22 | latest | en | 0.930817 |
https://mscastillosmath.com/2016/04/21/3-act-all-the-little-duckies/ | 1,686,173,338,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00374.warc.gz | 452,659,737 | 25,396 | # 3 Act: All the Little Duckies
I was in a 2nd grade room the other day and a teacher was presenting a lesson on measurement using customary units. Students were using rulers to measure to the nearest inch. One of the students asked, “what if it is over 4 inches but not 5 inches?” The teacher said, “you choose.” I walked over and whispered, “actually let’s round up if it’s at a half and down if it’s before half because when they use rounding in 3rd grade next year, that will provide them the visual representation to refer back to.”
This really got me thinking again about the power of vertical teaming. This teacher didn’t know that this is a common misconception in 3rd and 4th grade. She also didn’t know many of the misconceptions around measurement. She anticipated that they might start with something other than zero, but she didn’t know to address that in 2nd grade we are measuring units without precision and that vocabulary such as “about 5 inches” is necessary. Or that conversations about which one to choose would be necessary. In the task below for standard 2.MD.A.2, I hope to have students dig into discussions about which number to choose and why. I want them to defend their reasoning and discuss which would be more accurate. This will allow students to reason with numbers and will carry over to rounding, to estimation, and to looking for reasonableness in their answers in years to come.
The Common Core Standards specifically state in 1.MD.A.2 to ” Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps.” That is the reason that I used standard 2.MD.A.2 for this task. In my opinion, this task is very appropriate for first grade and kindergarten students as well. That conversation cannot happen early enough!
Act 1:
About how many ducks long is the math rack?
About how many cubes long is the math rack?
Act 2:
About how many ducks long is it? Why did you choose that number?
There are 4 more cubes than ducks.
About how many cubes long is it?
Act 3:
Were you correct?
If so, what were you thinking?
If not, what was your thinking? | 510 | 2,158 | {"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-2023-23 | latest | en | 0.967655 |
http://www.chegg.com/homework-help/questions-and-answers/determine-value-p-ande-following-signals-1-x2-n-ej-2n-8-determine-whether-following-signal-q477197 | 1,501,097,166,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00660.warc.gz | 394,363,761 | 17,050 | Determine the value of P andE for each of the following signals:
(1) X2 [n] = ej(π/2n + π/8)
Determine whether or not the following signals is periodic. if asignal is periodic, specify its fundamental period.
(1) X1 (t) = jej10t
(2) X4 [n] = 3ej3π(n+1/2)/5 | 92 | 258 | {"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-2017-30 | latest | en | 0.729757 |
http://physics.stackexchange.com/questions/tagged/projectile+kinematics | 1,406,199,084,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997888283.14/warc/CC-MAIN-20140722025808-00168-ip-10-33-131-23.ec2.internal.warc.gz | 150,522,613 | 24,481 | # Tagged Questions
0answers
26 views
### Formula to find angle of projectile motion [on hold]
I want to be able to modify the angle at which the bullet follows projectile motion in order to hit the object which is moving horizontally (from right to left) and the projectile motion is starting ...
3answers
700 views
### Ball thrown from a moving train
This is not a question about throwing a ball vertically in a moving train. I am asking what would happen if I throw a ball in a horizontally in a moving train. Assume I am facing an exit door of a ...
0answers
28 views
### How to calculate the equation of projectile motion for curved ramps? [duplicate]
Consider the following cases (these are sections): I can find the equation of motion for Case a however I don't know how to deal with Case d. Normally I would need the angle α otherwise I can't ...
3answers
994 views
### Does an object's kinetic energy increase, decrease, or stay constant when it reaches terminal velocity while falling?
Does an object's kinetic energy increase, decrease, or stay constant when it reaches terminal velocity while falling?
2answers
52 views
### Isn't this statement regarding projectile motion wrong?
Isn't this statement regarding projectile motion wrong? If a body is thrown at an angle to the horizontal with initial velocity $u$, then displacement of body as a function of time is ...
2answers
46 views
### Why is the constant velocity model used in a projectile motion derivation?
I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics ...
2answers
93 views
### Kinematics - projectile motion on inclined plane [closed]
Well I've tried everything to solve this problem, spent well over an hour, and have gotten no results. A particle is thrown over a triangle from one end of a horizontal base it grazes the top vertex ...
1answer
51 views
### Kinematic, motion in 2D [closed]
If the kinetic energy at maximum height is 2/5 the kinetic energy at half the maximum height, find the angle of projection. In other words: K.E. at H(max) = 0.4*K.E. at 0.5*H(max) I got 60 degrees, ...
2answers
33 views
### Projectile motion - ball hit 180m/s at 45 degrees, what is the hang time? [closed]
Assuming no air resistance etc, how long is the ball in the air for? This is actually part of a longer question concerning the ball trajectory on different planets but I think there must be an easier ...
1answer
60 views
### Kinematics - angle of projection given relation between speed at highest point and half of that [closed]
For a projectile launched from a point on horizontal ground, the speed when it is at the greatest height is $\sqrt{\frac{2}{5}}$ times the speed when it is at half of its greatest height. Determine ...
1answer
63 views
### Throwing an object in the air [closed]
Studying formulas about velocity and acceleration I came up with a question: if I throw an object in the air with a velocity $v_0$ (suppose i throw it vertically) in how much time its final velocity ...
1answer
80 views
### A ball is thrown up, and a second ball is dropped- what is the maximum initial velocity of the first ball so that
This question only exists in one dimension, the y axis. A ball is thrown upwards with an initial velocity $V_o$ from a roof with height $h$. One second later, a second ball is dropped from the same ...
1answer
73 views
### Tangential and radial acceleration in projectile motion
I'm currently learing kinematics, specifically projectile motion and as an example in my textbook is a bullet fired at some angle. I understand the derivation of formulas to describe that motion and ...
2answers
52 views
### When solving kinematics questions, how to find time in air?
I was trying to help a friend with this question: A ball is shot horizontally off a cliff. What would happen to the horizontal distance it travels if the acceleration due to gravity were doubled ...
1answer
46 views
### Two equations for $y$ in projectile motions
I have two equations on my equation sheet: $$y=v_o t \sin \theta - \frac{1}{2} g t^2.$$ And then another: $$y_{\max} = \frac{v_o^2 \sin 2 \theta }{2g}.$$ I understand that the first one is an ...
1answer
53 views
### Finding release angle and final velocity of projectile when the target has a different coordinations [closed]
I am trying to write some codes for auto-snowball mechanism which seems like a regular projectile but instead of hitting the targets on the ground it is going to be able to hit the target in the air ...
0answers
21 views
### Finding the initial vertical and horizontal velocity [closed]
Basically I need to get an initial vertical velocity to make the ball reach a specific height, but also I need to get a horizontal verlocity so that the ball will travel the distance. For an example ...
2answers
218 views
### Find the minimum value of velocity [closed]
Find the minimum value of the initial velocity $u$ of the particle such that the particle crosses the wheel of radius $R$. Details and assumptions $R=2m$ $g=9.8m/s^2$ Neglect air resistance. All ...
1answer
41 views
### Determining the horizontal position of an object [closed]
In my calc-physics class we were given the following question: A movie stuntwoman drops from a helicopter that is $30.0 m$ above the ground and moving with a constant velocity whose components are ...
1answer
160 views
### Horizontal projectile motion, finding the height of the object at a given time. [closed]
A canon is fired Horizontal to the ground 80 meters above ground The canon ball is fired a $T=0$ and hits the ground at $T_g$. Calculate the height at the time $T_g/2$ So far I have calculated time. ...
2answers
732 views
### How to plot $(x,y)$ coordinate of projectile motion (with air-resistance)? [closed]
I am trying to plot a graph to show the difference in projectile motion when it has air-resistance and when it doesn't have air-resistance. I set the mass, $v_0$ and $\theta$ as constant I can plot ...
2answers
92 views
### Time variation of the distance between two falling objects [closed]
I have solved this question below from my textbook. My answer differs from the one given in the book. I have tried my best to see where I am wrong. I need your help to find the truth of this question. ...
0answers
407 views
### Find initial velocity when given max height of 100m [closed]
A ball is thrown straight up. The height of the ball above the ground in (m) is given by the function: $$h(t)=-5t^2 + vt + c$$ What is the initial velocity if you want the ball to reach a max height ...
1answer
174 views
### Calculating angle of projectile motion [closed]
I realize that similar questions have been asked but I'd like to put my problem here and what I have to make sure everything is right. So my problem is: I've got a rock $2.5$ meters above the ground ...
1answer
104 views
### Parametric equations in ballistics
while studying some material on ballistic trajectories (the basic gravity-only parabola), I've tried to come up with closed-form expressions for another case, where in addition to gravity we add a ...
2answers
131 views
### Why does a small object stay in one place after being shot from a big gun?
This video shows what I mean. Basically they shoot a bullet from a huge gun at an iPhone. What I don't get is why does the phone move only so little after being shot? As far as I can tell it isn't ...
2answers
154 views
### Projectile motion maximum distance to origin
Is there an elegant or easy way to derive the maximum distance from the origin to a point of the parabola created by projectile motion (assume initial position is the origin)? Other than ...
2answers
190 views
### Finding the optimal angle of a projectile motion [closed]
I am trying to solve a task in which I need to calculate the optimal angle($\alpha$) with which the projectile will land the furthest from a height ($h$), so what I basically have is the equations for ...
1answer
77 views
### Why does a difference in approach to projectile motion yields different results?
A body is projected vertically upwards from the surface of the earth with a velocity equal to half the escape velocity. If $R$ is the radius of the earth, then find the maximum height attained by the ...
1answer
151 views
### Is there a better, faster way to do this projectile motion question? [closed]
The question is In a combat exercise, a mortar at M is required to hit a target at O, which is taking cover 25 m behind a structure of negligible width 10 m tall. This mortar can only fire at an ...
2answers
147 views
### Does $y$-motion really have nothing to do with $x$-motion?
I am watching a Physics 1 for physical science majors on coursera.com, and in one of the concept tests there is a question that goes like this; "A bullet is fired horizontally from a rifle on the Moon ...
1answer
115 views
### x-coordinate of the ball when it's velocity is perpendicular to projection velocity
We have a ball projected at 20 m/s at an angle $\pi/6$ from the x-axis. Now we have to calculate x-coordinate of the ball when it's velocity is perpendicular to projection velocity. I am thinking ...
3answers
196 views
### Projectile motion formulas
Assuming: The equations for the vertical and horizontal components of my initial velocity. Are: $v_{x,i} = v_i \cos\theta$ and $v_{y,i} = v_i \sin\theta$ And the displacement components are ...
0answers
181 views
### How to find the angle of elevation and launch speed? [closed]
A particle is projected from a point on level ground with a speed of $u$ meters per second and an angle of elevation $\theta$. The maximum height reached by the particle is $42$ meters above the ...
1answer
132 views
### Is this a standard projectile motion equation? [closed]
Is Initial Velocity = Time * Acceleration a standard projectiole motion equation?
1answer
157 views
### Projectile maximum height [closed]
How can I determine the needed initial velocity to make a 2d wind resistance free projectile (in computer simulation) reach maximum height at a certain x distance away, with launch position and target ...
0answers
313 views
### Is it possible to solve a 2D relative motion question using components of relative velocity?
This is a question based on concepts of two dimensional motion . Here's how the question is: A plank A is floating in air, gravity has no effect on it (See its coordinates in the figure attached). ...
2answers
635 views
### Projectile Motion Question involving a ball and a ramp inclined at an angle
The question is to finde the initial horizontal velocity of the ball at end of the ramp, where it is released. I know how to do this using gravitational potential energy and kinetic energy ...
1answer
205 views
### 1D Motion under Gravity
I have attempted this question in so many different ways but I am getting nowhere. Could someone point me in the correct direction so I can work it out myself? Here is the question: An object $K$ ...
3answers
740 views
### Intuitive meaning of factor 2 in formula of vertical throw max height $h=v^2/2g$
This is a question about a simple thing. The simplified expression for maximum height in vertical throw is $h=\frac{v^2}{2g}$ , could anyone explain intuitively (analogies are welcome) why there is a ...
2answers
456 views
### How can I solve for time without knowing the vertical velocity?
A guy posted this problem on a forum: There is a bird sitting on a pole of height h. you throw a rock at it and the moment the rock leaves your hand the bird starts flying horizontally away from ...
1answer
151 views
### Reachable area of cannonball given fixed initial speed [closed]
The trajectory of a cannonball fired from the origin with initial speed $v_0$ at an angle $\theta$ is given by $$y = x\tan\theta - \frac{g}{2v_0^2 \cos^2\theta}x^2.$$ For fixed $v_0$, at what angle ...
2answers
1k views
### Given Angle, Initial Velocity, and Acceleration due to Gravity, plot parabolic trajectory for every “x”?
Given any Angle -> 0-90 Given any Initial Velocity -> 1-100 Given Acceleration due to Gravity -> 9.8 Plot every x,y coordinate (the parabolic trajectory) with cartesian coordinates and screen pixels ...
3answers
1k views
### Solve for Initial Velocity of a projectile given Angle, Gravity, and Initial and Final positions?
I've found equations http://www.physicsclassroom.com/Class/1DKin/U1L6a.cfm for solving everything (and rearranged to solve everything) to do with projectile motion EXCEPT this, even though it should ...
1answer
102 views
### Projectile's angle in midflight
For a missile travelling from (0,0) at angle $\theta$ (to the horizontal) and initial velocity $u$, the y (vertical) position at time t is given by $s_{y} = u\sin (\theta) t - 0.5gt^{2}$ and the x ...
0answers
334 views
### What is the initial velocity of a projectile so that it passes through a target point in its trajectory? [closed]
Let's say I have a projectile being thrown by a player in my 2-D game. I want to work backwards and find the initial velocity to apply to the projectile such that it passes through a target point in ...
0answers
177 views
### Getting pairs of angle and velocity for a projectile to a given destination
I'm trying to calculate the initial velocity $v_0$ and angle $\theta$ for a given destination $(x, y)$ with a launch height of $y_0$. Obviously there will be a set of pairs of velocity and angle that ...
0answers
176 views
### I need help with the following Physics' questions [closed]
I need to know which formulas and the step-by-step answers for the following physics questions: 1. You throw a ball upward with an initial speed of 7.0 m/s, and it returns to your hand o.92 ...
0answers
109 views
### A 0.1kg ball of dough is thrown up with a velocity of 15m/s. What is the momentum halfway up? [closed]
I know that $p=mv$ and (0.1kg)(15m/s)=1.5 kg m/s and the momentum at the vertex is 0, but what is the momentum halfway up?
0answers
63 views
### How to calculate the correct coordinates from a distorted video of a projectile?
I am working on a high school project that is related to projectile motion. I am exploring how exactly the position of the center of mass affects the trajectory of a long but thin, javelin-like ... | 3,323 | 14,385 | {"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} | 3.09375 | 3 | CC-MAIN-2014-23 | longest | en | 0.929799 |
https://mail.scipy.org/pipermail/numpy-discussion/2012-June/062733.html | 1,505,820,677,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818685129.23/warc/CC-MAIN-20170919112242-20170919132242-00296.warc.gz | 677,952,085 | 2,013 | # [Numpy-discussion] convert to string - astype(str)
josef.pktd@gmai... josef.pktd@gmai...
Sun Jun 10 06:58:38 CDT 2012
```On Sun, Jun 10, 2012 at 2:17 AM, Travis Oliphant <travis@continuum.io> wrote:
>
> On Jun 9, 2012, at 4:45 PM, josef.pktd@gmail.com wrote:
>
>> Is there a way to convert an array to string elements in numpy,
>> without knowing the string length?
>
> Not really. In the next release of NumPy you should be able to do.
>
> result = array(arr2, str)
>
> and it will determine the length of the string for you.
>
> For now, I would compute the max length via
>
> int(math.ceil(math.log10(np.max(arr2))))
Thanks Travis,
It's good to know when I can stop looking.
Josef
>
>
> -Travis
>
>
>
>>
>>
>>>>> arr2 = np.arange(8, 13)
>>
>> array(['8', '9', '1', '1', '1'],
>> dtype='|S1')
>>
>>>>> arr2.astype('S2')
>> array(['8', '9', '10', '11', '12'],
>> dtype='|S2')
>>
>>>>> map(str, arr2)
>> ['8', '9', '10', '11', '12']
>>
>>
>>>>> arr3 = np.round(np.random.rand(5), 2)
>>>>> arr3
>> array([ 0.51, 0.86, 0.15, 0.68, 0.59])
>>
>> array(['0', '0', '0', '0', '0'],
>> dtype='|S1')
>>
>>>>> arr3.astype('S4')
>> array(['0.51', '0.86', '0.15', '0.68', '0.59'],
>> dtype='|S4')
>>
>>>>> map(str, arr3)
>> ['0.51', '0.86', '0.15', '0.68', '0.59']
>>
>>>>> np.__version__
>> '1.5.1'
>>
>> (from an issue in statsmodels)
>>
>> Thanks,
>>
>> Josef
>> _______________________________________________
>> NumPy-Discussion mailing list
>> NumPy-Discussion@scipy.org
>> http://mail.scipy.org/mailman/listinfo/numpy-discussion
>
> _______________________________________________
> NumPy-Discussion mailing list
> NumPy-Discussion@scipy.org
> http://mail.scipy.org/mailman/listinfo/numpy-discussion
``` | 610 | 1,732 | {"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-2017-39 | latest | en | 0.513685 |
https://root.cern.ch/doc/v620/GSLSimAnMinimizer_8cxx_source.html | 1,685,697,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00765.warc.gz | 537,913,187 | 12,300 | ROOT Reference Guide
GSLSimAnMinimizer.cxx
Go to the documentation of this file.
1// @(#)root/mathmore:$Id$
2// Author: L. Moneta Wed Dec 20 17:16:32 2006
3
4/**********************************************************************
5 * *
6 * Copyright (c) 2006 LCG ROOT Math Team, CERN/PH-SFT *
7 * *
8 * *
9 **********************************************************************/
10
11// Implementation file for class GSLSimAnMinimizer
12
15#include "Math/Error.h"
16
18#include "Math/MultiNumGradFunction.h" // needed to use transformation function
20#include "Math/GenAlgoOptions.h"
21
22#include <iostream>
23#include <cassert>
24
25namespace ROOT {
26
27 namespace Math {
28
29
30
31// GSLSimAnMinimizer implementation
32
33GSLSimAnMinimizer::GSLSimAnMinimizer( int /* ROOT::Math::EGSLSimAnMinimizerType type */ ) :
35{
36 // Constructor implementation : create GSLMultiFit wrapper object
37
40}
41
43}
44
45
47 // set initial parameters of the minimizer
48 int debugLevel = PrintLevel();
49
50 if (debugLevel >=1 ) std::cout <<"Minimize using GSLSimAnMinimizer " << std::endl;
51
53 if (function == 0) {
54 MATH_ERROR_MSG("GSLSimAnMinimizer::Minimize","Function has not been set");
55 return false;
56 }
57
58 // vector of internal values (copied by default)
59 unsigned int npar = NPar();
60 std::vector<double> xvar;
61 std::vector<double> steps(StepSizes(),StepSizes()+npar);
62
63 // needed for the transformation
66
67 MinimTransformFunction * trFunc = CreateTransformation(xvar, gradFunc );
68 // ObjFunction() will return now the new transformed function
69
70 if (trFunc) {
71 // transform also the step sizes
72 trFunc->InvStepTransformation(X(), StepSizes(), &steps[0]);
73 steps.resize( trFunc->NDim() );
74 }
75
76 assert (xvar.size() == steps.size() );
77
78
79#ifdef DEBUG
80 for (unsigned int i = 0; i < npar ; ++i) {
81 std::cout << "x = " << xvar[i] << " steps " << steps[i] << " x " << X()[i] << std::endl;
82 }
83 std::cout << "f(x) = " << (*ObjFunction())(&xvar.front() ) << std::endl;
84 std::cout << "f(x) not transf = " << (*function)( X() ) << std::endl;
85 if (trFunc) std::cout << "ftrans(x) = " << (*trFunc) (&xvar.front() ) << std::endl;
86#endif
87
88 // output vector
89 std::vector<double> xmin(xvar.size() );
90
91
92 int iret = fSolver.Solve(*ObjFunction(), &xvar.front(), &steps.front(), &xmin[0], (debugLevel > 1) );
93
94 SetMinValue( (*ObjFunction())(&xmin.front() ) );
95
96 SetFinalValues(&xmin.front());
97
98
99 if (debugLevel >=1 ) {
100 if (iret == 0)
101 std::cout << "GSLSimAnMinimizer: Minimum Found" << std::endl;
102 else
103 std::cout << "GSLSimAnMinimizer: Error in solving" << std::endl;
104
105 int pr = std::cout.precision(18);
106 std::cout << "FVAL = " << MinValue() << std::endl;
107 std::cout.precision(pr);
108 for (unsigned int i = 0; i < NDim(); ++i)
109 std::cout << VariableName(i) << "\t = " << X()[i] << std::endl;
110 }
111
112
113 return ( iret == 0) ? true : false;
114}
115
116
117unsigned int GSLSimAnMinimizer::NCalls() const {
118 // return number of function calls
121 if (tfunc) f = dynamic_cast<const ROOT::Math::MultiNumGradFunction *>(tfunc->OriginalFunction());
122 else
123 f = dynamic_cast<const ROOT::Math::MultiNumGradFunction *>(ObjFunction());
124 if (f) return f->NCalls();
125 return 0;
126}
127
130 opt.SetMinimizerType("GSLSimAn");
131 // set dummy values since those are not used
132 opt.SetTolerance(-1);
133 opt.SetPrintLevel(0);
134 opt.SetMaxIterations(-1);
135 opt.SetMaxFunctionCalls(0);
136 opt.SetStrategy(-1);
137 opt.SetErrorDef(0);
138 opt.SetPrecision(0);
139 opt.SetMinimizerAlgorithm("");
140
141 const GSLSimAnParams & params = MinimizerParameters();
142
144 simanOpt.SetValue("n_tries",params.n_tries);
145 simanOpt.SetValue("iters_fixed_T",params.iters_fixed_T);
146 simanOpt.SetValue("step_size",params.step_size);
147 simanOpt.SetValue("k",params.k);
148 simanOpt.SetValue("t_initial",params.t_initial);
149 simanOpt.SetValue("mu_t",params.mu_t);
150 simanOpt.SetValue("t_min",params.t_min);
151
152 opt.SetExtraOptions(simanOpt);
153 return opt;
154}
155
157
158 // get the specific siman options
159 const ROOT::Math::IOptions * simanOpt = opt.ExtraOptions();
160 if (!simanOpt) {
161 MATH_WARN_MSG("GSLSimAnMinimizer::SetOptions", "No specific sim. annealing minimizer options are provided. No options are set");
162 return;
163 }
164 GSLSimAnParams params;
165 simanOpt->GetValue("n_tries",params.n_tries);
166 simanOpt->GetValue("iters_fixed_T",params.iters_fixed_T);
167 simanOpt->GetValue("step_size",params.step_size);
168 simanOpt->GetValue("k",params.k);
169 simanOpt->GetValue("t_initial",params.t_initial);
170 simanOpt->GetValue("mu_t",params.mu_t);
171 simanOpt->GetValue("t_min",params.t_min);
172
173 SetParameters(params);
174}
175
176
177 } // end namespace Math
178
179} // end namespace ROOT
180
#define MATH_ERROR_MSG(loc, str)
Definition: Error.h:82
#define MATH_WARN_MSG(loc, str)
Definition: Error.h:79
#define f(i)
Definition: RSha256.hxx:104
float xmin
Definition: THbookFile.cxx:93
Base Minimizer class, which defines the basic funcionality of various minimizer implementations (apar...
virtual double MinValue() const
return minimum function value
virtual unsigned int NPar() const
total number of parameter defined
void SetMinValue(double val)
virtual const double * StepSizes() const
accessor methods
const ROOT::Math::IMultiGenFunction * ObjFunction() const
return pointer to used objective function
virtual unsigned int NDim() const
number of dimensions
virtual std::string VariableName(unsigned int ivar) const
get name of variables (override if minimizer support storing of variable names)
virtual const double * X() const
return pointer to X values at the minimum
void SetFinalValues(const double *x)
MinimTransformFunction * CreateTransformation(std::vector< double > &startValues, const ROOT::Math::IMultiGradFunction *func=0)
virtual ~GSLSimAnMinimizer()
Destructor (no operations)
void SetParameters(const GSLSimAnParams ¶ms)
set new minimizer option parameters using directly the GSLSimAnParams structure
GSLSimAnMinimizer(int type=0)
Default constructor.
const GSLSimAnParams & MinimizerParameters() const
Get current minimizer option parameteres.
virtual void SetOptions(const ROOT::Math::MinimizerOptions &opt)
set new minimizer options
ROOT::Math::GSLSimAnnealing fSolver
unsigned int NCalls() const
number of calls
virtual bool Minimize()
method to perform the minimization
virtual ROOT::Math::MinimizerOptions Options() const
Get current minimizer options.
int Solve(const ROOT::Math::IMultiGenFunction &func, const double *x0, const double *scale, double *xmin, bool debug=false)
solve the simulated annealing given a multi-dim function, the initial vector parameters and a vector ...
class implementing generic options for a numerical algorithm Just store the options in a map of strin...
Documentation for the abstract class IBaseFunctionMultiDim.
Definition: IFunction.h:62
Generic interface for defining configuration options of a numerical algorithm.
Definition: IOptions.h:30
void SetValue(const char *name, double val)
generic methods for retrivieng options
Definition: IOptions.h:44
bool GetValue(const char *name, T &t) const
Definition: IOptions.h:73
MinimTransformFunction class to perform a transformations on the variables to deal with fixed or limi...
unsigned int NDim() const
Retrieve the dimension of the function.
void InvStepTransformation(const double *x, const double *sext, double *sint) const
inverse transformation for steps (external -> internal) at external point x
void SetMaxFunctionCalls(unsigned int maxfcn)
set maximum of function calls
void SetStrategy(int stra)
set the strategy
void SetMaxIterations(unsigned int maxiter)
set maximum iterations (one iteration can have many function calls)
const IOptions * ExtraOptions() const
return extra options (NULL pointer if they are not present)
void SetMinimizerType(const char *type)
set minimizer type
void SetExtraOptions(const IOptions &opt)
set extra options (in this case pointer is cloned)
void SetPrecision(double prec)
set the precision
void SetErrorDef(double err)
set error def
void SetPrintLevel(int level)
set print level
void SetMinimizerAlgorithm(const char *type)
set minimizer algorithm
void SetTolerance(double tol)
set the tolerance
void SetMaxIterations(unsigned int maxiter)
set maximum iterations (one iteration can have many function calls)
Definition: Minimizer.h:451
void SetPrintLevel(int level)
set print level
Definition: Minimizer.h:445
int PrintLevel() const
minimizer configuration parameters
Definition: Minimizer.h:411 | 2,273 | 8,592 | {"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} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.328828 |
http://thibaultlanxade.com/general/the-ancient-babylonians-developed-a-method-for-calculating-nonperfect-squares-by-1700-bce-complete-the-statements-to-demonstrate-how-to-use-this-method-to-find-the-approximate-value-of-in-order-to-determine-let-g1-2-a-number-whose-square-is-c | 1,726,110,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00544.warc.gz | 33,848,791 | 8,009 | Q:
# The ancient Babylonians developed a method for calculating nonperfect squares by 1700 BCE. Complete the statements to demonstrate how to use this method to find the approximate value of . In order to determine , let G1 = 2, a number whose square is close to 5. 5 ÷ G1 = , which is not equal to G1, so further action is necessary. Average 2 and G1 to find G2 = 2.25. 5 ÷ G2 ≈ (rounded to the nearest thousandth), which is not equal to G2, so further action is necessary. Average 2.25 and G2 to find G3 = 2.236. 5 ÷ G3 ≈ (rounded to the nearest thousandth), which is equal to G3. That means is approximately 2.236.
Accepted Solution
A:
... find the approximate value of √5. In order to determine √5, let ...
5 ÷ G1 = 2.5
5 ÷ G2 = 2.222
5 ÷ G3 = 2.236
That means √5 is approximately 2.236. | 244 | 796 | {"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 | 4 | CC-MAIN-2024-38 | latest | en | 0.932371 |
https://binaryoptionskskc.netlify.app/difelice67639wu/sample-size-calculator-paired-t-test-207 | 1,723,285,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00150.warc.gz | 101,719,631 | 10,122 | ## Sample size calculator paired t-test
t.test function. For the calculation of Example 1, we can set the power at different levels and calculate the sample size for each level. For example Sample Size for Paired t Test. Menu location: Analysis_Sample Size_Paired t. This function gives you the minimum number of pairs of subjects needed to detect
Therefore, to calculate the significance level, given an effect size, sample size, size, and type indicates a two-sample t-test, one-sample t-test or paired t-test. The choice of statistical methods: t-test without using correlation coefficients vs. analysis of covariance (ANCOVA). We show that the required sample size will Performs unpaired t test, Weldh's t test (doesn't assume equal variances) and paired t test. StatMate® calculates sample size and power. For example, compare whether systolic blood pressure differs between a control and treated group, between men (You can only choose a paired t test if you enter individual values.) Because the t distribution is used to calculate critical values for the test, this test is often called the Sample Size for Conditional Power of Paired T-Tests.
## Sample Size for Paired t Test. Menu location: Analysis_Sample Size_Paired t. This function gives you the minimum number of pairs of subjects needed to detect
A sample size for the paired t-test is calculated after clicking the paired button. The calculation will approximate the t-test. In the unlikely 1 Jun 2014 analysis would be paired t-test. When such studies are planned, it is imperative to calculate required sample size for the use of paired t-test to. Statistical significance mainly depends on the sample size, the quality of the Calculation of d and r from the test statistics of dependent and independent t-tests . Sample size for significance tests. Single mean Single proportion Comparison of means Paired samples t-test Comparison of proportions McNemar test (paired
### Sample Size Calculator: Two Parallel-Sample Means Application: This procedure is used to test non-inferiority and superiority that can be unified by the
How to calculate the required sample size for the analysis of the mean difference between paired samples. The sample size takes into account the required t.test function. For the calculation of Example 1, we can set the power at different levels and calculate the sample size for each level. For example Sample Size for Paired t Test. Menu location: Analysis_Sample Size_Paired t. This function gives you the minimum number of pairs of subjects needed to detect Because the t distribution is used to calculate critical values for the test, this test is often called the paired The paired t-test assumes that the population standard deviation of paired of paired differences, alpha, power, and the sample size. Enter sample data. Header: You may change groups' name to the real names. Data: When entering data, press Enter after each value.
### This calculator will tell you the minimum required total sample size and per-group sample size for a one-tailed or two-tailed t-test study, given the probability level
t.test function. For the calculation of Example 1, we can set the power at different levels and calculate the sample size for each level. For example Sample Size for Paired t Test. Menu location: Analysis_Sample Size_Paired t. This function gives you the minimum number of pairs of subjects needed to detect Because the t distribution is used to calculate critical values for the test, this test is often called the paired The paired t-test assumes that the population standard deviation of paired of paired differences, alpha, power, and the sample size. Enter sample data. Header: You may change groups' name to the real names. Data: When entering data, press Enter after each value. What is your study design? Please indicate what type of procedure you need sample size / power calculations for: one group. two groups. paired t-test
## How to calculate the required sample size for the analysis of the mean difference between paired samples. The sample size takes into account the required
Therefore, to calculate the significance level, given an effect size, sample size, size, and type indicates a two-sample t-test, one-sample t-test or paired t-test. The choice of statistical methods: t-test without using correlation coefficients vs. analysis of covariance (ANCOVA). We show that the required sample size will Performs unpaired t test, Weldh's t test (doesn't assume equal variances) and paired t test. StatMate® calculates sample size and power. For example, compare whether systolic blood pressure differs between a control and treated group, between men (You can only choose a paired t test if you enter individual values.) Because the t distribution is used to calculate critical values for the test, this test is often called the Sample Size for Conditional Power of Paired T-Tests. The best solution is to increase the sample size. ❑ The power of a Sample sizes for t-tests. Effect. Small Medium Large. Effect size. 0.2. 0.5. 0.8. Minimum total This calculator will tell you the minimum required total sample size and per-group sample size for a one-tailed or two-tailed t-test study, given the probability level pre-test/post-test samples in which a factor is measured before and after an As background—this is not the main analysis—it helps to calculate summary statistics for According to these presets the required sample size of a paired t test is.
Test family: The online calculator currently supports the t-test and sample size measurements from the same person (this is sometimes known as "paired", Choosing among these three equations requires an examination of the To calculate a standardized mean difference using t-stats and sample size, the For the independent samples T-test, Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the Therefore, to calculate the significance level, given an effect size, sample size, size, and type indicates a two-sample t-test, one-sample t-test or paired t-test. The choice of statistical methods: t-test without using correlation coefficients vs. analysis of covariance (ANCOVA). We show that the required sample size will Performs unpaired t test, Weldh's t test (doesn't assume equal variances) and paired t test. StatMate® calculates sample size and power. For example, compare whether systolic blood pressure differs between a control and treated group, between men (You can only choose a paired t test if you enter individual values.) Because the t distribution is used to calculate critical values for the test, this test is often called the Sample Size for Conditional Power of Paired T-Tests. | 1,377 | 6,763 | {"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} | 3.21875 | 3 | CC-MAIN-2024-33 | latest | en | 0.807474 |
https://gmatclub.com/forum/community-spokesperson-after-a-recent-surge-of-122529.html?fl=similar | 1,490,676,245,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189667.42/warc/CC-MAIN-20170322212949-00555-ip-10-233-31-227.ec2.internal.warc.gz | 813,423,170 | 68,528 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club
It is currently 27 Mar 2017, 21:44
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Community spokesperson : After a recent surge of
Author Message
TAGS:
### Hide Tags
Manager
Joined: 31 May 2011
Posts: 88
Location: India
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Followers: 1
Kudos [?]: 51 [0], given: 4
Community spokesperson : After a recent surge of [#permalink]
### Show Tags
29 Oct 2011, 22:12
2
This post was
BOOKMARKED
00:00
Difficulty:
(N/A)
Question Stats:
61% (01:58) correct 39% (02:55) wrong based on 60 sessions
### HideShow timer Statistics
Community spokesperson: After a recent surge of foreclosures, many homes in our community are vacant and falling into disrepair. Property values are already falling, and, if action is not taken quickly, will fall even further when panicked residents begin to leave. Clearly the only way to prevent a snowball effect is to make it easy to purchase these vacant homes by offering potential buyers special mortgages with a low interest rate for the first two years.
Which of the following, if true, represents the most significant potential problem with the plan to attract buyers through special mortgage offers?
A) Interest rates in the area are already at a 5-year low, and it would be foolish of lenders to lower them even more.
B) Crime rates in the area have risen by 1% in the last year, making it less likely that potential buyers will be interested in moving to the area.
C) Most residents have lived in the area for over 20 years and do not want to leave their homes.
D) Low interests rates will likely attract buyers who will be unable to make payments when the interest rate goes up after the first two years.
E) There are few mortgage brokers in the area, making it unlikely that there will be enough personnel to assist a rush of potential home buyers.
If you have any questions
New!
Manager
Joined: 28 Sep 2011
Posts: 89
Followers: 0
Kudos [?]: 31 [0], given: 15
### Show Tags
29 Oct 2011, 22:49
If E is true, then the plan will not work. I pick E.
OA?
Manager
Joined: 31 May 2011
Posts: 88
Location: India
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Followers: 1
Kudos [?]: 51 [0], given: 4
### Show Tags
29 Oct 2011, 23:25
DexDee wrote:
If E is true, then the plan will not work. I pick E.
OA?
I will post the OA after few more discussions. But the OA is not E.
Intern
Joined: 15 Sep 2011
Posts: 6
Followers: 3
Kudos [?]: 24 [2] , given: 8
### Show Tags
30 Oct 2011, 09:36
2
KUDOS
IMO D
_________________
plz........ kudos plz......
Manager
Joined: 21 Aug 2010
Posts: 188
Location: United States
GMAT 1: 700 Q49 V35
Followers: 2
Kudos [?]: 100 [0], given: 141
### Show Tags
30 Oct 2011, 10:58
Initially I thought E.
If E is not the OA then it could be D
_________________
-------------------------------------
Last edited by g106 on 01 Nov 2011, 10:15, edited 1 time in total.
Director
Joined: 28 Jul 2011
Posts: 560
Location: United States
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 3
Kudos [?]: 225 [0], given: 16
### Show Tags
30 Oct 2011, 11:13
D looks the best choice
_________________
Manager
Joined: 31 May 2011
Posts: 88
Location: India
GMAT Date: 12-07-2011
GPA: 3.22
WE: Information Technology (Computer Software)
Followers: 1
Kudos [?]: 51 [0], given: 4
### Show Tags
30 Oct 2011, 20:26
Folks please try to explain your answers (like reason for selection and rejection of certain option) as this is a discussion forum.
BTW the OA is D
Intern
Joined: 27 Jan 2011
Posts: 28
Followers: 0
Kudos [?]: 8 [0], given: 0
### Show Tags
31 Oct 2011, 09:02
D
Manager
Joined: 29 Jun 2011
Posts: 164
WE 1: Information Technology(Retail)
Followers: 3
Kudos [?]: 21 [0], given: 29
### Show Tags
31 Oct 2011, 15:14
IMO-D.
As per the plan the buyers would be given special mortgages with a low interest rate for the first two years.Hence the scheme could fail if the buyers are unable to pay after interest rates are increased after 2 yrs.
Director
Status: Prep started for the n-th time
Joined: 29 Aug 2010
Posts: 704
Followers: 6
Kudos [?]: 173 [0], given: 37
### Show Tags
01 Nov 2011, 00:12
+1 for D.
The only other contender was C, but most does not mean all. Hence D wins.
Crick
Manager
Status: Next engagement on Nov-19-2011
Joined: 12 Jan 2011
Posts: 84
Location: New Delhi, India
Schools: IIM, ISB, & XLRI
WE 1: B.Tech (Information Technology)
Followers: 2
Kudos [?]: 23 [0], given: 5
### Show Tags
01 Nov 2011, 09:52
Sudhanshuacharya wrote:
Community spokesperson: After a recent surge of foreclosures, many homes in our community are vacant and falling into disrepair. Property values are already falling, and, if action is not taken quickly, will fall even further when panicked residents begin to leave. Clearly the only way to prevent a snowball effect is to make it easy to purchase these vacant homes by offering potential buyers special mortgages with a low interest rate for the first two years.
Which of the following, if true, represents the most significant potential problem with the plan to attract buyers through special mortgage offers?
According to the stem by decreasing interest rates for first two years will help to prevent snowball effect. We have to find something which represents potential problem with the plan
A) Interest rates in the area are already at a 5-year low, and it would be foolish of lenders to lower them even more.
Reducing interest rates will hamper lenders but has nothing to do with plan
B) Crime rates in the area have risen by 1% in the last year, making it less likely that potential buyers will be interested in moving to the area.
Off track and has nothing to do with proposed plan
C) Most residents have lived in the area for over 20 years and do not want to leave their homes.
Off track and has nothing to do with proposed plan
D) Low interests rates will likely attract buyers who will be unable to make payments when the interest rate goes up after the first two years.
This correlates proposed plan, potential buyers and drawback of reducing interest rates.
E) There are few mortgage brokers in the area, making it unlikely that there will be enough personnel to assist a rush of potential home buyers.
Off track and has nothing to do with proposed plan
Ans: 'D'
Time: 1.5 Mins
_________________
Preparation for final battel:
GMAT PREP-1 750 Q50 V41 - Oct 16 2011
GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row
MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39
My ongoing plan: http://gmatclub.com/forum/550-to-630-need-more-to-achieve-my-dream-121613.html#p989311
Appreciate by kudos !!
Intern
Joined: 06 Jul 2011
Posts: 26
Followers: 1
Kudos [?]: 5 [0], given: 8
### Show Tags
01 Nov 2011, 12:13
I don't think it's D. The plan is to attract potential buyers. Who cares if they can't pay anymore after 2 years? The question doesnt ask or state anything about long-term effects. They'd be able to sell those houses today and thats all that matters for the question.
For the lack of other reasonable choices I'd choose E. I don't really like it because in my opinion, they could still sell the houses and it'd just take a little longer if the brokers are busy, but again i think that's the most reasonable one.
Intern
Joined: 01 Apr 2011
Posts: 10
Followers: 0
Kudos [?]: 0 [0], given: 0
### Show Tags
06 Nov 2011, 18:57
I totalt agree! The main question is are we looking for a long term benefit or a short one? I believe a short one is asked for. E is a really bad choice but it's better since it gives explanations to the specific question? Anybody know how to distinguise between reason and specific wording in the question?
Posted from my mobile device
Manager
Joined: 08 Jul 2008
Posts: 147
Followers: 0
Kudos [?]: 3 [0], given: 1
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
08 Nov 2011, 06:35
According to the question "Which of the following, if true, represents the most significant potential problem with the plan to attract buyers through special mortgage offers?". E is better choice than D
Manager
Status: One last try =,=
Joined: 11 Jun 2010
Posts: 143
Followers: 3
Kudos [?]: 85 [0], given: 32
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
08 Nov 2011, 07:47
Sudhanshuacharya wrote:
Community spokesperson: After a recent surge of foreclosures, many homes in our community are vacant and falling into disrepair. Property values are already falling, and, if action is not taken quickly, will fall even further when panicked residents begin to leave. Clearly the only way to prevent a snowball effect is to make it easy to purchase these vacant homes by offering potential buyers special mortgages with a low interest rate for the first two years.
Which of the following, if true, represents the most significant potential problem with the plan to attract buyers through special mortgage offers?
A) Interest rates in the area are already at a 5-year low, and it would be foolish of lenders to lower them even more.
B) Crime rates in the area have risen by 1% in the last year, making it less likely that potential buyers will be interested in moving to the area.
C) Most residents have lived in the area for over 20 years and do not want to leave their homes.
D) Low interests rates will likely attract buyers who will be unable to make payments when the interest rate goes up after the first two years.
E) There are few mortgage brokers in the area, making it unlikely that there will be enough personnel to assist a rush of potential home buyers.
The scope of this CR question is the potential problem of the special mortgage offer. During my first read, I can eliminate choices A, B, and C.
Only D and E left, and I took D. I think not having enough personnel may not be the potential problem of the plan. Choice D, on the other hand, can result in raising the default rate when the number of buyers who won't be able to repay increases. So, that can be considered the potential issue of the plan.
_________________
There can be Miracles when you believe
Intern
Joined: 01 Apr 2011
Posts: 10
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
09 Nov 2011, 16:11
The problem is the wording of the question! There is no potential problem with attracting the buyers with low interests rates, they will be attracted! The problems come later but are not associated with selling houses!
Posted from GMAT ToolKit
BSchool Forum Moderator
Status: Flying over the cloud!
Joined: 17 Aug 2011
Posts: 906
Location: Viet Nam
GMAT Date: 06-06-2014
GPA: 3.07
Followers: 74
Kudos [?]: 636 [0], given: 44
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
13 Nov 2011, 09:42
+ 1 for D
_________________
Rules for posting in verbal gmat forum, read it before posting anything in verbal forum
Giving me + 1 kudos if my post is valuable with you
The more you like my post, the more you share to other's need
CR: Focus of the Week: Must be True Question
Manager
Status: SC SC SC SC SC.... Concentrating on SC alone.
Joined: 20 Dec 2010
Posts: 239
Location: India
Concentration: General Management
GMAT Date: 12-30-2011
Followers: 3
Kudos [?]: 61 [0], given: 47
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
13 Nov 2011, 10:16
+1 D
_________________
D- Day December 30 2011. Hoping for the happiest new year celebrations !
Aiming for 700+
Kudo me if the post is worth it
Senior Manager
Status: D-Day is on February 10th. and I am not stressed
Affiliations: American Management association, American Association of financial accountants
Joined: 12 Apr 2011
Posts: 270
Location: Kuwait
Schools: Columbia university
Followers: 5
Kudos [?]: 287 [0], given: 52
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
13 Nov 2011, 14:49
I go for D because:
A) Interest rates in the area are already at a 5-year low, and it would be foolish of lenders to lower them even more.
foolish is not a word we can rely on because it might not foolish as it has positive effect since it would increase the purchases.
B) Crime rates in the area have risen by 1% in the last year, making it less likely that potential buyers will be interested in moving to the area.
we can't trust percents in CR, we dont know how big the number is with 1% or how low. it is not representative.
C) Most residents have lived in the area for over 20 years and do not want to leave their homes.
this is irrelevant and there is an assurance word that we can't trust "DO NOT WANT".
D) Low interests rates will likely attract buyers who will be unable to make payments when the interest rate goes up after the first two years.
this answer serves best as it indicates a direct problem with the plan or conclusion that people would not be able to make payments.
E) There are few mortgage brokers in the area, making it unlikely that there will be enough personnel to assist a rush of potential home buyers.
totally irrelevant and out of scope
hope that helps good luck
_________________
Sky is the limit
Intern
Joined: 28 Feb 2014
Posts: 3
Concentration: Finance
Schools: Said'16, HEC Sept'16
GPA: 4
Followers: 0
Kudos [?]: 5 [0], given: 0
Re: Community spokesperson : After a recent surge of [#permalink]
### Show Tags
26 Mar 2015, 03:34
Am I the only one who chose answer B? If people are afraid of crime rates and will not buy homes for that reason then it doesn't matter whether mortgages are with low interest rates or not. People won't buy them in any case, so the plan won't work.
Re: Community spokesperson : After a recent surge of [#permalink] 26 Mar 2015, 03:34
Similar topics Replies Last post
Similar
Topics:
3 After several decades of surging demand, there has been a 10% fall in 3 20 Nov 2014, 11:30
12 After the recent court rulings, commercial shark fishing and 13 23 Jul 2014, 17:06
2 Community spokesperson: After a recent surge of 4 30 May 2013, 18:25
The recent surge in fear over the virulence of the Ebola 6 19 Feb 2009, 08:13
1 Railroad spokesperson: Of course it is a difficult task to 7 14 Apr 2008, 11:48
Display posts from previous: Sort by | 3,993 | 15,005 | {"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} | 3.734375 | 4 | CC-MAIN-2017-13 | longest | en | 0.908549 |
https://www.theschoolrun.com/subject/worksheets/maths/all | 1,638,196,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358774.44/warc/CC-MAIN-20211129134323-20211129164323-00453.warc.gz | 1,119,597,010 | 21,582 | # all Maths worksheets by Subject
### Multiplication square
A multiplication number square will help your child spot number patterns and begin to learn their multiplication tables and square numbers. Download TheSchoolRun's free, colour-coded multiplication tables to help them practise their multiplication facts and times tables at home.
### Complete the multiplication square: 7, 11 and 12 times tables
Year 4 times tables practice: complete the multiplication square with missing numbers from the 7, 11 and 12 times table.
### Complete the multiplication square: 6 and 9 times tables
Look at the multiplication square. Some of the numbers from the 6 and 9 times tables are missing – can you add them back in? What patterns can you identify on the square?
### Complete the multiplication square: 3, 4 and 8 times tables
Using your knowledge of the 3, 4 and 8 times tables, complete the multiplication square with the missing numbers. How fast can you complete the square?
### Complete the multiplication square: 2, 5 and 10 times tables
Complete the multiplication square with missing numbers from the 2, 5 and 10 times table.
### Weigh items to compare them
How good are you at estimating what something weighs? Let's try this activity, which involves estimating the weight of common household items then checking the weight on your kitchen scales.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Race to find 2D and 3D shapes
Do you know your 2D and 3D shapes? Let's race to see who can find the most items in 30 seconds!
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Make your own one pound shop
Let's play shops! This fun game will help your child recognise the value of different denominations of coins, as well as supporting their understanding of simple addition and subtraction problems.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Estimating height and length
Gather a selection of toys or household items that are different lengths and heights. Then ask your child to put them in length / height order and compare them using terms such as “longer / shorter / taller” and “double the length / half the height”.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
What did you do today? Let's talk about our daily routine. What is the first thing you do every day? What do you do in the morning, afternoon and night?
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The numbers 8 and 9
Playdough sausages, drawing an octopus, counting passengers on a bus – discover lots of activities to help your child learn the numbers 8 and 9.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The numbers 6 and 7
Are you a whizz at the numbers 1 to 5? Let’s have a look at 6 and 7.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The numbers 3 and 4
Do you know the numbers 3 and 4? Try these fun activities to help you recognise these numbers.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The numbers 1 and 2
The first numbers you will come across at school are the numbers 1 and 2. Here are lots of activities to help you learn to spot these numbers.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The number 10
The number ten is the first number with two digits! It is made up of the digits 1 and 0. Here are some activities to help you recognise the number 10.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### The number 5
After 1, 2, 3 and 4 comes the number 5! Can you spot the number 5 on a clock? Can you write the number 5 on a birthday card?
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Spot the 2D shapes
Shapes are all around us! Do you recognise any of these shapes? Can you find them in your house? What shapes can you see in this picture of a house?
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Solve simple number problems
Now you can count and recognise numbers, you can start to solve puzzles using the numbers 1 to 6.
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members.
### Solve problems using objects
Practise solving real-world problems to help you with your maths work when you get to school. Collect some teddies and some cookie counters (or some real cookies!) and do some problem-solving!
Subscribe now now to instantly download this content, plus gain access to 1000s of worksheets, learning packs and activities exclusively available to members. | 1,230 | 5,814 | {"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} | 3.4375 | 3 | CC-MAIN-2021-49 | latest | en | 0.896027 |
https://www.flashcardmachine.com/stats-chapter-1and2.html | 1,597,298,575,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738960.69/warc/CC-MAIN-20200813043927-20200813073927-00281.warc.gz | 667,231,692 | 6,718 | # Shared Flashcard Set
## Details
Stats chapter 1 and 2
Fain and Tammy
27
Education
09/05/2011
## Additional Education Flashcards
Term
Terms
Definition
Definitions/ Info
Term
Random Sample
Definition
Draw a sample from a larger population to represent the whole population. Put names in a hat. Make sure every member/element has equal chance of being selected.
Term
Randomly Assign
Definition
1/2 of random sample goes to treatment group and the other 1/2 goes to the control group.
Term
Population
Definition
entire collection of events (scores, incomes, speeds, etc.) that you're studying
Term
Sample
Definition
small number from larger population. Allows us to infer something about characteristics of population.
Term
External Validity
Definition
1st aspect of randomness. Does the sample reflect the population? Ex. Small town from NE wouldn't rep US hispanic culture well.
Term
Random Assignment
Definition
2nd aspect of randomness. After the subjects have been selected, subjects then have to be randomly assigned to treatment or control.
Term
Internal Validity
Definition
Are the results the result of the differences in the way we treated our groups (hope so) and not a result of WHO we placed in each group.
Term
Variable
Definition
property of an object/event that can take on different values (hair color, self-confidence, gender, personal control, treatment groups)
Term
Independent Variable
Definition
Researchers decide what these will be (group memberships like gender groups or teaching style, etc.)
Term
Dependent Variable
Definition
Researchers have NO control over these. (resulting self-esteem scores, personal control, etc.)
Term
Discrete Variables
Definition
limited number of values (gender, high school grades)
Term
Continuous Variables
Definition
any value between lowest and highest points on a scale (age, self-esteem)
Term
Quantitative Data
Definition
aka. Measurement Data.Numerical data (weights, test scores) other "how much" tests
Term
Measurement Data
Definition
aka. Quantitative Data. Numerical data (weights, test scores) other "how much" tests
Term
Categorical Data
Definition
aka. Qualitative/Frequency Data. (no numbers)
Term
Frequency Data
Definition
aka. Qualitative/Categorical Data. (no numbers)
Term
Qualitative Data
Definition
aka. Categorical/Frequency Data. (no numbers)
Term
Descriptive Statistics
Definition
Describing a set of data (means, graphs, extreme scores, oddly shaped distributions)
Term
Exploratory Data Analysis (EDA)
Definition
Joh Tukey showed necessity of paying close attention to examining data in close detail before invoking more tech involved procedures
Term
Inferential Statistics
Definition
do after descriptive statistics, after we have a basic understanding of the numbers.
Term
Parameter
Definition
a measure that refers to an entire population (average self-esteem score)
Term
Statistic
Definition
same measure as a parameter, calculated from sample of data we have collected.
Term
Nominal Scale
Definition
labels categorical data. (gender, political parties)
Term
Ordinal Scale
Definition
simplest true scale. Orders people, objects or events along a continuum (ranks in military)
Term
Interval Scale
Definition
allows us to speak of differences between scale points (same different between 10-15 degree C as there is between 15-20 degree C)
Term
Ratio Scale
Definition
has a true 0 point (true absence). Allows us to speak of ratios/fractions. (length, volume, time)
Supporting users have an ad free experience! | 770 | 3,547 | {"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} | 2.734375 | 3 | CC-MAIN-2020-34 | latest | en | 0.845399 |
https://www.nossaciencia.com.br/tempur-luxeadapt-sfwz/distance-between-2-lines-in-3d-calculator-94b4eb | 1,618,961,534,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00398.warc.gz | 1,025,018,320 | 24,190 | West Paces Ferry Apartments, Roll R Package, Weber Baby Q 1000, Bipolar And Pip, Without You The Kid Laroi Chords, Effects Of Single Parent Families On Children's Education, Stihl Ms271 Motor, Travis Johnson Man U, Guava Farm Near Me, " /> West Paces Ferry Apartments, Roll R Package, Weber Baby Q 1000, Bipolar And Pip, Without You The Kid Laroi Chords, Effects Of Single Parent Families On Children's Education, Stihl Ms271 Motor, Travis Johnson Man U, Guava Farm Near Me, " />
distance between 2 lines in 3d calculator Posts
quarta-feira, 9 dezembro 2020
We know that slopes of two parallel lines are equal. Your feedback and comments may be posted as customer voice. connecting the north end of one line to the south end of the other. Study the figure below: can you explain how to derive this formula? Then d(P,L) = |h−1,−2,1i×h5,0,1i| h5,0,1i = |h−2,6,10i| √ 26 = √ 140 is the distance between P and L. Question to the reader: what is the equation of the plane which contains the point P and the line L? To find the distance between two 2 points 3 points straight or parallel lines with the x and y coordinates value follow some simple steps of the distance between two points calculator: Input: Very first, select the type of points from the drop-down menu among which you want to calculate the distance. And how to calculate that distance? This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces.. Distance Formula and Its Use in 3D Geometry. Theory. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. So far my approach has been as follow: I have chosen some reference z values and recalculated all the deltas for both lines to these reference z values. It is a good idea to find a line vertical to the plane. We considered the Distance of a Point to a Line and the Distance of a Point to a Segment in Algorithm 2. The distance formula is derived from the Pythagorean theorem. It equals the perpendicular distance from any point on one line to the other line.. Next, enter the x, y, and z coordinates of the two points. We have also provided a solved example for you to understand the calculations. Distance between two lines in 3D: Let and denote the center positions and directional unit vectors of the two segments of lengths . This means, you can calculate the shortest distance between the point and a point of the plane. PS: this is what I'm working on, and i need to find out where the distance between the angled lines equals the horizontal one. The distance | P 1 P 2 | between the points P 1 (x 1, y 1, z 1) and P 2 (x 2, y 2, z 2) equals . The distance between the intersection points A´ 1 and A´ 2 is at the same time the distance between given lines, thus: Distance between two skew lines Through one of a given skew lines lay a plane parallel to another line and calculate the distance between any point of that line and the plane. We want to find the w(s,t) that has a minimum length for all s and t. This can be computed using calculus [Eberly, 2001]. Additional features of distance from a point to a line 3D calculator. Example 1: Find a) the parametric equations of the line passing through the points P 1 (3, 1, 1) and P 2 (3, 0, 2). If you got a point and a plane in the Euclidean space, you can calculate the distance between the point and the plane. Distance between two lines is equal to the length of the perpendicular from point A to line (2). We now consider the distance between both infinite lines and finite line segments. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. Mathematics - Mathematical rules and laws - numbers, areas, volumes, exponents, trigonometric functions and more ; Related Documents . Calculate Shortest Distance Between Two Lines Line passing through the point A(a1,b1,c1) This calculator determines the distance (also called metric) between two points in a 1D, 2D, 3D and 4D Euclidean, Manhattan, and Chebyshev spaces. By using this website, you agree to our Cookie Policy. Let be a vector between points on the two lines. 2. The given distance between two points calculator is used to find the exact length between two points (x1, y1) and (x2, y2) in a 2d geographical coordinate system.. Using the 3D Distance Formula Calculator To start, leave the Dimensions setting at 3. I now need to insert a line of a specific length in between those. Parallel Lines in 3D Geometry. Analytical geometry line in 3D space. To do it we must write the implicit equations of the straight line: $$r:\left\{ \begin{array}{l} 2x-y-7=0 \\ x-z-2=0 \end{array} \right. Distance Formula in Three Dimensions. Also, the solution given here and the Eberly result are faster than Teller'… The shortest distance between the lines is the distance which is perpendicular to both the lines given as compared to any other lines that joins these two skew lines. The distance formula is derived from the Pythagorean theorem. Distance between two 3D lines Parametric line equation: L 1: x = + t: y = + t: z = + t: L 2: x = + s: Line equation: L 1: x + = 1. Distance between Two Points is a length between point (x 1, y 1) & point (x 2, y 2) on the straight line. This library used for manipulating multidimensional array in a very efficient way. Shortest distance between two lines(d) We are considering the two line in space as line1 and line2. Sponsored Links . We are going to calculate the distance between the straight lines:$$$r:x-2=\dfrac{y+3}{2}=z \qquad r':x=y=z$ First we determine its relative position. I'm sure there is a simple way to find out where the horizontal distance between the two is the length I'm after but I was unable to figure it out. Alternatively, see the other Euclidean distance calculators: The formula for three-dimension distance is: Approximately 12.083 units separate the two points. b) Find a point on the line that is located at a distance of 2 units from the point (3, 1, 1). [6] 2019/11/19 09:52 Male / Under 20 years old / High-school/ University/ Grad student / A little / Purpose of use share | cite | improve this question | follow | edited Jul 2 '15 at 23:43. If they intersect, then at that line of intersection, they have no distance -- 0 distance -- between them. This website uses cookies to ensure you get the best experience. as much as possible , or whether it is possible to calculate the distance (pixel) between the two edges ( lines) ? The distance between two parallel lines in the plane is the minimum distance between any two points lying on the lines. d = ((2 - 1) 2 + (1 - 1) 2 + (2 - 0) 2) 1/2 = 2.24. For the normal vector of the form (A, B, C) equations representing the planes are: This is a 3D distance formula calculator, which will calculate the straight line or euclidean distance between two points in three dimensions. For any other combinations of endpoints, just supply the coordinates of 2 endpoints and click on the "GENERATE WORK" button. To find that distance first find the normal vector of those planes - it is the cross product of directional vectors of the given lines. [1] 2020/11/06 02:20 Male / 40 years old level / A teacher / A researcher / Very /, [2] 2020/08/07 18:28 Male / Under 20 years old / An engineer / Useful /, [3] 2020/06/16 04:22 Female / Under 20 years old / High-school/ University/ Grad student / Useful /, [4] 2020/05/11 03:49 Male / Under 20 years old / High-school/ University/ Grad student / Very /, [5] 2020/03/18 08:20 Female / Under 20 years old / High-school/ University/ Grad student / Very /, [6] 2019/11/19 09:52 Male / Under 20 years old / High-school/ University/ Grad student / A little /, [7] 2019/06/14 01:37 Male / Under 20 years old / High-school/ University/ Grad student / Very /, [8] 2019/05/21 18:35 Male / Under 20 years old / High-school/ University/ Grad student / Very /, [9] 2019/02/21 01:59 Female / 20 years old level / High-school/ University/ Grad student / Very /, [10] 2018/11/11 12:18 Male / Under 20 years old / High-school/ University/ Grad student / Not at All /. Parallel Lines in 3D Geometry. Distance Between Two Parallel Lines. 3D Distance Calculator. For example, the equations of two parallel lines The distance between two parallel planes is understood to be the shortest distance between their surfaces. Next, enter the x, y, and z coordinates of the two points. I get the picture from the camera using OPENCV, and using functions Canny ROI and get the following result ( Picture ) . Free distance calculator - Compute distance between two points step-by-step. You may translate everything into C++. Therefore, distance between the lines (1) and (2) is |(–m)(–c1/m) + (–c2)|/√(1 + m2) or d = |c1–c2|/√(1+m2). DISTANCE POINT-LINE (3D). Good day! Free distance calculator - Compute distance between two points step-by-step. How to find the distance between two points. Such a line is given by calculating the normal vector of the plane. In 3D geometry, the distance between two objects is the length of the shortest line segment connecting them; this is analogous to the two-dimensional definition. The top table holds the X, Y, & Z for the first point, the lower holds the X, Y, & Z for the second. Online space geometric calculator to find the shortest distance between given two lines in space, each passing through a point and parallel to a vector. In the case of non-parallel coplanar intersecting lines, the distance between them is zero.For non-parallel and non-coplanar lines (), a shortest distance between nearest points can be calculated. The shortest distance between two parallel lines is equal to determining how far apart lines are. To find the distance between two points, take the coordinates of two points such as (x 1, y 1) and (x 2, y 2) Use the distance formula (i.e) square root of (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 Calculate the horizontal and vertical distance between two points. The literal longest distance possible connecting the two lines in a straight line, i.e. The longest distance between one line and another measured parallel to the shortest distance between those same lines. The distance between two points in a three dimensional - 3D - coordinate system can be calculated as d = ((x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2)1/2 (1) Click Calculate Distance, and the tool will place a marker at each of the two addresses on the map along with a line between them. I have a question. The distance between two parallel lines in the plane is the minimum distance between any two points lying on the lines. To start, leave the Dimensions setting at 3. Additional features of distance from a point to a line 3D calculator. I will be very grateful for the help Distance between 2 Skew Lines. Distance from a point to a line 3D. This can be done by measuring the length of a line that is perpendicular to both of them. Consider two lines L1: and L2: . In three-dimensional geometry, one of the most crucial elements is a straight line. Given two line segments in space, with center positions r1,r2, ! Suggested Videos. Thank you for your questionnaire.Sending completion, Volume of a tetrahedron and a parallelepiped, Shortest distance between a point and a plane. This online calculator can find the distance between a given line and a given point. The line1 is passing though point A (a 1 ,b 1 ,c 1 ) and parallel to vector V 1 and The line2 is passing though point B(a 2 ,b 2 ,c 2 ) and parallel to vector V 2 . Section formula in 3D. This website uses cookies to ensure you get the best experience. It’s an online Geometry tool requires coordinates of 2 points in the two … Vector Form We shall consider two skew lines L 1 and L 2 and we are to calculate the distance between them. (Hint: use the Pythagorean theorem twice.) It equals the perpendicular distance from any point on one line to the other line.. 3. P = (2,3,1) is a point in space and Lis the line ~r(t) = (1,1,2)+t(5,0,1). vectors euclidean-geometry 3d. Apparently, selecting P and [2 3.166] from R to S line has lower distance of 1.1666. Therefore, two parallel lines can be taken in the form y = mx + c1… (1) and y = mx + c2… (2) Line (1) will intersect x-axis at the point A (–c1/m, 0) as shown in figure. The distance between them will appear just above the map in both miles and kilometers. The distance between two parallel lines is equal to the perpendicular distance between the two lines. The below mathematical formula is being used in this calculator to generate the work or validate the custom input values of assignment or homework problems on finding the distance between 2 points. Afterwards, visit our other calculators and tools. The intersection points are the points on each of the given skew lines closest to the other. Online space geometric calculator to find the shortest distance between given two lines in space, each passing through a point and parallel to a vector. One sometimes has to compute the minimum distance separating two geometric objects; for example, in collision avoidance algorithms. By using this website, you agree to our Cookie Policy. Thus the distance d betw… To find a vector, P=(Px,Py,Pz), perpendicular to both vectors (O and P), we need to solve the two simultaneous equations, O.P=0 and V.P=0. See, 3D Distance Calculator: Straight Line Distance Between Points, Hours Calculator: See How Many Hours are Between Two Times, Net Worth by Age Calculator for the United States in 2020, Average, Median, Top 1%, and all United States Net Worth Percentiles in 2020, Net Worth Percentile Calculator for the United States in 2020, Stock Total Return and Dividend Reinvestment Calculator (US), Least to Greatest Calculator: Sort in Ascending Order, Bond Pricing Calculator Based on Current Market Price and Yield, Income Percentile by Age Calculator for the United States in 2020, S&P 500 Return Calculator, with Dividend Reinvestment, Income Percentile Calculator for the United States in 2020, Household Income Percentile Calculator for the United States in 2020, Height Percentile Calculator for Men and Women in the United States, Bitcoin Return Calculator with Inflation Adjustment, Age Difference Calculator: Compute the Age Gap, Years Between Dates Calculator: Years between two dates, Month Calculator: Number of Months Between Dates, Average, Median, Top 1%, and all United States Household Income Percentiles in 2020, d: the distance between the two points (the hypotenuse), x1, y1, z1: the x, y, and z coordinates of point 1, x2, y2, z2: the x, y, and z coordinates of point 2. asked Jul 2 '15 at 23:28. tjvg1991 tjvg1991. And we considered the Distance of a Point to a Plane in Algorithm 4. In simple terms, Euclidean distance is the shortest between the 2 points irrespective of the dimensions. This is an example of minimizing a function of two variables over a square domain. Distance between points (4, 3) and (3, -2) is 5.099 Distance between two points calculator uses coordinates of two points A(xA, yA) A (x A, y A) and B(xB, yB) B (x B, y B) in the two-dimensional Cartesian coordinate plane and find the length of the line segment ¯¯¯¯¯ ¯AB A B ¯. Perpendicular distance from a point and the Eberly result are faster than Teller'… we know that slopes two. ; for example, the solution given here and the Eberly result are than. Pixel ) between the two line segments planes is understood to be the shortest between.: use the Pythagorean theorem twice. plane is the minimum distance between two (. To calculate the straight line, i.e ; Related Documents parallel, they must,! Study the figure below: can you explain how to find the Euclidean distance between two skew lines L and... Of a flight, drive, or whether it is possible to calculate the between! Next, enter the x, y, distance between 2 lines in 3d calculator using functions Canny and! Now because setting of JAVASCRIPT of the two lines in the plane is the minimum separating. Areas, volumes, exponents, trigonometric functions and more ; Related Documents point to Segment. The shortest distance between two parallel lines here and the Eberly result are faster than Teller'… we that... Another measured parallel to the length of a point to the length of the browser OFF... Planes are not parallel, they have no distance -- between them appear. Following is the minimum distance between them lower distance of 1.1666 2 and... Easily determine the distance formula is derived from the point and the distance of 1.1666,! 8 8 silver badges 26 26 bronze badges and comments may be posted as customer voice to,... Has to Compute the minimum distance between two parallel lines distance POINT-LINE 3D! Library used for manipulating multidimensional array in a straight distance between 2 lines in 3d calculator, i.e S line has lower distance a! To understand the calculations points on the lines is equal to the shortest distance between them the browser is.. The normal vector of the two lines in the direction of the plane will use the NumPy....: P 2: Make a Shortcut to this calculator on your Home Screen numbers, areas volumes... You get the picture from the point to a line is given by calculating the normal vector of two. -- between them at that line of intersection, they must intersect, then that! A very efficient way efficient way can be done by measuring the length of the perpendicular distance from Pythagorean! More geometric approach, and using functions Canny ROI and get the picture the. Simple terms, Euclidean distance between the point to the distance between the two lines in direction! 3.5 ) and ( -5.1, -5.2 ) distance between 2 lines in 3d calculator 2D space 3, 3.5 ) and -5.1... Shortest distance between them will appear just above the map in both miles and kilometers by! Know that slopes of two variables over a square domain be a vector between points the picture from Pythagorean! Positions r1, r2, trigonometric functions and more ; Related Documents and on. Units separate the two edges ( lines ) distance between 2 lines in 3d calculator find the distance between them three Dimensions that..., they have no distance -- 0 distance -- between them distance POINT-LINE ( 3D.... Parallel, they must intersect, eventually line and the plane is the minimum distance separating geometric...: the formula for three-dimension distance is: Approximately 12.083 units separate the two points in three Dimensions the of! To be the shortest distance between their surfaces distance button and we 'll show you distance! Manipulating multidimensional array in a very efficient way Eberly result are faster than Teller'… we that... Two line segments in space as line1 and line2 longest distance possible the! Lines in 3D space, the solution given here and the Eberly result are faster than Teller'… we that. Which will calculate the shortest distance between two lines in 3D: let and denote the center positions directional! The center positions and directional unit vectors of the perpendicular distance from a point to a line is equal length. Of distance from the drop-down menu endpoints, just supply the coordinates of the two lines ( d we... Or Euclidean distance calculators: the formula for three-dimension distance is: Approximately 12.083 units separate the two lines i... Will use the Pythagorean theorem twice. online calculator can find the shortest distance between a given.. Distance ( pixel ) between the point to a line is given by calculating the normal vector of perpendicular. Or Euclidean distance between the 2 points irrespective of the two lines ( d ) we to. Have also provided a solved example for you to understand the calculations calculating! Over a square domain this library used for manipulating multidimensional array in very... And [ 2 3.166 ] from R to S line has lower distance of a and. Of 2 endpoints and click on the lines minimum distance between 2 cities as well the. A very efficient way, trigonometric functions and more ; Related Documents explain how derive. Z coordinates of the Dimensions setting at 3 planes that contain these lines Pythagorean.. If you got a point to a Segment in Algorithm 2, just supply the coordinates of the two (! Vector of the other we may derive a formula using this website uses cookies to ensure you get best. These lines and kilometers can find the distance between two points formula directly find... Shortcut to this calculator on your Home Screen two line in space as line1 and line2 at.!, the shortest distance between two lines in a straight line three-dimensional geometry, one of other. To find the Euclidean distance between them a parallelepiped, shortest distance between them will appear above!: Approximately 12.083 units separate the two lines in 3D: let and denote the positions... They must intersect, then at that line of a point and the is! Point of the other Euclidean distance, we use a more geometric approach, and using functions Canny and! Equations of two variables over a square domain far apart lines are equal unit vectors of perpendicular... Possible to calculate the straight line 'll show you the distance of a point to line... Happens ( z2+delta ), leave the Dimensions setting at 3 useful for estimating the mileage a. Points irrespective of the perpendicular distance between two points uses cookies to ensure get. Have no distance -- between them ; for example, the solution given here the! Minimum distance between them connecting the north end of one line to the plane three-dimension. -5.1, -5.2 ) in 2D space are considering the two line segments the perpendicular! Using OPENCV, and end up with the same result and get the best experience are. The Compute distance button and we 'll show you the distance between two parallel lines is in plane... Array in a very efficient way twice. example, the solution given here and the distance between.. Approach, and end up with the same result plane in the direction of the Dimensions in \mathbb R3! Distance between their surfaces - Compute distance button and we 'll show you distance. And distance between 2 lines in 3d calculator ; Related Documents laws - numbers, areas, volumes, exponents, trigonometric functions more! Another measured parallel to the south end of the browser is OFF,.: the formula for three-dimension distance is: Approximately 12.083 units separate the two segments of.... Efficient way the Pythagorean theorem formula for three-dimension distance is the minimum distance the..., exponents, trigonometric functions and more ; Related Documents distance between 2 lines in 3d calculator on one line and plane! Best experience line of a point to a line is equal to the other..... Numpy library Compute the minimum distance between them to insert a line and another measured parallel the. Your questionnaire.Sending completion, Volume of a line vertical to the perpendicular distance between two lines... To select the dimension from the Pythagorean theorem if you got a point and a in. Collision avoidance algorithms, hit the Compute distance between parallel planes is understood to the. Z2+Delta ) anybody could help me that would be awesome and z coordinates of 2 endpoints and click the. Above the map in both miles and kilometers this website uses cookies to ensure get... Array in a straight line, i.e formula calculator to start, leave the setting! Dimensions setting at 3 formula using this website uses cookies to ensure you get the result... Same lines given two line in space as line1 and line2 at 3 formula using this approach and this... 2D space 8 8 silver badges 26 26 bronze badges efficient way the figure below: can you explain to... Line or Euclidean distance calculators: the formula for three-dimension distance is distance... Comments may distance between 2 lines in 3d calculator posted as customer voice distance formula calculator, which will calculate the distance the... Improve this question | follow | edited Jul 2 '15 at 23:43 the is! Endpoints, just supply the coordinates of the two points, and z coordinates of endpoints. A more geometric approach, and end up with the same result from the drop-down menu two lines and also. Also, the shortest between the two lines limited now because setting of JAVASCRIPT of the perpendicular distance from Pythagorean. Your Home Screen how to find the shortest distance between parallel planes is understood to the... Comments may be posted as customer voice understand the calculations to start, leave Dimensions! Is possible to calculate the closest distance between two parallel lines in \mathbb R^3 is. ( d ) we are to calculate the closest distance between two parallel is... The 2 points irrespective of the given skew lines L 1 and L 2 and we are to the. Share | cite | improve this question | follow | edited Jul 2 '15 at 23:43 easily the! | 5,833 | 24,836 | {"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} | 3.84375 | 4 | CC-MAIN-2021-17 | longest | en | 0.900571 |
http://www.lynda.com/SolidWorks-tutorials/Working-complex-calculations/143606/157804-4.html | 1,409,486,394,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500959239.73/warc/CC-MAIN-20140820021559-00228-ip-10-180-136-8.ec2.internal.warc.gz | 453,398,823 | 34,101 | # Working with complex calculations
## Video: Working with complex calculations
The Equation Editor gives us the ability to
## Working with complex calculations
The Equation Editor gives us the ability to create equations based on variables and input values. Equations can use the results from one equation as their input values, and some very complicated dynamic values can be calculated. These values can then easily be linked to sketch dimensions. In this case here, I'm going to create a dimension from the far-side of the part, to the center line of the part. And create a double-dimension. I'm going to type in, instead of 10, I'm going to say, equals and define a global variable called L And then I'm going to type in, yes, I want that to be 10, so I'm going to overwrite this value, here.
And you click OK again, creating a global variable called, L, and I'm giving it the value of 10. Now instead of creating a new global variable, what I'm going to do now is I'm going to actually create a dimension from this part, side of the part, to the center line. Again with a double dimension. But, instead of creating a new global variable, again I'm going to hit Equals. I'm going to go and grab one of those global variables I already created, which is L, and I'll say, hey, this is divided by 2. So it's going to create a little equation there. And then go ahead and click OK, and click OK again. So now I have the same value I had last time, but this time, I'm actually just using a formula to figure that value out.
The neat thing about this is if I change one of these values, for instance, if I change this to 12. The other variable automatically updates because it's being equation driven from the original value. From here you can continue on to add new equations or variables, or if you close this out, jump over here to equations and you go to manage equations, you can see I already have my global variable here. It's equal to twelve. And I have a couple other equations already predefined. At this point here I could say, maybe I want to define a global variable called hole. And I want to make it derived from maybe a function of sine of 30 degrees.
And I could add, maybe add 1 inch to it. And that will equate to 1.5, or whatever variables you wanted to use. It can do pretty complicated math here for you. You can create a really long string of wild equations, or you can even link those equations to other variables or other globals. Creating dynamic equation driven models in Solidworks is easy and it's a great tool to figure out complex shapes that would otherwise be very difficult to calculate or draw.
Show transcript
#### This video is part of
SolidWorks 2014 Essential Training
97 video lessons · 6466 viewers
Author
Expand all | Collapse all
1. ### Introduction
1m 51s
1. Welcome
1m 7s
2. Using the exercise files
44s
2. ### 1. Touring the Interface
31m 13s
1. Launching SolidWorks for the first time
3m 55s
2. Accessing and customizing the Ribbon
4m 14s
3. Touring the shortcut bar and identifying essential keys
7m 27s
4. Saving, renaming, and managing files
10m 28s
5. Working with the new view cube, or View Selector
2m 36s
6. New features in SolidWorks 2013 and 2014
2m 33s
3. ### 2. Getting Started with 3D
14m 11s
1. Understanding the 3D world
2m 31s
3m 15s
3. The virtual, parametric prototyping environment
1m 56s
4. The FeatureManager and feature-based modeling
3m 43s
5. History-based modeling and the rollback bar
2m 46s
4. ### 3. Basic Solid Modeling
28m 32s
1. Starting a new sketch
6m 50s
2. The six steps used in almost all modeling features
52s
3. The Line and Centerline tools
3m 25s
4. Using the Circle tool
1m 51s
5. Adding and removing relationships and dimensions
6m 56s
6. Understanding relationship types
3m 58s
7. System options, units, and templates
4m 40s
5. ### 4. More Drawing Tools
18m 28s
1. Drawing rectangles
5m 31s
2. Creating arcs in a sketch
4m 8s
3. Drawing splines in a sketch
4m 57s
4. Sketching polygons
3m 52s
6. ### 5. Sketch Editing Tools
36m 5s
1. Trimming and extending portions of a sketch
3m 54s
2. Creating offset geometry
3m 13s
3. Moving, copying, rotating, and scaling elements
3m 13s
4. Erasing, undoing, and redoing actions
2m 24s
5. Using the mirror tools
2m 24s
6. Creating repeating patterns in a sketch
4m 55s
7. Using construction lines to build robust sketches
3m 25s
8. Applying fillets and chamfers to a sketch
2m 32s
9. Working with slots
3m 46s
4m 1s
11. Using the Convert Entities command
2m 18s
7. ### 6. Reference Geometry
9m 33s
1. Working with planes
5m 28s
2. Placing and using axes
2m 22s
3. Placing a coordinate system
1m 43s
8. ### 7. Building 3D Geometry
17m 50s
1. Extruding a sketch into a 3D object
4m 36s
2. Using Revolve to create 3D parts
2m 42s
3. Using Loft to create complex shapes
4m 40s
4. Refining a loft shape with guide curves
2m 22s
5. Using the sweep to create wire and pipe shapes
3m 30s
9. ### 8. Removing Material
20m 23s
1. Modifying parts using the Extruded Cut tool
5m 42s
2. Working with the Revolved Cut tool
6m 19s
3. Using the Lofted Cut tool
3m 32s
4. Cutting holes and grooves with the Swept Cut tool
4m 50s
10. ### 9. Refining Geometry
21m 5s
1. Using fillets and chamfers to smooth corners
5m 58s
2. Creating repeating rectangular patterns
3m 16s
3. Creating a circular pattern
2m 27s
4. Mirroring objects
4m 0s
5. Using the Shell and Draft tools
3m 52s
6. Scaling parts
1m 32s
11. ### 10. Blocks
9m 39s
1. Working with reusable sketches and blocks
2m 47s
2. Creating blocks
3m 51s
3. Designing with blocks
3m 1s
12. ### 11. Assembly: Putting It All Together
29m 45s
1. Understanding the tools for beginning a new assembly
4m 46s
2. The basic steps in creating an assembly
3m 18s
3. Mating parts together in an assembly
6m 43s
4. Working with subassemblies
2m 9s
5. Linear and circular assembly patterns
4m 56s
3m 32s
7. Using Toolbox
4m 21s
15m 8s
1. Mating parts with coincident, parallel, and distance mates
4m 35s
2. Mating parts with width mates
5m 53s
3. Mating parts with path mates
2m 5s
4. Mating parts by aligning planes
2m 35s
14. ### 13. Hole Wizard
10m 20s
1. Getting started with the Hole Wizard
4m 38s
2. Positioning holes in layout sketches
5m 42s
15. ### 14. In-Context Modeling
15m 27s
1. Linking sketches to other parts
4m 28s
6m 48s
3. Using the Hole Wizard in context
4m 11s
16. ### 15. Creating Threads on Parts
17m 15s
7m 17s
2. Using a helix and Swept Path to create a thread
4m 2s
5m 56s
17. ### 16. Equations and Design Tables
17m 25s
1. Using equations to drive a sketch
5m 5s
2. Working with complex calculations
2m 6s
3. Integrating Microsoft Excel to manage design tables
7m 10s
4. Building assemblies using part configurations
3m 4s
18. ### 17. Part Drawings
23m 17s
1. Working with drawing templates
6m 49s
2. Setting up drawing options and sheet properties
3m 43s
3. Choosing the correct projection angle
2m 21s
4. Adding model views to a drawing
10m 24s
19. ### 18. Dimensioning
16m 8s
1. Creating general dimension notations
6m 37s
2. Creating ordinate and running dimensions
3m 0s
3. Dimensioning holes and curved features
3m 8s
4. Using the autodimension tools
3m 23s
20. ### 19. Adding General Annotations
14m 38s
1. Creating holes and callouts
5m 8s
2. Adding center marks and centerlines to a drawing
3m 46s
2m 57s
4. Making drawing revisions
2m 47s
21. ### 20. Assembly Drawings
11m 42s
2m 10s
2. Including a bill of materials
1m 42s
3. Adding balloons to specify parts on an assembly drawing
1m 39s
4. Adding a title block and sheet properties
2m 8s
5. Building an exploded view for an assembly drawing
4m 3s
22. ### Conclusion
1m 2s
1. Next steps
1m 2s
### Start learning today
Sometimes @lynda teaches me how to use a program and sometimes Lynda.com changes my life forever. @JosefShutter
@lynda lynda.com is an absolute life saver when it comes to learning todays software. Definitely recommend it! #higherlearning @Michael_Caraway
@lynda The best thing online! Your database of courses is great! To the mark and very helpful. Thanks! @ru22more
Got to create something yesterday I never thought I could do. #thanks @lynda @Ngventurella
I really do love @lynda as a learning platform. Never stop learning and developing, it’s probably our greatest gift as a species! @soundslikedavid
@lynda just subscribed to lynda.com all I can say its brilliant join now trust me @ButchSamurai
@lynda is an awesome resource. The membership is priceless if you take advantage of it. @diabetic_techie
One of the best decision I made this year. Buy a 1yr subscription to @lynda @cybercaptive
guys lynda.com (@lynda) is the best. So far I’ve learned Java, principles of OO programming, and now learning about MS project @lucasmitchell
Signed back up to @lynda dot com. I’ve missed it!! Proper geeking out right now! #timetolearn #geek @JayGodbold
Share a link to this course
### What are exercise files?
Exercise files are the same files the author uses in the course. Save time by downloading the author's files instead of setting up your own files, and learn by following along with the instructor.
### Can I take this course without the exercise files?
Yes! If you decide you would like the exercise files later, you can upgrade to a premium account any time.
How to use exercise files.
Learn by watching, listening, and doing, Exercise files are the same files the author uses in the course, so you can download them and follow along Premium memberships include access to all exercise files in the library.
Exercise files
How to use exercise files.
This course includes free exercise files, so you can practice while you watch the course. To access all the exercise files in our library, become a Premium Member.
Are you sure you want to mark all the videos in this course as unwatched?
This will not affect your course history, your reports, or your certificates of completion for this course.
Congratulations
You have completed SolidWorks 2014 Essential Training.
Become a member to add this course to a playlist
Join today and get unlimited access to the entire library of video courses—and create as many playlists as you like.
Become a member to like this course.
Join today and get unlimited access to the entire library of video courses.
Exercise files
Learn by watching, listening, and doing! Exercise files are the same files the author uses in the course, so you can download them and follow along. Exercise files are available with all Premium memberships. Learn more
How to use exercise files.
Thanks for contacting us.
You’ll hear from our Customer Service team within 24 hours.
Please enter the text shown below:
The classic layout automatically defaults to the latest Flash Player.
To choose a different player, hold the cursor over your name at the top right of any lynda.com page and choose Site preferencesfrom the dropdown menu.
• Mark video as unwatched
• Mark ALL videos as unwatched
Exercise files
Access exercise files from a button right under the course name.
Mark videos as unwatched
Remove icons showing you already watched videos if you want to start over.
Make the video wide, narrow, full-screen, or pop the player out of the page into its own window.
Interactive transcripts
Click on text in the transcript to jump to that spot in the video. As the video plays, the relevant spot in the transcript will be highlighted.
## Are you sure you want to delete this note?
Thanks for signing up.
We’ll send you a confirmation email shortly.
• new course releases
• general communications
• special notices
Keep up with news, tips, and latest courses with emails from lynda.com.
• new course releases | 3,217 | 11,659 | {"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-2014-35 | longest | en | 0.935045 |
http://www.cplusplus.com/forum/beginner/229816/ | 1,539,646,376,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583509958.44/warc/CC-MAIN-20181015225726-20181016011226-00361.warc.gz | 430,010,004 | 8,563 | ### Help with set/get functions!
Okay this is for an assignment that I have gotten a decent start on. I am confused in a few sections but mostly with "if" statements and get/ set functions. I have to make a class program that has hours minutes and seconds displayed all as two digit numbers. This is the instructions that I am most stuck on "
• Ensure that the hour value is in the range 0 – 23; if it is not, set the hour to 12.
• Ensure that the minute value is in the range 0 – 59; if it is not, set the minute to 0. • Ensure that the second value is in the range 0 – 59; if it is not, set the second to 0. Provide a set and get function for each data member."
I think my code while not perfect will work I am stuck on the IF statements that have no body in my code. If you have any advice for me let me know!
also confused on how to make the time be displayed as 00:00:00, all i can get is 0:0:0.
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657`` `````` #include using namespace std; class time { private: int hr; int min; int sec; public: time(int h, int m, int s); }; time::time(int h, int m, int s) { hr = h; min = m; sec = s; } int main() { cout << " Please enter each integer as two digits." << endl; cout << " Enter hours:"; int h; cin>> h; cout << " Enter minutes"; int m; cin>> m; cout << "Enter Seconds"; int s; cin>> s; if((h >= 24 || h < 0)) if((m >=59 || m <0)) if ((s >= 59 || s <0)) cout <
Hello jaredv11,
Welcome to the forum.
The class definition is fine although I would choose a name other than "time" to avoid any possible conflices with "time" that is used in other header files.
In main you input the hours, minutes and seconds, but never create n instance of the class to hold these values.
A set function like:
``1234`` ``````time::SetHour(int hour) { hr = hour; }``````
Or you could adjust this to set all three variables. The "set" functions are mainly used after an instance of the class has been made when you need to either set or change a value.
The opposite is the "Get" function, not that hard, and can be written as simple as:
`int time::getHour() { return hr; }`.
The if statements may be useful at some point, but I think you could do without them.
Work something up and post the new code.
Hope that helps,
Andy
Hello jaredv11,
I forgot to mention. You have defined your variables as ints. These do not store leading zeros, so you will need to take care of this when you output to the screen.
Andy
Hi Handy Andy,
Thank you for your response! I am currently working on it, I have a question about storing my variables as ints. I don't think I've been taught any other way at this point. Im not sure if i know how to change the way i out put the time to save leading variables. If I hold the inputed values will then save leading 0s if they are apart of the input?
Thanks
Hello jaredv11,
"int"s are fine to store the parts of time. "int"s along with any other type of variable, e.g., short, unsigned int, float, best not to use, double, better than a float, or long do not store leading zeros. If you enter a leading zero that will work, but it will be dropped when stored. The only way to keep a leading zero is to input to a string, but if you need to do any type of calculation you will have to change it back to a numeric variable. That is more work that you need to do.
It looks like you would understand an if/else statement better.
``1234`` ``````if (h < 10) std::cout << '0' << timeClass.GetHour() << ':'; else std::cout << timeClass.GetHour() << ':';``````
The same format for minutes and seconds.
I defined my instance of the class as "timeClass" just to stay away form "time".
Hope that helps,
Andy
Andy,
Thank you so much for you help. I finally finished the project, I ended up using "while" statements to keep the variables within the parameters.
However, my directions wanted me to use a constructor to keep the variables in the usable range if someone could help me build a constructor or point me in the right direction that would be much appreciated.
Heres where I ended up.
#include <iostream>
#include <fstream>
using namespace std;
class Time
{
private:
int hr;
int min;
int sec;
public:
Time(int h, int m, int s);
};
Time::Time(int h, int m, int s)
{
hr = h;
min = m;
sec = s;
}
int main()
{
signed int h;
signed int s;
signed int m;
cout << " Please enter each integer as two digits." << endl;
cout << " Enter hours:";
cin>> h;
cout << " Enter minutes:";
cin>> m;
cout << "Enter Seconds:";
cin>> s;
while (h < 0 || h > 23)
{
cout << h << " are not valid hours.\n";
cout << "Please enter hours 0-23: ";
cin >> h;
}
while (m < 0 || m > 59)
{
cout << m << " are not valid minutes.\n";
cout << "Please enter minutes 0-59: ";
cin >> m;
}
while (s < 0 || s > 59)
{
cout << s << " are not valid seconds.\n";
cout << "Please enter seconds 0-59: ";
cin >> s;
}
if (h < 10)
std::cout << '0' << h << ':';
else
std::cout << h << ':';
if (m < 10)
std::cout << '0' << m << ':';
else
std::cout << m << ':';
if (s < 10)
std::cout << '0' << s << ':';
else
std::cout << s << ':';
return 0;
}
@ jaredv11, yes you had a great start but you seem to ignore some important aspect(s) of the program specification.
`Provide a set and get function for each data member.`
and the controls
``123`` ``````Ensure that the hour value is in the range 0 – 23; if it is not, set the hour to 12. Ensure that the minute value is in the range 0 – 59; if it is not, set the minute to 0. Ensure that the second value is in the range 0 – 59; if it is not, set the second to 0``````
should be done inside the class.
Complete this code
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141`` ``````#include #include using namespace std; class Time { private: int hr; int min; int sec; public: Time(); // nice to have this too Time(const int &h, const int &m, const int &s); int getHour() const; int getMinutes() const; int getSeconds() const; void setHour(const int &hour); void setMinutes(const int &mins); void setSeconds(const int &secs); void showTime(); }; Time::Time(const int &h, const int &m, const int &s) { /* hr = h; min = m; sec = s;*/ setHour(h); // see setHour below. setMinutes(m); setSeconds(s); } Time::Time(){}; // getters int Time::getHour() const { return hr; } // Do same for minutes and seconds // Ensure that the hour value is in the range 0 – 23; if it is not, set the hour to 12. void Time::setHour(const int &hour) { if(hour >= 0 && hour <= 23){ hr = hour; } else{ hr = 12; } } // Ensure that the minute value is in the range 0 – 59; if it is not, set the minute to 0 void Time::setMinutes(const int &mins) { // do the control here } // Ensure that the second value is in the range 0 – 59; if it is not, set the second to 0 void Time::setSeconds(const int &secs) { // do the control here } /* if hr, sec and min are < 10, append "0" to it otherwise use it as is see http://www.cplusplus.com/reference/string/to_string/ */ void Time::showTime() { string time; string h, m, s; if(hr < 10){ h = "0" + to_string(hr); }else{ h = to_string(hr); } if(min < 10){ m = "0" + to_string(min); }else{ m = to_string(min); } if(sec < 10){ s = "0" + to_string(sec); }else{ s = to_string(sec); } cout << "Time is : " << h << ":" << m << ":" << s << endl; } int main() { // test show time Time t = Time(1,3,4); t.showTime(); Time t1 = Time(12,9,34); t1.showTime(); Time t2 = Time(-2, -2, 23); t2.showTime(); cout << " Please enter each integer as two digits." << endl; cout << " Enter hours: "; int h; cin>> h; cout << " Enter minutes: "; int m; cin>> m; cout << "Enter Seconds: "; int s; cin>> s; Time t3 = Time(h,m,s); t3.showTime(); // or using your setters Time t4; // this will call Time::Time(){}; t4.setHour(h); t4.setMinutes(m); t4.setSeconds(s); t4.showTime(); return 0; }``````
If you do all right, your output should be like this
``` Time is : 01:03:04 Time is : 12:09:34 Time is : 12:00:23 Please enter each integer as two digits. Enter hours: 12 Enter minutes: -1 Enter Seconds: 12 Time is : 12:00:12 Time is : 12:00:12 ```
Hope that helps
Last edited on
Hello jaredv11,
PLEASE ALWAYS USE CODE TAGS (the <> formatting button) when posting code.
http://www.cplusplus.com/articles/jEywvCM9/
http://www.cplusplus.com/articles/z13hAqkS/
Hint: You can edit your post, highlight your code and press the <> formatting button.
You can use the preview button at the bottom to see how it looks.
Watch your use of blank lines. To many makes the code hard to follow.
``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384`` ``````#include #include //using namespace std; // <--- Best not to use. class Time { private: int hr; int min; int sec; public: Time(int h, int m, int s); // <--- Needs Set and Get functions here. }; Time::Time(int h, int m, int s) { hr = h; min = m; sec = s; } // <--- Needs Set and Get functions definitions here. int main() { int h; // <--- signed is not need. "int" is already a signed type. int s; int m; std::cout << " Please enter each integer as two digits." << std::endl; std::cout << " Enter hours:"; std::cin >> h; std::cout << " Enter minutes:"; std::cin >> m; std::cout << "Enter Seconds:"; std::cin >> s; // <--- These while loops are good start, but in the wrong place. Need to be in the ctor or set functions of // the class. As is each while loop should follow each corresponding "cin". while (h < 0 || h > 23) { std::cout << h << " are not valid hours.\n"; std::cout << "Please enter hours 0-23: "; std::cin >> h; } while (m < 0 || m > 59) { std::cout << m << " are not valid minutes.\n"; std::cout << "Please enter minutes 0-59: "; std::cin >> m; } while (s < 0 || s > 59) { std::cout << s << " are not valid seconds.\n"; std::cout << "Please enter seconds 0-59: "; std::cin >> s; } // <--- These if statements are fine, but you should not be printing from the variables used for input, but the // class variables. This is where the get functions of the class would be used. if (h < 10) std::std::cout << '0' << h << ':'; else std::std::cout << h << ':'; if (m < 10) std::std::cout << '0' << m << ':'; else std::std::cout << m << ':'; if (s < 10) std::std::cout << '0' << s << ':'; else std::std::cout << s << ':'; return 0; }``````
``123456789`` ``````void Time::showTime(std::ostream& fout) { fout << std::setfill('0'); fout << "Time is : " << std::setw(2) << hour << ":" << std::setw(2) << minutes << ":" << std::setw(2) << seconds << '\n'; fout << std::setfill(' '); // Be sure to reset the fill character before returning. }`````` | 3,179 | 10,905 | {"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-2018-43 | latest | en | 0.872402 |
http://gamedev.stackexchange.com/tags/animation/hot | 1,440,840,388,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064420.15/warc/CC-MAIN-20150827025424-00184-ip-10-171-96-226.ec2.internal.warc.gz | 100,637,913 | 14,341 | # Tag Info
4
I'm using Hooke's Law here as the definition of a spring. () Given the derivatives of position and velocity, are velocity and force respectively, we can construct a differential equation for the stretching of the spring. Which is just a damped harmonic oscillator, and since we already know that only the under-damped case need analysis, we can obtain a ...
3
It looks like it's supposed to represent a motion blur or at least achieve a similar effect: to convey the feeling of movement and speed as the sword is swung in a arc. In 3D we sometimes go for similar effects with geometry or particle-effect based "trails" behind slashes. It looks the way it does for a variety of reasons, many of them due to the ...
3
If you call coroutines this way they will be executed in order, one after another. You should fire them up with StartCoroutine(Aiming()); public IEnumerator Aiming() { yield return StartCoroutine(LoadBow()); yield return StartCoroutine(DrawBow()); aimStates = AimStates.Aim; } ...
2
You will have to manually add all the frames for your animation. You don't need tweening unless you want to rotate or move the sprite from point A to B and have it automagically interpolate (be)'tween the key frames. For walking, jumping, shooting animations, etc. You have to do it manually. There's no magic for that. Here is a little guide to help you ...
1
You could disable the gravity on the Rigidbody and then just make gravity yourself by writing a script and using GetComponent<Rigidbody>().AddForce(-Vector3.up * Time.deltaTime * YOUR GRAVITY VALUE) and just change YOUR GRAVITY VALUE when ever you want the object to be slowed down.
1
I found this tutorial on the OpenGL wiki a month ago. It explains the basics of skeletal animations; but I hope it is good enough for what you are trying to do, if not, you can also search skeletal animation opengl on google, because there is much more on the opengl wiki about this topic.
1
You currently have a typo in the line you posted: this.canvascanvas.Height should be: this.canvas.canvasHeight Alternatively, you can use canvas.width and canvas.height as shown in this StackOverflow answer. Debugging tip When this type of code snippets don't work, try just outputting their values to the console with console.log(...). In your case, ...
1
If anyone could offer any ideas as to 'force' the game to complete the animation first before continuing I would be really greatful. It looks like you haven't implemented any way to tell if the animation is complete. You are using a shooting flag that appears to be a bit confused about it's purpose. The shooting flag gets set when the user clicks the ...
1
In your Animator Component, change Update Mode to Unscaled Time. Now the Animator can play even when TimeScale is 0. Useful for pause menus and similar. Docs
1
Ah yeah. Fire Emblem, a great game indeed. If you take a really close look to the critic attack of Lyn you will actually got it. The first slash is going from down-right to up-left. That will be the first slashing image starting from the left of the sheet combined with the second. So, we can say that the second slashing image is a continuation of the first ...
1
The solution suggested by Sebastian would work, but you'll end up with a lot of code just for a "fade out/fade in" animation. Libgdx has built-in functionnalities for that in Scene2D, you should use them. (Actions, Stage, Actor) That would make your code as simple as that : myPlayer.addAction(Actions.sequence(Actions.fadeOut(0.15f), ...
1
Why not set the alpha value of the sprites directly via sprite.setAlpha(float)? Of course you would have to call that by your rendering method and calculate the amount of alpha alteration using deltaTime. https://libgdx.badlogicgames.com/nightlies/docs/api/com/badlogic/gdx/graphics/g2d/Sprite.html#setAlpha-float-
1
The trajectory of the particles is a Bézier curve. A Bézier curve is a smooth curve along four points: pointA: startpoint pointB: startdirection+acceleration pointC: enddirection+accelleration pointD: endpoint Basically it generates a curve through from point A to point D, using the vectors towards A-B and C-D to determine the path. See C# (Monogame) ...
1
As the comment says offset only effects the children, i.e the text inside the button. What you need to look at is this piece of code: buttonStyle.up = new TextureRegionDrawable(settingsAtlas.findRegion("achievements")); buttonStyle.down = new TextureRegionDrawable(settingsAtlas.findRegion("achievements")); Since you set the same image for up and down you ...
1
It's unfortunate that I have to answer my own question, but I'll post the fix that I worked out in case it helps someone else. int current_image = 0; int current_image_add = 1; These two variables inside of the animate() function were being reset every time animation() was called, meaning that it never really cycled through anything because it continued ...
1
You could set a boolean for example to false before yield and then after the time had passed set it back to true and then just use if statement to check if the boolean is true/false. What about efficiency, I think there's not much of a difference. From what I understand the delay from coroutine affects only something you change before and after the delay so ...
Only top voted, non community-wiki answers of a minimum length are eligible | 1,186 | 5,405 | {"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} | 2.5625 | 3 | CC-MAIN-2015-35 | longest | en | 0.921438 |
http://webprod3.leeds.ac.uk/catalogue/dynmodules.asp?Y=201920&F=P&M=GEOG-3020 | 1,590,654,398,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00341.warc.gz | 138,051,940 | 4,711 | ## GEOG3020 Numerical Methods and Statistics
### 10 creditsClass Size: 40
Module manager: Prof Andy Baird
Email: a.j.baird@leeds.ac.uk
Taught: Semester 1 View Timetable
Year running 2019/20
### Pre-requisites
SOEE1160 Comp & Prog in Geosciences
### This module is mutually exclusive with
SOEE2250 Numerical Methods & Statistics
This module is not approved as a discovery module
### Objectives
Students are introduced to the most common numerical methods and their implementation in Python. They also learn how to handle and report data with uncertainties in an appropriate manner. They are required to think and write about the role of simulation in physical geography.
Learning outcomes
On completion of this module students should be able to:
1. Design and implement computer programs to solve numerical problems that would be impossible or time-consuming to complete by hand, and understand their limitations.
2. Derive expressions for simple numerical methods.
3. Solve mathematical problems via recall or use of the appropriate numerical method for finding the roots or optima of functions, solving linear systems of equations, interpolating values, performing numerical integration and differentiation, and solving initial-value and boundary-value problems.
4. State the advantage and disadvantages of different numerical methods and, where appropriate, conditions required for convergence.
5. Handle and report data with uncertainties in an appropriate manner.
6. Evaluate the role of numerical methods in environmental models and the need to consider data uncertainty when setting up and testing such models.
### Syllabus
Numerical methods
1. Errors in Numerical Methods
2. Finding Roots
3. One-Dimensional Optimisation
4. Linear Systems - Direct Methods
5. Linear Systems - Iterative Methods
6. Interpolation
7. Numerical Integration
8. Numerical Differentiation
9. Initial-Value Problems
10. Boundary-Value Problems.
Statistics
1. Error Representation
2. Error Propagation
3. Statistical Analysis
4. Normal Distribution
5. Least-Squares Fitting.
The wider context: the use of simulation in physical geography.
### Teaching methods
Delivery type Number Length hours Student hours Lecture 11 2.00 22.00 Practical 10 2.00 20.00 Private study hours 58.00 Total Contact hours 42.00 Total hours (100hr per 10 credits) 100.00
### Private study
Students will be expected to spend time reviewing course material, completing the problems sets and practicals, and practising computer programming. They will also spend time reading on the role of simulation in physical geography.
### Opportunities for Formative Feedback
Problems sets will be given out in the lectures and reviewed in subsequent lectures. Students will receive oral and written feedback on their computer programs.
### Methods of assessment
Coursework
Assessment type Notes % of formal assessment In-course Assessment In-class computer programming assignment coupled with a short report on the role of simulation in physical geography (1000 words, short-answer format) 40.00 Total percentage (Assessment Coursework) 40.00
Normally resits will be assessed by the same methodology as the first attempt, unless otherwise stated
Exams
Exam type Exam duration % of formal assessment Standard exam (closed essays, MCQs etc) 1 hr 30 mins 60.00 Total percentage (Assessment Exams) 60.00
Students who fail the module will be required to resit any failed component. | 723 | 3,457 | {"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-2020-24 | latest | en | 0.828619 |
https://math.stackexchange.com/questions/1231810/what-are-the-zeros-of-the-j-function | 1,660,263,149,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571536.89/warc/CC-MAIN-20220811224716-20220812014716-00141.warc.gz | 360,981,525 | 64,271 | What are the zeros of the j-function?
Recall that, for a complex number $\tau$ with positive imaginary part, the $j$-invariant is given by $j(\tau)=1728 \frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}$ where
$g_2(\tau)=60 \sum_{(m,n)\neq(0,0)}(m+n\tau)^{-4}$
and $g_3$ is unimportant. Clearly the zeros of $j$ are the same as the zeros of $g_2$ (and, if we are only considering the zeros, we can forget about the 60 and just consider the sum.)
Is there a way to directly see that $g_2(\rho)=0$, where $\rho=exp(\frac{2}{3} \pi i)$? (e.g. is there an algebraic trick to see that the terms in the sum cancel each other out?)
If not is there a reasonably simple indirect proof?
I have since worked out the answer to this. So in case anybody comes across the question, I will sketch it below.
Observe $\rho$ has the minimal polynomial $x^2+x+1=0$, so if $\Lambda$ is the lattice $\mathbb{Z}+\rho\mathbb{Z}$, we have $\rho \Lambda = \rho\mathbb{Z}+\rho^2\mathbb{Z}=\rho\mathbb{Z}-\mathbb{Z}-\rho\mathbb{Z}=\Lambda$.
But clearly $g_2(a z)=a^4 g_2(z)$, so $g_2(\rho)=\rho^4 g_2(\rho)$, which gives $g_2(\rho)=0$.
Of course, this doesn't tell us that these are all the zeros (upto $SL_2(\mathbb{Z})$-invariance), but it does at least find the zero.
• Heh, $-(1+\rho)\Bbb Z$ isn't the same as $-\Bbb Z-\rho\Bbb Z$ but your equality still holds. :-) Apr 14, 2015 at 0:17 | 484 | 1,368 | {"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} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.872372 |
https://devtalk.blender.org/t/using-python-to-bake-displacement-for-cycles-how-does-cycles-displace-a-point/7500 | 1,680,060,359,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00759.warc.gz | 238,462,163 | 15,930 | # Using Python to bake displacement for Cycles— How does Cycles displace a point?
EDIT: it appears my problem is color-space conversion in Cycles.
Hi everybody. I’ve been trying to create a script that bakes displacement from one ‘source’ mesh to a ‘target’ mesh by comparing the position of each vertex. How I’m doing it is explained below and I’ll share the code if anyone asks.
Problem:
I’m trying to bake displacement for Cycles to use, but I don’t really understand what data Cycles expects, nor do I understand what it does with the data. I can’t reproduce the displacement Cycles produces with Python.
Cycles displaces the geometry almost correctly, but there’s a noticeable margin of error on the vertices that transform along something other than the world x,y, z axes (diagonal transformations). I think somehow the normal of the face or normal is being used by Cycles, since the margin of error is worse in some test-meshes than in others. The problem is that I simply don’t know what Cycles is doing! It obviously isn’t just moving the vertices in global space (v.co += disp) because when I do that in Python, it’s exactly correct. Any help in this matter would be appreciated.
I think it’s possible that Cycles is performing some sort of color-space transformation to the vertex-color data. Any ideas on how to test this, or work around?
( I tried looking through the source code for Cycles, but I didn’t understand it (maybe I just wasn’t looking in the right place). This is the part of the source code I was looking through- nowhere do I see the code that actually changes the location of anything, it looks like it’s being done somewhere else. I’ve been poking around elsewhere and can’t find anything relevant.)
Here’s how I’m doing it:
1. I’m generate the data by subtracting the location of the target vertex from the source.
2. Then, I take the square root of each (x, y, z) component of the displacement vector. I think this is what is going on: right now it represents a translation on x, y, and z, and what I want is a vector that goes straight through the rectangular space of that transformation. Or to put it another way, x and y, x and z, y and z all make the legs of right triangles, and I take the square root to get the hypotenuse. Scaliing by the magic number 1.05 seems to help some of the problems in Cycles, but it’s not consistent.
3. Finally, I scale each vector until the maximum value in either direction, positive or negative, is 1. This way I can fit the data into vertex colors without it being clamped. I use a layer for positive coordinates and negative coordinates. The negative coordinates need to have their signs flipped, so they can be stored as colors.
If I skip the square root step, the result is bizarrely exaggerated, especially along the x, y and z axes. The direction is correct, but the magnitude is wrong. Taking the square root fixes this, for the most part.
My shader is simple, the two layers of vertex colors are subtracted ( pos - neg) and fed into a vector displacement node set to object space. Midlevel = 0, and the scale is the reciprocal of the value I scaled the vectors by in step 2 (right now I just have to do it manually, the value is printed in the console).
Here’s the code:
``````import bpy
import mathutils
import math
sign = lambda x: x and (1, -1)[x < 0]
def InactiveObject():
"""Returns the first selected object that is not Active"""
obReturn = None
if (len (bpy.context.selected_objects) != 0):
for ob in bpy.context.selected_objects:
if (ob != bpy.context.active_object):
obReturn = ob
break
return obReturn
def calcDisplacement(dict, matSource, vSource, matTarget, vTarget):
disp = dict
posOffset = 0
negOffset = 0
for i in range(len(vSource)):
v1 = (matTarget @ vTarget[i].co)
v2 = (matSource @ vSource[i].co)
value = v2 - v1
disp[i] = value
if (value.x > posOffset):
posOffset = value.x
elif (value.y > posOffset):
posOffset = value.y
elif (value.z > posOffset):
posOffset = value.z
if (value.x < negOffset):
negOffset = value.x
elif (value.y < negOffset):
negOffset = value.y
elif (value.z < negOffset):
negOffset = value.z
return (disp, posOffset, negOffset)
def scaleDisplacement(dispData, obSource, obTarget):
disp = dispData[0]
posOffset = dispData[1]
negOffset = dispData[2]
scale = 0
if ((posOffset != 0) and (negOffset != 0)):
if ( abs(posOffset) > abs(negOffset) ):
scale = 1/posOffset
else:
scale = -1/(negOffset)
print ("Scale: %s" % scale)
dispScaled = disp
for key, value in dispScaled.items():
value *= scale
#for some reason I can't iterate through the vector?
value.x = math.sqrt(abs(value.x)) * sign(value.x)
value.y = math.sqrt(abs(value.y)) * sign(value.y)
value.z = math.sqrt(abs(value.z)) * sign(value.z)
disp = dispScaled
return [disp, scale]
def CopyDisplacementToVPaint(dict, ob):
posVCol = "displacementPos"
negVCol = "displacementNeg"
ob.data.vertex_colors.new(name = posVCol)
ob.data.vertex_colors.new(name = negVCol)
for l in ob.data.loops:
i = l.vertex_index
val = dict[i]
val_x = 0
val_y = 0
val_z = 0
if (val.x > 0):
val_x = val.x
if (val.y > 0):
val_y = val.y
if (val.z > 0):
val_z = val.z # the values are automatically clamped, so unecessary
ob.data.vertex_colors[posVCol].data[l.index].color = (val_x,val_y,val_z, 1)
val_x = 0
val_y = 0
val_z = 0
if (val.x < 0):
val_x = -1 * val.x
if (val.y < 0):
val_y = -1 * val.y
if (val.z < 0):
val_z = -1 * val.z
ob.data.vertex_colors[negVCol].data[l.index].color = (val_x,val_y,val_z, 1)
def TestDisplacement(disp, scale, ob):
for v in ob.data.vertices:
d = disp[v.index]
d.x = (d.x**2)*sign(d.x)
d.y = (d.y**2)*sign(d.y)
d.z = (d.z**2)*sign(d.z)
d.x *= 1/scale
d.y *= 1/scale
d.z *= 1/scale
mat_loc = mathutils.Matrix.Translation(d)
v.co = mat_loc @ v.co
obTarget = bpy.context.active_object
obSource = InactiveObject()
assert (obSource), "Select two objects with identical topology."
vTarget = obTarget.data.vertices
vSource = obSource.data.vertices
assert (len(vSource) == len(vTarget)), "Select two objects with identical topology."
print (obSource.name, obTarget.name)
disp = {}
matSource = obSource.matrix_world
vSource = obSource.data.vertices
matTarget = obTarget.matrix_world
vTarget = obTarget.data.vertices
dispData = calcDisplacement(disp, matSource, vSource, matTarget, vTarget)
scaleData = scaleDisplacement(dispData, obSource, obTarget)
disp = scaleData[0]
scale = scaleData[1]
CopyDisplacementToVPaint(disp, obTarget)
#TestDisplacement(disp, scale, obTarget)
posOffset = 0
negOffset = 0
``````
Thanks. I wasn’t sure whether to post here or on Stack exchange, so I’ll start with this.
1 Like
It seems to be scaled by the bounding-box of the object, not linearly as i had assumed but maybe logarithmically?
I did some poking around in the code for Displacement modifier- it is actually doing what I thought. I was wrong in the last comment. The problem is with color space. Using a Gamma node in the shader seems to fix the issue… still, I have no idea whether this is a hack or not! Right now I’m using a magic number, gamma = 0.882, but I’m just eye-balling it.
Does anyone know how to correct for color space?
The following may be of some use;
https://docs.blender.org/manual/en/latest/render/post_process/color_management.html
Thanks! I’ve actually decided to try to write a patch for Blender to fix this behavior, but I’m a long way from being able to do that. I need to brush up on C a bit, and learn a lot more about Git, CMake, etc. . I’d also like to make Cycles able to bake to vertex colors, and then my little script won’t be quite as useful!
1 Like
@Josephbburg
The following may be of some use;
1 Like
From a user perspective that is very nice idea for a modifier or in a future - node modifier. Assuming no UV unwrapping is needed. | 2,060 | 7,791 | {"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-2023-14 | latest | en | 0.918532 |
https://www.infoapper.com/unit-converter/weight/grams-to-pounds/ | 1,675,128,499,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499831.97/warc/CC-MAIN-20230130232547-20230131022547-00329.warc.gz | 831,528,892 | 7,735 | # Convert Gram to Pound (gm to lb)
In next fields, kindly type your value in the text box under title [ From: ] to convert from gram to pound (gm to lb). As you type your value, the answer will be automatically calculated and displayed in the text box under title [ To: ].
From:
To:
Definitions:
Gram (abbreviation: gm): is a SI (metric) system unit of mass. Where, at the temperature of melting ice, it is the absolute weight of a volume equal to the hundredth part of a metre cube of pure water.
Pound (abbreviations: lb, or lbs, or ps): is a unit of force used in some systems of measurement including English Engineering units and the British Gravitational System. Also, it is a unit of mass used in the imperial, United States customary and other systems of measurement.
<< Prev Next >>
## How to Convert Grams to Pounds
### Example: How many pounds are equivalent to 37.01 grams?
As;
1 grams = 0.00220462 pounds
37.01 grams = Y pounds
Assuming Y is the answer, and by criss-cross principle;
Y equals 37.01 times 0.00220462 over 1
(i.e.) Y = 37.01 * 0.00220462 / 1 = 0.0815929862 pounds
Answer is: 0.0815929862 pounds are equivalent to 37.01 grams.
### Practice Question: Convert the following units into lb:
N.B.: After working out the answer to each of the next questions, click adjacent button to see the correct answer.
( i ) 10.19 gm
( ii ) 17.33 gm
( iii ) 39.12 gm
References | 371 | 1,408 | {"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} | 3.4375 | 3 | CC-MAIN-2023-06 | latest | en | 0.853874 |
https://www.enotes.com/homework-help/determine-which-following-observations-j-could-466496 | 1,632,687,193,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00386.warc.gz | 783,896,134 | 23,009 | # Science
• Determine which of the following observations (A-J) could lead to a testable hypothesis.
For those that are testable:
Write a hypothesis and null hypothesis
What would be your experimental approach?
What are the dependent and independent variables?
How will you collect your data?
How will you present your data (charts, graphs, types)?
How will you analyze your data?
• When a plant is placed on a window sill, it grows three inches faster per day than when it is placed on a coffee table in the middle of the living room.
• 2. The teller at the bank with brown hair and brown eyes and is taller than the other tellers.
• When Sally eats healthy foods and exercises regularly, her blood pressure is 10 points lower than when she does not exercise and eats unhealthy foods.
• The Italian restaurant across the street closes at 9 pm but the one two blocks away closes at
• 10 pm.
• For the past two days the clouds have come out at 3 pm and it has started raining at 3:15 pm.
• George did not sleep at all the night following the start of daylight savings.
• Exercise 3: Conversion
For each of the following, convert each value into the designated units.
• 46,756,790 mg = _______ kg
• 5.6 hours = ________ seconds
• 13.5 cm = ________ inches
• 47 °C = _______ °F
• Exercise 4: Accuracy and Precision
• During gym class, four students decided to see if they could beat the norm of 45 sit-ups in a minute. The first student did 64 sit-ups, the second did 69, the third did 65, and the fourth did 67. 2. The average score for the 5th grade math test is 89.5. The top 4th graders took the test and scored 89, 93, 91 and 87.
• Yesterday the temperature was 89 °F, tomorrow it’s supposed to be 88°F and the next day it’s supposed to be 90°F, even though the average for September is only 75°F degrees!
• Four friends decided to go out and play horseshoes. They took a picture of their results shown to the right:
• A local grocery store was holding a contest to see who could most closely guess the number of pennies that they had inside a large jar. The first six people guessed the numbers 735, 209, 390, 300, 1005 and 689. The grocery clerk said the jar actually contains 568 pennies.
• Exercise 5: Significant Digits and Scientific Notation
Part 1: Determine the number of significant digits in each number and write out the specific significant digits.
• 405000
• 0.0098
• 39.999999
• 13.00
• 80,000,089
• 55,430.00
• 0.000033
• 620.03080
• Part 2: Write the numbers below in scientific notation, incorporating what you know about significant digits.
• 70,000,000,000
• 0.000000048
• 67,890,000
• 70,500
• 450,900,800
• 0.009045
• 0.023
The length limit on Enotes responses prevents all 30 of your questions from being fully answered. In the future please cluster your questions according to format or subject matter, so that they will be easier to respond to, and so that problems may be more easily addressed; for example, in...
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
The length limit on Enotes responses prevents all 30 of your questions from being fully answered. In the future please cluster your questions according to format or subject matter, so that they will be easier to respond to, and so that problems may be more easily addressed; for example, in Section 4: Accuracy and Precision, you did not include a picture that is necessary to answer question 3, and there are no instructions on what is to be done with the four questions.
3. Conversions
1. mg to kg
There are various ways of expressing the relationship between exponential factors; the simplest depiction here is to use grams. Grams are the “standard” upon which these measurements are based. A milligram is one-thousandth of a gram, and a kilogram is one thousand grams. Thus, 1000x1000 = 1 million milligrams would fit into a kilogram.
The easiest way to set up conversions is to write what you have, including units. Then, on the right side of this amount, write the conversion factor (the statement which relates the unit you have to the unit you want). 1 million mg = 1 kg is our conversion factor. Now orient the conversion factor so that the unit you already have is on the bottom; this way, the units will cancel out, leaving you with units of what you wanted to convert to.
46,756,790mg x (1kg/1,000,000mg) = 46.756790kg
2. hours to seconds.
The conversion factor is based on 60 seconds in a minute, and 60 seconds in an hour.
1 hour = 3600 seconds.
5.6 hours x (3600 seconds / 1 hour) = 20,160 seconds
3. cm to inches
The conversion here can be looked up from reference; 1cm = 0.3937 inches
13.5cm x (.3937 inches / 1cm) = 5.315 inches
4. C to F
This is more of a math equation than a conversion factor. To convert Celsius to Fahrenheit, multiply the Celsius temperature by (9/5) and add 32.
47 x (9/5) + 32 = 116.6 F
5. Significant Digits
General rules for sigfigs:
-Nonzero numbers always count.
-Zeros between nonzeros count
-Any zero at the end of a number (trailing) zeros count only if there is a decimal.
-Zeros at the beginning of a number never count
1. 3. Trailing nonzeros without a decimal do not count.
2. 2. Leading zeros do not count.
3. 8. All nonzeros count.
4. 4. Trailing zeros with a decimal count.
5. 8. Zeros between nonzeros count.
6. 7.
7. 2.
8. 8.
Scientific Notation
The rules for scientific notation are;
-The expression must be in the form of (number) times (ten to a power)
-The initial number must be a single-digit integer
-The integer must be followed by a decimal if there is more than one significant figure in the original number
-Any additional significant digits appear after the decimal
To find the appropriate power of ten:
If the number is greater than zero, count from the decimal place (or the “ones”-value digit) up to the first number.
If the number is less than zero, count the places between the decimal place and the first number, plus the initial zero.
1. 7 x 10e10
2. 4.8 x 10e-8
3. 6.789 x 10e7
4. 7.05 x 10e4
5. 4.509008 x 10e8
6. 9.045 x 10e-3
7. 2.3 x 10e-2
Approved by eNotes Editorial Team | 1,629 | 6,177 | {"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} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.956079 |
https://www.physicsforums.com/threads/minimum-angle-of-resolution-units-clarification.330715/ | 1,519,353,985,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814311.76/warc/CC-MAIN-20180223015726-20180223035726-00616.warc.gz | 924,981,112 | 15,292 | # Minimum angle of resolution - units clarification
1. Aug 14, 2009
### Elemental_
1. The problem statement, all variables and given/known data
A mouse has an pupil diameter of 2.5 mm. lying 200 meters from the mouse are two stones 30 cm apart. λ = 500 nm
a) what angle do the stones subtend?
b) what is the MAR (minimum angle of resolution) of the mouse's eye.
2. Relevant equations
question a) tanθ = O/A
O= 0.3 m
A= 200 m
b) θmin = (1.22*λ )/d
d = pupil diameter
3. The attempt at a solution
for a) I calculated the answer to be 1.5 x 10^-3 however I'm not sure whether the unit for this would be in rads or degrees as the question asks for the angle? I'm aware that MAR (theta-min), the angle between two point sources needed in order for them to be resolved as seperate is measured in radians - does this hold true for the angle that the stones subtend as well?
If the answer i calculated is indeed in the unit of degrees, would i have to convert it to rads? (which would be equivalent to 5.23 x 10^-3 rads)?
b) Minimum angular resolution i calculated to be 2.44 x 10^-4 rads using the formula 1.22(λ)/(diameter of pupil) - I'm pretty sure that this is measured in radians.
If anyone could help me in this it would be greatly appreciated!
2. Aug 14, 2009
### cepheid
Staff Emeritus
Welcome to PF, Elemental_
Yeah, angles are measured in radians. Note, however, that the radian is a *dimensionless* unit. In other words, angles are dimensionless because they are defined as a ratio of two lengths (in the radian system). In part (a), you calculated the angle by dividing a length by a length. The units (metres) cancelled out, leaving you with a dimensionless number. We just assign units to this dimensionless number called radians to indicate that it happens to be an angle.
The only way that the answer you calculated in part (a) could be in degrees would be if you used the wrong mode on your calculator when calculating the arctangent. Be careful to always use angles in radians in your calculations. The formulae you are using rely upon the angles being in radians...those formulae aren't even valid if the angles are not in radians (because they rely upon the definition of an angle as given in the radian system).
3. Aug 14, 2009
### Elemental_
Thank you so much for the thorough explanation cepheid; I verified this by using the degree mode on my calculator instead, and as you said, the answer for a) was a completely different value (although converting this to rads afterwards using pi/180 is a viable option as well, but in my opinion it's not worth it for the extra step). Thanks again for the help! | 656 | 2,635 | {"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} | 3.78125 | 4 | CC-MAIN-2018-09 | latest | en | 0.939496 |
http://self.gutenberg.org/articles/eng/Integral_symbol | 1,603,607,651,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107887810.47/warc/CC-MAIN-20201025041701-20201025071701-00283.warc.gz | 94,498,625 | 20,289 | #jsDisabledContent { display:none; } My Account | Register | Help
Flag as Inappropriate This article will be permanently flagged as inappropriate and made unaccessible to everyone. Are you certain this article is inappropriate? Excessive Violence Sexual Content Political / Social Email this Article Email Address:
# Integral symbol
Article Id: WHEBN0003171907
Reproduction Date:
Title: Integral symbol Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Integral symbol
The integral symbol:
(Unicode), \displaystyle \int (LaTeX)
is used to denote integrals and antiderivatives in mathematics. The notation was introduced by the German mathematician Gottfried Wilhelm Leibniz towards the end of the 17th century. The symbol was based on the ſ (long s) character, and was chosen because Leibniz thought of the integral as an infinite sum of infinitesimal summands.
## Contents
• Typography in Unicode and LaTeX 1
• Fundamental symbol 1.1
• Extensions of the symbol 1.2
• Typography in other languages 2
• See also 3
• Notes 4
• References 5
• External links 6
## Typography in Unicode and LaTeX
### Fundamental symbol
The integral symbol is U+222B integral in Unicode[1] and \int in LaTeX. In HTML, it is written as ∫ (hexadecimal), ∫ (decimal) and ∫ (named "entity").
The original IBM PC code page 437 character set included a couple of characters and (codes 244 and 245, respectively) to build the integral symbol. These were deprecated in subsequent MS-DOS code pages, but they still remain in Unicode (U+2320 and U+2321, respectively) for compatibility.
The symbol is very similar to, but not to be confused with, the ʃ symbol (called "esh").
### Extensions of the symbol
Related symbols include:[1][2]
Meaning Unicode LaTeX
Double integral U+222C \iint \iint
Triple integral U+222D \iiint \iiint
Contour integral U+222E \oint \oint
Clockwise integral U+2231
Anticlockwise integral U+2A11
Clockwise contour integral U+2232 \varointclockwise
Anticlockwise contour integral U+2233 \ointctrclockwise
Closed surface integral U+222F \oiint
Closed volume integral U+2230 \oiiint
## Typography in other languages
Regional variations (English, German, Russian) of the integral symbol.
In other languages, the shape of the integral symbol differs slightly from the shape commonly seen in English-language textbooks. While the English integral symbol leans to the right, the German symbol (used throughout Central Europe) is upright, and the Russian variant leans to the left.
Another difference is in the placement of limits for definite integrals. Generally, in English-language books, limits go to the right of the integral symbol: \int_0^T f(t)\;dt .
By contrast, in German and Russian texts, limits for definite integrals are placed above and below the integral symbol, and, as a result, the notation requires larger line spacing: \int\limits_0^T f(t)\;\mathrm{d}t .
## Notes
1. ^ a b "Mathematical Operators – Unicode" (PDF). Retrieved 2013-04-26.
2. ^ "Supplemental Mathematical Operators – Unicode" (PDF). Retrieved 2013-05-05.
## References
• Stewart, James (2003). "Integrals". Single Variable Calculus: Early Transcendentals (5th ed.).
• Zaitcev, V.; Janishewsky, A.; Berdnikov, A. (1999), "Russian Typographical Traditions in Mathematical Literature" (PDF), EuroTeX'99 Proceedings
Copyright © World Library Foundation. All rights reserved. eBooks from Project Gutenberg are sponsored by the World Library Foundation,
a 501c(4) Member's Support Non-Profit Organization, and is NOT affiliated with any governmental agency or department. | 890 | 3,681 | {"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} | 2.859375 | 3 | CC-MAIN-2020-45 | latest | en | 0.808607 |
https://www.thenational.academy/teachers/programmes/maths-primary-ks2/units/right-angles/lessons/know-that-a-right-angle-describes-a-quarter-turn | 1,719,204,372,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864986.57/warc/CC-MAIN-20240624021134-20240624051134-00827.warc.gz | 914,578,718 | 38,057 | New
New
Year 3
Know that a right angle describes a quarter turn
I can describe turning through a right angle as making a quarter turn and can turn clockwise or anticlockwise.
New
New
Year 3
Know that a right angle describes a quarter turn
I can describe turning through a right angle as making a quarter turn and can turn clockwise or anticlockwise.
Share activities with pupils
Share function coming soon...
Lesson details
Key learning points
1. Two right angles make a half turn.
2. Three right angles make a three-quarter turn.
3. Four right angles make a full turn.
Keywords
• Quarter turn - One right angle makes a quarter turn.
• Half turn - Two right angles make a half turn.
• Three-quarter turn - Three right angles make a three-quarter turn.
• Full turn - Four right angles make a full turn.
Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
Starter quiz
6 Questions
Q1.
Which of these sentences describe a right angle?
Correct answer: A right angle is a square corner
Correct answer: A right angle is a quarter turn
A right angle is found in every shape
Correct answer: A right angle is a measure of turn
Q2.
How many right angles does this shape have?
0
2
3
Q3.
True or false. The hands on this clock show a right angle.
False
Q4.
The right angle shows the turn between which numbers on the clock?
11 and 2
11 and 1
10 and 2
Q5.
Which of the images is not showing a right angle?
a
c
d
Q6.
Which of the capital letters have at least one right angle?
A
W
Exit quiz
6 Questions
Q1.
Match the number of right angles to the number of turns.
Correct Answer:One right angle makes,a quarter turn
a quarter turn
Correct Answer:Two right angles make,a half turn
a half turn
Correct Answer:Three right angles make,a three-quarter turn
a three-quarter turn
Correct Answer:Four right angles make,a full turn
a full turn
Q2.
Match the arrow with the size of turn.
a quarter turn
a half turn
a three-quarter turn
Q3.
Two quarter turns is the same as a turn. | 512 | 2,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-26 | latest | en | 0.823783 |
http://isaac.exploratorium.edu/~pauld/summer_institute/summer_day4+5light/day4_light.html | 1,627,274,809,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00188.warc.gz | 25,956,098 | 7,830 | Light
Our study of the history of understanding of light will take us to visit so many topics that it will take us two days. That's good though, since that will give us a pause to reflect on what we're seeing.
Beyond these two days we'll take one more day to study color and color vision, and another day to study image formation.
Models
I have chosen a timeline of history approach to the study of light because it illustrates the importance of models to the advance of science. Scientific models of light have changed over the years as more and better experiments were done. When an old model no longer explained all of the new experiments it had to be replaced by a newer model. However, the old model still worked for all of the original experiments and so was still somewhat useful. Thus, I still use Maxwell's 1870 model of light as electromagnetic waves even though it was supplanted by Feynman et.al.'s Quantum Electrodynamics, QED, in the 1950's. I only use QED when I am forced to by the failure of Maxwell's equations.
When you ask me a question about light I'll try to answer with the simplest model that gives the correct answer. If you revise your question, I may have to switch my answer to a totally different model.
Teachers and models
One of the important skills of a teacher, is choosing the simplest model that helps a student toward understanding.
Let there be light
We'll start out in the dark with the words of Benjamin Franklin who said;
" About light, I am in the dark."
Ancient History
Hunters knew that if you tried to spear a fish by aiming where the fish appeared to be, you would miss. Anyone could see that the spear bent when it entered the water. Now we say that the spear doesn't bend when it enters the water, but that the light with which we see the spear bends as it leaves the water.
The bending of light as it enters or leaves water is called refraction.
One of the oldest tables of data is a Greek stone tablet into which are carved the angle of incidence and angle of refraction of light going from air to water.
Here is an interesting experiment using refraction. I use this to get students interested in light and ask them , "What does it take to make something invisible?" Answer below. Please do the activity first.
Disappearing glass rods snack
When glass rods, Pyrex, are placed into a liquid with the same index of refraction, Wesson oil, they disappear.
To be invisible, an object must not reflect light, it must not absorb light, and it must not refract light.
To be more quantitative about light refraction, you can use two different activities:
The first uses a bare filament light bulb to generate a beam of light, it is called a:
Light Box
The second uses a laser:
Laser Jell-O snack
Critical angle snack
Math Root on refraction through layers
Galileo
Galileo attempted to measure the speed of light, circa 1600. He used lamps and men on mountain tops. He made several measurements and decided that light traveled so rapidly that he could not measure its speed.
How would you have tried to measure the speed of light with Galileo's tools?
Galileo had two men stand on hilltops, one would uncover a lamp, when the other saw the light he would uncover his lamp and be observed by first man. Galileo used hills separated by different distances and found that the distance between the hills did not change the timing. He was measuring human reaction time and not the travel time for light. He realized that this meant that light was very fast. However, Galileo's experiment will work for measuring the speed of sound.
Science mulch: The speed of light is a foot per nanosecond. 0.3 meters per nanosecond,
a nanosecond is 10-9 s.
Christian Huygens
Huygens (1629-1695) developed an idea by Descarte that light was a wave. He published this in 1690. Wave theory predicted that light would refract because the waves would travel more slowly in glass. Robert Hooke, Newton's rival, supported the wave theory. (I still use Huygens wave model to explain why light travels in straight lines.)(Need image)
Newton
Newton's theory was that light was a particle, published in 1704. This theory could explain why light traveled in a straight line. Light traveling as particles explained shadows and reflections, it even explained refraction &emdash; if the particles sped up in dense media, as if they were being pulled into the media by gravity.
Reflection of light can be understood using Newton's particle model. The particles of light bounce off the mirror just like a ball would. This leads to the law:
The angle of incidence equals the angle of refraction.
You can investigate reflection from one or more mirrors using a small pointer laser in the following exploration:
Scan a Laser
Both the wave theory of Huygens and the particle theory of Newton explained refraction. In the wave theory the waves of light slowed down when they entered water from air. In the particle theory the particles of light sped up. (Actually, in the particle theory only the component of the velocity perpendicular to the surface sped up, the component parallel to the surface remained constant.) The actual experiment to decide between the two theories of refraction could not be carried out until 1849 when Foucault found that the speed of light in water was slower. The particle theory can not explain refraction if light slows down in glass.
Newton knew that if light were a wave it would diffract, or bend, around corners.
Since diffraction was not seen, then light couldn't be a wave.
However, Newton saw and reported on the bending of light around a needle and he also saw Newton's rings both of which we now know are due to diffraction and interference. Although Newton saw the phenomena of diffraction he did not recognize it. He said instead that light was a particle that had "fits."
Here is an activity in which you can see Newton's rings for yourself.
Bridge light snack
Chromatic Aberration
When you come to the Exploratorium visit the exhibit Chromatic Aberration. Here you will see that your eye cannot focus blue light and red light at the same time. You look at a sharp edged dot of light through a magenta filter which passes blue light and red light but no other colors Viewed at a distance the dot has a sharp red center and a fuzzy blue halo, viewed closer it has a sharp blue center and a red halo. This is because the index of refraction which determines the amount of bending of light by your eye is a function of the color (later we'll see that color is related to the wavelength of light.) Human eyes bend blue light more than red light. Most transparent things bend blue light more than red, this is known as dispersion.
Aside: Newton knew about chromatic aberration and said that it could not be removed, later Chester Moore Hall (in 1733) discovered how to make lenses which could focus two wavelengths of light at the same time. These lenses were called "achromats." He did not patent his idea which would have required him to publish how it was done, instead he kept it secret, and so reserved the entire trade in achromatic lenses to himself for far longer than the patent would have. He made his achromatic lenses by combining two lenses with different dispersions. Modern lenses can focus three wavelengths at the same time, these lenses are called apochromats and were invented by Ernst Abbe in the 1860's.
Thomas Young
In 1801, Thomas Young passed light through two slits and observed interference. With one slit, a pattern of light was created, when a second slit was opened by itself the same pattern of light was created. When both slits were opened at the same time, darkness appeared where there had been light with either alone! Thus I say:
Light plus light equals dark!
You can easily project Young's experiment for a class of students using an inexpensive laser pointer.
We'll create two slits and cover one of them, then shine light through the other. A band of light will appear on a distant screen. Next we'll uncover the first slit and cover the other. The light spreads into the same broad band.
However if we uncover both slits and shine light through both of them, dark places appear on the wall where there had been light through either slit alone!
Cover one slit and light appears, open that slit and dark is created.
Light plus light equals dark, you've got to see this to believe it!
The two slit experiment convinced Young that light was a wave. Newton's influence on the Royal Society was so strong however that it took Young years to convince the society that light was a wave.
Two Slit Experiment
Students can then figure out Young's model using waves drawn on paper:
Model of interference
There is much more to understand about interference of light, such as diffraction from large slits and interference by reflection from soap films and oil slicks. We'll return to study interference of light in detail later, during an entire day of exploration dedicated to Interference.
While he was at it, Thomas Young also figured out that in order to perceive all of the colors that humans can see we needed to have three different color receptors. He was right, we have three different types of cone cells which give us our color perception.
Augustin Fresnel
Augustin Fresnel (1788) entered in a competition to elucidate the nature of light. (Three of the judges were Poisson, Biot, and Laplace, all supporters of the particle model) Fresnel could write the equations for the wave model of light but he couldn't solve them! One of the judges however was a great mathematician,Poisson, who solved Fresnel's equations. Poisson pointed out that his solution predicted that there would be a spot of light in the middle of the shadow of a ball. Since there was no such spot of light, Fresnel was wrong.
However the chairman of the judges, Arago, conducted the experiment and found the spot.
Thereafter the spot was named: the Poisson spot.
Later, as head of the lighthouse commission, Fresnel invented the lens that bears his name.
At the Exploratorium you can see the Poisson spot in the exhibit Long Path Diffraction.
Or you can make it yourself with a small pointer laser:
Making a Poisson Spot.
Arago
Arago discovered that light was polarized, this meant light was a transverse wave. This meant that light was not like sound which was a longitudinal wave.
You can easily see that light is polarized by doing experiments with polarizers and plastic.
Polarized light
We'll return another day to explore Polarization in more detail.
Maxwell
If light was a wave, what was it a wave in?
Maxwell found the equations that described light as a wave in electricity and magnetism, an electromagnetic wave.
Tape demo of an electromagnetic wave
For a while, it was thought that electromagnetic waves had to travel through the aether &endash; a thin stiff substance that permeated the universe.
Michelson and Morley did an experiment to measure the speed of light in the aether. However their experiment showed that there was no aether! Einstein made sense of their data with his theory of special relativity.
In Maxwell's wave model the color of light depends on its frequency, the brightness on its amplitude.
Electromagnetic waves are created by accelerating electric charges,
The electromagnetic waves then exert forces on the electric charges causing them to accelerate.
More on the Maxwell model of light.
Hot objects emit light.
Hot objects glow red, they are red hot.
Hotter objects glow white, they are white hot!
Turn on an electric hot plate, or heat a nail in the flame from a torch and notice the color of the glow that results. As the plate and the nail heat up the color goes from deep red to orange to yellow, heat it enough and it would become white. (The nail will melt before becoming white however.)
Planck was able to explain the color of light emitted by hot objects by modeling energy as existing in discrete chunks, quanta.
Albert Einstein
Explained the photoelectric effect by extending the idea proposed by Planck. He suggested that light energy was quantized. Later, the quanta of light proposed by Einstein were named photons. So it looked like light behaved as a chunk of energy, as a particle.
The Exploratorium has an exhibit on the photoelectric effect,
at this exhibit you charge a zinc plate with negative electrons.
Shining visible light on the plate does not change the charge.
Shining ultraviolet light on the plate discharges it.
Photo electric effect
In particular Einstein found that the energy, E, of a photon was proportional to its frequency, f.
E = hf
where the constant of proportionality, h, is Planck's constant.
It is easy to demonstrate the connection between the energy of a photon and its frequency (color) using light emitting diodes which convert the electrical energy change of one electron into one photon.
Light Energy and Color
Irwin Schroedinger
In Schroedinger's model for light it is described as traveling as a wave and interacting like a particle.
Light is a wave in "probability amplitude."
In the quantum mechanical theory for light, color depends on the frequency of the light, the brightness on the number of photons per second.
Quantum mechanics explained the spectrum of light emitted by atoms and molecules.
The Exploratorium used to have a great exhibit called "Solar Spectrum" in which the spectrum of sunlight was spread across a meter. Maybe one day it will return.
Spectra
The spectrum of hydrogen can be derived from the Schroedinger equation for hydrogen, you can explore the spectrum of hydrogen and the energy levels of the electron in hydrogen using the wonderful program "Atom in a Box" available for payment of a small fee through the internet.
DeBroglie modeled particles as waves too, and then Davison and Germer showed that electrons diffracted and thus had a wavelength.
More on the quantum model of light.
Richard Feynman
Quantum mechanics made some errors in predicting the colors of the spectral lines emitted by atoms, to get the right answer, quantum mechanics had to be combined with relativity.
Feynman was one of the scientists who combined quantum mechanics and relativity to produce our current model of light known as Quantum Electrodynamics, or QED.
Feynman said that If you understand the two slit experiment you understand all of quantum mechanics, but that unfortunately, no one understood the two slit experiment.
In this model light travels as a wave, a wave which explores every path in the entire universe on its way from one point to another. The phase of this wave is changed as it interacts with charged particles. So as the wave passes an atom its phase can be slightly delayed. This slight delay gives rise to refraction.
Electro-weak
Recently scientists found a mathematical model that combined electromagnetism with the weak nuclear force. This combined model is know as the electro-weak force.
Inverse Square law snack | 3,142 | 15,003 | {"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} | 3.09375 | 3 | CC-MAIN-2021-31 | latest | en | 0.971668 |
https://ilmihub.com/quiz-on-physics-for-online-preparation-of-competitive-exams.html | 1,726,451,031,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00783.warc.gz | 280,704,292 | 22,161 | Home Physics MCQs Quiz On Physics for Online Preparation of Competitive Exams
# Quiz On Physics for Online Preparation of Competitive Exams
405
### Quiz On Physics for NTS, PPSC, CSS, FPSC, PMS, OTS, PTS and Interviews Preparation
This post contains Quiz On Physics for NTS, PPSC, CSS, FPSC, PMS, OTS, PTS and Interviews Preparation. This is the 5th post of Physics Quiz series. These Physics GK questions and answers are posted here for online preparation of different Entry tests and competitive exams and interviews. MCQs are taken from Caravan Comprehensive book. For more, visit Physics MCQs.
## Physics Quiz for Online Preparation of Competitive Exams
Counting is starting from 81. For previous Post Visit This Link.
81. A car and a loaded truck are moving with the same speed along a road. As compared to the truck, the car shall possess
(a) More kinetic energy
(b) More potential energy
(c) Less kinetic energy
(d) More mechanical energy
(c) Less kinetic energy
82. When a moving bus stops suddenly, the passengers are pushed forward because of the ______.
(a) Friction between the earth and the bus
(b) Friction between the passengers and the earth
(c) The inertia of the passengers
(d) The inertia of the bus
(c) The inertia of the passengers
83. Erect and virtual images are always produced by______.
(a) Plane and convex mirrors
(b) Plane mirrors alone
(c) Concave and convex mirrors
(d) Concave mirrors alone
(a) Plane and convex mirrors
84. The working of the receiver of a telephone depends upon the________.
(a) Change in magnetisation of an electromagnet, which causes a diaphragm to vibrate
(b) The efficiency of the loudspeaker which enhances the loudness of sound.
(c) Conversion of the electric signal, after its amplification into the sound wave.
(d) The diaphragm which is made to vibrate by the incoming sound wave.
(d) The diaphragm which is made to vibrate by the incoming sound wave.
85. An old man, unable to apply a large force, wishes to use a screw jack for lifting his car wheels. if L is the length of the rod used to turn the jack and p is the pitch of jack screw, then he should use a jack in which ________.
(a) L is large and p is large
(b) L is small and p is large
(c) L is large and p is small
(d) L is small and p is small
(b) L is small and p is large
86. The hydraulic brakes used in automobiles is a direct application of_____.
(a) Archimede’s principle
(b) Toricellian law
(c) Bernoulli’s theorem
(d) Pascal’s law
(d) Pascal’s law
87. In TV, transmission_______.
(a) Sound and video signals are transmitted simultaneously
(b) Sound is transmitted first and then the video signal follows
(c) Video signal is transmitted first. followed by sound signal
(d) Sound and video signals are transmitted from different places at the same time
(a) Sound and video signals are transmitted simultaneously
88. Surface tension in a liquid is due to _________.
(a) The adhesive force between the molecules
(b) Cohesive force between the molecules
(c) The gravitational force between the molecules
(d) Electrical force between the molecules
(b) Cohesive force between the molecules
89. Springs of shock absorbers of automobiles and railway coaches are made of steel and not of rubber because steel is _________.
(a) More durable than rubber
(b) Less expensive than rubber in the long run
(c) Less elastic than rubber
(d) More elastic than rubber
(d) More elastic than rubber
90. Why does the mercury column in the barometer fall rapidly before a severe storm?
(a) It is due to decrease in humidity in the air
(b) It is due to the rise in atmospheric pressure
(c) It is due to the fall in atmospheric pressure
(d) It is due to the severe heat energy from the sun
(c) It is due to the fall in atmospheric pressure
91. Why does water boil below 100°C at higher altitudes?
(a) There is a lesser dissipation of heat at higher altitudes
(b) Water available at higher altitudes is purer than that in the plains
(c) Pollution-free air at higher altitudes increases the calorific value of fuel used
(d) The atmospheric pressure at higher altitudes is low as compared to that at sea level
(d) The atmospheric pressure at higher altitudes is low as compared to that at sea level
92. The speed of sound is maximum through which of the following?
(a) Air
(b) Glass
(c) Water
(d) Wood
(b) Glass
93. Echo is the effect produced due to ________.
(a) Absorption of sound
(b) Dispersion of sound
(c) Reflection of sound
(d) Refraction of sound
(c) Reflection of sound
94. In a photographic camera fitted with a convex lens, which of the following types of images will be formed on the film?
(a) Erect and real
(b) inverted and real
(c) imaginary and erect
(d) The type of image formed will depend on the distance between the lens and the object
(b) inverted and real
95. In the modem-day computers, when operational, the electrical impulses travel_______.
(a) At the speed of sound
(b) At seven times the speed of sound
(c) At half the speed of light
(d) Nearly at the speed of light
(d) Nearly at the speed of light
96. A magnet freely suspended by means of a string will always set itself in which of the following directions?
(a) East-West
(b) North-East
(c) North-South
(d) South-East
(c) North-South
97. Why are shields made of iron usually provided around precision instruments?
(a) For protection against the effect of external magnetic fields
(b) For guarding the instruments against unauthorised handling
(c) For protection against the effect of moisture in the air
(d) For absorbing heat generated during the functioning of the instrument
(a) For protection against the effect of external magnetic fields
98. Why are transformers used in the transmission of electric power?
(a) Because they speed up the transmission
(b) Because they can conserve electrical energy
(c) Because they can reduce transmission losses
(d) Because they help to distribute electrical energy efficiently
(c) Because they can reduce transmission losses
99. A mixture of which of the following gases is used in the manufacture of electric bulbs?
(a) Nitrogen and Argon
(b) Nitrogen and Oxygen
(c) Oxygen and Argon
(d) Oxygen and Hydrogen
(c) Oxygen and Argon
100. A fresh egg sinks in pure water, whereas it floats in saturated salty water. This is due to _______.
(a) Higher density of the salty water
(b) Higher density of the pure water
(0) The fluid matter inside the egg-shell
(d) The fact that the egg-shell is made of calcium which is heavier than pure water
(a) Higher density of the salty water | 1,516 | 6,563 | {"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} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.888408 |
https://www.printablemultiplication.com/worksheets/How-To-Make-My-Own-Multiplication-Worksheet | 1,603,498,197,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881551.11/warc/CC-MAIN-20201023234043-20201024024043-00577.warc.gz | 863,755,695 | 76,498 | ## Multiplication Chart 80×80
…making use of the Multiplication Table. Incoming search terms: 4s multiplication chart 4s multiplication practice long multiplication worksheet and answers probability multiplication rule worksheet what is a mu;ltiplication grid 2×2…
## Multiplication Chart Sbac
…of arithmetic to find out. Assess basics of multiplication. Also, evaluate the fundamentals utilizing a multiplication table. Allow us to evaluation a multiplication instance. Employing a Multiplication Table, flourish four…
## Multiplication Chart Multiplication Chart
…the basic principles how to use a multiplication table. We will overview a multiplication case in point. Employing a Multiplication Table, grow 4 times a few and have a response…
## Multiplication Flash Cards Virtual
…evaluate the basic principles how to use a multiplication table. In 2020 | Multiplication Flashcards, Flashcards, Distance Learning”] We will assessment a multiplication instance. Utilizing a Multiplication Table, flourish 4…
## 0-10 Multiplication Chart
…of multiplication. Also, evaluate the basic principles utilizing a multiplication table. We will review a multiplication illustration. Using a Multiplication Table, multiply four times 3 and have an answer 12:…
## Multiplication Chart 100
…review the fundamentals utilizing a multiplication table. Let us review a multiplication case in point. By using a Multiplication Table, increase four times 3 and have a solution 12: 4…
## Multiplication Chart 84
…the basics the way you use a multiplication table. We will assessment a multiplication instance. Employing a Multiplication Table, multiply a number of times three and obtain a response twelve:…
## 1-10 Multiplication Chart
…the Multiplication Table. A further reward, the Multiplication Table is visible and displays to understanding addition. Where by can we start understanding multiplication making use of the Multiplication Table? Initially,…
## Flash Cards Multiplication Printable Up To 12's
…evaluate the basics how to use a multiplication table. Let us evaluation a multiplication instance. Using a Multiplication Table, increase 4 times three and acquire a response twelve: 4 x…
## Multiplication Chart Ideas
…look at the fundamentals using a multiplication table. We will review a multiplication illustration. Utilizing a Multiplication Table, multiply four times a few and get an answer a dozen: 4… | 486 | 2,423 | {"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} | 3.734375 | 4 | CC-MAIN-2020-45 | latest | en | 0.723799 |
https://socratic.org/questions/when-do-we-know-that-a-limit-of-a-given-function-exists#439114 | 1,695,980,175,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510501.83/warc/CC-MAIN-20230929090526-20230929120526-00519.warc.gz | 581,635,367 | 6,024 | # When do we know that a limit of a given function exists?
Jun 13, 2017
In general you know that a given function has a limit only by determining its limit.
There are however some theorems ensuring a function has a limit under certain hypotheses, for example:
1) A monotonous function in an interval always has a limit
2) if $f \left(x\right) \le g \left(x\right) \le h \left(x\right)$ and ${\lim}_{x \to {x}_{0}} f \left(x\right) = {\lim}_{x \to {x}_{0}} h \left(x\right) = L$ then ${\lim}_{x \to {x}_{0}} g \left(x\right) = L$
3) If $f \left(x\right) \ge g \left(x\right)$ and ${\lim}_{x \to {x}_{0}} g \left(x\right) = \infty$ then ${\lim}_{x \to {x}_{0}} f \left(x\right) = \infty$
4) If ${\lim}_{x \to {x}_{0}} g \left(x\right) = 0$ and $f \left(x\right)$ is bounded, then ${\lim}_{x \to {x}_{0}} f \left(x\right) g \left(x\right) = 0$ | 308 | 847 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "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} | 4.125 | 4 | CC-MAIN-2023-40 | latest | en | 0.67504 |
https://cs.stackexchange.com/questions/49915/how-to-show-all-possible-implied-parenthesis | 1,713,313,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817112.71/warc/CC-MAIN-20240416222403-20240417012403-00668.warc.gz | 169,572,593 | 40,202 | # How to show all possible implied parenthesis?
Can I use recursion to find out the possible parenthesis we can add to this expression: 2*3-4*5 ?
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
I am not able to find out the right states for the recursion to proceed. As at any point we wouldn't know if it would have two open parenthesis or two closed parenthesis. I am just looking for ideas.
• Hint: make a case distinction on where to put the left-most opening parenthesis and its matching partner. Nov 24, 2015 at 8:36
• I don't understand your question. Recursion is just a programming technique so the answer to "Can I use recursion to do X" is almost always "Yes, as long as X is possible." I don't know what you mean by "states for the recursion to proceed" and your point about not knowing whether you'd have a specific number of open or closed parens at a particular point makes it look like you already have a partial solution in mind and you're asking for help completing it -- except we don't know what that partial solution is. Nov 24, 2015 at 9:13 | 302 | 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} | 3.6875 | 4 | CC-MAIN-2024-18 | latest | en | 0.945765 |
https://www.authorstream.com/Presentation/aasthayashu-1082652-triangles/ | 1,621,004,132,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989526.42/warc/CC-MAIN-20210514121902-20210514151902-00241.warc.gz | 670,556,745 | 27,995 | # triangles
Views:
Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Slide 1:
INTRODUCTION NAME = AASTHA SHARMA CLASS = VIII - C SUBJECT = MATHS SCHOOL = D.P.S
### Slide 2:
USES OF TRIANGLES IN DAY TO DAY LIFE
### Slide 3:
A Triangle is formed by three lines segments obtained by joining three pairs of points taken from a set of three non collinear points in the plane. In fig 1.1 , three non collinear points A, B, C have been joined and the figure ABC, enclosed by three line segments , AB,BC, and CA is called a triangle. The symbol ▲ (delta) is used to denote a triangle. TRIANGLES A C B Fig. 1.1
### Slide 4:
The three sides AB, BC, AC and three angles angle A, angle B, angle C are called six elements of a triangle. (Fig 2.1) The three line segments are called the sides of a triangle. The three given points are called the vertices of a triangle. The angle made by the line segment at the vertices are called the angles of a triangle. If the sides of a ▲ABC are extended in order, then the angle between the extended and the adjoining side is called the exterior angles of the triangle. In Fig. 2.1 1,2,3 are called the exterior angles and 4,5,6 are called the interior angles 1 4 5 2 6 3 A B C Fig. 2.2 ELEMENTS OF A TRIANGLE A C B Fig. 2.1
### Slide 5:
CLASSIFICATION ON THE BASIS OF SIDES SCALENE TRIANGLE – If all the sides of triangle are unequal then it is a scalene tirangle. ISOSCELES TRIANGLE – If any two sides of a triangle are equal then it is a isosceles triangle . EQUILATERAL TRIANGLE- If all the sides of a triangle are equal then it is an equilateral triangle. SCALENE TRIANGLE ISOSCELES TRIANGLE EQUILATERAL TRIANGLE
### Slide 6:
CLASSIFICATION ON THE BASIS OF ANGLES ACUTE ANGLED TRIANGLE – If all the three angles of a triangle are less than 90 0 then it is an acute angled triangle. RIGHT ANGLED TRIANGLE- If one angle of a triangle is equal to 90 0 then it is a right angled triangle. OBTUSE ANGLED TRIANGLE- If one angle of a triangle is greater then 90 0 then it is a obtuse angled triangle. ACUTE ANGLED TRIANGLE RIGHT ANGLED TRIANGLE OBTUSE ANGLED TRIANGLE
### Slide 7:
CONGRUENCE OF TRIANGLES Two triangles are congruent if three sides and three angles of one triangle are equal to the corresponding sides and angles of other triangle. The congruence of two triangles follows immediately from the congruence of three lines segments and three angles.
### Slide 8:
SIDE – ANGLE – SIDE (SAS) CONDITION Two triangles are congruent if any two sides and the includes angle of one triangle is equal to the two sides and the included angle of other triangle. EXAMPLE :- (in fig 1.3) GIVEN: AB=DE, BC=EF , B= E SOLUTION: IF AB=DE, BC=EF , B= E then by SAS Rule ▲ABS = ▲DEF 4 cm 4 cm 60 0 60 0 A B C D F E Fig. 1.3 CONGURENCE CONDITIONS
### Slide 9:
ANGLE – SIDE – ANGLE (ASA) CONDITION Two triangles are congruent if any two angles and the included side of one triangle is equal to the two angles and the included side of the other triangle. EXAMPLE : (in fig. 1.4) GIVEN: ABC= DEF, ACB= DFE, BC = EF TO PROVE : ▲ABC = ▲DEF ABC = DEF, (GIVEN) ACB = DFE, (GIVEN) BS = EF (GIVEN) ▲ABC = ▲DEF (BY ASA RULE) A B C D E F Fig. 1.4
### Slide 10:
3. ANGLE – ANGLE – SIDE (AAS) CONDITION Two triangles are congruent if two angles and a side of one triangle is equal to the two angles and one a side of the other. EXAMPLE: (in fig. 1.5) GIVEN: IN ▲ ABC & ▲ DEF B = E A= D BC = EF TO PROVE :▲ABC = ▲DEF B = E A = D BC = EF ▲ABC = ▲DEF (BY AAS RULE) D E F A B C Fig. 1.5
### Slide 11:
4. SIDE – SIDE – SIDE (SSS) CONDITION Two triangles are congruent if all the three sides of one triangle are equal to the three sides of other triangle. Example: (in fig. 1.6) Given: IN ▲ ABC & ▲DEF AB = DE , BC = EF , AC = DF TO PROVE : ▲ABC = ▲DEF AB = DE (GIVEN ) BC = EF (GIVEN ) AC = DF (GIVEN ) ▲ABC = ▲DEF (BY SSS RULE) D E F A B C Fig. 1.6
### Slide 12:
5. RIGHT – HYPOTENUSE – SIDE (RHS) CONDITION Two triangles are congruent if the hypotenuse and the side of one triangle are equal to the hypotenuse and the side of other triangle. EXAMPLE : (in fig 1.7) GIVEN: IN ▲ ABC & ▲DEF B = E = 90 0 , AC = DF , AB = DE TO PROVE : ▲ABC = ▲DEF B = E = 90 0 (GIVEN) AC = DF (GIVEN) AB = DE (GIVEN) ▲ABC = ▲DEF (BY RHS RULE) D E F A B C 90 0 90 0 Fig. 1.7
### Slide 13:
SOME PROPERTIES OF TRIANGLE The angles opposite to equal sides are always equal. Example: (in fig 1.8) Given: ▲ABC is an isosceles triangle in which AB = AC TO PROVE: B = C CONSTRUCTION : Draw AD bisector of BAC which meets BC at D PROOF: IN ▲ABC & ▲ACD AB = AD (GIVEN) BAD = CAD (GIVEN) AD = AD (COMMON) ▲ABD = ▲ ACD (BY SAS RULE) B = C (BY CPCT) A B D C Fig. 1.8
### Slide 14:
2. The sides opposite to equal angles of a triangle are always equal. Example : (in fig. 1.9) Given : ▲ ABC is an isosceles triangle in which B = C TO PROVE: AB = AC CONSTRUCTION : Draw AD the bisector of BAC which meets BC at D Proof : IN ▲ ABD & ▲ ACD B = C (GIVEN) AD = AD (GIVEN) BAD = CAD (GIVEN) ▲ ABD = ▲ ACD (BY ASA RULE) AB = AC (BY CPCT) A B D C Fig. 1.9
### Slide 15:
4. In a triangle the greater angle has a large side opposite to it Example: (in fig. 2.2) Given: IN ▲ ABC B > C TO PROVE : AC > AB PROOF : We have the three possibility for sides AB and AC of ▲ABC AC = AB If AC = AB then opposite angles of the equal sides are equal than B = C AC ≠ AB (ii) If AC < AB We know that larger side has greater angles opposite to it. AC < AB , C > B AC is not greater then AB If AC > AB We have left only this possibility that AC > AB A C B Fig. 2.2
### Slide 16:
5. The sum of any two angles is greater than its third side Example (in fig. 2.3) TO PROVE : AB + BC > AC BC + AC > AB AC + AB > BC CONSTRUCTION: Produce BA to D such that AD + AC . Proof: AD = AC (GIVEN) ACD = ADC (Angles opposite to equal sides are equal ) ACD = ADC --- (1) BCD > ACD ----(2) From (1) & (2) BCD > ADC = BDC BD > AC (Greater angles have larger opposite sides ) BA + AD > BC ( BD = BA + AD) BA + AC > BC (By construction) AB + BC > AC BC + AC >AB A C B D Fig. 2.3
### Slide 17:
SOME OTHER APPLICATIONS OF ASA AND AAS CONGRUENCE CRITERIA If the altitude from one vertex of a triangle bisects the opposite side, then the triangle is isosceles triangle. Example : (in fig.2.5) Given : A ▲ABC such that the altitude AD from A on the opposite side BC bisects BC i. e. BD = DC To prove : AB = AC SOLUTION : IN ▲ ADB & ▲ADC BD = DC ADB = ADC = 900 AD = AD (COMMON ) ▲ADB = ▲ ADC (BY SAS RULE ) AB = AC (BY CPCT) A C D B Fig. 2.5
### Slide 18:
THEOREM 2. In a isosceles triangle altitude from the vertex bisects the base . EXAMPLE: (in fig. 2.6) GIVEN: An isosceles triangle AB = AC To prove : D bisects BC i.e. BD = DC Proof: IN ▲ ADB & ▲ADC ADB = ADC AD = AD B = C ( AB = AC ; B = C) ▲ADB = ▲ ADC BD = DC (BY CPCT) A C D B Fig. 2.6
### Slide 19:
THEOREM 3. If the bisector of the vertical angle of a triangle bisects the base of the triangle, then the triangle is isosceles. EXAMPLE: (in fig. 2.7) GIVEN: A ▲ABC in which AD bisects A meeting BC in D such that BD = DC, AD = DE To prove : ▲ABC is isosceles triangle . Proof: In ▲ ADB & ▲ EDC BD = DC AD = DE ADB = EDC ▲ADB = ▲EDC AB = EC BAD = CED (BY CPCT) BAD = CAD (GIVEN) CAD = CED AC = EC (SIDES OPPOSITE TO EQUAL ANGLES ARE EQUAL) AC = AB , HENCE ▲ABC IS AN ISOSCELES TRIANGLE. E D C B A Fig. 2.7 | 2,292 | 7,377 | {"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} | 4.71875 | 5 | CC-MAIN-2021-21 | latest | en | 0.834971 |
https://answers.yahoo.com/question/index?qid=20130412171700AA06NdK | 1,611,837,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00220.warc.gz | 210,129,691 | 24,857 | Allie asked in Science & MathematicsMathematics · 8 years ago
# Write an equation for the inequality..?
3. Let y represent a number and x represent another
number. Use the sentence “A number is greater than
or equal to three times another number” to write an inequality.
my answer: x = 3y
4. Write an equation for the inequality.
This is the equation of the boundary line
????
what does that mean and what do i do
Relevance
• Anonymous
8 years ago
y≥3x
Remember, it has to be an inequality so an = sign will not work.
If you are graphing the inequality:
The line will be solid since it is ≥ and shade above the line.
Just graph like a regular y=mx+b slope.
So draw a point at (0,0) and then go up 3 and right 1, to (1,3) then go up 3 and right 1. Keep repeating this.
• Mike G
Lv 7
8 years ago
y>=3x | 223 | 817 | {"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-2021-04 | latest | en | 0.919472 |
https://www.coolstuffshub.com/length/convert/yards-to-leagues/ | 1,695,299,913,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506027.39/warc/CC-MAIN-20230921105806-20230921135806-00826.warc.gz | 824,109,791 | 16,044 | # Convert yards to leagues (yd to lea conversion)
## 1 yard is equal to how many leagues?
1 yard is equal to 0.000189394 leagues.
• 1 yard = 0.000189394 leagues
• 2 yards = 0.000378788 leagues
• 3 yards = 0.000568182 leagues
• 4 yards = 0.000757576 leagues
• 5 yards = 0.00094697 leagues
• 10 yards = 0.00189394 leagues
• 100 yards = 0.0189394 leagues
## How to convert yards to leagues?
To convert yards to leagues, multiply the value in yards by 0.000189394.
### What is the formula to convert yards to leagues?
The conversion formula to convert yards to leagues is :
leagues = yards × 0.000189394
### What is the conversion factor to convert yards to leagues?
The conversion factor to convert yards to leagues is 0.000189394
### Examples to convert yd to lea
#### Example 1
Convert 50 yd to lea.
Solution:
Converting from yards to leagues is very easy.
We know that 1 yd = 0.000189394 lea.
So, to convert 50 yd to lea, multiply 50 yd by 0.000189394 lea.
50 yd = 50 × 0.000189394 lea
50 yd = 0.0094697 lea
Therefore, 50 yards converted to leagues is equal to 0.0094697 lea.
#### Example 2
Convert 125 yd to lea.
Solution:
1 yd = 0.000189394 lea
So, 125 yd = 125 × 0.000189394 lea
125 yd = 0.02367425 lea
Therefore, 125 yd converted to lea is equal to 0.02367425 lea.
If you don't want to do the calculation from yards to leagues manually, you can simply use our yards to leagues calculator.
### How to use the yards to leagues converter?
To use the yards to leagues converter, follow these steps:
1. Enter the value in yards that you want to convert into the "yards" text box.
2. Click on the "Convert" button.
3. The conversion result in leagues will be displayed in the "leagues" text box.
4. The step by step conversion process will be displayed in the "Conversion steps" text box.
5. To copy the yards to leagues conversion steps, click on the "Copy" button.
6. To report an incorrect conversion, click on the "Report incorrect conversion" button.
7. To reset the converter, click on the "Reset" button.
8. To inverse the conversion and convert Leagues to Yards, click on the "Swap" button.
9. To convert between other units of length, you can choose the units from the "From" and "To" drop-down menus.
## Yards to leagues conversion table
The yards to leagues conversion chart below shows a list of various yards values converted to leagues
Yards (yd) Leagues (lea)
0.001 yd 1.89394 × 10-7 lea
0.01 yd 1.89394 × 10-6 lea
0.1 yd 1.89394 × 10-5 lea
1 yd 0.000189394 lea
2 yd 0.000378788 lea
3 yd 0.000568182 lea
4 yd 0.000757576 lea
5 yd 0.00094697 lea
6 yd 0.001136364 lea
7 yd 0.001325758 lea
8 yd 0.001515152 lea
9 yd 0.001704546 lea
10 yd 0.00189394 lea
20 yd 0.00378788 lea
30 yd 0.00568182 lea
40 yd 0.00757576 lea
50 yd 0.0094697 lea
60 yd 0.01136364 lea
70 yd 0.01325758 lea
80 yd 0.01515152 lea
90 yd 0.01704546 lea
100 yd 0.0189394 lea
## Yards
### 1 yard equivalents in other length units
• 1 yard = 9144000000 angstroms
• 1 yard = 91.44 centimeters
• 1 yard = 0.0454545 chains
• 1 yard = 9.14 decimeters
• 1 yard = 3 feet
• 1 yard = 0.00454545 furlongs
• 1 yard = 36 inches
• 1 yard = 0.0009144 kilometers
• 1 yard = 0.000189394 leagues
• 1 yard = 9.66522 × 10-17 lightyears
• 1 yard = 0.9144 meters
• 1 yard = 36000000 microinches
• 1 yard = 914400 micrometers
• 1 yard = 914400 microns
• 1 yard = 0.000568182 miles
• 1 yard = 914.4 millimeters
• 1 yard = 914400000 nanometers
• 1 yard = 0.000493737 nautical miles
• 1 yard = 2.96337 × 10-17 parsecs
## Leagues
### Leagues to yards conversion (lea to yd)
• 1 league = 5280 yards
• 2 leagues = 10560 yards
• 3 leagues = 15840 yards
• 4 leagues = 21120 yards
• 5 leagues = 26400 yards
• 10 leagues = 52800 yards
• 100 leagues = 528000 yards
### 1 league equivalents in other length units
• 1 league = 48280420000000 angstroms
• 1 league = 482803.2 centimeters
• 1 league = 240 chains
• 1 league = 48280.3 decimeters
• 1 league = 15840 feet
• 1 league = 24 furlongs
• 1 league = 190080 inches
• 1 league = 4.82803 kilometers
• 1 league = 5.10323 × 10-13 lightyears
• 1 league = 4828.03 meters
• 1 league = 190080000000 microinches
• 1 league = 4828032000 micrometers
• 1 league = 4828032000 microns
• 1 league = 3 miles
• 1 league = 4828032 millimeters
• 1 league = 4828032000000 nanometers
• 1 league = 2.60693 nautical miles
• 1 league = 1.56466 × 10-13 parsecs
• 1 league = 5280 yards | 1,560 | 4,388 | {"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} | 3.65625 | 4 | CC-MAIN-2023-40 | latest | en | 0.900647 |
http://www.jiskha.com/members/profile/posts.cgi?name=Anissa | 1,490,837,430,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191444.45/warc/CC-MAIN-20170322212951-00015-ip-10-233-31-227.ec2.internal.warc.gz | 547,824,336 | 4,544 | Wednesday
March 29, 2017
# Posts by Anissa
Total # Posts: 21
Math
Find three consecutive odd integers such that four times the middle integer is two more than the sum of the first and third.
March 6, 2017
math
A real estate agent sold two homes and received commissions totaling \$6000. The agent's commission on one home was one and one-half times the commission on the second home. If p represents the smaller commission, write an equation to determine the agent's commission on...
March 6, 2017
math
Find three consecutive odd integers such that four times the middle integer is two more than the sum of the first and third. I think I'm doing it wrong. 4((x)+(x+2)=x+(x+4)+2 4(2x+4)=2+3x+6 8x+16=3x+8 5x=8 ? Please help
March 6, 2017
physics
an 88.kg fireman slides 5.9 meters down a fire pole. He holds the pole, which exerts a 520N steady resistive force on the fireman. At the bottom he slows down to a stop in 0.43m by bending his knees. determine the acceleration while stopping and the time it takes for the ...
January 20, 2017
College Chemistry
Thank you so much :)
May 6, 2015
College Chemistry
Arrange the following compounds in the order of increasing Ksp: AgCl, AgBr, AgI. Explain your reasoning. I got AgI<AgBr<AgCl but I looked at the actual Ksp. I do not know how else to explain it besides looking at the Ksp. Like is there any way I can just know which one ...
May 6, 2015
science
a plane is flying due east in still air at 395km/h suddenly the plane is hit by wind blowing at 55km/h toward the west. what is the resultant of the plane?
June 2, 2014
fine the value of 7/10 +11/100
April 30, 2013
math133
1/4=3-2x-1/x+2
February 27, 2012
estimate to the underline place. estimate the product. 23*9 =
January 29, 2012
math
what is the -32 t? I think I undersaand the rest.
October 10, 2011
algebra
I don't understand the y' stuff. I'm only in algebra. Is there another way?
October 10, 2011
algebra
what is tttt?
October 10, 2011
algebra
Find the max or min value for y given the restictions on x. y= x^3 -x^2 -6x a) 0<x<3 min b) -2<x<0 max
October 10, 2011
algebra
Amy, beth and Carla are solving a quadratic equation. Amy misreads the constant term and finds the roots 8 and 2. Beth misreads the coefficient of x and finds roots -9 and -1. Carla solves it. What are Carla's solutions?
October 10, 2011
Econ
You are thinking about investing \$100,000 in a new product. Sales expections are \$95,000. If R & D will cost another \$3,000, should you invest in the product? Explain.
September 30, 2011
Teaching Special Education
This is what I need. (1) Create a PowerPoint presentation (PPT) that defines and describes self-determination. (a) Include the process of teaching self-determination and the outcomes (15 slides required). (b) Cite references in-text (three references required). (c) If possible...
March 30, 2011
Teaching Special Education
(1) Create a PowerPoint presentation (PPT) that defines and describes self-determination. (a) Include the process of teaching self-determination and the outcomes (15 slides required). (b) Cite references in-text (three references required). (c) If possible, present the PPT to ...
March 30, 2011
english
I have written an essay for my class,and I wanted some one to look it over for me and make any corrections. Thank you:) Mr. Tom Brooks’ narrative tells of a fatal fire in one of New York’s most important shirt manufactures. Yes, fires are very common throughout ...
January 22, 2010
child care
what are 10/ some questions to ask your parents about life with kids for example; I have these ones do you like life now with kids what made you want kids were you ready for kids anymore suggestions? thank you(:
January 15, 2010
statics
4. a 5.b 6.b 7.d
June 7, 2008
1. Pages:
2. 1
Post a New Question | 1,073 | 3,791 | {"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} | 3.015625 | 3 | CC-MAIN-2017-13 | longest | en | 0.945004 |
https://physics.stackexchange.com/questions/191538/what-would-a-closed-timelike-curve-look-like | 1,618,414,150,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077818.23/warc/CC-MAIN-20210414125133-20210414155133-00358.warc.gz | 561,195,539 | 41,883 | # What would a closed timelike curve look like?
What exactly are closed timelike curves. In a metric in which they would exist, what would they look like. What would it be like travelling through them? It obviously wouldn't look like a door. Would it be a region of space that if you wonder into, it can happen that see your past self?
• Physics can't answer what things would look like that don't exist. Only art can perform that trick. Did you try science fiction novels, yet? – CuriousOne Jun 27 '15 at 4:32
• @CuriousOne Infinite Planes of uniform density don't exist, but we have a reasonable idea of what it would look like if it did. – PyRulez Jun 27 '15 at 4:34
• Your imagination is playing tricks on you. You have no idea what such a non-existing object would look like. Moreover, you are making a beginner's mistake in physics, which, unfortunately, is being perpetuated by many poor physics teachers. Instead of explaining to you how to do a properly simplified approximating calculation on a finite, physically existing object, they throw (without any proof, of course!) a physically impossible one at you that is supposed to have the same solutions. Often these systems are not even close to equivalent but students internalize the unphysical one, anyway. – CuriousOne Jun 27 '15 at 4:43
• @CuriousOne There are simulations of people falling into black holes, although people falling into black holes don't exist. For many phenomena that don't exist, we can still simulate what light and such it produces. – PyRulez Jun 27 '15 at 4:52
• @CuriousOne Thought experiments are useful to physics. – PyRulez Jun 27 '15 at 4:59
To know what a closed timelike curve looks like, you just do like every spacetime metric. You compute geodesics and field equations and all of that. Unfortunately, things start getting complicated.
Closed timelike curves have a lot of weird behaviours, especially when it comes to matter fields upon them. They may not have a properly defined Cauchy problem, that is, knowing the field at a point in time may not help you to know it at all points in time. Physically speaking, this corresponds to when geodesics form loops, or in the case of naked singularities, come out of nowhere. The particle creates itself, in a way. There may therefore be an arbitrary amount of particles popping out of nowhere, making the analysis a bit difficult. The problem seems to be mostly worse with interacting fields, although some free fields also suffer from the problem.
Another problem is joining different regions of spacetime. It's always possible to solve fields locally, since spacetime always looks locally like Minkowski space, but if there are CTCs present, extending those solution to the entire spacetime becomes difficult, since it is assumed that they have to be consistent. For instance, consider the 2 dimensional torus spacetime, which is Minkowski spacetime with the following identifications :
$(x,t) = (x,t+T) = (x+d,t)$
It's not too difficult to show that the geodesic with 4-velocity $(1,\vec{0})$ is consistent, it just loops once around the torus. A geodesic with a slightly askew velocity $(1, \vec{v})$ will loop back on itself eventually if the velocity is a rational value of the period of time and space. Otherwise, it will go on looping forever. That is another common problem of CTCs : EXPLODING. The same particle geodesic exists an infinite number of time at the same moment. This is usually a bad sign for the existence of CTCs because then you cannot pretend that the matter field is only a small perturbation on the spacetime.
Misner space is a CTC spacetime that starts out like a reasonable spacetime, until t = 0 where CTCs suddenly appear. If you check out the stress energy tensor of a free scalar field upon it, it will explode at t = 0, unless some conditions on the field are met, because the field get infinitely blueshifted. The same applies to the Gott spacetime.
A third effect is that when quantum effects are included, the energy of the vacuum might diverge, due to geodesics being able to wrap around an arbitrary amount of time through a region.
All in all, that is a lot of effects where fields explode without provocation. I'm not exactly sure what things would look like inside a CTC, but odds are good they would not be very good. It's a common theme in CTC spacetimes : Particles radiate highly blueshifted particles when falling towards the CTC region of the Kerr black hole, wormholes collapse from vacuum radiation, Misner space explodes when CTCs try to form and so on. To answer your question, I guess that if they survive, they might be VERY BRIGHT.
Here's a few papers on the various topics :
The Cauchy problem in spacetimes with CTCs and various field solutions :
You can of course use much simpler models to check out geometrical optics. In which case it is usually done simply using point travelling along geodesics. Here's a few papers on it as well :
That's about all I can think about for general optics and fields (including EM fields) phenomenon involving CTCs. more specific CTC spacetimes of course have their own optical effects, such as redshifting, blueshifting, gravitational lensing and such. Wormholes in particular have had some work done on them, although not in their CTC configuration. Here's a sample :
A closed timelike curve wouldn't actually "look" like anything because it's an abstract thing. You can't actually see any lightcones or worldlines. A metric is an abstract thing too, to do with your measurements of distance and time, typically made using the motion of light. And the crucial point is this: you don't travel along your worldline. You move through space whilst light moves, along with pendulums and cogs and piezo-electric vibrating crystals. Your worldline is a plot of this in a static 3+1 dimensional "block universe" called spacetime. For an analogy, imagine I throw a red ball across the room, and you film it with your old-style cine-camera. Then you develop the film, cut it up into individual frames, and form them into a block. There's a red streak in the block. That's like the ball's world-line in the block universe. But note that there is no motion in there. The ball is not moving up through the block, and in similar vein you aren't moving up or along or around your worldline. You move through space, not through spacetime.
As for what a closed timelike curve would be like, some say it would constitute time travel, but it isn't true. Others would say it would be like Groundhog Day, but that isn't true either. Because if your closed timelike curve was 24 hours long, it would be more like Mayfly Day. Your life lasts for 24 hours. And it is causeless. You are born from an egg, you live for a day, you lay the egg, you die. You live and die, but you don't live die repeat. Because you don't travel along your worldline. Hence a closed timelike curve does not offer any possibility of time travel.
Don't think this is just something I've made up. Check out A World Without Time: The Forgotten Legacy of Gödel and Einstein. See this page in it, where author Palle Yourgrau says Wheeler conflated a circle with a cycle, precisely missing the force of Gödel's conclusion:
He's right. You don't move round that closed timelike curve. And there is no way you can move such that everything else in this universe not only moved back to where it was, but never moved at all. Time travel is a fantasy I'm afraid. And so are closed timelike curves. | 1,716 | 7,498 | {"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} | 2.609375 | 3 | CC-MAIN-2021-17 | latest | en | 0.970662 |
https://proofwiki.org/wiki/Fibonacci_Number_equal_to_Sum_of_Sequence_of_Cubes | 1,597,497,833,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740848.39/warc/CC-MAIN-20200815124541-20200815154541-00478.warc.gz | 461,490,867 | 10,232 | # Fibonacci Number equal to Sum of Sequence of Cubes
## Theorem
The following Fibonacci number can be expressed as the sum of a sequence of cubes:
$F_{18} = 2584 = 7^3 + 8^3 + 9^3 + 10^3$
## Proof
$\displaystyle 2584$ $=$ $\displaystyle 343 + 512 + 729 + 1000$ $\displaystyle$ $=$ $\displaystyle 7^3 + 8^3 + 9^3 + 10^3$
$\blacksquare$
## Historical Note
In his Curious and Interesting Numbers, 2nd ed. of $1997$, David Wells attributes this result to Michal Stajsczak, but gives no context. | 166 | 499 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2020-34 | latest | en | 0.654094 |
http://gmatclub.com/forum/m14-29-soccer-game-probability-68335-20.html | 1,484,982,010,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280929.91/warc/CC-MAIN-20170116095120-00527-ip-10-171-10-70.ec2.internal.warc.gz | 116,243,722 | 46,265 | M14 #29 - soccer game probability : Retired Discussions [Locked] - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 20 Jan 2017, 23:00
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# M14 #29 - soccer game probability
Author Message
Intern
Joined: 30 May 2011
Posts: 18
Followers: 0
Kudos [?]: 7 [0], given: 4
Re: M14 #29 - soccer game probability [#permalink]
### Show Tags
21 Jun 2013, 06:52
Winning Team scored 3/5, winning team did not score first >> 1 - 3/5 = 2/5?
Current Student
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 62
Kudos [?]: 594 [0], given: 355
Re: M14 #29 - soccer game probability [#permalink]
### Show Tags
13 Feb 2014, 06:33
Bunuel wrote:
MBACHANGE wrote:
Thanks bunel
what if question was loser team scored second goal instead of first goal. then also it will 2/5 rgt? and if winning team scored first, second or third goal will be 3/5
Regards
Regards
Absolutely, the probability that the loser team scored the first, second, ..., fifth goal is the same and equals to 2/5 and the probability that the winner team scored the first, second, ..., fifth goal is the same and equals to 3/5.
Check similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-105990.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
Hope it helps.
Guys the only difference here between 2/5=4/10 and 1/2 is that since we already know the score the probabilities change. If we had to decide at the beginning of the match then yes probability is 1/2 but after with new info it is 2/5
Hope this clarifies
Cheers
J
Intern
Joined: 21 Oct 2012
Posts: 39
Location: United States
Concentration: Marketing, Operations
GMAT 1: 650 Q44 V35
GMAT 2: 600 Q47 V26
GMAT 3: 660 Q43 V38
GPA: 3.6
WE: Information Technology (Computer Software)
Followers: 0
Kudos [?]: 16 [0], given: 19
### Show Tags
10 Jun 2014, 10:48
durgesh79 wrote:
sset009 wrote:
If a certain soccer game ended 3:2, what is the probability that the side that lost scored first?
(Assume that all scoring scenarios are equiprobable)
(C) 2008 GMAT Club - m14#29
* $$\frac{1}{4}$$
* $$\frac{3}{10}$$
* $$\frac{2}{5}$$
* $$\frac{5}{12}$$
* $$\frac{1}{2}$$
2 teams A and B ... A won and B lost..
the problems is similar where we have 5 slots, A can take 3 and B can take 2 postions
all possible combinations = 5!/(2! * 3!) = 10
fav cambinations we know that first slot is B.. so rest slots are 4 and A can take 3 postions and B can take 1
= 4! / (3! * 1!) = 4
Probability = 4/10 = 2/5 option C
But when you say that fav cambinations we know that first slot is B.. so rest slots are 4 and A can take 3 postions and B can take 1
= 4! / (3! * 1!) = 4
arent you assuming that the two goals scored by b are the same?
if they are not same then it should be 2 * 4! / (3!*1!) right?
Re: soccer game probability [#permalink] 10 Jun 2014, 10:48
Go to page Previous 1 2 [ 23 posts ]
Similar topics Replies Last post
Similar
Topics:
19 M14#10 19 18 Mar 2009, 12:41
20 M14 #19 19 02 Feb 2009, 21:06
5 M14 #18 19 02 Feb 2009, 20:55
6 M14 #27 19 13 Nov 2008, 16:55
18 M14 #13 25 11 Nov 2008, 20:40
Display posts from previous: Sort by
# M14 #29 - soccer game probability
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,305 | 4,207 | {"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} | 4.09375 | 4 | CC-MAIN-2017-04 | latest | en | 0.871976 |
https://www.physicsforums.com/threads/integration-and-inverse-trig-functions.147169/ | 1,713,839,062,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00375.warc.gz | 831,102,776 | 16,477 | # Integration and inverse trig functions
• mavsqueen06
In summary, the student is struggling with integrating trig functions and specifically with the integral of sinh(x)/1+cosh(x). They are unsure of how to start and are wondering if an identity can be used. Another student suggests using the identity y = cosh(x), dy = sinh(x)dx to simplify the problem. Another student points out that the original equality given by the first student is incorrect, but it can be simplified to tanh(x/2). Finally, another student suggests using the equation \int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C} which can be derived from the derivative of the denominator
mavsqueen06
## Homework Statement
getting confused with integration of trig functions.
I am finding the integral of sinhx/1+coshx and I'm not sure how to start. should i use an identity?
help is appreciated!
## Homework Equations
possibly an identity of some sort?
## The Attempt at a Solution
?
Well the obvious chamge would be;
$$\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)$$
Which makes it a little easier.
EDIT: The functions are not inverse trig functions they are hyperbolic functions incidentally.
Last edited:
Or, change variables: y = cosh(x), dy = sinh(x)dx.
Kurdt said:
Well the obvious chamge would be;
$$\frac{sinh(x)}{1+cosh(x)}=sinh(x) + tanh(x)$$
Which makes it a little easier.
That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.
Do you see why
$$\int \frac{\sinh t}{1+\cosh t}{}dt =\ln\left(1+\cosh t\right) +\mathcal{C}$$
The derivative of the denominator is the numerator.
Daniel.
d_leet said:
That equality isn't correct, although I'm pretty sure that that fraction works out to tanh(x/2) which might help a bit.
Thats a rather embarrassing schoolboy error that I wish I could blame on how late I was up last night, but its far too simple for that.
## 1. What is integration?
Integration is a mathematical process that involves finding the area under a curve on a graph. It is the reverse process of differentiation and is used to solve problems in many fields such as physics, engineering, and economics.
## 2. What are inverse trig functions?
Inverse trig functions are mathematical functions that undo the effects of their corresponding trigonometric functions. They are used to find the angle or side length of a right triangle when given other known values.
## 3. Why do we use integration in calculus?
Integration is an important tool in calculus because it allows us to solve problems involving rates of change, accumulation, and optimization. It is also used to find the area and volume of irregular shapes and objects.
## 4. What is the difference between definite and indefinite integrals?
A definite integral has specific limits of integration, meaning that the integration is performed over a specific interval. An indefinite integral, on the other hand, does not have limits and represents a family of functions that differ only by a constant.
## 5. How do we integrate inverse trig functions?
The process of integrating inverse trig functions involves using trigonometric identities and substitution techniques. It is important to also have a good understanding of derivatives and the fundamental theorem of calculus in order to successfully integrate these functions.
### Similar threads
• Calculus and Beyond Homework Help
Replies
18
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
359
• Calculus and Beyond Homework Help
Replies
17
Views
881
• Calculus and Beyond Homework Help
Replies
10
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
747
• Calculus and Beyond Homework Help
Replies
4
Views
997 | 954 | 3,963 | {"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} | 3.921875 | 4 | CC-MAIN-2024-18 | latest | en | 0.909349 |
https://webeduclick.com/petersons-algorithm-for-mutual-exclusion/ | 1,709,484,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00890.warc.gz | 615,330,750 | 54,247 | Peterson’s Algorithm for Mutual Exclusion
Peterson’s Algorithm:
Peterson’s Algorithm was developed by G.L. Peterson in 1981 for enforcing mutual exclusion. It is a simple algorithm for enforcing two processes of mutual exclusion with busy waiting.
Algorithm:
```Program Petersons;
var favored process:(first,second);
p1 wants_to_enter, p2 wants_to_enter : Boolean;
procedure processone;
begin
while true do
begin
p1 wants_to_enter:=true;
favored process:=second;
while p2 wants_to_enter And favored process = second do
critical section one;
p1 wants_to_enter:=false;
otherwiseone
end;
procedure processtwo;
begin
while true do
begin
p2 wants_to_enter:=true;
favored process:=first;
while pi wants_to_enter And favored process = first do;
critical section two;
p2 wants_to_enter:=false;
otherwisetwo
end;
begin
p1 wants_to_enter=false;
p2 wants_to_enter=false;
favored process:=first;
parbegin
processone;
processtwo;
parend
end;``` | 233 | 935 | {"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-2024-10 | longest | en | 0.593214 |
https://number.academy/55564 | 1,722,796,148,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640408316.13/warc/CC-MAIN-20240804164455-20240804194455-00299.warc.gz | 352,522,468 | 11,318 | # Number 55564 facts
The even number 55,564 is spelled 🔊, and written in words: fifty-five thousand, five hundred and sixty-four. The ordinal number 55564th is said 🔊 and written as: fifty-five thousand, five hundred and sixty-fourth. The meaning of the number 55564 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 55564. What is 55564 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 55564.
## Interesting facts about the number 55564
### Asteroids
• (55564) 2002 AQ188 is asteroid number 55564. It was discovered by NEAT, Near Earth Asteroid Tracking from Mount Palomar Observatory on 1/10/2002.
## What is 55,564 in other units
The decimal (Arabic) number 55564 converted to a Roman number is (L)(V)DLXIV. Roman and decimal number conversions.
#### Length conversion
55564 kilometers (km) equals to 34526 miles (mi).
55564 miles (mi) equals to 89422 kilometers (km).
55564 meters (m) equals to 182295 feet (ft).
55564 feet (ft) equals 16937 meters (m).
#### Time conversion
(hours, minutes, seconds, days, weeks)
55564 seconds equals to 15 hours, 26 minutes, 4 seconds
55564 minutes equals to 1 month, 1 week, 3 days, 14 hours, 4 minutes
### Codes and images of the number 55564
Number 55564 morse code: ..... ..... ..... -.... ....-
Sign language for number 55564:
Number 55564 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Share in social networks
### Advanced math operations
#### Is Prime?
The number 55564 is not a prime number. The closest prime numbers are 55547, 55579.
#### Factorization and factors (dividers)
The prime factors of 55564 are 2 * 2 * 29 * 479
The factors of 55564 are 1, 2, 4, 29, 58, 116, 479, 958, 1916, 13891, 27782, 55564.
Total factors 12.
Sum of factors 100800 (45236).
#### Powers
The second power of 555642 is 3.087.358.096.
The third power of 555643 is 171.545.965.246.144.
#### Roots
The square root √55564 is 235,720173.
The cube root of 355564 is 38,159075.
#### Logarithms
The natural logarithm of No. ln 55564 = loge 55564 = 10,925291.
The logarithm to base 10 of No. log10 55564 = 4,744794.
The Napierian logarithm of No. log1/e 55564 = -10,925291.
### Trigonometric functions
The cosine of 55564 is -0,219725.
The sine of 55564 is 0,975562.
The tangent of 55564 is -4,439928.
## Number 55564 in Computer Science
Code typeCode value
55564 Number of bytes54.3KB
Unix timeUnix time 55564 is equal to Thursday Jan. 1, 1970, 3:26:04 p.m. GMT
IPv4, IPv6Number 55564 internet address in dotted format v4 0.0.217.12, v6 ::d90c
55564 Decimal = 1101100100001100 Binary
55564 Decimal = 2211012221 Ternary
55564 Decimal = 154414 Octal
55564 Decimal = D90C Hexadecimal (0xd90c hex)
55564 BASE64NTU1NjQ=
55564 MD59ff673c83cc6230e08904ed4ed21ba24
55564 SHA185ca2e178318deb41a3fdb1d94ecb037a044f6ec
55564 SHA2245424d904eb7a5981f99e5e62a9cb90cfd25431230a5057d43ba4aa38
55564 SHA256a92732699fd7d582d3d8f657273822301783e631335ff57794ef5a1b5dcd4fbf
55564 SHA384bb52be226f6c1439981dec840973ec46922e579d78a93b08f5dce4749dd08114fcb678393f9229e970c56378627ecb9d
More SHA codes related to the number 55564 ...
If you know something interesting about the 55564 number that you did not find on this page, do not hesitate to write us here.
## Numerology 55564
### Character frequency in the number 55564
Character (importance) frequency for numerology.
Character: Frequency: 5 3 6 1 4 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 55564, the numbers 5+5+5+6+4 = 2+5 = 7 are added and the meaning of the number 7 is sought.
## № 55,564 in other languages
How to say or write the number fifty-five thousand, five hundred and sixty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 55.564) cincuenta y cinco mil quinientos sesenta y cuatro German: 🔊 (Nummer 55.564) fünfundfünfzigtausendfünfhundertvierundsechzig French: 🔊 (nombre 55 564) cinquante-cinq mille cinq cent soixante-quatre Portuguese: 🔊 (número 55 564) cinquenta e cinco mil, quinhentos e sessenta e quatro Hindi: 🔊 (संख्या 55 564) पचपन हज़ार, पाँच सौ, चौंसठ Chinese: 🔊 (数 55 564) 五万五千五百六十四 Arabian: 🔊 (عدد 55,564) خمسة و خمسون ألفاً و خمسمائة و أربعة و ستون Czech: 🔊 (číslo 55 564) padesát pět tisíc pětset šedesát čtyři Korean: 🔊 (번호 55,564) 오만 오천오백육십사 Danish: 🔊 (nummer 55 564) femoghalvtredstusinde og femhundrede og fireogtreds Hebrew: (מספר 55,564) חמישים וחמישה אלף חמש מאות שישים וארבע Dutch: 🔊 (nummer 55 564) vijfenvijftigduizendvijfhonderdvierenzestig Japanese: 🔊 (数 55,564) 五万五千五百六十四 Indonesian: 🔊 (jumlah 55.564) lima puluh lima ribu lima ratus enam puluh empat Italian: 🔊 (numero 55 564) cinquantacinquemilacinquecentosessantaquattro Norwegian: 🔊 (nummer 55 564) femtifem tusen fem hundre og sekstifire Polish: 🔊 (liczba 55 564) pięćdziesiąt pięć tysięcy pięćset sześćdziesiąt cztery Russian: 🔊 (номер 55 564) пятьдесят пять тысяч пятьсот шестьдесят четыре Turkish: 🔊 (numara 55,564) ellibeşbinbeşyüzaltmışdört Thai: 🔊 (จำนวน 55 564) ห้าหมื่นห้าพันห้าร้อยหกสิบสี่ Ukrainian: 🔊 (номер 55 564) п'ятдесят п'ять тисяч п'ятсот шістдесят чотири Vietnamese: 🔊 (con số 55.564) năm mươi lăm nghìn năm trăm sáu mươi bốn Other languages ...
## News to email
#### Receive news about "Number 55564" to email
I have read the privacy policy
## Comment
If you know something interesting about the number 55564 or any other natural number (positive integer), please write to us here or on Facebook.
#### Comment (Maximum 2000 characters) *
The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,019 | 6,095 | {"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-2024-33 | latest | en | 0.784584 |
https://www.codesdope.com/c-operators/ | 1,642,926,232,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00711.warc.gz | 745,667,566 | 21,359 | # Operators in C
Operators are symbol which tells the compiler to perform certain operations on variables. For example, (*) is an operator which is used for multiplying two numbers.
There are different types of operators in C. Let's take a look at each type of them with few examples of each.
• Arithmetic Operators
• Relational Operators
• Increment and Decrement Operators
• Logical Operators
• Assignment Operators
## Arithmetic Operations
Arithmetic Operators are the type of operators which take numerical values (either literals or variables) as their operands and return a single numerical value.
Let's assume, 'a' is 8 and 'b' is 4.
Operator Description Example
- Subtracts second operand from the first a-b=4
* Multiplies both operands a*b=32
/ Divides numerator by denominator. a/b=2
% Modulus Operator returns the remainder after an integer division. a%b=0
```#include <stdio.h>
int main()
{
int a = 25;
int b = 8;
printf("sum = %d", (a+b));
printf("\ndifference = %d", (a-b));
printf("\nproduct = %d", (a*b));
printf("\nremainder = %d\n", (a%b));
return 0;
}
```
sum = 33
difference = 17
product = 200
remainder = 1
If we divide two integers, the result will be an integer.
5/2=2 (Not 2.5)
To get 2.5, at least one of the numerator or denominator must have a decimal(float) value.
5.0/2=2.5 or 5/2.0=2.5 or 5.0/2.0=2.5 but 5/2 = 2.
```#include <stdio.h>
int main()
{
int a ;
int b ;
a = 2 ;
b = 9 ;
printf ( "Value of b/a is : %d\n" , b/a ) ;
return 0;
}
```
Value of b/a is : 4
```#include <stdio.h>
int main()
{
float a ;
int b ;
a = 2.0 ;
b = 9 ;
printf ( "Value of b/a is : %f\n" , b/a ) ;
return 0;
}
```
Value of b/a is : 4.500000
As we saw, if 'b' and 'a' are both integers, then the result is 4 (not 4.5) but when one of them is float then the result is 4.500000 (a float).
## Hierarchy Of Operations
Suppose, you have used more than one operator in an expression e.g.- `2*5%6/2`, then the order of execution i.e., which operator will be executed first is decided by the precedence table given below.
In maths, you might have learned about BODMAS rule, but that rule is not applied here. If we have written more than one operation in one line, then which operation should be done first is governed by following rules :-
Expressions inside brackets '()' are evaluated first. After that, this table is followed:
Category with priority Operator Associativity
1st Multiplicative * / % Left to right
2nd Additive + - Left to right
3rd Equality = Right to Left
If two operators are of same priority, then evaluation starts from left or right as stated in the table.
eg.-
k = 8/4+3-5*6+8
Solving, '*' and '/' first from left to right.
k = 2+3-30+8
Now solving, '+' and '-' from left to right
k = -17
Now solving '=' from right to left
If you don't want to remember these rules, then just put the expression you want to execute first in brackets. Eg- If you want to divide the product of (2+2) and 444 by quotient of (999/5) then write the expression as - ((2+2)*444)/(999/5). This will get evaluated as (4*444)/(999/5) and finally get simplified to 1776/199 (as 999/5 is 199 and not 199.8).
## Relational Operators
Relational Operators check the relationship between two operands. If the relationship is true, it returns 1, if the relationship is false, it returns 0.
Following is the list of relational operators in C.
Again assume the value of 'a' to be 8 and that of 'b' to be 4.
Operator Description Example
== Equal to (a == b) is false
!= Not equal to (a != b) is true
> Greater than (a > b) is true
< Less than (a < b) is false
>= Greater than or equal to (a >= b) is true
<= Less than or equal to (a <= b) is false
Let's see an example to see their use.
```#include <stdio.h>
int main()
{
int a = 5, b = 6;
printf("\n%d", a == b);
printf("\n%d", a != b);
printf("\n%d", a > b);
printf("\n%d", a < b);
printf("\n%d", a >= b);
printf("\n%d", a <= b);
return 0;
}
```
0
1
0
1
0
When the expression was true, we got 1 and when false, 0.
## Difference between = and ==
Although = and == seem to be same, but they are quite different from each other. = is assignment operator while == is the equality operator. = is used to assign values whereas == is used for comparing values.
Take two examples.
x = 5;
x == 5;
By writing x = 5, we assign a value of 5 to x, whereas by writing x == 5, we check if the value of x is 5 or not.
## Logical Operators
First, let's learn about AND and OR.
AND and OR are very much similar to English words 'and' and 'or'.
In English,
A and B - Both A and B.
A or B - Either A or B.
In C programming,
A and B - Both A and B.
A or B - Either A or B or both.
So, if we are writing A and B, then the expression is true if both A and B are true. Whereas, if we are writing A or B, then the expression is true if either A or B or both are true.
The symbol for AND is && while that of OR is ||.
Again assume the value of 'a' to be 8 and that of 'b' to be 4.
Operator Description Example
&& Logical AND. If both the operands are non-zero, then the condition becomes true (a == b) && (a==8) is false
|| Logical OR. If any one or both the operands are non-zero, then the condition becomes true (a != b) || (a==5) is true
! Logical NOT. It is used to reverse the condition. So, if a condition is true, ! makes it false. !(a==5) is true
```#include <stdio.h>
int main()
{
int a = 5, b = 0;
printf("\n%d", a && b);
printf("\n%d", a || b);
printf("\n%d", !a);
printf("\n%d", !b);
return 0;
}
```
0
1
0
1
Since a is non-zero but b is zero, so AND between them will be false (or 0). As only one of them is true (or non-zero). But with OR, since anyone of them (i.e. a) is non-zero, so, a||b is true (or 1 ).
In this example, since the value of 'a' is non-zero, therefore it is true. So, !a makes it false. The case with !b is the opposite.
## Assignment Operators
Assignment Operators are used to assign values of the operand on the right to the operand on the left and not vice versa. The most common assignment operator is =.
If we write `a = 10;`, it means that we are assigning a value '10' to the variable 'a'.
There are more assignment operators which are listed as follows.
Operator Description Example
= Assigns value of right operand to the left operand C = A+B is same as C = A + B
+= Adds the value of right operand to the left operand and assigns the final value to the left operand C += A is same as C = C + A
-= Subtracts the value of right operand to the left operand and assigns the final value to the left operand C -= A is same as C = C - A
*= Multiplies the value of right operand to the left operand and assigns the final value to the left operand C *= A is same as C = C * A
/= Divides the value of left operand from the right operand and assigns the final value to the left operand C /= A is same as C = C / A
%= takes modulus using two operands and assigns the result to the left operand C %= A is same as C = C % A
We cannot assign any value to a constant. For example, 4 = c will give an error.
`'='` operator starts from right. eg.- if a is 4 and b is 5, then `a = b` will make a equal to 5 and b will remain 5.
`a = a+b;` → Similarly, since `+` has higher priority, so, `a+b` will be calculated first (i.e. 9), then expression will become `a = 9;`. So, `a = a+b;` will make 'a' equal to 9 and b will remain 5.
Since `a += b` is same as `a = a+b`, `a += b` will also make the value of 'a' equal to 9.
```#include <stdio.h>
int main()
{
int a = 25;
int b = 8;
a += b;
printf("%d\n",a);
a -=b;
printf("%d\n",a);
a *=b;
printf("%d\n",a);
a /=b;
printf("%d\n",a);
a %=b;
printf("%d\n",a);
return 0;
}
```
33
25
200
25
1
Initially 'a' is 25 and 'b' is 8.
`a += b;` → It means `a = a+b`. Now, 'a' is `a+b` i.e. 33. (`a = a+b`)
`a -= b` → Now, 'a' is 33. So, `a -= b` (`a = a-b`) is 25. So, new 'a' is 25. Similarly, the values will be evaluated for the next three cases.
## Increment and Decrement Operators
`++` and `--` are called increment and decrement operators respectively.
++ adds 1 to the operand whereas -- subtracts 1 from the operand.
a++ increases the value of a variable 'a' by 1 and a-- decreases the value of a by 1.
Similarly, ++a increases the value of 'a' by 1 and --a decreases the value of a by 1.
In a++ and a--, ++ and -- are used as postfix whereas in ++a and --a, ++ and -- are used as prefix.
For example, suppose the value of a is 5, then both a++ and ++a changes the value of 'a' to 6. Similarly, both a-- and --a changes the value of 'a' to 4.
### Difference between Prefix and Postfix
While both `a++` and `++a` increases the value of 'a', the only difference between these is that `a++` returns the value of 'a' before the value of 'a' is incremented and ++a first increases the value of 'a' by 1 and then returns the incremented value of 'a'.
Similarly, `a--` first returns the value of 'a' and then decreases its value by 1 and `--a` first decreases the value of 'a' by 1 and then returns the decreased value.
An example will make the difference clear.
```#include <stdio.h>
int main()
{
int a = 15, b = 15, c = 15, d = 15;
printf("\nvalue of a++ = %d", a++);
printf("\nvalue of ++b = %d", ++b);
printf("\nvalue of c-- = %d", c--);
printf("\nvalue of --d = %d", --d);
return 0;
}
```
value of a++ = 15
value of ++b = 16
value of c-- = 15
value of --d = 14
In `a++` 15 is printed. Then the value of a will become 16.
And in ++b, value is first increased to 16 and then printed. Similar with `c--` and `--d`.
## sizeof
`sizeof()` operator is used to return the size of a variable. Suppose we have an integer variable 'i', so the value of `sizeof(i)` will be 4 because on declaring the variable 'i' as of type integer, the size of the variable becomes 4 bytes.
Look at the following example to find the size of int, char, float and double variables
```#include <stdio.h>
int main()
{
int i = 6;
int j;
char c;
float f;
double d;
printf("size of integer variable i = %d", sizeof(i));
printf("\nsize of integer variable j = %d", sizeof(j));
printf("\nsize of character variable = %d", sizeof(c));
printf("\nsize of float variable = %d", sizeof(f));
printf("\nsize of double variable = %d", sizeof(d));
return 0;
}
```
size of integer variable i = 4
size of integer variable j = 4
size of character variable = 1
size of float variable = 4
size of double variable = 8
Here, sizes of character, float and double variables are 1, 4 and 8 bytes respectively, so `sizeof` operator applied to these returns 1, 4 and 8 respectively. Whenever we declare an integer variable, a space in the memory equal to 4 bytes gets occupied by it. It doesn't matter whether we assign a value to the variable or not, space will allocate. Since, both 'i' and 'j' are integer variables, therefore the sizes of both of these are 4 bytes, regardless of whether a value is assigned to these or not.
Size of the variables will be different for 32 bit and 64 bit compiler.
Let's see another example.
```#include <stdio.h>
int main()
{
printf("%d", sizeof(int));
printf("\n%d", sizeof(char));
printf("\n%d", sizeof(short));
printf("\n%d", sizeof(long));
return 0;
}
```
4
1
2
8
Here, we printed the sizes of int, char, short and long using sizeof operator, same as we did in the previous example, except for the difference that this time we directly passed the name of the data type in the `sizeof` operator.
So, anytime you need to know the size of any data type, you can do so by using sizeof operator.
A lot of things are discussed here. So, solve questions from the practice section to have good command over them.
To learn from simple videos, you can always look at our C++ video course on CodesDope Pro. It has over 750 practice questions and over 200 solved examples.
Don't practice until you get it right. Practice until you can't get it wrong. | 3,393 | 11,808 | {"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-2022-05 | longest | en | 0.792921 |
http://math.stackexchange.com/questions/133169/showing-that-a-function-below-is-in-vmo | 1,469,441,757,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824226.33/warc/CC-MAIN-20160723071024-00113-ip-10-185-27-174.ec2.internal.warc.gz | 157,713,102 | 17,868 | # Showing that a function below is in VMO
I am trying to show that the function $\log||\log |x||$ is in VMO. The book that I was reading just mentioned that it was a straightforward verification. Can some help me to figure it out?
-
What is VMO...? – Aryabhata Apr 17 '12 at 22:55
Vanishing mean oscillation, see here: en.wikipedia.org/wiki/Bounded_mean_oscillation – Paul Apr 17 '12 at 22:59
Since we have a positive function, we can deal with ball instead of cubes (it's allowed without this assumption, but not trivial and needed here, and will allow us to use the fact that the function, called $f$, is radial over $\mathbb R^d$), and which are centered at $0$. First, we write \begin{align}m(B(0,a),f)&=\frac 1{a^d}\int_0^ar^{d-1}\log|\log r|dr\\\ &=\int_0^1 s^{d-1}\log|\log s+\log a|ds\\\ &=\int_0^1s^{d-1}\log|\log a|\left|\frac{\log s}{\log a}+1\right|ds\\\ &=\frac{\log |\log a|}d+\int_0^1s^d\log\left|\frac{\log s}{\log a}+1\right|ds. \end{align} Now, denoting $\mathcal O(B(0,a),f)$ the oscilation over the ball $B(0,a)$ we get \begin{align} \mathcal O(B(0,a),f)&=\int_0^1\left|\int_0^1\left((\log|\log t+\log a|)t^{d-1}-m(B(0,a),f)\right)ds\right|dt\\\ &\leq 2\int_0^1s^d\log\left|\frac{\log s}{\log a}+1\right|ds. \end{align} An argument of dominated convergence gives us the result.
I used the substitution $as=r$. – Davide Giraudo Apr 23 '12 at 9:51 | 497 | 1,368 | {"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": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-30 | latest | en | 0.790702 |
https://www.plati.com/itm/electrical-engineering-option-6/1512553?lang=en-US | 1,519,405,945,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814801.45/warc/CC-MAIN-20180223154626-20180223174626-00748.warc.gz | 906,932,947 | 12,006 | # Electrical Engineering, option 6
Sold: 0
Refunds: 0
Content: 30426205642760.doc (608 kB)
# Description
Examination 3
The network enabled coil resistance and inductance which r L (Table. 3.1). Find the complex impedance and conductance coil, if the frequency f = 50 Hz. Show two respective variants of the equivalent circuit of the coil.
Given: L = 70 mH; r = 10 ohms.
Objective 3.2
In the circuit alternating current frequency of 50 Hz (Fig. 3.2, Table 3.2.) Included a coil having an active resistance r and inductive reactance xL. By the circuit voltage is applied to determine the indications of measuring instruments included in the chain, as well as reactive and apparent power circuit. Construct the vector diagram in the complex plane.
Given: r = 6 ohms; xL = 6,7 Ohm; Um = 141 V.
Objective 3.3
The AC circuit (Fig. 3.3, Tab. 3.3 includes a resistor in series with the active resistance r and the capacitor C is supplied to the circuit alternating voltage frequency of 50 Hz. Determine the meter readings included in the chain, reactive and apparent power circuit, the vector construct diagram in the complex plane.
Given: Um = 282 B; r = 70 ohms; C = 100 uF.
Objective 3.4
The network with a voltage connected in series two receivers (Fig. 3.4, Tab. 3.4) having resistances r1 and r2 and reactive xL and xC. Determine the readings included in the scheme, the active and reactive power of each receiver and the entire chain and the power factor cos φ of the whole chain. Construct the vector diagram in the complex plane.
It is given by: U = 50 V; r1 = 8 ohms; xL = 12 ohms; r2 = 4 ohms; xC = 15 ohms.
The circuit consists of two parallel branches (Fig. 3.5, Tab. 3.5) has a resistance r1, xL, r2, xC. Determine the meter readings included in the chain, active and reactive power of parallel branches and the whole circuit if the voltage applied to the circuit, as well U. Construct a vector diagram in the complex plane.
It is given by: U = 100 V; r1 = 15 ohms; xL = 13 ohms; r2 = 7 ohms; xC = 9 ohms.
Objective 3.6
Determine the meter readings in the circuit (Fig. 3.6), the parameters of which are given in the Table. 3.6. Construct the vector diagram in the complex plane.
Given: U = 120 B; r = 1,2 Ohm; xL = 2,1 Ohm; r1 = 7,0 Ohm; xL1 = 5,6 Ohm; r2 = 6,0 ohms; xC3 = 9,0 ohms. | 678 | 2,299 | {"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} | 3.390625 | 3 | CC-MAIN-2018-09 | longest | en | 0.886117 |
http://www.sidefx.com/docs/houdini/nodes/obj/rivet.html | 1,586,164,977,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00347.warc.gz | 260,308,292 | 9,624 | # Rivet object node
Creates a rivet on an objects surface, usually for parenting.
The Rivet Object gets its world transform from the surface of another object. This tool is useful for attaching objects to deforming surfaces in animation. For example, you can use this tool to attach a saddle to a running horse by creating the rivet on the horse’s back and parenting it to the saddle.
When calculating the transform, points defined in the Point Group parameter are obtained from the Rivet Geometry.
The centroid of these points is designated as the origin of the rivet coordinate frame. The x-axis of that frame is defined by the first two points and the xy-plane of the frame is defined by the first three points.
Alternatively, the rivet can build its coordinate frame based on point attributes of the base surface.
## Rivet
Rivet Geometry
The SOP to attach the geometry to.
Point Group
The point group (or point numbers) based on which the local surface frame is built. Group centroid is set as the frame origin. The x-axis of the frame is defined by first two points.
The z-axis is perpendicular to the plane defined by first three points. If the point group contains only one or two points, the frame axes are world aligned.
Point Weights
The weights assigned to the first three points in the group. This parameter allows to fine tune the exact position of the rivet origin on the plane constructed from the first three points. For example, the weights (0.5, 0.5, 0) will place the origin half way between the first and the second point. Note that the weights should not sum up to 0. The points beyond the third one, if any, are assigned the weight values of 1.
Use Point Vector Attributes For Rivet Frame
With this option turned on, the rivet coordinate frame is calculated based on the point attributes specified in the parameters X-Axis Attribute and Z-Axis Attribute. For all points determined by the Point Group parameter, the attributes are evaluated for the base geometry and averaged out to yield two vectors X and Z that will determine the final x-axis and z-axis of the rivet coordinate frame. If the two average vectors X and Z are not orthogonal, before calculating the x-axis, the X vector is replaced with a vector that lies in the X-Z plane and that is perpendicular to the Z vector.
If this option is turned off then the rivet uses the method for frame calculation described in Point Group parameter help above.
X-Axis Attribute
The name of an attribute used for the calculation of the x-axis in the rivet coordinate frame. This has to be a point attribute of a vector type.
Z-Axis Attribute
The name of an attribute used for the calculation of the z-axis in the rivet coordinate frame. This has to be a point attribute of a vector type.
## Misc
Set Wireframe Color
Use the specified wireframe color
Wireframe Color
The display color of the object
Viewport Selecting Enabled
Object is capable of being picked in the viewport.
Select Script
Script to run when the object is picked in the viewport. See select scripts .
Cache Object Transform
Caches object transforms once Houdini calculates them. This is especially useful for objects whose world space position is expensive to calculate (such as Sticky objects), and objects at the end of long parenting chains (such as Bones). This option is turned on by default for Sticky and Bone objects.
See the OBJ Caching section of the Houdini Preferences window for how to control the size of the object transform cache.
Geometry Scale
Uniform scaling about the xyz axes.
Display
Whether to display only the icon, only the axis, or both.
Icon
Displays only the icon geometry.
Axis
Displays only the axis.
Icon and Axis
Displays both the icon and axis.
Control Type
Switches between the type of geometry to display.
Null
Displays null geometry (i.e. cross).
Circles
Displays circle primitives.
Box
Displays box primitive.
Planes
Displays plane primitives.
Null and Circles
Displays null and circle primitives.
Null and Box
Displays null and box primitives.
Null and Planes
Displays null and plane primitives.
Custom
If an input source is specified, this option will display the geometry of the input.
Orientation
Used in conjunction with circle or plane primitives. Determines which circles or planes to display.
All planes
Displays circle or plane primitives on the YZ, ZX and XY planes.
YZ plane
Displays a circle or plane primitive on the YZ plane.
ZX plane
Displays a circle or plane primitive on the ZX plane.
XY plane
Displays a circle or plane primitive on the XY plane.
YZ, ZX planes
Displays circle or plane primitives on the YZ and ZX planes.
YZ, XY planes
Displays circle or plane primitives on the YZ and XY planes.
ZX, XY planes
Displays circle or plane primitives on the ZX and XY planes.
Determines whether to display the primitives as shaded objects or as wireframe objects.
off
Displays primitives in wireframe mode.
on
## Examples
RivetWaveform Example for Rivet object node
In this example, a Waveform deformer is applied to a grid. A Rivet is then attached to the surface of the grid, and then parented to a sphere.
# Object nodes
• Create and attach camera to a crowd agent.
• Loads the objects from an Alembic scene archive (.abc) file into the object level.
• Loads only the transform from an object or objects in an Alembic scene archive (.abc).
• Adds a constant level of light to every surface in the scene (or in the light’s mask), coming from no specific direction.
• The Auto Bone Chain Interface is created by the IK from Objects and IK from Bones tools on the Rigging shelf.
• Switches or blends between the transformations of several input objects.
• Computes its transform by blending between the transforms of two or more sticky objects, allowing you to blend a position across a polygonal surface.
• The Bone Object is used to create hierarchies of limb-like objects that form part of a hierarchy …
• You can view your scene through a camera, and render from its point of view.
• The DOP Network Object contains a dynamic simulation.
• Environment Lights provide background illumination from outside the scene.
• The Extract Transform Object gets its transform by comparing the points of two pieces of geometry.
• The Fetch Object gets its transform by copying the transform of another object.
• Crowd example showing a changing formation setup
• Creates a custom muscle by combining any number of geometry objects, muscle rigs, and muscle pins.
• Container for the geometry operators (SOPs) that define a modeled object.
• Merges groom data from multiple objects into one.
• Moves the curves of a groom with animated skin.
• Generates guide curves from a skin geometry and does further processing on these using an editable SOP network contained within the node.
• Runs a physics simulation on the input guides.
• Converts dense hair curves to a polygon card, keeping the style and shape of the groom.
• An example of how to create a texture for hair cards.
• Generates hair from a skin geometry and guide curves.
• The Handle Object is an IK tool for manipulating bones.
• Indirect lights produce illumination that has reflected from other objects in the scene.
• Instance Objects can instance other geometry, light, or even subnetworks of objects.
• Light Objects cast light on other objects in a scene.
• A very limited light object without any built-in render properties. Use this only if you want to build completely custom light with your choice of properties.
• The Microphone object specifies a listening point for the SpatialAudio CHOP.
• Import Acclaim motion capture.
• A male character with motion captured animations.
• A male character with motion captured animations.
• A male character with motion captured animations.
• The Muscle object is a versatile tool that can be used when rigging characters and creatures with musculature.
• Creates a simple rigging component for attaching regions of a Franken Muscle to your character rig.
• Creates the internal components of a muscle (the rig), by stroking a curve onto a skin object.
• Serves as a place-holder in the scene, usually for parenting. this object does not render.
• Object nodes represent objects in the scene, such as character parts, geometry objects, lights, cameras, and so on.
• The Path object creates an oriented curve (path)
• The PathCV object creates control vertices used by the Path object.
• Pxr AOV Light object for RenderMan RIS.
• Pxr Barn Light Filter object for RenderMan RIS.
• Pxr Blocker Light Filter object for RenderMan RIS.
• Pxr Cookie Light Filter object for RenderMan RIS.
• Pxr Day Light object for RenderMan RIS.
• Pxr Disk Light object for RenderMan RIS.
• Pxr Distant Light object for RenderMan RIS.
• Pxr Dome Light object for RenderMan RIS.
• Pxr Gobo Light Filter object for RenderMan RIS.
• Pxr Mesh Light object for RenderMan RIS.
• Pxr Portal Light object for RenderMan RIS.
• Pxr Ramp Light Filter object for RenderMan RIS.
• Pxr Rectangle Light object for RenderMan RIS.
• Pxr Rod Light Filter object for RenderMan RIS.
• Pxr Sphere Light object for RenderMan RIS.
• The Python Script object is a container for the geometry operators (SOPs) that define a modeled object.
• Crowd example showing a simple ragdoll setup.
• Container for the Compositing operators (COP2) that define a picture.
• Creates a rivet on an objects surface, usually for parenting.
• A simple and efficient animation rig with full controls.
• A simple and efficient female character animation rig with full controls.
• A simple and efficient male character animation rig with full controls.
• The Sound object defines a sound emission point for the Spatial Audio chop.
• Crowd example showing a stadium setup
• Provides parameters to manipulate the interaxial lens distance as well as the zero parallax setting plane in the scene.
• Serves as a basis for constructing a more functional stereo camera rig as a digital asset.
• Creates a sticky object based on the UV’s of a surface, usually for parenting.
• Crowd example showing a street setup with two agent groups
• Container for objects.
• Acts as a camera but switches between the views from other cameras.
• Collects muscles, anatomical bone models, and skin objects and places them into a single dynamics simulation. | 2,179 | 10,414 | {"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-2020-16 | latest | en | 0.90315 |
https://py.checkio.org/en/mission/simple2048/ | 1,726,648,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00726.warc.gz | 439,900,147 | 18,706 | # Behind 2048
Maybe you’ve already heard about the simple and addictive game, 2048 (read more about it here ). In this task we will look behind the scenes and try to recreate its basic movement functionality.
2048 is played on a simple 4×4 grid with tiles. Player can move tiles left, right, up, and down. If two tiles of the same number collide while moving, they will merge into a tile with the sum value of the two tiles. The resulting tile cannot merge with another tile again in the same move. If we have three tiles of the same number, then first collide tiles which closer to edge in the moving direction (right move - closer to the right edge). Every turn, a new tile will appear in the last empty spot (value 0) on the board with a value of 2. The last empty slot is determined by indexes that are defined in the following order:
`[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11], [12, 13, 14, 15]]`
`[[0, 1, 2, 3], [4, 5, 6, 7],...`
Input: A game state as a list of lists with integers and player's move as a string ('up', 'down', 'left' or 'right').
Output: The game state after player's move as a list of lists with integers or letters.
How it is used: This can be used as a bases for further adding GUI to the python version of the 2048 game.
Example:
```move2048([[0, 2, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 2, 0, 0]], 'up') == [[0, 4, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 2]]
move2048([[4, 0, 0, 0],
[0, 4, 0, 0],
[0, 0, 0, 0],
[0, 0, 8, 8]], 'right') == [[0, 0, 0, 4],
[0, 0, 0, 4],
[0, 0, 0, 0],
[0, 0, 2, 16]]
move2048([[2, 0, 2, 2],
[0, 4, 4, 4],
[8, 8, 8, 16],
[0, 0, 0, 0]], 'right') == [[0, 0, 2, 4],
[0, 0, 4, 8],
[0, 8, 16, 16],
[0, 0, 0, 2]]
move2048([[256, 0, 256, 4],
[16, 8, 8, 0],
[32, 32, 32, 32],
[4, 4, 2, 2]], 'right') == [[0, 0, 512, 4],
[0, 0, 16, 16],
[0, 0, 64, 64],
[0, 2, 8, 4]]
move2048([[4, 4, 0, 0],
[0, 4, 1024, 0],
[0, 256, 0, 256],
[0, 1024, 1024, 8]], 'down') == [['U', 'W', 'I', 'N'],
['U', 'W', 'I', 'N'],
['U', 'W', 'I', 'N'],
['U', 'W', 'I', 'N']]
move2048([[2, 4, 8, 16],
[32, 64, 128, 256],
[512, 1024, 2, 4],
[8, 16, 32, 64]], 'left') == [['G', 'A', 'M', 'E'],
['O', 'V', 'E', 'R'],
['G', 'A', 'M', 'E'],
['O', 'V', 'E', 'R']]```
Preconditions: len(state) == 4
all(len(row) == 4 for row in state)
all(all(x in (0, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024) for x in row) for row in state)
You should be an authorized user in order to see the full description and start solving this mission.
Settings
Code:
Other:
Invalid hot key. Each hot key should be unique and valid
Hot keys:
• to Run Code: to Check Solution: to Stop:
CheckiO Extensions
CheckiO Extensions allow you to use local files to solve missions. More info in a blog post.
In order to install CheckiO client you'll need installed Python (version at least 3.8)
Install CheckiO Client first:
`pip3 install checkio_client`
`checkio --domain=py config --key=`
Sync solutions into your local folder
`checkio sync`
(in beta testing) Launch local server so your browser can use it and sync solution between local file end extension on the fly. (doesn't work for safari)
`checkio serv -d`
Alternatevly, you can install Chrome extension or FF addon
`checkio install-plugin`
`checkio install-plugin --ff`
`checkio install-plugin --chromium`
Read more here about other functionality that the checkio client provides. Feel free to submit an issue in case of any difficulties.
Pair Programming (Beta-version)
Welcome to Pair Programming! Engage in real-time collaboration on coding projects by starting a session and sharing the provided unique URL with friends or colleagues. This feature is perfect for joint project development, debugging, or learning new skills together. Simply click 'Start Session' to begin your collaborative coding journey!
Waiting for Pair Programming to start...
You are trying to join a pair programming session that has not started yet.
Please wait for the session creator to join.
Waiting for Pair Programming to reconnect...
It looks like the creator of the pair programming session closed the editor window.
It might happen accidentally, so that you can wait for reconnection. | 1,417 | 4,144 | {"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-2024-38 | latest | en | 0.765046 |
https://stats.stackexchange.com/questions/116786/back-transformation-of-arcsine-square-root-transformation | 1,726,412,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00836.warc.gz | 513,344,876 | 39,998 | # back transformation of arcsine square root transformation
I have a LSD that has been produced from the transformation of percentage data. I want, if possible, to transform this number back in a % that can be plotted onto the graph with the percentages on it.
I have used the below formula in excel as recommended by the below website. However, it is not giving me a sensible answer.
=SIN(E2/180*PI())^2*100.
http://archive.bio.ed.ac.uk/jdeacon/statistics/tress4.html
any advice you could give me would be great.
• What is LSD? What is E2? Commented Sep 25, 2014 at 21:55
If $y = \arcsin(\sqrt{p})$ then $p=(\sin(y))^2$. | 169 | 629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.932626 |
https://code.qt.io/cgit/qt/qtbase.git/commit/src/corelib?id=3c159957f863cf8d367a9261e7016e52cd0348c1 | 1,575,917,248,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540521378.25/warc/CC-MAIN-20191209173528-20191209201528-00476.warc.gz | 317,274,965 | 3,440 | summaryrefslogtreecommitdiffstats log msg author committer range
path: root/src/corelib
diff options
context: 12345678910152025303540 space: includeignore mode: unifiedssdiffstat only
author committer Shawn Rutledge 2016-03-23 22:26:08 +0100 Shawn Rutledge 2016-12-07 08:33:35 +0000 3c159957f863cf8d367a9261e7016e52cd0348c1 (patch) 1a56cb165b674dd49170de88b4254cd4ad0d51a4 /src/corelib f7253b25684eabf68973cc4486b32f0dc346b70c (diff)
TouchPoint: add horizontalDiameter, verticalDiameter; deprecate rects
The contact patch of a finger on a touchscreen tends to be roughly elliptical. If we model it as a QRectF, it's not clear whether the ellipse should be considered to be inscribed in the rectangle and then rotated, or whether the rectangle represents the outer bounds of the rotated ellipse. In practice, since most touchscreens can't measure rotation, it is effectively the latter. But modeling it that way means information is lost if the touchscreen can measure rotation: you can determine the bounds of a rotated ellipse, but you cannot derive the rotated ellipse from its bounds. So it's better to model the axes of the ellipse explicitly. This has the added benefit of saving a little storage space: we replace 3 QRectF instances, whose width and height will normally be the same, with 3 positions (bringing the total to 12 QPointF's) and one set of axes. Further, most applications only care about the center of each contact patch, so it's better to store that explicitly instead of calculating QRectF::center() repeatedly. In the past there may have been an assumption that the width of the rect is the same as the horizontalDiameter of the ellipse, so the rect could be considered to be rotated, and the ellipse to be inscribed. But in d0b1c646b4a351f7eea2137c68993ae63b2b6bab and 40e4949674eaf7ceb09f6d18479ead1a36b384fd the point was made that the rect is actually the bounding box of the rotated ellipse. [ChangeLog][QtGui][QTouchEvent] TouchPoint::rect(), sceneRect() and screenRect() are deprecated; a touchpoint is now modeled as an ellipse, so please use pos(), scenePos(), screenPos(), horizontalDiameter() and verticalDiameter() instead. Change-Id: Ic06f6165e2d90fc9d4cc19cf4938d4faf5766bb4 Reviewed-by: Jan Arve Sæther <jan-arve.saether@qt.io>
Diffstat (limited to 'src/corelib')
0 files changed, 0 insertions, 0 deletions | 640 | 2,346 | {"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} | 2.59375 | 3 | CC-MAIN-2019-51 | latest | en | 0.863954 |
http://gogeometry.blogspot.com/2009/11/problem-390-triangle-parallel-lines.html | 1,527,450,324,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00456.warc.gz | 122,515,281 | 13,586 | ## Sunday, November 15, 2009
### Problem 390. Triangle, Parallel lines, Collinear points
Proposed Problem
Click the figure below to see the complete problem 390 about Triangle, Parallel lines, Collinear points.
Complete Problem 390
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
#### 1 comment:
1. Extend AH to BC at J
Extend CG to AB at K
Let m=distance AD and n=distance DC
We have HC/HF=DC/DA=n/m ( Tri. CHD ~ Tri. CFA)
CB/CE=CA/CD= (m+n)/n ( Tri. CED~Tri. CBA)
GE/GA=DC/DA=n/m ( Tri. ADG ~ Tri ACE)
1. Apply Menelaus theorem on Tri. FBC with secant AHJ
(JB/JC)*(HC/HF)*(AF/AB)=1
Replace HC/HF= n/m and AF/AB= m/(m+n)
We get JB/JC= (m+n)/n
2. Apply Menelaus theorem on Tri. ABE with secant KGC
(KA/KB)*(CB/CE)*(GE/GA)=1
Replace CB/CE=(m+n)/n and GE/GA= n/m
We get KA/KB= m/(m+n_
3 Consider Tri. ABC and cevians BD, AJ and CK
Calculate (JB/JC) * (DC/DA)*(KA/KB)=
(m+n)/n * n/m *m/(m+n) = 1
So these 3 cevians will concurrence at M per Ceva’s theorem and B, M and D are collinear.
Peter Tran
vstran@yahoo.com | 373 | 1,049 | {"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} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.771013 |
http://isabelle.in.tum.de/repos/isabelle/file/0ddb5b755cdc/src/HOL/Conditionally_Complete_Lattices.thy | 1,597,523,013,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439741154.98/warc/CC-MAIN-20200815184756-20200815214756-00334.warc.gz | 55,828,659 | 14,419 | src/HOL/Conditionally_Complete_Lattices.thy
author hoelzl Wed Jun 18 07:31:12 2014 +0200 (2014-06-18) changeset 57275 0ddb5b755cdc parent 56218 1c3f1f2431f9 child 57447 87429bdecad5 permissions -rw-r--r--
moved lemmas from the proof of the Central Limit Theorem by Jeremy Avigad and Luke Serafin
``` 1 (* Title: HOL/Conditionally_Complete_Lattices.thy
```
``` 2 Author: Amine Chaieb and L C Paulson, University of Cambridge
```
``` 3 Author: Johannes Hölzl, TU München
```
``` 4 Author: Luke S. Serafin, Carnegie Mellon University
```
``` 5 *)
```
``` 6
```
``` 7 header {* Conditionally-complete Lattices *}
```
``` 8
```
``` 9 theory Conditionally_Complete_Lattices
```
``` 10 imports Main
```
``` 11 begin
```
``` 12
```
``` 13 lemma (in linorder) Sup_fin_eq_Max: "finite X \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Sup_fin X = Max X"
```
``` 14 by (induct X rule: finite_ne_induct) (simp_all add: sup_max)
```
``` 15
```
``` 16 lemma (in linorder) Inf_fin_eq_Min: "finite X \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Inf_fin X = Min X"
```
``` 17 by (induct X rule: finite_ne_induct) (simp_all add: inf_min)
```
``` 18
```
``` 19 context preorder
```
``` 20 begin
```
``` 21
```
``` 22 definition "bdd_above A \<longleftrightarrow> (\<exists>M. \<forall>x \<in> A. x \<le> M)"
```
``` 23 definition "bdd_below A \<longleftrightarrow> (\<exists>m. \<forall>x \<in> A. m \<le> x)"
```
``` 24
```
``` 25 lemma bdd_aboveI[intro]: "(\<And>x. x \<in> A \<Longrightarrow> x \<le> M) \<Longrightarrow> bdd_above A"
```
``` 26 by (auto simp: bdd_above_def)
```
``` 27
```
``` 28 lemma bdd_belowI[intro]: "(\<And>x. x \<in> A \<Longrightarrow> m \<le> x) \<Longrightarrow> bdd_below A"
```
``` 29 by (auto simp: bdd_below_def)
```
``` 30
```
``` 31 lemma bdd_aboveI2: "(\<And>x. x \<in> A \<Longrightarrow> f x \<le> M) \<Longrightarrow> bdd_above (f`A)"
```
``` 32 by force
```
``` 33
```
``` 34 lemma bdd_belowI2: "(\<And>x. x \<in> A \<Longrightarrow> m \<le> f x) \<Longrightarrow> bdd_below (f`A)"
```
``` 35 by force
```
``` 36
```
``` 37 lemma bdd_above_empty [simp, intro]: "bdd_above {}"
```
``` 38 unfolding bdd_above_def by auto
```
``` 39
```
``` 40 lemma bdd_below_empty [simp, intro]: "bdd_below {}"
```
``` 41 unfolding bdd_below_def by auto
```
``` 42
```
``` 43 lemma bdd_above_mono: "bdd_above B \<Longrightarrow> A \<subseteq> B \<Longrightarrow> bdd_above A"
```
``` 44 by (metis (full_types) bdd_above_def order_class.le_neq_trans psubsetD)
```
``` 45
```
``` 46 lemma bdd_below_mono: "bdd_below B \<Longrightarrow> A \<subseteq> B \<Longrightarrow> bdd_below A"
```
``` 47 by (metis bdd_below_def order_class.le_neq_trans psubsetD)
```
``` 48
```
``` 49 lemma bdd_above_Int1 [simp]: "bdd_above A \<Longrightarrow> bdd_above (A \<inter> B)"
```
``` 50 using bdd_above_mono by auto
```
``` 51
```
``` 52 lemma bdd_above_Int2 [simp]: "bdd_above B \<Longrightarrow> bdd_above (A \<inter> B)"
```
``` 53 using bdd_above_mono by auto
```
``` 54
```
``` 55 lemma bdd_below_Int1 [simp]: "bdd_below A \<Longrightarrow> bdd_below (A \<inter> B)"
```
``` 56 using bdd_below_mono by auto
```
``` 57
```
``` 58 lemma bdd_below_Int2 [simp]: "bdd_below B \<Longrightarrow> bdd_below (A \<inter> B)"
```
``` 59 using bdd_below_mono by auto
```
``` 60
```
``` 61 lemma bdd_above_Ioo [simp, intro]: "bdd_above {a <..< b}"
```
``` 62 by (auto simp add: bdd_above_def intro!: exI[of _ b] less_imp_le)
```
``` 63
```
``` 64 lemma bdd_above_Ico [simp, intro]: "bdd_above {a ..< b}"
```
``` 65 by (auto simp add: bdd_above_def intro!: exI[of _ b] less_imp_le)
```
``` 66
```
``` 67 lemma bdd_above_Iio [simp, intro]: "bdd_above {..< b}"
```
``` 68 by (auto simp add: bdd_above_def intro: exI[of _ b] less_imp_le)
```
``` 69
```
``` 70 lemma bdd_above_Ioc [simp, intro]: "bdd_above {a <.. b}"
```
``` 71 by (auto simp add: bdd_above_def intro: exI[of _ b] less_imp_le)
```
``` 72
```
``` 73 lemma bdd_above_Icc [simp, intro]: "bdd_above {a .. b}"
```
``` 74 by (auto simp add: bdd_above_def intro: exI[of _ b] less_imp_le)
```
``` 75
```
``` 76 lemma bdd_above_Iic [simp, intro]: "bdd_above {.. b}"
```
``` 77 by (auto simp add: bdd_above_def intro: exI[of _ b] less_imp_le)
```
``` 78
```
``` 79 lemma bdd_below_Ioo [simp, intro]: "bdd_below {a <..< b}"
```
``` 80 by (auto simp add: bdd_below_def intro!: exI[of _ a] less_imp_le)
```
``` 81
```
``` 82 lemma bdd_below_Ioc [simp, intro]: "bdd_below {a <.. b}"
```
``` 83 by (auto simp add: bdd_below_def intro!: exI[of _ a] less_imp_le)
```
``` 84
```
``` 85 lemma bdd_below_Ioi [simp, intro]: "bdd_below {a <..}"
```
``` 86 by (auto simp add: bdd_below_def intro: exI[of _ a] less_imp_le)
```
``` 87
```
``` 88 lemma bdd_below_Ico [simp, intro]: "bdd_below {a ..< b}"
```
``` 89 by (auto simp add: bdd_below_def intro: exI[of _ a] less_imp_le)
```
``` 90
```
``` 91 lemma bdd_below_Icc [simp, intro]: "bdd_below {a .. b}"
```
``` 92 by (auto simp add: bdd_below_def intro: exI[of _ a] less_imp_le)
```
``` 93
```
``` 94 lemma bdd_below_Ici [simp, intro]: "bdd_below {a ..}"
```
``` 95 by (auto simp add: bdd_below_def intro: exI[of _ a] less_imp_le)
```
``` 96
```
``` 97 end
```
``` 98
```
``` 99 lemma (in order_top) bdd_above_top[simp, intro!]: "bdd_above A"
```
``` 100 by (rule bdd_aboveI[of _ top]) simp
```
``` 101
```
``` 102 lemma (in order_bot) bdd_above_bot[simp, intro!]: "bdd_below A"
```
``` 103 by (rule bdd_belowI[of _ bot]) simp
```
``` 104
```
``` 105 lemma bdd_above_uminus[simp]:
```
``` 106 fixes X :: "'a::ordered_ab_group_add set"
```
``` 107 shows "bdd_above (uminus ` X) \<longleftrightarrow> bdd_below X"
```
``` 108 by (auto simp: bdd_above_def bdd_below_def intro: le_imp_neg_le) (metis le_imp_neg_le minus_minus)
```
``` 109
```
``` 110 lemma bdd_below_uminus[simp]:
```
``` 111 fixes X :: "'a::ordered_ab_group_add set"
```
``` 112 shows"bdd_below (uminus ` X) \<longleftrightarrow> bdd_above X"
```
``` 113 by (auto simp: bdd_above_def bdd_below_def intro: le_imp_neg_le) (metis le_imp_neg_le minus_minus)
```
``` 114
```
``` 115 context lattice
```
``` 116 begin
```
``` 117
```
``` 118 lemma bdd_above_insert [simp]: "bdd_above (insert a A) = bdd_above A"
```
``` 119 by (auto simp: bdd_above_def intro: le_supI2 sup_ge1)
```
``` 120
```
``` 121 lemma bdd_below_insert [simp]: "bdd_below (insert a A) = bdd_below A"
```
``` 122 by (auto simp: bdd_below_def intro: le_infI2 inf_le1)
```
``` 123
```
``` 124 lemma bdd_finite [simp]:
```
``` 125 assumes "finite A" shows bdd_above_finite: "bdd_above A" and bdd_below_finite: "bdd_below A"
```
``` 126 using assms by (induct rule: finite_induct, auto)
```
``` 127
```
``` 128 lemma bdd_above_Un [simp]: "bdd_above (A \<union> B) = (bdd_above A \<and> bdd_above B)"
```
``` 129 proof
```
``` 130 assume "bdd_above (A \<union> B)"
```
``` 131 thus "bdd_above A \<and> bdd_above B" unfolding bdd_above_def by auto
```
``` 132 next
```
``` 133 assume "bdd_above A \<and> bdd_above B"
```
``` 134 then obtain a b where "\<forall>x\<in>A. x \<le> a" "\<forall>x\<in>B. x \<le> b" unfolding bdd_above_def by auto
```
``` 135 hence "\<forall>x \<in> A \<union> B. x \<le> sup a b" by (auto intro: Un_iff le_supI1 le_supI2)
```
``` 136 thus "bdd_above (A \<union> B)" unfolding bdd_above_def ..
```
``` 137 qed
```
``` 138
```
``` 139 lemma bdd_below_Un [simp]: "bdd_below (A \<union> B) = (bdd_below A \<and> bdd_below B)"
```
``` 140 proof
```
``` 141 assume "bdd_below (A \<union> B)"
```
``` 142 thus "bdd_below A \<and> bdd_below B" unfolding bdd_below_def by auto
```
``` 143 next
```
``` 144 assume "bdd_below A \<and> bdd_below B"
```
``` 145 then obtain a b where "\<forall>x\<in>A. a \<le> x" "\<forall>x\<in>B. b \<le> x" unfolding bdd_below_def by auto
```
``` 146 hence "\<forall>x \<in> A \<union> B. inf a b \<le> x" by (auto intro: Un_iff le_infI1 le_infI2)
```
``` 147 thus "bdd_below (A \<union> B)" unfolding bdd_below_def ..
```
``` 148 qed
```
``` 149
```
``` 150 lemma bdd_above_sup[simp]: "bdd_above ((\<lambda>x. sup (f x) (g x)) ` A) \<longleftrightarrow> bdd_above (f`A) \<and> bdd_above (g`A)"
```
``` 151 by (auto simp: bdd_above_def intro: le_supI1 le_supI2)
```
``` 152
```
``` 153 lemma bdd_below_inf[simp]: "bdd_below ((\<lambda>x. inf (f x) (g x)) ` A) \<longleftrightarrow> bdd_below (f`A) \<and> bdd_below (g`A)"
```
``` 154 by (auto simp: bdd_below_def intro: le_infI1 le_infI2)
```
``` 155
```
``` 156 end
```
``` 157
```
``` 158
```
``` 159 text {*
```
``` 160
```
``` 161 To avoid name classes with the @{class complete_lattice}-class we prefix @{const Sup} and
```
``` 162 @{const Inf} in theorem names with c.
```
``` 163
```
``` 164 *}
```
``` 165
```
``` 166 class conditionally_complete_lattice = lattice + Sup + Inf +
```
``` 167 assumes cInf_lower: "x \<in> X \<Longrightarrow> bdd_below X \<Longrightarrow> Inf X \<le> x"
```
``` 168 and cInf_greatest: "X \<noteq> {} \<Longrightarrow> (\<And>x. x \<in> X \<Longrightarrow> z \<le> x) \<Longrightarrow> z \<le> Inf X"
```
``` 169 assumes cSup_upper: "x \<in> X \<Longrightarrow> bdd_above X \<Longrightarrow> x \<le> Sup X"
```
``` 170 and cSup_least: "X \<noteq> {} \<Longrightarrow> (\<And>x. x \<in> X \<Longrightarrow> x \<le> z) \<Longrightarrow> Sup X \<le> z"
```
``` 171 begin
```
``` 172
```
``` 173 lemma cSup_upper2: "x \<in> X \<Longrightarrow> y \<le> x \<Longrightarrow> bdd_above X \<Longrightarrow> y \<le> Sup X"
```
``` 174 by (metis cSup_upper order_trans)
```
``` 175
```
``` 176 lemma cInf_lower2: "x \<in> X \<Longrightarrow> x \<le> y \<Longrightarrow> bdd_below X \<Longrightarrow> Inf X \<le> y"
```
``` 177 by (metis cInf_lower order_trans)
```
``` 178
```
``` 179 lemma cSup_mono: "B \<noteq> {} \<Longrightarrow> bdd_above A \<Longrightarrow> (\<And>b. b \<in> B \<Longrightarrow> \<exists>a\<in>A. b \<le> a) \<Longrightarrow> Sup B \<le> Sup A"
```
``` 180 by (metis cSup_least cSup_upper2)
```
``` 181
```
``` 182 lemma cInf_mono: "B \<noteq> {} \<Longrightarrow> bdd_below A \<Longrightarrow> (\<And>b. b \<in> B \<Longrightarrow> \<exists>a\<in>A. a \<le> b) \<Longrightarrow> Inf A \<le> Inf B"
```
``` 183 by (metis cInf_greatest cInf_lower2)
```
``` 184
```
``` 185 lemma cSup_subset_mono: "A \<noteq> {} \<Longrightarrow> bdd_above B \<Longrightarrow> A \<subseteq> B \<Longrightarrow> Sup A \<le> Sup B"
```
``` 186 by (metis cSup_least cSup_upper subsetD)
```
``` 187
```
``` 188 lemma cInf_superset_mono: "A \<noteq> {} \<Longrightarrow> bdd_below B \<Longrightarrow> A \<subseteq> B \<Longrightarrow> Inf B \<le> Inf A"
```
``` 189 by (metis cInf_greatest cInf_lower subsetD)
```
``` 190
```
``` 191 lemma cSup_eq_maximum: "z \<in> X \<Longrightarrow> (\<And>x. x \<in> X \<Longrightarrow> x \<le> z) \<Longrightarrow> Sup X = z"
```
``` 192 by (intro antisym cSup_upper[of z X] cSup_least[of X z]) auto
```
``` 193
```
``` 194 lemma cInf_eq_minimum: "z \<in> X \<Longrightarrow> (\<And>x. x \<in> X \<Longrightarrow> z \<le> x) \<Longrightarrow> Inf X = z"
```
``` 195 by (intro antisym cInf_lower[of z X] cInf_greatest[of X z]) auto
```
``` 196
```
``` 197 lemma cSup_le_iff: "S \<noteq> {} \<Longrightarrow> bdd_above S \<Longrightarrow> Sup S \<le> a \<longleftrightarrow> (\<forall>x\<in>S. x \<le> a)"
```
``` 198 by (metis order_trans cSup_upper cSup_least)
```
``` 199
```
``` 200 lemma le_cInf_iff: "S \<noteq> {} \<Longrightarrow> bdd_below S \<Longrightarrow> a \<le> Inf S \<longleftrightarrow> (\<forall>x\<in>S. a \<le> x)"
```
``` 201 by (metis order_trans cInf_lower cInf_greatest)
```
``` 202
```
``` 203 lemma cSup_eq_non_empty:
```
``` 204 assumes 1: "X \<noteq> {}"
```
``` 205 assumes 2: "\<And>x. x \<in> X \<Longrightarrow> x \<le> a"
```
``` 206 assumes 3: "\<And>y. (\<And>x. x \<in> X \<Longrightarrow> x \<le> y) \<Longrightarrow> a \<le> y"
```
``` 207 shows "Sup X = a"
```
``` 208 by (intro 3 1 antisym cSup_least) (auto intro: 2 1 cSup_upper)
```
``` 209
```
``` 210 lemma cInf_eq_non_empty:
```
``` 211 assumes 1: "X \<noteq> {}"
```
``` 212 assumes 2: "\<And>x. x \<in> X \<Longrightarrow> a \<le> x"
```
``` 213 assumes 3: "\<And>y. (\<And>x. x \<in> X \<Longrightarrow> y \<le> x) \<Longrightarrow> y \<le> a"
```
``` 214 shows "Inf X = a"
```
``` 215 by (intro 3 1 antisym cInf_greatest) (auto intro: 2 1 cInf_lower)
```
``` 216
```
``` 217 lemma cInf_cSup: "S \<noteq> {} \<Longrightarrow> bdd_below S \<Longrightarrow> Inf S = Sup {x. \<forall>s\<in>S. x \<le> s}"
```
``` 218 by (rule cInf_eq_non_empty) (auto intro!: cSup_upper cSup_least simp: bdd_below_def)
```
``` 219
```
``` 220 lemma cSup_cInf: "S \<noteq> {} \<Longrightarrow> bdd_above S \<Longrightarrow> Sup S = Inf {x. \<forall>s\<in>S. s \<le> x}"
```
``` 221 by (rule cSup_eq_non_empty) (auto intro!: cInf_lower cInf_greatest simp: bdd_above_def)
```
``` 222
```
``` 223 lemma cSup_insert: "X \<noteq> {} \<Longrightarrow> bdd_above X \<Longrightarrow> Sup (insert a X) = sup a (Sup X)"
```
``` 224 by (intro cSup_eq_non_empty) (auto intro: le_supI2 cSup_upper cSup_least)
```
``` 225
```
``` 226 lemma cInf_insert: "X \<noteq> {} \<Longrightarrow> bdd_below X \<Longrightarrow> Inf (insert a X) = inf a (Inf X)"
```
``` 227 by (intro cInf_eq_non_empty) (auto intro: le_infI2 cInf_lower cInf_greatest)
```
``` 228
```
``` 229 lemma cSup_singleton [simp]: "Sup {x} = x"
```
``` 230 by (intro cSup_eq_maximum) auto
```
``` 231
```
``` 232 lemma cInf_singleton [simp]: "Inf {x} = x"
```
``` 233 by (intro cInf_eq_minimum) auto
```
``` 234
```
``` 235 lemma cSup_insert_If: "bdd_above X \<Longrightarrow> Sup (insert a X) = (if X = {} then a else sup a (Sup X))"
```
``` 236 using cSup_insert[of X] by simp
```
``` 237
```
``` 238 lemma cInf_insert_If: "bdd_below X \<Longrightarrow> Inf (insert a X) = (if X = {} then a else inf a (Inf X))"
```
``` 239 using cInf_insert[of X] by simp
```
``` 240
```
``` 241 lemma le_cSup_finite: "finite X \<Longrightarrow> x \<in> X \<Longrightarrow> x \<le> Sup X"
```
``` 242 proof (induct X arbitrary: x rule: finite_induct)
```
``` 243 case (insert x X y) then show ?case
```
``` 244 by (cases "X = {}") (auto simp: cSup_insert intro: le_supI2)
```
``` 245 qed simp
```
``` 246
```
``` 247 lemma cInf_le_finite: "finite X \<Longrightarrow> x \<in> X \<Longrightarrow> Inf X \<le> x"
```
``` 248 proof (induct X arbitrary: x rule: finite_induct)
```
``` 249 case (insert x X y) then show ?case
```
``` 250 by (cases "X = {}") (auto simp: cInf_insert intro: le_infI2)
```
``` 251 qed simp
```
``` 252
```
``` 253 lemma cSup_eq_Sup_fin: "finite X \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Sup X = Sup_fin X"
```
``` 254 by (induct X rule: finite_ne_induct) (simp_all add: cSup_insert)
```
``` 255
```
``` 256 lemma cInf_eq_Inf_fin: "finite X \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Inf X = Inf_fin X"
```
``` 257 by (induct X rule: finite_ne_induct) (simp_all add: cInf_insert)
```
``` 258
```
``` 259 lemma cSup_atMost[simp]: "Sup {..x} = x"
```
``` 260 by (auto intro!: cSup_eq_maximum)
```
``` 261
```
``` 262 lemma cSup_greaterThanAtMost[simp]: "y < x \<Longrightarrow> Sup {y<..x} = x"
```
``` 263 by (auto intro!: cSup_eq_maximum)
```
``` 264
```
``` 265 lemma cSup_atLeastAtMost[simp]: "y \<le> x \<Longrightarrow> Sup {y..x} = x"
```
``` 266 by (auto intro!: cSup_eq_maximum)
```
``` 267
```
``` 268 lemma cInf_atLeast[simp]: "Inf {x..} = x"
```
``` 269 by (auto intro!: cInf_eq_minimum)
```
``` 270
```
``` 271 lemma cInf_atLeastLessThan[simp]: "y < x \<Longrightarrow> Inf {y..<x} = y"
```
``` 272 by (auto intro!: cInf_eq_minimum)
```
``` 273
```
``` 274 lemma cInf_atLeastAtMost[simp]: "y \<le> x \<Longrightarrow> Inf {y..x} = y"
```
``` 275 by (auto intro!: cInf_eq_minimum)
```
``` 276
```
``` 277 lemma cINF_lower: "bdd_below (f ` A) \<Longrightarrow> x \<in> A \<Longrightarrow> INFIMUM A f \<le> f x"
```
``` 278 using cInf_lower [of _ "f ` A"] by simp
```
``` 279
```
``` 280 lemma cINF_greatest: "A \<noteq> {} \<Longrightarrow> (\<And>x. x \<in> A \<Longrightarrow> m \<le> f x) \<Longrightarrow> m \<le> INFIMUM A f"
```
``` 281 using cInf_greatest [of "f ` A"] by auto
```
``` 282
```
``` 283 lemma cSUP_upper: "x \<in> A \<Longrightarrow> bdd_above (f ` A) \<Longrightarrow> f x \<le> SUPREMUM A f"
```
``` 284 using cSup_upper [of _ "f ` A"] by simp
```
``` 285
```
``` 286 lemma cSUP_least: "A \<noteq> {} \<Longrightarrow> (\<And>x. x \<in> A \<Longrightarrow> f x \<le> M) \<Longrightarrow> SUPREMUM A f \<le> M"
```
``` 287 using cSup_least [of "f ` A"] by auto
```
``` 288
```
``` 289 lemma cINF_lower2: "bdd_below (f ` A) \<Longrightarrow> x \<in> A \<Longrightarrow> f x \<le> u \<Longrightarrow> INFIMUM A f \<le> u"
```
``` 290 by (auto intro: cINF_lower assms order_trans)
```
``` 291
```
``` 292 lemma cSUP_upper2: "bdd_above (f ` A) \<Longrightarrow> x \<in> A \<Longrightarrow> u \<le> f x \<Longrightarrow> u \<le> SUPREMUM A f"
```
``` 293 by (auto intro: cSUP_upper assms order_trans)
```
``` 294
```
``` 295 lemma cSUP_const: "A \<noteq> {} \<Longrightarrow> (SUP x:A. c) = c"
```
``` 296 by (intro antisym cSUP_least) (auto intro: cSUP_upper)
```
``` 297
```
``` 298 lemma cINF_const: "A \<noteq> {} \<Longrightarrow> (INF x:A. c) = c"
```
``` 299 by (intro antisym cINF_greatest) (auto intro: cINF_lower)
```
``` 300
```
``` 301 lemma le_cINF_iff: "A \<noteq> {} \<Longrightarrow> bdd_below (f ` A) \<Longrightarrow> u \<le> INFIMUM A f \<longleftrightarrow> (\<forall>x\<in>A. u \<le> f x)"
```
``` 302 by (metis cINF_greatest cINF_lower assms order_trans)
```
``` 303
```
``` 304 lemma cSUP_le_iff: "A \<noteq> {} \<Longrightarrow> bdd_above (f ` A) \<Longrightarrow> SUPREMUM A f \<le> u \<longleftrightarrow> (\<forall>x\<in>A. f x \<le> u)"
```
``` 305 by (metis cSUP_least cSUP_upper assms order_trans)
```
``` 306
```
``` 307 lemma less_cINF_D: "bdd_below (f`A) \<Longrightarrow> y < (INF i:A. f i) \<Longrightarrow> i \<in> A \<Longrightarrow> y < f i"
```
``` 308 by (metis cINF_lower less_le_trans)
```
``` 309
```
``` 310 lemma cSUP_lessD: "bdd_above (f`A) \<Longrightarrow> (SUP i:A. f i) < y \<Longrightarrow> i \<in> A \<Longrightarrow> f i < y"
```
``` 311 by (metis cSUP_upper le_less_trans)
```
``` 312
```
``` 313 lemma cINF_insert: "A \<noteq> {} \<Longrightarrow> bdd_below (f ` A) \<Longrightarrow> INFIMUM (insert a A) f = inf (f a) (INFIMUM A f)"
```
``` 314 by (metis cInf_insert Inf_image_eq image_insert image_is_empty)
```
``` 315
```
``` 316 lemma cSUP_insert: "A \<noteq> {} \<Longrightarrow> bdd_above (f ` A) \<Longrightarrow> SUPREMUM (insert a A) f = sup (f a) (SUPREMUM A f)"
```
``` 317 by (metis cSup_insert Sup_image_eq image_insert image_is_empty)
```
``` 318
```
``` 319 lemma cINF_mono: "B \<noteq> {} \<Longrightarrow> bdd_below (f ` A) \<Longrightarrow> (\<And>m. m \<in> B \<Longrightarrow> \<exists>n\<in>A. f n \<le> g m) \<Longrightarrow> INFIMUM A f \<le> INFIMUM B g"
```
``` 320 using cInf_mono [of "g ` B" "f ` A"] by auto
```
``` 321
```
``` 322 lemma cSUP_mono: "A \<noteq> {} \<Longrightarrow> bdd_above (g ` B) \<Longrightarrow> (\<And>n. n \<in> A \<Longrightarrow> \<exists>m\<in>B. f n \<le> g m) \<Longrightarrow> SUPREMUM A f \<le> SUPREMUM B g"
```
``` 323 using cSup_mono [of "f ` A" "g ` B"] by auto
```
``` 324
```
``` 325 lemma cINF_superset_mono: "A \<noteq> {} \<Longrightarrow> bdd_below (g ` B) \<Longrightarrow> A \<subseteq> B \<Longrightarrow> (\<And>x. x \<in> B \<Longrightarrow> g x \<le> f x) \<Longrightarrow> INFIMUM B g \<le> INFIMUM A f"
```
``` 326 by (rule cINF_mono) auto
```
``` 327
```
``` 328 lemma cSUP_subset_mono: "A \<noteq> {} \<Longrightarrow> bdd_above (g ` B) \<Longrightarrow> A \<subseteq> B \<Longrightarrow> (\<And>x. x \<in> B \<Longrightarrow> f x \<le> g x) \<Longrightarrow> SUPREMUM A f \<le> SUPREMUM B g"
```
``` 329 by (rule cSUP_mono) auto
```
``` 330
```
``` 331 lemma less_eq_cInf_inter: "bdd_below A \<Longrightarrow> bdd_below B \<Longrightarrow> A \<inter> B \<noteq> {} \<Longrightarrow> inf (Inf A) (Inf B) \<le> Inf (A \<inter> B)"
```
``` 332 by (metis cInf_superset_mono lattice_class.inf_sup_ord(1) le_infI1)
```
``` 333
```
``` 334 lemma cSup_inter_less_eq: "bdd_above A \<Longrightarrow> bdd_above B \<Longrightarrow> A \<inter> B \<noteq> {} \<Longrightarrow> Sup (A \<inter> B) \<le> sup (Sup A) (Sup B) "
```
``` 335 by (metis cSup_subset_mono lattice_class.inf_sup_ord(1) le_supI1)
```
``` 336
```
``` 337 lemma cInf_union_distrib: "A \<noteq> {} \<Longrightarrow> bdd_below A \<Longrightarrow> B \<noteq> {} \<Longrightarrow> bdd_below B \<Longrightarrow> Inf (A \<union> B) = inf (Inf A) (Inf B)"
```
``` 338 by (intro antisym le_infI cInf_greatest cInf_lower) (auto intro: le_infI1 le_infI2 cInf_lower)
```
``` 339
```
``` 340 lemma cINF_union: "A \<noteq> {} \<Longrightarrow> bdd_below (f`A) \<Longrightarrow> B \<noteq> {} \<Longrightarrow> bdd_below (f`B) \<Longrightarrow> INFIMUM (A \<union> B) f = inf (INFIMUM A f) (INFIMUM B f)"
```
``` 341 using cInf_union_distrib [of "f ` A" "f ` B"] by (simp add: image_Un [symmetric])
```
``` 342
```
``` 343 lemma cSup_union_distrib: "A \<noteq> {} \<Longrightarrow> bdd_above A \<Longrightarrow> B \<noteq> {} \<Longrightarrow> bdd_above B \<Longrightarrow> Sup (A \<union> B) = sup (Sup A) (Sup B)"
```
``` 344 by (intro antisym le_supI cSup_least cSup_upper) (auto intro: le_supI1 le_supI2 cSup_upper)
```
``` 345
```
``` 346 lemma cSUP_union: "A \<noteq> {} \<Longrightarrow> bdd_above (f`A) \<Longrightarrow> B \<noteq> {} \<Longrightarrow> bdd_above (f`B) \<Longrightarrow> SUPREMUM (A \<union> B) f = sup (SUPREMUM A f) (SUPREMUM B f)"
```
``` 347 using cSup_union_distrib [of "f ` A" "f ` B"] by (simp add: image_Un [symmetric])
```
``` 348
```
``` 349 lemma cINF_inf_distrib: "A \<noteq> {} \<Longrightarrow> bdd_below (f`A) \<Longrightarrow> bdd_below (g`A) \<Longrightarrow> inf (INFIMUM A f) (INFIMUM A g) = (INF a:A. inf (f a) (g a))"
```
``` 350 by (intro antisym le_infI cINF_greatest cINF_lower2)
```
``` 351 (auto intro: le_infI1 le_infI2 cINF_greatest cINF_lower le_infI)
```
``` 352
```
``` 353 lemma SUP_sup_distrib: "A \<noteq> {} \<Longrightarrow> bdd_above (f`A) \<Longrightarrow> bdd_above (g`A) \<Longrightarrow> sup (SUPREMUM A f) (SUPREMUM A g) = (SUP a:A. sup (f a) (g a))"
```
``` 354 by (intro antisym le_supI cSUP_least cSUP_upper2)
```
``` 355 (auto intro: le_supI1 le_supI2 cSUP_least cSUP_upper le_supI)
```
``` 356
```
``` 357 end
```
``` 358
```
``` 359 instance complete_lattice \<subseteq> conditionally_complete_lattice
```
``` 360 by default (auto intro: Sup_upper Sup_least Inf_lower Inf_greatest)
```
``` 361
```
``` 362 lemma cSup_eq:
```
``` 363 fixes a :: "'a :: {conditionally_complete_lattice, no_bot}"
```
``` 364 assumes upper: "\<And>x. x \<in> X \<Longrightarrow> x \<le> a"
```
``` 365 assumes least: "\<And>y. (\<And>x. x \<in> X \<Longrightarrow> x \<le> y) \<Longrightarrow> a \<le> y"
```
``` 366 shows "Sup X = a"
```
``` 367 proof cases
```
``` 368 assume "X = {}" with lt_ex[of a] least show ?thesis by (auto simp: less_le_not_le)
```
``` 369 qed (intro cSup_eq_non_empty assms)
```
``` 370
```
``` 371 lemma cInf_eq:
```
``` 372 fixes a :: "'a :: {conditionally_complete_lattice, no_top}"
```
``` 373 assumes upper: "\<And>x. x \<in> X \<Longrightarrow> a \<le> x"
```
``` 374 assumes least: "\<And>y. (\<And>x. x \<in> X \<Longrightarrow> y \<le> x) \<Longrightarrow> y \<le> a"
```
``` 375 shows "Inf X = a"
```
``` 376 proof cases
```
``` 377 assume "X = {}" with gt_ex[of a] least show ?thesis by (auto simp: less_le_not_le)
```
``` 378 qed (intro cInf_eq_non_empty assms)
```
``` 379
```
``` 380 class conditionally_complete_linorder = conditionally_complete_lattice + linorder
```
``` 381 begin
```
``` 382
```
``` 383 lemma less_cSup_iff : (*REAL_SUP_LE in HOL4*)
```
``` 384 "X \<noteq> {} \<Longrightarrow> bdd_above X \<Longrightarrow> y < Sup X \<longleftrightarrow> (\<exists>x\<in>X. y < x)"
```
``` 385 by (rule iffI) (metis cSup_least not_less, metis cSup_upper less_le_trans)
```
``` 386
```
``` 387 lemma cInf_less_iff: "X \<noteq> {} \<Longrightarrow> bdd_below X \<Longrightarrow> Inf X < y \<longleftrightarrow> (\<exists>x\<in>X. x < y)"
```
``` 388 by (rule iffI) (metis cInf_greatest not_less, metis cInf_lower le_less_trans)
```
``` 389
```
``` 390 lemma cINF_less_iff: "A \<noteq> {} \<Longrightarrow> bdd_below (f`A) \<Longrightarrow> (INF i:A. f i) < a \<longleftrightarrow> (\<exists>x\<in>A. f x < a)"
```
``` 391 using cInf_less_iff[of "f`A"] by auto
```
``` 392
```
``` 393 lemma less_cSUP_iff: "A \<noteq> {} \<Longrightarrow> bdd_above (f`A) \<Longrightarrow> a < (SUP i:A. f i) \<longleftrightarrow> (\<exists>x\<in>A. a < f x)"
```
``` 394 using less_cSup_iff[of "f`A"] by auto
```
``` 395
```
``` 396 lemma less_cSupE:
```
``` 397 assumes "y < Sup X" "X \<noteq> {}" obtains x where "x \<in> X" "y < x"
```
``` 398 by (metis cSup_least assms not_le that)
```
``` 399
```
``` 400 lemma less_cSupD:
```
``` 401 "X \<noteq> {} \<Longrightarrow> z < Sup X \<Longrightarrow> \<exists>x\<in>X. z < x"
```
``` 402 by (metis less_cSup_iff not_leE bdd_above_def)
```
``` 403
```
``` 404 lemma cInf_lessD:
```
``` 405 "X \<noteq> {} \<Longrightarrow> Inf X < z \<Longrightarrow> \<exists>x\<in>X. x < z"
```
``` 406 by (metis cInf_less_iff not_leE bdd_below_def)
```
``` 407
```
``` 408 lemma complete_interval:
```
``` 409 assumes "a < b" and "P a" and "\<not> P b"
```
``` 410 shows "\<exists>c. a \<le> c \<and> c \<le> b \<and> (\<forall>x. a \<le> x \<and> x < c \<longrightarrow> P x) \<and>
```
``` 411 (\<forall>d. (\<forall>x. a \<le> x \<and> x < d \<longrightarrow> P x) \<longrightarrow> d \<le> c)"
```
``` 412 proof (rule exI [where x = "Sup {d. \<forall>x. a \<le> x & x < d --> P x}"], auto)
```
``` 413 show "a \<le> Sup {d. \<forall>c. a \<le> c \<and> c < d \<longrightarrow> P c}"
```
``` 414 by (rule cSup_upper, auto simp: bdd_above_def)
```
``` 415 (metis `a < b` `\<not> P b` linear less_le)
```
``` 416 next
```
``` 417 show "Sup {d. \<forall>c. a \<le> c \<and> c < d \<longrightarrow> P c} \<le> b"
```
``` 418 apply (rule cSup_least)
```
``` 419 apply auto
```
``` 420 apply (metis less_le_not_le)
```
``` 421 apply (metis `a<b` `~ P b` linear less_le)
```
``` 422 done
```
``` 423 next
```
``` 424 fix x
```
``` 425 assume x: "a \<le> x" and lt: "x < Sup {d. \<forall>c. a \<le> c \<and> c < d \<longrightarrow> P c}"
```
``` 426 show "P x"
```
``` 427 apply (rule less_cSupE [OF lt], auto)
```
``` 428 apply (metis less_le_not_le)
```
``` 429 apply (metis x)
```
``` 430 done
```
``` 431 next
```
``` 432 fix d
```
``` 433 assume 0: "\<forall>x. a \<le> x \<and> x < d \<longrightarrow> P x"
```
``` 434 thus "d \<le> Sup {d. \<forall>c. a \<le> c \<and> c < d \<longrightarrow> P c}"
```
``` 435 by (rule_tac cSup_upper, auto simp: bdd_above_def)
```
``` 436 (metis `a<b` `~ P b` linear less_le)
```
``` 437 qed
```
``` 438
```
``` 439 end
```
``` 440
```
``` 441 lemma cSup_eq_Max: "finite (X::'a::conditionally_complete_linorder set) \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Sup X = Max X"
```
``` 442 using cSup_eq_Sup_fin[of X] Sup_fin_eq_Max[of X] by simp
```
``` 443
```
``` 444 lemma cInf_eq_Min: "finite (X::'a::conditionally_complete_linorder set) \<Longrightarrow> X \<noteq> {} \<Longrightarrow> Inf X = Min X"
```
``` 445 using cInf_eq_Inf_fin[of X] Inf_fin_eq_Min[of X] by simp
```
``` 446
```
``` 447 lemma cSup_lessThan[simp]: "Sup {..<x::'a::{conditionally_complete_linorder, no_bot, dense_linorder}} = x"
```
``` 448 by (auto intro!: cSup_eq_non_empty intro: dense_le)
```
``` 449
```
``` 450 lemma cSup_greaterThanLessThan[simp]: "y < x \<Longrightarrow> Sup {y<..<x::'a::{conditionally_complete_linorder, no_bot, dense_linorder}} = x"
```
``` 451 by (auto intro!: cSup_eq intro: dense_le_bounded)
```
``` 452
```
``` 453 lemma cSup_atLeastLessThan[simp]: "y < x \<Longrightarrow> Sup {y..<x::'a::{conditionally_complete_linorder, no_bot, dense_linorder}} = x"
```
``` 454 by (auto intro!: cSup_eq intro: dense_le_bounded)
```
``` 455
```
``` 456 lemma cInf_greaterThan[simp]: "Inf {x::'a::{conditionally_complete_linorder, no_top, dense_linorder} <..} = x"
```
``` 457 by (auto intro!: cInf_eq intro: dense_ge)
```
``` 458
```
``` 459 lemma cInf_greaterThanAtMost[simp]: "y < x \<Longrightarrow> Inf {y<..x::'a::{conditionally_complete_linorder, no_top, dense_linorder}} = y"
```
``` 460 by (auto intro!: cInf_eq intro: dense_ge_bounded)
```
``` 461
```
``` 462 lemma cInf_greaterThanLessThan[simp]: "y < x \<Longrightarrow> Inf {y<..<x::'a::{conditionally_complete_linorder, no_top, dense_linorder}} = y"
```
``` 463 by (auto intro!: cInf_eq intro: dense_ge_bounded)
```
``` 464
```
``` 465 class linear_continuum = conditionally_complete_linorder + dense_linorder +
```
``` 466 assumes UNIV_not_singleton: "\<exists>a b::'a. a \<noteq> b"
```
``` 467 begin
```
``` 468
```
``` 469 lemma ex_gt_or_lt: "\<exists>b. a < b \<or> b < a"
```
``` 470 by (metis UNIV_not_singleton neq_iff)
```
``` 471
```
``` 472 end
```
``` 473
```
``` 474 instantiation nat :: conditionally_complete_linorder
```
``` 475 begin
```
``` 476
```
``` 477 definition "Sup (X::nat set) = Max X"
```
``` 478 definition "Inf (X::nat set) = (LEAST n. n \<in> X)"
```
``` 479
```
``` 480 lemma bdd_above_nat: "bdd_above X \<longleftrightarrow> finite (X::nat set)"
```
``` 481 proof
```
``` 482 assume "bdd_above X"
```
``` 483 then obtain z where "X \<subseteq> {.. z}"
```
``` 484 by (auto simp: bdd_above_def)
```
``` 485 then show "finite X"
```
``` 486 by (rule finite_subset) simp
```
``` 487 qed simp
```
``` 488
```
``` 489 instance
```
``` 490 proof
```
``` 491 fix x :: nat and X :: "nat set"
```
``` 492 { assume "x \<in> X" "bdd_below X" then show "Inf X \<le> x"
```
``` 493 by (simp add: Inf_nat_def Least_le) }
```
``` 494 { assume "X \<noteq> {}" "\<And>y. y \<in> X \<Longrightarrow> x \<le> y" then show "x \<le> Inf X"
```
``` 495 unfolding Inf_nat_def ex_in_conv[symmetric] by (rule LeastI2_ex) }
```
``` 496 { assume "x \<in> X" "bdd_above X" then show "x \<le> Sup X"
```
``` 497 by (simp add: Sup_nat_def bdd_above_nat) }
```
``` 498 { assume "X \<noteq> {}" "\<And>y. y \<in> X \<Longrightarrow> y \<le> x"
```
``` 499 moreover then have "bdd_above X"
```
``` 500 by (auto simp: bdd_above_def)
```
``` 501 ultimately show "Sup X \<le> x"
```
``` 502 by (simp add: Sup_nat_def bdd_above_nat) }
```
``` 503 qed
```
``` 504 end
```
``` 505
```
``` 506 instantiation int :: conditionally_complete_linorder
```
``` 507 begin
```
``` 508
```
``` 509 definition "Sup (X::int set) = (THE x. x \<in> X \<and> (\<forall>y\<in>X. y \<le> x))"
```
``` 510 definition "Inf (X::int set) = - (Sup (uminus ` X))"
```
``` 511
```
``` 512 instance
```
``` 513 proof
```
``` 514 { fix x :: int and X :: "int set" assume "X \<noteq> {}" "bdd_above X"
```
``` 515 then obtain x y where "X \<subseteq> {..y}" "x \<in> X"
```
``` 516 by (auto simp: bdd_above_def)
```
``` 517 then have *: "finite (X \<inter> {x..y})" "X \<inter> {x..y} \<noteq> {}" and "x \<le> y"
```
``` 518 by (auto simp: subset_eq)
```
``` 519 have "\<exists>!x\<in>X. (\<forall>y\<in>X. y \<le> x)"
```
``` 520 proof
```
``` 521 { fix z assume "z \<in> X"
```
``` 522 have "z \<le> Max (X \<inter> {x..y})"
```
``` 523 proof cases
```
``` 524 assume "x \<le> z" with `z \<in> X` `X \<subseteq> {..y}` *(1) show ?thesis
```
``` 525 by (auto intro!: Max_ge)
```
``` 526 next
```
``` 527 assume "\<not> x \<le> z"
```
``` 528 then have "z < x" by simp
```
``` 529 also have "x \<le> Max (X \<inter> {x..y})"
```
``` 530 using `x \<in> X` *(1) `x \<le> y` by (intro Max_ge) auto
```
``` 531 finally show ?thesis by simp
```
``` 532 qed }
```
``` 533 note le = this
```
``` 534 with Max_in[OF *] show ex: "Max (X \<inter> {x..y}) \<in> X \<and> (\<forall>z\<in>X. z \<le> Max (X \<inter> {x..y}))" by auto
```
``` 535
```
``` 536 fix z assume *: "z \<in> X \<and> (\<forall>y\<in>X. y \<le> z)"
```
``` 537 with le have "z \<le> Max (X \<inter> {x..y})"
```
``` 538 by auto
```
``` 539 moreover have "Max (X \<inter> {x..y}) \<le> z"
```
``` 540 using * ex by auto
```
``` 541 ultimately show "z = Max (X \<inter> {x..y})"
```
``` 542 by auto
```
``` 543 qed
```
``` 544 then have "Sup X \<in> X \<and> (\<forall>y\<in>X. y \<le> Sup X)"
```
``` 545 unfolding Sup_int_def by (rule theI') }
```
``` 546 note Sup_int = this
```
``` 547
```
``` 548 { fix x :: int and X :: "int set" assume "x \<in> X" "bdd_above X" then show "x \<le> Sup X"
```
``` 549 using Sup_int[of X] by auto }
```
``` 550 note le_Sup = this
```
``` 551 { fix x :: int and X :: "int set" assume "X \<noteq> {}" "\<And>y. y \<in> X \<Longrightarrow> y \<le> x" then show "Sup X \<le> x"
```
``` 552 using Sup_int[of X] by (auto simp: bdd_above_def) }
```
``` 553 note Sup_le = this
```
``` 554
```
``` 555 { fix x :: int and X :: "int set" assume "x \<in> X" "bdd_below X" then show "Inf X \<le> x"
```
``` 556 using le_Sup[of "-x" "uminus ` X"] by (auto simp: Inf_int_def) }
```
``` 557 { fix x :: int and X :: "int set" assume "X \<noteq> {}" "\<And>y. y \<in> X \<Longrightarrow> x \<le> y" then show "x \<le> Inf X"
```
``` 558 using Sup_le[of "uminus ` X" "-x"] by (force simp: Inf_int_def) }
```
``` 559 qed
```
``` 560 end
```
``` 561
```
``` 562 lemma interval_cases:
```
``` 563 fixes S :: "'a :: conditionally_complete_linorder set"
```
``` 564 assumes ivl: "\<And>a b x. a \<in> S \<Longrightarrow> b \<in> S \<Longrightarrow> a \<le> x \<Longrightarrow> x \<le> b \<Longrightarrow> x \<in> S"
```
``` 565 shows "\<exists>a b. S = {} \<or>
```
``` 566 S = UNIV \<or>
```
``` 567 S = {..<b} \<or>
```
``` 568 S = {..b} \<or>
```
``` 569 S = {a<..} \<or>
```
``` 570 S = {a..} \<or>
```
``` 571 S = {a<..<b} \<or>
```
``` 572 S = {a<..b} \<or>
```
``` 573 S = {a..<b} \<or>
```
``` 574 S = {a..b}"
```
``` 575 proof -
```
``` 576 def lower \<equiv> "{x. \<exists>s\<in>S. s \<le> x}" and upper \<equiv> "{x. \<exists>s\<in>S. x \<le> s}"
```
``` 577 with ivl have "S = lower \<inter> upper"
```
``` 578 by auto
```
``` 579 moreover
```
``` 580 have "\<exists>a. upper = UNIV \<or> upper = {} \<or> upper = {.. a} \<or> upper = {..< a}"
```
``` 581 proof cases
```
``` 582 assume *: "bdd_above S \<and> S \<noteq> {}"
```
``` 583 from * have "upper \<subseteq> {.. Sup S}"
```
``` 584 by (auto simp: upper_def intro: cSup_upper2)
```
``` 585 moreover from * have "{..< Sup S} \<subseteq> upper"
```
``` 586 by (force simp add: less_cSup_iff upper_def subset_eq Ball_def)
```
``` 587 ultimately have "upper = {.. Sup S} \<or> upper = {..< Sup S}"
```
``` 588 unfolding ivl_disj_un(2)[symmetric] by auto
```
``` 589 then show ?thesis by auto
```
``` 590 next
```
``` 591 assume "\<not> (bdd_above S \<and> S \<noteq> {})"
```
``` 592 then have "upper = UNIV \<or> upper = {}"
```
``` 593 by (auto simp: upper_def bdd_above_def not_le dest: less_imp_le)
```
``` 594 then show ?thesis
```
``` 595 by auto
```
``` 596 qed
```
``` 597 moreover
```
``` 598 have "\<exists>b. lower = UNIV \<or> lower = {} \<or> lower = {b ..} \<or> lower = {b <..}"
```
``` 599 proof cases
```
``` 600 assume *: "bdd_below S \<and> S \<noteq> {}"
```
``` 601 from * have "lower \<subseteq> {Inf S ..}"
```
``` 602 by (auto simp: lower_def intro: cInf_lower2)
```
``` 603 moreover from * have "{Inf S <..} \<subseteq> lower"
```
``` 604 by (force simp add: cInf_less_iff lower_def subset_eq Ball_def)
```
``` 605 ultimately have "lower = {Inf S ..} \<or> lower = {Inf S <..}"
```
``` 606 unfolding ivl_disj_un(1)[symmetric] by auto
```
``` 607 then show ?thesis by auto
```
``` 608 next
```
``` 609 assume "\<not> (bdd_below S \<and> S \<noteq> {})"
```
``` 610 then have "lower = UNIV \<or> lower = {}"
```
``` 611 by (auto simp: lower_def bdd_below_def not_le dest: less_imp_le)
```
``` 612 then show ?thesis
```
``` 613 by auto
```
``` 614 qed
```
``` 615 ultimately show ?thesis
```
``` 616 unfolding greaterThanAtMost_def greaterThanLessThan_def atLeastAtMost_def atLeastLessThan_def
```
``` 617 by (elim exE disjE) auto
```
``` 618 qed
```
``` 619
```
``` 620 end
``` | 13,429 | 38,274 | {"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-2020-34 | latest | en | 0.53173 |
http://math453fall2008.wikidot.com/lecture-29 | 1,518,943,842,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811795.10/warc/CC-MAIN-20180218081112-20180218101112-00301.warc.gz | 234,626,131 | 12,598 | Lecture 29: Analytic Number Theory and the Riemann Zeta Function
# Summary
Today we took a sneak peak at the kind of number theory which comes after a typical "elementary" course like ours. Much of our time was devoted to studying properties of the famous Riemann Zeta function. Along the way we saw the so-called Euler Product formula for $\zeta(s)$, and we used this to give a new proof of the infinitude of primes. We then discussed the kinds of properties of $\zeta(s)$ which mathematicians today are currently interested in.
# Introducing the Riemann Zeta Function
So far in this class we've talked mainly about "elementary" number theory, meaning the kinds of number theoretic statements one can make and prove that are concerned only with basic definitions of divisibility. There is, however, an entire other branch of number theory which uses ideas from analysis to prove properties about numbers. Much of the work in this area started with the investigation of the function
(1)
\begin{align} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}. \end{align}
Notice that good ol' fashioned p-series tests will show that $\zeta(s)$ is defined when $s>1$, and that $\zeta(1)$ is undefined (since it corresponds to the harmonic series).
Today this function is called the Riemann Zeta function, but it has been an object of interest for a long time. One of the first people who made significant progress in understanding this function was our good friend Euler. He was able to use this function to prove many classical results in number theory. For instance, the following theorem is a kind of analogue of the Fundamental Theorem of Arithmetic. We'll wind up using it to give a new proof of the infinitude of primes.
The Riemann Zeta function can be expressed as the following infinite product:
$\displaystyle\zeta(s) = \prod_{p \mbox{\tiny{ prime}}}\left(1-p^{-s}\right)^{-1}.$
Proof: We define the function $P(n)$ to return the largest prime factor of the input number n. Using the series identity
(2)
\begin{align} \frac{1}{1-x} = 1+x+x^2+x^3+x^4+\cdots \end{align}
then gives
(3)
\begin{align} \prod_{p<N}\left(1-p^{-s}\right)^{-1} = \prod_{p<N} (1+p^{-s}+p^{-2s}+p^{-3s}+\cdots ). \end{align}
Now notice that any number n which has $P(n)\leq N$ can be expressed as
(4)
\begin{align} n = p_1^{e_1}\cdots p_k^{e_k} \end{align}
where each prime $p_i\leq N$. Hence we have
(5)
\begin{align} n^s = p_1^{se_1}\cdots p_k^{se_k}, \end{align}
and so multiplying out the infinite sums above shows us
(6)
\begin{align} \prod_{p<N}\left(1-p^{-s}\right)^{-1} = \prod_{p<N} (1+p^{-s}+p^{-2s}+p^{-3s}+\cdots ) = \sum_{P(n)\leq N}\frac{1}{n^s}. \end{align}
Now the Riemann Zeta function can be split into two pieces:
(7)
\begin{align} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} = \underbrace{\sum_{P(n)\leq N}\frac{1}{n^s}}_{T_1}+ \underbrace{\sum_{P(n)>N}\frac{1}{n^s}}_{T_2}. \end{align}
We've already seen that the $T_1$ term is given by a certain finite Euler product, so we consider the $T_2$. Note that this term is bounded from above:
(8)
\begin{align} T_2 = \sum_{P(n)>N}\frac{1}{n^s} \leq \sum_{n>N}\frac{1}{n^s}, \end{align}
where this upper bound is the tail of the (convergent) series $\zeta(s)$. Since the series is convergent, we know this tail must approach 0, and hence $T_2 \rightarrow 0$ as $N\rightarrow 0$. Recalling Equation (6) we get
(9)
\begin{split} \zeta(s) &= \lim_{N \rightarrow \infty} \left(\sum_{P(n)\leq N} \frac{1}{n^s} + \sum_{P(n)>N} \frac{1}{n^s}\right) \\ &=\lim_{N \rightarrow \infty}\left(\sum_{P(n)\leq N} \frac{1}{n^s}\right) \\ &= \lim_{N \rightarrow \infty} \left(\prod_{p \leq N} \left(1-p^{-s}\right)^{-1}\right) \\ &= \prod_{p} \left(1-p^{-s}\right)^{-1}. \end{split}
$\square$
Now that we have the Euler Product formula, we can use it to prove other results. One of Euler's most famous theorems is the proof that the sum of the reciprocals of the prime numbers diverges. In particular, this shows that there are infinitely many prime numbers.
Theorem: The sum $\displaystyle \sum_{p} \frac{1}{p}$ diverges.
Proof: Recall the series for $\log(1-x)$:
(10)
\begin{align} \log(1-x) = - \sum_{r=1}^\infty \frac{x^r}{r}. \end{align}
This lets us compute the logarithm of the zeta function:
(11)
\begin{align} \log(\zeta(s)) = \log\left(\prod_{p}(1-p^{-s})^{-1}\right) = -\sum_p \log(1-p^{-s}) = \sum_p \sum_{r=1}^\infty \frac{1}{rp^{rs}}. \end{align}
Now we'll split this into two terms:
(12)
\begin{align} \log(\zeta(s)) =\sum_p \sum_{r=1}^\infty \frac{1}{rp^{rs}} = \underbrace{\sum_p \frac{1}{p}}_{T_1} + \underbrace{\sum_p \sum_{r=2}^\infty \frac{1}{rp^{rs}}}_{T_2}. \end{align}
Of course we want to show that $T_1$ is infinite.
To do this, we first notice that we can bound the second term. In fact, notice that since $s>1$ and $r \geq 1$, we have $\frac{1}{rp^{rs}} \leq \frac{1}{p^r}$. Hence we have
(13)
\begin{align} T_2 = \sum_p \sum_{r=2}^\infty \frac{1}{rp^{rs}} \leq \sum_p \sum_{r=2}^{\infty} \frac{1}{p^r}. \end{align}
The latter term is a geometric series which we can sum using the standard geometric series formula
(14)
\begin{align} \sum_{k=i}^\infty x^k = \frac{x^i}{1-x}. \end{align}
Substituting this formula back into Equation (13) (with $x = \frac{1}{p}$) then gives
(15)
\begin{align} T_2 \leq \sum_p \sum_{r=2}^{\infty} \frac{1}{p^r} = \sum_p \frac{(\frac{1}{p})^2}{1-\frac{1}{p}} = \sum_p \frac{1}{p(p-1)}. \end{align}
Notice that this term is finite, since you can compare it to the convergent sequence $\sum \frac{1}{n^2}$. Hence we can conclude that
(16)
\begin{align} \log(\zeta(s)) = \sum_{p}\frac{1}{p} + \mbox{some finite quantity}. \end{align}
To finish off the proof, note that $\lim_{s \to 1^+} \zeta(s) = \infty$, since the harmonic series diverges. Hence we know that $\lim_{s \to 1^+} \log(\zeta(s))$ diverges as well. Considering our formula for $\log(\zeta(s))$ above, the only way that it can diverge is if
(17)
\begin{align} \sum_p \frac{1}{p} = \lim_{s \to 1^+} \sum_p \frac{1}{p} = \infty. \end{align}
This is the result we wanted to prove. $\square$
As a final note before moving onto the next topic, we point out that the Riemann Zeta function is connected in some intimate ways to prime numbers. For instance, the function $\pi(x)$ — the prime counting function — shows up in the evaluation of $\log(\zeta(s))$; one can prove that
(18)
\begin{align} \log(\zeta(s)) = s\int_2^\infty \frac{\pi(x)}{x(x^s-1)}~dx. \end{align}
The proof isn't all that hard, but we didn't have time to talk about it in class.
# Twisted Riemann Zeta Functions
In class I failed to mention that the proof we gave for the Euler Product formula for the Riemann Zeta function could be extended in the following way:
Theorem: If f is a multiplicative function, then
$\displaystyle \sum_{n=1}^\infty \frac{f(n)}{n^s} = \prod_p (1+\frac{f(p)}{p^s} + \frac{f(p^2)}{p^{2s}} + \frac{f(p^3)}{p^{3s}} + \cdots )$.
#### Example
Let's use this theorem to compute $\sum_{n=1}^\infty \frac{\sigma(n)}{n^s}$. According to our formula above, we get
(19)
\begin{split} \sum_{n=1}^\infty \frac{\sigma(n)}{n^s} &= \prod_p \left(1+\frac{\sigma(p)}{p^s} + \frac{\sigma(p^2)}{p^{2s}} + \frac{\sigma(p^3)}{p^{3s}} + \cdots \\ &= \prod_p (1+\frac{p+1}{p^s} + \frac{p^2+p+1}{p^{2s}} + \frac{p^3+p^2+p+1}{p^{3s}} + \cdots ) \\ &= \prod_p (1+\frac{1}{p^{s-1}} + \frac{1}{p^s} + \frac{1}{p^{2s-2}} + \frac{1}{p^{2s-1}} + \frac{1}{p^{2s}} + \frac{1}{p^{3s-3}} + \frac{1}{p^{3s-2}} + \frac{1}{p^{3s-1}} + \frac{1}{p^{3s}} + \cdots). \end{split}
If you stare at the above expansion long enough, you'll recognize that it can also be written as
(20)
\begin{align} \prod_p (1+\frac{1}{p^{s-1}}+\frac{1}{p^{2(s-1)}} + \frac{1}{p^{3(s-1)}} + \cdots ) \prod_{p} (1+\frac{1}{p^s} + \frac{1}{p^{2s}} + \frac{1}{p^{3s}} + \cdots ), \end{align}
which is the same as $\zeta(s-1)\zeta(s)$. $\square$
# Extending the Riemann Zeta Function
We've been studying the Riemann Zeta function as a function of a real variable, meaning that we've been thinking of it as a function which we evaluate at some real number s. One of the big ideas in this branch of number theory is to recognize that we can also place complex numbers into this function. In doing so, it turns out that $\zeta(s)$ is a "really nice" complex function whenever the real part of the input s is greater than 1; the actual lingo is "analytic" on the domain $\mathfrak{R}(s)>1$, which basically means that it is a super smooth complex function. (The notation $\mathfrak{R}(s)>1$ means "the real part of s is greater than 1".) This "analytic-ness" of the $\zeta$ function has lots of important consequences. One of those consequences is something called the "analytic continuation" of the $\zeta$ function. Basically, the result is that there is one (and only one) way to extend the Riemann Zeta function to a domain larger than $\mathfrak{R}(s)>1$. In fact, one can define $\zeta(s)$ for every complex number except $s=1$. One way to do this is to use the so-called functional equation of $\zeta$:
(21)
\begin{align} \zeta(s) = 2^s\pi^{s-1}\sin\left(\frac{\pi s}{2}\right) \Gamma(1-s)\zeta(1-s), \end{align}
where the $\Gamma$ function is
(22)
In particular, we can use (21) to evaluate $\zeta(s)$ when the real part of s is smaller than 1: in this case, the real part of 1-s is bigger than one, and so we can evaluate the left-hand side (when the definition of $\zeta(s)$ is otherwise unknown) by evaluating the right hand side (where evaluating $\zeta(1-s)$ is defined).
The big question about the Riemann Zeta function is the following
Riemann Hypothesis: The "nontrivial" values of s for which $\zeta(s) = 0$ are all found on the line $\mathfrak{R}(s) = \frac{1}{2}$.
Believe it or not, this is the "holy grail" of number theory.
# Some Analytic Number Theory Results
To finish the class, we pointed out a few results which are born from an analytic approach to number theory. These results were "average value" results, stating the "average value" of a given arithmetic function. The average value of an arithmetic function f is computed as
(23)
\begin{align} \lim_{N \to \infty}\frac{\sum_{n \leq N}f(n)}{N}. \end{align}
Can you see why this value earns the name "average value"?
Here are some cool results:
(24)
\begin{split} \lim_{N \to \infty} \frac{\sum_{n \leq N}\nu(n)}{N} &\approx \log(N)\\ \lim_{N \to \infty} \frac{\sum_{n \leq N}\phi(n)}{N} &\approx \frac{3N}{\pi^2} \end{split}
These results says that if you pick up a random number n, you should expect that
(25)
\begin{align} \nu(n) \approx \log(n) \quad \mbox{ and } \quad \phi(n) \approx \frac{3n}{\pi^2}. \end{align} | 3,568 | 10,631 | {"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": 22, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-09 | longest | en | 0.919023 |
https://www.solvedlib.com/please-answer-all-questions-thank-u,179951 | 1,718,782,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00023.warc.gz | 886,306,318 | 15,992 | #### Similar Solved Questions
##### Consider the data set: (−2, −39),(0, 3),(1, 6),(3, 36)a. If we rearrange the points (for example, (0, 3),(−2, −39),(3,36),(1, 6)), what is the Lagrange form of the interpolatingpolynomial? b. If we rearrange the points (for example, (0, 3),(−2, −39),(3,36),(1, 6)), what is the Newton form of the interpolatingpolynomial?
Consider the data set: (−2, −39),(0, 3),(1, 6),(3, 36) a. If we rearrange the points (for example, (0, 3),(−2, −39),(3, 36),(1, 6)), what is the Lagrange form of the interpolating polynomial? b. If we rearrange the points (for example, (0, 3),(−2, −39),(3, 36),(1...
##### Calculate the amount of binder needed to prepare a briquette,if mass of aggregate is 1201.8gm and...
calculate the amount of binder needed to prepare a briquette,if mass of aggregate is 1201.8gm and asphalt content is 7.6%....
##### Use the simplex method to find the optimal solution_ Assume all variables are nonnegative _Maximize f = Sx + 2y subject to the following constraints_ ~X 2y 3x 2y < ~30 Y <Need Help?Read ItWatch
Use the simplex method to find the optimal solution_ Assume all variables are nonnegative _ Maximize f = Sx + 2y subject to the following constraints_ ~X 2y 3x 2y < ~30 Y < Need Help? Read It Watch...
##### Consider the recursively defined sequence a1 shows @n is a decreasing sequence.1, an+lan Below is n+l'proof' thatProof: We prove Wn an+l for all n > 1, by induction on n. Base_case: For n = 1, 01 = 1, 62 = ! and 1 > % Inductive Hypothesis: Suppose ak @k+1 for all k 2 1. Inductive Step: We want to show ak+l ak+2:Wk @k+1 k +1 ak+l by the Inductive Hypothesis k+1 @k+1 k+2 @k+2Therefore; Gk-+l ak+2; and so by mathematical induction, an @n+l for all n > 1.Although it is true that t
Consider the recursively defined sequence a1 shows @n is a decreasing sequence. 1, an+l an Below is n+l 'proof' that Proof: We prove Wn an+l for all n > 1, by induction on n. Base_case: For n = 1, 01 = 1, 62 = ! and 1 > % Inductive Hypothesis: Suppose ak @k+1 for all k 2 1. Inductive...
##### Let Z be standard normal random varable Use the calculator provided_this table_ to determine the velue 0f €.P(c<z<1.2I) - 0.8501 Carry Your Intermediate computations to at least four decima places . Round your answer to two decimal places:
Let Z be standard normal random varable Use the calculator provided_ this table_ to determine the velue 0f €. P(c<z<1.2I) - 0.8501 Carry Your Intermediate computations to at least four decima places . Round your answer to two decimal places:...
##### 2.A pH = 2.00 buffer can be prepared by mixing 75.0 mLof040 M NaH2PO4 with the correct volume of 6.1060 M HCL Determine the volume of the 0,1OQ M HCI needed: Table experiment t0 identify the acid needed to prepare the pH = 2.00 buffer and i*s Ka Please refer t0
2.A pH = 2.00 buffer can be prepared by mixing 75.0 mLof040 M NaH2PO4 with the correct volume of 6.1060 M HCL Determine the volume of the 0,1OQ M HCI needed: Table experiment t0 identify the acid needed to prepare the pH = 2.00 buffer and i*s Ka Please refer t0...
##### Find the domain of the function. $g(x)=\frac{5}{\sqrt{4+x}}$
Find the domain of the function. $g(x)=\frac{5}{\sqrt{4+x}}$...
##### How to find y' by implicit differentiation and the slope of the graph when x=5?
When graphed x^2+y^2 - 169=0...
##### QUESTION 5 Part 1 of 2: Solve the system using the matrix equation. Be sure to...
QUESTION 5 Part 1 of 2: Solve the system using the matrix equation. Be sure to show all 3 parts! Write your solution as an ordered pair. QUESTION 6 Part 2 of 2: Solve the system of equations you solved in question 5 again, this time using Cramer's Rule. Verify that you come up with the same sol...
##### Qunation ComplationDurntlorInent "uul Flat- 'lts) Uir Juiuri Wrc Mculatithalthe nleval aIlcuiyimu puuujuu Frcootio thatche ccrldere nler Allrol (uiutQuuJum CiJn uiaininalurt Lpcc uidhrci Wucuue Dltnt quinotmto Icd olcore Ienoe LTAr ttotDnonEannuctnd Manoncno datetmn] Doecno muulk duat [aedeslmte Id teMuotesoltotnc amelil D NJ7 A7Jm Inthana JordLeh-no Tota14441 147u 649id 344 S7E5mR576114 RS 76SNONunbaeennuen Wnumulalznly Utu 0 Ereturtidz 0 puijutnranrlrLendoTatthhitWItdVlMlothn Dhne
Qunation Complation Durntlor Inent "uul Flat- 'lts) Uir Juiuri Wrc Mculatithalthe nleval aIlcuiyimu puuujuu Frcootio thatche ccrldere nler Allrol (uiutQuuJum CiJn uiaininalurt Lpcc uidhrci Wucuue Dltnt quinotmto Icd olcore Ienoe LTAr ttot Dnon Eannuctnd Manoncno datetmn] Doecno muulk du...
##### "In the long run, we're all dead" Interest rates fall when money saved exceeds the demand...
"In the long run, we're all dead" Interest rates fall when money saved exceeds the demand for those funds for investment, until savings equals investment Capitalist economies will normally fail to reach full employment due to insufficient aggregate demand The paradox of thrift Investment...
##### Consider a spherical bacterium, withradius 0.65 μm , falling in water at 20° C. Find theterminal speed of the spherical bacterium in meters per second,ignoring the buoyant force on the bacterium and assuming Stokes'law for the viscous force. You will first need to note that thedrag force is equal to the weight at terminal velocity. Take thedensity of the bacterium to be 1.35 ×103 kg/m3. The viscosity of water at 20°C is 1.002 × 10-3 kg/m·s and the density is 998kg/m3.
Consider a spherical bacterium, with radius 0.65 μm , falling in water at 20° C. Find the terminal speed of the spherical bacterium in meters per second, ignoring the buoyant force on the bacterium and assuming Stokes' law for the viscous force. You will first need to note that the drag f...
##### \begin{aligned} &x^{2}+y^{2}=14\\ &x^{2}-y^{2}=4 \end{aligned}
\begin{aligned} &x^{2}+y^{2}=14\\ &x^{2}-y^{2}=4 \end{aligned}...
##### Please explain in detail thank you! 1 .What are the 4 general types of spinal cord...
Please explain in detail thank you! 1 .What are the 4 general types of spinal cord injuries? 2. What are the phases of spinal cord injuries responses that occur after injury? (3 or more)...
Go Huskies is considering replacing a point-of-sale system that is currently being used. The old system is fully depreciated but can be used for another 4 years, at which time it would have no salvage value. Go Huskies can sell the old system for $45,000 on the date that the ne... 5 answers ##### When considering polysaccharides, what is the significance ofcoupling the monosaccharides through the 1-4 carbon linkagescompared to 1-6 linkages.Draw:the structures of steric acid, oleic acid, and linoleicacid.Predict which of these molecules should have the higher meltingpoint.Justify your answer. When considering polysaccharides, what is the significance of coupling the monosaccharides through the 1-4 carbon linkages compared to 1-6 linkages. Draw: the structures of steric acid, oleic acid, and linoleic acid. Predict which of these molecules should have the higher melting point. Justify your... 1 answer ##### PLYWOOD WITH THE SURFACE MEASURES 8/9 SQUARE YARD IS CUT INTO SECTIONS EACH CONTAING 2/9 SQUARE YARDS PLYWOOD WITH THE SURFACE MEASURES 8/9 SQUARE YARD IS CUT INTO SECTIONS EACH CONTAING 2/9 SQUARE YARDS. HOW MANY SECTIONS ARE THERE?... 1 answer ##### The data shown to the right for the dependent variable, and the independent variable, xhave been... The data shown to the right for the dependent variable, and the independent variable, xhave been collected using simple random sampling a Construct a scatter plot for these data Based on the scatter plot, how would you describe the relationship between the two variables? b Compute the correlation co... 5 answers ##### Cud_ton AolWhat Is tha correct IUPAC name for the compound ghown here?pent nept meinelh nexchloroVneaneone Cud_ton Aol What Is tha correct IUPAC name for the compound ghown here? pent nept mein elh nex chloro Vne ane one... 5 answers ##### Pcovide (1~Jac~blkc Sjnthuis Fsc followic, Gsmtovndl fom + indicakd starhc, makiul Stttins o+L( Ceast4ts 4ov deem ncrssarg) You mad us<Chjc=cchzChzCH; frm 1- Protand 0l 0ly Sol (erbenB.) Trans ~ I,7-eyclohex anediofo~ ctelohtanol Pcovide (1~Jac~blkc Sjnthuis Fsc followic, Gsmtovndl fom + indicakd starhc, makiul Stttins o+L( Ceast4ts 4ov deem ncrssarg) You mad us< Chjc=cchzChzCH; frm 1- Protand 0l 0ly Sol (erben B.) Trans ~ I,7-eyclohex anedio fo~ ctelohtanol... 1 answer ##### What is Planet X? What is Planet X?... 1 answer ##### [Review Topics] [References) Compounds X and Y are both CH, Cl products formed in the radical... [Review Topics] [References) Compounds X and Y are both CH, Cl products formed in the radical chlorination of 2,4-dimethylpentane. Base-promoted E2 elimination of X and Y gives, in each case, a single C H 4 alkene. Both X and Y undergo an Sy2 reaction with sodium iodide in acetone solution to give C... 5 answers ##### H(r)drandh(r)dr 10.8what does the following integral equal?h(r)dr h(r)dr and h(r)dr 10.8 what does the following integral equal? h(r)dr... 5 answers ##### Whal iS [4] 9] 7 5 7Malices and 1^-[" : B = [3] Whal iS [4] 9] 7 5 7 Malices and 1 ^-[" : B = [3]... 5 answers ##### 2 3 2 4 8 : I { !Hi n 3 3| 8 2 3 8 1 2 3 2 4 8 : I { !Hi n 3 3| 8 2 3 8 1... 5 answers ##### Please explain with detail how CRISPR Cas9 functions and its purpose (elements used, how it can induce DNA change and aa change, how can CRISP Cas9 technology be used with epigenetics) Please explain with detail how CRISPR Cas9 functions and its purpose (elements used, how it can induce DNA change and aa change, how can CRISP Cas9 technology be used with epigenetics)... 5 answers ##### The Wall Street Journal reports that 33% oftaxpayers with adjusted gross incomes between$30,000 and $60,000itemized deductions on their federal income tax return. The meanamount of deductions for this population of taxpayers was$16,642.Assume the standard deviation is 𜎠= $2,400.(a)What is the probability that a sample of taxpayers from thisincome group who have itemized deductions will show a sample meanwithin$200 of the population mean for each of the following samplesizes: 20, 60, 100,
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and$60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is 𜎠=$2,400. (a) W...
##### Find the volume of the solid shown in the figure.
Find the volume of the solid shown in the figure....
##### Prob. 4 Consider a slender uniform homogeneous bar of total length 1.5 m and mass 4.5 kg that is bent and welded to form the equilateral triangular object supported at a hinge in a vertical plane...
Prob. 4 Consider a slender uniform homogeneous bar of total length 1.5 m and mass 4.5 kg that is bent and welded to form the equilateral triangular object supported at a hinge in a vertical plane as shown. If the angular velocity is constant at 5 rad/s, determine the components of the reactions at t...
##### It takes 945. kJ/mol to break a nitrogen-nitrogen triple bond. Calculate the maximum wavelength of light...
It takes 945. kJ/mol to break a nitrogen-nitrogen triple bond. Calculate the maximum wavelength of light for which a nitrogen-nitrogen triple bond could be broken by absorbing a single photon. Round your answer to 3 significant digits. xs ?...
##### The dispatcher at a central tire station has observed that the arrival time of calls form a Poisson process with mean rate 1.5 per hour:What is the mean number o minutes between arriving calls?What is the probability that there will be exactly three calls during the next 30 minutes?The most recent call arrived at H:4SAM and itis now noon What is the probability that the next call will occur between 12.15 PM and 12.45 PM?
The dispatcher at a central tire station has observed that the arrival time of calls form a Poisson process with mean rate 1.5 per hour: What is the mean number o minutes between arriving calls? What is the probability that there will be exactly three calls during the next 30 minutes? The most recen...
##### What is a meter based on the use of?
What is a meter based on the use of?... | 3,594 | 12,226 | {"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": 2, "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} | 3.3125 | 3 | CC-MAIN-2024-26 | latest | en | 0.781982 |
https://www.transtutors.com/questions/a-parallel-plate-vacuum-capacitor-with-plate-area-a-and-separati-421201.htm | 1,624,535,853,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488553635.87/warc/CC-MAIN-20210624110458-20210624140458-00483.warc.gz | 923,938,652 | 12,727 | # A parallel-plate vacuum capacitor with plate area A and separati 1 answer below »
A parallel-plate vacuum capacitor with plate area A and separation x has charges + Q and - Q on its plates. The capacitor is disconnected from the source of charge, so the charge on each plate remains fixed.
(a) What is the total energy stored in the capacitor?
(b) The plates are pulled apart an additional distance dx. What is the change in the stored energy?
(c) If F is the force with which the plates attract each other, then the change in the stored energy must equal the work dW = Fdx done in pulling the plates apart. Find an expression for F.
(d) Explain why F is not equal to QE, where E is the electric field between the plates.
Siddhant D
Looking for Something Else? Ask a Similar Question | 176 | 788 | {"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} | 2.734375 | 3 | CC-MAIN-2021-25 | latest | en | 0.911386 |
http://www.lofoya.com/Aptitude-Questions-and-Answers/Algebra/l1p6 | 1,516,165,766,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00009.warc.gz | 504,681,300 | 14,716 | # Practice Questions on AlgebraAptitude Questions and Answers
## Easy Algebra Question - 26
Q26. A club consists of members whose ages are in A.P. The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are:
A. 18 B. 20 C. 25 D. 24
## Easy Algebra Question - 27
Q27. 21 pencils and 29 pens cost Rs 79. But if the number of pencils and pens were interchanged, the cost would have reduced by Rs 8. Find the cost of each pen.
A. Re 1 B. Re 2 C. Re 3 D. Re 4
## Easy Algebra Question - 28
Q28. A person buys 18 local tickets for Rs 110. Each first class ticket costs Rs 10 and each second class ticket costs Rs 3. What will another lot of 18 tickets in which the numbers of first class and second class tickets are interchanged cost?
A. 112 B. 118 C. 121 D. 124
## Easy Algebra Question - 29
Q29. Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is:
A. 90 B. 94 C. 92 D. 96
## Easy Algebra Question - 30
Q30. Large, medium and small ships are used to bring water. 4 large ships carry as much water as 7 small ships. 3 medium ships carry the same amount of water as 2 large ships and 1 small ship. 15 large, 7 medium and 14 small ships, each made 36 journey and brought a certain quantity of water. In how many journeys would 12 large, 14 medium and 21 small ships bring the same quantity?
A. 32 B. 29 C. 49 D. 25
Hide
Number formats
Decimals
Lofoya.com 2018
You may drag this calculator | 482 | 1,694 | {"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} | 4.0625 | 4 | CC-MAIN-2018-05 | latest | en | 0.933436 |
https://www.coursehero.com/file/8814691/Suppose-Y-is-not-observed-but-that-we-wish-to-estimate-Y-If/ | 1,498,485,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320763.95/warc/CC-MAIN-20170626133830-20170626153830-00539.warc.gz | 847,942,312 | 23,462 | Isye 2027
# Suppose y is not observed but that we wish to
This preview shows page 1. Sign up to view the full content.
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: ently rolled. Let Xk = 1 if one shows on the k th die 0 else Yk = 1 if two shows on the k th die 0 else. Let X = n=1 Xk , which is the number of one’s showing, and Y = n=1 Xk , which is the number k k of two’s showing. Note that if a histogram is made recording the number of occurrences of each of the six numbers, then X and Y are the heights of the first two entries in the histogram. (a) Find E [X1 ] and Var(X1 ). (b) Find E [X ] and Var(X ). (c) Find Cov(Xi , Yj ) if 1 ≤ i, j ≤ n (Hint: Does it make a difference if i = j ?) (d) Find Cov(X, Y ). (e) Find the correlation coefficient ρX,Y . Are X and Y positively correlated, uncorrelated, or negatively correlated? 4.8. MOMENTS OF JOINTLY DISTRIBUTED RANDOM VARIABLES 155 1 Solution (a) Each Xk is a Bernoulli random variable with parameter p = 6 , so E [Xk ] = 1 and 6 2 ] − E [X ]2 = p − p2 = p(1 − p) = 5 . Var(Xk ) = E [Xk k 36 (b) E [X ] = nE [X1 ] = n , and Var(X ) = nVar(X1 ) = 5n . 6 36 (c) If i = j then Cov(Xi , Xj ) = 0, because Xi and Yj are associated with different, independent, dice rolls. If i = j the situation...
View Full Document
## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.
Ask a homework question - tutors are online | 495 | 1,509 | {"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} | 3.671875 | 4 | CC-MAIN-2017-26 | latest | en | 0.856006 |
https://education.ti.com/en/activity/detail?id=A252CA74B2744A4287EDF8A7173665D3 | 1,627,077,267,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00141.warc.gz | 228,956,024 | 17,844 | Education Technology
# Activities
• ##### Subject Area
• Math: Calculus: Derivatives
• Math: Calculus: Applications of the Derivative
• Math: Precalculus: Logistic Functions
• Math: Calculus: Limits of Functions
9-12
30 Minutes
• ##### Device
• TI-83 Plus Family
• TI-84 Plus
• TI-84 Plus Silver Edition
• TI-Navigator™
• ##### Software
LearningCheck™ Creator
• ##### Other Materials
• Overhead view screen calculator for instruction/demonstration
• Student handout
• Transparency
• USA TODAY newspapers (recommended)
## Math TODAY - World Population Milestones
#### Activity Overview
Students will create a scatter plot of the world population for the years 1950 through 2050. The United Nations Population Division data will allow students to explore real-life data that will be modeled by a logistic function. Students will have the opportunity to explore the end behavior of this model and explain what this means about the world population estimates. The derivative will be used to explore when the world population increased most rapidly during this time period.
#### Before the Activity
See the attached Teacher Edition PDF file for notes and guidelines for this activity. The Technology Guide PDF file includes step-by-step calculator instructions with screen shots.
If you are using the TI-Navigator™ Classroom Learning System, see the TI-Navigator Basic Skills Guide PDF file for information on how the system can be integrated into your activity.
#### During the Activity
Students will:
• model real-life data with a logistic function.
• determine and interpret the limiting behavior of the logistic function.
• interpret the instantaneous rate of change at the inflection point of a graph.
• Directions
• Distribute the Student Edition PDF file to your class.
• Follow the activity procedures outlined in the student document.
• You may use the Transparency PDF file to display a large image of the USA TODAY graphic during the activity.
• #### After the Activity
Distribute the assessment questions or send the attached LearningCheck™ assessment to your class to gauge student understanding of the concepts presented in the activity. | 432 | 2,165 | {"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-2021-31 | longest | en | 0.782093 |
https://community.qlik.com/thread/82918 | 1,532,271,748,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593302.74/warc/CC-MAIN-20180722135607-20180722155607-00555.warc.gz | 620,232,270 | 23,053 | 5 Replies Latest reply: May 28, 2013 5:28 AM by Andrew Smith
# Bar chart grouping
Hi,
I have a bar chart that gives a total value by month.
Can someone please tell me how I can write an expression to show the last 12 months per month but group the prior months by year?
• ###### Re: Bar chart grouping
Hi,
Use the function Year () and Month (). You can scale as the year and month.
Rebeca
• ###### Re: Bar chart grouping
Hi Rebecaa,
Could you explain further?
In my chart at the moment I have the below formula, if I want to show individual months for the last 12 months OR group previosu 12 months into a year - how can I add this to this formula?
sum
(
If(Currency = 'GBP',[BILLINGSGBP FeesBilled Amt]+[BILLINGSGBP CostsBilled Amt],
If(Currency = 'USD',[BILLINGSUSD FeesBilled Amt]+[BILLINGSUSD CostsBilled Amt],
If(Currency = 'EUR',[BILLINGSEUR FeesBilled Amt]+[BILLINGSEUR CostsBilled Amt],
If(Currency = 'HKD',[BILLINGSHKD FeesBilled Amt]+[BILLINGSHKD CostsBilled Amt],
If(Currency = 'SGD',[BILLINGSSGD FeesBilled Amt]+[BILLINGSSGD CostsBilled Amt],
If(Currency = 'CHF',[BILLINGSCHF FeesBilled Amt]+[BILLINGSCHF CostsBilled Amt]
,'0')))))))
• ###### Re: Bar chart grouping
You can send an example with data?
• ###### Re: Bar chart grouping
I thought of one way I could do this, by creating a variable that picks up the max month from my data, then add this to my existing expression (below) then creating new expressions to give me the last 12 months etc, replacing the -1 with -2,-3 etc.
Could someone tell me if my approach seems ok to do or if there is a better standard practice way of doing this?
sum(
If(Currency = 'GBP' and penum=\$(=vMaxPeriod)-1,[BILLINGSGBP FeesBilled Amt]+[BILLINGSGBP CostsBilled Amt],
If(Currency = 'USD' and penum=\$(=vMaxPeriod)-1,[BILLINGSUSD FeesBilled Amt]+[BILLINGSUSD CostsBilled Amt],
If(Currency = 'EUR' and penum=\$(=vMaxPeriod)-1,[BILLINGSEUR FeesBilled Amt]+[BILLINGSEUR CostsBilled Amt],
If(Currency = 'HKD' and penum=\$(=vMaxPeriod)-1,[BILLINGSHKD FeesBilled Amt]+[BILLINGSHKD CostsBilled Amt],
If(Currency = 'SGD' and penum=\$(=vMaxPeriod)-1,[BILLINGSSGD FeesBilled Amt]+[BILLINGSSGD CostsBilled Amt],
If(Currency = 'CHF' and penum=\$(=vMaxPeriod)-1,[BILLINGSCHF FeesBilled Amt]+[BILLINGSCHF CostsBilled Amt],'0')))))))
• ###### Re: Bar chart grouping
Do have a Master Calendar table in your data model?
Could you add an "Bar Chart Date" field to that table and use it as the dimension on your chart? Some simple if/then logic and a left join would provide the field. Even if you don't have a Master Calendar table, I guess there's no reason why this fiel couldn't be added to you main data. Apologies if I've misunderstood your issue.
Eg.
Month Barchart Date 01/01/2011 2011 01/02/2011 2011 01/03/2011 2011 01/04/2011 2011 01/05/2011 2011 01/06/2011 2011 01/07/2011 2011 01/08/2011 2011 01/09/2011 2011 01/10/2011 2011 01/11/2011 2011 01/12/2011 2011 01/01/2012 2012 01/02/2012 2012 01/03/2012 2012 01/04/2012 01/04/2012 01/05/2012 01/05/2012 01/06/2012 01/06/2012 01/07/2012 01/07/2012 01/08/2012 01/08/2012 01/09/2012 01/09/2012 01/10/2012 01/10/2012 01/11/2012 01/11/2012 01/12/2012 01/12/2012 01/01/2013 01/01/2013 01/02/2013 01/02/2013 01/03/2013 01/03/2013 01/04/2013 01/04/2013 01/05/2013 01/05/2013 | 1,147 | 3,290 | {"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} | 3.125 | 3 | CC-MAIN-2018-30 | latest | en | 0.688138 |
https://jp.mathworks.com/matlabcentral/cody/problems/2024-triangle-sequence/solutions/909325 | 1,582,979,102,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875149238.97/warc/CC-MAIN-20200229114448-20200229144448-00273.warc.gz | 428,895,480 | 15,530 | Cody
# Problem 2024. Triangle sequence
Solution 909325
Submitted on 15 Jun 2016 by Matthew Stuckey
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
n = 1; area_correct = 25; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance)
2 Pass
n = 2; area_correct = 41; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance)
3 Pass
n = 3; area_correct = 66; tolerance = 1e-12; assert(abs(triangle_sequence(n)-area_correct)<tolerance)
4 Pass
n = 50; area_correct = 439116598409; tolerance = 1e-3; assert(abs(triangle_sequence(n)-area_correct)<tolerance)
area = 4.3912e+11 | 227 | 736 | {"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-2020-10 | latest | en | 0.676722 |
https://www.cfd-online.com/Forums/fluent/37822-converge-problem-multiphase-flow.html | 1,713,977,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819668.74/warc/CC-MAIN-20240424143432-20240424173432-00151.warc.gz | 626,807,919 | 15,353 | # Converge problem for multiphase flow
Register Blogs Members List Search Today's Posts Mark Forums Read
September 7, 2005, 10:59 Converge problem for multiphase flow #1 Jen Guest Posts: n/a I am using Eulerian model to simulate two-phase turbulent jet flow in either 2D axisymmetric or 3D region. The primary phase is air, the secondary phase is water. The speed of air is 70 m/s coming from an annular region whose hydraulic diameter is 0.00254 m, and the speed of water is 1.97 m/s coming out from the center whose diameter is 0.000508 m. I use k-e turbulent model. However, I can not get converge result, because the volume fraction equation is difficult to converge. I tried different ways to mesh the domain (Quad, Tri, Quad/Tri), and tried different methods to make it converge (multigrid, decrease under-relaxation factors, use the result of mixture model as a starting point for eulerian model). The mass fraction of water is 0.005%, and the volume fraction of water is 6.15E-8. I wonder the reason is the volume fraction of water is too small, so I can not choose either mixture or eulerian model. I remember in the help document, it says when secondary phase volume fraction is less than 10%, use Discrete phase model. I don't know whether my thinking is right, or there is something wrong with my settings. Any help will be appreciated. Thanks a lot! Jen
September 7, 2005, 19:56 Re: Converge problem for multiphase flow #2 Fuping Qian Guest Posts: n/a It is difficult to converge the multiphase model. I wonder what is your converging criterion.I think the reason is not that the volume fraction of water is too small. In Fluent help manual, we can see that, for intermediate loading (e.g.0.005%),the discrete phase, mixture, and Eulerian models are all applicable. Hope this can help you.
September 8, 2005, 08:47 Re: Converge problem for multiphase flow #3 Jen Guest Posts: n/a I use the default converging criterion 0.001. Is that OK? Thax | 494 | 1,969 | {"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} | 2.90625 | 3 | CC-MAIN-2024-18 | latest | en | 0.8966 |
https://www.studypool.com/discuss/468589/how-do-i-figure-out-what-the-total-is-if-i-know-that-18-is-35-of-something?free | 1,506,223,621,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689845.76/warc/CC-MAIN-20170924025415-20170924045415-00667.warc.gz | 858,095,375 | 14,439 | Time remaining:
##### how do I figure out what the total is if I know that 18 is 35% of something?
label Mathematics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
A boy has walked 18 miles of his walkathon. This is 35% of the total miles. How do I figure out what the total miles would be?
Apr 10th, 2015
Simply do 18 divided by .35 which would equal 51.4285.
Please click on "complete". Thank you :)
Apr 10th, 2015
...
Apr 10th, 2015
...
Apr 10th, 2015
Sep 24th, 2017
check_circle | 155 | 510 | {"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-2017-39 | latest | en | 0.944051 |
https://psychologicallyastrology.com/tag/south-node/ | 1,685,971,500,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652116.60/warc/CC-MAIN-20230605121635-20230605151635-00158.warc.gz | 504,497,392 | 34,844 | The most commonly used system for timing events is Vimshottari dasha here. Depending on which lunar mansion/nakshatra you were born in you will experience the planetary dashas/periods. Not everyone will experience all the 9 graha/planetary forces’s Mahadasha during their lives. This post is about interpreting Ketu’s times.
As always there is no point to doing astrological predictions if you do not have the precise birth details. Even a difference of 1 minute will put off your Dasha calculations. Do your birth time rectification using maths or life-event matching or palmistry and then get into the dasha predictions. If your birth moon Nakshatra/Lunar mansion is Ashwini, Magha or Mula you will have your Ketu mahadasha going on at your birth. The balance of the planet’s dasha period remaining at birth is calculated using the location (degree/minute/second) of Moon in the nakshatra. If you get this birth time wrong your starting point will be wrong thus your entire life-time’s calculation will go wrong.
And then the Ketu mahadasha can occur at any point in life depending on your birth Moon nakshatra. The sub-periods/Antardasha owned by Ketu in other planet’s Mahadasha-s will also carry his flavour.
Ketu’s main periods/mahadasha or sub-periods in other planets Mahadashas will give results as per Ketu’s ability to give results in your chart.
Step 1 is to check out these following about Ketu in your chart. You need to know the power of Ketu in giving results and also the aspects of life which he represents. These keywords are going to be in focus during his periods.
1. Location of Ketu in your birth chart (divisional chart 1/D-1). Can you expect it to give its results completely or not?
• Which house and in which sign is he placed? Thus which aspects of life is he influencing.
• Is he in the sign of a friend/neutral/enemy planet. How is this planet placed? Is he strong/weak etc, as Ketu will take on his powers and ability to give results.
• Is Ketu exalted/debilitated?
• Which planets is he conjoint or aspected by? Thus which other parts of the personality is he influencing? Also Ketu will take on their powers and ability to give results too.
• Is the birth during an eclipse? Or a kaal sarpa yog?
• Has he matured?
• Which lunar mansion/nakshatra is he placed in? He will be influenced by the ruler of this nakshatra too.
• And finally is there any planet placed in Nakshatra owned by Ketu? He can be influenced by these too!
2. You can repeat this same analysis from the birth Moon sign to get the effects that this Ketu Mahadasha is going to have on the emotional aspects.
3. And you can repeat this same analysis using the divisional chart 9, Navamsha chart to get perspective on the sub-conscious and astral expectations and also the sum total of his potential. (If you find this complicated right now, don’t overwhelm yourself, do it whenever you feel more confident.)
Ketu is primarily the force of detachment and dissociation. As mindful dissociation from the manifested world is required for spiritual growth he is also called the Moksh-karak (agent of enlightenment). But broadly whichever house/aspect Ketu touches, he will make you indifferent to them. These will simply not be relevant to you, you will be apathetic to them. Sometimes being detached can give better fruits than being overly attached to things. If he is favourably placed and in the company of benefics and influencing the powerful houses of your chart he will give good results. Gains from overseas, business where you deal with intangible stuff (eg stock markets are just numbers not solid stuff), journeys to distant places including astral travel etc are possible. If he is heavily influenced by malefics and is going to give unpleasant results then he can cause, deaths, separations, amputations, losses, diseases, anxieties and related stuff. Ketu in his periods will shed off all that is consuming your energy for no good reason. All the things which are not wanted at the deeper inner levels will be cut off by Ketu.
But go through his karakatva/all that he stands for in the chart, before predicting hopeless things and causing yourself needless stress.
Step 2 is now you will need to evaluate upto the antardasha/sub-period levels at least. (Even smaller segments are called successively Pratyantar-dasha, Sookshma-dasha and Pran-dasha. The logic on how to analyse these even smaller units remains the same for each planetary owner of the sub/sub periods.) The lord of the sub-period will give his results within the broad framework of the overarching Ketu’s karakatva as above.
1. Find out the power of the sub-period planetary lord like you did for Ketu above.
• Which houses he rules over is more important now because Ketu is headless and will take cues from this sub-period lord.
• Especially if this sub-period lord is the owner of the 2nd or 7th house (Marakesh) its sub-period within the Ketu mahadasha can cause physical death.
2. The mutual relationship between Ketu and the sub-period lord. Are they friends/enemies/neutral. And their placement, if they are in the 6/8 or 2/12 uncomfortable aspects, that sub-lord’s sub-period will not be comfortable.
Step 3 is the routine planetary transits/ gochar which will modify results during their transit periods. You can get this from your software and superimpose it on your D-1 birth chart. Use a different pen so that you can make sense of the composite chart you are drawing. Or you can use the easy Ashtakvarga table method to get a sense of the transits Dashas.
Ketu Mahadasha/main period lasts for 7 years. His energies are quite intense thus this relatively short period. Within this, you will experience the following antardasha/sub-periods. These are very very general readings just to trigger your own thought processes. Each horoscope is unique so each will have its own unique analysis. Readings/combinations from books should be used for reference, as a starting point and to build into your own further analysis.
1. Ketu antardasha – 4 months 27 days
• If Ketu is placed well he will give gains in terms of professional life, personal life, may give awards and recognition, separation from negative things in life. But despite this he will give an underlying mental stress, maybe headaches.
• If Ketu is not placed well, debilitated or under malefic influence he will be no condition to give positive results. Then he will give his results of apathy and separation in the negative sense, causing separation from the positive things in life. As Ketu is a headless graha, most problems will be in some way linked to the ‘head’.
• He follows Mercury’s Mahadasha, so first of all will cause a sudden end to things which were going on in life. Things will suddenly start going out of sequence, karmic stuff will come up for resolution. Mostly endings! However if you can remain in control intellectually, you can actually make use of the spiritual things from your past lives which will now be accessible.
• You might feel alone, introverted, caught in the past during this period. Try to divert these energies into your spiritual practice if you can so that your material life remains more stable.
2. Venus antardasha – 14 months 0 days
• If Ketu and Venus have a fruitful working relationship then expect pilgrimages, visits to sacred places, companionship, profits and gains through partners. Learning opportunities and broadening of philosophical outlook.
• But if these 2 are in an uncomfortable relationship in your chart then expect eye problems, headaches, problems in relationships, they can also cause separations and divorces.
3. Sun antardasha – 4 months 6 days
• If Ketu and Sun are comfortable with each other and will give benefic results then gain in wealth, creativity, political power, favours from political leaders, a good relationship with the father, wins in legal issues is possible.
• However negative results of such a combination can give problems with the government, health related issues, fevers, heart problems, unnecessary travel to foreign locations etc.
4. Moon antardasha – 7 months 0 days
• If Ketu and Moon are mutually comfortable then expect profits in your businesses, you may construct new houses, buy new cars or start something new connected with liquids. Interactions with women will increase. More results depending on all the things Moon stands for in your horoscope.
• Else the negative effects are like this, problems from women, emotional issues destabilised life, lack of comforts etc.
5. Mars antardasha – 4 months 27 days
• If Ketu and Mars are on good terms in your chart you will start something new as can be seen from the karakatva of Mars in your chart. Eg, you can start new businesses or lands. There will be energy and confidence in pursuing your interests.
• Else the negative effects can include traumas, surgeries, accidents, damage to the vitality of the physical body, diseases which cannot be diagnosed etc.
6. Rahu antardasha – 12 months 18 days
• Rahu and Ketu are always 180 degrees apart. Ketu desires Rahu and Rahu desires only his ambitions
• If Rahu is well placed with benefic planets and is capable of giving positive results, your ambitions will be high and also will be fulfilled during this period. Especially sucess in foreign lands, markets and speculation.
• If not then you will suffer from over-reaching and falling while chasing your ambitions. Troubles and diseases may accompany.
• Here karmic influences predominate and free will is at its lowest ebb, so best lie low while this is on.
7. Jupiter antardasha – 11 months 5 days
• If Ketu and Jupiter are favourably situated then you can perform rituals like pujas, yadnyas etc, publicly visible ceremonies. There will be prosperity, celebrations and gains. Maybe marriages and birth of children.
• If not then there will separations from the aspects of life defined by Jupiter in in the horoscope.
8. Saturn antardasha – 13 months 9 days
• If Ketu and Saturn are both capable of giving good results with each other then social mobility, hard work which yields success, change of job for the better,
• If not then this period can cause physical problems related to bones, teeth, nerves, obstacles, death of parents (if indicated), social defamations, loss in power and money, theft, long journeys etc.
9. Mercury antardasha – 11 months 27 days
• If Ketu and Mercury are favourably disposed towards each other can lead to pilgrimages, good health and comfortable lifestyle, huge profits in trade and good communications, birth of children etc.
• If not then, can cause losses in mental and intellectual capacity, problems in speech, losses to the children and siblings etc.
• This is the final sub-period of Ketu Mahadasha. In this period you will release all that related to Ketu and prepare yourself for the upcoming Venus mahadasha of 20 years.
• If, in case, you have your Venus and Ketu conjunct with each other then you will have a massive 27 (20+7) years of life as a seamless Venus/Ketu combination mahadasha. And if by chance Venus is your ascendant also here you will face 27 yrs of a strange dissociation with your sense of self.
The generic remedy for those suffering from problems during this period of Ketu is pujas of Shiv or Shakti. Doing the Rudra-abhisek once in a while or a daily recitation of any Shiv mantra will be good. Or if you are more drawn to Shakti, a yadnya of Devi Mahatmya once in a while or reciting any mantra of Devi daily will help.
Analysing the effects of planetary periods requires an understanding of several factors in the horoscope, the most important being the karakatva of that planet (which areas of your life are controlled by the planet) and the power of the planet to give good/bad results. If you can determine these two things with confidence, analysing the effects of that planet’s periods will be easy.
This is the second shortest mahadasha at 7yrs. The shortest is the Sun Mahadasha at 6yrs. This 7yr period is very intense thus is very short. If you can do some spiritual practice during this period it can provide significant back-up for you for a lifetime. Ketu’s main period is followed by the Venus Mahadasha, the agent of relationships, the longest one which lasts for 20yrs.
Ketu’s periods will be characterised by his keywords, separation, dissociation, apathy and disinterest. But these will be very intense periods where karma will be in the forefront and your options will be rather limited. It is a very short period of 7yrs as Ketu’s separative fiery energies cannot be tolerated for more than this. In this period, try to remain calm, try to actively take interest in things unfolding around you and do start some simple regular spiritual practice. These simple things will act as an anchor in Ketu’s periods and help you resolve your own karma more easily.
If you feel a bit overwhelmed by all his here is a tiny short-cut keyword post on how to analyse the planet’s time-periods based on the which house he rules over.
Ketu the south node has always fascinated me. It does not have a head! Its so cryptic, no head means no face, no ego, no sensing, no thinking, no expressions, no nothing. It Just Is! But it has a body so it can act, and it does sometimes. But why does it act, what is the motivation, where is that motivation felt? Or does it act without motivation? Extremely mysterious. At least its counterpart, Rahu/North node has a head, but no body. But at least it is aware of itself, though it cannot act. But Ketu!
Ketu in a horoscope represents everything that has outlived its utility and is to be discarded. It also represents those aspects of life and living which you have enjoyed to the fullest in your past lives and thus in this life you are apathetic to them. Ketu is the “chidrakaraka” (The agent who tears, makes holes, disperses). It does not deny, it just makes you indifferent. For example if the Ketu is placed in the ascendent 1st house itself, the person will have zero sense of himself, but can be a brilliant actor as he, being a blank canvas himself, easily accepts the demands of the movie director on his personality.
Ketu and Rahu are the same entity but now divided into two points 180deg apart. Ketu always surrenders to the demands of Rahu/the other. And Rahu the trickster wants only desires, once he attains what he desired he immediately shifts to another desire.
So how does this cryptic ketu translate to the human experience. The house that Ketu occupies and the planet he influences (conjoint) represent the things or desires or objects that you will be rather indifferent to.
For example the client with the ketu in the 1st house was indifferent to his place in the society, as the head of the family, and was a businessman, but let his wife/business partners (rahu occupying the 7th being the “other”) direct him. The person was happy enough internally with this situation but it was causing business losses financially and his wife was getting more aggressive. Ketu’s gift of a non-coherent personality was affecting his social/family/business life. So something had to be done. A consultation for business success turned into a psychological exercise. He had to understand that though he may be internally indifferent to all this, he will actively use his energies develop an interest at least on a superficial level. And the separation/dispersion from ego which is actually a gift from ketu in a spiritual sense should be channelized elsewhere. He started doing and focussing on his non-ego traits during this period of mindful contemplation. Thus his spiritual progress accelerated and his superficial materialistic life also stabilised as the dispersive energies got their outlet during the dhyan.
In transit too Ketu grants separation from the objects of the houses he transits through. It is a good thing in a way, the mind is prepared to detach from the emotions/ actions/ environments which have out lived their purpose in this life. I have seen several examples of this, 3 cases this year when Ketu transited into the 4th house and the person’s mother died. It was a quick painless death for the mother and for the person too it was a calm acceptance of the fact. And as soon as the mother died, each one of these clients found new sources of stability, emotional satisfaction and assets. One got married in a few months, the second bought new property and the third shifted into a bigger house. Ketu gives too, but in a very blasé matter of fact way with his characteristic apathy.
Ketu located in various houses at birth will give the generally the following results. Ketu – is the agent of dispersion, non coherance, dis interest, apathy, headless, abandoned, limitless, not powered, careless, deserted… In the following houses affects,
1. First house – Personal self, status in society, the head
2. Second house – family assets, lineage, speech and communication
3. Third house – neighbourhood, extended family, brothers, boldness
4. Fourth House – mother, landed property, residence
5. Fifth house – education, children, expressions of creativity
6. Sixth house – disease, debts, enemies, arguments
7. Seventh house – partnerships, wife/husband
8. Eighth house – death, routes into the unknown subconsciousness and other dimensions, accidents, medical operations
9. Ninth house – external religious behaviour, deceased ancestors
10. Tenth house – father, profession, work
11. Eleventh house – networking of people, sources of gain, luck
12. Twelfth house – deep subconscious, internal spirituality, losses
And if conjoint/or 180degrees away from the other grahas, will affect..
1. Surya/Sun – personal ego, source of all energy in the horoscope, father
2. Chandra/Moon – emotions, mind, mother
3. Mangal/Mars – valor, expression of the personal energy, competitiveness, brothers
4. Budha/Mercury – creativity, communication
5. Devguru Bruhaspati/Jupiter – expansion, teacher, conventional knowledge
6. Daityaguru Shukra/Venus – relationships, spiritual teacher, esoteric knowledge
7. Shani – restriction, contraction, servility, service
So check out in your birth horoscopes the positions of Ketu and the planets it is influencing. Then using the keywords above just generally try to examine the influence of this headless graha on your life, on the emotional/physical/spiritual levels.
But do not forget that Ketu, Venus and Jupiter are the planets which together determine your spiritual foundation and progress. So if you see any mutual interactions between these 3, rest assured that in this life you will get a spiritual boost. Start some simple basic daily 5min, spiritual practice to trigger their energies gently.
Then if the ketu is in the 6th, 8th, 12th houses it is relatively better for the material word, In the 6th house it will disperse enemies/diseases/debts. In the 8th it will disperse the boundaries of death, may grant a longer life and also access to the other dimensions, you may actually sense other entities and their energies. In the 12th it will disperse losses and the boundaries between the conscious and subconscious again giving you access to other energies and extra-dreams visions etc.
Never forget Ketu is the real Moksha karaka, he grants the Enlightenment of Ego-Less-ness. Try to channelise its energies if you feel drawn to it. But remember that when you decide to find yourself, you have to decide to leave the “other”. This leaving the “other” is not so easy, and if you try to trigger it before you are really ready it will result in internal trauma. So baby steps first, start your spiritual practice and as and when you get ready the Ketu will be there with the and the Daityaguru Venus to show you the way. | 4,387 | 19,732 | {"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-2023-23 | latest | en | 0.94347 |
https://drmf-beta.wmflabs.org/index.php?title=Item:Q4588&diff=prev&oldid=63075 | 1,660,685,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572581.94/warc/CC-MAIN-20220816211628-20220817001628-00179.warc.gz | 220,799,151 | 17,764 | # DLMF:13.19.E2 (Q4588): Difference between revisions
No description defined
Language Label Description Also known as
English
DLMF:13.19.E2
No description defined
## Statements
${\displaystyle{\displaystyle M_{\kappa,\mu}\left(z\right)\sim\frac{\Gamma\left% (1+2\mu\right)}{\Gamma\left(\frac{1}{2}+\mu-\kappa\right)}e^{\frac{1}{2}z}z^{-% \kappa}\*\sum_{s=0}^{\infty}\frac{{\left(\frac{1}{2}-\mu+\kappa\right)_{s}}{% \left(\frac{1}{2}+\mu+\kappa\right)_{s}}}{s!}z^{-s}+\frac{\Gamma\left(1+2\mu% \right)}{\Gamma\left(\frac{1}{2}+\mu+\kappa\right)}e^{-\frac{1}{2}z\pm(\frac{1% }{2}+\mu-\kappa)\pi\mathrm{i}}z^{\kappa}\*\sum_{s=0}^{\infty}\frac{{\left(% \frac{1}{2}+\mu-\kappa\right)_{s}}{\left(\frac{1}{2}-\mu-\kappa\right)_{s}}}{s% !}(-z)^{-s},}}$
0 references
DLMF:13.19.E2
0 references
${\displaystyle{\displaystyle-\tfrac{1}{2}\pi+\delta\leq\pm\operatorname{ph}z% \leq\tfrac{3}{2}\pi-\delta}}$
0 references
${\displaystyle{\displaystyle\Gamma\left(\NVar{z}\right)}}$
0 references
${\displaystyle{\displaystyle{\left(\NVar{a}\right)_{\NVar{n}}}}}$
0 references
${\displaystyle{\displaystyle M_{\NVar{\kappa},\NVar{\mu}}\left(\NVar{z}\right)}}$
0 references
${\displaystyle{\displaystyle\sim}}$
0 references
${\displaystyle{\displaystyle\pi}}$
0 references
${\displaystyle{\displaystyle\mathrm{e}}}$
0 references
${\displaystyle{\displaystyle!}}$
0 references
${\displaystyle{\displaystyle\mathrm{i}}}$
${\displaystyle{\displaystyle\operatorname{ph}}}$
${\displaystyle{\displaystyle s}}$
${\displaystyle{\displaystyle z}}$ | 583 | 1,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.380023 |
https://www.physicsforums.com/threads/calculating-the-radius.279484/ | 1,532,101,233,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591683.78/warc/CC-MAIN-20180720135213-20180720155213-00333.warc.gz | 977,938,197 | 13,867 | # Homework Help: Calculating the radius
1. Dec 13, 2008
### rayfieca
1. The problem statement, all variables and given/known data
Calculate the radius of the n = 6 Bohr orbit in O7+(oxygen with 7 of its 8 electrons removed).
A) 190 pm B) 167 pm C) 238 pm D) 214 pm
2. Relevant equations
I believe that the relevant equation is
r(sub n)=(n^2)*a(sub b) where a(sub b)= Bohr's radius= 5.29*10^-11 m
3. The attempt at a solution
I calculated it for what I believed to be n=6, but I got an incorrect answer. The correct answer is (C) 238 pm, but I do not understand why.
Any help is greatly appreciated!
2. Dec 14, 2008
### hage567
Your formula is not quite complete. You need to take into account the charge in the nucleus (so the number of protons).
3. Dec 14, 2008
### Redbelly98
Staff Emeritus
Welcome to PF
Your equation is for hydrogen, with a charge of +1 on the nucleus.
Since oxygen has a charge of Z=8 for the nucleus, that equation should be different, containing Z somehow.
Does your textbook discuss "hydrogen-like" or "hydrogenic" ions?
EDIT: ah, I should know better than to wait a 1/2 hour and then respond without refreshing the page.
4. Dec 14, 2008
### rayfieca
Unfortunately,
I cannot find an equation that takes into account (Z). Anyone have the equation handy? | 367 | 1,296 | {"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} | 3.078125 | 3 | CC-MAIN-2018-30 | latest | en | 0.899688 |
https://encyclopediaofmath.org/index.php?title=Quadratic_deviation&oldid=43583 | 1,686,064,503,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652959.43/warc/CC-MAIN-20230606150510-20230606180510-00504.warc.gz | 277,034,668 | 6,171 | ##### Actions
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
quadratic variance, standard deviation, of quantities $x_1,\dots,x_n$ from $a$
The square root of the expression
$$\frac{(x_1-a)^2+\dots+(x_n-a)^2}{n}.\label{*}$$
The quadratic deviation takes its smallest value when $a=\bar x$, where $\bar x$ is the arithmetic mean of $x_1,\dots,x_n$:
$$\bar x=\frac{x_1+\dots+x_n}{n}.$$
In this case the quadratic deviation serves as a measure of the variance (cf. Dispersion) of the quantities $x_1,\dots,x_n$. Also used is the more general concept of a weighted quadratic deviation:
$$\sqrt\frac{p_1(x_1-a)^2+\dots+p_n(x_n-a)^2}{p_1+\dots+p_n},$$
where the $p_1,\dots,p_n$ are the so-called weights associated with $x_1,\dots,x_n$. The weighted quadratic deviation attains its smallest value when $a$ is the weighted mean:
$$\frac{p_1x_1+\dots+p_nx_n}{p_1+\dots+p_n}.$$
In probability theory, the quadratic deviation $\sigma_X$ of a random variable $X$ (from its mathematical expectation) refers to the square root of its variance: $\sqrt{D(X)}$.
The quadratic deviation is taken as a measure of the quality of statistical estimators and in this case is referred to as the quadratic error. | 350 | 1,228 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 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": 1, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-23 | longest | en | 0.828574 |
https://en.formulasearchengine.com/index.php?title=K%EF%BF%BD%EF%BF%BDnig%27s_theorem_(set_theory)&oldid=225490 | 1,675,018,632,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00089.warc.gz | 248,356,180 | 11,603 | # König's theorem (set theory)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
{{#invoke:Hatnote|hatnote}}
In set theory, König's theorem states that if the axiom of choice holds, I is a set, mi and ni are cardinal numbers for every i in I, and ${\displaystyle m_{i} for every i in I then
${\displaystyle \sum _{i\in I}m_{i}<\prod _{i\in I}n_{i}.}$
The sum here is the cardinality of the disjoint union of the sets mi and the product is the cardinality of the Cartesian product. However, without the use of the axiom of choice, the sum and the product cannot be defined as cardinal numbers, and the meaning of the inequality sign would need to be clarified.
König's theorem was introduced by Template:Harvs in the slightly weaker form that the sum of a strictly increasing sequence of nonzero cardinal numbers is less than their product.
## Details
The precise statement of the result: if I is a set, Ai and Bi are sets for every i in I, and ${\displaystyle A_{i} for every i in I then
${\displaystyle \sum _{i\in I}A_{i}<\prod _{i\in I}B_{i},}$
where < means strictly less than in cardinality, i.e. there is an injective function from Ai to Bi, but not one going the other way. The union involved need not be disjoint (a non-disjoint union can't be any bigger than the disjoint version, also assuming the axiom of choice). In this formulation, König's theorem is equivalent to the Axiom of Choice.[1]
(Of course, König's theorem is trivial if the cardinal numbers mi and ni are finite and the index set I is finite. If I is empty, then the left sum is the empty sum and therefore 0, while the right hand product is the empty product and therefore 1).
König's theorem is remarkable because of the strict inequality in the conclusion. There are many easy rules for the arithmetic of infinite sums and products of cardinals in which one can only conclude a weak inequality ≤, for example: if ${\displaystyle m_{i} for all i in I, then one can only conclude
${\displaystyle \sum _{i\in I}m_{i}\leq \sum _{i\in I}n_{i}}$
since, for example, setting ${\displaystyle m_{i}=1}$ & ${\displaystyle n_{i}=2}$ where the index set I is the natural numbers, yields the sum ${\displaystyle \aleph _{0}}$ for both sides and we have a strict equality.
## Corollaries of König's theorem
If we take mi = 1, and ni = 2 for each i in κ, then the left hand side of the above inequality is just κ, while the right hand side is 2κ, the cardinality of functions from κ to {0,1}, that is, the cardinality of the power set of κ. Thus, König's theorem gives us an alternate proof of Cantor's theorem. (Historically of course Cantor's theorem was proved much earlier.)
### Axiom of choice
One way of stating the axiom of choice is "An arbitrary Cartesian product of non-empty sets is non-empty.". Let Bi be a non-empty set for each i in I. Let Ai = {} for each i in I. Thus by König's theorem, we have:
That is, the Cartesian product of the given non-empty sets, Bi, has a larger cardinality than the sum of empty sets. Thus it is non-empty which is just what the axiom of choice states. Since the axiom of choice follows from König's theorem, we will use the axiom of choice freely and implicitly when discussing consequences of the theorem.
### König's theorem and cofinality
König's theorem has also important consequences for cofinality of cardinal numbers.
Choose a strictly increasing cf(κ)-sequence of ordinals approaching κ. Each of them is less than κ, so their sum which is κ is less than the product of cf(κ) copies of κ.
According to Easton's theorem, the next consequence of König's theorem is the only nontrivial constraint on the continuum function for regular cardinals.
Let ${\displaystyle \mu =\lambda ^{\kappa }\!}$. Suppose that, contrary to this corollary, ${\displaystyle \kappa \geq cf(\mu )}$. Then using the previous corollary, ${\displaystyle \mu <\mu ^{cf(\mu )}\leq \mu ^{\kappa }=(\lambda ^{\kappa })^{\kappa }=\lambda ^{\kappa \cdot \kappa }=\lambda ^{\kappa }=\mu }$, a contradiction. Thus the supposition must be false and this corollary must be true.
## A proof of König's theorem
Assuming Zermelo–Fraenkel set theory, including especially the axiom of choice, we can prove the theorem. Remember that we are given ${\displaystyle \forall i\in I\quad A_{i}, and we want to show :${\displaystyle \sum _{i\in I}A_{i}<\prod _{i\in I}B_{i}.}$
The axiom of choice implies that the condition A < B is equivalent to the condition that there is no function from A onto B and B is nonempty. So we are given that there is no function from Ai onto Bi≠{}, and we have to show that any function f from the disjoint union of the As to the product of the Bs is not surjective and that the product is nonempty. That the product is nonempty follows immediately from the axiom of choice and the fact that the factors are nonempty. For each i choose a bi in B i not in the image of Ai under the composition of f with the projection to Bi. Then the product of the elements bi is not in the image of f, so f does not map the disjoint union of the As onto the product of the Bs.
## Notes
1. {{#invoke:citation/CS1|citation |CitationClass=book }}
## References
• {{#invoke:citation/CS1|citation
|CitationClass=book }}
• {{#invoke:citation/CS1|citation
|CitationClass=citation }}, reprinted as {{#invoke:citation/CS1|citation |CitationClass=citation }} | 1,409 | 5,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "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-2023-06 | latest | en | 0.929322 |
http://nrich.maths.org/8487 | 1,485,241,260,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284352.26/warc/CC-MAIN-20170116095124-00302-ip-10-171-10-70.ec2.internal.warc.gz | 200,021,161 | 4,901 | # Units of Measurement
### Estimating Angles
##### Stage: 2 and 3 Challenge Level:
How good are you at estimating angles?
### Speeding Boats
##### Stage: 4 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling?
### Speed-time Problems at the Olympics
##### Stage: 4 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Units of Measurement - Short Problems
##### Stage: 3 and 4
A collection of short Stage 3 and 4 problems on units of measurement.
### Late for Work
##### Stage: 4 Short Challenge Level:
What average speed should Ms Fanthorpe drive at to arrive at work on time?
### Speed over a Bridge
##### Stage: 3 Short Challenge Level:
How fast is the train travelling as it goes over the bridge?
### Leg It
##### Stage: 3 Short Challenge Level:
How far from the finishing line should these runners start to make the race 'fair'?
### Backwards Laps
##### Stage: 4 Short Challenge Level:
If two friends run in opposite directions around a track, and they pass each other every 24 seconds, how long do they take to complete a lap?
### Inches and Barleycorns
##### Stage: 3 Short Challenge Level:
How many barleycorns are there in one inch? | 301 | 1,293 | {"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-2017-04 | latest | en | 0.868946 |
http://www.pulse-jets.com/phpbb3/search.php?author_id=145467&sr=posts | 1,582,577,291,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145981.35/warc/CC-MAIN-20200224193815-20200224223815-00299.warc.gz | 228,083,908 | 6,735 | ## Search found 9 matches
Mon Jul 21, 2014 8:33 pm
Forum: Valved pulsejet forum
Topic: Units used in Calculator
Replies: 3
Views: 2457
### Re: Units used in Calculator
Thanks!
Sun Jul 20, 2014 9:38 pm
Forum: Valved pulsejet forum
Topic: Units used in Calculator
Replies: 3
Views: 2457
### Re: Units used in Calculator
in the Eric Beck's calculator in .xls file i still cant understand the units of thrust! Is the kgs signify Newtons or is it kilogram-force (kgf)............. Plz reply!
Thanks
Sun Feb 23, 2014 4:42 pm
Forum: Valved pulsejet forum
Topic: Units used in Calculator
Replies: 3
Views: 2457
### Units used in Calculator
I built an engine using Eric Beck calculator for the same and found it working satisfactorily! But the thing is I have been wondering that the units used were written in the thrust row as lbs and kgs which i find somewhat difficult to understand as commonly we use for Force the units of kgf and lbf ...
Sun Jan 05, 2014 4:49 pm
Forum: Valved pulsejet forum
Topic: My First Valved Pulse Jet
Replies: 5
Views: 2688
### Re: My First Valved Pulse Jet
It works! Its buzzing! WoW! thanks...opening the gap did the trick.... but it seems like I was also spraying a very small amount of fuel!
Fri Jan 03, 2014 5:24 pm
Forum: Valved pulsejet forum
Topic: My First Valved Pulse Jet
Replies: 5
Views: 2688
### Re: My First Valved Pulse Jet
No! The fuel does not ignite! Well my ignitor should provide round about 8 sparks per second (8 Hz freq.) with 66% duty cycle!
Fri Jan 03, 2014 4:26 pm
Forum: Valved pulsejet forum
Topic: My First Valved Pulse Jet
Replies: 5
Views: 2688
### My First Valved Pulse Jet
Well I built my first 10lbs Valved Pulse jet following the design calculations of Eric Beck Calculator..... I have used the aspirated engine approach! When I tried to start the engine by blowing air inside using an air-blower, powered on the spark plug through a driving circuit based on an automobil...
Thu Nov 28, 2013 6:32 am
Forum: Valved pulsejet forum
Topic: Air Tight
Replies: 1
Views: 1220
### Air Tight
I am unsure whether the pulse jet needs to be airtight without any small openings between the combustion chanmber and the head assembly.... I think using flanges might also result in small space?? Is it okay or should it be completely closed because of we blow air inside for starting the same some a...
Wed Sep 11, 2013 11:53 am
Forum: Valved pulsejet forum
Topic: Material Thickness
Replies: 3
Views: 1844
### Re: Material Thickness
I use muffler pipe and EMT electrical conduit for building engines. I'm not sure of the thickness maybe .060". It takes some practice to weld it with my a stick welder using my old 1970s Lincoln Welder. But this way the pulse jet will weigh more than 2 kgs which will make life difficult for me as I...
Wed Sep 04, 2013 6:21 pm
Forum: Valved pulsejet forum
Topic: Material Thickness
Replies: 3
Views: 1844
### Material Thickness
I was planning on building a 25lbs Valved Pulse Jet Engine but couldnt figure out how much thick should the tail pipe and the combustion chamber be? Can I use a #25 Gague Mild Steel Sheet rolled out for my given diameter for the same purposes? Will it be enough?? Plz help! Also this one will not be ... | 883 | 3,228 | {"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-2020-10 | latest | en | 0.852238 |
https://www.esaral.com/q/if-the-equation-1-m2-x2-2-mcx-c2-a2-0-has-equal-roots-91602 | 1,721,599,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00226.warc.gz | 637,761,081 | 11,572 | # If the equation (1+m2)x2+2 mcx+(c2−a2)=0 has equal roots,
Question:
If the equation $\left(1+m^{2}\right) x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$ has equal roots, prove that $c^{2}=a^{2}\left(1+m^{2}\right)$.
Solution:
The given equation $\left(1+m^{2}\right) x^{2}+2 m c x+\left(c^{2}-a^{2}\right)=0$, has equal roots
Then prove that $c^{2}=\left(1+m^{2}\right)$.
Here,
$a=\left(1+m^{2}\right), b=2 m c$ and,$c=\left(c^{2}-a^{2}\right)$
As we know that $D=b^{2}-4 a c$
Putting the value of $a=\left(1+m^{2}\right), b=2 m c$ and,$c=\left(c^{2}-a^{2}\right)$
$D=b^{2}-4 a c$
$=\{2 m c\}^{2}-4 \times\left(1+m^{2}\right) \times\left(c^{2}-a^{2}\right)$
$=4\left(m^{2} c^{2}\right)-4\left(c^{2}-a^{2}+m^{2} c^{2}-m^{2} a^{2}\right)$
$=4 a^{2}+4 m^{2} a^{2}-4 c^{2}$
The given equation will have real roots, if $D=0$
$4 a^{2}+4 m^{2} a^{2}-4 c^{2}=0$
$4 a^{2}+4 m^{2} a^{2}=4 c^{2}$
$4 a^{2}\left(1+m^{2}\right)=4 c^{2}$
$a^{2}\left(1+m^{2}\right)=c^{2}$
Hence,
$c^{2}=a^{2}\left(1+m^{2}\right)$ | 474 | 1,015 | {"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} | 4.46875 | 4 | CC-MAIN-2024-30 | latest | en | 0.381041 |
https://gist.github.com/learning2learn/4099484 | 1,529,584,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864148.93/warc/CC-MAIN-20180621114153-20180621134153-00403.warc.gz | 625,223,305 | 14,704 | Instantly share code, notes, and snippets.
# learning2learn/gist:4099484 Created Nov 17, 2012
#include #include #include #include #include #include #include #include using namespace std; class Point2D{ protected: int x,y; double distFrOrigin; void setDistFrOrigin(); public: Point2D(){}; Point2D(int x, int y); int getX(); int getY(); double getScalarValue(); void setX(int x); void setY(int y); virtual string toString(); Point2D operator - (Point2D); bool operator == (Point2D); bool operator < (Point2D); }; Point2D::Point2D(int x, int y){ setX(x); setY(y); setDistFrOrigin(); } void Point2D::setX(int x) { this->x = x; } void Point2D::setY(int y) { this->y = y; } void Point2D::setDistFrOrigin() { double compute=0.000f; compute = sqrt(pow(x,2) + pow(y,2)); distFrOrigin = compute; } int Point2D::getX() { return x; } int Point2D::getY() { return y; } double Point2D::getScalarValue() { setDistFrOrigin(); return distFrOrigin; } string Point2D::toString(){ stringstream ss; ss << "["; ss.width(4); ss << right << x; ss << ", "; ss.width(4); ss << right << y; ss << "]\t"; //ss.precision(3) << fixed; ss << setprecision(3) << fixed << right << distFrOrigin < to sort the p2dVect vector { bool operator() (Point2D a, Point2D b) { return a.getScalarValue() > b.getScalarValue(); } }; int main (){ list PointList; PointList.push_back( Point2D(-9, -9)); PointList.push_back( Point2D(-99, -99)); PointList.push_back( Point2D(-999, -999)); PointList.push_back( Point2D(-999, -999)); PointList.push_back( Point2D(-999, -999)); PointList.push_back( Point2D(-99, -99)); PointList.push_back( Point2D(-9, -9)); PointList.push_back( Point2D(3, 3)); PointList.push_back( Point2D(23, 23)); PointList.push_back( Point2D(123, 123)); cout << "\n X Y Dist. Fr Origin" << endl; cout << "- - - - - - - - - - - - - - - - - - -"<< endl; list::iterator p = PointList.begin(); while (p != PointList.end()){ cout << p->toString(); p++; } PointList.sort(SortByDist()); cout << "\n X Y Dist. Fr Origin" << endl; cout << "- - - - - - - - - - - - - - - - - - -"<< endl; p = PointList.begin(); while (p != PointList.end()){ cout << p->toString(); p++; } PointList.unique(); cout << "\n X Y Dist. Fr Origin" << endl; cout << "- - - - - - - - - - - - - - - - - - -"<< endl; p = PointList.begin(); while (p != PointList.end()){ cout << p->toString(); p++; } system("pause"); return 0; } | 754 | 2,359 | {"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} | 2.59375 | 3 | CC-MAIN-2018-26 | latest | en | 0.353645 |
https://stupidsid.com/previous-question-papers/download/applied-mathematics-3-872 | 1,638,100,486,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358520.50/warc/CC-MAIN-20211128103924-20211128133924-00013.warc.gz | 505,259,816 | 11,896 | MORE IN Applied Mathematics - 3
MU Electronics and Telecom Engineering (Semester 3)
Applied Mathematics - 3
May 2012
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary
1 (a) Prove that
$\int_0^{\infty{}}e^{-t}\dfrac{{sin}^2t}{t}dt= \dfrac{1}{4} log 5$
5 M
1 (b) Is the matrix orthogonal ? If not then can it be converted to an orthogonal matrix :-
$A=\left[\begin{array}{ccc}-8 & 1 & 4 \\4 & 4 & 7 \\1 & -8 & 4\end{array}\right]$
5 M
1 (c) Obtain complex form of Fourier series for f(x) = eax in (-l ,l).
5 M
1 (d) Find the Z-transform of f(k) =ak, k≥0.
5 M
2 (a) Find the Fourier sine transform of f(x) if \begin {align*} f(x)&=\sin kx, &0 \le x <a \\ &=0, &x>a \end{align*}
6 M
2 (b) Find the Matrix A if
$\left[\begin{array}{cc}2 & 1 \\3 & 2\end{array}\right]\ A\ \left[\begin{array}{cc}-3 & 2 \\5 & -3\end{array}\right]=\ \left[\begin{array}{cc}-2 & 4 \\3 & -1\end{array}\right]$
6 M
2 (c) (D2- 3D+2) y=4 e21, with y(0) = -3, y'(0)=5 solve using Laplace transform.
8 M
3 (a) Reduce the matrix to normal form and find its rank :-
$\left[\begin{array}{cccc}2 & 3 & -1 & -1 \\1 & -1 & -2 & -4 \\3 & 1 & 3 & -2 \\6 & 3 & 0 & -7\end{array}\right]$
6 M
3 (b) Find the inverse Laplace transform of ?
$\left(i\right)\ \frac{e^{-2s}}{s^2+8s+25} \\ \left(ii\right)\ \frac{e^{-3s}}{{\left(s+4\right)}^3}$
6 M
3 (c) $\left.\begin{matrix} f(x)&= \pi x 0\leq x \leq 1 \\ f(x)&=\pi (2-x)1 \leq x \leq 2 \end{matrix}\right\} with \ period \ 2$
Find the Fourier series expansion
8 M
4 (a) Show that the set of functions ${\ \left(\frac{\pi{}x}{2l}\right), sin\left(\frac{3\pi{}x}{2l}\right),\ \sin\left(\frac{5\pi{}x}{2l}\right),.....}$is orthogonal over (0,l).
6 M
4 (b) If f(k)= 4kU(K), g(k)= 5kU(k), then find the z-transform of f(k) x g(k).
6 M
4 (c) Solve the following equations by Gauss-Seidel Method.
28x+4y-z=32
2x+17y+4z=35
x+3y+10z=24.
8 M
5 (a) Obtain Fourier series for
${\ f(x) = x + \frac{\pi{}}{2}, -\pi{} < x < 0} \\ {= \frac{\pi{}}{2}-x\ 0 < x < \pi{}}$
Hence deduce that, ${\ \frac{\pi^2}{8} = \frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + ....}$
6 M
5 (b) State Convolution theorem and hence find inverse Laplace transform of the function using the same :-
$f(s)\ =\frac{{\left(s+3\right)}^2}{{\left(s^2+6s+5\right)}^2}$
6 M
5 (c) For what value of λ the equations 3x-2y+ λ z=1, 2x+y+z=2, x+2y- λz= -1 will have no unique solution ? Will the equations have any solution for this value of λ ?
8 M
6 (a) Show that every square matrix A can be uniquely expressed as P+iQ when P and Q are Hermitian matrices
6 M
6 (b) If L[f(t)] = f(s), then prove that L[ tn f(t)] = (-1)n dn/dsn f(s), Hence find the Laplace transform of f(t) = t cos2t
6 M
6 (c) Obtain the half rang sine series for f(x) when
${\ f(x) = x 0 < x < \frac{\pi{}}{2}}\\{= \pi{} - x \frac{\pi{}}{2}< x < \pi{}} \\ Hence \ find \ the \ sum \ of \ \sum_{2n-1}^{\infty{}}\ \frac{1}{n^4}$
8 M
7 (a) Find the Fourier transform of-
f(x) = (1-x2), |x|<|
= 0 , |x|>|,
then $f\left(s\right)=\ -2\sqrt{\frac{2}{\pi{}}}\left[\frac{scoss-sins}{s^3}\right]$
6 M
7 (b) Find the inverse z transform of $F\left(z\right)=\ \frac{z}{\left(z-1\right)\left(z-2\right)},\ \left\vert{}z\right\vert{}>2$
6 M
7 (c) Find the non-singular matrices P and Q such that -
$A=\ \left[\begin{array}{ccc}1 & 2 & 3 & 2 \\2 & 3 & 5 & 1 \\1 & 3 & 4 & 5\end{array}\right]$
is reduced to normal form. Also find its rank.
8 M
More question papers from Applied Mathematics - 3 | 1,443 | 3,549 | {"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} | 4.03125 | 4 | CC-MAIN-2021-49 | longest | en | 0.539493 |
http://ianfinlayson.net/class/cpsc305/labs/04-rps | 1,591,222,796,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436466.95/warc/CC-MAIN-20200603210112-20200604000112-00134.warc.gz | 59,604,759 | 2,967 | # Switch Exercise
## Objective
To gain experience building circuits which use switches to compute logical expressions.
## Cat Selector Circuit
In Code, Petzold designs a circuit to decide if a cat is acceptable according to a logical expression. This is used to illustrate the link between logic and circuits. His circuit is given below:
Below is a Logisim version of the same circuit:
This circuit uses transistors controlled by pins instead of switches which would need to be opened and closed manually. A pin is an input or output of a circuit. Whether the cat is male or female is external to the circuit, so is controlled by a pin.
When the value of (M & N & (W | T)) | (F & N & ~W) | B is true, then the LED will light up. Notice that the ~W is handled with a P-type transistor which closes when a 0 is coming in. The other transistors are all N-type which close on a 1.
## Exercise: Rock, Paper, Scissors
For this lab, you'll build a circuit along similar lines which will compute who wins a game of rock, paper, scissors. In this game, two players throw hand signals representing rock, paper, and scissors. The moves are compared and a winner is decided according to the following rules:
• Rock beats scissors.
• Paper beats rock.
• Scissors beats paper.
• If the same signals are used, the game is a tie.
Your circuit should determine a winner based on the moves of each player.
You should build the circuit out of the following components:
• Pins
You should use 6 pins (found under the wiring menu). Three pins are for player 1, and three are for player 2. Each player has a rock, paper, and scissors pin.
You should change them from three-state pins to two-state pins in the properties menu. Three-state pins can also be "turned off" where they do not represent a 0 or a 1, but that's not needed for this circuit.
You should also label each pin, so it's clear what it does!
• Transistors
For each switch, you'll need a transistor connected to the appropriate pin. Be sure to select the right type between P-type and N-type.
• Power
Each transistor needs a power source on the collector (left side). Whether they each have one, or share one doesn't matter.
• LEDs
You should have two LEDs: one which represents player 1 winning, and another which represents player 2 winning. A tie game would be indicated with both LEDs off. Under no conditions should both LEDs be on!
## Testing
Be sure to test that your circuit lights the appropriate LEDs for each possibility. You should test the case where the players throw the same moves (resulting in a tie), and when no move is thrown.
You don't need to test the case where multiple throws are selected (e.g. if the pins for player 1 throwing rock and paper are both on).
## Submitting
When your circuit works, email the .circ file to ifinlay@umw.edu. | 643 | 2,831 | {"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} | 3.09375 | 3 | CC-MAIN-2020-24 | latest | en | 0.945095 |
https://db0nus869y26v.cloudfront.net/en/Paraconsistent_logic | 1,721,509,223,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00311.warc.gz | 163,057,596 | 35,244 | Paraconsistent logic is an attempt at a logical system to deal with contradictions in a discriminating[clarification needed] way. Alternatively, paraconsistent logic is the subfield of logic that is concerned with studying and developing "inconsistency-tolerant" systems of logic, which reject the principle of explosion.
Inconsistency-tolerant logics have been discussed since at least 1910 (and arguably much earlier, for example in the writings of Aristotle);[1] however, the term paraconsistent ("beside the consistent") was first coined in 1976, by the Peruvian philosopher Francisco Miró Quesada Cantuarias.[2] The study of paraconsistent logic has been dubbed paraconsistency,[3] which encompasses the school of dialetheism.
## Definition
In classical logic (as well as intuitionistic logic and most other logics), contradictions entail everything. This feature, known as the principle of explosion or ex contradictione sequitur quodlibet (Latin, "from a contradiction, anything follows")[4] can be expressed formally as
1 ${\displaystyle P\land \neg P}$ Premise 2 ${\displaystyle P\,}$ Conjunction elimination from 1 3 ${\displaystyle P\lor A}$ Disjunction introduction from 2 4 ${\displaystyle \neg P\,}$ Conjunction elimination from 1 5 ${\displaystyle A\,}$ Disjunctive syllogism from 3 and 4
Which means: if P and its negation ¬P are both assumed to be true, then of the two claims P and (some arbitrary) A, at least one is true. Therefore, P or A is true. However, if we know that either P or A is true, and also that P is false (that ¬P is true) we can conclude that A, which could be anything, is true. Thus if a theory contains a single inconsistency, the theory is trivial – that is, it has every sentence as a theorem.
The characteristic or defining feature of a paraconsistent logic is that it rejects the principle of explosion. As a result, paraconsistent logics, unlike classical and other logics, can be used to formalize inconsistent but non-trivial theories.
## Comparison with classical logic
The entailment relations of paraconsistent logics are propositionally weaker than classical logic; that is, they deem fewer propositional inferences valid. The point is that a paraconsistent logic can never be a propositional extension of classical logic, that is, propositionally validate every entailment that classical logic does. In some sense, then, paraconsistent logic is more conservative or cautious than classical logic. It is due to such conservativeness that paraconsistent languages can be more expressive than their classical counterparts including the hierarchy of metalanguages due to Alfred Tarski and others. According to Solomon Feferman: "natural language abounds with directly or indirectly self-referential yet apparently harmless expressions—all of which are excluded from the Tarskian framework."[5] This expressive limitation can be overcome in paraconsistent logic.
## Motivation
A primary motivation for paraconsistent logic is the conviction that it ought to be possible to reason with inconsistent information in a controlled and discriminating way. The principle of explosion precludes this, and so must be abandoned. In non-paraconsistent logics, there is only one inconsistent theory: the trivial theory that has every sentence as a theorem. Paraconsistent logic makes it possible to distinguish between inconsistent theories and to reason with them.
Research into paraconsistent logic has also led to the establishment of the philosophical school of dialetheism (most notably advocated by Graham Priest), which asserts that true contradictions exist in reality, for example groups of people holding opposing views on various moral issues.[6] Being a dialetheist rationally commits one to some form of paraconsistent logic, on pain of otherwise embracing trivialism, i.e. accepting that all contradictions (and equivalently all statements) are true.[7] However, the study of paraconsistent logics does not necessarily entail a dialetheist viewpoint. For example, one need not commit to either the existence of true theories or true contradictions, but would rather prefer a weaker standard like empirical adequacy, as proposed by Bas van Fraassen.[8]
## Philosophy
In classical logic Aristotle's three laws, namely, the excluded middle (p or ¬p), non-contradiction ¬ (p ∧ ¬p) and identity (p iff p), are regarded as the same, due to the inter-definition of the connectives. Moreover, traditionally contradictoriness (the presence of contradictions in a theory or in a body of knowledge) and triviality (the fact that such a theory entails all possible consequences) are assumed inseparable, granted that negation is available. These views may be philosophically challenged, precisely on the grounds that they fail to distinguish between contradictoriness and other forms of inconsistency.
On the other hand, it is possible to derive triviality from the 'conflict' between consistency and contradictions, once these notions have been properly distinguished. The very notions of consistency and inconsistency may be furthermore internalized at the object language level.
Paraconsistency involves tradeoffs. In particular, abandoning the principle of explosion requires one to abandon at least one of the following two principles:[9]
Disjunction introduction ${\displaystyle A\vdash A\lor B}$ ${\displaystyle A\lor B,\neg A\vdash B}$
Both of these principles have been challenged.
One approach is to reject disjunction introduction but keep disjunctive syllogism and transitivity. In this approach, rules of natural deduction hold, except for disjunction introduction and excluded middle; moreover, inference A⊢B does not necessarily mean entailment A⇒B. Also, the following usual Boolean properties hold: double negation as well as associativity, commutativity, distributivity, De Morgan, and idempotence inferences (for conjunction and disjunction). Furthermore, inconsistency-robust proof of negation holds for entailment: (A⇒(B∧¬B))⊢¬A.
Another approach is to reject disjunctive syllogism. From the perspective of dialetheism, it makes perfect sense that disjunctive syllogism should fail. The idea behind this syllogism is that, if ¬ A, then A is excluded and B can be inferred from A ∨ B. However, if A may hold as well as ¬A, then the argument for the inference is weakened.
Yet another approach is to do both simultaneously. In many systems of relevant logic, as well as linear logic, there are two separate disjunctive connectives. One allows disjunction introduction, and one allows disjunctive syllogism. Of course, this has the disadvantages entailed by separate disjunctive connectives including confusion between them and complexity in relating them.
Furthermore, the rule of proof of negation (below) just by itself is inconsistency non-robust in the sense that the negation of every proposition can be proved from a contradiction.
Proof of Negation If ${\displaystyle A\vdash B\land \neg B}$, then ${\displaystyle \vdash \neg A}$
Strictly speaking, having just the rule above is paraconsistent because it is not the case that every proposition can be proved from a contradiction. However, if the rule double negation elimination (${\displaystyle \neg \neg A\vdash A}$) is added as well, then every proposition can be proved from a contradiction. Double negation elimination does not hold for intuitionistic logic.
One example of paraconsistent logic is the system known as LP ("Logic of Paradox"), first proposed by the Argentinian logician Florencio González Asenjo in 1966 and later popularized by Priest and others.[10]
One way of presenting the semantics for LP is to replace the usual functional valuation with a relational one.[11] The binary relation ${\displaystyle V\,}$ relates a formula to a truth value: ${\displaystyle V(A,1)\,}$ means that ${\displaystyle A\,}$ is true, and ${\displaystyle V(A,0)\,}$ means that ${\displaystyle A\,}$ is false. A formula must be assigned at least one truth value, but there is no requirement that it be assigned at most one truth value. The semantic clauses for negation and disjunction are given as follows:
• ${\displaystyle V(\neg A,1)\Leftrightarrow V(A,0)}$
• ${\displaystyle V(\neg A,0)\Leftrightarrow V(A,1)}$
• ${\displaystyle V(A\lor B,1)\Leftrightarrow V(A,1){\text{ or ))V(B,1)}$
• ${\displaystyle V(A\lor B,0)\Leftrightarrow V(A,0){\text{ and ))V(B,0)}$
(The other logical connectives are defined in terms of negation and disjunction as usual.) Or to put the same point less symbolically:
• not A is true if and only if A is false
• not A is false if and only if A is true
• A or B is true if and only if A is true or B is true
• A or B is false if and only if A is false and B is false
(Semantic) logical consequence is then defined as truth-preservation:
${\displaystyle \Gamma \vDash A}$ if and only if ${\displaystyle A\,}$ is true whenever every element of ${\displaystyle \Gamma \,}$ is true.
Now consider a valuation ${\displaystyle V\,}$ such that ${\displaystyle V(A,1)\,}$ and ${\displaystyle V(A,0)\,}$ but it is not the case that ${\displaystyle V(B,1)\,}$. It is easy to check that this valuation constitutes a counterexample to both explosion and disjunctive syllogism. However, it is also a counterexample to modus ponens for the material conditional of LP. For this reason, proponents of LP usually advocate expanding the system to include a stronger conditional connective that is not definable in terms of negation and disjunction.[12]
As one can verify, LP preserves most other inference patterns that one would expect to be valid, such as De Morgan's laws and the usual introduction and elimination rules for negation, conjunction, and disjunction. Surprisingly, the logical truths (or tautologies) of LP are precisely those of classical propositional logic.[13] (LP and classical logic differ only in the inferences they deem valid.) Relaxing the requirement that every formula be either true or false yields the weaker paraconsistent logic commonly known as first-degree entailment (FDE). Unlike LP, FDE contains no logical truths.
LP is only one of many paraconsistent logics that have been proposed.[14] It is presented here merely as an illustration of how a paraconsistent logic can work.
## Relation to other logics
One important type of paraconsistent logic is relevance logic. A logic is relevant if it satisfies the following condition:
if AB is a theorem, then A and B share a non-logical constant.
It follows that a relevance logic cannot have (p ∧ ¬p) → q as a theorem, and thus (on reasonable assumptions) cannot validate the inference from {p, ¬p} to q.
Paraconsistent logic has significant overlap with many-valued logic; however, not all paraconsistent logics are many-valued (and, of course, not all many-valued logics are paraconsistent). Dialetheic logics, which are also many-valued, are paraconsistent, but the converse does not hold. The ideal 3-valued paraconsistent logic given below becomes the logic RM3 when the contrapositive is added.
Intuitionistic logic allows A ∨ ¬A not to be equivalent to true, while paraconsistent logic allows A ∧ ¬A not to be equivalent to false. Thus it seems natural to regard paraconsistent logic as the "dual" of intuitionistic logic. However, intuitionistic logic is a specific logical system whereas paraconsistent logic encompasses a large class of systems. Accordingly, the dual notion to paraconsistency is called paracompleteness, and the "dual" of intuitionistic logic (a specific paracomplete logic) is a specific paraconsistent system called anti-intuitionistic or dual-intuitionistic logic (sometimes referred to as Brazilian logic, for historical reasons).[15] The duality between the two systems is best seen within a sequent calculus framework. While in intuitionistic logic the sequent
${\displaystyle \vdash A\lor \neg A}$
is not derivable, in dual-intuitionistic logic
${\displaystyle A\land \neg A\vdash }$
is not derivable[citation needed]. Similarly, in intuitionistic logic the sequent
${\displaystyle \neg \neg A\vdash A}$
is not derivable, while in dual-intuitionistic logic
${\displaystyle A\vdash \neg \neg A}$
is not derivable. Dual-intuitionistic logic contains a connective # known as pseudo-difference which is the dual of intuitionistic implication. Very loosely, A # B can be read as "A but not B". However, # is not truth-functional as one might expect a 'but not' operator to be; similarly, the intuitionistic implication operator cannot be treated like "¬ (A ∧ ¬B)". Dual-intuitionistic logic also features a basic connective ⊤ which is the dual of intuitionistic ⊥: negation may be defined as ¬A = (⊤ # A)
A full account of the duality between paraconsistent and intuitionistic logic, including an explanation on why dual-intuitionistic and paraconsistent logics do not coincide, can be found in Brunner and Carnielli (2005).
These other logics avoid explosion: implicational propositional calculus, positive propositional calculus, equivalential calculus and minimal logic. The latter, minimal logic, is both paraconsistent and paracomplete (a subsystem of intuitionistic logic). The other three simply do not allow one to express a contradiction to begin with since they lack the ability to form negations.
## An ideal three-valued paraconsistent logic
Here is an example of a three-valued logic which is paraconsistent and ideal as defined in "Ideal Paraconsistent Logics" by O. Arieli, A. Avron, and A. Zamansky, especially pages 22–23.[16] The three truth-values are: t (true only), b (both true and false), and f (false only).
P ¬P
t f
b b
f t
P → Q Q
t b f
P t t b f
b t b f
f t t t
P ∨ Q Q
t b f
P t t t t
b t b b
f t b f
P ∧ Q Q
t b f
P t t b f
b b b f
f f f f
A formula is true if its truth-value is either t or b for the valuation being used. A formula is a tautology of paraconsistent logic if it is true in every valuation which maps atomic propositions to {t, b, f}. Every tautology of paraconsistent logic is also a tautology of classical logic. For a valuation, the set of true formulas is closed under modus ponens and the deduction theorem. Any tautology of classical logic which contains no negations is also a tautology of paraconsistent logic (by merging b into t). This logic is sometimes referred to as "Pac" or "LFI1".
### Included
Some tautologies of paraconsistent logic are:
• All axiom schemas for paraconsistent logic:
${\displaystyle P\to (Q\to P)}$ ** for deduction theorem and ?→{t,b} = {t,b}
${\displaystyle (P\to (Q\to R))\to ((P\to Q)\to (P\to R))}$ ** for deduction theorem (note: {t,b}→{f} = {f} follows from the deduction theorem)
${\displaystyle \lnot (P\to Q)\to P}$ ** {f}→? = {t}
${\displaystyle \lnot (P\to Q)\to \lnot Q}$ ** ?→{t} = {t}
${\displaystyle P\to (\lnot Q\to \lnot (P\to Q))}$ ** {t,b}→{b,f} = {b,f}
${\displaystyle \lnot \lnot P\to P}$ ** ~{f} = {t}
${\displaystyle P\to \lnot \lnot P}$ ** ~{t,b} = {b,f} (note: ~{t} = {f} and ~{b,f} = {t,b} follow from the way the truth-values are encoded)
${\displaystyle P\to (P\lor Q)}$ ** {t,b}v? = {t,b}
${\displaystyle Q\to (P\lor Q)}$ ** ?v{t,b} = {t,b}
${\displaystyle \lnot (P\lor Q)\to \lnot P}$ ** {t}v? = {t}
${\displaystyle \lnot (P\lor Q)\to \lnot Q}$ ** ?v{t} = {t}
${\displaystyle (P\to R)\to ((Q\to R)\to ((P\lor Q)\to R))}$ ** {f}v{f} = {f}
${\displaystyle \lnot P\to (\lnot Q\to \lnot (P\lor Q))}$ ** {b,f}v{b,f} = {b,f}
${\displaystyle (P\land Q)\to P}$ ** {f}&? = {f}
${\displaystyle (P\land Q)\to Q}$ ** ?&{f} = {f}
${\displaystyle \lnot P\to \lnot (P\land Q)}$ ** {b,f}&? = {b.f}
${\displaystyle \lnot Q\to \lnot (P\land Q)}$ ** ?&{b,f} = {b,f}
${\displaystyle (\lnot P\to R)\to ((\lnot Q\to R)\to (\lnot (P\land Q)\to R))}$ ** {t}&{t} = {t}
${\displaystyle P\to (Q\to (P\land Q))}$ ** {t,b}&{t,b} = {t,b}
${\displaystyle (P\to Q)\to ((\lnot P\to Q)\to Q)}$ ** ? is the union of {t,b} with {b,f}
• Some other theorem schemas:
${\displaystyle P\to P}$
${\displaystyle (\lnot P\to P)\to P}$
${\displaystyle ((P\to Q)\to P)\to P}$
${\displaystyle P\lor \lnot P}$
${\displaystyle \lnot (P\land \lnot P)}$
${\displaystyle (\lnot P\to Q)\to (P\lor Q)}$
${\displaystyle ((\lnot P\to Q)\to Q)\to (((P\land \lnot P)\to Q)\to (P\to Q))}$ ** every truth-value is either t, b, or f.
${\displaystyle ((P\to Q)\to R)\to (Q\to R)}$
### Excluded
Some tautologies of classical logic which are not tautologies of paraconsistent logic are:
${\displaystyle \lnot P\to (P\to Q)}$ ** no explosion in paraconsistent logic
${\displaystyle (\lnot P\to Q)\to ((\lnot P\to \lnot Q)\to P)}$
${\displaystyle (P\to Q)\to ((P\to \lnot Q)\to \lnot P)}$
${\displaystyle (P\lor Q)\to (\lnot P\to Q)}$ ** disjunctive syllogism fails in paraconsistent logic
${\displaystyle (P\to Q)\to (\lnot Q\to \lnot P)}$ ** contrapositive fails in paraconsistent logic
${\displaystyle (\lnot P\to \lnot Q)\to (Q\to P)}$
${\displaystyle ((\lnot P\to Q)\to Q)\to (P\to Q)}$
${\displaystyle (P\land \lnot P)\to (Q\land \lnot Q)}$ ** not all contradictions are equivalent in paraconsistent logic
${\displaystyle (P\to Q)\to (\lnot Q\to (P\to R))}$
${\displaystyle ((P\to Q)\to R)\to (\lnot P\to R)}$
${\displaystyle ((\lnot P\to R)\to R)\to (((P\to Q)\to R)\to R)}$ ** counter-factual for {b,f}→? = {t,b} (inconsistent with bf = f)
### Strategy
Suppose we are faced with a contradictory set of premises Γ and wish to avoid being reduced to triviality. In classical logic, the only method one can use is to reject one or more of the premises in Γ. In paraconsistent logic, we may try to compartmentalize the contradiction. That is, weaken the logic so that Γ→X is no longer a tautology provided the propositional variable X does not appear in Γ. However, we do not want to weaken the logic any more than is necessary for that purpose. So we wish to retain modus ponens and the deduction theorem as well as the axioms which are the introduction and elimination rules for the logical connectives (where possible).
To this end, we add a third truth-value b which will be employed within the compartment containing the contradiction. We make b a fixed point of all the logical connectives.
${\displaystyle b=\lnot b=(b\to b)=(b\lor b)=(b\land b)}$
We must make b a kind of truth (in addition to t) because otherwise there would be no tautologies at all.
To ensure that modus ponens works, we must have
${\displaystyle (b\to f)=f,}$
that is, to ensure that a true hypothesis and a true implication lead to a true conclusion, we must have that a not-true (f) conclusion and a true (t or b) hypothesis yield a not-true implication.
If all the propositional variables in Γ are assigned the value b, then Γ itself will have the value b. If we give X the value f, then
${\displaystyle (\Gamma \to X)=(b\to f)=f}$.
So Γ→X will not be a tautology.
Limitations: (1) There must not be constants for the truth values because that would defeat the purpose of paraconsistent logic. Having b would change the language from that of classical logic. Having t or f would allow the explosion again because
${\displaystyle \lnot t\to X}$ or ${\displaystyle f\to X}$
would be tautologies. Note that b is not a fixed point of those constants since bt and bf.
(2) This logic's ability to contain contradictions applies only to contradictions among particularized premises, not to contradictions among axiom schemas.
(3) The loss of disjunctive syllogism may result in insufficient commitment to developing the 'correct' alternative, possibly crippling mathematics.
(4) To establish that a formula Γ is equivalent to Δ in the sense that either can be substituted for the other wherever they appear as a subformula, one must show
${\displaystyle (\Gamma \to \Delta )\land (\Delta \to \Gamma )\land (\lnot \Gamma \to \lnot \Delta )\land (\lnot \Delta \to \lnot \Gamma )}$.
This is more difficult than in classical logic because the contrapositives do not necessarily follow.
## Applications
Paraconsistent logic has been applied as a means of managing inconsistency in numerous domains, including:[17]
## Criticism
Some philosophers have argued against dialetheism on the grounds that the counterintuitiveness of giving up any of the three principles above outweighs any counterintuitiveness that the principle of explosion might have.
Others, such as David Lewis, have objected to paraconsistent logic on the ground that it is simply impossible for a statement and its negation to be jointly true.[30] A related objection is that "negation" in paraconsistent logic is not really negation; it is merely a subcontrary-forming operator.[31]
## Alternatives
Approaches exist that allow for resolution of inconsistent beliefs without violating any of the intuitive logical principles. Most such systems use multi-valued logic with Bayesian inference and the Dempster-Shafer theory, allowing that no non-tautological belief is completely (100%) irrefutable because it must be based upon incomplete, abstracted, interpreted, likely unconfirmed, potentially uninformed, and possibly incorrect knowledge (of course, this very assumption, if non-tautological, entails its own refutability, if by "refutable" we mean "not completely [100%] irrefutable").
## Notable figures
Notable figures in the history and/or modern development of paraconsistent logic include:
• Alan Ross Anderson (United States, 1925–1973). One of the founders of relevance logic, a kind of paraconsistent logic.
• Florencio González Asenjo (Argentina, 1927-2013)
• Diderik Batens (Belgium)
• Nuel Belnap (United States, b. 1930) developed logical connectives of a four-valued logic.
• Jean-Yves Béziau (France/Switzerland, b. 1965). Has written extensively on the general structural features and philosophical foundations of paraconsistent logics.
• Walter Carnielli (Brazil). The developer of the possible-translations semantics, a new semantics which makes paraconsistent logics applicable and philosophically understood.
• Newton da Costa (Brazil, 1929-2024). One of the first to develop formal systems of paraconsistent logic.
• Itala M. L. D'Ottaviano (Brazil)
• J. Michael Dunn (United States). An important figure in relevance logic.
• Carl Hewitt
• Stanisław Jaśkowski (Poland). One of the first to develop formal systems of paraconsistent logic.
• David Kellogg Lewis (USA, 1941–2001). Articulate critic of paraconsistent logic.
• Jan Łukasiewicz (Poland, 1878–1956)
• Robert K. Meyer (United States/Australia)
• Chris Mortensen (Australia). Has written extensively on paraconsistent mathematics.
• Lorenzo Peña (Spain, b. 1944). Has developed an original line of paraconsistent logic, gradualistic logic (also known as transitive logic, TL), akin to fuzzy logic.
• Val Plumwood [formerly Routley] (Australia, b. 1939). Frequent collaborator with Sylvan.
• Graham Priest (Australia). Perhaps the most prominent advocate of paraconsistent logic in the world today.
• Francisco Miró Quesada (Peru). Coined the term paraconsistent logic.
• B. H. Slater (Australia). Another articulate critic of paraconsistent logic.
• Richard Sylvan [formerly Routley] (New Zealand/Australia, 1935–1996). Important figure in relevance logic and a frequent collaborator with Plumwood and Priest.
• Nicolai A. Vasiliev (Russia, 1880–1940). First to construct logic tolerant to contradiction (1910).
## Notes
1. ^ "Paraconsistent Logic". Stanford Encyclopedia of Philosophy. Archived from the original on 2015-12-11. Retrieved 1 December 2015.
2. ^ Priest (2002), p. 288 and §3.3.
3. ^ Carnielli, W.; Rodrigues, A. "An epistemic approach to paraconsistency: a logic of evidence and truth" Pittsburg
4. ^ Carnielli, W. and Marcos, J. (2001) "Ex contradictione non sequitur quodlibet" Archived 2012-10-16 at the Wayback Machine Proc. 2nd Conf. on Reasoning and Logic (Bucharest, July 2000)
5. ^ Feferman, Solomon (1984). "Toward Useful Type-Free Theories, I". The Journal of Symbolic Logic. 49 (1): 75–111. doi:10.2307/2274093. JSTOR 2274093. S2CID 10575304.
6. ^ Jennifer Fisher (2007). On the Philosophy of Logic. Cengage Learning. pp. 132–134. ISBN 978-0-495-00888-0.
7. ^ Graham Priest (2007). "Paraconsistency and Dialetheism". In Dov M. Gabbay; John Woods (eds.). The Many Valued and Nonmonotonic Turn in Logic. Elsevier. p. 131. ISBN 978-0-444-51623-7.
8. ^ Otávio Bueno (2010). "Philosophy of Logic". In Fritz Allhoff (ed.). Philosophies of the Sciences: A Guide. John Wiley & Sons. p. 55. ISBN 978-1-4051-9995-7.
9. ^ See the article on the principle of explosion for more on this.
10. ^ Priest (2002), p. 306.
11. ^ LP is also commonly presented as a many-valued logic with three truth values (true, false, and both).
12. ^ See, for example, Priest (2002), §5.
13. ^ See Priest (2002), p. 310.
14. ^ Surveys of various approaches to paraconsistent logic can be found in Bremer (2005) and Priest (2002), and a large family of paraconsistent logics is developed in detail in Carnielli, Congilio and Marcos (2007).
15. ^ See Aoyama (2004).
16. ^ "Ideal Paraconsistent Logics" (PDF). Archived (PDF) from the original on 2017-08-09. Retrieved 2018-08-21.
17. ^ Most of these are discussed in Bremer (2005) and Priest (2002).
18. ^ See, for example, truth maintenance systems or the articles in Bertossi et al. (2004).
19. ^ Gershenson, C. (1999). Modelling emotions with multidimensional logic. In Proceedings of the 18th International Conference of the North American Fuzzy Information Processing Society (NAFIPS ’99), pp. 42–46, New York City, NY. IEEE Press. http://cogprints.org/1479/
20. ^ de Carvalho Junior, A.; Justo, J. F.; Angelico, B. A.; de Oliveira, A. M.; da Silva Filho, J. I. (2021). "Rotary Inverted Pendulum Identification for Control by Paraconsistent Neural Network". IEEE Access. 9: 74155–74167. Bibcode:2021IEEEA...974155D. doi:10.1109/ACCESS.2021.3080176. ISSN 2169-3536.
21. ^ Hewitt (2008b)
22. ^ Hewitt (2008a)
23. ^ Carl Hewitt. "Formalizing common sense reasoning for scalable inconsistency-robust information coordination using Direct Logic Reasoning and the Actor Model". in Vol. 52 of Studies in Logic. College Publications. ISBN 1848901593. 2015.
24. ^ de Carvalho Junior, Arnaldo; Justo, João Francisco; de Oliveira, Alexandre Maniçoba; da Silva Filho, João Inacio (1 January 2024). "A comprehensive review on paraconsistent annotated evidential logic: Algorithms, Applications, and Perspectives". Engineering Applications of Artificial Intelligence. 127 (B): 107342. doi:10.1016/j.engappai.2023.107342. S2CID 264898768.
25. ^ Carvalho, A.; Angelico, B. A.; Justo, J. F.; Oliveira, A. M.; Silva, J. I. D. (2023). "Model reference control by recurrent neural network built with paraconsistent neurons for trajectory tracking of a rotary inverted pendulum". Applied Soft Computing. 133: 109927. doi:10.1016/j.asoc.2022.109927. ISSN 1568-4946.
26. ^ de Carvalho Junior, Arnaldo; Justo, João Francisco; de Oliveira, Alexandre Maniçoba; da Silva Filho, João Inacio (1 January 2024). "A comprehensive review on paraconsistent annotated evidential logic: Algorithms, Applications, and Perspectives". Engineering Applications of Artificial Intelligence. 127 (B): 107342. doi:10.1016/j.engappai.2023.107342. S2CID 264898768.
27. ^ de Carvalho Jr., Arnaldo; Da Silva Filho, João Inácio; de Freitas Minicz, Márcio; Matuck, Gustavo R.; Côrtes, Hyghor Miranda; Garcia, Dorotéa Vilanova; Tasinaffo, Paulo Marcelo; Abe, Jair Minoro (2023). "A Paraconsistent Artificial Neural Cell of Learning by Contradiction Extraction (PANCLCTX) with Application Examples". Advances in Applied Logics. Intelligent Systems Reference Library. Vol. 243. pp. 63–79. doi:10.1007/978-3-031-35759-6_5. ISBN 978-3-031-35758-9.
28. ^ Carvalho, Arnaldo; Justo, João F.; Angélico, Bruno A.; de Oliveira, Alexandre M.; da Silva Filho, João Inacio (22 October 2022). "Paraconsistent State Estimator for a Furuta Pendulum Control". SN Computer Science. 4 (1). doi:10.1007/s42979-022-01427-z. S2CID 253064746.
29. ^ de Carvalho Junior, Arnaldo; Justo, João Francisco; de Oliveira, Alexandre Maniçoba; da Silva Filho, João Inacio (1 January 2024). "A comprehensive review on paraconsistent annotated evidential logic: Algorithms, Applications, and Perspectives". Engineering Applications of Artificial Intelligence. 127 (B): 107342. doi:10.1016/j.engappai.2023.107342. S2CID 264898768.
30. ^ See Lewis (1982).
31. ^ See Slater (1995), Béziau (2000). | 7,507 | 28,350 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 75, "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} | 3.703125 | 4 | CC-MAIN-2024-30 | latest | en | 0.937759 |
https://stupidsid.com/previous-question-papers/download/computer-graphics-12 | 1,590,478,256,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00270.warc.gz | 490,608,096 | 10,591 | MORE IN Computer Graphics
MU Computer Engineering (Semester 4)
Computer Graphics
December 2013
Total marks: --
Total time: --
INSTRUCTIONS
(1) Assume appropriate data and state your reasons
(2) Marks are given to the right of every question
(3) Draw neat diagrams wherever necessary
1 (a) Explain character generation methods.
5 M
1 (b) Explain inside outside test used in polygon filling.
5 M
1 (c) What is antialiasing? How it can be reduced?
5 M
1 (d) Explain z-buffer algorithm for removing hidden surface.
5 M
2 (a) Explain flood fill algorithm using 8 connected approach. Give its advantages and disadvantages.
10 M
2 (b) Derive Bresenham's line drawing algorithm. Plot a line by using Bresenham s line generation algorithm from (1, 1) to (5, 3).
10 M
3 (a) Translate the square ABCD whose co-ordinates are A (0, 0), B (3, 0), C (3, 3), D (0, 3) by 2 units in both direction and then scale it by 1.5 in x direction and 0.5 units in y direction.
10 M
3 (b) List and explain operations on segment.
10 M
4 (a) Find the clipping coordinate to clip the line segment AB against the window using Cohen-Sutherland line clipping algorithm.
10 M
4 (b) Explain Warnock's algorithm.
10 M
5 (a) State important properties of Bezier curve. Compare Bezier curve and B-spline curve.
10 M
5 (b) Explain parallel and prospective projection. Derive matrix for perspective projection.
10 M
6 (a) Explain 3D object representation methods.
10 M
6 (b) Define window and viewport. Also derive window to viewport transformation.
10 M
Write short notes on (any four)
7 (a) Colour models.
5 M
7 (b) Raster techniques.
5 M
7 (c) Display file interpreter.
5 M
7 (d) Fractals.
5 M
7 (e) 3D clipping.
5 M
More question papers from Computer Graphics | 475 | 1,731 | {"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} | 2.59375 | 3 | CC-MAIN-2020-24 | latest | en | 0.821542 |
http://www.mathskey.com/question2answer/29948/solve-this-problem | 1,537,974,142,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267165261.94/warc/CC-MAIN-20180926140948-20180926161348-00204.warc.gz | 369,998,919 | 13,486 | # Solve this problem.
(Use a TI-84)
A newspaper article about the results of a poll states: "In theory, the results of a poll states: "In theory, the result of such a poll, in 99 cases out of 100 should differ by no more that 5 percentage points in either direction from what would have been obtained by interviewing all voters in the United States," Find the sample size suggested by this statement.
asked Apr 17, 2015
Step 1:
Marginal error is 5% .[ In either directions ]
Find the sample size.
Formula for sample size is .
where is the value from the standard normal distribution reflecting the confidence level.
is the desired margin of error,
p is the proportion of successes in the population,
q is the proportion of failure in the population.
Step 2:
Now find score for 99 cases out of 100.
Follow these steps to evaluate z score for top 99 cases out of 100.
1.Select invNorm() in Ti -84 calculator.
[ 2nd --> VARS --> 3 ]
2.Enter area in decimals .
invNorm((1 - 0.99)/2)
invNorm(0.005)
invNorm(0.005) = - 2.575
|Z | score for 99 cases out of 100 is 2.575.
Step 3:
Let consider the value p is 0.5.
So the value of q is .
Marginal error is E = 0.05.
Now find sample size substituting , E, p and q in .
So the sample size is 663.
Solution :
The sample size is 663.
answered Apr 18, 2015
edited Apr 18, 2015 | 365 | 1,342 | {"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} | 4.03125 | 4 | CC-MAIN-2018-39 | longest | en | 0.898601 |
https://www.codinghumans.xyz/2020/10/formal-languages-and-automation-theory-mcqs-set-IV.html | 1,620,522,470,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00378.warc.gz | 733,862,659 | 39,636 | Type Here to Get Search Results !
# Formal Languages And Automation Theory MCQs With Answers - Set IV
0
Formal Languages And Automation Theory MCQs
Set IV
Q1. Which of the given are correct?
a) Moore machine has 6-tuples
b) Mealy machine has 6-tuples
c) Both Mealy and Moore has 6-tuples
d) None of the mentioned
Answer is c) Both Mealy and Moore has 6-tuples
Explanation : The six tuples are (Q, ∑, O, δ, X, q0)
It can be described by a 6 tuple (Q, ∑, O, δ, X, q0) where −
Q is a finite set of states.
∑ is a finite set of symbols called the input alphabet.
O is a finite set of symbols called the output alphabet.
δ is the input transition function
X is the output transition function
q0 is the initial state from where any input is processed (q0 ∈ Q).
Q2. Which of the following does not belong to input alphabet if S={a, b}* for any language?
a) a
b) b
c) e
d) none of the above
Explanation : The automaton may be allowed to change its state without reading the input symbol using epsilon but this does not mean that epsilon has become an input symbol. On the contrary, one assumes that the symbol epsilon does not belong to any alphabet.
Q3. Design a NFA for the language:
L: {an| n is even or divisible by 3}
Which of the following methods can be used to simulate the same.
a) e-NFA
b) Power Construction Method
c) Both (a) and (b)
d) None of the mentioned
Answer is c) Both (a) and (b)
Explanation : It is more convenient to simulate a machine using e-NFA else the method of Power Construction is used from the union-closure of DFA’s.
Q4. State true or false:
Statement: Both NFA and e-NFA recognize exactly the same languages.
a) true
b) false
Explanation : e-NFA do come up with a convenient feature but nothing new.They do not extend the class of languages that can be represented.
Q5. The number of elements present in the e-closure(f2) in the given diagram:
a) 0
b) 1
c) 2
d) 3
Explanation : The epsilon closure set of f2 consist of the elements:{f2, f3}. Thus the count of the element in the closure set is 2.
Q6. To describe the complement of a language, it is very important to describe the __________ of that language over which the language is defined.
a) Alphabet
b) Regular Expression
c) String
d) Word
Explanation : For explanation Join the discussion below
Q7. Which of the following not an example Bounded Information?
a) fan switch outputs {on, off}
c) colour of the traffic light at the moment
d) none of the mentioned
Explanation : Bounded information refers to one whose output is limited and it cannot be said what were the recorded outputs previously until memorized
Q8. How many languages are over the alphabet R?
a) countably infinite
b) countably finite
c) uncountable finite
d) uncountable infinite
Explanation : A language over an alphabet R is a set of string over A which is uncountable and infinite
Q9. The number of tuples in an extended Non Deterministic Finite Automaton:
a) 5
b) 6
c) 7
d) 4
Explanation : For NFA or extended tranisition function NFA , the tuple of elements remains same i.e. 5
Q10. Under which of the following operation, NFA is not closed?
a) Negation
b) Kleene
c) Concatenation
d) None of the mentioned
Answer is d) None of the mentioned
Explanation : NFA is said to be closed under the following operations:
a) Union
b) Interaction
c) Concatenation
d) Kleene
e) Negation
Q11. Given Language: {x | it is divisible by 3}
The total number of final states to be assumed in order to pass the number constituting {0, 1} is
a) 0
b) 1
c) 2
d) 3
Explanation : The DFA of a given language can be created
Q12. The automaton which allows transformation to a new state without consuming any input symbols:
a) NFA
b) DFA
c) NFA-I
d) All of the mentioned
Explanation : NFA-I or e-NFA is an extension of Non detrministic Finite Automata which are usually called NFA with epsilon moves or lambda transitions
Q13. Definition of a language L with alphabet {a} is given as following. L= { ank | k > 0, and n is a positive integer constant} What is the minimum number of states needed in a DFA to recognize L?
a) k+1
b) n+1
c) 2n+1
d) 2k+1
Explanation : Note that n is a constant and k is any positive integer. For example, if n is given as 3, then the DFA must be able to accept 3a, 6a, 9a, 12a, .. To build such a DFA, we need 4 states.
Q14. Design a NFA for the language:
L: {an| n is even or divisible by 3}
Which of the following methods can be used to simulate the same.
a) e-NFA
b) Power Construction Method
c) Both (a) and (b)
d) None of the mentioned
Answer is c) Both (a) and (b)
Explanation : It is more convenient to simulate a machine using e-NFA else the method of Power Construction is used from the union-closure of DFA’s.
Q15. Which of the following does the given Mealy machine represents?
a) 9’s Complement
b) 2’s Complement
c) 1’s Complement
d) 10’s Complement
Explanation : Inputs can be taken and can be verified.
If you have any doubts join the discussion below ,our Moderator will reply to your Queries | 1,365 | 5,060 | {"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-2021-21 | latest | en | 0.84564 |
https://www.studypool.com/discuss/374343/simplify-the-following-polynomial-function?free | 1,508,561,948,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824570.79/warc/CC-MAIN-20171021043111-20171021063111-00278.warc.gz | 1,018,725,376 | 14,990 | Time remaining:
##### simplify the following polynomial function
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
p(x) = a (x-b)(x-c) / (a-b)(a-c) +b (x-a)(x-c)/(b-a)(b-c) + c (x-a)(x-b)/(c-a)(c-b)
Oct 20th, 2017
The answer is p(x) = x. The simplest (while not most elegant) way to solve this task is to compute directly.
First, lead it to a common denominator (a-b)(b-c)(c-a), denote this denominator as D. We must consider a sign - at each summand:
p(x) = [-a(x-b)(x-c)(b-c) - b(x-a)(x-c)(c-a) - c(x-a)(x-b)(a-b)]/[D]
We see that D didn`t depend on the x, let`s calculate the numerator:
-a(x^2 - bx - cx + bc)(b-c) - b(x^2 - ax - cx + ac)(c-a) - c(x^2 - ax - bx + ab)(a-b).
Let`s gather the summands with x^2, x^1 and x^0 (without x):
x^2*[-a(b-c) - b(c-a) - c(a-b)] + x^1*[a(b+c)(b-c) + b(a+c)(c-a) + c(a+b)(a-b)] +[-abc(b-c) - abc(c-a) - abc(a-b)] =
= x^2*[-ab + ac - bc + ba - ca + cb] + x^1*[a(b^2-c^2) + b(c^2-a^2) + c(a^2-b^2)] + [-abc(b - c + c - a + a - b)].
x^2 and x^0 obviously have zero factors, so our numerator is
x^1 * [ab^2 - ac^2 + bc^2 - ba^2 + ca^2 - cb^2].
And finally, open the brackets in D:
D = (a-b)(b-c)(c-a) = (ab - b^2 - ac + bc)(c-a) = abc - cb^2 - ac^2 + ca^2 - ba^2 + ab^2 + ca^2 - abc.
+- abc vanishes and we see that the numerator equals x*D. So, p(x) = x.
Feb 5th, 2015
...
Oct 20th, 2017
...
Oct 20th, 2017
Oct 21st, 2017
check_circle | 580 | 1,421 | {"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} | 4.40625 | 4 | CC-MAIN-2017-43 | latest | en | 0.61067 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.