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https://www.instasolv.com/question/the-work-done-on-a-particle-of-mass-m-by-a-force-x-k-being-a-constan-z7w576 | 1,611,260,894,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527850.55/warc/CC-MAIN-20210121194330-20210121224330-00426.warc.gz | 815,313,755 | 11,460 | The work done on a particle of mass...
Question
# The work done on a particle of mass ( m ) by a force, ( Kleft[frac{x}{left(x^{2}+y^{2}right)^{3 / 2}} hat{mathbf{i}}+frac{y}{left(x^{2}+y^{2}right)^{3 / 2}} hat{mathbf{j}}right](K ) being a constant of appropriate dimensions), when the particle is taken from the point ( (a, 0) ) to the point ( (0, a) ) along a circular path of radius ( a ) about the origin in the ( x ) - ( y ) plane is (a) ( frac{2 K pi}{a} )(b) ( frac{K pi}{a} )(c) ( frac{K pi}{2 a} )(d) 0
NEET/Medical Exams
Physics
Solution
121
4.0 (1 ratings) | 210 | 569 | {"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-2021-04 | latest | en | 0.691516 |
https://psychologslask.com.pl/2018/Jan/20_2839.html | 1,660,286,311,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571584.72/warc/CC-MAIN-20220812045352-20220812075352-00598.warc.gz | 436,095,816 | 6,638 | calculate grade for m10 grinder
# calculate grade for m10 grinder
• ### Understanding the Difference between Chain Grades and How ...
May 07, 2015· Chain Grades. One of the safety measures implemented was to place chain in Grades based on the ultimate breaking strength of that chain. This number is what we see today G30, G43, G70, G80 G100 and the common chain grades. The number after each letter is N/mm 2.
• ### Bolt Depot Bolt Grade Markings and Strength Chart
Class Stainless markings vary. Most stainless is nonmagnetic. Usually stamped A2.
• ### IXL | Compare and convert metric units of length | 3rd ...
Improve your math knowledge with free questions in "Compare and convert metric units of length" and thousands of other math skills.
• ### How to Calculate Power Based on Work and Time dummies
Sometimes, it isn't just the amount of work you do but the rate at which you do work that's important. In physics, the concept of power gives you an idea of how much work you can expect in a certain amount of time. Power in physics is the amount of work done divided by the [.]
• ### Calculate Weight of a Steel Plate | Chapel Steel
Calculate the weight of a steel plate with Chapel's conversion calculator. ... Weight Calculators. Weight of a Steel Plate. Please enter values then click on Calculate. ... Inches to Millimeters Millimeters to Inches Pounds to Kilograms Kilograms to Pounds Metric Tons to Pounds Metric Tons to Tons Fahrenheit to Celsius Celsius to Fahrenheit ft ...
• ### Circumference Calculator | Math Goodies
Our circumference calculator provides circumference of a circle by entering its radius. Use our circumference calculator to determine the area of a circle. For more help on the circumference of a circle, try our Circumference Lesson.
• ### Diameter / Decimal / Metric Chart for Small Diameters ...
Grades ASTM Grade 4 or 5 1 A307, Grade A 4 or 5 2 5 2 8 5 A325, A449 9 5+ A193, B7 and B16 10 or 12 8 A490, A354, Grade 8D 10 or 12 A540, B21 through B24 Common DIN Numbers for Metric Fasteners DIN Hex Cap Screws 931 Coarse thread pitch partially threaded (specify grade)
• ### Homewyse Calculator: Cost to Grind Large Tree Stump
The cost to Grind a Large Tree Stump starts at 157 per stump, but can vary significantly with site conditions and options. Get fair costs for your SPECIFIC project requirements. See typical tasks and time to grind a large tree stump, along with per unit costs and material requirements. See professionally prepared estimates for large stump grinding work.
• ### Understanding Metric Fasteners Fastener Mart
Understanding Metric Fasteners. = Pitch (distance from thread to thread), in millimeters 20 = Length, in millimeters (see below) This example illustrates how a coarse threaded screw may appear. M12 x 25 = 12 mm diameter, coarse thread is assumed ( mm), 25 mm long M12 is a matching coarse threaded nut (To avoid confusion,...
• ### Fastener, Bolt and Screw Design Torque and Force ...
The following design resources are for design screws and bolts for the proper torque, stress, strain, preload and other engineering critical design parameters. Thread Stress Area Calculator and Equation. Calculating Assembly Torque per ISO 68 ISO 724. Screw Stress Area 100 ksi Greater. Bolt Stress Area less Than 100 ksi.
• ### Best Mixer Grinder for Idli Dosa Batter for Indian ...
Mar 24, 2019· Elgi Ultra Grind – Best Wet Grinder for Dosa Batter in India. If you are a south Indian, you will never underestimate the essence of dosas and idlis. The Elgi Ultra Grind wet grinder is an excellent option for you. The grinder comes with a certification of AISI 304 grade stainless steel in construction.
• ### How to Calculate the Hydraulic Grade Line
Oct 23, 2013· How to Calculate the Hydraulic Grade Line October 23, 2013 by Bernie Roseke,, PMP 1 Comment The Hydraulic Grade Line is a graph of the pressure and gravitational heads plotted along the position of the pipeline or channel.
• ### Stonemill Essentials Sea Salt Grinder: calories, nutrition ...
Personalized health review for Stonemill Essentials Sea Salt Grinder: 0 calories, nutrition grade (C plus), problematic ingredients, and more. Learn the good bad for 250,000+ products.
• ### GRADE SLOPE CONTROLS FOR PAVERS AND MILLING .
GRADE SLOPE CONTROLS FOR PAVERS AND MILLING MACHINES. MOBAMATIC: MOBA Mobile Automation AG Kapellenstraße 15 ... » Slope display even at grade control SLOPE SENSOR » Very accurate grade position sensor (± 0,5 mm) for milling applications ... raged by the sensor to calculate an average elevation. By averaging and filtering, small ...
• ### FUTEK Bolt Torque Calculator | Bolt Torque Calculators
To find your resistance: Use a multimeter to determine the resistance across the +Signal and Signal wires. This value is your bridge resistance. Typical Futek sensors have . | 1,106 | 4,851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2022-33 | latest | en | 0.844579 |
https://www.eduzip.com/question/5498 | 1,642,932,070,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304217.55/warc/CC-MAIN-20220123081226-20220123111226-00605.warc.gz | 788,702,722 | 9,899 | 6 years ago in Quantitative Aptitude
# If 32x-y=32x-y= √ 27 , then the value of 3x-y will be ?
[A] <span style="white-space: nowrap; font-size:larger"> √<span style="text-decoration:overline;"> 3 </span></span>
[B] 3
[C] 1/<span style="white-space: nowrap; font-size:larger"> √<span style="text-decoration:overline;"> 27 </span></span>
[D] 1/<span style="white-space: nowrap; font-size:larger"> √<span style="text-decoration:overline;"> 3 </span></span>
## Overall Stats
Attempted 6
Correct 2
Incorrect 1
Viewed 3
Nazia Sayed - 4 years ago
Nazia Sayed from Mumbai, India is saying 3 is correct answer
Sayed badruzama - 4 years ago
Mahantesh Janjeer - 4 years ago
Mahantesh Janjeer from Bengaluru, India is saying 1/ 27 is correct answer
• [A] 26.9%
• [B] 27.3%
• [C] 27.8%
• [D] 26.5%
• [A] 27
• [B] 212
• [C] 2
• [D] 25
• [A] 26 cm
• [B] 13 cm
• [C] 19.5 cm
• [D] 20.5 cm
• [A] 59.96
• [B] 599.6
• [C] 5.996
• [D] 0.5996
• [A] Rs. 800
• [B] Rs. 900
• [C] Rs. 600
• [D] RS. 700 | 420 | 992 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-05 | latest | en | 0.500867 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=128&t=41148&view=print | 1,604,058,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910204.90/warc/CC-MAIN-20201030093118-20201030123118-00683.warc.gz | 381,982,699 | 2,078 | Page 1 of 1
### Final Temperature at Constant Pressure, and then at Constant Volume
Posted: Fri Feb 01, 2019 3:52 pm
Hey guys,
I am working on problem 4C.3, which asks to calculate the final temperature and the change in enthalpy when 765 J of energy is transferred as heat to 0.820 mol Kr (g) at 298 L and 1.00 atm.
I know how to calculate the final temperature and enthalpy with this information but I am unsure of how to calculate that with constant pressure and then with constant volume. If someone could explain to get me started it would be much appreciated!
### Re: Final Temperature at Constant Pressure, and then at Constant Volume
Posted: Sat Feb 02, 2019 11:09 am
Hi Mariah!
I'm assuming that you mean T = 298 K and not V = 298 L. As such, the volume won't be constant, but the pressure will be. So I don't think you need to worry about there being a constant volume since we don't know if the gas does any work. We know that no work is done ON the gas, but by adding heat, we could make the gas DO work.
Hope that helps!
### Re: Final Temperature at Constant Pressure, and then at Constant Volume
Posted: Mon Feb 04, 2019 2:09 pm
For this question you need to know that for an ideal gas Cp(specific heat at constant pressure) = (5/2) R and at constant volume is Cv = (3/2) R where R is equal to 8.312 J*K-1*mol-1.
You can find these values on the constant and formulas sheet.
Hope this helps! | 373 | 1,414 | {"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} | 3.515625 | 4 | CC-MAIN-2020-45 | latest | en | 0.943067 |
https://www.physicsforums.com/members/physics_fool.148994/recent-content | 1,627,548,484,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00615.warc.gz | 977,157,611 | 8,542 | # Recent content by physics_fool
1. ### Finding the coefficient of kinetic friction
Ok...nvm. I got it...thanks anyways:smile:
2. ### Finding the coefficient of kinetic friction
That's the question as it was given... but for your second question, isn't that to find the net force...not the force of friction?
3. ### Finding the coefficient of kinetic friction
Homework Statement A 50 kg block slides down a 32° ramp with an acceleration of 3.2 m/s2. What is the coefficient of kinetic friction between the block and the ramp? Homework Equations \mukinetic=\frac{f_{kinetic}}{F_{normal}} The Attempt at a Solution I have gone through many... | 157 | 644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-31 | latest | en | 0.928825 |
http://hydraraptor.blogspot.com/2007/10/die-swell-revisited.html | 1,566,719,792,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027323221.23/warc/CC-MAIN-20190825062944-20190825084944-00530.warc.gz | 96,135,400 | 14,387 | ## Tuesday, 2 October 2007
### Die swell revisited
My machine is back up and running again after replacing the thermistor and MSP430 micro. I added some high temperature insulation to keep the thermistor wires away from the heater wires in future! I salvaged it from the 10A shunt of an old multimeter that I scrapped.
I should really recalibrate the temperature measurement but after finding temperature is not too critical I can't be bothered at the moment. I might wait until I after I have blown it up again!
In my previous article Equations of Extrusion I put forward a theory for the significance of the Y axis crossing point on this graph, i.e. the minimum filament diameter at zero flow rate, was that it was the size of the hole in the nozzle.
My explanation was that perhaps I had inadvertently opened out the hole in the extruder nozzle by drilling too far from the back. While my extruder was off the machine for its new thermistor I had the opportunity to inspect the end of the nozzle.
As you can see the hole is still the correct size, 0.5mm, so my theory was wrong. My revised explanation for the minimum filament size is that HDPE is so viscous that there is a minimum pressure to make it flow through a small hole, below which it does not flow at all. The minimum diameter is then the hole size plus the die swell at that minimum pressure.
I used a fine stiff wire as a probe to try to get an idea of the depth of the hole. I think it is no more than 1mm, so I am at a loss to explain why I get variable die swell and other people do not. Perhaps my HDPE is different, I know mine is translucent whereas Forrest Higgs' is opaque white.
1. Okay, let's compare notes. I was getting pretty consistently 0.8 mm with the last extruder barrel.
With the new one I did a direct drill without pecking and I'm pretty sure that I've got a 0.5 mm orifice. It's been drilled through a dimple in a 0.127 mm copper sheet (0.005 inches).
I let the extruder barrel heat up and the HDPE extrude by thermal expansion only and got a HDPE thread without stretching of about 0.635 mm.
I also ran the system at several mm/sec extrusion rate and got 0.584 mm. I'm assuming that the smaller diameter is due to gravity driven stretching of the filament near the extruder before it had a chance to cool.
As best as I can conjure the difference between our respective systems lies in the fact that your extrusion hole, at ~1 mm is just a shade under 8 times deeper than mine.
Mind, this doesn't set aside the potential for differences between the feedstock HDPEs that we are using.
2. The thermal expansion during warmup gives about 0.9mm filament on my system.
I think I understand this now. The deeper the hole and / or the faster the feed, the more pressure required, the more die swell.
When the hole is as short as yours, i.e. shorter than its diameter, the die swell becomes insignificant compared to the hole size.
3. That's what seems to be implied. I'm not sure, though. As you said some time ago, die swell seems to be a pretty complicated phenomena. | 705 | 3,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-35 | latest | en | 0.96472 |
https://www.resurrectionofgavinstonemovie.com/what-does-funding-gap-mean/ | 1,675,545,924,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500154.33/warc/CC-MAIN-20230204205328-20230204235328-00515.warc.gz | 982,819,798 | 10,277 | Resurrectionofgavinstonemovie.com
Live truth instead of professing it
# What does funding gap mean?
## What does funding gap mean?
A term sometimes applied to a period when federal agencies lack authority to obligate or spend funds because their appropriations for that period have not been enacted. Spending gaps occur most frequently at the beginning of a fiscal year, but agencies occasionally run out of money later in the year.
## How do you calculate funding gap?
To calculate its gap ratio, a business must divide the total value of its interest-sensitive assets by the total value of its interest-sensitive liabilities. Once it has this quotient, the business may represent it as a decimal or as a percentage.
What is a positive funding gap?
A positive gap, or one greater than one, is the opposite, where a bank’s interest rate sensitive assets exceed its interest rate sensitive liabilities. A positive gap means that when rates rise, a bank’s profits or revenues will likely rise. There are two types of interest rate gaps: fixed and variable.
What should I offer investors in return?
Angel investors typically want from 20 to 25 percent return on the money they invest in your company. Venture capitalists may take even more; if the product is still in development, for example, an investor may want 40 percent of the business to compensate for the high risk it is taking.
### What does a negative funding gap mean?
A negative gap is a situation where a financial institution’s interest-sensitive liabilities exceed its interest-sensitive assets. A negative gap is not necessarily a bad thing, because if interest rates decline, the entity’s liabilities are repriced at lower interest rates. In this scenario, income would increase.
### What is funding & gapping in banking?
INTRODUCTION. “Funds gap” or “gap” is positive when the peso amount of sensitive assets exceeds that of sensitive liabilities. The gap is negative if sensitive liabilities exceed sensitive assets. When sensitive assets are equal to sensitive liabilities, we have a zero fund gap.
What is the gap ratio?
It is the ratio of a company’s rate sensitive assets to the liabilities to see how much profit is recognized.
What is a negative funding gap?
A negative gap is a situation where a financial institution’s interest-sensitive liabilities exceed its interest-sensitive assets. A negative gap is not necessarily a bad thing, because if interest rates decline, the entity’s liabilities are repriced at lower interest rates.
#### What is RSA and RSL?
• RSA = all the assets that mature or are repriced within the. gapping period (maturity bucket) • RSL = all the liabilities that mature or are repriced within. the gapping period (maturity bucket)
#### What ROI do investors look for?
7%-10%
For stock market investments, anywhere from 7%-10% is usually considered a good ROI, and many investors use the S&P to guide their investment strategy. There are other types of investments you can make and those have different expectations, such as: Government bonds can produce a return of around 5%.
How do you do a gap analysis?
The four steps of a gap analysis are:
1. Identify the current situation. Define what is important for you in your department or organization.
2. Set S.M.A.R.T goals of where you want to end up. S.M.A.R.T.
3. Analyze gaps from where you are to where you want to be.
4. Establish a plan to close existing gaps.
What is Viability Gap Funding?
Viability Gap Funding (VGF) Means a grant one-time or deferred, provided to support infrastructure projects that are economically justified but fall short of financial viability.
## What is a funding gap?
A funding gap occurs when there are not enough funds to finance operations or future development projects. Funding gaps are common for early-stage companies as it is difficult to accurately estimate future operating expenses and profit margins are narrow.
## What is the meaning of the viability of VGF?
Viability literally means ability to survive successfully. VGF is an economic instrument (or scheme) of Government of India ,launched in 2004 with the motive of supporting projects which come under public-private partnerships(PPP) model. Basically, it is a grant to support projects that are economically justified but are not financially viable.
Why do infrastructure projects have poor financial viability?
The lack of financial viability usually arises from long gestation periods and the inability to increase user charges to commercial levels. Infrastructure projects also involve externalities that are not adequately captured in direct financial returns to the project sponsor. | 920 | 4,673 | {"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} | 2.53125 | 3 | CC-MAIN-2023-06 | latest | en | 0.934573 |
https://fyhao.com/2012/03/computer-and-it/overview-for-functional-programming/ | 1,702,122,574,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100909.82/warc/CC-MAIN-20231209103523-20231209133523-00217.warc.gz | 298,512,922 | 11,170 | # Overview for Functional Programming
In computer science, functional programming is a programming paradigm that treats computation as the evaluation of mathematical functions and avoids state and mutable data. It emphasizes the application of functions, in contrast to the imperative programming style, which emphasizes changes in state.
In practice, the difference between mathematical functions and notion of a “function” used in imperative programming is that the imperative function can have side effect, it changes the program state. Means the same language expression can have different results over time, but functional programming is not, the data is immutable, unchangeable, the same language expression always lead to same result.
Functional programming promotes higher order function, that it is able to pass function as a parameter to other function.
In JavaScript,
``` var len = function(list) { return list.length; } var sum = function(list) { var total = 0; for(var i = 0; i < list.length; i++) { total += list[i]; } return total; }; var avg = function(list) { var total = c(list, sum); return total / c(list, len); }; var c = function(list, m) { return m(list); }; var a = [1,2,3,4,5]; c(a, sum); // 15 c(a, avg); // 3 ```
Many languages support Closures, the popular one is JavaScript, F#, Erlang, Scala (JVM Language). Java 7 is adding support declaration of Lambda function. Excitingly, PHP 5.3.0 started to support anonymous function now.
Before Java 7 is introduced, we also can declare anonymous classes (a class without assigning reference). From inside the method of anonymous classes to access the variable outside the scope of the anonymous class, the variable should declared as final. Means, the variable must be immutable, unchangeable, otherwise it caused side effect.
``` final int a = 0; new SomeClass() { void someMethod() { System.out.println("a: " + a); } } ```
But sometimes we want to achieve some simple computation, we may just define an anonymous function passed as an argument to other function.
## Author: fyhao
Jebsen & Jessen Comms Singapore INTI University College Bsc (Hon) of Computer Science, Coventry University
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 494 | 2,254 | {"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-2023-50 | longest | en | 0.795714 |
http://wiki.zcubes.com/Manuals/calci/GAMMAINV | 1,643,239,084,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305006.68/warc/CC-MAIN-20220126222652-20220127012652-00674.warc.gz | 64,369,917 | 7,221 | # Manuals/calci/GAMMAINV
• is the probability value associated with gamma distribution
• & are the values of the shape and rate parameters.
• gives accurate value of the solution.
• is any positive integer.
• GAMMAINV(),returns the inverse of the gamma cumulative distribution.
## Description
• This function gives the inverse value of Cumulative Gamma Probability Distribution.
• This distribution is the Continuous Probability Distribution on the positive real line and it is of the reciprocal of a variable distributed according to the gamma distribution with two parameters & .
• It is used in Bayesian statistics.
• In , is the probability value associated with Gamma Distribution, is called shape parameter and is the rate parameter of the distribution.
• If , then .
• GAMMAINV use the iterating method to find the value of .
• Suppose the iteration has not converged after 100 searches, then the function gives the error result.
• This function will give the error result when
1.Any one of the arguments are non-numeric
2. or
3. or
## Examples
1. =GAMMAINV(0.65189,2,5) = 11.1335534510
2. =GAMMAINV(0.006867292,5,7) = 8.155481331
3. =GAMMAINV(0.1543869,9,3) = 18.0467153645
4. =GAMMAINV(1,9,3) = 82.51739521528073
5. =GAMMAINV(1.1,9,3) = NAN, because
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GAMMA Distribution | 346 | 1,303 | {"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} | 3.0625 | 3 | CC-MAIN-2022-05 | latest | en | 0.750163 |
https://programmer.group/gym102956-partial-solution.html | 1,656,459,906,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103619185.32/warc/CC-MAIN-20220628233925-20220629023925-00337.warc.gz | 517,679,124 | 11,015 | # Gym102956 partial solution
Keywords: gym
Take more.
Thank you@ hht2005 Help.
#### B. Beautiful Sequence Unraveling
Define a good sequence with a length of \ (n \) \ (a \) is a sequence in which there is no position \ (1 \ le I < n \) such that \ (\ max\{a_1,\ldots,a_i\}=\min\{a_{i+1},\ldots,a_n \} \).
Count the good sequence with length of \ (n \) and value range of \ ([1,k] \), modulo \ (p \) and output.
$$1\le n\le 400$$，$$1\le k\le 10^8$$，$$998244353\le p\le 10^9+9$$，$$p\in\text{prime}$$.
Solution
$$k$$ is yours, if you consider the direct coolness related to \ (k \).
Considering avoiding \ (k \), it is not difficult to find that there are only \ (n \) different numbers in the legal sequence, which is directly discretized.
Consider that \ (f_{i,j} \) represents a sequence with a length of \ (I \) and the number of good sequences with a value range of \ ([1,j] \).
A good sequence is not easy to find, but considering a bad sequence \ (a \), set its maximum position \ (P \) so that \ (\ max\{a_1,\ldots,a_p\}=\min\{a_{p+1},\ldots,a_n \} \), it is easy to find a big property: \ (a_{p+1},\ldots,a_n \) is a good sequence. If it is not a good sequence, \ (P \) is not the maximum.
Enumerating the maximum position \ (p \) and enumerating its value \ (m \) can get the DP formula:
$f_{i,j}=j^i-\sum^{i-1}_{p=1}\sum^j_{m=1}(m^p-(m-1)^p)(f_{i-p,j-m+1}-f_{i-p,j-m})$
\ (O(n^3) \) can be achieved using prefixes and optimizations.
Consider setting \ (g_, I \) to represent the sequence with length of \ (n \), and the number of good sequences with value range of \ ([1,i] \):
$g_i=f_{n,i}-\sum^{i-1}_{j=1}\binom i j g_j$
So the answer is: \ (\ displaystyle \ sum ^ n {I = 1} \ BINOM k i g u I \).
Combination number calculation, time complexity \ (O(n^3) \).
Code
const int N=400;
int n,K,mod,f[N+10][N+10],ans,g[N+10],sum[N+10][N+10][N+10];
x+=y;
if(x>=mod) x-=mod;
}
void Del(int &x,int y) {
x-=y;
if(x<0) x+=mod;
}
int fpow(int x,int y) {
int ret=1;
for(;y;y>>=1) {
if(y&1) ret=1ll*ret*x%mod;
x=1ll*x*x%mod;
}
return ret;
}
int C(int n,int m) {
int ret=1,fm=1;
for(int i=n;i>=n-m+1;i--) ret=1ll*ret*i%mod;
for(int i=1;i<=m;i++) fm=1ll*fm*i%mod;
return 1ll*ret*fpow(fm,mod-2)%mod;
}
//n!/(n-m)!/m!
int po[N+10][N+10],inv[N+10][N+10];
void init() {
for(int i=1;i<=n;i++) {
po[i][0]=inv[i][0]=1;
inv[i][1]=fpow(po[i][1]=i,mod-2);
for(int j=2;j<=n;j++) po[i][j]=1ll*po[i][j-1]*i%mod,inv[i][j]=1ll*inv[i][j-1]*inv[i][1]%mod;
}
}
int main() {
init();
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++) {
f[i][j]=po[j][i];
for(int t=1;t<=j;t++) {
int A=po[t][i],B=po[t-1][i];
int sum1=(sum[i-1][j-t][t]-sum[i-1][j-t+1][t]+mod)%mod;
int sum2=(sum[i-1][j-t+1][t-1]-sum[i-1][j-t][t-1]+mod)%mod;
}
for(int t=1;t<=n;t++) sum[i][j][t]=(sum[i-1][j][t]+1ll*f[i][j]*inv[t][i]%mod)%mod;
}
for(int i=1;i<=n;i++) {
g[i]=f[n][i];
for(int j=1;j<i;j++) Del(g[i],1ll*C(i,j)*g[j]%mod);
}
write((ans+mod)%mod),enter;
return WDNMD;
}
#### C. Brave Seekers of Unicorns
Defining a good sequence satisfies the following conditions:
• The sequence is not empty.
• There are no three consecutive elements exclusive or and \ (= 0 \).
• Sequence increment.
• The value range of the sequence is \ ([1,n] \).
Given \ (n \), count the good sequence, and take the module for the answer \ (998244353 \).
Solution
Let \ (f_i \) represent the number of all schemes ending with \ (I \), including:
$f_i=\sum^{i-1}_{j=1}f_j-[j\oplus i<j]f_{j\oplus i}$
$$f_j$$ can be prefixed and taken away by one wave. Considering the later, it can be deformed into \ (\ sum_ {K < K \ oplus I < I} f_k \).
Considering the case of \ (i\oplus j\oplus k=0 \), there are only two \ (1 \) at most in a bit. Enumerate the highest \ (1 \) bit \ (d \) of \ (i \) in the binary system. It is determined that the \ (d \) bit of \ (j \) is \ (1 \) and the \ (d \) bit of \ (K \) is \ (0 \), so the \ (d \) bit can be arbitrarily taken in the future.
Then the range of \ (k \) is \ ([2^d,2^{d+1}-1] \), which can also be prefixed and combined.
Time complexity \ (O(n\log n) \).
Code
const int N=1e6,mod=998244353;
ll f[N+10],n,sum[N+10];
int main() {
scanf("%d",&n);
f[1]=1;sum[1]=1;
for(int i=2;i<=n;i++) {
f[i]=sum[i-1]+1;
int j;
for(j=20;j>=0;j--)
if(i&(1<<j)) break;
j--;
for(;j>=0;j--) if(i&(1<<j)) f[i]=(f[i]-sum[(1<<(j+1))-1]+sum[(1<<j)-1]+mod)%mod;
sum[i]=(f[i]+sum[i-1])%mod;
}
printf("%lld\n",sum[n]%mod);
}
#### D. Bank Security Unification
Take a subsequence from a sequence \ (a \) with a length of \ (n \), so that the sum of bitwise and values of two adjacent elements is the largest.
$$2\le n\le 10^6$$，$$0\le a_i\le 10^{12}$$.
Solution
We want the bitwise and maximum, that is, we want the binary digits of two adjacent elements to be as same as possible.
Consider that \ (f_i \) represents the maximum answer of the subsequence with the last bit of \ (I \), and the naive transfer is \ (O(n^2) \).
Consider pruning. If we consider the current one, where do we want to transfer from, it is obviously the same bit. If we choose these bits later, they must be better than those in the front.
Therefore, the record \ (las_{i,0/1} \) indicates where the last one in the \ (I \) bit is \ (0 / 1 \), which is only transferred from the corresponding \ (Las \) each time.
Time complexity \ (O(n\log a_i) \).
Code
const int N=1e6;
int n;
ll f[N+10],g[N+10],ans;
int las[44][2];
int main() {
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%lld",&f[i]);
g[2]=(f[1]&f[2]);
for(int i=1;i<=2;i++)
for(int j=39;j>=0;j--)
if(f[i]&(1ll<<j)) las[j][1]=i;
else las[j][0]=i;
for(int i=3;i<=n;i++) {
for(int j=39;j>=0;j--) {
int zz=(bool)(f[i]&(1ll<<j));
g[i]=max(g[i],g[las[j][zz]]+(f[las[j][zz]]&f[i]));
}
for(int j=39;j>=0;j--)
if(f[i]&(1ll<<j)) las[j][1]=i;
else las[j][0]=i;
}
for(int i=1;i<=n;i++) ans=max(ans,g[i]);
printf("%lld\n",ans);
}
#### E. Brief Statements Union
Given \ (n \) numbers, \ (k \) restrictions, such as \ (a_l \ and a {L + 1} \ and \ ldots \ and a {r} = x \).
For a certain restriction, if the restriction can be deleted so that there is a sequence satisfying the condition, it is said to be good.
Output which sequences are good and which sequences are bad.
$$1\le n,k\le 10^6$$，$$0\le x_i\le 10^{18}$$.
Solution
Considering by bit, the remaining restrictions are divided into two categories: \ (1 \) restrictions and \ (0 \) restrictions.
Override the \ (1 \) limit and consider the number of conflicts with the \ (0 \) limit:
• Is \ (0 \): all OK.
• For \ (1 \): the conflict is an insider, knife.
• $$> 1$$: not at all.
$$0$$ limit is considered.
Consider how to delete the \ (1 \) limit. For each point that is only covered by the \ (1 \) limit once, we can delete this \ (1 \) limit to meet several \ (0 \) limits.
For a certain \ (0 \) restriction, we will remove the \ (1 \) restriction that can be deleted, take out the \ (1 \) restriction that legalizes the \ (0 \) restriction, and cross such a set of all \ (0 \) restrictions.
\ (O(n\log x_i) \) can be achieved by using difference.
Code
const int N=1e6;
int n,K;
int ql[N+10],qr[N+10];
ll qx[N+10],f[N+10],g[N+10];
int nxt[N+10],pre[N+10],ans[N+10];
int vio[N+10],tot,id,pos[N+10],seg[N+10];
int main() {
scanf("%d %d",&n,&K);
for(int i=1;i<=K;i++) scanf("%d %d %lld",&ql[i],&qr[i],&qx[i]);
nxt[n+1]=n+1,pre[0]=0;
for(int k=0;k<60;k++) {
for(int i=1;i<=n;i++) f[i]=g[i]=0;
for(int i=1;i<=K;i++)
if((1ll<<k)&qx[i])
f[ql[i]]++,f[qr[i]+1]--,
g[ql[i]]+=i,g[qr[i]+1]-=i;
for(int i=1;i<=n;i++) f[i]+=f[i-1],g[i]+=g[i-1];
for(int i=n;i>=1;i--) nxt[i]=f[i]?nxt[i+1]:i;
id=tot=0;
for(int i=1;i<=K;i++)
if(!((1ll<<k)&qx[i]))
if(nxt[ql[i]]>qr[i])
vio[++tot]=i;
if(tot==0) {
for(int i=1;i<=K;i++) ans[i]++;
continue;
}
if(tot==1) ans[vio[1]]++;
for(int i=1;i<=n;i++)
if(f[i]==1&&!pos[g[i]])
seg[pos[g[i]]=++id]=g[i];
for(int i=1;i<=n;i++) pre[i]=(f[i]==1)?i:pre[i-1];
for(int i=n;i>=1;i--) nxt[i]=(f[i]==1)?i:nxt[i+1];
int l=0,r=id;
for(int i=1;i<=tot;i++) {
int u=vio[i];
if(nxt[ql[u]]<=qr[u]) l=max(l,pos[g[nxt[ql[u]]]]);
else l=id+1;
if(pre[qr[u]]>=ql[u]) r=min(r,pos[g[pre[qr[u]]]]);
else r=0;
}
for(int i=l;i<=r;i++) ans[seg[i]]++;
for(int i=1;i<=K;i++) pos[i]=0;
}
for(int i=1;i<=K;i++)
if(ans[i]==60) putchar('1');
else putchar('0');
puts("");
}
#### F. Border Similarity Undertaking
Given a letter rectangle, find out how many sub rectangles it has to satisfy that all the surrounding characters are the same.
$$1\le n,m\le 2000$$.
Solution
Consider matrix divide and conquer, that is, divide and conquer the length and width of a matrix.
Divide and conquer each matrix, and consider calculating the number of matrices across the centerline.
This process can preprocess the extension distance of a point up, down, left and right, which can be dealt with violently.
The left and right sides of the center line are processed respectively to process the number of rectangles that can be formed on the left and right sides.
The code is more detailed and complex, with time complexity \ (O(nm\log n) \).
Code
const int N=2000;
int n,m;
char ch[N+10][N+10];
int up[N+10][N+10],dn[N+10][N+10],lef[N+10][N+10],rig[N+10][N+10];
int al[N+10][N+10],ar[N+10][N+10];
ll ans;
void solve(int l1,int r1,int l2,int r2) {
if(l1==r1||l2==r2) return;
if(r1-l1+1<=r2-l2+1) {
int mid=(r2+l2)>>1,L,R;
for(int i=l1;i<=r1;i++)
for(int j=l1;j<=r1;j++)
al[i][j]=ar[i][j]=0;
for(int i=l1;i<=r1;i++) {
for(int j=mid;j>=l2&&j>=lef[i][mid];j--)
al[i][min(dn[i][j],r1)]++,
al[i][max(up[i][j],l1)]++;
for(int j=r1-1;j>=i;j--) al[i][j]+=al[i][j+1];
for(int j=l1+1;j<=i;j++) al[i][j]+=al[i][j-1];
for(int j=mid+1;j<=r2&&j<=rig[i][mid+1];j++)
ar[i][min(dn[i][j],r1)]++,
ar[i][max(up[i][j],l1)]++;
for(int j=r1-1;j>=i;j--) ar[i][j]+=ar[i][j+1];
for(int j=l1+1;j<=i;j++) ar[i][j]+=ar[i][j-1];
}
for(int i=l1;i<=r1;i++)
for(int j=i+1;j<=r1;j++) {
if(ch[i][mid]!=ch[j][mid]) continue;
if(ch[i][mid]!=ch[i][mid+1]) continue;
if(ch[j][mid]!=ch[j][mid+1]) continue;
L=lef[i][mid]>=lef[j][mid]?al[i][j]:al[j][i];
R=rig[i][mid+1]<=rig[j][mid+1]?ar[i][j]:ar[j][i];
ans+=1ll*L*R;
}
solve(l1,r1,l2,mid);
solve(l1,r1,mid+1,r2);
}
else {
int mid=(l1+r1)>>1,L,R;
for(int i=l2;i<=r2;i++)
for(int j=l2;j<=r2;j++)
al[i][j]=ar[i][j]=0;
for(int i=l2;i<=r2;i++) {
for(int j=mid;j>=l1&&j>=up[mid][i];j--)
al[i][min(rig[j][i],r2)]++,
al[i][max(lef[j][i],l2)]++;
for(int j=r2-1;j>=i;j--) al[i][j]+=al[i][j+1];
for(int j=l2+1;j<=i;j++) al[i][j]+=al[i][j-1];
for(int j=mid+1;j<=r1&&j<=dn[mid+1][i];j++)
ar[i][min(rig[j][i],r2)]++,
ar[i][max(lef[j][i],l2)]++;
for(int j=r2-1;j>=i;j--) ar[i][j]+=ar[i][j+1];
for(int j=l2+1;j<=i;j++) ar[i][j]+=ar[i][j-1];
}
for(int i=l2;i<=r2;i++)
for(int j=i+1;j<=r2;j++) {
if(ch[mid][i]!=ch[mid][j]) continue;
if(ch[mid][i]!=ch[mid+1][i]) continue;
if(ch[mid][j]!=ch[mid+1][j]) continue;
L=up[mid][i]>=up[mid][j]?al[i][j]:al[j][i];
R=dn[mid][i]<=dn[mid][j]?ar[i][j]:ar[j][i];
ans+=1ll*L*R;
}
solve(l1,mid,l2,r2);
solve(mid+1,r1,l2,r2);
}
}
int main() {
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
cin>>ch[i][j];
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) {
up[i][j]=i>1&&ch[i][j]==ch[i-1][j]?up[i-1][j]:i;
lef[i][j]=j>1&&ch[i][j]==ch[i][j-1]?lef[i][j-1]:j;
}
for(int i=n;i>=1;i--)
for(int j=m;j>=1;j--) {
dn[i][j]=i<n&&ch[i][j]==ch[i+1][j]?dn[i+1][j]:i;
rig[i][j]=j<m&&ch[i][j]==ch[i][j+1]?rig[i][j+1]:j;
}
solve(1,n,1,m);
printf("%lld\n",ans);
}
#### G. Biological Software Utilities
Find the number of trees with perfect matching \ (n \) points\ (1\le n\le 10^6\).
Solution
Bind two points together. There are \ (n!!=1\times 3\times \ldots \times (n-1) \) methods.
There are \ (\ frac{n}{2})^{n/2-2} \) kinds of trees that can be composed of points tied together.
There are \ (4^{n/2-1} \) kinds of interconnection between points.
Multiply, no, time complexity \ (O(n) \).
Code
const int N=1e6,mod=998244353;
int n;
int ans=1;
int fpow(int x,int y) {
if(y<0) return 1;
int ret=1;
for(;y;y>>=1) {
if(y&1) ret=1ll*ret*x%mod;
x=1ll*x*x%mod;
}
return ret;
}
int main() {
scanf("%d",&n);
if(n%2) {puts("0");return 0;}
for(int i=1;i<=n;i+=2) ans=1ll*ans*i%mod;
ans=1ll*ans*fpow(n/2,n/2-2)%mod*fpow(4,n/2-1)%mod;
printf("%d\n",ans);
}
#### H. Bytelandia States Union
Go from \ ((x_1,y_1) \) to \ ((x_2,y_2) \) to find the shortest path.
There are different schemes from \ ((x,y) \) to the four directions. See the original topic for details.
$$1\le T\le 5\times 10^4$$，$$1\le x_1,y_1,x_2,y_2\le 10^9$$.
Solution
Fraud problem, I can responsibly tell you that the cost of a path \ ((x_1,y_1),(x_2,y_2),(x_3,y_3),\ldots,(x_k,y_k) \) is \ (x ^ 2_ky ^ 2_k-x ^ 2_1y ^ 2_1-x_k ^ 2-x_k ^ 2 + \ sum ^ {K} {I = 1} x ^ 2_i + y ^ 2_i \)? Can ask hht2005 , anyway, I won't.
To minimize the path, you only need to minimize the following values, so just get as close to the line \ (x=y \) as possible.
The time complexity of each time is \ (O(1) \).
Code
void solve() {
int x1,y1,x2,y2;
scanf("%lld %lld %lld %lld",&x1,&y1,&x2,&y2);
ans=(p2(1ll*x2*y2%mod)-p2(1ll*x1*y1%mod)+mod)%mod;
ans=(ans-p2(x2)-p2(y2)+mod)%mod;
if(x1>x2) swap(x1,x2),swap(y1,y2);
if(y2>y1)
if(x1<y1) {
int nx=min(y1,x2);
if(nx==x2) {
}
else {
int nxt=min(x2,y2);
if(nxt==y2) {
}
else {
}
}
}
else {
int ny=min(x1,y2);
if(ny==y2) {
}
else {
int nxt=min(x2,y2);
if(nxt==x2) {
}
else {
}
}
}
else {
}
printf("%lld\n",(ans%mod+mod)%mod);
}
signed main() {
ll t;scanf("%lld",&t);
while(t--) solve();
}
#### I. Binary Supersonic Utahraptors
Alice and Bob are playing games again. Alice has black and white items, and Bob also has black and white items.
There are \ (k \) rounds. In each round, Alice gives Bob $$s_i$$ items first, and Bob returns \ (s_i \) items.
Alice wants the difference between Alice's black items and Bob's white items to be as small as possible, while Bob wants to maximize the above items.
Find the last value.
$$1\le n,m,k\le 3\times 10^5$$.
Solution
People will be invincible. Alice and Bob can't decide all this.
The answer is the difference between the total number of black items and \ (n \).
Time complexity \ (O(1) \).
Code
#include<bits/stdc++.h>
using namespace std;
int n,m,k,r,y;
int main() {
scanf("%d %d %d",&n,&m,&k);
for(int i=1,x;i<=n;i++) {
scanf("%d",&x);
if(x==1) r++;
}
for(int i=1,x;i<=m;i++) {
scanf("%d",&x);
if(x==1) r++;
}
printf("%d\n",abs(n-r));
}
Find the smallest independent set covering all edges of a graph, \ (2\le n\le 3\times 10^5 \), \ (1\le m\le 3\times 10^5 \).
Solution
To cover all edges, two adjacent points must have different states.
Considering the coloring of bipartite graphs, if the original graph has odd rings, there must be no solution, otherwise take the smaller point set of that color.
Time complexity \ (O(n+m) \).
Code
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
const int N=3e5;
int n,m,vis[N+10],cnt[4],ans;
vector<int> G[N+10];
bool dfs(int u,int col) {
vis[u]=col;cnt[col]++;
bool fl=1;
for(auto v:G[u])
if(!vis[v]) fl&=dfs(v,3-col);
else if(vis[v]==col) return 0;
return 1;
}
int main() {
scanf("%d %d",&n,&m);
for(int i=1,x,y;i<=m;i++) {
scanf("%d %d",&x,&y);
G[x].pb(y),G[y].pb(x);
}
for(int i=1;i<=n;i++)
if(!vis[i]) {
cnt[1]=cnt[2]=0;
if(!dfs(i,1)) {
puts("-1");
return 0;
}
ans+=min(cnt[1],cnt[2]);
}
printf("%d\n",ans);
}
#### K. Bookcase Solidity United
There are \ (n \) boards, the \ (I \) board has stability \ (a_i \), and the boards are arranged from high to low.
Place an iridium ball on the board, the \ (I \) board will be broken after being placed with more than or equal to \ (a_i \), and \ (\ lfloor\frac k 2\rfloor \) will fall onto the next board.
Find out at least how many iridium balls are needed to break the front \ (i \) board.
$$1\le n\le 70$$，$$1\le a_i\le 150$$.
Solution
Let \ (f_{l,r,k} \) represent the minimum number of balls to break the board of \ ([l,r] \) and leave \ (k \) balls.
There is an obvious shift: \ (f {L, R, K} + \ max (a {R + 1} - K, 0) \ to f {L, R + 1, \ max (k, a {R + 1}) / 2} \), which means inheriting several balls to hit the following ones.
At the same time, we consider piecewise smashing, \ (f {L, R, K 1 + K 2} = f {L, mid, K 1} + F {mid + 1, R, K 2} \).
Direct violent transfer is \ (O(n^3m^2) \).
Code
#include<bits/stdc++.h>
using namespace std;
const int N=200,M=200;
int n,a[N+10],m;
int f[N+10][N+10][M+10];
void chkmin(int &x,int y) {
if(x>y) x=y;
}
int main() {
scanf("%d",&n);
memset(f,0x3f,sizeof f);
for(int i=1;i<=n;i++) scanf("%d",&a[i]),f[i][i][a[i]/2]=a[i],m=max(m,a[i]);
for(int len=2;len<=n;len++)
for(int l=1;l+len-1<=n;l++) {
int r=l+len-1;
for(int i=0;i<=m;i++) chkmin(f[l][r][max(i,a[r])/2],f[l][r-1][i]+max(0,a[r]-i));
for(int i=0;i<=m;i++)
for(int j=l;j<r;j++)
for(int k=0;k<=i;k++)
chkmin(f[l][r][i],f[l][j][k]+f[j+1][r][i-k]);
}
for(int i=1;i<=n;i++) {
int ans=INT_MAX;
for(int j=0;j<=m;j++) ans=min(ans,f[1][i][j]);
printf("%d ",ans);
}
puts("");
return 0;
}
#### M. Brilliant Sequence of Umbrellas
Construct an increasing sequence with a value range of \ ([1,n] \) with a length of at least \ (\ lceil \frac2 3 \sqrt{n}\rceil \), so that the \ (\ gcd \) between two pairs is also increased.
$$1\le n\le 10^{12}$$.
Solution
Observe the example, consider constructing a sequence with Coprime every two numbers, and then take the product of two adjacent terms of the sequence as the original sequence.
This must satisfy the \ (\ gcd \) increment.
Code
typedef long long ll;
const int N=1e6;
ll n,a[N+10],cnt,b[N+10];
int main() {
scanf("%lld",&n);
b[++cnt]=1,b[++cnt]=1;
int lim=ceil(2.0*sqrt(n)/3);
for(ll i=2;cnt<=lim+1&&b[cnt]*i<=n;i++)
if(__gcd(b[cnt-1],i)==1)
b[++cnt]=i;
printf("%lld\n",cnt-1);
for(int i=1;i<cnt;i++) printf("%lld ",b[i]*b[i+1]);
puts("");
}
#### N. Best Solution Unknown
There are \ (n \) individuals playing the game. Everyone has a strength value. The one with a large strength value will defeat the one with a small strength value. All of them have a chance to win.
Winning a person can increase the strength value of \ (1 \), and each person can only play with adjacent people.
Every time you pick a game at random, ask who is likely to win.
$$1\le n\le 10^6$$，$$1\le a_i\le 10^9$$.
Solution
Consider whether a person can win for a large section, draw the maximum value, and find that the maximum value is divided into two parts, one is left and the other is right. If you want to win the mountain over the maximum value.
Then we can divide and conquer, find the maximum value to divide and conquer, and record at least how much to the left or right each time before we can win.
When divide and conquer, use ST table to speed up the calculation of the maximum value.
Code
const int N=1e6;
int a[N+10],n,Log[N+10],Max[N+10][30],id[N+10][30];
bool nb[N+10];
int query(int l,int r) {
int k=Log[r-l+1];
int ret=max(Max[l][k],Max[r-(1<<k)+1][k]);
if(ret==Max[l][k]) return id[l][k];
else return id[r-(1<<k)+1][k];
}
void dfs(int l,int r,int c) {
if(l>r) return;
if(l==r) {
if(a[l]>=c) nb[l]=1;
return;
}
int id=query(l,r);
if(a[id]<c) return;
nb[id]=1;
dfs(l,id-1,max(c,a[id]-id+1+l));
dfs(id+1,r,max(c,a[id]+id+1-r));
}
int main() {
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
Log[1]=0;
for(int i=2;i<=n;i++) Log[i]=Log[i/2]+1;
for(int i=1;i<=n;i++) Max[i][0]=a[i],id[i][0]=i;
for(int j=1;j<=20;j++)
for(int i=1;i+(1<<(j-1))<=n;i++) {
Max[i][j]=max(Max[i][j-1],Max[i+(1<<(j-1))][j-1]);
if(Max[i][j]==Max[i][j-1]) id[i][j]=id[i][j-1];
else id[i][j]=id[i+(1<<(j-1))][j-1];
}
dfs(1,n,0);
int cnt=0;
for(int i=1;i<=n;i++) if(nb[i]) cnt++;
printf("%d\n",cnt);
for(int i=1;i<=n;i++) if(nb[i]) printf("%d ",i);
puts("");
}
Posted by blacksnday on Mon, 01 Nov 2021 03:04:56 -0700 | 7,353 | 19,552 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2022-27 | longest | en | 0.740269 |
https://www.concepts-of-physics.com/mechanics/water-bottle-in-free-fall.php | 1,709,468,976,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476374.40/warc/CC-MAIN-20240303111005-20240303141005-00383.warc.gz | 715,770,564 | 5,458 | # Water Bottle in Free Fall
By
The concept of weight and weightlessness is often not easy. We define weight $$W$$ of a body as the gravitational force that the earth exerts on the body. However, it is measured by measuring the effect of some kind of contact force. When we stand on a weighing machine, our legs press the machine downward and it is this contact force $$F$$ that causes compression in the spring inside the balance. The compression in the spring is displayed as a weight. The connecting link is Newton's first and third laws. As the body pushes the machine down by a force $$F$$, the machine pushes it up by the same force. As the body is in equilibrium, $$F=W$$.
If the weighing machine (together with the body on it) is in free fall, the contact force $$F$$ becomes zero. This is expressed by saying that the body has become weightless. If you use a freely falling frame, all phenomenon can be described by assuming $$g=0$$ and then applying Newton's laws.
## Procedure
Take a bottle of 1 litre capacity. Make a small hole in its wall near the bottom. Put a piece of empty pen refill in the hole. Hold the bottle in your hand, close the hole by a finger and fill water in it. Remove the finger and see that water comes out of the hole and goes in a parabolic path. Now leave the bottle from a height and catch it before it falls on the floor. As long as it is falling, there will not be any parabolic stream. Once you catch it, the stream will reappear.
## Discussion
During the free fall, the water in the bottle is in state of weightlessness and does not push through the hole.
Weightlessness shows up in this demo through the vanishing of pressure difference between the water at top surface and just inside the hole. When the bottle is stationary, this pressure at the hole is $$p_h=p_0+h\rho g$$, where $$p_0$$ is pressure at the surface (atmospheric pressure), $$h$$ is the height of the water column and $$\rho$$ is the density of water. When in free fall, in the frame attached to the bottle we can substitute $$g=0$$ and the pressure just inside the hole become $$p_h=p_0$$. The pressure outside the hole is also equal to $$p_0$$. Thus there is no pressure difference for the water to flow out of the hole.
## Related
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or | 526 | 2,282 | {"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 | 4 | CC-MAIN-2024-10 | latest | en | 0.947562 |
http://answers.opencv.org/question/2348/missing-region-in-disparity-map/ | 1,547,752,540,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00558.warc.gz | 15,844,198 | 13,481 | # missing region in disparity map
i am currently working on stereo processing using opencv2.3 and a Pointgrey Bumblebee2 stereocamera as an input device. Acquiring images is done via the libdc1394.
My code for rectification and stereo processing is the following:
void StereoProcessing::calculateDisparityMap(const Mat &left, const Mat &right, Mat &disparity_map)
Mat map11, map12, map21, map22, left_rectified, right_rectified, disp16;
// Computes the undistortion and rectification transformation maps
initUndistortRectifyMap(this->camera_matrix1,
this->distance_coefficients1,
this->R1,
this->P1,
this->output_image_size,
CV_16SC2,
map11,
map12);
initUndistortRectifyMap(this->camera_matrix2,
this->distance_coefficients2,
this->R2,
this->P2,
this->output_image_size,
CV_16SC2,
map21,
map22);
// creates rectified images
remap(left, left_rectified, map11, map12, INTER_LINEAR);
remap(right, right_rectified, map21, map22, INTER_LINEAR);
// calculates 16-bit disparitymap
this->stereo_bm(left_temp, right_temp, disp16);
disp16.convertTo(disparity_map, CV_8U, 255 / (this->stereo_bm.state->numberOfDisparities * 16.0));
}
This works fine except for a black left border in the disparity map, which is the following:
The input images are these two - unrectified as you can see ;) :
So my question is now: Is this normal behaviour? Or do you see any mistake i have done so far? As another information, the rectification works fine.
edit retag close merge delete
Sort by ยป oldest newest most voted
I think it's a normal behaviour that region with x-coordinates between 0 and max_disparity is not reconstructed.
Suppose on a rectified left image you have a point with x-coordinate x0, where 0 < x0 < max_disparity. Potential matches for this point on the right rectified image have x-coordinate in < x0-max_disparity; x0 > range. But if x0 < max_disparity part of this range has negative x-coordinate and is not visible on the rectified right image. So disparity cannot be calculated.
And one advice: for better results use StereoSGBM algorithm. It usually gives much better results than simple block matching (StereoBM) algorithm.
more
Is there any way to solve this problem using the StereoBM Algorithm
( 2014-03-14 08:57:24 -0500 )edit
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https://brokerewkxqj.netlify.app/riedel2365noxu/binary-indexed-tree-pdf-108.html | 1,726,542,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00138.warc.gz | 122,863,450 | 10,511 | Binary indexed tree pdf
Fenwick tree and segment tree are very powerful data structures, but you will seldom see them alone in a real life applications. They are always used to implement other algorithms efficiently. For instance, BIT is used to implement the Arithmetic Binary Indexed Tree or Fenwick Tree - GeeksforGeeks Dec 11, 2014 · An alternative solution is Binary Indexed Tree, which also achieves O(Logn) time complexity for both operations. Compared with Segment Tree, Binary Indexed Tree requires less space and is easier to implement.. Representation. Binary Indexed …
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Binary Indexed Tree Test. View as PDF · Submit solution · All submissions · Best submissions
Binary Index Trees - CS Department A binary index tree is the perfect data structure to allow us to update which candies have been eaten and answer many queries of this nature in sequence. One way to solve the problem is as follows: 1) Start with an empty binary index tree. 2) Whenever an item is eaten add … Binary Tree Data Structure - GeeksforGeeks Binary Tree Data Structure A tree whose elements have at most 2 children is called a binary tree. Since each element in a binary tree can have only 2 children, we typically name them the left and right child. Binary Indexed Trees Binary Indexed Trees Nick Haliday, Ryan Jian September 28, 2012 1 Introduction A binary indexed tree, also known as a Fenwick tree, is a data structure used to e ciently calculate and update cumulative frequency tables, or pre x sums. Binary indexed trees typically only show up in Gold problems, however they could start appearing more often in Data Structures - Binary Indexed Tree (Fenwick) (Arabic ...
CHAPTER 27 Binary Search Trees and Hash Tables
CHAPTER 27 Binary Search Trees and Hash Tables CHAPTER 27 Binary Search Trees and Hash Tables 27.1 Binary Search Trees The binary tree is a binary search tree if it satisfies the following property: for any node n, the keys in the left subtree elements (indexed 0 - 999), the hash function might simply return the low-order three digits of the geeksforgeeks.pdf/Advanced-Data-Structure.json at master ... Dismiss Join GitHub today. GitHub is home to over 40 million developers working together to host and review code, manage projects, and build software together.
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Recursion Problem 1 - University of KwaZulu-Natal
Binary Indexed Tree or Fenwick Tree - GeeksforGeeks
Dec 31, 2015 · SOME DISTRIBUTIONS ON FINITE ROOTED BINARY TREES UCDMS RESEARCH REPORT NO. UCDMS2015/2, SCHOOL OF MATHEMATICS AND (indexed by their addresses) of a ranked plane tree and choose a random leaf node according the increasing binary tree lifting (see [14, Ex. 17, p. 132] and the references therein). The bijection, Tb# n 3^t# 6.006 Introduction to Algorithms Spring 2008 For ... Large array indexed by time does the trick. This will not work for arbitrary precision The new requirement necessitates a design amendment. 2. Lecture 3 Ver 2.0 Scheduling and Binary Search Trees 6.006 Spring 2008 Binary Search Trees (BST) 49 49 79 79 49 46 79 49 46 41 64 insert 49 Walk down tree to find desired time 2. Add in nodes Binary Search Trees Complexity Of Dictionary Operations ... remove(index) (indexed binary search tree) Complexity Of Dictionary Operations get(), put() and remove() Data Structure Worst Case Expected Hash Table O(n) O(1) Binary Search Tree O(n) O(log n) Balanced Binary Search Tree indexed binary tree, not indexed binary search tree). Can’t use hash tables for either of these applications. What are Binary search trees? - Quora
Section 2 -- Binary Tree Problems Here are 14 binary tree problems in increasing order of difficulty. Some of the problems operate on binary search trees (aka "ordered binary trees") while others work on plain binary trees with no special ordering. The next section, Section 3, shows the solution code in C/C++. Lecture 5: Scheduling and Binary Search Trees Lecture 5 Scheduling and Binary Search Trees 6.006 Fall 2011 What if times are in whole minutes? Large array indexed by time does the trick. This will not work for arbitrary precision time or verifying width slots for landing. Key Lesson: Need fast insertion into sorted list. 3 Competitive Programmer’s Handbook - CSES Chapter 1 Introduction Competitive programming combines two topics: (1) the design of algorithms and (2) the implementation of algorithms. The design of algorithms consists of problem solving and mathematical thinking. The Strong Law of Large Numbers and the Entropy Ergodic ... This paper is devoted to the strong law of large numbers and the entropy ergodic theorem for nonhomogeneous M-bifurcating Markov chains indexed by a M-branch Cayley tree, which generalizes the | 1,196 | 5,429 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.911465 |
https://tomblackson.com/PHI_319_420/lecture15.html | 1,696,390,685,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00013.warc.gz | 608,858,609 | 4,690 | # Philosophy, Computing, and Artificial Intelligence
PHI 319. Semantics and Donkey Sentences.
## Building Meaning With The Lambda Calculus
We can use the lambda calculus to associate truth-conditions with a sentence.
### The Lambda Calculus
λ is the eleventh letter of the Greek alphabet.
Alonzo Church and Stephen Kleene developed the λ-calculus in the 1930’s to think about the computation necessary to convert a function and its argument to a value.
Here is a formula in the lambda calculus
λx.MAN(x)
In this formula, the prefix λx binds the occurrence of the variable x in the one-place predicate MAN(x).
The formula λx.MAN(x) may be understood as a one-place function. The lambda binds a variable that holds a place for a substitution of the argument in the one-place predicate. In the expression
λx.MAN(x)@vincent
the symbol @ separates the function from the argument. The left hand expression
λx.MAN(x)
is the function. The right hand expression
vincent
is the argument. The substitution process (the replacement of a bound variable in a function with an argument) is called β-reduction. In the example, what results from β-reduction is the sentence
MAN(vincent)
### The Meaning of Words as Computations
We can think of the meanings of the words in the lexicon as computations, and we can represent these computations with the lambda calculus. Here is a simple example:
'every' : λP.λQ.∀x(P@x → Q@x)
'some' or 'a' : λP.λQ.∃x(P@x ∧ Q@x)
'no' : λP.λQ.∀x(P@x → ¬Q@x)
'boxer' : λy.BOXER(y)
'walks' : λx.WALKS(x)
Here is the representation tree for the noun phrase 'every boxer.' It shows how the computation for 'every' takes the computation for 'boxer' to produce the computation for 'every boxer.'
``` every boxer (NP)
λP.λQ.∀x(P@x → Q@x)@λy.BOXER(y)
/ \
/ \
/ \
every (DET) boxer (NOUN)
λP.λQ.∀x(P@x → Q@x) λy.BOXER(y) ```
The expression at the root of the tree is
λP.λQ.∀x(P@x → Q@x)@λy.BOXER(y)
By substitution, this reduces to
λQ.∀x(λy.BOXER(y)@x → Q@x)
This reduces to
λQ.∀x(BOXER(x) → Q@x)
If this noun phrase is given a verb phrase, the result is a sentence. Here is the representation tree (with substitutions completed) for the sentence 'every boxer walks'
``` every boxer walks (S)
∀x(BOXER(x) → WALKS(x))
/ \
/ \
/ \
every boxer (NP) walks(VP)
λQ.∀x(BOXER(x) → Q@x) λx.WALKS(x)
/ \
/ \
/ \
every (DET) boxer (NOUN)
λP.λQ.∀x(P@x → Q@x) λy.BOXER(y)
```
Here is the representation tree for the sentence 'vincent loves mia '
``` vincent loves mia (S)
LOVE(vincent, mia)
/ \
/ \
/ \
vincent (NP) loves mia (VP)
λP.P@vincent λz(LOVE(z,mia))
/ \
/ \
/ \
loves (TV) mia (NP)
λX.λz.(X@λx.LOVE)(z,x)) λP.P@mia
```
### The Meanings of Words as Types
It is sometimes useful to think of lambda substitution in terms of types. There are two basic types, e and t. The first is the type of entities in the domain of the model. The second is the type for truth-values (true and false). Compound types are built from these two basic types.
One place predicates, for example, have the type <e, t>. When a one-place predicate is given an expression of type e, it returns an expression with type t.
To understand more clearly how this works, think again about the lambda expression
λx.MAN(x)@vincent
By β-conversion, this expression reduces to
MAN(vincent)
In terms of types, the lambda expression
``` λx.MAN(x)@vincent
<e, t> e```
reduces to
``` MAN(vincent)
t```
Here is a grammar whose lexical entries are associated with formulas in the lambda calculus
S → VP NP
NP → NAME
NP → DET N
VP → IV
VP → TV NP
NAME → mia: λP.P@mia
NAME → vincent: λP.P@vincent
DET → every: λP.λQ.∀x(P@x → Q@x)
DET → some: λP.λQ.∃x(P@x ∧ Q@x)
N → man: λx.MAN(x)
N → woman: λx.WOMAN(x)
IV → laughs: λx.LAUGHS(x)
TV → loves: λX.λz.(X@λx.LOVE)(z,x))
Here are the representation trees showing the type conversions
``` S : t
/ \
/ \
NP : (e → t) → t VP : e → t
NP : (e → t) → t
|
|
NAME : (e → t) → t
NP : (e → t) → t
/ \
/ \
DET : (e → t) → (e → t) → t N : e → t
VP : e → t
|
|
IV : e → t
VP : e → t
/ \
/ \
TV : e → (e → t) NP : (e → t) → t
```
Here is the representation tree for a sentence of the grammar ('every man laughs'):
```
every man laughs (S)
∀x(MAN(x) → LAUGHS(x))
t
/ \
/ \
every man (NP) laughs (VP)
λQ.∀x(BOXER(x) → Q@x) λx.LAUGHS(x)
(e→t)→t e→t
/ \
/ \
every (DET) man (NOUN)
λP.λQ.∀x(P@x → Q@x) λy.MAN(y)
(e →t)→(e→t)→t e→t
```
## Problems with the Semantics
Discourse Representation Theory (DRT) is a possible solution to these problems. This theory, however, is beyond the scope of the course. This semantics is not without its problems. Here are two of the more famous problems.
### Indefinite NPs in Donkey sentences
When Peter Geach called attention to the phenomenon in language (now known as "donkey anaphora"), he used sentences about farmers and donkeys to construct his counterexample.
The word anaphora is from ἀναφορά, meaning "carrying back."
One problem concerns the truth conditions for "donkey" sentences.
The truth-conditions for the following two sentences
If John owns a donkey, he feeds it.
Every farmer who owns a donkey feeds it.
are
∀x((donkey(x) ∧ own(john, x)) → feeds(john,x))
∀x∀y((farmer(x) ∧ donkey(y) ∧ own(x, y)) → feeds(x, y))
In neither is the indefinite NP an existentially quantified expression.
In the first, a donkey (located in the antecedent of a conditional) is a universally quantified expression with wide scope over the material conditional.
In the second, a donkey (located in relative clause that modifies a universally quantified NP) is also a universally quantified expression taking wide scope.
### Indefinite NPs in Discourse
Another problem concerns the way indefinite NPs function in discourse.
The sentence A dog came in can be part of a discourse
A dog came in. It sat down.
but the translation
∃x(dog(x) ∧ came_in(x)). sat_down(x).
does not make sense becase the the variable in sat_down(x) is free.
Here is another argument in terms of discourse.
In classical logic, ∃xφ and ¬∀¬xφ are equivalent. ¬(φ ∧ ψ) and φ → ¬ψ are also equivalent. So
∃x(dog(x) ∧ came_in(x))
¬∀x(dog(x) → ¬came_in(x))
are logically equivalent, but the sentences
A dog came in.
Not every dog failed to come in.
are not interchangeable. Only the first can be extended in conversation
A dog came in. It sat down.
Not every dog failed to come in. It sat down.
The second discourse does not make sense.
## What we Accomplished in this Lecture
We looked at how to use the lambda calculus to generate the logical form of a sentence. | 2,033 | 7,272 | {"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-2023-40 | latest | en | 0.795966 |
http://www.physicsforums.com/showthread.php?p=3719281 | 1,408,823,834,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500826679.55/warc/CC-MAIN-20140820021346-00123-ip-10-180-136-8.ec2.internal.warc.gz | 547,584,319 | 12,023 | # Where is the center of the universe?
by thetexan
Tags: universe
Astronomy
PF Gold
P: 23,218
Quote by Drakkith They don't say there isn't an edge. They say that the model doesn't require there to be an edge for it to work. As far as I know at least.
There is some truth to that! You can put it in terms of Occam Razor commandment---Thou shalt not make thy models unnecessarily complicated.
We see no evidence of a boundary, so why put it in the model? Plus it would be a mathematical headache. If there is a boundary forces would be unbalanced and expansion would appear asymmetrical unless you put the earth at center. But putting earth at center is "Un-Copernican"
The thing a lot of newcomers don't realize is that cosmology is a mathematical science that is primarily concerned with a math MODEL of the universe and that model runs according to the 1915 GR equation (Einstein Field Equation) which is our current law of gravity. If you have a model which must run according to an equation you don't have a lot of freedom to mess around.
The GR equation describes the evolving geometry of all space. We know it changes and the law of gravity describes how. On page one of the book you are given a manifold which is all space thru all time, and a metric or distance function that describes the geometry---what paths are straight and how distances relate to areas and volumes and how distances change thru time etc.
Personally I find the idea of a boundary UNINTUITIVE. The universe is supposed to be ALL SPACE. So what would a boundary be separating space from?
The GR equation is by far the most accurate law of gravity we've ever had---it gives more precise numbers than the older Newton law. It has been tested repeatedly and checks out every time. And it is a law both of gravity and evolving geometry because they are the same thing (how massive objects influence geometry).
So there is a COMMON SENSE reason to base models of the universe's geometry on the GR equation. Its a no-brainer in fact.
So if you accept to base your model cosmo on the accepted law of gravity/geometry, then the idea of a boundary has no standing. How to implement? Maybe by having the average density of galaxies gradually peter out so that our region is surrounded by a large "void". But then expansion would most likely be decelerating so let's fix that with a larger cosmological constant, and so on. It is not naturally a part of the picture. Just giving the idea of a boundary a meaning seems likely to lead to headaches and a more complex picture.
The discussion gets over into talking about what's called the "cosmological principle". On large scale the universe seems on average uniform. Matter seems uniformly distributed throughout space, at any one time. So we infer looking back in time. At each epoch matter was uniformly distributed at some average density that prevailed at that time.
So people say "homogeneous and isotropic" which basically just means evenly distributed on largescale average.
That makes the model simple and we see no evidence to the contrary so since the job is to get the simplest model with the best fit, and no evidence to contrary, evenness is assumed.
Maybe, as Drakkith suggests, it is at heart an Occam thing--the ancient tradition in mathematical sciences of keeping the model simple.
PF Gold
P: 28
Quote by marcus Personally I find the idea of a boundary UNINTUITIVE. The universe is supposed to be ALL SPACE. So what would a boundary be separating space from?
Precisely. "Bubble theory" as a model for the uber-universe, does not require any membranes around the bubbles but rather they may simply be viewed as cosmological villages in the vast darkness, collections of local activity activated by some localized phenomenon such as a big bang and separated from any other such areas of activity by lots and lots of nothing, or something or what have you. And, of course, the mass and energy making up the village are reorganized every 14 billion years or so, at least in this mind model. It's always been there and always will be there following the various conservation laws.
But, no boundary or membrane encapsulating the universe or universes is needed since the vastness of infinite space pretty much takes care of the isolation question on its own.
Best,
RD
P: 38
Quote by marcus So if you accept to base your model cosmo on the accepted law of gravity/geometry, then the idea of a boundary has no standing. How to implement? Maybe by having the average density of galaxies gradually peter out so that our region is surrounded by a large "void". But then expansion would most likely be decelerating so let's fix that with a larger cosmological constant, and so on. It is not naturally a part of the picture. Just giving the idea of a boundary a meaning seems likely to lead to headaches and a more complex picture.
Astronomy Sci Advisor PF Gold P: 23,218 it gets into semantics and I lose interest. most "other universe" talk seems vacuous, devoid of empirical content. I try to keep language simple and consistent with professional usage---universe is all space and all physical existence. That seems to be how it is used in 99% of the cosmology research papers that come out daily on the preprint archive. Have a look for yourself. http://arxiv.org/list/astro-ph.CO/recent Essentially nothing about "multiverse" in the run of mill professional literature. More confined to popular media where they stimulate the imagination in order to sell books. Talk about string theory and God and multiverses and stuff.
P: 38 You obviously don't get my point. Call our accelerated expaning bubble of galaxies 'A', picture it in space and surround it with 'A2', 'A3', ...
Mentor
P: 11,844
Quote by voxilla You obviously don't get my point. Call our accelerated expaning bubble of galaxies 'A', picture it in space and surround it with 'A2', 'A3', ...
What about it? Even if it's possible, there's no way for us to know at the moment. Our current theory is difficult enough as it is without trying to complicate it with unknowable stuff.
P: 53
Quote by Chalnoth The problem is that explosions are messy. Really messy.
I think that sometimes are not that messy :)
P: 4,800
Quote by minio I think that sometimes are not that messy :)
I believe that you only get this nice picture in a few, very specific wavelengths.
P: 38
Quote by Drakkith What about it? Even if it's possible, there's no way for us to know at the moment. Our current theory is difficult enough as it is without trying to complicate it with unknowable stuff.
At least it can explain the acceleration of expansion, because of attraction to surrounding A2, A3, ... without need for dark stuff, isn't that a simplification ?
P: 366 How would you see an outer edge or for that matter an inner edge, from our view point we only see signals between objects?
Mentor
P: 11,844
Quote by voxilla At least it can explain the acceleration of expansion, because of attraction to surrounding A2, A3, ... without need for dark stuff, isn't that a simplification ?
Absolutely not. And it doesn't even explain the accelerating expansion.
P: 38 One day I may make a GPU simulation out of it, to show how it can work.
P: 38
Quote by marcus it gets into semantics and I lose interest. most "other universe" talk seems vacuous, devoid of empirical content. I try to keep language simple and consistent with professional usage---universe is all space and all physical existence. That seems to be how it is used in 99% of the cosmology research papers that come out daily on the preprint archive. Have a look for yourself. http://arxiv.org/list/astro-ph.CO/recent Essentially nothing about "multiverse" in the run of mill professional literature. More confined to popular media where they stimulate the imagination in order to sell books. Talk about string theory and God and multiverses and stuff.
You might as well say that the cosmos revolve around the earth. That kind of perspective amounts to the very same thing. It could well be true but without proof in either favor, it is speculation and only serves to hinder progress. Nobody will ever find a way to prove it if we refuse to consider the possibilities.
Existing models are useful for practical applications but contemplating what is beyond our knowledge domain is paramount to discovery.
Astronomy Sci Advisor PF Gold P: 23,218 http://edge.org/response-detail/805/...annot-prove-it or google "Steinhardt annual question 2005"
P: 366 Just to reinforce some previous comments. The entire Universe has no center, for it to have a center would also preclude a leading edge. This would violate the Cosmological principle and also undermine relativity by applying different and preferential reference frames. The BB was not a ballistic explosion in a pre-existing space and is entirely background independant. To try to assume external vantage points "outside" the Universe is pointless and does not provide any helpful understanding IMO. Now there are edges to the Universe, but these are not spatial; they are temporal. When I stand and look up into the sky I am on the temporal edge of the Universe. I hope this helps and am happy to discuss this further as sometimes it can help for a layperson to explain this. (My head still hurts if I think about it too much.) Cosmo
P: 4,800
Quote by marcus http://edge.org/response-detail/805/...annot-prove-it or google "Steinhardt annual question 2005"
I've always found that response to anthropic arguments to be rather pathetic.
Mentor
P: 11,844
Quote by Fuzzy Logic You might as well say that the cosmos revolve around the earth. That kind of perspective amounts to the very same thing. It could well be true but without proof in either favor, it is speculation and only serves to hinder progress. Nobody will ever find a way to prove it if we refuse to consider the possibilities. Existing models are useful for practical applications but contemplating what is beyond our knowledge domain is paramount to discovery.
We have overwhelming evidence of at least a single universe and zero evidence of more than a single universe. I think in this case we should stick to a single universe model until something tells us otherwise. | 2,220 | 10,228 | {"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} | 2.828125 | 3 | CC-MAIN-2014-35 | latest | en | 0.953787 |
https://math.answers.com/Q/How_many_millions_are_in_25_billion | 1,702,253,799,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102697.89/warc/CC-MAIN-20231210221943-20231211011943-00170.warc.gz | 430,152,535 | 45,529 | 0
# How many millions are in 25 billion?
Updated: 9/17/2023
Wiki User
7y ago
The answer depends on whether you're using the U.S. or British billion.
The U.S. billion is a thousand millions, so 25 billion is 25,000 millions.
The British billion is a million millions, so 25 billion is 25,000,000 millions.
The answer depends upon where you are:
• In those countries that use the long scale (based on powers of a million), like Europe:
1 billion = 1 million million = 1,000,000 million
→ 25 billion = 25,000,000 million (25 million million)
• In those countries that use the short scale (based on powers of a thousand plus one), like USA:
1 billion = 1 thousand million = 1,000 million
→ 25 billion = 25,000 million (25 thousand million).
Wiki User
7y ago
Wiki User
10y ago
two thousand five hundred | 218 | 813 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-50 | latest | en | 0.898931 |
https://becalculator.com/166-5-cm-to-inch-converter.html | 1,718,669,971,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00202.warc.gz | 107,230,638 | 16,100 | # 166.5 cm to inch converter
## FAQs on 166.5 cm to inch
### How many inches are in a cm?
If you want to convert 166.5 cm to a number in inches, first you need to determine how many inches one centimeter equals.
This is how I will be specific: one centimeter is equivalent to 0.3937 inches.
### How can you convert 1 cm into inches?
To convert 1 centimeter into inches, you need to multiply 1cm by the conversion rate of 0.3937.
This makes it much easier to convert 166.5 cm to inches.
Therefore, 1 cm to inches = 1 x 0.3937 = 0.3937 inches.
This will allow you to answer the following question easily and quickly.
• What is one centimeter into inches?
• What is cm into inches conversion?
• How many inches is equal to 1 cm?
• What does 1 cm equal in inches?
### Definition:Centimeter
Centimeter is an International Standard Unit of Length. It is equal to one hundredth of a meter. It’s approximately equivalent to 39.37 inches.
### Definition of Inch
Anglo-American units of length are in inches. 12 inches equals one foot, and 36 inches equals one yard. According to the modern standard, one inch equals 2.54 cm.
### How do u convert 166.5 cm to inches?
By the above, you have fully grasped cm to inches.
Here is the exact algorithm:
Value in inches = value in cm × 0.3937
So, 166.5 cm to inches = 166.5 cm × 0.3937 = 6.555105 inches
This formula can be used to answer the related questions:
• What’s the formula to convert inches from 166.5 cm?
• How do I convert cm to inches?
• How to change inches from cm?
• How to calculate cm to inches?
• How big are 166.5 cm to inches?
cm inch 165.7 cm 6.523609 inch 165.8 cm 6.527546 inch 165.9 cm 6.531483 inch 166 cm 6.53542 inch 166.1 cm 6.539357 inch 166.2 cm 6.543294 inch 166.3 cm 6.547231 inch 166.4 cm 6.551168 inch 166.5 cm 6.555105 inch 166.6 cm 6.559042 inch 166.7 cm 6.562979 inch 166.8 cm 6.566916 inch 166.9 cm 6.570853 inch 167 cm 6.57479 inch 167.1 cm 6.578727 inch 167.2 cm 6.582664 inch 167.3 cm 6.586601 inch
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https://courses.lumenlearning.com/wm-accountingformanagers/chapter/significant-variance/ | 1,718,418,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00845.warc.gz | 156,965,416 | 18,279 | ## Significant Variance
### Learning Outcomes
• Determine if a variance is significant
How do we know if a variance is significant? Looking at history can be a good start. Let’s take a look at some examples and review variance analysis:
You might want to save these videos to favorites as we will be working with them again in the next few learning outcomes!
Now back to our old standby. What if Simply Yoga had the following information available to you?
• 2010: Sales $7000, wages$3500
• 2011: Sales $7000, wages$3400
• 2012: Sales $7000, wages$3600
• 2013: Sales $7000, wages$5000
What do you notice? Would you consider the wage variances between 2010 and 2012 to be significant? What about 2013? With stable sales, each variance would need to be examined to insure that payroll was prepared correctly, but the variance in 2013 when comparing wages year to year is significant. This could happen if there was a teacher shortage, and a higher wage was needed to get enough staff. Can you think of any other reasons that a wage difference this significant might happen?
Let’s look at another example with our old friends at Hupana Running Company:
Hupana Running Company’s direct labor information is as follows:
Year Direct Labor Budgeted Direct Labor Actual
2010 $20,500$20,400
2011 $21,000$21,155
2012 $22,000$28,500
2013 $23,000$23,900
Which years would you not worry about from a variance perspective and which would be of concern?
When we look at various years, we can see that small variances happen. It might relate to unanticipated overtime in the case of small unfavorable variances or a particularly quick worker in the case of small favorable variances. But look at 2012. This is a variance that needs to be investigated. Maybe there was a machine breakdown that left employees standing around for a week, or perhaps high turnover led to a period of time needed to train new employees, thus slowing production.
What is another answer for the higher labor in 2012? We don’t see a very important number that could help answer this question: sales. What if we had a banner year, and sales were through the roof? Then this variance makes perfect sense.
Whatever the case, it is important to look at history and determine if a variance is significant or a normal variance. | 532 | 2,295 | {"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} | 3.109375 | 3 | CC-MAIN-2024-26 | latest | en | 0.941393 |
https://www.ncertpoint.com/2021/12/how-many-watts-does-it-take-to-run-a-well-pump.html | 1,709,584,714,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00228.warc.gz | 908,743,923 | 29,438 | # How many watts does it take to run a well pump?
Consumer reports estimates a well pump uses roughly 700 watts. Multiply it by 3 to obtain a reasonable approximation of the beginning wattage for the pump and you run roughly 2100 which still offers plenty of wiggle space in terms of power.
### Correspondingly, will a 5000 watt generator power a well pump?
Homeowners may frequently power most domestic appliances with between 3000 and 6500 watts. If your house has a modest furnace and city water, you may normally assume that 3000-5000 watts will fulfil your requirements. If you have a bigger furnace and/or a well pump, you will likely require a 5000 to 6500 watt generator.
### Subsequently, question is, what big generator do I need to power a water pump?
In the example of a 1 HP conventional submersible well pump; a 4-5 KW generator is needed for the well pump even though 1 KW is the amount of electricity the pump requires while it is operating.
### Also Know, how many watts does a 1/2 HP well pump use?
The instruction booklet for the transfer switch specifies that a shallow well 1/2 hp pump needs 1000 operating watts + 2350 starting watts.
### How many amps does a 1 HP well pump draw?
1 HP submersible pump motor will draw a current of 2 amperes if it is single phase motor at a voltage of 230 volts, and if it is a 3phase motor at a voltage of 440 volts, it will draw a current of 1.7 amperes, maximum.
### What will a 9000 watt generator run?
Operating a sump pump, freezer, refrigerator, and furnace will use up to 4000 watts. Add in some lights, a small television, and a microwave oven, and a 7500-watt unit is working close to its continuous capacity. A larger 9,000-watt standby unit could probably handle the additional load of a 1-ton air conditioner.
### What size generator do I need to run a refrigerator and freezer?
For example, if you want a generator to run a refrigerator and a freezer, the wattage (table 2) of the refrigerator would be 800 and the freezer would be 1,000. To select the correct size generator, you decide if both refrigerator and freezer are to start at the same time. If so, you would need (1,800 X 4) 7,200 watts.
### What will a 10000 watt generator run?
A 10,000 watt generator is ideal for running a refrigerator and other kitchen appliances, even at the same time. You can also use one to power a furnace, large window air conditioning unit, and even clothes washers and dryers.
### What size generator do I need to run my AC?
run a few smaller items. 15,000 btu air conditioner. run a few smaller items. 15,000 btu air conditioner. Buying a Portable RV Generator. Size of Air Conditioner Approximate Starting Watts Approximate Running Watts 13,500 btu 2800-3000W 1500-2000W 15,000 btu 3300-3500W 1300-1800W
### Will a 7500 watt generator run my house?
A 7500-watt generator is a great size for most homeowners. With a 7500-watt generator, you can power up most household appliances including your refrigerator, hot water heater, well pump, freezer, light, and oven. A 7500- watt generator will get you through your next power outage in comfort.
### How many amps does a 2 HP well pump draw?
Generator
Joe Generator Wattage Guide Household & Office Running Wattage Starting Wattage Pump, Well, 2 HP 3750 7500 Pump, Well, 3 HP 5000 10000 Pump, Well, 5 HP 7500 15000 Pump, Well, 7 – 1/2 HP 10000 20000
5.0 amps
### How many watts does a 1 hp water pump use?
How Much Electricity Do You Need to Produce? Survival Appliances Rated Watts Surge Watts Sump Pump 800 2000 Water Pump (1 HP) 1900 5700 Water Pump (2 HP) 2500 7500
### How many amps does a 1.5 HP well pump draw?
Member. A 230v 1.5 HP submersible pump is metering about 11.4 amps per leg. The motor nameplate states running current to be 10.6A. Metered starting amperage only increases about 1A above running amperage.
### Will a 2000 watt generator run a well pump?
A 2000 watt generator could power a 500 watt deep freezer and one element of the electric stove. A 2000 watt generator could start up the sump pump and then keep it running during a storm, and once running and drawing 800 watts, you could use the same generator to power the 1000 watt water well pump.
### How many amps does a 3/4 HP well pump draw?
Re: Amp draw on a 3/4HP pump Running amps will be 6 to 8 amps.
### How many amps does a deep well pump draw?
For example, if the pump wattage is 2,200 watts, and the voltage is 110 volts, the current is 2,200 /110 = 20 Amps. This is the value of the amps drawn by the well pump.
### How many watts does a septic pump use?
Appliance Wattage Guide – Calculate How Much Energy You Use TOOL OR APPLIANCE RUNNING (RATED) (RATED) WATTS ADDITIONAL STARTING WATTS Refrigerator/ Freezer 700 2200 Sump Pump – 1/3 HP 800 1300 Sump Pump – 1/2 HP 1050 2200 Water Well Pump – 1/3 HP 1000 2000 | 1,282 | 4,844 | {"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-2024-10 | latest | en | 0.920962 |
https://lavillatte.com/gcf-and-lcm-worksheets/ | 1,618,753,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038492417.61/warc/CC-MAIN-20210418133614-20210418163614-00634.warc.gz | 469,438,031 | 11,779 | # Gcf And Lcm Worksheets
Worksheets for the greatest common factor () and least common multiple () of numbers. you can create free printable worksheets for finding the greatest common factor () and least common multiple () of up to different numbers. the worksheets can be made in or formats, and are customizable with lots of options you can choose the number ranges for the and separately, the number of problems, workspace, font size, border, and border color.
Grade factoring worksheets these math worksheets cover factoring to prime numbers as well as determining the greatest common factor () or least common multiple () of two numbers. sample grade factoring worksheet. Factoring and worksheets our grade factoring worksheets provide students practice in factoring and finding prime factors of numbers and finding the greatest common factor and the the least common multiple of pairs of numbers.
find the of these numbers. guided lesson explanation yeah i get it. Printable worksheets and lessons. and step-by-step lesson- we go over both skills in detail.you will learn when they are used and how to find them. guided lesson - the skill is stepped up just a bit because we introduce finding the and between numbers.
## List of Gcf And Lcm Worksheets
Guided lesson explanation - yeah, i get it i went a little overboard with the explanation on this, but kids usually get it. Worksheet on prime factorization,, and find the of the numbers by listing multiples.,.,.,.,.,.,. Below are two grade math worksheets with word problems needing the use of greatest common factors () or least common multiples () to solve.
mixing and word problems encourages students to read and think about the questions, rather than simply recognizing a pattern to the solutions. these worksheets are files. Finding the greatest common factor () and least common multiple () find the of each.
), ), ), ), ), ), ), ), ), ), find the of each. Using least common multiple and greatest common factor worksheets is fine but we can do better. been teaching factor trees for years with the same mixed bag level of understanding. put together a list of my favorite strategies for teaching and.
### 1. Puzzle Maneuvering Middle Teachers Pay Gcf Lcm Worksheets
Some of the worksheets for this concept are least common multiple, least common multiples, least common multiple, greatest common factor es, greatest common factor, greatest common factor,, factors multiples primes prime factors and. Displaying top worksheets found for - and practice.
### 2. Placing Points Grid Grade Math Worksheets Practice Worksheet Greatest Common Factor Multiple Gcf Lcm
Id language school subject ( mathematics ) grade age - main content, and divisibility rules other contents addition and subtraction of fractions add to my workbooks embed in my website or add to classroom. Displaying top worksheets found for - rd grade and.
### 3. Factors Multiples Mazes Riddles Coloring Page Number Math Activity Gcf Lcm Worksheets
On these printable worksheets, students will identify multiples. and word problems worksheet ) lily has collected u.s. stamps and international stamps. she wants to display them in identical groups of u.s. and international stamps, with no stamps left over.
Id language school subject math grade age - main content integers other contents add to my workbooks download file embed in my website or add to classroom. The apps, sample questions, videos and worksheets listed below will help you learn and word problems access lesson plan resources for and word problems sample questions related to and word problems and continue reading.
Multiple choice questions in - displaying top worksheets found for this concept. some of the worksheets for this concept are grade chapter multiple choice math test name date, least common multiples, ace your math test reproducible work, quiz on , grade math practice test, multiple choice es, math fraction operation study guide, math mammoth grade b.
### 4. Lessons Gcf Lcm Worksheets
At the gym, swims every days, runs every days and cycles every days. if she did all three activities today, in how many days will she do all three activities again on the same day. Divide the product of numbers by provides the of those numbers. answers key.
### 5. Ten Problems Variables Worksheet Grade Lesson Planet Gcf Lcm Worksheets
This worksheet and quiz will help students measure their aptitude with and word problems. the questions on the quiz will require students to use their mathematical knowledge. quiz. A. and word problems. e. share skill. worksheet find the greatest common factor of the following numbers.
### 6. Interactive Worksheet Gcf Lcm Worksheets
Ns.b) use the distributive property to factor out the greatest common factor from an addition expression with two whole numbers (ns.c). Greatest common factor () and least common multiple () word problems grade math word problems worksheet read and answer each question.
### 7. Column Grade Word Problem Worksheet Worksheets Factoring Greatest Common Factor Answers Finding Problems Gcf Lcm
Example find the of and using listing method., prime factor tree worksheet., prime factor tree worksheet., prime factor tree worksheet., prime factor tree worksheet. to link to this greatest common factor worksheets page, copy the following code to your site more topics.
these math worksheets cover factoring to prime numbers as well as determining the greatest common factor or least common multiple of two numbers. you can use the generator to make worksheets either in or format both are easy to print. Worksheets math grade factoring.
factoring, and worksheets. our grade factoring worksheets provide students practice in factoring and finding prime factors of numbers and finding the greatest common factor and the the least common multiple of pairs of numbers., and worksheets - a total of worksheets with answer keys.
### 8. 5 Ways Factor Trees Middle School Math Gcf Lcm Worksheets
Find the of these numbers, b. find the of these numbers, a. needs. plastic containers. for her cookies.,.,. the two stacks will be the same height at. inches., ,.,. the tours leave at the same time every minutes. (answer choice c)., ,,. , ,. . This activity is a fun puzzle worksheet that will have students finding the and and matching the correct answer to be given a clue of what to draw in a grid.
if all the answers are correct they will reveal a surprise picture. this worksheet is suited for your core group and. subjects. With this download you get a fun activity worksheet, differentiated practice worksheets that have the same problems, and a common core practice worksheet.
worksheet is a fun puzzle worksheet that will have students finding the and and matching the correct answer to be.
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Some of the worksheets for this concept are and word problems, of monomials, and word problems, lesson applying and to fraction operations, greatest common factor, math mammoth grade b sample, least common multiple, greatest common factor in diagrams. | 2,879 | 14,794 | {"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} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.925609 |
http://aa.springer.de/papers/9348003/2300993/appa.htm | 1,653,063,213,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662533972.17/warc/CC-MAIN-20220520160139-20220520190139-00699.warc.gz | 1,067,137 | 4,644 | Astron. Astrophys. 348, 993-999 (1999)
## Appendix A: the travel of the hot bubbles towards the solar surface
### A.1. The velocity of the bubbles
Accepting that a hot bubble forms in the solar core, we can ask how can such a bubble survive when it starts to rise, when accelerated by the buoyant force. At first sight, it may seem that such a bubble can easily loose its surplus inner energy on its pathway by turbulent viscosity and other forms of dissipation. Moreover, heat expansion may quickly cool the bubble and so it may dissolve with its surroundings. Nevertheless, quantitative estimates show that the bubbles may easily keep their identity on travelling towards the solar surface and that if their energy surplus is significant enough, they can even reach the surface layers, where they are certainly disintegrate due to a developing shock wave ("sonic boom", Grandpierre 1981).
Let us see some equations determining the conditions of the bubbles on their way towards the surface. Following Gorbatsky (1964), if the hot bubble's surplus energy is dominated by radiation, and their initial energy surplus when they start to rise is , then
where is the temperature in the bubble and R is the radius of the bubble, and
where is the pressure outside the bubble.
The more exact calculation modifies the value of this coefficient to 0.40. Using a value for , and , one can get for and . Of course, a hotter bubble can contain the same amount of internal energy with a smaller size.
Now let us calculate the velocity of a bubble! Following Gorbatsky (1964), let us assume the following plausible assumptions:
1. the bubble keeps its spherical form
2. the region between the bubble and its environment is thin (very narrow turbulent wake)
3. the density and temperature may be regarded uniform within the bubble
4. the pressure within the bubble equals with the pressure outside it.
On its travel the bubble meets with resistance, therefore it is necessary to put a term in the equation of motion of the bubble describing it. Since the molecular viscosity is extremely low, it is enough to take into account only the turbulent drag here:
where is the coefficient of the turbulent drag, v is the velocity of the bubble. The quantity represents the so-called "induced-mass" term occuring for a body moving in a hydrodynamic medium (see Landau, Lifshitz, 1959, Sect. 11) in case of a spherical bubble. Using the equality , a consequence of the assumption 3, Eq. (16) can be arranged into a more suitable form:
The initial conditions for these equation are: at , and . As the bubble becomes accelerated to a large enough velocity, the turbulent drag and the gravitational force will balance the buoyant force and the velocity becomes steady,
Now accepting that (since for a movement with a constant speed the potential flow v and its potential does not depend on time, and therefore the pressure distribution becomes , see Problem 2, Sect. 10, Landau & Lifshitz 1959, p. 25), and substituting from the condition of hydrostatic equilibrium ,
The acceleration "a" can be estimated with , , , , , so and the bubble reach the constant speed during . The obtained velocity of the hot bubbles is much larger then the average speed of convective cells in the solar convective zone.
It has to note here that recent experimental and computational results suggest that for extremely high Rayleigh numbers the turbulent convection turns to a thready flow. "The flow is driven entirely (in the limit of infinite Ra) by these threads. The heat flux is carried by flows that maintain their identity...and can cross a convecting layer with little mixing between them. The width of the threads, in spite of entrainment, decreases with Rayleigh number instead of increasing as one might have expected on the basis of the simple `higher Ra means more turbulence means more mixing' line of argument" (Spruit, 1997). | 827 | 3,914 | {"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-2022-21 | latest | en | 0.922433 |
https://math.answers.com/Q/How_do_you_write_52_percent_as_a_decimal | 1,716,522,939,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00364.warc.gz | 318,832,894 | 46,995 | 0
# How do you write 52 percent as a decimal?
Updated: 9/16/2023
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Q: How do you write 52 percent as a decimal?
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### How do you write 0.52 as a percent?
To convert any decimal to percent, simply multiply the decimal by 100.Putting your example into the equation we get: 0.52 × 100 = 52(%)
### What is 52 percent in decimal?
Well, 52 percent = 52/100ths = 0.52
0.52
### What is 0.52 as a decimal and percent?
You just wrote it in a decimal format 0.52 and the percent is 52%
### How do you do 52 percent of 83?
Convert 52% into a decimal. 52%=.52 .52•83=43.16
### What is 52 in percent?
To convert any decimal to percent, simply multiply the decimal by 100. So, 5.2 × 100 = 520%
56% = 0.56
### How do you change 13 25 to a decimal and a percent?
13/25 to a decimal and a percent = 0.52; 52%
### What percent of 104 is 52?
52 is what percent of 104:= 52 / 104= 0.5Converting decimal to a percentage:0.5 * 100 = 50%
### What 52 is what percent of 104?
52 is what percent of 104:= 52 / 104= 0.5Converting decimal to a percentage:0.5 * 100 = 50%
### How do you write 15.5 percent as a decimal?
To write 15.5 percent as a decimal, you divide the percent value by 100. So, 15.5 percent as a decimal is 0.155.
### How do you write 4 percent as decimal?
Well how you write 4 percent as a decimal is 0.04 | 465 | 1,438 | {"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-2024-22 | latest | en | 0.879019 |
https://www.tutorialspoint.com/finding-distance-between-two-points-in-a-2-d-plane-using-javascript | 1,685,528,007,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646457.49/warc/CC-MAIN-20230531090221-20230531120221-00725.warc.gz | 941,191,768 | 9,716 | # Finding distance between two points in a 2-D plane using JavaScript
## Problem
We are required to write a JavaScript function that takes in two objects both having x and y property specifying two points in a plane.
Our function should find and return the distance between those two points.
## Example
Following is the code −
Live Demo
const a = {x: 5, y: -4};
const b = {x: 8, y: 12};
const distanceBetweenPoints = (a = {}, b = {}) => {
let distance = 0;
let x1 = a.x,
x2 = b.x,
y1 = a.y,
y2 = b.y;
distance = Math.sqrt((x2 - x1) * 2 + (y2 - y1) * 2);
return distance;
};
console.log(distanceBetweenPoints(a, b));
## Output
6.164414002968976
Advertisements | 194 | 668 | {"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} | 2.53125 | 3 | CC-MAIN-2023-23 | latest | en | 0.805022 |
https://www.my-mooc.com/en/mooc/hydraulics-shui-li-xue-tsinghuax-hyd1-1x/ | 1,708,496,238,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00834.warc.gz | 928,970,133 | 91,114 | list 10 sequences
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Hydraulics is one of the basic courses of civil engineering, hydraulic engineering, environmental engineering, architecture, and engineering physics. This science focuses on the laws of fluid dynamics and their interactions with the boundaries.
The main objective of this course is to develop the following:
• Understanding of the basic concepts and related theories of hydraulics
• Innovation consciousness and scientific literacy
• Analysis and the ability to solve the hydraulic problems met in practice
The course content focuses on basic theories:
• Physical and mechanical properties of fluid
• Hydrostatics
• Hydrokinematics
• Hydrodynamics
• Dimensional analysis and similitude
• Flow resistance and energy loss
Hydraulics will expose you to important and interesting applications of mathematics and mechanics in engineering. You’ll participate in experiments and view videos in this class. If you take this course, you will not only gain expertise, but also have fun and know more about the nature about hydraulics in a fun and engaging way.
《水力学》是水利、土木、环境、建筑和工程物理等相关专业的技术基础课，它以水为主要对象研究流体运动的规律以及流体与边界的相互作用。本课程的主要任务是培养学生在三个方面的知识与能力：(1) 掌握水力学基本概念和基本理论；(2) 培养创新意识和科学素养；(3) 培养分析和解决工程实际中水力学问题的能力和实验技能。课程内容包括：基础部分、专题部分和实验部分。基础部分：1.流体的物理力学性质；2.静力学；3.运动学；4.动力学基础；5.量纲分析和相似理论；6.流动阻力和能量损失。专题部分：1.有旋流动和有势流动；2.边界层理论基础与绕流运动；3.孔口、管嘴出流有压管流。实验部分：包括常规教学实验和设计型（创新）实验。
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## Prerequisite
• Calculus, or higher mathematics
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## Syllabus
• Basic concepts and related theories of hydraulics
• Innovation consciousness and scientific literacy
• How to analyze and solve the hydraulic problems met in practice
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## Instructors
Li Ling
Hydraulic Engineering
Tsinghua University
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## Content Designer
Tsinghua University is a Chinese university in Beijing, considered to be one of the most prestigious in the People's Republic of China. When it was founded in 1911, it was a preparatory school for students wishing to do postgraduate studies in American universities.
In 1925, Tsinghua developed into a university and now offers four-year undergraduate degrees (bachelor's degree) and graduate degrees (master's degree and doctorate).
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5/5 | 927 | 3,721 | {"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-2024-10 | latest | en | 0.660687 |
https://mmerevise.co.uk/a-level-maths-revision/the-modulus-function/ | 1,726,888,091,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00096.warc.gz | 354,608,101 | 56,725 | The Modulus Function
A LevelAQAEdexcelOCR
The Modulus Function
The modulus of a number is the size of the number, whether it is negative or positive, for example the modulus of $6$ is $6$ and the modulus of $-6$ is $6$.
A LevelAQAEdexcelOCR
Modulus Notation
The following notation is relevant to the modulus:
• The modulus of a number, e.g. $x$ is written as $|x|$.
• Generally, $|x| = x$ for $x \geq 0$, and $|x| = -x$ for $x <0$
• Functions also have a modulus: e.g. if $f(x) = -2$, then $|f(x)| = 2$
• $|f(x)| = f(x)$ when $f(x) \geq 0$ and $|f(x)| = -f(x)$ when $f(x)<0$
• If the modulus is inside the function, e.g. $f(|x|)$, then you apply the modulus to the $x$-value before applying the function, i.e. $f(|-4|) = f(4)$
A LevelAQAEdexcelOCR
Graphs of Modulus Functions – Straight Lines
There are $3$ types of modulus graphs that you may be asked to draw:
1. $y = |f(x)|$ – all negative values of $f(x)$ are made positive, by reflecting the negative section of the graph of $f(x)$ in the $x$-axis. This restricts the range to $|f(x)| \geq 0$ (or a subset within $|f(x)| \geq 0$, e.g. $|f(x)| \geq 2$.
2. $y = f(|x|)$ – the negative $x$-values give the same result as the corresponding positive $x$-values, so the graph of $f(x)$ for $x \geq 0$ is reflected in the $y$-axis, for negative $x$-values.
3. $y = |f(-x)|$ – the $x$-values swap sign (i.e. from positive to negative of from negative to positive), so the graph of $f(x)$ is reflected in the $y$-axis. Then, all negative values of $f(x)$ are made positive by reflecting the negative section of the graph of $f(x)$ in the $x$-axis. The range is restricted, as with $y = |f(x)|$.
The best and easiest way to draw these graphs is to plot the graph of $y = f(x)$ first, and then reflect it in the appropriate axis or axes.
Example: For $f(x) = 2x-1$, sketch the graphs of
\begin{aligned} y &= |f(x)| \\ y &= f(|x|) \\ y &= f|(-x)| \end{aligned}
A LevelAQAEdexcelOCR
A LevelEdexcel
Graphs of Modulus Functions – Quadratics and Cubics etc.
For modulus graphs where the function is a quadratic or cubic etc. the same rules apply as for straight lines – however, sketching them will be a little bit harder.
Example: For $f(x) = x^2-2x$, sketch the graphs of
\begin{aligned} y &= |f(x)| \\ y &= f(|x|) \\ y &= |f(-x)| \end{aligned}
A LevelEdexcel
Solving Modulus Equations Graphically
To solve modulus equations of the form $|f(x)| = n$ or $|f(x)| = |g(x)|$, you can solve them graphically, using the following method:
Step 1: Sketch the graphs of $y = |f(x)|$ and $y = n$, on the same pair of axes.
Step 2: Work out the ranges of $x$ for which $f(x) \geq 0$ and $f(x) < 0$ from the graph.
e.g. $f(x) \geq 0$ for $x \leq \textcolor{red}{a}$ or $x \geq \textcolor{blue}{b}$ and $f(x) < 0$ for $\textcolor{red}{a} < x < \textcolor{blue}{b}$
Step 3: Use step 2 to write $2$ new equations, one that holds for each range of $x$:
$f(x) = n$ for $x \leq \textcolor{red}{a}$ or $x \geq \textcolor{blue}{b}$
$- f(x) = n$ for $\textcolor{red}{a} < x < \textcolor{blue}{b}$
Step 4: Solve each equation in turn and check that the solutions are valid, and remove any that are outside the range of $x$ for that equation.
Step 5: Check that the solutions look correct, by looking at the graph.
Note: Use the same method for $|f(x)| = |g(x)|$, by replacing $n$ with $g(x)$.
A LevelAQAEdexcelOCR
Solving Modulus Equations Algebraically
For equations of the form $|f(x)| = n$ and $|f(x)| = g(x)$ you can solve them algebraically instead of graphically – if you feel that you understand the topic well enough.
Example: Solve $|2x+2| = x+4$
Step 1: Solve for positive values:
\begin{aligned} 2x + 2 &= x+4 \\ \textcolor{red}{x} &= \textcolor{red}{2} \end{aligned}
Step 2: Solve for negative values:
\begin{aligned} -(2x + 2) &= x+4 \\ 3x &= -6 \\ \textcolor{red}{x} &= \textcolor{red}{-2} \end{aligned}
Step 3: Combine the solutions:
The solutions are $\textcolor{red}{x = 2}$ and $\textcolor{red}{x = -2}$
For equations of the form $|f(x)| = |g(x)|$, it is easier to do solve them algebraically, using the following rule:
“If $|a| = |b|$, then $a^2 = b^2$
So if $|f(x)| = |g(x)|$, then $[f(x)]^2 = [g(x)]^2$
Example: Solve $|x-1| = |2x+3|$
Step 1: Square both sides:
\begin{aligned} |x-1| &= |2x+3| \\ (x-1)^2 &= (2x+3)^2 \end{aligned}
Step 2: Expand and simplify:
\begin{aligned} x^2 - 2x + 1 &= 4x^2 + 12x + 9 \\ 3x^2 + 14x + 8 &= 0 \\ (3x+2)(x+4) &= 0 \end{aligned}
Step 3: So, the solutions are:
$\textcolor{red}{x = - \dfrac{2}{3}}$ and $\textcolor{red}{x = -4}$
A LevelAQAEdexcelOCR
Note:
You can also solve modulus inequalities using these methods. The graphical method of solving inequalities will be helpful, since there will often be a quadratic involved. Another rule that will be helpful is:
$|x-a| < b \, \iff \, a - b < x < a+b$
A LevelAQAEdexcelOCR
Example 1: Solving Modulus Equations Graphically – Straight Lines
Solve $|2x+3| = x+2$
[3 marks]
Step 1: Sketch the graphs of $y = |2x+3|$ and $y = x+2$ on the same pair of axes.
Step 2: Work out the ranges of $x$ for which $f(x) \geq 0$ and $f(x) < 0$ from the graph:
$2x+3 \geq 0$ for $x \geq - \dfrac{3}{2}$ and $2x+3 < 0$ for $x < - \dfrac{3}{2}$
Step 3: Use step 2 to write $2$ new equations, one that holds for each range of $x$:
(1) $2x+3 = x+2$ for $x \geq - \dfrac{3}{2}$
(2) $- (2x+3) = x+2$ for $x < - \dfrac{3}{2}$
Step 4: Solve each equation in turn and check that the solution are valid, and remove any that are outside the range of $x$ for that equation.
Solving (1): $x=-1$ (this is valid since $-1 \geq - \dfrac{3}{2}$)
Solving (2): $3x = -5 \Rightarrow x = - \dfrac{5}{3}$ (this is valid since $- \dfrac{5}{3} < - \dfrac{3}{2}$)
Step 5: Check that the solutions look correct, by looking at the graph. The two solutions appear to be correct.
A LevelAQAEdexcelOCR
A LevelEdexcel
Example 2: Solving Modulus Equations Graphically – Quadratics and Cubics etc.
Solve $|x^2 - 4| = 3$
[4 marks]
Step 1: Sketch the graphs of $y = |x^2-4|$ and $y = 3$ on the same pair of axes.
Step 2: Work out the ranges of $x$ for which $f(x) \geq 0$ and $f(x) < 0$ from the graph:
$x^2 - 4 \geq 0$ for $x \leq -2$ or $x \geq 2$
and $x^2 - 4 < 0$ for $-2 < x < 2$
Step 3: Use step 2 to write $2$ new equations, one that holds for each range of $x$:
(1) $x^2 - 4 = 3$ for $x \leq -2$ or $x \geq 2$
(2) $- (x^2 - 4) = 3$ for $-2 < x < 2$
Step 4: Solve each equation in turn and check that the solution are valid, and remove any that are outside the range of $x$ for that equation.
Solving (1): $x^2 = 7 \Rightarrow x = \sqrt{7}$ and $x = - \sqrt{7}$ (this is valid since $- \sqrt{7} \leq - 2$ and $\sqrt{7} \geq 2$)
Solving (2): $x^2 - 1 = 0 \Rightarrow x = 1$ and $x = -1$ (this is valid since $1$ and $-1$ both lie within $-2 < x < 2$)
Step 5: Check that the solutions look correct, by looking at the graph. The four solutions appear to be correct.
A LevelEdexcel
The Modulus Function Example Questions
Question 1: For the function $f(x) = 3x^2 + 4x - 9$, find the following:
a) $f(-2)$
b) $|f(-2)|$
c) $f(|-2|)$
d) $-|f(2)|$
[4 marks]
A Level AQAEdexcelOCR
a) $f(-2) = 3(-2)^2 + 4(-2) - 9 = -5$
b) $|f(-2)| = |-5| = 5$
c) $f(|-2|) = f(2) = 3(2)^2 + 4(2) - 9 = 11$
d) $- |f(2)| = - |11| = - 11$
Gold Standard Education
Question 2:
a) For the function $f(x) = 3x-3$, $x \in \mathbb{R}$, sketch the graphs of:
i) $y = |f(x)|$
ii) $y = f(|x|)$
iii) $y = |f(-x)|$
b) Hence, or otherwise, solve the equation $|3x - 3| = \dfrac{3}{2}$
[7 marks]
A Level AQAEdexcelOCR
a)i) The negative section needs to be reflected in the $x$-axis:
ii) For the negative $x$-values, reflect the line in the $y$-axis:
iii) Reflect $f(x)$ in the $y$-axis, and then reflect the negative section in the $x$-axis:
b)
Firstly, sketch the graphs of $y = |3x - 3|$ (using part a)i)) and $y = \dfrac{3}{2}$ on the same pair of axes:
$3x - 3 \geq 0$ when $x \geq 1$ and $3x - 3 < 0$ when $x < 1$
So, we can form two equations:
(1) $3x - 3 = \dfrac{3}{2}$ for $x \geq 1$
(2) $-(3x-3) = \dfrac{3}{2}$ for $x < 1$
Then, we can solve these:
Solving (1): $3x = \dfrac{9}{2} \Rightarrow x = \dfrac{3}{2}$ for $x \geq 1$ (this is valid since $\dfrac{3}{2} \geq 1$)
Solving (2): $3x = \dfrac{3}{2} \Rightarrow x = \dfrac{1}{2}$ for $x < 1$ (this is valid since $\dfrac{1}{2} < 1$)
From looking at the graph, both solutions seem to be correct.
Gold Standard Education
Question 3: Solve the equation $2|-x-3| = x+4$
[3 marks]
A Level AQAEdexcelOCR
$2|-x-3| = x-4$
Therefore $2(-x-3) = x+4$ or $-2(-x-3) = x+4$
Then, solve these:
\begin{aligned} 2(-x-3) &= x + 4 \\ -2x - 6 &= x + 4 \\ 3x &= - 10 \\ x &= - \dfrac{10}{3} \end{aligned}
\begin{aligned}-2(-x-3) &= x + 4 \\ 2x + 6 &= x + 4 \\ x &= - 2 \end{aligned}
So, $x = - \dfrac{10}{3}$ or $x = - 2$
Gold Standard Education
Question 4: Solve the equation $|2x+4| = -2|x+1|$
[3 marks]
A Level AQAEdexcelOCR
Actually, we need not do any calculation at all. On the left side of the equation, we have $|2x + 4|$, which is defined to be always positive.
However, the right side of the equation is always negative (as we have $-2$ multiplied by an always-positive term).
We can conclude, then, that there could only be a solution if both graphs meet at the $x$-axis.
However, this is not the case as the first graph touches the $x$-axis at $x=-2$ while the second graph touches the $x$-axis at $x=-1$.
Hence, there are no solutions.
Gold Standard Education
Question 5: Solve the equation $|2x+2| = |x-2|$
[4 marks]
A Level AQAEdexcelOCR
\begin{aligned} |2x+2| &= |x-2| \\ (2x+2)^2 &= (x-2)^2 \\ 4x^2 + 8x + 4 &= x^2 - 4x + 4 \\ 3x^2 + 12x &= 0 \\ x^2 + 4x &= 0 \\ x(x+4) &= 0 \end{aligned}
Hence, $x = 0$ or $x = -4$
Gold Standard Education
A Level
A Level
A Level | 3,719 | 9,880 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 194, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-38 | latest | en | 0.725568 |
https://www.statsmodels.org/v0.10.2/generated/statsmodels.tsa.statespace.structural.UnobservedComponentsResults.test_heteroskedasticity.html | 1,716,930,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059148.63/warc/CC-MAIN-20240528185253-20240528215253-00839.warc.gz | 848,023,574 | 4,607 | # statsmodels.tsa.statespace.structural.UnobservedComponentsResults.test_heteroskedasticity¶
method
UnobservedComponentsResults.test_heteroskedasticity(method, alternative='two-sided', use_f=True)
Test for heteroskedasticity of standardized residuals
Tests whether the sum-of-squares in the first third of the sample is significantly different than the sum-of-squares in the last third of the sample. Analogous to a Goldfeld-Quandt test. The null hypothesis is of no heteroskedasticity.
Parameters
methodstring {‘breakvar’} or None
The statistical test for heteroskedasticity. Must be ‘breakvar’ for test of a break in the variance. If None, an attempt is made to select an appropriate test.
alternativestring, ‘increasing’, ‘decreasing’ or ‘two-sided’
This specifies the alternative for the p-value calculation. Default is two-sided.
use_fboolean, optional
Whether or not to compare against the asymptotic distribution (chi-squared) or the approximate small-sample distribution (F). Default is True (i.e. default is to compare against an F distribution).
Returns
outputarray
An array with (test_statistic, pvalue) for each endogenous variable. The array is then sized (k_endog, 2). If the method is called as het = res.test_heteroskedasticity(), then het[0] is an array of size 2 corresponding to the first endogenous variable, where het[0][0] is the test statistic, and het[0][1] is the p-value.
Notes
The null hypothesis is of no heteroskedasticity. That means different things depending on which alternative is selected:
• Increasing: Null hypothesis is that the variance is not increasing throughout the sample; that the sum-of-squares in the later subsample is not greater than the sum-of-squares in the earlier subsample.
• Decreasing: Null hypothesis is that the variance is not decreasing throughout the sample; that the sum-of-squares in the earlier subsample is not greater than the sum-of-squares in the later subsample.
• Two-sided: Null hypothesis is that the variance is not changing throughout the sample. Both that the sum-of-squares in the earlier subsample is not greater than the sum-of-squares in the later subsample and that the sum-of-squares in the later subsample is not greater than the sum-of-squares in the earlier subsample.
For $$h = [T/3]$$, the test statistic is:
$H(h) = \sum_{t=T-h+1}^T \tilde v_t^2 \Bigg / \sum_{t=d+1}^{d+1+h} \tilde v_t^2$
where $$d$$ = max(loglikelihood_burn, nobs_diffuse)` (usually corresponding to diffuse initialization under either the approximate or exact approach).
This statistic can be tested against an $$F(h,h)$$ distribution. Alternatively, $$h H(h)$$ is asymptotically distributed according to $$\chi_h^2$$; this second test can be applied by passing asymptotic=True as an argument.
See section 5.4 of [1] for the above formula and discussion, as well as additional details.
TODO
• Allow specification of $$h$$
References
1(1,2)
Harvey, Andrew C. 1990. Forecasting, Structural Time Series Models and the Kalman Filter. Cambridge University Press. | 736 | 3,041 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-22 | latest | en | 0.858276 |
http://www.physicsforums.com/showpost.php?p=3411314&postcount=2 | 1,411,120,386,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657131238.51/warc/CC-MAIN-20140914011211-00017-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 732,798,782 | 2,807 | View Single Post
Sci Advisor P: 6,112 Using the prime number theorem, this can be estimated. π(n) ~ n/ln(n), so π(2n) - π(n) ~ 2n/ln(2n) - n/ln(n) which is approximately n/ln(n). | 67 | 179 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2014-41 | latest | en | 0.713126 |
http://www.besthomeloanratestoday.com/how-banks-calculate-interest-on-loans/ | 1,604,074,044,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911027.72/warc/CC-MAIN-20201030153002-20201030183002-00596.warc.gz | 123,579,937 | 8,681 | # How Banks Calculate Interest On Loans
### Contents
Interest Rates: AER and APR explained – MoneySavingExpert – Therefore the savings rate is what the bank pays you for borrowing your money.. Rough compound interest calculation rule of thumb for maths nerds: divide 72 by. The APR takes into account not just the interest on the loan but also other.
Education Loan EMI Calculator – Education Loan Interest. – Benefits of using the Education Loan EMI Calculator.. Just make sure that you consider the bank’s interest payment calculations, whether it is calculated at a daily reducing balance, or on a quarterly reducing balance. Calculate your Education Loan EMI.
SBI’s NRE Account: Features, How To Avail Loans, Interest Rates – State Bank of India (SBI), the largest lender. be withdrawn for making local payments in rupees, said SBI. 4. Interest earned on NRE accounts is exempt from Indian Income tax. 5. Rupee loan is.
Loan Calculator – Calculate EMI, Affordability, Tenure & Interest Rate – Loan tenure calculator loan Tenure. Interest Rate calculator interest rate. How much interest would I have to pay for an item (electronic gadgets, furniture, household appliances etc.) that I Banks charge more than just the interest rate on loans. When obtaining a loan, lenders charge various.
Not allowed to issue FDs and give loans, how payments banks function and offer higher interests on deposits? – and sanction loans, and may accept only demand deposits through savings bank accounts and current accounts. Despite the bars on these key banking activities, some payments banks started their.
How do banks calculate interest on loans – How. :: GoFTP Answers – How banks calculate home vaules? Please tell us which questions below are the same as this one: How do banks calculate interest on loans?
3 Ways to Calculate Bank Interest on Savings – wikiHow – · To calculate bank interest on savings, use the formula for calculating the effect of compound interest on your bank balance. In this formula, “P” stands for the principal, “r” is the annual rate of interest, and n is the.
How to Pay Off Student Loan Debt While Still Saving and Investing – Not doing so can lead to penalties, extra interest. said for the safety of liquid cash in the bank. That’s why building an emergency fund is a must-have before you can go gung-ho on paying off.
Formulas and Examples to Calculate Interest on Savings – To use compound interest, you need to adjust several numbers. Change the annual rate to a monthly rate: 5 percent divided by 12 months becomes 0.004167. Also, convert the number of periods to 12. To calculate for more than one year, you’d use 12 per year.
How do lenders determine your personal loan eligibility? – After evaluating your profile against a set parameters, lenders employ risk-based pricing to determine the rate of interest. for loans or credit cards. Your employment experience is the period of. | 611 | 2,924 | {"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-2020-45 | latest | en | 0.922994 |
https://mchenrysoftware.com/970960q.htm | 1,685,398,011,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00512.warc.gz | 437,171,939 | 4,756 | # Questions related to SAE paper 970960
The following are our responses to questions we have received on SAE paper 970960 "Effects of Restitution in the Application of Crush Coefficients"
Questions | #1 | #2 | #3 | #4 | #5 | Top | End | Homesite |
• Question#1: If one performs a revised SMAC calculation, all springs would initially have the (same?) input value of stiffness, K1. When each spring starts to unload, the maximum dynamic crush is known for that spring. At that time, K2, rho and gamma can be calculated for each spring using the formulas you present. But re-calculating K1 would make no sense, since K1 is not used for the rest of the time-history. So how does Equation 8 in SAE paper 970960 get implemented to define K1?
Answer #1: The four fitted constants are calculated only once, before the start of an application run. When a given radial spring starts to unload, the residual crush for that spring is calculated from equations (17) and (24), or the equivalent equation (9). Equation (8) was presented only for the purpose of showing that the value of K1 is independent of the extent of restored energy and, thereby, of the restitution coefficient.
Questions | #1 | #2 | #3 | #4 | #5 | Top | End |
• Question#2: Equation 8 of SAE paper 970960, specifying the loading stiffness, is identical to Equation 10 of SAE paper 910119 (except for notation). Of course, this follows because both demand a consistency between the loading energy in SMAC and the loading energy in CRASH.
1. Answer #2: The derivation of equation (8) is presented in appendix 2 in the form of equation (26). While the manner of derivation is different, the result is equivalent to equation (10) in SAE #910119.
Questions | #1 | #2 | #3 | #4 | #5 | Top | End |
• Question#3: Figure 6a of SAE paper 970960 shows an original SMAC specification of A,B and KV. Since original SMAC does not use A and B, I assume that A and B were specified as target values used by the sophisticated analyst to select the KV value. Likewise, on Figure 8 was KV selected based on A, B and the calculated deflections?
Answer #3: In figure 6A, the listed values of A, B, (/)1, ()1, and ()1 were used to calculate the value for Kv (by application of either equations (2) through (5) or equation (8)). In Figures 7, 8 and 9, the same A and B values as in Figure 6A were applied with different inputs for the restitution properties:
Fig.6A Fig. 7, 8 & 9 A 317 317 LB/IN B 56 56 LB/IN2 (m)1 30 30 IN ()1 0.2000 0.1500 - (/)1 0.8000 0.8775 - K1 54.74 63.7 LB/IN2 K2 54.74 95.5 LB/IN2
Note that a major point in the paper is the fact that a given set of values for A and B can serve to represent a wide range of restitution behavior.
Questions | #1 | #2 | #3 | #4 | #5 | Top | End |
• Question#4: Clarification #7 of Appendix C of SAE paper 970960 states that the elastic range (full dimensional recovery) is equal to A/B. This implies disturbingly large elastic deformations. For example, the first Figure A3, A/B is 357/13=27.5 inches. Do you see this as a serious problem?
1. Answer#4: Clarification #7 refers specifically to the CRASH (EDCRASH) computer program in which the implied elastic deformation range, in terms of full dimensional recovery is equal to A/B. In prior publications (e.g., SAE papers 910119, 920607, etc.) "A" has been referred to as the "zero residual crush force" or the "stiffness coefficient which represents the maximum force per unit width of the contact area which produces no crush." Therefore, the quantity A/B must be recognized as effectively being the elastic deformation range in terms of full dimensional recovery.
With the revised form of restitution modeling presented in SAE #970960, the corresponding elastic deformation range, in terms of full dimensional recovery, is given by equation (26) with set to zero. In that case, the maximum deformation for elastic behavior, in terms of full dimensional recovery, is defined as:
For the A, B values of the first Figure A3, which were fitted by Monk and Guenther (Monk, M.W., Guenther, D.A., "Update of CRASH II Computer Model Damage Tables, Vol. I", DOT HS-806446, March 1983), the following results can be obtained:
A = 357 LB/IN
B = 13 LB/IN2
()1 = 20 INCHES
()1 = 0.225
(/)1 = 0.750
K1 = 58.6 LB/IN2
K2 = 47.5 LB/IN2
While 12.93 inches of elastic deformation, in terms of full dimensional recovery, also seems large, it is certainly more believable than the 27.5 inches of elastic deformation in terms of full dimensional recovery in CRASH (EDCRASH) for the same (questionable) fitted values of A and B.
The purpose for inclusion of Figure A3 was to demonstrate that (1) unusual A, B fits such as Category 4, Rear, can be accommodated with crush properties that appear to be reasonable and (2) force equilibrium is possible in collisions between vehicles with widely different values for A and B.
As indicated in the paper, progress towards a rigorous and complete validation is data-limited at the present time.
Questions | #1 | #2 | #3 | #4 | #5 | Top | End |
• Question#5: If you extrapolate your V line for the revised SMAC in Figure 8 of the paper SAE 97-0960 to zero residual crush you come up with a V of approximately 15 MPH. Am I misinterpreting this graph or using it incorrectly?
Answer#5: In Figure 8 of SAE #970960 (revised SMAC) the maximum deformation corresponding to a zero residual deformation, from equation (9):
For = 5.308 Inches, the coefficient of restitution, from equation (1):
Since the calculated value for is > 1.00, is set equal to the maximum value of 1.0.
Thus, extension of the slope of the V line in Figure 8 will include a change in the slope of the V line at that value of residual deflection where the calculated equals 1.0.
From equation (1):
=6.410 Inches
From equation (9) the corresponding value of residual deflection is:
The intercept is then obtained using equation (19) and the coefficient of restitution (1.0):
= (2.00)(6.74) = 13.48 MPH
This means that, for the given fitted properties, the residual deformation would be zero in a 6.74 MPH SAE barrier crash. The corresponding V would be 13.48 MPH.
Another item to note is that because of the selected form of restitution control in the original NHTSA SMAC (EDSMAC), the V line has an intercept through (0,0). In particular, in original SMAC the residual deflection approaches ~95% of the maximum deflection at small deflections (see Figure 6 of the paper). In real-life, the residual deflection should become a very small portion of the maximum deflection at small deflections (i.e, the ratio of should approach 0.0). | 1,702 | 6,599 | {"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-2023-23 | latest | en | 0.929408 |
https://www.airmilescalculator.com/distance/kcz-to-ogn/ | 1,606,854,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141681524.75/warc/CC-MAIN-20201201200611-20201201230611-00661.warc.gz | 508,437,262 | 7,775 | # Distance between Kōchi (KCZ) and Yonaguni Jima (OGN)
Flight distance from Kōchi to Yonaguni Jima (Kōchi Airport – Yonaguni Airport) is 899 miles / 1447 kilometers / 781 nautical miles. Estimated flight time is 2 hours 12 minutes.
## Map of flight path from Kōchi to Yonaguni Jima.
Shortest flight path between Kōchi Airport (KCZ) and Yonaguni Airport (OGN).
## How far is Yonaguni Jima from Kōchi?
There are several ways to calculate distances between Kōchi and Yonaguni Jima. Here are two common methods:
Vincenty's formula (applied above)
• 898.864 miles
• 1446.581 kilometers
• 781.091 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 899.388 miles
• 1447.425 kilometers
• 781.547 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Kōchi Airport
City: Kōchi
Country: Japan
IATA Code: KCZ
ICAO Code: RJOK
Coordinates: 33°32′45″N, 133°40′8″E
B Yonaguni Airport
City: Yonaguni Jima
Country: Japan
IATA Code: OGN
ICAO Code: ROYN
Coordinates: 24°28′0″N, 122°58′40″E
## Time difference and current local times
There is no time difference between Kōchi and Yonaguni Jima.
JST
JST
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 143 kg (316 pounds).
## Frequent Flyer Miles Calculator
Kōchi (KCZ) → Yonaguni Jima (OGN).
Distance:
899
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
899
Round trip? | 485 | 1,668 | {"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-2020-50 | latest | en | 0.789222 |
http://interactivepython.org/courselib/static/pythonds/Recursion/pythondsConvertinganIntegertoaStringinAnyBase.html | 1,553,567,608,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204768.52/warc/CC-MAIN-20190326014605-20190326040605-00098.warc.gz | 112,419,148 | 9,690 | # 4.5. Converting an Integer to a String in Any Base¶
Suppose you want to convert an integer to a string in some base between binary and hexadecimal. For example, convert the integer 10 to its string representation in decimal as "10", or to its string representation in binary as "1010". While there are many algorithms to solve this problem, including the algorithm discussed in the stack section, the recursive formulation of the problem is very elegant.
Let’s look at a concrete example using base 10 and the number 769. Suppose we have a sequence of characters corresponding to the first 10 digits, like convString = "0123456789". It is easy to convert a number less than 10 to its string equivalent by looking it up in the sequence. For example, if the number is 9, then the string is convString[9] or "9". If we can arrange to break up the number 769 into three single-digit numbers, 7, 6, and 9, then converting it to a string is simple. A number less than 10 sounds like a good base case.
Knowing what our base is suggests that the overall algorithm will involve three components:
1. Reduce the original number to a series of single-digit numbers.
2. Convert the single digit-number to a string using a lookup.
3. Concatenate the single-digit strings together to form the final result.
The next step is to figure out how to change state and make progress toward the base case. Since we are working with an integer, let’s consider what mathematical operations might reduce a number. The most likely candidates are division and subtraction. While subtraction might work, it is unclear what we should subtract from what. Integer division with remainders gives us a clear direction. Let’s look at what happens if we divide a number by the base we are trying to convert to.
Using integer division to divide 769 by 10, we get 76 with a remainder of 9. This gives us two good results. First, the remainder is a number less than our base that can be converted to a string immediately by lookup. Second, we get a number that is smaller than our original and moves us toward the base case of having a single number less than our base. Now our job is to convert 76 to its string representation. Again we will use integer division plus remainder to get results of 7 and 6 respectively. Finally, we have reduced the problem to converting 7, which we can do easily since it satisfies the base case condition of $$n < base$$, where $$base = 10$$. The series of operations we have just performed is illustrated in Figure 3. Notice that the numbers we want to remember are in the remainder boxes along the right side of the diagram.
Figure 3: Converting an Integer to a String in Base 10
ActiveCode 1 shows the Python code that implements the algorithm outlined above for any base between 2 and 16.
Notice that in line 3 we check for the base case where n is less than the base we are converting to. When we detect the base case, we stop recursing and simply return the string from the convertString sequence. In line 6 we satisfy both the second and third laws–by making the recursive call and by reducing the problem size–using division.
Let’s trace the algorithm again; this time we will convert the number 10 to its base 2 string representation ("1010").
Figure 4: Converting the Number 10 to its Base 2 String Representation
Figure 4 shows that we get the results we are looking for, but it looks like the digits are in the wrong order. The algorithm works correctly because we make the recursive call first on line 6, then we add the string representation of the remainder. If we reversed returning the convertString lookup and returning the toStr call, the resulting string would be backward! But by delaying the concatenation operation until after the recursive call has returned, we get the result in the proper order. This should remind you of our discussion of stacks back in the previous chapter.
Self Check
Write a function that takes a string as a parameter and returns a new string that is the reverse of the old string.
Write a function that takes a string as a parameter and returns True if the string is a palindrome, False otherwise. Remember that a string is a palindrome if it is spelled the same both forward and backward. For example: radar is a palindrome. for bonus points palindromes can also be phrases, but you need to remove the spaces and punctuation before checking. for example: madam i’m adam is a palindrome. Other fun palindromes include:
• kayak
• aibohphobia
• Live not on evil
• Reviled did I live, said I, as evil I did deliver
• Go hang a salami; I’m a lasagna hog.
• Able was I ere I saw Elba
• Kanakanak – a town in Alaska
• Wassamassaw – a town in South Dakota
Next Section - 4.6. Stack Frames: Implementing Recursion | 1,058 | 4,767 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2019-13 | latest | en | 0.912409 |
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2022 WAEC CHEMSTRY PRACTICAL VERIFIED ANSWERS
*2022 WAEC CHEMISTRY PRACTICAL RUNZ*
(1a)
Volume of burrete = 50.00cm³
Volume of pipette = 25.00cm³
Indicator = Methyl orange
TABULATE
Final burette readings (cm³); 37.50 | 22.50 | 37.50 | 44.50
Initial burette readings(cm³); 0.00 | 0.00 | 0.00 | 0.00
Volume f Acid used (cm³); 37.50 | 22.50 | 22.50 | 22.50
Volume of A used = 1st + 2nd + 3rd/3
= 22.50+22.50+22.50/3
Average volume of A used = 22.50cm³
(1bi)
Concentration of A ( CA )?
2.03g in 500cm³ of solution
Volume = 500cm³/1000 = 0.500dm³
In g/dm³ = 2.03/0.5 = 4.06g/dm³
To find concentration in mol/dm³
Conc. of A in mol/dm³ = Conc. of A in g/dm³/Molar mass
Conc. of A in mol/dm³ = 4.06/36.5
= 0.1112mol/dm³
Molar mass of A (HCL) = 1+35.5 = 36.5g/mol
(1bii)
Number of mole of the Acid in average titre
Na = CaVa
= 0.1112 × 22.50
= 2.52moles
(1biii)
Nb= ? Na= 3, Ca= 0.1112mol/dm³, Cb = 0.12mol/dm³ , Va = 22.50cm³, Vb = 25.00cm³
CaVa/CbVb= nA/nB
0.1112×22.50/0.12×25 = 3/nB
nB × 2.502 = 9
nB = 9/2.502 =
nB = 3.597moles
(1biv)
Mole ratio of Acid to base
nA:nB
= 3:3
= 1:1
============================
(2)
(2a)
Test; C + 10cm³ of distilled water + filter
Observation; C dissolves partially to give a light green solution. Blue filtrate and green residue
Inference; C is a mixture of soluble and insoluble salt
(2bi)
Test; 2cm³ of filtrate + NaOH in drops in excess
Observation; Blue precipitate remains
Inference; CU²+ is present
(2bii)
Test; 2bi + warm
Observation; a colourless gas with a chocking smell which turns moist red litmus blue and forms dense white fumes with hydrogen chloride gas
Inference; NH³ from NH⁴+
(2biii)
Test; 2cm³ of filtrate + BaCl² + excess HCl
Observation; white precipitate, precipitate Remains
Inference; SO²- CO3²- or SO4²- is present, SO4²- confirmed
(2ci)
Test; Residue + HNOg
Observations; effervescence of a colourless and odourless gas which turns like water milky and turns moist blue to litmus red
Inference; CO2 from CO3²-
(2cii)
Test; 2ci + NH3 in drops in excess
Observations; blue precipitate. Precipitate dissolved to give a deep blue solution
Inference; CU2+ present
CU2+ confirmed
===============================
(3)
(3ai)
The value will increase
(3aii)
The occur as a result of the decrease in the concentration of base due to the added volume of water
(3bi)
There will be no visible reaction because copper is less than Zinc in the electrochemical series
(3bii)
It absorb water and become sticky because it is hygroscopic
(3biii)
The solution turns pink
(3c)
When NaOH is added to the solution of zn³+, a white precipitate is formed which later dissolve in excess NaOH due to the formation of zinc hydroxide
==============================
Categories: Waec | 1,058 | 2,979 | {"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-2024-33 | latest | en | 0.632462 |
https://www.axisandallies.org/forums/user/count_zeppelin/topics | 1,717,006,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059384.83/warc/CC-MAIN-20240529165728-20240529195728-00869.warc.gz | 551,947,835 | 20,600 | This thread appears to be long dead, but I never get tired of math questions! :-)
To answer the original question: The statistics of when a bomber gets shot down are determined by something called the geometric distribution, which you can find plenty of info about at http://en.wikipedia.org/wiki/Geometric_distribution. The geometric distribution describes the first occurrence of an event when you make repeated, independent trials. For example, the number of times you have to flip a coin before getting heads; the number of times you have to be dealt a poker hand before getting a royal flush; and the number of times you have to roll a die before getting a 1 are all described by geometric distributions. Once you know the probability of success in a single event, the geometric distribution tells you how long you have to wait before seeing the first success.
Some of the information for the geometric distribution:
–-The probability of getting your first success (in our example “success” is the bomber getting shot down!) on the nth turn is (1/6)*(5/6)^(n-1). This is because, in order to be shot down on the nth turn, the bomber must first survive the preceding n-1 turns, which has a probability of (5/6)^(n-1), and must then be shot down on turn n, which has a probability of 1/6.
–-The probability of getting your first success by the nth turn is 1 - (5/6)^n. This is just 1 minus the probability of the bomber surviving the first n turns. (It’s also the sum of the probabilities of getting shot down on turn 1, 2, …, up to n, but that’s the long way to do it!)
Given this, a number of other facts can be calculated (I’ll spare you the details…) The question “When will the bomber, statistically speaking, be shot down?” has three different answers: the mean, median, and mode. Here’s what each of those is:
—The mode is the single turn on which the bomber is most likely to be shot down. This may be surprising, but it’s the very first turn! The probability of getting shot down is the same (1/6) on all turns, provided the bomber makes it to that turn , but first it must survive all the turns before it. Hence, the first turn is the most likely turn-of-death since the bomber doesn’t have to survive any prior turns in order to reach it.
–-The median is the first turn for which the bomber has a 50% chance of getting shot down before it. This is the fourth turn, because the probability of getting shot down on one of the first four turns is 52% whereas the probability of surviving the first four turns is 48%. So it’s about 50-50 whether the bomber survives at least 4 turns.
–-The mean , also known as the expected value, is the average number of turns that the bomber will survive. This is 6 turns. It turns out that you can actually calculate this by taking the reciprocal of the 1/6 probability of getting shot down on a given turn, although the reason why that works is slightly less obvious than it might seem.
So, which of the mean, median, and mode is the most useful? When should we expect to lose the bomber? Well, the mode is the least useful for answering this sort of question. One can make a case for either the mean or the median, but for the types of calculation that people usually have in mind, the mean is the way to go. For example, if you’re trying to figure out the average amount of economic damage inflicted minus damage received from each bombing raid, then the statistic you want is the average, aka the mean.
Of course, the question of whether bombing raids are worth doing involves a lot more than calculating the average IPC’s lost and destroyed. My test: If you want to decide whether bombing a given country with your country is useful, ask yourself, “Would I destroy \$10 of mine if I also got to destroy \$10 of theirs?” If the answer is “Heck yes!” then SBR’s are useful. For example, Japan bombing Russia generally makes sense, as does US or UK bombing Germany. A weaker power bombing a stronger power is usually not so smart. | 918 | 4,008 | {"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-2024-22 | latest | en | 0.917726 |
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# Search results
Found 450 matches
Speeds and feeds - Spindle speed
The phrase speeds and feeds or feeds and speeds refers to two separate velocities in machine tool practice, cutting speed and feed rate. They are often ... more
Force between two nearby magnetized surfaces (relative to flux density)
The Gilbert model assumes that the magnetic forces between magnets are due to magnetic charges near the poles. This model produces good approximations that ... more
Force between two nearby magnetized surfaces
The Gilbert model assumes that the magnetic forces between magnets are due to magnetic charges near the poles. This model produces good approximations that ... more
Worksheet 324
The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from –15ºC to 40ºC . (a) What is its change in length between these temperatures? Assume that the bridge is made entirely of steel.
Strategy
Use the equation for linear thermal expansion to calculate the change in length , ΔL . Use the coefficient of linear expansion, α ,for steel from Table 13.2, and note that the change in temperature, ΔT , is 55ºC
Thermal Expansion - Linear
(b) convert the change in temperature if Kelvin and Fahrenheit degrees. **
**this section is not included in the Reference material
Celsius <-> Kelvin
Celsius <-> Fahrenheit
Discussion
Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Mean anomaly - function of gravitational parameter
In celestial mechanics, the mean anomaly is an angle used in calculating the position of a body in an elliptical orbit in the classical two-body problem. ... more
In aerodynamics, wing loading is the total weight of an aircraft divided by the area of its wing. The stalling speed of an aircraft in straight, level ... more
Menelaus' theorem ( transversal line passes inside triangle )
Menelaus’ theorem, named for Menelaus of Alexandria, is a theorem about triangles in plane geometry. Given a triangle ABC, ... more
Ceva's theorem (lines from vertices to the opposite sides of a triangle)
Ceva’s theorem is a theorem about triangles in Euclidean plane geometry. Given a triangle ABC, let the lines AO, BO and CO ... more
Force between two magnetic poles
The Gilbert model assumes that the magnetic forces between magnets are due to magnetic charges near the poles. This model produces good approximations that ... more
Epicyclic gearing (overal gear ratio)
An epicyclic gear train consists of two gears mounted so that the center of one gear revolves around the center of the other. A carrier connects the ... more
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Category | 637 | 3,016 | {"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} | 3.265625 | 3 | CC-MAIN-2020-16 | latest | en | 0.910813 |
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Viewing 3 posts - 1 through 3 (of 3 total)
• Author
Posts
• #6493
RuslanY
Participant
Hello Nazca-design Team!
Thanks for the extremely handy design tool. I am a beginner nazca user and want to understand this better for experimentation in our university laboratory. I have several questions about creating simple optical elements: What is a way there to make L1, L2 and L3 the same length?
This is a fairly simple case, but in practice it is more difficult to create interconnects. As for example in this structure:
In red rectangles, elements that are created using cells and are added multiple times. These elements must be connected by waveguides, while adjusting the length of L1 with a certain step. It would be convenient to do as shown by the dotted line. If I add each section of the waveguide manually, then I encounter problems as the direction of bend:
What else can I try in such cases?
#6510
Ronald
Keymaster
Dear Ruslan,
Good to hear Nazca works well for you.
Starting with your last image: Nazca uses outward pointing connections by definition. If you define your own block/cell, make sure you define pins pointing outwards from the circuit element. This would never lead to confusion of bend directions. Bend directions follow the standard mathematical convention of positive angles for counter clockwise rotation. Moreover, Nazca avoids quite nasty global layout states, e.g. no global cross section or global flip states, nor mixed in- and outward pin/port directions as mentioned above.
For the dotted L1 line case, you can just add an upward bend to the two blocks you connect this particular shape to. Connect the two bends with `ubend_p2p(length=L)`, where L is the length of the straight section, like in a sliding trumpet, used to extend the shape.
For length extraction in two port elements/cells you can use the nd.trace module, for geometrical path lengths. To optimize the length of a section you can use a root or minimizer solver, as discussed in this trail_cellpost. The trial cell decorator shown there will be part of the nd.bb_util module in Nazca >0.5.13. (`@nd.bb_util.trial_cell`).
If you want to trace through interconnects and blocks/cells you can resort to the pathfinder module. That would need a dedicated tutorial to explain how to use it though. It can handle geometrical, optical and electrical path information.
Ronald
#6515
RuslanY
Participant
Dear Ronald,
Thank you for your answers! The recommendations are really useful and I can succeed.
Viewing 3 posts - 1 through 3 (of 3 total)
• You must be logged in to reply to this topic. | 596 | 2,657 | {"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} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.896984 |
http://math.stackexchange.com/questions/204157/pointwise-limit-on-a-complete-metric-space | 1,469,493,697,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00273-ip-10-185-27-174.ec2.internal.warc.gz | 154,970,244 | 19,250 | # pointwise limit on a complete metric space
Let $\{f_n: X\rightarrow \mathbb{R}\}$ be a sequence of continuous real-valued functions on a complete metric space, $X$. Suppose this sequence has a pointwise limit, $f$. How easy is it to see that there must be some point of $X$ at which $f$ is continuous?
I tried to use Baire Category with my open sets the sets at which the oscillation of f is bounded above by some number decreasing to zero with te index of the sets. But I couldn't show these were dense. Am I on the right track? Thanks for reading and any help would be greatly appreciated. I know that this issue has been studied and that the proof of this fact is definitely in a number of books that I don't have access to so I would be perfectly happy with an answer that just indicates how trivial or non trivial this fact is.
-
Not on an arbitrary metric space, but this may help: mathoverflow.net/questions/32033/… – Byron Schmuland Sep 28 '12 at 19:48
The points of continuity of a pointwise limit of continuous functions (= Baire class 1 function: en.wikipedia.org/wiki/Baire_function) form a comeager $G_\delta$. See Kechris, Classical Descriptive Set Theory, Theorem 24.24, page 193: books.google.com/… – commenter Sep 28 '12 at 20:16
Thanks for the references, guys – Leray Hirsch Sep 29 '12 at 7:53
$\newcommand{\cl}{\operatorname{cl}}$Let $V$ be any non-empty open set in $X$.
Fix $\epsilon>0$. For $n,k\in\Bbb N$ let $G_n(k)=\{x\in V:|f_n(x)-f_{n+k}(x)|>\epsilon\}$, and let $G_n=\bigcup_{k\in\Bbb N}G_n(k)$; $G_n$ is open in $X$. Let $x\in V$; there is an $n(x)\in\Bbb N$ such that $|f_n(x)-f(x)|<\frac{\epsilon}2$ whenever $n\ge n(x)$, so $|f_{n(x)}(x)-f_{n(x)+k}(x)|<\epsilon$ for all $k\in\Bbb N$, and therefore $x\notin G_{n(x)}$. Thus, $\bigcap_{n\in\Bbb N}G_n=\varnothing$. But $V$ is a Baire space, so there must be some $n\in\Bbb N$ such that $G_n$ is not dense in $V$; let $U=V\setminus\cl_XG_n$.
Fix $x\in U$. Since $f_n$ is continuous, there is a $\delta>0$ such that $|f_n(y)-f_n(x)|<\epsilon$ for all $y\in B(x,\delta)$, the open ball in $X$ of radius $\delta$ centred at $x$, and we may further assume that $\cl_XB(x,\delta)\subseteq U$. Thus, for each $y\in B(x,\delta)$ we have
$$|f(y)-f(x)|\le|f(y)-f_n(y)|+|f_n(y)-f_n(x)|+|f_n(x)-f(x)|<3\epsilon\;,$$
so the oscillation of $f$ on $\operatorname{cl}_XB(x,\delta)$ is at most $3\epsilon$.
Thus, for any non-empty open $V\subseteq X$ and any $\epsilon>0$ there is a non-empty open $W$ such that $\cl_XW\subseteq V$ and the oscillation of $f$ on $W$ is at most $\epsilon$. Moreover, we can make the diameter of $W$ as small as we like. It follows that for any non-empty open $V_0\subseteq X$ there is a sequence $\langle V_n:n\in\Bbb N\rangle$ of non-empty open subsets of $X$ such that for each $n\in\Bbb N$ we have $\cl_XV_{n+1}\subseteq V_n$, the diameter of $V_{n+1}$ is less than $2^{-n}$, and the oscillation of $f$ on $V_{n+1}$ is at most $2^{-n}$. $X$ is a Baire space, so $$\{x\}=\bigcap_{n\in\Bbb N}\cl_XV_n=\bigcap_{n\in\Bbb N}V_n\subseteq V_0$$ for some $x\in X$, and clearly $f$ is continuous at $x$. Thus, the set of points of continuity of $f$ is dense in $X$.
-
+1 Nice answer! – user38268 Sep 29 '12 at 1:15
Thanks, Brian!! I guess when I said "variation" I meant oscillation. – Leray Hirsch Sep 29 '12 at 7:51
@Leray: You’re welcome! – Brian M. Scott Sep 29 '12 at 7:52
@BrianM.Scott nice answer! but I fail to see why after picking $n$ (so that $G_n$ is not dense) we might assume that $|f_n(x)-f(x)| < \epsilon$ for $x\in U$. From the definition, I only see that there is a $k$ such that $|f_n(x)-f_k(x)| < \epsilon$... – Yul Otani Aug 24 '13 at 17:36
@Yul: No, you get $|f_n(x)-f_{n+k}(x)|\le\epsilon$ for all $k\in\Bbb N$, and hence in the limit $|f_n(x)-f(x)|\le\epsilon$ as well. (The strictness of the inequality at the end of the displayed line comes from the middle term.) – Brian M. Scott Aug 24 '13 at 19:05 | 1,346 | 3,929 | {"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} | 3.515625 | 4 | CC-MAIN-2016-30 | latest | en | 0.841148 |
https://laszukdawid.com/2016/09/ | 1,596,875,685,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737319.74/warc/CC-MAIN-20200808080642-20200808110642-00555.warc.gz | 393,968,119 | 41,987 | This blog needs some lightening up and what a better way to do that than post some graphs? Exactly! I like to collect data on various things and then make a graph of them. Things are much better when presented with axis and some number around it.
Not so long time ago I had birthday. Although it happens rather regularly, the last one was special. For the first time I’ve enabled notification on Facebook. It appears that people (a.k.a. friends) are rather nice and are willing to write something positive about you on that day. Here is to memory of that special day!
Summary:
I’ve scraped Facebook’s posts and messages which were within 48 h from the Birthday 00:01 AM. Sent times are aggregated on the graph and text content is summarised in word stats.
Green graph displays wishes density as a function of time obtained using Gaussian Kernel Distribution Estimation. Blue dots show cumulative wish count. Both are normalised to highest value being 1. Statistically speaking, green and blue shows (up to constants) probability density function and cumulative density function respectively.
What is nice about this graph is that it generally shows at what time of the day my peers are active. It seems that majority is active in the morning 10-11 am and after 8 pm, which is reassuring that they are very normal average people.
Additional insight gives extracted content. Although there’s not enough data, I think it nicely hints on my background. I’ll leave interpretation, but will also point up that the percentage of wish to total falls closely to the ratio of people who read posts, i.e. 12 – 16 % (even though this is not related as birthday notification goes to everyone).
Wishes:
Happy Birthday: 34 (38.20%)
Wszystkiego najlepszego/All the best: 11 (12.36%)
(tylko) Najlepszego/(only) The best: 10 (11.24%)
Sto lat/100+ years: 16 (17.98%)
Emoji/Emoticons:
Smileys: 25 (28.09%)
Kisses: 13 (14.61%)
Hearths: 3 (3.37%)
Reference:
Dawid: 20 (22.47%)
You: 17 (19.1%)
Boy: 2 (2.23%)
Total number of exclamation mark (!):
!: 82
!!: 11
!!!: 4
!!!!: 2
!!!!!: 1
# Take it easy
tl;dr: I’m going less quality, more quantity.
I like to write. I write lots, but usually don’t publish it. Whenever I think about something and write it down, I then rethink what I thought and want to rewrite what I wrote. It takes me ages to write something very precise and something that I wouldn’t be immediately ashamed of. This blog was meant to be the place for that content, for things I wouldn’t quickly regret. And, although, I’m rather OK with the content, I’m deeply disappointed with the frequency. Thus: change.
This blog initially was meant to be my academical window. Something that when people look me up (for whatever reason), they’d see something over which I would have control. However, my view on World has yet again changed. Despite my passion for research and likeness for academia free-thinking diverse environment, I’ve decided to leave it. There is so much great research being done in industry and private sector that I think I’m shooting myself in foot by sticking only to that audience.
Plan for this blog is changing to present cool things that happen and fun stuff I’m working on. I enjoy working and I like to write. If the content is not up to standard quality, shame. But not having a content at all is just disgraceful. How can I show people what I’m working on, if I don’t show them anything at all.
In Python’s mantra: It’s easier to ask for forgiveness than for permission.
# On the Phase Coupling of Two Components Mixing in Empirical Mode Decomposition
Another of my papers [1] (see About) has been published recently. Unfortunately, I cannot present it here in a full scope. Below I’m presenting an abstract from the paper and invite interested people in contacting me. This topic is really interesting and there is plenty to be done.
On the Phase Coupling of Two Components Mixing in Empirical Mode Decomposition
Abstract:
This paper investigates frequency mixing effect of Empirical Mode Decomposition (EMD) and explores whether it can be explained by simple phase coupling between components of the input signal. The input is assumed to be a linear combination of harmonic oscillators. The hypothesis was tested assuming that phases of input signals’ components would couple according to Kuramoto’s model. Using a Kuramoto’s model with as many oscillators as the number of intrinsic mode functions (result of EMD), the model’s parameters were adjusted by a particle swarm optimisation (PSO) method. The results show that our hypothesis is plausible, however, a different coupling mechanism than the simple sine-coupling Kuramoto’s model are likely to give better results.
In a very, very brief, but figure-enhanced summary: we present that for certain range of frequencies ratio a mix in frequency appears which can be explained as a difference of original frequencies. Figure below presents Fourier transform F of correlation function between each pair of IMFs for different values of frequency f. All values were normalised, such that for given f the maximum is equal to one. Additionally, on the top projection of the figure two lines have been drawn – F1 = 13-f (dashed line) and F2 = 2(13 – f) (dash-dotted line).
[1] D. Laszuk, O. J. Cadenas, and S. J. Nasuto (July, 2016) On the Phase Coupling of Two Components Mixing in Empirical Mode Decomposition, Advances in Data Science and Adaptive Analysis, vol. 8, no. 1. [Link] | 1,267 | 5,471 | {"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-2020-34 | latest | en | 0.93554 |
http://nedrilad.com/Tutorial/topic-24/Virtual-Reality-Concepts-and-Technologies-409.html | 1,505,970,938,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687642.30/warc/CC-MAIN-20170921044627-20170921064627-00660.warc.gz | 250,256,125 | 5,482 | Game Development Reference
In-Depth Information
Certain applications do not make do with a Boolean response and require finding
the accurate minimum distance between a pair of polyhedrons. This value is also the
distance between the origin and the Minkowski difference of two polytopes, defined
as: A
.If A and B are convex, even this polyhedron is shown
as convex. The GJK method (Gilbert et al., 1988) aims at determining the distance
between A
B
={
x
y ;
x
A ,
y
B
}
B . This
simplex is a point, a segment, a triangle or a tetrahedron formed by the vertices of
A
B and the origin by iteration, without explicitly calculating A
B .
In practice, as the Minkowski difference is not constructed, every vertex of A
B
is not known explicitly, but is systematically defined as the difference between a vertex
of A and a vertex of B . The shortest distance between simplex S i and the origin is
calculated at step i. Let V i be the point of this simplex, closest to the origin. The
simplex is then reduced by selecting the surface, edge or vertex of the simplex on
which the V i is located. Or S i , the minimum simplex found. We then find the support
point w i of A
B (in fact, the difference of the current supporting points of A and B as
per the direction and its opposite respectively). Point w i is then added to the simplex,
and we reiterate the algorithm on the convex envelop of the union of simplex S i and
point w i . Point V i and simplex S i are found in a single step: It is enough to take the
smallest sub-set { w j } of vertices of S i such as:
λ j ×
V i =
w j
and λ j > 0,
j
where V i is the closest point between the origin and the convex envelop of this set
{ w j }. This set is determined by doing an exhaustive search, starting with the set of one,
then two and then three vertices of S i . It is sufficient to choose one vertex of A
Bas
a simplex at the beginning of the algorithm. The algorithm ends when no new vertex
close to the origin can be added to the simplex. The minimum distance is then the
distance between this simplex and the origin. Figure 17.2 illustrates some calculation
steps of the GJK algorithm. This algorithm has had several upgrades, mainly the ISA-
GJK, making it better for numerical calculations (Bergen, 1999). There are some other
approaches to calculate the distance between polytopes, but we will not describe them
here as these algorithms require pre-processing the objects before they can be applied:
The DK method (Dobkin & Kirkpatrick, 1990) consists of constructing a series of
increasingly rough approximations of the polytopes in advance.
Then the distance between the rough approximations is calculated and then ascer-
tained more and more accurately by refining the approximation of polytopes. The LC
method (Lin & Canny, 1991), which was later called Voronoi marching, is based on
a section of the space surrounding each polytope in Voronoi regions. i.e. two points
selected respectively on the two polytopes; these two points are on a support element
(vertex, edge or face) of the polytope. These two points are separated by a minimum
distance, if they are included respectively in the Voronoi region of the support of the
other point. This criterion makes it possible to find the solution points and determine
the distance between the polytopes, only by searching the minimum distance.
The advantage of being able to determine the distance between two polytopes is
being able to anticipate their approach and ensure that this proximity is not converted
sooner or later into interpenetration. A method avoiding distance calculation involves
“enlarging'' all the objects, i.e. creating a safety zone around them and using this zone | 864 | 3,688 | {"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-2017-39 | latest | en | 0.934386 |
https://camnangtienganh.vn/us/19-how-many-hours-is-9-to-6pm-hienthithang-hienthinam/ | 1,702,156,292,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00656.warc.gz | 170,456,739 | 52,183 | # 19 How Many Hours Is 9 To 6pm 12/2023
Below is the best information and knowledge about how many hours is 9 to 6pm compiled and compiled by the Cẩm Nang Tiếng Anh team, along with other related topics such as: how many hours is 9 to 4, how many hours is 9 to 7, how many hours is 9 to 5, how many hours is 10am to 6pm, how many hours is 8am to 6pm, 9 to 3 is how many hours, 7am to 6pm is how many hours, 11am to 6pm is how many hours
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The most popular articles about how many hours is 9 to 6pm
## 1. How Many Hours Is 9am To 6pm? – DateDateGo
How Many Hours Is 9am To 6pm? – DateDateGo How many hours from 9am to 6pm? There are 9 hours from 9am to 6pm. You May Also Want To Calculate. How Many Hours Is 1pm To 9am?
How many hours from 9am to 6pm? There are 9 hours from 9am to 6pm.
## 2. 9am to 6pm is how many hours? – Hours Calculator – iamrohit.in
9am to 6pm is how many hours? – Hours Calculator – iamrohit.in 9 hours · 540 minutes.
This simple online utility tool makes it easy to calculate the difference in hours and minutes between two given times. To calculate the hours and minutes contained in a time period you need to know its beginning and end time.
## 3. 9am to 6pm is how many hours – calculator.name
9am to 6pm is how many hours – calculator.name How many hours between 9am and 6pm? The time from 9am to 6pm is 9 hours. Time duration calculator is to find out how many hours are there from 9 am (october …
How many hours between 9am and 6pm? The time from 9am to 6pm is 9 hours
## 4. How many hours is 9am to 6pm? – Answers
How many hours is 9am to 6pm? – Answers 9 hours. User Avatar · Wiki User. ∙ 2013-05-09 17:54:02. This answer is: 👍 Helpful (0) 👎 Not Helpful (0). Add a Comment. User Avatar.
Q: How many hours is 9am to 6pm?
## 5. Well, your blog post says 9am-6pm, which is 9 hours, lunch …
Well, your blog post says 9am-6pm, which is 9 hours, lunch … Well, your blog post says 9am-6pm, which is 9 hours, lunch included. But I wouldn’t expect anything less if you’re going to do 4 days per week. I’d much …
Well, your blog post says 9am-6pm, which is 9 hours, lunch included. But I wouldn’t expect anything less if you’re going to do 4 days per week. I’d much rather work five days a week, 11am-5pm or 12pm-6pm, without a lunch break. Can you really put off your personal life every day from 9-6? I think th…
## 6. How many hours between two days – Online calculator
How many hours between two days – Online calculator This calculator helps you calculate how many hours between two days, for example, between Monday 8 a.m and Wednesday, 6 p.m..
If you enter the day and time for each of the two time points, it will show you how many hours (and minutes, if you wish) between these. In my example, there will be 58 hours from Monday 8 a.m to Wednesday 6 p.m. It also shows the opposite for convenience – how many hours from Wednesday 6 p.m to nex…
## 7. Top 6 11Pm To 6Pm Is How Many Hours 27655 People Liked …
Top 6 11Pm To 6Pm Is How Many Hours 27655 People Liked … How many hours is it from 7am to 3pm? How many hours sleep will I get tonight? How many hours is 9/5 in a day? How do I calculate my hours? Why …
Message Here
## 8. Hours Calculator Between Two Times – DQYDJ
Hours Calculator Between Two Times – DQYDJ Hours Calculator: See How Many Hours are Between Two Times. Time. Written by: PK. Below is an hours calculator. Use the easy-to-enter inputs and enter a …
Below is an hours calculator. Use the easy-to-enter inputs and enter a starting time and an ending time, then hit the ‘Calculate’ button. We’ll add up the number of hours (up to 23 hours and 59 minutes maximum).Hours Calculator Between Two TimesUsing the Hours CalculatorTo use the tool to find the h…
## 9. Half Day Leave – HRSINGAPORE
Half Day Leave – HRSINGAPORE Hi HR practitioners,. Our half day working hours is 9 am to 1 pm and 1 pm to 6 pm. Is this common? An employee highlighted that she did not …
Our working hours are 9 am to 6 pm, with one hour lunch. Hence actual working hours are 8 hours per day. Morning leave is between 9 am and 1 pm (4 hours) and afternoon leave is between 2 pm and 6 pm (4 hours).
## 10. How many hours, minutes and seconds between two times
How many hours, minutes and seconds between two times Time Duration Calculator – How many hours, minutes and seconds between two times – Clock math Calculator – online calculators.
Clock Math Calculator
From a selected clock time, add or subtract hours, minutes and seconds
Enter clock time information at “From:”
Enter hours and minutes. Select am or pm.
Enter the number of hours, minutes and seconds you wish to add to or subtract from the clock time.
## 11. How many work hours is 9am to 6pm – 44Bars.com
How many work hours is 9am to 6pm – 44Bars.com If you’re also wondering “how many hours am I working?” and you are on a 9AM to 5PM job， then the question is how many hours is 9 to 5.…How many hours?
Convert both times to 24 hour format， adding 12 to any pm hours. 8:55am becomes 8:55 hours (start time) …If the start minutes are greater than the end minutes… …Subtract end time minutes from start time minutes… …Subtract the hours… …Put(not add) the hours and minutes together – 6:45 (6 hours and 45…
## 12. Hours Calculator
Hours Calculator This hours calculator computes the number of hours and minutes between two times. A full version can calculate the hours between two times on different …
A 24-hour clock typically uses the numbers 0-23, where 00:00 indicates midnight, and a day runs from midnight to midnight over the course of 24 hours. This time format is an international standard, and is often used to avoid the ambiguity resulting from the use of a 12-hour clock. The hours from 0-1…
## 13. What Is The Ideal Number Of Working Hours Per Day To Be …
What Is The Ideal Number Of Working Hours Per Day To Be … 8 hours? 10 hours? 6 hours? How long should we work every day to respect our … When working from 9 a.m. to 6 p.m. in an office, with a break from 12 to 1 …
Because of the accumulation of cerebral fatigue, working all this time is connected to lower results in both cognitive and intellectual tests. Being in the office for more than 8 hours a day is associated with poorer overall health an with a 40% higher risk of developing heart disease or stress…
## 14. How many hours is 9AM to 6PM? – Newsbasis.com
How many hours is 9AM to 6PM? – Newsbasis.com The traditional American business hours are 9:00 a.m. to 5:00 p.m., Monday to Friday, representing a workweek of five eight-hour days comprising 40 hours in …
and you are on a 9AM to 5PM job, then the question is how many hours is 9 to 5. The answer is exactly eight hours….How many hours?
## 15. 9 hours before 6pm – Calendar Maniacs
9 hours before 6pm – Calendar Maniacs Here we will show the exact answer to 9 hours before 6pm. In other words, what time was it 9 hours before 6pm?
We of course took into account that there are twenty-four hours in a day, which include twelve hours in the am
and twelve hours in the pm. Below is the answer to what time it was 9 hours before 6pm.
## 16. Quick Answer: How Many Hours Is It From 9 30am To 4pm?
Quick Answer: How Many Hours Is It From 9 30am To 4pm? and you are on a 9AM to 5PM job, then the question is how many hours is 9 to 5. The answer is exactly eight hours. … There are 9 hours from 9am to 6pm.
45 minutes is 45 minutes * (1 hour / 60 minutes) = 45/60 hours = 0.75 hours. 45 seconds is 45 seconds * (1 hour / 3600 seconds) = 45/3600 hours = 0.0125 hours. Adding them all together we have 2 hours + 0.75 hours + 0.0125 hours = 2.7625 hours.
## 17. How Many Hours Since Tuesday At 6pm? – DateTimeGo
How Many Hours Since Tuesday At 6pm? – DateTimeGo Find out how many hours since 6pm Tuesday. Calculate how many hours has it been since Tuesday at 6pm.
There are 4 hours and 39 minutes from Sunday, October 2, 10:39 PM to Tuesday, September 27, 6:00 PM.
## 18. How many hours a week do you normally work? – IndiaBIX
How many hours a week do you normally work? – IndiaBIX This is the HR interview questions and answers on “How many hours a week do you … Post your comments +9 -1 … But normally, my duty was 10am to 6pm.
Time is important for me, but when my work comes I look more towards finishing my work efficiently and completely rather looking to how many hours I worked and then it’s okay even if it takes more time than usual, but there’ll be the true satisfaction of completing the whole work successfully.
## 19. Differences in Average Working Hours Around the World
Differences in Average Working Hours Around the World How many hours should you expect to work in a graduate career? … remains the traditional “9 to 5” – but that it’s increasingly common to work beyond this.
According to official statistics on working hours in Canada, employed Canadians worked an average of 36.6 hours per week in 2012, with significant variation depending on age, gender and location. For example, in Quebec the average working hours were 35.4 hours, compared to 39 hours in Alberta – a cl…
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Synthetic: Cẩm Nang Tiếng Anh US | 2,545 | 9,208 | {"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-2023-50 | latest | en | 0.955788 |
https://overlordoftheuberferal.com/tag/sine/ | 1,591,293,786,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347445880.79/warc/CC-MAIN-20200604161214-20200604191214-00380.warc.gz | 464,338,279 | 16,957 | # White Rites
The blancmange curve is an interesting fractal formed by summing a series of zigzags. It
takes its name from its resemblance to the milk-pudding known as a blancmange
(blanc-manger in French, meaning “white eating”):
Blancmange curve
In successive zigzags, the number of zags doubles as their height halves, i.e. z(i) = z(i-1) * 2, h(i) = h(i-1) / 2. If all the zigzags are represented at once, the construction looks like this:
Zigzags 1 to 10
Zigzags 1 to 10 (animated)
Here is a step-by-step construction, with the total sum of zigzags in white, the present zigzag in red and the previous zigzag in green:
Blancmange curve stage 1
Stage 2
Stage 3
Stage 4
Stage 5
Stage 6
Stage 7
Stage 8
Stage 9
Stage 10
Blancmange curve (animated)
It’s easy to think of variants on the standard blancmange curve. Suppose the number of zags triples as their height is divided by three, i.e. z(i) = z(i-1) * 3, h(i) = h(i-1) / 3:
Blancmange curve for z(i) = z(i-1) * 3, h(i) = h(i-1) / 3
Continue reading “White Rites”… | 337 | 1,038 | {"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-2020-24 | latest | en | 0.931864 |
http://www.slader.com/textbook/9781285057095-calculus-10th-edition/ | 1,542,311,029,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742906.49/warc/CC-MAIN-20181115182450-20181115204450-00449.warc.gz | 505,785,745 | 21,609 | # Calculus, 10th Edition Calculus, 10th Edition
ISBN: 9781285057095 / 1285057090
### Chapter P
Preparation For Calculus
P.1 Graphs and Models Exercises p.8 P.2 Linear Models and Rates of Change Exercises p.16 P.3 Functions and Their Graphs Exercises p.27 P.4 Fitting Models to Data Exercises p.34 Review Exercises p.37 Problem Solving p.39
### Chapter 1
Limits And Their Properties
1.1 A Preview of Calculus Exercises p.47 1.2 Finding Limits Graphically and Numerically Exercises p.55 1.3 Evaluating Limits Analytically Exercises p.67 1.4 Continuity and One-Sided Limits Exercises p.79 1.5 Infinite Limits Exercises p.88 Review Exercises p.91 Problem Solving p.93
### Chapter 2
Differentiation
2.1 The Derivative and the Tangent Line Problem Exercises p.103 2.2 Basic Differentiation Rules and Rates of Change Exercises p.114 2.3 Product and Quotient Rules and Higher-Order Derivatives Exercises p.125 2.4 The Chain Rule Exercises p.136 2.5 Implicit Differentiation Exercises p.145 2.6 Related Rates Exercises p.153 Review Exercises p.157 Problem Solving p.159
### Chapter 3
Applications Of Differentiation
3.1 Extrema on an Interval Exercises p.167 3.2 Rolle's Theorem and the Mean Value Theorem Exercises p.174 3.3 Increasing and Decreasing Functions and the First Derivative Test Exercises p.183 3.4 Concavity and the Second Derivative Test Exercises p.192 3.5 Limits at Infinity Exercises p.202 3.6 A Summary of Curve Sketching Exercises p.212 3.7 Optimization Problems Exercises p.220 3.8 Newton's Method Exercises p.229 3.9 Differentials Exercises p.236 Review Exercises p.238 Problem Solving p.241
### Chapter 4
Integration
4.1 Antiderivatives and Indefinite Integrals Exercises p.251 4.2 Area Exercises p.263 4.3 Riemann Sums and Definite Integrals Exercises p.273 4.4 The Fundamental Theorem of Calculus Exercises p.288 4.5 Integration by Substitution Exercises p.301 4.6 Numerical Integration Exercises p.310 Review Exercises p.312 Problem Solving p.315
### Chapter 5
Logarithmic, Exponential, And Other Transcendental Functions
5.1 The Natural Logarithmic Function: Differentiation Exercises p.325 5.2 The Natural Logarithmic Function: Integration Exercises p.334 5.3 Inverse Functions Exercises p.343 5.4 Exponential Functions: Differentiation and Integration Exercises p.352 5.5 Bases Other than e and Applications Exercises p.362 5.6 Inverse Trigonometric Functions: Differentiation Exercises p.372 5.7 Inverse Trigonometric Functions: Integration Exercises p.380 5.8 Hyperbolic Functions Exercises p.390 Review Exercises p.393 Problem Solving p.395
### Chapter 6
Differential Equations
6.1 Slope Fields and Euler's Method Exercises p.403 6.2 Differential Equations: Growth and Decay Exercises p.412 6.3 Separation of Variables and the Logistic Equation Exercises p.421 6.4 First-Order Linear Differential Equations Exercises p.428 Review Exercises p.431 Problem Solving p.433
### Chapter 7
Applications Of Integration
7.1 Area of a Region Between Two Curves Exercises p.442 7.2 Volume: The Disk Method Exercises p.453 7.3 Volume: The Shell Method Exercises p.462 7.4 Arc Length and Surfaces of Revolution Exercises p.473 7.5 Work Exercises p.483 7.6 Moments, Centers of Mass, and Centroids Exercises p.494 7.7 Fluid Pressure and Fluid Force Exercises p.501 Review Exercises p.503 Problem Solving p.505
### Chapter 8
Integration Techniques, L'Hopital's Rule, And Improper Integrals
8.1 Basic Integration Rules Exercises p.512 8.2 Integration by Parts Exercises p.521 8.3 Trigonometric Integrals Exercises p.530 8.4 Trigonometric Substitution Exercises p.539 8.5 Partial Fractions Exercises p.549 8.6 Integration by Tables and Other Integration Techniques Exercises p.555 8.7 Indeterminate Forms and L'Hopital's Rule Exercises p.564 8.8 Improper Integrals Exercises p.575 Review Exercises p.579 Problem Solving p.581
### Chapter 9
Infinite Series
9.1 Sequences Exercises p.592 9.2 Series and Convergence Exercises p.601 9.3 The Integral Test and p-Series Exercises p.609 9.4 Comparisons of Series Exercises p.616 9.5 Alternating Series Exercises p.625 9.6 The Ratio and Root Tests Exercises p.633 9.7 Taylor Polynomials and Approximations Exercises p.644 9.8 Power Series Exercises p.654 9.9 Representation of Functions by Power Series Exercises p.662 9.10 Taylor and Maclaurin Series Exercises p.673 Review Exercises p.676 Problem Solving p.679
### Chapter 10
Conics, Parametric Equations, And Polar Coordinates
10.1 Conics and Calculus Exercises p.692 10.2 Plane Curves and Parametric Equations Exercises p.703 10.3 Parametric Equations and Calculus Exercises p.711 10.4 Polar Coordinates and Polar Graphs Exercises p.722 10.5 Area and Arc Length in Polar Coordinates Exercises p.731 10.6 Polar Equations of Conics and Kepler's Laws Exercises p.739 Review Exercises p.742 Problem Solving p.745
### Chapter 11
Vectors And The Geometry Of Space
11.1 Vectors in the Plane Exercises p.755 11.2 Space Coordinates and Vectors in Space Exercises p.763 11.3 The Dot Product of Two Vectors Exercises p.773 11.4 The Cross Product of Two Vectors in Space Exercises p.781 11.5 Lines and Planes in Space Exercises p.790 11.6 Surface in Space Exercises p.802 11.7 Cylindrical and Spherical Coordiantes Exercises p.809 Review Exercises p.811 Problem Solving p.813
### Chapter 12
Vector-Valued Functions
12.1 Vector-Valued Functions Exercises p.821 12.2 Differentiation and Integration of Vector-Valued Functions Exercises p.830 12.3 Velocity and Acceleration Exercises p.838 12.4 Tangent Vectors and Normal Vectors Exercises p.848 12.5 Arc Length and Curvature Exercises p.860 Review Exercises p.863 Problem Solving p.865
### Chapter 13
Functions Of Several Variables
13.1 Introduction to Functions of Several Variables Exercises p.876 13.2 Limits and Continuity Exercises p.887 13.3 Partial Derivatives Exercises p.896 13.4 Differentials Exercises p.905 13.5 Chain Rules for Functions of Several Variables Exercises p.913 13.6 Directional Derivatives and Gradients Exercises p.924 13.7 Tangent Planes and Normal Lines Exercises p.933 13.8 Extrema of Functions of Two Variables Exercises p.942 13.9 Applications of Extrema Exercises p.949 13.10 Lagrange Multipliers Exercises p.958 Review Exercises p.960 Problem Solving p.963
### Chapter 14
Multiple Integration
14.1 Iterated Integrals and Area in the Plane Exercises p.972 14.2 Double Integrals and Volume Exercises p.983 14.3 Change of Variables: Polar Coordinates Exercises p.991 14.4 Center of Mass and Moments of Inertia Exercises p.1000 14.5 Surface Area Exercises p.1007 14.6 Triple Integrals and Applications Exercises p.1017 14.7 Triple Integrals in Other Coordinates Exercises p.1025 14.8 Change of Variables: Jacobians Exercises p.1032 Review Exercises p.1034 Problem Solving p.1037
### Chapter 15
Vector Analysis
15.1 Vector Fields Exercises p.1049 15.2 Line Integrals Exercises p.1061 15.3 Conservative Vector Fields and Independence of Path Exercises p.1072 15.4 Green's Theorem Exercises p.1081 15.5 Parametric Surfaces Exercises p.1091 15.6 Surface Integrals Exercises p.1104 15.7 Divergence Theorem Exercises p.1112 15.8 Stoke's Theorem Exercises p.1119 Review Exercises p.1120 Problem Solving p.1123
##### Can you find your fundamental truth using Slader as a completely free Calculus solutions manual?
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https://pt.coursera.org/learn/computer-vision-basics/reviews | 1,669,754,650,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710711.7/warc/CC-MAIN-20221129200438-20221129230438-00228.warc.gz | 484,883,907 | 99,531 | Voltar para Computer Vision Basics
## Comentários e feedback de alunos de Computer Vision Basics da instituição Universidade de Buffalo
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## Sobre o curso
By the end of this course, learners will understand what computer vision is, as well as its mission of making computers see and interpret the world as humans do, by learning core concepts of the field and receiving an introduction to human vision capabilities. They are equipped to identify some key application areas of computer vision and understand the digital imaging process. The course covers crucial elements that enable computer vision: digital signal processing, neuroscience and artificial intelligence. Topics include color, light and image formation; early, mid- and high-level vision; and mathematics essential for computer vision. Learners will be able to apply mathematical techniques to complete computer vision tasks. This course is ideal for anyone curious about or interested in exploring the concepts of computer vision. It is also useful for those who desire a refresher course in mathematical concepts of computer vision. Learners should have basic programming skills and experience (understanding of for loops, if/else statements), specifically in MATLAB (Mathworks provides the basics here: https://www.mathworks.com/learn/tutorials/matlab-onramp.html). Learners should also be familiar with the following: basic linear algebra (matrix vector operations and notation), 3D co-ordinate systems and transformations, basic calculus (derivatives and integration) and basic probability (random variables). Material includes online lectures, videos, demos, hands-on exercises, project work, readings and discussions. Learners gain experience writing computer vision programs through online labs using MATLAB* and supporting toolboxes. * A free license to install MATLAB for the duration of the course is available from MathWorks....
## Melhores avaliações
AO
28 de abr de 2020
Lays a good foundation for Computer Vision. There should be more programming examples as some of the labs were beyond the scope of what was taught in the videos, especially the last one.
ZA
12 de nov de 2022
My Message is going to anyone who demand to up his skill in Digital World, You Ought to Enroll this course..
Also if You are a Beginner for Computer Vision world it will Make Like AI
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## 1 — 25 de 499 Avaliações para o Computer Vision Basics
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28 de abr de 2020 | 918 | 3,349 | {"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} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.862864 |
https://johnmayhk.wordpress.com/2008/04/10/%e4%bb%a3%e5%a0%82%e6%b1%82%e5%85%b6%e8%ac%9b/ | 1,481,162,928,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542323.80/warc/CC-MAIN-20161202170902-00063-ip-10-31-129-80.ec2.internal.warc.gz | 869,739,227 | 23,382 | # Quod Erat Demonstrandum
## 2008/04/10
### 代堂求其講
Filed under: Junior Form Mathematics — johnmayhk @ 7:31 下午
1. 表明自己不懂足球，球技差劣。再引入足球由古至今的設計。
2. 指出其中一種設計是以若干正五邊形和正六邊形組成。
3. 故弄玄虛地順帶提及 1985 年發現的 $C_{60}$ 結構（誤導學生以為阿 sir 連 Chem 都識）。
4. 問學生如何找出足球面上五邊形及六邊形的數目。
5. 煞有介事地指出他們的數學書上有條式可以解決這問題。
6. 重提數學書中的歐拉公式：$V + F = E + 2$
7. 完成上面 6 點吹水後，入戲肉：（一）證明歐拉公式*（二）利用歐拉公式，證明足球面上有 12 個正五邊形及 20 個正六邊形，順道驗證 $C_{60}$ 為何是 $60$
8. 完成 7，看看時間已差不多，吹水地談歐拉公式和地圖五色問題有關，又順帶吹一吹地圖四色問題。
*『證明歐拉公式』，第一步是立體平面化，第二步是三角形化，第三步是去三角形化，想法來自《從 1 到無窮大》一書，但相信提及的是個頗粗疏的證法，不過對我和中三的同學應該可以接受的。
## 11 則迴響 »
1. 小弟甘拜下風，John Sir果然是前輩。
我的代堂話題暫時只有
1. 三次數學危機
2. 楊輝三角
3. 歐氏幾何與非歐幾何淺談
4. 少量數學趣題
「歐拉公式」的證明前幾年還記得(因中二學生要做有關「五個正多面體」的Project)，現在只記得第一步，不過前輩提起，小弟立刻去重溫一下。
迴響 由 Kam — 2008/04/10 @ 9:53 下午 | 回覆
2. I wonder why Euler’s formula is stated while the elementary Euler’s theorem (number theoretic) is never mentioned in the text book? The proof is even simpler.
迴響 由 koopa — 2008/04/11 @ 8:09 上午 | 回覆
Dunno if you can answer me something. My Internet explorer can’t visit all of the sites that includes password, nor can it perform a pop-up thing… I checked, I allow the pop-up already, and the security is low… dunno why.
I’m forced to use firefox
迴響 由 Opmux — 2008/04/11 @ 5:11 下午 | 回覆
4. To Kam sir
Good, the topics suggested sound interesting. However, I really have no idea on the delivery of topic like non-Euclidean geometry, will it be too advanced to lower form students?
To koopa
According to the curriculum of HK primary school (2000 onwards), the topic of prime number is not required (it is a suggested enrichment topic in P.4), and I was told in a meeting with some representatives from Curriculum Development Institute that it is likely that the topic of prime number will be removed. Hence, it is difficult to imagine that the elementary Euler’s theorem (number theoretic) will be introduced in the future primary or secondary mathematics curricula. Easy or difficult? It depends. Of course, for mathematics competition players or mathematics lovers, they are just a piece of cake.
To Edmund
迴響 由 johnmayhk — 2008/04/11 @ 5:55 下午 | 回覆
5. John sir…
之前上堂見過的borane, 結構也很"有趣"
http://en.wikipedia.org/wiki/Borane
那些closo, nido很考腦筋
有空看看 ^^
迴響 由 小R — 2008/04/12 @ 7:22 下午 | 回覆
6. @John Sir
有關「歐氏幾何和非歐幾何淺談」，對於初中學生，我只會介紹一下歐氏的公理、公設系統，然後以平行線公理來介紹出非歐幾何。當中可以讓學生認識到世事，包括數學知識上面有極多可能性。而且可以使他們思考到凡事都要三思，用心就能夠看到事情的真假。(又走到德育和哲學課)
再者，由非歐幾何中，又可以介紹一下相對論的宇宙觀，學生最愛聽這些科幻題材。
迴響 由 Kam — 2008/04/13 @ 12:53 上午 | 回覆
7. To 小R
非常好的資料！始終覺得，如果找到其他領域可以運用純數學（係認真地運用，不是用來做幾個統計圖嚟『做樣』），攪純數便多一層意義。人覺得數學冇用，好可能佢地學的數學太少。
To Kam sir
「非歐幾何」我所知不多，經你一提，我也想可好好學習了。謝謝！
迴響 由 johnmayhk — 2008/04/13 @ 3:42 下午 | 回覆
8. 我也很想做學生，等你來代課。因為題目看來很有趣，kam的講題也有趣。
另，我也想知一些中國數學，例如那個幾何的機械化證明（吳法），不知我這個小學層次的數學白痴會否聽得懂，根據過去經驗，純數字講解很快發夢，如果是牽涉人事就應該可以的。
迴響 由 tseyanyan — 2008/10/15 @ 11:33 上午 | 回覆
9. 茵茵老師，待我有時間，一定會介紹一下，等我。
迴響 由 johnmayhk — 2008/10/16 @ 11:18 上午 | 回覆
10. @John sir
sorry~現在才看到回覆…
謝謝欣賞boranes
懶R覺得, 從這些結構去思考圖形, 比起靠公式之類易理解
本身這些原子的排法是照各自排斥較少的方式
e.g. tetrahedron (e.g. CH4), bond angle可以從John sir教過的AM
笨R相信, 有效運用數學, 首先要理解數學怎樣表現現實
很多人擅於運用數學工具, 但不明白當中意義, 也不會解說
最近在論壇看到很多人亂用數學, 很灰 orz
雖然笨R自己quantitative的方式不太行…
迴響 由 小R — 2008/10/18 @ 8:14 下午 | 回覆 | 1,664 | 3,181 | {"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": 4, "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-2016-50 | latest | en | 0.486769 |
https://www.roelpeters.be/logarithmic-scale-in-r-with-ggplot2/ | 1,719,320,256,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865972.21/warc/CC-MAIN-20240625104040-20240625134040-00095.warc.gz | 855,348,332 | 21,096 | Home ยป Logarithmic scale in R with ggplot2
Logarithmic scale in R with ggplot2
2020: the year that every data scientist became a virologist. For the past weeks we’ve been numbed with statistics and plots about the coronavirus. A recurring feature of these plots is that they often have a logarithmic axes. Here’s how to achieve this in R’s ggplot2.
Lets’s say you are trying to plot daily new COVID-19 cases in the United States using ggplot2 (until 2020-03-28). That would look like this.
ggplot(usa,aes(x = DATE, y = NEW_CASES)) +
geom_line() +
geom_point() +
t +
ylab('DAILY NEW CASES (LINEAR SCALE)')
However, epidemics tend to grow exponentially. Given this property, it often makes sense to plot this on a logarithmic scale, and not on a linear one. In ggplot2, we can do this fairly easy using one of the following functions.
Of course, you can do exactly the same for the x axis. Furthermore, the library also allows for logarithmic tick marks using annotation_logticks().
ggplot(usa,aes(x = DATE, y = NEW_CASES)) +
geom_line() +
geom_point() +
t +
ylab('DAILY NEW CASES (LINEAR SCALE)') +
scale_y_log10() +
annotation_logticks(sides = 'l')
By the way, if you’re having trouble understanding some of the code and concepts, I can highly recommend “An Introduction to Statistical Learning: with Applications in R”, which is the must-have data science bible. If you simply need an introduction into R, and less into the Data Science part, I can absolutely recommend this book by Richard Cotton. Hope it helps!
Wanna know how I made these charts so crisp in ggplot2? Have a look at this blog post.
Say thanks, ask questions or give feedback
Technologies get updated, syntax changes and honestly… I make mistakes too. If something is incorrect, incomplete or doesn’t work, let me know in the comments below and help thousands of visitors. | 450 | 1,853 | {"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} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.862311 |
http://medical-dictionary.thefreedictionary.com/median+strips | 1,503,100,987,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105195.16/warc/CC-MAIN-20170818233221-20170819013221-00632.warc.gz | 281,680,498 | 13,827 | # median
(redirected from median strips)
Also found in: Dictionary, Thesaurus, Legal, Financial, Encyclopedia.
## median
[me´de-an]
1. situated in the median plane or in the midline of a body or structure.
2. any value that divides the probability distribution of a random variable in half; that is, the probability of observing a value above the median and the probability of observing a value below the median are both less than or equal to one-half. For a finite population or sample, the median is the middle value of an odd number of values (arranged in ascending value) or any value between the two middle values of an even number of values; in the latter case it is conventional to use the average of the two middle values.
## me·di·an
(mē'dē-ăn),
1. middle; lying in the midline. Synonym(s): medianus
2. The middle value in a set of measurements; like the mean, a measure of central tendency.
[L. medianus, middle]
## median
/me·di·an/ (me´de-in)
1. situated in the median plane or in the midline of a body or structure.
2. the value of the middle item of a series when the items are arranged in numerical order. Symbol m.
## median
(mē′dē-ən)
1. Relating to, located in, or extending toward the middle.
2. Anatomy Of, relating to, or situated in or near the plane that divides a bilaterally symmetrical animal into right and left halves; mesial.
3. Statistics Relating to or constituting the middle value in a distribution.
n.
1. A median point, plane, line, or part.
2. Statistics The middle value in a distribution, above and below which lie an equal number of values.
3. Mathematics
a. A line that joins a vertex of a triangle to the midpoint of the opposite side.
b. The line that joins the midpoints of the nonparallel sides of a trapezoid.
## median (med)
[mē′dē·ən]
Etymology: L, medius, middle
(in statistics) the number representing the middle value of the scores in a sample. In an odd number of scores arrayed in ascending order, it is the middle score. In an even number of scores so arrayed, it is the average of the two central scores.
## median
EBM
The exact middle value in a set of values that has been arranged in order from highest to lowest—i.e., there are as many values greater and less than the median. Where there is an even number of values, the median is designated as halfway between the two middle values.
Statistics
The midpoint of data after being ranked in a distribution, above and below which lie an equal number of values; a measure of central location, which divides a set of data into two equal parts; the data value at rank 0.5 x (n + 1).
The median is a better measure than the mean of the centre of a data distribution when the data are not symmetrically (normally) distributed, as it is not affected as severely as the mean by the outliers and non-symmetry typical of biological data. The median appears as a line in the box of a box-and-whisker plot and divides the middle two quartiles. Medians are compared by non-parametric statistical procedures.
## me·di·an
(mē'dē-ăn)
1. Central; middle; lying in the midline.
2. The middle value in a set of measurements; like the mean, a measure of central tendency.
## median
1. Situated in or towards the MEDIAN PLANE of the body.
2. In statistics, the middle value when observations are ranked in order of magnitude.
## median
1. (of a structure or character) describing a location along the line of bilateral symmetry
2. (in statistics), the middle value in a frequency distribution, on either side of which lie values with equal total frequencies.
## median
middle value in range of measurements; mean, median and mode are identical in normally distributed data
## me·di·an
(mē'dē-ăn)
1. Central; middle; lying in the midline.
2. The middle value in a set of measurements.
## median (mē´dēən),
adj 1. pertaining to the middle.
n 2. a measure of central tendency attained by a calculation or count that separates all cases in a ranked distribution into halves. The median may be used as an average score.
median lethal dose,
n See dose, lethal, median.
median line,
median mandibular point,
n See point, median mandibular.
median nerve,
n one of the terminal branches of the brachial plexus that extends along the radial portions of the forearm and the hand and supplies various muscles and the skin of these parts.
median retruded relation,
median rhomboid glossitis,
median sagittal plane,
n See plane, median sagittal.
## median
1. situated in the median plane or in the midline of a body or structure.
2. the perpendicular line that divides the area of a frequency curve into two equal halves.
median calving date
the number of days between the first calving in the herd and the 50th percentile calving; an excellent measure of fertility status of seasonally calving herds; in dairy herds the target is 18 days.
median eminence
part of the hypophysis.
median nerve
see Table 14.
median nerve block
the anesthetic agent is injected on the medial aspect of the forelimb, just distal to the elbow. An area encircling most of the fetlock and pastern is desensitized.
median nerve injury
results in overextension and dropping of the carpus.
References in periodicals archive ?
And as all gardeners know, the success of any landscape plan depends on the caring people who take special pride in keeping all these areas, including the city's 22 acres of planted median strips, a wonderful part of Eugene's public space.
gutters, curb, sidewalks and median strips abut the existing, the existing shall be saw cut.
In order to make Capital city more beautiful and attractive, the CDA has also decided to beautify all entry points, median strips and green belts along the major avenues with colourful flowering plants and fascinating landscaping.
The Fire Department has also received several calls about fires in the median strips on city streets.
During visit to different parks and green areas along the major road sides of the sectors, the Chairman directed Environment Wing to remove rank vegetation, shrubs and bushes; beautification and aesthetic uplift of green belts, median strips, roundabout and intersections.
Median strips would be decorated with flowers to aesthetically boost the city.
They have planted Joshuas in street median strips and around basins that collect stormwater.
Under this plan, traffic signals and signage will be improved, lane marking will be carried out, median strips will be decorated with flowers to beautify the city.
Since the oleander is everywhere, including and especially on freeway median strips, it is downgraded to a commonplace, uninspiring species.
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Code Listing by sri hari bhupala haribhakta
Code 1-3 of 3
### decimation matrix 1.0 - sri hari bhupala haribhaktaTools / Build Tools
This function gives the decimation matrix which after post mulitpling it with the given signal gives the downsampled version of the original signal...
D = decimmtx(x,N)
decimates the given signal by a factor of N.
[D,y] = decimmtx(x,N)
gives D as well as the decimated signal y
### elementary signals 1.0 - sri hari bhupala haribhaktaTools / Build Tools
The signals presented here can be used in demonstrating how the signal varies when the operation is made on either independent variable or dependent variable. The parameters can be varied to have a feel of the variation.
Example(1):...
### convlution matrix in four lines 1.0 - sri hari bhupala haribhaktaTools / Build Tools
CONV_MTX Convolution matrix.
If 'x' is a column vector of length 'nx', then conv_mtx(x,nh) gives a toeplitz
matrix 'X' of size (nx+nh-1) times (nh). 'nh' is the length of the column vector
'h' with which 'x' is convolved. Thus... | 601 | 2,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2018-43 | longest | en | 0.664603 |
http://www.solutioninn.com/in-a-survey-169-respondents-say-that-they-never-use | 1,503,094,543,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105187.53/warc/CC-MAIN-20170818213959-20170818233959-00377.warc.gz | 681,688,316 | 7,465 | # Question: In a survey 169 respondents say that they never use
In a survey, 169 respondents say that they never use a credit card, 1227 say that they use it sometimes, and 2834 say that they use it frequently. What is the probability that a randomly selected person uses a credit card frequently? Is it unlikely for someone to use a credit card frequently? How are all of these results affected by the fact that the responses were obtained by those who decided to respond to the survey posted on the Internet by America Online?
Consider an event to be “unlikely” if its probability is less than or equal to 0.05. (This is equivalent to the same criterion commonly used in inferential statistics, but the value of 0.05 is not absolutely rigid, and other values such as 0.01 are sometimes used instead.)
View Solution:
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Views649 | 184 | 836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-34 | latest | en | 0.962947 |
https://www.doubtnut.com/question-answer-physics/a-perfect-gas-at-27c-is-heated-at-constant-pressure-to-327c-if-original-volume-of-gas-at-27c-is-v-th-11797062 | 1,674,987,818,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499710.49/warc/CC-MAIN-20230129080341-20230129110341-00067.warc.gz | 750,534,017 | 79,918 | Home
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# A perfect gas at 27^(@)C is heated at constant pressure to 327^(@)C. If original volume of gas at 27^(@)C is V, then volume at 327^(@)C is
Khareedo DN Pro and dekho sari videos bina kisi ad ki rukaavat ke!
V3V2VV//2
Solution : From Charles law V prop T <br> :. (V_2)/(V_1)=(T_2)/(T_1)=(327+273)/(27+273) = 600/300=2 <br> implies V_(2) = 2V | 181 | 483 | {"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} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.67874 |
https://numberworld.info/530000 | 1,582,949,609,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148375.36/warc/CC-MAIN-20200229022458-20200229052458-00378.warc.gz | 476,955,358 | 4,069 | # Number 530000
### Properties of number 530000
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 53
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
g5ig
sin(530000)
0.68380798173863
cos(530000)
0.72966200676103
tan(530000)
0.93715717058375
ln(530000)
13.180632285528
lg(530000)
5.7242758696008
sqrt(530000)
728.01098892805
Square(530000)
### Number Look Up
Look Up
530000 which is pronounced (five hundred thirty thousand) is a great figure. The cross sum of 530000 is 8. If you factorisate the number 530000 you will get these result 2 * 2 * 2 * 2 * 5 * 5 * 5 * 5 * 53. The figure 530000 has 50 divisors ( 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 53, 80, 100, 106, 125, 200, 212, 250, 265, 400, 424, 500, 530, 625, 848, 1000, 1060, 1250, 1325, 2000, 2120, 2500, 2650, 4240, 5000, 5300, 6625, 10000, 10600, 13250, 21200, 26500, 33125, 53000, 66250, 106000, 132500, 265000, 530000 ) whith a sum of 1307394. 530000 is not a prime number. The number 530000 is not a fibonacci number. The figure 530000 is not a Bell Number. The figure 530000 is not a Catalan Number. The convertion of 530000 to base 2 (Binary) is 10000001011001010000. The convertion of 530000 to base 3 (Ternary) is 222221000122. The convertion of 530000 to base 4 (Quaternary) is 2001121100. The convertion of 530000 to base 5 (Quintal) is 113430000. The convertion of 530000 to base 8 (Octal) is 2013120. The convertion of 530000 to base 16 (Hexadecimal) is 81650. The convertion of 530000 to base 32 is g5ig. The sine of the figure 530000 is 0.68380798173863. The cosine of the number 530000 is 0.72966200676103. The tangent of the number 530000 is 0.93715717058375. The square root of 530000 is 728.01098892805.
If you square 530000 you will get the following result 280900000000. The natural logarithm of 530000 is 13.180632285528 and the decimal logarithm is 5.7242758696008. I hope that you now know that 530000 is very great number! | 804 | 2,077 | {"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} | 3.546875 | 4 | CC-MAIN-2020-10 | latest | en | 0.74003 |
https://www.mylotto-app.com/two-dimensional-lotto-winner-apps/europe/lotto-winner-for-la-primitiva-lottery/ | 1,720,848,122,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00044.warc.gz | 731,550,252 | 21,792 | Select Page
Lotto Winner for La Primitiva Lottery
Best lotto code generator
La Primitiva Lottery app
This app helps players to choose numbers with an algorithm, UNIQUE IN THE WORLD for La Primitiva Lottery draws.
Generating two-dimensional and Even/Odd numbers.
Best code generator in the WORLD (Two-Dimensional)!
IF YOU ARE ON THIS PAGE MEANS THAT EVERYTHING HAPPENS FOR A REASON!
Increase your winning chances with this app!
La Primitiva two-dimensional app generating numbers contains an algorithm, UNIQUE IN THE WORLD that acts like real lottery draws. As in reality, every number is chosen in a certain time. This is how it happens with the app and it generates the time as well as the numbers, being the closest simulation to the real lottery draw.
You can choose how many even and odd numbers the app to generate and play.
La Primitiva lotto winning numbers
Play the official La Primitiva lottery by selecting 6 numbers from a guess range of 1-49. A reintegro number, which decides the La Primitiva jackpot, is automatically added to your official Spanish lottery ticket. theLotter offers 4 or 8-line tickets, and 8, 9, 10 or 11-number systematic forms online that create all possible combinations of your selected numbers for more ways to win. The probability of this happening is 1 in 13,983,816.
La Primitiva Lottery – Poisson approximation
Probabilities associated with La Primitiva Lottery are determined by calculations based on equally likely outcomes but their interpretation is empirical, as follows. If an experiment (such as a weekly Lotto drawing) is repeated under similar conditions many times (mathematically the requisite number of times must approach infinity) then the probability of an event is the long run proportion of experimental repetitions on which the event occurs. (This is called a law of large numbers. It is a mathematical result only).
As illustration, the probability of obtaining a fourth place prize is 0.0009686. This means that in 26,000 Lotto draws, each particular choice of six numbers will yield a fourth place prize (matching exactly 4 numbers) on approximately 0.0009686 of those plays, that is 26000 x 0.000986 = 25 times. There is no guarantee that this will happen 25 times. The actual number of times is random and subject to what is called the Poisson approximation. The figure of 25 represents an expected number of winnings and is a useful measure by which to compare the performance of different strategies.
Tips – how to generate, play & win at La Primitiva Lottery
Odd numbers are those which end in 1, 3, 5, 7 and 9 while even numbers are those which end in 0, 2, 4, 6 and 8. You are more likely to win if your odd-even number combination is either 2-4, 3-3 or 4-2. More than 70% of the winning combinations in all the lotteries 6/49 in the past two years are in these category.
For further details or any questions please send me an email at: [email protected]
Disclaimer:
This application has no link in no way with La Primitiva Lottery organism and it does not allow to participate in the official game of La Primitiva Lottery.
TIPS:
Generate almost equal numbers odd-even, until you see your favorite number. Mathematically, these numbers you have to play about 25 times. It is not easy but it is the best way to win!
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Are you making plans to get a roof repairer for your private home development venture? Below 5 major suggestions that let you discover the high-quality one for your region.No. 1 - Check roofers' compensation and liability coverage.Before some thing else, ask your... | 1,251 | 5,854 | {"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-2024-30 | latest | en | 0.919763 |
http://www.mathisfunforum.com/viewtopic.php?id=4373&p=183 | 1,397,827,746,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00115-ip-10-147-4-33.ec2.internal.warc.gz | 533,395,149 | 6,767 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #4551 2014-02-16 00:42:04
bobbym
Offline
### Re: 10 second questions
Hi;
Sorry, wrote down the wrong number for 4651.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4552 2014-02-17 00:16:22
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solution #4652 is correct. Neat work!
#4653. Find the values of x, y, and z if
.
#4654. Find the value of
.
Character is who you are when no one is looking.
## #4553 2014-02-17 01:53:34
bobbym
Offline
### Re: 10 second questions
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4554 2014-02-17 22:19:53
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4653 and #4654 are correct. Marvelous!
#4655. Find the value of
.
#4656. Simplify :
.
Character is who you are when no one is looking.
## #4555 2014-02-17 22:36:57
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4556 2014-02-18 00:30:36
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4655 and #4656 are correct. Fantastic!
#4657. If
and
, then find the value of
.
#4658. Find the value of
.
Character is who you are when no one is looking.
## #4557 2014-02-18 00:35:24
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4558 2014-02-18 13:46:11
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4657 and #4658 are correct. Neat work!
#4659. 5004 ÷ 139 - 6 = ?
#4660. 7500 + (1250 ÷ 50) = ?
Character is who you are when no one is looking.
## #4559 2014-02-18 14:01:41
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4560 2014-02-19 00:24:06
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4659 and #4660 are correct. Good work!
#4661. (8 ÷ 88) x 8888088 = ?
#4662. The value of 1001 ÷ 11 of 13 is _________.
Character is who you are when no one is looking.
## #4561 2014-02-20 00:35:12
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
#4663. Find the value of
.
#4664. Find the value of x if
Character is who you are when no one is looking.
## #4562 2014-02-20 01:07:10
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4563 2014-02-21 00:19:44
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4663 and #4664 are correct. Neat work!
#4665. A boy was asked to write the value of 25 x 92. He wrote it as 2592. What is the difference between the obtained and the actual value?
#4666. 2 - [2 - {2 - 2(2 + 2)}] = ?
Character is who you are when no one is looking.
## #4564 2014-02-21 00:23:45
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4565 2014-02-22 00:17:31
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4665 and #4666 are correct. Good work!
#4667. Find the value of 25 - 5[2 + 3{2 - 2(5 - 3) + 5} - 10] ÷ 4.
#4668. 3640 ÷ 14 x 16 + 340 = ?
Character is who you are when no one is looking.
## #4566 2014-02-22 03:02:29
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4567 2014-02-22 16:20:02
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4667 and #4668 are correct. Neat work!
#4669. 100 x 10 - 100 + 2000 ÷ 100 = ?
#4670.
Character is who you are when no one is looking.
## #4568 2014-02-22 16:22:40
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4569 2014-02-23 00:17:21
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4669 and #4670 are correct. Brilliant!
#4671.
#4672.
Character is who you are when no one is looking.
## #4570 2014-02-23 00:54:29
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4571 2014-02-24 00:31:40
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4671 and #4672 are correct. Excellent!
Simplify:
#4673.
#4674.
Character is who you are when no one is looking.
## #4572 2014-02-24 00:40:12
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #4573 2014-02-25 00:32:04
ganesh
Moderator
Offline
### Re: 10 second questions
Hi bobbym,
The solutions #4673 and #4674 are correct. Remarkable!
#4675. Evaluate :
.
#4676.
Character is who you are when no one is looking.
## #4574 2014-02-25 00:36:12
John E. Franklin
Star Member
Offline
### Re: 10 second questions
igloo myrtilles fourmis
## #4575 2014-02-25 01:02:36
bobbym
Offline
### Re: 10 second questions
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 2,218 | 7,139 | {"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-2014-15 | longest | en | 0.883769 |
http://forum.arduino.cc/index.php?topic=98510.msg739115 | 1,508,569,343,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824618.72/warc/CC-MAIN-20171021062002-20171021082002-00436.warc.gz | 135,082,571 | 8,683 | Go Down
### Topic: How do I know if address is in hex or dec? (Read 1 time)previous topic - next topic
#### Whitworth
##### Mar 26, 2012, 09:53 pm
I'm learning how to read and write data using the wire interface at the moment, but as my question says I'm unsure if the code is interpretting the address as a hex value or decimal, for example:
Code: [Select]
`void setup(){ Wire.begin(); // join i2c bus (address optional for master) Serial.begin(9600); // start serial for output writeTo(DEVICE, 0x16, 8); } // end of setup`
Where writeTO is:
Code: [Select]
`void writeTo(int device, byte address, byte val) { Wire.beginTransmission(device); //start transmission to device Wire.write(address); // send register address Wire.write(val); // send value to write Wire.endTransmission(); //end transmission } // end of writeTo`
So, i'm using writeTo(DEVICE, 0x16, , how does it know that the 0x16 is hex and not decimal?
#### dxw00d
#1
##### Mar 26, 2012, 10:00 pm
The '0x' part tells the compiler that you are using a hex constant.
#2
##### Mar 26, 2012, 10:01 pm
0x16 tells the code it is hex (the 0x part).
0x00 is a byte (8 bits) as in
byte flag1 = 0x01;
0x0000 is a int or word (16 bits), as in
0x01234567 is a long (32 bits), as in
unsigned long currentTime = millis();
B11110000, B indicates binary.
'Nothing' preceding a number indicates it is decimal.
Designing & building electrical circuits for over 25 years. Screw Shield for Mega/Due/Uno, Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at my website.
#### dxw00d
#3
##### Mar 26, 2012, 10:06 pmLast Edit: Mar 26, 2012, 10:10 pm by dxw00d Reason: 1
Quote
B11110000, B indicates binary.
This is something the Arduino IDE does. There is also 0b11110000.
A leading 0 indicates an octal constant, I believe.
#### AWOL
#4
##### Mar 26, 2012, 10:12 pm
Quote
'Nothing' preceding a number indicates it is decimal.
...and a leading zero indicates it is octal
"Pete, it's a fool looks for logic in the chambers of the human heart." Ulysses Everett McGill.
Do not send technical questions via personal messaging - they will be ignored.
I speak for myself, not Arduino.
#### Whitworth
#5
##### Mar 26, 2012, 10:20 pm
Thank you, it seems obvious now. If I had the register map below:
and I wanted to write FS_SEL=3 on register 0x16 then am I correct in saying:
FS_SEL =3 makes the line equal in binary:
00011000
which is equal to 24 in decimal, so i'd want:
writeTo(DEVICE, 0x16, 24); ?
#6
##### Mar 27, 2012, 02:06 am
Or,
writeTo(DEVICE, 0x16, B00011000);
Or
writeTo(DEVICE, 0x16, 0x18);
Makes it a lot more straightforward which bits are being manipulated.
24, you can't tell where the bits are without doing the conversions.
Designing & building electrical circuits for over 25 years. Screw Shield for Mega/Due/Uno, Bobuino with ATMega1284P, & other '328P & '1284P creations & offerings at my website.
#### RandallR
#7
##### Mar 27, 2012, 02:17 pm
You could also write:
writeTo(DEVICE, 0x16, (3<<3));
That is the value 3 shifted 3 bit to the left.
I assume that you know that you will also be writing a 0 to DLPF_CFG when you do this.
#8
##### Mar 27, 2012, 02:39 pmLast Edit: Mar 27, 2012, 02:42 pm by Graynomad Reason: 1
Quote
which is equal to 24 in decimal, so i'd want:
writeTo(DEVICE, 0x16, 24); ?
While technically this is correct it's unreadable, a day later you'll be wondering WTF the value means.
So I would use the B00011000 or (3<<3) options mentioned. Even better define a value to shift by
#define FS_SEL 3
(3<<FS_SEL)
You will notice that is how most AVR direct port manipulation is written. Straight away you know that the FS_SEL field is being set to 3.
If the DLPF_CFG field is <> 0 then
(3<<FS_SEL) | (dlpf_cfg_val << DLPF_CFG)
or
(3<<FS_SEL) | dlpf_cfg_val
because it's bit offset is 0 anyway.
______
Rob
Rob Gray aka the GRAYnomad www.robgray.com
#### Whitworth
#9
##### Mar 27, 2012, 05:37 pm
Thanks Graynomad, that approach makes a lot of sense I'll be using that way in new code I write.
#### GoForSmoke
#10
##### Mar 27, 2012, 05:40 pm
If you are changing different registers then you might want to make a function to read the register and only change the bits you want changed. Perhaps you will need args for what to & and what to | with the original data before writing back. | 1,318 | 4,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-43 | latest | en | 0.763515 |
http://wordsdomination.com/hex.html | 1,542,394,257,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743110.55/warc/CC-MAIN-20181116173611-20181116195611-00276.warc.gz | 362,565,643 | 12,441 | # hex
share
## Exampleshex's examples
• Buy hex, Business Industrial items on eBay. Find great deals on eBay Motors, Home Garden items and get what you want now!. — “hex items - Get great deals on Business Industrial, eBay”,
• In mathematics and computer science, hexadecimal (also base- 16, hexa, or hex) is a numeral system with a radix, or base, of 16. It uses six*** distinct symbols, most often the symbols 0–9 to represent values Since each hexadecimal digit represents four binary digits ( bits). — “Hexadecimal”, schools-
• Shop for hex at Target. Find products like hexbug, hex bug and more. Choose from Jonah Hex (2 Discs) (Includes Digital Copy) (Blu-ray/DVD) (Widescreen), Jonah Hex - Lilah Adult Wig and other products. — “hex : Target Search Results”,
• In mathematics, hexadecimal or simply hex is a numeral system with a radix or base of 16 usually written using the symbols 0–9 and A–F or a–f. For example, the decimal numeral 79 whose binary representation is 01001111 can be written as 4F in hexadecimal (4 = 0100, F = 1111). — “Hexadecimal - Definition”,
• Definition of hex from Webster's New World College Dictionary. Meaning of hex. Pronunciation of hex. Definition of the word hex. Origin of the word hex. — “hex - Definition of hex at ”,
• A simple and easy to understand explanationof the hexadecimal number system is given. — “Hexadecimal Numbers Explained”,
• Template:Numeral systems In mathematics and computer science, hexadecimal, base-Template:Num, or simply hex, is a numeral system with a radix, or base, of 16, usually written using the symbols 0–9 and A–F, or a–f. Its primary purpose is to represent. — “Hexadecimal - CryptoDox”,
• All about HEXADECIMAL and How is It used on Computers Programmers often refer to it as simply, Hex (which by itself should mean only six, such as the number of sides that a "Hex wrench" has, but it almost always means Hexadecimal if the subject is computer software rather than hardware or your ride). — “What is HEXADECIMAL and Where, How and Why is It used on”,
• hex (plural hexes) An evil spell or curse. A witch. (rare) A spell (now rare but still found in compounds such as hex sign and to hex (third-person singular simple present hexes, present participle hexing, simple. — “hex - Wiktionary”,
• hexadecimal adj. Of, relating to, or based on the number 16: the hexadecimal number system. Of or relating to six***ths Each half byte (four bits) is assigned a hex digit as shown in the following chart with its decimal and binary equivalents. — “hexadecimal: Definition from ”,
• Hex definition, to bewitch; practice witchcraft on: See more. — “Hex | Define Hex at ”,
• Hex is a web design business with more than 10 years of online experience. No frills, no fluff, just plain old websites that look and work great. — “Hex Singapore - Web Design Services. We Build Websites The”, hex.sg
• Greek - six 1. n.An evil curse or spell 2. prefix meaning six 3. v.To charm or cast a spell upon 4. A six pack of anything 5. Short for hexadecimal. — “Urban Dictionary: hex”,
• In mathematics and computer science, hexadecimal (also base 16, or hex) is a positional numeral system with a radix, or base, of 16. It uses six*** distinct symbols, most often the symbols 0–9 to represent values zero to nine, Each hexadecimal digit represents four binary digits (bits) (also. — “Hexadecimal - Wikipedia, the free encyclopedia”,
• Definition of hex in the Online Dictionary. Meaning of hex. Pronunciation of hex. Translations of hex. hex synonyms, hex antonyms. Information about hex in the free online English dictionary and encyclopedia. — “hex - definition of hex by the Free Online Dictionary”,
• I may be dead, but I still know what's going on! I'm also trying to help save the world from evil - all before lunch! Read my blog. Voodoo rituals, fallen angels, lesbian ghosts, evil fairies you've got to be a true HEX-pert to keep track of it all. How well have you been paying attention?. — “BBC America - Hex”,
• Hex is a board game played on a hexagonal grid, theoretically of any size and several possible shapes, but traditionally as a 11x11 rhombus. Other popular dimensions are 13x13 and 19x19 as a result of the game's relationship to the older game of Go. — “Hex (board game) - Wikinfo”,
• Harry Potter role playing game. Get sorted, attend Hogwarts, play quidditch, and more. — “Hogwarts Extreme: An Interactive Harry Potter RPG”,
• With Jemima Rooper, Jamie Davis, Amber Sainsbury, Colin Salmon. I have to admit that I was not expecting much from Hex. However after watching the first episode I was impressed and looking forward to the next. I am a fan of shows that feature the occult and Hex did not disappoint me. — “Hex (TV Series 2004–2005) - IMDb”,
• HEX : - Incense Oils & Potions Jewelry Candles Ritual Soaps Witches' Herbal Dolls & Poppets Bags & Pouches Ritual Tools Spell Kits HEX Mugs Amulets Books Miscellaneous Crystals & Stones Art & Statuary ecommerce, open source, shop, online. — “HEX, Old World Witchery”,
## Blogs & Forumblogs and forums about hex
• “Making Connections One Blog at a Time In Generation Hex, Marla Alupoaicei and Dillon Burroughs explore the history, culture, and practices of Wicca”
— Generation Hex Blog Tour,
• “Hex.Ro Forum - Statistics Center”
— Hex.Ro Forum - Statistics Center, hex.ro
• “Posted by Hex in Essential Tips. As web usage has been increasing throughout the years, How to Make an Awesome Corporate Blog http://j.mp/9v732n 2010-10-07”
Hex Blog, hex.sg
• “Hex Folk. Staff. Contributors. Submit. Guidelines. Style Guide. Blog. 19 Nov 2010. Review: Asa © 2010 Hex Press | Site Created by Arrowyn Craban / Verdandi Design | Use of this website constitutes”
— Blog " Hex Magazine,
• “Try and submit a link to hex.io and it had better be legit! You may not create a HEX.IO link for it. Detection rates are around 80 to 90%, with false”
— HEX!O Redirection Intelligence " Blog Archive " HEX!O, hex.io
• “Hex-Rays forum. Decompilation and binary program ***ysis in general Please ensure you read any forum rules as you navigate around the board”
— Hex-Rays forum • User Control Panel • Login, hex-
• “Though Josh Brolin has been acting since the '80s ('Goonies,' anyone?), it's only in the past couple of years that the talented Angeleno has Please keep your comments relevant to Josh Brolin Interview Jonah Hex blog entry. Email addresses are never displayed, but they are required to”
— Josh Brolin Talks 'Jonah Hex,' Fart Jokes and Megan Fox - The, | 1,616 | 6,480 | {"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-2018-47 | latest | en | 0.856616 |
https://blog.doublehelix.csiro.au/jumping-puzzle/ | 1,709,220,153,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00514.warc.gz | 129,203,079 | 22,308 | # Blog
## Jumping puzzle
By
,
Time to jump start your brain for the day. Can you solve this jumping puzzle?
### Assembling the pieces
1. First make the playing pieces. Cut along the thick black lines to get three boy strips and three girl strips
2. Fold each strip along the dotted lines to make a triangle and hold it together with a piece of sticky tape.
3. The remaining piece of paper has a line of seven circles – this is the board.
### The puzzle
1. Put one boy piece on each of the three stripy circles, and one girl piece on each of the umbrella circles. Leave the middle circle empty.
2. The aim of the game is to swap the girls and the boys.
3. There are two ways a piece can move. They can move into the next circle if it is empty. Or they can jump over a different piece into an empty circle.
4. Girls can only jump over boys. Boys can only jump over girls.
5. Girls can only move towards the boys’ side and boys can only move towards the girls’ side. No-one can move backwards.
6. There are only two solutions to the puzzle. Good luck!
### What’s happening?
This jumping puzzle is very easy to get wrong – you can make a mistake on the second move that makes it impossible! Here are a few clues to get you started.
You can only jump over one piece at a time. So two boys or two girls in a row are very hard to move past. Try to keep things going girl, boy, girl, boy.
You only have one empty space to move pieces into. Try watching where the space moves as you play through the puzzle. It needs to visit the whole board, but it’ll spend more time in the middle.
For a bigger hint, here’s a list of moves that should work:
Move 1 girl
Move 2 boys
Move 3 girls
Move 3 boys
Move 3 girls
Move 2 boys
Move 1 girl
But you’ll have to work out which ones to move!
If you’re after more maths activities for kids, subscribe to Double Helix magazine!
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1. y2/3 = 6x + 4lnx - 2
I have to give my answer in the form of y2 = g(x)
I find it tricky to convert y2/3 to y2.. Can someone show me how to do it?
Thanks
2. (Original post by sabre2th1)
y2/3 = 6x + 4lnx - 2
I have to give my answer in the form of y2 = g(x)
I find it tricky to convert y2/3 to y2.. Can someone show me how to do it?
Thanks
Take both side of equation on the 3rd power
For RHS use
where a=6x b=4lnx and c=-2
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# GMAT Question of the Day (Feb 14): Coordinate Geometry and Sentence Correction
- Feb 14, 02:00 AM Comments [0]
Math (DS)
The vertices of a triangle have coordinates $(x, 1)$ , $(5, 1)$ , and $(5, y)$ where $x < 5[/latex] and [latex]y > 1$ . What is the area of the triangle?
1. $x = y$
2. Angle at the vertex $(x, 1)$ = angle at the vertex $(5, y)$
Question Discussion & Explanation
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Verbal (SC)
The aristocratic values expressed in the writings of Marguerite Yourcenar place her within the French classical tradition, as does her passionate interest in history, particularly Roman history.
(A) as does
(B) so do
(C) as do
(D) so is the case with
(E) similarly, does
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# If Sally can paint a house in 4 hours, and John can paint
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If Sally can paint a house in 4 hours, and John can paint [#permalink]
### Show Tags
02 Nov 2012, 09:46
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### HideShow timer Statistics
If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
A. 2 hours and 24 minutes
B. 3 hours and 12 minutes
C. 3 hours and 44 minutes
D. 4 hours and 10 minutes
E. 4 hours and 33 minutes
[Reveal] Spoiler: OA
Last edited by Bunuel on 02 Nov 2012, 13:06, edited 1 time in total.
Renamed the topic and edited the question.
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### Show Tags
02 Nov 2012, 10:16
Hello...
Sally can paint a house in 4 hours; Which means in 1 hour she can paint 1/4th of the house.
John can paint the same house in 6 hours. Which means in 1 hour she can paint 1/6th of the house.
Together in 1 hour they can paint: - 1/4 + 1/6 = 5/12th of the house.
Total Hours for painting the house together will be 12/5 = 2.4 Hours.
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### Show Tags
02 Nov 2012, 10:25
1
KUDOS
shrinivas280390 wrote:
Hello...
Sally can paint a house in 4 hours; Which means in 1 hour she can paint 1/4th of the house.
John can paint the same house in 6 hours. Which means in 1 hour she can paint 1/6th of the house.
Together in 1 hour they can paint: - 1/4 + 1/6 = 5/12th of the house.
Total Hours for painting the house together will be 12/5 = 2.4 Hours.
Thanks a lot I really made a stupid mistake :/
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### Show Tags
02 Nov 2012, 12:11
MariaF wrote:
shrinivas280390 wrote:
Hello...
Sally can paint a house in 4 hours; Which means in 1 hour she can paint 1/4th of the house.
John can paint the same house in 6 hours. Which means in 1 hour she can paint 1/6th of the house.
Together in 1 hour they can paint: - 1/4 + 1/6 = 5/12th of the house.
Total Hours for painting the house together will be 12/5 = 2.4 Hours.
Thanks a lot I really made a stupid mistake :/
It'd be interesting how to use elimination techniques solving this problem?
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### Show Tags
02 Nov 2012, 12:17
MariaF wrote:
Guys, help me please to solve this easy problem. I think I just make a stupid mistake
If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
A. 2 hours and 24 minutes
B. 3 hours and 12 minutes
C. 3 hours and 44 minutes
D. 4 hours and 10 minutes
E. 4 hours and 33 minutes
Time taken will be = x*y/x+y = 4*6/4+6 [ A]
Manager
Joined: 25 Jun 2012
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### Show Tags
02 Nov 2012, 12:22
RJSPO wrote:
MariaF wrote:
Guys, help me please to solve this easy problem. I think I just make a stupid mistake
If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
A. 2 hours and 24 minutes
B. 3 hours and 12 minutes
C. 3 hours and 44 minutes
D. 4 hours and 10 minutes
E. 4 hours and 33 minutes
Time taken will be = x*y/x+y = 4*6/4+6 [ A]
Nice trick, I have to remember it, x*y/(x+y)
and if it would be three workers is it equal x*y*z/(x+y+z)?
Manager
Joined: 31 Oct 2012
Posts: 61
Schools: Mays '16, Simon '16
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### Show Tags
02 Nov 2012, 12:30
RJSPO wrote:
MariaF wrote:
Guys, help me please to solve this easy problem. I think I just make a stupid mistake
If Sally can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
A. 2 hours and 24 minutes
B. 3 hours and 12 minutes
C. 3 hours and 44 minutes
D. 4 hours and 10 minutes
E. 4 hours and 33 minutes
Time taken will be = x*y/x+y = 4*6/4+6 [ A]
Nice trick, I have to remember it, x*y/(x+y)
and if it would be three workers is it equal x*y*z/(x+y+z)?
No, for 3 workers it will be quite a complicated formula : x*y*z/xy+yz+zx
Better to use reciprocal formula for other cases
Senior Manager
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Re: If Sally can paint a house in 4 hours, and John can paint [#permalink]
### Show Tags
14 Nov 2012, 03:13
$$\frac{1}{4}+\frac{1}{6}=\frac{10}{24}=\frac{5}{12}$$
$$Rt=W -> \frac{5houses}{12hrs}xt=1$$
$$t=\frac{12}{5}=2\frac{2}{5}hours=2 hours and 24 minutes$$
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Re: If Sally can paint a house in 4 hours, and John can paint [#permalink]
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06 Jan 2016, 11:35
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Re: If Sally can paint a house in 4 hours, and John can paint [#permalink] 06 Jan 2016, 11:35
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### comment on
Need Help??
There is no trick.
In the problem, each envelope can contain any number. The requirement of a distribution of numbers that has a continuous probability distribution with non-zero density everywhere is on the one you create for your algorithm. That number is made up out of thin air and has no connection to either of the numbers in the envelopes. The trick is that it gives us a chance of telling the two numbers apart.
The probability density function that you are trying to describe, "uniform over the real numbers", is not a valid probability density function. It is a classic result both in real analysis and probability that it can't be. (The real analysis statement is that no such measure exists, the probability theory statement is that no such probability distribution exists. Those statements are the same.)
This is why you have to be very careful in the wording to even get a well-defined problem. Given any two numbers and the algorithm, there is a well-defined probability that you're right, and that probability is over 0.5. Prior to the numbers and algorithm, the probability of your being right is undefined and undefinable.
Again I'd like to point people to Bently Preece's excellent explanation. The goal of the question is to come up with a single strategy that will give better than even odds for every game in a particular class of games. And the described strategy does that.
Now I haven't explained why this particular probability distribution can't exist. The technical reason is that the axioms of probability require the probability of a countable disjoint union to be the sum of the probabilities, and that the probability of the whole set is 1. But the real line is a countable disjoint union of intervals, each of which has probability 0 (for the reason that you explained), so the probability of the whole set is a countable sum of 0's, which is 0. Contradicting the requirement that it be 1.
Now it is easy to object that one could just define probabilities differently. And it is true, one could. But any way you define it, you'll have a lot of subtleties around infinity, and attempting to think about a uniform probability distribution on the real numbers will be one place that you'll encounter lots of them.
BTW in the USA this fact is generally covered in a real analysis class in advanced undergraduate or beginning graduate school mathematics.
In reply to Re^2: Spooky math problem by tilly
in thread Spooky math problem by tilly
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Ok, so you should find your 15,10, and 5 rep maxes and then decrement down by 5lbs from each to find your weekly workout weight. Let's say you're doing side raises. I dunno about ya'll, but I'm not exactly using alotta weight for 15 reps on these. Let's say I'm using 20, would I wanna use 10 for 2 days, 15, for 2 days, and 20 for 2 days. And, if I'm getting the hst principle wrong, feel free to fill me in
2. I didnt find the maxes beforehand, since they constantly change with developing strength levels. Naturally ramp up the weight to a weight that feels right when you change the reps.
You say 2 days in there, you mean two weeks right?
•
3. nah, yer supposed to raise the weight on each workout, meaning there's 6 days in there ya know. Anyways, if I can do a weight more than the given number of reps in a given exercise should I? I wish the hst site explained it a lil better.
4. I had the same dilemma, so i just used the same weight (20 pounds) for 3 workouts in a row until i was able to increase. worked fine for me as I have gotten great results from HST.
5. Cool, thanks man. I'll give this workout a try, I gotta do something new.
•
6. Originally posted by goin_big
nah, yer supposed to raise the weight on each workout, meaning there's 6 days in there ya know.
What are you talking about? Are you seriously contemplating a 6 workout cycle?
7. I have no idea what hst you are doing, but each 2 week cycle is divided up into 6 workouts.......you're confusing me
8. Hm. I am talking about hypertrophy specific training. You may want to read the article explaing HST again goin_big.
You do not understand how to use HST. After you re-read it come back and ask questions
9. yes I do. Monday, wednesday, friday, you do a full body workout, increasing the weight each day until on the 6th day which would be the second friday you would be at your max weight for that rep scheme and so on and so forth, I dunno what you're talking about....
10. • Repetitions will decrease every 2 weeks in the following order: 15 reps for 2 weeks Þ 10 reps for 2 weeks Þ 5 reps for 2 weeks Þ then continue with your 5 rep max for 2 weeks or begin 2 weeks of negatives. 15¹s can be skipped when you are about to start over after the first 8 week cycle. If you are feeling strain-type injuries coming on don't skip the 15s.
From the guide explaining HST on the HST site.
11. Yes I know this, I have no idea what point you're trying to get to here, 6-8 weeks, divided into 6 workouts every 2 weeks, it's so easy.........nm, I know the whole routine, if you think I don't, whatever.
12. Originally posted by goin_big
yes I do. Monday, wednesday, friday, you do a full body workout, increasing the weight each day until on the 6th day which would be the second friday you would be at your max weight for that rep scheme and so on and so forth, I dunno what you're talking about....
Then why do you say THIS, which is totally wrong. Re-Read my last post.
EDIT
AHh we talked on chat, and I realised what you are saying, which is alot less wrong, its not particularly useful what you are thinking
Originally posted by ex_banana-eater
Hm. I am talking about hypertrophy specific training. You may want to read the article explaing HST again goin_big.
You do not understand how to use HST. After you re-read it come back and ask questions
goin_big
Spend some time reading the info at the HST site.
Then come here & ask some questions once you have a good understanding of HST.
That way you`ll be doing yourself a favor as well as everyone else here
14. I read the entire site already nelson, me and banana already talked about this in chat. He was misreading what I was writing. I believe he thought that I thought the routine was a 2 week cycle while I meant that it was a 6-8 week cycle divided into 2 week blocks. Once again people are assuming, what a lovely idea it is to assume that I was wrong when i wasn't.
15. I didn`t make any assumptions, rather it was clear from your posts that you didn`t have a grasp of the concept behind HST.
Maybe you just have trouble expressing yourself which led me to make this invalid assumption.
16. well anyways, I got an answer to the original question, that was really the only thing I was wondering.
17. Call me a sissy if you want but I can't even imagine doing legs 3 days a week. And legs are my favorite bodypart to train. A decent squatter will be loading their back and knees with 350-500 lbs 3 times a week. What is really scary is they will follow it up with stiff-legged deads sometimes with, in many cases close to the same poundages. While I know this protocol works well for some people (as do ALL training methods) I can't imagine it being a long-term use method for intermediate level lifters. And please don't flame me to hell if you use it and it is working wonders for you. I don't doubt that for a minute. I do however think there are many, many protocols much more effective. As most people that are famliar with me know I'm not much of a fan of volume training, but I would bet money on a mid-level (9-12 sets BP) volume routine delivering better results. Then again, I'd bet money on 6 sets a bodypart with mid level intensity beating them both for the AVERAGE trainee. Yes, probably you. Yes, I am biased.
18. Just to shed a little light on this, the idea is to increase the weight in EACH of the 6 workouts in a phase. The thing about 5 lb increments is just an example. If you are doing leg presses, 5 lb increments are nothing. The best strategy, especially for smaller bodyparts like arms is to use percentages of the 6th workout, not fixed numbers. I try to stick to 65% or so for the first workout, and go up from there. So your increase on lighter weight exercises may be 2.5 lbs, or 1 lb.
IA, the idea seems fairly sound (I am a HIT devotee myself, just looking for a change of pace). The recommendation is 14-16 sets per whole body workout (my norm is 14 for a 1/2 split workout, so the volume increase isn't a big deal), but working out 3 times a week is new for me (normally twice) so I will be supplementing with 15g of anti-c's (BCAA & Glut) a day to help with recovery.
The biggest problem I see with HST is you don't have a firm handle if you are getting stronger at any point during the 6 weeks, whereas with HIT, I know exactly where I stand at EVERY workout.
19. i came into the same problem when i was reading about hst and trying to decide on a routine.. like all my isolation exercises would be just too low to increase weight every session.. so what i decided to do is just stick with all compound movements instead.. i do Squat, Deadlift, Incline/Flat Bnech, Bent Over Rows, Military Press, Shrugs, Barbell Curls, Close Grip Bench.. i can do a pretty go amount of weight with Hammers and OverHead Tri Extensions.. so on my 2nd workout of the week i put these in.. i just started my routine.. so nothing to report yet
20. Originally posted by Sanosuke
i came into the same problem when i was reading about hst and trying to decide on a routine.. like all my isolation exercises would be just too low to increase weight every session.
If you're doing DB laterals and your 10 RM is 30lbs going backwards by 5 your workout would look like this.
5/10/15/20/25/30
The first three workouts would be to easy what you could do is this double up on poundages each workout.
20/20/25/25/30/30
21. yeah but say your max is like 20 for 10 reps.. then you be going to the negatives.. so thats y id just stick with compounds if your a weakling like me
22. If you're weak I would recommend building up some strength before starting HST you'll get better gains!
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https://www.sciforums.com/threads/john-t-nordbergs-theory.134047/#post-3053618 | 1,695,718,510,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510179.22/warc/CC-MAIN-20230926075508-20230926105508-00889.warc.gz | 1,078,293,780 | 15,926 | # John T. Nordberg's theory...
Discussion in 'Physics & Math' started by curious45, Mar 21, 2013.
1. ### curious45Registered Member
Messages:
42
What do mathematically and physics minded folks make of this guys theories?
The posits what he calls a "Ball-of-Light Particle Model", where rather than special relativity, he defines time = c (time and the speed of light are the same). General relativity is supposed to be covered by a "euclidean form of GR".
He claims to have unified all the forces.
He has a lot of video clip explainations.
There's an experiment which he claims his model experimentally predicts. (I cant post links)
I don't really have enough physics and math comprehension to understand it all, but on the surface, it does seem that making time = c would be mathematically equivilant to some of SR, and perhaps simpler?
Would love somebody with more knowledge and skill to give feedback on these ideas?
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3. ### arfa branecall me arfValued Senior Member
Messages:
7,832
Yes, well.
I can say after quick look at his "theory" that it falls over at the first experimental test.
Since if gravity was a vector cross-product (of electric and magnetic "force" vectors), then wires would (appear to) get heavier as the current increases. This is not what we observe.
And a vector cross product can't possibly represent gravity in three or more dimensions. Can you see why?
Finally, trying to "extend" our notion of time so it becomes equivalent to the speed of light doesn't make sense either; speed is a change in distance over time (so time is "defined" inside speed or velocity). He's claiming that time is equal to change in distance divided by . . . time ?? It's self referential, so doesn't give a "new" definition at all, nor does it make any physical sense.
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5. ### curious45Registered Member
Messages:
42
Ill take your word for it that this is an implication, re gravity. I couldn't find much on people weighing electricity (especially high voltage where I suppose youd see the most difference, if this is true) except for this:
"However, you can weigh changes that act as indicators of electricity. In particular, you can insert two pieces of silver, two electrodes, into a solution of silver nitrate and weigh the pieces before and after arranging what we now call a `direct current' to flow, the weight of one electrode decreases and the other increases by the same amount."
Actually I cant see why a vector cross-product couldnt operate in three dimensions. Its a number that goes up, couldn't that be a spherical "wave" in effect like he describes?
So far as the time thing, as I understand it, he defines speed as distance traveled by the thing divided by the distance travelled by light.
Not that it makes conceptual sense to me, but then neither does most of theoretical physics.
BTW, does standard physics have an explaination for his sphere current magnetism experiment? (the magnetic current "reversing" etc)
Thats probably the aspect I am most interested in, as he presents that as "proof".
7. ### eramSciengineerValued Senior Member
Messages:
1,877
He has bungled up the concept of four-velocity.
8. ### arfa branecall me arfValued Senior Member
Messages:
7,832
This isn't relevant to his claim that gravity is a vector cross product. It is relevant to the theory that ions in solution migrate under the influence of an electric field.
Of course a cross product 'operates' in three dimensions, unfortunately a cross product of vectors has one dimension.
As to any "proof", it would need to explain why wires with a current flowing through them don't get noticeably heavier (in fact they do get very slightly heavier but not because of cross products of electric and magnetic vectors)
9. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Are you thinking of the dot product (scalar product)?
10. ### arfa branecall me arfValued Senior Member
Messages:
7,832
No, the cross product is orthogonal, or normal, to the plane the two vectors lie on, and has the same dimension as each, namely one dimension.
11. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
It can point in any direction in 3D space, and have any magnitude, so I don't see the problem?
12. ### arfa branecall me arfValued Senior Member
Messages:
7,832
I don't follow this. If you have two vectors in the plane and take their cross product, the result is another vector pointing in exactly one direction.
We got taught that cross products only make sense in three dimensions (fairly obvious), but then you find out that you can use them to find the distance from a point to a line, which is obviously two dimensions, go figure.
13. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
I may have misunderstood from the beginning
Is Nordberg's cross product supposed to represent the value of the gravitational field at a single location, or the whole field of a gravitating object?
14. ### eramSciengineerValued Senior Member
Messages:
1,877
Nobody knows. Not even him.
15. ### arfa branecall me arfValued Senior Member
Messages:
7,832
From what I've read in the link above, he's using the standard definition (is there another one?) by taking $\vec E \times \vec B$.
Apparently the resulting one-dimensional vector represents gravity. This is unlikely since (the working version of) gravity is described by a tensor with 10 independent components. The only place gravity is one dimensional is where F = mg, which as we all know is a first order approximation.
"Everything is linear to first order" is a well known incantation physicists use to thwart higher order demons and spirits.
16. ### Romain4chessRegistered Member
Messages:
2
I think what he is trying to say (at least how i get it) is that the cross product of the E and B fields tangential to the surface of the sphere has a resulting G vector that always points towards the center. Remember that this is a sphere and not just a plane piece of paper. Look at it as you have a perfectly rounded onion. Now cut this onion into extremely thin horizontal slices (lets call these the E fields) then cut the onion into extremely thin vertical slices (lets call these the B fields). The Cross product of these then give a G field towards the center (like sticking a needle towards the center of the onion from any point on the outside). So any atom within the sphere of E and B fields will experience a G force towards the center. Its a pretty fascinating topic. I don't know if its true but at least he looks at fusion in a way that seem pretty logical ( i.e you have a fixed point, and you push or accelerate all atoms towards that point). Its the same thing a fusor does and the same thing our sun does. I have been thinking of ways to do such a thing myself, but no luck really. Which is why if he is right I personally believe it will bring us much closer to fusion energy.
17. ### Romain4chessRegistered Member
Messages:
2
Also according to his theory a wire is not expected to get heavier when current flows through it. Remember that the proposed G field will point towards the center of a wire, not downwards to the earth center. So if a current flows through a wire and creates B in a circular path around the wire, this then intersects with E and creates a G vector on the wire from all directions. The fact that G acts in all directions and towards the center will not cause the wire to get heavier. Instead it will just crush the wire (as John said). I'm not saying this guy is right I'm just saying he is actually making some sense. Remember he states that this is the basics of the Z pinch.. and we know that the Z pinch works
18. ### Write4UValued Senior Member
Messages:
19,458
Question: Does resistance not suggest that the energy flowing through the wire has mass, which makes the wire heavier, when it flows through it?
19. ### Engell79Registered Senior Member
Messages:
110
John T. Nordberg has created countless Youtube vids with pseudo sicence, i woulndt waste my time on him even if i got paid to do so.
20. ### Write4UValued Senior Member
Messages:
19,458
Ty for the reference.
But it does not answer my reasonable question from an objective mathematical viewpoint.
Last edited: Sep 26, 2016
21. ### danshawenValued Senior Member
Messages:
3,950
"Self referential" is exactly what our definition of time is. How could you expect any different when the only meaning an equation can capture is a proportional relationship? When you get to the "root" of the proportional relationship that is time, you are still presented with the riddle of what is different between something traveling a linear path at c and something that may be undergoing quantum spin at speeds that are greater.
No proportional relationship (your math) can get any deeper into the meaning of time or the speed of light, no many how many math or physics PhDs you have.
22. ### Write4UValued Senior Member
Messages:
19,458
What do you mean by getting "deeper" into the meaning of time or the speed of light? The point is that the universe functions with mathematical precision. That is the deterministic part of the Universe. It cannot not function mathematically, the mathematical function IS the fundamental essence of spacetime. There are no deeper concepts necessary.
But I disagree with the proposition that Time is self-referential. IMO, Time is change-referential.
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23. ### danshawenValued Senior Member
Messages:
3,950
Proportional relativistic change self-referential, yes.
It is true that no deeper concepts (other than proportional ones) are necessary to capture all that is relativity and its associated math. But as I repeatedly pointed out here before, relativity doesn't really "capture" the nature of the quantum spin flips of paired electrons, or the speed with which the process proceeds, any more than the nature (derivation of the) fudge factor for the universal gravitational constant G is completely "captured" by Newton's or Einstein's GR field equations. It is a constant in a proportional relationship. Where does it come from? Is no "deeper" concept, other than that suggested by proportional mathematics, "necessary"? Really?
If you have no concept of the temporal dependence of quantum entangled paired electron or photon spin flips, you don't understand the fundamental basis of time at all. The proportional relationship that is time literally depends on whichever energy transfer event(s) occurs faster, and despite what relativity tells us, it isn't the propagation of light.
Gödel tried telling us this about the nature of mathematical (proportional) reasoning. It's either incomplete because the description itself is incomplete, or inconsistent because your proportional view of things caused you to divide something by zero. Few really listened.
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# Econ 281 Chapter3b - Indifference Curves 3 dimensional...
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1 Indifference Curves 3 dimensional graphs are difficult to graph and understand In practice, consumer preference is graphed using 2 goods on the X and Y axis and INDIFFERENCE CURVES Each INDIFFERENCE CURVE plots all the goods combinations that yield the same utility; that a person is indifferent between
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2 U=√2 U=2 x y Consider the utility function U=(xy) 1/2 . Each indifference curve below shows all the baskets of a given utility level. Consumers are indifferent between baskets along the same curve. 0 1 2 4 1 2
3 U=√2 U=2 x y From the indifference curves, we know that: A ≈ B, C ≈ D C A, C B, D A and D B 0 1 2 4 1 2 A B C D
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4 1). Completeness => each basket lies on only one indifference curve 2). Transitivity => indifference curves do not cross 3). Negative Slope => when a consumer likes both goods (MU a and MU b are positive), the indifference curve is downward sloping 4). Thin curves => indifference curves are not “thick”
5 x y A B IC 1 IC 2 C B A. (different indifference curves) A ≈ C (same indifference curve) B ≈ C (same indifference curve) Therefore: B ≈ A by transitivity Contradiction!
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6 IC 1 x y Y: Preferred to A A X: Less preferred More of any good is more preferred and less of a good is less preferred, so an indifference curve cannot extend into areas X or Y; it must slope downward
7 x y Since more is better, baskets B and C should be preferred to basket A, yet they all lie on the same indifference curve, implying indifference. •• 0 1 2 4 1 2 A B C
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8 University Of Alberta IC 1 IC 2 IC 3 North East C B A Example of “more is better” violation
Example : Goods people don’t care about Complete? Transitive?
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Econ 281 Chapter3b - Indifference Curves 3 dimensional...
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Ask a homework question - tutors are online | 713 | 2,638 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-05 | latest | en | 0.734013 |
http://www.numbersaplenty.com/1987 | 1,597,117,338,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738727.76/warc/CC-MAIN-20200811025355-20200811055355-00506.warc.gz | 153,134,202 | 3,315 | Search a number
1987 is a prime number
BaseRepresentation
bin11111000011
32201121
4133003
530422
613111
75536
oct3703
92647
101987
111547
121197
13b9b
14a1d
158c7
hex7c3
1987 has 2 divisors, whose sum is σ = 1988. Its totient is φ = 1986.
The previous prime is 1979. The next prime is 1993. The reversal of 1987 is 7891.
It can be divided in two parts, 19 and 87, that multiplied together give a triangular number (1653 = T57).
1987 is nontrivially palindromic in base 13.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1987 - 23 = 1979 is a prime.
1987 is an undulating number in base 13.
1987 is a lucky number.
It is a zygodrome in base 2.
It is not a weakly prime, because it can be changed into another prime (1907) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a good prime.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 993 + 994.
It is an arithmetic number, because the mean of its divisors is an integer number (994).
21987 is an apocalyptic number.
1987 is a deficient number, since it is larger than the sum of its proper divisors (1).
1987 is an equidigital number, since it uses as much as digits as its factorization.
1987 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 504, while the sum is 25.
The square root of 1987 is about 44.5757781760. The cubic root of 1987 is about 12.5718528487.
The spelling of 1987 in words is "one thousand, nine hundred eighty-seven". | 476 | 1,610 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-34 | latest | en | 0.910857 |
https://metanumbers.com/210503 | 1,638,315,272,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964359082.76/warc/CC-MAIN-20211130232232-20211201022232-00526.warc.gz | 468,169,393 | 7,363 | # 210503 (number)
210,503 (two hundred ten thousand five hundred three) is an odd six-digits composite number following 210502 and preceding 210504. In scientific notation, it is written as 2.10503 × 105. The sum of its digits is 11. It has a total of 2 prime factors and 4 positive divisors. There are 209,160 positive integers (up to 210503) that are relatively prime to 210503.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 11
• Digital Root 2
## Name
Short name 210 thousand 503 two hundred ten thousand five hundred three
## Notation
Scientific notation 2.10503 × 105 210.503 × 103
## Prime Factorization of 210503
Prime Factorization 181 × 1163
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 210503 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 210,503 is 181 × 1163. Since it has a total of 2 prime factors, 210,503 is a composite number.
## Divisors of 210503
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 211848 Sum of all the positive divisors of n s(n) 1345 Sum of the proper positive divisors of n A(n) 52962 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 458.806 Returns the nth root of the product of n divisors H(n) 3.9746 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 210,503 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 210,503) is 211,848, the average is 52,962.
## Other Arithmetic Functions (n = 210503)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 209160 Total number of positive integers not greater than n that are coprime to n λ(n) 104580 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 18791 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 209,160 positive integers (less than 210,503) that are coprime with 210,503. And there are approximately 18,791 prime numbers less than or equal to 210,503.
## Divisibility of 210503
m n mod m 2 3 4 5 6 7 8 9 1 2 3 3 5 6 7 2
210,503 is not divisible by any number less than or equal to 9.
## Classification of 210503
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (210503)
Base System Value
2 Binary 110011011001000111
3 Ternary 101200202102
4 Quaternary 303121013
5 Quinary 23214003
6 Senary 4302315
8 Octal 633107
10 Decimal 210503
12 Duodecimal a199b
16 Hexadecimal 33647
20 Vigesimal 16653
36 Base36 4ifb
## Basic calculations (n = 210503)
### Multiplication
n×y
n×2 421006 631509 842012 1052515
### Division
n÷y
n÷2 105252 70167.7 52625.8 42100.6
### Exponentiation
ny
n2 44311513009 9327706422933527 1963510185146776234081 413324784503951837602752743
### Nth Root
y√n
2√n 458.806 59.4866 21.4198 11.6052
## 210503 as geometric shapes
### Circle
Radius = n
Diameter 421006 1.32263e+06 1.39209e+11
### Sphere
Radius = n
Volume 3.90718e+16 5.56835e+11 1.32263e+06
### Square
Length = n
Perimeter 842012 4.43115e+10 297696
### Cube
Length = n
Surface area 2.65869e+11 9.32771e+15 364602
### Equilateral Triangle
Length = n
Perimeter 631509 1.91874e+10 182301
### Triangular Pyramid
Length = n
Surface area 7.67498e+10 1.09928e+15 171875
## Cryptographic Hash Functions
md5 de5f8f8fd8496ef7be2e1da67d2cdf6c 8efe11231bbe322f0bfa8a9982399f64b7f312b3 0a8f4ac402f0f00f515aa7f502ea6424ea6abdfa5a437ce616b3136478f44a4f 605a210d5fb92ad6acac84a36f674f18b5d0ea5fdda84fec519b451e5a740ecf07f3c53b380a3d47b17b63a9f7fd753405e7931935a5a3e122533d7d2cefd7bd 8610608a45dfe768c070b45b30e37d7d991ff609 | 1,477 | 4,258 | {"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} | 3.71875 | 4 | CC-MAIN-2021-49 | latest | en | 0.809679 |
https://www.mersenneforum.org/showthread.php?p=596820 | 1,675,797,532,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500628.77/warc/CC-MAIN-20230207170138-20230207200138-00103.warc.gz | 899,287,123 | 12,902 | mersenneforum.org Best msieve poly scores
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2022-01-01, 15:36 #199 swellman Jun 2012 2×52×7×11 Posts 2022 Record GNFS Polynomial Scores Code: Digits Score Found by Date Number 145 1.394e-11 fivemack 04-16 ? 146 1.232e-11 Max0526 02-17 (136*10^208-1)/9 147 1.034e-11 Max0526 03-17 (10^215+54*10^107-1)/9 148 9.335e-12 Max0526 02-17 (59*10^229+13)/9 149 7.912e-12 Max0526 03-17 (56*10^225-11)/9 150 7.312e-12 Max0526 03-19 (77*10^216-23)/9 151 6.514e-12 lorgix/spin 11-17 547^97-1 152 5.193e-12 fivemack 04-16 ? 153 4.496e-12 RichS 09-17 434040:3059 154 3.821e-12 Max0526 02-17 (17*10^222-53)/9 155 3.922e-12 Jayson King 11-12 RSA-155 156 2.893e-12 Max0526 03-17 (65*10^224-11)/9 157 2.776e-12 Max0526 02-17 (49*10^243-31)/9 158 2.430e-12 Max0526 02-17 (86*10^228-23)/9 159 2.051e-12 fivemack 04-16 9436:1335 160 1.673e-12 Max0526 02-17 (13*10^242-31)/9 161 1.471e-12 unconnected 02-16 933436:12438 162 1.263e-12 RichD 04-16 P202+1 163 1.098e-12 charybdis 06-21 2360:1756 164 9.561e-13 Gimarel 08-18 11040:10088 165 8.531e-13 Max0526 09-17 (34*10^236-7)/9 166 7.11e-13 chris2be8 04-14 3408:1399 167 6.902e-13 fivemack 04-16 8154:2379 168 5.32e-13 sashamkrt 10-13 xyyx130_119 169 4.622e-13 charybdis 04-21 2360:1737 170 3.910e-13 RichD 02-16 P226+1 (wblipp) 171 3.730e-13 Max0526 02-17 (4*10^261-13)/9 172 3.517e-13 charybdis 07-20 HCN 10+9_610L 173 3.187e-13 Gimarel 11-21 434040:3116 174 2.248e-13 charybdis 03-21 4788:12575 175 2.263e-13 Gimarel 12-15 4788:5241 176 1.771e-13 charybdis 06-20 785232:11494 177 1.545e-13 richs/spin 11-18 829332:3673 178 1.583e-13 EdH 04-20 HCN 12-7_945 179 1.226e-13 VBCurtis/spin 03-18 13*2^914-1 180 1.007e-13 CADO team 03-18 RSA-180 181 9.461e-14 Vebis/Gimarel 07-18 xyyx 145_89 182 8.175e-14 Gimarel 01-22 829332:i3712 183 6.382e-14 charybdis 08-20 HCN 10+7_302 184 5.874e-14 charybdis 10-20 HCN 7+3_317 185 4.799e-14 charybdis/spin 02-21 2_2750M 186 4.380e-14 Gimarel 01-21 HCN 7+5_346 187 3.728e-14 Max0526 10-18 R(1980) 188 3.293e-14 Gimarel 09-18 xyyx150_83 189 2.857e-14 Gimarel 01-21 4788:12574 190 2.261e-14 CADO team 05-18 RSA-190 191 2.372e-14 charybdis/spin 01-21 2_2390M 192 1.806e-14 Gimarel/spin 02-19 L1532 193 1.558e-14 swellman 07-21 2,2630M 194 1.221e-14 RichD 07-15 4788:5236 195 1.044e-14 RichD 03-16 4788:5246 196 1.035e-14 Gimarel 10-20 3366:2224 197 8.249e-15 Gimarel 08-21 2,2870M 198 7.107e-15 Gimarel 01-16 E-M 199 5.487e-15 fivemack 12-19 F1459 200 5.370e-15 CADO team 05-18 RSA-200 201 4.934e-15 Gimarel 04-21 3366:2226 202 4.117e-15 Gimarel 03-21 3408:1693 204 3.496e-15 Plutie 09-21 3,748+ 205 2.215e-15 Gimarel/spin 05-18 276:2122 206_5 2.111e-15 VBCurtis/spin 03-19 xyyx139_123 206_6 1.953e-15 swellman/spin 03-19 xyyx139_123 207_5 1.763e-15 Max0526/spin 04-19 2,2330L 207_6 1.940e-15 vebis/spin 04-19 2,2330L 208_5 1.439e-15 E. Branger 08-17 (64*10^329-1)/9 210_5 1.188e-15 vebis/spin 05-19 2,2330M 210_6 1.491e-15 CADO team 05-18 RSA-210 211_5 1.007e-15 Gimarel/spin 10-19 2,2210M 211_6 1.219e-15 VBCurtis 09-19 2,2210M 212_5 6.698e-16 Gimarel 06-13 10,770M 212_6 9.426e-16 Shi Bai et al 07-12 RSA-704 (CADO) 213_6 1.329e-15 Gimarel/spin 12-20 276:2140 216_5 3.61e-16 Gimarel 08-13 3,766+ 216_6 4.66e-16 frmky 08-13 3,766+ 217_5 4.024e-16 swellman/spin 01-20 2,1165+ 217_6 7.093e-16 VBCurtis-Gimarel/spin 12-19 2,1165+ 218_5 2.637e-16 frmky? 05-16 2,1285- 220_6 4.178e-16 charybdis 08-21 4788:12596 221_6 4.075e-16 Gimarel 12-21 2,2246M 230_6 8.065e-17 Sam Gross 08-18 RSA-230 232_6 7.690e-17 Bai et al 10-15 RSA-768 240_6 2.957e-17 Boudot et al 11-19 RSA-240 250_6 7.008e-18 Boudot et al 02-20 RSA-250 Please PM me with any corrections. Records found in 2022 are in bold. Last fiddled with by swellman on 2022-02-05 at 00:44
2022-01-01, 17:48 #200 swellman Jun 2012 2×52×7×11 Posts 2022 Record GNFS Polynomial Scores - Another View Degree 5 Code: Digits Score Found by Date Number 145 1.394e-11 fivemack 04-16 ? 146 1.232e-11 Max0526 02-17 (136*10^208-1)/9 147 1.034e-11 Max0526 03-17 (10^215+54*10^107-1)/9 148 9.335e-12 Max0526 02-17 (59*10^229+13)/9 149 7.912e-12 Max0526 03-17 (56*10^225-11)/9 150 7.312e-12 Max0526 03-19 (77*10^216-23)/9 151 6.514e-12 lorgix/spin 11-17 547^97-1 152 5.193e-12 fivemack 04-16 ? 153 4.496e-12 RichS 09-17 434040:3059 154 3.821e-12 Max0526 02-17 (17*10^222-53)/9 155 3.922e-12 Jayson King 11-12 RSA-155 156 2.893e-12 Max0526 03-17 (65*10^224-11)/9 157 2.776e-12 Max0526 02-17 (49*10^243-31)/9 158 2.430e-12 Max0526 02-17 (86*10^228-23)/9 159 2.051e-12 fivemack 04-16 9436:1335 160 1.673e-12 Max0526 02-17 (13*10^242-31)/9 161 1.471e-12 unconnected 02-16 933436:12438 162 1.263e-12 RichD 04-16 P202+1 163 1.098e-12 charybdis 06-21 2360:1756 164 9.561e-13 Gimarel 08-18 11040:10088 165 8.531e-13 Max0526 09-17 (34*10^236-7)/9 166 7.11e-13 chris2be8 04-14 3408:1399 167 6.902e-13 fivemack 04-16 8154:2379 168 5.32e-13 sashamkrt 10-13 xyyx130_119 169 4.648e-13 swellman 06-22 72708:i1788 170 4.083e-13 unconnected 02-22 54444_269 171 3.730e-13 Max0526 02-17 (4*10^261-13)/9 172 3.517e-13 charybdis 07-20 HCN 10+9_610L 173 3.187e-13 Gimarel 11-21 434040:3116 174 2.329e-13 bur 09-22 1992:1695 175 2.263e-13 Gimarel 12-15 4788:5241 176 1.771e-13 charybdis 06-20 785232:11494 177 1.545e-13 richs/spin 11-18 829332:3673 178 1.583e-13 EdH 04-20 HCN 12-7_945 179 1.226e-13 VBCurtis/spin 03-18 13*2^914-1 180 1.007e-13 CADO team 03-18 RSA-180 181 9.461e-14 Vebis/Gimarel 07-18 xyyx 145_89 182 8.175e-14 Gimarel 01-22 829332:i3712 183 6.382e-14 charybdis 08-20 HCN 10+7_302 184 5.874e-14 charybdis 10-20 HCN 7+3_317 185 4.799e-14 charybdis/spin 02-21 2_2750M 186 4.390e-14 Gimarel 03-22 11040:i10202 187 3.728e-14 Max0526 10-18 R(1980) 188 3.293e-14 Gimarel 09-18 xyyx150_83 189 2.857e-14 Gimarel 01-21 4788:12574 190 2.456e-14 Gimarel 12-22 11,759L 191 2.372e-14 charybdis/spin 01-21 2_2390M 192 1.861e-14 Gimarel 03-22 HCN 3+2,1674L 193 1.558e-14 swellman 07-21 2,2630M 194 1.361e-14 Gimarel 04-22 HCN 5+2,1090M 195 1.072e-14 swellman 06-22 HCN 11-3,825M 196 1.035e-14 Gimarel 10-20 3366:2224 197 8.249e-15 Gimarel 08-21 2,2870M 198 7.107e-15 Gimarel 01-16 E-M 199 5.571e-15 swellman 11-22 HCN 11-5,277 200 5.370e-15 CADO team 05-18 RSA-200 201 4.934e-15 Gimarel 04-21 3366:2226 202 4.117e-15 Gimarel 03-21 3408:1693 203 3.781e-15 Gimarel 10-22 xyyx147_104 204 3.496e-15 Plutie 09-21 3,748+ 205 2.215e-15 Gimarel/spin 05-18 276:2122 206 2.111e-15 VBCurtis/spin 03-19 xyyx139_123 207 1.763e-15 Max0526/spin 04-19 2,2330L 208 1.679e-15 Gimarel 11-22 HP2(4496):i314 210 1.188e-15 vebis/spin 05-19 2,2330M 211 1.007e-15 Gimarel/spin 10-19 2,2210M 212 6.698e-16 Gimarel 06-13 10,770M 216 3.61e-16 Gimarel 08-13 3,766+ 217 4.024e-16 swellman/spin 01-20 2,1165+ 218 2.637e-16 frmky? 05-16 2,1285- Degree 6 Code: Digits Score Found by Date Number 206 1.953e-15 swellman/spin 03-19 xyyx139_123 207 1.940e-15 vebis/spin 04-19 2,2330L 208 1.858e-15 swellman 10-22 HP2(4496):i314 210 1.491e-15 CADO team 05-18 RSA-210 211 1.219e-15 VBCurtis 09-19 2,2210M 212 9.426e-16 Shi Bai et al 07-12 RSA-704 (CADO) 213 1.329e-15 Gimarel/spin 12-20 276:2140 216 4.66e-16 frmky 08-13 3,766+ 217 7.093e-16 VBCurtis-Gimarel/spin 12-19 2,1165+ 220 4.178e-16 charybdis 08-21 4788:12596 221 4.075e-16 Gimarel 12-21 2,2246M 225 2.347e-16 Gimarel 11-22 2,1108+ 230 8.065e-17 Sam Gross 08-18 RSA-230 232 7.690e-17 Bai et al 10-15 RSA-768 240 2.957e-17 Boudot et al 11-19 RSA-240 250 7.008e-18 Boudot et al 02-20 RSA-250 Please PM me with any corrections. Records found in 2022 are in bold. Last fiddled with by swellman on 2022-12-21 at 14:26
2022-02-05, 23:56 #201 unconnected May 2009 Moscow, Russia 1011010100002 Posts New record poly for 170 digits number.
2022-07-04, 12:42 #202 swellman Jun 2012 2·52·7·11 Posts C195 Record 11-3_825M Of the Homogeneous Cunningham project. Code: n: 102321453509007438552518035673503689807237071382340769983097912090046069867427160104968782313205511591594280004987619773186120222377338832107141591151925451767603846299026958761993890626201339301 skew: 120110430.39 c0: -26717595915537577622988596689067944188263507700 c1: 277644615868194727166497853430607747043 c2: 18235103670163734605438695652900 c3: -84687056017688046474981 c4: -1627919593310272 c5: 1314180 Y0: -43497919484087675136316845101798495909 Y1: 47080517401981302402397 # MurphyE (Bf=8.590e+09,Bg=4.295e+09,area=2.684e+16) = 1.025e-08 # cownoise/msieve scores as 1.072e-014 Last fiddled with by swellman on 2022-07-04 at 12:46
2022-11-23, 12:37 #203 swellman Jun 2012 1111000010102 Posts C199 Record A new GNFS poly record for a C199 from 11-5,277 of the Homogeneous Cunningham project. Code: n: 5713756417392562663734372416521875819197574873591645665428848917297162453507490818079472541955120036706610908031772953520305096398798780918779670328441212324600228650424280176741157527919825437554659 skew: 80585229.11 c0: 132203115349697707126210650369442246549683208087 c1: 15179763894189484421156406573185245034164 c2: -210886048633803559196429051461753 c3: -6486886650241898705846194 c4: 55013523060473376 c5: 249923520 Y0: -174117411110133051509764609118960724108 Y1: 2963900250561308037230819 # MurphyE (Bf=8.590e+09,Bg=4.295e+09,area=2.684e+16) = 6.908e-09 # skew 80585229.11, size 1.600e-019, alpha -8.747, combined = 5.571e-15 rroots = 5
2023-01-02, 15:57 #204 swellman Jun 2012 74128 Posts 2023 Record GNFS Polynomials Degree 5 Code: Digits Score Found by Date Number 145 1.394e-11 fivemack 04-16 ? 146 1.232e-11 Max0526 02-17 (136*10^208-1)/9 147 1.034e-11 Max0526 03-17 (10^215+54*10^107-1)/9 148 9.335e-12 Max0526 02-17 (59*10^229+13)/9 149 7.912e-12 Max0526 03-17 (56*10^225-11)/9 150 7.312e-12 Max0526 03-19 (77*10^216-23)/9 151 6.514e-12 lorgix/spin 11-17 547^97-1 152 5.193e-12 fivemack 04-16 ? 153 4.496e-12 RichS 09-17 434040:3059 154 3.821e-12 Max0526 02-17 (17*10^222-53)/9 155 3.922e-12 Jayson King 11-12 RSA-155 156 2.893e-12 Max0526 03-17 (65*10^224-11)/9 157 2.776e-12 Max0526 02-17 (49*10^243-31)/9 158 2.430e-12 Max0526 02-17 (86*10^228-23)/9 159 2.051e-12 fivemack 04-16 9436:1335 160 1.673e-12 Max0526 02-17 (13*10^242-31)/9 161 1.471e-12 unconnected 02-16 933436:12438 162 1.263e-12 RichD 04-16 P202+1 163 1.098e-12 charybdis 06-21 2360:1756 164 9.561e-13 Gimarel 08-18 11040:10088 165 8.531e-13 Max0526 09-17 (34*10^236-7)/9 166 7.11e-13 chris2be8 04-14 3408:1399 167 6.902e-13 fivemack 04-16 8154:2379 168 5.32e-13 sashamkrt 10-13 xyyx130_119 169 4.648e-13 swellman 06-22 72708:i1788 170 4.083e-13 unconnected 02-22 54444_269 171 3.730e-13 Max0526 02-17 (4*10^261-13)/9 172 3.517e-13 charybdis 07-20 HCN 10+9_610L 173 3.187e-13 Gimarel 11-21 434040:3116 174 2.329e-13 bur 09-22 1992:1695 175 2.263e-13 Gimarel 12-15 4788:5241 176 1.771e-13 charybdis 06-20 785232:11494 177 1.545e-13 richs/spin 11-18 829332:3673 178 1.583e-13 EdH 04-20 HCN 12-7_945 179 1.226e-13 VBCurtis/spin 03-18 13*2^914-1 180 1.007e-13 CADO team 03-18 RSA-180 181 9.461e-14 Vebis/Gimarel 07-18 xyyx 145_89 182 8.175e-14 Gimarel 01-22 829332:i3712 183 6.382e-14 charybdis 08-20 HCN 10+7_302 184 5.874e-14 charybdis 10-20 HCN 7+3_317 185 4.799e-14 charybdis/spin 02-21 2_2750M 186 4.390e-14 Gimarel 03-22 11040:i10202 187 3.728e-14 Max0526 10-18 R(1980) 188 3.293e-14 Gimarel 09-18 xyyx150_83 189 2.857e-14 Gimarel 01-21 4788:12574 190 2.456e-14 Gimarel 12-22 11,759L[l 191 2.372e-14 charybdis/spin 01-21 2_2390M 192 1.861e-14 Gimarel 03-22 HCN 3+2,1674L 193 1.558e-14 swellman 07-21 2,2630M 194 1.361e-14 Gimarel 04-22 HCN 5+2,1090M 195 1.265e-14 EdH/spin 01-23 HCN 11+2,289 196 1.035e-14 Gimarel 10-20 3366:2224 197 8.249e-15 Gimarel 08-21 2,2870M 198 7.107e-15 Gimarel 01-16 E-M 199 5.571e-15 swellman 11-22 HCN 11-5,277 200 5.370e-15 CADO team 05-18 RSA-200 201 4.934e-15 Gimarel 04-21 3366:2226 202 4.117e-15 Gimarel 03-21 3408:1693 203 3.781e-15 Gimarel 10-22 xyyx147_104 204 3.496e-15 Plutie 09-21 3,748+ 205 2.215e-15 Gimarel/spin 05-18 276:2122 206 2.111e-15 VBCurtis/spin 03-19 xyyx139_123 207 1.763e-15 Max0526/spin 04-19 2,2330L 208 1.679e-15 Gimarel 11-22 HP2(4496):i314 210 1.188e-15 vebis/spin 05-19 2,2330M 211 1.007e-15 Gimarel/spin 10-19 2,2210M 212 6.698e-16 Gimarel 06-13 10,770M 216 3.61e-16 Gimarel 08-13 3,766+ 217 4.024e-16 swellman/spin 01-20 2,1165+ 218 2.637e-16 frmky? 05-16 2,1285- Degree 6 Code: Digits Score Found by Date Number 206 1.953e-15 swellman/spin 03-19 xyyx139_123 207 1.940e-15 vebis/spin 04-19 2,2330L 208 1.858e-15 swellman 10-22 HP2(4496):i314 210 1.491e-15 CADO team 05-18 RSA-210 211 1.219e-15 VBCurtis 09-19 2,2210M 212 9.426e-16 Shi Bai et al 07-12 RSA-704 (CADO) 213 1.329e-15 Gimarel/spin 12-20 276:2140 216 4.66e-16 frmky 08-13 3,766+ 217 7.093e-16 VBCurtis-Gimarel/spin 12-19 2,1165+ 220 4.178e-16 charybdis 08-21 4788:12596 221 4.075e-16 Gimarel 12-21 2,2246M 225 2.347e-16 Gimarel 11-22 2,1108+ 230 8.065e-17 Sam Gross 08-18 RSA-230 232 7.690e-17 Bai et al 10-15 RSA-768 240 2.957e-17 Boudot et al 11-19 RSA-240 250 7.008e-18 Boudot et al 02-20 RSA-250 Please PM me with any corrections. Records found in 2023 are in bold. Last fiddled with by swellman on 2023-01-27 at 20:27
Similar Threads Thread Thread Starter Forum Replies Last Post drone84 Msieve 4 2017-06-28 09:18 VBCurtis Msieve 0 2016-04-11 21:33 Andi47 Msieve 1 2011-03-28 04:30 Batalov Msieve 54 2010-01-13 19:45 Jeff Gilchrist Msieve 5 2008-12-29 23:07
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Tue Feb 7 19:18:52 UTC 2023 up 173 days, 16:47, 1 user, load averages: 1.15, 1.05, 0.94 | 6,958 | 13,446 | {"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} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.279687 |
https://www.mathalino.com/reviewer/plane-trigonometry/sum-and-difference-two-angles | 1,713,090,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00284.warc.gz | 836,510,832 | 10,740 | # Sum and Difference of Two Angles
Go to the derivation of sum and difference of two angles if you want information on where these formulas came from.
1. $\sin (\alpha + \beta) = \sin \alpha \, \cos \beta + \cos \alpha \, \sin \beta$
2. $\sin (\alpha - \beta) = \sin \alpha \, \cos \beta - \cos \alpha \, \sin \beta$
3. $\cos (\alpha + \beta) = \cos \alpha \, \cos \beta - \sin \alpha \, \sin \beta$
4. $\cos (\alpha - \beta) = \cos \alpha \, \cos \beta + \sin \alpha \, \sin \beta$
5. $\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \, \tan \beta}$
6. $\tan (\alpha - \beta) = \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \, \tan \beta}$ | 239 | 670 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.385707 |
https://www.arispeaks.com/geography-how-to-calculate-gradient/ | 1,575,838,535,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540514893.41/warc/CC-MAIN-20191208202454-20191208230454-00318.warc.gz | 623,944,416 | 10,517 | # Geography How To Calculate Gradient
What is Slope calculator? Slope calculator is a free online tool used to calculate the slope/gradient when coordinates of two points are given.
However, understanding where and when habitat will be impacted is not so simple, as coastal geography and the level of human development. These areas appear to have a more significant elevational.
Kinematics 2.1.1 Define displacement, velocity, speed and acceleration. Displacement Displacement is the distance moved in a particular direction.
In cycling terms, “gradient” simply refers to the steepness of a section of road. A flat road is said to have a gradient of 0%, and a road with a higher gradient (e.g. 10%) is steeper than a road with a lower gradient (e.g. 5%).
You might want to check out the Google sites below which is organised by level and topics for more Geography resources.
Key stage 1 – years 1 and 2. The principal focus of mathematics teaching in key stage 1 is to ensure that pupils develop confidence and mental fluency with whole numbers, counting and place value.
The simplest way to think about slope is to consider two points in a three. In physical geography, the orientation of the land surface is defined by its aspect, which is. A simple way to compute the slope of the river is to dissolve all the stream.
To calculate the rate of change we first need to know the change. Here we use this equation: The rate of change can be calculated from the above graph by finding the gradient of the trend line,
These are basic questions that doctors, trainers, fitness buffs and weight-watchers would all like to have answered, and now researchers from Texas have derived a fundamental equation to calculate.
In this study for Geology, Frances Westall and colleagues examine some of the oldest rocks on Earth — in the Barberton Greenstone Belt, South Africa (older than 3.3. physical weathering rates and.
Gabbro An intrusive igneous rock that develops from mafic magma and whose mineral crystals are coarse. Mineralogically this rock is identical to basalt. Gaia Hypothesis The Gaia hypothesis states that the temperature and composition of the Earth’s surface are actively controlled by life on the planet.
Nov 26, 2011. Often slope is calculated as a ratio of "rise over run" in which run is the. In civil engineering applications and physical geography, the slope is.
3. An extract from topographical map 2528DA CULLINAN. Orthophoto map 2528 DA 16 CULLINAN.
The online Gradient Calculator is able to help calculate the gradient of a straight line which is the slope.
SAS software and R were used to perform the tests and calculate the estimates. were enriched using a 44/67% discontinuous Percoll (GE Lifesciences, Pittsburgh PA) gradient. Splenic lymphocytes were.
These are basic questions that doctors, trainers, fitness buffs and weight-watchers would all like to have answered, and now researchers from Texas have derived a fundamental equation to calculate.
In this study for Geology, Frances Westall and colleagues examine some of the oldest rocks on Earth — in the Barberton Greenstone Belt, South Africa (older than 3.3. physical weathering rates and.
Make even the trickiest of maths topics accessible with the help of these CIMT teaching materials. For years, CIMT have been my first port of call when looking for high-quality practice questions, but not enough maths teachers seem to be familiar with them!
“We can then use this information to identify potential PV installation programmes based on geography, cost. models of the earth’s surface to determine the size, aspect and gradient of each roof in.
The following variables were recorded at each transect: start and end time, transect length, distance from the start of the transect to the dominant debris line, beach gradient. GIS was used to.
For attribution, the original author(s), title, publication source (PeerJ) and either DOI or URL of the article. now is known about the ‘cannabis reservoir’ relative to the geography of the United.
Hotels Near Atomic Cowboy A kid-friendly Western ski town. a trendy boutique hotel with easy access to a sugary 3-mile beach and an ideal base for kayaking, hiking, snorkeling and stand-up paddleboarding, among other. THE SPIKE. It was late-afternoon. Forty-nine of us, forty-eight men and one woman, lay on the green waiting for the spike to open. We were
Course Summary Study for a test or get caught up with your middle school geography class with the fun material in this course.
Ncert Solutions Of Social Science Class 9 Economics fiscal and social responsibility, and strategic planning, to create pathways to success for students and workforce and economic development for the community. What sets you apart from other. Free Download NCERT solutions for class 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 for all subjects like maths, science, english, hindi, sanskrit for free
SAS software and R were used to perform the tests and calculate the estimates. were enriched using a 44/67% discontinuous Percoll (GE Lifesciences, Pittsburgh PA) gradient. Splenic lymphocytes were.
Based on geography, a market is analyzed across North America. Mitcorp, Yateks, 3R, Coantec, Gradient Lens, AIT, Wohler, SENTECHAMAR NARAIN. Reasons to Get this Report: Additional global industrial.
TileStache is a Python-based server application that can serve up map tiles based on rendered geographic data. You might be familiar with TileCache the venerable open source WMS server from MetaCarta. TileStache is similar, but we hope simpler and.
subject to the constraint {eq}displaystyle , g(x, y, z, t)= x^2 + y^2 + z^2 + t^2 -225=0 , {/eq}. We calculate the gradient vectors: {eq}displaystyle , nabla f (x, y, z, t) = left langle frac.
-To construct contour maps, cross-sections and calculate gradients using. One common problem in geography is the need to portray data that have been.
Introduced or glacial relict-Phylogeography of the cryptogenic ascidian Molgula manhattensis De Kay, 1843 (Ascidiacea, Pleurogona)
Geography is unique in bridging the social sciences (human geography) with the natural sciences (physical geography). Human geography concerns the understanding of the dynamics of cultures, societies and economies, and physical geography concerns the understanding of the dynamics of physical landscapes and the environment.
Would Ecologists Work In Museums The group aims to inform decisions about conservation and the rights of local communities, and their work has led to the protection. a conservation ecologist with the Field Museum, said at the. to ask big questions, explore new ideas, and discover how and why things work. Richard Pyle, Bishop Museum, Hawaii, Documenting the Global Biodiversity.
Based on geography, a market is analyzed across North America. Mitcorp, Yateks, 3R, Coantec, Gradient Lens, AIT, Wohler, SENTECHAMAR NARAIN. Reasons to Get this Report: Additional global industrial.
To calculate the rate of change we first need to know the change. Here we use this equation: The rate of change can be calculated from the above graph by finding the gradient of the trend line,
A number of methods for calculating gradient and aspect has been proposed for a. geographical information system database at a grid spacing of 30 m.
Length of human pregnancies can vary naturally by as much as five weeks Date: August 6, 2013 Source: Oxford University Press (OUP) Summary: The length of.
In mathematics, the slope or gradient of a line is a number that describes both the direction and the steepness of the line. Slope is often denoted by the letter m; there is no clear answer to the question why the letter m is used for slope, but it might be from the "m for multiple" in the equation of a straight line "y = mx + b" or "y = mx + c". Slope is calculated by finding the ratio of.
The grade (also called slope, incline, gradient, mainfall, pitch or rise) of a physical feature, landform or constructed line refers to the tangent of the angle of that surface to the horizontal.It is a special case of the slope, where zero indicates horizontality.A larger number indicates higher or steeper degree of "tilt". Often slope is calculated as a ratio of "rise" to "run", or as a.
R Social Science Project Meyer, W. G. (2016). Estimating: the science of uncertainty. Paper presented at PMI® Global Congress 2016—EMEA, Barcelona, Spain. Newtown Square, PA: Project. To analyze the economic impacts of policies in particularly important fields such as knowledge production, human capital supply, promotion of R&D investment. such policies. This project aims to. 14 May 2018 Building Nations
Heres what i would do: Overlay the roads layer with a dtm layer, this way, all the vertices of the segments would get a Z coordinate; Calculate.
“We can then use this information to identify potential PV installation programmes based on geography, cost. models of the earth’s surface to determine the size, aspect and gradient of each roof in.
The relationship between wire length, width, area and resistance. In this investigation I am going to investigate the relationship between wire length, width and resistance.
Coordinates: 31°54’S 26°53’E. Queenstown is a town inthe Eastern Cape in. South Africa. It lies on the Komani River, which forms part of the Great Kei system of rivers.
For attribution, the original author(s), title, publication source (PeerJ) and either DOI or URL of the article. now is known about the ‘cannabis reservoir’ relative to the geography of the United.
Coastal management. A level geography. Methodology. Primary and secondary data sources. Fieldwork techniques.
High Levels of Diversity Uncovered in a Widespread Nominal Taxon: Continental Phylogeography of the Neotropical Tree Frog Dendropsophus minutus Marcelo Gehara1,2*, Andrew J. Crawford3,4, Victor G. D. Orrico5, Ariel Rodrı´guez1, Stefan Lo¨tters6, Antoine Fouquet7, Lucas S. Barrientos3, Francisco Brusquetti8, Ignacio De la Riva9, Raffael Ernst10, Giuseppe Gagliardi Urrutia11, Frank Glaw12.
In contrast, coarse particles forming a non-fluidized rough substrate are lifted at a critical upward force due to the pore pressure gradient, according to their individual masses, which provides a.
The Gradient (also called Slope) of a straight line shows how steep a straight line is. Calculate. To calculate the Gradient: Divide the change in height by the.
However, understanding where and when habitat will be impacted is not so simple, as coastal geography and the level of human development. These areas appear to have a more significant elevational.
The gradient is the slope of a linear equation, represented in the simplest form as y = mx + b. In Earth Science, the gradient is usually used to measure how. | 2,406 | 10,760 | {"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-2019-51 | latest | en | 0.937431 |
https://www.jiskha.com/display.cgi?id=1334101409 | 1,503,338,902,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109470.15/warc/CC-MAIN-20170821172333-20170821192333-00453.warc.gz | 897,109,551 | 4,383 | # ms.sue
posted by .
can you give me an easy way how to find percentages these ways:
_____% of ___?_____ = ___?____
__?___ of ________=_________
• ms.sue -
Change the percent to a decimal and multiply.
20% of 50 = 0.2 * 50 = 10
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More Similar Questions | 762 | 2,671 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2017-34 | latest | en | 0.692142 |
http://au.metamath.org/mpegif/drnguc1p.html | 1,531,792,980,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589537.21/warc/CC-MAIN-20180717012034-20180717032034-00251.warc.gz | 32,862,646 | 8,089 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > drnguc1p Structured version Unicode version
Theorem drnguc1p 20093
Description: Over a division ring, all nonzero polynomials are unitic. (Contributed by Stefan O'Rear, 29-Mar-2015.)
Hypotheses
Ref Expression
drnguc1p.p Poly1
drnguc1p.b
drnguc1p.z
drnguc1p.c Unic1p
Assertion
Ref Expression
drnguc1p
Proof of Theorem drnguc1p
StepHypRef Expression
1 simp2 958 . 2
2 simp3 959 . 2
3 eqid 2436 . . . . . 6 coe1 coe1
4 drnguc1p.b . . . . . 6
5 drnguc1p.p . . . . . 6 Poly1
6 eqid 2436 . . . . . 6
73, 4, 5, 6coe1f 16609 . . . . 5 coe1
873ad2ant2 979 . . . 4 coe1
9 drngrng 15842 . . . . 5
10 eqid 2436 . . . . . 6 deg1 deg1
11 drnguc1p.z . . . . . 6
1210, 5, 11, 4deg1nn0cl 20011 . . . . 5 deg1
139, 12syl3an1 1217 . . . 4 deg1
148, 13ffvelrnd 5871 . . 3 coe1 deg1
15 eqid 2436 . . . . 5
1610, 5, 11, 4, 15, 3deg1ldg 20015 . . . 4 coe1 deg1
179, 16syl3an1 1217 . . 3 coe1 deg1
18 eqid 2436 . . . . 5 Unit Unit
196, 18, 15drngunit 15840 . . . 4 coe1 deg1 Unit coe1 deg1 coe1 deg1
20193ad2ant1 978 . . 3 coe1 deg1 Unit coe1 deg1 coe1 deg1
2114, 17, 20mpbir2and 889 . 2 coe1 deg1 Unit
22 drnguc1p.c . . 3 Unic1p
235, 4, 11, 10, 22, 18isuc1p 20063 . 2 coe1 deg1 Unit
241, 2, 21, 23syl3anbrc 1138 1
Colors of variables: wff set class Syntax hints: wi 4 wb 177 wa 359 w3a 936 wceq 1652 wcel 1725 wne 2599 wf 5450 cfv 5454 cn0 10221 cbs 13469 c0g 13723 crg 15660 Unitcui 15744 cdr 15835 Poly1cpl1 16571 coe1cco1 16574 deg1 cdg1 19977 Unic1pcuc1p 20049 This theorem is referenced by: ig1peu 20094 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1555 ax-5 1566 ax-17 1626 ax-9 1666 ax-8 1687 ax-13 1727 ax-14 1729 ax-6 1744 ax-7 1749 ax-11 1761 ax-12 1950 ax-ext 2417 ax-rep 4320 ax-sep 4330 ax-nul 4338 ax-pow 4377 ax-pr 4403 ax-un 4701 ax-inf2 7596 ax-cnex 9046 ax-resscn 9047 ax-1cn 9048 ax-icn 9049 ax-addcl 9050 ax-addrcl 9051 ax-mulcl 9052 ax-mulrcl 9053 ax-mulcom 9054 ax-addass 9055 ax-mulass 9056 ax-distr 9057 ax-i2m1 9058 ax-1ne0 9059 ax-1rid 9060 ax-rnegex 9061 ax-rrecex 9062 ax-cnre 9063 ax-pre-lttri 9064 ax-pre-lttrn 9065 ax-pre-ltadd 9066 ax-pre-mulgt0 9067 ax-addf 9069 ax-mulf 9070 This theorem depends on definitions: df-bi 178 df-or 360 df-an 361 df-3or 937 df-3an 938 df-tru 1328 df-ex 1551 df-nf 1554 df-sb 1659 df-eu 2285 df-mo 2286 df-clab 2423 df-cleq 2429 df-clel 2432 df-nfc 2561 df-ne 2601 df-nel 2602 df-ral 2710 df-rex 2711 df-reu 2712 df-rmo 2713 df-rab 2714 df-v 2958 df-sbc 3162 df-csb 3252 df-dif 3323 df-un 3325 df-in 3327 df-ss 3334 df-pss 3336 df-nul 3629 df-if 3740 df-pw 3801 df-sn 3820 df-pr 3821 df-tp 3822 df-op 3823 df-uni 4016 df-int 4051 df-iun 4095 df-br 4213 df-opab 4267 df-mpt 4268 df-tr 4303 df-eprel 4494 df-id 4498 df-po 4503 df-so 4504 df-fr 4541 df-se 4542 df-we 4543 df-ord 4584 df-on 4585 df-lim 4586 df-suc 4587 df-om 4846 df-xp 4884 df-rel 4885 df-cnv 4886 df-co 4887 df-dm 4888 df-rn 4889 df-res 4890 df-ima 4891 df-iota 5418 df-fun 5456 df-fn 5457 df-f 5458 df-f1 5459 df-fo 5460 df-f1o 5461 df-fv 5462 df-isom 5463 df-ov 6084 df-oprab 6085 df-mpt2 6086 df-of 6305 df-1st 6349 df-2nd 6350 df-riota 6549 df-recs 6633 df-rdg 6668 df-1o 6724 df-oadd 6728 df-er 6905 df-map 7020 df-en 7110 df-dom 7111 df-sdom 7112 df-fin 7113 df-sup 7446 df-oi 7479 df-card 7826 df-pnf 9122 df-mnf 9123 df-xr 9124 df-ltxr 9125 df-le 9126 df-sub 9293 df-neg 9294 df-nn 10001 df-2 10058 df-3 10059 df-4 10060 df-5 10061 df-6 10062 df-7 10063 df-8 10064 df-9 10065 df-10 10066 df-n0 10222 df-z 10283 df-dec 10383 df-uz 10489 df-fz 11044 df-fzo 11136 df-seq 11324 df-hash 11619 df-struct 13471 df-ndx 13472 df-slot 13473 df-base 13474 df-sets 13475 df-ress 13476 df-plusg 13542 df-mulr 13543 df-starv 13544 df-sca 13545 df-vsca 13546 df-tset 13548 df-ple 13549 df-ds 13551 df-unif 13552 df-0g 13727 df-gsum 13728 df-mnd 14690 df-submnd 14739 df-grp 14812 df-minusg 14813 df-mulg 14815 df-subg 14941 df-cntz 15116 df-cmn 15414 df-abl 15415 df-mgp 15649 df-rng 15663 df-cring 15664 df-ur 15665 df-drng 15837 df-psr 16417 df-mpl 16419 df-opsr 16425 df-psr1 16576 df-ply1 16578 df-coe1 16581 df-cnfld 16704 df-mdeg 19978 df-deg1 19979 df-uc1p 20054
Copyright terms: Public domain W3C validator | 2,486 | 4,488 | {"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} | 3.078125 | 3 | CC-MAIN-2018-30 | latest | en | 0.134428 |
https://onlinetyari.com/blog/solve-number-system-questions/ | 1,656,652,672,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00395.warc.gz | 488,678,703 | 26,309 | # Tips to Solve Number System Questions in CDS 2016
0
3429
Tips to Solve Number System Questions in CDS 2016: Combined Defence Services Examination is a national level entrance exam conducted by UPSC twice every year. UPSC CDS (I) was conducted in the month of February, 2016. UPSC has released the official information on the exam dates and application process. This year, UPSC CDS (II) will be conducted on 23rd of October.
CDS (II) examination for Indian Army Academy, Indian Naval Academy and Air Force Academy comprises of three papers i.e. English, Mathematics and General Knowledge. For a complete in-depth analysis of the exam pattern of UPSC CDS (II), follow the link. Hence, we can say that the mathematics section is a key component to ensure a good score for the CDS (II) examination.
One of the most important topics of CDS Elementary mathematics is number systems. In this article, we will be focusing on the tips to solve number system questions easily. But first, lets have a look at the CDS (I) exam analysis of the number system questions.
## Analysis of Number System Questions in CDS (I) 2016 Paper
The UPSC CDS (I) exam was conducted on 14th February. Here, we have mentioned a table which highlights the questions pattern for the number system topic of CDS exam.
Topic No. of questions Basics of number system 4 LCM and HCF 3 Surds and Indices 2 BODMAS 3 Factors 3
From the above information, we can conclude that there was a fair distribution of questions for each of the topics in number systems. However, we can not rely on the previous paper of CDS exam completely. It is important for candidates to prepare for each of the topics mentioned in the syllabus for UPSC CDS (II) Elementary Mathematics.
## Tips to Solve Number System Questions in CDS
Weight age of number system in CDS exam is around 15% every year. Number system is the fundamental concept of any topic in mathematics. We are surrounded by numbers in our daily life like calculations, counting, analysis, etc. To help you in your preparation, we have mentioned some important tips to solve number system questions in CDS exam. Have a look at these:
1. To learn the number system concepts, make use of NCERT books of class 9th, 10th, 11th and 12th. This is the best way to learn the basic fundamentals.
2. Learn the types of numbers concepts, so as to solve easy questions without pen and paper. Memorize them and keep revising them till the date of your examination.
3. The best way to solve number system questions quickly is to work on your basics i.e. LCM and HCF and divisibility rules. Make your basics strong and half of your work is done.
4. Try to solve at least 50 questions of this topic ranging from easy to intermediate levels of difficulty.
5. Focus on questions related to finding unit digit place of the product of numbers. Try to solve around 50 questions of this particular topic.
6. Lastly, go to the Remainder theorem then learn the shortcuts on how to calculate number of zeros in a given factorial. Try to solve around 50 problems on remainder theorem. Let us try to learn an application of remainder theorem with an example:
Find the remainder when (123 x 453 x 656) is divided by 13.
Solution: Step 1: Find out the remainders individually.
Remainder of 123/13 = 6
Remainder of 453/13 = 11
Remainder of 656/13 = 6
Step 2: Multiply these individual remainders to find final remainder
Remainder of (123 x 453 x 656)/13= Remainder of ( ) =Remainder of ( ) = 6.
We hope all these preparation tips to solve number system questions will prove beneficial for the upcoming UPSC CDS (II) Examination. If there any doubts or queries, let us know in the comments section below. | 876 | 3,711 | {"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-2022-27 | longest | en | 0.936135 |
http://freetofindtruth.blogspot.com/2017/01/58-115-58-year-old-steve-ray-to-replace.html | 1,532,006,294,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590901.10/warc/CC-MAIN-20180719125339-20180719145339-00051.warc.gz | 142,097,860 | 22,784 | ## Sunday, January 8, 2017
### 58 115 | 58-year-old Steve Ray to replace Charles Brotman for Inauguration Day ceremonies, January 20, 2017
Notice they're having a 58-year old do it! Unreal!
Again, this past election was the 58th Presidential Election in U.S. history, and Inauguration Day, January 20, 2017, is a date with '58' numerology.
1/20/2017 = 1+20+20+17 = 58
https://en.wikipedia.org/wiki/United_States_presidential_election,_2016
United = 3+5+9+2+5+4 = 28
States = 1/10+2+1+2+5+1/10 = 12/30
United States = 40/58
Washington = 5+1+1/10+8+9+5+7+2+6+5 = 49/58
On top of that, notice the name of the 58-year old is Steve Ray, having gematria of '115'. Here is Donald Trump, taking office with the 115th Congress, after running a 511-day campaign.
Donald Trump began his Presidential Campaign June 16, 2015, exactly 511-days before election day, November 8, 2016.
https://en.wikipedia.org/wiki/Donald_Trump_presidential_campaign,_2016
June 16 can be written 16/6. The White House = 166; Secret Society = 166
http://www.cnn.com/2017/01/08/politics/donald-trump-charles-brotman-inauguration-announcer/index.html
Let us also examine the Washington Post's reporting on the switch. | 391 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-30 | latest | en | 0.918573 |
http://poj.org/showmessage?message_id=342534 | 1,638,183,546,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358705.61/warc/CC-MAIN-20211129104236-20211129134236-00635.warc.gz | 69,025,975 | 3,067 | Online JudgeProblem SetAuthorsOnline ContestsUser
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## 数组大小一定要注意，苦逼的过了几遍，最后居然是数组太小.....
Posted by 20122430139 at 2014-03-17 14:57:05 on Problem 1458
```#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define MAX 1001 //刚开始是100，就错了
#define Max(x,y) (x)>(y)?(x):(y)
int main()
{
char string1[MAX];
char string2[MAX];
//char string1a[26];
//char string1b[26];
int string[MAX][MAX];
int max1,max2,i,j;
while(scanf("%s%s",string1,string2)!=EOF)
{
memset(string, 0, sizeof(string));
//gets(string1);
//gets(string2);
//memset(c,0,sizeof(string));
max1=strlen(string1);
max2=strlen(string2);
for(i=1;i<=max1;i++)
for(j=1;j<=max2;j++)
{
if(string1[i-1]==string2[j-1])
{
string[i][j]=string[i-1][j-1]+1;
}
else
string[i][j] = Max(string[i][j-1], string[i-1][j]);
}
printf("%d\n",string[max1][max2]);
}
return 0;
} ```
Followed by: | 356 | 1,018 | {"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} | 2.6875 | 3 | CC-MAIN-2021-49 | latest | en | 0.22656 |
http://ballistipedia.com/index.php?title=Precision_Models&oldid=28 | 1,579,864,417,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250619323.41/warc/CC-MAIN-20200124100832-20200124125832-00224.warc.gz | 19,824,583 | 5,912 | # Precision Models
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
# Correction Factors
## Rayleigh correction factor
$c_{R}(n) = 4^n \sqrt{\frac{n}{\pi}} \frac{ N!(N-1)!} {(2N)!}$ To avoid overflows this is better calculated using log-gammas, as in the following spreadsheet formula:
EXP(LN(SQRT(N/PI())) + N*LN(4) + GAMMALN(N+1) + GAMMALN(N) - GAMMALN(2N+1))
## Gaussian correction factor
$\frac{1}{c_{G}(n)} = \sqrt{\frac{2}{n-1}}\,\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)} \, = \, 1 - \frac{1}{4n} - \frac{7}{32n^2} - \frac{19}{128n^3} + O(n^{-4})$ The third-order approximation is adequate. The following spreadsheet formula gives a more direct calculation:
=EXP(LN(SQRT(2/(N-1))) + GAMMALN(N/2) - GAMMALN((N-1)/2))
## Bessel correction factor
$c_{B}(n) = \frac{n}{n-1}$ | 303 | 844 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-05 | latest | en | 0.696565 |
https://artofproblemsolving.com/wiki/index.php?title=Talk:Gmaas&diff=prev&oldid=103324 | 1,606,924,981,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141711306.69/warc/CC-MAIN-20201202144450-20201202174450-00156.warc.gz | 195,142,233 | 10,298 | # Difference between revisions of "Talk:Gmaas"
Follow the following steps to summon Gmaas:
1. Draw a circle and circumscribe it with a regular hexagon
2. Write the numerical value of $17^{36}$ along the edge of the circle
3. Write the numerical value of $33^{29}$ along the edge of the hexagon
4. Write the numerical value of $cot(0)$ 5 centimeters above the hexagon
5. Write the numerical value of $tan(90)$ 5 centimeters below the hexagon
6. Write the numerical value of $ln(0)$ inside the circle
7. Write the numerical value of the melting point of water inside the circle. Include units and write 15 significant digits
8. Write the numerical value of the gravitational constant inside the circle. Include units and write 25 significant digits.
9. Carry the paper you used to write those numbers in your hand
10. Travel at the speed of light with that paper and Gmaas will be summoned | 217 | 898 | {"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} | 2.703125 | 3 | CC-MAIN-2020-50 | latest | en | 0.776225 |
http://www.math.wpi.edu/Course_Materials/MA1022D01/areaapprox/node2.html | 1,544,599,125,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00610.warc.gz | 432,765,944 | 2,571 | # Exercises
1. Suppose that the the velocity of an object traveling in one dimension is given by for . Let be the position of the object at time , assuming that .
1. As explained in your text, the position of the object at is equal to the area under the velocity curve from to . Use the midpoint rule with subdivision to approximate the position of the object at .
2. Let be the right endpoint rule and the left endpoint rule approximations with subdivisions to the position of the object at . Explain why the following relation holds for any value of .
(Hint - look at the results of leftbox and rightbox commands.)
3. Evaluate and . Based on your results, do you think your approximation using the midpoint rule with subintervals was within of the exact value for the area? Explain why or why not.
2. Consider the function
on the interval .
1. Use the error bound formula to find the smallest value of that guarantees that approximates the area to within . That is, find the smallest value of that guarantees that .
2. The value of given by the error bound is usually conservative. That is, in practice the desired accuracy can be achieved with a smaller value of . Given that
find the smallest value of such that . | 254 | 1,222 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-51 | latest | en | 0.906753 |
https://www.systutorials.com/docs/linux/man/l-cgelsy/ | 1,685,854,602,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00742.warc.gz | 1,118,559,132 | 9,720 | # cgelsy (l) - Linux Manuals
## NAME
CGELSY - computes the minimum-norm solution to a complex linear least squares problem
## SYNOPSIS
SUBROUTINE CGELSY(
M, N, NRHS, A, LDA, B, LDB, JPVT, RCOND, RANK, WORK, LWORK, RWORK, INFO )
INTEGER INFO, LDA, LDB, LWORK, M, N, NRHS, RANK
REAL RCOND
INTEGER JPVT( * )
REAL RWORK( * )
COMPLEX A( LDA, * ), B( LDB, * ), WORK( * )
## PURPOSE
CGELSY computes the minimum-norm solution to a complex linear least squares problem:
minimize || A X - B ||
using a complete orthogonal factorization of A. A is an M-by-N matrix which may be rank-deficient.
Several right hand side vectors b and solution vectors x can be handled in a single call; they are stored as the columns of the M-by-NRHS right hand side matrix B and the N-by-NRHS solution matrix X.
The routine first computes a QR factorization with column pivoting:
R11 R12 ]
R22 ]
with R11 defined as the largest leading submatrix whose estimated condition number is less than 1/RCOND. The order of R11, RANK, is the effective rank of A.
Then, R22 is considered to be negligible, and R12 is annihilated by unitary transformations from the right, arriving at the complete orthogonal factorization:
T11 0 Z
]
The minimum-norm solution is then
Zaq inv(T11)*Q1aq*B ]
]
where Q1 consists of the first RANK columns of Q.
This routine is basically identical to the original xGELSX except three differences:
o The permutation of matrix B (the right hand side) is faster and
more simple.
o The call to the subroutine xGEQPF has been substituted by the
the call to the subroutine xGEQP3. This subroutine is a Blas-3
version of the QR factorization with column pivoting.
o Matrix B (the right hand side) is updated with Blas-3.
## ARGUMENTS
M (input) INTEGER
The number of rows of the matrix A. M >= 0.
N (input) INTEGER
The number of columns of the matrix A. N >= 0.
NRHS (input) INTEGER
The number of right hand sides, i.e., the number of columns of matrices B and X. NRHS >= 0.
A (input/output) COMPLEX array, dimension (LDA,N)
On entry, the M-by-N matrix A. On exit, A has been overwritten by details of its complete orthogonal factorization.
LDA (input) INTEGER
The leading dimension of the array A. LDA >= max(1,M).
B (input/output) COMPLEX array, dimension (LDB,NRHS)
On entry, the M-by-NRHS right hand side matrix B. On exit, the N-by-NRHS solution matrix X.
LDB (input) INTEGER
The leading dimension of the array B. LDB >= max(1,M,N).
JPVT (input/output) INTEGER array, dimension (N)
On entry, if JPVT(i) .ne. 0, the i-th column of A is permuted to the front of AP, otherwise column i is a free column. On exit, if JPVT(i) = k, then the i-th column of A*P was the k-th column of A.
RCOND (input) REAL
RCOND is used to determine the effective rank of A, which is defined as the order of the largest leading triangular submatrix R11 in the QR factorization with pivoting of A, whose estimated condition number < 1/RCOND.
RANK (output) INTEGER
The effective rank of A, i.e., the order of the submatrix R11. This is the same as the order of the submatrix T11 in the complete orthogonal factorization of A.
WORK (workspace/output) COMPLEX array, dimension (MAX(1,LWORK))
On exit, if INFO = 0, WORK(1) returns the optimal LWORK.
LWORK (input) INTEGER
The dimension of the array WORK. The unblocked strategy requires that: LWORK >= MN + MAX( 2*MN, N+1, MN+NRHS ) where MN = min(M,N). The block algorithm requires that: LWORK >= MN + MAX( 2*MN, NB*(N+1), MN+MN*NB, MN+NB*NRHS ) where NB is an upper bound on the blocksize returned by ILAENV for the routines CGEQP3, CTZRZF, CTZRQF, CUNMQR, and CUNMRZ. If LWORK = -1, then a workspace query is assumed; the routine only calculates the optimal size of the WORK array, returns this value as the first entry of the WORK array, and no error message related to LWORK is issued by XERBLA.
RWORK (workspace) REAL array, dimension (2*N)
INFO (output) INTEGER
= 0: successful exit
< 0: if INFO = -i, the i-th argument had an illegal value
## FURTHER DETAILS
Based on contributions by
A. Petitet, Computer Science Dept., Univ. of Tenn., Knoxville, USA
E. Quintana-Orti, Depto. de Informatica, Universidad Jaime I, Spain
G. Quintana-Orti, Depto. de Informatica, Universidad Jaime I, Spain | 1,198 | 4,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-23 | longest | en | 0.780346 |
https://byjus.com/question-answer/a-catapult-consists-of-two-parallel-rubber-strings-each-of-length-10-text-cm-and/ | 1,716,699,801,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00414.warc.gz | 119,879,414 | 34,644 | 1
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Question
A catapult consists of two parallel rubber strings, each of length 10 cm and cross-sectional area 10 mm2 . When stretched by 5 cm, it can throw a stone of mass 100 gm to a vertical height of 25 m. Determine the Young's modulus of elasticity of rubber.
A
107 N/m2
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B
108 N/m2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
109 N/m2
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D
1010 N/m2
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Solution
The correct option is B 108 N/m2 Given, for the catapult L=10 cm,ΔL=5 cm,A=10 mm2 For the stone, m=100 gm,h=25 m A stretched catapult has elastic potential energy stored in it (Strain energy stored in both the rubber strings) U=2×(12×Stress×Strain×Volume) =2×(12×(Strain×Y)×Strain×Volume) ⇒U=2×(12×Y×(ΔLL)2×(A×L)) ⇒U=2×(12YA ΔL2L) This energy, when imparted to the stone, it flies off to a height 25 m. Energy possessed by the stone =mgh. Now, U=mgh ⇒YA ΔL2L=mgh ⇒Y=mgh LA ΔL2=(10−1 kg)×10 (m/s2)×(25 m)×(10×10−2 m)(10×10−6 m2)×(5×10−2 m)2 ∴Y=108 N/m2
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Join BYJU'S Learning Program | 468 | 1,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.756478 |
https://www.cuemath.com/jee/basic-examples-set-5-ellipses/ | 1,621,328,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989756.81/warc/CC-MAIN-20210518063944-20210518093944-00191.warc.gz | 752,229,803 | 20,255 | # Basic Examples On Ellipses Set-5
Go back to 'Ellipse'
Example – 12
Let P be a point on the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,\,\,0 < b < a.$$ Let the line parallel to the y-axis passing through P meet the circle $${x^2} + {y^2} = {a^2}$$ at the point Q such that P and Q are on the same side of the x-axis. For two positive real numbers r and s, find the locus of the point R on PQ such $$PR:RQ = r:s$$ as P varies over the ellipse.
Solution: The circle $${x^2} + {y^2} = {a^2}$$ is the auxiliary circle of the given ellipse.
We can assume point P to be $$(a\cos \theta ,\,b\sin \theta )$$ so that Q will be $$(a\cos \theta ,\,b\sin \theta )$$ The point R (h, k) divides the segment PQ internally in the ratio r : s. Thus,
$h = \frac{{ar\cos \theta + as\cos \theta }}{{r + s}},\,\,k = \frac{{ar\sin \theta + bs\sin \theta }}{{r + s}}$
We need to eliminate $$\theta$$ from these two relations to obtain a relation between h and k. Thus,
$\cos \theta = \frac{h}{a},\,\,\,\,\,\sin \theta = \frac{{k(r + s)}}{{ar + bs}}$
Squaring and adding the two gives us the required relation:
$\frac{{{h^2}}}{{{a^2}}} + \frac{{{{(r + s)}^2}{k^2}}}{{{{(ar + bs)}^2}}} = 1$
The required locus of R is
$\frac{{{x^2}}}{{{a^2}}} + \frac{{{{(r + s)}^2}{y^2}}}{{{{(ar + bs)}^2}}} = 1$
which is an ellipse, as might have been expected.
Example - 13
Consider the ellipse $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$$ Let F be its focus (ae, 0) and S be its vertex (a, 0). Consider any point P on the ellipse whose eccentric angle is $$\phi ,$$ while $$\angle SFP = \theta .$$ Prove that
$\tan \frac{\theta }{2} = \sqrt {\frac{{1 + e}}{{1 - e}}} \tan \left( {\frac{\phi }{2}} \right)$
Solution:
We have, from the figure,
$\cos \theta = \frac{{ - FT}}{{PF}}$
Since F is the focus, note that PF will simply be e times the distance of P from the directrix $$x = \frac{a}{e}.$$ Thus,
$$PF = e\left( {\frac{a}{e} - a\cos \phi } \right)$$
\begin{align}&\qquad\qquad\;\; = a - ae\cos \phi \\&\Rightarrow \quad \cos\theta = \frac{{a(\cos \phi - e)}}{{a(1 - e\cos \phi )}} = \frac{{\cos \phi - e}}{{1 - e\cos \phi }}\qquad\qquad...\left( 1 \right)\end{align}
Thus,
\begin{align}&\qquad\quad{\tan ^2}\left( {\frac{\theta }{2}} \right) = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = \frac{{(1 + e)(1 - \cos \phi )}}{{(1 - e)(1 + \cos \phi )}}\left\{ {Using{\rm{ }}\left( 1 \right)} \right\}\\&{\rm{}} \qquad\qquad\qquad\quad= \frac{{(1 + e)}}{{(1 - e)}}{\tan ^2}\left( {\frac{\phi }{2}} \right)\\&\Rightarrow \quad\; \tan \left( {\frac{\theta }{2}} \right) = \sqrt {\frac{{1 + e}}{{1 - e}}} \tan \left( {\frac{\phi }{2}} \right)\end{align} | 1,000 | 2,679 | {"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": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-21 | latest | en | 0.738499 |
https://aast.edu/en/colleges/coe/alex/dept/course-details.php?course_id=1393&unit_id=66 | 1,720,973,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00884.warc.gz | 67,490,115 | 15,560 | # Measurements & Instrumentation
• Electronics & Communications Engineering |
#### Description
Measurements of errors, Accuracy, Precision, Resolution, Sensitivity. Statistical analysis (Mean, Deviation, Standard Deviation, Variance). Units and standards of measurement. Electromechanical indicating instruments. Analog Instruments (DC Ammeter (Ayrton Shunt), DC Voltmeter, Ohmmeter (Series type, Shunt Type), AC- Instruments with Rectifiers, Bridge measurements (AC Bridges, DC Bridges), Digital instruments for measuring basic parameters, oscilloscope techniques.
#### Program
Bachelor in Electronics and Communications Engineering
#### Objectives
• Understanding the basic measurement techniques such as accuracy, precision, standards. To study rnthe operation and construction of analog, electronic and digital multi-meters.
#### Textbook
Data will be available soon!
#### Course Content
content serial Description
1Definitions, The importance of electronic measurements for engineers, Types of rnerrors.
2Statistical analysis.
3Review on the fundamental and derived units, Classification of standards, rnElectrical standards, IEEE standards.
4Permanent magnet moving coil.
5DC voltmeters, sensitivity, Use the sensitivity method for the design of DC rnvoltmeter, Analyze a circuit taken into consideration in loading effect.
6Series type and shunt type ohmmeters, Calibration of DC instruments .
77th Week Exam
8Alternating current indicating instruments, AC voltmeters with full wave rnrectifiers and half Wave rectifiers.
9DC bridges and sources of error, AC bridges.
10AC voltmeters using rectifiers.
11True RMS – Responding Voltmeter.
12Component measuring instruments, Basic Q-meter circuits: a- Direct rnconnection b- Series connection c- Parallel connection Sources of errorâ€.
13Oscilloscope measurements (phase shift, period and voltages).
14Oscilloscope block diagram.
15Oscilloscope techniques, Special oscilloscopes, (a) storage oscilloscope, (b) rnsampling oscilloscope (c) Digital storage oscilloscopeâ€.
16Final Examination
#### Markets and Career
• Generation, transmission, distribution and utilization of electrical power for public and private sectors to secure both continuous and emergency demands.
• Electrical power feeding for civil and military marine and aviation utilities.
• Electrical works in construction engineering. | 480 | 2,370 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.72245 |
https://anadat-r.davidzeleny.net/doku.php/en:numecolr:pca | 1,597,172,393,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738819.78/warc/CC-MAIN-20200811180239-20200811210239-00225.warc.gz | 211,335,423 | 9,062 | # Analysis of community ecology data in R
David Zelený
### Others
en:numecolr:pca
This script is part of supporting materials, coming with the book of Borcard et al. 2011.
To import the function definition directly into R, use the following:
```source ('http://www.davidzeleny.net/anadat-r/doku.php/en:numecolr:pca?do=export_code&codeblock=1')
```
PCA
```PCA <- function(Y, stand=FALSE)
#
# Principal component analysis (PCA) with option for variable standardization
#
# stand = FALSE : center by columns only, do not divide by s.d.
# stand = TRUE : center and standardize (divide by s.d.) by columns
#
# Author: Pierre Legendre, May 2006
{
Y = as.matrix(Y)
obj.names = rownames(Y)
var.names = colnames(Y)
size = dim(Y)
Y.cent = apply(Y, 2, scale, center=TRUE, scale=stand)
Y.cov = cov(Y.cent)
Y.eig = eigen(Y.cov)
k = length(which(Y.eig\$values > 1e-10))
U = Y.eig\$vectors[,1:k]
F = Y.cent %*% U
U2 = U %*% diag(Y.eig\$value[1:k]^(0.5))
G = F %*% diag(Y.eig\$value[1:k]^(-0.5))
rownames(F) = obj.names
rownames(U) = var.names
rownames(G) = obj.names
rownames(U2) = var.names
axenames <- paste("Axis",1:k,sep=" ")
colnames(F) = axenames
colnames(U) = axenames
colnames(G) = axenames
colnames(U2) = axenames
#
# Fractions of variance
varY = sum(diag(Y.cov))
eigval = Y.eig\$values[1:k]
relative = eigval/varY
rel.cum = vector(length=k)
rel.cum[1] = relative[1]
for(kk in 2:k) { rel.cum[kk] = rel.cum[kk-1] + relative[kk] }
#
out <- list(total.var=varY, eigenvalues=eigval, rel.eigen=relative,
rel.cum.eigen=rel.cum, U=U, F=F, U2=U2, G=G, stand=stand,
obj.names=obj.names, var.names=var.names, call=match.call() )
class(out) <- "PCA"
out
}
print.PCA <-
function(x, ...)
{
cat("\nPrincipal Component Analysis\n")
cat("\nCall:\n")
cat(deparse(x\$call),'\n')
if(x\$stand) cat("\nThe data have been centred and standardized by column",'\n')
cat("\nTotal variance in matrix Y: ",x\$total.var,'\n')
cat("\nEigenvalues",'\n')
cat(x\$eigenvalues,'\n')
cat("\nRelative eigenvalues",'\n')
cat(x\$rel.eigen,'\n')
cat("\nCumulative relative eigenvalues",'\n')
cat(x\$rel.cum.eigen,'\n')
invisible(x)
}
biplot.PCA <-
function(x, scaling=1, plot.axes=c(1,2), color.obj="black", color.var="red", ...)
# scaling = 1 : preserves Euclidean distances among the objects
# scaling = 2 : preserves correlations among the variables
{
#### Internal function
larger.frame <- function(mat, percent=0.07)
# Produce an object plot 10% larger than strictly necessary
{
range.mat = apply(mat,2,range)
z <- apply(range.mat, 2, function(x) x[2]-x[1])
range.mat[1,]=range.mat[1,]-z*percent
range.mat[2,]=range.mat[2,]+z*percent
range.mat
}
####
if(length(x\$eigenvalues) < 2) stop("There is a single eigenvalue. No plot can be produced.")
if(length(which(scaling == c(1,2))) == 0) stop("Scaling must be 1 or 2")
par(mai = c(1.0, 0.75, 1.0, 0.5))
if(scaling == 1) {
# Distance biplot, scaling type = 1: plot F for objects, U for variables
# This projection preserves the Euclidean distances among the objects
lf.F = larger.frame(x\$F[,plot.axes])
biplot(x\$F[,plot.axes],x\$U[,plot.axes],col=c(color.obj,color.var), xlim=lf.F[,1], ylim=lf.F[,2], arrow.len=0.05, asp=1)
title(main = c("PCA biplot","scaling type 1"), family="serif", line=3)
} else {
# Correlation biplot, scaling type = 2: plot G for objects, U2 for variables
# This projection preserves the correlation among the variables
lf.G = larger.frame(x\$G[,plot.axes])
biplot(x\$G[,plot.axes],x\$U2[,plot.axes],col=c(color.obj,color.var), xlim=lf.G[,1], ylim=lf.G[,2], arrow.len=0.05, asp=1)
title(main = c("PCA biplot","scaling type 2"), family="serif", line=3)
}
invisible()
}
# mite.hel = decostand(mite, "hel")
# mite.hel.D = dist(mite.hel)
# mite.correlog = mantel.correlog(mite.hel.D, XY=mite.xy, nperm=99, cutoff=FALSE)
# mite.correlog
# mite.correlog\$mantel.res
# plot(mite.correlog)``` | 1,272 | 3,848 | {"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-2020-34 | latest | en | 0.491127 |
http://cran.ma.ic.ac.uk/web/packages/lessSEM/vignettes/log-likelihood-gradients.html | 1,718,700,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.70/warc/CC-MAIN-20240618073942-20240618103942-00808.warc.gz | 8,682,422 | 4,274 | # Ableitung der Log-Likelihood
$L(\pmb\theta) = \underbrace{k\ln(2\pi)}_{1} + \underbrace{\ln(|\pmb\Sigma(\pmb\theta)|)}_{2} + \underbrace{(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))}_{3}$
Wir wollen nach $$\pmb \theta$$ ableiten.
## Element 1
Es gilt $$\frac{\partial}{\partial \theta_j} k\ln(2\pi)= 0$$
## Element 2
Es gilt:
$\frac{\partial}{\partial \theta_j}\ln(|\pmb\Sigma(\pmb\theta)|) = \frac{1}{|\pmb\Sigma(\pmb\theta)|}\frac{\partial}{\partial \theta_j}|\pmb\Sigma(\pmb\theta)|$
$\frac{\partial}{\partial \theta_j}|\pmb\Sigma(\pmb\theta)| = |\pmb\Sigma(\pmb\theta)|tr(\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta))$ und somit:
$\frac{\partial}{\partial \theta_j}\ln(|\pmb\Sigma(\pmb\theta)|) = \frac{1}{|\pmb\Sigma(\pmb\theta)|}|\pmb\Sigma(\pmb\theta)|tr(\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)) = tr(\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta))$
Wir brauchen also die Ableitung der modell-implizierten Kovarianzmatrix nach den Parametern: $$\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)$$. Dabei gilt: $$\pmb\Sigma(\pmb\theta) = \pmb F (\pmb I - \pmb A)^{-1} \pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T$$.
### Fall 1: Der Parameter $$\theta_j$$ ist in $$\pmb S$$.
Dann gilt: Außer $$\pmb S$$ kann alles andere als Konstante behandelt werden. Es folgt:
$\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta) = \pmb F (\pmb I - \pmb A)^{-1} \frac{\partial}{\partial \theta_j}\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T$ wobei $$\frac{\partial}{\partial \theta_j}\pmb S$$ eine sparse Matrix mit einsen an den Stellen ist, an denen $$\theta_j$$ vorkommt.
Zusammenfassung:
$\frac{\partial}{\partial \theta_j}\ln(|\pmb\Sigma(\pmb\theta)|) = tr(\pmb\Sigma(\pmb\theta)^{-1}\pmb F (\pmb I - \pmb A)^{-1} \frac{\partial}{\partial \theta_j}\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T)$
Achtung: Wenn die Person Missings hat, kann man die Matrix $$\pmb F$$ so anpassen, dass die entsprechenden Zeilen und Spalten herausfallen.
### Fall 2: Der Parameter $$\theta_j$$ ist in $$\pmb A$$.
Dann gilt: Außer $$\pmb A$$ kann alles andere als Konstante behandelt werden. Zudem gilt: $$\frac{\partial}{\partial a_i}\pmb A^{-1} = \pmb A^{-1}\frac{\partial \pmb A}{\partial a_i} \pmb A^{-1}$$ (https://math.stackexchange.com/questions/4074265/derivative-involving-inverse-matrix?noredirect=1&lq=1). Es folgt:
$\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta) = \pmb F[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}][\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T] + \pmb F(\pmb I - \pmb A)^{-1} \pmb S[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}]^T\pmb F^T$
Zusammenfassung:
$\frac{\partial}{\partial \theta_j}\ln(|\pmb\Sigma(\pmb\theta)|) = tr(\pmb\Sigma(\pmb\theta)^{-1}[\pmb F[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}][\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T] + \pmb F(\pmb I - \pmb A)^{-1} \pmb S[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}]^T\pmb F^T])$
### Fall 3: Der Parameter $$\theta_j$$ ist in $$\pmb m$$, wobei $$\pmb m$$ die Mittelwertstruktur des SEM ist.
Dann gilt: Die Ableitung ist $$0$$.
Hinweis: Element 2 ist unabhängig vom Datensatz!
## Element 3
$\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))$
Es gilt:
\begin{aligned} &\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))\\ =& [\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\frac{\partial}{\partial \theta_j}[\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))] \\ =& [\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}[(\pmb x - \pmb \mu(\pmb\theta))] \end{aligned}
mit $$\pmb\mu (\pmb\theta) = \pmb F(\pmb I - \pmb A)^{-1}\pmb m$$ wobei $$\pmb m$$ die Mittelwertstruktur des SEMs ist.
### Fall 1: Der Parameter $$\theta_j$$ ist in $$\pmb S$$.
Dann gilt: Außer $$\pmb S$$ kann alles andere als Konstante behandelt werden. Es folgt: $$[\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T] = 0$$ und somit
\begin{aligned} &[\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}[(\pmb x - \pmb \mu(\pmb\theta))] \\ =&(\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta)) \end{aligned}
Es gilt (https://math.stackexchange.com/questions/4074265/derivative-involving-inverse-matrix?noredirect=1&lq=1): $\frac{\partial}{\partial \theta_j} \pmb \Sigma(\pmb\theta)^{-1} = -\pmb \Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}\pmb \Sigma(\pmb\theta)\Sigma(\pmb\theta)^{-1}$ und somit:
\begin{aligned} &\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))\\ =&(\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta))\\ =& (\pmb x - \pmb \mu(\pmb\theta))^T[-\pmb \Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}\pmb \Sigma(\pmb\theta)\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta))\\ =& (\pmb x - \pmb \mu(\pmb\theta))^T[-\pmb \Sigma(\pmb\theta)^{-1}\pmb F (\pmb I - \pmb A)^{-1} \frac{\partial}{\partial \theta_j}\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta))\\ \end{aligned}
Hinweis: Der letzte Schritt wurde bei Element 2 besprochen.
### Fall 2: Der Parameter $$\theta_j$$ ist in $$\pmb A$$.
$$\pmb A$$ findet sich auch in der Mittelwertstruktur wieder. Hier gilt
\begin{aligned} &\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))\\ =& [\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}[(\pmb x - \pmb \mu(\pmb\theta))] \end{aligned}
mit $$[\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))] = [- \frac{\partial}{\partial \theta_j}\pmb \mu(\pmb\theta))] = -\frac{\partial}{\partial \theta_j}\pmb F(\pmb I - \pmb A)^{-1}\pmb m = -\pmb F(\pmb I - \pmb A)^{-1}\frac{\partial (\pmb I - \pmb A)}{\partial \theta_j}(\pmb I - \pmb A)^{-1}\pmb m$$
Es folgt: \begin{aligned} &\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))\\ =& 2*[-\pmb F(\pmb I - \pmb A)^{-1}\frac{\partial (\pmb I - \pmb A)}{\partial \theta_j}(\pmb I - \pmb A)^{-1}\pmb m]^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T[\frac{\partial}{\partial \theta_j}\pmb\Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta))\\ =& 2*[-\pmb F(\pmb I - \pmb A)^{-1}\frac{\partial (\pmb I - \pmb A)}{\partial \theta_j}(\pmb I - \pmb A)^{-1}\pmb m]^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) \\ &+ (\pmb x - \pmb \mu(\pmb\theta))^T[-\pmb \Sigma(\pmb\theta)^{-1}[\pmb F[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}][\pmb S ((\pmb I - \pmb A)^{-1})^T \pmb F^T] \\ &+ \pmb F(\pmb I - \pmb A)^{-1} \pmb S[(\pmb I - \pmb A)^{-1} \frac{\partial\pmb A}{\partial \theta_j}(\pmb I - \pmb A)^{-1}]^T\pmb F^T]\pmb \Sigma(\pmb\theta)^{-1}](\pmb x - \pmb \mu(\pmb\theta))\\ \end{aligned}
Hinweis: Der letzte Schritt wurde bei Element 3 besprochen.
### Fall 3: Der Parameter $$\theta_j$$ ist in $$\pmb m$$.
Dann gilt: Außer $$\pmb\mu (\pmb\theta) = \pmb F(\pmb I - \pmb A)^{-1}\pmb m$$ kann alles andere als Konstante behandelt werden.
\begin{aligned} &\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))\\ =& [\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\frac{\partial}{\partial \theta_j}[\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta))] \\ =& [\frac{\partial}{\partial \theta_j}(\pmb x - \pmb \mu(\pmb\theta))^T]\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}\frac{\partial}{\partial \theta_j}[(\pmb x - \pmb \mu(\pmb\theta))] \\ =& (-\pmb F(\pmb I - \pmb A)^{-1}\pmb e)^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) + (\pmb x - \pmb \mu(\pmb\theta))^T\pmb\Sigma(\pmb\theta)^{-1}(-\pmb F(\pmb I - \pmb A)^{-1}\pmb e)\\ =& 2*(- \pmb F(\pmb I - \pmb A)^{-1}\pmb e)^T\pmb\Sigma(\pmb\theta)^{-1}(\pmb x - \pmb \mu(\pmb\theta)) \end{aligned} wobei $$\pmb e = \begin{bmatrix} 0 & 0 & ... & 1 & ... &0\end{bmatrix}^T$$ ein Vektor ist, der eine eins an der Stelle hat, an der $$\theta_j$$ in $$\pmb m$$ sitzt. | 4,273 | 9,853 | {"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} | 3.484375 | 3 | CC-MAIN-2024-26 | latest | en | 0.165161 |
https://cs.stackexchange.com/questions/103611/reduction-between-these-two-languages/103631 | 1,624,109,348,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487648194.49/warc/CC-MAIN-20210619111846-20210619141846-00522.warc.gz | 190,979,752 | 38,465 | # Reduction between these two languages
I'm given $$L_\cap=\{\langle M_1\rangle\#\langle M_2\rangle\mid L(M_1)\cap L(M_2)\neq\emptyset\}$$ and $$L_U=\{\langle M\rangle\#w|M \text{ accepts } w\}$$.
How can I reduce the former to the latter: $$L_\cap\leq_RL_U$$? My idea was to build a machine $$M$$ simulating $$M_1$$ and $$M_2$$ in series. $$M$$ accepts a word if and only if both $$M_1$$ and $$M_2$$ do. So we feed all words to the machine taking $$M$$ and words until we find one, otherwise keep running. But the proplem is the machine is not halting guaranteed since it can run forever. What will be the correct solution?
• You write in words that you want to reduce $L_U$ to $L_\cap$ but in symbols that you want to reduce the other way. Which is it? – David Richerby Jan 30 '19 at 15:21
• @DavidRicherby Actually both direction. But I got stuck by one, the one as in the symbol. I had the idea now. Could you help me to see whether is correct or not? For $L_\cap\leq_R L_U$, first test the correctness whether they are two valid machine codes. If no then reject right away. If yes, rewrite a program called $A$ needing empty input and with a big while loop running through all words and in the loop let the words are fed to both $M_1$ and $M_2$ in series: Just simulate them together. If the word is accepted then return accept, otherwise check the next word (in alphabetic order). – Upc Jan 30 '19 at 15:41
• @DavidRicherby Now feed this new program $A$ with $\lambda$ to the oracle machine can solve $L_U$. If accept then output accept, otherwise reject. – Upc Jan 30 '19 at 15:43
• @XavierYang, please update in the question since the description in the comment is much better what is in the question. – John L. Jan 30 '19 at 15:44
Your solution is almost there: I do not think it is necessary to construct a machine that always halts. The problem running $$M$$ on input $$w$$ does not always halt either.
Running $$M_1$$ and $$M_2$$ over all input strings $$x$$ is tricky: indeed,if one of the machines does not halt on a particular input, then the new machine $$M$$ will not consider the next input. The solution for that is usually called dove tiling, where more and more inputs are simulated step by step.
This can be avoided (or better: hidden) by assuming non-deterministic machines. $$M$$ guesses an input $$x$$ and simulates both $$M_1$$ and $$M_2$$ on that input $$x$$. Now $$M$$ halts (on the empty input) whenever the languages of $$M_1$$ and $$M_2$$ intersect. | 695 | 2,484 | {"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": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-25 | latest | en | 0.874417 |
https://stats.stackexchange.com/questions/415903/quantify-measure-of-oscillation-and-amplitude-in-time-series-data | 1,722,711,477,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00514.warc.gz | 440,221,951 | 39,623 | Quantify measure of oscillation and amplitude in time-series data
Let's define a term, score, as a metric that measures the extent of oscillation by incorporating information about amplitude and frequency. High score means that big oscillations happen frequently. Low score means that small oscillations happen intermittently.
Compare Set 1 and Set 2. They have the same amplitude, but has different frequency, making set 1 to have lower score=2.1 then set 2 score=4.
I want to come up with a metric that allows me to quantify the score. I was thinking of taking derivatives, sum them up, and divide by the total number of points:
But I don't think this is going to work because the absolute difference amplitudes of two points depends on when I sampled them. Furthermore, my data has uneven sampling rate, which makes $$\Delta t_i$$ non-uniform.
How should I do?
$$y_t = \beta_0 + \sum_{frq=0}^n \beta_{frq} \cdot f_{frq}(t)$$
where $$f_{frq}(t) = sin(2\pi \cdot frq \cdot t)$$
the size of coefficients $$\beta$$ could then be used to approximate the score you are interested in. This approach also overcomes the irregular sampling as you only need to evaluate each sine wave at the points where you have data. | 297 | 1,217 | {"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": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2024-33 | latest | en | 0.949522 |
http://www.slidesearchengine.com/slide/digital-signal-processing-and-control-system-under-matlab-environment | 1,527,010,314,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864837.40/warc/CC-MAIN-20180522170703-20180522190703-00111.warc.gz | 461,544,699 | 10,880 | # Digital Signal Processing and Control System under MATLAB Environment
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Information about Digital Signal Processing and Control System under MATLAB Environment
Technology
Published on February 16, 2014
Author: PSJamwal10
Source: slideshare.net
## Description
A practical report on Digital Signal Processing and Control System under MATLAB Environment present in M.Tech 1st Semester during session 2013-14.
A Practical Report on DIGITAL SIGNAL PROCESSING & CONTROL SYSTEM Under MATLAB Environment Submitted To: Mr. Asim Ali Khan Associate Professor Mr. Manpreet Singh Manna Associate Professor Submitted By: Paramjeet Singh Jamwal PG/ICE/136321 M.Tech First Semester DEPARTMENT OF ELECTRICAL AND INSTRUMENTATION ENGINEERING SANT LONGOWAL INSTITUTE OF ENGINEERING & TECHNOLOGY LONGOWAL - 148106 JAN 2014 1|Page info4eee | MATLAB
S.No. 1. 2. 3. 4. 5. 6. 7. Command Clc clear all close all zeros(N) zeros(M,N) ones(N) ones(M,N) 8. subplot(m,n,p) 9. stem(y) 10. 11. 12. stem(x,y) xlabel(‘text’) ylabel(‘text’) Input(‘How many assignments’) sum(A) prod(A) mean(A) length(A) inv(A) A’ rand(N) rand(N,M) magic(N) 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. max(A) min(A) Function Clear command window and homes the cursors Removes all variables, globals, functions and MEX links. Closes all the open figure windows. An N-by-N matrix of zeros. An M-by-N matrix of zeros. An N-by-N matrix of ones. An M-by-N matrix of ones. Breaks the Figure window into an m-by-n matrix of small axes, selects the p-th axes for the current plot, and returns the axis handle. Plots the data sequence y as stems from the x axis terminated with circles for the data value. If y is a matrix then each column is plotted as a separate series. Plots the data sequence y at the values specified in x. Adds text beside the x-axis on the current axis. Adds text beside the y-axis on the current axis. To get user defined value To find sum of matrix ‘A’ column wise. To find product of ‘A’(column wise). Find mean of ‘A’. To find length of matrix ‘A’. To find inverse of matrix ‘A’ if possible. It gives transpose of matrix ‘A’. It generates N-N random matrix. It generates N-M random matrix. It generates N-N matrix whose sum of all rows and columns are equal. It shows largest element present in each column. It shows smallest element present in each column. 2|Page info4eee | MATLAB
1. Write a program to generate unit step signal. Program: n=input('enter lowest index= '); m=input('enter highest index= '); t=n:m; x=ones(1,m-n+1); stem(t,x); axis([n m 0 1.5]); title('Unit Step Signal') xlabel('Value of n') ylabel('Amplitude') Command Window: >> psj_unitstep enter lowest index= 7 enter highest index= 17 Figure Window: 3|Page info4eee | MATLAB
2. Write a program to generate ramp signal. Program: n=input('enter lowest index= '); m=input('enter highest index= '); t=n:m; x=0:m-n; stem(t,x); axis([n m 0 m-n+1]); title('Unit Ramp Signal') xlabel('Value of n') ylabel('Amplitude') Command Window: >> psj_unitramp enter lowest index= 7 enter highest index= 17 Figure Window: 4|Page info4eee | MATLAB
3. Write a program to generate unit impulse signal. Program: n=input('enter the value of impluse point= '); t=n-2:n+2; x=[zeros(1,2) ones(1,1) zeros(1,2)]; stem(t,x); axis([n-2 n+2 0 1.5]); title('Unit Impulse Signal') xlabel('Value of n') ylabel('Amplitude') Command Window: >> psj_unitimpulse enter the value of impluse point= 7 Figure Window: 5|Page info4eee | MATLAB
4. Write a program to find the convolution of two numbers. Program: x=input('Enter the value of x= '); h=input('Enter the value of h= '); lx=length(x); lh=length(h); z(1,1:lx+lh-1)=zeros; l=max(lx,lh); for i=1:l y(1,2*i-1)=x(1,i)*h(1,i); z(1,2*i-1)=z(1,2*i-1)+y(1,2*i-1); end for j=2:l for i=j:l y(1,2*i-j)=x(1,i)*h(1,i-j+1)+x(1,i-j+1)*h(1,i); z(1,2*i-j)=z(1,2*i-j)+y(1,2*i-j); end end disp('The Convolution of x & h is'); z Command Window: >> psj_convolution Enter the value of x= [2,-1,0,0,1,0,-1] Enter the value of h= [1,2,2,1,0,-1,0] The Convolution of x & h is z= 2 3 2 0 0 0 2 -1 -2 -2 0 1 0 6|Page info4eee | MATLAB
5. Write a program to generate cosine wave of different amplitude. Program: t=0:0.01:10; c1=0.5*cos(2*pi*t); c2=cos(4*pi*t); c3=(2/3)*cos(6*pi*t); c=1+c1+c2+c3; subplot(4,1,1),plot(t,c1); xlabel('time axis----->'),ylabel('Amplitude----->'); axis([0 10 -0.7 0.7]); title('0.5*cos(2*pi*t)') subplot(4,1,2),plot(t,c2); xlabel('time axis----->'),ylabel('Amplitude----->'); title('cos(4*pi*t)') axis([0 10 -1.2 1.2]); subplot(4,1,3),plot(t,c3); xlabel('time axis----->'),ylabel('Amplitude----->'); title('(2/3)*cos(6*pi*t)') subplot(4,1,4),plot(t,c); xlabel('time axis----->'),ylabel('Amplitude of c----->'); title('1+0.5*cos(2*pi*t)+cos(4*pi*t)+(2/3)*cos(6*pi*t)') Command Window: >> psj_cosine 7|Page info4eee | MATLAB
Figure Window: 8|Page info4eee | MATLAB
6. Inbuilt function for Fourier transform in MATLAB. S.No. Command FFT(X) 1. FFT(X,N) FFT(X,N,DIM) FFT2(X) 2. FFT2(X,MROWS,NCOLS) FFTN(X) 3. FFTN(X,SIZ) IFFT(X) IFFT(X,N) IFFT(X,N,DIM) 4. IFFT(..., 'symmetric') IFFT(..., 'nonsymmetric') IFFT2(F) IFFT2(F,MROWS,NCOLS) 5. IFFT2(..., 'symmetric') IFFT2(..., 'nonsymmetric') IFFTN(F) 6. IFFTN(F,SIZ) Function Discrete Fourier transform (DFT) of vector X. For matrices, the FFT operation is applied to each column. For N-D arrays, the FFT operation operates on the first non-singleton dimension. N-point FFT padded with zeros if X has less than N points and truncated if it has more. FFT operation across the dimension DIMs. Two-dimensional Fourier transform of matrix X. If X is a vector, the result will have the same orientation. Pads matrix X with zeros to size MROWS-byNCOLS before transforming. N-dimensional discrete Fourier transform of the N-D array X. If X is a vector, the output will have the same orientation. Pads X so that its size vector is SIZ before performing the transform. If any element of SIZ is smaller than the corresponding dimension of X, then X will be cropped in that dimension. Inverse discrete Fourier transform of X. N-point inverse transform. Inverse discrete Fourier transform of X across the dimension DIM. IFFT to treat X as conjugate symmetric along the active dimension. This option is useful when X is not exactly conjugate symmetric merely because of round-off error. IFFT to make no assumptions about the symmetry of X. Two-dimensional inverse Fourier transform of matrix F. If F is a vector, the result will have the same orientation. Pads matrix F with zeros to size MROWS-byNCOLS before transforming. IFFT2 to treat F as conjugate symmetric in two dimensions so that the output is purely real. This option is useful when F is not exactly conjugate symmetric merely because of round-off error. IFFT2 to make no assumptions about the symmetry of F. N-dimensional inverse discrete Fourier transform of the N-D array F. If F is a vector, the result will have the same orientation. Pads F so that its size vector is SIZ before performing the transform. If any element of SIZ is 9|Page info4eee | MATLAB
IFFTN(..., 'symmetric') IFFTN(..., 'nonsymmetric') 7. DFTMTX(N) CONJ(DFTMTX(N))/N S = SPECTROGRAM(X) S = SPECTROGRAM(X,WINDOW) 8. S= SPECTROGRAM(X,WINDOW,NOV ERLAP) S= SPECTROGRAM(X,WINDOW,NOV ERLAP,NFFT) S= SPECTROGRAM(X,WINDOW,NOV ERLAP,NFFT,Fs) [S,F,T] = SPECTROGRAM(...) [S,F,T] = smaller than the corresponding dimension of F, then F will be cropped in that dimension. IFFTN to treat F as multidimensionally conjugate symmetric so that the output is purely real. This option is useful when F is not exactly conjugate symmetric merely because of round-off error. IFFTN to make no assumptions about the symmetry of F. N-by-N complex matrix of values around the unitcircle whose inner product with a column vector of length N yields the discrete Fourier transform of the vector. If X is a column vector of length N, then DFTMTX(N)*X yields the same result as FFT(X); however, FFT(X) is more efficient. The inverse discrete Fourier transform matrix. Spectrogram of the signal specified by vector X in the matrix S. By default, X is divided into eight segment with 50% overlap, each segment is windowed with a Hamming window. The number of frequency points used to calculate the discrete Fourier transforms is equal to the maximum of 256 or the next power of two greater than the length of each segment of X. If X cannot be divided exactly into eight segments, X will be truncated accordingly. when WINDOW is a vector, divides X into segments of length equal to the length of WINDOW, and then windows each segment with the vector specified in WINDOW. If WINDOW is an integer, X is divided into segments of length equal to that integer value, and a Hamming window of equal length is used. If WINDOW is not specified, the default is used. the number of samples each segment of X overlaps. NOVERLAP must be an integer smaller than WINDOW if WINDOW is an integer. NOVERLAP must be an integer smaller than the length of WINDOW if WINDOW is a vector. If NOVERLAP is not specified, the default value is used to obtain a 50% overlap. specifies the number of frequency points used to calculate the discrete Fourier transforms. If NFFT is not specified, the default NFFT is used. Fs is the sampling frequency specified in Hz. If Fs is specified as empty, it defaults to 1 Hz. If it is not specified, normalized frequency is used. a vector of frequencies F and a vector of times T at which the spectrogram is computed. F has length equal to the number of rows of S. T has length k and its value corresponds to the center of each segment. where F is a vector of frequencies in Hz computes 10 | P a g e info4eee | MATLAB
SPECTROGRAM(X,WINDOW,NOV ERLAP,F,Fs) [S,F,T,P] = SPECTROGRAM(...) SPECTROGRAM(...) 9. [X,T] = INSTDFFT(XHAT,LOWB,UPPB) 10 [XHAT,OMEGA] = NSTDFFT(X,LOWB,UPPB) the spectrogram at those frequencies using the Goertzel algorithm. The specified frequencies in F are rounded to the nearest DFT bin commensurate with the signal's resolution. P is a matrix representing the Power Spectral Density (PSD) of each segment. For real signals, SPECTROGRAM returns the one-sided modified periodogram estimate of the PSD of each segment; for complex signals and in the case when a vector of frequencies is specified, it returns the two-sided PSD. The PSD estimate for each segment on a surface in the current figure with no output arguments. It uses SURF(f,t,10*log10(abs(P)) where P is the fourth output argument. A trailing input string, FREQLOCATION, controls where MATLAB displays the frequency axis. This string can be either 'xaxis' or 'yaxis'. Setting this FREQLOCATION to 'yaxis' displays frequency on the y-axis and time on the x-axis. The default is 'xaxis' which displays the frequency on the x-axis. If FREQLOCATION is specified when output arguments are requested, it is ignored. the inverse nonstandard FFT of XHAT, on a powerof-2 regular grid (non necessarily integers) on the interval [LOWB,UPPB]. Output arguments are X the recovered signal computed on the time interval T given by T = LOWB + [0:n-1]*(UPPB-LOWB)/n, where n is the length of XHAT. Outputs are vectors of length n. The length of XHAT must be a power of 2. a nonstandard FFT of signal X sampled on a powerof-2 regular grid (non necessarily integers) on the interval [LOWB,UPPB]. Output arguments are XHAT the shifted FFT of computed on the interval OMEGA given by OMEGA = [-n:2:n-2]/(2*(UPPBLOWB)) where n is the length of X. Outputs are vectors of length n. Length of X must be a power of 2. 11 | P a g e info4eee | MATLAB
7. Write a program to find the Fourier transform of cosine function. Program: t=0:0.01:10; x=cos(2*pi*t); y=fft(x); z=ifft(y); subplot(3,1,1); plot(x); title('Cosine Wave') subplot(3,1,2); plot(y); title('FFT of Cosine Wave') subplot(3,1,3); plot(z); title('IFFT') Command Window: >> psj_fourier Figure Window: 12 | P a g e info4eee | MATLAB
8. Write a program to find the DFT of given sequence. Program: h=input('enter sequence= '); N=input('enter N='); n=length(h); x=[h,zeros(1,N-n)]; y=zeros(N,1); for k=1:N, for n=1:N, y(k)=y(k)+x(n)*exp(-i*2*pi*(k-1)*(n-1)/N); end end disp('DFT of x='); y Command Window: >> psj_dft enter sequence= [1 2 4 8 16 32 64 128] enter N=8 DFT of x= y= 1.0e+002 * 2.5500 0.4864 + 1.6607i -0.5100 + 1.0200i -0.7864 + 0.4607i -0.8500 - 0.0000i -0.7864 - 0.4607i -0.5100 - 1.0200i 0.4864 - 1.6607i 13 | P a g e info4eee | MATLAB
9. Write a program to find the IDFT of given sequence. Program: h=input('enter sequence = '); N=input('enter N = '); n=length(x); x=[h,zeros(1,N-n)]; y=zeros(N,1); for n=1:N, for k=1:N, y(k)=y(k)+(1/N)*x(n)*exp(i*2*pi*(k-1)*(n-1)/N); end end disp('IDFT of Sequence = '); y Command Window: >> psj_idft enter sequence = [36 -4+9.656i -4+4i -4+1.656i -4 -4-1.656i -4-4i -4-9.656i] enter N = 8 IDFT of Sequence = y= 1.0000 2.0003 + 0.0000i 3.0000 + 0.0000i 4.0003 5.0000 - 0.0000i 5.9997 - 0.0000i 7.0000 + 0.0000i 7.9997 - 0.0000i 14 | P a g e info4eee | MATLAB
10. Write a program to find the DFT of given sequence using matrix. Program: x=input('Enter the sequence = '); n=length(x); w=dftmtx(n); y=[w*x']; w; disp('DFT of Sequence'); y Command Window: >> psj_dftumtx Enter the sequence = [1 2 3 4 4 3 2 1] DFT of Sequence y= 20.0000 -5.8284 - 2.4142i 0 -0.1716 - 0.4142i 0 -0.1716 + 0.4142i 0 -5.8284 + 2.4142 15 | P a g e info4eee | MATLAB
11. Write a program to find the IDFT using matrix. Program: x=input('enter the sequence = '); n=length(x); w=conj(dftmtx(n))/n; y=[inv(w)*x']/n; w; disp('IDFT of Sequence'); y Command Window: >> psj_idftumtx enter the sequence = [36 -4+9.656i -4+4i -4+1.656i -4 -4-1.656i -4-4i -49.656i] IDFT of Sequence y= 1.0000 + 0.0000i 2.0003 + 0.0000i 3.0000 - 0.0000i 4.0003 - 0.0000i 5.0000 + 0.0000i 5.9997 + 0.0000i 7.0000 - 0.0000i 7.9997 + 0.0000i 16 | P a g e info4eee | MATLAB
12. Write a program to generate DFT Matrix without using direct command. Program: N=input('Enter the value of N = '); x=zeros(N,N); for j=1:N, n=N*(j-1); for k=1:N, x(k+n)=exp(-i*2*pi*(j-1)*(k-1)/N); end end disp('Resultant matrix '); x Command Window: >> psj_dftmtx Enter the value of N = 4 Resultant matrix x= 1.0000 1.0000 1.0000 1.0000 1.0000 0.0000 - 1.0000i -1.0000 - 0.0000i -0.0000 + 1.0000i 1.0000 -1.0000 - 0.0000i 1.0000 + 0.0000i -1.0000 - 0.0000i 1.0000 -0.0000 + 1.0000i -1.0000 - 0.0000i 0.0000 - 1.0000i 17 | P a g e info4eee | MATLAB
13. Write a program to find the step response and impulse response of transfer function. Program: sys=tf([8 18 31],[1 6 14 24]) subplot(2,1,1) step(sys) subplot(2,1,2) impulse(sys) Command Window: >> psj_sni Transfer function: 8 s^2 + 18 s + 31 ----------------------s^3 + 6 s^2 + 14 s + 24 Figure Window: 18 | P a g e info4eee | MATLAB
14. Design a system to convert Fahrenheit to degree Celsius. Setup: 19 | P a g e info4eee | MATLAB
15. Design a system for torque converter. Setup: -------------------------------------A Group -------------------------------------- 20 | P a g e info4eee | MATLAB
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MATLAB Licensed Toolboxes. ... Control System: ... Analog and digital signal processing tools: | 5,182 | 17,319 | {"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-2018-22 | latest | en | 0.653367 |
https://metanumbers.com/255155900000000 | 1,623,572,732,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487607143.30/warc/CC-MAIN-20210613071347-20210613101347-00371.warc.gz | 373,183,434 | 8,282 | ## 255155900000000
255,155,900,000,000 (two hundred fifty-five trillion one hundred fifty-five billion nine hundred million) is an even fifteen-digits composite number following 255155899999999 and preceding 255155900000001. In scientific notation, it is written as 2.551559 × 1014. The sum of its digits is 32. It has a total of 17 prime factors and 162 positive divisors. There are 102,062,320,000,000 positive integers (up to 255155900000000) that are relatively prime to 255155900000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 15
• Sum of Digits 32
• Digital Root 5
## Name
Short name 255 trillion 155 billion 900 million two hundred fifty-five trillion one hundred fifty-five billion nine hundred million
## Notation
Scientific notation 2.551559 × 1014 255.1559 × 1012
## Prime Factorization of 255155900000000
Prime Factorization 28 × 58 × 2551559
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 17 Total number of prime factors rad(n) 25515590 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,155,900,000,000 is 28 × 58 × 2551559. Since it has a total of 17 prime factors, 255,155,900,000,000 is a composite number.
## Divisors of 255155900000000
162 divisors
Even divisors 144 18 9 9
Total Divisors Sum of Divisors Aliquot Sum τ(n) 162 Total number of the positive divisors of n σ(n) 6.36644e+14 Sum of all the positive divisors of n s(n) 3.81488e+14 Sum of the proper positive divisors of n A(n) 3.9299e+12 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59736e+07 Returns the nth root of the product of n divisors H(n) 64.9268 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,155,900,000,000 can be divided by 162 positive divisors (out of which 144 are even, and 18 are odd). The sum of these divisors (counting 255,155,900,000,000) is 636,643,795,131,960, the average is 3,929,899,969,950,.37.
## Other Arithmetic Functions (n = 255155900000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 102062320000000 Total number of positive integers not greater than n that are coprime to n λ(n) 12757790000000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7937162729336 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 102,062,320,000,000 positive integers (less than 255,155,900,000,000) that are coprime with 255,155,900,000,000. And there are approximately 7,937,162,729,336 prime numbers less than or equal to 255,155,900,000,000.
## Divisibility of 255155900000000
m n mod m 2 3 4 5 6 7 8 9 0 2 0 0 2 6 0 5
The number 255,155,900,000,000 is divisible by 2, 4, 5 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (255155900000000)
Base System Value
2 Binary 111010000001000000011100110010000010011100000000
3 Ternary 1020110102122222101021000110112
4 Quaternary 322001000130302002130000
5 Quinary 231420433240400000000
6 Senary 2302401000253250452
8 Octal 7201003462023400
10 Decimal 255155900000000
12 Duodecimal 2474ab0a793a28
16 Hexadecimal e8101cc82700
20 Vigesimal 14i70aif0000
36 Base36 2ig102xh4w
## Basic calculations (n = 255155900000000)
### Multiplication
n×i
n×2 510311800000000 765467700000000 1020623600000000 1275779500000000
### Division
ni
n⁄2 1.27578e+14 8.5052e+13 6.3789e+13 5.10312e+13
### Exponentiation
ni
n2 65104533304810000000000000000 16611805789468769879000000000000000000000000 4238600256837114500369136100000000000000000000000000000000 1081503863273505103744737253817990000000000000000000000000000000000000000
### Nth Root
i√n
2√n 1.59736e+07 63426.2 3996.7 760.959
## 255155900000000 as geometric shapes
### Circle
Radius = n
Diameter 5.10312e+14 1.60319e+15 2.04532e+29
### Sphere
Radius = n
Volume 6.95834e+43 8.18128e+29 1.60319e+15
### Square
Length = n
Perimeter 1.02062e+15 6.51045e+28 3.60845e+14
### Cube
Length = n
Surface area 3.90627e+29 1.66118e+43 4.41943e+14
### Equilateral Triangle
Length = n
Perimeter 7.65468e+14 2.81911e+28 2.20971e+14
### Triangular Pyramid
Length = n
Surface area 1.12764e+29 1.95772e+42 2.08334e+14
## Cryptographic Hash Functions
md5 34c7be2db83e3d3d5d1045bb561035a1 c17f022677a2a6947931eb51d3717c8c420ea288 664749438a8cc37402d130525723ddd89fa71586070c4c6c017dbfef1658cf00 c3be41c22a69ff3427e777aaba4169a768b5583f4421439ec9b8f6c3f7bd9bee772e4b57d67a962780403dc77c09d958a648abfa5692931f75d9b32bdcd79694 73ce99a89ee1620527f24830607ffae75bba63c8 | 1,781 | 4,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.76241 |
https://www.studiestoday.com/concept-cubes-and-cube-roots-cubes-and-cube-roots-basic-concepts-part-36747.html | 1,726,830,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00697.warc.gz | 920,713,242 | 22,966 | # CBSE Class 8 Mathematics Cubes And Cube Roots Basic Notes Set A
Download CBSE Class 8 Mathematics Cubes And Cube Roots Basic Notes Set A in PDF format. All Revision notes for Class 8 Mathematics have been designed as per the latest syllabus and updated chapters given in your textbook for Mathematics in Class 8. Our teachers have designed these concept notes for the benefit of Class 8 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 8 Mathematics for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 8 Mathematics given on studiestoday
## Revision Notes for Class 8 Mathematics Chapter 7 Cubes and Cube Roots
Class 8 Mathematics students should refer to the following concepts and notes for Chapter 7 Cubes and Cube Roots in Class 8. These exam notes for Class 8 Mathematics will be very useful for upcoming class tests and examinations and help you to score good marks
### Chapter 7 Cubes and Cube Roots Notes Class 8 Mathematics
CBSE basic concepts on Cubes and Cube Roots. Prepared by HOD Mathematics of one of the best CBSE schools in Delhi. Based on CBSE and CCE guidelines. The students should read and remember these concepts to gain perfection which will help him to get more marks in CBSE examination.
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### CBSE Class 8 Mathematics Chapter 7 Cubes and Cube Roots Notes
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We have provided all notes for each topic of Class 8 Mathematics Chapter 7 Cubes and Cube Roots as per latest CBSE syllabus | 1,493 | 7,104 | {"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 | 0.883706 |
https://www.reddit.com/r/askmath/comments/1o196j/a_deck_of_shuffled_cards_mixed_front_and_back_has/ | 1,516,455,263,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889617.56/warc/CC-MAIN-20180120122736-20180120142736-00551.warc.gz | 970,386,351 | 17,630 | ×
This is an archived post. You won't be able to vote or comment.
[–] 3 points4 points (0 children)
If you want to see how many times larger it is, divide by 1080. It'll be about 1.03 * 1086. What this means is, If you take the number of atoms in the universe and square it, that number will still be about a million times smaller than the number of possible permutations of cards.
[–] 2 points3 points (4 children)
104! is incorrect. It's 52! * 252 -- 84 orders of magnitude lower.
[–][deleted] (3 children)
[deleted]
[–] 1 point2 points (2 children)
So let's go from there:
52! x 252 = 3.6 x 1083
1080 is 1/3600 of that, or 0.028%
Two ways to put it:
• If you used every atom in the known universe to represent one possible permutation, you would only be 0.028% done when you run out of atoms.
• If you used more universes to keep going, you would need 3600 universes to represent all possible permutations.
[–][deleted] (1 child)
[deleted]
[–] 0 points1 point (0 children)
We're talking about cards mixed front and back, meaning some cards may be facing up and the others facing down.
• If we ignore the orientation and only care about the value of cards, there are 52! permutations.
• If, on the opposite, we ignore the card values and only care about whether they're facing up or down, 52 cards with 2 possibilities for each, that gives you 252 permutations.
• When we mix both together, we have 52! permutations for card value, and for each permutation there are 252 permutations for card orientation, giving you a total of 52! x 252 permutations of both value and orientation. | 429 | 1,610 | {"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} | 3.984375 | 4 | CC-MAIN-2018-05 | latest | en | 0.904625 |
http://www.mathworks.com/help/simbio/ug/using-rate-rules-when-the-rate-of-change-is-determined-by-another-species.html?nocookie=true | 1,398,398,515,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00535-ip-10-147-4-33.ec2.internal.warc.gz | 725,560,607 | 7,513 | Accelerating the pace of engineering and science
# Documentation Center
• Trial Software
## Create a Rate Rule for the Rate of Change That Is Determined by Another Species
A species from one reaction can determine the rate of another reaction if it is in the second reaction rate equation. Similarly, a species from a reaction can determine the rate of another species if it is in the rate rule that defines that other species.
Suppose you have a SimBiology model with three species and one reaction as defined next.
Reaction a –> b Rate equation v = -k * a Rate rule dc/dt = k2 * a Species a = 10 mole; b = 0 mole; c = 5 mole Parameters k1 = 1 second-1; k2 = 1 second-1
The solution for the species in the reaction are:
```
and
```
With the rate rule dc/dt = k2*a dependent on the reaction, dc/dt = k2(aoe-k1t), and the solution is:
`c = co + k2ao/k1(1 - e-k1t)`
Enter the following commands to set up a SimBiology model accordingly and simulate it.
```m = sbiomodel('m'); | 258 | 986 | {"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} | 3.046875 | 3 | CC-MAIN-2014-15 | longest | en | 0.904355 |
http://mizar.uwb.edu.pl/version/current/html/proofs/diff_2/41 | 1,571,853,282,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-43/segments/1570987835748.66/warc/CC-MAIN-20191023173708-20191023201208-00062.warc.gz | 128,318,366 | 1,321 | let x0, x1 be Real; :: thesis: [!sin,x0,x1!] = ((2 * (cos ((x0 + x1) / 2))) * (sin ((x0 - x1) / 2))) / (x0 - x1)
[!sin,x0,x1!] = ((sin x0) - (sin x1)) / (x0 - x1)
.= (2 * ((cos ((x0 + x1) / 2)) * (sin ((x0 - x1) / 2)))) / (x0 - x1) by SIN_COS4:16 ;
hence [!sin,x0,x1!] = ((2 * (cos ((x0 + x1) / 2))) * (sin ((x0 - x1) / 2))) / (x0 - x1) ; :: thesis: verum | 190 | 355 | {"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-2019-43 | latest | en | 0.323614 |
https://cs.stackexchange.com/questions/79674/residual-graph-of-a-graph-with-bidirectional-edges | 1,576,117,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540534443.68/warc/CC-MAIN-20191212000437-20191212024437-00206.warc.gz | 333,846,042 | 30,333 | # Residual graph of a graph with bidirectional edges?
Let's suppose we have a directed graph $G$ which has at least a pair of vertices $v,w$ such that $(v,w) \in E, (w,v) \in E$.
$e.g:$
In the example, there is an edge going from $C$ to $A$ and viceversa.
So, my question is, what would be the way to model this graph with a residual graph?
The residual graph is not a graph modeling method. It represents how we can change the flow on edges of a graph $G$ in order to increase the total flow when we compute the maximum flow.
The residual capacity used when you construct the residual graph is defined as $$c_f(u,v) = \begin{cases} c(u,v) - f(u,v) & \text{if (u,v) \in E} \\ f(v,u) & \text{if (v,u) \in E} \\ 0 & \text{otherwise} \end{cases}\$$
So, we cannot have both $(u,v)$ and $(v,u)$ in $E$ even though the antiparallel edges do not contradict the main network flow properties.
Your graph contains antiparallel edges which you should get rid of before you run a maximum flow algorithm on that graph, e.g., Ford-Fulkerson algorithm.
You could transform this graph into equivalent one with no antiparallel edges as following. You choose one antiparallel edge and "split" it into two edges. For example take $AC$ and introduce a new vertex $F$ and two new edges $AF$ and $FC$ with weights equal to $5$, i.e., $w(AF)=w(FC) = 5$. Similarly for every pair of antiparallel edges in the graph.
• If the edges represent capacities (as is usual with flow networks), there is nothing wrong with antiparallel edges. Were you thinking that the edges reflect the flow, rather than the capacities? Those are two different things, and it seems the question isn't clear about what its graph is supposed to represent. – D.W. Aug 4 '17 at 5:45
• @D.W. They are capacities. I remove antiparallel edges in order compute residual capacities. Residual capacity is defined as $cap(u,v)-f(u,v)$ if $(u,v) \in E$ and $f(v,u)$ if $(v,u) \in E$ (Ref: Cormen, and etc.). – fade2black Aug 4 '17 at 6:30
• If they are capacities, antiparallel edges are fine; you can still compute the residual graph. You define $c_f(u,v) = c(u,v)-f(u,v)$ if $f(u,v)>0$, else $c_f(u,v) = f(v,u)$ if $f(v,u)>0$, else $c_f(u,v)=0$. – D.W. Aug 4 '17 at 16:37 | 637 | 2,225 | {"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} | 3.78125 | 4 | CC-MAIN-2019-51 | latest | en | 0.908968 |
https://discuss.codecademy.com/t/try-it-x-problem/78496 | 1,537,912,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162385.84/warc/CC-MAIN-20180925202648-20180925223048-00220.warc.gz | 488,166,530 | 4,614 | # Try It! (x problem)
#1
Hello,
I have done this exercise however I'm stuck with a problem. It asks us to "Use filter() and a lambda expression to print out only the squares that are between 30 and 70 (inclusive)." So aren't we supposed to do the second version I show below instead of the correct version which in the first code.
Because It asks us to "only the squares that are between 30 and 70 ". Then why does it accept the "x"s instead of "x**2" ?
``````1)
squares=[x**2 for x in range(1,11)]
print filter(lambda x: x>=30 and x<=70, squares)
&
2)
squares=[x**2 for x in range(1,11)]
print filter(lambda x: x**2>=30 and x**2<=70, squares)``````
#2
This version doesn't have any data to filter. Consider:
``````>>> squares=[x**2 for x in range(1,11)]
>>> squares
[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
>>> print (list(filter(lambda x: x**2>=30 and x**2<=70, squares)))
[]
>>>``````
Now examine the above filter. 1**2 is 1, 4**2 is 16, 9**2 is 81, &c.
None of the values is in the prescribed range.
For future reference, consider that Python recognizes an inequality expression as a range (in a lambda)...
``30 < x < 70``
I left out the `=` since neither 30 nor 70 are perfect squares.
#3
This topic was automatically closed 7 days after the last reply. New replies are no longer allowed. | 397 | 1,306 | {"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} | 3.234375 | 3 | CC-MAIN-2018-39 | latest | en | 0.900055 |
https://www.programming-idioms.org/idiom/202/sum-of-squares/5854/ruby | 1,675,359,451,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500035.14/warc/CC-MAIN-20230202165041-20230202195041-00575.warc.gz | 959,471,847 | 7,323 | # Idiom #202 Sum of squares
Calculate the sum of squares s of data, an array of floating point values.
``s = data.sum{ _1**2 }``
``s = data.sum{|i| i**2}``
``````(defn square [x] (* x x))
(def s (reduce + (map square data)))``````
``````(defn square [x] (* x x))
(def s (transduce (map square) + data))``````
``````(defn square [x] (* x x))
(def s (->> data (map square) (reduce +)))``````
``using System.Linq;``
``var s = data.Sum(x => x * x);``
``var s = data.map((v) => v * v).reduce((sum, v) => sum + v);``
``s = sum( data**2 )``
``import "math"``
``````var s float64
for _, d := range data {
s += math.Pow(d, 2)
}``````
``def s = data.sum { it ** 2 }``
``sumOfSquares = sum . map (^2)``
``s = data.reduce((a, c) => a + c ** 2, 0)``
``import java.util.Arrays;``
``double s = Arrays.stream(data).map(i -> i * i).sum();``
``uses math;``
``````var
data: array of double;
...
s := SumOfSquares(data);
...``````
``use List::Util qw(sum);``
``my \$s = sum map { \$_ ** 2 } @data;``
``s = sum(i**2 for i in data)``
``let s = data.iter().map(|x| x.powi(2)).sum::<f32>();``
Bart | 395 | 1,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-06 | longest | en | 0.404048 |
https://ladywiththepants.com/d450e3d39c4ff3231b141cef23486137.html | 1,653,341,524,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00043.warc.gz | 404,870,942 | 2,879 | # Is negative number can be odd?
## Is negative number can be odd?
An odd number is an integer when divided by two, either leaves a remainder or the result is a fraction. One is the first odd positive number but it does not leave a remainder 1. Since odd numbers are integers, negative numbers can be odd.
## Is even number or not?
All the numbers ending with 0,2,4,6 and 8 are even numbers. For example, numbers such as 14, 26, 32, 40 and 88 are even numbers. If we divide a number into two groups with an equal number of elements in each, then the number is an even number.
## Do even numbers include negatives?
An even number is an integer that can be divided exactly or evenly by \color{red}2. By performing mental math, it's obvious that the numbers below, including the negative numbers, are even because they are all divisible by 2.
## What does even number mean?
Definition of even number : a whole number that is able to be divided by two into two equal whole numbers The numbers 0, 2, 4, 6, and 8 are even numbers.
## What is even numbers with examples?
Even Number: Any number which is exactly divisible by 2 is called an even number. i.e. if a number when divided by 2 leaves no remainder, then the number is called an even number. Examples of Even numbers: 2, 4, 6, 8, 10, 12, 14, 42, 100, 398, 996 etc.
## What is even and odd numbers?
What are even and odd numbers? Even numbers are divisible by 2 without remainders. They end in 0, 2, 4, 6, or 8. Odd numbers are not evenly divisible by 2 and end in 1, 3, 5, 7, or 9. You can tell whether a number is odd or even regardless of how many digits it has by looking at the final digit.
## What are even and odd numbers with examples?
All the numbers ending with 1,3,5,7 and 9 are odd numbers. For example, numbers such as 11, 23, 35, 47 etc. For example, numbers such as 14, 26, 32, 40 and 88 are even numbers.
Is Zamasu stronger than Goku?
Is Yondu the most powerful?
Is Yamcha the strongest human?
How powerful is the Hulk World-Breaker?
## Are old Tupperware containers safe to use?
Should you throw away old Tupperware? If your Tupperware container is old, you should use it for other purposes and no longer store or reheat food. Plastic containers that are cracked or warped are not safe since they might trap bacteria, and scratched surfaces can also leak harmful chemicals when microwaved.
## Does Oliver Queen have a kid in Arrow?
Oliver Queen's biological son Connor Hawke first appeared in Green Arrow #0 (October 1994) and was created by Kelley Puckett and Jim Aparo. His mother was Sandra “Moonday” Hawke, a woman Ollie had had a brief relationship with in his useless playboy days.
## Does Oliver Queen have a son in Arrow?
Ben Lewis is joining the cast of the CW's Arrow for a season-long arc after being introduced in Monday night's season premiere as the future adult version of William “Will” Clayton, the son of Oliver Queen and the late Samantha Clayton. | 749 | 2,955 | {"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-2022-21 | latest | en | 0.945178 |
https://kurst-wear.com/qa/how-many-cm-are-in-a-ruler.html | 1,627,441,928,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153521.1/warc/CC-MAIN-20210728025548-20210728055548-00262.warc.gz | 371,561,754 | 17,966 | # How Many Cm Are In A Ruler?
## How many centimeter does a ruler have?
30 centimetersThe ruler is the most common measuring device.
In the United States, most rulers have the imperial measurements along one long edge while the other long edge shows the metric measurements.
A standard ruler is 12-inches long and 30 centimeters in length..
## What is 4 by 6 inches in cm?
Size table of standard photosSize (cm)Accurate size (mm)Size in inches10 x 15 cm102 x 152 mm4″ x 6″13 x 18 cm127 x 178 mm5″ x 7″15 x 21 cm152 x 216 mm6″ x 8,5″18 x 24 cm180 x 240 mm7″ x 9,5″6 more rows
## Where is CM in ruler?
A ruler can be defined as a tool or device used to measure length and draw straight lines. A ruler or measuring tape can be used to measure lengths in both metric and customary units. Here, the ruler is marked in centimeters (cms) along the top and in inches along the bottom.
## What size is a 4 by 6 photo?
4×6: 4×6 prints measure approximately 4” x 5 ⅞”. This is the standard size in the photofinishing industry because this print size mirrors the aspect ratio of most digital cameras viewfinder. 4×6 prints are perfect for framed photos, cards and for a physical backup of any of your favorite digital images.
## Is 1 cm the same as 1 inch?
The term centimetre is abbreviated as “cm” where one centimetre is equal to the one-hundredth of a meter. In short, 1 centimetre = 0.01 meter = 10 millimeter = 0.3937 inches. The relationship between inch and cm is that one inch is exactly equal to 2.54 cm in the metric system.
## How many cm exactly is an inch?
2.54 cmWe know that 1 inch = 2.54 cm.
## What size is 2.5 cm in inches?
Convert 2.5 Centimeters to Inches2.5 Centimeters (cm)0.984252 Inches (in)1 cm = 0.393701 in1 in = 2.540000 cm
## What height is 5 foot 6?
5 foot and 12 inches is the same as 6 foot 0 inches….4 feet 0 inches= 121.92 centimeters5 feet 5 inches= 165.10 centimeters5 feet 6 inches= 167.64 centimeters5 feet 7 inches= 170.18 centimeters29 more rows
## What size is 18 inches in cm?
Quick lookup: Inches to cmInches118cm2.5445.72
## How long is an inch on your finger?
One inch (2.5 cm) is roughly the measurement from the top knuckle on your thumb to your thumb tip. Measure yours to see how close it is to 1 inch.
## What size is 7 cm in inches?
Centimeters to Inches tableCentimetersInches5 cm1.97 in6 cm2.36 in7 cm2.76 in8 cm3.15 in16 more rows
## How many cm is 12 inch ruler?
Inches to CM Conversioninch112cm2.5430.48
## How long is a ruler?
12 in or 30 cm in length is useful for a ruler to be kept on a desk to help in drawing. Shorter rulers are convenient for keeping in a pocket. Longer rulers, e.g., 18 in (46 cm) are necessary in some cases. Rigid wooden or plastic yardsticks, 1 yard long, and meter sticks, 1 meter long, are also used.
## Who big is a centimeter?
Equivalence to other units of length1 centimetre= 10 millimetres= 0.01 metres= 0.393700787401574803149606299212598425196850 inches(There are exactly 2.54 centimetres in one inch.)
## What size is 2 cm?
2 CM equals to 0.7874015748031495 Inches.
## Is 6 foot a good height for a guy?
6′1″/185cm is the ideal male height. It’s somewhat tall without being too tall. … The ideal height range for men though is 6′1″ to 6′5″. 6′0″/183 is seen as the shortest height in the acceptable range, and most women will be completely fine with a guy who is six feet tall and most men won’t consider him short or small.
## Is a Dollar Bill 6 inches long?
SIZE: US currency bills are are 2.61 inches wide and 6.14 inches long; they are . 0043 inches thick and weigh 1 gram. … PRODUCTION OF DOLLAR BILLS: It costs the US government 4.2 cents to produce a U.S. bill.
## How long is 1 cm on a ruler?
Each centimeter is labeled on the ruler (1-30). Example: You take out a ruler to measure the width of your fingernail. The ruler stops at 1 cm, meaning that your nail is precisely 1 cm wide. So if you counted five lines from 9 cm, for instance, you’d get 9.5 cm (or 95 mm).
## Which is bigger 1 cm or 1 inch?
When converting from a smaller to a larger unit, you will always have fewer of the larger units. A centimeter is smaller than an inch, so a given length will have more centimeters than inches.
## What size is 50 cm in inches?
Convert 50 Centimeters to Inchescmin50.0019.68550.0119.68950.0219.69350.0319.69796 more rows | 1,294 | 4,381 | {"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} | 3.046875 | 3 | CC-MAIN-2021-31 | latest | en | 0.882893 |
http://grado33.ml/ryfew/homework-help-precalculus-gad.php | 1,550,557,332,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247489425.55/warc/CC-MAIN-20190219061432-20190219083432-00341.warc.gz | 114,950,318 | 7,139 | Skip Nav
# Precalculus Homework Help: Answers & Questions
## We Can Help with Your Pre-Calculus
❶Go to chapter Working with Complex Numbers: Lesson 7 - How to Write a Linear Equation.
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Expiration is not a valid, future date. Year Expiration Year is required. Zip Code Zip code is required. Secure Server tell me more. Lesson 1 - Functions: Lesson 2 - Transformations: How to Shift Graphs on a Plane. Lesson 5 - How to Compose Functions. Lesson 6 - Inverse Functions. Lesson 8 - What is Function Notation: Lesson 9 - Discontinuous Functions: Lesson 10 - Monotonic Function: Lesson 1 - What Is an Exponential Function? Lesson 2 - Exponential Growth vs.
Lesson 3 - What is a Logarithm? Lesson 4 - How to Graph Logarithms: Lesson 5 - How to Evaluate Logarithms. Lesson 6 - Logarithmic Properties. Lesson 7 - Practice Problems for Logarithmic Properties. Lesson 8 - How to Solve Exponential Equations. Lesson 9 - How to Solve Logarithmic Equations. Lesson 10 - Logarithmic Function: Lesson 1 - What is an Inequality? Lesson 2 - How to Graph 1- and 2-Variable Inequalities. Lesson 4 - Graphing Inequalities: Lesson 1 - What are the Different Types of Numbers?
Lesson 3 - What is a Linear Equation? Lesson 4 - Linear Equations: Intercepts, Standard Form and Graphing. Lesson 7 - How to Write a Linear Equation. Lesson 8 - What is a System of Equations? Lesson 1 - What is a Parabola? Lesson 3 - What is a Function? Lesson 5 - How to Factor Quadratic Equations: Lesson 6 - Factoring Quadratic Equations: Polynomial Problems with a Non-1 Leading Coefficient. Lesson 7 - How to Complete the Square. Lesson 8 - Completing the Square Practice Problems.
Lesson 12 - Graphing Circles: Identifying the Formula, Center and Radius. Lesson 3 - How to Simplify Expressions with Exponents. Lesson 4 - Rational Exponents.
Lesson 5 - Simplifying Expressions with Rational Exponents. Lesson 1 - What are Piecewise Functions? Lesson 3 - How to Graph Piecewise Functions.
Lesson 4 - Translating Piecewise Functions. Lesson 5 - How to Evaluate Composite Functions. Simply post your question and get it answered by professional tutor within 30 minutes. Usually, this class is a preparation period before a calculus class and is based on every aspect of algebra, which was studied during high school years. Sometimes pre-calculus is divided into algebra and trigonometry, which can be an additional headache for any student.
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What we could find: | 1,157 | 5,249 | {"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-2019-09 | longest | en | 0.864938 |
https://fr.mathworks.com/matlabcentral/cody/problems/44480-the-average-of-the-second-largest-values/solutions/1409575 | 1,580,081,021,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694071.63/warc/CC-MAIN-20200126230255-20200127020255-00082.warc.gz | 453,921,206 | 15,835 | Cody
# Problem 44480. The average of the second largest values
Solution 1409575
Submitted on 8 Jan 2018 by Erik Luiten
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
x = [1 2 3 4;8 7 6 5;9 0 1 4]; y_correct = (3 + 7 + 4) / 3; assert(isequal(ave_2ndM(x),y_correct))
t = 3 7 4
2 Pass
x = [2 4 ;5 9; 0 5;88 52]; y_correct = (2 + 5 + 0 + 52) / 4; assert(isequal(ave_2ndM(x),y_correct))
t = 2 5 0 52
3 Pass
x = [97 93 68 58 24 74 55 49 40 24 63 53 46 37 19]; y_correct = (93 + 55 + 53) / 3; assert(isequal(ave_2ndM(x),y_correct))
t = 93 55 53
4 Pass
x = [ 82 65 14 40 44 73 81 18 53 2 15 46 40 42 99 66 44 84 66 17 52 83 81 63 11 98 9 7 30 38]; y_correct = 61.5; assert(isequal(ave_2ndM(x),y_correct))
t = 65 73 46 66 81 38 | 393 | 855 | {"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-2020-05 | latest | en | 0.471385 |
https://www.free-online-calculator-use.com/five-number-summary-calculator.html | 1,713,084,985,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00147.warc.gz | 718,379,220 | 33,007 | Five Number Summary Calculator with Step-by-Step Solution
This calculator will find an entered data set's minimum value, first quartile, median, third quartile, maximum value, and interquartile range.
Plus, the calculator generates the step-by-step solution and a visual representation of the results.
Five Number Summary Calculator
Calculate the five-number summary of a data set, and see the step-by-step solution.
Special Instructions
Selected Data Record:
A Data Record is a set of calculator entries that are stored in your web browser's Local Storage. If a Data Record is currently selected in the "Data" tab, this line will list the name you gave to that data record. If no data record is selected, or you have no entries stored for this calculator, the line will display "None".
DataData recordData recordSelected data record: None
Load or Clear Sample Data Set:
To quickly see how the 5-Number Summary Calculator works, tap the "Sample" button. To clear sample entries, tap the "Clear" button.
Enter or paste data set:Enter or paste data set:Enter or paste data set:Enter or paste data set:
Data set:
Enter each element of the data set (or paste a copied data set) into this text box. Be sure each number is separated by a space, a comma, a line return, or any combination of the three.
Elements:# of elements:Number of elements in set:Numbers of elements in set:
Number of elements in set:
This is the total number of elements detected in the data set field.
Minimum:Minimum:Minimum:Minimum:
Minimum:
This is the minimum or smallest value in the data set.
Q1:1st quartile:1st quartile (lower):First quartile (Q1/lower):
1st quartile (lower):
This is the 1st quartile of the data set based on the interpolation method, which is just one of many methods that can be used. The larger the data set, the closer the results of all methods will be to each other.
Median:Median:Median (middle value):Median (middle value):
Median:
This is the median of the data set based on the interpolation method of calculating quartiles, which is just one of many methods that can be used. The larger the data set, the closer the results of all methods will be to each other.
Q3:3rd quartile:Third quartile (upper):Third quartile (Q3/upper):
Third quartile (upper):
This is the third quartile of the data set based on the interpolation method of calculating quartiles, which is just one of many methods that can be used. The larger the data set, the closer the results of all methods will be to each other.
Maximum:Maximum:Maximum:Maximum:
Maximum:
This is the maximum or largest value in the data set.
Inter range:Interquartile range:Interquartile range:Interquartile range:
Interquartile range:
This is the interquartile range, which results from of subtracting Q1 from Q3.
If you would like to save the current entries to the secure online database, tap or click on the Data tab, select "New Data Record", give the data record a name, then tap or click the Save button. To save changes to previously saved entries, simply tap the Save button. Please select and "Clear" any data records you no longer need.
Learn
What a 5-Number Summary is and How to Calculate it.
What is a five-number summary?
The 5-number summary gives you a quick snapshot of the distribution of a data set using five descriptive statistics.
The five descriptive statistics are:
• Minimum (smallest)
• 1st quartile (lower quartile)
• Median (middle value)
• 3rd quartile (upper quartile)
• Maximum (largest)
The summary can then be used to compare data sets, especially when converted into visual representations called box plots or box and whisker plots.
How to Calculate Five Number Summary
To illustrate how to calculate a 5-number summary, let's use the following example data set:
90, 130, 400, 200, 350, 70, 325, 250, 150, 275, 270, 150, 130, 59, 200, 450, 300, 220, 100, 200, 400, 200, 250, 95, 180, 170, 150
I used the following steps to calculate the Five Number Summary. Please note that the method I use for calculating quartiles is only one of many different methods that can be used.
Sort Data Set In Descending Order
PositionValue
159
270
390
495
5100
6130
7130
8150
9150
10150
11170
12180
13200
14200
15200
16200
17220
18250
19250
20270
21275
22300
23325
24350
25400
26400
27450
Find the Minimum and Maximum
Based on the table above, the minimum value in the data set is 59, and the maximum value in the data set is 450.
Find the 1st Quartile (Lower)
To find the position of the 1st quartile (lower), we solve for the following equation, where n is equal to the number of values in the set (27):
Q1 Position = 0.25(n + 1) = 0.25(27+ 1) = 0.25 x 28 = 7
Since 7 is an integer, the 1st quartile is the value in the 7th position, which is 130.
Find the Median (Middle)
To find the position of the middle value we solve for the following equation, where n is equal to the number of values in the set (27):
Median Position = 0.50(n + 1) = 0.50(27+ 1) = 0.50 x 28 = 14
Since 14 is an integer, the median is the value in the 14th position, which is 200.
Find the 3rd Quartile (Upper)
To find the position of the 3rd quartile, we solve for the following equation, where n is equal to the number of values in the set (27):
Q3 Position = 0.75(n + 1) = 0.75(27+ 1) = 0.75 x 28 = 21
Since 21 is an integer, the 3rd quartile is the value in the 21st position, which is 275.
Below are the color-coded results of my 5 Number Summary calculations for the entered set of data, followed by my attempt to create a rough visual presentation of the summary (hopefully somewhat similar to an official Box and Wisper Plot).
5 Number Summary
Minimum59
Q1 (Lower)130
Median (Middle)200
Q3 (Upper)275
Maximum450
500 -410 -320 -230 -140 -
But what about a data set that does not yield integers for positions? In that case, we can use interpolation to find the median and quartiles. To show how interpolating is used, let's remove the last value from the earlier example:
90, 130, 400, 200, 350, 70, 325, 250, 150, 275, 270, 150, 130, 59, 200, 450, 300, 220, 100, 200, 400, 200, 250, 95, 180, 170
I used the following steps to calculate the Five Number Summary. Please note that the method I use for calculating quartiles is only one of many different methods that can be used.
Sort Data Set In Descending Order
PositionValue
159
270
390
495
5100
6130
7130
8150
9150
10170
11180
12200
13200
14200
15200
16220
17250
18250
19270
20275
21300
22325
23350
24400
25400
26450
Find the Minimum and Maximum
Based on the table above, the minimum value in the data set is 59, and the maximum value in the data set is 450.
Find the 1st Quartile (Lower)
To find the position of the 1st quartile (lower), we solve for the following equation, where n is equal to the number of values in the set (26):
Q1 Position = 0.25(n + 1) = 0.25(26+ 1) = 0.25 x 27 = 6.75
Since 6.75 is not an integer, we can use interpolation to find the 1st quartile that is located somewhere between the values located at the 6th and 7th positions, which are 130 and 130, respectively.
Step #1: Find the difference between the two values between which Q1 is located:
130 - 130 = 0
Step #2: Get the decimal part of the Q1 position and multiply it by the result in step #1:
0.75 x 0 = 0
Step #3: Add the result in step # 2 to the smallest number in step #1:
0 + 130 = 130
Q1 = 130
Find the Median (Middle)
To find the position of the middle value we solve for the following equation, where n is equal to the number of values in the set (26):
Median Position = 0.50(n + 1) = 0.50(26+ 1) = 0.50 x 27 = 13.5
Since 13.5 is not an integer, we can use interpolation to find the median that is located somewhere between the values located at the 13th and 14th positions, which are 200 and 200, respectively.
Step #1: Find the difference between the two values between which Q1 is located:
200 - 200 = 0
Step #2: Get the decimal part of the median position and multiply it by the result in step #1:
0.5 x 0 = 0
Step #3: Add the result in step # 2 to the smallest number in step #1:
0 + 200 = 200
Median = 200
Find the 3rd Quartile (Upper)
To find the position of the 3rd quartile, we solve for the following equation, where n is equal to the number of values in the set (26):
Q3 Position = 0.75(n + 1) = 0.75(26+ 1) = 0.75 x 27 = 20.25
Since 20.25 is not an integer, we can use interpolation to find the 1st quartile that is located somewhere between the values located at the 20th and 21st positions, which are 275 and 300, respectively.
Step #1: Find the difference between the two values between which Q1 is located:
300 - 275 = 25
Step #2: Get the decimal part of the Q3 position and multiply it by the result in step #1:
0.25 x 25 = 6.25
Step #3: Add the result in step # 2 to the smallest number in step #1:
6.25 + 275 = 281.25
Q3 = 281.25
Below are the color-coded results of my 5 Number Summary calculations for the entered set of data, followed by my attempt to create a rough visual presentation of the summary (hopefully somewhat similar to an official Box and Wisper Plot).
5 Number Summary
Minimum59
Q1 (Lower)130
Median (Middle)200
Q3 (Upper)281.25
Maximum450
500 -410 -320 -230 -140 -
Move the slider to left and right to adjust the calculator width. Note that the Help and Tools panel will be hidden when the calculator is too wide to fit both on the screen. Moving the slider to the left will bring the instructions and tools panel back into view.
Also note that some calculators will reformat to accommodate the screen size as you make the calculator wider or narrower. If the calculator is narrow, columns of entry rows will be converted to a vertical entry form, whereas a wider calculator will display columns of entry rows, and the entry fields will be smaller in size ... since they will not need to be "thumb friendly". | 2,668 | 9,856 | {"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} | 3.5 | 4 | CC-MAIN-2024-18 | longest | en | 0.845036 |
http://forums.worden.com/keeploggedin.aspx?g=posts&t=68418 | 1,547,740,000,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658988.30/warc/CC-MAIN-20190117143601-20190117165551-00038.warc.gz | 88,012,846 | 20,608 | How to use the formula expression？ Rate this Topic:
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yx25606
Posted : Sunday, April 15, 2018 12:22:13 PM
Registered User
Joined: 10/5/2016
Posts: 71
Y=SMA(X,N,M) ≠avg(x,n)
Y=(M*X+(N-M)*Y')/N
How to use the formula expression？How to have values from the first day?
yx25606
Posted : Sunday, April 15, 2018 12:24:57 PM
Registered User
Joined: 10/5/2016
Posts: 71
Y=SMA(X,N,M) Is not equal to avg(x,n)
Y=(M*X+(N-M)*Y')/N
How to use the formula expression？How to have values from the first day?
Bruce_L
Posted : Monday, April 16, 2018 12:31:17 PM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
If Y' is the previous value of the moving average, X is whatever is being averaged, M = 1, and N is the period of the moving average, then you can use the following (this is Wilder's smoothing which is a type of exponential moving average).
XAVG(X, P)
Where P = (2 * N) - 1
If Y' is the previous value of the moving average, X is whatever is being averaged, M = 2 AND N is the period of the moving average plus one, then you can use the following (this is a normal exponential moving average average).
XAVG(X, P)
Where P is the period of the moving average.
The moving average is calculated entirely from the values on the chart, so it will not start plotting until the algoritm being used has finished initializing.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
yx25606
Posted : Monday, April 16, 2018 1:36:25 PM
Registered User
Joined: 10/5/2016
Posts: 71
Thank you！
XAVG(c,2*(10-1))?Can't seem to pass。
5 days average numerical, must start from the fifth day is in front of the four days?
Bruce_L
Posted : Monday, April 16, 2018 1:41:23 PM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
The second argument has to be an numeric integer and can't be a formula. So you have to write it as follows (and it would be ((2 * 10 - 1) and not (2 * (10 - 1)) as far as the value to use goes.
XAVG(C, 19)
Or as:
`XAVGC19`
You are going to have four or five blank spaces before the moving average starts plotting if the period used is 5. So in the case of the above, you would have 18 or 19 blank spaces before the moving average starts plotting.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
yx25606
Posted : Monday, April 16, 2018 2:45:57 PM
Registered User
Joined: 10/5/2016
Posts: 71
Thank you！
Understand, but the problem was not solved, can have a look at the breakthrough in the sent to your email? No function can be done?
Bruce_L
Posted : Monday, April 16, 2018 2:53:44 PM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
I think I am just not understanding what you are saying. It looks like we already replied to the only email we received from you on the subject.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
yx25606
Posted : Monday, April 16, 2018 3:16:08 PM
Registered User
Joined: 10/5/2016
Posts: 71
May line has a problem, I again hair once, figure can see?
Bruce_L
Posted : Monday, April 16, 2018 3:35:40 PM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
I have poured over the email along with each of the three attachments and I have no other ideas as to what is being asked (several other people have looked at it as well to try and brainshtorm about what you are asking).
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
yx25606
Posted : Monday, April 16, 2018 4:03:34 PM
Registered User
Joined: 10/5/2016
Posts: 71
Basic like this, don't do the first day there will be an average value。Thank you very much！
You are going to have four or five blank spaces before the moving average starts plotting if the period used is 5. So in the case of the above, you would have 18 or 19 blank spaces before the moving average starts plotting.
Bruce_L
Posted : Monday, April 16, 2018 4:15:56 PM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
Just guessing again, but the Personal Criteria Formula Language doesn't have a way to reference the previously calculated value in a formula. So while you can use a Custom PCF Cumulative Indicator which can add or subtract a calculation from a running total, there is no way to reference the previous value of the running total to use in the calculations.
There also isn't a way to use nested IIF() functions in order to test how many bars of data are available and then use different calculations based on the number of bars. The moment you get to a test which asks for a number of bars exceeding the number of available bars, the IIF() function will return an error instead of returning the true or false result.
Say you are trying to return a 5 period simple moving average of C with the data series starting on the first bar. You can't just return C for the first bar, (C + C1) / 2 for the second bar (C + C1 + C2) / 3 for the third bar, (C + C1 + C2 + C3) / 4 for the fourth bar, and (C + C1 + C2 + C3 + C4) / 5 for all of the rest of the bars. The concept is sound, but there isn't a way to implement it in the Personal Criteria Formula Language.
-Bruce
Personal Criteria Formulas
TC2000 Support Articles
yx25606
Posted : Monday, April 16, 2018 11:00:12 PM
Registered User
Joined: 10/5/2016
Posts: 71
Delayed so much of your time, thank you very much！
Bruce_L
Posted : Tuesday, April 17, 2018 9:22:32 AM
Worden Trainer
Joined: 10/7/2004
Posts: 64,504
You're welcome.
-Bruce
Personal Criteria Formulas
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https://number.academy/40245 | 1,656,861,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104244535.68/warc/CC-MAIN-20220703134535-20220703164535-00479.warc.gz | 464,318,750 | 12,018 | Number 40245
Number 40,245 spell 🔊, write in words: forty thousand, two hundred and forty-five . Ordinal number 40245th is said 🔊 and write: forty thousand, two hundred and forty-fifth. The meaning of number 40245 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 40245. What is 40245 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 40245.
What is 40,245 in other units
The decimal (Arabic) number 40245 converted to a Roman number is (XL)CCXLV. Roman and decimal number conversions.
Weight conversion
40245 kilograms (kg) = 88724.1 pounds (lbs)
40245 pounds (lbs) = 18255.0 kilograms (kg)
Length conversion
40245 kilometers (km) equals to 25008 miles (mi).
40245 miles (mi) equals to 64769 kilometers (km).
40245 meters (m) equals to 132036 feet (ft).
40245 feet (ft) equals 12267 meters (m).
40245 centimeters (cm) equals to 15844.5 inches (in).
40245 inches (in) equals to 102222.3 centimeters (cm).
Temperature conversion
40245° Fahrenheit (°F) equals to 22340.6° Celsius (°C)
40245° Celsius (°C) equals to 72473° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
40245 seconds equals to 11 hours, 10 minutes, 45 seconds
40245 minutes equals to 3 weeks, 6 days, 22 hours, 45 minutes
Zip codes 40245
• Zip code 40245 Louisville, Kentucky, Jefferson, USA a map
Codes and images of the number 40245
Number 40245 morse code: ....- ----- ..--- ....- .....
Sign language for number 40245:
Number 40245 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
Mathematics of no. 40245
Multiplications
Multiplication table of 40245
40245 multiplied by two equals 80490 (40245 x 2 = 80490).
40245 multiplied by three equals 120735 (40245 x 3 = 120735).
40245 multiplied by four equals 160980 (40245 x 4 = 160980).
40245 multiplied by five equals 201225 (40245 x 5 = 201225).
40245 multiplied by six equals 241470 (40245 x 6 = 241470).
40245 multiplied by seven equals 281715 (40245 x 7 = 281715).
40245 multiplied by eight equals 321960 (40245 x 8 = 321960).
40245 multiplied by nine equals 362205 (40245 x 9 = 362205).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 40245
Half of 40245 is 20122,5 (40245 / 2 = 20122,5 = 20122 1/2).
One third of 40245 is 13415 (40245 / 3 = 13415).
One quarter of 40245 is 10061,25 (40245 / 4 = 10061,25 = 10061 1/4).
One fifth of 40245 is 8049 (40245 / 5 = 8049).
One sixth of 40245 is 6707,5 (40245 / 6 = 6707,5 = 6707 1/2).
One seventh of 40245 is 5749,2857 (40245 / 7 = 5749,2857 = 5749 2/7).
One eighth of 40245 is 5030,625 (40245 / 8 = 5030,625 = 5030 5/8).
One ninth of 40245 is 4471,6667 (40245 / 9 = 4471,6667 = 4471 2/3).
show fractions by 6, 7, 8, 9 ...
Calculator
40245
Is Prime?
The number 40245 is not a prime number. The closest prime numbers are 40241, 40253.
Factorization and factors (dividers)
The prime factors of 40245 are 3 * 5 * 2683
The factors of 40245 are 1 , 3 , 5 , 15 , 2683 , 8049 , 13415 , 40245
Total factors 8.
Sum of factors 64416 (24171).
Powers
The second power of 402452 is 1.619.660.025.
The third power of 402453 is 65.183.217.706.125.
Roots
The square root √40245 is 200,611565.
The cube root of 340245 is 34,269201.
Logarithms
The natural logarithm of No. ln 40245 = loge 40245 = 10,602741.
The logarithm to base 10 of No. log10 40245 = 4,604712.
The Napierian logarithm of No. log1/e 40245 = -10,602741.
Trigonometric functions
The cosine of 40245 is 0,364121.
The sine of 40245 is 0,931352.
The tangent of 40245 is 2,557808.
Properties of the number 40245
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 40245 in Computer Science
Code typeCode value
40245 Number of bytes39.3KB
Unix timeUnix time 40245 is equal to Thursday Jan. 1, 1970, 11:10:45 a.m. GMT
IPv4, IPv6Number 40245 internet address in dotted format v4 0.0.157.53, v6 ::9d35
40245 Decimal = 1001110100110101 Binary
40245 Decimal = 2001012120 Ternary
40245 Decimal = 116465 Octal
40245 Decimal = 9D35 Hexadecimal (0x9d35 hex)
40245 BASE64NDAyNDU=
40245 MD529c08ec2725d8cc323fdd68147a40703
40245 SHA1047153d6eebfca3358be26419533b4823b99f8fb
40245 SHA25604a113f1db1042f411e5021d2e621030e43e9b67c2d828c75dbafbf08e781831
40245 SHA3842de7663dfa2644a0fe6ba1566d864204081f100584ff6bf328c4ce2ce8d892783d909a7dd79fceb95cca155415bdfbc3
More SHA codes related to the number 40245 ...
If you know something interesting about the 40245 number that you did not find on this page, do not hesitate to write us here.
Numerology 40245
Character frequency in number 40245
Character (importance) frequency for numerology.
Character: Frequency: 4 2 0 1 2 1 5 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 40245, the numbers 4+0+2+4+5 = 1+5 = 6 are added and the meaning of the number 6 is sought.
Interesting facts about the number 40245
Asteroids
• (40245) 1998 WO7 is asteroid number 40245. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from Lincoln Laboratory, Socorro on 11/23/1998.
Number 40,245 in other languages
How to say or write the number forty thousand, two hundred and forty-five in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 40.245) cuarenta mil doscientos cuarenta y cinco German: 🔊 (Anzahl 40.245) vierzigtausendzweihundertfünfundvierzig French: 🔊 (nombre 40 245) quarante mille deux cent quarante-cinq Portuguese: 🔊 (número 40 245) quarenta mil, duzentos e quarenta e cinco Chinese: 🔊 (数 40 245) 四万零二百四十五 Arabian: 🔊 (عدد 40,245) أربعون ألفاً و مئتان و خمسة و أربعون Czech: 🔊 (číslo 40 245) čtyřicet tisíc dvěstě čtyřicet pět Korean: 🔊 (번호 40,245) 사만 이백사십오 Danish: 🔊 (nummer 40 245) fyrretusinde og tohundrede og femogfyrre Dutch: 🔊 (nummer 40 245) veertigduizendtweehonderdvijfenveertig Japanese: 🔊 (数 40,245) 四万二百四十五 Indonesian: 🔊 (jumlah 40.245) empat puluh ribu dua ratus empat puluh lima Italian: 🔊 (numero 40 245) quarantamiladuecentoquarantacinque Norwegian: 🔊 (nummer 40 245) førti tusen, to hundre og førti-fem Polish: 🔊 (liczba 40 245) czterdzieści tysięcy dwieście czterdzieści pięć Russian: 🔊 (номер 40 245) сорок тысяч двести сорок пять Turkish: 🔊 (numara 40,245) kırkbinikiyüzkırkbeş Thai: 🔊 (จำนวน 40 245) สี่หมื่นสองร้อยสี่สิบห้า Ukrainian: 🔊 (номер 40 245) сорок тисяч двiстi сорок п'ять Vietnamese: 🔊 (con số 40.245) bốn mươi nghìn hai trăm bốn mươi lăm Other languages ...
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If you know something interesting about the number 40245 or any natural number (positive integer) please write us here or on facebook. | 2,390 | 6,919 | {"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-2022-27 | latest | en | 0.659018 |
http://www.jiskha.com/display.cgi?id=1355968046 | 1,498,418,397,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320570.72/warc/CC-MAIN-20170625184914-20170625204914-00401.warc.gz | 575,615,592 | 4,110 | # math... Ms, Sue can you help?
posted by .
Apparently, I got this question wrong on a test and I'd like to know what error I made... can you help?
Here's what I did:
2b^2+b=6
b^2+1/2b=3
b^2+1/2b+(1/2)^2=3+(1/2)^2
(b+1/2)^2=3+1/4
b+1/2=+/- <13/4>
b=-1/2+/-<13>/2
b=(-1+/-<13>)/2
<> means its in a radical and +/- means plus or minus
• math... Ms, Sue can you help? -
The error is in your third line
you take 1/2 of the middle coefficient, and then square it
(1/2)(1/2) = 1/4 --- (1/4)^2 = 1/16
3rd line:
b^2 + (1/2)b + 1/16 = 3 + 1/16
(b+1/4)^2 = 49/16
b + 1/4 = ± 7/4
b = -1/4 ± 7/4 = 6/4 or -8/4
b = -2 or 3/2
since we have rational answers, we know it would have factored to
(x+2)(2b-3)
= 2b^2 + b - 6 , the original. | 336 | 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} | 4.09375 | 4 | CC-MAIN-2017-26 | latest | en | 0.921523 |
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