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http://nrich.maths.org/public/leg.php?code=31&cl=2&cldcmpid=1889 | 1,448,593,284,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447906.82/warc/CC-MAIN-20151124205407-00133-ip-10-71-132-137.ec2.internal.warc.gz | 170,233,215 | 10,264 | Search by Topic
Resources tagged with Addition & subtraction similar to Counters:
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Challenge level:
Difference
Stage: 2 Challenge Level:
Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.
A Square of Numbers
Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
Domino Numbers
Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
Stage: 1 and 2 Challenge Level:
Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.
Dodecamagic
Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
Code Breaker
Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
Which Symbol?
Stage: 2 Challenge Level:
Choose a symbol to put into the number sentence.
A Dotty Problem
Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
Cycling Squares
Stage: 2 Challenge Level:
Can you make a cycle of pairs that add to make a square number using all the numbers in the box below, once and once only?
Prompt Cards
Stage: 2 Challenge Level:
These two group activities use mathematical reasoning - one is numerical, one geometric.
How Much Did it Cost?
Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
Reach 100
Stage: 2 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
The Pied Piper of Hamelin
Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
On Target
Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
Hubble, Bubble
Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
The Puzzling Sweet Shop
Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
Totality
Stage: 1 and 2 Challenge Level:
This is an adding game for two players.
All Seated
Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
One Million to Seven
Stage: 2 Challenge Level:
Start by putting one million (1 000 000) into the display of your calculator. Can you reduce this to 7 using just the 7 key and add, subtract, multiply, divide and equals as many times as you like?
A-magical Number Maze
Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
Polo Square
Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
Zargon Glasses
Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
Asteroid Blast
Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
Neighbours
Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
Money Bags
Stage: 2 Challenge Level:
Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. How many pennies did Ram put in each bag?
Five Coins
Stage: 2 Challenge Level:
Ben has five coins in his pocket. How much money might he have?
Bean Bags for Bernard's Bag
Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
Twenty Divided Into Six
Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
Spell by Numbers
Stage: 2 Challenge Level:
Can you substitute numbers for the letters in these sums?
Stage: 2 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
Arranging the Tables
Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
Criss Cross Quiz
Stage: 2 Challenge Level:
A game for 2 players. Practises subtraction or other maths operations knowledge.
Pouring the Punch Drink
Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
Cubes Within Cubes
Stage: 2 Challenge Level:
We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used?
Seven Square Numbers
Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
Worms
Stage: 2 Challenge Level:
Place this "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, what other totals can you make?
Plants
Stage: 1 and 2 Challenge Level:
Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
Number Differences
Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
Page Numbers
Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
Prison Cells
Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
Sealed Solution
Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
Finding Fifteen
Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
Shapes in a Grid
Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
Picture a Pyramid ...
Stage: 2 Challenge Level:
Imagine a pyramid which is built in square layers of small cubes. If we number the cubes from the top, starting with 1, can you picture which cubes are directly below this first cube?
Train Carriages
Stage: 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
Play to 37
Stage: 2 Challenge Level:
In this game for two players, the idea is to take it in turns to choose 1, 3, 5 or 7. The winner is the first to make the total 37.
More Plant Spaces
Stage: 2 and 3 Challenge Level:
This challenging activity involves finding different ways to distribute fifteen items among four sets, when the sets must include three, four, five and six items.
How Old?
Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information? | 2,217 | 9,191 | {"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-2015-48 | longest | en | 0.904629 |
https://artofproblemsolving.com/wiki/index.php?title=1952_AHSME_Problems/Problem_35&oldid=78372 | 1,723,059,644,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640707024.37/warc/CC-MAIN-20240807174317-20240807204317-00714.warc.gz | 81,992,451 | 11,696 | # 1952 AHSME Problems/Problem 35
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
With a rational denominator, the expression $\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}$ is equivalent to:
$\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad \textbf{(B)}\ \frac {\sqrt {6} - 2 + \sqrt {10}}{6} \qquad \textbf{(C)}\ \frac{2+\sqrt{6}+\sqrt{10}}{10} \qquad\\ \textbf{(D)}\ \frac {2 + \sqrt {6} - \sqrt {10}}{6} \qquad \textbf{(E)}\ \text{none of these}$
## Solution
Let $k=\sqrt{2}+\sqrt{3}$ Then $\frac{\sqrt{2}}{k-\sqrt{5}}\implies \frac{\sqrt{2}(k+\sqrt{5})}{k^2-\sqrt{5}^2}\implies\frac{\sqrt{2}k+\sqrt{10}}{k^2-5}\implies \frac{\sqrt{2}(\sqrt{2}+\sqrt{3})+\sqrt{10}}{(\sqrt{2}+\sqrt{3})^2-5}\implies \frac{2+\sqrt{6}+\sqrt{10}}{2\sqrt{6}}\implies\frac{2\sqrt{6}+6+\sqrt{60}}{12}\implies \frac{\sqrt{6}+3+\sqrt{15}}{6}\fbox{A}$ | 392 | 888 | {"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} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.443962 |
https://openi.nlm.nih.gov/detailedresult.php?img=PMC4440753_pone.0126620.g003&req=4 | 1,521,852,057,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257649508.48/warc/CC-MAIN-20180323235620-20180324015620-00712.warc.gz | 657,983,245 | 11,930 | New operational matrices for solving fractional differential equations on the half-line. Bhrawy AH, Taha TM, Alzahrani EO, Baleanu D, Alzahrani AA - PLoS ONE (2015) Bottom Line: An upper bound of the absolute errors is obtained for the approximate and exact solutions.Several numerical examples are implemented for FDEs and systems of FDEs including linear and nonlinear terms.We demonstrate the high accuracy and the efficiency of the proposed techniques. View Article: PubMed Central - PubMed Affiliation: Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia; Department of Mathematics, Faculty of Science, Beni-Suef University, Beni-Suef, Egypt. ABSTRACTIn this paper, the fractional-order generalized Laguerre operational matrices (FGLOM) of fractional derivatives and fractional integration are derived. These operational matrices are used together with spectral tau method for solving linear fractional differential equations (FDEs) of order ν (0 < ν < 1) on the half line. An upper bound of the absolute errors is obtained for the approximate and exact solutions. Fractional-order generalized Laguerre pseudo-spectral approximation is investigated for solving nonlinear initial value problem of fractional order ν. The extension of the fractional-order generalized Laguerre pseudo-spectral method is given to solve systems of FDEs. We present the advantages of using the spectral schemes based on fractional-order generalized Laguerre functions and compare them with other methods. Several numerical examples are implemented for FDEs and systems of FDEs including linear and nonlinear terms. We demonstrate the high accuracy and the efficiency of the proposed techniques. No MeSH data available. © Copyright Policy Related In: Results - Collection License getmorefigures.php?uid=PMC4440753&req=5 .flowplayer { width: px; height: px; } pone.0126620.g003: Graph of the absolute error function for N = 10, α = 0 and ν = λ = 0.5, for Example 6. Mentions: In Table 7, we list the results obtained by the fractional-order generalized Laguerre generalized collocation (FGLC) method with various choices of α, N = 10, and ν = λ = 0.5. The present method is compared with the shifted Chebyshev spectral tau (SCT) method given in [48]. As we see from Table 7, it is clear that the result obtained by the present method for each choice of the parameter α is superior to that obtained by SCT method. Fig 3 shows the absolute error function at N = 10, α = 0 and ν = λ = 0.5. The obtained results of this example show that the present method is very accurate by selecting a few number of fractional-order generalized Laguerre generalized functions.
New operational matrices for solving fractional differential equations on the half-line.
Bhrawy AH, Taha TM, Alzahrani EO, Baleanu D, Alzahrani AA - PLoS ONE (2015)
Related In: Results - Collection
Show All Figures
getmorefigures.php?uid=PMC4440753&req=5
pone.0126620.g003: Graph of the absolute error function for N = 10, α = 0 and ν = λ = 0.5, for Example 6.
Mentions: In Table 7, we list the results obtained by the fractional-order generalized Laguerre generalized collocation (FGLC) method with various choices of α, N = 10, and ν = λ = 0.5. The present method is compared with the shifted Chebyshev spectral tau (SCT) method given in [48]. As we see from Table 7, it is clear that the result obtained by the present method for each choice of the parameter α is superior to that obtained by SCT method. Fig 3 shows the absolute error function at N = 10, α = 0 and ν = λ = 0.5. The obtained results of this example show that the present method is very accurate by selecting a few number of fractional-order generalized Laguerre generalized functions.
Bottom Line: An upper bound of the absolute errors is obtained for the approximate and exact solutions.Several numerical examples are implemented for FDEs and systems of FDEs including linear and nonlinear terms.We demonstrate the high accuracy and the efficiency of the proposed techniques.
View Article: PubMed Central - PubMed
Affiliation: Department of Mathematics, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia; Department of Mathematics, Faculty of Science, Beni-Suef University, Beni-Suef, Egypt.
ABSTRACT
In this paper, the fractional-order generalized Laguerre operational matrices (FGLOM) of fractional derivatives and fractional integration are derived. These operational matrices are used together with spectral tau method for solving linear fractional differential equations (FDEs) of order ν (0 < ν < 1) on the half line. An upper bound of the absolute errors is obtained for the approximate and exact solutions. Fractional-order generalized Laguerre pseudo-spectral approximation is investigated for solving nonlinear initial value problem of fractional order ν. The extension of the fractional-order generalized Laguerre pseudo-spectral method is given to solve systems of FDEs. We present the advantages of using the spectral schemes based on fractional-order generalized Laguerre functions and compare them with other methods. Several numerical examples are implemented for FDEs and systems of FDEs including linear and nonlinear terms. We demonstrate the high accuracy and the efficiency of the proposed techniques.
No MeSH data available. | 1,186 | 5,333 | {"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-2018-13 | longest | en | 0.84068 |
https://www.airmilescalculator.com/distance/ksu-to-hov/ | 1,701,199,692,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099942.90/warc/CC-MAIN-20231128183116-20231128213116-00559.warc.gz | 727,884,476 | 22,053 | # How far is Ørsta from Kristiansund?
The distance between Kristiansund (Kristiansund Airport, Kvernberget) and Ørsta (Ørsta–Volda Airport, Hovden) is 85 miles / 137 kilometers / 74 nautical miles.
The driving distance from Kristiansund (KSU) to Ørsta (HOV) is 117 miles / 188 kilometers, and travel time by car is about 3 hours 55 minutes.
85
Miles
137
Kilometers
74
Nautical miles
## Distance from Kristiansund to Ørsta
There are several ways to calculate the distance from Kristiansund to Ørsta. Here are two standard methods:
Vincenty's formula (applied above)
• 85.293 miles
• 137.266 kilometers
• 74.118 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 85.042 miles
• 136.862 kilometers
• 73.899 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).
## How long does it take to fly from Kristiansund to Ørsta?
The estimated flight time from Kristiansund Airport, Kvernberget to Ørsta–Volda Airport, Hovden is 39 minutes.
## Flight carbon footprint between Kristiansund Airport, Kvernberget (KSU) and Ørsta–Volda Airport, Hovden (HOV)
On average, flying from Kristiansund to Ørsta generates about 38 kg of CO2 per passenger, and 38 kilograms equals 83 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Kristiansund to Ørsta
See the map of the shortest flight path between Kristiansund Airport, Kvernberget (KSU) and Ørsta–Volda Airport, Hovden (HOV).
## Airport information
Origin Kristiansund Airport, Kvernberget
City: Kristiansund
Country: Norway
IATA Code: KSU
ICAO Code: ENKB
Coordinates: 63°6′42″N, 7°49′28″E
Destination Ørsta–Volda Airport, Hovden
City: Ørsta
Country: Norway
IATA Code: HOV
ICAO Code: ENOV
Coordinates: 62°10′48″N, 6°4′26″E | 576 | 2,029 | {"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-2023-50 | latest | en | 0.773537 |
http://igoogledrive.blogspot.com/2012/09/Alternate-for-Excel-function-AVERAGEIFS-in-Google-Spreadsheet.html | 1,516,130,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886639.11/warc/CC-MAIN-20180116184540-20180116204540-00510.warc.gz | 154,842,354 | 17,557 | ## Thursday, September 20, 2012
Question:
I have a spreadsheet that works perfecctly in excel but when I upload to google one field shows an error "Name".
The formula I'm using is: =AVERAGEIFS(F3:F500,G3:G500,"w")
This works in excel and whait it does is looks in column G for a cell with a w in it, it then works out the average for the figures in column F when next to a "w" in G
Solution:
Here are some ways by which you can achieve the same thing in Google Spreadsheet, that can be acheived by AVERAGEIFS in EXCEL:
=SUMIF(G3:G;"w";F3:F)/COUNTIF(G3:G;"w")
OR
=AVERAGE(FILTER(F3:F;G3:G="w"))
OR
=AVERAGE(QUERY(F3:G;"select F where G = 'w'"))
I hope the above solution will help you, and if you need more help then please do comment below on this blog itself, I will try to help you out.
If this blog post was helpful to you, and if you think you want to help me too and make my this blog survive then please donate here: http://igoogledrive.blogspot.com/2012/09/donate.html
Thanks,
Kishan,
#### 1 comment:
1. The second formula works perfectly! It can also be transposed and works the same way for row-based conditional averaging. Great solution!! | 324 | 1,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-05 | longest | en | 0.928106 |
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# Faster Than the Speed of Life
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posted on Aug, 29 2009 @ 08:21 AM
I have been thinking again lately and somethineg has crossed my mind. I really would like to know your thoughts on this.
It is undeniable that we can measure the speed of almost everything. I am not sure if we can measure the speed the Earth is flying through the Universe, but we can measure the speed that it rotates and almost every other speed associated with it.
We can measure the speed of a race car, an olympic athlete, a regular person walking, a turle, or anything that we want to.
There are parameters set up for this diagnosis of speed. formulas used to determine the speed relative to stop. But we usually all miss the most important one; the Speed of Life.
All of the measurements have to have an environment to occur in. Life. The rate at which we progress.
When you measure a light beam you have to have a basis for it. If it were going 2 miles an hour, it would be much different, but it would still be moving through time.
At the speed light travels now it can still be measured, so it must still be traveling through the parameters that we live by.
Some may say there are things that move faster, but we can't see them. I can buy that, but they too are moving slower then we live from one instance to the next.
Theoretically, if an object (void of any speed or mass consequences) were to move faster than we live - from moment to moment - what would happen to it.
Would it go back in time, would it cease to exist? If it were so fast that when it left it's start point it moved faster than the smallest measurement of time, what would happen to its existence? You would think that a new measurement of time would have been made by the speed of this object, but if it were to move faster than that too, what would happen to it?
That is what I don't understand. Where would this object be if it moved faster than time elapses? Would it vanish into another dimension, would it create a solid line the size of itself from one side of the universe to the other?
If any of you can help me out with this thought, please, don't hesitate to respond.
[edit on 29-8-2009 by esteay812]
posted on Aug, 29 2009 @ 09:36 AM
I look forward to the responses you will get on this very intriguing question. I am no expert by any means; however, I will give you my thoughts -- which may seem baffling.
I personally do not believe in time. It is an artificial abstract. The reason is, is that all life takes place "in the here and now". It always has. It has never, ever, been anything but now.
Trust me. I have been around for over 60 years, and for every moment of those years it was always "now".
Think about it. Somehow, all of eternity is encapsulated in now.
edit to add: So you can't measure the speed of it, because there isn't any speed to it. It isn't moving. It just is.
[edit on 8/29/2009 by wayno]
posted on Aug, 29 2009 @ 09:47 AM
Awesome perception, one that does have ties in many of my ideas. Being Now and always now may be what allows things to be measured. The happen in the present, one moment after the next.
The speed at which they occur still happens slower than the speed of now. So another question that could be posed is; how do we measure the speed of now. This would possibly be the cresting point from the reality into the next, or dimension if-you-will.
The only thing that makes Present, past, and future real is the fact that we perceive it. "I think therfore I am". We preceive it, therfore it is. Though we only experience things in the now and always the now another clue to the validity of time is the fact that we age, grow, and everntually die.
If it was always now, wouldn't we live in the same state forever - eternally?
posted on Aug, 29 2009 @ 09:53 AM
Life ?
HMm i just posted about this... LIFE is not YOU.. and i do not mean that in a bad way.. i just think people for get what the word means.. most people use the word LIFE to describe there own REALITY when its not.
Life is a function it has no "speed" NONE because YOU my little friend make the universe WORK
there is no measurement for the recreation of the universe and thats LIFES job
like the tree, like that cat, like the dog, and YOU all one of the same thing
; )
Do not worry about it. in time you will understand
posted on Aug, 29 2009 @ 09:57 AM
its all IN YOU .. NOT NOW
just wanted to correct you on that..
I know its hard to wrap ones head around it but how your mind works is the very same way the universe was created..
Just a thought ! think im kidding? Look up
your mind has NO shape NONE zip ...
What els has no shape? THE UNIVERSE
and i love the fact we are one of the same, for every living being there is a universe
posted on Aug, 29 2009 @ 10:00 AM
we can and have measured the speed our earth moves with, and the number is staggering, i think its about 20.000 km/per hour
posted on Aug, 29 2009 @ 10:06 AM
I don't think you understand. The speed at which we live is obviously real. If it weren't there would be no basis for speed measurements of any kind. All of these measurements fall within the parameters of our observance. If an object moved faster than our perception or if it moved faster than time, in present moments, what would happen to it?
Maybe you will understand one day, but don't worry if you don't.
posted on Aug, 29 2009 @ 10:40 AM
Originally posted by Tetragrammaton
we can and have measured the speed our earth moves with, and the number is staggering, i think its about 20.000 km/per hour
Yes, I've read reports like this before. I don't recall however the specifics. Is that the speed that the earth travels around the sun? Our whole solar system is moving in space. Is that speed calculated in? Our whole galaxy is moving thru space. Is that speed known and calculated as well?
This thing never ends. Its all relative. And as symetricA says its all happening in us anyway.
posted on Aug, 29 2009 @ 11:32 AM
The speed of life is 6. Six what? Possibly a nice zen contemplation but it doesn't make sense.
By definition speed is the rate at which something moves from one location in space to another. Speed is the ratio between distance and time. Unless you are talking about how far someone travels in their lifetime, the term "speed of life" is meaningless. There is no numerator for the denominator.
posted on Aug, 29 2009 @ 11:51 AM
No i Do not think you understand.. The "speed" word.. here in this topic
The reason for infinity is the understanding of ones own reality that is why we have a thing called ODD numbers and EVEN numbers..
You can not measure infinity as it is a part of the very REAL thing you call life..
We can do equations that can work out the average life span of a Human yes "but this is not the same as your topic" because you equate life as "spirit"
Hence you are mixing them up
So if you wish to have a hypothesis about the SPEED of LIFE i suggest you get more educated in what LIFE in fact is..
It is endless .. and by that i mean YOU the start of everything and the END of everything you know starts and ends with you..
You can not measure perspectives FACT
now .. be for you get all humpty dumpty sat on a wall with me please take the time to read what i said.
and just for the record math is my fav subject even tho my english stinks
posted on Aug, 29 2009 @ 11:56 AM
Thank you phage
its NOT possible to measurer ones perspective.. Tho we can measure the life span of "something" that is alive ect, this is not the same as trying to measure LIFE in on its self
You can quantify it yes as a component of reality in essence 1
In math the logic of this would be to put life in a box as we do with elements 1 blah blah
but yes phage is correct
not possible in the way you are saying or trying to
was just adding to what phage said, "was not directed at phage" but the OP kinda thing ; )
posted on Aug, 31 2009 @ 06:32 PM
I would just like to throw in that we wouldn't even notice the passing of time without the movement of objects. A poster earlier said that everything is in the here and now, and I agree with him on that. Time is in our heads. It is not an illusion, it is there, but it is there only to us who think it is there. I wonder if someone who has learned about the existence of time can 'unlearn' it so to speak?
new topics
top topics
0 | 2,051 | 8,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-13 | latest | en | 0.982597 |
http://medical-dictionary.thefreedictionary.com/Rayleigh+equation | 1,555,966,515,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578582584.59/warc/CC-MAIN-20190422195208-20190422221208-00220.warc.gz | 116,244,502 | 11,100 | # Rayleigh equation
## Ray·leigh e·qua·tion
(rā'lē),
a ratio of red to green required by each observer to match spectral yellow.
Synonym(s): Rayleigh test
## Rayleigh,
Lord John W.S., English physicist and Nobel laureate, 1842-1919.
rayl - unit of acoustic impedance.
Rayleigh equation - a ratio of red to green required by each observer to match spectral yellow. Synonym(s): Rayleigh test
Rayleigh test - Synonym(s): Rayleigh equation
## Rayleigh equation
A colour equation representing a match of yellow (usually 589 nm) with a mixture of red (usually 670 nm) and green (usually 535 nm). It is used to differ-entiate certain types of colour deficiencies. The anomaloscope is built on this principle. See defective colour vision.
Mentioned in ?
References in periodicals archive ?
In this present paper, we investigate the existence and uniqueness of the periodic solutions of the following Rayleigh equation with two delays
As we know, Rayleigh equation can be derived from many fields, such as physics, mechanics and engineering technique fields.
In this direction, many researchers (see [4-9]) continued to discuss the Rayleigh equation and got some new results on the T-periodic solutions of Eq.
Consider the following Rayleigh equation with two delays
1) is a very simple version of Rayleigh equation with two delays, all the results in [1,3-9] and the references therein cannot be applicable to Eq.
New results on the periodic solutions for a kind of Rayleigh equation with two deviating arguments, Math.
Peng, Periodic solutions for p-Laplacian neutral Rayleigh equation with a deviating argument, Nonlinear Analysis (2007), doi:10.
15]N of the residual nitrate increases in proportion to the logarithm of the residual nitrate fraction (Kendall 1998), which can be expressed using the classic Rayleigh equation (Mariotti et al.
Using the Rayleigh equation we calculated a linear relationship between N[O.
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Open / Close | 445 | 1,953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2019-18 | latest | en | 0.884664 |
https://www.kdnuggets.com/2021/05/ensemble-methods-explained-plain-english-bagging.html | 1,695,446,331,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00121.warc.gz | 946,593,606 | 31,138 | Ensemble Methods Explained in Plain English: Bagging
Understand the intuition behind bagging with examples in Python.
By Claudia Ng, Senior Data Scientist
In this article, I will go over a popular homogenous model ensemble method — bagging. Homogenous ensembles combine a large number of base estimators or weak learners of the same algorithm.
The principle behind homogenous ensembles is the idea of “wisdom of the crowd” — the collective predictions of many diverse models is better than any set of predictions made by a single model. There are three requirements to achieve this:
1. The models must be independent;
2. Each model performs slightly better than random guessing;
3. All individual models have similar performance on their own.
When these three requirements are satisfied, adding more models should improve the performance of your ensemble.
Ensemble methods help to reduce variance and combat overfitting to your train dataset, thus allowing your model to better learn generalized patterns rather than overfitting to the noise in your train dataset.
How Bagging Works
In bagging, a large number of independent weak models are combined to learn the same task with the same goal. The term “bagging” comes from `bootstrap + aggregating`, whereby each weak learner is trained on a random subsample of data sampled with replacement (bootstrapping), and then the models’ predictions are aggregated.
Bootstrapping guarantees independence and diversity, because each subsample of data is sampled separately with replacement and we are left with different subsets to train our base estimators.
The base estimators are weak learners that perform only slightly better than random guessing. An example of such a model is a shallow decision tree limited to a maximum depth of three. The predictions from these models are then combined through averaging.
Bagging can be applied to both classification and regression problems. For regression problems, the final predictions will be an average (soft voting) of the predictions from base estimators. For classification problems, the final predictions will be the majority vote (hard voting).
Diagram of Bagging Algorithms
Implementing Bagging Algorithms with Scikit-Learn
You can build your own bagging algorithm using `BaggingRegressor` or `BaggingClassifier` in the Python package Scikit-Learn.
To begin, instantiate your base estimator and enter this as your base estimator in `BaggingRegressor` or `BaggingClassifier`. Below are an example of a bagging regressor with a linear regression as the base estimator, and an example of a bagging classifier with a decision tree classifier as the base estimator. The default number of estimators is 10.
``````from sklearn.linear_model import LinearRegression
from sklearn.ensemble import BaggingRegressor
reg_lr = LinearRegression()
reg_bag = BaggingRegressor(base_estimator=reg_lr)
reg_bag.fit(X_train, y_train)
====================================================================
from sklearn.tree import DecisionTreeClassifier
from sklearn.ensemble import BaggingClassifier
clf_dt = DecisionTreeClassifier(max_depth=3)
clf_bag = BaggingClassifier(base_estimator=clf_dt)
clf_bag.fit(X_train, y_train)``````
Random Forest
Random forest® is a popular example of a bagging algorithm. It uses averaging to ensemble a number of individual decision trees trained on a subset of the train dataset.
Using scikit-learn’s random forest algorithm in Python, you can specify tree-specific parameters. The following are some important hyperparameters to tune so that it is optimized for your dataset:
• `n_estimators`: number of trees to train to be aggregated. Usually between 100 to 500 trees is enough and generally, more trees will improve your model (with diminishing returns) but it will also be more computationally expensive;
• `max_depth` : the maximum depth of a tree. A deeper tree will help to reduce the bias but at the expense of increasing variance. The aggregation of multiple trees in the random forest algorithm can help to combat this, but you should still be careful;
• `max_features`: the maximum number of features to consider at each split. A good starting point is usually the square root of the number of features.
Another important concept in the random forest algorithm is the out-of-bag (OOB) score. When performing bootstrapping, there will be instances left out of the subsamples used to train the estimators. These out-of-sample instances can be used to evaluate the model to obtain an out-of-bag (OOB) score, in essence like a pseudo-validation set for the random forest model. To obtain the OOB score, set `oob_score=True` when initializing your random forest object.
``````rf = RandomForestClassifier(n_estimators=100, oob_score=True)
rf.fit(X_train, y_train)
print(rf.oob_score_)``````
Note that scikit-learn’s random forest algorithms will return an error if there are null values in your features, so remember to fill null values with pandas’ `fillna` before calling fit on your data, otherwise it will throw an error.
Pros and Cons of Bagging
• Reduces variance: Given that the sampling is done truly randomly with bootstrapping, bagging usually helps to reduce variance and combat overfitting.
• Easy to parallelize: Estimators are independent, so models can be built in parallel with bagging.
• Greater stability and robustness: The high number of estimators aggregated together help to provide more stability and robustness to the predictions.
• Difficult to interpret: The final predictions of a bagging algorithm are based on the mean predictions from the base estimators. While this improves accuracy, it becomes more difficult to interpret the model.
Summary
Bagging is based on the idea of collective learning, where many independent weak learners are trained on bootstrapped subsamples of data and then aggregated via averaging. It can be applied to both classification and regression problems.
The random forest algorithm is a popular example of a bagging algorithm. When tuning hyperparameters in the random forest algorithm for your dataset, the important three areas to pay attention to are: i) number of trees (`n_estimators`), ii) prune the trees (start with `max_depth` but also explore samples required in a node and/or for splitting), iii) the maximum number of features to consider at each split (`max_features`).
Bagging is a very powerful concept and example of model ensemble methods. Hope this inspires you to try out a bagging algorithm next time you approach a predictive modeling problem!
Bio: Claudia Ng is a Senior Data Scientist.
Original. Reposted with permission.
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• MATH IS EVERYWHERE!
• Ruby is in the 10th grade and loves math and realizes that math is everywhere! She wants to show you her day to day life and how math relates to everything she does.
• After school one day Ruby and her friends went out for pizza. The pizza cost \$20. There were four fewer people contributing to the pizza than the original plan. Each person had to now give 25 cents more. Ruby is now wondering how many people were supposed to come originally?
• To get home Ruby has to use the ferry to cross the body of water.
• The ferry is due to cruise south. After 1 hour Ruby realizes that the ferry is actually S50°W. Ruby wonders if he was travelling at a speed of 100km/h how far off course are they?
• 50°
• .
• N
• After Ruby finally arrives home she realizes her mom is trying to figure something out so she goes to ask if she need help
• Hey Mom! What're you doing?
• Hey Ruby! I was just trying to figure out what dimensions would create the largest fencing around this pool. But one side doesn't need fencing because of the shed
• Hmmm since one side of the pool is against the shed and we only have 400m of fencing I wonder what dimensions would maximize the area
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Video Marketing | 339 | 1,506 | {"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-2019-26 | latest | en | 0.964907 |
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Exit Ticket.mov
# Multiplying Fractions
Unit 2: Number System
Lesson 2 of 12
## Big Idea: Using multiplication of rational numbers to solve problems involving accumulating resources measured in fractional quantities.
Print Lesson
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Standards:
Subject(s):
Math, Fractions, Number Sense and Operations, direct variation, mixed numbers, rational numbers, Unit Rates, pre
40 minutes
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P. 3994. A 60 kg boy notices a nice fruit on a tree. Even with stretched arms he cannot reach it, he is a 0.5 m below the fruit. The boy crouches down - his centre of mass descends 0.4 m - and then he leaps up. Thus he just reaches the fruit. Assume that he exerts a constant force on the ground while he touches it.
a) What is this force at least?
b) What is the average power of the boy?
(4 points)
Deadline expired on 11 October 2007.
Statistics on problem P. 3994.
282 students sent a solution. 4 points: 113 students. 3 points: 10 students. 2 points: 61 students. 1 point: 55 students. 0 point: 38 students. Unfair, not evaluated: 5 solutions.
• Problems in Physics of KöMaL, September 2007
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# Thread: Convert this to non-recursive
1. ## Convert this to non-recursive
```public static int f(int a, int b)
{
Z.h1(a, b); //// arbitrary code
if (Z.h2(a, b))
return Z.h3(a, b);
else
{
Z.h4(a, b); ////arbitrary code
return f(Z.h5(a, b), Z.h6(a, b)); //// recursive call
}
}```
-h1 and h4 --------------------> void functions
-h2 ------------------------------> boolean function
-h3, h5 and h6 -------------->return int
-h1 ... h6-----------------------> will not call f
My Attempt to convert the code above to non-recursive:
```
public static int f(int a, int b)
{
while(true)
{
Z.h1(a, b); //// arbitrary code
if (Z.h2(a, b)) //// base case
return Z.h3(a, b);
else
{
Z.h4(a, b); /////arbitrary code
int temp = Z.h5(a, b);
b = Z.h6(a, b));
a=temp;
}
}
}```
I'm not sure if I'm right.
Any ideas?
2. ## Re: Convert this to non-recursive
It looks fine to me. Did you test it out to make sure the outputs match up? | 421 | 1,604 | {"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-2020-10 | latest | en | 0.77656 |
https://radacad.com/fun-with-dax-blackjack-game?replytocom=44539 | 1,579,958,779,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00015.warc.gz | 607,350,883 | 37,621 | # Fun with DAX – Blackjack Game
I thought it might be fun to try something a little different with this blog and use a combination of bookmarks with a little DAX magic to come up with a game that plays a hand of Blackjack. Power BI and DAX are not the ideal engine to build a game of Blackjack with, but it was an interesting challenge and I thought I would share my approach that takes advantage of quite a few novel techniques.
First of all, let’s look at the finished result. The PBIX file allows you to play a single hand that involves you playing the House. The first view of the table shows the first card for both you and the house. You can then click the hit button to deal new cards until you decide to hold (or go bust), at which point the house cards are revealed with a dynamic message that shows if you won or lost. There is no mechanism to bet or play additional hands. The game is dynamic enough to understand if you win, lose or draw based on your decision on when to hold.
### Data Model
The data model for the game consists of 6 unrelated data-tables. Three of the tables control the side of the game for Player 1 (you), while the other three data-tables control the game for the house. In the diagram below, the three data tables on the left-hand side (‘P1’, ‘Player 1’ and ‘P1 Turn’) control the game for Player 1, while the three data tables on the left control the game for the House.
The P1 and P2 data-tables at the top are effectively the shuffled cards as dealt in order of ‘Turn’. They both have the same pattern and structure with 10 rows and three columns. The reason there are two tables is so we can have a random set of cards for Player 1 (the P1 data-table) and a different set of random cards for the House (P2).
The ‘Turn’ column in each data-table is numbered 1 to 10 and controls the order of the deal. The ‘Cards’ column contains a random number between 1 and 13 (1 = Ace, 2 = 2 … 11 = J, 12 = Q, 13 = K), while the ‘Suit’ column is a random number between 1 and 4 to allow us to deal cards from any of the four suits (♠ ♣ ♥ ♦). These tables are generated in Power Query using a random number technique to ensure that different values were generated for each row of both tables.
My first stab at creating a random number in Power Query ended up providing the same number for each row, so I passed the value of the previous step as a way to force Power Query to generate a new value for each row
The M code for the P1 table starts by generating a list of numbers {1..10} which is then converted to a table with two extra columns added containing the random numbers needed for Card (1 to 13) and Suit (1 to 4).
The M code for the P2 table
While the P2 table is a copy of P1, it generates a different set of random numbers.
The P1 and P2 tables are hidden from the final model and are used as the basis for the next two tables.
The DAX fun begins in the next data-table.
For Player 1, we generate a data-table based on the P1 table from the previous step. There will be 1-row per turn and a series of columns that help determine if the player wins the game.
The Turn, Cards and Suit columns are derived from the P1 table unchanged, but four new columns are added. No rows are added or filtered.
Firstly the ‘Card Value’ column converts the ‘Cards’ column back to the numeric value associated with the card. Ace = 1, 2 = 2…. J, Q & K all equal 10. This value is used to determine your score to compare in the game. While an Ace card can represent two possible values (1 or 11), this is handled later in the code that generates the ‘Score’ column.
The ‘Card Face’ column uses the UNICHAR function to combine both the ‘Cards’ and ‘Suit’ columns to a Unicode value that represents one of the 52 playing cards.
Then a ‘Card Colour’ column is generated to return a 1 or 0 that will be used by conditional formatting to set the font to be red for heart and diamond suits, and black for club and spade suits.
Finally, a ‘Score’ column is added that performs a cumulative sum over the ‘Card Value’ column based on incrementing turns. A nested variable scope is introduced to handle the scenario whereby an Ace card can have a numeric value of either 1 or 11. The logic first tests to see if there has been an Ace in any of the previous turns and counts the Aces so far. The B1 and B2 variables then provide the two possible best scores based on the combination of Aces dealt to this point. The SWITCH statement analyses both values stored in the B1 and B2 variables to determine the best possible score under 21 to return. If both B1 and B2 are over 21, then a value of 99 is returned which represents a bust.
The last of the three data-tables in the model used to control Player 1 is the ‘P1 Turn’ data-table. This is simply a 10-row single column table with numbers 1 through 10. The purpose of this table is to use in conjunction with a slicer that will help keep track of what turn Player 1 is up to. A series of 5 bookmarks will be used to snapshot a slicer selected in 5 different states. A series of bookmarks will be taken with this slicer having a different value selected which is how the game can keep track of the progress through the game.
A [P1 Card Filter] calculated measure keeps track of the selected value over the above slicer which is used as a filter on the table-visual that is used to reveal cards for Player 1.
A [P1 Score] calculated measure uses the [P1 Card Fitler] measure to determine the current score for Player 1 depending on how many cards they have had dealt.
Both the [P1 Card Filter] and [P1 Score] calculated measures have equivalent versions to handle the scoring for the House, although are slightly different in that the house only has 1 hand, or shows their final state.
The DAX to generate the data-table to control the scoring for the house is as follows
This table is similar to the ‘Player 1’ table in that it carries forward the three columns (Turn, Cards & Score) from the ‘P2’ table generated in Power Query. The difference is that a new column called ‘Hit or Hold’ is added to control the rule that says the House must keep dealing until they hit 17. Once the cumulative ‘Score’ is at 17 or greater the house will stop dealing.
The final RETURN statement filters the overall data-table so it only has as many rows as the first hold – which could be a bust score.
The cards can only be in 2 states for the House. State 1 is that only the first card is showing, while the other state is that all cards are showing.
### Report View
There are two report pages. The first is simply a welcome screen with a button linked to a bookmark that takes you to the second report page where the game takes place.
1. A standard bookmark button that is linked to an action that jumps back to a bookmark showing the initial welcome screen.
2. There are five identical images added that link to slightly different bookmarks. Each image sets a different value on the slicers at item (10) and then hides itself to reveal the next image. The [P1 Card Fitler] calculated measure watches the slicer at (10) and is used as a filter on the data-table at (5) to control how many cards to show.
3. A single image that is linked to a final bookmark that reveals the house cards AND the [result] measure to show at that point, if Player 1 or the House has the best score.
4. A simple card visual that shows the current score for Player 1 that is based on how many cards have been dealt
5. A table visual with two columns to show the cards for Player 1. The first column has the Turn number. This is hidden by making the font colour the same as the background. The second column shows the UNICHAR card and has conditional formatting to set the font to red or black depending on the value for ‘Card Colour’. Finally, this table has uses the [P1 Card Filter] calculated measure to determine how many rows to display which is based on the slicer at (10)
6. A table visual with two columns two show the cards for the House. The first column is hidden by making the font colour the same as the background. The second column initially shows only one card. When the Hold image is clicked (3), all the cards for the House are revealed showing the pre-determined values. Conditional formatting is applied to control the colour of the cards to match the suit and the table visual uses the [House Card Filter] calculated measure to control which cards should be shown based on the slicer at (9).
7. A simple card visual that keeps track of the score for the house based on which cards have been revealed.
8. A card visual that shows the result. The result will be dynamic and will show if Player 1 has won, lost or drawn.
9. A slicer to control the turn number for the House. This slicer is hidden at all times and is initially set to value 1. The final bookmark sets this slicer to 2 which causes all the House cards to be shown.
10. A slicer to control the turn number for Player 1. This slicer is also hidden at all times and is stepped through from value 1 through to 5 by each of the images at (2).
The game is dynamic enough that with the same set of cards, depending on how many times the Player hits and then holds, they can win, lose or draw.
There is no concept of betting and it’s only good enough for one hand. It does showcase a variety of features available in Power BI. The refresh button needs to be clicked to generate a new set of cards using the randomiser in Power Query. Unfortunately the “Publish to Web” feature offers no ability to randomise the cards, so you end up playing the same hand over and over. | 2,249 | 9,638 | {"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} | 3.25 | 3 | CC-MAIN-2020-05 | latest | en | 0.928405 |
http://oeis.org/A167006 | 1,582,536,662,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145910.53/warc/CC-MAIN-20200224071540-20200224101540-00525.warc.gz | 102,052,319 | 3,930 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A167006 G.f.: exp( Sum_{n>=1} x^n/n * Sum_{k=0..n} binomial(n^2, n*k) ). 13
1, 2, 6, 66, 4258, 1337374, 1933082159, 11353941470188, 291885138650054688, 29463501750534915665304, 12844314786465829040693498639, 21675661852919288704454219459892060, 156969579902607123047763327413679853875703 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS Logarithmic derivative yields A167009. Equals row sums of triangle A209196. LINKS Seiichi Manyama, Table of n, a(n) for n = 0..57 EXAMPLE G.f.: A(x) = 1 + 2*x + 6*x^2 + 66*x^3 + 4258*x^4 + 1337374*x^5 +... log(A(x)) = 2*x + 8*x^2/2 + 170*x^3/3 + 16512*x^4/4 + 6643782*x^5/5 + 11582386286*x^6/6 +...+ A167009(n)*x^n/n +... PROG (PARI) {a(n)=polcoeff(exp(sum(m=1, n, sum(k=0, m, binomial(m^2, k*m))*x^m/m)+x*O(x^n)), n)} for(n=0, 20, print1(a(n), ", ")) CROSSREFS Cf. A167009, A209196, A155200. Cf. variants: A206848, A228809. Sequence in context: A006517 A217630 A091458 * A087331 A097419 A219037 Adjacent sequences: A167003 A167004 A167005 * A167007 A167008 A167009 KEYWORD nonn AUTHOR Paul D. Hanna, Nov 17 2009 STATUS approved
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Last modified February 24 04:30 EST 2020. Contains 332197 sequences. (Running on oeis4.) | 591 | 1,618 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2020-10 | latest | en | 0.493133 |
http://m.csmonitor.com/1998/0611/061198.feat.feat.2.html | 1,477,111,635,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718423.65/warc/CC-MAIN-20161020183838-00461-ip-10-171-6-4.ec2.internal.warc.gz | 154,213,994 | 9,730 | Close X
Math Chat: Crossing a Rickety Bridge at Night By Flashlight
Old bridge-crossing challenge (Tiku Majum-der and Conrad Weiser)
"There are four people who need to cross a river at night. There is a bridge that can only hold up to two people at a time. There is one flashlight that must be used when crossing. (It is extremely dark, and someone must bring the flashlight back to the others; no throwing anything, no halfway crosses, etc.).
"The four people take different amounts of time to cross the river. If two people cross together, they travel at the slower person's rate. The times are 10 minutes, 5 minutes, 2 minutes, and 1 minute for each of the four individuals." How fast can they complete the passage? Can they do it in 19 minutes? in 17 minutes?
They certainly can do it in 19 minutes, by having 1-minute "Speedy" escort each of the other three across the bridge. That's 2 + 5 + 10 = 17 minutes crossing, plus Speedy's two 1-minute return trips, for a total of 19 minutes.
Amazingly, Aubrey Dunne, James Fahs, Hany Farid, Jocelyn Granger, Heather Anne Harrison, Chad Heeter, Sam Ragucci, Erik Randolph, John Robertson, and Mike Bevan found a 17-minute solution. First 1 and 2 cross, 1 returns, 5 and 10 cross, 2 returns, 1 and 2 cross, for a total of 2 + 1 + 10 + 2 + 2 = 17 minutes.
Princeton freshmen Amanda Fulmer, David Greco, Michael Lindahl, and Graham Meyer explain that this method wins essentially because 2 - 1 &lt; 5 - 2. If it took the second person 4 minutes instead of 2, the best strategy would be to have Speedy escort the other three, because 4 - 1 &gt; 5 - 4.
Kenneth Eggert suggests the more imaginative solution of having Speedy carry each of the other three across in a total time of 5 minutes flat.
New intelligence challenge | 466 | 1,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-44 | latest | en | 0.907391 |
https://medium.com/@LeonFedden/the-no-free-lunch-theorem-62ae2c3ed10c | 1,563,457,561,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00126.warc.gz | 453,309,288 | 24,896 | # The No Free Lunch Theorem (or why you can’t have your cake and eat it)
Oct 6, 2017 · 6 min read
In this series I’ll be taking a random walk through the algorithms and ideas that I learn from my Natural Computing class that I’m taking at Goldsmiths, University of London.
Please find part two, part three and part four of the series here!
Where possible, I’ll be combining the topical material with a deep learning / connectionist slant to complement the work I’ll be doing elsewhere this year.
The No Free Lunch Theorem (NFLT) is named after the phrase, there ain’t no such thing as a free lunch. As a pretext to this article, it is worth exploring the meaning of this phrase. There is/ain’t no such thing as free lunch has origins in the mid-nineteenth century onwards, whereby bar and saloon owners would attract drinkers in with free food on condition that they brought a drink, whereby the working class customer would likely be better off not drunk and with the money in their pocket.
There are a few things we need to review before getting into the guts of the NFLT. If you are like me and generally baulk at mathematics, feel free to skip to the higher level section further down this article.
Firstly, we are dealing with comparison of optimisation problems, sometimes called ‘cost functions’ or ‘energy functions’, which are represented as a mapping between points in a search space and the points in the space of possible ‘fitness’ or ‘cost’ values. Note that NFLT also stands for search algorithms, but the specifics will not be covered in this article.
Next, there are a few conditionals of NFLT:
1. The search space the optimiser iterates through will be finite.
2. The space of possible cost values will also be finite.
These two conditions are automatically met for optimisation algorithms ran on any computer; even if the potential values are floating point variables there is still a discrete range of values due to how floating point values are represented on a computer. Because the size of the two spaces are are both finite, this means the size of the set of all possible problems are finite.
Probability theory was utilised so that David H. Wolpert and William G. Macready could generalise their results for both stochastic and deterministic algorithms.
Here, ‘dₘ’ is the set of size ‘m’ of the cost values ‘y’. The ‘y’ cost values are associated to the aforementioned mapping (using the cost function ‘f’) of the search space and the cost values. This above statement gives us the probability of getting a sequence of cost values from the algorithm ‘a’ iterated ‘m’ times using the cost function ‘f’. Using this, Wolpert and Macready went on to state their first theorem:
The above theorem (the proof found in No Free Lunch Theorems for Optimisation) shows a few things. For the pair of algorithms ‘a₁’ and ‘a₂’ in the above equation, we can firstly, and crucially, note that the average of performance is independent of the algorithm. This theorem is demonstrating what a given algorithm gains on one of class problems it is equally offset by its performance on the remaining problems.
In their paper, Wolpert and Macready give a second NFLT, which is analogous to the first but shows that objective functions that vary with time. For brevity and my own sanity I’ve omitted this from the article. This second theorem means if an algorithm outperforms another with certain kinds of cost function dynamics, then the reverse must be true on the set of all other cost function dynamics.
The NLFT are a set of mathematical proofs and general framework that explores the connection between general-purpose algorithms that are considered “black-box” and the problems they solve.
The No Free Lunch Theorems state that any one algorithm that searches for an optimal cost or fitness solution is not universally superior to any other algorithm.
This is due to the huge potential problem space for the application of the general-purpose algorithm; if an algorithm is particularly adept at solving one class of problem, and the fitness surface that comes with it, then then it has to perform worse on the remaining average of problems. Wolpert and Macready wrote in their paper, No Free Lunch Theorems for Optimisation:
“If an algorithm performs better than random search on some class of problems then in must perform worse than random search on the remaining problems.”
In short because of the NFLT proof, we can state: A superior black-box optimisation strategy is impossible.
There is another issue to consider, which is the generality versus the specificity of the algorithm. This means we must weigh the tradeoff between an efficient algorithm which is very specific in the space of problems it can address, or the general algorithm which a jack of all trades, but a master of none.
In the real world, we need to decide on engineering solutions to build practical models that solve real problems. So, understanding NFLT, how might we identify favourable approaches that tackle our problems?
For certain pattern recognition problems, we tend to find that certain algorithms perform better than others which is a consequence of the algorithm’s fitness or cost function fit to the particular problem. For our particular given problem, in order to find the best algorithm, NFLT should remind us that we need to focus on the particular problem at hand, the assumptions, the priors (extra information), the data and the cost.
So in short, how well the algorithm will do is determined by how aligned the algorithm is with the actual problem at hand. Wolpert and Macready wrote in an earlier paper, No Free Lunch Theorems for Search, the following:
Ultimately, of course, the only important question is, how do I find good solutions for my given cost function ‘f’? The proper answer to this question is to start with the given ‘f’, determine certain salient features of it, and then construct a search algorithm, a, specifically tailored to match those features.
Don’t worry, you can put your deep learning hat back on.
What the NFLT is trying to tell us is, we are generally not going to find off the shelf algorithms that fit perfectly to our data. We are going to have to architect the algorithm to better fit the data — for example, make a Neural Network recurrent to better fit a time series, or make an Multi Layer Perceptron into a Convolutional Neural Network to better understand the spacial information in the input data. The beauty of neural nets is that they are modifiable in architecture, allowing us to engineer new solutions specific to the problem at hand, but then become specialised as you settle on a good configuration.
I should add, you can put on your Genetic Algorithm hat or whatever hat you choose to wear — there are always ways of adapting whatever algorithm you use — e.g changing how you encode the DNA in the GA or how the selection is done.
Ultimately, the lesson to take away today is that randomly selecting a good algorithm without making any structural assumptions on the problem is similar to the proverbial “needle in the haystack”. It will serve us well to consider the problem and the data, and act accordingly, engineering a solution with a good fit to the task at hand. | 1,487 | 7,250 | {"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-2019-30 | latest | en | 0.930743 |
http://www.enotes.com/homework-help/what-axis-symmetery-parabola-y-16x-2-16x-440035 | 1,462,117,793,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860116587.6/warc/CC-MAIN-20160428161516-00100-ip-10-239-7-51.ec2.internal.warc.gz | 481,065,248 | 11,682 | # What is the axis of symmetry of the parabola y = 16x^2 - 16x
Posted on
The axis of symmetry of the parabola y = 16x^2 - 16x has to be determined.
For a parabola of the form y - k = a(x - h)^2 the axis of symmetry is x = h.
y = 16x^2 - 16x
=> y = 16(x^2 - x + 1/4) - 4
=> y + 4 = 16*(x - 1/2)^2
The axis of symmetry of the parabola is x = 1/2 | 146 | 350 | {"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-2016-18 | longest | en | 0.832113 |
https://itprospt.com/qa/191352/30-find-the-sum-of-1-3-5-59 | 1,686,395,014,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00358.warc.gz | 373,622,241 | 10,930 | 1
30. Find the sum of 1 + 3 + 5 + ... + 59.
Question
30. Find the sum of 1 + 3 + 5 + ... + 59.
30. Find the sum of 1 + 3 + 5 + ... + 59.
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https://quantumcomputing.stackexchange.com/questions/24270/if-a4-b4-ab-i-what-is-a-good-circuit-for-sqrt-a-sqrt-b?noredirect=1 | 1,709,588,449,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00247.warc.gz | 476,175,118 | 41,884 | # TL/DR
What is a good circuit for: $$\frac{1}{2}\begin{pmatrix} -i & i & 1 & 1 \\ 1 & 1 & -i & i \\ i & -i & 1 & 1 \\ 1 & 1 & i & -i\end{pmatrix},$$
as this may be a useful matrix for taking square roots of other unitaries?
EDIT - there are some significant errors in my question, and I retract the setup. For example the initial circuit below is improper in at least three ways, as I didn't phase the ancillae, I got the IQFT wrong, and I didn't uncompute properly.
But I don't want to delete the question as @CraigGidney's answer is still valid to the original question, and his pointers to his posts are instructive, and deserves the check!
# Separate Square Roots
In more detail, given two unitary operators $$A,B$$ acting on an $$n$$-qubit state $$|\psi\rangle$$, where $$A$$ and $$B$$ are inverses of each other, e.g., $$AB=BA=I$$, and both $$A$$ and $$B$$ are of order $$4$$, e.g., $$A^4=B^4=I$$, we wish to find a good circuit for $$\sqrt A\sqrt B$$, as this might be part of a product formula for Hamiltonian simulation.
Initially we can construct the square roots for each of $$A$$ and $$B$$ separately, noting that because the eigenvalues of $$A$$ and $$B$$ are the fourth roots of unity, e.g., $$\pm i, \pm 1$$, we can use two ancillae for phase estimation with $$S$$ gates that rotate the ancillae and take the roots:
The last Hadamard gates are included to emphasize that the ancillae all revert back to $$|0\rangle$$.
# Parallel Circuit
Because $$[\sqrt A, \sqrt B]=0$$, i.e. the above circuits commute, we can execute them in parallel with four ancillae (two at the top and two at the bottom):
With the above parallel circuit, we have six controlled applications each of $$A$$ and $$B$$, sixteen Hadamard gates, and eight $$S$$/$$CS$$ gates.
# Initial Serial Circuit
But in the first circuits the ancillae revert back to $$|0\rangle$$, and we can also execute them in series:
This only uses two ancillae, with the same number of gates as, but double the depth of, the parallel circuit.
# Simplified Serial Circuit
However, recall that $$H^2=I$$, and we also were given that $$AB=A^2 B^2=I$$. Thus much of the above cancels out, leaving:
This simplified serial circuit uses two ancillae and six $$S$$ and one $$CZ$$ gate, but with ten $$H$$ gates and significantly only three controlled applications each of $$A$$ and $$B$$.
# Question
Quirk tells us that the remaining highlighted circuit in the middle of the last figure is equal to the matrix mentioned in the intro:
$$\frac{1}{2}\begin{pmatrix} -i & i & 1 & 1 \\ 1 & 1 & -i & i \\ i & -i & 1 & 1 \\ 1 & 1 & i & -i\end{pmatrix}.$$
I suspect the highlighted circuit could be simplified further still. Can the above circuit/matrix be simplified any more, using, say, Clifford+$$T$$ gates?
Indeed, the $$S$$ gates, the $$CZ$$ gate, etc. are all Clifford gates anyways (although cube roots/fourth roots/etc. would use non-Clifford gates).
• Mark, would it be possible for you to elaborate a bit about the application of this?
– Lior
Feb 27, 2022 at 6:46
What is a good circuit for [...]
Here's a circuit that does it.
Congrats, you've rediscovered that phase kickback can be used to implement powers of operations! But it's not just limited to cases where the eigenvalues are multiples of i, such as the fractional fourier transform. It works for anything (but only efficiently if you can implement big powers of the operation). For example, you can use it to turn incrementing into a continuous operation:
• Thanks, how did you go from my 13-gate circuit to your six-gate circuit so quickly? Was it intuitive, or did you have it in your back-pocket? Or did you auto-synthesize it somehow? Feb 26, 2022 at 21:27
• @MarkS I opened Quirk and made a manual version that matched. Then I mirrored it so I had it then its inverse, so the overall operation was identity. Then I iteratively manually optimized one half while preserving the fact that the whole thing was an identity. I was verifying its identity-ness by applying it to entangled qubits and making sure they were in the same state at the end. Then the optimized half was the result. An alternative automated method that would have worked was using Cirq's KAK decomposition functionality, which can give min-CZ versions of any two qubit unitary. Feb 26, 2022 at 21:30
• @MarkS Ah, I made have made a mistake when copying the unitary matrix. I tried it again and got a different circuit. I started from the circuit link you gave this time, instead of trying to resynthesize the matrix, so it should be correct. I updated the answer. Feb 26, 2022 at 21:39
• @MarkS Specific things I used were: S commutes with CZ, CP conjugated by a Clifford U can be rewritten into C(UPU^dagger), etc. Feb 26, 2022 at 21:40 | 1,270 | 4,753 | {"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": 36, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-10 | longest | en | 0.918206 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-early-transcendentals-8th-edition/appendix-a-numbers-inequalities-and-absolute-values-a-exercises-page-a9/40 | 1,531,774,020,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589455.35/warc/CC-MAIN-20180716193516-20180716213516-00099.warc.gz | 899,417,211 | 12,507 | ## Calculus: Early Transcendentals 8th Edition
$68≤ F≤86$
The interval on the Celsius scale corresponds to the temperature range. $20≤C≤30$ can be calculated as follows: Given: $C=\frac{5}{9}(F-32)$ $9C=5(F-32)$ We get, $F=\frac{9C}{5}+32$ Further, $20≤C≤30$ $\frac{9}{5}(20)≤ \frac{9}{5}C≤\frac{9}{5}(30)$ $36≤ \frac{9}{5}C≤54$ $36+32≤ \frac{9}{5}C+32≤54+32$ $68≤ \frac{9}{5}C+32≤86$ $68≤ F≤86$ Therefore, the interval on the Celsius scale corresponds to the temperature range $68≤ F≤86$. | 201 | 490 | {"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} | 3.984375 | 4 | CC-MAIN-2018-30 | latest | en | 0.607033 |
https://sec.oercommons.org/EN/browse?f.mccrs_alignment=MCCRS.Math.Content.HSF-LE.A.1c | 1,631,983,799,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056548.77/warc/CC-MAIN-20210918154248-20210918184248-00425.warc.gz | 557,774,151 | 15,166 | Updating search results...
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This task gives students an opportunity to work with exponential functions in a real world context involving continuously compounded interest. They will study how the base of the exponential function impacts its growth rate and use logarithms to solve exponential equations.
Subject:
Mathematics
Functions
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
Only Sharing Permitted
CC BY-NC-ND
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This lesson unit is intended to help teachers assess how well students are able to interpret exponential and linear functions and in particular to identify and help students who have the following difficulties: translating between descriptive, algebraic and tabular data, and graphical representation of the functions; recognizing how, and why, a quantity changes per unit intervale; and to achieve these goals students work on simple and compound interest problems.
Subject:
Algebra
Measurement and Data
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Only Sharing Permitted
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This lesson unit is intended to help teachers assess how well students are able to: articulate verbally the relationships between variables arising in everyday contexts; translate between everyday situations and sketch graphs of relationships between variables; interpret algebraic functions in terms of the contexts in which they arise; and reflect on the domains of everyday functions and in particular whether they should be discrete or continuous.
Subject:
Functions
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
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CC BY
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This is a direct task suitable for the early stages of learning about exponential functions. Students interpret the relevant parameters in terms of the real-world context and describe exponential growth.
Subject:
Mathematics
Functions
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
Unrestricted Use
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This problem provides an opportunity to experiment with modeling real data. Populations are often modeled with exponential functions and in this particular case we see that, over the last 200 years, the rate of population growth accelerated rapidly, reaching a peak a little after the middle of the 20th century and now it is slowing down.
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الرياضيات
وظائف
نوع المادة:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
المؤلف:
Illustrative Mathematics
05/01/2012
Unrestricted Use
CC BY
Rating
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The purpose of this task is to give students an opportunity to explore various aspects of exponential models (e.g., distinguishing between constant absolute growth and constant relative growth, solving equations using logarithms, applying compound interest formulas) in the context of a real world problem with ties to developing financial literacy skills.
Subject:
Mathematics
Functions
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics | 715 | 3,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2021-39 | latest | en | 0.862526 |
https://math.answers.com/math-and-arithmetic/What_is_25_percent_off_of_something | 1,725,793,039,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650976.41/warc/CC-MAIN-20240908083737-20240908113737-00295.warc.gz | 369,667,391 | 47,445 | 0
# What is 25 percent off of something?
Updated: 9/18/2023
Wiki User
11y ago
To get 25% off of something:
Example:
25% off of 170
= 25% discount applied to 170
= 170 - (25% * 170)
= 170 - (0.25 * 170)
= 170 - 42.5
= 127.50
Wiki User
11y ago | 102 | 255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.805313 |
http://www.onemathematicalcat.org/algebra_book/online_problems/dist_law.htm | 1,582,890,026,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875147154.70/warc/CC-MAIN-20200228104413-20200228134413-00065.warc.gz | 207,977,799 | 8,305 | Practice with the Distributive Law
PRACTICE WITH THE DISTRIBUTIVE LAW
• PRACTICE (online exercises and printable worksheets)
• This page gives an in-a-nutshell discussion of the concepts.
Want more details, more exercises? Read the full text!
One definition of the English verb to distribute is to spread out.
The distributive law is one of the most frequently used tools in algebra,
and tells how multiplication gets ‘spread out’ when it interacts
with addition—it gives a different order of operations that can be used.
The basic statement of the distributive law is deceptively simple.
Many important tools, however, are consequences of this law.
One of the most famous is ‘FOIL’, which we'll see is a memory device
for correctly multiplying expressions of the form $\,(a + b)(c + d)\,$.
THE DISTRIBUTIVE LAW
For all real numbers $\,a\,$, $\,b\,$, and $\,c\,$: $$a(b + c) = ab + ac$$
The distributive law offers two different orders of operation that always give the same result:
• The expression $\,a(b + c)\,$ specifies this order:
add $\,b\,$ to $\,c\,$; then multiply $\,a\,$ by this sum.
• The expression $\,ab + ac\,$ specifies this order:
multiply $\,a\,$ and $\,b\,$; multiply $\,a\,$ and $\,c\,$; then add these results.
One effective visual way to understand the statement of the distributive law is to use areas (see below).
Form a rectangle with height $\,a\,$ and width $\,b + c\,$.
The area of this rectangle is height times width: $a(b + c)\,$.
However, the area can also be found by summing the areas of the two smaller rectangles: $\,ab + ac\,$.
Thus, $\,a(b + c) = ab + ac\,$.
It helps some students first learning to use the distributive law to draw the following arrows:
EXAMPLES:
Simplify: $\,a(b - c)$
Answer: $ab - ac$
Do not change the order of the letters:
write ‘$\,ab - ac\,$’, not (say) ‘$\,ba - ac\,$’.
Simplify: $(-a + b)(-c)$
Answer: $ac - bc$
You must write your answers in the simplest possible way
to be recognized as correct in the exercise below.
Simplify: $-(-a + c)$
Answer: $a - c$
Simplify: $2a(4b - 3c)$
Answer: $8ab - 6ac$
Master the ideas from this section
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
For example, the answer to ‘$\,a(b - c)\,$’ must be input as ‘$\,ab - ac\,$’ .
Although (for example) ‘$\,ba - ca\,$’ is a correct answer, it is not recognized as correct. | 670 | 2,403 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2020-10 | latest | en | 0.846149 |
https://www.enjoytutorials.com/python-programming/python-sum-function-tutorial/ | 1,659,966,067,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570827.41/warc/CC-MAIN-20220808122331-20220808152331-00211.warc.gz | 691,874,015 | 70,117 | # Python sum() Function Tutorial
• Post author:
• Post category:Python
In this section, we will learn what the sum() function is and how to use it in Python.
## sum() Function in Python
The Python sum() function is used to sum the items of an iterable object.
For example, if the target iterable object is a list with integer numbers, then calling this function will add those numbers together and return the sum as a result.
## Python sum() Function Syntax:
`sum(iterable, initialValue)`
## Python sum() Function Parameters
The function takes two arguments:
• The first argument is the iterable object that we want to sum its elements.
• The second argument is the initial number that we want the elements of the iterable object to be added to. This value is optional and, if ignored, the default value, which is 0 will be used instead.
## Python sum() Function Return Value
The return value of this method is the sum of all items in the target iterable object.
## Example: sum() function in python
```tuple1 = (1,2,3,4,5,6,7)
print(sum(tuple1))```
Output:
`28`
## Example: python sum list
```list1 = [1,2,3,4,5,6,7]
print(sum(list1))```
Output:
`28`
## Example: sum() function and Python Dictionary
```dictionary = {
0: "John",
1: "Doe",
2: 200
}
print(sum(dictionary))```
Output:
`3`
Note that invoking the sum() function in a dictionary will sum up its keys! | 344 | 1,394 | {"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-2022-33 | latest | en | 0.689833 |
http://www.cppblog.com/liangairan/articles/135001.html | 1,722,956,098,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640492117.28/warc/CC-MAIN-20240806130705-20240806160705-00771.warc.gz | 37,463,500 | 11,789 | # 永远也不完美的程序
• 随笔 - 39
• 文章 - 61
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• 积分 - 270167
• 排名 - 94
### 最新评论
1. 怎样获得顶点的TBN
uniform vec3 lightpos; //传入光源的模型坐标吧
uniform vec4 eyepos;
varying vec3 lightdir;
varying vec3 halfvec;
varying vec3 norm;
varying vec3 eyedir;
attribute vec3 rm_Tangent;
void main(void)
{
vec4 pos = gl_ModelViewMatrix * gl_Vertex;
pos = pos / pos.w;
//把光源和眼睛从模型空间转换到视图空间
vec4 vlightPos = (gl_ModelViewMatrix * vec4(lightpos, 1.0));
vec4 veyePos = (gl_ModelViewMatrix * eyepos);
lightdir = normalize(vlightPos.xyz - pos.xyz);
vec3 eyedir = normalize(veyePos.xyz - pos.xyz);
//模型空间下的TBN
norm = normalize(gl_NormalMatrix * gl_Normal);
vec3 vtangent = normalize(gl_NormalMatrix * rm_Tangent);
vec3 vbinormal = cross(norm,vtangent);
//将光源向量和视线向量转换到TBN切线空间
lightdir.x = dot(vtangent, lightdir);
lightdir.y = dot(vbinormal, lightdir);
lightdir.z = dot(norm , lightdir);
lightdir = normalize(lightdir);
eyedir.x = dot(vtangent, eyedir);
eyedir.y = dot(vbinormal, eyedir);
eyedir.z = dot(norm , eyedir);
eyedir = normalize(eyedir);
halfvec = normalize(lightdir + eyedir);
gl_FrontColor = gl_Color;
gl_TexCoord[0] = gl_MultiTexCoord0;
gl_Position = ftransform();
}
// vertex shaderuniform vec3 lightpos; //传入光源的模型坐标吧uniform vec4 eyepos;varying vec3 lightdir;varying vec3 halfvec;varying vec3 norm;varying vec3 eyedir;attribute vec3 rm_Tangent;void main(void){ vec4 pos = gl_ModelViewMatrix * gl_Vertex; pos = pos / pos.w; //把光源和眼睛从模型空间转换到视图空间 vec4 vlightPos = (gl_ModelViewMatrix * vec4(lightpos, 1.0)); vec4 veyePos = (gl_ModelViewMatrix * eyepos); lightdir = normalize(vlightPos.xyz - pos.xyz); vec3 eyedir = normalize(veyePos.xyz - pos.xyz); //模型空间下的TBN norm = normalize(gl_NormalMatrix * gl_Normal); vec3 vtangent = normalize(gl_NormalMatrix * rm_Tangent); vec3 vbinormal = cross(norm,vtangent); //将光源向量和视线向量转换到TBN切线空间 lightdir.x = dot(vtangent, lightdir); lightdir.y = dot(vbinormal, lightdir); lightdir.z = dot(norm , lightdir); lightdir = normalize(lightdir); eyedir.x = dot(vtangent, eyedir); eyedir.y = dot(vbinormal, eyedir); eyedir.z = dot(norm , eyedir); eyedir = normalize(eyedir); halfvec = normalize(lightdir + eyedir); gl_FrontColor = gl_Color; gl_TexCoord[0] = gl_MultiTexCoord0; gl_Position = ftransform();}
uniform float shiness;
uniform vec4 ambient, diffuse, specular;
uniform sampler2D bumptex;
uniform sampler2D basetex;
float amb = 0.2;
float diff = 0.2;
float spec = 0.6;
varying vec3 lightdir;
varying vec3 halfvec;
varying vec3 norm;
varying vec3 eyedir;
void main(void)
{
vec3 vlightdir = normalize(lightdir);
vec3 veyedir = normalize(eyedir);
vec3 vnorm = normalize(norm);
vec3 vhalfvec = normalize(halfvec);
vec4 baseCol = texture2D(basetex, gl_TexCoord[0].xy);
//Normal Map里的像素normal定义于该像素的切线空间
vec3 tbnnorm = texture2D(bumptex, gl_TexCoord[0].xy).xyz;
tbnnorm = normalize((tbnnorm - vec3(0.5))* 2.0);
float diffusefract = max( dot(lightdir,tbnnorm) , 0.0);
float specularfract = max( dot(vhalfvec,tbnnorm) , 0.0);
if(specularfract > 0.0){
specularfract = pow(specularfract, shiness);
}
gl_FragColor = vec4(amb * ambient.xyz * baseCol.xyz
+ diff * diffuse.xyz * diffusefract * baseCol.xyz
+ spec * specular.xyz * specularfract ,1.0);
}
//fragment shaderuniform float shiness;uniform vec4 ambient, diffuse, specular;uniform sampler2D bumptex;uniform sampler2D basetex;float amb = 0.2;float diff = 0.2;float spec = 0.6;varying vec3 lightdir;varying vec3 halfvec;varying vec3 norm;varying vec3 eyedir;void main(void){ vec3 vlightdir = normalize(lightdir); vec3 veyedir = normalize(eyedir); vec3 vnorm = normalize(norm); vec3 vhalfvec = normalize(halfvec); vec4 baseCol = texture2D(basetex, gl_TexCoord[0].xy); //Normal Map里的像素normal定义于该像素的切线空间 vec3 tbnnorm = texture2D(bumptex, gl_TexCoord[0].xy).xyz; tbnnorm = normalize((tbnnorm - vec3(0.5))* 2.0); float diffusefract = max( dot(lightdir,tbnnorm) , 0.0); float specularfract = max( dot(vhalfvec,tbnnorm) , 0.0); if(specularfract > 0.0){ specularfract = pow(specularfract, shiness); } gl_FragColor = vec4(amb * ambient.xyz * baseCol.xyz + diff * diffuse.xyz * diffusefract * baseCol.xyz + spec * specular.xyz * specularfract ,1.0);}
(上为底纹理和法线纹理，下为它们与某破壁模型合作的效果，纹理from planetpixelemporium.com)
(我想是游戏最常用的用途：砖墙。我想是最常用的NormalMap,from NEHE)
(自己把墙壁BaseMap放入Photoshop的normalMapFilter里弄的NormalMap，呃.....)
posted on 2010-11-29 17:58 狂烂球 阅读(2305) 评论(0) 编辑 收藏 引用 所属分类: 图形编程
只有注册用户登录后才能发表评论。 【推荐】100%开源！大型工业跨平台软件C++源码提供，建模，组态！ 相关文章: | 1,653 | 4,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-33 | latest | en | 0.146472 |
https://www.rdocumentation.org/packages/nloptr/versions/1.2.1/topics/bobyqa | 1,591,433,858,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00382.warc.gz | 857,238,764 | 5,928 | # bobyqa
0th
Percentile
##### Bound Optimization by Quadratic Approximation
BOBYQA performs derivative-free bound-constrained optimization using an iteratively constructed quadratic approximation for the objective function.
##### Usage
bobyqa(x0, fn, lower = NULL, upper = NULL, nl.info = FALSE,
control = list(), ...)
##### Arguments
x0
starting point for searching the optimum.
fn
objective function that is to be minimized.
lower, upper
lower and upper bound constraints.
nl.info
logical; shall the original NLopt info been shown.
control
list of options, see nl.opts for help.
...
additional arguments passed to the function.
##### Details
This is an algorithm derived from the BOBYQA Fortran subroutine of Powell, converted to C and modified for the NLOPT stopping criteria.
##### Value
List with components:
par
the optimal solution found so far.
value
the function value corresponding to par.
iter
number of (outer) iterations, see maxeval.
convergence
integer code indicating successful completion (> 0) or a possible error number (< 0).
message
character string produced by NLopt and giving additional information.
##### Note
Because BOBYQA constructs a quadratic approximation of the objective, it may perform poorly for objective functions that are not twice-differentiable.
##### References
M. J. D. Powell. The BOBYQA algorithm for bound constrained optimization without derivatives,'' Department of Applied Mathematics and Theoretical Physics, Cambridge England, technical reportNA2009/06 (2009).
cobyla, newuoa
• bobyqa
##### Examples
# NOT RUN {
fr <- function(x) { ## Rosenbrock Banana function
100 * (x[2] - x[1]^2)^2 + (1 - x[1])^2
}
(S <- bobyqa(c(0, 0, 0), fr, lower = c(0, 0, 0), upper = c(0.5, 0.5, 0.5)))
# }
Documentation reproduced from package nloptr, version 1.2.1, License: LGPL-3
### Community examples
Looks like there are no examples yet. | 461 | 1,912 | {"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} | 2.703125 | 3 | CC-MAIN-2020-24 | latest | en | 0.707271 |
https://www.physicsforums.com/threads/diverging-lens.518238/ | 1,511,241,535,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806316.80/warc/CC-MAIN-20171121040104-20171121060104-00520.warc.gz | 861,765,421 | 15,078 | # Diverging Lens
1. Jul 31, 2011
### fromthepast
1. The problem statement, all variables and given/known data
A lens forms an image of an object. The object is 16 cm from the lens. The image is 12 cm
from the lens on the same side as the object. (a) What is the focal length of the lens? Is the
lens converging or diverging? (b) If the object is 8.5 mm tall, how tall is the image? Is it erect or inverted? Draw a principle-ray diagram.
2. Relevant equations
1/f = 1/s + 1/s'
3. The attempt at a solution
I got 6.8cm for the focal length and .638cm for the height. When I draw my principle-ray diagram, the object and image are to the left of the second focal point, which is to the left of the lens. Shouldn't the 2nd focal point be between the object and image on the left side of the lens?
2. Jul 31, 2011
### fromthepast
Anybody?
3. Jul 31, 2011
### ehild
The image is at the same side as the object. Is the image real or virtual?
ehild
4. Jul 31, 2011
### fromthepast
virtual
5. Jul 31, 2011
### ehild
Your result for the focal length is not correct. The virtual image distance has minus sign in the lens formula. Calculate again.
ehild
6. Jul 31, 2011
### fromthepast
-48cm?
7. Jul 31, 2011
### ehild
Yes, the focal length is -48 cm. The lens is diverging so the focal length is negative.
ehild | 398 | 1,329 | {"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} | 3.484375 | 3 | CC-MAIN-2017-47 | longest | en | 0.867886 |
https://jitely.info/?post=7170 | 1,632,355,531,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057403.84/warc/CC-MAIN-20210922223752-20210923013752-00657.warc.gz | 347,717,244 | 3,784 | # What are edges in math
## Geometric solids
Here we give you an overview of the topic Geometric solids. In our video we introduce you to different geometric bodies and show you how you can recognize them. Check it out right now!
### What are geometric bodies?
A geometric body consists of different surfaces and is a 3D object. You can geometric solids Take it in your hand and fill it with air or water. These body in the mathematics have special names and properties that we will introduce to you here.
Mathematical solids you can examine and calculate. Have along geometric body corners, edge and Surfaces. Some also have a tip.
Now let's look at each one body in maths more precisely.
### The geometric solid cube
The cube is a body, which consists of square surfaces. As geometric body it looks exactly like a dice.
And how many edges does a cube have? A cube in geometry Has
• 12 edges,
• 8 corners,
• 6 surfaces.
### The geometric body cuboid
The cuboid as body in the mathematics consists of rectangles.
How many areas does a cuboid have? Has the cuboid as a body in math
• 6 surfaces,
• 12 edges,
• 8 corners.
To the Properties of bodies belongs the surface area or the volume. In our extra video on the volume of cuboids and cubes | 283 | 1,256 | {"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-2021-39 | latest | en | 0.92261 |
https://programmer.help/blogs/detailed-explanation-of-huffman-tree.html | 1,679,912,495,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948620.60/warc/CC-MAIN-20230327092225-20230327122225-00658.warc.gz | 524,714,150 | 4,460 | # 1: Definition
1: Given n weights as n leaf nodes, a binary tree is constructed. If the weighted path length (wpl) of the tree reaches the minimum, such a binary tree is called the optimal binary tree, also known as Huffman tree. Other books are translated as Huffman tree. 2: Huffman tree is the tree with the shortest weighted path length, and the node with larger weight is closer to the root.
## Basic concepts
1: Path and path length: the path between child or grandson nodes that can be reached from one node down in a tree is called path. The number of branches in a path is called the path length. If the specified number of layers of the root node is 1, the path length from the root node to the L-th layer node is L-1 2: Node weight and weighted path length: if a node in the tree is assigned a value with a certain meaning, this value is called the weight of the node. The weighted path length of a node is the product of the path length from the root node to the node and the weight of the node 3: Weighted path length of the tree: the weighted path length of the tree is specified as the sum of the weighted path lengths of all leaf nodes, which is recorded as WPL(weighted path length). The binary tree with greater weight and closer to the root node is the optimal binary tree. The smallest WPL is Huffman tree
# 2: Create Huffman tree
## Idea:
1: Sort from small to large. Each data is a node, and each node can be regarded as the simplest binary tree 2: Take out the two binary trees with the smallest weight of the root node 3: Form a new binary tree. The weight of the root node of the new binary tree is the sum of the weight of the root node of the previous two binary trees 4: Then the new binary tree is sorted again according to the weight of the root node and repeated Step 1-2-3-4 until all the data in the sequence are processed to obtain a Huffman tree
For example: 13, 7, 8, 3, 29, 6, 1 Sort 1, 3, 6, 7, 8, 13, 29
Preorder traversal: 67, 29, 38, 15, 7, 8, 23, 10, 4, 1, 3, 6, 13
```package com.atgguigu.huffmanTree;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
public class HuffmanTree {
public static void main(String[] args) {
int[] arr = {13, 7, 8, 3, 29, 6, 1};
Node node = createHuffmanTree(arr);
preOrder(node);
}
public static Node createHuffmanTree(int[] arr){
//Put the array into the Node collection for easy management
List<Node> nodes = new ArrayList<>();
for (int value : arr) {
}
while (nodes.size() > 1){
// 1: Sort
Collections.sort(nodes);
// 2: Take out the two binary trees with the smallest weight of the root node
Node leftNode = nodes.get(0);
Node rightNode = nodes.get(1);
// 3: Form a new binary tree. The weight of the root node of the new binary tree is the sum of the weight of the root node of the previous two binary trees
Node parent = new Node(leftNode.value+ rightNode.value);
parent.left = leftNode;
parent.right = rightNode;
nodes.remove(leftNode);
nodes.remove(rightNode);
// 4: The new binary tree is sorted again according to the weight of the root node
}
return nodes.get(0);
}
//Preorder traversal
public static void preOrder(Node node){
if(node != null){
node.preOrder();
}else {
System.out.println("You play with me? The empty tree is still coming");
}
}
}
//Node class
//Comparable < node > set sorting
class Node implements Comparable<Node>{
int value; //Node weight
Node left; //Left child node
Node right; //Right child node
public Node(int value){
this.value = value;
}
@Override
public String toString() {
return "Node{" +
"value=" + value +
'}';
}
//Comparable < node > set sorting
@Override
public int compareTo(Node o) {
//Indicates sorting from small to large
//return -(this.value-o.value); Indicates sorting from large to small
return this.value-o.value;
}
//Preorder traversal
public void preOrder(){
System.out.print("==>"+this);
if(this.left != null){
this.left.preOrder();
}
if(this.right != null){
this.right.preOrder();
}
}
}
```
Tags: Java data structure
Posted on Fri, 29 Oct 2021 02:53:46 -0400 by Mattyspatty | 1,055 | 4,498 | {"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-2023-14 | longest | en | 0.824464 |
http://www.unt.edu/UNT/departments/CC/Benchmarks/sprsum97/resamp.htm | 1,464,752,193,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464054526288.74/warc/CC-MAIN-20160524014846-00201-ip-10-185-217-139.ec2.internal.warc.gz | 867,477,336 | 4,157 | # Statistical Computing Tips
## Sampling with Replacement in SPSS: Creating Bootstrap Confidence Intervals for the Correlation Coefficient
Resampling based approaches to estimating a statistic and determining the properties of the estimator are becoming increasingly identified with what is sometimes referred to as "modern methods" of data analysis (Fox and Long, 1990). Resampling, using subsamples of an original sample, allows one to incorporate all of the original characteristics of the observed empirical distribution into the significance estimates and confidence intervals for the statistic under consideration. Efron (1979) proposed a resampling plan, which he called the "bootstrap." The bootstrap technique has received considerable attention over the years since its popularization by Efron (an overview of these developments can be found in Efron and Tibshirani, 1993). In a bootstrap resampling scheme, the initial sample of observations is treated as if they constitute the population under study. By randomly resampling with replacement from this proxy population, the sampling error of the original population can be estimated and confidence intervals constructed for most statistics that are being evaluated.
For the present illustrative example, 11 pairs of observations constitute the sample/population. From these 11 pairs of data, sampling with replacement is used to create a new sample of size 11 (a bootstrap sample). In this new sample, the data pair (.18, .20) might appear only once or could appear multiple times in the sample of 11 data pairs. A Pearson's correlation coefficient is calculated for each of these bootstrap samples. The standard deviation of these bootstrap correlations is calculated thus giving an estimate of the standard error of the correlation coefficient. The calculation of the 2.5th and 97.5th percentiles of the distribution of the bootstrap correlations will give an estimate of the 95th confidence intervals for the correlation coefficient (percentile method). A bias adjustment for the percentile method (bias corrected percentile method) is discussed in Efron and Tibshirani (1993); here, we discuss only the unadjusted percentile method.
Thompson (1993) discusses using the bootstrap methodology in conjunction with traditional statistical significance testing to explore result replicability. Thompson's data set (page 370)is used as our example:
### Data Set (From Thompson, 1993):
Y X
1.00 .18 .20
2.00 .54 1.88
3.00 -.49 -.76
4.00 ..92 .42
5.00 .22 .32
6.00 .75 -.56
7.00 .66 1.55
8.00 -2.65 -1.21
9.00 -.51 -.66
10.00 .47 -.96
11.00 -.09 -.21
This data set produces the following statistics:
### Normal Theory Significance and CI:
r = 0.56, p = .073, 95% CI = (-0.06, .868)
The SPSS syntax included here uses the SPSS INPUT PROGRAM to generate 1000 samples (n=11 per sample)of randomly sampled case id's (sampling with replacement). The MATCH FILES procedure is used to copy data from the original file (the x,y pairs) into the working data file.
```**** Bootstrap Confidence Intervals for
**** the Correlation Coefficient
**** create 1000 bootstrap samples of size
**** n=11, use sampling with replacement
input program.
loop samp=1 to 1000.
+ loop #i=1 to 11.
+ compute id=trunc(uniform(11))+1.
+ end case.
+ end loop.
+ leave samp.
end loop.
end file.
end input program.
execute.
sort cases by id.
match files file=* /table='a:\thompson.sav' /by id.
sort cases by samp.
split file by samp.
execute.
**** calculate a correlation coefficient for each bootstrap sample
CORRELATIONS
/VARIABLES=y x
/PRINT=TWOTAIL SIG
/MISSING=PAIRWISE .```
Once the correlation output has been saved to an output text file, one removes the sample ids, correlation values, and pvalues from the output file of the 1000 bootstrap samples:
```SET WIDTH=80.
FILE TYPE NESTED FILE='a:\corr.out' RECORD=1-80 (A).
RECORD TYPE SAMP:'.
DATA LIST / sample 9-16.
RECORD TYPE X'.
DATA LIST RECORDS=3 / corr 13-18 // pvalue 16-19 .
END FILE TYPE.
FORMATS corr (F8.2) pvalue (F8.2) sample (F8.2) .
execute.```
Next, the lower 2.5th and upper 97.5th percentiles of the empirical distribution of correlation coefficients are calculated:
```FREQUENCIES VARIABLES=corr
/FORMAT=NOTABLE
/PERCENTILES= 2.5 97.5
/STATISTICS=STDDEV MEAN.
CORR
Mean .573 Std dev .162
Percentile Value Percentile Value
2.50 .179 97.50 .828
Valid cases 1000 Missing cases 0```
Our bootstrap 95th percentiles are (.179, .828). Since these intervals do ot include 0, this is taken to be a rejection of the null hypothesis, that the correlation coefficient is zero in the population Power estimation (Cohen, 1988) with the bootstrap is accomplished by counting the proportion of redrawn samples that lead to a statistically significant estimator (for a given alpha level):
```**** Probability to reject an assumed false
**** null hypothesis (simulated power).
do if pvalue<<=.05).
compute count=1.
else if pvalue>>(.05).
compute count=0.
end if.
execute.
FREQUENCIES
VARIABLES=count.
COUNT
Valid Cum
Value Label Value Frequency Percent Percent Percent
.00 523 52.3 52.3 52.3
1.00 477 47.7 47.7 100.0
- - -
Total 1000 100.0 100.0
Valid cases 1000 Missing cases 0```
Power estimate based on distributional assumptions (Using Cohen's power tables) = .460
Resampling based power estimate = .477
## References:
Cohen, J. (1988). Statistical Power Analysis for the Behavioral Sciences. Lawrence Erlbaum Associates, Hillsdale, New Jersey.
Efron, B & Tibshirani, R.J. (1993). An Introduction to the Bootstrap. Chapman and Hall, New York.
Fox, J & Long, J.S. (1990). Modern Methods of Data Analysis. Sage Publications, Newbury, Park, CA.
Thompson, B. (1993). The Use of Statistical Significance Tests in Research: Bootstrap and Other Alternatives. Journal of Experimental Education, 61(4), 361-377.
Previous Article Next Article | 1,495 | 5,999 | {"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-2016-22 | latest | en | 0.893054 |
https://checksheet.app/google-sheets-formulas/sqrtpi/ | 1,721,879,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518532.66/warc/CC-MAIN-20240725023035-20240725053035-00258.warc.gz | 133,459,331 | 10,949 | # SQRTPI
The `SQRTPI` function returns the square root of the product of a supplied number (value) and pi (3.14159265359). This function is useful in geometry and trigonometry calculations where the area and circumference of circles are used.
## Usage
Use the `SQRTPI` formula with the syntax shown below, it has 1 required parameter:
Parameters:
1. value (required):
The number to be multiplied by pi and then the square root taken.
## Examples
Here are a few example use cases that explain how to use the `SQRTPI` formula in Google Sheets.
### Calculate the circumference of a circle
To calculate the circumference of a circle, multiply the diameter by pi. The `SQRTPI` function can be used to simplify this calculation by taking the square root of the product of the diameter and pi: `=2*SQRTPI(diameter/2)`.
### Calculate the area of a circle
To calculate the area of a circle, square the radius and multiply by pi. The `SQRTPI` function can be used to simplify this calculation by taking the square root of the product of the radius squared and pi: `=SQRTPI(radius^2)`.
### Calculate the standard deviation of a set of values
The `SQRTPI` function can be used to calculate the standard deviation of a set of values. The formula for the standard deviation is the square root of the variance. The variance is the average of the squared differences from the mean. Therefore, the standard deviation can be calculated as follows: `=SQRTPI(AVERAGE((values - AVERAGE(values))^2))`.
## Common Mistakes
`SQRTPI` not working? Here are some common mistakes people make when using the `SQRTPI` Google Sheets Formula:
### Not multiplying the input by pi
One common mistake when using the `SQRTPI` function is forgetting to multiply the input by pi before taking the square root.
### Providing a negative input
The `SQRTPI` function only works with positive inputs. Providing a negative input will result in an error.
The following functions are similar to `SQRTPI` or are often used with it in a formula:
• `PI`
The `PI()` function returns the value of pi to 10 decimal places, which is approximately 3.1415926536. This function is often used in geometry and trigonometry calculations.
• `SIN`
The `SIN` function in Google Sheets returns the sine of a given angle in radians. Sine is a mathematical function that describes a smooth repetitive oscillation. It is commonly used in trigonometry, physics, and engineering to model phenomena such as waves, oscillations, and periodic motion.
• `COS`
The `COS` function in Google Sheets returns the cosine of an angle provided in radians. It is commonly used in trigonometry to calculate the cosine of an angle. The function takes one parameter, the angle in radians.
You can learn more about the `SQRTPI` Google Sheets function on Google Support. | 637 | 2,809 | {"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} | 3.875 | 4 | CC-MAIN-2024-30 | latest | en | 0.797873 |
https://math.stackexchange.com/questions/2716512/chain-rule-deriviation | 1,563,522,937,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00497.warc.gz | 474,673,384 | 35,385 | # Chain rule deriviation
Given gradient and hessian of $\phi$ and the first and second derivatives of $h$ I would like to find the gradient and hessian matrix of $f$
$f(x)=h(\phi(x))$
Where $f:R^n \longrightarrow R$
$\phi:R^n \longrightarrow R$
$h:R \longrightarrow R$
According to the chain rule: $\nabla f = h'(\phi(x))\nabla\phi(x)$ Is it correct and trivial and doesn't need any other proof? And how can I find the hessian of $f$?
• Well, yeah, the chain rule says exactly that, why should you need some other proof? And just compute all (mixed) second order partial derivatives to find the hessian – VanillaThunder Mar 31 '18 at 19:08
• you might have to use the chain rule again... – VanillaThunder Mar 31 '18 at 19:14
• @cbdes Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… – gimusi Apr 3 '18 at 14:35
Since
$$f(x_1,x_2,...,x_n)=h(\phi(x_1,x_2,...,x_n))$$
by chain rule we have that
$$\frac{\partial f}{\partial x_i}=h'(\phi(\vec x))\frac{\partial \phi}{\partial x_i}\implies \nabla f(\vec x) = h'(\phi(\vec x))\nabla\phi(\vec x)$$
For the Hessian note that
$$\frac{\partial^2 f}{\partial x_i^2}=h''(\phi(\vec x))\left(\frac{\partial \phi}{\partial x_i}\right)^2+h'(\phi(\vec x))\frac{\partial^2 \phi}{\partial x_i^2}$$
$$\frac{\partial^2 f}{\partial x_ix_j}=h''(\phi(\vec x))\frac{\partial \phi}{\partial x_i}\frac{\partial \phi}{\partial x_j}+h'(\phi(\vec x))\frac{\partial^2 \phi}{\partial x_ix_j}$$ | 504 | 1,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-30 | longest | en | 0.807844 |
https://matholympiad.org.bd/forum/viewtopic.php?p=14729 | 1,603,144,270,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107866404.1/warc/CC-MAIN-20201019203523-20201019233523-00475.warc.gz | 435,350,165 | 7,476 | ## BdMO National 2013: Higher Secondary 4
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm
### BdMO National 2013: Higher Secondary 4
If the fraction $\dfrac{a}{b}$ is greater than $\dfrac{31}{17}$ in the least amount while $b<17$, find $\dfrac{a}{b}$.
SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm
### Re: BdMO National 2013: Higher Secondary 4
Is it enough to find the successor term of $\frac{14}{17}$ in fairy sequence $f_{17}$?
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:
### Re: BdMO National 2013: Higher Secondary 4
By the definition of Farey sequence, yes. But that too involves manual search for a primitive solution of a linear diophantine equation.
Use $L^AT_EX$, It makes our work a lot easier!
SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm
### Re: BdMO National 2013: Higher Secondary 4
I got $\frac{11}{6}$. Is it the answer?
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
photon
Posts: 186
Joined: Sat Feb 05, 2011 3:39 pm
Location: dhaka
Contact:
### Re: BdMO National 2013: Higher Secondary 4
$\frac {11}{6}$ is the answer , I had found it by trial and error method .
Try not to become a man of success but rather to become a man of value.-Albert Einstein
Ismail Rifat
Posts: 1
Joined: Tue Aug 16, 2011 11:37 am
### Re: BdMO National 2013: Higher Secondary 4
I used Continued Fraction. And got $\frac{11}{6}$. Is it one from the right ways? | 575 | 1,684 | {"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 | 3 | CC-MAIN-2020-45 | latest | en | 0.786402 |
https://www.jiskha.com/display.cgi?id=1364856251 | 1,498,678,421,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323730.30/warc/CC-MAIN-20170628185804-20170628205804-00524.warc.gz | 884,377,260 | 3,790 | # algebra
posted by .
Consider making a four-digit I.D. number using the digits 3,5,8, and 0
a. How many I.D. numbers can be formed using each digit once?
b. How many can be formed using each digit once and not using 0 first?
c. How many can be formed if repetition is allowed and any digit can be first?
d. How many can be formed if repetition is allowed but 0 is not used first?
• algebra -
a) so we can use the 0 at the front
number of ways = 4x3x2x1 = 24
b) can't have zero at front
number of ways = 3x3x2x1 = 18
c) repeats are allowed and zero can be at front
number of ways = 4x4x4x4= 256
d) repeats allowed but no zero at front
number of ways = 3x4x4x4 = 192 | 211 | 669 | {"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-2017-26 | longest | en | 0.89275 |
https://en.wikipedia.org/wiki/Megacycle | 1,493,427,201,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123102.83/warc/CC-MAIN-20170423031203-00100-ip-10-145-167-34.ec2.internal.warc.gz | 781,407,644 | 11,629 | # Cycle per second
(Redirected from Megacycle)
A 1000 kilocycle military grade crystal resonator with an octal base
The cycle per second was a once-common English name for the unit of frequency now known as the hertz. The plural form was typically used, often written cycles per second, cycles/second, c.p.s., c/s, ~, or, ambiguously, just cycles. The term comes from the fact that sound waves have a frequency measurable in their number of vibrations, or cycles, per second.[1]
With the organization of the International System of Units in 1960, the cycle per second was officially replaced by the hertz, or reciprocal second. Symbolically, "cycle per second" units are "cycle/second", while hertz is "1/second" or ${\displaystyle {\text{s}}^{-1}}$.[2] This particular mandate has been so widely adopted as to render the old 'cycle per second' all but extinct.
For higher frequencies, kilocycles (kc), as an abbreviation of kilocycles per second were often used on components or devices. Other higher units like megacycle (Mc) and less commonly kilomegacycle (kMc) were used before 1960[3] and in some later documents.[4] These have modern equivalents such as kilohertz (kHz), megahertz (MHz), and gigahertz (GHz).
The rate at which aperiodic or stochastic events occur may be expressed in becquerels (as in the case of radioactive decay), not hertz, since although the two are mathematically similar by convention hertz implies regularity where becquerels implies the requirement of a time averaging operation. Thus, 1 Bq is 1 event per second on average whereas 1 hertz is 1 event per second on a regular cycle.
Cycle can also be a unit for measuring usage of reciprocating machines, especially presses, in which cases cycle refers to one complete revolution of the mechanism being measured (i.e. the shaft of a reciprocating engine). | 433 | 1,843 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-17 | latest | en | 0.946887 |
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### 2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare
2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare. We have economics answers for 2022 exam. These questions can be viewed here. This WAEC Economics paper for SSCE for 2022 will be taken as scheduled in the time table.
Furthermore, this post, will be giving you samples of current WAEC economics questions. Therefore, if you are to participate in the examination this year these questions will help you a lot. In fact, this is how I made my best grades during my time. I encourage you to read through and get prepared for this exam.
### THE OBJECTIVE OF POST:
Because people are asking about WAEC Economics questions and answers 2022/2023. They need expo and here they are. They are also asking for JAMB 2022 ECONOMICS questions and answers pdf, The solutions are the objectives of this post. This post therefore takes care of those students who plan to do well in the WAEC exams this year.
However, the best way to read this post is to read it alongside clicking the highlighted topics for referencing. Especially those of you aspiring for oversea admission and scholarships. So check out these related topics that follow.
## WAEC Objectives and Essay answers 2022
The questions below are the WAEC 2021 Economics Practice Questions. Go through them and be ready to score high in your WAEC 2021 Economics Examination.
1. If at 10K per kg, 1000kg of yam were purchased, the resultant point elasticity of demand is
A. 0.33
B. 0.0001
C. 1
D.10000
1. A situation in which all inputs are doubled and output also doubles is known as
A. constant proportions
B. increasing returns to scale
C. constant returns
D. constant returns to scale.
ANSWER: D (constant returns to scale)
1. Developments outside a given firm which reduce the firm’s costs are called
A. internal economics
B. external economics
C. external diseconomies
D. optimum effects
### 2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare
1. Scarcity in economics means that _
A. human wants are limitless
B. the economy has very few resources
C. the economy can scarcely produce anything
D. resources are limited.
2. Air is essential to life but commands no price! But Diamond is not essential to life but commands a high price! This is the paradox of __
A. thrift B. value C. abundance D. scarcity.
3. Economics of scale operate only when __
A. marginal cost is falling with input
B. average cost is falling with input
C. fixed cost is variable
D. variable cost is less than the fixed cost.
4. Efficiency in production involves _
A. reducing the size of the workforce
B. producing a given output with the lowest cost combination of factors of production
D. increasing the quantity of the fixed factor of production.
### HERE IS THE 2022/2023 WAEC ECONOMICS QUESTIONS AND ANSWERS
1. An effect of inflation is that it _
B. favours debtors at the expense of creditors
C. increases the real income of salary earners
D. increases the value of a country’s exports.
2. What could be the opportunity cost of a nuclear power station?
A. the running costs of the power station
B. a coal-fired power station
C. the current value of the power station
D. the cost of building the power station.
3. The market for a good was in equilibrium. A change occurred which resulted in a new equilibrium with a higher price for the good and a lower quantity traded.
What change would have caused this?
A. the demand curve moved to the left
B. the demand curve moved to the right
C. the supply curve moved to the left
D. the supply curve moved to the right.
4. A demand curve shows the relationship between the quantity demanded and __
A. a change in income
B. consumer tastes
C. the supply of the product
D. the price of the product
5. A government subsidises the production of pineapples. This is likely to _
A. increase the price of pineapples
B. raise the costs of supplying pineapples
C. raise revenue for the government
D. cause the supply of pineapples to increase at every price.
6. What indicates the existence of external costs in an economy?
A. An international trade deficit has caused the country to be in debt.
B. National companies have borrowed from foreign investors.
C. Private costs of production are less than social costs.
D. Private costs of production are more than social benefits.
### 2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare
1. What might be a disadvantage to a trade union when arguing for an increase in its members’ pay?
A. an increase in imports of a cheaper, similar product
B. the closure of a local training college resulting in fewer potential workers
C. the development of a new and profitable brand of the company’s product
D. the development of new techniques that increase productivity.
2. A German car manufacturer decided to produce its cars in a factory in China. What would not be a reason why they might have chosen to do this?
A. cheaper wage costs in China
B. the availability of raw materials
C. to gain external economies from skilled labour in China
D. to increase Germany self-sufficiency.
3. A government removed the quota on goods imported into the country.
What is the most likely result of this?
A. a decrease in demand for domestic production
B. a decrease in domestic unemployment
C. a decrease in exports
D. a decrease in the balance of trade deficit.
4. A modern corporation is owned by __
A. debenture holders B. ordinary shareholders C. preference shareholders D. creditors.
5. One of the most important factors that should be considered in the location of an industry is _
A. nearness to the financial centre
B. assured patronage by government functionaries
C. availability of inputs and market
6. What is the term used to describe a policy aimed at promoting the local production of goods which are usually imported?
A. deregulation B. import substitution C. tariff reduction D. backward integration.
7. Progressive tax structure is designed to _
A. take more from the income of the poor
B. take more from the income of the rich
C. take equal proportion of income from both the rich and the poor
D. reduce the problems emanating from tax imposition.
### 2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare
1. Which of the following best describes the production function?
A. it indicates the best output to produce
B. it relates naira inputs to naira outputs
C. it relates physical outputs to physical inputs
D. it indicates the best way to combine factors to produce any given output.
2. Which of the following reward is associated with entrepreneurship as a factor of production?
A. salaries B. profits C. interest D. rents.
3. In a market economy, the question of what, how and for whom to produce are solved by the __
A. elected representative of the people
B. planning committee
C. price mechanism
D. government.
4. At any given level of output, a firm’s total variable cost equals __
A. total cost less marginal cost
B. total cost less total fixed cost
C. total cost less average cost
D. average variable cost and marginal variable cost.
Age groups (years) Distribution (%)
Above 60 30
15 – 60 45
0 – 14 25
### HERE IS THE 2021/2022 WAEC ECONOMICS QUESTIONS AND ANSWERS
1. From the above, the estimated dependency ratio of the population shown above is _ A. 11: 9 B.9: 11 C. 7: 3 D. 3: 7
2. Which of the following factors is NOT responsible for the rural /urban drift in Nigeria?
A. the infrastructural facilities in the cities
B. declining fertility of rural farmlands
C. rural electrification programme
D. higher living standards in urban areas.
3. The necessity of choice is due to the fact that _
A. human wants are inelastic
B. customers like to maximize satisfaction
C. resources are abundant
D. consumers are selective.
4. What is meant by labour supply?
A. number of people in working population
B. number of men and hours they work
C. number of hours during which the middle-aged persons work
D. number of workforce multiplied by the hours they worked .
5. Any payment to a factor of production in excess of what is necessary to keep its present employment is known as __
A. real income B. profit C. economic rent D. real wage
6. The market where there are many differentiated products is called __
A. monopoly B. perfect competition C. monopolistic competition D. Oligopoly.
7. One of the advantages of large scale production is that
A. there is a rise in the cost of administration.
B. consumers sacrifice their individual tastes.
C. the firm can use labour-saving machinery.
D. the demand for a firm’s products becomes localized.
8. The location of timber and plywood industries in West Africa is mainly influenced by the availability of
A. transport.
B. water.
C. raw materials.
D. labour supply.
9. Malthus’ population theory states that
A. high death rate may lead to low productivity.
B. population may outgrow the means of subsistence.
C. people will always decide to have children.
D. migration may leave some parts of the world barren.
10. Which of the following is not likely to be an effect of a growing population?
A. Rise in demand
B. Unemployment
C. Fall in standard of living
D. High per capita income
11. Personal savings are generally low in West Africa because of
A. the level of income of the people.
B. the refusal of banks to grant loans.
C. overpopulation.
D. cheaper foreign currencies.
12. Devaluation of currency in a country is likely to lead to
A. increasing population.
B. increasing imports.
C. exports becoming cheaper.
D. reduced exports.
13. Which of the following is not a benefit derived from the petroleum industry?
A. Increased foreign exchange earnings
B. Establishment of refineries and petrochemical industries
C. Employment of a greater proportion of the population
D. Development of airports, seaports and other social infrastructure
14. An efficient weapon used in resolving disputes between employees is
A. co-operation.
B. collective bargaining.
C. display of placards.
D. legal action.
### WAEC ECONOMICS QUESTIONS AND ANSWERS
1. A country’s balance of payment is in deficit when
A. a country’s payments for imports of invisible goods are greater than her receipts from exports of invisible goods.
B. the total receipts from her export of visible and invisible goods are greater than her payments for visible and invisible imports.
C. it can record a surplus on the current account of her balance of payments accounts.
D. the total payments for visible and invisible imports are greater than the total receipts from her exports of visible and invisible goods.
2. Which of the following features best describes peasant agriculture in West Africa? It
A. specializes in the production of one crop.
B. involves the use of small farm holdings.
C. is a capital-intensive system of farming.
D. is mostly associated with tree crops
### 2022/2022 WAEC Economics Essay/Objective Questions and Answers – How to prepare
PAPER 2 [Essay]
1. (a) Define mobility of labour.
(b) Describe any four factors influencing the supply of labour.
ANS: (b) i. the size of the population;
ii. age distribution of the population;
iii. working hours;
iv. certain beliefs /practices;
v. level of wages;
vi. number of public holidays;
vii. official retirement age.
1. (a) Define money.
(b) State the three motives for holding money.
(c) Mention two determinants of each of the motives for holding money.
ANS: Transactions motive
(i) size of income;
(ii) interval between wage payments;
(iii) availability of credit;
(iv) family size.
Precautionary motive
(i) size of income;
(ii) interval between wage payments;
(iii) availability of credit;
(iv) perception of risks.
Speculative motive
(i) the rate of interest;
(ii) the degree of risk aversion.
1. (a) With the aid of a diagram, explain a minimum price.
(b) State any five measures by which a minimum price for an agricultural produce can be made effective.
2. (a) What is a supply schedule?
(b) Using an example, show how a market supply schedule of a product is obtained from individual supply schedules.
(c) State three examples of exceptional demand.
ANS: (c) Exceptional demand:
i. Fixed demand or perfectly inelastic demand e.g. salt
ii. Perfectly elastic demand
iii. Expectation of future increase in price.
iv. Articles of ostentation.
v. Giffen goods.
1. (a) With examples, distinguish between direct and indirect tax.
(b) Explain any four problems of tax collection in any West African Country.
ANS: (a) A direct tax is a tax on incomes and properties. Examples include persona income tax, company tax, death duties, inheritance tax, capital gains taxes, etc.
Whereas, an indirect tax is tax on goods and services. Examples include sales tax, import and export duties, excise tax, purchase tax, value added tax, etc.
(b) (i) absence of reliable records of business activities and revenues collected.
(ii) the prevalence of subsistence production.
(iii) corruption on the part of tax officials.
(iv) high level of tax evasion and tax avoidance etc.
1. (a) What is demographic transition theory ?
(b) Explain the three stages of the theory.
ANS: (a) Demographic Transition Theory is concerned with the historical population growth of society. It explains the relationship between fertility and mortality on population growth and how developed countries in contemporary times have passed through three identical stages of population history.
(b) Stages of the Theory;
stage 1 (pre-transition phase/ stage)
stage 2 (transition phase/ stage)
And, stage 3 (post transition phase /stage)
### 2021/2022 WAEC ECONOMICS QUESTIONS AND ANSWERS
1. (a) Differentiate between shares and debentures.
(b) Identify any four problems encountered by farmers in raising capital
ANS: (a) A share is the smallest unit into which the capital of a company is divided. It is a unit of ownership of a business concern while a debenture is a loan capital or corporate bond. A debenture holder is a creditor to a company.
(b) (i) low level of loanable funds;
(ii) inability of firms to produce the required collateral facilities;
(iii) under-developed money and capital markets;
(iv) high cost of loans;
(v) fluctuation of share prices;
(vi) Government financial regulation.
1. (a) State and explain the law of comparative cost advantage.
(b) Give two limitations of the law as a theory of international trade.
ANS: (a)
(b) Some limitations of the law are:
– It assumes that there are no transport costs.
– And, it assumes free mobility of factors of production.
– It assumes that production conditions (technology and the resource base of countries) will remain the same over time.
2. (a) Why is scarcity a fundamental problem in Economics?
(b) Give a reason why Economics is a;
(i) Science;
(ii) Social Science.
ANS: (a) Economics seeks to study the relationship between ends and means. Ends are unlimited while the means are limited. Scarcity means resources are limited in relation to the ends. Economics is therefore concerned with allocating the limited resources among the competing and unlimited wants.
(b) (i) Economics is a science because it adopts the scientific method.
(ii) Economics is a social science because it studies human behaviour. e.g. If the price of a commodity rises, people will buy less, other things equal.
1. (a) What is a centrally planned economy?
(b) Outline any four features of a capitalist economy?
ANS: (a) A centrally planned economy is one in which there is public ownership of the means of production and in which the decisions as to what to produce, how to produce and for whom to produce are made by the government.
(b) Features of capitalist economy;
i. there is private ownership of resources;
ii. production decisions are taken mainly by private individuals and organizations;
iii. capital is provided by private individuals and organizations;
iv. there is the profit motive;
v. there is very limited government participation in economic activity. | 3,635 | 16,084 | {"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-2024-38 | latest | en | 0.914448 |
https://socratic.org/questions/5927ccb27c014943aa1aaeae#429934 | 1,660,677,112,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00493.warc.gz | 478,313,429 | 6,350 | # Question #aaeae
May 26, 2017
a) $3 {x}^{2} - \frac{8}{{x}^{3}} + \frac{1}{2 \sqrt{x}}$
b) $40 {\left(2 x - 1\right)}^{3}$
c)$\frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x - \cos \frac{x}{\sin} ^ 2 x$
#### Explanation:
a) $\frac{d}{\mathrm{dx}} \left[{x}^{3} + \frac{4}{x} ^ 2 + {x}^{\frac{1}{2}}\right] = \frac{d}{\mathrm{dx}} \left[{x}^{3}\right] + 4 \frac{d}{\mathrm{dx}} \left[{x}^{- 2}\right] + \frac{d}{\mathrm{dx}} \left[{x}^{\frac{1}{2}}\right]$
Using the power rule:
$= 3 {x}^{2} - 8 {x}^{-} 3 + \frac{1}{2} {x}^{- \frac{1}{2}}$
$= 3 {x}^{2} - \frac{8}{{x}^{3}} + \frac{1}{2 \sqrt{x}}$
b) $\frac{d}{\mathrm{dx}} \left[5 {\left(2 x - 1\right)}^{4}\right] = 5 \frac{d}{\mathrm{dx}} \left[{\left(2 x - 1\right)}^{2}\right]$
The chain rule states that:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
Let $y = {u}^{4}$ while $u = 2 x - 1$
Using the power rule:
$\frac{\mathrm{dy}}{\mathrm{du}} = 4 {g}^{3} , \setminus \quad \frac{\mathrm{du}}{\mathrm{dx}} = 2$
$5 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = 40 {\left(2 x - 1\right)}^{3}$
c) $\frac{d}{\mathrm{dx}} \left[\frac{{x}^{2} + 4}{\sin} x\right] = \frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] + 4 \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$
Solve for $\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right]$ first.
Using the product rule:
$\frac{d}{\mathrm{dx}} \left[f \left(x\right) \cdot g \left(x\right)\right] = f \left(x\right) \frac{\mathrm{dg}}{\mathrm{dx}} + g \left(x\right) \frac{\mathrm{df}}{\mathrm{dx}}$
$\frac{d}{\mathrm{dx}} \left[{x}^{2} \cdot \frac{1}{\sin} x\right] = \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \cdot \frac{1}{\sin} x + \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] \cdot {x}^{2}$
$= \frac{2 x}{\sin} x + \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] \cdot {x}^{2}$
Solve for $\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$
Using the chain rule, let:
$y = \frac{1}{u}$ while $u = \sin x$
$\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{u} ^ 2$
$\frac{\mathrm{du}}{\mathrm{dx}} = \cos x$
$\frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = - \cos \frac{x}{u} ^ 2$
Sub back $u = \sin x$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \cos \frac{x}{\sin} ^ 2 x$
We can also use the reciprocal rule to solve for $\frac{1}{\sin} x$
$\frac{d}{\mathrm{dx}} \left[\frac{1}{f} \left(x\right)\right] = - \frac{\mathrm{df}}{\mathrm{dx}} \cdot \frac{1}{f} ^ 2 \left(x\right)$
Finally, $\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] = \frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x$
Now solve for part 2
$\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$ has already been solved
$\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] = - \cos \frac{x}{\sin} ^ 2 x$
Now, sub back in.
$\frac{d}{\mathrm{dx}} \left[{x}^{2} / \sin x\right] + 4 \frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right] = \frac{2 x}{\sin} x - \frac{{x}^{2} \cos x}{\sin} ^ 2 x - \cos \frac{x}{\sin} ^ 2 x$ | 1,378 | 3,061 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 31, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-33 | latest | en | 0.263856 |
https://sudonull.com/post/105949-Orthogonal-a-model-of-the-world-with-an-alternative-theory-of-relativity | 1,639,043,619,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00179.warc.gz | 591,421,644 | 7,658 | # Orthogonal - a model of the world with an alternative theory of relativity
In 2011-2013 Australian writer Greg Egan has published the Orthogonal trilogy (The Clockwork Rocket, The Ethernal Flame, The Arrows of Time). The books describe an amazing world in which there are no fluids and electric charges, four-eyed intelligent creatures live, able to change shape and multiply by fission, using air not for chemical reactions, but for cooling their body, and light - for transmitting nerve impulses. The speed of light in this world is variable: violet photons move much faster than red ones. Therefore, stars do not look like white dots, but like rainbow stripes
Even in the first book, the heroes found out that the reason for this behavior of light lies in the properties of the space-time of their universe: unlike our world, which is Minkowski space, their spatial and temporal coordinates are completely equal. Any body moves along its path in four-dimensional space-time with constant speed, uniform motion there looks like a straight line, and accelerated - like an arc. For example, a flight of a spaceship to another star and vice versa can be represented as follows:
We consider the thrust of the ship during acceleration and braking to be constant, and in this case the trajectory of its movement in space-time will be an arc of a circle. In a finite time, the ship will reach infinite (by the clock of a stationary observer) speed, and the main part of the flight will take place in zero time. At the same time, the time for the passengers of the ship will go as usual, and it can be measured along the length of the trajectory in the figure. When the ship returns to the starting point, it turns out that only a few years have passed on the home planet, while centuries could have passed for the passengers of the ship. Moreover, if the acceleration / deceleration phases last a little longer, the ship may return at the same moment when it started, or maybe even earlier:
True, the Universe will have to somehow solve the paradoxes that arise in this case, and these decisions may turn out to be unexpected for the inhabitants of the planet.
Photons obey the same laws as other bodies. They differ in their own vibrations. All photons are the same, and in their own frame of reference they have the same frequency (and zero wavelength). But when the observer sees photons moving at different speeds, then their frequencies seem different to him. For red light (the slowest), the frequency and wavelength is minimal, and for violet it is maximum:
Here, the red line is the trajectory of the red photon, and the violet is the trajectory of the violet, respectively. The eye sees photons whose trajectories are between these lines.
The heroes of the book see light in a speed range from 76/144 to 192/144 of the speed of blue light (blue photons are those that fly in spacetime at an angle of 45 degrees to the observer, that is, their apparent speed in space is equal to any speed reference systems in space-time). Thus, the observer sees only those photons whose trajectory lies between the two cones:
Half the angle at the apex of the inner (red) cone is 27 degrees, and the outer (purple) is 54 degrees. If the trajectory of a star intersects this space, then the star can be seen:
Here, a slowly moving star was considered. If the star’s speed becomes greater, then the trajectory will consist of two parts:
Soon after the start of the trilogy, strange stars began to appear in the sky - Hartlers (Hurtler - a pest?). They arose as violet dots, from which iridescent stripes quickly diverged in two directions:
Heroes suggested (and correctly) that hartlers are stars whose trajectory is perpendicular to the trajectory of their world. That is, each such star exists only at one moment in time, but at the same time occupies its entire trajectory. In space-time, it could look like this:
But if you look closely at this picture, it turns out that half of the trajectory directed towards the hartler’s movement is formed by photons that fly backward in time in the hartler’s frame of reference! Judging by the contents of the third book, stars do not emit such photons, so in reality the picture should look like this:
I was wondering what such a universe would look like if you navigate it in a very maneuverable ship. To do this, I decided to write a game with the simplest plot - there is the Universe, there are several stars in it that need to be visited and redeemed (just by flying nearby).
The first question was what form of space-time to choose. The heroes of the trilogy quickly came to the conclusion that the Universe should be finite, but for a long time doubted which one. As a result, they came to the conclusion that this should be a four-dimensional sphere (i.e. a sphere in 5-dimensional space). True, for some reason they needed to have regions with negative curvature in it (otherwise there would be some problems with entropy), that is, the sphere should be distorted in shape. But for simplicity, I took the usual homogeneous sphere.
The position and speed of the ship are described by a pair of perpendicular vectors. Vector P determines the current position in time space, V is the drift direction (drift speed is always the same). In addition, we need three vectors X, Y, Z, which determine the orientation of the ship in space (and the image on the screen). All vectors are taken in 5-dimensional space, have a length of 1 and are perpendicular to each other. Thus, the ship is described by the 5 * 5 orthogonal matrix.
It turns out that all the movements and maneuvers of the ship in this representation are just rotations of the matrix in coordinate planes. General drift - rotation in the (P, V) plane (vectors X, Y, Z remain unchanged), acceleration and braking in 3D - rotations in the (V, Z) plane, lateral accelerations - rotations in (V, X) and in ( V, Y), change of orientation of the ship - rotation in (X, Z) and (Y, Z). Rotation speeds are determined by the general parameters of the game, and they can be changed on the control panel.
The star trajectory is also a pair of perpendicular vectors (P0, V0). At any time T (according to the clock of the star itself), its position will be P1 = P0 * cos (T) + V0 * sin (T), and the drift velocity will be V1 = V0 * cos (T) -P0 * sin (T). To get an image of a star, we need to determine the parameters of a photon emitted from the point (P1, V1) and reaching our point (P, V): what color it will be and from which side it will arrive. To do this, it is enough for us to connect the points P and P1 with an arc of a large circle and see which way it goes from point P and which side it enters into P1.
For simplicity, we assume that a photon cannot fly more than a quarter of a circle. In fact, the book says nothing about the image of a star visible from the night side of the planet, or about phantom stars from the opposite side of the universe, the light from which was focused in the vicinity of the world of heroes of the trilogy. This means that it suffices to consider the case when the angle between the vectors P and P1 is sharp, i.e. (P, P1)> 0. It turns out that, firstly, the condition (P, V1)> 0 must be fulfilled - otherwise the star would have to emit a photon into the past, and secondly, (P1, V)> 0 - otherwise the photon will fly to us from the future, and without special means we can’t see him.
After that, it suffices to project the vector P1 onto the space (V, X, Y, Z) (tangent to the sphere at the point P). Let the vector S = (v, x, y, z) be obtained. Then the length L of the vector S corresponds to the distance traveled by the photon (more precisely, equal to its sine), the quantity v / L is the cosine of the angle between the path of the photon and our path in space-time, which determines the color of the photon, and (x, y, z) - the direction from which the photon arrived - and we can depict it by the usual methods.
It turns out that catching stars is far from easy. The simplest case is when the star is close, and our speeds do not differ very much (as was the first picture with cones). With the help of outboard engines, we can easily eliminate lateral speeds. The star on the screen from the rainbow strip will turn into a point, our trajectories in space-time will be in the same plane, and we will fly to the star () or from it - it's hard to tell in advance).
A natural desire is to aim at a star and begin to accelerate. But what will happen?
It can be seen that the portion of the trajectory that we see is getting farther - in fact, we are starting to see the star’s increasingly distant past. In addition, the visible star moves away and becomes smaller and dimmer. And if we have at least a slight lateral displacement, then we will see that the red part of the trajectory contracts, stopping on the green, and then on the blue part of the spectrum.
At this point, you can slow down a little and wait until the star becomes larger in size. And then start to catch her. But what happens if we miss?
From the side where we looked, the purple part of the track suddenly changes to red, and we will see that it is moving away. And the star itself will be on the opposite side from us. Therefore, we need to deploy the ship and move to the star again.
But catching a close star is not very difficult. Problems begin when the star is far away and we see a long and thin rainbow trail. Where to look for a star, where to fly is completely impossible to understand. Sometimes I succeed. More often than not.
If you want to experiment with this space, you can get the exe file here (for .NET 3.5), and the source here .
A more detailed description of the laws of physics of this world (up to the quantum level) is on the site of the author of the trilogy: www.gregegan.net | 2,191 | 9,827 | {"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-2021-49 | latest | en | 0.937648 |
https://discourse.julialang.org/t/fast-exponential-of-a-large-matrix/116998 | 1,723,439,115,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00623.warc.gz | 153,574,900 | 7,907 | # Fast Exponential of a large matrix
I have to perform the exponential of a matrix. I have compared the built-in `exp` function to the `exponential!` function from the ExponentialUtilites package. The latter is supposed to be much faster than the built-in function. However, I don’t see this!
``````A = randn(2^12, 2^12)
@time exp(im*A);
71.589389 seconds (2.02 M allocations: 1.880 GiB, 0.10% gc time, 2.65% compilation time)
A = randn(2^12, 2^12)
@time exponential!(im*A);
71.524002 seconds (3.93 M allocations: 1.741 GiB, 0.25% gc time, 5.79% compilation time)
``````
Why is this so? And is there any other way to calculate the exponential of a large matrix efficiently, like using multi-threading?
1 Like
If you’re willing to sacrifice accuracy maybe an idea could be to truncate its series expansion: Matrix exponential - Wikipedia
Wait, can’t this be done without allocating?
I’m not an expert, but… if we just think about a taylor expansion:
x + x^2 + x^3 + x^4 times some coefficients.
This sum can be split into an accumulator where the final result will be stored and some chunk of memory were the calculation is being done. The calculation is you just keep multiplying by the same matrix and can be done iteratively: each step is multiplying some NxN matrix corresponding to the output of the previous step, times x which is also an NxN matrix. This has known size and can be allocated beforehand.
Alternatively, isn’t this the kind of thing that BLAS should be really good at doing? Why not just use that?
If you have a Hermitian `A` you should use `cis(A)` instead.
Do you need the whole matrix, or do you just need to multiply it by a vector? In the latter case there are faster possibilities.
3 Likes
Hi, @stevengj! Yes, the matrix is Hermitian. And yes, you are right; I just need to multiply `exp(im*A)` by a vector. Could you please tell me what other possibilities you are referring to? Thanks!
Hi @tom-plaa! I am not sure how much accuracy I will lose by doing a series truncation. But I would prefer to calculate `exp(im*A)` since it’s important to preserve the unitarity.
`expv` which uses krylov methods is likely much better here.
4 Likes
@Oscar_Smith this is wonderful! Thanks!
It’s well known that Taylor series are a terrible way to compute matrix exponentials in general: Taking gradients of a matrix exponential - #8 by stevengj
3 Likes
Unfortunately, it looks like ExponentialUtilities.jl can’t currently exploit this: cisv(t, A, b) to compute exp(im*A*t)*b for Hermitian A? · Issue #176 · SciML/ExponentialUtilities.jl · GitHub
It might be worth trying KrylovKit.jl instead, since it supports this case in its `exponentiate` function, by passing an imaginary t and a Hermitian A.
3 Likes
@stevengj it seems `expv` in `ExponentialUtilities` is much faster than `exponentiate` in `KrylovKit`
``````@time expv(im, B, phi);
0.491426 seconds (46 allocations: 1.065 MiB)
@time exponentiate(B, im, phi);
11.420562 seconds (973 allocations: 26.749 MiB, 0.10% gc time)
``````
Check the accuracy of the result. In my experience ExponentialUtilities is faster than KrylovKit with default settings. However KrylovKit tries very hard to give you the most accurate result (by using restart IIRC) while ExponentialUtilities just gives you less accurate results.
@Oscar_Smith How to reduce the number of allocations for `expv`? So in my work, I have to repeatedly apply the `exp(im*H)`, where H is Hermitian, on a state, which gets updated in each run. Because of large memory allocations, I end up getting much slower computation with `expv`, than with `exponential!`. Here is a minimal working code for the latter
``````function run_sim(t_final, U, initial_state, final_state)
for t in 1:t_final
final_state .= U * initial_state
end
final_state
end
function main(n)
d = 2^n
initial_state = rand(ComplexF64, d)
final_state = initial_state
t_final = 10
A = rand(ComplexF64, (d, d))
H = (A .+ A') / 2
U = exponential!(im*H)
for _ in 1:10
final_state .= run_sim(t_final, U, initial_state, final_state)
end
end
@time main(10);
1.326368 seconds (122 allocations: 145.607 MiB)
``````
Here is the same code using `expv`
``````function run_sim(t_final, H, initial_state, final_state)
for t in 1:t_final
final_state .= expv(im, H, initial_state)
end
final_state
end
function main(n)
d = 2^n
initial_state = rand(ComplexF64, d)
final_state = initial_state
t_final = 10
A = rand(ComplexF64, (d, d))
H = (A .+ A') / 2
for _ in 1:10
final_state .= run_sim(t_final, H, initial_state, final_state)
end
end
@time main(10);
5.373086 seconds (4.91 k allocations: 101.364 MiB)
``````
You can reuse some memory by using caches but unfortunately the docs are broken for that. Here is how some of it works:
``````state = rand(ComplexF64, d)
krylovdim = 30 # default
# 1. create a Krylov subspace cache
KS = ExponentialUtilities.KrylovSubspace{ComplexF64}(length(state), min(krylovdim, length(state)))
# 2. populate Krylov space
ExponentialUtilities.arnoldi!(KS, H, state; ishermitian=true)
# 3. compute exponential, stores result back in state
ExponentialUtilities.expv!(state, -im*Δt, KS)
# now repeat 2. and 3.
``````
I think this still allocates some memory but you cannot really work around it because you have complex valued states. But I am not quite sure about this.
1 Like
It does reduce the allocations but is still large.
`````` 5.003996 seconds (1.51 k allocations: 58.343 MiB)
`````` | 1,507 | 5,449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-33 | latest | en | 0.923252 |
http://forums.freshershome.com/printthread.php?s=9aeff628bc60984070cbc21a65d2f2fd&t=4974&pp=12&page=1 | 1,585,395,901,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370491857.4/warc/CC-MAIN-20200328104722-20200328134722-00293.warc.gz | 71,104,867 | 2,510 | # Brain Teasers............... 10 points for the best?
• 04-06-2008, 12:19 AM
Sriram I
Brain Teasers............... 10 points for the best?
Sinbad the sailor is known to have travelled around the world seven times. During which round (which time) of travel did he die of a mysterious disease?
-----------------------------------------------------------------------------
There are 7999 card castles made every year around the world. each castle is of 15 storeys. each card castle is made up of 1853 cards. every card is either an ace or an heart but not a diamond nor a spade. every storey of the card castle requires 100 cards which reduces by 6 for every floor made up. my question that will not be to you is not that how many cards are there in a pack but is what is the probability that u get to make sixteenth floor of the castle?
-------------------------------------------------
if a train travels at 340 miles per hour and a man is travelling in the opposite direction outside the train at 10 kms per hour at what speed are the passengers travelling
LOTS OF FUN
HOPE U ENJOY IT
• 08-04-2009, 12:58 AM
sahil100
1st: sinbad died during 8th trip
2nd: probability=0 because 15th floor was made by 4 cards and 16th would be made by 4-6=-2(not possible)
3rd:passengers at 544 kmph(fastest train ?)
• 07-17-2013, 03:51 PM
prakashrao.n | 344 | 1,339 | {"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-2020-16 | latest | en | 0.946049 |
https://chemistry.stackexchange.com/questions/34239/does-the-bohr-model-violate-the-uncertainty-principle/34241#34241 | 1,642,834,022,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00433.warc.gz | 234,107,483 | 33,812 | # Does the Bohr model violate the uncertainty principle?
In a book, it is stated that one reason for the failure of Bohr's theory was the fact that it violates the uncertainty principle. Is this fact true? How so?
• There is an issue of semantics here. The Bohr model predates the Uncertainty principle, so one can't really speak to the impropriety 'violation' implies. However, Bohr's model is not compatible with the Uncertainty principle, and the two really can not be rigorously combined to model the quantum mechanics of atoms. Jul 21 '15 at 21:29
$$\Delta x \Delta p \geq \frac{\hbar}{2}$$
When you think about it, the whole of classical mechanics violates this principle. Think of a particle with mass $m$, a defined position $x_0$ at time $0$, a defined velocity $v_0$ at time $0$, and a constant acceleration $a$. The laws of kinematics state that at any time $t$, the position $x_t = x_0 + v_0t + \frac{1}{2}at^2$, and the momentum $p_t = m(v_0 + at)$, with zero uncertainty in both quantities, i.e. $\Delta x \Delta p = 0$. That works in most real-life cases because on a macroscopic level, quantum effects are negligible.
In the Bohr model of the atom the energy (thus its momentum) of the electron and the radius of its orbit (thus its position) are precisely defined quantities. This is a direct violation of the Uncertainty Principle. Since the energy levels of the hydrogen atom had been shown to be experimentally well defined by Bohr’s model ($\Delta$p ~ 0), we are left with the conclusion that we know very little about the exact location of the electrons in space ($\Delta$x ~ infinity). | 401 | 1,611 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | latest | en | 0.920057 |
https://community.airtable.com/t/updating-status-with-formula-field/46297 | 1,656,343,905,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103334753.21/warc/CC-MAIN-20220627134424-20220627164424-00541.warc.gz | 220,405,538 | 6,179 | # Updating status with formula field
Hello, I am horrible at formulas and it’s really becoming a frustration as I continue to learn Airtable. I am wanting to create a status in the formula field to then automate an email trigger.
1. The formula needs to update to either 60-days or 30-days based on the contract end date.
2. I already have fields for the start date and end date so the formula can work with either.
Seems simple enough but I am a struggle bus.
What formula do I need to enter? If we are 60-days out from the contract ending, update the status to 60-days, and if we a 30-days from the contract ending, update the status to 30-days.
Start with IF(AND(, correct?
IF(AND({Contract End Date} …blank stare…, “60-days”), repeat blank stare, “30-days”)
Help
See the formula below which should do as you asked using 4 nested IFs
``````IF(
{Contract End Date},
IF(
{Contract End Date} > TODAY(),
IF(
DATETIME_DIFF({Contract End Date}, TODAY(), "days") <= 30,
"30 days",
IF(
DATETIME_DIFF({Contract End Date}, TODAY(), "days") <= 60,
"60 days",
"more than 60 days"
)
),
"contract end date passed"
),
"no contract end date"
)
``````
If you don’t want any text to display if its neither within 30 days nor 60 days then simplify that formula to:
``````IF(
{Contract End Date},
IF(
{Contract End Date} > TODAY(),
IF(
DATETIME_DIFF({Contract End Date}, TODAY(), "days") <= 30,
"30 days",
IF(
DATETIME_DIFF({Contract End Date}, TODAY(), "days") <= 60,
"60 days"
)
)
)
)
``````
3 Likes
You just saved me! This makes zero sense right now, but it will over the next few months as I keep using Airtable. Thank you so much! I am beyond appreciative of your help!!
Working from outside in:
1. The first IF() makes sure the End Date field isn’t empty. This prevents the field from showing errors
2. The second IF() makes sure that the End Date hasn’t happened yet. If its already occurred then you will get a negative number for “the difference between days”
3. The third and 4th IF()s are straightforward. They are presented in this order because checking for if a number is less than 60 first may include numbers that are also less than 30. “A number less than 30 must be less than 60”, so logically we want to know “If it is not less than 30, is it less than 60?”
1 Like
This topic was solved and automatically closed 3 days after the last reply. New replies are no longer allowed. | 611 | 2,392 | {"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-2022-27 | latest | en | 0.890112 |
https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Circuitry/Book%3A_I_Direct_Current_(DC)/14%3A_Magnetism_and_Electromagnetism | 1,586,467,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371876625.96/warc/CC-MAIN-20200409185507-20200409220007-00249.warc.gz | 773,164,615 | 21,446 | # 14: Magnetism and Electromagnetism
• 14.1: Permanent Magnets
Centuries ago, it was discovered that certain types of mineral rock possessed unusual properties of attraction to the metal iron. One particular mineral, called lodestone, or magnetite, is found mentioned in very old historical records (about 2500 years ago in Europe, and much earlier in the Far East) as a subject of curiosity. Later, it was employed in the aid of navigation, as it was found that a piece of this unusual rock would tend to orient itself in a north-south direction if left free to
• 14.2: Electromagnetism
The discovery of the relationship between magnetism and electricity was, like so many other scientific discoveries, stumbled upon almost by accident. The Danish physicist Hans Christian Oersted was lecturing one day in 1820 on the possibility of electricity and magnetism being related to one another, and in the process demonstrated it conclusively by experiment in front of his whole class! By passing an electric current through a metal wire suspended above a magnetic compass, Oersted was able to
• 14.3: Magnetic Units of Measurement
If the burden of two systems of measurement for common quantities (English vs. metric) throws your mind into confusion, this is not the place for you! Due to an early lack of standardization in the science of magnetism, we have been plagued with no less than three complete systems of measurement for magnetic quantities.
• 14.4: Permeability and Saturation
• 14.5: Electromagnetic Induction
While Oersted’s surprising discovery of electromagnetism paved the way for more practical applications of electricity, it was Michael Faraday who gave us the key to the practical generation of electricity: electromagnetic induction. Faraday discovered that a voltage would be generated across a length of wire if that wire was exposed to a perpendicular magnetic field flux of changing intensity.
• 14.6: Mutual Inductance
If two coils of wire are brought into close proximity with each other so the magnetic field from one links with the other, a voltage will be generated in the second coil as a result. This is called mutual inductance: when voltage impressed upon one coil induces a voltage in another. | 471 | 2,226 | {"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-2020-16 | latest | en | 0.948881 |
https://pylops.readthedocs.io/en/v1.18.0/gallery/plot_avo.html | 1,669,637,337,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710503.24/warc/CC-MAIN-20221128102824-20221128132824-00585.warc.gz | 520,490,891 | 8,764 | # AVO modelling¶
This example shows how to create pre-stack angle gathers using the pylops.avo.avo.AVOLinearModelling operator.
import matplotlib.pyplot as plt
import numpy as np
from scipy.signal import filtfilt
import pylops
from pylops.utils.wavelets import ricker
plt.close("all")
np.random.seed(0)
Let’s start by creating the input elastic property profiles
nt0 = 501
dt0 = 0.004
ntheta = 21
t0 = np.arange(nt0) * dt0
thetamin, thetamax = 0, 40
theta = np.linspace(thetamin, thetamax, ntheta)
# Elastic property profiles
vp = 1200 + np.arange(nt0) + filtfilt(np.ones(5) / 5.0, 1, np.random.normal(0, 80, nt0))
vs = 600 + vp / 2 + filtfilt(np.ones(5) / 5.0, 1, np.random.normal(0, 20, nt0))
rho = 1000 + vp + filtfilt(np.ones(5) / 5.0, 1, np.random.normal(0, 30, nt0))
vp[201:] += 500
vs[201:] += 200
rho[201:] += 100
# Wavelet
ntwav = 41
wavoff = 10
wav, twav, wavc = ricker(t0[: ntwav // 2 + 1], 20)
wav_phase = np.hstack((wav[wavoff:], np.zeros(wavoff)))
# vs/vp profile
vsvp = 0.5
vsvp_z = np.linspace(0.4, 0.6, nt0)
# Model
m = np.stack((np.log(vp), np.log(vs), np.log(rho)), axis=1)
fig, axs = plt.subplots(1, 3, figsize=(9, 7), sharey=True)
axs[0].plot(m[:, 0], t0, "k", lw=6)
axs[0].set_title("Vp")
axs[0].set_ylabel(r"$t(s)$")
axs[0].invert_yaxis()
axs[0].grid()
axs[1].plot(m[:, 1], t0, "k", lw=6)
axs[1].set_title("Vs")
axs[1].invert_yaxis()
axs[1].grid()
axs[2].plot(m[:, 2], t0, "k", lw=6, label="true")
axs[2].set_title("Rho")
axs[2].invert_yaxis()
axs[2].grid()
axs[2].legend()
Out:
<matplotlib.legend.Legend object at 0x7f770c9d4320>
We create now the operators to model the AVO responses for a set of elastic profiles
# constant vsvp
PPop_const = pylops.avo.avo.AVOLinearModelling(
theta, vsvp=vsvp, nt0=nt0, linearization="akirich", dtype=np.float64
)
# depth-variant vsvp
PPop_variant = pylops.avo.avo.AVOLinearModelling(
theta, vsvp=vsvp_z, linearization="akirich", dtype=np.float64
)
We can then apply those operators to the elastic model and create some synthetic reflection responses
dPP_const = PPop_const * m.ravel()
dPP_const = dPP_const.reshape(nt0, ntheta)
dPP_variant = PPop_variant * m.ravel()
dPP_variant = dPP_variant.reshape(nt0, ntheta)
fig, axs = plt.subplots(1, 2, figsize=(10, 5), sharey=True)
axs[0].imshow(
dPP_const,
cmap="gray",
extent=(theta[0], theta[-1], t0[-1], t0[0]),
vmin=dPP_const.min(),
vmax=dPP_const.max(),
)
axs[0].set_title("Data with constant VP/VS")
axs[0].axis("tight")
axs[1].imshow(
dPP_variant,
cmap="gray",
extent=(theta[0], theta[-1], t0[-1], t0[0]),
vmin=dPP_variant.min(),
vmax=dPP_variant.max(),
)
axs[1].set_title("Data with variable VP/VS")
axs[1].axis("tight")
plt.tight_layout()
Finally we can also model the PS response by simply changing the linearization choice as follows
PSop = pylops.avo.avo.AVOLinearModelling(
theta, vsvp=vsvp, nt0=nt0, linearization="ps", dtype=np.float64
)
We can then apply those operators to the elastic model and create some synthetic reflection responses
dPS = PSop * m.ravel()
dPS = dPS.reshape(nt0, ntheta)
fig, axs = plt.subplots(1, 2, figsize=(10, 5), sharey=True)
axs[0].imshow(
dPP_const,
cmap="gray",
extent=(theta[0], theta[-1], t0[-1], t0[0]),
vmin=dPP_const.min(),
vmax=dPP_const.max(),
)
axs[0].set_title("PP Data")
axs[0].axis("tight")
axs[1].imshow(
dPS,
cmap="gray",
extent=(theta[0], theta[-1], t0[-1], t0[0]),
vmin=dPS.min(),
vmax=dPS.max(),
)
axs[1].set_title("PS Data")
axs[1].axis("tight")
plt.tight_layout()
Total running time of the script: ( 0 minutes 0.961 seconds)
Gallery generated by Sphinx-Gallery | 1,264 | 3,563 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-49 | latest | en | 0.338066 |
http://teachbridge.com/articles-resources/articles/ellen-pomer/ep-bil-notes/count-1/ | 1,603,946,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107902745.75/warc/CC-MAIN-20201029040021-20201029070021-00634.warc.gz | 109,297,463 | 13,566 | Session 1: Introduction to Counting Part 1
Posted on Sep 18, 2014
From Thursday September 18th, 16:15 ET, BILlies Retreat
Ellen Caitlin Pomer has been teaching online for nearly 17 years. She is co-author of the acclaimed ‘Standard Bidding with SAYC’ and founder of Bridge Forum (www.bridge-forum.com). She returns to bridge teaching in the BIL, where she has taught many topics, and is delighted to be back teaching ‘Introduction to Counting’. She is also available for private sessions. She can be reached at epomer@outlook.com. Please advise others that if they wish to receive these notes, that they should email me with their full name and BBO I.D. Enjoy!
Please note that due to the holy Jewish holidays, there will be no session on Thursday, September 25th and will resume Thursday, October 2rd at 4:15pm Eastern. To all Jewish members who celebrate these holidays, I wish you and yours a Happy, Healthy Jewish New Year.
We will look at a variety of issues which are involved with counting.
Bridge is a game of counting so let’s break it down as to when you count.
a) When declarer does not have what we call a ‘cold’ contract (i.e. more losers than s/he can afford to make the contract), counting may help. We will see examples where this is true.
b) Defenders count declarer’s hand. If, for example West opens 1NT and the contract becomes 3NT, and declarer has shown 17 HCPs (High Card Points) and declarer must have the missing A, as partner has signalled s/he doesn’t like hearts, (to be discussed below) South now knows that partner must have the ♣K (as declarer can’t have the A and the missing ♣K). When in, North should feel free to lay down his ♣A from AQJ as partner should have the Club King, and in doing so, you can defeat the contract if you get your tricks in time.
c) Defenders give count to one another, but when? When on defense, and your partner leads, you give attitude. Thus partner leads the 2 and you hold the JT63, play the Ten (lower of two equally ranking cards) to say you like the suit. But when declarer plays a suit, give count, thus traditionally high-low with an even number of cards (8652) and low-high with an odd number of cards (J32).
What if the suit has been played one round? Now we are giving remainder count. Thus with 852 remaining in the above example, we now play low-high, thus the 2; and with J3 left from the above J32, we play the Jack if it makes sense to do so.
Remember it is very important to give your partner count on defense but you must be the judge. If giving count or attitude in a specific situation only helps declarer, LIE!
More Tips for Counting
a) The bidding at the table gives enormous help. For example, if an opponent, say East, opens a weak two 2♠, and North-South land up in 4, declarer already starts with a good count on the opponents cards, knowing one opponent has six in a suit.
b) The lead is also a key for declarer. Say you are in 4♠, and your opponent, who passed in first seat, leads the .AK. Later s/he shows up with the Club King. There is no way s/he can have the missing ♣K or s/he would have opened the bidding.
Mike Lawrence, a member of BBO, wrote: ‘How to Read Your Opponents’Cards’ many years ago and has two CDs on counting: ‘Counting at Bridge’ and ‘Counting at Bridge 2’. I highly recommend you consider purchasing Lawrence’s first CD, with software by BBO’s founder, Fred Gitelman, from the BBO ‘Online Store’.
Below you will find the four hands we counted for your review, while hand 4 is ‘homework; which we will discuss at our next session.
Reading the Opponents’ Cards: Counting Shape
Hand 1
N
North
42
AQ975
AJ
K1075
S
South
AKQ73
K3
KQ2
AQ4
W
West
N
North
E
East
S
South
1
Pass
4NT
Pass
5
Pass
7NT
All Pass
The bidding could be better! The 5♠ bid, as we will see, with 4NT as RKC Blackwood shows 2 controls — the, two Aces and the Queen of hearts.. Even though South is heading toward notrump, you respond to 4NT based on the last suit bid unless, thus 5♠ showing two Aces and the Q.
Here is the full deal:
N
North
42
AQ975
AJ
K1075
W
West
6
J4
1098653
J962
E
East
J10975
10862
74
83
S
South
AKQ73
K3
KQ2
AQ4
You cash your winners outside of clubs — 3 spades, 3 hearts, 3 diamonds — so you need to win 4 clubs. As you cash your winners, we count one opponent’s hand, say West, here. We find out that West follows to one spade, two hearts, and East follows to only two diamonds: We know that West’s shape is 1-2-6-4. While West has four clubs and East only two, it is twice as likely that West has the ♣J and he does. It is not a done deal that West has the club Jack, but this is an informed decision. With a club finesse to the Ten, declarer has 13 tricks.
Hand 2
N
North
KQJ10
832
KQJ
J53
S
South
A8754
Q6
952
AQ9
W
West
N
North
E
East
S
South
1
1
Pass
4
All Pass
Here is the full deal:
N
North
KQJ10
832
KQJ
J53
W
West
93
107
A853
108762
E
East
62
AKJ954
1074
K4
S
South
A8754
Q6
952
AQ9
Thus far East has shown 6 hearts and 2 spades and by playing on diamonds, declarer knows West’s shape: 2-6-3-2. (How does declarer know that East has only 3 diamonds? Note that on the run of extra spades to get more information (a common technique), East played a diamond. Counting the hand we know East has two clubs and one must be the ♣K as East needs the king for his opening bid given East has the A. So the issue is not which hand holds the ♣K — that we have come to know — but how many clubs does East holds? Now we know to lead a club to the Queen, cashing the ♣A, dropping the ♣K.
Hand 3
N
North
Q1043
QJ9
K103
872
S
South
AK72
652
AJ4
AKQ
W
West
N
North
E
East
S
South
Pass
Pass
2NT
Pass
3
Pass
3
Pass
4
All Pass
West cashes AK and follows with the 3. East ruffs and exits with a club.
Here is the full deal:
N
North
Q1043
QJ9
K103
872
W
West
65
AK1073
2
J9654
E
East
J98
84
Q98765
103
S
South
AK72
652
AJ4
AKQ
You win, cash trump and two more clubs. When East discards on the third round, you know that West started with 2-5-1-5 shape, so you can lead to dummy’s K and finesse the J with certainty
Hand 4
N
North
84
AKJ
AJ873
Q74
S
South
10
93
K1065
KJ10952
W
West
N
North
E
East
S
South
2
Dbl
4
5
All Pass
West cashes ♠AK. You ruff and…
Here is the full deal:
N
North
84
AKJ
AJ873
Q74
W
West
AK9652
74
Q42
83
E
East
QJ73
Q108652
9
A6
S
South
10
93
K1065
KJ10952
… knock out the ♣A. East does best to return a club. To count the hand, play AK and ruff the J. Now you know West started with two hearts and two clubs. Presumably he had six spades for his opening bid so he must have three diamonds. This means that West has only one diamond. Play the K in case the singleton is the Q and when it is not finesse the J
5. Homework
N
North
A873
Q10
AJ82
A76
S
South
QJ1096
963
K107
K8
W
West
N
North
E
East
S
South
Pass
1NT
2
3
Pass
4
All pass
West leads 8 and East cashes AK. West plays 2 on East’s second heart. East leads a third heart and South trumps with the ♠Q while West discards a small club. Declarer now leads trump with West holding one trump and East, the remainder. What line of play do you now take to give yourself the best chance of making this contract given you already have 3 losers? | 2,147 | 7,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-45 | latest | en | 0.967252 |
http://www.patheos.com/blogs/janetheactuary/2017/08/much-childs-life-worth.html | 1,537,916,012,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267162563.97/warc/CC-MAIN-20180925222545-20180926002945-00347.warc.gz | 392,873,748 | 18,330 | # How much is a child’s life worth?
How much is a child’s life worth? August 7, 2017
Today’s Tribune article on children left in hot cars features a proposal to require that all new cars include some sort of visual and audio warning device so that parents won’t forget to take their babies out of the car. Whether it’s a recognition that a rear door was opened just before the start of the trip, but not after the end of the drive, or a recognition of a “breath” in the backseat, advocates, or, specifically, U.S. Rep. Jan Schakowsky, say(s), “My car reminds me when I get out of the car and the keys are in there — how can we not remind the driver to check the backseat?”
As it is, an average of 37 children have died from being left behind in cars every year since 1998. Statistics from Kidsandcars.org show no discernible pattern except that the numbers have been increasing. (The Trib article suggests this is due to global warming, but this seems improbable; maybe the advent of smartphones caused increasing numbers of parents to be distracted, e.g., by having a phone call on the way to daycare/work.) They also show a breakdown, that 55% percent of deaths were due to a child being forgotten (or a parent committing murder but successfully convincing everyone it was an accident), 28% were children who “got in on their own” (e.g., were playing unsupervised but then either locked themselves in or fell asleep) and 13% were intentional (whether that includes the “I thought my kid would be fine” or just the homocides, I don’t know).
But do the math.
There were 17.5 million new cars sold last year.
Assume that this enhancement technology costs \$1,000 per car.
Assume that all 38 * 55% percent of the kids who die are saved by this technology.
That’s a cost of 833 million per life saved.
That’s an extreme case. We could say that the tech costs \$100; then it’s 83 million per life saved. If we assume that the go-in-on-their-own lives are also saved, that brings the cost-per-life down, too, and we’re at about \$50 million.
And, again, this is all for something that’s the result of human error in the first place. And some of those lives saved have to be balanced out by the marginal cost meaning that, for some number of people, this will tip the scales to keeping an older, less safe car for longer.
If this weren’t a quick lunchtime post, I’d dig up more background, on whether there are already metrics in place for this sort of thing (though this particular legislation doesn’t seem to care), or whether any of this technology exists in the first place to give some idea of its cost. (It seems not to, though, or I imagine it would have been mentioned; back-up cameras, the mandate for which is coming up, are currently available as an option.)
Plus, there’s always the issue of risk compensation — if people think of this as a guarantee, and take for granted that they’ll be warned, and drop any other “don’t forget the kid” measures they had previous adopted, then it might not have the hoped-for effect.
On the other hand, perhaps it will turn out to be a low-cost item. After all, the solution for kids being stuck in trunks, turned out to be a low tech glow-in-the-dark trunk release.
And if it weren’t mandatory, how much would you pay for a warning device, if you have small children of your own, or if you don’t? What if there were an aftermarket device, or an app paired with the carseat?
But as it is, I’ll just leave it as a discussion item.
Image: By Carin Araujo, http://www.prtc.net/~carin (Stock.xchng #197853) [Copyrighted free use], via Wikimedia Commons
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Everyday maths 2
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# 2.2 Rounding to approximate an answer
You might round in order to approximate an answer. At the coffee shop, you might want to buy a latte for £2.85, a cappuccino for £1.99 and a tea for £0.99. It is natural to round these amounts up to £3, £2 and £1 in order to arrive at an approximate cost of £6 for all three drinks.
## Activity 4: Approximation
Find an approximate answer to each of the following.
1. Cost of 3 kg of bananas at 79p per kg.
2. The total cost of 3 books at £1.99, £2.85 and £4.90.
3. Change from a £20 note when £3.84 has been spent.
4. Approximate cost of one carton of fruit juice when a pack of 4 are sold together for 99p. | 241 | 860 | {"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-2024-22 | latest | en | 0.930696 |
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Home Brew Forums > White house beer
11-15-2012, 05:43 AM #681
sweetcell
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Quote:
Originally Posted by OhReally One way to calculate is to take the difference in gravities and multiply by 131.25. (OG-FG) x 131.25 = ABV 1.060-1.018 = 0.042 x 131.25 = 5.51% ABV Cheers
being lazy, i let a website do the math for me
http://pint.com.au/calculators/alcohol/ tells me 1.060 --> 1.018 = 5.45% ABV.
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What hops should I grow? Hop grower's comparison table. Looking for cheap honey?
Drinking: galaxy/conan IPA, a farmhouse with ECY08 & brett blend
Aging: imperial chocolate stout, sour cherry mead, rye sour ECY20/ECY34 split, oud bruin & a few other sours, acerglyn, a BDSA
11-15-2012, 01:17 PM #682
alestateyall
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Quote:
Originally Posted by mtnagel I brewed the White House ale from Northern Brewer on 10/31. Used a yeast starter and had healthy fermentation for a few days and no more bubbles for at least a week. The temperature was at the low end for the yeast, which is 64F, which I was around or a little higher for the first several days then it got cold and basement dropped to 63F. I wanted to check the gravity but I broke my hydrometer this weekend Got to my LHBS on Monday and got a new one and the gravity is at 1.020. I measured Tuesday and got the same 1.020. Northern said the FG should be about 1.007 - 1.014. There was also a healthy head of krausen on top too when I first took the gravity on Monday and it was less, but still there on Tuesday. I did move it to my dining room, which keeps it around 68F on Monday evening. There has been very slow air lock activity. I wanted to move the beer to secondary so I could reuse the White House ale yeast to brew a milk stout which I plan to put in a used whiskey barrel. Due to vacations coming up (thanksgiving and xmas), I really wanted to brew yesterday reusing the yeast but I decided to just use the dry yeast that came with the kit because I figured my fermentation was stuck and I didn't want to use questionable yeast. So I'll probably just leave the honey ale in primary until I get back from vacation so it will probably be in primary for about 4 weeks. I will probably take a couple gravity readings before bottling just to see where it's at, but I can't imagine it wouldn't be done by then.
1.020 is a bit high but it is not the end of the world. Just make sure to read the gravity again and confirm the gravity has stopped changing. If the gravity is stable you can safely bottle at 1.020 and the beer will still be great.
You don't need to secondary. Leaving the beer in primary longer has the same effect.
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11-15-2012, 07:01 PM #683
mtnagel
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Thanks for the info. I only was going to move to secondary so I could reuse the yeast in another beer that I needed to brew now. Now that I'm not reusing the yeast, it will primary for about a month, just like the 10 gallons of White House honey porter I made.
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-Matt
11-15-2012, 10:35 PM #684
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Sent you a completely unrelated PM colincbn.
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11-16-2012, 12:16 AM #685
MtnHiBrewin
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I rarely move anything to a secondary. Every now and then, but not often. My White House Porter stayed in the primary 19 days, and turned out great.
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11-16-2012, 12:38 AM #686
sloose
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Quote:
Originally Posted by MtnHiBrewin I rarely move anything to a secondary. Every now and then, but not often. My White House Porter stayed in the primary 19 days, and turned out great.
My only reasons for secondary are 1) to free up a primary fermenter and 2) I'm rather clumsy when it comes to the trub cake while racking so the less of it there is the clearer my beer gets. Not that I really care about clarity.
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Bottled&Drinking: Founders All Day IPA Clone, Centennial blonde, Vienna Hefeweizen, Belgian wit, blizzard imperial IPA,
Conditioning: hard local cider
Primary: more local hard cider!
Secondary:
On Deck: English bitter, English IPA
11-16-2012, 04:37 PM #687
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SO, fun day here, Trying to make this honey porter and i miss read the recipe (i thought there was only 3.3 pounds of malt, not 2 cans of 3.3 pounds ) so in a frantic search for a solution i used approximately 1.5 pounds of brown sugar. may be an interesting result. any ideas on what i can expect. I am more than happy to share the results if anyone is interested.
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11-16-2012, 05:03 PM #688
sweetcell
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Quote:
Originally Posted by Ben_D in a frantic search for a solution i used approximately 1.5 pounds of brown sugar. may be an interesting result. any ideas on what i can expect.
the beer will be quite dry (the opposite of sweet), and it will be low on maltiness. you will likely taste the alcohol a little more. the brown sugar should give a subtle caramel flavor, especially if you happened to used dark brown sugar (the typical stuff you bake with is light brown sugar).
3.3 pounds of LME has the same potential as approx 2.75 lb of brown sugar. so technically you needed to add another 1.25 pounds to get back to the same point. however it was a good thing that you limited yourself to 1.5, going with 2.75 would have been too much. my suggestion: if you can, get a pound or a pound and a half of dried extract, boil it in a little water to sterilize it, and add it to your fermenter. liquid will also work, but unless your LHBS sells it in bulk it can be annoying to open a can, use some, then need to re-seal and keep for another brew day. also, it's bad for freshness if you aren't going to brew again soon. DME does a better job of sitting on the shelf.
whatever you do (including leaving it as-is), let us know how it turns out.
__________________
.
What hops should I grow? Hop grower's comparison table. Looking for cheap honey?
Drinking: galaxy/conan IPA, a farmhouse with ECY08 & brett blend
Aging: imperial chocolate stout, sour cherry mead, rye sour ECY20/ECY34 split, oud bruin & a few other sours, acerglyn, a BDSA
11-17-2012, 07:33 PM #689
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Thanks for the feedback brotha. It's bubbling nicely now and tasted good (sweet with a good roasted flavor, i had to use 40l crystal malt btw) before i put in the yeast so hopefully it maintains some of that flavor. I'm going to just let it go and see what happens. Call me hopeful. I will keep updating.
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11-18-2012, 12:06 AM #690
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I am trying my version tonight. It is good. I taste honey up front then a malty sweet finish. Mild fruity hops flavor. I don't get any alcohol taste. It is a pretty heavy beer.
OG 1060
FG 1017
I used wildflower honey from a local bee keeper. The honey was awesome by itself.
I used caramel 60 for the "amber crystal malt."
It seems darker than the pictures from the video. iBrewmaster estimates 10 SRM which seems about right.
__________________ | 2,194 | 8,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-52 | latest | en | 0.941247 |
https://gre-test-prep.com/math-lectures/solve-simultaneous-solve-onlin.html | 1,550,491,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247486480.6/warc/CC-MAIN-20190218114622-20190218140622-00173.warc.gz | 582,039,754 | 11,919 | Algebra Tutorials!
Monday 18th of February
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Depdendent Variable
Number of equations to solve: 23456789
Equ. #1:
Equ. #2:
Equ. #3:
Equ. #4:
Equ. #5:
Equ. #6:
Equ. #7:
Equ. #8:
Equ. #9:
Solve for:
Dependent Variable
Number of inequalities to solve: 23456789
Ineq. #1:
Ineq. #2:
Ineq. #3:
Ineq. #4:
Ineq. #5:
Ineq. #6:
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Ineq. #8:
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• absolute Maximum and Minimum Function radicals | 938 | 3,746 | {"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-2019-09 | latest | en | 0.841922 |
http://www.hongjun.sg/2011/01/creative-answer-to-mid-term-chemistry.html | 1,519,233,326,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813691.14/warc/CC-MAIN-20180221163021-20180221183021-00053.warc.gz | 446,956,874 | 29,920 | Skip to main content
### Creative answer to a mid-term Chemistry question from University of Washington
A very creative answer demonstrating true application of science.
As quoted from a Facebook note:
The following is an actual question given on a University of Washington chemistry mid-term exam. The answer by one student was so "profound" that the professor shared it with colleagues, via the Internet, which is, of course, why we now have the pleasure of enjoying it as well...
Here is the "Bonus Question" on the exam: "Is Hell exothermic (gives off heat) or endothermic (absorbs heat)?"
Most of the students wrote proofs of their beliefs using Boyle's Law (gas cools when it expands and heats when it is compressed) or some variant. One student, however, wrote the following:
First, we need to know how the mass of Hell is changing in time. So we need to know the rate at which souls are moving into Hell and the rate at which they are leaving. I think that we can safely assume that once a soul gets to Hell, it will not leave. Therefore, no souls are leaving.
As for how many souls are entering Hell, let's look at the different Religions that exist in the world today. Most of these religions state that if you are not a member of their religion, you will go to Hell. Since there is more than one of these religions and since people do not belong to more than one religion, we can project that all souls go to Hell.
With birth and death rates as they are, we can expect the number of souls in Hell to increase exponentially.
Now, we look at the rate of change of the volume in Hell because Boyle's Law states that in order for the temperature and pressure in Hell to stay the same, the volume of Hell has to expand proportionately as souls are added. This gives two possibilities:
1. If Hell is expanding at a slower rate than the rate at which souls enter Hell, then the temperature and pressure in Hell will increase until all Hell breaks loose.
2. If Hell is expanding at a rate faster than the increase of souls in Hell, then the temperature and pressure will drop until Hell freezes over.
So which is it? If we accept the postulate given to me by Teresa (a girlfriend of mine during my Freshman year) that, "it will be a cold day in Hell before I sleep with you", and take into account the fact that I slept with her last night, then number 2 must be true, and thus I am sure that Hell is exothermic and has already frozen over.
The corollary of this theory is that since Hell has frozen over, it follows that it is not accepting any more souls and is therefore, extinct...leaving only Heaven, thereby proving the existence of a divine being which explains why, last night, Teresa kept shouting "Oh my God."
THIS STUDENT RECEIVED THE ONLY "A."
### Ho Ching named 5th most powerful and is mistaken as first lady by Forbes
Forbes named Singapore Prime Minister wife and CEO of Temasek Holdings, Ho Ching, as the 5th most powerful woman in the world. Ho Ching is mistaken as Singapore's first lady! OMG!
I wonder how can Forbes makes such a blunder. For a complete list, refer to here.
### How to stop FortiClient from starting automatically?
Installed FortiClient recently but the challenge in disabling the application/service from running automatically on every start-up annoyed me. Attempt to stop 'FortiClient Service Scheduler' only return 'Parameter is incorrect' error message.
An article on Technet help solve my trouble. To stop FortiClient from starting automatically, try the following:
Shut down FortiClient from the system tray.
Run net stop fortishield on command prompt.
Run msconfig.
On msconfig, switch to the Services tab. Clear the FortiClient Service Scheduler check box and click Apply.Run services.msc on command prompt to open up show all available services.Look for FortiClient Service Scheduler. Switch Startup type to Manual.Restart your computer. FortiClient should not be running automatically the next time round. Hope it helps.
### Access blocked websites, stream US / UK stuff
For people who like to stream US/UK stuff but don't want to pay VPN, try Hola. They have support for Windows, Android, Chrome, Firefox and Apple.
Alternatively, try TunnelBear VPN. It's free for first 500MB of data each month. They have support for Windows, iOS devices, Mac and Android. | 941 | 4,331 | {"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-2018-09 | latest | en | 0.963372 |
https://math.answers.com/other-math/Find_the_sum_of_the_reciprocals_of_two_real_numbers_given_that_these_numbers_have_a_sum_of_150_and_a_product_of_40 | 1,701,733,732,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00565.warc.gz | 442,304,955 | 45,943 | 0
# Find the sum of the reciprocals of two real numbers given that these numbers have a sum of 150 and a product of 40?
Updated: 4/28/2022
Wiki User
13y ago
1/x + 1/y = (y+x)/xy
But y + x = sum = 150, and xy = product = 40
So sum of reciprocals = 150/40 = 3.75
Wiki User
13y ago | 104 | 287 | {"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-2023-50 | latest | en | 0.874347 |
https://www.mathworks.com/matlabcentral/cody/problems/37-pascal-s-triangle/solutions/163656 | 1,511,530,717,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00022.warc.gz | 832,786,831 | 11,708 | Cody
# Problem 37. Pascal's Triangle
Solution 163656
Submitted on 18 Nov 2012 by Vue Sicker
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 0; correct = [1]; assert(isequal(pascalTri(n),correct))
y = 1
2 Pass
%% n = 1; correct = [1 1]; assert(isequal(pascalTri(n),correct))
y = 1 1
3 Pass
%% n = 2; correct = [1 2 1]; assert(isequal(pascalTri(n),correct))
y = 1 2 1
4 Pass
%% n = 3; correct = [1 3 3 1]; assert(isequal(pascalTri(n),correct))
y = 1 3 3 1
5 Pass
%% n = 10; correct = [1 10 45 120 210 252 210 120 45 10 1]; assert(isequal(pascalTri(n),correct))
y = 1 10 45 120 210 252 210 120 45 10 1 | 283 | 745 | {"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-2017-47 | latest | en | 0.515946 |
https://medium.com/math-for-data-science/math-for-data-science-bits-wilp-program-58417c654665?source=post_internal_links---------1---------------------------- | 1,674,815,662,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494976.72/warc/CC-MAIN-20230127101040-20230127131040-00755.warc.gz | 425,691,104 | 35,462 | # Math for Data Science — Lecture 01 Basic Matrix Operations via Python
In this lecture, we will study Matrix Operations (Initialization, Vector, Multiplication, Transpose, Special Matrix) via Python.
Matrix Definition
Matrix Notation
Matrix Initialization
`import numpy as npprint(np.zeros((2,3)))print(np.random.randint(low = 5,high=12, size=(2, 3)))A = np.matrix([[1, 2], [3, 4], [5,6]])print("Type of Matrix A:{}, having {} rows and {} columns".format(type(A),A.shape[0],A.shape[1]))`
Vectors
• Row vector is having 1 row and N columns
• Column vector is having N rows and 1 column
Matrix Equality
Matrix Multiplication
In below Image 09, matrix multiplication is shown both via python code and hand as well.
• Matrix multiplication isn’t commutative (AB != BA)
Matrix Transpose
Rules of Matrix Transpose
Special Matrix
Reduce size of Sparse Matrix using CSR (Compressed Sparse Row)
Properties of Determinants
WORK IN PROGRESS !! Jupyter Notebook/.py will be uploaded soon..
Our Next lecture is about using Elementary row operations to find Row Echelon Form (REF), Reduced Row Echelon Form (RREF) and Rank of Matrix (https://medium.com/math-for-data-science/math-for-data-science-lecture-02-elementary-row-operations-via-python-46885a33a84c).
Meanwhile, Please feel free to clap if you liked the article. Also, please subscribe to my YouTube Channel https://www.youtube.com/channel/UC4yh4xPxRP0-bLG_ldnLCHA?sub_confirmation=1
--
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## More from Math for data science
Idea is to understand math for data science from intuition and code (mainly python) | 393 | 1,577 | {"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-2023-06 | latest | en | 0.777426 |
https://phys.libretexts.org/Bookshelves/Relativity/General_Relativity_(Crowell)/01%3A_Geometric_Theory_of_Spacetime | 1,721,777,357,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518130.6/warc/CC-MAIN-20240723224601-20240724014601-00654.warc.gz | 390,691,045 | 30,597 | # 1: Geometric Theory of Spacetime
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“I always get a slight brain-shiver, now [that] space and time appear conglomerated together in a gray, miserable chaos.” – Sommerfeld
• 1.1: Time and Causality
You may have heard that relativity is a theory that can be interpreted using non-Euclidean geometry. The invariance of betweenness is a basic geometrical property that is shared by both Euclidean and non-Euclidean geometry. We say that they are both ordered geometries. With this geometrical interpretation in mind, it will be useful to think of events not as actual notable occurrences but merely as an ambient sprinkling of points at which things could happen.
• 1.2: Experimental Tests of the Nature of Time
In 1971, Hafele and Keating brought atomic clocks aboard commercial airliners and went around the world, once from east to west and once from west to east. Hafele and Keating observed that there was a discrepancy between the times measured by the traveling clocks and the times measured by similar clocks that stayed at the lab in Washington. The result was that the east-going clock lost 59 ns , while the west going one gained 273 ns. This establishes that time is not universal and absolute.
• 1.3: Non-simultaneity and Maximum Speed of Cause and Effect
Instantaneous communication is impossible. There must be some maximum speed at which signals can propagate — or, more generally, a maximum speed at which cause and effect can propagate — and this speed must for example be greater than or equal to the speed at which radio waves propagate. It is also evident from these considerations that simultaneity itself cannot be a meaningful concept in relativity.
• 1.4: Ordered Geometry
Euclid's familiar geometry of two-dimensional space has the following axioms, which are expressed in terms of operations that can be carried out with a compass and unmarked straightedge.
• 1.5: The Equivalence Principle (Part 1)
A central principle of relativity known is the equivalence principle: - that is, accelerations and gravitational fields are equivalent. There is no experiment that can distinguish one from the other.
• 1.6: The Equivalence Principle (Part 2)
Earlier, we saw experimental evidence that the rate of flow of time changes with height in a gravitational field. We can now see that this is required by the equivalence principle. By the equivalence principle, there is no way to tell the difference between experimental results obtained in an accelerating laboratory and those found in a laboratory immersed in a gravitational field.
• 1.E: Geometric Theory of Spacetime (Exercises)
This page titled 1: Geometric Theory of Spacetime is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Benjamin Crowell via source content that was edited to the style and standards of the LibreTexts platform. | 2,374 | 7,194 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.196649 |
http://www.vvz.ethz.ch/Vorlesungsverzeichnis/lerneinheit.view?lerneinheitId=113880&semkez=2016W&lang=de | 1,606,307,275,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182776.11/warc/CC-MAIN-20201125100409-20201125130409-00167.warc.gz | 171,884,203 | 3,788 | Ab 2. November 2020 findet das Herbstsemester 2020 online statt. Ausnahmen: Veranstaltungen, die nur mit Präsenz vor Ort durchführbar sind.
Bitte beachten Sie die per E-Mail kommunizierten Informationen der Dozierenden.
# 401-4531-66L Topics in Rigidity Theory
Semester Herbstsemester 2016 Dozierende M. Burger Periodizität einmalige Veranstaltung Lehrsprache Englisch
Kurzbeschreibung The aim of this course is to give detailed proofs of Margulis' normal subgroup theorem and his superrigidity theorem for lattices in higher rank Lie groups. Lernziel Understand the basic techniques of rigidity theory. Inhalt This course gives an introduction to rigidity theory, which is a set of techniques initially invented to understand the structure of a certain class of discrete subgroups of Lie groups, called lattices, and currently used in more general contexts of groups arising as isometries of non-positively curved geometries. A prominent example of a lattice in the Lie group SL(n, R) is the group SL(n, Z) of integer n x n matrices with determinant 1. Prominent questions concerning this group are: - Describe all its proper quotients.- Classify all its finite dimensional linear representations.- More generally, can this group act by diffeomorphisms on "small" manifolds like the circle? - Does its Cayley graph considered as a metric space at large scale contain enough information to recover the group structure?In this course we will give detailed treatment for the answers to the first two questions; they are respectively Margulis' normal subgroup theorem and Margulis' superrigidity theorem. These results, valid for all lattices in simple Lie groups of rank at least 2 --like SL(n, R), with n at least 3-- lead to the arithmeticity theorem, which says that all lattices are obtained by an arithmetic construction. Literatur - R. Zimmer: "Ergodic Theory and Semisimple groups", Birkhauser 1984.- D. Witte-Morris: "Introduction to Arithmetic groups", available on Arxiv- Y. Benoist: "Five lectures on lattices in semisimple Lie groups", available on his homepage.- M.Burger: "Rigidity and Arithmeticity", European School of Group Theory, 1996, handwritten notes, will be put online. Voraussetzungen / Besonderes For this course some knowledge of elementary Lie theory would be good. We will however treat Lie groups by examples and avoid structure theory since this is not the point of the course nor of the techniques. | 561 | 2,434 | {"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-2020-50 | longest | en | 0.751368 |
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# What Is a Ballpark Estimate?
Article Details
• Written By: Malcolm Tatum
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2003-2018
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A ballpark estimate is a term used to identify an approximation of an outcome that is based on the information that is readily available to the person or group who is making the estimate. Unlike a guess, a ballpark estimate takes into consideration data that can be verified with relative ease and employs expertise in making rational projections based on that data, coming up with an answer that is reasonable in light of all known factors. By contrast, a guess is tends to rely more on subjective understandings and less on verifiable data.
The fanciful name for the ballpark estimate comes from the allusion to many types of sports that are played in a ballpark. The concept has to do with coming up with an estimate that, like the ball that is struck with a bat and lands somewhere within the contained area of the park, is within reasonable distance from the goal. While not considered an actual quotation or covenant, the estimate serves as a guideline for determining whether to move forward with a project, or to abandon it in favor of some other activity.
With a ballpark estimate, the data available for analysis and consideration in forming the estimate provides the essential range of values necessary to come up with a reasonable answer. Within the scope of that estimate, it is understood that should other relevant information appear during the project, the outcome could be changed in some manner. For example, a car insurance agent may present a ballpark estimate to a potential client, based on the known details about the customer. If the agent finds out later that the customer has several tickets over the last couple of years and an accident or two that was not reported at the time the estimate was requested, this will make an impact on the final premium that the agent extends to that client.
While a ballpark estimate is not the final word, it will take into consideration enough information to make a reasonable guess at the final outcome. In actual practice, an estimate of this type may be very close to the outcome, especially if no previously unknown factors arise that have any real bearing on the result. At other times, the appearance of a significant amount of information that was not taken into consideration previously can render the estimate more or less useless. Typically, a ballpark estimate is not considered binding and serves only as a guideline until it is possible to determine the outcome more precisely.
## Recommended
Reminiscence Post 2 I once tried to get an online insurance quote, and the site asked me to answer a few general questions. It actually said I would receive a ballpark figure in a few minutes, based on my answers. If I was happy with that car insurance estimate, I could fill out a more detailed application and a real life insurance agent would contact me with a more specific set of numbers. I guess a lot of people were more interested in getting ballpark estimate from several different insurance companies than getting specific numbers from agents. I know there are times when all I want from a repairman or contractor is a rough estimate. Inaventu Post 1 I remember when I needed to get some repairs done on my house, I called a general contractor and he started listing all of the things that needed attention. It was a much longer list that I anticipated. I asked him to give me a ball park estimate on the cost of repairs, and he said \$20,000 or so. That's a mighty big ballpark, but at least I knew how much to borrow from the bank if he did everything on that list. The actual cost turned out to be much less, but at least the ballpark figure prepared me for the worst. | 815 | 4,009 | {"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-2018-51 | longest | en | 0.953504 |
https://fenicsproject.discourse.group/t/boundary-term-appears-to-be-missing-in-fenics-navier-stokes-example/14360 | 1,721,256,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00003.warc.gz | 221,736,231 | 9,481 | # Boundary term appears to be missing in fenics Navier Stokes example (?)
Dear all,
In this example on a numerical solution of the Navier-Stokes equations with Fenics, a boundary term appears to be missing (?)
Consider the second step in the section ‘Variational formulation’, and the second step in the incremental pressure correction scheme (IPCS). One obtains the equation
-\nabla \cdot u^{\star} / \Delta t + \nabla^2 p^{n+1} - \nabla^2 p^n = 0
then one multiplies by a test function q, integrates, and obtains
-\langle \nabla \cdot u^{\star}, q \rangle_\Omega / \Delta t + \langle \nabla^2 p^{n+1} - \nabla^2 p^n , q \rangle_\Omega = 0, where the subscript \Omega denotes integration within the manifold \Omega.
The second term is then integrated by parts
\langle \nabla^2 p^{n+1} -\nabla^2 p^n , q \rangle_\Omega = - \langle \nabla (p^{n+1} - p^n) , \nabla q \rangle_\Omega + \langle q [\nabla (p^{n+1} - p^n) ] \cdot \hat n \rangle_{\partial \Omega},
where \partial \Omega is the boundary of \Omega and \hat n its unit normal vector pointing outside \Omega.
Thus the variational problem to solve should be
-\langle \nabla \cdot u^{\star}, q \rangle_\Omega / \Delta t - \langle \nabla (p^{n+1} - \nabla) p^n , \nabla q \rangle_\Omega + \langle q\nabla[ (p^{n+1} - p^n) ] \cdot \hat n \rangle_{\partial \Omega} = 0
This is the same equation as (3.34) here, modulo the boundary term, which is missing in there. Why? I don’t see any reason why that term would vanish.
Thank you
Best,
Hi,
So the last term in your last equation is basically grad(p).n = 0. Take right wall of a box as an example, its unit normal vector is (1,0), so the gradient of pressure in x direction must be 0 if grad(p).n = 0 holds. That being said, p is a constant in x direction on the right wall, which means the total force acting upon right wall is balanced (inside and outside of the right wall). So the box is rigid, not deformable, as I interpreted it. Hope it makes sense to you.
Ive written up notes summarizing bcs on splitting schemes at
https://computationalphysiology.github.io/oasisx/splitting_schemes.html
Hello,
I am sorry, but your reply is totally unclear to me, see below
It is not: the test function appears under the gradient operator too.
The right wall is a line joining the two points (L,0) and (L,h), and it has x=const so it does not make sense to state that a field is ‘constant in the x directon on the right wall’.
I’m sorry, is it \nabla (pq) \cdot \vec{n} or q \nabla p \cdot \vec{n}? I guess it should be the latter.
If it is \nabla p \cdot \vec{n} = 0 with \vec{n} = (1, 0), we have \frac{\partial p}{\partial x} = 0 at x=1 (right wall). This is telling you p at both sides of the right wall are identical so that p has no gradient along x direction at x=1, though p might be a function of y. For example,
10.001 | 10.001
9.021 | 9.021
-0.203 | -0.203
2.189 | 2.189
where vertical line | is the right wall.
Forget about it if this does not make sense to you.
Integration by parts here is wrong.
You are right, I corrected my original post.
My original question still stands: q in the boundary term \langle q [\nabla (p^{n+1} - p^n )] \cdot \hat{n}\rangle_{\partial \Omega} vanishes on the parts of \partial \Omega where p is known, i.e., on the right wall. However, in general it does not vanish elsewhere, so I don’t see why this term is missing as a whole in the Fenics example.
I guess this is not a Dirichlet BC where the field is prescribed (or is given numbers directly) so that test function vanishes. In addition at least p^{n+1} is not known.
I did, but I don’t see the answer in there.
See specifically:
https://computationalphysiology.github.io/oasisx/splitting_schemes.html#essential-boundary-conditions
and the following section.
Note that \phi ln those notes represent the difference between the old and new pressure
I could not find the answer in there. It says “However, if \partial \Omega_N \neq 0, then we need to consider the open boundary conditions for the pressure correction schemes”. In the channel flow example of my original post \partial \Omega \neq 0, so we are in this case.
However, still I don’t see why the boundary terms are absent in the code for the channel flow of my original post. I wonder whether this is because in the pressure-correction scheme, one assumes that p^{n+1} takes a known value on \partial \Omega, and thus q= 0 on \partial \Omega (?)
The text states that at all Dirichlet conditions for velocity, one tend to use dphi/dn=0 on this boundary.
For the remaining boundary the one where we have outflow, you set phi=0. Thus you have no boundary term.
So what is the physical meaning of \nabla (p^{n+1} - p^n) \cdot \vec{n} = 0? pressure gradient normal to the wall stays the same? Could you show us an example? Thanks!
Thank you for your reply. Of course, if \partial \phi / \partial n= 0 on the boundaries where one has Dirichlet conditions for the velocity, then that boundary term vanishes.
That is not stated in the screenshot that you posted.
For the sake of other users, and in an effort to improve this beautiful Fenics project, I would like to point out some clarity issues on this point:
• The only place in this webpage where the condition \partial \phi / \partial n= 0 on \partial \Omega is stated is in Section “Essential boundary conditions,” paragraph “Assume that \partial \Omega_N = 0 […] Thus we use that \partial \phi / \partial n = 0 on \partial \Omega.” I find this very unclear, because, as it is, the text implies that the condition \partial \phi / \partial n = 0 on \partial \Omega holds only within the conditions of that paragraph, i.e., in the case where \partial \Omega_N = 0 . Also, “Thus we use that \partial \phi / \partial n = 0 on \partial \Omega” does not specify that this condition holds only for boundaries where Dirichlet boundary conditions for the velocity are imposed.
• Something about this boundary term should be said in the fenics example program or webpage on Navier Stokes equations, it is very confusing to have to go through all this to find out eventually that that tern vanishes because an unmentioned boundary condition was hypothesized.
I hope that this will help!
That is stated in the text i first referenced to, that references the second screenshot
As you refer to.
If you continue looking through this text, you will get the full formulation.
I.e. The equations in:
https://computationalphysiology.github.io/oasisx/splitting_schemes.html#navier-stokes-equation
Im open for suggestions on how to improve the text though.
Please note that you are looking at an 8 year old tutorial, while the webpage i am referencing is maintained (by me).
I Also have an updated version of the splitting scheme at:
https://jsdokken.com/dolfinx-tutorial/chapter2/ns_code2.html
which writes out the assumptions on the boundary condition explicitly. | 1,829 | 6,892 | {"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-2024-30 | latest | en | 0.816482 |
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In this project, you will manage charts to meet objectives 5.1 through 5.3 of the Excel MOS Associate exam. To complete the project, you will create charts 5.1, modify charts 5.2, and format charts 5.3. Note: This project was created for PC users since the MOS Exam is completed in the PC user environment.
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1
Start Excel. Download and open the file named Student_MOS19_XL_PROJ5_Charts.xlsx. Save the file as Last_First_MOS19_XL_PROJ5_Charts.
0
2
Edit the chart to include the 4th Quarter data series.
8
3
Change the Rows/Columns so that the Departments are the Legend entries.
8
4
Resize the chart to the range B14:G30, ensuring that the chart is just inside the cells’ borders.
6
5
Insert a title for the chart and enter Government Expenses.
6
6
Add Alt Text to the chart and enter the description, Chart shows a comparison of government expenses for the first through fourth quarters.
10
7
Insert a 3-D Pie Chart comparing the Department Totals using Departments as the legend. The chart title will display as Department Total.
8
8
Move the Department Total chart to a new sheet and rename the sheet Pie Chart.
6
9
Apply Chart Style 3, and then apply Colorful Palette 3.
8
10
On the Expenses worksheet use Quick Analysis to apply Data Bars to the range F4:F11. Note, Mac users, apply Solid Blue Data Bars to the range.
8
11
Ensure that the chart is not selected. Insert a Right Block Arrow shape pointing to the Capital Projects data for the 4th Quarter of the Government Expenses chart. Position the shape in row 18 and resize the shape so that the Height is 0.20 and the Width is 1.07.
Hint:
8
12
On the Pie Chart sheet, insert a Text Box in the Department Total chart. Position the text box in the pie shape to the right of the 29% data label. In the text box, type Largest percent, apply Bold to the text, and then resize the Text Box to Height 0.30 and Width 1.2.
12
13
Change the Chart Layout to 1. Use the snipping tool to take a rectangular snip of the Pie Chart Area including the title. Copy and then paste the snip in cell I2 of the Expenses worksheet. Resize the snip so that the upper left corner sits just inside the cell borders of I2, Height 5.91 and Width 6.67.
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Template:Essay-like Quantum mechanics is most often used to describe matter on the scale of molecules, atoms, or elementary particles. However, some phenomena, particularly at low temperatures, show quantum behavior on a macroscopic scale. The best-known examples of macroscopic quantum phenomena are superfluidity and superconductivity; another example is the quantum Hall effect. Since 2000 there has been extensive experimental work on quantum gases, particularly Bose–Einstein Condensates.
Between 1996 to 2003 four Nobel prizes were given for work related to macroscopic quantum phenomena.[1] Macroscopic quantum phenomena can be observed in superfluid helium and in superconductors,[2] but also in dilute quantum gases and in laser light. Although these media are very different, their behavior is very similar as they all show macroscopic quantum behavior.
Quantum phenomena are generally classified as macroscopic when the quantum states are occupied by a large number of particles (typically Avogadro's number) or the quantum states involved are macroscopic in size (up to km size in superconducting wires).
## Consequences of the macroscopic occupation
Fig.1 Left: only one particle; usually the small box is empty. However, there is a certain chance of that the particle is in the box. This chance is given by Eq.(15). Middle: a few particles. There are usually some particles in the box. We can define an average, but the actual number of particles in the box has large fluctuations around this average. Right: large number of particles. The fluctuations around the average are small.
The concept of macroscopically-occupied quantum states is introduced by Fritz London.[3][4] In this section it will be explained what it means if the ground state is occupied by a very large number of particles. We start with the wave function of the ground state written as
with Ψ₀ the amplitude and ${\displaystyle \varphi }$ the phase. The wave function is normalized so that
The physical interpretation of the quantity
depends on the number of particles. Fig.1 represents a container with a certain number of particles with a small control volume ΔV inside. We check from time to time how many particles are in the control box. We distinguish three cases:
1. There is only one particle. In this case the control volume is empty most of the time. However, there is a certain chance to find the particle in it given by Eq.(15). The chance is proportional to ΔV. The factor ΨΨ is called the chance density.
2. If the number of particles is a bit larger there are usually some particles inside the box. We can define an average, but the actual number of particles in the box has relatively large fluctuations around this average.
3. In the case of a very large number of particles there will always be a lot of particles in the small box. The number will fluctuate but the fluctuations around the average are relatively small. The average number is proportional to ΔV and ΨΨ is now interpreted as the particle density.
In quantum mechanics the particle probability flow density Jp (unit: particles per second per m²) can be derived from the Schrödinger equation to be
with q the charge of the particle and ${\displaystyle {\vec {A}}}$ the vector potential. With Eq.(13)
If the wave function is macroscopically occupied the particle probability flow density becomes a particle flow density. We introduce the fluid velocity vs via the mass flow density
The density (mass per m³) is
so Eq.(17) results in
This important relation connects the velocity, a classical concept, of the condensate with the phase of the wave function, a quantum-mechanical concept.
## Superfluidity
{{#invoke:main|main}}
Fig.2 Lower part: vertical cross section of a column of superfluid helium rotating around a vertical axis. Upper part: Top view of the surface showing the pattern of vortex cores. From left to right the rotation speed is increased resulting in an increasing vortex-line density.
Below the lambda-temperature helium shows the unique property of superfluidity. The fraction of the liquid that forms the superfluid component is a macroscopic quantum fluid. The helium atom is a neutral particle so q=0. Furthermore the particle mass m=m₄ so Eq.(20) reduces to
For an arbitrary loop in the liquid this gives
Due to the single-valued nature of the wave function
with n integer, we have
The quantity
is the quantum of circulation. For a circular motion with radius r
In case of a single quantum (n=1)
When superfluid helium is put in rotation Eq.(25) will not be satisfied for all loops inside the liquid unless the rotation is organized around vortex lines as depicted in Fig.2. These lines have a vacuum core with a diameter of about 1 Å (which is smaller than the average particle distance). The superfluid helium rotates around the core with very high speeds. Just outside the core (r = 1 Å) the velocity is as large as 160 m/s. The cores of the vortex lines and the container rotate as a solid body around the rotation axes with the same angular velocity. The number of vortex lines increases with the angular velocity as shown in the upper half of the figure. Note that the two right figures both contain six vortex lines, but the lines are organized in different stable patterns.[5]
## Superconductivity
{{#invoke:main|main}}
### Fluxoid quantization
For superconductors the bosons involved are the so-called Cooper pairs which are quasiparticles formed by two electrons.[6] Hence m = 2me and q = -2e where me and e are the mass of an electron and the elementary charge. It follows from Eq.(20) that
Integrating Eq.(27) over a closed loop gives
As in the case of helium we define the vortex strength
and use the general relation
where Φ is the magnetic flux enclosed by the loop. The so-called fluxoid is defined by
In general the values of κ and Φ depend on the choice of the loop. Due to the single-valued nature of the wave function and Eq.(28) the fluxoid is quantized
The unit of quantization is called the flux quantum
The flux quantum plays a very important role in superconductivity. The earth magnetic field is very small (about 50 μT), but it generates one flux quantum in an area of 6 by 6 μm. So, the flux quantum is very small. Yet it was measured to an accuracy of 9 digits as shown in Eq.(33). Nowadays the value given by Eq.(33) is exact by definition.
Fig. 3. Two superconducting rings in an applied magnetic field
a: thick superconducting ring. The integration loop is completely in the region with vs=0;
b: thick superconducting ring with a weak link. The integration loop is completely in the region with vs=0 except for a small region near the weak link.
In Fig. 3 two situations are depicted of superconducting rings in an external magnetic field. One case is a thick-walled ring and in the other case the ring is also thick-walled, but is interrupted by a weak link. In the latter case we will meet the famous Josephson relations. In both cases we consider a loop inside the material. In general a superconducting circulation current will flow in the material. The total magnetic flux in the loop is the sum of the applied flux Φa and the self-induced flux Φs induced by the circulation current
### Thick ring
The first case is a thick ring in an external magnetic field (Fig. 3a). The currents in a superconductor only flow in a thin layer at the surface. The thickness of this layer is determined by the so-called London penetration depth. It is of μm size or less. We consider a loop far away from the surface so that vs=0 everywhere so κ=0. In that case the fluxoid is equal to the magnetic flux (Φv=Φ). If vs=0 Eq.(27) reduces to
Taking the rotation gives
Using the well-known relations ${\displaystyle {\vec {\nabla }}\times {\vec {\nabla }}\varphi =0}$ and ${\displaystyle {\vec {\nabla }}\times {\vec {A}}={\vec {B}}}$ shows that the magnetic field in the bulk of the superconductor is zero as well. So, for thick rings, the total magnetic flux in the loop is quantized according to
Fig. 4. Schematic of a weak link carrying a superconducting current is. The voltage difference over the link is V. The phases of the superconducting wave functions at the left and right side are assumed to be constant (in space, not in time) with values of φ1 and φ2 respectively.
Weak links play a very important role in modern superconductivity. In most cases weak links are oxide barriers between two superconducting thin films, but it can also be a crystal boundary (in the case of high-Tc superconductors). A schematic representation is given in Fig. 4. Now consider the ring which is thick everywhere except for a small section where the ring is closed via a weak link (Fig. 3b). The velocity is zero except near the weak link. In these regions the velocity contribution to the total phase change in the loop is given by (with Eq.(27))
The line integral is over the contact from one side to the other in such a way that the end points of the line are well inside the bulk of the superconductor where vs=0. So the value of the line integral is well-defined (e.g. independent of the choice of the end points). With Eqs.(31), (34), and (38)
Without proof we state that the supercurrent through the weak link is given by the so-called DC Josephson relation[7]
The voltage over the contact is given by the AC Josephson relation
The names of these relations (DC and AC relations) are misleading since they both hold in DC and AC situations. In the steady state (constant ${\displaystyle \Delta \varphi ^{*}}$) Eq.(41) shows that V=0 while a nonzero current flows through the junction. In the case of a constant applied voltage (voltage bias) Eq.(41) can be integrated easily and gives
Substitution in Eq.(40) gives
This is an AC current. The frequency
is called the Josephson frequency. One μV gives a frequency of about 500 MHz. By using Eq.(44) the flux quantum is determined with the high precision as given in Eq.(33).
The energy difference of a Cooper pair, moving from one side of the contact to the other, is ΔE = 2eV. With this expression Eq.(44) can be written as ΔE = which is the relation for the energy of a photon with frequency ν.
The AC Josephson relation (Eq.(41)) can be easily understood in terms of Newton's law, (or from one of the London equation's[8]). We start with Newton's law
${\displaystyle {\vec {F}}=m{\mathrm {d} }{\vec {v}}_{s}/{\mathrm {d} }t.}$
Substituting the expression for the Lorentz force
${\displaystyle {\vec {F}}=q({\vec {E}}+{\vec {v}}_{s}\times {\vec {B}})}$
and using the general expression for the co-moving time derivative
${\displaystyle {\mathrm {d} }{\vec {v}}_{s}/{\mathrm {d} }t=\partial {\vec {v}}_{s}/\partial t+(1/2){\vec {\nabla }}v_{s}^{2}-{\vec {v}}_{s}\times ({\vec {\nabla }}\times {\vec {v}}_{s})}$
gives
${\displaystyle (q/m)({\vec {E}}+{\vec {v}}_{s}\times {\vec {B}})=\partial {\vec {v}}_{s}/\partial t+(1/2){\vec {\nabla }}v_{s}^{2}-{\vec {v}}_{s}\times ({\vec {\nabla }}\times {\vec {v}}_{s}).}$
Eq.(20) gives
${\displaystyle 0={\vec {\nabla }}\times {\vec {v}}_{s}+(q/m){\vec {\nabla }}\times {\vec {A}}={\vec {\nabla }}\times {\vec {v}}_{s}+(q/m){\vec {B}}}$
so
${\displaystyle (q/m){\vec {E}}=\partial {\vec {v}}_{s}/\partial t+(1/2){\vec {\nabla }}v_{s}^{2}.}$
Take the line integral of this expression. In the end points the velocities are zero so the ∇v2 term gives no contribution. Using
${\displaystyle \int {\vec {E}}\cdot {\mathrm {d} }{\vec {l}}=-V}$
and Eq.(38), with q = -2e and m =2me, gives Eq.(41).
### DC SQUID
{{#invoke:main|main}}
Fig. 5. Two superconductors connected by two weak links. A current and a magnetic field are applied.
Fig. 6. Dependence of the critical current of a DC-SQUID on the applied magnetic field
Figure 5 shows a so-called DC SQUID. It consists of two superconductors connected by two weak links. The fluxoid quantization of a loop through the two bulk superconductors and the two weak links demands
If the self-inductance of the loop can be neglected the magnetic flux in the loop Φ is equal to the applied flux
with B the magnetic field, applied perpendicular to the surface, and A the surface area of the loop. The total supercurrent is given by
Substitution of Eq(45) in (47) gives
Using a well known geometrical formula we get
Since the sin-function can vary only between −1 and +1 a steady solution is only possible if the applied current is below a critical current given by
Note that the critical current is periodic in the applied flux with period Φ₀. The dependence of the critical current on the applied flux is depicted in Fig. 6. It has a strong resemblance with the interference pattern generated by a laser beam behind a double slit. In practice the critical current is not zero at half integer values of the flux quantum of the applied flux. This is due to the fact that the self-inductance of the loop cannot be neglected.[9]
### Type II superconductivity
{{#invoke:main|main}}
Fig. 7. Magnetic flux lines penetrating a type-II superconductor. The currents in the superconducting material generate a magnetic field which, together with the applied field, result in bundles of quantized flux.
Type-II superconductivity is characterized by two critical fields called Bc1 and Bc2. At a magnetic field Bc1 the applied magnetic field starts to penetrate the sample, but the sample is still superconducting. Only at a field of Bc2 the sample is completely normal. For fields in between Bc1 and Bc2 magnetic flux penetrates the superconductor in well-organized patterns, the so-called Abrikosov vortex lattice similar to the pattern shown in Fig. 2.[10] A cross section of the superconducting plate is given in Fig. 7. Far away from the plate the field is homogeneous, but in the material superconducting currents flow which squeeze the field in bundles of exactly one flux quantum. The typical field in the core is as big as 1 tesla. The currents around the vortex core flow in a layer of about 50 nm with current densities on the order of 15Template:E A/m². That corresponds with 15 million ampère in a wire of one mm².
## Dilute quantum gases
The classical types of quantum systems, superconductors and superfluid helium, were discovered in the beginning of the 20th century. Near the end of the 20th century, scientists discovered how to create very dilute atomic or molecular gases, cooled first by laser cooling and then by evaporative cooling.[11] They are trapped using magnetic fields or optical dipole potentials in ultrahigh vacuum chambers. Isotopes which have been used include rubidium (Rb-87 and Rb-85), strontium (Sr-87, Sr-86, and Sr-84) potassium (K-39 and K-40), sodium (Na-23), lithium (Li-7 and Li-6), and hydrogen (H-1). The temperatures to which they can be cooled are as low as a few nanokelvin. The developments have been very fast in the past few years. A team of NIST and the University of Colorado has succeeded in creating and observing vortex quantization in these systems.[12] The concentration of vortices increases with the angular velocity of the rotation, similar to the case of superfluid helium and superconductivity.
## References and footnotes
1. These Nobel prizes were for the discovery of super-fluidity in helium-3 (1996), for the discovery of the fractional quantum Hall effect (1998), for the demonstration of Bose–Einstein condensation (2001), and for contributions to the theory of superconductivity and superfluidity (2003).
2. D.R. Tilley and J. Tilley, Superfluidity and Superconductivity, Adam Hilger, Bristol and New York, 1990
3. Fritz London Superfluids (London, Wiley, 1954-1964)
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12. {{#invoke:Citation/CS1|citation |CitationClass=journal }} | 3,899 | 16,094 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 51, "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-2019-43 | longest | en | 0.927236 |
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multidimensional signals 多维信号(2)
多维信号
To utilize coding over finite fields for multidimensional signals, we start with prime elements and study the algebraic integer ring of cyclotomic field Q(e2πi/n) modulo prime ideals for arbitrary integer n using the property that prime elements generate prime ideals. 为实现面向多维信号有限域上的编码,本文从素元出发,利用素元生成的理想为素理想,研究对任意正整数n,分圆域Q(e2πi/n)的代数整数环模素理想所得的剩余域. 短句来源 Applying the principle of generating surrogate data to generate surrogate time series of measured multivariate time series and using linear and generalized redundancy as test statistics,we discuss the scheme of resisting noise in a proposed quantitative method to detect nonlinearity in multidimensional signals. 采用多变量时间序列替代数据生成原理,生成实测多变量时间序列的多组替代时间序列. 综合线性冗余和广义冗余两种检验统计量,对一种定量检验多维信号非线性方法的抗噪声能力进行分析. 短句来源
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SAMPLING THEOREMS FOR MULTIDIMENSIONAL NONBANDLIMMITED SIGNALS 多维非带限信号的抽样定理 短句来源 recombination of signals. 信号重组。 短句来源 of broadcasting signals. 主要实现了对异地广播信号的实时监测,录音,数据存储,异常告警等功能。 短句来源 Blind separation of multidimensional mixed signals based on fast independent component analysis 基于快速独立分量分析的多维混合信号盲分离 短句来源 On the Multidimensional Turbo Codes 多维Turbo码研究 短句来源
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multidimensional signals
Extension to the case of multidimensional signals is considered. However, it has been difficult to take advantage of this rich source of information concerning the signal, especially for multidimensional signals. Kramer's sampling theorem for multidimensional signals and its relationship with Lagrange-type interpolations Extending the notion of second-order correlations, we define thecumulants of stationary non-Gaussian random fields, and demonstrate their potential for modeling and reconstruction of multidimensional signals and systems. We present an example of adaptive optimization of the sampling grid for a magnetic field that is associated with a spectrum of multidimensional signals having fractal similarity. 更多
A new method of multilevel coding based on multidimensional signal constellations and corresponding decoding algorithm is presented in this paper. The method solves the problem that the decoding complexity grows approximately exponentally with the increase of coding gains in previously known trellis-coded modulation systems. As examples, Some coding systems based on the four dimensional lattice D4 and the eight dimensional lattice E8 are given. The results show that the required normalized (each two dimensional)... A new method of multilevel coding based on multidimensional signal constellations and corresponding decoding algorithm is presented in this paper. The method solves the problem that the decoding complexity grows approximately exponentally with the increase of coding gains in previously known trellis-coded modulation systems. As examples, Some coding systems based on the four dimensional lattice D4 and the eight dimensional lattice E8 are given. The results show that the required normalized (each two dimensional) decoding complexity of the new coding scheme grows almost linearly with the increase of coding gains. 本文提出了一种对多维信号星座进行多级编码的新方法和相应的译码算法。这种新方法解决了以前网格编码调制系统中译码复杂度大致上随编码增益指数增长的问题。作为例子,我们给出了在4维格D_4和8维格E_8上的一些编码系统。结果表明:新的编码方式所需要的规范化(每2维)译码复杂度几乎随编码增益线性地增长。 An approach for designing multilevel trellis coded modulation schemes based on multidimensional signal constellations is described through two examples. We discussed some related problems in detail, such as, how to construct chains of lattice partitioning and constellations, and how to select binary convolutional codes. 本文通过两个实例描述了基于多维信号星座的多级网格编码调制方式的设计方法,并详细讨论了格分割链和星座的构造以及二元卷积码的选择等有关问题。 For a multicomponent system a mathematical model for multidimensional signals and its processing approach (transformation of coordinates) have been established. Combined measurement (PCM) system for proers consisting of non-contact sensors (on-line non-contact process combined measurement system) have been designed and proposed in the paper. On-line PCM system is a new mode of process analyzer system. It provides a stable, fast response and a very high accuracy that is bard to attain in conventional discrete... For a multicomponent system a mathematical model for multidimensional signals and its processing approach (transformation of coordinates) have been established. Combined measurement (PCM) system for proers consisting of non-contact sensors (on-line non-contact process combined measurement system) have been designed and proposed in the paper. On-line PCM system is a new mode of process analyzer system. It provides a stable, fast response and a very high accuracy that is bard to attain in conventional discrete and selective measurement of multicomponent analyzer system. The principle of PCM system can be widely utilized in other fields for on-line continuous industrial analysis. 为多元组分建立了多维信号空间数学模型及其处理方法（坐标变换），设计了包含非接触传感器的过程组合测量系统──在线非接触过程组合测量（ｏｎ－ｌｉｎｅ）ＰＣＭ系统．作为一种新型的分析系统，它具有稳定、快速响应及很高的精确度，其原理也可应用于工业在线连续分析的其他领域中． << 更多相关文摘
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2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社 | 1,542 | 5,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2019-43 | latest | en | 0.567936 |
http://math.berkeley.edu/programs/undergraduate/major/teaching | 1,371,656,590,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708835190/warc/CC-MAIN-20130516125355-00088-ip-10-60-113-184.ec2.internal.warc.gz | 154,162,030 | 8,152 | # Course Requirements: Major with a Teaching Concentration
### Requirements for the Major with a Teaching Concentration
The new teaching concentration is designed to increase the number and quality of math teachers. It requires the completion of three new courses: Math 151, 152 and 153. It also includes a modification to the typical pure math major sequence.
Following are the required courses for the teaching concentration:
Statistics 20 or 25 Probability and Statistics
Mathematics 1A-1B Calculus
Mathematics 53 Multivariable Calculus
Mathematics 54 Linear Algebra & Differential Equations
Mathematics 55 Discrete Mathematics
Mathematics 110 Linear Algebra
Mathematics 113 Abstract Algebra
Any two of:
Mathematics 128A Numerical Analysis,
Mathematics 130 Classical Geometry, or
Mathematics 135 Set Theory
Mathematics 151 Mathematics for the Secondary School Curriculum I
Mathematics 152 Mathematics for the Secondary School Curriculum II
Mathematics 153 Mathematics for the Secondary School Curriculum III
Mathematics 160 History of Mathematics
In addition, students are encouraged (but not required) to take Mathematics 104 Analysis, Mathematics 115 Number Theory, and Mathematics 185 Complex Analysis.
We will accept Computer Science 70 in lieu of Mathematics 55 for students with a double major in Computer Science or Electrical Engineering and Computer Science.
Following is a brief description of the Mathematics 151-153 series:
Mathematics 151 treats fractions, rational numbers, basic number theory and the Euclidean algorithm, rigid motions, dilations, geometry of similar triangles, and linear equations and their graphs.
Mathematics 152 treats linear inequalities and their graphs, simultaneous linear equations, functions (quadratic, polynomial, rational, exponential, and logarithmic), basic Euclidean geometry, and discussion of axiomatization.
Mathematics 153 treats trigonometric functions, the least upper bound axiom, limits and n-th room, area and volume, basic mensuration formulas, the theory of calculus up to the abstract definitions of exp and log. | 416 | 2,086 | {"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-2013-20 | latest | en | 0.822642 |
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Posted by | · · · · | Blog
[In Part 1 of this series, we discussed how to use Autofilters to manage large sets of data, enabling you to work with only the records that are relevant. In this article we will explore how the Subtotal function in Excel can leverage the power of filters even more.]
### Part Two: Subtotals
If you’ve used Excel at all, then you are probably familiar with how to sum a column or a row of numbers. What you may not know is that the Subtotal() function allows for much more than basic mathematical manipulation of data, and can be combined with additional functions to perform those manipulations in a more dynamic manner.
Below is a screenshot illustrating just a few of the additional functions that can be paired with Subtotal(), with each option numbered to serve as a clear visual reference for the examples that follow.
Now, back to our data.
In this sheet, we see that there are 5 records, only one of which is male. Because we used the Sum() function in cell C9 (“=SUM(C4:C8)”) we can easily see that the total of the Financial Outcome column (cells C4 through C8) is \$1,750.
In this screenshot, after filtering, we see that the same records have been filtered on Gender, leaving only the record in row 4 displayed. But the Sum() function in cell C9 is still reporting that the sum of cells C4 through C8 is (still) \$1,750:
But instead of using the Sum() function, we can use the Subtotal() function to see what happens. Instead of just “summing” the values in those cells, we ask it to do it selectively. This is what the Subtotal() function does.
By using the formula “=SUBTOTAL(9,C4:C8)” instead of always getting \$1,750, the “total” will change and it will show you the sum of only the records that are being displayed based on the filters. So, to see the sum of all Financial Outcomes for all Males, the results looks like this:
To see the total for all females, there is no need to change the formula, simply change the filters and now you will have this:
And to see the total for all Hispanic Females:
This is a very powerful feature that is easy to learn and use.
But wait, there’s more! (And no, it’s not a Ginzu knife!)
Often, instead of summing a column, you just want to know how many records there are. By adding additional variables to the Subtotal() function, we can easily do that.
Using the example immediately above, the logical flow of the function ” =SUBTOTAL(9,C4:C8)” is as follows:
• Function: Subtotal
• Type of Subtotal: 9 (sum of all records being displayed)
• Range to Sum: Cells C4 through C8
To COUNT the records instead of summing the dollar amount, we would simply change the “9” above to a “3” so that instead of:
=SUBTOTAL(9,C4:C8)
…it should look like this:
=SUBTOTAL(3,C4:C8)
Which gives you this end result:
There are lots of other functions that the Subtotal can perform…:
…but the Total (9) and the CountA (3) are the most frequently useful ones. (Note: the difference between Count and CountA is that Count counts every cell in the range, CountA only counts the cells that are not blank. This will make sense when you care about this distinction.)
I hope this will give you a little bit of insight into how these two simple functions can really help you take a fresh look at your data. Excel has some extraordinary features, but the Autofilters and the Subtotal functions, when used together, can be a great yet simple entry into the uses of this amazing tool.
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Index > Main > Perlin noise, the making of
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Joined: 25 Sep 2003
Posts: 2139
Location: Estonia
Lets start off easy, this sample is a Java implementation of what I call "Perlin in a square" 'cuz of its formatting. Has some weird stuff, that I don't understand, but I get the jist.
The problem is there just isn't a good way to solve it in ASM. Or maybe in any language. It has got mixed integer, float stuff and LUT that you can't SIMD properly. Well it really is fast (compared to previous versions), but all this code is for ONE PIXEL!
Okay, so what else can we find on the Internet (Google...). There's a FAQ that describes somewhat, what's behind. This guy is in the middle of making a really impressive game (billions of purely generated planets) and he has optimized it a bit (though SSE4.1 has got rounding that makes the conversions a bit easier).
It wasn't until this code that I started to ponder about SSE implementation (more functions here).
To get the very true nature of this noise, there's one sentence that I came up with:
Pick some pseudo-random values, space them evenly on a pixel grid and interpolate other pixels using weighed average.
For the sake of clarity, lets deal only with 2D now!
Lets meet the words:
1) random - we're playing with a range of 0..255 (black to white)
2) evenly - random pixel every 2^n (where n<N) pixels, where a generated image is 2^N*2^N pixels
3) weighed - if it were linear, the image would look like pyramids, but we'll use what's called the "ease curve" on the FAQ.
I'm not very goot at C, but the tests that I made with Intel's otherwise perfect SIMDer compiler concluded that Perlin's noise function is too big of a nut to crack. So I figure I need to do it manually. Here's the Intel's version of hand-made (robotbastards code) intrinsics code. All inlined because you can't put SSE in a function (i.e. its really hard). Ugly heh?
Code:
```;438 lines - prologue and epilogue stripped
movaps xmm2, XMMWORD PTR _2il0floatpacket\$3(rip) ;137.10
movaps xmm1, xmm0 ;137.10
subps xmm1, XMMWORD PTR _2il0floatpacket\$1(rip) ;137.10
movdqa xmm6, XMMWORD PTR [rsp+192] ;137.10
movaps xmm11, XMMWORD PTR [rsp+304] ;137.10
movaps xmm8, XMMWORD PTR [rsp+256] ;137.10
cvtps2dq xmm3, xmm1 ;137.10
movdqa xmm13, XMMWORD PTR _2il0floatpacket\$8(rip) ;137.10
cvtdq2ps xmm1, xmm3 ;137.10
pand xmm3, XMMWORD PTR _2il0floatpacket\$2(rip) ;137.10
paddd xmm3, XMMWORD PTR [rsp+272] ;137.10
movaps xmm9, XMMWORD PTR _2il0floatpacket\$9(rip) ;137.10
subps xmm0, xmm1 ;137.10
movaps xmm12, XMMWORD PTR _2il0floatpacket\$10(rip) ;137.10
movaps xmm10, XMMWORD PTR _2il0floatpacket\$12(rip) ;137.10
mulps xmm2, xmm0 ;137.10
subps xmm2, XMMWORD PTR _2il0floatpacket\$4(rip) ;137.10
movaps XMMWORD PTR [rsp+336], xmm0 ;137.10
movaps xmm1, xmm0 ;137.10
mulps xmm1, xmm0 ;137.10
mulps xmm2, xmm0 ;137.10
addps xmm2, XMMWORD PTR _2il0floatpacket\$5(rip) ;137.10
mulps xmm1, xmm0 ;137.10
mulps xmm2, xmm1 ;137.10
movdqa xmm1, XMMWORD PTR [rsp+240] ;137.10
movdqa xmm5, xmm1 ;137.10
movaps XMMWORD PTR [rsp+352], xmm2 ;137.10
movdqa xmm4, xmm1 ;137.10
movdqa XMMWORD PTR [rsp+368], xmm3 ;137.10
movaps xmm1, xmm0 ;137.10
subps xmm1, XMMWORD PTR _2il0floatpacket\$7(rip) ;137.10
movaps XMMWORD PTR [rsp+384], xmm1 ;137.10
movdqa xmm1, xmm14 ;137.10
pand xmm5, xmm13 ;137.10
cvtdq2ps xmm7, xmm5 ;137.10
movaps xmm2, xmm7 ;137.10
movdqa XMMWORD PTR [rsp+400], xmm1 ;137.10
movdqa xmm1, xmm14 ;137.10
cmpltps xmm2, xmm9 ;137.10
pand xmm4, xmm13 ;137.10
movdqa XMMWORD PTR [rsp+416], xmm1 ;137.10
movdqa xmm1, xmm14 ;137.10
pand xmm6, xmm13 ;137.10
movdqa XMMWORD PTR [rsp+432], xmm1 ;137.10
movdqa xmm1, xmm14 ;137.10
movaps xmm3, xmm7 ;137.10
cmpltps xmm3, xmm12 ;137.10
movaps xmm15, xmm3 ;137.10
movdqa XMMWORD PTR [rsp+448], xmm1 ;137.10
movaps xmm1, xmm11 ;137.10
andps xmm1, xmm2 ;137.10
andnps xmm2, xmm0 ;137.10
orps xmm1, xmm2 ;137.10
movaps xmm2, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
andps xmm15, xmm0 ;137.10
cmpeqps xmm2, xmm7 ;137.10
cmpeqps xmm7, xmm10 ;137.10
orps xmm2, xmm7 ;137.10
movaps xmm7, xmm11 ;137.10
andps xmm7, xmm2 ;137.10
andnps xmm2, xmm8 ;137.10
orps xmm7, xmm2 ;137.10
movdqa xmm2, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm3, xmm7 ;137.10
orps xmm15, xmm3 ;137.10
pand xmm2, xmm5 ;137.10
cvtdq2ps xmm3, xmm2 ;137.10
pand xmm5, xmm14 ;137.10
cvtdq2ps xmm2, xmm5 ;137.10
xorps xmm5, xmm5 ;137.10
xorps xmm7, xmm7 ;137.10
cmpeqps xmm3, xmm5 ;137.10
movaps xmm5, xmm1 ;137.10
andps xmm5, xmm3 ;137.10
subps xmm7, xmm1 ;137.10
andnps xmm3, xmm7 ;137.10
movaps xmm7, XMMWORD PTR [rsp+320] ;137.10
orps xmm5, xmm3 ;137.10
xorps xmm1, xmm1 ;137.10
cmpeqps xmm2, xmm1 ;137.10
movaps xmm3, xmm15 ;137.10
andps xmm3, xmm2 ;137.10
xorps xmm1, xmm1 ;137.10
subps xmm1, xmm15 ;137.10
andnps xmm2, xmm1 ;137.10
orps xmm3, xmm2 ;137.10
movaps xmm3, xmm7 ;137.10
cvtdq2ps xmm1, xmm4 ;137.10
movaps xmm15, xmm1 ;137.10
movaps xmm2, xmm1 ;137.10
cmpltps xmm15, xmm9 ;137.10
andps xmm3, xmm15 ;137.10
andnps xmm15, xmm0 ;137.10
orps xmm3, xmm15 ;137.10
cmpltps xmm2, xmm12 ;137.10
movaps xmm15, xmm2 ;137.10
andps xmm15, xmm0 ;137.10
movaps xmm0, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
cmpeqps xmm0, xmm1 ;137.10
cmpeqps xmm1, xmm10 ;137.10
orps xmm0, xmm1 ;137.10
movaps xmm1, xmm7 ;137.10
andps xmm1, xmm0 ;137.10
andnps xmm0, xmm8 ;137.10
orps xmm1, xmm0 ;137.10
movdqa xmm0, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm2, xmm1 ;137.10
orps xmm15, xmm2 ;137.10
pand xmm0, xmm4 ;137.10
cvtdq2ps xmm2, xmm0 ;137.10
pand xmm4, xmm14 ;137.10
cvtdq2ps xmm1, xmm4 ;137.10
xorps xmm0, xmm0 ;137.10
xorps xmm4, xmm4 ;137.10
cmpeqps xmm2, xmm0 ;137.10
movaps xmm0, xmm3 ;137.10
andps xmm0, xmm2 ;137.10
subps xmm4, xmm3 ;137.10
andnps xmm2, xmm4 ;137.10
orps xmm0, xmm2 ;137.10
xorps xmm2, xmm2 ;137.10
cmpeqps xmm1, xmm2 ;137.10
movaps xmm3, xmm15 ;137.10
andps xmm3, xmm1 ;137.10
xorps xmm2, xmm2 ;137.10
subps xmm2, xmm15 ;137.10
andnps xmm1, xmm2 ;137.10
movaps xmm2, XMMWORD PTR [rsp+384] ;137.10
orps xmm3, xmm1 ;137.10
movaps xmm4, xmm11 ;137.10
movaps xmm1, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
movaps xmm3, XMMWORD PTR [rsp+288] ;137.10
subps xmm0, xmm5 ;137.10
mulps xmm0, xmm3 ;137.10
movaps XMMWORD PTR [rsp+464], xmm5 ;137.10
cvtdq2ps xmm5, xmm6 ;137.10
movaps xmm0, xmm5 ;137.10
movaps xmm15, xmm5 ;137.10
cmpltps xmm0, xmm9 ;137.10
andps xmm4, xmm0 ;137.10
andnps xmm0, xmm2 ;137.10
orps xmm4, xmm0 ;137.10
cmpltps xmm15, xmm12 ;137.10
movaps xmm0, xmm15 ;137.10
andps xmm0, xmm2 ;137.10
cmpeqps xmm1, xmm5 ;137.10
cmpeqps xmm5, xmm10 ;137.10
orps xmm1, xmm5 ;137.10
movaps xmm5, xmm11 ;137.10
andps xmm5, xmm1 ;137.10
andnps xmm1, xmm8 ;137.10
orps xmm5, xmm1 ;137.10
movdqa xmm1, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm15, xmm5 ;137.10
orps xmm0, xmm15 ;137.10
pand xmm1, xmm6 ;137.10
cvtdq2ps xmm15, xmm1 ;137.10
pand xmm6, xmm14 ;137.10
cvtdq2ps xmm5, xmm6 ;137.10
xorps xmm1, xmm1 ;137.10
xorps xmm6, xmm6 ;137.10
cmpeqps xmm15, xmm1 ;137.10
movaps xmm1, xmm4 ;137.10
andps xmm1, xmm15 ;137.10
subps xmm6, xmm4 ;137.10
andnps xmm15, xmm6 ;137.10
orps xmm1, xmm15 ;137.10
xorps xmm4, xmm4 ;137.10
cmpeqps xmm5, xmm4 ;137.10
movaps xmm6, xmm0 ;137.10
andps xmm6, xmm5 ;137.10
xorps xmm4, xmm4 ;137.10
subps xmm4, xmm0 ;137.10
andnps xmm5, xmm4 ;137.10
orps xmm6, xmm5 ;137.10
movdqa xmm5, XMMWORD PTR [rsp+368] ;137.10
pand xmm5, xmm13 ;137.10
movaps xmm6, xmm7 ;137.10
cvtdq2ps xmm0, xmm5 ;137.10
movaps xmm15, xmm0 ;137.10
movaps xmm4, xmm0 ;137.10
cmpltps xmm15, xmm9 ;137.10
andps xmm6, xmm15 ;137.10
andnps xmm15, xmm2 ;137.10
orps xmm6, xmm15 ;137.10
cmpltps xmm4, xmm12 ;137.10
movaps xmm15, xmm4 ;137.10
andps xmm15, xmm2 ;137.10
movaps xmm2, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
cmpeqps xmm2, xmm0 ;137.10
cmpeqps xmm0, xmm10 ;137.10
orps xmm2, xmm0 ;137.10
movaps xmm0, xmm7 ;137.10
andps xmm0, xmm2 ;137.10
andnps xmm2, xmm8 ;137.10
orps xmm0, xmm2 ;137.10
andnps xmm4, xmm0 ;137.10
movdqa xmm0, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
orps xmm15, xmm4 ;137.10
pand xmm0, xmm5 ;137.10
cvtdq2ps xmm4, xmm0 ;137.10
pand xmm5, xmm14 ;137.10
cvtdq2ps xmm2, xmm5 ;137.10
xorps xmm0, xmm0 ;137.10
xorps xmm5, xmm5 ;137.10
cmpeqps xmm4, xmm0 ;137.10
movaps xmm0, xmm6 ;137.10
andps xmm0, xmm4 ;137.10
subps xmm5, xmm6 ;137.10
andnps xmm4, xmm5 ;137.10
orps xmm0, xmm4 ;137.10
xorps xmm4, xmm4 ;137.10
cmpeqps xmm2, xmm4 ;137.10
movaps xmm5, xmm15 ;137.10
andps xmm5, xmm2 ;137.10
xorps xmm4, xmm4 ;137.10
subps xmm4, xmm15 ;137.10
movdqa xmm15, XMMWORD PTR [rsp+400] ;137.10
andnps xmm2, xmm4 ;137.10
movaps xmm4, XMMWORD PTR [rsp+336] ;137.10
orps xmm5, xmm2 ;137.10
pand xmm15, xmm13 ;137.10
movaps xmm2, xmm11 ;137.10
movaps xmm5, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
subps xmm0, xmm1 ;137.10
mulps xmm0, xmm3 ;137.10
cvtdq2ps xmm8, xmm15 ;137.10
movaps xmm6, xmm8 ;137.10
movaps xmm0, XMMWORD PTR [rsp+464] ;137.10
subps xmm1, xmm0 ;137.10
mulps xmm1, XMMWORD PTR [rsp+352] ;137.10
cmpltps xmm6, xmm12 ;137.10
cmpeqps xmm5, xmm8 ;137.10
movaps XMMWORD PTR [rsp+464], xmm0 ;137.10
movaps xmm0, xmm8 ;137.10
movaps xmm1, xmm6 ;137.10
andps xmm1, xmm4 ;137.10
cmpltps xmm0, xmm9 ;137.10
andps xmm2, xmm0 ;137.10
andnps xmm0, xmm4 ;137.10
orps xmm2, xmm0 ;137.10
movaps xmm0, XMMWORD PTR [rsp+208] ;137.10
cmpeqps xmm8, xmm10 ;137.10
orps xmm5, xmm8 ;137.10
movaps xmm8, xmm11 ;137.10
andps xmm8, xmm5 ;137.10
andnps xmm5, xmm0 ;137.10
orps xmm8, xmm5 ;137.10
movdqa xmm5, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm6, xmm8 ;137.10
orps xmm1, xmm6 ;137.10
pand xmm5, xmm15 ;137.10
cvtdq2ps xmm8, xmm5 ;137.10
pand xmm15, xmm14 ;137.10
cvtdq2ps xmm6, xmm15 ;137.10
xorps xmm5, xmm5 ;137.10
xorps xmm15, xmm15 ;137.10
cmpeqps xmm8, xmm5 ;137.10
movaps xmm5, xmm2 ;137.10
andps xmm5, xmm8 ;137.10
subps xmm15, xmm2 ;137.10
andnps xmm8, xmm15 ;137.10
orps xmm5, xmm8 ;137.10
xorps xmm2, xmm2 ;137.10
cmpeqps xmm6, xmm2 ;137.10
movaps xmm8, xmm1 ;137.10
andps xmm8, xmm6 ;137.10
xorps xmm2, xmm2 ;137.10
subps xmm2, xmm1 ;137.10
movdqa xmm1, XMMWORD PTR [rsp+416] ;137.10
andnps xmm6, xmm2 ;137.10
orps xmm8, xmm6 ;137.10
pand xmm1, xmm13 ;137.10
movaps xmm2, xmm7 ;137.10
cvtdq2ps xmm15, xmm1 ;137.10
movaps xmm6, xmm15 ;137.10
movaps xmm8, xmm15 ;137.10
cmpltps xmm6, xmm9 ;137.10
andps xmm2, xmm6 ;137.10
andnps xmm6, xmm4 ;137.10
orps xmm2, xmm6 ;137.10
movaps xmm6, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
cmpltps xmm8, xmm12 ;137.10
andps xmm4, xmm8 ;137.10
cmpeqps xmm6, xmm15 ;137.10
cmpeqps xmm15, xmm10 ;137.10
orps xmm6, xmm15 ;137.10
movaps xmm15, xmm7 ;137.10
andps xmm15, xmm6 ;137.10
andnps xmm6, xmm0 ;137.10
orps xmm15, xmm6 ;137.10
movdqa xmm6, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm8, xmm15 ;137.10
orps xmm4, xmm8 ;137.10
pand xmm6, xmm1 ;137.10
cvtdq2ps xmm8, xmm6 ;137.10
pand xmm1, xmm14 ;137.10
cvtdq2ps xmm6, xmm1 ;137.10
xorps xmm1, xmm1 ;137.10
xorps xmm15, xmm15 ;137.10
cmpeqps xmm8, xmm1 ;137.10
movaps xmm1, xmm2 ;137.10
andps xmm1, xmm8 ;137.10
subps xmm15, xmm2 ;137.10
andnps xmm8, xmm15 ;137.10
orps xmm1, xmm8 ;137.10
xorps xmm2, xmm2 ;137.10
cmpeqps xmm6, xmm2 ;137.10
movaps xmm8, xmm4 ;137.10
andps xmm8, xmm6 ;137.10
xorps xmm2, xmm2 ;137.10
subps xmm2, xmm4 ;137.10
andnps xmm6, xmm2 ;137.10
orps xmm8, xmm6 ;137.10
movaps xmm6, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
movaps xmm4, xmm11 ;137.10
subps xmm1, xmm5 ;137.10
mulps xmm1, xmm3 ;137.10
movdqa xmm1, XMMWORD PTR [rsp+432] ;137.10
movaps XMMWORD PTR [rsp+400], xmm5 ;137.10
pand xmm1, xmm13 ;137.10
movaps xmm5, XMMWORD PTR [rsp+384] ;137.10
cvtdq2ps xmm8, xmm1 ;137.10
movaps xmm2, xmm8 ;137.10
movaps xmm15, xmm8 ;137.10
cmpltps xmm2, xmm9 ;137.10
andps xmm4, xmm2 ;137.10
andnps xmm2, xmm5 ;137.10
orps xmm4, xmm2 ;137.10
cmpltps xmm15, xmm12 ;137.10
movaps xmm2, xmm15 ;137.10
andps xmm2, xmm5 ;137.10
cmpeqps xmm6, xmm8 ;137.10
cmpeqps xmm8, xmm10 ;137.10
orps xmm6, xmm8 ;137.10
andps xmm11, xmm6 ;137.10
andnps xmm6, xmm0 ;137.10
orps xmm11, xmm6 ;137.10
movdqa xmm6, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
andnps xmm15, xmm11 ;137.10
orps xmm2, xmm15 ;137.10
pand xmm6, xmm1 ;137.10
cvtdq2ps xmm6, xmm6 ;137.10
pand xmm1, xmm14 ;137.10
xorps xmm8, xmm8 ;137.10
cvtdq2ps xmm1, xmm1 ;137.10
xorps xmm11, xmm11 ;137.10
cmpeqps xmm6, xmm8 ;137.10
movaps xmm8, xmm4 ;137.10
andps xmm8, xmm6 ;137.10
subps xmm11, xmm4 ;137.10
andnps xmm6, xmm11 ;137.10
orps xmm8, xmm6 ;137.10
xorps xmm4, xmm4 ;137.10
cmpeqps xmm1, xmm4 ;137.10
movaps xmm6, xmm2 ;137.10
andps xmm6, xmm1 ;137.10
xorps xmm4, xmm4 ;137.10
subps xmm4, xmm2 ;137.10
movdqa xmm2, XMMWORD PTR [rsp+448] ;137.10
andnps xmm1, xmm4 ;137.10
orps xmm6, xmm1 ;137.10
pand xmm2, xmm13 ;137.10
movaps xmm1, xmm7 ;137.10
cvtdq2ps xmm6, xmm2 ;137.10
movaps xmm4, xmm6 ;137.10
cmpltps xmm4, xmm9 ;137.10
movaps xmm9, xmm6 ;137.10
andps xmm1, xmm4 ;137.10
andnps xmm4, xmm5 ;137.10
cmpltps xmm9, xmm12 ;137.10
orps xmm1, xmm4 ;137.10
movaps xmm4, XMMWORD PTR _2il0floatpacket\$11(rip) ;137.10
andps xmm5, xmm9 ;137.10
cmpeqps xmm4, xmm6 ;137.10
cmpeqps xmm6, xmm10 ;137.10
orps xmm4, xmm6 ;137.10
andps xmm7, xmm4 ;137.10
andnps xmm4, xmm0 ;137.10
movdqa xmm0, XMMWORD PTR _2il0floatpacket\$13(rip) ;137.10
orps xmm7, xmm4 ;137.10
andnps xmm9, xmm7 ;137.10
orps xmm5, xmm9 ;137.10
pand xmm0, xmm2 ;137.10
cvtdq2ps xmm4, xmm0 ;137.10
pand xmm2, xmm14 ;137.10
xorps xmm0, xmm0 ;137.10
cvtdq2ps xmm2, xmm2 ;137.10
movaps xmm6, xmm1 ;137.10
cmpeqps xmm4, xmm0 ;137.10
andps xmm6, xmm4 ;137.10
xorps xmm0, xmm0 ;137.10
subps xmm0, xmm1 ;137.10
andnps xmm4, xmm0 ;137.10
orps xmm6, xmm4 ;137.10
xorps xmm0, xmm0 ;137.10
cmpeqps xmm2, xmm0 ;137.10
movaps xmm1, xmm5 ;137.10
andps xmm1, xmm2 ;137.10
xorps xmm0, xmm0 ;137.10
subps xmm0, xmm5 ;137.10
movaps xmm5, XMMWORD PTR [rsp+464] ;137.10
andnps xmm2, xmm0 ;137.10
movaps xmm0, XMMWORD PTR [rsp+400] ;137.10
orps xmm1, xmm2 ;137.10
subps xmm6, xmm8 ;137.10
mulps xmm6, xmm3 ;137.10
subps xmm8, xmm0 ;137.10
mulps xmm8, XMMWORD PTR [rsp+352] ;137.10
subps xmm0, xmm5 ;137.10
mulps xmm0, XMMWORD PTR [rsp+224] ;137.10
divps xmm5, XMMWORD PTR _2il0floatpacket\$14(rip) ;137.10
```
So has anyone got ideas or half-baked code to start from? Might even be in C.
Can anyone tell me why is this different or is it?
_________________
My updated idol http://www.agner.org/optimize/
31 Jul 2008, 22:26
tom tobias
Joined: 09 Sep 2003
Posts: 1320
Location: usa
tom tobias
Thank you, Madis, for teaching us about Perlin Noise, a complete novelty for me, at least. I was impressed with the last link's list of potential applications in 1,2,3 & 4 dimensions. I don't know if one could say the code above is "ugly", but it is certainly unintuitive. To address your question about whether or not hugo elias' applications of Perlin Noise are distinctive from Perlin's own applications, why not send them both an inquiry? Since Perlin teaches at NYU, and Elias' web page has not been updated for the past five years, it could well be the case that these guys are not going to be offended by your inquiry--they may even be willing to share their code, or offer some assistance to you, on one of your many interesting projects....
01 Aug 2008, 09:00
Joined: 25 Sep 2003
Posts: 2139
Location: Estonia
tom tobias wrote:
send them both an inquiry?
Yeah, I was planning on doing that exact thing, but I thought people on this board will be less offended and might come to an asm realization of the code quicker. I'm just fishing here for an easy way out, but if there really aren't any helping leads soon, I will contact them about this optimization challenge
Actually I'm quite amazed that there is so little info about this great algorithm. Its variants are used here and here and look at the astonishing images a few bytes are capable of doing here. Perlin is behind all of these!
01 Aug 2008, 10:35
bitRAKE
Joined: 21 Jul 2003
Posts: 3417
Location: vpcmipstrm
bitRAKE
I'd do the whole thing with fixed point fractions [-1,1)=[\$8000,\$7FFF). Or DWORDs, but I don't think they are needed for a few octaves. Something like:
Code:
``` mov ebx,WIDTH - 1
mov edx,0
.0:
mov ecx,OCTAVES - 1
xor edx,ebx ; changed bits from last itteration
.4: bt edx,ecx
jnc .5
; select new random number for octave[ecx]
; reset interpolation for octave[ecx]
.5: dec ecx
jns .4
; interpolate octaves in parallel
; accumulate points (weighted?)
; output value
mov edx,ebx
dec ebx
jns .0 ```
...for the 1D Perlin Noise. The octaves would be in XMM registers. Type of interpolation would determine needed registers - 12 DWORD octaves would look really smooth. EBX is the (X) coordinate and the output is (Y). WIDTH and OCTAVES should be a power of two.
_________________
15 Aug 2008, 03:43
Joined: 25 Sep 2003
Posts: 2139
Location: Estonia
About octaves I've calculated the maximum to achieve is:
Octaves = log(2,Image_side_in_pixels)
so 12 octaves would be an overkill even for 4096x4096px image.
I've used 256x256 as the image and 6 or 7 octaves.
Okay, but atleast we're getting somewhere. With float (as Perlin itself did) you have all the precision you need and I think its not always needed. When you go integers, you start experiencing rounding errors (unless you are really careful). The fixed point would be the way to go and then you can even change the precision on the fly without losing penalty to SSE flt=>int int=>flt conversions.
The ideal I'd like to see is the same number format all through the algorithm. If its not possible, then I have to kill all the latency by hiding them between faster non-dependent instructions.
15 Aug 2008, 07:27
Joined: 25 Sep 2003
Posts: 2139
Location: Estonia
Okay something solid for you. Too much theory already spoken :S
Code:
```format binary as "BMP"
db 0x42, 0x4D, 0x36, 0x00, 0x03, 0x00, 0x00, 0x00, 0x00, 0x00, 0x36, 0x00
db 0x00, 0x00, 0x28, 0x00, 0x00, 0x00, 0x00, 0x01, 0x00, 0x00, 0x00, 0x01
db 0x00, 0x00, 0x01, 0x00, 0x18, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00
db 0x03, 0x00, 0x13, 0x0B, 0x00, 0x00, 0x13, 0x0B, 0x00, 0x00, 0x00, 0x00
db 0x00, 0x00, 0x00, 0x00, 0x00, 0x00
B32 equ and 0FFFFFFFFh ;clip to 32-bit space!!! FASM uses 64-bit
macro RND a
{
x=(a shl 13 xor a) B32
x=((((x*x) B32)*15731) B32+789221)*x+1376312589
a=(x shr 16) and 255
}
repeat 256*256
z=0
rept 7 octave ;FASM doesn't allow 7 to be var instead
{
a=% and (not ((1 shl octave -1)*0101h))
RND a
z=z+a
}
a=z/7
db a,a,a ;R=G=B
end repeat
```
The above code is a sample I'm playing with. You need to manually change the 7 mentioned throughout the code (two places) to change the behaviour.
Description: The sample output and histogram Filesize: 46.52 KB Viewed: 3922 Time(s)
_________________
My updated idol http://www.agner.org/optimize/
15 Aug 2008, 11:25
bitRAKE
Joined: 21 Jul 2003
Posts: 3417
Location: vpcmipstrm
bitRAKE
A little less blocky:
Code:
```repeat 256*256
z=0
rept 7 octave {
a=% and (not ((1 shl octave -1)*0101h))
RND a
a=a shr ((octave-1)/3)
z=z+a
}
a=z/5
db a,a,a ;R=G=B
end repeat ```
Description: Filesize: 28.14 KB Viewed: 3906 Time(s)
_________________
15 Aug 2008, 15:08
Joined: 25 Sep 2003
Posts: 2139
Location: Estonia
Actually the blocks should be removed with interpolation which I haven't done in FASM-macro-assisted generator.
If my calcs are correct this non-blocky code adds ~20% to the time, but since I'm dealing with .1 seconds, I cannot be accurate enough. I will make some more tests and smooth the output.
PS. Sorry for long delays on this topic. I meant to research this theme a lot, but work and vacation intercepted. I will get back to this...
15 Aug 2008, 18:11
tom tobias
Joined: 09 Sep 2003
Posts: 1320
Location: usa
tom tobias
I will get back to this...
15 Aug 2008, 22:14
bitRAKE
Joined: 21 Jul 2003
Posts: 3417
Location: vpcmipstrm
bitRAKE
I had an idea for some Perlin type noise, but it is not so good.
_________________ | 8,827 | 34,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2022-27 | latest | en | 0.947159 |
https://help.scilab.org/docs/6.0.0/ja_JP/sinc.html | 1,717,072,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00660.warc.gz | 252,785,745 | 3,793 | Change language to:
English - Français - Português - Русский
Please note that the recommended version of Scilab is 2024.1.0. This page might be outdated.
See the recommended documentation of this function
# sinc
sinc関数
### 呼び出し手順
`t=sinc(x)`
x
t
### 説明
`x`がベクトルまたは行列の場合, `t=sinc(x)`はベクトルまたは行列となります. ただし, `x(i)~=0`の場合に`t(i)=sin(x(i))/x(i)`, `x(i)==0`の場合に`t(i)=1` です.
### 例
```x=linspace(-10,10,3000);
plot2d(x,sinc(x))```
```[X,Y] = meshgrid(-10:0.25:10,-10:0.25:10);
f = sinc(sqrt((X).^2+(Y).^2));
mesh(X,Y,f);```
### 参照
• sin — 正弦関数
• cos — 余弦関数
Report an issue << sin Trigonometry sind >>
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https://www.clutchprep.com/chemistry/practice-problems/69833/consider-the-following-reaction-no2-g-no-g-1-2o2-g-the-following-data-were-colle | 1,618,583,259,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00399.warc.gz | 810,953,754 | 36,001 | Problem: Consider the following reaction: NO2(g) → NO(g) + 1/2O2(g) The following data were collected for the concentration of NO2 as a function of time:A. What is the average rate of the reaction between 10 and 20 s?B. What is the average rate of the reaction between 50 and 60 s?C. What is the rate of formation of O2 between 50 and 60 s?
FREE Expert Solution
We are being asked to determine the average of the reaction for the reaction:
NO2(g) → NO(g) + 1/2O2(g)
Recall that for a reaction aA bB, the rate of a reaction is given by:
$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{a}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\Delta t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{b}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{B}\mathbf{\right]}}{\mathbf{\Delta t}}}$
where:
Δ[A] = change in concentration of reactants or products (in mol/L or M), [A]final – [A]initial
Δt = change in time, tfinal – tinitial
Since NO2 is a reactant, the rate with respect to NO2 is negative (–) since we’re losing reactants.
100% (424 ratings)
Problem Details
Consider the following reaction: NO2(g) → NO(g) + 1/2O2(g) The following data were collected for the concentration of NO2 as a function of time:
A. What is the average rate of the reaction between 10 and 20 s?
B. What is the average rate of the reaction between 50 and 60 s?
C. What is the rate of formation of O2 between 50 and 60 s? | 459 | 1,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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-2021-17 | latest | en | 0.849202 |
http://www.ck12.org/earth-science/Erosion-by-Wind/rwa/The-Mystery-of-the-Sailing-Stones/r1/ | 1,490,866,582,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218193288.61/warc/CC-MAIN-20170322212953-00657-ip-10-233-31-227.ec2.internal.warc.gz | 482,281,830 | 29,198 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Erosion by Wind
## Wind erodes sediments to form terrain features especially in the desert.
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The Mystery of the Sailing Stones
### The Mystery of the Sailing stones
Credit: Jon Sullivan
Source: http://en.wikipedia.org/wiki/File:Death_8_bg_082303.jpg
License: CC BY-NC 3.0
“Sailing stones” rest at the far end of a smooth track on a dry lakebed. How they get there has baffled observers for at least a century. No one has ever seen the rocks move, but scientists are getting closer to figuring out how they do it.
#### How Do the Rocks Move?
Many observations have been made over the years:
• The rocks have very different sizes. Many are 1 to 2 lbs, many are as large as 20 to 30 lbs. Amazingly, one is estimated to be 700 lbs!
• Stones may move short distances, but one moved more than 600 feet in one winter!
• Stone sliding is seasonal: the rocks do not move in summer, but they do not move every winter.
• The tracks may change directions. Big rocks may move farther than small rocks, but not always.
Over the decades many hypotheses have been proposed, including the following. Do the rocks move
1. by sliding downhill?
2. by people or animals?
3. during an earthquake?
4. from magnetic attraction?
5. by strong winds?
6. in an ice floe?
#### Explore More
Read the following article to learn more about the racing rocks of Racetrack Playa in Death Valley National Park. Then answer the following questions.
1. Hypotheses 1, 2, 3 and 4 above were discounted easily. How would you test each one of these?
2. What must happen for wind to be able to move such large rocks?
3. What is the current hypothesis to explain how the racing rocks move?
4. What technological devices are now being used to study the racing rocks and what do they do?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Please to create your own Highlights / Notes | 551 | 2,266 | {"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-2017-13 | latest | en | 0.921964 |
http://www.maa.org/press/periodicals/mathematics-magazine/mathematics-magazine-february-2008 | 1,496,024,474,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612008.48/warc/CC-MAIN-20170529014619-20170529034619-00113.warc.gz | 714,193,410 | 21,731 | # Mathematics Magazine - February 2008
### ARTICLES
Seth Zimmerman and Chungwu Ho
pp. 3-15
Most of us were shown that equals 2 long before we were troubled by convergence, or even knew what it meant. We might have taken special delight in seeing the golden mean emerge from the snake-like . This paper investigates such infinitely nested radicals, considering general forms like , where a and b are rational. Under what conditions do these converge? Can any given real number be a limit? What patterns do families of these radicals create on the number line? All of these questions are considered here, most of them resolved by proof.
Synthetic Partial Fraction Decompositions
William A. Adkins and Mark Davidson
pp. 16-26
The partial fraction decomposition of a general rational function over the real numbers has been routinely treated in calculus texts, where the procedure is normally taught, via the technique of undetermined coefficients. As has been observed numerous times, there is an alternative algorithmic method for partial fraction decompositions that primarily involves repeated division by polynomials. The goal of this article is to present the partial fraction decomposition algorithm in a format that is amenable to recursive hand calculations in calculus or differential equations classes. These are calculations that are done on the coefficients of the polynomials involved, and hence we refer to the method as synthetic partial fraction decomposition, due to the use of synthetic division by linear and quadratic polynomials. A Brief History of Impossibility
Jeff Suzuki
pp. 27-38
Polynomial Root Squeezing
Matthew Boelkins, Justin From, and Samuel Kolins
pp.39-44
Given a real polynomial with distinct real zeros, the Polynomial Root Dragging Theorem states that if one or more zeros of the polynomial are moved to the right, then all of the critical numbers also move to the right with none of the critical numbers moving as much as the root that is moved most. But what happens if some of the roots of the polynomial are dragged in opposing directions, either toward or away from each other? The Polynomial Root Squeezing Theorem shows that when two zeros of a polynomial are squeezed together, the outermost critical numbers move inward. We then apply the Root Squeezing Theorem to prove results about which polynomials have derivatives with minimum span; that is, the distance from their derivatives’ smallest to greatest zeros is the least possible.
### NOTES
Paint it Black - A Combinatorial Yawp
Arthur T. Benjamin, Jennifer J. Quinn, James A. Sellers, and Mark A. Shattuck
pp. 45-50
We begin by solving a problem posed from the Monthly. Show
. We prove this combinatorially by pairing up objects that have an odd value of r with objects that have an even value of r. The proof, and its generalizations, lead to many new and interesting identities.
Integration by Parts and Infinite Series
Shelby J. Kilmer
pp. 51-54
Sometimes tabular integration by parts doesn’t terminate, and when this happens, an infinite series is formed. Using this method we derive infinite series for the sine, cosine, and exponential functions and several for pi. We also give criteria for the convergence of such a series.
What Fraction of a Soccer Ball is Covered with Pentagons?
P.K. Aravind
pp. 55-57
Many soccer balls have their surface covered with a pattern consisting of twelve pentagons and twenty hexagons. We use elementary geometric methods to calculate the fraction of the area of the soccer ball covered by the pentagons.
Euler’s Triangle Inequality via Proofs Without Words
Roger B. Nelsen
pp. 58-61
Euler’s triangle inquality states that , where R and r denote, respectively, the circumradius and the inradius of a triangle. In this Note we use “proofs without words ” to prove three simple lemmas that can be combined with the arithmetic mean – geometric mean inequality to reduce the proof of Euler’s triangle inequality to simple algebra.
### PROOF WITHOUT WORDS
Double Angle Formula via Area
James D. Currie
pp.62
A triangle labeled two different ways verifies the double angle formula for sines.
The Cauchy–Schwarz Inequality
Sidney Kung
A rectangular area is partitioned two different ways to demonstrate the Cauchy–Schwarz Inequality for vectors in two dimensions.
pp. 63-69
pp. 70
### REVIEW
67th Annual William Lowell Putnam Exam Solutions
pp. 72 | 954 | 4,398 | {"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-2017-22 | latest | en | 0.938122 |
https://gfbrandenburg.wordpress.com/2015/03/19/which-groups-are-over-or-under-represented-in-regular-dc-public-schools-and-the-dc-charter-schools/ | 1,532,366,134,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00479.warc.gz | 649,297,581 | 25,340 | ## Which groups are over- or under-represented in regular DC public schools and the DC charter schools?
I’ve been taking a careful look at where various groups of students in Washington DC are enrolled, using various official DCPS and OSSE documents, to see which groups are under- or over-represented in my city’s regular public schools and charter schools.
Here is the bottom line:
#### Black students are over-represented in the charter schools, but every other single group I calculated (special education students, English-language learners, students eligible for free or reduced-price lunches, Hispanic students, students deemed officially At Risk, white or Asian students, multiracial students and even students who got ‘proficient’ or ‘advanced’ on the DC-CAS) are over-represented in the regular DC public schools.
I will present my results below in the form of circle graphs, and I made it so that the area of each circle is proportional to the total number of students in each group. I also showed where the division line between the public school population and the charter school population would be if the subgroup’s population was split evenly among the two school systems. If you want to take a closer look at any of the graph, merely click on it.
## All Publicly-Funded DC Students
First, the overall student population. As you can see, the regular DC public schools enrolled 57.2% of all publicly-funded students last year, while the charter schools enrolled 42.8%, and the total population was 76,659 students.
## African-American students
Next, let’s look at where Black (African-American) students are enrolled. You may be surprised to find that they are over-represented in the DC charter schools, and under-represented in the regular DC public schools. They are the ONLY group to be divided in that manner.The dotted circle represents the entire population of publicly-funded students; last year, Black students represented 71.6% of that entire enrollment, or 54.894 students. Only 52.7% of those (Black) students were enrolled in the regular DC public schools, which is a smaller percentage than the percentage seen in the orange slice of the pie chart just above this paragraph. And 47.3% of all African-American students in publicly funded schools are in the charter sector, which is a larger percentage than the percentage shown in the green slice in the previous pie chart.
## Students in Poverty
Next, let’s look at students who are eligible for free or reduced-price lunch, but keep in mind that this statistic is no longer very accurate, since for the last two years, quite a few schools have been permitted to designate every single one of their students as eligible, no matter what the family income might be, if the school as a whole fits various conditions. For what it’s worth, here is the graph:
As you can see, 70.8% of all students in regular public schools and in the charter sector (or 54,270 students) are eligible for free or reduced-price lunches, which is a measure of family poverty. Of those students, 60.4% are in the regular public schools, which means that students in poverty are somewhat over-represented in the regular DC public schools.
## Special Education Students
Here, the green-and-orange circle is fairly small, because according to the official statistics, only about one student in eight (or 12.5%, or 9,592) is officially considered to be in special education. Once again, the outer dotted circle shows the size of the entire DC publicly-funded student population, both public and charter, and you can see that special education students are over-represented in the regular public school sector and under-represented in the charter school sector.
## Students Learning English as a Second Language
Here we are looking at students who are learning English as a second language (the acronyms for this status change nearly every year – ESL, ESOL, ELL, etc). Not all of them are of Hispanic origin! There are probably over a hundred different languages spoken by the parents and families of students enrolled in DC’s public and charter schools.Once again, this group of English-language-learners is over-represented in the regular DC public school sector; as a whole, they make up about 9.2% of the entire student population.
## Hispanic Students
Here we are looking at the entire population of Hispanic students (quite a few of whom are fully fluent in English!). Once again, this subgroup is over-represented in the DC public school system and under-represented in the charter school sector. They compose about one student in seven of the grand total (14.4%), and of that group, 65.0% are enrolled in the regular DC public schools.As a reminder, the percentage of students enrolled in the regular DC public school system is only about 57%, so this is about an eight percentage-point gap.
## At-Risk Students
You may recall that this is a fairly new, official subgroup of students, comprising students who are homeless, are over-age, on welfare or food stamps, and various other indications of risk. I have heard from some charter advocates that these numbers are not always accurate. I don’t have access to school-level official records; I only have what DCPS and the charter sector provide to OSSE and then OSSE disseminates.
Notice here that the dotted line for equal division is very close to the dark line separating the orange wedge (regular public schools) from the green wedge (charter schools). So while this group of students is over-represented in the regular public schools, it’s not by a lot (1.4 percentage points). Note that At-Risk students make up nearly half (45.2%) of the entire DC publicly-funded student body!
## White or Asian Students
You may wonder why I combined the populations of white (caucasian) and Asian students. Simple answer: both groups are pretty small; in fact, many schools have none at all of either group. In cases where there were some students enrolled, but there were too few for them to be enumerated, that meant there were fewer than 10 students. I arbitrarily decided to make that number five (5) at each such school. I also omitted Native American students altogether because their numbers were so tiny you couldn’t even see the circle if I were to draw one.
Unlike in some cities, where white students are self-segregating themselves into charter schools, DC’s self-segregation pattern is different in that about half-a-dozen elementary schools in upper Northwest are located in overwhelmingly affluent and predominantly white neighborhoods. For various reasons, white families there are sending their students to those local public schools (and to Deal and Wilson) in greater and greater numbers, so the proportions of white students at those schools has been pretty steadily increasing for the past 10 years or so. By contrast, there is only a very small number of charter schools that have any significant number of white or asian students.
As a result of these historical patterns, you can see that white and asian students are very strongly over-represented in the regular DC public school sector: 78.8%. White or asian students make up about 10% of all publicly-funded students in DC.
## Multi-Racial Students
This is another relatively small group, comprising less than two percent of all publicly-funded preK-12 students in DC. It is strongly over-represented in the regular DC public school sector.
## Students Who Scored ‘Proficient’ or ‘Advanced’ on the DC-CAS
You may be surprised to discover that students who “passed” the DC-CAS in 2014, that is, who were marked ‘proficient’ or ‘advanced’ on that test, are slightly over-represented in the regular DC public school sector. Caveat: in my calculations, I averaged the percentages of those ‘passing’ in reading and those ‘passing’ in math, to come up with a single number for each school. I then calculated how many students that was at each school who ‘passed’, and then added all those totals together, for each school in each sector, and then compared those totals to the entire population in that sector. (Not by hand! I used Excel!)
So there you have it: for all of the subgroups I calculated, every single one was somewhat or slightly or strongly over-represented in the regular DC public school sector, except for the total numbers of African-American students.
Published in: on March 19, 2015 at 3:30 pm Comments (8)
1. Is there any way to get data on retention/dismissal rates?
Like
2. Do you mean holding students back in a grade, and/or suspension rates?
Like
3. LearnDC has data on “mobility” between schools. It doesn’t get into expulsions vs. voluntary transfers, but there is lots of anecdotal information about how schools “counsel out” kids and call them voluntary transfers but they are expulsions in all but name.
Like
• I’m playing with that mobility data right now. The vast majority of the charters admit nobody at all after school starts; they all lose students, anywhere from 1% to 20%. The regular public schools on the average gain about as many as they lose.
In addition, the official suspension rates at the charter schools are much higher than the official suspension rates at the regular public schools.
Like
4. Guy, if you control for the very affluent and very white ward 3 schools, what happens to this information. Charters are located generally in poorer parts of the city due to cost of real estate and how does that skew the data.
It would also be interesting to see the racial and poverty characteristics of students who move school mid year. Doubt OSSE will share the data on that, but you can be surprised.
Like
• I’m positive that the kids who move from one school to another mid-year are predominantly poor, at risk, and black or hispanic. I don’t see how to show that via data though.
Like
5. You can find charter school data at data.dcpcsb.org.
Like | 2,105 | 9,910 | {"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-2018-30 | latest | en | 0.96073 |
https://www.cradle-cfd.com/media/column/a100 | 1,721,400,275,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00740.warc.gz | 624,749,298 | 7,625 | # Case Study – Learn about Thermo-Fluid Analyses Optimization No. 10: Optimal shape of a louver (2)
#### Optimal shape of a louver (2)
For the optimal design for a louver, the previous column explained about pressure loss and the relation among the length, angle, and pitch of the louver. This time, we will create a model and set conditions.
We perform a two-dimensional flow analysis assuming the flow in the louver as two-dimensional. For the height direction, we create a one-pitch model and set periodic boundary conditions because the flow behavior repeats per pitch. In addition, we set a flow path of 1,000 mm before and after the louver. So, the model will be a band-like plate as shown in Figure 2.1. Figure 2.2 is a zoomed part of the louver. Then, we set a stationary wall condition on the portion corresponding to the louver surface, and set a free slip wall condition on other surfaces. For the surfaces shown in the figure with the caption “Periodic boundary condition”, we set periodic boundary conditions so that the number 11 and 12, and the number 21 and 22 will be the pairs, respectively. We set the velocity 5 m/s as the fixed velocity for the inlet, and set 0 Pa as the fixed static pressure for the outlet.
Figure 2.1 CFD model of a louver
Figure 2.2 Zoomed louver
The number of design variables is two: the length and angle of the louver slat; therefore, the number of samples is defined as (2+1)×(2+2)=12. Here, let us set the parameters in EOopti as shown in Figure 2.2: Louver length 20 – 100 mm, louver angle 15 – 60 degrees.
Figure 2.3 Condition setting in EOopti
We prepare design of experiments, create a model based on the planned experimental conditions, and obtain pressure loss using SC/Tetra. Next, we obtain loss coefficient ζ and length per pitch L/t. Here, the length per pitch is 1/sinθ where θ is the louver angle. Loss coefficient ζ is derived by the following equation:
In the equation above, Pinlet is the pressure at the inlet, from the following factors: Density of air 1.206 kg/m3 , inflow velocity 5 m/s.
Figure 2.4 and 2.5 are examples of an analysis result output from SC/Tetra.
Figure 2.4 and 2.5 show velocity distribution and static pressure distribution, respectively, with the condition of 50 mm for the louver length and 30 degrees for the louver angle. The calculation was performed for one pitch; however, the figures show the results for several pitches using the periodic copy function of Postprocessor.
In Figure 2.4, we can see that the flow direction varies along the louver, the flow separates at the edge of the louver, and the velocity increases between the louvers. In Figure 2.5, we can see that, at the edge of the louver, static pressure increases where the flow meets, and decreases where the flow separates. Between the louvers, static pressure decreases as velocity increases, then, static pressure increases. In addition, we can see that static pressure distribution is almost even at the outlet of the louver.
Figure 2.4 Velocity distribution
Figure 2.5 Static pressure distribution
In the next column, we will see the exploring results of optimal solution.
[Reference] Machine Engineering Handbook – Fluid engineering (in Japanese)
User's Guide Optimization (Option)
Professor Gaku Minorikawa | Faculty of Science and Engineering,
Department of Mechanical Engineering, Hosei University
Certified environmental measurer (noise and vibration)
• 1992 Joined EBARA CORPORATION
• 1999 Became an assistant at Hosei University Faculty of Engineering
• 2001 Obtained Doctor of Engineering at Tokyo Institute of Technology
• 2004 Became Assistant Professor at Hosei University Faculty of Engineering
• 2010 Became Professor at Hosei University Faculty of Science and Engineering | 871 | 3,761 | {"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-2024-30 | latest | en | 0.882109 |
http://www.eduplace.com/math/mathsteps/5/b/5.primefact.ideas.html | 1,542,443,341,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743351.61/warc/CC-MAIN-20181117082141-20181117104141-00244.warc.gz | 424,232,074 | 3,188 | ## Prime Factors
Making sure your students' work is neat and orderly will help prevent them from losing factors when constructing factor trees. Have them check their prime factorizations by multiplying the factors to see if they get the original number.
Prerequisite Skills and Concepts: Students will need to know and be able to use exponents. They also will find it helpful to know the rules of divisibility for 2, 3, 4, 5, 9 and 10.
Write the number 48 on the board.
• Ask: Who can give me two numbers whose product is 48?
Students should identify pairs of numbers like 6 and 8, 4 and 12, or 3 and 16. Take one of the pairs of factors and create a factor tree for the prime factorization of 48 on the board or on an overhead transparency as shown below.
• Ask: How many factors of two are there? (4)
How do I express that using an exponent?
Students should say to write it as "24" If they don't, remind them that the exponent tells how many times the base is taken as a factor. Finish writing the prime factorization on the board as .
• Say: When we write a composite number as the product of prime numbers, we have written the prime factorization for the number. In this case, the prime factorization of 48 is .
Next, find the prime factorization for 48 using a different set of factors.
• Ask: What do you notice about the prime factorization of 48 for this set of factors?
Students should notice that the prime factorization of 48 = for both of them.
• Say: There is a theorem in mathematics that says when we factor a number into a product of prime numbers, it can only be done one way, not counting the order of the factors
Illustrate what is meant by that by showing them 12 = or 12 = .
• Say: Now let's try one on your own. Find the prime factorization of 60 by creating a factor tree for 60
Have someone come to the board and show how to find the prime factorization of 60. | 446 | 1,889 | {"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} | 4.75 | 5 | CC-MAIN-2018-47 | latest | en | 0.954289 |
https://www.slideserve.com/delphine/seismomath | 1,544,571,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823705.4/warc/CC-MAIN-20181211215732-20181212001232-00065.warc.gz | 1,016,944,667 | 13,188 | SeismoMath!
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# SeismoMath! - PowerPoint PPT Presentation
SeismoMath!. Math Colloquium #7 Nancy Ikeda April 13, 2010. Problem. Q: How can earthquake forecasting models be tested? Most often, researchers have to just wait to see if their predicted earthquake occurs. Solution. A: Use a Monte Carlo simulation
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### SeismoMath!
Math Colloquium #7
Nancy Ikeda
April 13, 2010
Problem
• Q: How can earthquake forecasting models be tested?
• Most often, researchers have to just wait to see if their predicted earthquake occurs
Solution
• A: Use a Monte Carlo simulation
• Create a realistic synthetic earthquake catalog to test the forecasting method
• The synthetic catalog represents the null hypothesis: there is nothing “causing” the earthquakes and they occur “randomly”
Synthetic EQ Project
• Collaborated with Dr. David Bowman, Chair, Dept. of Geological Sciences at CSUF (my advisor)
• Worked in “The Field” at CSUF
• Used Macintosh computers
• Programmed in IDL
Earthquake Catalog
• Location
• Magnitude
• Time
• Depth
Location
• Must be in the form of latitude and longitude
• Should be located near a tectonic plate boundary
• Aftershocks should be located near the mainshock
Location
http://mineralsciences.si.edu/tdpmap/
Location
Actual Locations
CA Region, 1980-2000
Synthetic Locations
CA Region, 1980-2010
Location
• Probability map
• 5 km x 5 km cells
• Find the number of EQ in each cell with some aftershocks removed (declustered)
• Use random number generator to select a cell
• Randomly offset the earthquake from the center of the cell
Location
• Aftershocks are located near the mainshock
• They are placed a random distance and direction from the mainshock
• Distance is based on mainshock magnitude
Location
Background only
BG + Aftershocks
Magnitude
• Gutenberg-Richter (GR) Law
N = 10a - bm or log N = a - bm
Number of events (cumulative)
Global Catalog 1984 - 2003
Magnitude
Magnitude
• tapered GR distribution
[Kagan and Jackson, 2000; Kagan, 2002]
• has an exponential taper applied to the cumulative number of events (for higher magnitudes)
Magnitude
Used Felzer et al’s [2002] inverse transform technique to generate a random magnitude:
Since Kagan’s formula is in the form of a cumulative distribution, it follows that it will take on values between 0 and 1.
Magnitude
To generate a magnitude from a random number r, we must solve this equation for m.
But how?!?!
Magnitude
• Use the Lambert W function, W(x)
• It is the inverse of the function
f(x) = x·ex
Thus, for x = yey, then y = W(x)
Magnitude
Now, with , if x = yey, then y = W(x):
Magnitude
• Halley’s Method was used (similar to Newton’s Method)
• Forx ≥ e, W(x) can be approximated by ln x – ln(ln x)
• For x < e, an approximation of the function for argument values near 0 had to be found
Magnitude
• fit a quartic curve to the Lambert W function
• y = -0.0285x4 + 0.1892x3 – 0.508x2 + 0.9138x
• R2 = 0.99995
• 5 iterations
• Then plug into magnitude formula
Time
• Earthquakes occur randomly in time
• Aftershocks occur after large EQs
• Aftershocks decay over time
California, 1980-2000
Time
• Epidemic-Type Aftershock Sequence (ETAS) model
Total Eqs in CA
M ≥ 3
Time
• To use the formula, time and magnitude have to be plugged in
• All of the parameters had to be approximated also: K, , c, p,
Time
• An estimate formwas calculated
• Tried to fit the other parameters
• K = [0.04, 0.09]
• a = [0.4, 0.8]
• C = 0.02 (about 30 minutes)
• P = [1.5, 1.75]
• Picked parameter values for a region
• Each aftershock sequence has a new set of parameters based on selected regional parameters
Time
Time vs Magnitude
For background EQs
Time vs Magnitude
For All Synthetic EQs
Depth
• found the average depth of events for a region
• And the average depth of events in the 5 km x 5 km cell
• Assigned events a depth based on the cell average, following a normal distribution
• If a cell had no previous events, it was assigned the average depth for the region
Running the Program
• Load in file for real data (ANSS)
• 1984 - 2003
• Minimum magnitude = 3.5
• Depth = 40 km
• Load in region boundary data (including ETAS parameters)
• Select earthquakes from a region
• Estimate m
• Create location probability map
Running the Program
• Create background earthquakes
• New m is generated for each year
• Use poissonian distribution for day of event
• Assign random time on day
• Assign location based on a-value map
• Assign magnitude
• Run ETAS on each background event
• New ETAS parameters are generated for each background event
• ETAS parameters are fixed for each aftershock sequence
• Run daily to determine number of aftershocks per day
• Assign aftershocks a time, location and magnitude
Running the Program
• Run ETAS on all aftershocks individually
• New set of parameters are used again
• This continues until the end of the catalog
• Index the events
• Create final catalog
• Originally 40 years
• Cut out the first 10 years
• Cut out any events that happened after 40 years
• Write events to a file
Global Synthetic Catalog
Magnitude Distributions
Real Catalog
1984 - 2003
Synthetic Catalog
1980 - 2010
Global Synthetic Catalog
Time vs Magnitude
Real Catalog
Synthetic Catalog
What’s left/next?
• Use synthetic catalog to test the accelerating moment release (AMR) method
• Write a paper on the use of the Lambert W function for generating magnitudes
• Find even more realistic formulas and start over using Matlab (instead of IDL)
References
Corless, R.M., Gonnet, G. H., Hare, D.E.G., Jeffrey, D. J., and D.E. Knuth, On the Lambert W Function, Advances in Computational Mathematics, vol. 5, p. 329-359, 1996.
Felzer, K.R., Becker, T. W., Abercrombie, R. E., Ekstrom, G., and J. R. Rice, Triggering of the 1999 Mw 7.1 Hector Mine earthquake by aftershocks of the 1992 Mw 7.3 Landers earthquake, JGR, v. 107, B9, 2190, 2002.
Helmstetter, A., and D. Sornette, Sub-critical and Super-critical Regimes in Epidemic Models of Earthquake Aftershocks, JGR, 107, B10, 2237, 2002.
http://mathworld.wolfram.com/LambertW-Function.html
http://mineralsciences.si.edu/tdpmap/
Kagan, Y. Y., Universality of the Seismic Moment-frequency Relation, Pure and Applied Geophysics, 155, p. 537-573, 1999.
Kagan, Y. Y., and D. D. Jackson, Probabilistic earthquake forecasting, GJI, v. 143, p. 438-453, 2000.
Kagan, Y. Y., Seismic moment distribution revisited: I. Statistical results, GJI, v. 148, p. 520-541, 2002.
Ogata, Y., Seismicity Analysis through Point-process Modeling: A Review, Pure and Applied Geophysics, 155, p. 471-507, 1999.
Ogata, Y., and J. Zhuang, Space-time ETAS models and an improved extension, Tectonophysics, 413, p. 13-23, 2006. | 2,028 | 7,297 | {"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-2018-51 | latest | en | 0.843683 |
https://search.r-project.org/CRAN/refmans/concstats/html/concstats_concstats.html | 1,701,886,939,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100602.36/warc/CC-MAIN-20231206162528-20231206192528-00474.warc.gz | 566,770,659 | 2,237 | concstats_concstats {concstats} R Documentation
## A set of Market Structure, Concentration, and Inequality Measures
### Description
A convenience function which calculates a selected set of different market structure, inequality and concentration measures more or less commonly used, e.g. k-firm ratios, Entropy, HHI, Palma ratio, and others in a one step procedure to provide a first overview.
### Usage
concstats_concstats(x, na.rm = TRUE, digits = NULL)
### Arguments
x A non-negative numeric vector. na.rm A logical vector that indicates whether NA values should be excluded or not. Must be either TRUE or FALSE. Defaults to TRUE. If set to FALSE the computation yields NA if vector contains NA values. digits A non-null value for digits specifies the minimum number of significant digits to be printed in values. The default is NULL and will use base R print option. Significant digits defaults to 7.
### Details
concstats_concstats computes a set of different and selected structural, inequality, and concentration measures in a one step procedure. The resulting data frame contains eight measures: number of firms with market share, numbers equivalent, the cumulative share of the top (top 3 and top 5) firm(s) in percentage, the hhi index, the entropy index, and the palma ratio. However, all measures can be computed individually or in groups.
### Value
A data frame of numeric measures with default settings.
### Note
The vector of market shares should be in a decimal form corresponding to the total share of individual firms/units. The vector should sum up to 1.
### See Also
concstats_mstruct(), concstats_comp(), concstats_inequ()
### Examples
# a vector of market shares
x <- c(0.35, 0.4, 0.05, 0.1, 0.06, 0.04)
# a selected set of different structural, concentration, and inequality
# measures
concstats_concstats(x, digits = 2)
[Package concstats version 0.1.6 Index] | 446 | 1,911 | {"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} | 2.515625 | 3 | CC-MAIN-2023-50 | latest | en | 0.705517 |
https://lists.osgeo.org/pipermail/postgis-users/2014-July/039399.html | 1,713,057,003,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00785.warc.gz | 335,857,453 | 3,987 | # [postgis-users] Finding records withing XX distance from a geometry in the same table
Alexander W. Rolek a.rolek at gmail.com
Thu Jul 10 09:01:01 PDT 2014
```The 2 step query is an interesting idea. Would I need to do that in 2
queries, or can that be accomplished in a single statement?
On Thu, Jul 10, 2014 at 8:58 AM, Rémi Cura <remi.cura at gmail.com> wrote:
> Hey,
> I'm glad you found out.
> I never tried but you could probably try to index on the result of the
> cast :
> CREATE INDEX ON ... USING GIST ((CAST geom AS GEOGRAPHY))
> .
>
> However if your distance are not that big (compaired to earth curve),
> stick to geometry.
> If your distances are big, go geography.
> You could also simply do it in 2 steps, a first step with geometry and
> maybe 2 times the distance to be sure on all the table, a second step more
> precise with geography on only the result of the previous filter.
>
> Cheers,
> Rémi-C
>
>
>
> 2014-07-10 17:47 GMT+02:00 Alexander W. Rolek <a.rolek at gmail.com>:
>
> Remi -
>>
>> Thanks again for the quick response. I dug into this a bit more last
>> night, and apparently the issue was the distance I was using with the
>> ST_DWithin call. When I had a projection under 2230, the unit of measure
>> was feet. With 4326 it appears to be degrees. I kept the distance of 100 on
>> both tests, so the 2230 project returned quick, but a distance of 100 on a
>> 4326 projection is very large so it was returning my whole table. When I
>> changed the distance to something small (0.0009) my query is now fast. ;-)
>>
>> My next challenge will be figuring out the best way to convert between
>> degrees & meters. I understand the geography type is better for queries
>> like this, but I'm already using the geometry type for other queries. Is it
>> common to have both geometry and geography for a record in a table? I know
>> I can cast using ::geography but the queries are dramatically slower.
>>
>> Alex
>>
>>
>> On Thu, Jul 10, 2014 at 1:11 AM, Rémi Cura <remi.cura at gmail.com> wrote:
>>
>>> Hey,
>>> If I take into account your last mail,
>>> you probably have forget to transform your data.
>>>
>>> If you have a table in srid 4326
>>> , you can't use a transform in your querry if you want it to uses
>>> indexes.
>>> OR you have to buil and index like this :
>>> CREATE INDEX ON sdgis.parcels4326 USING GIST (ST_Transform(geom, 4326));
>>>
>>> So can you confirm thatl your geom have the correct srid in the correct
>>> table?
>>>
>>> SELECT DISTINCT ST_SRID(geom)
>>> FROM sdgis.parcels ;
>>>
>>> and
>>> SELECT DISTINCT ST_SRID(geom)
>>> FROM sdgis.parcels4326 ;
>>>
>>> , run it with "explain analyse" (Select the text of your querry, then
>>> press shift+F7).
>>> You should see a graphic explaining what it happening. You should look
>>> for sequential scans.
>>>
>>> You can also try the same querry without CTE :
>>>
>>> SELECT p.*
>>> FROM (
>>> SELECT geom
>>> FROM sdgis.parcels4326 AS parcels
>>> WHERE apn = '3500600300
>>> ) AS mip
>>> INNER JOIN sdgis.parcels4326 AS p ON
>>> (ST_DWITHIN(mip.geom,p.geom,100)=TRUE);
>>>
>>>
>>> Of course vacuum analyze both table before running the querry.
>>> It would be difficult to help you more without detailed information
>>> (tables declarations, querry used, result of explain analyse), because
>>> slowness can come from several reasons .
>>>
>>> Here is a link of requirements for people having slow querry on postgres
>>> mailing list (https://wiki.postgresql.org/wiki/Slow_Query_Questions).
>>> I'm not guru so I couldn't use all this informations but that give you
>>> an idea of how many reasons of slow query they can be
>>>
>>> Cheers,
>>> Rémi-C
>>>
>>>
>>>
>>>
>>> ""
>>> I tried the query that Remi suggested (thank you!) and it works and
>>> doesn't work. I have the exact same table in two different projections
>>> (2230, 4326). When I run the ST_Dwithin query against the 2230 table, I get
>>> results in under 100 ms. When I run the same query against the 4326
>>> projection it still takes around 270 seconds! I have confirmed that I have
>>> the gist index on the 4326 table. Any ideas why the 4326 projection would
>>> be dramatically slower?
>>>
>>> ""
>>>
>>>
>>> 2014-07-09 18:38 GMT+02:00 Alexander W. Rolek <a.rolek at gmail.com>:
>>>
>>> Remi -
>>>>
>>>> Thanks for the quick response. Sorry to respond to you directly, but my
>>>> message settings are set on digest so I can't respond to the thread yet.
>>>>
>>>> I went through your steps, and have built out my query, but it's still
>>>> taking around 270 seconds to run the query, and it's returning ever record
>>>> in my table. Here's my query:
>>>>
>>>> WITH my_input_polygon AS (
>>>> SELECT geom
>>>> FROM sdgis.parcels4326 AS parcels
>>>> WHERE apn = '3500600300'
>>>> )
>>>> SELECT p.*
>>>> FROM my_input_polygon AS mip
>>>> INNER JOIN sdgis.parcels4326 AS p ON
>>>> (ST_DWITHIN(mip.geom,p.geom,100)=TRUE);
>>>>
>>>> I also have indexes on the apn and the geom columns (see attached
>>>> screen shots)
>>>>
>>>> Any ideas what I'm missing here?
>>>>
>>>> Thanks again for the help
>>>>
>>>> --
>>>> Alexander W. Rolek
>>>>
>>>
>>>
>>
>>
>> --
>> Alexander W. Rolek
>> 303-829-9989
>>
>
>
--
Alexander W. Rolek
303-829-9989
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``` | 1,477 | 5,394 | {"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-2024-18 | latest | en | 0.935419 |
https://www.enotes.com/homework-help/there-any-value-x-that-satisfies-x-2-3x-2-gt-0-4x-451073 | 1,498,344,340,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320362.97/warc/CC-MAIN-20170624221310-20170625001310-00590.warc.gz | 854,178,028 | 12,504 | # Is there any value of x that satisfies x^2 + 3x + 2 >= 0 and 4x^2 + 5x + 1 >= 0
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
The values of x that satisfy `x^2 + 3x + 2 >= 0` and `4x^2 + 5x + 1 >= 0` have to be determined.
`x^2 + 3x + 2 >= 0`
=> `x^2 + 2x + x + 2 >= 0`
=> `x(x + 2) +1(x + 2) >= 0`
=> `(x + 1)(x + 2) >= 0`
This is the case when:
• `x + 1 >= 0` and `x + 2 >= 0`
=> `x >= -1` and `x >= -2`
=> `x >= -1`
• `x + 1 <= 0` and `x + 2 <= 0`
=> `x <= -1` and `x <= -2`
=> `x <= -2`
The solution for this inequality is `(-oo, -2]U[-1, oo)`
`4x^2 + 5x + 1 >= 0`
=> `4x^2 + 4x + x + 1 >= 0`
=> `4x(x + 1) + 1(x + 1) >= 0`
=> `(4x + 1)(x + 1) >= 0`
This is true when
• `4x + 1 >= 0` and `x + 1 >= 0`
=> `x >= -1/4` and `x >= -1`
=> `x >= -1/4`
• `4x + 1 <= 0` and `x + 1 <= 0`
=> `x <= -1/4` and `x <= -1`
=> `x <= -1`
The solution of this inequality is `(-oo, -1]U[-1/4, oo)`
The intersection of `(-oo, -2]U[-1, oo)` and `(-oo, -1]U[-1/4, oo)` is `(-oo, -2]U[-1/4, oo)`
The values of x that satisfy both the inequalities lie in `(-oo, -2]U[-1/4, oo)` | 544 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2017-26 | longest | en | 0.670298 |
https://pick3master333.com/2013/08/page/3/ | 1,586,159,462,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00247.warc.gz | 565,325,863 | 20,705 | # Archive | August 2013
You are browsing the site archives by date.
## 6 Consecutive Doubles Hold the Key to C100 Success
I am now thinking that the absolute best time to play the C100 numbers is right after you see 6 back to back doubles. This does not always happen in any one particular state but you can go to us-lotteries.com and look for this in other states. Just look around for 6 consecutive doubles. Once you […]
## I Hit 532 Box Last Night Using the C100 System
532 hit last night in New York. I don’t usually post my winnings, but I think this time it’s appropriate so you can see just how powerful the C100 system is. I took my own advice and after seeing those 6 back to back doubles (6D), I decided to play. I played all 100 C100 […]
## 6D in New York: Expect a Winning Cycle Next
Today is August 20, 2013. New York just came out of a very clear cut 6 draw losing cycle starting on August 16 with 884 and ending on August 19 with 991. I say “clear cut” because these 6 Non-C100 number were all doubles. Let’s call this losing cycle 6D (6 back to back doubles): 991 […]
## The NEW Updated C100 System
The NEW Updated C100 System: Available Tomorrow August 20! *Now you need less money to get started. I give you the exact amount on the report. *The 3 Cycles in the Game (NEW! Use this secret knowledge to know when to play and when not to play). *How to Set up Your Own Pick 3 […]
## \$225 Profit for July in NY
See the Excel document below showing a profit of \$225 in New York in July 2013 using the C100 system. Click on the link to open the document. \$225 in NY-JULY By keeping a +1 and -1 count for each win and loss, you end up with a +18 count. Now, each time you win […]
## 2 Discoveries from the C100 System
To all of you who did not purchase my C100 system; the numbers with a yellow highlight on the Excel sheet below are hits from my C100 system. This system is based on playing 100 numbers box at 5dimes for 0.25 cents (\$25). The gray numbers are numbers not on the list, therefore loses. As you […]
## Two C100 Points
Two Points: 1. Each time you lose \$25, you have to win twice (\$12.50 x 2 = \$25) to make up for that one loss. Meaning that the wins must outnumber the losses by at least 3x to see a profit. This realization is a step in the right direction. 2. Avoid playing when you […] | 613 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-16 | latest | en | 0.904233 |
https://www.teachoo.com/4723/739/Ex-9.5--2---Show-homogeneous--y--=-x-y---x/category/Ex-9.5/ | 1,558,508,458,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256764.75/warc/CC-MAIN-20190522063112-20190522085112-00498.warc.gz | 965,733,311 | 10,360 | 1. Chapter 9 Class 12 Differential Equations
2. Serial order wise
3. Ex 9.5
Transcript
Ex 9.5, 2 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. ๐ฆ^โฒ=(๐ฅ+๐ฆ)/๐ฅ Step 1: Find ๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = (๐ฅ + ๐ฆ)/๐ฅ Step 2. Putting F(x, y) = ๐๐ฆ/๐๐ฅ and find F(๐x, ๐y) So, F(x, y) = (๐ฅ + ๐ฆ)/๐ฅ F(๐x, ๐y) = (๐๐ฅ +๐๐ฆ)/๐๐ฅ = (๐(๐ฅ +๐ฆ))/๐๐ฅ = (๐ฅ + ๐ฆ)/๐ฅ = F(x, y) = ๐ยฐF(x, y) Therefore F(x, y) Is a homogenous function of degree zero. Hence ๐๐ฆ/๐๐ฅ is a homogenous differential equation Step 3: Solving ๐๐ฆ/๐๐ฅ by putting y = vx Put y = vx. differentiating w.r.t.x ๐๐ฆ/๐๐ฅ = x ๐๐ฃ/๐๐ฅ+๐ฃ๐๐ฅ/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ + v Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐๐ฆ/๐๐ฅ = (๐ฅ + ๐ฆ)/๐ฅ ๐ฅ ( ๐๐ฃ)/๐๐ฅ + v = (๐ฅ + ๐ฃ๐ฅ)/๐ฅ ๐ฅ ( ๐๐ฃ)/๐๐ฅ + v = 1+๐ฃ ๐ฅ (๐ฅ ๐๐ฃ)/๐๐ฅ = 1+๐ฃโ๐ฃ ๐ฅ ( ๐๐ฃ)/๐๐ฅ = 1 ( ๐๐ฃ)/๐๐ฅ = 1/๐ฅ Integrating both sides โซ1โใ๐๐ฃ=โซ1โใ๐๐ฅ/๐ฅ ใ ใ v = log|๐ฅ|+๐ Putting v = ๐ฆ/๐ฅ ๐ฆ/๐ฅ = log|๐ฅ| + c y = x log|๐| + cx
Ex 9.5 | 925 | 1,316 | {"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-2019-22 | latest | en | 0.640318 |
https://www.mastguru.com/a-person-incurs-a-loss-of-5-be-selling-a-watch-for-rs-1140-at-what-price-should-the-wat/822 | 1,582,656,537,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00082.warc.gz | 812,672,803 | 11,191 | # A person incurs a loss of 5% be selling a watch for Rs. 1140. At what price should the watch be sold to earn 5% profit.
• Rs.1200
• Rs.1230
• Rs.1260
• Rs.1290
Similar Questions :
## 1. 100 oranges are bought at the rate of Rs. 350 and sold at the rate of 48 per dozen. The percentage of profit is
• \begin{aligned} 12\frac{2}{7} \% \end{aligned}
• \begin{aligned} 13\frac{2}{7} \% \end{aligned}
• \begin{aligned} 14\frac{2}{7} \%\end{aligned}
• \begin{aligned} 15\frac{2}{7} \% \end{aligned}
## 2. In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit
• 70%
• 80%
• 90%
• None of above
• Rs. 70
• Rs. 72
• Rs. 74
• Rs. 76
• 51:52
• 52:53
• 53:54
• 54:55
## 5. A material is purchased for Rs. 600. If one fourth of the material is sold at a loss of 20% and the remaining at a gain of 10%, Find out the overall gain or loss percentage
• \begin{aligned} 4\frac{1}{2} \end{aligned}
• \begin{aligned} 3\frac{1}{2} \end{aligned}
• \begin{aligned} 2\frac{1}{2} \end{aligned}
• \begin{aligned} 1\frac{1}{2} \end{aligned}
Read more from - Profit and Loss Questions Answers | 430 | 1,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-10 | latest | en | 0.794768 |
http://a1vbcode.com/snippet-4011.asp | 1,679,867,258,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00094.warc.gz | 1,358,325 | 4,235 | Find Code: All Words Any of the Words Exact Phrase Home : Code : Forums : Submit : Mailing List : About : Contact
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Secant method to approximate root of a function Author: BasuDip Website: http://www.geocities.com/basudip_in/vb6code/ Submitted: 12/24/2006 Version: VB6 Compatibility: VB6 Category: Mathematics Views: 16353 Compute a root of a given equation (function of 1 variable) using Secant method. This is my second program on Methods of Numerical Analysis. Declarations: 'none Code: Option Explicit Dim Xp As Single, Xn As Single Private Sub Form_Load() Xp = 1: Xn = 2 ' Initial Values End Sub Rem Modify here to type in the equation and set the initial values ****** Private Function myFunc(ByVal X As Single) As Single myFunc = X * X - Sqr(X) - 2 ' Let F(x)=x*x-x^0.5-2 End Function Private Function SecantRootX(ByVal X0 As Single, _ ByVal X1 As Single) As Single Dim F0 As Single, F1 As Single If X0 = X1 Then SecantRootX = X0 Exit Function End If F0 = myFunc(X0) F1 = myFunc(X1) SecantRootX = (X0 * F1 - X1 * F0) / (F1 - F0) End Function Private Sub Form_Activate() Dim nextXroot As Single, xR As Single, fR As Single, iCount As Integer iCount = 1 Print "=================================================================" Print "X0= " & FormatNumber(Xp, 5) & vbTab & vbTab & "|" & vbTab & vbTab & "F(X0)= " & FormatNumber(myFunc(Xp), 5) Print "------------------------------------------------------------------------------------------------------------" Print "X1= " & FormatNumber(Xn, 5) & vbTab & vbTab & "|" & vbTab & vbTab & "F(X1)= " & FormatNumber(myFunc(Xn), 5) Do While Abs(Xp - Xn) > 0 nextXroot = SecantRootX(Xp, Xn) xR = FormatNumber(nextXroot, 5) fR = FormatNumber(myFunc(nextXroot), 5) Print "------------------------------------------------------------------------------------------------------------" iCount = iCount + 1 Print "X" & iCount & "= " & xR & vbTab & vbTab & "|" & vbTab & vbTab & "F(X" & iCount & ")= "; fR If fR = 0 Then Exit Do DoEvents Xp = Xn: Xn = nextXroot Loop Print "==================================================================" Print " Thus the root of the equation is " & vbTab & FormatNumber(nextXroot, 5) & " ." End Sub | 662 | 2,341 | {"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-2023-14 | longest | en | 0.337982 |
https://isabelle.in.tum.de/repos/isabelle/diff/a5a9c433f639/src/Modal/s43.thy | 1,618,995,892,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039526421.82/warc/CC-MAIN-20210421065303-20210421095303-00618.warc.gz | 401,853,742 | 3,434 | src/Modal/s43.thy
changeset 0 a5a9c433f639
```--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/src/Modal/s43.thy Thu Sep 16 12:20:38 1993 +0200
@@ -0,0 +1,79 @@
+(* Title: 91/Modal/S43
+ ID: \$Id\$
+ Author: Martin Coen
+ Copyright 1991 University of Cambridge
+
+This implements Rajeev Gore's sequent calculus for S43.
+*)
+
+S43 = Modal0 +
+
+consts
+ S43pi :: "[sobj=>sobj, sobj=>sobj, sobj=>sobj,\
+\ sobj=>sobj, sobj=>sobj, sobj=>sobj] => prop"
+ "@S43pi" :: "[sequence, sequence, sequence, sequence, sequence,\
+\ sequence] => prop" ("S43pi((_);(_);(_);(_);(_);(_))" [] 5)
+
+rules
+(* Definition of the star operation using a set of Horn clauses *)
+(* For system S43: gamma * == {[]P | []P : gamma} *)
+(* delta * == {<>P | <>P : delta} *)
+
+ lstar0 "|L>"
+ lstar1 "\$G |L> \$H ==> []P, \$G |L> []P, \$H"
+ lstar2 "\$G |L> \$H ==> P, \$G |L> \$H"
+ rstar0 "|R>"
+ rstar1 "\$G |R> \$H ==> <>P, \$G |R> <>P, \$H"
+ rstar2 "\$G |R> \$H ==> P, \$G |R> \$H"
+
+(* Set of Horn clauses to generate the antecedents for the S43 pi rule *)
+(* ie *)
+(* S1...Sk,Sk+1...Sk+m *)
+(* ---------------------------------- *)
+(* <>P1...<>Pk, \$G |- \$H, []Q1...[]Qm *)
+(* *)
+(* where Si == <>P1...<>Pi-1,<>Pi+1,..<>Pk,Pi, \$G * |- \$H *, []Q1...[]Qm *)
+(* and Sj == <>P1...<>Pk, \$G * |- \$H *, []Q1...[]Qj-1,[]Qj+1...[]Qm,Qj *)
+(* and 1<=i<=k and k<j<=k+m *)
+
+ S43pi0 "S43pi \$L;; \$R;; \$Lbox; \$Rdia"
+ S43pi1
+ "[| (S43pi <>P,\$L'; \$L;; \$R; \$Lbox;\$Rdia); \$L',P,\$L,\$Lbox |- \$R,\$Rdia |] ==> \
+\ S43pi \$L'; <>P,\$L;; \$R; \$Lbox;\$Rdia"
+ S43pi2
+ "[| (S43pi \$L';; []P,\$R'; \$R; \$Lbox;\$Rdia); \$L',\$Lbox |- \$R',P,\$R,\$Rdia |] ==> \
+\ S43pi \$L';; \$R'; []P,\$R; \$Lbox;\$Rdia"
+
+(* Rules for [] and <> for S43 *)
+
+ boxL "\$E, P, \$F, []P |- \$G ==> \$E, []P, \$F |- \$G"
+ diaR "\$E |- \$F, P, \$G, <>P ==> \$E |- \$F, <>P, \$G"
+ pi1
+ "[| \$L1,<>P,\$L2 |L> \$Lbox; \$L1,<>P,\$L2 |R> \$Ldia; \$R |L> \$Rbox; \$R |R> \$Rdia; \
+\ S43pi ; \$Ldia;; \$Rbox; \$Lbox; \$Rdia |] ==> \
+\ \$L1, <>P, \$L2 |- \$R"
+ pi2
+ "[| \$L |L> \$Lbox; \$L |R> \$Ldia; \$R1,[]P,\$R2 |L> \$Rbox; \$R1,[]P,\$R2 |R> \$Rdia; \
+\ S43pi ; \$Ldia;; \$Rbox; \$Lbox; \$Rdia |] ==> \
+\ \$L |- \$R1, []P, \$R2"
+end
+
+ML
+
+local
+
+ val S43pi = "S43pi";
+ val SS43pi = "@S43pi";
+
+ val tr = LK.seq_tr1;
+ val tr' = LK.seq_tr1';
+
+ fun s43pi_tr[s1,s2,s3,s4,s5,s6]=
+ Const(S43pi,dummyT)\$tr s1\$tr s2\$tr s3\$tr s4\$tr s5\$tr s6;
+ fun s43pi_tr'[Abs(_,_,s1),Abs(_,_,s2),Abs(_,_,s3),
+ Abs(_,_,s4),Abs(_,_,s5),Abs(_,_,s6)] =
+ Const(SS43pi,dummyT)\$tr' s1\$tr' s2\$tr' s3\$tr' s4\$tr' s5\$tr' s6;
+in
+val parse_translation = [(SS43pi,s43pi_tr)];
+val print_translation = [(S43pi,s43pi_tr')]
+end``` | 1,353 | 3,298 | {"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-2021-17 | latest | en | 0.510349 |
https://www.teacherspayteachers.com/Product/Subtraction-Data-Probes-With-Without-and-Mixed-pages-involving-Borrowing-2142840 | 1,513,074,194,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948515313.13/warc/CC-MAIN-20171212095356-20171212115356-00706.warc.gz | 800,361,329 | 26,165 | Total:
\$0.00
# Subtraction Data Probes - With, Without, and Mixed pages involving Borrowing!
Resource Type
Common Core Standards
Product Rating
3.9
9 ratings
File Type
Compressed Zip File
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Product Description
I am an elementary Special Education teacher of a class of nine students, each with significant, and ranging disabilities. We spend most of our time working on foundational skills including math. I spent a lot of time searching for straightforward data probes that would simply test the student’s ability to subtract. Too often the skills progressed quicker than my students did, or it tested a wide range of skills when we needed to work on just one.
I designed this product to be a series of systematic data probes, beginning with simple subtraction, and (eventually) working up to triple digit subtraction with and without multi-step borrowing. I have found that to show true mastery of subtraction after learning both how to subtract, and then to borrow, the last step is to complete a page with mixed problems (some borrowing and some without borrowing) discerning when and when not to borrow.
In this packet you will find 12 probes, each with three trials. The probes range from single digit subtraction with single digit answers to triple digit subtraction that both involves and doesn’t involving borrowing. Each trial includes 10 problems to allow for clear and easy percentages. There are data sheets for each digit of skill, single, double, and triple, a whole class data sheet, as well as a blank sheet. The probes are as follows:
- 1 Single Digit Subtraction Probe
o Single Digit Subtraction, with Single Digit Answers
- 3 “Under 20” Subtraction Probes
o (Under 20) Double Digit Minus Single Digit, without any borrowing
o (Under 20) Double Digit Minus Single Digit, with all borrowing
o (Under 20) Double Digit Minus Single Digit, with and without borrowing
- 3 Double Digit Subtraction Probes
o Double Digit Subtraction, without any borrowing
o Double Digit Subtraction, with all borrowing
o Double Digit Subtraction, with and without borrowing
- 3 Triple Digit Subtraction Probes
o Triple Digit Subtraction, without any borrowing
o Triple Digit Subtraction, with all borrowing
o Triple Digit Subtraction, with and without borrowing
Since creating this product, my subtraction data probes are on hand, easy to use, and very straightforward. Having these as a resource creates consistent and thorough data. I can show true trends by using all three trials and the student’s mastery of the skills are obvious.
I hope this alleviates some stress from your life and makes your data tracking more efficient and quality! Enjoy!
-Jordan Nichols
My SpEd Life
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Teaching Duration
N/A
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\$3.00 | 626 | 2,832 | {"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-2017-51 | latest | en | 0.899312 |
https://femci.gsfc.nasa.gov/cte/index.html | 1,723,090,065,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00889.warc.gz | 209,561,091 | 3,455 | Explanation of the Coefficient of Thermal Expansion (CTE) For Use in NASTRAN
Dr. William CaseApril 14, 1994
By definition, thermal strain at temperature T is the change in length of a member, due to a change in temperature, ( T - Tref ), divided by the original length l of that member. Denoting thermal strain at temperature T as eT
(1).
The thermal strain is zero at the reference temperature, Tref .
Also, by definition, the coefficient of thermal expansion (CTE or ) at temperature T is is defined as
(2).
Therefore,
(3),
where t is a dummy variable of integration. Define to be the average value of alpha over the temperature range from Tref to T.
Then,
(4).
Then, the thermal strain can be written as
(5).
Generally, thermal expansion data is given as a curve of vs. T. Based on equations (2) and (5), it is seen that
(6),
whereas
(7).
NASTRAN requires that be input, not . However, the two are equal when is constant. An example of the calculation of , along with the NASTRAN material data input, is given below.
Temp, deg K 103 Delta L/L 106 Alpha 106 Alpha-bar 0 -4.90 0.0 16.7 10 -4.90 0.1 17.3 20 -4.90 0.4 17.9 30 -4.89 1.4 18.6 40 -4.86 3.3 19.2 50 -4.83 5.7 19.9 60 -4.75 8.1 20.4 70 -4.66 10.3 20.9 80 -4.54 12.2 21.3 90 -4.41 13.9 21.7 100 -4.27 15.4 22.1 120 -3.93 17.6 22.7 140 -3.56 19.4 23.3 160 -3.16 21.0 23.8 180 -2.73 22.2 24.2 200 -2.27 23.2 24.4 220 -1.80 23.9 24.7 240 -1.32 24.4 24.9 260 -0.83 24.8 25.1 280 -0.33 25.2 25.3 300 0.18 25.5 25.4
NASTRAN Temperature Dependent Material Data Input for Magnesium
(using degrees Kelvin for temperature)
```MAT1 11 6.5E6 2.4E6 .066 25.35E-6 293.
MATT1 115
TABLEM1 115 +T1151
+T1151 0. 16.7E-6 10. 17.3E-6 20 17.9E-6 30. 18.6E-6 +T1152
+T1152 40. 19.2E-6 50. 19.9E-6 60. 20.4E-6 70. 20.9E-6 +T1153
+T1153 80. 21.3E-6 90. 21.7E-6 100. 22.1E-6 120. 22.7E-6 +T1154
+T1154 140. 23.3E-6 160. 23.8E-6 180. 24.2E-6 200. 24.4E-6 +T1155
+T1155 220. 24.7E-6 240. 24.9E-6 260. 25.1E-6 280. 25.3E-6 +T1156
+T1156 300. 25.4E-6 ENDT
```
Dr. William Case
April 14, 1994
HOME | FEMCI BOOK | PRESENTATIONS | REFERENCES | WORKSHOP | LINKS | ABOUT | MINUTES | PRIVACY | 973 | 2,347 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2024-33 | latest | en | 0.691954 |
https://math.stackexchange.com/questions/530135/prove-that-the-min-and-max-of-2-continuous-function-are-continuous | 1,720,944,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514551.8/warc/CC-MAIN-20240714063458-20240714093458-00802.warc.gz | 344,295,471 | 37,705 | # Prove that the min and max of 2 continuous function are continuous
Prove that if $$f$$ and $$g$$ are continuous functions the so are $$\min\{f(x),g(x)\}$$ and $$\max\{f(x),g(x)\}$$
I know this is true when $$f$$ and $$g$$ are not intersect each other, then I can compare them. However, I don't know how to prove it's true when they are intersect.
• You write this is in calculus/proof-writing. Do you know $\epsilon, \delta$ limits and continuity? Is this how you are expecting to write this up? Commented Oct 17, 2013 at 19:15
• You can prove that $\min,\,\max \colon \mathbb{R}^2 \to \mathbb{R}$ are continuous. Then the continuity of compositions of continuous functions does the rest. Commented Oct 17, 2013 at 19:16
• A different hint: $\max(f,g)=\frac12(f+g+|f-g|)$.
– dls
Commented Oct 17, 2013 at 19:20
• @mixedmath yes, I'm expecting a ϵ-δ proof. Commented Oct 17, 2013 at 19:21
• @DanielFischer, I don't think I have learn that technique yet Commented Oct 17, 2013 at 19:22
Let $h(x) = \min\{f(x),g(x)\}$. Suppose $x_0$ is such that $f(x_0) = g(x_0)$. We want to show $h$ is continuous at $x_0$.
Take $\epsilon > 0$, then there is a $\delta_f$ so that $|f(x) - f(x_0)| < \epsilon$ for $|x-x_0| < \delta_f$, and similarly for $g$ and some $\delta_g$ (with the same $\epsilon$).
Use this, and the fact that $h(x_0) = f(x_0) = g(x_0)$ to show that $|h(x) - h(x_0)| < \epsilon$ whether $h(x) = f(x)$ or $h(x) = g(x)$ as long as $|x-x_0| < \delta$ for some $\delta$.
• Also, what is your reasoning for setting $x_0$ s.t. $f(x_0)=g(x_0)$? Does this prove the statement generally, or only for the case in which we have an $x_0$ with this property? Commented Jun 15, 2015 at 13:42
• @baronVT 'and the fsct that $h(x_0)=f(x_0)=g(x_0)$'? I don't see how you may say this Commented May 17, 2016 at 12:06 | 643 | 1,810 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.923202 |
http://finance.yahoo.com/mbview/threadview/?bn=d9b6ee08-22cd-3a9f-9f64-dea01be8d612&tid=970250119000-190c45e6-8422-3d15-8139-cd1fc38f13b9&tls=nm%2C%2C273 | 1,394,854,621,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678695683/warc/CC-MAIN-20140313024455-00067-ip-10-183-142-35.ec2.internal.warc.gz | 54,835,342 | 25,581 | Fri, Mar 14, 2014, 11:37 PM EDT - U.S. Markets closed
# Corrections Corporation of America Message Board
• grunka grunka Sep 29, 2000 1:55 PM Flag
## Conversion
Today is the 10th trading day prior to the first
conversion period. Based on the closing price of the common
these ten days, the average is under \$1 so we should
get 24.46 common shares for each Pref B, Correct?
SortNewest | Oldest | Most Replied Expand all replies
• At no time in the past 10 trading days did PZN
trade under one dollar. So how could the average be
under \$1. ? I show the following
closing prices
(based on Yahoo) for the past ten
1.9375, 1.9375, 1.8125, 1.8125, 1.7500, 1.1875, 1.1875,
1.1250, 1.000, 1.3125 = 15.0625 total. divided by 10 =
1.50625 = average price per share. 24.46 divided by
1.50625 = 16.2395 shares for each Preferred B
share
converted . I don't know how PZN will handle the fractions.
• 4 Replies to bensabia
• Bensabia had the "conversion" price bang on back on Friday, if the closing prices noted in his post # 15472 are "official". If so. it looks like \$1.50 is valid.
Thanks ben, good call!
• I was basing it on the adjusted closing cost, do you know for a fact that thats not the figure that will be used for the average, or will PZN use whatever figure that is to their advantage.
• Ben, I believe you are correct. Therefore in the
first conversion period a PZN.B shr will become
16.239004149 shrs of common.
100 shrs of PZN.B will
become 1623.9 shrs of the common. Upon conversion,
fractional shrs will be credited in cash, and the account
should show 1623 shrs and .9x1.50625 (\$1.36) in cash.
That's my take, anyway.
Question is, what will
PZN trade at during the ten days preceeding the
second conversion period? More than a \$1.50 average?
Those meaning to convert will have to decide the answer
before 10/13.
• I've been out of touch with the board since last
week or so when I sold some PZN for tax loss
purposes..can you give me your assessment as to what the
general opinion is regarding the conversion and when is
the window open ?
Would appreciate, still have
substantial position.
• The first half of conversion period was pre-dividend. If PZN closes at \$1.125 today I figure the conversion price is \$1.4375 for the first period.
• It looks like today's close will be \$1.3125. This
should mean a conversion price of \$1.50.
However,
just for fun, I checked Yahoo's historical quote data
and found that for the period prior to the dividend
there is an adjusted price listed which totals \$4.4626
for the first five days of the conversion period. If
this price is used the conversion price becomes
\$1.0275.
Note: the press release says the average close on the | 763 | 2,702 | {"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-2014-10 | latest | en | 0.928083 |
https://www.vedantu.com/question-answer/choose-the-correct-option-about-alternating-class-11-physics-cbse-5f89eb30b39c3957d6e37257 | 1,718,320,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861517.98/warc/CC-MAIN-20240613221354-20240614011354-00677.warc.gz | 965,795,833 | 32,628 | Courses
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# Choose the correct option about Alternating voltage – A) It is independent of timeB) It varies directly with timeC) It varies inversely with time.D) It varies with time.
Last updated date: 13th Jun 2024
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Hint: Alternating voltage is the changing sinusoidal form of voltage which is the voltage in Alternating current. The voltage changes its polarity with time. The number of times the polarity changes depends on the frequency of the Alternating voltage.
Alternating current involves the current flow in the form of a sinusoidal wave. It’s direction changes with time as the sine wave. Alternating voltage is the voltage involved in the alternating current. Since the current changes with the time the alternating current voltage should also change with time. Actually, it is the reverse. An alternating voltage introduces an alternating current in the circuit.
Let us consider the alternating form of voltage. The equation is given by –
$V={{V}_{0}}\sin \omega t$
Where, V is the voltage at an instant ‘t’, ${{V}_{0}}$ is the maximum amplitude of voltage, $\omega$ is the angular frequency and ‘t’ is the time.
From the equation we can conclude that the Voltage varies with time and is not directly proportional. It changes as a function of sine.
Therefore, Alternating Voltage varies as a function of time.
The correct answer is option D.
Note:
- The current lags or leads the voltage in phase with respect to the elements used in the circuit. A resistor only circuit does not show variation in phase between the voltage and current, whereas a inductor and capacitor circuit shows phase variations.
- The alternating current and alternating voltage are related to each other in various ways depending on the elements in the circuit. | 401 | 1,872 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | latest | en | 0.887046 |
https://republicofsouthossetia.org/question/tom-compared-2-12-and-7-6-by-first-comparing-each-fraction-to-1-2-and-1-step-1-2-12-is-less-than-15805698-51/ | 1,652,680,150,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00142.warc.gz | 579,512,516 | 14,291 | ## Tom compared 2/12 and 7/6 by first comparing each fraction to 1/2 and 1. Step 1: 2/12 is less than 1/2. Step 2: 7/6 is less than 1 but more
Question
Tom compared 2/12 and 7/6 by first comparing each fraction to 1/2 and 1. Step 1: 2/12 is less than 1/2. Step 2: 7/6 is less than 1 but more than 1/2. Step 3: So, 2/12 < 7/6 In which step did Tom mess up?
in progress 0
3 months 2022-02-16T15:13:02+00:00 1 Answer 0 views 0 | 178 | 426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-21 | latest | en | 0.946509 |
https://www.geeksforgeeks.org/practice-problems-finite-automata-set-2/?ref=rp | 1,696,070,556,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00170.warc.gz | 832,161,879 | 35,933 | Open In App
# Practice problems on finite automata | Set 2
Que-1: Draw a deterministic and non-deterministic finite automate which either starts with 01 or end with 01 of a string containing 0, 1 in it, e.g., 01010100 but not 000111010. Explanation – Draw a DFA and NFA of same language whose strings only reach to the final state containing either 01 at start or at the end. If anything else is come then come out to the final state then it does not accept. NFA of the given string is as follows: DFA of the given string is as follows: Here, q0 shows the initial state, q1, q2 are the transition states, and q3, q4, q5, q6, q7 are the transition and final states. Que-2: Draw a deterministic and non-deterministic finite automate which starts with 01 and ends with 01 of a string containing 0, 1 in it, e.g., 01000101 but not 000111001. Explanation – Draw a DFA and NFA of same language whose strings only reach to the final state containing 01 at start and at the end. If anything else is come then come out to the final state then it does not accept. NFA of the given string is as follows: DFA of the given string is as follows:
Here, q0 shows the initial state, q1, q2, q3 are the transition states, and q4, q5 are the transition and final states. Que-3: Draw a deterministic finite automata which recognize a string containing binary representation 0, 1 in the form of multiple 2, e.g., 1010 but not 01101. Explanation – Draw a DFA whose strings only reach to the final state containing 0 at the end that means number is multiple of 2. If anything else is come then come out to the final state then it does not accept. DFA of the given string is as follows: Here, q0 shows the initial and final state, q1 is the transition states. Que-4: Draw a deterministic finite automata which recognize a string containing binary representation 0, 1 in the form of multiple 3, e.g., 1001 but not 1000. Explanation – Draw a DFA whose string only reach to the final state containing binary number is multiple of 3. If anything else is come then come out to the final state then it does not accept. DFA of the given string is as follows: Here, q0 shows the initial and final state, q1, q2 are the transition states. Read – Practice problems on finite automata | 562 | 2,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2023-40 | latest | en | 0.894523 |
http://www.programmingwithbasics.com/2017/04/geeksforgeeks-solution-for-max-value.html | 1,500,892,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00625.warc.gz | 538,406,126 | 44,137 | # Geeksforgeeks Solution For " Max value "
GeeksforGeeks Solution For Hard Domain .Below You Can Find The Solution Of School Basic ,Easy ,Medium . Or Hackerrank Solution You Can Also Direct Submit Your Solution to Geeksforgeeks Same Problem .You Need to login then you can submit you answers
Problem :- Max value
Solution :-
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,sum,res=0;
cin>>n;
int arr[n];
for(i=0;i<n;i++)
cin>>arr[i];
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
sum=abs((arr[i]-i)-(arr[j]-j));
if(res<sum)
res=sum;
}
}
cout<<res<<endl;
}
return 0;
}
Output:- | 203 | 637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-30 | longest | en | 0.175201 |
https://codeforces.com/problemset/problem/1366/E | 1,675,663,120,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00445.warc.gz | 190,496,467 | 13,832 | E. Two Arrays
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given two arrays $a_1, a_2, \dots , a_n$ and $b_1, b_2, \dots , b_m$. Array $b$ is sorted in ascending order ($b_i < b_{i + 1}$ for each $i$ from $1$ to $m - 1$).
You have to divide the array $a$ into $m$ consecutive subarrays so that, for each $i$ from $1$ to $m$, the minimum on the $i$-th subarray is equal to $b_i$. Note that each element belongs to exactly one subarray, and they are formed in such a way: the first several elements of $a$ compose the first subarray, the next several elements of $a$ compose the second subarray, and so on.
For example, if $a = [12, 10, 20, 20, 25, 30]$ and $b = [10, 20, 30]$ then there are two good partitions of array $a$:
1. $[12, 10, 20], [20, 25], [30]$;
2. $[12, 10], [20, 20, 25], [30]$.
You have to calculate the number of ways to divide the array $a$. Since the number can be pretty large print it modulo 998244353.
Input
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 2 \cdot 10^5$) — the length of arrays $a$ and $b$ respectively.
The second line contains $n$ integers $a_1, a_2, \dots , a_n$ ($1 \le a_i \le 10^9$) — the array $a$.
The third line contains $m$ integers $b_1, b_2, \dots , b_m$ ($1 \le b_i \le 10^9; b_i < b_{i+1}$) — the array $b$.
Output
In only line print one integer — the number of ways to divide the array $a$ modulo 998244353.
Examples
Input
6 3
12 10 20 20 25 30
10 20 30
Output
2
Input
4 2
1 3 3 7
3 7
Output
0
Input
8 2
1 2 2 2 2 2 2 2
1 2
Output
7 | 587 | 1,593 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-06 | latest | en | 0.609766 |
https://www.jiskha.com/display.cgi?id=1332289320 | 1,511,054,554,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00278.warc.gz | 790,709,863 | 3,797 | Physics
posted by .
1.Can an object reverse its direction of travel while maintaining a constant acceleration? If so, give an example. If not, explain why not.
2. If you were standing in a bus moving at constant velocity, would you have to lean in some special way to compensate for the bus's motion? EXPLAIN. WHat if the bus were at a constant acceleration?eXPLAIN.
• Physics -
1. Yes
example: throwing a ball up and catching it on the way down
2. No
Next part: yes, you would have to lean if the bus were accelerating, to keep yourself from tipping over due to the static friction force on your feet.
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More Similar Questions | 456 | 2,186 | {"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} | 2.984375 | 3 | CC-MAIN-2017-47 | latest | en | 0.930198 |
http://www.yourarticlelibrary.com/project-management/planning-expenditure-of-a-project-with-calculation-project-management/95020/ | 1,503,331,099,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886109157.57/warc/CC-MAIN-20170821152953-20170821172953-00494.warc.gz | 705,967,784 | 14,407 | Planning Expenditure of a Project (With Calculation) | Project Management
In this article we would like to discuss the steps for planning expenditure of a project, along with the preparation of the cash flow as per schedule of activities-by means of an illustration.
1. Project Data Base with Scheduling:
Project: Construction of a building—Project 14.
Project Cash: Rs. 7, 50,000.
Project Schedule: Target to complete in 32 weeks.
Scheduling: estimated data:
2. The Initial Network Construction with the Project Data Base:
Important points for revision:
1. Legend : for the network diagram
2. EST of head event = EST of tail event + tij of the relevant activity, but when a number of activities land up on a head event from different tail events with activities of different tij -s then the EST of the head event is with highest time units : e.g. activities E, F and H converge in event 7.
... The EETs are from event 3 = 12 + 10 = 22
from event 4 = 13 + 2 = 15
from event 6 = 17 + 4 = 21
Hence EST of event 7 is with the highest figure of 22.
3. For LFT, it is backward working; when different activities emanate from a single tail event, the LFT of the tail event is with the least time units.
3. Expenditure Planning as per Activities:
We have defined the project in terms of activities and the project cost as per such activities.
As we know the time elements per activity, we can prepare the expenditure plan for the project per activity and within activity per week, as shown below:
We find from the table that the budgeted expenses are categorised per activity and, within each activity, the requirement of funds per month. For easier understanding we have discussed the plan in a simple manner but, according to the circumstances and the need, the activities can be further split into sub-activities and grouped as per the projected breakdown structures, such as.
Plumbing may be sub-divided as:
Fund requests as per activities:
The group responsible for plumbing activities is to request for funds Rs. 25,000 in the beginning of the second month and Rs. 35,000 in the third month and meet the expenses, progressively, as per the actual works.
The detailed system of codification of the activity and the collection of actual expenditures against the relevant codes becomes the part of the project cost system outside the stream of the organisation’s normal financial accounting.
The critical path in a network is the sequence of critical activities showing the longest path in the network from the starting event to the final event of the project. We also know that Time, in implementation of a project, is very much related to the project cost.
While reduction in duration may reduce the administrative cost it may cause extra cost on account of more labour, overtime, extra machines etc. In general, reduction of activity duration causes reduction in indirect costs but increases other direct costs.
There may be a situation to complete the project earlier than what has been envisaged in a network plan for the project showing the critical path, completed on the basis of normal activity duration and consuming the normal resources.
It is possible, to a certain extent, to reduce the activity duration by employing additional resources and, as such, additional direct cost. Of course, such reduction is possible only up to a limit which is the minimum duration.
This process of shortening the activity duration is called ‘crashing’. The ‘time cost trade-off’ represents adjustments of project schedule with a view to reduce the total projected time even at extra cost and the process is known as project crashing. The relation between time, i.e. the activity duration, and cost (or the activity cost) can be presented graphically by Activity Cost Slope curve as explained hereafter.
The terms used in activity cost slope are:
Normal Time Tn = the minimum time required to complete an activity, under normal conditions.
Normal Cost Cn = the lowest direct cost estimated to complete an activity, in normal time.
Crash Time Tc = the minimum time required to complete an activity by using extra resources, i.e. by extra cost.
Crash Cost Cc = the direct cost that is estimated in completing the activity, by crash time.
Activity Cost Slope (ACS) represents the additional direct cost incurred per unit of time saved in completing an activity and is expressed as:
ACS=Cc – Cn /Tn-Tc
ACS may be shown graphically:
Steps for crashing:
We have just discussed crashing of an activity. When we follow the similar process for the entire project to establish total possible crashing (and the relevant increase in the project cost) it is called Project Crashing.
The process can best be elaborated with an example:
We recall the same illustration discussed under Project data base scheduling, i.e.
a. Project duration of 32 weeks
b. Project cost of Rs. 7,50,000
c. Critical path as shown below with
d. Critical activities A, B, E, I and J.
We bring forward the network construction again:
The possibilities of crashing the project are explored and the revised schedule of operation with the time crashed, extra cost and the crash cost as estimated are produced below:
The total indirect cost estimated is, say, Rs. 3, 20,000; i.e. @ Rs. 10,000 per week.
Step 1:
We know, firstly, that the Critical Path is the longest one and, as such, reducing the duration of non-critical activity will not help. Hence, we would like to deal with crashing the critical activities. We review the table of critical activities which can be crashed.
Study of the critical activities suggest that activities I and J cannot be reduced.
But other critical activities can be reduced at extra cost as detailed below:
The duration of the non-critical activities F and H also cannot be reduced.
Step 2:
Crashing is done gradually from the cheapest one to crash to the costliest one, in sequence. The cheapest being the activity B, we start crashing B till the critical path does not change, i.e. we crash one week, review the critical path, and then we crash the second week (which is the maximum possible to crash).
Review again, after crashing B for two weeks:
(a) Project duration will be 32 weeks – 2 weeks = 30 weeks.
(b) Project cost Rs. 750 + Extra cost 3 + 3 = Rs. 7, 56,000.
(c) The network after B crashed by two weeks will appear as follows:
(d) The critical path is not disturbed, i.e. A, B, E, I and J. The next crashing is E, being cheaper than A and the maximum possible reduction is by one week. Therefore, activity duration will be 9 weeks. This reduction of E by one week will change the timing of the events (7), (8) and (9) as 19, 19 + 6 = 25, and 25 + 4 = 29, respectively.
But, again, this will change the time of event 6 also, as, going backward from (7), it will be 19 minus 3 (activity duration of H). This will change the activity H also as critical and to a second critical path, when we crash the non-critical activity G by 1 week to 6 weeks and review the position.
The network will appear as per diagram produced below with two different critical paths:
Project duration now becomes 29 weeks (i.e. the effect of the reduction of E in the longest path).
Project cost increased further by Rs. 9,000, i.e. 7,56,000 + 9,000 = Rs. 765,000 (E costing Rs. 4,000 and G costing Rs. 5,000 as extra cost for reduction by one week.)
Now to explore the possibility of reduction of the duration of activity common to both the critical paths, i.e. activity A and by one week possible at an extra cost of Rs. 5,000.
The status is reviewed after crashing the activity A as follows:
Project duration – 28 weeks
Project cost – Rs. 7, 65,000 + 5,000 = Rs. 7, 70,000
Network diagram:
The Critical Paths are (a) A, B, E, I and J and (b) A, D, G, H, I and J.
Project cost revision after crashing:
We have reached the limit of possible crashing [as the data table with estimations for crashing indicates that activities F and H cannot be crashed].
We can tabulate our findings with the complete project crashing:
From the above table the management may review the overall cost aspect of crashing and take a decision on the course of crashing, that is, if the management is considered with the total minimum cost (which, in the illustration above, is Rs. 10, 50,000) then a decision should be to crash A, B, E and G as above and decide on the total project schedule estimate of 28 weeks.
5. Expenditure Planning After Crashing:
When the management would like to review the position after possible crashing, the overall effect of both direct and indirect costs and the project duration are worked out.
The changed scenario is assessed and a final decision of the course of crashing is made. With these estimated changes, a new expenditure plan is developed which is illustrated following the same example of Project 14, with the management decision to crash A, B, E and G. The fund request per activity will also change accordingly.
Notes:
1. The Direct Cost (D/C) has been apportioned between the weeks on a pro rata basis (i.e. on the basis of activity duration) although, in reality, it can be different. Major part of direct cost for an activity of 9 weeks duration may be required to be spent in first two weeks.
Therefore, in Expenditure Planning, such possibilities are to be considered on the basis of the usual practice in the prevalent business for the purpose of planning the apportionment of such expenses.
2. The Indirect Cost (I/C) normally vary with the time. Such expenses are apportioned to different activities concurrently taking place, in proportion to the time consumed by different activities during the same week or months etc.
For instance, activities during the weeks 09 to 12 have been envisaged as:
Total indirect expenses for four weeks 09 to 12 = 4 x Rs. 10,000 = Rs. 40,000. Such expenses are pro-rata to B, C, D, E and G in the ratio of 1 : 4 : 1 : 3 : 3.
6. Cash Forecast:
From the estimated expenses per activity and per week we can prepare the cash forecast as follows:
Notes to cash forecast:
1. Withdrawal of fund, projected at the beginning is in excess of the fund requirement as a safety measure to meet emergency expenses, the excess of Rs. 35,000. The same amount is retained with the project management as an imprest cash till the completion of the project, thereby the withdrawals are estimated to be at the level of estimated expenses.
2. The normal financial accounting carried out by the organisation’s central accounts. Monitoring the actual expenditure on the project and controlling the costs is carried out by project management as the expenditure details are collected on the basis of activity code.
The code is followed as planned with the integrated breakdown structure. This enables to exercise the cost control, locating the functionaries responsible for such cost and cost control.
3. While the direct costs are allocated, the indirect costs are appropriated to the respective activity in cost/cash forecast on the basis of estimated activity duration and in case of actual on the basis of actual time spent on activity times the budgeted rate of indirect cost.
Accordingly, every functionary in the project team shall have to accumulate the actual time spent on activities which is considered for controlling the project schedule as well as the indirect costs.
4. At every month end, the expense statement is prepared showing per activity the forecasted expenses as well as the actual expenses incurred for the same month. The variances of such expenses are analysed and corrective actions taken when required. | 2,552 | 11,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2017-34 | latest | en | 0.935548 |
http://www.ee.nmt.edu/~wedeward/EE212/SP03/ee212.html | 1,516,171,713,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886830.8/warc/CC-MAIN-20180117063030-20180117083030-00379.warc.gz | 439,060,038 | 2,177 | EE 212: Circuits & Signals II
Spring 2003 Schedule: MWF 10:00am - 10:50am in MSEC 103
Instructor: Kevin Wedeward, Office: Workman 221, Phone: 835-5708, e-mail: wedeward@nmt.edu, web-page: www.ee.nmt.edu/~wedeward/
Office Hours: MWF 09:00am - 10:00am, MWF 11:00am - 12:00pm and by appointment
Associated Lab: EE212L (Circuits & Signals II Lab)
Course Objective: Develop an understanding of linear circuits and their analysis using
• frequency-domain techniques
Course Prerequisite: EE211 (Circuits & Signals I)
Required Text: "Elementary Linear Circuit Analysis, 2nd ed." by Leonard S. Bobrow
Topics:
1. Sinusoidal analysis using frequency-based approach (chapter 8)
2. Power definitions for waveforms (chapter 9)
3. Frequency response (chapter 10)
4. Laplace Transform (chapter 11)
5. Two-port networks (chapter 12)
6. Fourier Series (chapter 13)
Homework: Homework will be assigned, collected, and graded on a weekly basis. You are encouraged to work with other students so long as the written work turned in is your own.
1. HW1 due beginning of class (BOC) W 01/29/03: 8.6 (assume in steady-state), 8.16 (use Euler's formula), 8.19 (write answer in both rectangular and exponential (polar) form, sketch answers in complex plane)
2. HW2 due BOC F 02/07/03: 8.25, 8.27, 8.28, 8.42, 8.46
3. HW3 due BOC W 02/19/03: 9.2, 9.3, 9.11, 9.16, 9.17, 9.19
4. HW4 due BOC M 03/03/03: 9.28, 9.30, 10.6, 10.7
5. HW5 due BOC W 03/19/03: 10.11, 10.14, 10.16; use matlab to plot frequency response and frequency response as Bode plot for 10.14
6. HW6 due M 3/24: 10.15, 10.17 (include phase response); use matlab to plot frequency response and frequency response as Bode plot for both
7. HW7 due F 3/28: 10.20, 10.21, 10.26
8. HW8 due W 4/9: 10.40, 10.46, 10.47, 10.55
9. HW9 due W 4/16: 11.1, 11.2, 11.3, 11.4, 11.7
10. HW10 due F 4/25: 11.12a,d, 11.13a, 11.14b,c, 11.15, 11.16
11. HW11 due F 5/9: 11.22, 11.38, 11.40, 11.46, 11.49
• Homework: 15%
• Three Exams: 55%
• Final Exam: 30%
• Scheduled for Tuesday, May 13 at 1:30pm in MSEC 103 (review sheet)
1. Chapter 8
2. Chapter 9
3. Chapter 10
4. Chapter 11
Matlab Examples:
1. example 1: frequency response
2. example 2: frequency response | 821 | 2,214 | {"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-2018-05 | latest | en | 0.840888 |
https://able2know.org/topic/208160-3 | 1,701,932,931,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00525.warc.gz | 102,309,294 | 11,043 | 8
# Can an object be accelerating and yet -not- moving?
dalehileman
1
Fri 15 Feb, 2013 03:18 pm
@Finn dAbuzz,
Quote:
Do you believe that a decision made through intuition has simply invoked a random decision generator?
Don't think that's what Max meant but it did sound that way didn't iot
Quote:
….isn't very effective for problems much more complex than "Do I move…...into that clearing ahead?"
Disagree, think it's much more profound. But don't want to start a war
Quote:
or "Which person do I ask to watch my baggage while I go to the restroom?"
Now this might require some pretty sharp judgement depending on many factors not accessible to the conscious--nonetheless as you say, comes quick
Quote:
but something is going on in the mind…..beyond spinning a wheel……...
Yes Buzz, much much more
0 Replies
maxdancona
1
Fri 15 Feb, 2013 04:15 pm
@Finn dAbuzz,
No. A decision made through intuition is often worse than a random decision generator. Intuition encompasses our fears, our desires and our prejudices.
People acting on intuition are acting irrationally.
I avoid making a decision based on intuition on anything important except for the time where a quick decision is more important than a logical one. By that I mean that if I can't come up for a logically defensible reason for a feeling I have, I ignore the feeling.
So many irrational things that people do and believe are based on what they feel is right even though they can't support it logically.
Falco
1
Fri 15 Feb, 2013 04:43 pm
@raprap,
What do you mean by a boundary condition? I've always thought of it as an instantaneous condition. Take a simple graph paper with a slope, for example, where there is plot of velocity with respect to time, with a nonzero slope passing the x-intercept, where the acceleration remains constant throughout the run. Therefore, when the instantaneous velocity is zero (hitting the x-intercept), the object in motion is changing from from one direction to another direction (+ to - or - to +) depending on if the slope is positive or negative. It can be from positive direction of motion to the negative direction, or vice versa. Only at such intercept, where v is instantaneously zero and such non-zero slope is constant therefore at that instantaneous movement there is a positive or negative non-zero acceleration, arbitrary to the sign of the slope.
raprap
1
Sat 16 Feb, 2013 03:28 am
Falco
5
Sat 16 Feb, 2013 09:47 am
@raprap,
I understand the definition of boundary condition, but I think it is inaccurate in describing an instance where an object is instantaneously at rest, v(t)=0, but is either on the verge of starting to move or is turning around and changing direction. At such an instantaneous moment, the velocity is zero (ds/dt=0), but the acceleration is nonzero (dv/dt 0).
There aren't any boundaries involved or finding the area under the a line involved. Using a position vs. time graph you simply need to find the equation to a tangent line and find the value at the point of interest to find the instantaneous vleocity, and in a similar manner the tangent of a velocity-time graph represents instantaneous acceleration.
Take the trajectory of the ball for example,
a = dv/dt = -g (using the convention that upwards is positive),
v = -g\int{dt} + C
v = -gt + v0
where v0 is the initial velocity, and we let this to be positive since it was tossed upwards.
Plotting as a function of t, v becomes smaller, until at some point, -gt + v0 = zero. But with constant acceleration of -g. In the curve of velocity vs. time. At some time, t, the instantaneous velocity is zero. The acceleration is the slope of the velocity curve at time t.
Still fail to see how the boundary condition applies, unless I'm missing something.
dalehileman
1
Sat 16 Feb, 2013 12:48 pm
@Falco,
Evidently then the math establishes it. Yet Her "creation" is not necessarily tied to the math (man's creation) while Intuition stubbornly insists on that moment of zero acceleration
engineer
1
Sat 16 Feb, 2013 02:24 pm
@dalehileman,
If an object at zero velocity has zero acceleration, it can never move. All stationary objects that are set into motion have an acceleration applied to them while they are stationary.
georgeob1
1
Sat 16 Feb, 2013 02:35 pm
@Falco,
Falco wrote:
Still fail to see how the boundary condition applies, unless I'm missing something.
Your issue here is a semantical one, not mathematical. Substitute " initial condition" for "boundary contition" and you may feel better. Raprap (I believe) referred to the fundamental theorem of calculus (the integral as the anti derivative). The issues here are simply the constants associated with the two integrations of the (variable or constant) acceleration. One is the initial position, the other is the initial velocity.
dalehileman
1
Sat 16 Feb, 2013 02:51 pm
@engineer,
Quote:
If an object at zero velocity has zero acceleration, it can never move.
I'm neither physicist nor mathematician, can only respond it's counterintuitive
Quote:
All stationary objects that are set into motion have an acceleration applied to them while they are stationary.
Being a rank amateur at this sort of thing capable of only the most literal interpretation, it seems that this sentence has several possible meanings so perhaps ought to be reworded for the benefit of the Intellectually Disadvantaged (me)
georgeob1
1
Sat 16 Feb, 2013 04:15 pm
@dalehileman,
Think of it this way - imagine a stationary object that is not moving. Now imagine that you push it - i.e. apply a force to it - at the instant you apply the force the object starts to move - it is accelerating and the acceleration is proportional to the force applied and inversely proportional to the mass of the object (F = M x A, or eqivalently, A = F/M).
dalehileman
1
Sat 16 Feb, 2013 05:21 pm
@georgeob1,
Quote:
Think of it this way - imagine a stationary object that is not moving.
Yes that's usually how I imagine a stationary object
Quote:
Now imagine that you push it …….starts to move - it is accelerating……... (F = M x A, or eqivalently, A = F/M).
Of course, that's how I had always understood it. But that doesn't explain how it's accelerating when it's stationary
aspvenom
1
Sat 16 Feb, 2013 05:25 pm
@dalehileman,
Acceleration is the rate of change of velocity. Velocity can be zero and changing(in the ball's case going from positive velocity to negative velocity at the peak of the trajectory), in which case there is an acceleration.
In other words, if magnitude of velocity is zero, and the direction is changing, then the object is accelerating.
In addition to this, consider an object being spun around at a constant speed (note I did not say velocity) in a circle. Because the direction is changing, the object is accelerating, angular acceleration in this case. Actually in this case of an object traveling around in a circle, the magnitude of its average velocity is always zero, because it has no displacement, returning always to its starting point.
georgeob1
1
Sat 16 Feb, 2013 05:32 pm
@dalehileman,
dalehileman wrote:
But that doesn't explain how it's accelerating when it's stationary
Thanks for pointing out my redundant phrase.
What contradiction can there be in the idea of an accelerating object that has a zero velocity? Engineer provided a perfectly clear example with a vertically thrown ball on the verge of reversing its trajectory. Clearly gravity is still operating on the ball at the moment when it is no longer moving, so the idea of an unopposed force applied to an object with zero velocity shouldn't be a problem.
I can tell you, from my own experience, that an aircraft in vertical flight will slow down and (if one is gentle on the controls), stop, and the nose will fall vertically down. Moreover one experiences near zero "g" at that moment confirming the acceleration of the aircraft cancelling out some or most of the gravity acceleration.
dalehileman
1
Sat 16 Feb, 2013 06:51 pm
@georgeob1,
Quote:
What contradiction can there be in the idea of an accelerating object that has a zero velocity?
Intuition (mine anyhow) insists that when it's stopped it isn't accelerating
georgeob1
1
Sat 16 Feb, 2013 07:22 pm
@dalehileman,
Do you understand or acknowledge that the slope of a line can be increasing even though, at that moment, it is (say) horizontal? There are infinitely many points on the line between the present one and any nearby point, just as there are infinitely many moments in time between now and a second later. Moreover there are infinitely many real numbers between any two coordinate values or measures of time with which to describe them. All involve continuous processes and measures that are fundamental to mathematics and science. There are, at a subatomic level, minimum quantities (quanta) of mass, energy and momentum, but not time and space. Even there the nature of mass & energy becomes complex and the descriptions of it as involving waves or mass is essentially metaphorical.
How about a falling rock accelerating under the force of gravity. If it's velocity at some instant is (say) v, is it accelerating to a greater speed due to gravity ? All of modern mechanics is based on the observation that this is true. What difference is there if the initial velocity, v, is zero?
0 Replies
IRFRANK
1
Sat 16 Feb, 2013 07:57 pm
@maxdancona,
Quote:
By that I mean that if I can't come up for a logically defensible reason for a feeling I have, I ignore the feeling.
I like that. Helps one avoid mistakes.
0 Replies
maxdancona
1
Sat 16 Feb, 2013 09:31 pm
@aspvenom,
aspvenom wrote:
Because the direction is changing, the object is accelerating, angular acceleration in this case.
The math here is really not as not nearly as difficult as you all are making it.
An object going at constant speed around a circle is not experiencing angular acceleration. It is experiencing linear acceleration (i.e. acceleration in a direction). Angular acceleration is the result of a torque meaning that the speed would be changing.
aspvenom
1
Sat 16 Feb, 2013 09:36 pm
@maxdancona,
Well, there isn't linear velocity, only angular... I was just trying to make it sensible to Dale. I have failed at that.
Falco
1
Sat 16 Feb, 2013 09:41 pm
@georgeob1,
georgeob1 wrote:
Falco wrote:
Still fail to see how the boundary condition applies, unless I'm missing something.
Your issue here is a semantical one, not mathematical. Substitute " initial condition" for "boundary contition" and you may feel better. Raprap (I believe) referred to the fundamental theorem of calculus (the integral as the anti derivative). The issues here are simply the constants associated with the two integrations of the (variable or constant) acceleration. One is the initial position, the other is the initial velocity.
That makes more sense. I appreciate your taking the time in explaining that.
maxdancona
1
Sat 16 Feb, 2013 10:50 pm
@aspvenom,
You are incorrect aspvenom.
The linear velocity is the speed and the direction of the object at any given time. An object going at a constant speed in a circular path absolutely has a speed and a direction at any given time.
So yes, this object has a non-zero linear velocity.
The linear acceleration is the change in the linear velocity. Since the direction the object is moving is changing, there is absolutely a linear acceleration.
The angular velocity is the angular speed (i.e. the number of radians the object is travelling around the circle per second) with a direction along the axis of rotation. There is an angular velocity for this object at constant speed around a circular path, but it isn't changing because at constant speed the object travels the same number of radians each second, and since the direction is the axis of rotation it doesn't change.
And of course, since the angular velocity isn't changing, the angular acceleration for constant speed circular motion is zero. If the speed of the object is changing, then that would mean a non-zero angular acceleration.
Actually, this is rather simple math. These answers are things you probably learned and forgot in high school.
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Transient fields - Question by puzzledperson | 2,900 | 12,549 | {"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} | 3.03125 | 3 | CC-MAIN-2023-50 | latest | en | 0.959414 |
https://trustconverter.com/en/area-conversion/circular-thous/circular-thous-to-ares.html | 1,603,676,141,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107890108.60/warc/CC-MAIN-20201026002022-20201026032022-00209.warc.gz | 575,594,917 | 9,234 | # Circular thous to Ares Conversion
Circular thou to are conversion allow you make a conversion between circular thou and are easily. You can find the tool in the following.
### Area Conversion
to
input
= 5.06708 × 10-12
= 5.06708E-12
= 5.06708e-12
= 1.01342 × 10-11
= 1.01342E-11
= 1.01342e-11
= 1.52012 × 10-11
= 1.52012E-11
= 1.52012e-11
= 2.02683 × 10-11
= 2.02683E-11
= 2.02683e-11
= 2.53354 × 10-11
= 2.53354E-11
= 2.53354e-11
### Quick Look: circular thous to ares
circular thou 1 circ th 2 circ th 3 circ th 4 circ th 5 circ th 6 circ th 7 circ th 8 circ th 9 circ th 10 circ th 11 circ th 12 circ th 13 circ th 14 circ th 15 circ th 16 circ th 17 circ th 18 circ th 19 circ th 20 circ th 21 circ th 22 circ th 23 circ th 24 circ th 25 circ th 26 circ th 27 circ th 28 circ th 29 circ th 30 circ th 31 circ th 32 circ th 33 circ th 34 circ th 35 circ th 36 circ th 37 circ th 38 circ th 39 circ th 40 circ th 41 circ th 42 circ th 43 circ th 44 circ th 45 circ th 46 circ th 47 circ th 48 circ th 49 circ th 50 circ th 51 circ th 52 circ th 53 circ th 54 circ th 55 circ th 56 circ th 57 circ th 58 circ th 59 circ th 60 circ th 61 circ th 62 circ th 63 circ th 64 circ th 65 circ th 66 circ th 67 circ th 68 circ th 69 circ th 70 circ th 71 circ th 72 circ th 73 circ th 74 circ th 75 circ th 76 circ th 77 circ th 78 circ th 79 circ th 80 circ th 81 circ th 82 circ th 83 circ th 84 circ th 85 circ th 86 circ th 87 circ th 88 circ th 89 circ th 90 circ th 91 circ th 92 circ th 93 circ th 94 circ th 95 circ th 96 circ th 97 circ th 98 circ th 99 circ th 100 circ th are 5.067075 × 10-12 a 1.013415 × 10-11 a 1.5201225 × 10-11 a 2.02683 × 10-11 a 2.5335375 × 10-11 a 3.040245 × 10-11 a 3.5469525 × 10-11 a 4.05366 × 10-11 a 4.5603675 × 10-11 a 5.067075 × 10-11 a 5.5737825 × 10-11 a 6.08049 × 10-11 a 6.5871975 × 10-11 a 7.093905 × 10-11 a 7.6006125 × 10-11 a 8.10732 × 10-11 a 8.6140275 × 10-11 a 9.120735 × 10-11 a 9.6274425 × 10-11 a 1.013415 × 10-10 a 1.06408575 × 10-10 a 1.1147565 × 10-10 a 1.16542725 × 10-10 a 1.216098 × 10-10 a 1.26676875 × 10-10 a 1.3174395 × 10-10 a 1.36811025 × 10-10 a 1.418781 × 10-10 a 1.46945175 × 10-10 a 1.5201225 × 10-10 a 1.57079325 × 10-10 a 1.621464 × 10-10 a 1.67213475 × 10-10 a 1.7228055 × 10-10 a 1.77347625 × 10-10 a 1.824147 × 10-10 a 1.87481775 × 10-10 a 1.9254885 × 10-10 a 1.97615925 × 10-10 a 2.02683 × 10-10 a 2.07750075 × 10-10 a 2.1281715 × 10-10 a 2.17884225 × 10-10 a 2.229513 × 10-10 a 2.28018375 × 10-10 a 2.3308545 × 10-10 a 2.38152525 × 10-10 a 2.432196 × 10-10 a 2.48286675 × 10-10 a 2.5335375 × 10-10 a 2.58420825 × 10-10 a 2.634879 × 10-10 a 2.68554975 × 10-10 a 2.7362205 × 10-10 a 2.78689125 × 10-10 a 2.837562 × 10-10 a 2.88823275 × 10-10 a 2.9389035 × 10-10 a 2.98957425 × 10-10 a 3.040245 × 10-10 a 3.09091575 × 10-10 a 3.1415865 × 10-10 a 3.19225725 × 10-10 a 3.242928 × 10-10 a 3.29359875 × 10-10 a 3.3442695 × 10-10 a 3.39494025 × 10-10 a 3.445611 × 10-10 a 3.49628175 × 10-10 a 3.5469525 × 10-10 a 3.59762325 × 10-10 a 3.648294 × 10-10 a 3.69896475 × 10-10 a 3.7496355 × 10-10 a 3.80030625 × 10-10 a 3.850977 × 10-10 a 3.90164775 × 10-10 a 3.9523185 × 10-10 a 4.00298925 × 10-10 a 4.05366 × 10-10 a 4.10433075 × 10-10 a 4.1550015 × 10-10 a 4.20567225 × 10-10 a 4.256343 × 10-10 a 4.30701375 × 10-10 a 4.3576845 × 10-10 a 4.40835525 × 10-10 a 4.459026 × 10-10 a 4.50969675 × 10-10 a 4.5603675 × 10-10 a 4.61103825 × 10-10 a 4.661709 × 10-10 a 4.71237975 × 10-10 a 4.7630505 × 10-10 a 4.81372125 × 10-10 a 4.864392 × 10-10 a 4.91506275 × 10-10 a 4.9657335 × 10-10 a 5.01640425 × 10-10 a 5.067075 × 10-10 a
A circular thou is a unit of area, equal to the area of a circle with a diameter of one thou.
Name of unitSymbolDefinitionRelation to SI unitsUnit System
circular thoucirc th
π4 mil2
5.067075×10−10 m2
Imperial/US
#### conversion table
circular thousarescircular thousares
1= 5.067075E-126= 3.040245E-11
2= 1.013415E-117= 3.5469525E-11
3= 1.5201225E-118= 4.05366E-11
4= 2.02683E-119= 4.5603675E-11
5= 2.5335375E-1110= 5.067075E-11
The are (/ˈɑːr/ or /ˈɛər/) is a unit of area, equal to 100 square metres (10 m × 10 m), used for measuring land area. It was defined by older forms of the metric system, but is now outside of the modern International System of Units (SI).
Name of unitSymbolDefinitionRelation to SI unitsUnit System
area
≡ 100 m2
≡ 100 m2
Metric system SI
### conversion table
arescircular thousarescircular thous
1= 197352515997.896= 1184115095987.3
2= 394705031995.787= 1381467611985.2
3= 592057547993.668= 1578820127983.1
4= 789410063991.559= 1776172643981
5= 986762579989.4410= 1973525159978.9
### Conversion table
circular thousares
1= 5.067075 × 10-12
197,352,515,997.89= 1
### Legend
SymbolDefinition
exactly equal
approximately equal to
=equal to
digitsindicates that digits repeat infinitely (e.g. 8.294 369 corresponds to 8.294 369 369 369 369 …) | 2,268 | 4,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.525195 |
https://discourse.processing.org/t/rubix-cube-randomization/6896 | 1,660,371,400,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00026.warc.gz | 212,343,080 | 8,432 | # Rubix Cube Randomization
I’ve created this Rubix Cube Test Sketch, and it works totally fine for first. When I started creating a Randomizer Function to randomize all the little Tiles on each side, it went kind of buggy, and I can’t figure out why it is… it’d be really nice to get some help, and I’m really thankfull on getting help and learning. When you look at the Code, you’ll notice that I made this function “randomize” in the class “cube”, and that’s where the problem is. It always changed up the tiles, but then they kinda all go black except of a few?!
``````Cube cube;
void setup() {
size(600, 450);
cube = new Cube();
cube.randomize();
}
void draw() {
background(0);
stroke(0);
for (int i = 0; i < cube.sides.length; i++) {
for (int j = 0; j < cube.sides[i].tiles.length; j++) {
color temp = check(cube.sides[i].tiles[j].col);
fill(temp);
rect(cube.sides[i].tiles[j].self.x*50, cube.sides[i].tiles[j].self.y*50, 50, 50);
}
}
}
color check(String c) {
color col = color(0);
switch (c) {
case "orange":
col = color(244, 182, 66);
break;
case "white":
col = color(255);
break;
case "green":
col = color(0, 255, 0);
break;
case "yellow":
col = color(251, 255, 63);
break;
case "red":
col = color(255, 0, 0);
break;
case "blue":
col = color(0, 0, 255);
break;
}
return col;
}
class Cube {
Side[] sides = new Side[6];
Cube() {
for (int i = 0; i < 6; i++) {
PVector input = new PVector(0, 0);
String temp = "";
switch (i) {
case 0:
temp = "orange";
input = new PVector(0, 3);
break;
case 1:
temp = "white";
input = new PVector(3, 0);
break;
case 2:
temp = "green";
input = new PVector(3, 3);
break;
case 3:
temp = "yellow";
input = new PVector(3, 6);
break;
case 4:
temp = "red";
input = new PVector(6, 3);
break;
case 5:
temp = "blue";
input = new PVector(9, 3);
break;
}
sides[i] = new Side(temp, input);
}
}
void randomize() {
int numberOfTiles = sides.length * sides[0].tiles.length;
for (int i = 0; i < numberOfTiles; i++) {
int s1 = round(random(0, 5));
int s2 = round(random(0, 5));
if (s2 == s1) {
while (s2 == s1) {
s2 = round(random(0, 5));
}
}
int t1 = round(random(0, 8));
int t2 = round(random(0, 8));
if (t2 == t1) {
while (t2 == t1) {
t2 = round(random(0, 8));
}
}
println(s1, t1, ":", s2, t2);
String tempCol = sides[s1].tiles[t1].col;
PVector tempS = new PVector(sides[s1].tiles[t1].self.x,sides[s1].tiles[t1].self.y);
sides[s2].tiles[t2].self = tempS;
sides[s2].tiles[t2].col = tempCol;
}
}
}
class Side {
Tile[] tiles = new Tile[9];
PVector self;
Side(String col, PVector input) {
self = input;
int y = 0;
for (int i = 0; i < tiles.length; i++) {
PVector s = new PVector(input.x+(i%3), input.y+y);
tiles[i] = new Tile(col, s);
if (i%3 == 2) {
y++;
}
}
}
}
class Tile {
String col;
PVector self;
Tile(String c, PVector s) {
col = c;
self = new PVector(0,0);
self = s;
}
}
``````
1 Like
Your problem is complicated! Break it down into smaller steps and make sure each one works on its own. Normally you would randomize a cube by making ~20 random turns. But that’s a lot! Can you just make one turn? Hint: What would a completed cube that has had a single turn made look like?
You might then realize that the way you are randomizing your cube is, well, wrong. You don’t want to shuffle the tiles around! That’s like pulling all the stickers off and rearranging them! … And that might leave you with a cube you can’t even solve!
Don’t think about it in terms of moving tiles. Thing of it in terms of rotating sides! Figure out a way to rotate one face. Which tiles go where? Then another. Then the other 4. How about making them rotate 180 degrees instead of 90? How about -90? How can your user pick a face to rotate?
Once you have face rotations working, randomizing the cube is easy…
2 Likes
Oh, yes, that was kind of a problem, really… But I still don’t understand what the randomizer failure was just there. Anyways, I’m thankfull you helped me, again. Hope you have a great day!
So I changed a lot of things up now, and with the “working” randomizer, I went to ‘https://ruwix.com/online-rubiks-cube-solver-program/’ to check if it works, and if it was solvable. But the Website Solver told me it was not, and I had to “add every edge once” or something. Here is the current code:
``````Cube cube;
void setup() {
size(600, 450);
cube = new Cube();
cube.randomize();
}
void draw() {
background(0);
stroke(0);
for (int i = 0; i < cube.sides.length; i++) {
for (int j = 0; j < cube.sides[i].tiles.length; j++) {
color temp = check(cube.sides[i].tiles[j].col);
fill(temp);
rect(cube.sides[i].tiles[j].self.x*50, cube.sides[i].tiles[j].self.y*50, 50, 50);
}
}
}
color check(String c) {
color col = color(0);
switch (c) {
case "orange":
col = color(244, 182, 66);
break;
case "white":
col = color(255);
break;
case "green":
col = color(0, 255, 0);
break;
case "yellow":
col = color(251, 255, 63);
break;
case "red":
col = color(255, 0, 0);
break;
case "blue":
col = color(0, 0, 255);
break;
}
return col;
}
void keyPressed() {
if (key == 'q') {
cube.turnMiddleRight(0,3);
} else if (key == 'w') {
cube.turnMiddleRight(3,6);
} else if (key == 'e') {
cube.turnMiddleRight(6,9);
} else if (key == 'a') {
cube.turnMiddleLeft(0,3);
} else if (key == 's') {
cube.turnMiddleLeft(3,6);
} else if (key == 'd') {
cube.turnMiddleLeft(6,9);
} else if (key == 't') {
cube.turnTopDown(0,7);
} else if (key == 'z') {
cube.turnTopDown(1,8);
} else if (key == 'u') {
cube.turnTopDown(2,9);
} else if (key == 'g') {
cube.turnTopUp(0,7);
} else if (key == 'h') {
cube.turnTopUp(1,8);
} else if (key == 'j') {
cube.turnTopUp(2,9);
}
}
class Cube {
Side[] sides = new Side[6];
Cube() {
for (int i = 0; i < 6; i++) {
PVector input = new PVector(0, 0);
String temp = "";
switch (i) {
case 0:
temp = "green";
input = new PVector(0, 3);
break;
case 1:
temp = "yellow";
input = new PVector(3, 0);
break;
case 2:
temp = "orange";
input = new PVector(3, 3);
break;
case 3:
temp = "white";
input = new PVector(3, 6);
break;
case 4:
temp = "blue";
input = new PVector(6, 3);
break;
case 5:
temp = "red";
input = new PVector(9, 3);
break;
}
sides[i] = new Side(temp, input);
}
}
void turnMiddleLeft(int start, int end) {
for (int i = start; i < end; i++) {
String tempCol1 = sides[5].tiles[i].col;
sides[5].tiles[i].col = sides[0].tiles[i].col;
String tempCol2 = sides[4].tiles[i].col;
sides[4].tiles[i].col = tempCol1;
tempCol1 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol2;
sides[0].tiles[i].col = tempCol1;
}
}
void turnMiddleRight(int start, int end) {
for (int i = start; i < end; i++) {
String tempCol1 = sides[0].tiles[i].col;
sides[0].tiles[i].col = sides[5].tiles[i].col;
String tempCol2 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol1;
tempCol1 = sides[4].tiles[i].col;
sides[4].tiles[i].col = tempCol2;
sides[5].tiles[i].col = tempCol1;
}
}
void turnTopUp(int start, int end) {
for (int i = start; i < end; i+=3) {
String tempCol1 = sides[5].tiles[i].col;
sides[5].tiles[i].col = sides[1].tiles[i].col;
String tempCol2 = sides[3].tiles[i].col;
sides[3].tiles[i].col = tempCol1;
tempCol1 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol2;
sides[1].tiles[i].col = tempCol1;
}
}
void turnTopDown(int start, int end) {
for (int i = start; i < end; i+=3) {
String tempCol1 = sides[1].tiles[i].col;
sides[1].tiles[i].col = sides[5].tiles[i].col;
String tempCol2 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol1;
tempCol1 = sides[3].tiles[i].col;
sides[3].tiles[i].col = tempCol2;
sides[5].tiles[i].col = tempCol1;
}
}
void randomize() {
//turnMiddleRight(0, 3); // Turns Middle Top Row to the Right
//turnMiddleLeft(0,3); // Turns Middle Top Row to the Left
//turnMiddleRight(3, 6); // Turns Middle Middle Row to the Right
//turnMiddleLeft(3,6); // Turns Middle Middle Row to the Left
//turnMiddleRight(6, 9); // Turns Middle Bottom Row to the Right
//turnMiddleLeft(6,9); // Turns Middle Bottom Row to the Left
//turnTopDown(0, 7); // Turns Top Left Row Downwards
//turnTopUp(0, 7); // Turns Top Left Row Upwards
//turnTopDown(1, 8); // Turns Top Middle Row Downwards
//turnTopUp(1, 8); // Turns Top Middle Row Upwards
//turnTopDown(2, 9); // Turns Top Right Row Downwards
//turnTopUp(2, 9); // Turns Top Right Row Upwards
int times = round(random(20,40));
for (int i = 0; i < times; i++) {
int dec = round(random(1,2));
if (dec == 1) {
dec = round(random(1,2));
if (dec == 1) {
//Left
dec = round(random(0,2));
int c = 0;
switch (dec) {
case 1:
c = 3;
break;
case 2:
c = 6;
break;
}
turnMiddleLeft(c,c+3);
} else {
//Right
dec = round(random(0,2));
int c = 0;
switch (dec) {
case 1:
c = 3;
break;
case 2:
c = 6;
break;
}
turnMiddleRight(c,c+3);
}
} else {
dec = round(random(1,2));
if (dec == 1) {
dec = round(random(0,2));
turnTopDown(dec,dec+7);
} else {
dec = round(random(0,2));
turnTopUp(dec,dec+7);
}
}
}
}
}
class Side {
Tile[] tiles = new Tile[9];
PVector self;
Side(String col, PVector input) {
self = input;
int y = 0;
for (int i = 0; i < tiles.length; i++) {
PVector s = new PVector(input.x+(i%3), input.y+y);
tiles[i] = new Tile(col, s);
if (i%3 == 2) {
y++;
}
}
}
}
class Tile {
String col;
PVector self;
Tile(String c, PVector s) {
col = c;
self = new PVector(0,0);
self = s;
}
}
``````
“Every edge must be added once.” is the error I go when I input a cube that didn’t have one of each edge pieces. That is, there are 12 cubes in a cube that have two sides colored - these are the 12 edge pieces. The two colored sides on them are always different, and there are four sides of each color on them in total.
It’s possible that your cube got scrambled wrong. Or - more likely - it’s possibly you entered the scrambled cube incorrectly into the checker.
Show up a picture of the scrambled cube you generated and we’ll see if we can point out the problem.
So, I have changed a little code because I found a little failure, but it still gives me the same Failure. I’m also adding two Pictures from my Cube, and two from the Website, so that you could see the scarmbled and standard Version. (tho, since I’m a “new User” I can only post 1 Image and only 2 Links per Post, I’m going to put them on my Website for you, on http://baipyrus.bplaced.net/images/)
``````Cube cube;
void setup() {
size(600, 450);
cube = new Cube();
//cube.randomize();
}
void draw() {
background(0);
stroke(0);
for (int i = 0; i < cube.sides.length; i++) {
for (int j = 0; j < cube.sides[i].tiles.length; j++) {
color temp = check(cube.sides[i].tiles[j].col);
fill(temp);
rect(cube.sides[i].tiles[j].self.x*50, cube.sides[i].tiles[j].self.y*50, 50, 50);
}
}
}
color check(String c) {
color col = color(0);
switch (c) {
case "orange":
col = color(244, 182, 66);
break;
case "white":
col = color(255);
break;
case "green":
col = color(0, 255, 0);
break;
case "yellow":
col = color(251, 255, 63);
break;
case "red":
col = color(255, 0, 0);
break;
case "blue":
col = color(0, 0, 255);
break;
}
return col;
}
void keyPressed() {
if (key == 'q') {
cube.turnMiddleRight(0,3);
} else if (key == 'w') {
cube.turnMiddleRight(3,6);
} else if (key == 'e') {
cube.turnMiddleRight(6,9);
} else if (key == 'a') {
cube.turnMiddleLeft(0,3);
} else if (key == 's') {
cube.turnMiddleLeft(3,6);
} else if (key == 'd') {
cube.turnMiddleLeft(6,9);
} else if (key == 't') {
cube.turnTopDown(0,7);
} else if (key == 'z') {
cube.turnTopDown(1,8);
} else if (key == 'u') {
cube.turnTopDown(2,9);
} else if (key == 'g') {
cube.turnTopUp(0,7);
} else if (key == 'h') {
cube.turnTopUp(1,8);
} else if (key == 'j') {
cube.turnTopUp(2,9);
}
}
class Cube {
Side[] sides = new Side[6];
Cube() {
for (int i = 0; i < 6; i++) {
PVector input = new PVector(0, 0);
String temp = "";
switch (i) {
case 0:
temp = "green";
input = new PVector(0, 3);
break;
case 1:
temp = "yellow";
input = new PVector(3, 0);
break;
case 2:
temp = "orange";
input = new PVector(3, 3);
break;
case 3:
temp = "white";
input = new PVector(3, 6);
break;
case 4:
temp = "blue";
input = new PVector(6, 3);
break;
case 5:
temp = "red";
input = new PVector(9, 3);
break;
}
sides[i] = new Side(temp, input);
}
}
void turnMiddleLeft(int start, int end) {
for (int i = start; i < end; i++) {
String tempCol1 = sides[5].tiles[i].col;
sides[5].tiles[i].col = sides[0].tiles[i].col;
String tempCol2 = sides[4].tiles[i].col;
sides[4].tiles[i].col = tempCol1;
tempCol1 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol2;
sides[0].tiles[i].col = tempCol1;
}
if (start == 0) {
//Turn Upper Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[1].tiles[i].col;
}
sides[1].tiles[2].col = tiles[0];
sides[1].tiles[5].col = tiles[1];
sides[1].tiles[8].col = tiles[2];
sides[1].tiles[1].col = tiles[3];
sides[1].tiles[7].col = tiles[5];
sides[1].tiles[0].col = tiles[6];
sides[1].tiles[3].col = tiles[7];
sides[1].tiles[6].col = tiles[8];
} else if (start == 6) {
//Turn Lower Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[3].tiles[i].col;
}
sides[3].tiles[6].col = tiles[0];
sides[3].tiles[3].col = tiles[1];
sides[3].tiles[0].col = tiles[2];
sides[3].tiles[7].col = tiles[3];
sides[3].tiles[8].col = tiles[6];
sides[3].tiles[5].col = tiles[7];
sides[3].tiles[2].col = tiles[8];
sides[3].tiles[1].col = tiles[5];
}
}
void turnMiddleRight(int start, int end) {
for (int i = start; i < end; i++) {
String tempCol1 = sides[0].tiles[i].col;
sides[0].tiles[i].col = sides[5].tiles[i].col;
String tempCol2 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol1;
tempCol1 = sides[4].tiles[i].col;
sides[4].tiles[i].col = tempCol2;
sides[5].tiles[i].col = tempCol1;
}
if (start == 0) {
//Turn Upper Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[1].tiles[i].col;
}
sides[1].tiles[6].col = tiles[0];
sides[1].tiles[3].col = tiles[1];
sides[1].tiles[0].col = tiles[2];
sides[1].tiles[7].col = tiles[3];
sides[1].tiles[8].col = tiles[6];
sides[1].tiles[5].col = tiles[7];
sides[1].tiles[2].col = tiles[8];
sides[1].tiles[1].col = tiles[5];
} else if (start == 6) {
//Turn Lower Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[3].tiles[i].col;
}
sides[3].tiles[2].col = tiles[0];
sides[3].tiles[5].col = tiles[1];
sides[3].tiles[8].col = tiles[2];
sides[3].tiles[1].col = tiles[3];
sides[3].tiles[7].col = tiles[5];
sides[3].tiles[0].col = tiles[6];
sides[3].tiles[3].col = tiles[7];
sides[3].tiles[6].col = tiles[8];
}
}
void turnTopUp(int start, int end) {
for (int i = start; i < end; i+=3) {
String tempCol1 = "";
if (start == 0) {
tempCol1 = sides[5].tiles[i+2].col;
sides[5].tiles[i+2].col = sides[1].tiles[i].col;
} else if (start == 2) {
tempCol1 = sides[5].tiles[i-2].col;
sides[5].tiles[i-2].col = sides[1].tiles[i].col;
} else {
tempCol1 = sides[5].tiles[i].col;
sides[5].tiles[i].col = sides[1].tiles[i].col;
}
String tempCol2 = sides[3].tiles[i].col;
sides[3].tiles[i].col = tempCol1;
tempCol1 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol2;
sides[1].tiles[i].col = tempCol1;
}
if (start == 0) {
//Turn Left Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[0].tiles[i].col;
}
sides[0].tiles[6].col = tiles[0];
sides[0].tiles[3].col = tiles[1];
sides[0].tiles[0].col = tiles[2];
sides[0].tiles[7].col = tiles[3];
sides[0].tiles[1].col = tiles[5];
sides[0].tiles[8].col = tiles[6];
sides[0].tiles[5].col = tiles[7];
sides[0].tiles[2].col = tiles[8];
} else if (start == 2) {
//Turn Right Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[4].tiles[i].col;
}
sides[4].tiles[2].col = tiles[0];
sides[4].tiles[5].col = tiles[1];
sides[4].tiles[8].col = tiles[2];
sides[4].tiles[1].col = tiles[3];
sides[4].tiles[7].col = tiles[5];
sides[4].tiles[0].col = tiles[6];
sides[4].tiles[3].col = tiles[7];
sides[4].tiles[6].col = tiles[8];
}
}
void turnTopDown(int start, int end) {
for (int i = start; i < end; i+=3) {
String tempCol1 = sides[1].tiles[i].col;
if (start == 0) {
sides[1].tiles[i].col = sides[5].tiles[i+2].col;
} else if (start == 2) {
sides[1].tiles[i].col = sides[5].tiles[i-2].col;
} else {
sides[1].tiles[i].col = sides[5].tiles[i].col;
}
String tempCol2 = sides[2].tiles[i].col;
sides[2].tiles[i].col = tempCol1;
tempCol1 = sides[3].tiles[i].col;
sides[3].tiles[i].col = tempCol2;
if (start == 0) {
sides[5].tiles[i+2].col = tempCol1;
} else if (start == 2) {
sides[5].tiles[i-2].col = tempCol1;
} else {
sides[5].tiles[i].col = tempCol1;
}
}
if (start == 0) {
//Turn Left Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[0].tiles[i].col;
}
sides[0].tiles[2].col = tiles[0];
sides[0].tiles[5].col = tiles[1];
sides[0].tiles[8].col = tiles[2];
sides[0].tiles[1].col = tiles[3];
sides[0].tiles[7].col = tiles[5];
sides[0].tiles[0].col = tiles[6];
sides[0].tiles[3].col = tiles[7];
sides[0].tiles[6].col = tiles[8];
} else if (start == 2) {
//Turn Right Part
String[] tiles = new String[9];
for (int i = 0; i < tiles.length; i++) {
tiles[i] = sides[4].tiles[i].col;
}
sides[4].tiles[6].col = tiles[0];
sides[4].tiles[3].col = tiles[1];
sides[4].tiles[0].col = tiles[2];
sides[4].tiles[7].col = tiles[3];
sides[4].tiles[1].col = tiles[5];
sides[4].tiles[8].col = tiles[6];
sides[4].tiles[5].col = tiles[7];
sides[4].tiles[2].col = tiles[8];
}
}
void L() {
turnTopDown(0, 7);
}
void L1() {
turnTopUp(0, 7);
}
void U() {
turnMiddleLeft(0, 3);
}
void U1() {
turnMiddleRight(0, 3);
}
void D() {
turnMiddleRight(6, 9);
}
void D1() {
turnMiddleLeft(6, 9);
}
void R() {
turnTopDown(2, 9);
}
void R1() {
turnTopUp(2, 9);
}
void randomize() {
//turnMiddleRight(0, 3); // Turns Middle Top Row to the Right
//turnMiddleRight(3, 6); // Turns Middle Middle Row to the Right
//turnMiddleLeft(3,6); // Turns Middle Middle Row to the Left
//turnMiddleLeft(6,9); // Turns Middle Bottom Row to the Left
//turnTopUp(0, 7); // Turns Top Left Row Upwards
//turnTopDown(1, 8); // Turns Top Middle Row Downwards
//turnTopUp(1, 8); // Turns Top Middle Row Upwards
//turnTopUp(2, 9); // Turns Top Right Row Upwards
int times = round(random(20, 40));
for (int i = 0; i < times; i++) {
int dec = round(random(1, 2));
if (dec == 1) {
dec = round(random(1, 2));
if (dec == 1) {
//Left
dec = round(random(0, 2));
int c = 0;
switch (dec) {
case 1:
c = 3;
break;
case 2:
c = 6;
break;
}
turnMiddleLeft(c, c+3);
} else {
//Right
dec = round(random(0, 2));
int c = 0;
switch (dec) {
case 1:
c = 3;
break;
case 2:
c = 6;
break;
}
turnMiddleRight(c, c+3);
}
} else {
dec = round(random(1, 2));
if (dec == 1) {
dec = round(random(0, 2));
turnTopDown(dec, dec+7);
} else {
dec = round(random(0, 2));
turnTopUp(dec, dec+7);
}
}
}
}
}
class Side {
Tile[] tiles = new Tile[9];
PVector self;
Side(String col, PVector input) {
self = input;
int y = 0;
for (int i = 0; i < tiles.length; i++) {
PVector s = new PVector(input.x+(i%3), input.y+y);
tiles[i] = new Tile(col, s);
if (i%3 == 2) {
y++;
}
}
}
}
class Tile {
String col;
PVector self;
Tile(String c, PVector s) {
col = c;
self = new PVector(0,0);
self = s;
}
}
``````
I see a problem with the scrambled cube:
On the far left side in the middle is a white tile.
On the far right side in the middle is a yellow tile.
But there is no edge piece that pairs a white tile with a yellow tile!
The white face and the yellow face are not adjacent!
Bonus problem: Your cube seems to have at least two green-red edges. Which edges are missing?
Some rotation must not be working right.
There are only really 6 operations you can do to a cube, often denoted by: L, R, U, D, F, and B.
That’s what the buttons down below are in the scrambler.
Does you cube change in the same way when you do each of these operations?
You need to do some debugging.
Thanks for these lots of help you gave me, I, again, fixed a lot of things so now I can press every button once (L,L’,R,R’,U,U’,D,D’,F,F’,B,B’) and the Cube is solvable! Though, somehow, when a lot of keys are getting pressed after one another, it gets confused again it is not solvable anymore … Got a lot of things done today, and i’m really thankfull you could help me! | 7,079 | 20,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-33 | latest | en | 0.513106 |
https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_13&diff=next&oldid=61749 | 1,669,706,030,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00245.warc.gz | 141,349,243 | 11,077 | # Difference between revisions of "1958 AHSME Problems/Problem 13"
## Problem
The sum of two numbers is $10$; their product is $20$. The sum of their reciprocals is:
$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$
## Solution
$x+y=10$
$xy=20$
$\frac1x+\frac1y=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}=\frac{10}{20}=\boxed{\frac12\textbf{ (B)}}$
(Which is a lot easier than finding the roots of $x^2-10x+20$.) | 195 | 495 | {"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": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-49 | latest | en | 0.641458 |
https://www.dataunitconverter.com/byte-per-second-to-yobibit-per-second | 1,713,846,934,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00286.warc.gz | 670,666,017 | 16,679 | # Bps to Yibps → CONVERT Bytes per Second to Yobibits per Second
expand_more
info 1 Bps is equal to 0.0000000000000000000000066174449004242213 Yibps
Sec
Min
Hr
Day
Sec
Min
Hr
Day
S = Second, M = Minute, H = Hour, D = Day
Input Bytes per Second (Bps) - and press Enter.
## Bytes per Second (Bps) Versus Yobibits per Second (Yibps) - Comparison
Bytes per Second and Yobibits per Second are units of digital information used to measure storage capacity and data transfer rate.
Bytes per Second is one of the very "basic" digital unit where as Yobibits per Second is a "binary" unit. One Byte is equal to 8 bits. One Yobibit is equal to 1024^8 bits. There are 151,115,727,451,828,646,838,272 Byte in one Yobibit. Find more details on below table.
Bytes per Second (Bps) Yobibits per Second (Yibps)
Bytes per Second (Bps) is a unit of measurement for data transfer bandwidth. It measures the number of Bytes that can be transferred in one Second. Yobibits per Second (Yibps) is a unit of measurement for data transfer bandwidth. It measures the number of Yobibits that can be transferred in one Second.
## Bytes per Second (Bps) to Yobibits per Second (Yibps) Conversion - Formula & Steps
The Bps to Yibps Calculator Tool provides a convenient solution for effortlessly converting data rates from Bytes per Second (Bps) to Yobibits per Second (Yibps). Let's delve into a thorough analysis of the formula and steps involved.
Outlined below is a comprehensive overview of the key attributes associated with both the source (Byte) and target (Yobibit) data units.
Source Data Unit Target Data Unit
Equal to 8 bits
(Basic Unit)
Equal to 1024^8 bits
(Binary Unit)
The formula for converting the Bytes per Second (Bps) to Yobibits per Second (Yibps) can be expressed as follows:
diamond CONVERSION FORMULA Yibps = Bps x 8 ÷ 10248
Now, let's apply the aforementioned formula and explore the manual conversion process from Bytes per Second (Bps) to Yobibits per Second (Yibps). To streamline the calculation further, we can simplify the formula for added convenience.
FORMULA
Yobibits per Second = Bytes per Second x 8 ÷ 10248
STEP 1
Yobibits per Second = Bytes per Second x 8 ÷ (1024x1024x1024x1024x1024x1024x1024x1024)
STEP 2
Yobibits per Second = Bytes per Second x 8 ÷ 1208925819614629174706176
STEP 3
Yobibits per Second = Bytes per Second x 0.0000000000000000000000066174449004242213
Example : By applying the previously mentioned formula and steps, the conversion from 1 Bytes per Second (Bps) to Yobibits per Second (Yibps) can be processed as outlined below.
1. = 1 x 8 ÷ 10248
2. = 1 x 8 ÷ (1024x1024x1024x1024x1024x1024x1024x1024)
3. = 1 x 8 ÷ 1208925819614629174706176
4. = 1 x 0.0000000000000000000000066174449004242213
5. = 0.0000000000000000000000066174449004242213
6. i.e. 1 Bps is equal to 0.0000000000000000000000066174449004242213 Yibps.
Note : Result rounded off to 40 decimal positions.
You can employ the formula and steps mentioned above to convert Bytes per Second to Yobibits per Second using any of the programming language such as Java, Python, or Powershell.
### Unit Definitions
#### What is Byte ?
A Byte is a unit of digital information that typically consists of 8 bits and can represent a wide range of values such as characters, binary data and it is widely used in the digital world to measure the data size and data transfer speed.
arrow_downward
#### What is Yobibit ?
A yobibit (Yib or Yibit) is a binary unit of digital information that is equal to 1,208,925,819,614,629,174,706,176 bits and is defined by the International Electro technical Commission(IEC). The prefix 'yobi' is derived from the binary number system and it is used to distinguish it from the decimal-based 'yottabit' (Yb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
## Excel Formula to convert from Bytes per Second (Bps) to Yobibits per Second (Yibps)
Apply the formula as shown below to convert from 1 Bytes per Second (Bps) to Yobibits per Second (Yibps).
A B C
1 Bytes per Second (Bps) Yobibits per Second (Yibps)
2 1 =A2 * 0.0000000000000000000000066174449004242213
3
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
## Python Code for Bytes per Second (Bps) to Yobibits per Second (Yibps) Conversion
You can use below code to convert any value in Bytes per Second (Bps) to Bytes per Second (Bps) in Python.
bytesperSecond = int(input("Enter Bytes per Second: "))
yobibitsperSecond = bytesperSecond * 8 / (1024*1024*1024*1024*1024*1024*1024*1024)
print("{} Bytes per Second = {} Yobibits per Second".format(bytesperSecond,yobibitsperSecond))
The first line of code will prompt the user to enter the Bytes per Second (Bps) as an input. The value of Yobibits per Second (Yibps) is calculated on the next line, and the code in third line will display the result.
## Frequently Asked Questions - FAQs
#### How many Yobibits(Yibit) are there in a Byte?expand_more
There are 0.0000000000000000000000066174449004242213 Yobibits in a Byte.
#### What is the formula to convert Byte to Yobibit(Yibit)?expand_more
Use the formula Yibit = Byte x 8 / 10248 to convert Byte to Yobibit.
#### How many Bytes are there in a Yobibit(Yibit)?expand_more
There are 151115727451828646838272 Bytes in a Yobibit.
#### What is the formula to convert Yobibit(Yibit) to Byte?expand_more
Use the formula Byte = Yibit x 10248 / 8 to convert Yobibit to Byte.
#### Which is bigger, Yobibit(Yibit) or Byte?expand_more
Yobibit is bigger than Byte. One Yobibit contains 151115727451828646838272 Bytes.
## Similar Conversions & Calculators
All below conversions basically referring to the same calculation. | 1,663 | 5,788 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.818591 |
http://cboard.cprogramming.com/cplusplus-programming/80947-operator-printable-thread.html | 1,480,841,164,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541220.56/warc/CC-MAIN-20161202170901-00001-ip-10-31-129-80.ec2.internal.warc.gz | 44,200,958 | 6,911 | # Or operator?
Show 80 post(s) from this thread on one page
Page 1 of 2 12 Last
• 07-13-2006
Cdrwolfe
Or operator?
Trying to get my Or operator to work and i guess it just isn't.
Code:
switch (rdm)
{
case 0: // Go Forward
{ if (Grid[++x][y] != 1||2||3)
{
Grid[x][y] = Output[j];
Pop[index][j] = 1;
cout << "Forward Picked" << endl;
}
else
{
Grid[--x][y];
cout << "Error" << endl;
}
}
break;
If case = 0, i want it to check the forward array in a 2d array and only procede if there isn't a 1,2 or 3 int in the element.
Can i use an Or operator in the If statement like that?
Thanks
Regards Wolfe
• 07-13-2006
twomers
>> if (Grid[++x][y] != 1||2||3
Code:
if ( Grid[++x][y] != 1 || Grid[++x][y] != 2 || Grid[++x][y] != 3 )
It's long, and convoluted, but necessary. I think there was a post about something similar recently.
EDIT:
This probably does work (the ++x thing for the array), but out of habit I would be more inclined to use Grid[x+1][y], I think it's just a habit and style thing.
• 07-13-2006
Cdrwolfe
Damn made a mistake elsewhere i think
if ( Grid[++x][y] != 1 || Grid[++x][y] != 2 || Grid[++x][y] != 3 )
Hopefully this should work but i have to change it to:
if ( Grid[++x][y] != 1 || Grid[x][y] != 2 || Grid[x][y] != 3 )
So it only increments once.
back in a mo
• 07-13-2006
twomers
Does [++x] actually increment it, or does it just 'peek' to the next one. I genuinally don't know. Just wondering. Oh, please use code tags, even for small snippets, or quite tags if your quoting code.
• 07-13-2006
anon
Are you sure this is what you mean? Wouldn't this be true always?
Example:
a = 5
if a != 1 (true) or ... -> true
a = 1
if a != 1 (false) or a != 2 (true) or ... -> true
May-be you mean, if a!=1 && a!=2 && a!= 3 ? ( !(a==1||a==2||a==3) )
Then may-be you could use:
if a > 3
and/or
if a >=1 && a <= 3
P.S The ++ operator changes the value, so you couldn't use it in multiple evaluations. The code would probably easier to read if you incremented the variable outside the evaluations.
By the way: grid[--x][y]; is a really weird way to decrement x.
• 07-13-2006
Cdrwolfe
Also added a j-- so it repets without losing a turn because of error
and i also removed the check for 3 as it wasn't needed.
here is the whole function for anyone interested.
Yes no strings Anon :D
Code:
void GetFillGrid(int (Output)[10], int (Pop)[PopSize][AAlgh], int (Grid)[X][Y] )
// Fill grid with an Individual and then store the sequence {Forward,Left,Right}
// as one of the population until whole population is created.
{
int x = 12;
int y = 12;
int rdm;
int index;
int j;
for (index = 0; index < PopSize; index++)
{
Grid[x][y] = Output[0];
for ( j = 1; j < PopSize; j++)
{
rdm = rand()%3;
switch (rdm)
{
case 0: // Go Forward
{ if (Grid[++x][y] != 1 && Grid[x][y] != 0 /*|| Grid[x][y] != 3*/)
{
Grid[x][y] = Output[j];
Pop[index][j] = 1;
cout << "Forward Picked" << endl;
}
else
{
Grid[--x][y];
j--;
cout << "Error" << endl;
}
}
break;
case 1: // Go Left
{ if (Grid[x][--y] != 1 && Grid[x][y] != 0 /*|| Grid[x][y] != 3*/)
{
Grid[x][y] = Output[j];
Pop[index][j] = 2;
cout << "Left Picked" << endl;
}
else
{
Grid[x][++y];
j--;
cout << "Error" << endl;
}
}
break;
case 2: // Go Right
{ if (Grid[x][++y] != 1 && Grid[x][y] != 0 /*|| Grid[x][y] != 3*/)
{
Grid[x][y] = Output[j];
Pop[index][j] = 3;
cout << "Right Picked" << endl;
}
else
{
Grid[x][--y];
j--;
cout << "Error" << endl;
}
}
break;
default:
break;
}
}
for (index = 0; index < 24; index++)
{
for (j = 0; j < 24; j++)
{
cout << Grid[index][j] << " ";
}
cout << endl;
cin.get();
}
}
}
• 07-13-2006
twomers
>> (int (Output)[10], int (Pop)[PopSize][AAlgh], int (Grid)[X][Y] )
Just our of curiosity, why do you have the variable name in ()'s? that could be:
Code:
(int Output[10], int Pop[PopSize][AAlgh], int Grid[X][Y] )
if I'm not much mistaken.
• 07-13-2006
Cdrwolfe
Grid is an array i beleive it needs to be in Brackets () in order for it to be called as an argument or parameters something like that.
could be wrong.
• 07-13-2006
twomers
Try taking them out, and see if it works. And post the answer here.
• 07-13-2006
Cdrwolfe
Weird i know i definately read about adding them so i did, but yes your right they are not needed, now i've got to find out where i read it :)
Thanks..
• 07-13-2006
SlyMaelstrom
Why are you starting x and y at 12 and incrementing from there? Are you not concerned about the first 12 elements? (actually 13 if you count that you increment right away) Those kinds of things should have comments expaining why you're doing that. Magic numbers are just magic numbers if we can't see the whole code.
• 07-13-2006
Cdrwolfe
Sorry.....
It's set to 12/12 becuase i need to make sure the sequence created doesn't go "out of bounds"
Here is the wholfe thing if you want a look.
Oh and input has to be over 10 int's
Code:
#include <iostream>
#include <stdio.h>
#include <string>
#define RAND_MAX 3
#include <time.h>
#define PopSize 10
#define AAlgh 12
#define X 24
#define Y 24
using namespace std;
using std::string;
using std::cout;
using std::endl;
using std::cin;
void GetUserInput(string &Input)
{
cout << "Please enter Amino Acid Sequence \"H,P\": " << endl;
cin >> Input;
cin.get();
}
void GetSetupString(string theInput, int Output[10])
// To declare an Array as a reference from Main, syntax is differant then other
// references, ie: Type (&ArryName)[Size], try finding that in any tutorial :)
// Input is now changed into the correct alphabet {1,0} and stored in output
// edit, don't need to reference Arrys.
{
int index;
for (index = 0; index < 10; ++index)
{
switch (theInput[index])
{
case 'H':
Output[index] = 1;
break;
case 'P':
Output[index] = 0;
break;
default:
break;
}
cout << Output[index] ;
}
cout << endl;
cin.get();
}
void GetCreateGrid(int Grid[X][Y])
{
// Fill each element in arry
int index;
int j;
for (index = 0; index < X; index++)
{
for ( j = 0; j < Y; j++)
{
Grid[index][j] = 8;
cout << Grid[index][j] << " " ;
}
cout << endl;
}
cout << "Start Filling Grid" << endl;
}
void GetFillGrid(int Output[10], int Pop[PopSize][AAlgh], int Grid[X][Y] )
// Fill grid with an Individual and then store the sequence {Forward,Left,Right}
// as one of the population until whole population is created.
{
int x;
int y;
int rdm;
int index;
int j;
int i;
for (index = 0; index < PopSize; index++)
{
// clear whole grid again?
x = 12;
y = 12;
Grid[x][y] = Output[0]; // Place first element in [12][12] so it is
// impossible for it to go out of bounds
for ( j = 1; j < PopSize; j++)
{
rdm = rand()%3;
switch (rdm)
{
case 0: // Go Forward
{ if (Grid[++x][y] != 1 && Grid[x][y] != 0 )
{
Grid[x][y] = Output[j];
Pop[index][j] = 1;
cout << "Forward Picked" << endl;
}
else
{
Grid[--x][y];
j--;
cout << "Error" << endl;
}
}
break;
case 1: // Go Left
{ if (Grid[x][--y] != 1 && Grid[x][y] != 0 )
{
Grid[x][y] = Output[j];
Pop[index][j] = 2;
cout << "Left Picked" << endl;
}
else
{
Grid[x][++y];
j--;
cout << "Error" << endl;
}
}
break;
case 2: // Go Right
{ if (Grid[x][++y] != 1 && Grid[x][y] != 0 )
{
Grid[x][y] = Output[j];
Pop[index][j] = 3;
cout << "Right Picked" << endl;
}
else
{
Grid[x][--y];
j--;
cout << "Error" << endl;
}
}
break;
default:
break;
}
}
for (i = 0; i < X; i++)
{
for (j = 0; j < Y; j++)
{
cout << Grid[i][j] << " ";
}
cout << endl;
}
cout << endl;
// cout << "Grid Sequence: " << Pop[index][j] << endl;
for (i = 0; i < X; i++)
{
for ( j = 0; j < Y; j++)
{
Grid[i][j] = 8;
cout << Grid[i][j] << " " ;
}
cout << endl;
}
cout << "Grid Cleared" << endl;
cin.get();
}
}
int main ()
{
srand ( time(NULL) );
string Input;
int Output[10];
int Pop[PopSize][AAlgh];
int Grid[X][Y];
GetUserInput(Input);
GetSetupString(Input, Output);
GetCreateGrid(Grid);
GetFillGrid(Output, Pop, Grid);
cout << "all functions done" << endl;
cin.get();
}
I'am sure there are many newb mistakes and better ways to do what i've done, but i only started really getting into c++ like 5-7 days ago so bear with me thanks.
Regards Wolfe
• 07-13-2006
Mario F.
Quote:
Originally Posted by Cdrwolfe
Weird i know i definately read about adding them so i did, but yes your right they are not needed, now i've got to find out where i read it :)
Thanks..
What you probably read is that to declare a pointer to an array you will in fact need to wrap the name of the array inside parenthesis because of the operator precedence rules.
Code:
int (*myarray)[5]; // declares a pointer to an array of 5 ints
int *myarray[5]; // declares an array of 5 pointers to int
• 07-13-2006
Cdrwolfe
Thanks Mario.
Regards Wolfe
• 07-14-2006
Cdrwolfe
Quick question?
does this code do what iu think it does?
Code:
if (Grid[x-1][y] == 1)
{
Pop[index][11] = ++fitness;
}
If shall we say [x] = 5, and [y] = 5, fitness = 0.
Then does ( Grid[x-1][y] == 1 ) equate to:
If Grid [4][5] contains a '1' then Pop[index][11] = Pop[index][1].
Is this right or wrong so far it just doesn't work as expected so i thought someone may know why.
Regards Wolfe
Show 80 post(s) from this thread on one page
Page 1 of 2 12 Last | 2,964 | 9,041 | {"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-2016-50 | latest | en | 0.858935 |
http://www.sandiegouniontribune.com/sdut-retirement-iq-test-022709-2009feb27-story.html | 1,508,742,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825700.38/warc/CC-MAIN-20171023054654-20171023074654-00366.warc.gz | 554,237,695 | 46,118 | Retirement: Test your financial planning IQ
The formula for a financially successful retirement used to be straightforward: Work for decades for one employer and then live happily ever after on the pension, Social Security and whatever personal savings you were able to amass.
With regular checks from your company and Uncle Sam, the amount of savings was important but not critical.
Today, with pensions vanishing and an economic crisis withering savings, it's increasingly up to individuals to take charge of their finances to fund retirements that can stretch for up to 30 years because of longer lifespans.
Is your retirement IQ up to the challenge? Take the test and find out. (Answers at bottom):
1. What percentage of your savings can you withdraw annually in retirement without risk of running out of money?
(a) 3 percent (b) 4 percent (c) 7 percent (d) 10 percent
2. Approximately what percentage of pre-retirement income is generally needed to maintain a person's current lifestyle in retirement?
(a) 45 to 60 percent (b) 60 to 75 percent (c) 75 to 99 percent (d) 100 percent or more
3. Working full-time for three years past one's anticipated retirement date and continuing to save 15 percent of salary could raise annual retirement income by how much?
(a) 7 percent (b) 12 percent (c) 17 percent (d) 22 percent
4. At what age will most of today's workers be eligible for full Social Security retirement benefits?
(a) 62 or 63 (b) 64 or 65 (c) 66 or 67 (d) 70
5. How much extra can workers 50 or older contribute to their retirement plans in 2009?
(a) \$6,000 extra, for a maximum \$21,000 (b) \$5,500 extra, for a maximum \$22,000 (c) \$7,000 extra, for a maximum \$24,000 (d) \$7,500 extra, for a maximum \$25,000.
6. The typical person age 50 and older with a 401(k) account with his or her current employer holds about how much in the account?
(a) \$47,000 (b) \$97,000 (c) \$147,000 (d) \$247,000
7. The number of workers age 65 and over is expected to grow by how much over the next decade?
(a) More than 20 percent (b) More than 40 percent (c) More than 60 percent (d) More than 80 percent
8. What percent of homeowners age 50 to 65 plan to use home equity to finance ordinary living expenses in retirement?
(a) 6 percent (b) 10 percent (c) 20 percent (d) 50 percent
9. A job layoff in one's 50s or 60s typically reduces total household wealth by what percent?
(a) 11 percent for married couples and 23 percent for single people (b) 16 percent for married couples and 28 percent for single people (c) 21 percent for married couples and 33 percent for single people (d) 31 percent for married couples and 43 percent for single people
10. What percent of U.S. workers are covered by traditional defined-benefit retirement plans (pensions)?
(a) 10 percent (b) 20 percent (c) 50 percent (d) 75 percent
–––
1. (b) The 4 percent rule advocated by many financial planners holds that if you withdraw no more than 4 percent of your portfolio in the first year of retirement and then increase that amount for inflation each year, your money should last at least 30 years. That rough guideline takes into consideration the role of expected earnings on your portfolio as well as inflation.
2. (c) Most employees will need an average of between 77 and 94 percent, according to Aon Consulting's 2008 Replacement Ratio Study. It's best to plan for the high side since health and medical costs are impossible to predict. Also, people tend to spend more money when they have more leisure time.
3. (d) Investment management company T. Rowe Price says retirement income goes up about 7 percent for each additional year of work, or around 22 percent after three years.
4. (c) Between 66 and 67, depending on date of birth. Retire before that and your benefits will be reduced by 20 to 30 percent. Check retirement benefits by year of birth at the Social Security site ( http://ssa.gov ) for your specifics. Knowing the right age is essential to making retirement plans that won't leave you short of money.
5. (b) Employees age 50 and up can make up to \$5,500 in catch-up contributions in 2009, added to a base contribution limit of \$16,500 for a maximum \$22,000.
6. (b) \$96,809 as of Feb. 26, according to the Employee Benefit Research Institute. Whether you're ahead or behind your neighbors and peers in retirement savings, don't lose sight of the long term – keep saving.
7. (d) The number of workers age 65 or older is predicted to soar by more than 80 percent by 2016 (from 2006 totals), according to the U.S. Bureau of Labor Statistics.
8. (a) 6 percent, according to a 2007 report by the Center for Retirement Research at Boston College. You should try to avoid tapping home equity for routine retirement expenses if possible. The housing crisis has called into question projections that home equity is likely to become an increasingly important source of retirement income.
9. (c) 21 percent for married couples and 33 percent for single people, according to a 2007 report by the Urban Institute. These statistics serve as a warning to build up emergency savings and not cut things too close in case of an unexpected job loss late in your working career.
10. (b) AARP says only 20 percent are covered, meaning most people have to look after their own finances with such vehicles as 401(k)s and IRAs for their nest eggs.
–––
SCORING SYSTEM:
0-3: Better bone up fast or you'll have to keep working till you drop.
4-5: You need to study some more.
6-7: Not bad, but the road to retirement affluence could still be bumpy for you.
8-9: If your planning matches your knowledge, you should be in good shape.
10: Congratulations! May you live long and prosper.
Copyright © 2017, The San Diego Union-Tribune | 1,397 | 5,751 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-43 | latest | en | 0.928857 |
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