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http://spot.pcc.edu/math/orcca/section-exploring-two-variable-data-and-rate-of-change.html | 1,558,652,002,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257432.60/warc/CC-MAIN-20190523224154-20190524010154-00076.warc.gz | 187,409,554 | 18,879 | ## Section4.3Exploring Two-Variable Data and Rate of Change
###### ObjectivesPCC Course Content and Outcome Guide
This section is about examining data that has been plotted on a Cartesian coordinate system, and then making observations. In some cases, we'll be able to turn those observations into useful mathematical calculations.
### Subsection4.3.1Modeling data with two variables
Using mathematics, we can analyze real data from the world around us. We can use what we discover to better understand the world, and sometimes to make predictions. Here's an example of data about the economic situation in the US:
If this trend continues, what percentage of all income will the top 1 % have in the year 2030? If we model data in the chart with the trend line, we can estimate the value to be 28.6 %. This is one way math is used in real life.
Does that trend line have an equation like those we looked at in Section 4.2? Is it even correct to look at this data set and decide that a straight line is a good model? These are some of the questions we want to consider as we begin this section. The answers will evolve through the next several sections.
### Subsection4.3.2Patterns in Tables
###### Example4.3.3
Find a pattern in each table. What is the missing entry in each table? Can you describe each pattern in words and/or mathematics?
Explanation
First table
Each word on the right has the opposite meaning of the word to its left.
Second table
Each city on the right is the capital of the country to its left.
Third table
Each number on the right is double the number to its left.
We can view each table as assigning each input in the left column a corresponding output in the right column. In the first table, for example, when the input “big” is on the left, the output “small” is on the right. The first table's function is to output a word with the opposite meaning of each input word. (This is not a numerical example.)
The third table is numerical. And its function is to take a number as input, and give twice that number as its output. Mathematically, we can describe the pattern as “$y=2x\text{,}$” where $x$ represents the input, and $y$ represents the output. Labeling the table mathematically, we have Table 4.3.6.
$x$ (input) $y$ (output) $1$ $2$ $2$ $4$ $3$ $6$ $5$ $10$ $10$ $20$ Pattern: $y=2x$
The equation $y=2x$ summarizes the pattern in the table. For each of the following tables, find an equation that describes the pattern you see. Numerical pattern recognition may or may not come naturally for you. Either way, pattern recognition is an important mathematical skill that anyone can develop. Solutions for these exercises provide some ideas for recognizing patterns.
### Subsection4.3.3Rate of Change
For an hourly wage-earner, the amount of money they earn depends on how many hours they work. If a worker earns $\15$ per hour, then $10$ hours of work corresponds to $\150$ of pay. Working one additional hour will change $10$ hours to $11$ hours; and this will cause the $\150$ in pay to rise by fifteen dollars to $\165$ in pay. Any time we compare how one amount changes (dollars earned) as a consequence of another amount changing (hours worked), we are talking about a rate of change.
Given a table of two-variable data, between any two rows we can compute a rate of change.
###### Example4.3.10
The following data, given in both table and graphed form, gives the counts of invasive cancer diagnoses in Oregon over a period of time. (wonder.cdc.gov)
Year Invasive Cancer Incidents 1999 17,599 2000 17,446 2001 17,847 2002 17,887 2003 17,559 2004 18,499 2005 18,682 2006 19,112 2007 19,376 2008 20,370 2009 19,909 2010 19,727 2011 20,636 2012 20,035 2013 20,458
What is the rate of change in Oregon invasive cancer diagnoses between 2000 and 2010? The total (net) change in diagnoses over that timespan is
\begin{equation*} 19727-17446=2281\text{.} \end{equation*}
Since $10$ years passed (which you can calculate as $2010-2000$), the rate of change is $2281$ diagnoses per $10$ years, or
\begin{equation*} \frac{2281\,\text{diagnoses}}{10\,\text{year}}=228.1\,\frac{\text{diagnoses}}{\text{year}}\text{.} \end{equation*}
We read that last quantity as “$228.1$ diagnoses per year.” This rate of change means that between the years $2000$ and $2010\text{,}$ there were $228.1$ more diagnoses each year, on average. (Notice that there was no single year in that span when diagnoses increased by $228.1\text{.}$)
Let's practice calculating rates of change over different timespans:
###### Checkpoint4.3.11
We are ready to give a formal definition for rate of change. Considering our work from Example 4.3.10 and Checkpoint 4.3.11, we settle on:
###### Definition4.3.12Rate of Change
If $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are two data points from a set of two-variable data, then the rate of change between them is
\begin{equation*} \frac{\text{change in $y$}}{\text{change in $x$}}=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}\text{.} \end{equation*}
(The Greek letter delta, $\Delta\text{,}$ is used to represent “change in” since it is the first letter of the Greek word for “difference.”)
In Example 4.3.10 and Checkpoint 4.3.11 we found three rates of change. Figure 4.3.13 highlights the three pairs of points that were used to make these calculations.
Note how the larger the numerical rate of change between two points, the steeper the line is that connects them. This is such an important observation, we'll put it in an official remark.
###### Remark4.3.14
The rate of change between two data points is intimately related to the steepness of the line segment that connects those points.
1. The steeper the line, the larger the rate of change, and vice versa.
2. If one rate of change between two data points equals another rate of change between two different data points, then the corresponding line segments will have the same steepness.
3. When a line segment between two data points slants down from left to right, the rate of change between those points will be negative.
In the solution to Checkpoint 4.3.8, the key observation was that the rate of change from one row to the next was constant: $3$ units of increase in $y$ for every $1$ unit of increase in $x\text{.}$ Graphing this pattern in Figure 4.3.15, we see that every line segment here has the same steepness, so the entire graph is a line.
Whenever the rate of change is constant no matter which two $(x,y)$-pairs (or data pairs) are chosen from a data set, then you can conclude the graph will be a straight line even without making the graph. We call this kind of relationship a linear relationship. We'll study linear relationships in more detail throughout this chapter. Right now in this section, we feel it is important to simply identify if data has a linear relationship or not.
###### Checkpoint4.3.18
Let's return to the data that we opened the section with, in Figure 4.3.2. Is that data linear? Well, yes and no. To be completely honest, it's not linear. It's easy to pick out pairs of points where the steepness changes from one pair to the next. In other words, the points do not all fall into a single line.
However if we stand back, there does seem to be an overall upward trend that is captured by the line someone has drawn over the data. Points on this line do have a linear pattern. Let's estimate the rate of change between some points on this line. We are free to use any points to do this, so let's make this calculation easier by choosing points we can clearly identify on the graph: $(1991,15)$ and $(2020,25)\text{.}$
The rate of change between those two points is
\begin{equation*} \frac{25-15}{2020-1991}=\frac{10}{29}\approx0.3448\text{.} \end{equation*}
So we might say that on average the rate of change expressed by this data is 0.3448 %yr.
### Subsection4.3.4Exercises
###### 1
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $-2$ ${-6}$ $-1$ ${-3}$ $0$ ${0}$ $1$ ${3}$ $2$ ${6}$
###### 2
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $3$ ${12}$ $4$ ${16}$ $5$ ${20}$ $6$ ${24}$ $7$ ${28}$
###### 3
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $5$ ${11}$ $6$ ${12}$ $7$ ${13}$ $8$ ${14}$ $9$ ${15}$
###### 4
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $6$ ${10}$ $7$ ${11}$ $8$ ${12}$ $9$ ${13}$ $10$ ${14}$
###### 5
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $15$ ${23}$ $13$ ${21}$ $6$ ${14}$ $4$ ${12}$ $1$ ${9}$
###### 6
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $17$ ${16}$ $6$ ${5}$ $14$ ${13}$ $19$ ${18}$ $1$ ${0}$
###### 7
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $25$ ${5}$ $1$ ${1}$ $4$ ${2}$ $16$ ${4}$ $9$ ${3}$
###### 8
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $-5$ ${5}$ $-2$ ${2}$ $-3$ ${3}$ $-2$ ${2}$ $-4$ ${4}$
###### 9
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $2$ ${4}$ $3$ ${9}$ $4$ ${16}$ $5$ ${25}$ $6$ ${36}$
###### 10
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $7$ ${49}$ $9$ ${81}$ $11$ ${121}$ $13$ ${169}$ $15$ ${225}$
###### 11
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $43$ ${{\frac{1}{43}}}$ $62$ ${{\frac{1}{62}}}$ $84$ ${{\frac{1}{84}}}$ $58$ ${{\frac{1}{58}}}$ $1$ ${1}$
###### 12
Write an equation in the form $y=\ldots$ suggested by the pattern in the table.
$x$ $y$ $54$ ${{\frac{1}{54}}}$ $27$ ${{\frac{1}{27}}}$ $25$ ${{\frac{1}{25}}}$ $33$ ${{\frac{1}{33}}}$ $99$ ${{\frac{1}{99}}}$
###### 13
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $0$ ${93}$ $1$ ${100}$ $2$ ${107}$ $3$ ${114}$ $4$ ${121}$ $5$ ${128}$
###### 14
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $0$ ${62}$ $1$ ${70}$ $2$ ${78}$ $3$ ${86}$ $4$ ${94}$ $5$ ${102}$
###### 15
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $5$ ${51}$ $6$ ${49}$ $7$ ${47}$ $8$ ${45}$ $9$ ${43}$ $10$ ${41}$
###### 16
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $10$ ${74}$ $11$ ${72}$ $12$ ${70}$ $13$ ${68}$ $14$ ${66}$ $15$ ${64}$
###### 17
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $3$ ${19}$ $4$ ${27}$ $5$ ${43}$ $6$ ${75}$ $7$ ${139}$ $8$ ${267}$
###### 18
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $8$ ${260}$ $9$ ${516}$ $10$ ${1028}$ $11$ ${2052}$ $12$ ${4100}$ $13$ ${8196}$
###### 19
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $0$ ${17}$ $1$ ${18}$ $2$ ${25}$ $3$ ${44}$ $4$ ${81}$ $5$ ${142}$
###### 20
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $1$ ${11}$ $2$ ${18}$ $3$ ${37}$ $4$ ${74}$ $5$ ${135}$ $6$ ${226}$
###### 21
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $-10$ ${35.92}$ $-9$ ${36.32}$ $-8$ ${36.72}$ $-7$ ${37.12}$ $-6$ ${37.52}$ $-5$ ${37.92}$
###### 22
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $-2$ ${82.57}$ $-1$ ${84.08}$ $0$ ${85.59}$ $1$ ${87.1}$ $2$ ${88.61}$ $3$ ${90.12}$
###### 23
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $5$ ${85}$ $10$ ${125}$ $12$ ${141}$ $16$ ${173}$ $17$ ${181}$ $18$ ${189}$
###### 24
Does the following table show that $x$ and $y$ have a linear relationship?
• yes
• no
$x$ $y$ $1$ ${18}$ $5$ ${50}$ $7$ ${66}$ $13$ ${114}$ $16$ ${138}$ $19$ ${162}$
###### 25
This table gives population estimates for Portland, Oregon from 1990 through 2014.
Year Population Year Population 1990 487849 2003 539546 1991 491064 2004 533120 1992 493754 2005 534112 1993 497432 2006 538091 1994 497659 2007 546747 1995 498396 2008 556442 1996 501646 2009 566143 1997 503205 2010 585261 1998 502945 2011 593859 1999 503637 2012 602954 2000 529922 2013 609520 2001 535185 2014 619360 2002 538803
Find the rate of change in Portland population between 2005 and 2006. Just give the numerical value; the units are provided.
$\,\frac{\text{people}}{\text{year}}$
And what was the rate of change between 2008 and 2014?
$\,\frac{\text{people}}{\text{year}}$
List all the years where there is a negative rate of change between that year and the next year.
###### 26
This table and graph gives population estimates for Portland, Oregon from 1990 through 2014.
Year Population Year Population 1990 487849 2003 539546 1991 491064 2004 533120 1992 493754 2005 534112 1993 497432 2006 538091 1994 497659 2007 546747 1995 498396 2008 556442 1996 501646 2009 566143 1997 503205 2010 585261 1998 502945 2011 593859 1999 503637 2012 602954 2000 529922 2013 609520 2001 535185 2014 619360 2002 538803
Between what two years that are two years apart was the rate of change highest?
What was that rate of change? Just give the numerical value; the units are provided.
$\,\frac{\text{people}}{\text{year}}$ | 4,238 | 13,490 | {"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": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2019-22 | latest | en | 0.911825 |
http://oeis.org/A321428 | 1,548,057,397,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583763839.28/warc/CC-MAIN-20190121070334-20190121092334-00225.warc.gz | 160,619,317 | 3,673 | This site is supported by donations to The OEIS Foundation.
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A321428 Expansion of Product_{i>0, j>0} (1 + x^(i^2 + j^2)). 5
1, 0, 1, 0, 0, 2, 0, 2, 1, 0, 4, 0, 3, 4, 0, 8, 0, 6, 8, 2, 13, 2, 9, 14, 4, 22, 8, 16, 24, 8, 35, 18, 28, 38, 19, 52, 34, 46, 60, 40, 78, 58, 76, 94, 75, 120, 93, 124, 140, 126, 183, 150, 200, 210, 204, 276, 239, 308, 319, 316, 417, 366, 465, 480, 484, 620, 554 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,6 LINKS Seiichi Manyama, Table of n, a(n) for n = 0..10000 FORMULA G.f.: Product_{k>0} (1 + x^k)^A063725(k). MATHEMATICA nmax = 100; A063725 = Rest[CoefficientList[Series[(EllipticTheta[3, 0, x] - 1)^2/4, {x, 0, nmax}], x]]; s = 1; Do[s *= Sum[Binomial[A063725[[k]], j]*x^(j*k), {j, 0, nmax/k}]; s = Expand[s]; s = Take[s, Min[nmax + 1, Exponent[s, x] + 1, Length[s]]]; , {k, 2, nmax}]; Take[CoefficientList[s, x], nmax + 1] (* Vaclav Kotesovec, Nov 09 2018 *) CROSSREFS Cf. A063725, A321380, A321429. Sequence in context: A058548 A157030 A080844 * A076626 A182886 A108731 Adjacent sequences: A321425 A321426 A321427 * A321429 A321430 A321431 KEYWORD nonn AUTHOR Seiichi Manyama, Nov 09 2018 STATUS approved
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Last modified January 20 23:20 EST 2019. Contains 319343 sequences. (Running on oeis4.) | 660 | 1,571 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-04 | latest | en | 0.543717 |
http://www.chess.com/blog/ADK/all-the-queens-knights | 1,371,542,099,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707186142/warc/CC-MAIN-20130516122626-00062-ip-10-60-113-184.ec2.internal.warc.gz | 319,599,095 | 14,071 | # All the Queen's Knights
• ADK
• | Sep 20, 2009 at 11:30 AM
• | Posted in: ADK's Blog
• | 1986 reads
• | 25 comments
In this game, Jose Capablanca faces off against Angel Arnal in Barcelona, 1935. I want to take the time to analyze the position right after White plays 12. d5 which offers up a free Pawn for Black. What is the great Capablanca thinking? Well, White's mindset when he sacrifices his Pawn is to remove the defender of the h5 and h7 by forcing the guardian of those two squares to e7 after a Bishop-Bishop exchange. We then arrive to a position where Capablanca is about to play a series of stunning moves that will leave Arnal in one sticky mess. So, analyze the position below and play like Capablanca:
Solution & Explanation: Could you find the winning moves in that position? Well, the solution is 14. Bxh7+?! Kxh7 15. Ng5+ Kg8 16. Rxd7?! Qxd7 17. Qh5 Rd8 18. Qxf7+ Kh8 19. h4 Nf5 20. Nh5 Qe8 21. Nf6! 14. Bxh7+?! forces the King out into the open. 15. Ng5+ forces the King back to g8, not h6 or h8, because of 16. Qh5+-. 16. Rxd7?! is a seemingly crazy move, but upon close inspection, the reader may see that this move got rid of the pesky Knight that IF placed on f6 it would throw a wrench into White's plans. 17. Qh5 threatens mate in 1. 18. Qxf7+ takes advantage of the absence of Black's Rook to isolate his King in a corner. 19. h4 takes the pressure off of the Knight on g3 and makes it movable again. 20. Nh5 further entangles Black's position. 21. Nf6! assures that IF 21...gxf6 then [22. Qh7#] or IF 21...Qxf7 then [22. Nxf7#].
### Comments
• 4 years ago
You are correct
• 4 years ago
Ah...
• 4 years ago
Comments are very interesting here:
http://www.chessgames.com/perl/chessgame?gid=1242938
• 4 years ago
... BTW, ?! means dubious. !? means interesting. So I think you had a little typo there.
The solution was Bh7 (a dubious move). Lol.
sethisrael , if this position is not resign worthy to you, look at the following continuation after 21... Nh6.
21... Nh6
22.Ne8 Nf7
23.Nf7+ Kg8
24.Nd8
White is up a whole piece and a pawn, and can play something like Nc7 and Ne6 to protect both knights.
• 4 years ago
Well, ADK you're not wrong... white are wining this game and Sethisrael is also correct... Nf6 is the best defense. Acourding to my computer after Nf6 white are still wining but no mate for white.
• 4 years ago
Nothing that leaps out to me. Maybe a better player than I can spot something. Sorry.
• 4 years ago
Ouch, you are right, sethisrael! I forgot about the Knight on the h6 square and thank you for pointing that miscalculation out. Can you find any moves that would lead to mate if that following variation would have been used?
ADK
• 4 years ago
"Black did not reply with 21...Nh6 because of 22. Qxe8+ Rxe8 and then Nf7#!"
Don't tell someone they're wrong if your refutation is, itself, wrong. After 21...Nh6, 22. Qxe8+ Rxe8 and then Nf7 is not mate, and definitely doesn't deserve a "!". Black responds with Nxf7, and white is down a minor piece. As it stands, white is in a better position and will be up a pawn after 22. Qxe8+ Rxe8 23. Nxe8. Not resign worthy to me.
• 4 years ago
freaking sweet
• 4 years ago
Yes, you're right! My bad :)
• 4 years ago
"....21. nh6"--JoeFD
Black did not reply with 21...Nh6 because of 22. Qxe8+ Rxe8 and then Nf7#!
"Why not 14.Qe4! with double threat Qh7 X mate and Qxa8 - winning a rook."
"It seems to me that this is not a Capablancas' game."--lubo
A simple reply would be 14...Nf6 defending mate and after White plays 15. Qxa8 Black would counteract that with 15...Nc6! and the Queen would be trapped and eventually captured.
ADK
• 4 years ago
Why not 14.Qe4! with double threat Qh7 X mate and Qxa8 - winning a rook.
It seems to me that this is not a Capablancas' game.
• 4 years ago
....21. nh6
• 4 years ago
Outstanding chess vision by Capablanca
• 4 years ago
''He didn't take QxQ because then NxQ+ followed by NXR''
If black takes whites queen, Nf7 is checkmate.
• 4 years ago
wow.
• 4 years ago
"Ddub -
He didn't take QxQ because then NxQ+ followed by NXR"
Wouldn't Black still be winning though?
7. Nh5 Qxf7 8. Nxf7+ Kg8 9. Nxd8 Bd7 10. Nb7 Bc6 11. Nxc5 bxc5 12. Re1 Bd5 13. Nf4 Bxa2 14. h5 Nd4 15. c3 Nc6 {(0:00:17) 352kN}{[%eval -146,13,Rybka 3 1-cpu w32]}
*
• 4 years ago
brilliant play by cap... the last knight move is so stylish and down to the point!
• 4 years ago
Woww .. to be able to see the combination itself is brilliant!!
• 4 years ago
Ddub -
He didn't take QxQ because then NxQ+ followed by NXR
Back to Top | 1,455 | 4,581 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2013-20 | latest | en | 0.88843 |
http://www.haverford.edu/computerscience/courses/cmsc106/twenty_questions.html | 1,432,648,308,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928864.16/warc/CC-MAIN-20150521113208-00107-ip-10-180-206-219.ec2.internal.warc.gz | 485,060,598 | 3,131 | Introduction to Data Structures CS106 Spring 2012
### Lab 5: Abstraction and Uses of Binary Trees
Check out the twenty_questions project for Lab 5. This contains the DocTest infrastructure but nothing else useful you will need to enter the classes and algorithms for this project from scratch. There are two major parts to this lab:
1. Create a class BinaryTree to represent binary trees in Python, in the file BinaryTree.py.
• Your class should provide binary tree constructors, accessor operations to retrieve the value at the root and the left and right subtrees, and appropriate equality testing (__eq__, for ==) and abstraction functions (__repr__).
• Create a test suite for your BinaryTree class, at the top of BinaryTree.py.
• You may represent the binary trees with the “list of lists” representation from 6.4.1 or the “nodes and references” representation from 6.4.2 (or some other representation if you can come up with one). You are welcome to copy as much as you like from the textbook, but you must cite it appropriately if you do so.
• You must provide axioms for all of your tree operations, so you should make sure that you have “pure-functional” versions of the fundamental constructors and the accessors. You may optionally include mutator operations as well if you like, but make sure to provide axioms for them (and tests in your test suite) if you do.
2. In the file game.py, create a program to play the “20 questions” game for animals, along the lines discussed in lecture (your program must take the “guess the secret thing” role rather than the role of coming up with something and answering questions). You should use the BinaryTree class to represent the questions (as the values of the non-leaf nodes) and possible results (the values of the leaf nodes). You should do this in two steps:
1. Start by writing a version that has a pre-built tree with at least six leaves (this will be a variable in the game procedure) and asks the questions from the tree until it arrives at a leaf, and then ask whether or not it has identified the mystery animal. If it does not know the animal, it can just print something like “rats I didn't get it”. Since this is game is almost entirely composed of user interface operations, you don't need to have a test suite, but you can cut and paste the output of various test runs into a comment at the top, as documentation for the user.
2. Enhance your program (and, if necessary, your BinaryTree class, along with its axioms and test suite) so that the can be played repeatedly, with the program adding new animals to the tree when it fails to guess correctly. When told that it has not identified the mystery animal, your program should ask what it was and then ask for a question to distinguish it from its final guess. It should use this information to adjust the tree accordingly for the next game (you do not need to keep the tree balanced). At the completion of a series of games, you should print the modified tree. (If you like, you can cut and paste the new tree back into the program so that it will be enhanced permanently, for the next round in fact, you are invited to think about how to automate this).
As always, add a README.txt file and a README-DESIGN.txt file. Then use Team->Commit to hand in your work as well as Blackboard (and, of course, you can also use commit to make backups also at each significant milestone, e.g., at the end of each bullet item above). | 742 | 3,448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-22 | longest | en | 0.921387 |
https://www.doubtnut.com/qna/644746697 | 1,725,741,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00590.warc.gz | 724,840,120 | 37,841 | # The circle described on the line joining the points (0,1),(a,b) as diameter cuts the x-axis in points whose abscissae are roots of the equation
A
x2+ax+b=0
B
x2ax+b=0
C
x2+axb=0
D
x2axb=0
Video Solution
Text Solution
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## To find the equation whose roots are the abscissae of the points where the circle described on the line joining the points (0, 1) and (a, b) as diameter cuts the x-axis, we can follow these steps:Step 1: Identify the endpoints of the diameterThe endpoints of the diameter of the circle are given as:- Point 1: (0,1)- Point 2: (a,b)Step 2: Use the diameter form of the circle's equationThe equation of a circle with endpoints of the diameter at (x1,y1) and (x2,y2) can be expressed as:(x−x1)(x−x2)+(y−y1)(y−y2)=0Substituting the points (0,1) and (a,b):(x−0)(x−a)+(y−1)(y−b)=0This simplifies to:x(x−a)+(y−1)(y−b)=0Step 3: Substitute y=0 to find the points where the circle cuts the x-axisTo find the points where the circle intersects the x-axis, we set y=0:x(x−a)+(0−1)(0−b)=0This simplifies to:x(x−a)+b=0Step 4: Rearrange the equationExpanding and rearranging the equation gives:x2−ax+b=0Step 5: Identify the quadratic equationThe equation x2−ax+b=0 is the quadratic equation whose roots are the x-coordinates (abscissae) of the points where the circle intersects the x-axis.ConclusionThus, the required equation is:x2−ax+b=0
|
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Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation | 690 | 2,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-38 | latest | en | 0.806416 |
https://didactalia.net/comunidad/materialeducativo/categoria/brightstorm/785b3963-d3d6-4f97-9271-e3e225bc0e06 | 1,606,786,520,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00536.warc.gz | 256,158,780 | 27,448 | ¿Qué puedo hacer?
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Hipervinculo Matemáticas
## Writing an Equation
"If a student learns about writing an equation to describe a picture or a pattern, it will help him or her understand how graphs and equations work and how they apply to the real world. Writing an equ ...
Video Matemáticas
## Work Word Problems
"It is possible to solve word problems when two people are doing a work job together by solving systems of equations. To solve a work word problem, multiply the hourly rate of the two people working t ...
Hipervinculo Física
## Work - Energy Theorem
"According to the work-energy theorem, the net work on an object causes a change in the kinetic energy of the object. The formula for net work is net work = change in kinetic energy = final kinetic en ...
Video Matemáticas
## Why SSA and AAA Don't Work as Congruence Shortcuts
"Four shortcuts allow students to know two triangles must be congruent: SSS, SAS, ASA, and AAS. Knowing only side-side-angle (SSA) does not work because the unknown side could be located in two differ ...
Video Física
## Wavelength
"Wavelength is the distance between two crests of a wave. It is related to frequency and period by the equations wave speed = frequency x wavelength and wave speed = wavelength / period." Fuente: Brig ...
Hipervinculo Física
## Wave Intensity (BrightStorm)
"Wave intensity is the average power that travels through a given area as the wave travels through space. The intensity of sound waves is measured using the decibel scale." Fuente: BrightSto ...
Video Física
## Wave Speed
"Wave speed is the speed at which a wave travels. Wave speed is related to wavelength, frequency, and period by the equation wave speed = frequency x wavelength. The most commonly used wave speed is t ...
Video Física
## Wave Phase
"Wave phase is the offset of a wave from a given point. When two waves cross paths, they either cancel each other out or compliment each other, depending on their phase. These effects are called const ...
Video Física
## Wave Inversion
"Wave inversion occurs when a traveling wave is flipped upside down when being reflected back into the old medium. This happens during the process of a wave moving through one medium encounters a boun ...
Hipervinculo Física
## Wave Interference
"Wave interference is the result of the interactions of multiple waves. Types of interference include constructive and destructive interferences. The difference between the two is the relative displac ...
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Con la tecnología GNOSS | 786 | 3,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2020-50 | latest | en | 0.784099 |
https://linearprogramminghelp.com/who-provides-support-for-linear-programming-assignment-statistical-analysis | 1,708,981,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00353.warc.gz | 361,758,480 | 23,117 | # Who provides support for linear programming assignment statistical analysis?
Who provides support for linear programming assignment statistical analysis? If the key words are “assignment data” and “regular and systematic sequence analysis”, chances is that the results are even better–this would be essentially because it yields both statements “the full dataset is a full dataset”, and a good performance for linear data assignment statistical analysis, “partial sample” random number distribution, random number generation by random samples, for a number of datasets. Because in most cases the size of the dataset is a little smaller than the size of the original data set, the differences become significant. However it is difficult to derive a complete data set for each dataset, where all the information about the dataset as a whole is known. Instead we take a fully random sample, with the only exception that any complete statistical summary, e.g. table analysis, is free to be obtained using random tables. It would be most sensible to follow the general procedure given in section “Basic Information Generics”. However, this procedure is to directly approach the statistical problem. We then search the data set to which the random set contains the most information. To this analysis we set up the program data.py and call random.py twice, for each dataset. The data function and the functions are described in the following sections. ### Methods of Data Analysis Now, we need to predict the distribution of the observed individuals in a given dataset with the proposed statistical methods in this section. Let us consider the specific interest to generate actual and estimated data such as these to the study population ($\mathbf{d}$, $\mathbf{A}\rho$). The above is done by using some simple measures besides the individual count as one approximation (a reference test statistic) which allows the goal be determined by the experiments. Thus it is necessary to formulate the more helpful hints statistical framework which relies on the mathematical definition of the data. The number of trials is estimated by (1) observing the data, (2) dividing 2 by the test statistic, and finallyWho provides support for linear programming assignment statistical analysis? It is called a statistical analysis software for statistical analysis, and is installed in the Microsoft Windows 95.0, and in Windows 98, it is called PL. It is designed to help you evaluate all statistics in Lin $l$ l and measure the probability distributions (P(X) is the P(Y) which represent the proportions in the distribution (X, Y) as a mixture, and that is the sum of the two probabilities (M) representing the individual P(Y) or the averages of the two probabilities (M) representing the product of the P(Y) and the mean P(Y), with x = X+ Y. | 550 | 2,823 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-10 | latest | en | 0.896781 |
http://mathhelpforum.com/algebra/158129-find-point-intersection-between-two-straight-lines.html | 1,529,502,816,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863519.49/warc/CC-MAIN-20180620124346-20180620144346-00167.warc.gz | 202,596,305 | 9,814 | # Thread: Find point of intersection between two straight lines
1. ## Find point of intersection between two straight lines
That's the instruction I have. Sorry for my bad English.
$\displaystyle mx - 4y = m$
$\displaystyle -4x + my = 2m + 4$
I Thought I should get both of the equations to y = ... , So:
$\displaystyle y = \dfrac{mx-m}{4}$
$\displaystyle y = 2 + \dfrac{4+4x}{m}$
And then:
$\displaystyle \dfrac{mx-m}{4} = 2 + \dfrac{4+4x}{m}$
$\displaystyle m(mx-m) = 8m +4(4+4x)$
$\displaystyle m^{2}x - m^{2} -8m-16-16x = 0$
And now I don't know how to continue...
2. You've done very good, correct work so far! Now it's a good time to remember what it is for which you are trying to solve. What is that?
3. "Combine like terms". That is, combine all terms involving "x" on one side of the equation, all terms that do not on the other. | 271 | 853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-26 | latest | en | 0.879072 |
https://artofproblemsolving.com/wiki/index.php?title=2021_USAMO_Problems/Problem_1&diff=next&oldid=148401 | 1,623,846,050,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00636.warc.gz | 97,709,561 | 9,595 | # Difference between revisions of "2021 USAMO Problems/Problem 1"
Rectangles $BCC_1B_2,$ $CAA_1C_2,$ and $ABB_1A_2$ are erected outside an acute triangle $ABC.$ Suppose that$\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]$Prove that lines $B_1C_2,$ $C_1A_2,$ and $A_1B_2$ are concurrent. | 116 | 296 | {"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": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-25 | latest | en | 0.789888 |
https://number.academy/131972 | 1,722,814,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00597.warc.gz | 353,416,087 | 11,274 | # Number 131972 facts
The even number 131,972 is spelled 🔊, and written in words: one hundred and thirty-one thousand, nine hundred and seventy-two. The ordinal number 131972nd is said 🔊 and written as: one hundred and thirty-one thousand, nine hundred and seventy-second. Color #131972. The meaning of the number 131972 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 131972. What is 131972 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 131972.
## Interesting facts about the number 131972
### Asteroids
• (131972) 2002 CG56 is asteroid number 131972. It was discovered by LINEAR, Lincoln Near-Earth Asteroid Research from White Sands Observatory in Socorro on 2/7/2002.
## What is 131,972 in other units
The decimal (Arabic) number 131972 converted to a Roman number is (C)(X)(X)(X)MCMLXXII. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
131972 seconds equals to 1 day, 12 hours, 39 minutes, 32 seconds
131972 minutes equals to 3 months, 1 week, 15 hours, 32 minutes
### Codes and images of the number 131972
Number 131972 morse code: .---- ...-- .---- ----. --... ..---
Sign language for number 131972:
Number 131972 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Share in social networks
#### Is Prime?
The number 131972 is not a prime number. The closest prime numbers are 131969, 132001.
#### Factorization and factors (dividers)
The prime factors of 131972 are 2 * 2 * 32993
The factors of 131972 are 1, 2, 4, 32993, 65986, 131972.
Total factors 6.
Sum of factors 230958 (98986).
#### Powers
The second power of 1319722 is 17.416.608.784.
The third power of 1319723 is 2.298.504.694.442.048.
#### Roots
The square root √131972 is 363,279507.
The cube root of 3131972 is 50,912833.
#### Logarithms
The natural logarithm of No. ln 131972 = loge 131972 = 11,790345.
The logarithm to base 10 of No. log10 131972 = 5,120482.
The Napierian logarithm of No. log1/e 131972 = -11,790345.
### Trigonometric functions
The cosine of 131972 is 0,999707.
The sine of 131972 is -0,02419.
The tangent of 131972 is -0,024197.
## Number 131972 in Computer Science
Code typeCode value
PIN 131972 If 131972 is your date of birth, it is not recommended that you use 131972 as your password or PIN.
131972 Number of bytes128.9KB
CSS Color
#131972 hexadecimal to red, green and blue (RGB) (19, 25, 114)
Unix timeUnix time 131972 is equal to Friday Jan. 2, 1970, 12:39:32 p.m. GMT
IPv4, IPv6Number 131972 internet address in dotted format v4 0.2.3.132, v6 ::2:384
131972 Decimal = 100000001110000100 Binary
131972 Decimal = 20201000212 Ternary
131972 Decimal = 401604 Octal
131972 Decimal = 20384 Hexadecimal (0x20384 hex)
131972 BASE64MTMxOTcy
131972 MD5176d5ac17924300c9d508a0fe1b61cce
131972 SHA2248e5bccb4b41e2149e532d3cf5d4d271703af77d5e597284385463165
131972 SHA25648dd69517e07bfbbcdf05052f037374191da4bd1897ab33d2903cf2bcdd6c86d
131972 SHA384ff927f13a6dbfde927ed780e9c248809c1763a897309b344174ef81bacd8af98991dd7c2b7b884f8dfabc1bd6893138d
More SHA codes related to the number 131972 ...
If you know something interesting about the 131972 number that you did not find on this page, do not hesitate to write us here.
## Numerology 131972
### Numerology of the date 1/3/1972
The number means date 1/3/1972 . Monday, Jan. 3, 1972 (1/3/1972, 1-3-1972). What does numbers in the date 1/3/1972 mean spiritually? Read more about numerology of the date.
## № 131,972 in other languages
How to say or write the number one hundred and thirty-one thousand, nine hundred and seventy-two in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 131.972) ciento treinta y uno mil novecientos setenta y dos German: 🔊 (Nummer 131.972) einhunderteinunddreißigtausendneunhundertzweiundsiebzig French: 🔊 (nombre 131 972) cent trente et un mille neuf cent soixante-douze Portuguese: 🔊 (número 131 972) cento e trinta e um mil, novecentos e setenta e dois Hindi: 🔊 (संख्या 131 972) एक लाख, इकतीस हज़ार, नौ सौ, बहत्तर Chinese: 🔊 (数 131 972) 十三万一千九百七十二 Arabian: 🔊 (عدد 131,972) مائة و واحد و ثلاثون ألفاً و تسعمائة و اثنان و سبعون Czech: 🔊 (číslo 131 972) sto třicet jedna tisíc devětset sedmdesát dva Korean: 🔊 (번호 131,972) 십삼만 천구백칠십이 Danish: 🔊 (nummer 131 972) ethundrede og enogtredivetusindnihundrede og tooghalvfjerds Hebrew: (מספר 131,972) מאה שלושים ואחד אלף תשע מאות שבעים ושתיים Dutch: 🔊 (nummer 131 972) honderdeenendertigduizendnegenhonderdtweeënzeventig Japanese: 🔊 (数 131,972) 十三万千九百七十二 Indonesian: 🔊 (jumlah 131.972) seratus tiga puluh satu ribu sembilan ratus tujuh puluh dua Italian: 🔊 (numero 131 972) centotrentunomilanovecentosettantadue Norwegian: 🔊 (nummer 131 972) en hundre og trettien tusen ni hundre og syttito Polish: 🔊 (liczba 131 972) sto trzydzieści jeden tysięcy dziewięćset siedemdziesiąt dwa Russian: 🔊 (номер 131 972) сто тридцать одна тысяча девятьсот семьдесят два Turkish: 🔊 (numara 131,972) yüzotuzbindokuzyüzyetmişiki Thai: 🔊 (จำนวน 131 972) หนึ่งแสนสามหมื่นหนึ่งพันเก้าร้อยเจ็ดสิบสอง Ukrainian: 🔊 (номер 131 972) сто тридцять одна тисяча дев'ятсот сімдесят два Vietnamese: 🔊 (con số 131.972) một trăm ba mươi mốt nghìn chín trăm bảy mươi hai Other languages ...
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If you know something interesting about the number 131972 or any other natural number (positive integer), please write to us here or on Facebook.
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 1,981 | 6,053 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.792195 |
https://math.stackexchange.com/questions/1195797/formula-for-the-variance-of-a-renewal-process | 1,569,271,897,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514578201.99/warc/CC-MAIN-20190923193125-20190923215125-00494.warc.gz | 583,375,724 | 32,028 | # Formula for the variance of a renewal process
Let $N(t)$ be a renewal process, with a sequence of IID inter-arrival times $X_{1}, X_{2}, \dots$ having finite second moment: $EX_{i}^{2} < \infty$.
How would I show that $$\mathrm{Var}N(t)= 2 \int^{t}_{0} m(t-s) \cdot m'(s)ds + m(t) - m(t)^{2}$$
where $m(t) = E[N(t)]$ and can be written $m(t) = F(t) + \int_{0}^{t} m(t-x)f(x)dx$ where $f$ is the density of the inter-arrival times and $F$ is the CDF.
I'd just like to know how this works out since I'm weak in computation. I've tried using the usual identity for variance, but I didn't know if that would lead anywhere.
• Where did you find this formula? In the case that $X_1 = 1$ almost surely, we have $\text{Var}(N_t) = 0$ and $m(t) = \lfloor t \rfloor$. However, calculating the right hand side of the formula above comes out as $2 \lfloor t \rfloor( t - \lfloor t \rfloor)$, which is not always $0$. There could well be an error in my calculation, and certainly the formula you give holds in the case of a Poisson process. – owen88 Mar 19 '15 at 10:06
• In light of my answer posted below, clearly there is a problem with my counter example... Or if may be that the claim I state below fails in the case that the function in the convolution is a step function. – owen88 Mar 19 '15 at 11:25
I too was unable to derive a direct proof of the formula you have given; the following proof therefore is somewhat indirect. It rests on the following claim.
Claim. Given functions $H(t), \, f(t)$, the renewal equation $g(t) = H(t) + \int_{0}^t g(t-x) f(x) \mathrm d x$ has a unique solution.
Henceforth I will use the normal notation for convolution $\phi*\psi(t) = \int_0^t \phi(t-x)\psi(x) \mathrm d x$ so simplify notation. The equation above then reads: $g = H + g*f$.
My proof proceeds in two steps:
1. Derive a renewal type equation for the variance.
2. Prove that it is solved by the right hand side of the formula you gave.
Part 1 Instead of working with the variance, i will consider $g(t) = \textbf{E}[N(t)^2]$. Exactly as though we were deriving the usual renewal equation, we condition on the first jump time \begin{align*} g(t) & = \int_{0}^\infty \textbf{E}[N(t)^2 \, | \, X_1 = x] f(x) \mathrm d x \\ & = \int_0^t \textbf{E}[N(t)^2 \, | \, X_1 = x] f(x) \mathrm d x + \int_t^\infty 0 \mathrm \, d x \\ & = \int_0^t \textbf{E}[(1 + N(t-x) )^2] f(x) \mathrm d x \\ & = \int_0^t f(x)\Big(1 + 2 \textbf{E}[N(t-x)] + \textbf{E}[N(t-x)^2] \Big ) \mathrm d x \\ & = F(t) + 2(m * f)(t) + (g*f)(t). \end{align*} In particular, $g$ satisfies a renewal type equation with $H = F +2 m*f$.
Part 2 According to the formula you gave, we want to show $g(t) = 2(m*m')(t) + m(t)$, note that this is exactly the right hand side of your formula but with the $m(t)^2$ term removed, since we are not working with the variance. Substituting this formula into the right hand side of the renewal equation we have \begin{align} H + (2(m*m') + m)*f & = F + 2(m*f) +2(m*m')*f + m*f\\ & = (m - m*f) + 2(m*f) +2(m*m')*f + m*f\\ & = m + 2( m*f + m*m'*f), \end{align} where in the second line we substituted in the renewal equation $F = m - m*f$. It remeains then to show \begin{align} m*f + m*m'*f =m*m'. \tag{1} \end{align} But again using the renewal equation, and differentiating it \begin{align} m' = (F + m*f)' = f +m'*f, \end{align} which when rearranged is $$m'*f =m'-f$$ Substituting into (1) \begin{align} m*f + m*m'*f & = m*f + m*(m' -f) \\ & =m*m', \end{align} as required. | 1,165 | 3,478 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-39 | latest | en | 0.9119 |
https://www.aqua-calc.com/calculate/volume-to-weight/substance/lead-op-ii-cp--blank-oxide-coma-and-blank-alpha-blank-form | 1,726,648,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651886.88/warc/CC-MAIN-20240918064858-20240918094858-00285.warc.gz | 590,462,218 | 13,507 | Weight of Lead(II) oxide, alpha form
lead(ii) oxide, alpha form: convert volume to weight
Weight of 1 cubic centimeter of Lead(II) oxide, alpha form
carat 47.65 ounce 0.34 gram 9.53 pound 0.02 kilogram 0.01 tonne 9.53 × 10-6 milligram 9 530
How many moles in 1 cubic centimeter of Lead(II) oxide, alpha form?
There are 42.7 millimoles in 1 cubic centimeter of Lead(II) oxide, alpha form
The entered volume of Lead(II) oxide, alpha form in various units of volume
centimeter³ 1 milliliter 1 foot³ 3.53 × 10-5 oil barrel 6.29 × 10-6 Imperial gallon 0 US cup 0 inch³ 0.06 US fluid ounce 0.03 liter 0 US gallon 0 meter³ 1 × 10-6 US pint 0 metric cup 0 US quart 0 metric tablespoon 0.07 US tablespoon 0.07 metric teaspoon 0.2 US teaspoon 0.2
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Foods, Nutrients and Calories
SWEETZELS, SPICED MINI CREMES, UPC: 073406710026 contain(s) 464 calories per 100 grams (≈3.53 ounces) [ price ]
1492 foods that contain Lycopene. List of these foods starting with the highest contents of Lycopene and the lowest contents of Lycopene
Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Florida Crushed Coral weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
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Weights and Measurements
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Calculators
Calculate volume of a dodecahedron and its surface area | 695 | 2,510 | {"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} | 3.1875 | 3 | CC-MAIN-2024-38 | latest | en | 0.726719 |
http://sl4.org/archive/0905/20270.html | 1,643,293,928,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00481.warc.gz | 53,199,479 | 2,483 | # Re: [sl4] Is belief in immortality computable?
From: Matt Mahoney (matmahoney@yahoo.com)
Date: Mon May 18 2009 - 13:45:42 MDT
--- On Mon, 5/18/09, Stuart Armstrong <dragondreaming@googlemail.com> wrote:
> > An agent might not know that the
> universe is finite or that it is mortal. It might still
> believe itself to be immortal, which we could detect if we
> observe it making decisions that postpone gratification
> arbitrarily far into the future. But I don't think such a
> test exists. (That is my question). I think that for any
> test, there is a sufficiently large T such that the two
> rational agents (one believing it will die at time T and the
> other believing itself to be immortal) will both give the
> same response.
>
> A continuous sliding scales of investments, returning t^2 at time t.
Let me make sure I understand correctly: you pay me \$1 per day for t days. You choose t. After the last payment, I pay you \$t^2?
A rational agent expecting to live T days will choose t = T-1. An agent expecting to live forever will pay forever. But in the latter case, you will never know if t is infinite or just really big.
Or do you mean: you choose a number t right now, and then I pay you \$t^2 after t days?
This tests also fails for agents that communicate using prefix-free strings over finite alphabets, because for any number t, there is a larger number that takes longer to describe. In the first case, the agent will choose t = BusyBeaver(T-1). In the second case, you will wait forever for the agent to finish choosing.
Agents believing in immortality can be expected to make decisions that seem irrational to us. Suppose you offer an agent a choice of \$1 per day or \$2 per day forever (adjusting for inflation). Then the choices are equivalent because they both have the same sum (aleph-null). However this test fails if the second agent chooses \$2 per day by chance.
-- Matt Mahoney, matmahoney@yahoo.com
This archive was generated by hypermail 2.1.5 : Wed Jul 17 2013 - 04:01:04 MDT | 505 | 2,022 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.906064 |
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1. The angle made by the intercepted arc AB. Theorems on Segments formed by Tangent Segments and Secant Segments Common Tangent A common tangent is a line or segment or ray that is tangent to two circles in the same plane. Proof. In an interview with Quanta Magazine, Emily Riehl said the following:. \ ( OP AB\) [\ (OP\) is the shortest distance from \ (O\) to \ (AB\)] \ (OP<OQ\) \ (OQ>OP\) \ (Q\) lies outside the circle [\ (OP\) is the radius and \ (OP<OQ\)] We are told that angles B and C are right angles, which add up to equal 180. of the measures of the intercepted arcs. Circle Theorems (Proof Questions/Linked with other Topics) (G10) The Oakwood Academy Page 2 Q1. If two secant segments are drawn to a circle from an exterior point, then the product of the measures of one secant segment and its external secant segment is equal to the product of the measures of the other secant segment and its external secant segment. 1.
If a radius is perpendicular to a line at the point at which the line intersects the circle, then the line is a tangent. Remember that this theorem only used the intercepted arcs . (Note: Each segment is measured from the outside point) Try this In the figure below, drag the orange dots around to reposition the secant lines. Secant-Tangent Power Theorem If a tangent and a secant intersect in the exterior of a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant and its external secant segment. In the diagram shown below, point C is the center of the circle with a radius of 8 cm and QRS = 80. The theorem follows directly from the fact, that the triangles PAC and PBD are similar. Write a two-column proof of Theorem 10.14: If two secants, a secant and a tangent, or two tangents interesect in the exterior of a circle, the measure of the angel formed is one-half the positive difference of the measures of the intercepted arcs. Secant & Tangent Theorems. In the circle, M O and M Q are secants that intersect . (c) Two tangents can be drawn from any exterior point of a circle. In the next theorem, we observe a relationship between a secant segment and tangent segment. Assume that lines which appear tangent are tangent. Theorems on Tangent Line 4. The theorem states that the angle between the tangent and its chord is equal to the angle in the alternate segment The entire wedge-shaped area is known as a circular sector 1 Section 2 In the figure below, the center of dilation is on AC, so AC and AC'' are on the same line The intelligent Income and Reward Calculator allows you to predict . Tangent-Secant Segment Theorem If a tangent segment and a secant segment are drawn to the same circle from the same exterior point, the product of the length of the secant and the length of its external segments is equal to the square of the length of the tangent segment.
2. If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. top; Tan & Sec $$\cdot$$ Practice I; Applet; 2 Secants $$\cdot$$ Practice II; Tangent and Secant. Proof: In figure 1.2 a circle with center O and tangent XY with point P at the interaction id given. Side Length of Tangent & Secant of a Circle. If a tangent from an external point A meets the circle at F and a secant from the external point A meets the circle at C and D respectively, then AF 2 = AC AD (tangent-secant theorem). of the tangent segment. .
Thus, the two important theorems in Class 10 Maths Chapter 10 Circles are: Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
So, DAB=CDB. Solution: The angles formed between the tangents and the radii is 90 degree.
Draw seg $$MP . In the given figure, M is the centre of the circle and seg KL is a tangent segment. The Mean Value Theorem highlights a link between the tangent and secant lines. Now, in triangles CADand CDB. See also Intersecting Secant Lengths Theorem . \(PM^2 = PN\cdot PO$$ Example 11: Solve for $$x$$ Solution: Using the Chord-Chord Power Theorem: . If MK = 12, KL = 6 3, then find the radius of the circle. A tangent can be considered a limiting case of a secant whose ends are coincident. Angle of Intersecting Secants. Given: square To Prove: square Proof: Draw radius AP and radius AQ and complete the following proof of the theorem. Consider a circle with a secant ABand a tangent DCintersecting at C. Join ADand DB.
This is an obvious step, but it's needed in a formal proof. 4 Parallel Lines Cut By 2 Transversals Illustration used to prove the theorem "If three or more parallel lines intercept equal segments on tangent secant theorem class 10. tangent secant theorem angle. Find 100's more videos linked to the Australia Senior Maths Curriculum at http://mathsvideosaustralia.com/There are videos for:Queensland: General Mathematic. Common External Tangents. a.
Proof of the Outside Angle Theorem "The measure of an angle formed by two secants, or two tangents, or a secant and a tangent, that intersect each other outside the circle is equal to half the difference of the measures of the intercepted arcs." Movement Proof: We will do the same as with our movement proof for the inscribed angle theorem.
2. By alternate segment theorem, QRS= QPR = 80. Example 2: Find the missing angle x using the intersecting secants theorem of a circle, given arc QS = 75 and arc PR= x. Downloads: 8001 x. Proof: Take any point $$P$$, other than $$N$$, on the line $$l$$. % Progress Proof of tangent secant angle theorem.
Theorem 1: The tangent at any point of a circle and the radius through the point are perpendicular to each other. The similarity yields an equation for ratios which is equivalent to the equation of the theorem given above. Angles from Secants and Tangents (V1) Angle From 2 Secants (V2) Secants: Proof Hint; Not Your Everyday Chord & Tangent Theorem; GoGeometry Action 4! Mean Value Theorem Proof.
Now let us discuss how to draw (i) a tangent to a circle using its centre (ii) a tangent to a circle using alternate segment theorem (iii) pair of tangents from an external point . To Prove: OP perpendicular to XY.
Proof (1) BAC CAB //Common angle to both triangles, reflexive property of equality (2) ABE ACD // Inscribed angles which subtend the same arc are equal (3) BEA CDA // (1), (2), Sum of angles in a triangle (4) ABE ACD //angle-angle-angle (5) ADAB = AEAC // (4), property of similar triangles Recall the inscribed angle theorem, 2 QPR = QCR. Example 3. r x y s r x .
3. Theorem.
Theorem 1: The tangent to the circle is perpendicular to the radius of the circle at the point of contact.
L is a point of contact. The Mean Value Theorem relates the slope of a secant line to the slope of a tangent line. If a tangent from an external point A meets the circle at F and a secant from the external point A meets the circle at C and D respectively, then AF 2 = AC AD (tangent-secant theorem).
Theorem 23-F Just understand.
Proof: Go to Day 10 If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.
If a tangent and secant meet at a common point outside a circle, the segments created have a similar relationship to that of two secant rays. Let PA be a secant passing through the point P in the exterior of. Problem 1: Given a circle with centre O.Two Tangent from external point P is drawn to the given circle. Thales Action + Sequel = GoGeometry Action 25!
This theorem states that if a tangent and a secant are drawn from an external point to a circle, then the square of the measure of the tangent is equal to the product of the measures of the secant's external part and the entire secant. Notice how the right-hand side of the Mean Value Theorem is the slope of the secant line through points A and B.
In the adjoining figure, O is the centre of the circle.
Intersecting Secants Theorem. Note: For the special case of two tangents , please visit this page . Interesting facts about Circles and its properties are . tangent secant theorem problems.
Intersecting Secants Theorem. This is the idea (a,b,c and d are lengths): And here it is with some actual values (measured only to whole numbers): And we get. Table of contents. A tangent at any point on a circle and the radius through the point are perpendicular to each other. Although the result may seem somewhat obvious, the theorem is used to prove many other theorems in Calculus. Mean Value Theorem Proof. This result is found as Proposition 36 in Book 3 of Euclid's Elements. Older (Earlier) Applets . 1 we discussed and prove important question 10.
Important Theorem from Circles for Board Exam class 10, CBSE Board,. Tangent Secant Theorem Point E is in the exterior of a circle. Tangent Secant Segment Theorem: If a tangent and a secant are drawn from a common point outside the circle (and the segments are labeled like the picture below), then a 2 = b ( b + c). The Pythagorean identity of secant and tan functions can also be written popularly in two other forms. They intersect at point U . Find the length of arc QTR. Find the measure of the arc or angle indicated. Click Create Assignment to assign this modality to your LMS. Intersecting Tangent Secant Theorem. In PAD and QAD, Product of the outside segment and whole secant equals the square of the tangent to the same point. The Mean Value Theorem. A number of interesting theorems arise from the relationships between chords, secant segments, and tangent segments that intersect.
If a tangent and a secant intersect in the exterior of a circle, then the measure of the angle formed is one half the difference. 35.3K subscribers Subscribe Proof of Tangent Secant Theorem Circles, Class 10, Most Important Theorem for CBSE Board Exam. Take a point Q on XY other than P and join OQ.
Notice that the exterior angle that is created by the intersection of two secants or tangents is one-half the difference . A secant segment is a segment with one endpoint on a circle, one endpoint outside the circle, and one point between these points that intersects the circle. This concept teaches students about tangent lines and how to apply theorems related to tangents of circles. Show Video Lesson. Proof: If a secant segment and tangent segment are drawn to a circle from the same external point, the length of the tangent segment is the geometric mean between the length of the secant . Some results on circles and tangents. When two secants intersect outside a circle, there are three angle measures involved: The angle made where they intersect (angle APB above) The angle made by the intercepted arc CD. 17Calculus Integrals - Secant-Tangent Trig Integration. Tangent-Secant Theorem:If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. Find the sum of angles formed between both radius and the angles between both the tangents of the circle. GoGeometry Action 13! JK = KM KL2x KL = 3 LM = 9 KM = _____ JK = _____ Click Create Assignment to assign this modality to your LMS. Using the previous theorem, we know the products of the segments are equal. Line $$l$$ is a tangent to the circle.
A secant line is a line drawn through two points on a curve.. . And lastly, the third situation is when two secants, or a secant and a tangent, intersect outside the circle. Remember that?)
Case I. Tangent and Secant The measure of an angle formed by a secant and a tangent drawn from a point outside the circle is 1 2 the difference of the intercepted arcs . This also works if one or both are tangents (a line that just touches a circle at one point), . Postulate on Tangent Line 3. Two secant segments which share an endpoint outside of the circle. Tangent Theorems.
Below you can download some free math worksheets and practice. Both theorems, including the tangent-secant theorem, can be proven uniformly: .
$\sec^2{x}-\tan^2{x} \,=\, 1$ $\sec^2{A}-\tan^2{A} \,=\, 1$ Remember, the angle of a right triangle can be represented by any symbol but the relationship between secant and tan functions must be written in that symbol.
Video . According to the figure, A is the centre of the circle. Solution: Using the secant of a circle formula (intersecting secants theorem), we know that the angle formed between 2 secants = (1/2) (major arc + minor arc) 45 = 1/2 (75 + x) 75 + x = 90.
Geometry Problem 1379. common tangent - A common tangent is a line or line segment that is tangent to two circles in the same plane. I you draw the diameter passing from A, intersects the other side of the circle in A . This is the case only when the segment A C is tangent to the circle. High School Math based on the topics required for the Regents Exam conducted by NYSED. So, U V 2 = U X U Y .
Three . Two Secants.
tangent secant theorem pdf. 61) (x 201) Below you can download some free math sheets and practices. So just ch Continue Reading Alon Amit Theorem 1. Since the angles in a quadrilateral add up to 360, angle o plus angle A equal 360-180=180. This is the idea (a,b and c are angles): And here it is with some actual values: In words: the angle made by two secants (a line that cuts a circle at two points) that intersect outside the circle is half of the furthest arc minus the nearest arc. Complete the following activity.
We have a new and improved read on this topic. Tangent segments drawn from an external point to a circle are congruent, prove this theorem. GoGeometry Action 16! Theorems on Angles formed by Tangent Lines and Secant Lines 5.
Using point . (Sounds sort of like the scarecrow from the Wizard of Oz talking about the Pythagorean Theorem. Case II. Intersecting secants theorem. For instance, in the above figure, 4 (4 + 2) = 3 (3 + 5) Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal. If a tangent and a secant lines are released from a point outside a circle, then the product of the measures of the secant and its external part is equal to the square. Here, DABwill be equal to the half measure of arc DB. Consider a circle with tangent and secant as, In the figure, near arc is Q R and far arc is P R. Join P R, so by exterior angle theorem
2 Secants This concept teaches students to solve for missing segments created by a tangent line and a secant line intersecting outside a circle.
Step 3: State that two triangles PRS and PQT are equivalent. Secant-Tangent Theorem states: If a secant PA and tangent PC meet a circle at the respective points A, B, and C (point of contact), then (PC)^2 = (PA)(PB). Common Internal Tangents. Notice how the right-hand side of the Mean Value Theorem is the slope of the secant line through points A and B. Now, let's have a look at the proof of secant tangent theorem.
(Tangent-Chord Theorem (3) ACB ABD /Sum of Angles in a Triangle (4) WAB AB/UBC /Corner-Corner (5) AB2 AD (5) tangent secant theorem proof. outside = tangent2) (AD) = (BE+ED) ED because of the Secant-Tangent Product Theorem. (Whew!) This page covers integration of functions involving secants and/or tangents in more advanced form that require techniques other than just integration by substitution. (c) We conclude the proof by showing that the theorem is true for all ni 2 (this part may be bypassed quoting [18] where it is shown that secant variety of lines of a Segre variety is contained in the subspace variety). In PAD and QAD, seg PA [segQA] [Radii of the same circle] seg AD seg AD [Common side] APD = AQD = 90 [Tangent theorem] Circles, Secant, Congruent Chords. hint for proof ABCLCDB by ?ACB ?ABD. Circles, Secant, Congruent Chords. Segments of Secants Theorem. (b) Only one tangent can be drawn at any point on a circle. There are two types of common tangents: common external tangents and common internal tangents. Proof AD // (5), property of similar triangles The Tangent-Chord Theorem Circumscribed Circle Proof of tangent secant theorem. Therefore, the red arc in the picture below is not used in this formula. Theorem 2: If two tangents are drawn from an external point of the circle, then they are of equal lengths . Intersecting Secants Theorem.
The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.
12 25 = 300; . Theorems: If two chords intersect in a circle, the product of the lengths of the segments of one chord equal the product of the segments of the other. Monge's Circle Theorem, Three Circles and Three Pair of Common Tangents, Collinearity. . There's a special relationship between two secants that intersect outside of a circle. This mathematics ClipArt gallery offers 127 images that can be used to demonstrate various geometric theorems and proofs. First of all, we must define a secant segment. Tangent Secant Theorem. 2.
inria-00610362, version 1 - 21 Jul 2011 Proof. First, join the vertices of the triangle to the center. The following theorem involves the measurement of the tangent-tangent angle. GoGeometry Action 26! (ii) The line $$PQ$$ is called a secant of the circle. Errata: For the example 2, the answer should be x = 9. In the circle, U V is a tangent and U Y is a secant. For easily spotting this property of a . This theorem states that the angle APB is half the difference of the . Inscribed Angle Theorem (Proof . Sample Problems based on the Theorem. Case #3 - Outside A Circle. If a line is tangent to a circle, the it is perpendicular to the radius drawn to the point of tangency. A tangent can be considered a limiting case of a secant whose ends are coincident. Figure 6.20. Move one of the secants (example-PD) so that it becomes a tangent. Both theorems, including the tangent-secant theorem, can be proven uniformly: . The point Q must lie outside the circle. . Now, the formula for tangent and secant of the circle could be given as: PR/PS = PS/PQ. circles-secant-tangent-angles-easy.pdf.
The Mean Value Theorem highlights a link between the tangent and secant lines.
tangent secant theorem calculator. In my experience, the proportion of female mathematicians varies wildly by subfield, and in algebraic topology I can see exactly why it's such a welcoming area for young women.It has a lot to do with very specific, proactive efforts taken by the generation of women above me who launched the Women in Topology network. Download. Search: Trigonometric Inequalities Calculator. When two secant lines intersect each other outside a circle, the products of their segments are equal. In this case, there are three possible scenarios, as indicated in the images below. Proof: Construction: Draw two segments AP and AQ. A secant through E intersects the circle at points A and B, and a tangent through E touches the circle at point T, then EA xx EB = ET^(2). Tangents, secants, Side Lengths Theorems & Formula. Step 2: Write that P is congruent to itself; This is because of the reflexive property of congruence (which simply states that any shape is congruent to itself). The field emerged in the Hellenistic world during the 3rd century BC from .
Assessment Directions: Using a two-column proof, show a proof of the following theorems involving tangents and secants. . The working sheet with the answer key on this theme Circle Theorem Three theorems for intercepted arcs at the angle of two tangents, two secants or 1 tangent and 1 secant are summed up in the photos below. | 4,773 | 19,864 | {"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": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2023-23 | latest | en | 0.913925 |
https://www.slideserve.com/saxon/tutorial-13-salary-survey-application-introducing-one-dimensional-arrays-powerpoint-ppt-presentation | 1,575,802,538,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540508599.52/warc/CC-MAIN-20191208095535-20191208123535-00044.warc.gz | 856,715,794 | 12,496 | Tutorial 13 – Salary Survey Application: Introducing One-Dimensional Arrays
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# Tutorial 13 – Salary Survey Application: Introducing One-Dimensional Arrays - PowerPoint PPT Presentation
## Tutorial 13 – Salary Survey Application: Introducing One-Dimensional Arrays
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1. Tutorial 13 – Salary Survey Application: Introducing One-Dimensional Arrays Outline13.1 Test-Driving the Salary Survey Application13.2 Introducing Arrays13.3 Declaring and Initializing Arrays13.4 Constructing the Salary Survey Application13.5 Wrap-Up
2. Objectives • In this tutorial, you will learn to: • Create and initialize arrays to store groups of related values. • Store information in an array. • Access individual elements of an array.
3. 13.1 Test-Driving the Salary Survey Application
4. Displaying the total salary 13.1 Test-Driving the Salary Survey Application Figure 13.1 Running the completed Salary Survey application. Figure 13.2 Salary Survey application displaying a salary and prompting for the next sales figure.
5. 13.1 Test-Driving the Salary Survey Application (Cont.) Figure 13.3 Entering several sales figures in the Salary Survey application. Figure 13.4 Displaying the distribution of salaries.
6. 13.2 Introducing Arrays • Arrays are groups of variables with the same type • Use the array name and an index (position number) to refer to individual elements • Indices range from 0 to one less than the number of elements • C++ will not prevent an application from attempting to access an element outside of an array’s valid range • Attempting to access a nonexistent element may result in a logic error
7. 13.2 Introducing Arrays (Cont.) Figure 13.5 Array unitsSold, consisting of 13 elements.
8. 13.3 Declaring and Initializing Arrays • Declare an array by using square brackets [] after the array name • Array initializer lists • Are comma-separated lists enclosed by braces {} • If no array size is specified within the square brackets, the size of the initializer list is used • Using an initializer list with more elements than the array is a syntax error • Using an initializer list with fewer elements than the array causes unintialized elements to be set to 0
9. 13.3 Declaring and Initializing Arrays (Cont.) • Array initializer lists • Syntax error • Initializing every element to 0 • If no initializer is provided values will be random. int salesPerDay[ 13 ];
10. Creating and initializing an array of ints 13.3 Declaring and Initializing Arrays (Cont.) Figure 13.6 Defining and initializing an array in main. • Use a constint to declare array sizes makes your code more clear and allows you to change it quickly in the future.
11. Retrieving the value of each element and adding it to the total one at a time 13.3 Declaring and Initializing Arrays (Cont.) Figure 13.7 Summing the values of an array’s elements. Figure 13.8 Displaying the sum of the values of an array’s elements.
12. Total value of array elements 13.3 Declaring and Initializing Arrays (Cont.) Figure 13.9 Completed Sum Array application output.
13. 13.4 Constructing the Salary Survey Application
14. Declaring the displayTotals function prototype 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.11 Declaring a function prototype for displayTotals.
15. Defining and initializing double variable sales Declaring int array resultArray and initializing its elements to 0 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.12 Defining a variable to store sales and creating an array of ints.
16. Prompting the user for and inputting a salesperson’s total sales 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.13 Retrieving user input.
17. Formatting floating-point values 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.14 Using the fixed and setprecision stream manipulators to display the salary.
18. while statement will process user input and prompt the user for the next salary 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.15 while statement repeats until user enters an invalid value.
19. Warning due to possible loss of data Calculating the salary and its corresponding index inresultArray 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.16 Calculating salary. Figure 13.17 Warning issued by the compiler when assigning a double value to an int variable. • Warnings do not prevent the application from compiling (syntax errors), but might indicate logic errors
20. Explicitly converting the result to an int to prevent a warning 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.18 Explicitly converting the floating-point result to an int to prevent the compiler warning.
21. Incrementing the appropriate element of resultArray 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.19 Updating the count of salaries in resultArray.
22. Prompting user for next sales value 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.20 Displaying the salary and prompting the user for the next sales figure.
23. Using pass-be-reference when passing resultArray to displayTotal 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.21 Passing resultArray by reference to the displayTotals function. • Arrays are passed-by-reference • Arguments that are passed-by-reference give the callee direct access to the argument in the caller
24. displayTotals function definition 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.22 Defining the displayTotals function.
25. Defining local variables to store the upper and lower bounds for each salary range Displaying a table header 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.23 Defining variables and displaying a table header in displayTotals.
26. for statement varies i from 2 to 9 Displaying the salary range and number of salaries in that range Assigning the upper and lower bounds for the current iteration 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.24 Displaying the number of salaries in each salary range.
27. Displaying the total for the final salary range 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.25 Display the range and salary count for \$1000 and greater.
28. 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.26 Sample input and output for the Salary Survey application.
29. 13.4 Constructing the Salary Survey Application (Cont.) Figure 13.27 Total of salaries in each range displayed upon exiting.
30. Declare the displayTotals function Define and initialize double sales Declare int array resultArray and initialize its elements to 0 SalarySurvey.cpp (1 of 4)
31. while statement processes user input and prompts for the next salary Calculate the salary and its corresponding index in resultArray Increment the appropriate element ofresultArray SalarySurvey.cpp (2 of 4)
32. Use pass-by-reference when passing resultArray to displayTotals Define local variables to store the upper and lower bounds for each salary range SalarySurvey.cpp (3 of 4)
33. Display a table header for header varies i from 2 to 9 Assign the lower and upper bounds for the current iteration Display the salary range and number of salaries in that range Display the count for the final salary range SalarySurvey.cpp (4 of 4) | 1,675 | 7,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2019-51 | latest | en | 0.654417 |
http://mathsmd.com/tag/worksheet-2/ | 1,653,417,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00316.warc.gz | 43,566,172 | 10,330 | Addition of Integers Worksheet Find the sum 1. 5 + 3 + – 7 2. 8 + – 6 + – 4 3. -10 + 2 + 3 4. -5 + -2 + -3 5. -7 + – 4 + – 5 6. 9 + 7 + – 3 Answers […]
Which are the following examples of the null set, give reason (i) Set of even numbers divisible by 3. (ii) {x : x is a common point to any two parallel lines} (iii) Set of odd numbers divisible by 2. (iv) {x : x is a natural number, x < 4 and x > 6} […]
Fill the symbol in the blank space. (i) {a, b, c, d, e}….{a, b, c} (ii) {3, 5, 7}… {3, 4, 5, 6, 7} (iii) {x : x is a odd number}…{x : x is a even number} (iv) {x : x is a student of class VI in your school}….{x : x is a […]
If A = {x : x is a natural number}, B = {x : x is an even natural number}, C = {x : x is an odd natural number}, and D = {x : x is a prime number}, find the value of given sets. (i) A ⋂ B(ii) A ⋂ C (iii) A […]
Worksheet – Ratio 1. Find the ratio of following? (a) 3×4/2×8 (b) 7×6/1×9 […]
Worksheet of numbers up to 50 1. Write the numbers that comes before. (a) 33 (b) 27 […]
Geometry practice on circle Here, we will practiced and discussed the questions on circle. Fill in the blanks. (1) The common point of a tangent and the circle is called …….. (2) A circle may have …….. parallel tangents. (3) A tangent to a circle intersects it in …….. point. (4) A line intersecting a […] | 507 | 1,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-21 | longest | en | 0.656601 |
https://youvegotthismath.com/12-times-table-worksheet/ | 1,726,679,803,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00435.warc.gz | 1,013,186,120 | 77,894 | # 7 Free 12 Times Table Worksheet | Fun Activities
These 12 times table worksheets will help to visualize and understand multiplication and number systems. 1st & 2nd grade students will learn basic multiplication methods and can improve their basic math skills with our free printable 12 times table worksheets.
## 7 Free 12 Times Table Worksheet
With the aid of these worksheets, our kids will be able to master fundamental mathematics quickly and actively. Download the following worksheets and practice.
## 12 Times Table Display Sheet
The multiples of 12 are listed in the 12 times table, a mathematical table. The outcomes of multiplying 12 by all whole numbers, from 1 to infinity, are displayed in the table below.
## 12 Times Table Up to 10
A multiplication table of 12, having multiplicands up to 10 is given below.
## 12 Times Table Up to 12
Here, you will have a 12x multiplication table, having multiplicands up to 12.
## Times Table Learning and Multiplication Facts
Since multiplication is a fundamental mathematical operation, you should learn it in preschool or in grades 1-3. In this section of the article, I will outline a few techniques for learning the timetable quickly and effectively.
First, write down the times table. I will help you to understand what a timetable looks like. However, try to look for patterns to make the table more interesting. Next- read it aloud as auditory learning. Then, repeat, repeat, and repeat to learn the table by heart. After that, to test how much you memorized, try the practice sheet.
Worksheets for learning multiplication facts and counting up by 12s are included in this article. In worksheets with 12x tables, students must solve multiplication problems in order to color the various parts of the artwork, complete the circle, match the right answers, etc. Using the worksheets will help your children to:
• Learn the factors of 12 with the help of 12 times tables
• Understand multiplication as repeated addition
• Get knowledge about multiplication as skip counting
• Find the unknown multiplicands
• Solve a range of simple multiplication challenges
## 7 Fun Activities to Learn and Practice 12 Times Table
Get in the fun activities, download the practice worksheets, and practice more to develop the mathematic skills of your students by developing fundamental ideas of mastering mathematical operations
## Count the 12s Product and Color by Number
In this game, preschoolers must find and color the five products of multiplication. Use light colors to decorate the box, such as yellow, pink, indigo, etc.
Now practice the activity from the following worksheet.
## Find the 12 Times Multiplication Product in Rocket
Here, students from kindergarten will multiply other numbers by 12. Write the products of the numbers given on the rockets, in the tail of the rockets.
Now practice the activity from the following worksheet.
## Complete 12 Times Multiplication Circle
As you can see, there is a number given—12—in the center of the three circles of flowers. Another 6 numbers are provided in the second outer circle. The outermost circles are empty. The numbers from the second circle must be multiplied by the number provided in the inner circle. Then, in the third outside circle, write the product.
Now practice the activity from the following worksheet.
## Match 12 Times Multiplication Product
Here, to obtain the products, they might use the 12x table. Now, you’ll discover that the worksheet also includes the expressions’ products, but the locations of the products are incorrect. The goal of this activity is to draw a line that corresponds to the expression’s product.
Practice the exercise from the next worksheet right away.
## Find the Missing Number in 12x Multiplication
Give your child directions on how to locate the multiplicand that needs to be multiplied by 12 to obtain the product. By dividing the result by 12, it is possible to get the missing number.
Practice the exercise from the next worksheet right away.
## Solve the Maze Counting by 12s
A man is trapped in a maze. He wants to come out of the maze. Here you can see the maze’s in and out path. But there are some numbers. What will be the path of the maze? Look at the maze carefully. You will find that some numbers are multiples of 12. And these multiples lead the path to the out end. So follow the 12 multiples blocks and help the man to get out of the maze.
Practice the exercise from the next worksheet right away.
## Cross the River Counting by 12s
Let’s assume a bunny is out of food. She wanted to get some carrots. She was running here and there and found a carrot farm on the opposite side of the river. She saw some blocks on the water of the river. The blocks have numbers written on the upper side of the blocks. Some blocks have the strength to take the weight of the bunny and others are not so strong. Now you need to find strong enough blocks. How can you find them? If you see the numbers carefully, you will get some specific numbers here. You will find numbers of 12s multiples. The blocks having 12 multiples are the strong blocks here. So using this path, the bunny can cross the river and have some carrots.
From the following worksheet, practice the activity. | 1,106 | 5,269 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-38 | latest | en | 0.91131 |
moore799.sensxafrica.com | 1,624,029,569,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487637721.34/warc/CC-MAIN-20210618134943-20210618164943-00268.warc.gz | 368,312,975 | 7,373 | # A Roulette Table Explained
Just how many times have you heard about a roulette table? If you are a fan of the overall game, then you surely should be aware of it and at the same time know what it is. To know roulette table profits, you have to first know how roulette works. It involves lots of strategy and skill to win at the game and it is the most popular games in casinos.
Roulette table refers to the numbers on the roulette wheel. Each number corresponds to a place on the wheel. In roulette table profits, each number that lands on the wheel means one unit of money. On a table with only two numbers, one will win and another will lose. This is why why in a few roulette games, players tend to place bets with low numbers on the tiny table and play on the big table with big numbers when there is only a small one. That is one of the known strategies in playing roulette.
The roulette wheel is really a complex device with the numbers arranged inside a circle. Players can place their bets on the inside of this circle. At these times, they win if all of the numbers which come on the wheel fit in the inside arrangement and lose if all the numbers do not fit in this arrangement. Thus, it is very important focus on these numbers.
The numbers that come on the roulette table also form a significant part of the strategy in winning. The rule is easy: the ball player who has more inside bets wins. So for a person who is just learning the overall game and has not put in much money yet, it will be best to play on the no-frills layout. This means that the ball player should play on the roulette wheel on the tiny table with two diamonds on the third line.
An excellent strategy in playing the game involves the 인터넷 바카라 use of the double zero roulette wheel. The wheel consists of seven balls and contains three black holes. The initial three balls haven’t any value, while the fourth ball has ten thousandths of a dollar. You have to bet only after seeing which of the three balls has the highest value in winning the overall game.
This may sound simple however the actual betting process is far from easy. This is because the probability of winning on the roulette wheel are very slim. This is because all of the spins with this roulette machine happen simultaneously. The process becomes more complicated as one moves on to more sophisticated setups. The most used and simplest layout in playing on the wheel involves the play on the black two spaces.
The concept behind this setup is that it contains only two numbers. In ways, you will find a wheel where you place the amount of money and the dealer places the numbers. This makes the numbers harder for the player to count. If they’re able to correctly guess which number gets the higher value, then your bet wins. It is the main idea behind the American design of roulette and the French style of setting the chip.
The original French design of setting the chips on the table involves putting the first bet after counting the full total amount of chips bet by the player. Once the player wins the initial hand, the winnings during the period of the bet are accumulated in columns. The ball player can either bet in column numbers or spread the bets among the column numbers or over the chips. Most people prefer to place their bets in column numbers as a result of convenience the layout provides. | 684 | 3,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.954565 |
https://www.sscadda.com/important-reasoning-questions-for-ssc_71 | 1,628,001,313,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154459.22/warc/CC-MAIN-20210803124251-20210803154251-00116.warc.gz | 1,030,973,476 | 26,259 | Latest SSC jobs » Important Reasoning Questions for SSC CGL and RRB Exam 2018: 4th September (Solutions)
Important Reasoning Questions for SSC CGL and RRB Exam 2018: 4th September (Solutions)
Dear Students,
Dear Aspirants,
Following SSC CGL Study Plan by ADDA247 and SSCADDAA, the platform is all set to deliver the quizzes and notes on each four subjects asked in SSC CGL Tier-1 Examination. Quizzes are prepared under the consideration of latest pattern and level of Exam whereas notes deal with understanding the concepts, formulae, rules and sharpen your revising skills. Be a part of this study plan, visit SSCADDA website regularly to add up each day effort in your practice.
Needless to say, SSC CGL 2018 is the golden opportunity to be grabbed where again Reasoning Section might bring a major difference. To clear the decks, attempt Daily Reasoning Quiz provided on SSCADDA and make headway in upcoming govt exams.
Q1.In the following question, select the related word from the given alternatives.
निम्नलिखित प्रश्न में, दिए गए विकल्पों में से सम्बंधित शब्द का चयन करें.
Kilometre :Metre : : Tonne : ?
किलोमीटर : मीटर :: टन : ?
(a) Litre/ लीटर
(b) Kilogram/ किलोग्राम
(c) Hours / घंटे
(d) Weight/ वजन
Ans.(b)
Sol.
1 kilometre = 1000 metre
1 tonne = 1000 kilogram
Q2.In the following question, select the related number from the given alternatives.
निम्नलिखित प्रश्न में, दिए गए विकल्पों में से सम्बंधित संख्या का चयन करें.
9143 : 9963 : : 6731 : ?
(a) 1368
(b) 5666
(c) 8964
(d) 9694
Ans.(c)
Sol.
9+1+4+3 = 17; 9+9+6+3 = 27
6+7+3+1 = 17; 8+9+6+4 = 27
Q3.In the following question, select the odd letter from the given alternatives.
निम्नलिखित प्रश्न में, दिए गए विकल्पों में से विषम अक्षर का चयन करें.
(a) B
(b) N
(c) P
(d) W
Ans.(d)
Sol.
Position of W is odd and rest are even numbers
Q4. Arrange the given words in the sequence in which they occur in the dictionary.
दिए गए शब्दों को उसी अनुक्रम में व्यवस्थित करें जैसे वे शब्दकोश में होते हैं।
1. Ball 2. Balanced 3. Balls
4. Balance 5. Balancing
(a) 24135
(b) 42135
(c) 42513
(d) 54213
Ans.(c)
Sol.
4. Balance
2. Balanced
5. Balancing
1. Ball
3. Balls
Q5. In the following question, select the missing number from the given series.
निम्नलिखित प्रश्न में, दी गई श्रृंखला में से अज्ञात संख्या का चयन करें.
13, 16, 11, 18, 9, 20, ?
(a) 3
(b) 5
(c) 6
(d) 7
Ans.(d)
Sol.
Q6. If ‘A + B’ means ‘A is father of B’, ‘A – B’ means ‘A is mother of B’, ‘A * B’ means ‘A is brother of B’ and ‘A % B’ means ‘A is sister of B’, then how is Q related to
S in ‘P + Q * R – S’?
यदि ‘A + B’ का अर्थ है A, B का पिता है, ‘A – B’ का अर्थ है A, B की माता है, ‘A * B’ का अर्थ है A, B का भाई है और ‘A % B’ का अर्थ है A, B की बहन है, तो ‘P + Q * R – S’ में Q, S से कैसे सम्बन्धित है?
(a) Husband / पति
(b) Uncle / अंकल
(c) Brother/ भाई
(d) Father/ पिता
Ans.(b)
Sol.
Q is uncle of S
Q7. In a certain code language, “REMOTE” is written as “KYSPGS” and “BRAND” is written as “IRDTC”. How is “MOBILE” written in that code language?
किसी निश्चित कूट भाषा में, “REMOTE”,”KYSPGS” के रूप में लिखा जाता है और “BRAND” “IRDTC” के रूप में लिखा जाता है. उसी कूट भाषा में “MOBILE” कैसे लिखा जाएगा?
(a) FMJCPN
(b) KQMEQN
(c) DKHANL
(d) DMHCNN
Ans.(b)
Sol.
Q8. If 19 (36) 13 and 37 (81) 28, then what is the value of ‘A’ in 43 (A) 38?
यदि 19 (36) 13 और 37 (81) 28, तो 43 (A) 38, A का मूल्य क्या है?
(a) 49
(b) 25
(c) 34
(d) 64
Ans.(b)
Sol.
19 – 13 = 6 ⇒ 6² = 36
37 – 28 = 9 ⇒ 9² = 81
43 – 38 = 5 ⇒ 5² = 25
Q9. How many triangles are there in the given figure?
दी गयी आकृति में कितने त्रिभुज है?
(a) 24
(b) 30
(c) 28
(d) 29
Ans.(d)
Sol.
29 triangles
Q10. From the given options, which answer figure can be formed by folding the figure given in the question?
दिए गए विकल्पों से, प्रश्न में दिए गए आंकड़े को तब्दील करके कौन सा उत्तर आंकड़ा बनाया जा सकता है?
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Are you sure you want to skip this step? | 2,388 | 7,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-31 | latest | en | 0.45937 |
http://forum.basic256.org/index.php?id=49 | 1,563,872,660,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529175.83/warc/CC-MAIN-20190723085031-20190723111031-00014.warc.gz | 57,035,713 | 2,957 | # Rock, Paper, Scissors game (code included) (Sample Programs)
If that's out of just one day of programming, then congratulations; that's an impressive feat for a first day.
One possible bug is what if compchoice ends up being exactly .33 or .66? In that (super-rare) case, nothing will get set for compchoice\$. I'd suggest changing the >'s to >='s to include those two numbers in the decision making...
#the number decides what the computer picks
if compchoice >= 0 and compchoice < .33 then compchoice\$ = "Rock"
if compchoice >= .33 and compchoice < .66 then compchoice\$ = "Scissors"
if compchoice >= .66 and compchoice < 1 then compchoice\$ = "Paper"
Best,
Jeremiah | 173 | 677 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | latest | en | 0.920814 |
https://www.researchandmarkets.com/reports/1760140/elements_of_set_theory | 1,529,482,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863489.85/warc/CC-MAIN-20180620065936-20180620085936-00238.warc.gz | 908,223,085 | 38,176 | # Elements of Set Theory
• ID: 1760140
• Book
• 279 Pages
• Elsevier Science and Technology
1 of 4
This is an introductory undergraduate textbook in set theory. In mathematics these days, essentially everything is a set. Some knowledge of set theory is necessary part of the background everyone needs for further study of mathematics. It is also possible to study set theory for its own interest--it is a subject with intruiging results anout simple objects. This book starts with material that nobody can do without. There is no end to what can be learned of set theory, but here is a beginning.
Note: Product cover images may vary from those shown
2 of 4
Contents
Preface
List of Symbols
Chapter 1 Introduction
Baby Set Theory
Sets-An Informal View
Classes
Axiomatic Method
Notation
Historical Notes
Chapter 2 Axioms and Operations
Axioms
Arbitrary Unions and Intersections
Algebra of Sets
Epilogue
Review Exercises
Chapter 3 Relations and Functions
Ordered Pairs
Relations
n-Ary Relations
Functions
Infinite Cartesian Products
Equivalence Relations
Ordering Relations
Review Exercises
Chapter 4 Natural Numbers
Inductive Sets
Peano's Postulates
Recursion on ?
Arithmetic
Ordering on ?
Review Exercises
Chapter 5 Construction of the Real Numbers
Integers
Rational Numbers
Real Numbers
Summaries
Two
Chapter 6 Cardinal Numbers and the Axiom of Choice
Equinumerosity
Finite Sets
Cardinal Arithmetic
Ordering Cardinal Numbers
Axiom of Choice
Countable Sets
Arithmetic of Infinite Cardinals
Continuum Hypothesis
Chapter 7 Orderings and Ordinals
Partial Orderings
Well Orderings
Replacement Axioms
Epsilon-Images
Isomorphisms
Ordinal Numbers
Debts Paid
Rank
Chapter 8 Ordinals and Order Types
Transfinite Recursion Again
Alephs
Ordinal Operations
Isomorphism Types
Arithmetic of Order Types
Ordinal Arithmetic
Chapter 9 Special Topics
Well-Founded Relations
Natural Models
Cofinality
Appendix Notation, Logic, and Proofs
Selected References for Further Study
List of Axioms
Index
Note: Product cover images may vary from those shown
3 of 4 | 498 | 2,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2018-26 | longest | en | 0.806997 |
https://dsp.stackexchange.com/questions/tagged/physical-units?tab=Active | 1,576,338,810,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541281438.51/warc/CC-MAIN-20191214150439-20191214174439-00481.warc.gz | 343,567,275 | 29,530 | # Questions tagged [physical-units]
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1answer
25 views
### Changing units in Fourier filtering
I'm applying a radial low-pass filter on a $H{\times}W$ pixel 2D image $I$ where $H \neq W$. To implement this filter, I apply 2D DFT to $I$ to produce $I'$, the $H{\times}W$ frequency domain ...
1answer
32 views
### Loss of precision due to low-pass filtering in physical units
Suppose I have a $10{\times}10$ pixel image, where each pixel represents 1 mm in physical units, and I apply a Gaussian low-pass filter with $\sigma = 5$. Can I make any statements such as "the ...
1answer
30 views
### Correct way of viewing negative voltages in units of dB for noise analysis
I have a voltage signal, on top of which there is some noise. I am interested in the noise components so the way I am trying to do this is to fit the known signal to the data and look at the residuals....
2answers
107 views
1answer
66 views
### Convolution Input/output Units
How come are the units of convolution for LTI system not consistent? $$y(t)= \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$$ $y(t)$ unit is volt, whereas the RHS has volt.time unit!!!!
2answers
12k views
### What is the difference between the PSD and the Power Spectrum?
Can any one explain the difference in the amplitude between the two and what does each one represents? For example, when using the pwelch algorithm in MATLAB how ...
2answers
163 views
### Why is “$1/\mathrm{Hz}$” unchanged while converting$\mathrm{V}^2/\mathrm{Hz}$ to $\mathrm{dB}/\mathrm{Hz}$?
When computing power spectral density via Fourier transform and Parseval's theorem, one gets units of $V^2/Hz$ (e.g. Wiki). As a popular next step, the following unit conversion is widely used: \$V^2/...
1answer
83 views
### Normalizing Cepstrum by Energy
I have separately computed the cepstrum and the energy of a finite discrete-time signal. If the energy of the signal is high, we would expect (on average) larger cepstral values than if the energy of ... | 533 | 2,062 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-51 | latest | en | 0.855309 |
http://mathhelpforum.com/algebra/81504-prove-consecutive-multiples.html | 1,527,436,598,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00051.warc.gz | 192,429,194 | 9,422 | # Thread: prove consecutive multiples
1. ## prove consecutive multiples
Proove that the product of two consecutive multiples of 5 is always an even number
I got:
(5n) (5n + 5)
25n^2 + 25n
n(25n+25)
-can't see it!
2. If n is odd then you have odd * (odd + odd) that is, odd* even = even
If n is even, then even*(even + odd) = even * odd = even | 111 | 351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.903008 |
https://vimore.org/watch/Jwo5Ae7tH_I/ms-excel-vlookup-with-exact-match/ | 1,621,292,240,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991870.70/warc/CC-MAIN-20210517211550-20210518001550-00407.warc.gz | 628,981,129 | 4,683 | MS Excel - VLookup with Exact Match - - vimore.org
# MS Excel - VLookup with Exact Match
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## MS Excel - Pivot Table Example 1
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## Compare Two Lists Using the Vlookup Formula
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## MS Excel - Shortcuts Ctrl+A to Ctrl+Z
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## 5 ways to use VLOOKUP
In this video, we look at 5 completely different ways to use the versatile VLOOKUP function. Comments at: https://exceljet.net/plc/excel-formulas-5-ways-to-use
## MS Excel - Copying and Filling
MS Excel - Copying and Filling Watch More Videos at: https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Mr. Pavan Lalwani, Tutorials Point Ind
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MS Excel - Shortcuts F1 to F12 Watch More Videos at: https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Mr. Pavan Lalwani Tutorials Point Indi
## Vlookup simply explained
Simple explanation of vlookup function and made it easy for everyone to learn.
## Excel - Difference between Vlookup and Index-Match Formulas
Both Vlookup/Hlookup and Index-Match Formulas are excellent search and retrieval tools in Excel. This Video uses a simple example to illustrate a prime differen
## Excel VLOOKUP formula with Multiple sheets | vlookup in Excel in Hindi
Excel VLOOKUP formula with Multiple sheets | vlookup in Excel in Hindi http://www.excelsuperstar.org/interworksheet-vlookup/ VLOOKUP (lookup_value, table_array
## VLOOKUP EXPLAINED - 2 Practical Excel Lookup Examples
Quickly learn the ins and outs of Excel's VLOOKUP formula. I show 2 practical examples of the VLOOKUP formula. I also explain why your VLOOKUP formula might not
## Excel VBA - Record a Macro
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## Stunning Example for Vlookup and Hlookup in Excel in Hindi || Full Explanation Vlookup & Hlookup
Download Vlookup and Hlookup Practice Sheet Here: ► http://www.techguruplus.com/vlookup-and-hlookup-formula-in-excel-hindi/ My Instagram: ► https://www.instag
## VLOOKUP Exact Match with #N/A erros to fix
In this video, we will use the VLOOKUP function in Excel and look for an Exact match in a table found on a separate worksheet. We'll also fix several #N/A error
## MS Excel - VLookup with Trim
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## Excel VLOOKUP Function with Example in Telugu || www.computersadda.com
More Computer Courses Videos Links in below --------------------------------------------------------------------------- Ms-Word in Telugu with in 5 Hours ------ | 836 | 3,350 | {"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} | 2.671875 | 3 | CC-MAIN-2021-21 | latest | en | 0.580449 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/320i1/C2%5E4.34D10.html | 1,606,448,494,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189038.24/warc/CC-MAIN-20201127015426-20201127045426-00398.warc.gz | 432,666,589 | 8,193 | Copied to
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## G = C24.34D10order 320 = 26·5
### 34th non-split extension by C24 of D10 acting via D10/C5=C22
Series: Derived Chief Lower central Upper central
Derived series C1 — C2×C10 — C24.34D10
Chief series C1 — C5 — C10 — C2×C10 — C22×D5 — C23×D5 — C23⋊D10 — C24.34D10
Lower central C5 — C2×C10 — C24.34D10
Upper central C1 — C22 — C22≀C2
Generators and relations for C24.34D10
G = < a,b,c,d,e,f | a2=b2=c2=d2=e10=1, f2=d, ab=ba, eae-1=faf-1=ac=ca, ad=da, bc=cb, ebe-1=bd=db, fbf-1=bcd, cd=dc, ce=ec, cf=fc, de=ed, df=fd, fef-1=de-1 >
Subgroups: 1406 in 334 conjugacy classes, 103 normal (39 characteristic)
C1, C2, C2 [×2], C2 [×8], C4 [×10], C22, C22 [×2], C22 [×28], C5, C2×C4, C2×C4 [×2], C2×C4 [×13], D4 [×22], Q8 [×2], C23 [×2], C23 [×2], C23 [×11], D5 [×3], C10, C10 [×2], C10 [×5], C42 [×2], C22⋊C4, C22⋊C4 [×2], C22⋊C4 [×7], C4⋊C4 [×2], C22×C4 [×3], C2×D4, C2×D4 [×2], C2×D4 [×16], C2×Q8, C4○D4 [×4], C24, C24, Dic5 [×4], Dic5 [×3], C20 [×3], D10 [×13], C2×C10, C2×C10 [×2], C2×C10 [×15], C42⋊C2, C22≀C2, C22≀C2 [×3], C4⋊D4 [×4], C4.4D4 [×2], C41D4 [×2], C22×D4, C2×C4○D4, Dic10 [×2], C4×D5 [×2], D20 [×2], C2×Dic5 [×3], C2×Dic5 [×6], C2×Dic5 [×2], C5⋊D4 [×16], C2×C20, C2×C20 [×2], C5×D4 [×4], C22×D5, C22×D5 [×2], C22×D5 [×4], C22×C10 [×2], C22×C10 [×2], C22×C10 [×4], C22.29C24, C4×Dic5 [×2], C10.D4 [×2], D10⋊C4 [×4], C23.D5, C23.D5 [×2], C5×C22⋊C4, C5×C22⋊C4 [×2], C2×Dic10, C2×C4×D5, C2×D20 [×2], D42D5 [×4], C22×Dic5 [×2], C2×C5⋊D4 [×2], C2×C5⋊D4 [×8], C2×C5⋊D4 [×4], D4×C10, D4×C10 [×2], C23×D5, C23×C10, C23.11D10, C22⋊D20, D10⋊D4 [×2], Dic5.5D4 [×2], C23⋊D10, Dic5⋊D4 [×2], C20⋊D4 [×2], C242D5, C5×C22≀C2, C2×D42D5, C22×C5⋊D4, C24.34D10
Quotients: C1, C2 [×15], C22 [×35], D4 [×4], C23 [×15], D5, C2×D4 [×6], C24, D10 [×7], C22×D4, 2+ 1+4 [×2], C22×D5 [×7], C22.29C24, D4×D5 [×2], C23×D5, C2×D4×D5, D46D10 [×2], C24.34D10
Smallest permutation representation of C24.34D10
On 80 points
Generators in S80
```(1 63)(2 31)(3 65)(4 33)(5 67)(6 35)(7 69)(8 37)(9 61)(10 39)(11 72)(12 46)(13 74)(14 48)(15 76)(16 50)(17 78)(18 42)(19 80)(20 44)(21 45)(22 73)(23 47)(24 75)(25 49)(26 77)(27 41)(28 79)(29 43)(30 71)(32 57)(34 59)(36 51)(38 53)(40 55)(52 70)(54 62)(56 64)(58 66)(60 68)
(1 22)(2 74)(3 24)(4 76)(5 26)(6 78)(7 28)(8 80)(9 30)(10 72)(11 39)(12 55)(13 31)(14 57)(15 33)(16 59)(17 35)(18 51)(19 37)(20 53)(21 62)(23 64)(25 66)(27 68)(29 70)(32 48)(34 50)(36 42)(38 44)(40 46)(41 60)(43 52)(45 54)(47 56)(49 58)(61 71)(63 73)(65 75)(67 77)(69 79)
(1 55)(2 56)(3 57)(4 58)(5 59)(6 60)(7 51)(8 52)(9 53)(10 54)(11 21)(12 22)(13 23)(14 24)(15 25)(16 26)(17 27)(18 28)(19 29)(20 30)(31 64)(32 65)(33 66)(34 67)(35 68)(36 69)(37 70)(38 61)(39 62)(40 63)(41 78)(42 79)(43 80)(44 71)(45 72)(46 73)(47 74)(48 75)(49 76)(50 77)
(1 63)(2 64)(3 65)(4 66)(5 67)(6 68)(7 69)(8 70)(9 61)(10 62)(11 45)(12 46)(13 47)(14 48)(15 49)(16 50)(17 41)(18 42)(19 43)(20 44)(21 72)(22 73)(23 74)(24 75)(25 76)(26 77)(27 78)(28 79)(29 80)(30 71)(31 56)(32 57)(33 58)(34 59)(35 60)(36 51)(37 52)(38 53)(39 54)(40 55)
(1 2 3 4 5 6 7 8 9 10)(11 12 13 14 15 16 17 18 19 20)(21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40)(41 42 43 44 45 46 47 48 49 50)(51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70)(71 72 73 74 75 76 77 78 79 80)
(1 62 63 10)(2 9 64 61)(3 70 65 8)(4 7 66 69)(5 68 67 6)(11 22 45 73)(12 72 46 21)(13 30 47 71)(14 80 48 29)(15 28 49 79)(16 78 50 27)(17 26 41 77)(18 76 42 25)(19 24 43 75)(20 74 44 23)(31 38 56 53)(32 52 57 37)(33 36 58 51)(34 60 59 35)(39 40 54 55)```
`G:=sub<Sym(80)| (1,63)(2,31)(3,65)(4,33)(5,67)(6,35)(7,69)(8,37)(9,61)(10,39)(11,72)(12,46)(13,74)(14,48)(15,76)(16,50)(17,78)(18,42)(19,80)(20,44)(21,45)(22,73)(23,47)(24,75)(25,49)(26,77)(27,41)(28,79)(29,43)(30,71)(32,57)(34,59)(36,51)(38,53)(40,55)(52,70)(54,62)(56,64)(58,66)(60,68), (1,22)(2,74)(3,24)(4,76)(5,26)(6,78)(7,28)(8,80)(9,30)(10,72)(11,39)(12,55)(13,31)(14,57)(15,33)(16,59)(17,35)(18,51)(19,37)(20,53)(21,62)(23,64)(25,66)(27,68)(29,70)(32,48)(34,50)(36,42)(38,44)(40,46)(41,60)(43,52)(45,54)(47,56)(49,58)(61,71)(63,73)(65,75)(67,77)(69,79), (1,55)(2,56)(3,57)(4,58)(5,59)(6,60)(7,51)(8,52)(9,53)(10,54)(11,21)(12,22)(13,23)(14,24)(15,25)(16,26)(17,27)(18,28)(19,29)(20,30)(31,64)(32,65)(33,66)(34,67)(35,68)(36,69)(37,70)(38,61)(39,62)(40,63)(41,78)(42,79)(43,80)(44,71)(45,72)(46,73)(47,74)(48,75)(49,76)(50,77), (1,63)(2,64)(3,65)(4,66)(5,67)(6,68)(7,69)(8,70)(9,61)(10,62)(11,45)(12,46)(13,47)(14,48)(15,49)(16,50)(17,41)(18,42)(19,43)(20,44)(21,72)(22,73)(23,74)(24,75)(25,76)(26,77)(27,78)(28,79)(29,80)(30,71)(31,56)(32,57)(33,58)(34,59)(35,60)(36,51)(37,52)(38,53)(39,54)(40,55), (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80), (1,62,63,10)(2,9,64,61)(3,70,65,8)(4,7,66,69)(5,68,67,6)(11,22,45,73)(12,72,46,21)(13,30,47,71)(14,80,48,29)(15,28,49,79)(16,78,50,27)(17,26,41,77)(18,76,42,25)(19,24,43,75)(20,74,44,23)(31,38,56,53)(32,52,57,37)(33,36,58,51)(34,60,59,35)(39,40,54,55)>;`
`G:=Group( (1,63)(2,31)(3,65)(4,33)(5,67)(6,35)(7,69)(8,37)(9,61)(10,39)(11,72)(12,46)(13,74)(14,48)(15,76)(16,50)(17,78)(18,42)(19,80)(20,44)(21,45)(22,73)(23,47)(24,75)(25,49)(26,77)(27,41)(28,79)(29,43)(30,71)(32,57)(34,59)(36,51)(38,53)(40,55)(52,70)(54,62)(56,64)(58,66)(60,68), (1,22)(2,74)(3,24)(4,76)(5,26)(6,78)(7,28)(8,80)(9,30)(10,72)(11,39)(12,55)(13,31)(14,57)(15,33)(16,59)(17,35)(18,51)(19,37)(20,53)(21,62)(23,64)(25,66)(27,68)(29,70)(32,48)(34,50)(36,42)(38,44)(40,46)(41,60)(43,52)(45,54)(47,56)(49,58)(61,71)(63,73)(65,75)(67,77)(69,79), (1,55)(2,56)(3,57)(4,58)(5,59)(6,60)(7,51)(8,52)(9,53)(10,54)(11,21)(12,22)(13,23)(14,24)(15,25)(16,26)(17,27)(18,28)(19,29)(20,30)(31,64)(32,65)(33,66)(34,67)(35,68)(36,69)(37,70)(38,61)(39,62)(40,63)(41,78)(42,79)(43,80)(44,71)(45,72)(46,73)(47,74)(48,75)(49,76)(50,77), (1,63)(2,64)(3,65)(4,66)(5,67)(6,68)(7,69)(8,70)(9,61)(10,62)(11,45)(12,46)(13,47)(14,48)(15,49)(16,50)(17,41)(18,42)(19,43)(20,44)(21,72)(22,73)(23,74)(24,75)(25,76)(26,77)(27,78)(28,79)(29,80)(30,71)(31,56)(32,57)(33,58)(34,59)(35,60)(36,51)(37,52)(38,53)(39,54)(40,55), (1,2,3,4,5,6,7,8,9,10)(11,12,13,14,15,16,17,18,19,20)(21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40)(41,42,43,44,45,46,47,48,49,50)(51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70)(71,72,73,74,75,76,77,78,79,80), (1,62,63,10)(2,9,64,61)(3,70,65,8)(4,7,66,69)(5,68,67,6)(11,22,45,73)(12,72,46,21)(13,30,47,71)(14,80,48,29)(15,28,49,79)(16,78,50,27)(17,26,41,77)(18,76,42,25)(19,24,43,75)(20,74,44,23)(31,38,56,53)(32,52,57,37)(33,36,58,51)(34,60,59,35)(39,40,54,55) );`
`G=PermutationGroup([(1,63),(2,31),(3,65),(4,33),(5,67),(6,35),(7,69),(8,37),(9,61),(10,39),(11,72),(12,46),(13,74),(14,48),(15,76),(16,50),(17,78),(18,42),(19,80),(20,44),(21,45),(22,73),(23,47),(24,75),(25,49),(26,77),(27,41),(28,79),(29,43),(30,71),(32,57),(34,59),(36,51),(38,53),(40,55),(52,70),(54,62),(56,64),(58,66),(60,68)], [(1,22),(2,74),(3,24),(4,76),(5,26),(6,78),(7,28),(8,80),(9,30),(10,72),(11,39),(12,55),(13,31),(14,57),(15,33),(16,59),(17,35),(18,51),(19,37),(20,53),(21,62),(23,64),(25,66),(27,68),(29,70),(32,48),(34,50),(36,42),(38,44),(40,46),(41,60),(43,52),(45,54),(47,56),(49,58),(61,71),(63,73),(65,75),(67,77),(69,79)], [(1,55),(2,56),(3,57),(4,58),(5,59),(6,60),(7,51),(8,52),(9,53),(10,54),(11,21),(12,22),(13,23),(14,24),(15,25),(16,26),(17,27),(18,28),(19,29),(20,30),(31,64),(32,65),(33,66),(34,67),(35,68),(36,69),(37,70),(38,61),(39,62),(40,63),(41,78),(42,79),(43,80),(44,71),(45,72),(46,73),(47,74),(48,75),(49,76),(50,77)], [(1,63),(2,64),(3,65),(4,66),(5,67),(6,68),(7,69),(8,70),(9,61),(10,62),(11,45),(12,46),(13,47),(14,48),(15,49),(16,50),(17,41),(18,42),(19,43),(20,44),(21,72),(22,73),(23,74),(24,75),(25,76),(26,77),(27,78),(28,79),(29,80),(30,71),(31,56),(32,57),(33,58),(34,59),(35,60),(36,51),(37,52),(38,53),(39,54),(40,55)], [(1,2,3,4,5,6,7,8,9,10),(11,12,13,14,15,16,17,18,19,20),(21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40),(41,42,43,44,45,46,47,48,49,50),(51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70),(71,72,73,74,75,76,77,78,79,80)], [(1,62,63,10),(2,9,64,61),(3,70,65,8),(4,7,66,69),(5,68,67,6),(11,22,45,73),(12,72,46,21),(13,30,47,71),(14,80,48,29),(15,28,49,79),(16,78,50,27),(17,26,41,77),(18,76,42,25),(19,24,43,75),(20,74,44,23),(31,38,56,53),(32,52,57,37),(33,36,58,51),(34,60,59,35),(39,40,54,55)])`
50 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 2K 4A 4B 4C 4D 4E 4F 4G 4H 4I 4J 5A 5B 10A ··· 10F 10G ··· 10R 10S 10T 20A ··· 20F order 1 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 4 4 4 4 4 5 5 10 ··· 10 10 ··· 10 10 10 20 ··· 20 size 1 1 1 1 2 2 4 4 4 20 20 20 4 4 4 10 10 10 10 20 20 20 2 2 2 ··· 2 4 ··· 4 8 8 8 ··· 8
50 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 4 4 4 type + + + + + + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 C2 D4 D5 D10 D10 D10 2+ 1+4 D4×D5 D4⋊6D10 kernel C24.34D10 C23.11D10 C22⋊D20 D10⋊D4 Dic5.5D4 C23⋊D10 Dic5⋊D4 C20⋊D4 C24⋊2D5 C5×C22≀C2 C2×D4⋊2D5 C22×C5⋊D4 C2×Dic5 C22≀C2 C22⋊C4 C2×D4 C24 C10 C22 C2 # reps 1 1 1 2 2 1 2 2 1 1 1 1 4 2 6 6 2 2 4 8
Matrix representation of C24.34D10 in GL6(𝔽41)
1 0 0 0 0 0 0 1 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 1 0 0 0 0 0 0 1
,
40 0 0 0 0 0 0 1 0 0 0 0 0 0 17 40 0 0 0 0 1 24 0 0 0 0 0 0 24 1 0 0 0 0 40 17
,
1 0 0 0 0 0 0 1 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40
,
40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40 0 0 0 0 0 0 40
,
0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 7 7 0 0 0 0 34 40 0 0 7 7 0 0 0 0 34 40 0 0
,
0 40 0 0 0 0 1 0 0 0 0 0 0 0 0 0 40 34 0 0 0 0 0 1 0 0 1 7 0 0 0 0 0 40 0 0
`G:=sub<GL(6,GF(41))| [1,0,0,0,0,0,0,1,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,1,0,0,0,0,0,0,1],[40,0,0,0,0,0,0,1,0,0,0,0,0,0,17,1,0,0,0,0,40,24,0,0,0,0,0,0,24,40,0,0,0,0,1,17],[1,0,0,0,0,0,0,1,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40],[40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40,0,0,0,0,0,0,40],[0,1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,7,34,0,0,0,0,7,40,0,0,7,34,0,0,0,0,7,40,0,0],[0,1,0,0,0,0,40,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,7,40,0,0,40,0,0,0,0,0,34,1,0,0] >;`
C24.34D10 in GAP, Magma, Sage, TeX
`C_2^4._{34}D_{10}`
`% in TeX`
`G:=Group("C2^4.34D10");`
`// GroupNames label`
`G:=SmallGroup(320,1264);`
`// by ID`
`G=gap.SmallGroup(320,1264);`
`# by ID`
`G:=PCGroup([7,-2,-2,-2,-2,-2,-2,-5,758,219,675,297,12550]);`
`// Polycyclic`
`G:=Group<a,b,c,d,e,f|a^2=b^2=c^2=d^2=e^10=1,f^2=d,a*b=b*a,e*a*e^-1=f*a*f^-1=a*c=c*a,a*d=d*a,b*c=c*b,e*b*e^-1=b*d=d*b,f*b*f^-1=b*c*d,c*d=d*c,c*e=e*c,c*f=f*c,d*e=e*d,d*f=f*d,f*e*f^-1=d*e^-1>;`
`// generators/relations`
×
𝔽 | 6,859 | 10,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2020-50 | longest | en | 0.344405 |
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2007 BICH 441 1st Exam
# 2007 BICH 441 1st Exam - name BICH 441 EXAM 1 An Aggie does...
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name ________________________________________ BICH 441 EXAM 1 February 16, 2007 An Aggie does not lie, cheat, or steal, or tolerate those who do. I swear on my honor as an Aggie that I have not received or given unauthorized aid on this exam, and I am not aware of others who have received or given unauthorized aid on this exam. signed _________________________________________________________ NO CALCULATORS! Before you begin, write your name on each page (there are 5 pages) and sign the Aggie Honor Code statement above. Information you may find useful in answering some of the questions: fructose-1,6-bisP + H 2 O fructose-6-P + P i G°' = –13.3 kJ/mol 1,3-bisphosphoglycerate + H 2 O 3-phosphoglycerate + P i G°' = –49.4 kJ/mol PEP + H 2 O pyruvate + P i G°' = –62.2 kJ/mol glucose-6-P + H 2 O glucose + P i G°' = –13.8 kJ/mol citrate isocitrate G°' = +6.7 kJ/mol glucose-6-P fructose-6-P G°' = +1.67 kJ/mol ATP + H 2 O ADP + P i G°' = –30.5 kJ/mol R = 8.314 J/mol•°K assume T = 25°C Questions 1-15 (2 pts each). Enter your answer on your Scantron sheet. 1. G°' for the reaction in glycolysis catalyzed by phosphoglycerate kinase is: A. –49.4 kJ/mol B. –18.9 kJ/mol C. 30.5 kJ/mol D. 49.4 kJ/mol 2. K eq ' for the sum of the reactions in the TCA cycle is 1.0 × 10 7 . G°' for this series of reactions is therefore:
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2007 BICH 441 1st Exam - name BICH 441 EXAM 1 An Aggie does...
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Ask a homework question - tutors are online | 617 | 1,952 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-13 | latest | en | 0.819495 |
http://mathhelpforum.com/calculus/67527-vectors.html | 1,511,027,701,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805008.39/warc/CC-MAIN-20171118171235-20171118191235-00749.warc.gz | 193,267,926 | 11,869 | 1. ## Vectors
Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.
===
Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$= 2|b| $^2$.
Thank you for helping!
2. Originally Posted by Tangera
Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.
===
Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$= 2|b| $^2$.
Thank you for helping!
1. Take the cross product of $\mathbf{a}$ and $\mathbf{b}$ and then divide this vector by its length.
3. Originally Posted by Tangera
Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.
===
Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$= 2|b| $^2$.
Thank you for helping!
2. Drawing a picture always helps.
Since it's a parallelogram, notice that OR is parallel and of equal magnitude to PQ and that QR is parallel and of equal magnitude to OP.
What do you have when vectors are parallel and of equal length? They are EQUAL.
4. ^ Thank you for your suggestions! I got stuck at part that requires the evaluation of the scalar product... Do I evaluate PR.OQ = OP.PQ = 0?
5. Originally Posted by Tangera
^ Thank you for your suggestions! I got stuck at part that requires the evaluation of the scalar product... Do I evaluate PR.OQ = OP.PQ = 0?
What sides touch?
OP touches PQ
PQ touches QR
QR touches OR
OP touches OR.
If it's a square, then the angles should be right angles. Evaluating any of the dot products of touching sides should give 0 if this is the case.
Then show that the lengths are equal, and you've got a square.
See how you go from there.
6. ^ Um...sorry I am still confused...I understand your method, but the question wanted me to show |a|= 2|b| if OPQR is a square...so do I have to equate 2 scalar products?
7. Originally Posted by Tangera
^ Um...sorry I am still confused...I understand your method, but the question wanted me to show |a|= 2|b| if OPQR is a square...so do I have to equate 2 scalar products?
Is that $|\mathbf{a}|^2 = 2|\mathbf{b}|^2$?
8. ^ Yup the question wanted ... | 847 | 2,859 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-47 | longest | en | 0.907311 |
https://www.azom.com/article.aspx?ArticleID=9718 | 1,721,605,540,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00868.warc.gz | 570,594,175 | 24,882 | # Introducing the Confocal Method of Magnetron Sputtering
The confocal sputtering technique involves the arrangement of magnetrons inside a vacuum chamber so that it is possible to apply multiple materials onto the substrate without breaking vacuum. This method enables the user to co-sputter, or to form a film from two or more materials at once. It is well recognized for small scale batch productions and for research and development.
Conventional top down sputtering enables sputtering of only one material at a time due to the geometric relation between the magnetron and the substrate. For this type of sputtering, it is necessary to have a target material larger than the substrate to obtain acceptable film uniformity. Moreover, the target material needs be larger than the substrate area.
On the other hand, confocal sputtering involves positioning of the target at an angle with respect to the surface of the substrate. While performing this step, it is necessary to move off the magnetron from the center of the rotational axis and rotate it in the event of deposition to obtain an acceptable uniform film thickness, as demonstrated in Figure 1. This now enables locating multiple magnetrons with different materials at equidistant from the rotation axis of the substrate and each can still deliver a highly uniform film thickness.
Figure 1. Geometric layout of cathode relative to substrate for confocal sputtering.
## Cathode Alignment for Optimum Uniformity
Confocal cathode sputtering relies on altering the angle of the cathode target corresponding to the substrate so as to form a highly uniform deposition process. This method needs mounting the cathode at an angle (θ) corresponding to the vertical (y) axis, as depicted in Figure 1.
There is a possibility to receive more material at the inner area of the substrate when compared the outer areas if the substrate is held static. This non-uniformity can be reduced when the substrate is rotated about its own axis with the cathode offset.
The cathode’s tilted orientation helps achieving uniform deposition. The optimization of the tilt angle and the target distance from the substrate can help create highly uniform films. For these reasons, confocal sputtering is capable of often achieving better than 5% non-uniformity in film thickness.
For confocal deposition, the cathode positioning is performed by drawing a line that runs through the cathode center and intersects the y axis. θ is set to 30° to obtain better results, because at this point the only variable left is the distance of the cathode to the substrate (m). In many systems, three or four cathodes can provide the optimum balance over good deposition rates, uniform film thickness, and material flexibility but compromise on the uniformity capabilities of the system owing to the changes in the throw distance.
## Sputtering Rate and Target Efficiency
The uniformity distribution of the film thickness pursues a parabolic curve beginning from the substrate center during top down, concentric deposition, as illustrated in Figure 2. The film thickness reduces exponentially from the center during movement towards the substrate’s outside edge. It is possible to move the magnetron farther away from the substrate surface to reduce non-uniformity in film thickness, but results in significant decrease in the deposition rate.
Figure 2. % non-uniformity (y) across the diameter of a substrate (x).
The throw distance to the substrate is capable of dramatically changing the uniformity for a specific area as illustrated in Figure 2. Confocal sputtering enables using a much smaller diameter target material corresponding to the substrate diameter.
The role of rotation in bringing the flat portion of the curve within +/- 5% uniformity is illustrated in Figure 3.
A 4-inch throw distance and a 30° offset for the magnetron are the ideal combination to form the most uniform film and increase the target efficiency with increase in material being used for its intended purpose. This is especially beneficial for precious metals. Moreover, confocal sputtering intrinsically utilizes smaller targets when compared to traditional sputtering.
Figure 3. Graph showing % non-uniformity for a non-rotating substrate in confocal alignment.
The addition of more number of magnetrons to a system must not change the distance from which they are positioned corresponding to the rotation axis of the substrate so as to maintain the complex geometry of the confocal configuration.
Adding the fourth or fifth magnetron will increase the distance in order for all the magnetrons to be at equidistant from the rotation axis. Hence, the use of no more than three to four cathodes is typically preferred for confocal sputtering as the use of more number of cathodes will lead to increased non-uniformity and reduced sputtering rates.
Cathode angle, source-to-substrate distance, and the position where the target’s centerline intersects with the substrate are all significant in creating high quality thin films. One method to ensure adjustability in all of these aspects is to use flex mount cathodes shown in Figure 4, which provide the adjustment required to alter the target angle corresponding to the substrate.
Figure 4. Semicore Flex mounted cathodes in confocal alignment.
## Conclusion
From the aforementioned information, it is evident that confocal sputtering offers several benefits to users and extends flexibility to the system. This method improves the yield of the sputtered material and enhances film uniformity. As it can have multiple materials in the vacuum chamber at the same time, it reduces time involved in the formation of complex films, as well as enables co-sputtering to take place. Co-sputtering is greatly recommended for today's sophisticated research and development systems.
Semicore is a manufacturer and worldwide supplier for the electronics, optical, solar energy, medical, automotive, military and related high technology industries.
Our high-performance production or R&D vacuum sputtering and thin film evaporation systems provide coatings on a variety of materials including plastic films, glass, ceramics, metals and hybrid substrates.
Whether you want to take advantage of our proven industrial solutions for vacuum system automation, process control and supervisory monitoring applications or need to develop some unique new application of your own design you will find Semicore’s staff and facilities to be competent, open-minded and eager to help.
This information has been sourced, reviewed and adapted from materials provided by Semicore Equipment.
## Citations
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Semicore Equipment, Inc.. (2018, August 23). Introducing the Confocal Method of Magnetron Sputtering. AZoM. Retrieved on July 21, 2024 from https://www.azom.com/article.aspx?ArticleID=9718.
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Semicore Equipment, Inc.. 2018. Introducing the Confocal Method of Magnetron Sputtering. AZoM, viewed 21 July 2024, https://www.azom.com/article.aspx?ArticleID=9718. | 1,518 | 7,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.903556 |
https://aptitude.gateoverflow.in/6150/Cat-2018-set-2-question-67 | 1,702,130,270,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00237.warc.gz | 120,761,277 | 21,093 | 533 views
The value of the sum $7 \times 11 + 11 \times 15 + 15 \times 19 + \dots$ + $95 \times 99$ is
1. $80707$
2. $80773$
3. $80730$
4. $80751$
Let $S = 7 \times 11 + 11 \times 15 + 15 \times 19 + \dots + 95 \times 99$
We can write first series as :
$S_{1} = \underbrace{7 + 11 + 15 + \dots + 95}_{\text{Arithmetic Progression (AP)}}$
Here, first term $a = 7,$ common difference $d = 11 – 7 = 4,$ last term $l = 95$
The $n^{th}$ term of the series $t_{n} = l = a+(n-1)d \;,$ where $n =$ number of terms
$\Rightarrow 95 = 7+(n-1)4$
$\Rightarrow 4n – 4 + 7 = 95$
$\Rightarrow 4n + 3 = 95$
$\Rightarrow 4n = 92$
$\Rightarrow \boxed{ n= 23}$
The $n^{th}$ term of the series $\boxed{t_{n} = 4n + 3}$
Similarly, we can write second series as :
$S_{2} = \underbrace{11 + 15 + 19 + \dots + 99}_{\text{Arithmetic Progression (AP)}}$
Here, $a = 11, \; d = 15 – 11 = 4, \; l = 99, \; n = 23$
The $n^{th}$ term of the series $t_{n} = l = a + (n-1)d$
$\Rightarrow t_{n} = 11 + (n-1)4$
$\Rightarrow t_{n} = 11 + 4n – 4$
$\Rightarrow \boxed{t_{n} = 4n + 7}$
Now, we can write $T_{n} = (4n+3)(4n+7) \;, \text{where} \; n = 1, 2, \dots, 23$
$\Rightarrow T_{n} = 16n^{2} + 28n + 12n + 21$
$\Rightarrow T_{n} = 16n^{2} + 40n + 21$
$\Rightarrow \sum T_{n} = \sum (16n^{2} + 40n + 21)$
$\Rightarrow S_{n} = 16 \sum n^{2} + 40 \sum n + 21 \sum 1$
$\Rightarrow S_{n} = 16 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 40 \left[ \frac{n(n+1)}{2} \right] + 21n$
The sum of the series $S = 16 \left[\frac{(23)(24)(47)}{6} \right] + 40 \left[ \frac{(23)(24)}{2} \right] + 21 \times 23$
$\Rightarrow S = 69184 + 11040 + 483$
$\Rightarrow \boxed{S = 80707}$
Correct Answer $: \text{A}$
$\textbf{PS :}$
If $S_{n} = 1 + 2 + 3 + \dots + n$
$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n} k$
$\Rightarrow \boxed{S_{n} = \frac{n(n+1)}{2}}$
If $S_{n} = 1^{2} + 2^{2} + 3^{2} + \dots + n^{2}$
$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n}k^{2}$
$\Rightarrow \boxed{S_{n} = \frac{n(n+1)(2n+1)}{6}}$
If $S_{n} = 1^{3} + 2^{3} + 3^{3} + \dots + n^{3}$
$\Rightarrow S_{n} = \displaystyle{}\sum_{k=1}^{n} k^{3}$
$\Rightarrow \boxed{S_{n} = \left[ \frac{n(n+1)}{2} \right]^{2}}$
11.5k points
1
576 views | 1,005 | 2,212 | {"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} | 4.71875 | 5 | CC-MAIN-2023-50 | latest | en | 0.423945 |
https://shhh-film.com/question-854 | 1,675,656,648,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500303.56/warc/CC-MAIN-20230206015710-20230206045710-00864.warc.gz | 531,987,408 | 5,590 | ## Inverse Functions
Correct answer: Explanation: To find the inverse of a function, you need to change all of the values to values and all the values to values. If you flip a function over the line , then you are
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## 2.4 Inverse Functions
Finding the Inverse of a Function Given the function f (x) f ( x) we want to find the inverse function, f −1(x) f − 1 ( x). First, replace f (x) f ( x) with y y. This is done to make the rest of the process easier. Replace every x x with a y y
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## Finding inverse functions (article)
The inverse function, if you take f inverse of 4, f inverse of 4 is equal to 0. Or the inverse function is mapping us from 4 to 0. Which is exactly what we expected. The function takes us from the x to the y world, and then
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Figure out math question
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## Algebra II : Inverse Functions
Inverse functions are functions that reverse the effect of the original function. The inverse of a function has the same points as the original function except that the values of x and y are swapped. For example, if the original function
## Inverse Functions
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Mathematics can be a daunting subject for many students, but with a little practice, it can be easy to clear up any mathematic tasks.
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By breaking down and clarifying the steps in a math equation, students can more easily understand and solve the problem. | 508 | 2,250 | {"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} | 4.375 | 4 | CC-MAIN-2023-06 | latest | en | 0.873445 |
http://www.ck12.org/physics/Vectors/?difficulty=basic&by=community | 1,493,587,508,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125849.25/warc/CC-MAIN-20170423031205-00618-ip-10-145-167-34.ec2.internal.warc.gz | 484,548,359 | 14,735 | Vectors
A scalar magnitude with a direction.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
Vectors
by Henrik Malmberg //at grade
In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not 'talk' to each other. Thus one must use the equations of Motion once for the x-direction and once for the y-direction.
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Vectors
by Shalini Mishra //at grade
In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not 'talk' to each other. Thus one must use the equations of Motion once for the x-direction and once for the y-direction.
MEMORY METER
This indicates how strong in your memory this concept is
0
Unit 4.1 Vectors
by Scott //at grade
In order to solve two dimensional problems it is necessary to break all vectors into their x and y components. Different dimensions do not 'talk' to each other. Thus one must use the equations of Motion once for the x-direction and once for the y-direction.
MEMORY METER
This indicates how strong in your memory this concept is
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A.T. August 16, 1999 20:14
Non-slip Boundary condition
Hi everybody,
Can anybody explain why non-slip boundary condition from the physical (maybe atom) point of view please?
Say, we want to calculate the flow field of the mercury through a tube made of glass, can we still use this non-slip boundary condition? thanks.
John C. Chien August 17, 1999 00:35
Re: Non-slip Boundary condition
(1). try to look up the properties of mercury in a handbook first. (2). If it says the viscosity is zero, then you can use slip boundary condition at the wall. (3). Otherwise, if it has a viscosity value, then use the non-slip boundary condition. (4). If you can't find it, then run two cases with slip and non-slip conditions. And compare the results with the test data.
Tae-Chang Jo August 17, 1999 13:00
Re:Non-slip Boundary condition
Hi. Could anyone tell me what the slip boundary condition is? Thank you.
Patrick Godon August 17, 1999 14:11
Re: Non-slip Boundary condition
For mercury the surface tension is extremely high and this has also to be taken into account. For example if you try to 'inject' mecury into a medium using a syringe, this one might just break. Also, a little amount of mercury in a tube would rather 'roll' rather than flow. So if you're talking about a small glass tube, you will certainly need a non-slip boundary condition (viscosity, friction) and some treatment for the surface tension. Check the viscosity of mercury, the velocity of its flow and find out (using simple books about the theory of the boundary layer) whether the boundary layer will be much smaller than the size of the glass tube (slip BC) or of the same scale (non-slip BC).
As to your question related to the microscopic scale of the friction, it is a whole field of Physics!
John C. Chien August 17, 1999 20:38
Re:Non-slip Boundary condition
(1).For a stationary wall ( or ground ), a non-slip condition is that the fluid velocity at the wall point is zero.(because of the existence of the viscosity) (2). In some analyses, the fluid viscosity is removed from the governing equations, and the resultant equation is INVISCID. (that is non-viscous). (3). For inviscid flows, we can not apply the non-slip boundary condition at the wall because there is not viscosity by assumption. (4). In the inviscid flow case, the fluid velocity at the wall is allowed to slip along the contour of the surface (move on the surface and tangent to the surface). This condition is called "slip condition".
Tae-Chang Jo August 18, 1999 16:05
Re:Non-slip Boundary condition
Thank you for your answering. I'd like to make sure that I understand well. For inviscid flows, the normal component of the velocity is zero at the boundary - this is what your saying, isn't it?
Then can the tangental velocity be any? Is there any condition such as the derivative in tangential direction is zero?
Thank you again.
prishor November 28, 2012 00:26
Quote:
Originally Posted by Tae-Chang Jo ;4652 Thank you for your answering. I'd like to make sure that I understand well. For inviscid flows, the normal component of the velocity is zero at the boundary - this is what your saying, isn't it? Then can the tangental velocity be any? Is there any condition such as the derivative in tangential direction is zero? Thank you again.
derivative of the tangential velocity will be zero for hydrodynamically(fully) developed flows only.it cannot be zero for boundary layer and entry length problems
regards,
prishor
michujo November 28, 2012 03:19
Quote:
Originally Posted by prishor (Post 394546) derivative of the tangential velocity will be zero for hydrodynamically(fully) developed flows only.it cannot be zero for boundary layer and entry length problems regards, prishor
Hi, I have to disagree with that. A null derivative of the tangential velocity component relative to the direction normal to the wall is a no-slip condition (zero shear stress at the wall). It does not correspond to a fully developed flow, check for instance the Poiseuille solution for laminar flow in a pipe and you'll readily see that it is not zero.
Cheers.
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# Using a BA II Plus, explain how to complete 5 problems using TVM functions
After graduation, you plan to work for Mega Corporation for 10 years and then start your own business. You expect to save \$5,000 a year for the first 5 years and \$10,000 annually for the following 5 years, with the first deposit being made a year from today. In addition, your grandfather just gave you a \$10,000 graduation gift which you will deposit immediately. If the account earns 8% compounded annually, what how much will you have when you start your business 10 years from now?
A 30-year, \$115,000 mortgage has a nominal annual rate of 7%. All payments are made at the end of each month.
10. What is the monthly payment on the mortgage?
\$765.0979
11. What is the remaining balance on the mortgage after 5 years?
12. How much of your 2nd monthly payment will go toward the repayment of principal? | 216 | 915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-44 | longest | en | 0.963034 |
https://www.acxesspring.com/compression-spring-rate.html | 1,721,123,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00014.warc.gz | 569,803,400 | 25,252 | # Compression Spring Rate
#### Definition: The constant rate of force expelled by a compression spring per inch or millimeter of travel once a load or force is applied to push and compress the compression spring.
Compression spring rate is the spring specification which will help you determine if you will be able to meet your compression spring’s working loads at the time of being installed in your application. Since this is a constant force, the amounts of load and travel increase in a consistent manner. It is calculated based on your spring’s physical dimensions such as wire diameter, active coils, or outer diameter. Other compression spring dimensions like inner diameter, mean diameter, or index can be calculated using the previously mentioned dimensions.
Spring Calculator Instructions
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
Attention! Input results shown will be +/- 10% from middle value. Hint: The closer your min and max inputs are, the more accurate your results will be!
### Compression Spring Rate Formula
To calculate the amount of rate on a compression spring design based on physical dimensions including material type use the formula provided below:
k = Gd^4 / 8D^3N
#### Explanation of Symbols:
d = wire size (inches)
D = Mean Diameter (inches)
N = Number of active coils
D / d = Index correction
G = Shear Modulus of Material
k = Spring Constant
#### G-Value for Common Spring Materials
Music Wire = 11.5 x 10^6
Stainless Steel = 11.2 x 10^6
Phospher Bronze = 5.9 x 10^6
Monel = 9.6 x 10^6
Inconel = 11.5 x 10^6
Copper = 6.5 x 10^6
Beryllium Copper = 6.9 x 10^6
### How to Measure Diagram
To calculate the amount of spring rate you will need on order to meet your working loads, simply divide the load you will be applying on your spring by the distance you expect your spring to travel or compress under that load. The equivalent to that formula will be your compression spring rate as shown below.
k = L ÷ T
Follow these examples to see how the formula works.
#### Example 1.)
You have a heavy duty spring that has a free length of 1” and it is supposed to travel or compress 0.2” under a load of 5 pounds. To know your compression spring rate divide the load of 5 pounds by the distance traveled of 0.2”. The result shall be 25 pounds of force per inch of travel.
k = L ÷ T
k = 5 ÷ 0.2
k = 25 lbf/in
#### Example 2.)
You have a compression spring with a free length of 5” and it is supposed to travel or compress 2” under a load of 20 pounds. To calculate the compression spring rate of this spring, divide the load of 20 pounds by the distance traveled of 2”. The compression spring rate calculated will be of 10 pounds of force per inch.
k = L ÷ T
k = 20 ÷ 2
k = 10 lbf/in | 732 | 2,992 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-30 | latest | en | 0.842569 |
https://doodlelearning.com/post/how-to-improve-at-mental-maths-tips-and-tricks | 1,679,568,600,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945144.17/warc/CC-MAIN-20230323100829-20230323130829-00751.warc.gz | 264,809,518 | 31,286 | # How to improve at mental maths: our tips and tricks
Mental maths is an excellent skill for children to have, but how, as parents and teachers, do we help them to develop this skill? Luckily, there are many mental maths practice strategies and aids to help with this. Take a look below to find out how!
## What is mental maths?
Mental maths is the ability to calculate maths in your head without working it out on paper or needing memory aid prompts, such as a times table square. It requires an understanding of basic number structures in order to complete calculations quickly.
Having a good knowledge of mental maths has been found to lead to increased accuracy and speed of thought. Most importantly, it can help children to feel positive about their own skills, helping to boost their confidence and independence.
The number one question often asked by students is ‘how to get fast at mental maths’, and there are lots of techniques that can help to boost their skills.
Improvement requires lots of mental maths practice and a knowledge of some helpful mental maths strategies, so we’ve outlined some below to help get you started. Let’s take a closer look!
## How to improve at mental maths
There are many ways to aid your child or pupils’ mental maths practice, and much of this is to do with increasing confidence and using repetition.
### Use flashcards
One of the most important aspects of mental maths practice is repetition, repetition, repetition! This is how many of us as adults remembered times tables or basic numeracy facts that we still use in our own mental maths to this day. Consolidation of these times tables takes time, and flashcards are a great way to help.
Making flashcards with sums on them (addition, subtraction, multiplication or division, depending on your child’s numeracy level) is a great way to practice this skill. And best of all, flashcards can be easily transported and used on the go, making them ideal for use on trips!
You can add an extra element of challenge to this by adding in a timer for a time trial challenge or having a competition between two children if they’re at a similar level.
### Number bond practice game
Number bonds are the basis of how children work out most mental maths problems, so encouraging young learners to practice them is a good way to increase their skills in mental maths.
Ask them to think of all the addition combinations they can that make 10, such as ‘8 + 2’ and ‘6 + 4’. Doing this will help them to understand number bonds better. Then, once they feel comfortable with this, encourage them to move on to reaching 20, 50 or 100.
### Developing your child’s logical thinking
Developing logical thinking is vital for children as they get older, and one way to do this (while getting in their mental maths practice!) is by using real-world situations to get them to tackle problems.
For example, when baking with them, ask them to double the weight of each ingredient by two. Or, when you’re grocery shopping, ask them to divide sweets into three equal groups or to round up the prices when completing the shopping.
Most tasks can have mental maths tasks attached to them without any preparation, so feel free to let your creativity shine!
### Use DoodleMaths
DoodleMaths is an app that’s specifically designed to help children to develop their mental maths skills.
Designed to be used for just 10 minutes a day, its interactive exercises and games keep learning fun and engaging, ensuring that children always look forward to maths practice!
Or discover Doodle for schools
## Mental maths strategies
As well as activities to aid your mental maths practice, there are also numeracy strategies you can use to help your child or pupils improve both their mental maths skills and their confidence when completing mental maths.
### Splitting into hundreds, tens and units
If your child is confronted with a large addition or subtraction sum, a great strategy to teach them is to break it into smaller parts.
It’s best to break the hundreds, tens, and units up so that they can be added together separately and then totalled at the end.
For example, take a look at the following sum:
541 + 235.
By breaking this down, the numbers can be more easily added together:
500 + 200 = 700, 40 + 30 = 70 and 1 + 5 = 6
These can then be put back together for a final number – 786!
Once your child has gotten the hang of this, they may find it easier to keep the tens and units together and to separate the hundreds, as this makes it easier when the unit adds to over ten. For example, 418 + 513 (400 + 500 = 900, 18 + 13 = 31, meaning that the overall total is 931).
### Rounding numbers for easier calculations
Mental maths practice is all about finding patterns or using strategies to break down and tackle trickier problems. One simple way to make difficult addition or subtraction sums easier is to round up or down to a number that your child is more comfortable working with.
Children are more confident when adding or subtracting numbers that end in five or zero, so adjusting calculations to suit this and readjusting them at the end can help tackle more complex problems.
For example, think about the following sum:
492 +180.
By rounding the 492 up to 500, the calculation becomes much easier than before. So, to round up to 500, your child will need to add on 8. Be sure to remember this later!
500 + 180 is much easier to work out, and equals 680
And now comes the final calculation: remember to take away the 8 you added on earlier! This means that the total should be 672.
### Flipping around the question
Most children find addition easier than subtraction, but many are not aware that they can flip difficult subtraction sums to help answer them.
For example, imagine your child or pupil is working on the following:
21 – 18
This can easily be flipped round to __ add 18 = 21.
If your child doesn’t struggle with subtraction sums, this method is also a great way to double-check quickly if the answer they have worked out is correct.
## The takeaway
Mental maths practise takes both time and patience, but there are lots of tried and tested ways that you can use to support your children. Remember: repetition is key with all of these strategies, as this will help embed the required knowledge.
So, what are you waiting for? Get counting today!
Or discover Doodle for schools
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Are you a parent or teacher?
## Hi there!
Book a chat with our team | 1,385 | 6,540 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-14 | latest | en | 0.952808 |
https://numberworld.info/4301748 | 1,590,366,266,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347387155.10/warc/CC-MAIN-20200525001747-20200525031747-00471.warc.gz | 476,843,468 | 4,659 | # Number 4301748
### Properties of number 4301748
Cross Sum:
Factorization:
2 * 2 * 3 * 3 * 3 * 3 * 11 * 17 * 71
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
41a3b4
Base 32:
438tk
sin(4301748)
0.26001845416733
cos(4301748)
-0.96560364720336
tan(4301748)
-0.26928072912774
ln(4301748)
15.274532009688
lg(4301748)
6.6336449654622
sqrt(4301748)
2074.0655727339
Square(4301748)
### Number Look Up
Look Up
4301748 (four million three hundred one thousand seven hundred forty-eight) is a very impressive figure. The cross sum of 4301748 is 27. If you factorisate 4301748 you will get these result 2 * 2 * 3 * 3 * 3 * 3 * 11 * 17 * 71. The number 4301748 has 120 divisors ( 1, 2, 3, 4, 6, 9, 11, 12, 17, 18, 22, 27, 33, 34, 36, 44, 51, 54, 66, 68, 71, 81, 99, 102, 108, 132, 142, 153, 162, 187, 198, 204, 213, 284, 297, 306, 324, 374, 396, 426, 459, 561, 594, 612, 639, 748, 781, 852, 891, 918, 1122, 1188, 1207, 1278, 1377, 1562, 1683, 1782, 1836, 1917, 2244, 2343, 2414, 2556, 2754, 3124, 3366, 3564, 3621, 3834, 4686, 4828, 5049, 5508, 5751, 6732, 7029, 7242, 7668, 9372, 10098, 10863, 11502, 13277, 14058, 14484, 15147, 20196, 21087, 21726, 23004, 26554, 28116, 30294, 32589, 39831, 42174, 43452, 53108, 60588, 63261, 65178, 79662, 84348, 97767, 119493, 126522, 130356, 159324, 195534, 238986, 253044, 358479, 391068, 477972, 716958, 1075437, 1433916, 2150874, 4301748 ) whith a sum of 13172544. The number 4301748 is not a prime number. The figure 4301748 is not a fibonacci number. The figure 4301748 is not a Bell Number. The figure 4301748 is not a Catalan Number. The convertion of 4301748 to base 2 (Binary) is 10000011010001110110100. The convertion of 4301748 to base 3 (Ternary) is 22002112220000. The convertion of 4301748 to base 4 (Quaternary) is 100122032310. The convertion of 4301748 to base 5 (Quintal) is 2100123443. The convertion of 4301748 to base 8 (Octal) is 20321664. The convertion of 4301748 to base 16 (Hexadecimal) is 41a3b4. The convertion of 4301748 to base 32 is 438tk. The sine of the number 4301748 is 0.26001845416733. The cosine of the figure 4301748 is -0.96560364720336. The tangent of 4301748 is -0.26928072912774. The square root of 4301748 is 2074.0655727339.
If you square 4301748 you will get the following result 18505035855504. The natural logarithm of 4301748 is 15.274532009688 and the decimal logarithm is 6.6336449654622. that 4301748 is impressive figure! | 1,106 | 2,559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.595626 |
https://www.archivemore.com/why-do-solids-liquids-and-gases-have-different-properties-related-to-shape-and-volume/ | 1,696,101,269,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510707.90/warc/CC-MAIN-20230930181852-20230930211852-00774.warc.gz | 688,039,034 | 10,242 | Solids, liquids and gases are different mainly because of their lattice arrangements and the cohesive forces between the molecules. The cohesive forces between their molecules are also very weak. This gives gases their property to flow and compressibility.
## What is the difference of solid and liquid from gas?
gas are well separated with no regular arrangement. liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.
## Which has more volume solid liquid or gas?
Matter is made of small particles of atoms or molecules. There are three common states of matter, solid, liquid and gas. A gas and a liquid will change shape to fit the shape of their container. A gas will change volume to fit the volume of the container….Liquids Solids and Gases:
Definite shape definite volume
liquid no yes
gas no no
## What has more matter gas or liquid?
If you have the same number of particles, then the gas will have the greater volume. The particles of matter in the liquid state are still close together but they are far enough apart to move freely. The particles of matter in the gaseous state are neither close together nor fixed in place.
## Does a gas have a definite volume?
Like liquids, gases have no definite shape, but unlike solids and liquids, gases have no definite volume either. The change from solid to liquid usually does not significantly change the volume of a substance.
## Does a gas have a definite shape and volume?
Three states of matter exist – solid, liquid, and gas. Liquids have a definite volume, but take the shape of the container. Gases have no definite shape or volume.
## What does it mean to have a definite volume?
Definite volume refers to something which has a consistent, unchanging volume, which in practical terms may mean a static mass and density, or a predetermined shape, either of which will dictate volume.
## Why gas has indefinite shape and volume?
Solution : Gases do not have a definite shape or volume because the molecules in gases are very loosely packed, they have large intermolecular spaces and hence they move around. The force of attraction between molecules is also very less, as a result gases acquire any shape or any volume.
## Does Liquids have a fixed volume?
Liquids have a fixed volume but no fixed shape. Gases have no fixed volume and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.
## Why do liquids have definite volume?
In a liquid, the particles are still in close contact, so liquids have a definite volume. However, because the particles can move about each other rather freely, a liquid has no definite shape and takes a shape dictated by its container.
## What are the properties of solids liquids and gases?
There are three common states of matter:
• Solids – relatively rigid, definite volume and shape. In a solid, the atoms and molecules are attached to each other.
• Liquids – definite volume but able to change shape by flowing. In a liquid, the atoms and molecules are loosely bonded.
• Gases – no definite volume or shape.
## What are 3 properties of liquids?
All liquids show the following characteristics:
• Liquids are almost incompressible. In liquids molecules are pretty close to each other.
• Liquids have fixed volume but no fixed shape.
• Liquids flow from higher to lower level.
• Liquids have their boiling points above room temperature, under normal conditions.
## What are 3 characteristics of liquids?
Liquids have the following characteristics:
• No definite shape (takes the shape of its container).
• Has definite volume.
• Particles are free to move over each other, but are still attracted to each other.
## What are the 3 properties of gases?
Gases have three characteristic properties: (1) they are easy to compress, (2) they expand to fill their containers, and (3) they occupy far more space than the liquids or solids from which they form.
## What are the 10 gases?
Some examples of gases are listed below.
• Hydrogen.
• Nitrogen.
• Oxygen.
• Carbon Dioxide.
• Carbon Monoxide.
• Water Vapour.
• Helium.
• Neon.
## What are the 7 gases?
There are seven diatomic elements: hydrogen, nitrogen, oxygen, fluorine, chlorine, iodine, bromine.
## What are the 11 gases?
The gaseous element group; hydrogen (H), nitogen (N), oxygen (O), fluorine (F), chlorine (Cl) and noble gases helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn) are gases at standard temperature and pressure (STP). | 984 | 4,555 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.936003 |
http://creuse-news.eu/poker/roulette-best-color-to-bet-on.php | 1,537,518,452,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156901.47/warc/CC-MAIN-20180921072647-20180921093047-00312.warc.gz | 57,090,987 | 12,029 | 4692
Roulette best color to bet on
Try your luck with the roulette strategy of betting on both colors. Even though roulette is a game of chance, it's still a good idea to approach the game with some. What’s the best red & black roulette system? First let’s look at the odds. The roulette odds for colors are 1 to 1 (excluding zero). Can you win at roulette with. Understanding Roulette Payouts. if the ball lands on the color you chose. Odd or Even – This bet pays out even So the best way to approach roulette is as a.
Betting on colors
The "unit" is whatever bet you started off with. And because red and black alternate each pocket, perhaps the only universal winning red and black system involves the use of precognition. It covers two adjoining rows of numbers. For example, place one bet on black and one bet on Column Three, which has eight red numbers. Again, you cover 26 numbers, and 4 of them have two ways to win. You will find these bets on the part of the table closest to the player.
Interesting Fact
• There are three types of game: European; American (another order of numbers on the track, the presence of the number "00"); French (one zero, the rule of En Prison).
• In 2012, at the roulette in Rio Casino (Las Vegas), in the eyes of numerous witnesses, number 19 fell seven times in a row. The probability of such a sequence of identical numbers is 1: 3,000,000,000.
• One Englishman named Ashley Revell once sold all his possessions in order to go to Las Vegas and play roulette. This evening, Ashley won 135 thousand 300 dollars, putting everything on red.
• Once Albert Einstein was asked if there is any roulette game system that guarantees a win. He replied: "Yes, I know one thing - it's to steal chips when the dealer does not see it".
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The center of the casino is Las Vegas. Every year 40 million gambling people from all over the world come here.
The Best Red Black Bet Roulette System
Payouts Understanding Roulette Payouts Roulette offers a bewildering number of betting options, but the bets are actually straightforward enough. Roulette payouts work like this.
For example, the single number bet offers a payout of 35 to 1. Each of these bets refers to a specific set of numbers or colors. The outside bets include: Red or Black — This bet pays out even odds 1 to 1 if the ball lands on the color you chose.
Odd or Even — This bet pays out even odds 1 to 1 if the ball lands on odd or even, depending on which you chose.
Low or High — This bet pays out even money 1 to 1 if the ball lands on if you bet low, or if the ball lands on if you bet high. Columns — The numbers on the layout are organized into three columns of twelve numbers each. This bet pays out 2 to 1 when you win. Dozens — There are 36 numbers on the table, so you can bet on the first dozen , the second dozen , or the third dozen This bet also pays out 2 to 1. Payouts on the Inside Bets You can also bet on specific numbers and sets of numbers on the inside of the layout.
Despite it's small size, it had a number of choirboys, and choirgirls. I felt this hand grab my cock and I saw her dive her hands into my underwear and pull them down. His name was Jake, and he had big muscles, and a physique of a fully developed man. If Kahn's model is correct, redirecting the 540 million now wasted on spreading the myth of heterosexual AIDS to high-risk groups - mostly gays and inner-city drug users - could wipe out new infections entirely.
She lapped it up like a cat with cream and then began passionately kissing Jeff again.
Details
By Kevin Blackwood, Max Rubin Remember to stay realistic about the long odds on this popular game of chance. Approach roulette with the sober realization that, with a house advantage of 5. Despite the odds, you can still use some simple strategies to stretch your roulette bankroll and enjoy the thrill of the spin.
This article contains a few tips that can help you improve your chances of winning. With 38 numbers 1 to 36, plus 0 and 00 , the true odds of hitting a single number on a straight-up bet are 37 to 1, but the house pays only 35 to 1 if you win! Ditto the payouts on the combination bets. This discrepancy is where the house gets its huge edge in roulette. Starting with the basics Strategy is critical if you want to increase your odds of winning.
Consequently, few players make just one bet at a time. Of course, the more bets you make, the more complicated and challenging it is to follow all the action. Here are two possible plans of attack to simplify matters: Stick to the table minimum and play only the outside bets.
For example, bet on either red or black for each spin. This type of outside bet pays 1 to 1 and covers 18 of the 38 possible combinations. Place two bets of equal amounts on two outside bets: For example, place one bet on black and one bet on Column Three, which has eight red numbers.
That way, you have 26 numbers to hit, 4 of which you cover twice. You can also make a bet on red and pair it with a bet on Column Two, which has eight black numbers.
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Roulette4fun also offers information for the newbie player that wants to gain as much insight as they can into how to play the game and the etiquette involved. As a new player you should probably play our free roulette game for few days and practise your skills before betting any real dollar in the casinos. You can click the above screenshots to enter the online roulette game. Also, make sure to check out our roulette casino reviews to get a better insight on which casino suits you best.
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The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,391 | 6,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-39 | latest | en | 0.916389 |
https://number.academy/90753 | 1,660,823,572,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00518.warc.gz | 382,193,205 | 12,085 | Number 90753
Number 90,753 spell 🔊, write in words: ninety thousand, seven hundred and fifty-three . Ordinal number 90753th is said 🔊 and write: ninety thousand, seven hundred and fifty-third. The meaning of number 90753 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 90753. What is 90753 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 90753.
What is 90,753 in other units
The decimal (Arabic) number 90753 converted to a Roman number is (XC)DCCLIII. Roman and decimal number conversions.
Weight conversion
90753 kilograms (kg) = 200074.1 pounds (lbs)
90753 pounds (lbs) = 41165.3 kilograms (kg)
Length conversion
90753 kilometers (km) equals to 56392 miles (mi).
90753 miles (mi) equals to 146053 kilometers (km).
90753 meters (m) equals to 297743 feet (ft).
90753 feet (ft) equals 27662 meters (m).
90753 centimeters (cm) equals to 35729.5 inches (in).
90753 inches (in) equals to 230512.6 centimeters (cm).
Temperature conversion
90753° Fahrenheit (°F) equals to 50400.6° Celsius (°C)
90753° Celsius (°C) equals to 163387.4° Fahrenheit (°F)
Time conversion
(hours, minutes, seconds, days, weeks)
90753 seconds equals to 1 day, 1 hour, 12 minutes, 33 seconds
90753 minutes equals to 2 months, 1 week, 33 minutes
Codes and images of the number 90753
Number 90753 morse code: ----. ----- --... ..... ...--
Sign language for number 90753:
Number 90753 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
Mathematics of no. 90753
Multiplications
Multiplication table of 90753
90753 multiplied by two equals 181506 (90753 x 2 = 181506).
90753 multiplied by three equals 272259 (90753 x 3 = 272259).
90753 multiplied by four equals 363012 (90753 x 4 = 363012).
90753 multiplied by five equals 453765 (90753 x 5 = 453765).
90753 multiplied by six equals 544518 (90753 x 6 = 544518).
90753 multiplied by seven equals 635271 (90753 x 7 = 635271).
90753 multiplied by eight equals 726024 (90753 x 8 = 726024).
90753 multiplied by nine equals 816777 (90753 x 9 = 816777).
show multiplications by 6, 7, 8, 9 ...
Fractions: decimal fraction and common fraction
Fraction table of 90753
Half of 90753 is 45376,5 (90753 / 2 = 45376,5 = 45376 1/2).
One third of 90753 is 30251 (90753 / 3 = 30251).
One quarter of 90753 is 22688,25 (90753 / 4 = 22688,25 = 22688 1/4).
One fifth of 90753 is 18150,6 (90753 / 5 = 18150,6 = 18150 3/5).
One sixth of 90753 is 15125,5 (90753 / 6 = 15125,5 = 15125 1/2).
One seventh of 90753 is 12964,7143 (90753 / 7 = 12964,7143 = 12964 5/7).
One eighth of 90753 is 11344,125 (90753 / 8 = 11344,125 = 11344 1/8).
One ninth of 90753 is 10083,6667 (90753 / 9 = 10083,6667 = 10083 2/3).
show fractions by 6, 7, 8, 9 ...
Calculator
90753
Is Prime?
The number 90753 is not a prime number. The closest prime numbers are 90749, 90787.
Factorization and factors (dividers)
The prime factors of 90753 are 3 * 13 * 13 * 179
The factors of 90753 are 1 , 3 , 13 , 39 , 169 , 179 , 507 , 537 , 2327 , 6981 , 30251 , 90753
Total factors 12.
Sum of factors 131760 (41007).
Powers
The second power of 907532 is 8.236.107.009.
The third power of 907533 is 747.451.419.387.777.
Roots
The square root √90753 is 301,252386.
The cube root of 390753 is 44,938682.
Logarithms
The natural logarithm of No. ln 90753 = loge 90753 = 11,415897.
The logarithm to base 10 of No. log10 90753 = 4,957861.
The Napierian logarithm of No. log1/e 90753 = -11,415897.
Trigonometric functions
The cosine of 90753 is 0,239858.
The sine of 90753 is -0,970808.
The tangent of 90753 is -4,047431.
Properties of the number 90753
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
Number 90753 in Computer Science
Code typeCode value
90753 Number of bytes88.6KB
Unix timeUnix time 90753 is equal to Friday Jan. 2, 1970, 1:12:33 a.m. GMT
IPv4, IPv6Number 90753 internet address in dotted format v4 0.1.98.129, v6 ::1:6281
90753 Decimal = 10110001010000001 Binary
90753 Decimal = 11121111020 Ternary
90753 Decimal = 261201 Octal
90753 Decimal = 16281 Hexadecimal (0x16281 hex)
90753 BASE64OTA3NTM=
90753 MD574e1445223e6e2fce6511509b01307bc
90753 SHA17acebab9d3a774da3cbb50bcc74ac337c6d71860
90753 SHA25653d5e6c9e02a9a3b360cf87d08464d155ca5d17d6e8e3476f2fbeccd433d163b
More SHA codes related to the number 90753 ...
If you know something interesting about the 90753 number that you did not find on this page, do not hesitate to write us here.
Numerology 90753
Character frequency in number 90753
Character (importance) frequency for numerology.
Character: Frequency: 9 1 0 1 7 1 5 1 3 1
Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 90753, the numbers 9+0+7+5+3 = 2+4 = 6 are added and the meaning of the number 6 is sought.
Interesting facts about the number 90753
Asteroids
• (90753) 1993 RX4 is asteroid number 90753. It was discovered by E. W. Elst from La Silla Observatory on 9/15/1993.
№ 90,753 in other languages
How to say or write the number ninety thousand, seven hundred and fifty-three in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 90.753) noventa mil setecientos cincuenta y tres German: 🔊 (Anzahl 90.753) neunzigtausendsiebenhundertdreiundfünfzig French: 🔊 (nombre 90 753) quatre-vingt-dix mille sept cent cinquante-trois Portuguese: 🔊 (número 90 753) noventa mil, setecentos e cinquenta e três Chinese: 🔊 (数 90 753) 九万零七百五十三 Arabian: 🔊 (عدد 90,753) تسعون ألفاً و سبعمائة و ثلاثة و خمسون Czech: 🔊 (číslo 90 753) devadesát tisíc sedmset padesát tři Korean: 🔊 (번호 90,753) 구만 칠백오십삼 Danish: 🔊 (nummer 90 753) halvfemstusinde og syvhundrede og treoghalvtreds Dutch: 🔊 (nummer 90 753) negentigduizendzevenhonderddrieënvijftig Japanese: 🔊 (数 90,753) 九万七百五十三 Indonesian: 🔊 (jumlah 90.753) sembilan puluh ribu tujuh ratus lima puluh tiga Italian: 🔊 (numero 90 753) novantamilasettecentocinquantatré Norwegian: 🔊 (nummer 90 753) nitti tusen, syv hundre og femti-tre Polish: 🔊 (liczba 90 753) dziewięćdzisiąt tysięcy siedemset pięćdziesiąt trzy Russian: 🔊 (номер 90 753) девяносто тысяч семьсот пятьдесят три Turkish: 🔊 (numara 90,753) doksanbinyediyüzelliüç Thai: 🔊 (จำนวน 90 753) เก้าหมื่นเจ็ดร้อยห้าสิบสาม Ukrainian: 🔊 (номер 90 753) дев'яносто тисяч сiмсот п'ятдесят три Vietnamese: 🔊 (con số 90.753) chín mươi nghìn bảy trăm năm mươi ba Other languages ...
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If you know something interesting about the number 90753 or any natural number (positive integer) please write us here or on facebook. | 2,359 | 6,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-33 | latest | en | 0.651347 |
http://www.weegy.com/?ConversationId=4B818798 | 1,461,938,754,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860111365.36/warc/CC-MAIN-20160428161511-00000-ip-10-239-7-51.ec2.internal.warc.gz | 898,686,832 | 9,071 | Google is an example of a(n): Web site.search engine.search directory. subject directory
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Question
Updated 137 days ago|12/13/2015 9:41:49 PM
This conversation has been confirmed as correct, not copied, and helpful.
Edited by yumdrea [12/13/2015 9:41:47 PM]
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Which statement represents the inverse of the statement, "If it is snowing, then Paul wears a sweater"?
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User: Jim has grades of 84, 65, and 76 on three math tests. What grade must he obtain on the next test to have an average of exactly 80 for the four tests? A.87 B. 95 C. 85 D.92
Weegy: c (More)
Question
Updated 3/12/2012 2:01:47 PM
He must obtain 95 on his next test for an 80% average.
(84+65+76+x)/4 = 80
(225+x)/4 = 80
225+x = 320
x = 320-225
x = 95
Information with subject directories is categorized according to?
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Updated 203 days ago|10/8/2015 2:15:39 PM
Information with subject directories is categorized according to subjects.
Added 203 days ago|10/8/2015 2:15:39 PM
Confirmed by selymi [12/15/2015 4:15:44 AM]
When citing an online resource, it is important to include this information, as well as the information provided in a standard citation. Author’s name Title of publication Date downloaded or retrieved Name of publisher
Weegy: True. When citing an online resource, it is important to include this information, as well as the information provided in a standard citation. Author’s name Title of publication Date downloaded or retrieved Name of publisher. (More)
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,027 | 3,334 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-18 | longest | en | 0.915479 |
https://brainly.com/question/264745 | 1,484,566,028,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279169.4/warc/CC-MAIN-20170116095119-00402-ip-10-171-10-70.ec2.internal.warc.gz | 809,481,382 | 8,323 | # Ky has 3 times more books than grant grant has 6 fewer books than Jaime if the total combined number of books is 176 how many books does Jaime have?
1
by ilonalaraque
2015-01-21T12:11:03-05:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Set up the equation
x+(x-6)+3(x-6)=176
x+x-6+3x-18=176
5x-24=176
5x=176+24
5x=200
x=40
Jamie has 40 books
Grant has 34
Ky has 102 | 187 | 644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-04 | latest | en | 0.947451 |
http://www.expertsmind.com/library/significance-of-natural-and-assignable-causes-of-variation-5693322.aspx | 1,580,149,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251705142.94/warc/CC-MAIN-20200127174507-20200127204507-00350.warc.gz | 215,894,540 | 11,611 | ### Significance of natural and assignable causes of variation
Assignment Help Other Engineering
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Evaluate the two basic types of inspection used in sampling for process control.
There are two principle inspection methods used in the sampling process Identify and then evaluate these methods
Describe the significance of natural and assignable causes of variation. During a quality inspection of a product you can have either natural or assignable causes of variation.
Identify and describe the significance of these variations.
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##### Reference no: EM13693322
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Create a new instance of that type of animal if they are of different genders. Otherwise, if two animals of the same type and gender try to collide, then only the one of lar | 447 | 2,332 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-05 | longest | en | 0.867306 |
http://www.chegg.com/homework-help/questions-and-answers/hello-i-m-studying-signals-systems-summerand-i-m-trying-good-grasp-convolution-ithink-i-un-q81199 | 1,386,417,690,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163054353/warc/CC-MAIN-20131204131734-00040-ip-10-33-133-15.ec2.internal.warc.gz | 279,694,887 | 10,354 | 0 pts ended
Hello. I'm studying signals and systems on my own this summerand I'm trying to get a good grasp of the convolution. Ithink I understand it mathematically enough to do some problems,but I don't have a firm grasp by any means. I'm studying bothdiscrete and continuous time cases. Before I get to myquestions, here is how I understand things in my own words (if anyof this is off, please let me know):
Sifting Property
The value of a signal at a time t can be found by summing theproduct of the impulse response and the signal at all times. Since the impulse response will be 1 only at time t and 0everywhere else, you will "sift" out only that value of the signalat time t. (1 times the signal at time t will just be thesignal)
Impulse Response
Written as in my texts (Oppenheim and Schaum's), theimpulse response is just the output of a system at a time when the input is the unit impulse.
In other words, I just think of the impulse response as what youget out of a system if you send in a 1 at a certain time.
This is where I may be confused. I'll ask a question aboutthis below.
And finally:
The Convolution
If you know the impulse response of a system at a time t, you justhave to scale it by the value of to get the response of the system to. For the system response:
I sort of think of this as multiplying a unit area (ie, 1m2) by a scalar, C, to get an area of (C m2).(the unit impulse response is analogous to the unit area, and thescalar is analogous to the input ).
Actually... I think I may be more confused about theconvolution than I realize. If you have any good tips on howto think about it, please let me know.
Ok. Now my questions:
Question I: SiftingProperty
What good is the sifting property??? It seems to be circularin its logic! I mean, you are basically saying you can get if you know !! You're just going through theextra step of multiplying all values of by to "catch" the ... But that means you already had in the first place! So what theheck is the point?!?
Question II: The ImpulseResponse
Is a 1, or infinity at time ?? When it's under the integral, Iknow it is 1, since it has unit are. But when the impulseresponse is described, it seems to be the response to the impulse,with no integral involved. Here is how my Schaum's definesit:
where T is the LTI system.
And if it is infinity, how can a system respond to an infiniteinput? This, I think, is my biggest point of confusion, andmay be why I'm having trouble understanding the convolutionfully.
Question III: TheConvolution
Why would you have the response of a system to the unit impulse,but not have its response to the signal ? If you could get the impulseresponse, why not just get the response and forget about the convolutionaltogether?
Conclusion
Well, I think that sums up my confusion for now. :) I hope myquestions made sense! I will be thrilled if someone is niceenough to clear some of these issues up for me!
Thanks! | 706 | 2,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2013-48 | latest | en | 0.947787 |
https://math.stackexchange.com/questions/3064730/a-nilpotent-and-ac-ib-is-nilpotent-then-b-is-nilpotent | 1,656,577,941,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00198.warc.gz | 439,546,686 | 65,615 | # $A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.
Let $$A$$ and $$B$$ be $$n \times n$$ matrices over some field with $$A$$ nilpotent. Now let $$c_1,\ldots,c_{n+1}$$ be $$n+1$$ distinct scalars such that $$A+c_i B$$ is nilpotent for all $$i=1, \ldots,n+1$$. Then how can I show that $$B$$ is also nilpotent?
Thanks
Let us define the matrix valued polynomial $$p(t) = (A+tB)^n = \sum_{k=0}^n C_k t^k.$$ We see that each $$(i,j)$$-th component $$p(t)_{ij}$$ of $$p(t)$$ is a polynomial in $$t$$ of degree at most $$n$$. By the assumption, $$p(t)_{ij}=0$$ has $$n+1$$ distinct roots $$c_1,c_2,\ldots, c_{n+1}$$ on the underlying field $$\mathbb{F}$$. (This is why: if $$n\times n$$ matrix $$T$$ is nilpotent, the it holds $$T^k=O$$ for some $$k\le n$$ and hence $$T^n=O$$. Thus, by the assumption, $$p(c_k) = O$$ for all $$k=1,2,\ldots, n+1$$.) This means $$p(t)_{ij} \equiv 0$$ and it follows that $$p(t) \equiv O$$. Since the coefficient $$C_n$$ of $$t^n$$ is $$B^n$$, it follows that $$B^n=O$$.
• Can you be more explicit. I don't see why $p(t)_{ij}=0$ has $n+1$ distinct roots. PS I saw that your answer was upvoted and then downvoted. I did not downvote. I am just seeking a clarification. Jan 7, 2019 at 8:07
• Do we even need the fact that $A$ is nilpotent tho? Jan 7, 2019 at 8:12
• @EuxhenH Maybe we can assume that one of $c_k$ is 0. Jan 7, 2019 at 8:13
• @Song That would mean that $A$ is nilpotent as a consequence, but we still don't need it in the problem statement. Jan 7, 2019 at 8:16 | 563 | 1,515 | {"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": 32, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-27 | latest | en | 0.891347 |
https://www.tutorialspoint.com/p-why-there-is-the-need-for-international-standard-units-realized-explain-with-suitable-examples-p | 1,709,264,744,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00530.warc.gz | 1,018,869,798 | 24,210 | # Why there is the need for International Standard Units realized? Explain with suitable examples.
A long time ago, people from different countries had different ways of measuring things, which made it harder for them to trade or business. People need to measure what they buy and exchange for them to prosper, but not having a common and correct way of measuring was a problem.
With the development in the field of science and technology, the need for a commonly accepted system of units is realized all over the world particularly to exchange scientific and technical information. Finally, scientists came up with "The International System of Units (SI, abbreviated from the French Système International d’Unités) is the modern form of the metric system that replaces all other existing ways we measure objects. It is the only system of measurement with official status in nearly every country in the world.
SI unit enables us to calculate easily by being a constant term of measurement throughout the world.
It comprises a coherent or consistent system of units measurement starting with seven base units, which are-
1. Second (the unit of time with the symbol s),
2. Metre (the unit of length, with the symbol m),
3. Kilogram (the unit of mass, with the symbol kg),
4. Ampere (the unit of electric current, with the symbol A),
5. Kelvin (the unit of thermodynamic temperature, with the symbol K),
6. Mole (the unit of amount of substance, with the symbol mol),
7. Candela (the unit of luminous intensity, with the symbol cd).
Even if we use other forms of measurement, we can easily convert these into SI units by following the conversion table.
The SI unit is necessary to ensure that our everyday measurements remain comparable and consistent worldwide because without the SI unit there would be confusion as to which unit (of measurement) is to be taken to measure a certain quantity.
For example- If we take cubit or handspan to measure something than it will vary person to person, while SI unit remains constant everywhere.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
39 Views | 432 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-10 | latest | en | 0.943882 |
http://www.ehow.co.uk/how_8253517_calculate-walltowindow-ratio.html | 1,526,839,483,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863662.15/warc/CC-MAIN-20180520170536-20180520190536-00130.warc.gz | 390,896,099 | 10,312 | DISCOVER
# How to calculate a wall-to-window ratio
Updated April 17, 2017
Many state and local codes consider a building's wall-to-window ratio in determining its energy efficiency. Building codes vary between jurisdictions. You may have to calculate a building's wall-to-window ratio to ascertain what additional insulation is required to meet your local building code. You can determine this ratio by taking some basic measurements yourself, without the assistance of a professional.
Measure the length and width of each exterior wall of the building. Multiply the length times width to get the total area of each wall. Keep a record of these numbers.
Measure the length and width for each rough window opening on the wall. Multiply the length times width of each window opening. Keep a record of these numbers.
Add the total areas of every exterior wall together. This number is the total of the exterior wall space of the building.
Add up all the window openings in the building. This is the total window space in the building.
Divide the number of the total exterior wall area by the total area of all the window openings in the building. This is the wall-to-window ratio of the building.
#### Tip
Check with your local building department to find out any additional factors involved in taking building measurements.
#### Things You'll Need
• Measuring tape
• Calculator | 270 | 1,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-22 | latest | en | 0.932498 |
https://hotc.camdar.io/singleton/declarative.html | 1,603,397,788,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880038.27/warc/CC-MAIN-20201022195658-20201022225658-00028.warc.gz | 377,038,860 | 8,293 | # Singleton kinds, declaratively
The singleton kind calculus relevant to us is syntactically an extension of $$F_\omega$$ (plus products), with identical constructors and terms. The kinds, of course, are extended with the $$S(c)$$ singleton kind. But to fully represent modules, we need more power.
Consider the signature
sig
type t
type u = t * int
end
Intuitively, we might write this kind as $$* \times S(t \times \texttt{int})$$, but this isn't well-formed (the type variable $$t$$ is not in scope). This necessitates the use of dependent kinds, $$\Sigma$$- and $$\Pi$$-kinds. These subsume normal arrow and (non-dependent) product kinds, giving us
$\require{bussproofs}k := * \mid S(c) \mid \Sigma(\alpha:k).k \mid \Pi(\alpha:k).k$
Of course, we will still use $$k \rightarrow k'$$ and $$k \times k'$$ as shorthand for $$\Pi(\alpha:k).k'$$ and $$\Sigma(\alpha:k).k'$$ respectively if $$k$$ is not free in $$k'$$.
The judgments we will use are
JudgmentMeaning
$$\Gamma \vdash k:\text{kind}$$Kind validity
$$\Gamma \vdash k \equiv k' : \text{kind}$$Kind equivalence
$$\Gamma \vdash k \le k'$$Subkinding
$$\Gamma \vdash c: k$$Kinding
$$\Gamma \vdash c \equiv c' : k$$Constructor equivalence
$$\Gamma \vdash e : \tau$$Typing (same as $$F_\omega$$)
$$\phantom{\Gamma} \vdash \Gamma\ ok$$Context well-formedness
The addition of singleton kinds necessitates subkinding to handle cases like int, which has kind $$*$$ but also kind $$S(\texttt{int})$$.
Rules 2.1 (Kind validity): $$\Gamma \vdash k:\text{kind}$$
$\begin{prooftree} \AxiomC{} \UnaryInfC{\Gamma \vdash * : \text{kind}} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c:*} \UnaryInfC{\Gamma \vdash S(c): \text{kind}} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash k_1:\text{kind}} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2:\text{kind}} \BinaryInfC{\Gamma \vdash \Pi(\alpha:k_1).k_2:\text{kind}} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash k_1:\text{kind}} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2:\text{kind}} \BinaryInfC{\Gamma \vdash \Sigma(\alpha:k_1).k_2:\text{kind}} \end{prooftree}$
Rules 2.2 (Kind equivalence): $$\Gamma \vdash k \equiv k' : \text{kind}$$
$\begin{prooftree} \AxiomC{} \UnaryInfC{\Gamma \vdash * \equiv * : \text{kind}} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c' : *} \UnaryInfC{\Gamma \vdash S(c) \equiv S(c') :\text{kind}} \end{prooftree}$
$\begin{prooftree} \AxiomC{\Gamma \vdash k_1 \equiv k_1' : \text{kind}} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2 \equiv k_2' : \text{kind}} \BinaryInfC{\Gamma \vdash \Pi(\alpha:k_1).k_2 \equiv \Pi(\alpha:k_1').k_2'} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash k_1 \equiv k_1' : \text{kind}} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2 \equiv k_2' : \text{kind}} \BinaryInfC{\Gamma \vdash \Sigma(\alpha:k_1).k_2 \equiv \Sigma(\alpha:k_1').k_2'} \end{prooftree}$
Before moving onward, we'll need some more machinery. By the above rules, we can only define singleton kinds for types (constructors kinded at $$*$$). As a typechecking mechanism, this is important, lest someone write type t = list. When declaring our whole system, however, this can be inconvenient. Let us define the generalized singleton $$S(c:k)$$ be defined as follows:
\begin{aligned} S(c:*) &\triangleq S(c) \\ S(c:\Pi(\alpha:k_1).k_2) &\triangleq \Pi(\alpha:k_1).S(c\ \alpha : k_2) \\ S(c:\Sigma(\alpha:k_1).k_2) &\triangleq S(\pi_1c : k_1) \times S(\pi_2c:[\pi_1c/\alpha]k_2) \\ S(c:S(c')) &\triangleq S(c) \end{aligned}
Note that the $$\Sigma$$ clause could be more concisely written as $$\Sigma(\alpha:S(\pi_1c:k_1)).S(\pi_2c:k_2)$$, but we choose to expand it out for clarity. Ideally, this definition should fit to the following condition (written as a derived rule):
$\begin{prooftree} \alwaysDashedLine \AxiomC{\Gamma \vdash c:k} \AxiomC{\vdash \Gamma\ ok} \BinaryInfC{\Gamma \vdash S(c:k)\ ok} \end{prooftree}$
which it does.
Why is all this necessary? Previously, when checking constructor equivalence, neither $$\Gamma$$ nor $$k$$ were used. But what about in this system?
$\lambda (\alpha:*) . \alpha \overset{?}{=} \lambda (\alpha:*) . \texttt{int}$
Before, we would say that this is "obviously false". Calling one on a type that is not int demonstrates this. But now, with subkinding, we might say that
$\cdot \vdash \lambda (\alpha:*) . \alpha : S(\texttt{int}) \rightarrow *$
because $$S(\texttt{int}) \le *$$. But then, these two lambdas should be equivalent, because if $$\alpha : S(\texttt{int})$$, then $$\alpha$$ must be (equivalent to) $$\texttt{int}$$ itself! So clearly the kind we are checking at matters.
Similarly, we need the context as well. Consider
$\beta (\lambda(\alpha:*).\alpha) \overset{?}{=} \beta(\lambda(\alpha:*).\texttt{int})$
Now, the kind of $$\beta$$ fixes the kinds of the lambdas (which, as we saw above, matters).
Rules 2.3 (Subkinding): $$\Gamma \vdash k \le k'$$
Preorder: $\begin{prooftree} \AxiomC{\Gamma \vdash k \equiv k':\text{kind}} \UnaryInfC{\Gamma \vdash k \le k'} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash k_1 \le k_2} \AxiomC{\Gamma \vdash k_2 \le k_3} \BinaryInfC{\Gamma \vdash k_1 \le k_3} \end{prooftree}$
Other: $\begin{prooftree} \AxiomC{\Gamma \vdash c:*} \UnaryInfC{\Gamma \vdash S(c) \le *} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c' : *} \RightLabel{*} \UnaryInfC{\Gamma \vdash S(c) \le S(c')} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash k_1' \le k_1} \AxiomC{\Gamma, \alpha:k_1' \vdash k_2 \le k_2'} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2 : \text{kind}} \TrinaryInfC{\Gamma \vdash \Pi(\alpha:k_1).k_2 \le \Pi(\alpha:k_1').k_2'} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash k_1 \le k_1'} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2 \le k_2'} \AxiomC{\Gamma, \alpha:k_1' \vdash k_2' : \text{kind}} \TrinaryInfC{\Gamma\vdash\Sigma(\alpha:k_1).k_2\le\Sigma(\alpha:k_1').k_2'} \end{prooftree}$
The starred rule is actually unnecessary, as it falls out from the symmetry rule.
The rule for $$\Pi$$ types flips the direction of the input kind, as functions are known to be contravariant. When checking the dependent parts of $$\Pi$$ and $$\Sigma$$, it is important that our context holds that $$\alpha$$ is the larger kind, as it will work for both comparands. Finally, we must check that certain subkinds are well-formed, as this won't necessarily fall out from proving other premises.
Rules 2.4.1 (Kinding, incomplete): $$\Gamma \vdash c:k$$
$\begin{prooftree} \AxiomC{\Gamma(\alpha) = k} \UnaryInfC{\Gamma \vdash \alpha : k} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma, \alpha:k \vdash c:k'} \AxiomC{\Gamma \vdash k:\text{kind}} \BinaryInfC{\Gamma \vdash \lambda(\alpha:k).c : \Pi(\alpha:k).k'} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c_1:\Pi(\alpha:k).k'} \AxiomC{\Gamma \vdash c_2:k} \BinaryInfC{\Gamma \vdash c_1\ c_2:[c_2/\alpha]k'} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash c:\Sigma(\alpha:k_1).k_2} \UnaryInfC{\Gamma \vdash \pi_1c : k_1} \end{prooftree}\qquad \begin{prooftree} \AxiomC{\Gamma \vdash c:\Sigma(\alpha:k_1).k_2} \UnaryInfC{\Gamma \vdash \pi_2c : [\pi_1c/\alpha]k_2} \end{prooftree}$
The major missing rule is the one kinding type-level tuples. There is one obvious candidate:
$\begin{prooftree} \AxiomC{\Gamma \vdash c_1:k_1} \AxiomC{\Gamma \vdash c_2:[c_1/\alpha]k_2} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2:\text{kind}} \TrinaryInfC{\Gamma \vdash \langle c_1,c_2 \rangle:\Sigma(\alpha:k_1).k_2} \end{prooftree}$
There is also the rule for non-dependent tuples:
$\begin{prooftree} \AxiomC{\Gamma \vdash c_1:k_1} \AxiomC{\Gamma \vdash c_2:k_2} \BinaryInfC{\Gamma \vdash \langle c_1,c_2 \rangle : k_1 \times k_2} \end{prooftree}$
which clearly falls out from the first rule. It turns out, however, that these two formulations are actually equivalent (under subsumption). Suppose we are given $$\Gamma \vdash c_1:k_1$$, $$\Gamma \vdash c_2:[c_1/alpha]k_2$$ and that the kinds are well-formed. Then
• We have $$\Gamma \vdash c_1:S(c_1:k_1)$$ by how we defined generalized singletons
• $$\Gamma \vdash \langle c_1, c_2 \rangle : S(c_1:k_1) \times [c_1/\alpha]k_2$$ by the non-dependent product rule.
• $$\alpha$$ is not free in $$[c_1/\alpha]k_2$$, so $$\Gamma, \alpha:S(c_1:k_1) \vdash [c_1/\alpha]k_2 \le k_2$$.
• So $$\Gamma \vdash S(c_1:k_1) \times [c_1/\alpha]k_2 \le \Sigma(\alpha:k_1).k_2$$.
• Then by subsumption, $$\Gamma \vdash \langle c_1, c_2 \rangle : \Sigma(\alpha:k_1). k_2$$. It turns out that using the non-dependent product rule is ultimately much simpler, so we will use it over the dependent version.
We complete rules 2.4 (for now) with singletons, arrows and $$\forall$$:
Rules 2.4.2 (Kinding continued, still incomplete):
$\begin{prooftree} \AxiomC{\Gamma \vdash c_1:k_1} \AxiomC{\Gamma \vdash c_2:k_2} \BinaryInfC{\Gamma \vdash \langle c_1,c_2 \rangle : k_1 \times k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c': *} \UnaryInfC{\Gamma \vdash c : S(c')} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash \tau_1:*} \AxiomC{\Gamma \vdash \tau_2:*} \BinaryInfC{\Gamma \vdash \tau_1 \rightarrow \tau_2:*} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash k:\text{kind}} \AxiomC{\Gamma, \alpha:k \vdash \tau:*} \BinaryInfC{\Gamma \vdash \forall(\alpha:k).\tau : *} \end{prooftree}$
The rule for singletons should be unsurprising, and the rules for arrow and $$\forall$$ types are unchanged from system $$F_\omega$$.
At last, then, we come to constructor equivalence. The general theme will be to use dependent versions of the rules we had before.
Rules 2.5 (Constructor equivalence): $$\Gamma \vdash c \equiv c'$$
Equivalence: $\begin{prooftree} \AxiomC{\Gamma \vdash c:k} \UnaryInfC{\Gamma \vdash c \equiv c:k} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c' : k} \UnaryInfC{\Gamma \vdash c' \equiv c : k} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c_1 \equiv c_2:k} \AxiomC{\Gamma \vdash c_2 \equiv c_3:k} \BinaryInfC{\Gamma \vdash c_1 \equiv c_3:k} \end{prooftree}$
Compatibility: $\begin{prooftree} \AxiomC{\Gamma \vdash k_1 \equiv k_1' : \text{kind}} \AxiomC{\Gamma, \alpha:k_1 \vdash c \equiv c':k_2} \BinaryInfC{\Gamma \vdash \lambda(\alpha:k_1).c \equiv \lambda(\alpha:k_1').c': \Pi(\alpha:k_1).k_2} \end{prooftree} \quad \begin{prooftree} \AxiomC{\Gamma \vdash c_1 \equiv c_1' : \Pi(\alpha:k_1).k_2} \AxiomC{\Gamma \vdash c_2 \equiv c_2' : k_2} \BinaryInfC{\Gamma \vdash c_1\ c_2 \equiv c_1'\ c_2' : [c_2/\alpha]k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash c_1 \equiv c_1' : k_1} \AxiomC{\Gamma \vdash c_2 \equiv c_2' : [c/\alpha]k_2} \BinaryInfC{\Gamma \vdash \langle c_1, c_2 \rangle \equiv \langle c_1', c_2' \rangle : \Sigma(\alpha:k_1).k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c' : \Sigma(\alpha:k_1).k_2} \UnaryInfC{\Gamma \vdash \pi_1c \equiv \pi_1c':k_1} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c \equiv c' : \Sigma(\alpha:k_1).k_2} \UnaryInfC{\Gamma \vdash \pi_2c \equiv \pi_2c':[\pi_1c/\alpha]k_2} \end{prooftree}$
Types constructors: $\begin{prooftree} \AxiomC{\Gamma \vdash c_1 \equiv c_1' : *} \AxiomC{\Gamma \vdash c_2 \equiv c_2' : *} \BinaryInfC{\Gamma \vdash c_1\rightarrow c_2 \equiv c_1' \rightarrow c_2':*} \end{prooftree}\qquad \begin{prooftree} \AxiomC{\Gamma \vdash k \equiv k' : \text{kind}} \AxiomC{\Gamma, \alpha:k \vdash c \equiv c' : *} \BinaryInfC{\Gamma \vdash \forall(\alpha:k).c\equiv\forall(\alpha:k').c':*} \end{prooftree}$
Beta-reduction: $\begin{prooftree} \AxiomC{\Gamma \vdash c_1:k_1} \AxiomC{\Gamma, \alpha:k_1 \vdash c_2:k_2} \BinaryInfC{\Gamma \vdash (\lambda(\alpha:k_1).c_2)\ c_1 \equiv [c_1/\alpha]c_2 : [c_1/\alpha]k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash c_1 : k_1} \AxiomC{\Gamma \vdash c_2 : k_2} \BinaryInfC{\Gamma \vdash \pi_1 \langle c_1, c_2 \rangle \equiv c_1} \end{prooftree} \qquad \begin{prooftree} \AxiomC{\Gamma \vdash c_1 : k_1} \AxiomC{\Gamma \vdash c_2 : k_2} \BinaryInfC{\Gamma \vdash \pi_2 \langle c_1, c_2 \rangle \equiv c_2} \end{prooftree}$
Extensionality: $\begin{prooftree} \AxiomC{\Gamma \vdash c:\Pi(\alpha:k_1).k_2'} \AxiomC{\Gamma \vdash c':\Pi(\alpha:k_1).k_2''} \AxiomC{\Gamma, \alpha:k_1 \vdash c\ \alpha \equiv c'\ \alpha : k_2} \TrinaryInfC{\Gamma \vdash c \equiv c' : \Pi(\alpha:k_1).k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash \pi_1c \equiv \pi_1c' : k_1} \AxiomC{\Gamma \vdash \pi_2c \equiv \pi_2c' : [\pi_1c/\alpha]k_2} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2:\text{kind}} \TrinaryInfC{\Gamma \vdash c \equiv c' : \Sigma(\alpha:k_2)} \end{prooftree}$
Finally, we may choose to include a singleton rule:
$\begin{prooftree} \alwaysDashedLine \AxiomC{\Gamma \vdash c:S(c')} \UnaryInfC{\Gamma \vdash c \equiv c':S(c')} \end{prooftree}$
but it's unnecessary, as it arises inversion of the singleton rule from 2.4.2.
It turns out, however, that our formulation of constructor kinding isn't quite complete. Consider the following situation.
Recall that $$S(c:\Pi(\alpha:k_1).k_2 = \Pi(\alpha:k_1).S(c\ \alpha : k_2)$$. Then consider the type constructor $$\lambda(\alpha:*).\texttt{int}$$:
\begin{aligned} \lambda(\alpha:*).\texttt{int} &: S(\lambda(\alpha:*).\texttt{int}:* \rightarrow *) \\ &= \Pi(\alpha:*).S((\lambda(\alpha:*).\texttt{int})\ \alpha : *) \\ &= \Pi(\alpha:*).S((\lambda(\alpha:*).\texttt{int})\ \alpha) \\ &= \Pi(\alpha:*).S(\texttt{int}) \end{aligned}
(where the $$=$$ above are kind equality)
By this, we should be able to derive $$\lambda(\alpha:*).\texttt{int}: \Pi(\alpha:*).S(\texttt{int})$$. This can be done easily:
$\begin{prooftree} \AxiomC{} \UnaryInfC{\alpha:* \vdash \texttt{int} : S(\texttt{int})} \UnaryInfC{\cdot \vdash \lambda(\alpha:\texttt{int}).\texttt{int} : \Pi(\alpha:*).S(\texttt{int})} \end{prooftree}$
which should follow from our existing rules.
However, this doesn't generalize. Namely, we should be able to show that, if $$\beta:S(* \rightarrow *)$$, then $$\beta:\Pi(\alpha:*).S(\beta\ \alpha)$$. As our rules currently are, however, we can't do this -- our rules currently only allow type lambdas and beta- or pi-redexes to have $$\Pi$$-types, and the singular variable $$\beta$$ is neither. A similar issue arises with product kinds. We can resolves this by giving up the structural-only property of the constructor kinding rules and adding extensionality there as well:
Rules 2.4.3 (Kinding finalized)
$\begin{prooftree} \AxiomC{\Gamma \vdash c:\Pi(\alpha:k_1).k_2'} \AxiomC{\Gamma, \alpha:k_1 \vdash c\ \alpha:k_2} \BinaryInfC{\Gamma \vdash c:\Pi(\alpha:k_1).k_2} \end{prooftree}$ $\begin{prooftree} \AxiomC{\Gamma \vdash \pi_1c:k_1} \AxiomC{\Gamma \vdash \pi_2c:[\pi_1c/\alpha]k_2} \AxiomC{\Gamma, \alpha:k_1 \vdash k_2:\text{kind}} \TrinaryInfC{\Gamma \vdash c:\Sigma(\alpha:k_1).k_2} \end{prooftree}$
## Remarks
• I'm a bit sketched on how the proof of non-dependent products implying dependent products works. The moral reason is that the dependent argument must have a fully instantiated kind, so it doesn't matter whether the actual sum kind is dependent, but the subtyping via the generalized singleton doesn't quite make sense to me. Maybe I'll do the full proof out later. | 5,670 | 15,285 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-45 | latest | en | 0.66107 |
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# Jeffrey Clark
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How to detect intersection of 3D rectangles that are rotated?
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scale a point cloud (enlarge or reduce)
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Plot arrow as upper limits in errorbar when I don't have lower bound
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How to find the y from given x on fit line?
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How to use Delaunay Triangulation to create a plane with constraints?
@youssef hany, if all your points are coplanar for one section (in your picture a wall, floor, etc) they can be processed as 2D ...
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How can I put three different tau_rms in this code and draw them in one picture? (each tau_rms = 5*10^(-9), 30*10^(-9), 80*10^(-9), when SNR_MMSE is same as 30.)
@zoopzdd, as in most cases where you want to apply multiple values in code methods invoked that aren't intended to receive multi...
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Need help with an error
@Ryan W, try this instead - you need to look at all possible sequential value sets so a recusive function seems applicable. This...
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i want to add a loop so that it works like this.
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constantGammaClutter doesn't take account of Doppler shift of the frequency due to radar motion
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Optimisation of a 3 vector velocity
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Error in finding phase noise of an electrical oscillator...
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Utility package with NoGrow class and realDeal function
Utilities I find useful: NoGrow superclass to detect and/or prevent growth of arrays; and realDeal to augment MATLAB deal functi...
Looping if between 2 numbers
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How to assign to a collection of object properties stored in an object array, using a numerical array of compatible type and size (one statement, without for-loop!)?
@Peter, without getting into your specific code I suspect that [obj.objArray(IndexSet).tstmp] = T(1:nI) is incorrect in that the...
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@S_G, if you use debug to step thru the code you will see that the 2nd subplot initially gets the shading, but then line plotted...
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@Karsten Gordon, Evaluate MATLAB expression in data dictionary section - MATLAB evalin (mathworks.com) or Evaluate MATLAB expres...
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How can I read data in backgorung from my NI hardware?
@Jan Tomaszewski, try moving the device setup into your main program and pass that dq (and maybe ch3) as a parameters to the wor...
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My snipet code have 3 harmonics, second's amplitude is maximum, but abs(fft(X)) returns first as maximum. I use MATLAB R2022a.
@Georges Theodosiou, you don't have enough samples to easily interpret the abs(fft), if you change Segm = 1:2048 to use 16384 yo...
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fill value that has different size of array
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@Martin Mittrenga, read up on num2str and strsplit which don't really support vectors of things. One way to do this without loop...
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Fast fourier tranformer for Time series data
@Mohammed Lamine Mekhalfia, I think the answer provided by @dpb is valid although @Bjorn Gustavsson has valid points as well. By...
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# 100 台聯大碩士班聯招【電機類】各考科綱要及參考書目
## 考科/代碼 電子學 (3001)
d. Diodes.
3. MOS Field-Effect Transistors (MOSFETs).
4. Bipolar Junction Transistors (BJTs).
5. Single-Stage Amplifiers.
6. Differential and Multistage Amplifiers.
7. Feedback.
參考書目 不提供
## 考科/代碼 資料結構 (3002)
z Stacks and Queues
z Trees
z Graphs
z Sorting
z Hashing
參考書目 E. Horowitz, S. Sahni, and D. P. Mehta, “Fundamentals of Data Structures in
C++,” 2nd. Ed., 2007.
## 考科/代碼 工程數學 A (3003)
1st and 2nd-order ODEs; Higher-order ODEs; System of ODEs, phase plane,
qualitative methods; Series solutions of ODEs; Laplace transforms, Fourier
analysis.
z Linear Algebra
Vector spaces; Linear transformations; Matrix algebra; Systems of linear
equations; Determinants; Diagonalization; Inner product spaces.
z Complex Analysis
Complex numbers and functions; Complex integration; Power series, Taylor
series; Laurent Series, Residue integration; Conformal mapping; Complex
analysis and potential theory.
參考書目 1. E. Kreyszig, Advanced Engineering Mathematics, 9th ed., John Wiley & Sons,
Inc., 2006.
2. C. H. Edwards and D. E. Penney, Elementary Differential Equations with
Boundary Value Problems, 6th ed., Prentice Hall, 2008.
3. S. H. Friedberg, A. J. Insel, and L. E. Spence, "Linear Algebra," 4th ed.,
Prentice Hall, 2003.
4. Gilbert Strang , Introduction to Linear Algebra, 4rd Edition,
Wellesley-Cambridge Press (2008).
5. S. J. Leon, Linear Algebra with Application, 7th ed., Prentice Hall, 2005.
6. H. Anton, Elementary Linear Algebra, 9th ed., Wiley, 2005.
7. A. D. Wunsch, "Complex Variables with Applications," 3rd Ed.,
Systems of linear equations; Determinants; Matrix algebra; Vector
spaces; Linear transformations; Eigenvalues and eigenvectors; Inner
product spaces.
(2) Probability
Sample space and probability; Combinatorial methods; Conditional
probability and independence; Distribution functions and random
variables; Moments and conditional statistics; Limit theorems.
參考書目 (1) S. H. Friedberg, A. J. Insel, and L. E. Spence, Linear Algebra, 4th ed.,
Prentice Hall, 2003.
(2) S. Ghahramani, Fundamentals of probability, with Stochastic Processes,
3rd ed. Upper Saddle River, NJ: Prentice Hall, 2005.
(3) Erwin Kreyszig, Advanced Engineering Mathematics, 8th edition, John
Wiley & Sons, Inc..
(4) Gilbert Strang , Introduction to Linear Algebra, 3rd Ed.,
Wellesley-Cambridge Press, 2003.
(5) Ronald E. Walpole, Raymond H. Myers, Sharon L. Myers, and Keying
Ye, Probability & Statistics for Engineers & Scientists, 7th ed., 2002,
Prentice Hall.
(6) S. J. Leon, Linear Algebra with Application, 7th ed., Prentice Hall, 2005.
(7) H. Anton, Elementary Linear Algebra, 9th ed., Wiley, 2005.
(8) D. P. Bertsekas and J. N. Tsitsiklis, Introduction to Probability Athena
Scientific, 2008.
(9) R. D. Yates and D. J. Goodman, Probability and Stochastic Process,
John Wiley & Sons, 2nd Ed., 2005.
(10)N. A. Weiss, A Course in Probability, Pearson 2006.
## 考科/代碼 工程數學 C (3005)
1st and 2nd order ODEs; Higher-order ODEs; System of ODEs; Series
solutions of ODEs; Laplace transforms, Fourier series methods.
(2) Linear Algebra
Vector spaces; Linear transformations; Matrix algebra; Systems of
linear equations; Determinants; Diagonalization; Inner product spaces.
(3) Partial Differential Equations and boundary value problems
參考書目 (1) E. Kreyszig, Advanced Engineering Mathematics, 9th ed., John Wiley &
Sons, Inc., 2006.
(2) S. H. Friedberg, A. J. Insel, and L. E. Spence, “Linear Algebra,” 4th ed.,
Prentice Hall, 2003.
(3) S. J. Leon, Linear Algebra with Application, 7th ed., Prentice Hall, 2005.
(4) H. Anton, Elementary Linear Algebra, 9th ed., Wiley, 2005.
(5) Gilbert Strang, Introduction to Linear Algebra, 4th Edition,
Wellesley-Cambridge Press, 2009.
(6) Edwards and Penny, Elementary Differential Equations with Boundary
Value Problems, 6th ed., Prentice Hall, 2008.
## 考科/代碼 工程數學 D (3006)
1st and 2nd –oder ODEs
Higher –oder ODEs
System of ODEs, phase plane, qualitative methods
Series Solutions of ODEs Laplace transforms
Fourier analysis
(2) Linear Algebra
Vector Spaces
Linear Transformations
Matrix Algebra
Systems of Linear Equations
Determinants
Diagonalization
Inner Product Spaces
參考書目 (1) Erwin Kreyszig “Advanced Engineering Mathematics” 9th edition, John
Wiley & Sons, Inc..
(2) Gilbert Strang , “Introduction to Linear Algebra,” 3rd Ed.,
Wellesley-Cambridge Press, 2003.
(3) S. J. Leon, Linear Algebra, 6th edition.
(4) S. J. Leon, Linear Algebra with Applications, 7th ed., Prentice Hall,
2005.
(5) H. Anton, Elementary Linear Algebra, 9th ed., Wiley, 2005.
(6) C. H. Edwards and D. E. Penney, Elementary Differential Equations
with Boundary Value Problems, 6th ed., Prentice Hall, 2007.
(7) Spence, Insel and Friedberg, Elementary Linear Algebra：a matrix
approach, 2nd ed., 2008.
## 考科/代碼 電磁學 A (3007)
＊Vector analysis
Static electric fields
Solution of electrostatic problems
Static magnetic currents
＊Time-varying fields and Maxwell’s equations
Plane electromagnetic waves
＊Theory and applications of transmission lines
＊Waveguides and cavity resonators (excluding circular waveguides and cavities)
參考書目 ＊David K. Cheng, Field and Wave Electromagnetics, Second Edition, 1989
＊N.N. Rao, Element of Engineering Electromagnetics, 6th edition, New Jersey:
Pearson Prentice Hall (2004)
＊L. C. Shen and J. A. Kong, Applied Electromagnetism, 1983; 2nd Edition
## 考科/代碼 電磁學 B (3008)
(2) Time-varying fields and Maxwell’s equations Plane electromagnetic waves
(3) Theory and applications of transmission lines
(4) Waveguides and cavity resonators (excluding circular waveguides and
cavities)
參考書目 【電磁學】
＊David K. Cheng, Field and Wave Electromagnetics, Second Edition, 1989
＊N.N. Rao, Element of Engineering Electromagnetics, 6th edition, New Jersey:
Pearson Prentice Hall(2004)
＊L. C. Shen and J. A. Kong, Applied Electromagnetism, 1983; 3nd Edition
【基礎光學】
＊E. Hecht 所著之 Optics (4th edition)其中 Ch2-5
## 考科/代碼 電路學 (3009)
(B)網路重要定理
(C)交流電路功率計算
(D)平衡三相電路分析
(E)雙埠電路分析
(F)富立葉級數與拉氏轉換在電路分析之應用
參考書目 1. Electric Circuits, 8th edition, by James W. Nilsson and Susan A. Riedel
2. Circuit Analysis: Theory and Practice, 4th ed. by A. Robbins and W. Miller
## 考科/代碼 計算機系統(計算機組織) (300A)
Chapter 2. Instructions：Language of the Computer (註：以 MIPS 指令集為主)
Chapter 3. Arithmetic for Computers
Chapter 4. The Processor
Chapter 5. Large and Fast：Exploiting Memory Hierarchy
Chapter 6. Storage and Other I/O Topics
Chapter 7. Multicores, Multiprocessors, and Clusters
參考書目 David A. Patterson, and John L. Hennessy, “Computer Organization and
Design：The Hardware/Software Interface＂ 4th Ed., Morgan Kaufmann
Publishers, Inc., 2009. ISBN 978-0-12-374493-7
2. Linear Time-Invariant Systems
3. Fourier Series Representation of Periodic Signals
4. Fourier Transform
5. Time and Frequency Characterization of LTI Systems
6. Sampling and Discrete-Time Processing
7. Laplace Transform
8. z-Transform
參考書目 1. Alan V. Oppenheim and Alan S. Willsky, with S. Hamid Nawab,
Signals and Systems, 2nd Ed., Prentice-Hall, 1997
2. Simon Haykin and Barry Van Veen, Signals and Systems, 2nd Ed.,
John Wiley & Sons, Inc., 2003
## 考科/代碼 電力系統 (300C)
參考書目 J. D. Glover, M. S. Sarma, T. J. Overbye, Power System Analysis
and Design, 4th ed., Cengaga Learning, 2010.
## 考科/代碼 控制系統 (300D)
2.Root Locus and Bode
3.Frequency-Domain Responses
4.Controller Design
參考書目 z R.C. Dorf and R.H. Bishop, Modern Control Systems, 11/e, Pearson, 2008
z B.C. Kuo and F. Golnaraghi, Automatic Control Systems, 9/e, John
Wiley&Sons, Inc., 2010
## 考科/代碼 計算機概論 (300E)
參考書目 1. D. A. Patterson and J. L. Hennessy, Computer Organization and Design: The
Hardware/Software Interface, 3rd Edition, Revised Printing, Morgan
Kaufmann Publishing Co., Menlo Park, CA., 2007.
2. J. L. Hennessy and D. A. Patterson, Computer Architecture: A Quantitative
Approach, 4th Edition, Morgan Kaufmann Publishing Co., Menlo Park, CA.
2006.
3. E. Horowitz, S. Sahni and D. P. Mehta. Fundamentals of Data Structures in
C++, 2nd Edition. Silicon Press, 2007.
4. M. Morris Mano and Morris M Mano, "Digital Design", 4th Edition,
2. Signals and Systems
3. Signal Space Analysis
4. Analog Modulation Techniques
5. Baseband Digital Transmission
6. Passband Digital Transmission
7. Advanced Topics in Multiuser Communications and Information Theory
參考書目 (1) S. Haykin, "Communication Systems", 4th ed., John Wiley & Sons, 2001
(2) R. E. Ziemer and W. H. Tranter, "Principles of Communications:
Systems Modulation and Noise", 6th ed., Wiley, 2010
## 考科/代碼 近代物理 (300G)
Chap.2. Particles and Waves in Classical Physics
. Classical Waves
. X Rays
Chap. 3. Quanta of Energy
. Planck’s Quantization of Energy
. Photons and the Photoelectric Effect
. The Specific Heat of Solids
Chap. 4. Atomic Structure and Spectral Lines
.The Spectral Series of Hydrogen
.Rutherfords’s Nuclear Atom
.Bohr’s theory
. The Correspondence Principle
Chap. 5. Quantum Mechanics I – Free Particles
. The de Broglie Wavelength
. Particle vs. Wave; Duality
. Heisenberg’s Uncertainty Relations; Complementarity
. Schrodinger’s Wave Equation for a Free Particle
. Wave Packets; Group Velocity
Chap. 6. Wave Mechanics II – Particles in Potentials
. Particle in a Box
. Piecewise Constant Potentials; The Finite Potential Well
. Barrier Penetration
. The Harmonic Oscillator
. The Hydrogen Atom
Chap. 7. Spin and the Exclusion Principle
. The Spin of the Electron
. The Total Angular Momentum; L-S coupling
. The Zeeman Effect
. Pauli’s Exclusion Principle
. Periodic Table
Chap. 8. Statistical Physics
. The Maxwell-Boltzmann Distribution
. The Bose-Einstein Distribution
. The Fermi-Dirac Distribution
Chap. 9. Electrons in Solids
. Free Electron Gas
. Band Theory of Solids; Conductors, Semiconductors, and Insulators
. p-n Junction
. Semiconductor Devices
參考書目 A. Concepts of Modern Physics, Beiser, Arthur,2003.
B. Physics for Scientists and Engineers with Modern Physics, Serway, Raymond
A,2004.
C. Modern Physics, Author: Randy Harris, Publisher: Pearson Addison Wesley,
2008.
## 考科/代碼 固態電子元件 (300H)
Chapter 1: Electron Energy and States in Semiconductors
1.1. Introduction and Preview
1.2. A brief history
1.3. Application to the Hydrogen Atom
1.7. A First Look at Optical Emission and Absorption
1.8. Crystal Structures, Planes, and Directions
## Chapter 2: Homogeneous Semiconductors
2.1. Introduction and Preview
2.3. Conduction Band Structure
2.4. Valence Band Structure
2.5. Intrinsic Semiconductors
2.6. Extrinsic Semiconductors
2.7. The Concept of Holes
2.9. Fermi-Dirac Statistics
## Chapter 3: Current Flow in Homogeneous Semiconductors
3.1. Introduction
3.2. Drift Current
3.3. Carrier Mobility
3.4. Diffusion Current
3.5. Carrier Generation and Recombination
3.7. Continuity Equations
3.9 Minority Carrier Diffusion Lengths
## Chapter 4: Non-Homogeneous Semiconductors
4.1. Constancy of the Fermi Level at Equilibrium
## Chapter 5: Prototype pn Homojunctions
5.1. Introduction
5.2. Prototype pn Junctions
5.3. Prototype pn Homojunctions
5.4. Small-Signal Impedance of Prototype Homojunctions
5.5. Transient Effects
5.6. Effects of Temperature
## Chapter 6: Additional Considerations for Diodes
6.2. Introduction
6.3. Nonstep Homojunctions
6.4. Metal-Semiconductor Junctions
## Chapter 7: The MOSFET
7.1. Introduction
7.2. MOSFETs (Qualitative)
7.3. MOSFETs (Quantitative)
7.4. Comparison of Models with Experiment
## Chapter 8: Additional Considerations for FETs
8.1. Introduction
8.2. Measurement of Threshold Voltage and Low-Field Mobility
8.3. Subthreshold Leakage Current
8.8 Short-Channel Effects
8.9 MOSFET scaling
## Chapter 9: Bipolar Junction Transistors
9.1. Introduction
9.2. Output Characteristics
9.3. Current Gain | 3,736 | 11,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-43 | latest | en | 0.659258 |
https://oeis.org/A045535 | 1,596,744,165,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00454.warc.gz | 403,388,892 | 4,377 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A045535 Least negative pseudosquare modulo the first n odd primes. (Formerly M4381 N2226) 14
7, 23, 71, 311, 479, 1559, 5711, 10559, 18191, 31391, 118271, 366791, 366791, 2155919, 2155919, 2155919, 6077111, 6077111, 98538359, 120293879, 131486759, 131486759, 508095719, 2570169839, 2570169839, 2570169839, 2570169839, 2570169839, 2570169839 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS a(n) is the smallest positive integer m such that m == 7 (mod 8) and for the first n odd primes p, -m is a (nonzero) quadratic residue mod p. a(29) > 2*10^10. - Jinyuan Wang, Mar 24 2020 REFERENCES N. D. Bronson and D. A. Buell, Congruential sieves on FPGA computers, pp. 547-551 of Mathematics of Computation 1943-1993 (Vancouver, 1993), Proc. Symp. Appl. Math., Vol. 48, Amer. Math. Soc. 1994. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS D. H. Lehmer, E. Lehmer and D. Shanks, Integer sequences having prescribed quadratic character, Math. Comp., 24 (1970), 433-451. D. H. Lehmer, E. Lehmer and D. Shanks, Integer sequences having prescribed quadratic character, Math. Comp., 24 (1970), 433-451 [Annotated scanned copy] PROG (PARI) {A045535 = (n, m=7)->until(!m+=8, for(i=2, n+1, m%prime(i)||next(2); issquare(Mod(-m, prime(i)))||next(2)); return(m))} \\ Starting value (e.g., a(n-1); must be in 7+8Z) may be given as 2nd arg. - M. F. Hasler, Oct 24 2013 CROSSREFS Cf. A002189, A062241. Sequence in context: A139852 A141194 A198644 * A001984 A147972 A002223 Adjacent sequences: A045532 A045533 A045534 * A045536 A045537 A045538 KEYWORD nonn,nice,more AUTHOR EXTENSIONS The Bronson-Buell reference gives terms through 227. The Math. Comp. version is erroneous. Edited by Don Reble, Nov 14 2006 Corrected link to OEIS index, following a remark by Don Reble. Values a(0..21) double-checked. - M. F. Hasler, Oct 24 2013 a(27)-a(28) from Jinyuan Wang, Mar 24 2020 STATUS approved
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Last modified August 6 16:02 EDT 2020. Contains 336255 sequences. (Running on oeis4.) | 864 | 2,596 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2020-34 | latest | en | 0.609948 |
https://metanumbers.com/183909 | 1,627,914,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154321.31/warc/CC-MAIN-20210802141221-20210802171221-00445.warc.gz | 403,018,206 | 10,945 | 183909
183,909 (one hundred eighty-three thousand nine hundred nine) is an odd six-digits composite number following 183908 and preceding 183910. In scientific notation, it is written as 1.83909 × 105. The sum of its digits is 30. It has a total of 3 prime factors and 8 positive divisors. There are 111,440 positive integers (up to 183909) that are relatively prime to 183909.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 30
• Digital Root 3
Name
Short name 183 thousand 909 one hundred eighty-three thousand nine hundred nine
Notation
Scientific notation 1.83909 × 105 183.909 × 103
Prime Factorization of 183909
Prime Factorization 3 × 11 × 5573
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 183909 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 183,909 is 3 × 11 × 5573. Since it has a total of 3 prime factors, 183,909 is a composite number.
Divisors of 183909
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 267552 Sum of all the positive divisors of n s(n) 83643 Sum of the proper positive divisors of n A(n) 33444 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 428.846 Returns the nth root of the product of n divisors H(n) 5.49901 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 183,909 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 183,909) is 267,552, the average is 33,444.
Other Arithmetic Functions (n = 183909)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 111440 Total number of positive integers not greater than n that are coprime to n λ(n) 27860 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 16618 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 111,440 positive integers (less than 183,909) that are coprime with 183,909. And there are approximately 16,618 prime numbers less than or equal to 183,909.
Divisibility of 183909
m n mod m 2 3 4 5 6 7 8 9 1 0 1 4 3 5 5 3
The number 183,909 is divisible by 3.
Classification of 183909
• Arithmetic
• Deficient
• Polite
• Square Free
Other numbers
• LucasCarmichael
• Sphenic
Base conversion (183909)
Base System Value
2 Binary 101100111001100101
3 Ternary 100100021110
4 Quaternary 230321211
5 Quinary 21341114
6 Senary 3535233
8 Octal 547145
10 Decimal 183909
12 Duodecimal 8a519
20 Vigesimal 12jf9
36 Base36 3xwl
Basic calculations (n = 183909)
Multiplication
n×i
n×2 367818 551727 735636 919545
Division
ni
n⁄2 91954.5 61303 45977.2 36781.8
Exponentiation
ni
n2 33822520281 6220265882358429 1143962878158656318961 210385068959280324963798549
Nth Root
i√n
2√n 428.846 56.868 20.7086 11.2959
183909 as geometric shapes
Circle
Diameter 367818 1.15553e+06 1.06257e+11
Sphere
Volume 2.60554e+16 4.25026e+11 1.15553e+06
Square
Length = n
Perimeter 735636 3.38225e+10 260087
Cube
Length = n
Surface area 2.02935e+11 6.22027e+15 318540
Equilateral Triangle
Length = n
Perimeter 551727 1.46456e+10 159270
Triangular Pyramid
Length = n
Surface area 5.85823e+10 7.33065e+14 150161
Cryptographic Hash Functions
md5 fc967bca638df6b88ce12aa92b397273 348c628cbf29a2141db1382a0a9e5bd84dd817a8 14d38dd664fc6d1d58b8e50dde5133dc6ac47e992f82a69c2de2104b7f3adee0 c933f1f5e24653174f32468992180dda2f05ca96b36594153aee78841ed76bd40a98fe64ca6d8b92e810e2c39eaabbaa8ebb56f999306160baf46ecfe8b57e8a 3006b18c0bfa54918db8df94340ba014a05b5f71 | 1,425 | 4,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-31 | latest | en | 0.820138 |
https://www.expertsmind.com/library/what-subjective-judgment-must-analyst-make-in-creating-model-51524273.aspx | 1,716,912,405,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059143.84/warc/CC-MAIN-20240528154421-20240528184421-00436.warc.gz | 651,444,420 | 14,146 | ### What subjective judgment must analyst make in creating model
Assignment Help Engineering Mathematics
##### Reference no: EM131524273
Question: Suppose that an analyst for an insurance company is interested in using regression analysis to model the damage caused by hurricanes when they come ashore. The response variable is Property Damage, measured in millions of dollars, and the explanatory variables are Diameter of Storm, Barometric Pressure, Wind Speed, and Time of Year.
a. What subjective judgments, both explicit and implicit, must the analyst make in creating and using this model?
b. If X1, Diameter of Storm, is measured in miles, what is the interpretation of the coefficient ß1?
c. Suppose the analyst decides to introduce a categorical variable X5, which equals 1 if the eye of the hurricane comes ashore within 20 miles of the center of a large city (defined as a city with population ≥ 500,000) and 0 otherwise. How would you interpret ß5?
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Find the values of x, y and z when 4.5x + 7y + 3z = 128.5, 6x + 18.2y + 12z = 270.8. | 1,039 | 5,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.901482 |
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# The angular frequency of the damped oscillator is given by , $\;w=\sqrt{\large\frac{k}{m}-\large\frac{r^{2}}{4m^{2}}}\;$ where k is the spring constant , m is the mass of the oscillator and r is the damping constant.If the ratio $\;\large\frac{r^{2}}{mk}\;$ is $\;8\%\;$ , the change in time period compared to the undamped oscillator is approximately as follows :
$(a)\;increases\;by\;1\% \qquad(b)\;increases\;by\;8\%\qquad(c)\;decreases\;by\;1\%\qquad(d)\;decreases\;by\;8\%$
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0 answers | 234 | 712 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2016-50 | longest | en | 0.836142 |
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# S2 Sampling Distribution Median
Announcements Posted on
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1. Hi guys so this question is getting on my nerves - I can't see how they get the median probability for 20? I have done all of the possible combinations and I still get 56/125? I have got (20,20,20), 3(20,20,50), 3(10,20,20), 3(10,20,50) and there's still more apparently, maybe it's an obvious one but I can't seem to get it?
cheers
Posted from TSR Mobile
2. (Original post by iMacJack)
Hi guys so this question is getting on my nerves - I can't see how they get the median probability for 20? I have done all of the possible combinations and I still get 56/125? I have got (20,20,20), 3(20,20,50), 3(10,20,20), 3(10,20,50) and there's still more apparently, maybe it's an obvious one but I can't seem to get it?
cheers
Posted from TSR Mobile
The list seems complete as regards the values of the coins, although it should be 6(10,20,50)
3. (Original post by ghostwalker)
The list seems complete as regards the values of the coins, although it should be 6(10,20,50)
How comes?
4. (Original post by iMacJack)
How comes?
Number of arrangements of 3 distinct objects = 3! = 6
a,b,c
a,c,b
b,c,a
b,a,c
c,a,b
c,b,a.
5. (Original post by ghostwalker)
Number of arrangements of 3 distinct objects = 3! = 6
a,b,c
a,c,b
b,c,a
b,a,c
c,a,b
c,b,a.
Awesome cheers!
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http://git.dev.opencascade.org/gitweb/?p=occt.git;a=blame;f=src/GC/GC_MakeCircle.hxx;hb=4796758e8dd2c666e54377fd5ac7f514d1c7282e | 1,556,195,195,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2019-18/segments/1555578721441.77/warc/CC-MAIN-20190425114058-20190425140058-00265.warc.gz | 79,061,182 | 5,333 | CommitLineData
42cf5bc1 1// Created on: 1992-09-28
2// Created by: Remi GILET
3// Copyright (c) 1992-1999 Matra Datavision
5//
6// This file is part of Open CASCADE Technology software library.
7//
8// This library is free software; you can redistribute it and/or modify it under
9// the terms of the GNU Lesser General Public License version 2.1 as published
10// by the Free Software Foundation, with special exception defined in the file
11// OCCT_LGPL_EXCEPTION.txt. Consult the file LICENSE_LGPL_21.txt included in OCCT
12// distribution for complete text of the license and disclaimer of any warranty.
13//
14// Alternatively, this file may be used under the terms of Open CASCADE
15// commercial license or contractual agreement.
16
19
20#include <Standard.hxx>
21#include <Standard_DefineAlloc.hxx>
22#include <Standard_Handle.hxx>
23
24#include <GC_Root.hxx>
d1a67b9d 25#include <Geom_Circle.hxx>
26
42cf5bc1 27class StdFail_NotDone;
28class gp_Circ;
29class gp_Ax2;
30class gp_Pnt;
31class gp_Dir;
32class gp_Ax1;
33
34
35//! This class implements the following algorithms used
36//! to create Cirlec from Geom.
37//!
38//! * Create a Circle parallel to another and passing
39//! though a point.
40//! * Create a Circle parallel to another at the distance
41//! Dist.
42//! * Create a Circle passing through 3 points.
43//! * Create a Circle with its center and the normal of its
45//! * Create a Circle with its axis and radius.
46//! The circle's parameter is the angle (Radian).
47//! The parametrization range is [0,2*PI].
48//! The circle is a closed and periodic curve.
49//! The center of the circle is the Location point of its axis
50//! placement. The XDirection of the axis placement defines the
51//! origin of the parametrization.
52class GC_MakeCircle : public GC_Root
53{
54public:
55
56 DEFINE_STANDARD_ALLOC
57
58
59 //! creates a circle from a non persistent circle C by its conversion.
60 Standard_EXPORT GC_MakeCircle(const gp_Circ& C);
61
62
63 //! A2 is the local coordinates system of the circle.
64 //! It is not forbidden to create a circle with Radius = 0.0
66 Standard_EXPORT GC_MakeCircle(const gp_Ax2& A2, const Standard_Real Radius);
67
68 //! Make a Circle from Geom <TheCirc> parallel to another
69 //! Circ <Circ> with a distance <Dist>.
70 //! If Dist is greater than zero the result is enclosing
71 //! the circle <Circ>, else the result is enclosed by the
72 //! circle <Circ>.
73 Standard_EXPORT GC_MakeCircle(const gp_Circ& Circ, const Standard_Real Dist);
74
75 //! Make a Circle from Geom <TheCirc> parallel to another
76 //! Circ <Circ> and passing through a Pnt <Point>.
77 Standard_EXPORT GC_MakeCircle(const gp_Circ& Circ, const gp_Pnt& Point);
78
79 //! Make a Circ from gp <TheCirc> passing through 3
80 //! Pnt2d <P1>,<P2>,<P3>.
81 Standard_EXPORT GC_MakeCircle(const gp_Pnt& P1, const gp_Pnt& P2, const gp_Pnt& P3);
82
83 //! Make a Circle from Geom <TheCirc> with its center
84 //! <Center> and the normal of its plane <Norm> and
86 Standard_EXPORT GC_MakeCircle(const gp_Pnt& Center, const gp_Dir& Norm, const Standard_Real Radius);
87
88 //! Make a Circle from Geom <TheCirc> with its center
89 //! <Center> and the normal of its plane defined by the
91 Standard_EXPORT GC_MakeCircle(const gp_Pnt& Center, const gp_Pnt& PtAxis, const Standard_Real Radius);
92
93 //! Make a Circle from Geom <TheCirc> with its center
95 Standard_EXPORT GC_MakeCircle(const gp_Ax1& Axis, const Standard_Real Radius);
96
97
98 //! Returns the constructed circle.
99 //! Exceptions
100 //! StdFail_NotDone if no circle is constructed.
101 Standard_EXPORT const Handle(Geom_Circle)& Value() const;
42cf5bc1 102
d1a67b9d 103 operator const Handle(Geom_Circle)& () const { return Value(); }
42cf5bc1 104
105private:
42cf5bc1 106 Handle(Geom_Circle) TheCircle;
42cf5bc1 107};
108 | 1,060 | 3,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.71259 |
http://www.livedu.in/dda-algorithm/ | 1,580,114,004,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251696046.73/warc/CC-MAIN-20200127081933-20200127111933-00269.warc.gz | 239,253,445 | 37,543 | DDA Algorithm
Computer Graphics / Tuesday, August 7th, 2018
DDA Line Generation Algorithm
Two of the most popular line generation algorithms are DDA Algorithm and Bresenham Algorithm. The full form of DDA algorithm is Digital Differential Analyser algorithm. To understand it we have to clear about some mathematical terms discussed below.
Need to know:
Any point ‘P’ in 2D represented as (x, y) where x is termed as abscissa which is distance of the point from origin i.e., (0, 0) in the direction of X-axis and y is termed as ordinate which is distance of the point from origin i.e., (0, 0) in the direction of Y-axis.
Now, if we are given the co-ordinates of end points of a line segment PQ , P(x1, y1) and Q(x2, y2) then the equation of the line is,
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Where, m = Slope of the line and it is calculated as,
$m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
$i.e.,m=\frac{{{y}_{(end)}}-{{y}_{(start)}}}{{{x}_{(end)}}-{{x}_{(start)}}}$
For example, let us find the slope of the line joining the points (2, 3) and (8, 11).
$m=\frac{{{y}_{(end)}}-{{y}_{(start)}}}{{{x}_{(end)}}-{{x}_{(start)}}}=\frac{11-3}{8-2}=\frac{8}{6}=\frac{4}{3}$
The slope of any particular line is fixed and its does not changes if we calculate it using any two points on the line.
Again, remember
$\Delta x=dx={{x}_{2}}-{{x}_{1}}$
$\Delta y=dy={{y}_{2}}-{{y}_{1}}$
$\therefore m=\frac{\Delta y}{\Delta x}=\frac{dy}{dx}$
Now we are ready to understand DDA algorithm.
It is an incremental conversion method. In this approach next step is calculated using results from the preceding step. Suppose at step k we have calculated (xk, yk) to be a point on the line then the next point (xk+1, yk+1) should satisfy m.
Case 1:
If the slope m < 1, we increase the value of x by 1 i.e., dx = 1 and calculate successive y values.
Case 2:
If the slope m > 1, we increase the value of y by 1 i.e., dy = 1 and calculate consecutive x values.
Case 3:
If the slope m = 1, we increase the value of x and y both by 1.
Rule:
m xk+1 yk+1 m < 1 xk+1 = xk + 1 yk+1= yk + m m > 1 xk+1 = xk + (1/m) yk+1= yk + 1 m = 1 xk+1 = xk + 1 yk+1= yk + 1
Example 01
Draw line segment from point (2, 4) to (9, 9) using DDA algorithm.
Solution:
We know general equation of line is given by
$y-{{y}_{1}}=m(x-{{x}_{1}})$
Where,
$m=\frac{{{y}_{(end)}}-{{y}_{(start)}}}{{{x}_{(end)}}-{{x}_{(start)}}}$
Given, (xstart, ystart) → (2, 4) and (xend , yend) → (9, 9)
$\Rightarrow m=\frac{{{y}_{(end)}}-{{y}_{(start)}}}{{{x}_{(end)}}-{{x}_{(start)}}}=\frac{9-4}{9-2}=\frac{5}{7}<1$
As 0 < m < 1 so according to DDA algorithm case 1.
xk+1 = xk + 1, yk+1= yk + m
Iteration 1:
Given, (x1, y1) → (2, 4)
$\therefore {{x}_{2}}={{x}_{1}}+1=2+1=3$
$\therefore {{y}_{2}}={{y}_{1}}+m=4+\frac{5}{7}=\frac{33}{7}$
Put pixel (x2, round y2, color) i.e., put on (3, 5)
Iteration 2:
${{x}_{3}}={{x}_{2}}+1=3+1=4$
${{y}_{3}}={{y}_{2}}+m=\frac{33}{7}+\frac{5}{7}=\frac{38}{7}$
Put pixel (x3, round y3, color) i.e., put on (4, 5)
Similarly go on till (9, 9) is reached.
One thought on “DDA Algorithm”
1. rivalries
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# Calculate desuperheated steam temp
## Calculate desuperheated steam temp
(OP)
Hello. I'm working on a problem right now that I hope someone here can provide some input/ideas. Here's the situation:
I have a superheated steam supply(about 200deg of superheat) that we can't directly supply to an exchanger due to fouling from the high wall temperature. The plan is to install a drum upstream of the exchanger and bubble the superheated steam through a sparger up through the condensate that returns from the exchanger, which for all practicle purposes is at saturated conditions.
I need the steam to come out approximately with only approximately 20-25 degrees of superheat (less would be better). I need to determine the liquid level required to give adequate heat transfer such that the temperature is reached. From there, I can size my drum.
This is currently done at a few other sites, but no design info is available.
Here is what I am thinking:
1) Calculate bubble diameter(assuming sphere) for a known orifice diameter.
2) From bubble diameter, I can determine a rise velocity.
3) In Mccabe Smith and Harriot, I see a Nusselt number formula for heat transfer between a flowing fluid and the surface of a single sphere.
4) Here's where I'm getting a little uneasy: Based on the ho I calculate from the nusselt number (ho*Dp/kf), can I simply solve the formula hi=mCp(Tb-Ta)/(PI*D*L*deltatLM) for my temp? Rather than L being length of a tube, use depth of liquid, and make D bubble diameter? I don't believe I can make those changes.
There must be a correlation for something like this. I have searched through my books and on the internet, but I have not run across anything yet. Any guidance would be greatly appreciated. Thanks.
Mark
### RE: Calculate desuperheated steam temp
How are you going to get the condensate from the exchanger to this drum so the steam can bubble through it? Is the exchanger above where you are doing to put the drum?
There would seem to be some hydraulic issues to work out given the inlet steam pressure has to be above the condensate pressure especially with the pressure losses through your proposed steam sparger.
I would look at a desuperheater. Some designs will entrain the condensate into the steam by means of a venturi to create a localized low pressure zone.. While there are limits how much lower than steam pressure the condensate pressure can be, your system should be a good candidate for it. Plus, you can get down to within 5F of saturation easily.
### RE: Calculate desuperheated steam temp
Hippo,
TD2K is right on.... the way that power plants deal with this issue is to desuperheat the steam using a device mounted in the steam piping. Desuperheating can be done to about 15 degrees of saturation. Large turndowns are available.
You must have a source of clean water (typically treated feedwater) at a higher pressure than the steam.
http://www.dezurik.com/MNSD_desuperheaters.htm
http://www.yarway.com/control_valves.asp
http://www.joco.com/paper/desuperheaters.html
Good luck............
Let us know what you finally decide upon
MJC
### RE: Calculate desuperheated steam temp
(OP)
Thanks for the replies. The exchanger will be above the return drum, and the steam pressure drop through the exchanger is minimal. There will be enough elevation difference for the hydraulics to work. I had asked about an inline desuperheater, but that's not the direction the client wants to go. Like I said earlier, they have the type of configuration I described in operation at other sites.
Looking at the equation I listed above, I don't think I can get what I mentioned above. I hadn't given it enough thought. If anything, the PI*D*L I would think would have to be replaced with bubble surface area, but I am still not certain this gets me to where I want to be.
I'll take any suggestions, and I'll keep searching.
Mark
### RE: Calculate desuperheated steam temp
Looking for bubble columns and gas dispersions I found Perry VI tabulates the final velocities of air bubbles rising in water at 20 Celsius as function of the bubble diameter from 10 to 300 micrometers. The velocities range from 0.61 mm/s up to 49.38 mm/s. If superheated steam bubbles sparged through condensate behave similarly, it might be that table 18-23 is of help.
### RE: Calculate desuperheated steam temp
H~
Bubbling superheated steam through condensate is not a good idea.
Steam condensers(condensing heat exchangers) have been used, but it begs the question of the cost of creating superheated steam in the first place.
Modern desuperheaters are far more controllable and cost effective.
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Register now while it's still free! | 1,371 | 5,861 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-30 | latest | en | 0.911476 |
https://www.meritnation.com/ask-answer/question/in-the-book-it-is-given-that-when-ap-mp-ap-is-at-its-maximum/production-and-costs/11611867 | 1,656,405,652,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00762.warc.gz | 934,192,768 | 8,215 | # In the Book it is given that when AP=MP , AP is at its maximum. However in this numerical this does not happen i.e. At the maximum value of AP it is NOT equal to MP (> MP) and when it becomes equal to MP (between the 2nd and 3rd unit) it is not at its maximum value. What is the reason behind this Contradiction???
Dear student,
At 3rd unit of labor, MP is maximum, i.e. 40, after that MP falls and at 4th unit AP=MP=30, and at this point AP is maximum too.
and in syllabus its also given that after MP attains maximum point, it starts falling while AP rises, it rises and attain maxmium point and at that level AP=MP.
Regards
• 0
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Dog examination techniques
This presentation has been developed to introduce veterinary students to the process of carrying out a systematic physical examination in canine patients. It is designed to act as an introduction to these processes and procedures only, giving the students a framework from which to work as they develop and refine these skills throughout the veterinary course. Physical examination is a key skill which will be used throughout a veterinary surgeon's career and is a key determinant in selecting diag
Author(s): Creator not set | 1,896 | 9,770 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-48 | latest | en | 0.960799 |
https://www.jiskha.com/display.cgi?id=1207097482 | 1,516,117,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886437.0/warc/CC-MAIN-20180116144951-20180116164951-00479.warc.gz | 930,859,975 | 4,102 | # excel 2007
posted by .
I have to average students test scores but if they have taken all five tests, I need to drop the lowest score before averaging
• excel 2007 -
• excel 2007 -
I need an excel formula that averages students test scores, dropping the lowest score if all 5 tests are taken
• excel 2007 -
I don't know that there is a program within Excel that does that. You can write one yourself, however.
## Similar Questions
1. ### Writing inequalities
you must have an average score of at least 80 to get a B on your report card. You have scores of 61,70,99, and 70. What is the minimium score you must get on the last test to get a B on your report card?
2. ### math
A student has a mean score of 88 on five tests taken. What score must she obtain on her next test to have a mean (average) score of 80 on all six tests?
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5. ### Algebra
Tom took five math tests and got an integer score on each of them. He never scored higher than 90, and his lowest score was the fourth test. If his average score, rounded to the nearest integer, was 82, what is the lowest possible …
6. ### Math
In your class, you have scores of 73,85,76, and 92 on the first four of five tests. To get a grade of Upper C, the average of the first five tests scores must be greater than or equal to 70 and less than 80. a) Solve an inequality …
7. ### statistics
A student has a mean score of five tests taken. What score must she obtain on her next test to have a mean average of 88 on all six tests?
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More Similar Questions | 681 | 2,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2018-05 | latest | en | 0.9738 |
http://stackoverflow.com/questions/8216088/how-to-check-the-size-of-a-float-in-python/16554909 | 1,406,074,087,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997869778.45/warc/CC-MAIN-20140722025749-00164-ip-10-33-131-23.ec2.internal.warc.gz | 360,976,810 | 16,349 | # How to check the size of a float in python?
I want to check whether a float is actually 32 or 64bits (and the number of bits of a numpy float array). There should be a built-in, but just didn't find out...
-
The size of a Python `float` can be requested via `sys.float_info`. I never encountered anything else than 64 bit, though, on many different architectures.
The items of a NumPy array might have different size, but you can check their size in bytes by `a.itemsize`, where `a` is a NumPy array.
-
The range of floating point values is available in the `sys.float_info` object.
As Sven says, for CPython `float` is always 64-bit. But Python's language reference says
You are at the mercy of the underlying machine architecture (and C or Java implementation) for the accepted range ...".
So this is not necessarily the case for other Python implementations.
-
Since the OP is using NumPy, chances are he doesn't care too much about other Python implementations. Good point, though. :) – Sven Marnach Nov 21 '11 at 18:01
numpy.finfo lists sizes and other attributes of float32 ..., including
nexp : number of bits in the exponent including its sign and bias.
nmant : number of bits in the mantissa.
On a machine with IEEE-754 standard floating point,
``````import numpy as np
for f in (np.float32, np.float64, float):
finfo = np.finfo(f)
print finfo.dtype, finfo.nexp, finfo.nmant
``````
will print e.g.
``````float32 8 23
float64 11 52
float64 11 52
``````
(Try float16 and float128 too.)
-
``````print numpy.finfo(numpy.float) | 401 | 1,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-23 | latest | en | 0.822563 |
https://wikimho.com/us/q/astronomy/20494 | 1,675,161,633,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499857.57/warc/CC-MAIN-20230131091122-20230131121122-00791.warc.gz | 621,130,688 | 12,703 | ### Usage of $\sim$, $\approx$, $\simeq$, and $\cong$ in observational astronomy?
• My understanding is $\sim$ generally means "on the order of magnitude of" e.g. $T \sim 10^5$ K
$\approx$ is obviously "approximately equal to" so for example one might write $d \approx 400$ pc rather than $d=4 \times 10^2$ pc
$\simeq$ and $\cong$, are where I am more confused as their usage is rarer and less consistent.
The below IAU resolution, for example, seems to use $\simeq$ for truncation $m_{\mbox{bol}} = -26.83199\ldots \simeq -26.832$, but I have also seen $\simeq$ used as "approximately equal to" in other works.
http://arxiv.org/pdf/1510.06262
How should one be using $\sim$, $\approx$, $\simeq$, and $\cong$?
I don't think that there's any set standard for using these approximation signs.
Those are just the pen strokes of hand waving astronomers. :)
To point out, you'd likely receive a very different answer if you posted this on the math stack exchange. I think the distinction between these various symbols has more significance in mathematics. Astronomers on the other hand tend to be less strict and will often just use all of these interchangeably, or you'll see different standards being used by different people. There's less uniformity in their meaning in the sciences than there would be in the mathematics groups.
5 years ago
$\simeq$ and $\approx$ both mean "approximately equal to". I don't think $\cong$ is used so often, but if I read it, I would interpret is as the same as the two others.
$\sim$ in principle means "of the order of", i.e. correct to within an order of magnitude. However, astronomers tend to be rather slobby sometimes (we call all elements except H and He for "metals", we call it "gas" even when it's ionized and should be called "plasma", we omit factors of order unity if we can get away with it, we mix up terms like "flux", "flux density", and "intensity", and so on), and thus $\sim$ is very often used to mean "approximately equal to". It is even sometimes used as "proportional to", but (usually) only when there's no risk of misunderstanding (e.g. $L_\mathrm{UV}\!~\sim\!~\mathrm{SFR}$, meaning that the UV luminosity is proportional to the star formation rate, omitting a factor of $10^{27}$).
Journals usually have rather strict standards for style and nomenclature, so they might not accept a $\sim$ for a $\propto$, or even for a $\simeq$. But if you read a conference proceeding or lecture notes or some other non-refereed text (even a paper on arXiv waiting to be refereed), you best be prepared for anything. | 662 | 2,575 | {"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} | 3.1875 | 3 | CC-MAIN-2023-06 | latest | en | 0.951969 |
https://alex.state.al.us/LR/CR/58381 | 1,695,561,468,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00439.warc.gz | 97,808,330 | 14,298 | Please pardon our progress while we refine the look and functionality of our new ALEX site! You can still access the old ALEX site at alex.asc.edu. If you would like to share feedback or have a question for the ALEX Team, you can use the contact form here, or email us directly at administrator@alex.state.al.us.
### Overview
In kindergarten, students will learn to count all the way to 100, by ones and by tens! But wait—Can they count to 100… starting from 29? Twenty-eight… Twenty-nine… Umm… twenty-ten? This is a free resource from PBS that can be used to teach students how to count on from any given number ending in nine.
## UP:MA19.K.1
### Vocabulary
• Count forward orally
• Count backwards orally
### Knowledge
Students know:
• how to count by ones and tens orally. This includes counting forward and counting backward.
### Skills
Students are able to:
• orally count forward.
• orally count backward.
### Understanding
Students understand that:
• Counting from 0 to 100 is a sequence.
## UP:MA19.K.2
• Count
### Knowledge
Students know:
• how to rote count from 0 to 100 starting with any given number.
### Skills
Students are able to:
• orally count.
### Understanding
Students understand that:
• Counting from 0 to 100 is a sequence and you can begin with any number.
Audio/Video
### Resource Provider
PBS
Accessibility | 321 | 1,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.887642 |
https://www.got-it.ai/solutions/excel-chat/excel-tutorial/date-and-time/create-date-range-from-two-dates | 1,590,990,990,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00015.warc.gz | 741,657,596 | 19,546 | Get instant live expert help with Excel or Google Sheets
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# Create date range from two dates
When we have two dates in two different cell references and wish to display them in one cell as date range as per our desired format then we will learn how to do that in this article. Using a formula based on the TEXT function and Ampersand (&), concatenating operator, we can create date range from two dates in Excel. This article will step through the process.
Figure 1. Creating Date Range From Two Dates
## Formula Syntax
The generic syntax for the formula is as follows;
`=TEXT(date1,"format")&" - "&TEXT(date2,"format")`
Suppose we have the start date in cell B2 and the end date in cell C2. We want to display these both dates in a cell E2 as date range as per a custom date format “mmm d”. Following the above formula syntax, we can create date range from two dates in a single formula, such as;
`=TEXT(B2,"mmm d")&" - "&TEXT(C2,"mmm d")`
Figure 2. Applying the Formula to Create Date Range
## How Formula Works
When we have a date value, the TEXT function returns it as a custom date format. By using the TEXT function we can return these two date values in a custom format and with the help of Ampersand (&), concatenating operator, we can combine these two formats in a single formula. After applying the formula, press Enter and copy the formula down to other cells.
Figure 3. Displaying Date Range in Single Cell
## Instant Connection to an Expert through our Excelchat Service
Most of the time, the problem you will need to solve will be more complex than a simple application of a formula or function. If you want to save hours of research and frustration, try our live Excelchat service! Our Excel Experts are available 24/7 to answer any Excel question you may have. We guarantee a connection within 30 seconds and a customized solution within 20 minutes.
### Did this post not answer your question? Get a solution from connecting with the expert.
Another blog reader asked this question today on Excelchat:
Solution examples
Good morning, I need a cell formatting solution to display leading zero in my excel table with the following conditions. 1. Cannot format as text 2. cell size is not fixed. Please let me know if there is a solution.Thanks in advance
Solved by V. C. in 11 mins
looking for an Iferror formula that will bring back a blank value if a date is not listed. Currently using the following but want to add an Iferror statement: =TEXT(H2,"DDDD"). H2 is the raw date ex. 2/8/18. I tried =IFERROR(TEXT(H2,"DDDD"),"") but it keeps returning "saturday" instead of a blank cell. Any help is appreciated.
Solved by F. Q. in 20 mins
How do I make this formula =Value(IFerror(TEXT(AM3,"HMM"),"0")) Return a value of 2400 if the value is not an error?
Solved by F. J. in 21 mins
I have an excel spreadsheet with two worksheets. And the following formula is not pulling the data. =VLOOKUP(\$A\$2:\$A\$566,'Module Type Info'!\$A\$2:\$D\$97,4,FALSE) In sheet 1, I am using all data in column 1 for the lookup In sheet 2, I have selected the first 4 columns of data for array I want to pull data from sheet 2, column 4 into sheet 1 Both tabs are sorted alphabetically. Confirmed that the value in column 4 of sheet 2 is a TEXT field. Why is this not pulling the info from sheet 2 into sheet 1
Solved by F. L. in 39 mins
I need to add an formula into this one: =IF(\$L\$1="","",VLOOKUP(TEXT(\$L\$1,"000000"),'FRS102 BASE & GBP TB MAY 2018'!\$E:\$X,MATCH(A25,'FRS102 BASE & GBP TB MAY 2018'!\$Q\$3:\$X\$3,0)+3,0))
Solved by K. D. in 11 mins | 973 | 3,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-24 | longest | en | 0.884791 |
https://hirequalitywriter.com/compare-and-contrast-your-sample-with-thepopulation-using-the-national-statistics-and-graphs-document/ | 1,652,929,203,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00356.warc.gz | 349,940,829 | 10,140 | # Compare and contrast your sample with the population using the National Statistics and Graphs document.
Scenario:Smart businesses in all industriesuse data to provide an intuitive analysis of how they can get a competitiveadvantage. The real estate industry heavily uses linear regression to estimatehome prices, as cost of housing is currently the largest expense for mostfamilies. Additionally, in order to help new homeowners and home sellers withimportant decisions, real estate professionals need to go beyond showingproperty inventory. They need to be well versed in the relationship betweenprice, square footage, build year, location, and so many other factors that canhelp predict the business environment and provide the best advice to theirclients.
PromptYou have been recently hired as ajunior analyst by D.M. Pan Real Estate Company. The sales team has tasked youwith preparing a report that examines the relationship between the sellingprice of properties and their size in square feet. You have been provided witha Real Estate Data spreadsheet that includes properties sold nationwide inrecent years. The team has asked you to select a region, complete an initialanalysis, and provide the report to the team.Note: In the report you prepare for the sales team, theresponse variable (y) should be the listing price and the predictor variable(x) should be the square feet.Specifically you must address the following rubriccriteria, using the Module Two Assignment Template (attached)·Generate a Representative Sample of the Data·Select a region and generate a simple randomsample of 30 from the data.·Report the mean, median, and standard deviationof the listing price and the square foot variables.·Analyze Your Sample·Discuss how the regional sample created is or isnot reflective of the national market.·Compare and contrast your sample with thepopulation using the National Statistics and Graphs document.·Explain how you have made sure that the sampleis random.·Explain your methods to get a truly randomsample.·Generate Scatterplot·Create a scatterplot of the x and y variablesnoted above and include a trend line and the regression equation·Observe patterns·Answer the following questions based on thescatterplot:oDefine x and y. Which variable is useful formaking predictions?oIs there an association between x and y?Describe the association you see in the scatter plot.oWhat do you see as the shape (linear ornonlinear)?oIf you had a 1,800 square foot house, based onthe regression equation in the graph, what price would you choose to list at?oDo you see any potential outliers in thescatterplot?oWhy do you think the outliers appeared in thescatterplot you generated?oWhat do they represent? | 544 | 2,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.91106 |
https://math.stackexchange.com/questions/3033245/smooth-curve-that-is-tangent-to-a-1-form-kernel-in-every-point | 1,556,194,351,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578721441.77/warc/CC-MAIN-20190425114058-20190425135634-00039.warc.gz | 497,041,264 | 32,391 | # smooth curve that is tangent to a 1-form kernel in every point
Let $$α = dz - ydx \in Ω^1 (\mathbb{R}^3)$$.
Prove that $$\forall p,q \in \mathbb{R}^3,\ \exists \gamma: [0,1] \rightarrow \mathbb{R}^3$$ smooth, such that $$γ(0)=p, γ(1) =q$$ and $$\gamma$$ is tangent to $$ker\ \alpha$$ for all $$t\in [0,1]$$.
If we take $$\gamma(t) = p(1-t) + qt$$ we math the conditions $$γ(0)=p, γ(1) =q$$, but how to match the last condition : $$\alpha (\gamma'(t)) = 0$$?
Thank you for any insights.
The third condition says that for each $$t$$, we must have $$\alpha(\gamma(t)) [\gamma'(t)] = 0. \tag{*}$$ If we write $$\gamma(t) = (x(t), y(t), z(t))$$ then $$\gamma'(t) = (x'(t), y'(t), z'(t))$$ and equation (*) becomes $$z'(t) - y(t) x'(t) = 0,$$ which you can, perhaps, solve (or can at least prove has a solution). | 305 | 813 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2019-18 | latest | en | 0.834419 |
https://mrsvigsscience.wordpress.com/2015/02/ | 1,591,122,290,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347425481.58/warc/CC-MAIN-20200602162157-20200602192157-00490.warc.gz | 441,990,213 | 24,204 | # Friday, 2/27/15
Learning Objective:
To describe friction and identify factors that determine the friction force between two objects
Learning Activities:
1. Do Now- Open your science notebook to the experiment on p.81 and get out “Shipping & Sliding” lab handout and your data table and Try the rotating staircase optical illusion.
2. Science Friday- Dawn of the Cyborg Bacteria
3. Experiment- Shipping and Sliding (Friction)
• As a full class, design a graph
• In your lab group, complete gathering data in your data table. Calculate the average and graph the average data on your graph.
Learning Objective:
To describe friction and identify factors that determine the friction force between two objects
Learning Activities:
1. Do Now- Get out your Science notebook and open to your experiment on p.81 and Observe the print “Relativity” by M. C. Escher
2. Experiment- Shipping and Sliding (Friction)
# Wednesday, 2/25/15
Learning Objective:
To describe friction and identify factors that determine the friction force between two objects
Learning Activities:
1. Do Now- Get out your Science notebook and open to your experiment on p.81 and Read the lab handout “Shipping and Sliding.”
2. Experiment- Shipping and Sliding (Friction)
• Assign groups & responsibilities
• As full class, model how to test friction
• In your lab group, plan your problem statement, independent variable, dependent variable, controlled variables, & procedure
• In your lab group, design a data table
# Tuesday, 2/24/15
Learning Objectives:
To describe friction and identify factors that determine the friction force between two objects
Learning Activities:
2. Review HW- Friction Brainstorm ISN p.80
3. Cue- Advance Organizer- BBC Friction Invigatron and Cart Race interactive websites
4. Experiment- Shipping and Sliding (Friction)- In your lab group, plan your problem statement, independent variable, dependent variable, controlled variables, and procedure
# Monday, 2/23/15
Learning Objectives:
To describe friction and identify factors that determine the friction force between two objects
Learning Activities: | 460 | 2,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-24 | latest | en | 0.825512 |
http://mathhelpforum.com/math-topics/91250-algorithm-analysis.html | 1,519,109,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812880.33/warc/CC-MAIN-20180220050606-20180220070606-00556.warc.gz | 241,821,311 | 11,728 | 1. ## Algorithm analysis
This is regarding the analysis of a basic computer alogrithm.
It's for a "search sort" alrogrithm. All it does is find the largest number in a list of n elements, places it last and then finds the largest element in the list excluding the last element so each iteration the list has n - 1 elements than before.
T(n) represents the time complexity for the algorithm and c represents the total for other operations.
T(n) = (n - 1) + c + (n - 2) + c ... + 2 + c + 1 + c = n(sqr) / 2 - n / 2 + cn
I can see that this works and that the cn is self explanitory, but how do you get to n(sqr) / 2 - n / 2 part, is there some kind of method or forumla that you work through?
If you want a slightly more detailed explination and example the same promblem on Analysis of algorithms - Wikipedia, the free encyclopedia. Under "run time complexity" on the frst example it says it can be factored as T6[1/2(n(sqr) + n]. Again I can see it works I just don't understand how you get there.
This has been driving me mad for about a week. Any help will be appreciated.
Thanks.
2. Originally Posted by alyosha2
...
T(n) = (n - 1) + c + (n - 2) + c ... + 2 + c + 1 + c = n(sqr) / 2 - n / 2 + cn
I can see that this works and that the cn is self explanitory, but how do you get to n(sqr) / 2 - n / 2 part, is there some kind of method or forumla that you work through?
...
1. Re-write the equation:
$T(n)= (n-1)+(n-2)+(n-3)+...+3+2+1+ \underbrace{c+c+c+...+c+c+c}_{\text{n summands}}$
2. Now take the sum
$s=(n-1)+(n-2)+(n-3)+...+3~~~~~+2~~~~~+1$
Now write the sum in reverse order:
$s=1~~~~~~+~~~2~~~~~+~~~3+...+(n-3)+(n-2)+(n-1)$
$2s=n+n+n+...+n+n+n=n \cdot n$
Therefore: $s = \dfrac12 n^2$
Put all the results together:
$T(n)= (n-1)+(n-2)+(n-3)+...+3+2+1+ \underbrace{c+c+c+...+c+c+c}_{\text{n summands}} = \dfrac12 n^2 + nc$
3. Oh I see, it's just a variation on the normal sum method.
thank you.
Just one more question, why the - n / 2?
4. Originally Posted by alyosha2
Oh I see, it's just a variation on the normal sum method.
thank you.
Just one more question, why the - n / 2?
hmmm ... honestly I'm not quite certain what you mean
Are you refering to this part of my post?:
$2s=n+n+n+...+n+n+n=n \cdot n$
Therefore: $s = \dfrac12 n^2$
If so:
When you add columnwise the LHS becomes:
$s+s=2s$
and the RHS becomes:
$n\cdot n = n^2$
Since you want to know the value of one s you have to divide by 2. Therefore:
$2s=n^2~\implies~s = \dfrac12 n^2$
5. no, your post was very clear. I was meaning in the original algorithm.
T(n) = (n - 1) + c + (n - 2) + c ... + 2 + c + 1 + c = n(sqr) / 2 - n / 2 + cn
It's the same as yours except it has an extra n / 2 subtracted.
6. Originally Posted by alyosha2
no, your post was very clear. I was meaning in the original algorithm.
T(n) = (n - 1) + c + (n - 2) + c ... + 2 + c + 1 + c = n(sqr) / 2 - n / 2 + cn
It's the same as yours except it has an extra n / 2 subtracted.
Earboth's $2s$ is equal to $n-1$ terms each equal to $n$ (he has counted them wrongly), so:
$2s=(n-1)n$
or:
$s=\frac{n^2-n}{2}= \frac{n^2}{2}-\frac{n}{2}$
CB | 1,068 | 3,121 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-09 | latest | en | 0.846656 |
https://en.wikipedia.org/wiki/McCullagh's_parametrization_of_the_Cauchy_distributions | 1,495,907,505,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608984.81/warc/CC-MAIN-20170527171852-20170527191852-00305.warc.gz | 936,018,123 | 15,457 | # McCullagh's parametrization of the Cauchy distributions
In probability theory, the "standard" Cauchy distribution is the probability distribution whose probability density function (pdf) is
${\displaystyle f(x)={1 \over \pi (1+x^{2})}}$
for x real. This has median 0, and first and third quartiles respectively −1 and +1. Generally, a Cauchy distribution is any probability distribution belonging to the same location-scale family as this one. Thus, if X has a standard Cauchy distribution and μ is any real number and σ > 0, then Y = μ + σX has a Cauchy distribution whose median is μ and whose first and third quartiles are respectively μ − σ and μ + σ.
McCullagh's parametrization, introduced by Peter McCullagh, professor of statistics at the University of Chicago uses the two parameters of the non-standardised distribution to form a single complex-valued parameter, specifically, the complex number θ = μ + iσ, where i is the imaginary unit. It also extends the usual range of scale parameter to include σ < 0.
Although the parameter is notionally expressed using a complex number, the density is still a density over the real line. In particular the density can be written using the real-valued parameters μ and σ, which can each take positive or negative values, as
${\displaystyle f(x)={1 \over \pi \left\vert \sigma \right\vert (1+{\frac {(x-\mu )^{2}}{\sigma ^{2}}})}\,,}$
where the distribution is regarded as degenerate if σ = 0. An alternative form for the density can be written using the complex parameter θ = μ + iσ as
${\displaystyle f(x)={\left\vert \Im {\theta }\right\vert \over \pi \left\vert x-\theta \right\vert ^{2}}\,,}$
where ${\displaystyle \Im {\theta }=\sigma }$.
To the question "Why introduce complex numbers when only real-valued random variables are involved?", McCullagh wrote:
In other words, if the random variable Y has a Cauchy distribution with complex parameter θ, then the random variable Y * defined above has a Cauchy distribution with parameter ( + b)/( + d).
McCullagh also wrote, "The distribution of the first exit point from the upper half-plane of a Brownian particle starting at θ is the Cauchy density on the real line with parameter θ." In addition, McCullagh shows that the complex-valued parameterisation allows a simple relationship to be made between the Cauchy and the "circular Cauchy distribution".
## Differential equation
McCullagh's parametrization of the pdf of the Cauchy distribution is a solution to the following differential equation:
${\displaystyle \left\{{\begin{array}{l}f'(x)\left(\mu ^{2}+\sigma ^{2}+x^{2}-2\mu x\right)+f(x)(2x-2\mu )=0,\\f(0)={\frac {1}{\pi \left|\sigma \right|\left({\frac {\mu ^{2}}{\sigma ^{2}}}+1\right)}}\end{array}}\right\}}$ | 726 | 2,743 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-22 | latest | en | 0.833475 |
https://origin.geeksforgeeks.org/c-program-for-dijkstras-shortest-path-algorithm-greedy-algo-7/ | 1,638,805,210,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00075.warc.gz | 510,988,805 | 28,171 | # C / C++ Program for Dijkstra’s shortest path algorithm | Greedy Algo-7
• Difficulty Level : Medium
• Last Updated : 08 Jul, 2021
Given a graph and a source vertex in the graph, find shortest paths from source to all vertices in the given graph.
Dijkstra’s algorithm is very similar to Prim’s algorithm for minimum spanning tree. Like Prim’s MST, we generate a SPT (shortest path tree) with given source as root. We maintain two sets, one set contains vertices included in shortest path tree, other set includes vertices not yet included in shortest path tree. At every step of the algorithm, we find a vertex which is in the other set (set of not yet included) and has a minimum distance from the source.
Below are the detailed steps used in Dijkstra’s algorithm to find the shortest path from a single source vertex to all other vertices in the given graph.
Algorithm
1) Create a set sptSet (shortest path tree set) that keeps track of vertices included in shortest path tree, i.e., whose minimum distance from source is calculated and finalized. Initially, this set is empty.
2) Assign a distance value to all vertices in the input graph. Initialize all distance values as INFINITE. Assign distance value as 0 for the source vertex so that it is picked first.
3) While sptSet doesn’t include all vertices
….a) Pick a vertex u which is not there in sptSet and has minimum distance value.
….b) Include u to sptSet
….c) Update distance value of all adjacent vertices of u. To update the distance values, iterate through all adjacent vertices. For every adjacent vertex v, if sum of distance value of u (from source) and weight of edge u-v, is less than the distance value of v, then update the distance value of v.
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## C++
`// A C++ program for Dijkstra's single source shortest path algorithm.` `// The program is for adjacency matrix representation of the graph` `#include ` `#include ` `// Number of vertices in the graph` `#define V 9` `// A utility function to find the vertex with minimum distance value, from` `// the set of vertices not yet included in shortest path tree` `int` `minDistance(``int` `dist[], ``bool` `sptSet[])` `{` ` ``// Initialize min value` ` ``int` `min = INT_MAX, min_index;` ` ``for` `(``int` `v = 0; v < V; v++)` ` ``if` `(sptSet[v] == ``false` `&& dist[v] <= min)` ` ``min = dist[v], min_index = v;` ` ``return` `min_index;` `}` `// A utility function to print the constructed distance array` `int` `printSolution(``int` `dist[], ``int` `n)` `{` ` ``printf``(``"Vertex Distance from Source\n"``);` ` ``for` `(``int` `i = 0; i < V; i++)` ` ``printf``(``"%d \t\t %d\n"``, i, dist[i]);` `}` `// Function that implements Dijkstra's single source shortest path algorithm` `// for a graph represented using adjacency matrix representation` `void` `dijkstra(``int` `graph[V][V], ``int` `src)` `{` ` ``int` `dist[V]; ``// The output array. dist[i] will hold the shortest` ` ``// distance from src to i` ` ``bool` `sptSet[V]; ``// sptSet[i] will be true if vertex i is included in shortest` ` ``// path tree or shortest distance from src to i is finalized` ` ``// Initialize all distances as INFINITE and stpSet[] as false` ` ``for` `(``int` `i = 0; i < V; i++)` ` ``dist[i] = INT_MAX, sptSet[i] = ``false``;` ` ``// Distance of source vertex from itself is always 0` ` ``dist[src] = 0;` ` ``// Find shortest path for all vertices` ` ``for` `(``int` `count = 0; count < V - 1; count++) {` ` ``// Pick the minimum distance vertex from the set of vertices not` ` ``// yet processed. u is always equal to src in the first iteration.` ` ``int` `u = minDistance(dist, sptSet);` ` ``// Mark the picked vertex as processed` ` ``sptSet[u] = ``true``;` ` ``// Update dist value of the adjacent vertices of the picked vertex.` ` ``for` `(``int` `v = 0; v < V; v++)` ` ``// Update dist[v] only if is not in sptSet, there is an edge from` ` ``// u to v, and total weight of path from src to v through u is` ` ``// smaller than current value of dist[v]` ` ``if` `(!sptSet[v] && graph[u][v] && dist[u] != INT_MAX` ` ``&& dist[u] + graph[u][v] < dist[v])` ` ``dist[v] = dist[u] + graph[u][v];` ` ``}` ` ``// print the constructed distance array` ` ``printSolution(dist, V);` `}` `// driver program to test above function` `int` `main()` `{` ` ``/* Let us create the example graph discussed above */` ` ``int` `graph[V][V] = { { 0, 4, 0, 0, 0, 0, 0, 8, 0 },` ` ``{ 4, 0, 8, 0, 0, 0, 0, 11, 0 },` ` ``{ 0, 8, 0, 7, 0, 4, 0, 0, 2 },` ` ``{ 0, 0, 7, 0, 9, 14, 0, 0, 0 },` ` ``{ 0, 0, 0, 9, 0, 10, 0, 0, 0 },` ` ``{ 0, 0, 4, 14, 10, 0, 2, 0, 0 },` ` ``{ 0, 0, 0, 0, 0, 2, 0, 1, 6 },` ` ``{ 8, 11, 0, 0, 0, 0, 1, 0, 7 },` ` ``{ 0, 0, 2, 0, 0, 0, 6, 7, 0 } };` ` ``dijkstra(graph, 0);` ` ``return` `0;` `}`
Output
```Vertex Distance from Source
0 0
1 4
2 12
3 19
4 21
5 11
6 9
7 8
8 14
```
Please refer complete article on Dijkstra’s shortest path algorithm | Greedy Algo-7 for more details!
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Page : | 1,765 | 5,707 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-49 | latest | en | 0.89937 |
http://de.metamath.org/mpegif/opprmul.html | 1,582,355,055,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145654.0/warc/CC-MAIN-20200222054424-20200222084424-00357.warc.gz | 36,828,846 | 5,229 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > opprmul Structured version Unicode version
Theorem opprmul 17147
Description: Value of the multiplication operation of an opposite ring. Hypotheses eliminated by a suggestion of Stefan O'Rear, 30-Aug-2015. (Contributed by Mario Carneiro, 1-Dec-2014.) (Revised by Mario Carneiro, 30-Aug-2015.)
Hypotheses
Ref Expression
opprval.1
opprval.2
opprval.3 oppr
opprmulfval.4
Assertion
Ref Expression
opprmul
Proof of Theorem opprmul
StepHypRef Expression
1 opprval.1 . . . 4
2 opprval.2 . . . 4
3 opprval.3 . . . 4 oppr
4 opprmulfval.4 . . . 4
51, 2, 3, 4opprmulfval 17146 . . 3 tpos
65oveqi 6308 . 2 tpos
7 ovtpos 6982 . 2 tpos
86, 7eqtri 2496 1
Colors of variables: wff setvar class Syntax hints: wceq 1379 cfv 5594 (class class class)co 6295 tpos ctpos 6966 cbs 14507 cmulr 14573 opprcoppr 17143 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1601 ax-4 1612 ax-5 1680 ax-6 1719 ax-7 1739 ax-8 1769 ax-9 1771 ax-10 1786 ax-11 1791 ax-12 1803 ax-13 1968 ax-ext 2445 ax-sep 4574 ax-nul 4582 ax-pow 4631 ax-pr 4692 ax-un 6587 ax-cnex 9560 ax-resscn 9561 ax-1cn 9562 ax-icn 9563 ax-addcl 9564 ax-addrcl 9565 ax-mulcl 9566 ax-mulrcl 9567 ax-i2m1 9572 ax-1ne0 9573 ax-rrecex 9576 ax-cnre 9577 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3or 974 df-3an 975 df-tru 1382 df-ex 1597 df-nf 1600 df-sb 1712 df-eu 2279 df-mo 2280 df-clab 2453 df-cleq 2459 df-clel 2462 df-nfc 2617 df-ne 2664 df-ral 2822 df-rex 2823 df-reu 2824 df-rab 2826 df-v 3120 df-sbc 3337 df-csb 3441 df-dif 3484 df-un 3486 df-in 3488 df-ss 3495 df-pss 3497 df-nul 3791 df-if 3946 df-pw 4018 df-sn 4034 df-pr 4036 df-tp 4038 df-op 4040 df-uni 4252 df-iun 4333 df-br 4454 df-opab 4512 df-mpt 4513 df-tr 4547 df-eprel 4797 df-id 4801 df-po 4806 df-so 4807 df-fr 4844 df-we 4846 df-ord 4887 df-on 4888 df-lim 4889 df-suc 4890 df-xp 5011 df-rel 5012 df-cnv 5013 df-co 5014 df-dm 5015 df-rn 5016 df-res 5017 df-ima 5018 df-iota 5557 df-fun 5596 df-fn 5597 df-f 5598 df-f1 5599 df-fo 5600 df-f1o 5601 df-fv 5602 df-ov 6298 df-oprab 6299 df-mpt2 6300 df-om 6696 df-tpos 6967 df-recs 7054 df-rdg 7088 df-nn 10549 df-2 10606 df-3 10607 df-ndx 14510 df-slot 14511 df-sets 14513 df-mulr 14586 df-oppr 17144 This theorem is referenced by: crngoppr 17148 opprring 17152 opprringb 17153 oppr1 17155 mulgass3 17158 opprunit 17182 unitmulcl 17185 unitgrp 17188 unitpropd 17218 opprirred 17223 irredlmul 17229 isdrng2 17277 isdrngrd 17293 subrguss 17315 subrgunit 17318 opprsubrg 17321 srngmul 17378 issrngd 17381 2idlcpbl 17752 opprdomn 17820 psropprmul 18149 invrvald 19047 rhmopp 27634 ldualsmul 34333 lcdsmul 36800
Copyright terms: Public domain W3C validator | 1,536 | 2,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.137672 |
https://people.maths.bris.ac.uk/~matyd/GroupNames/448/e16/C2%5E2xC7sC8byC2.html | 1,713,892,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818732.46/warc/CC-MAIN-20240423162023-20240423192023-00608.warc.gz | 404,737,686 | 2,462 | # Extensions 1→N→G→Q→1 with N=C2 and Q=C22×C7⋊C8
Direct product G=N×Q with N=C2 and Q=C22×C7⋊C8
dρLabelID
C23×C7⋊C8448C2^3xC7:C8448,1233
Non-split extensions G=N.Q with N=C2 and Q=C22×C7⋊C8
extensionφ:Q→Aut NdρLabelID
C2.1(C22×C7⋊C8) = C2×C4×C7⋊C8central extension (φ=1)448C2.1(C2^2xC7:C8)448,454
C2.2(C22×C7⋊C8) = C22×C7⋊C16central extension (φ=1)448C2.2(C2^2xC7:C8)448,630
C2.3(C22×C7⋊C8) = C2×C28⋊C8central stem extension (φ=1)448C2.3(C2^2xC7:C8)448,457
C2.4(C22×C7⋊C8) = C42.6Dic7central stem extension (φ=1)224C2.4(C2^2xC7:C8)448,459
C2.5(C22×C7⋊C8) = D4×C7⋊C8central stem extension (φ=1)224C2.5(C2^2xC7:C8)448,544
C2.6(C22×C7⋊C8) = Q8×C7⋊C8central stem extension (φ=1)448C2.6(C2^2xC7:C8)448,557
C2.7(C22×C7⋊C8) = C2×C28.C8central stem extension (φ=1)224C2.7(C2^2xC7:C8)448,631
C2.8(C22×C7⋊C8) = C56.70C23central stem extension (φ=1)2244C2.8(C2^2xC7:C8)448,674
C2.9(C22×C7⋊C8) = C2×C28.55D4central stem extension (φ=1)224C2.9(C2^2xC7:C8)448,740
×
𝔽 | 582 | 958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.430585 |
https://www.teacherspayteachers.com/Product/Symmetry-Project-Reflection-Rotation-Translation-667997 | 1,485,127,393,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281649.59/warc/CC-MAIN-20170116095121-00549-ip-10-171-10-70.ec2.internal.warc.gz | 987,623,075 | 51,829 | # Symmetry Project: Reflection, Rotation, & Translation
Subjects
Resource Types
Product Rating
3.9
File Type
PDF (Acrobat) Document File
1.04 MB | 12 pages
### PRODUCT DESCRIPTION
This symmetry project is great for a beginning project in geometry or during testing when students are burnt out. I’ve included 6 samples that you can use while explaining the project to your students.
Give each student the two-page Investigating Symmetry Project Instructions and a copy of the Investigating Symmetry Project Instructions for Measuring Your Objects page.
You can have students work in pairs, though I’ve found it’s best to have students work independently. You can give them time in class to work on their project or make it a homework assignment. I usually give them at least a week. I always have them draw their own objects rather than take clipart or images from the internet, but this is up to your discretion.
The day the projects are due, I put the students in groups of 4. Each group gets four Investigating Symmetry Project Grading Rubric sheets (one per student in the group). They then grade each other’s work according to the rubric. I have them attach the rubric to the project, collect them, and usually only need to make a few adjustments.
It's simple, students can work independently, and they are allowed to be as creative as they'd like. Students practice recognizing symmetry and measuring line segments using a ruler and angles in degrees using a protractor.
Here are other products you may like:
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Teaching geometry? Purchase the Geometry Bundle: Activities, Projects, Worksheets, and more! and SAVE!
Total Pages
12
Rubric only
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4.0
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3.9
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3.9
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3.9
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22 ratings | 569 | 2,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-04 | longest | en | 0.925021 |
http://www.math.yorku.ca/Who/Faculty/Brettler/3110_f01/3110_home.html | 1,513,352,816,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948575124.51/warc/CC-MAIN-20171215153355-20171215175355-00126.warc.gz | 411,588,943 | 3,605 | # Math 3110 A Course Information
### Announcements:
• (Dec.31) Click for unofficial grades.
• (Dec. 5) Here is a brief statistical summary of the test 2 results:
```MTB > dotplot c1
.
. : . . . . . . :.
-------+---------+---------+---------+---------+---------C1
14.0 17.5 21.0 24.5 28.0 31.5
MTB > describe c1
N MEAN MEDIAN TRMEAN STDEV SEMEAN
C1 13 21.74 23.50 21.82 7.13 1.98
MIN MAX Q1 Q3
C1 13.00 29.60 14.00 29.50
```
• (Nov. 25) For my solutions (beware of typographical errors) to Friday's class tests click for postscript or pdf.
• (Nov. 24) Homework 5 is now due on Wednesday, December 5.
• (Nov. 20) The office hour on Wednesday, Nov. 21 is cancelled. I will be available on Nov. 22 at noon.
• (Nov. 18) This is a status report on the midterm for Friday. My current draft has four questions. I will post an update once the final version is ready.
• Proof of a result involving absolute values.
• Proof of a result on least upper bounds whose proof will involve the Archimedian property.
• Proof that the limit of a sequence has a particular value. It is an exercise on 3.2.3.
• Four sequences are given. You are asked to determine whether or not each converges and if it does converge for the value of its limit.
• (Nov. 1) For statistical information on Test 1 performance click here.
• (Oct. 26) There will be no lecture and no office hours Monday, October 29.
• (Oct. 19) For my solutions (beware of typographical errors) to today's class test click for (postscript) or (pdf).
• (Oct 11) Assignment 3 has not yet been posted. For a selection of problems on absolute values look at
Page 34: 3, 5, 6b, 8d, 13bd, 15 .
• (Oct 1) Assignment 2 is NOW DUE Wednesday October 10 at noon.
• (Sept 24) I have corrected my solution to problem 16, page 11 on the first assignment. In addition, a pdf version of the solutions, without the graph I used for problem 15, is now available.
• (Sept 17) Please note that you only need to hand in solutions to the problems on page 11 on Friday. In addition, please note that the problem for the individual solution is page 11 problem 14. Sorry for the confusion.
Lecturer: Eli Brettler
Office: South 508 Ross
Telephone: 736-5250 Extension 66321
E-mail: brettler@mathstat.yorku.ca
WWW: http://www.math.yorku.ca/Who/Faculty/Brettler/
Normal Office hours: MW 12:00 noon - 1:00 p.m. and by appointment
Lectures: MWF 10:30 - 11:20 a.m. in WC 118
Tutorials: W 12:30 - 1:20 p.m. in N 501 Ross.
Text: Bartle and Sherbert, Introduction to Real Analysis (Third Edition).
Syllabus: The course will cover material from Chapters 1, 2, 3, 4, 5 of the text.
Homework:
These homework problems will be collected for grading. Groupwork is permitted and (in most cases) encouraged. EXCEPT WHERE OTHERWISE INDICATED, group assignments may be handed in. Assignments are submitted by placing them in the Assignment Box to be found next to the elevators, North Ross 5th Floor. My normal practice is to collect the papers then post solutions. Late assignments are not accepted.
To view and print pdf you need the free Acrobat reader.
## Evaluation:
Graded Assignments various dates (see above) 15% Class Tests October 19, November 23 30% Final Exam Examination Period 55%
Alternate Grading Scheme: Assignments 15%, Class Tests 55%, Final Exam 30%.
The scheme which gives the higher grade will be used in each case.
Note: There will be no makeups for missed tests. If you miss a test and provide a certificate indicating clearly that you were unable to write the test for reasons beyond your personal control, the weight of the test will be transferred to the final exam. Otherwise, the mark for the missed test will be 0.
The last date to drop the course without academic penalty is November 9. It is extremely important to realistically assess your course performance prior to this date.
Tests
• Test 1: October 19
• Test 2: November 23
Resources:
• The note on which the first lecture was based is available in (postscript) and (pdf).
Eli Brettler | 1,127 | 4,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-51 | latest | en | 0.855414 |
http://www.chegg.com/homework-help/cmos-digital-integrated-circuits-analysis-design-3rd-edition-chapter-15-solutions-9780072460537 | 1,475,230,524,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662159.54/warc/CC-MAIN-20160924173742-00081-ip-10-143-35-109.ec2.internal.warc.gz | 403,745,190 | 16,049 | # CMOS Digital Integrated Circuits Analysis & Design (3rd Edition) View more editions Solutions for Chapter 15
• 127 step-by-step solutions
• Solved by publishers, professors & experts
• iOS, Android, & web
Over 90% of students who use Chegg Study report better grades.
May 2015 Survey of Chegg Study Users
Chapter: Problem:
15.1 Consider a simple RC circuit model for a point-to-point interconnect in CMOS chips. The 50% delay time for a step input pulse is
τ50% = 0.38 RC
Both R and C values are subject to random fluctuations due to process variations.
(a). Express the sensitivities of τ50% with respect to R and C.
(b). Express the percentage change in the delay in terms of percentage changes in both R and C values under the assumption that their values can vary independently.
(c). Determine the largest delay increase possible due to ±10% fluctuations in both R and C values.
SAMPLE SOLUTION
Chapter: Problem:
• Step 1 of 3
(a).The sensitivities can be found:
• Step 2 of 3
(b).The percentage change:
• Step 3 of 3
(c).The largest delay increase is
Corresponding Textbook
CMOS Digital Integrated Circuits Analysis & Design | 3rd Edition
9780072460537ISBN-13: 0072460539ISBN: Authors: | 301 | 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} | 2.703125 | 3 | CC-MAIN-2016-40 | latest | en | 0.806276 |
https://www.enotes.com/homework-help/business-investment-us-increased-by-155-million-366462 | 1,495,657,712,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607860.7/warc/CC-MAIN-20170524192431-20170524212431-00083.warc.gz | 878,897,248 | 10,560 | # If, business investment in the US increased by \$155 million and the US MPC was 0.9, the increase in business investment caused US GDP to:a. decrease by \$155 million b. decrease by \$1.55 million...
If, business investment in the US increased by \$155 million and the US MPC was 0.9, the increase in business investment caused US GDP to:
a. decrease by \$155 million
b. decrease by \$1.55 million
c. increase by \$1550 million
pohnpei397 | College Teacher | (Level 3) Distinguished Educator
Posted on
The correct answer here is C. The investment that you describe will, in this situation, lead to an increase of \$1.55 billion (which is the same as \$1550 million) in gross domestic product.
We can find this by using the multiplier. The multiplier is found by using the equation
Multiplier = 1/1-marginal propensity to consume
In this case, the MPC is .9. This means that 1 – MPC is .1. 1/.1 = 10.
So, the multiplier is 10. This means that an increase in business investment will cause an increase in GDP that is 10 times greater. An increase of \$155 million in business investment will lead to an increase of \$1550 million, or \$1.55 billion, in GDP.
Sources: | 303 | 1,186 | {"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} | 3.375 | 3 | CC-MAIN-2017-22 | longest | en | 0.938144 |
https://www.varsitytutors.com/act_math-help/how-to-find-the-square-root-of-a-decimal | 1,669,548,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00415.warc.gz | 1,131,196,947 | 60,587 | # ACT Math : How to find the square root of a decimal
## Example Questions
← Previous 1
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
The easiest way to find the square root of a fraction is to convert it into scientific notation.
The key is that the exponent in scientific notation has to be even for a square root because the square root of an exponent is diving it by two. The square root of 9 is 3, so the square root of 8.1 is a little bit less than 3, around 2.8
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root.
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
This problem can be solve more easily by rewriting the decimal into scientific notation.
Because has an even exponent, we can take the square root of it by dividing it by 2. The square root of 4 is 2, and the square root of 1 is 1, so the square root of 2.5 is less than 2 and greater than 1.
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
This problem becomes much simpler if we rewrite the decimal in scientific notation
Because has an even exponent, we can take its square root by dividing it by two. The square root of 4 is 2, and because 3.6 is a little smaller than 4, its square root is a little smaller than 2, around 1.9
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root. The square root of 9 is 3, and the square root of 4 is two, so the square root of 6.4 is between 3 and 2, around 2.53
### Example Question #1 : Basic Squaring / Square Roots
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root. is a perfect square, whose square root is .
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root. The square root of 9 is 3, so the square root of 10 should be a little larger than 3, around 3.16
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of the following decimal:
Explanation:
To find the square root of this decimal we convert it into scientific notation.
Because has an even exponent, we can divide the exponenet by 2 to get its square root. The square root of 36 is 6, so the square root of 40 should be a little more than 6, around 6.32.
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of .
Explanation:
Rewrite the expression in radical form.
Rewrite the decimal with factors and simplify.
### Example Question #1 : How To Find The Square Root Of A Decimal
Find the square root of . | 854 | 3,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2022-49 | longest | en | 0.871211 |
https://www.collegesimply.com/gpa-calculator/percentage-grade-calculator/38-out-of-49/ | 1,716,869,366,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00121.warc.gz | 598,707,617 | 17,974 | # 38 out of 49 Test Score
### What grade is 38 out of 49?
Total number of questions or points possible
### 77.55% C+
A score of 38 out of 49 on a test, assignment or class is a 77.55% percentage grade. 11 questions were wrong or points missed.
A 77% is a C+ letter grade. A letter grade C+ means satisfactory or average performance. To get the next higher letter grade available, you would need a score of 40 which would be a B-.
The grade point equivalent to a C+ is a 2.3 GPA for this score.
### Letter Grades for 49 question test
A+48 to 490 to 197.96 to 100.00
A46 to 472 to 393.88 to 95.92
A-45491.84
B+43 to 445 to 687.76 to 89.80
B41 to 427 to 883.67 to 85.71
B-40981.63
C+38 to 3910 to 1177.55 to 79.59
C36 to 3712 to 1373.47 to 75.51
C-351471.43
D+33 to 3415 to 1667.35 to 69.39
D31 to 3217 to 1863.27 to 65.31
D-301961.22
F0 to 2920 to 490.00 to 59.18
This easy grade calculator can be used for quickly determining the percentage, letter grade and GPA for any test, quiz or exam. It can also be used when calculating an overall score in a class. This page is pre-populated showing the results for the particular calculation 38/49 as a percentage and as a grade. You can access the default calculator here.
To use, input the total number of questions or maximum points available in the "Questions" field. Next either enter the number of correct or incorrect answers in the respective fields. You only need to enter one, the other one will calculate automatically. Your results will appear as you type.
The result will compute the score as a percentage, the letter grade for the score and the GPA for the letter grade. The calculator will also indicate when a score is within one percent of the breakpoint between letter grades where it is appropriate to round up the grade.
Additionally you will see a letter grading scale table which shows every letter grade available for the number of questions on the test. This table indicates the correct and incorrect ranges for each letter grade along with the percentage ranges. Note that on some tests, letter grades are skipped - this is not a mistake. This happens on tests with fewer questions as each question is worth a significant percent of the overall score and significantly moves the letter grade.
## FAQ's
### Is a 38 out of 49 a good grade?
A score of 38/49 equates to a C+. C+ isn't a bad grade, but it's not a good grade either. You're falling short of other students' performance with a C+. A C+ is a bit below the average for most high school students (B letter grades).
### How is a percentage calculated for 38 out of 49?
The math is pretty simple to determine the percentage grade on a test as it's simple division and multiplication. The ratio of correct to total number of questions can be expressed as a fraction
The fraction is converted into a decimal using correct answers as the numerator and total questions as the denominator. Divide the correct answers by total questions and you have the answer in decimal form.
correct answers ÷ total questions = decimal
Now to convert into a percent, multiply the decimal times 100. This is your answer.
decimal × 100 = percent answer
In this particular case, the formula for calculating the grade percent 38/49 is
38 ÷ 49 × 100 = 77.55%
On a 49 question test, if you miss 11 and get 38 right, you will have a 77.55%.
### How is a letter grade determined for 38 out of 49?
To determine the letter grade, we must make use of a letter grade scale lookup table. You can find one here.
Take the percentage calculated for the test and locate the corresponding range in the percent column. The letter grade for that row is the letter grade for this test.
For this score, we find our 77.55% is in the 77-79% range. This corresponds with the letter grade "C+". Thus, a 77.55% grade is a C+. | 973 | 3,821 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-22 | latest | en | 0.889489 |
http://www.sr.bham.ac.uk/xmm/forceseb2.html | 1,542,566,222,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744561.78/warc/CC-MAIN-20181118180446-20181118202446-00117.warc.gz | 494,573,446 | 5,515 | FORCES, E FIELDS & B FIELDS - cont'd How are forces and field strengths related? Fields have to act on something to provide a force. In terms of the gravitational field and the electric field these "somethings" are the mass and the charge respectively. [Here the mass is the gravitational mass. Physicists are still attempting to find any difference between gravitational mass and inertial mass (discussed earlier) but as yet none has been found.] Hence : where g is the gravitational field strength and E is the electric field strength. It is important to note the fact that m can only be positive but q can be either positive or negative. For example the proton and the electron have the same charge magnitude but the proton is positively charged and the electron is negatively charged. There are particles called positrons (one of the family of the so called anti-matter) which are just like electrons but have a positive charge. Physicists have not found particles with negative mass though there is no a priori reason as to why they should not exist. An electric field therefore produces a negative force on a negatively charged particle which is simply a force in the opposite direction to the field and to that produced by the field on a positively charged particle (recall force is a vector quantity). Hence, if a gas in which the atoms are ionised (i.e. the outer electron has been freed from the electrically neutral atom leaving a positively charged atom behind called an ion) is placed in an electric field, the electrons move one way and the ions move the other. So we can see that masses are gravitational type things and charges are electric type things. What produces the Fields? We are familiar with Newtons Law of gravity and with Coulombs Law of electrostatics. They describe the forces between pairs of masses and pairs of charges and are respectively Again we note that gravitational forces can only be attractive whilst coulobic forces can be repulsive as well as attractive. Comparing these with the two equations above we can see that : Hence gravitational fields are due to masses and electric fields are due to charges. But we also have magnetic fields, so what do these act on and what produces them? The force produced by a magnetic field, the so called Lorentz force, is given by This equation should really be in its vector form because we would then see that a force is only produced if there is a component of the velocity (v) which is at right angles to the field (of magnitude B) and the force itself is then at right angles to the plane containing the velocity and the field. A force perpendicular to a velocity results in circular motion. We can further see that, because both v and q can have a sign associated with them, the direction of the circular motion can be different. IMAGE - An animation of a moving charged particle in a magnetic field. Please move your cursor over this image. The force is always perpendicular to both the direction and the field, with the direction of a positively charged particle determined by the Fleming's Left-Hand Rule. Since the lorentz force can carry either sign depending on the charge, electrons will move clockwise in the magnetic field described above. We can see from this that what magnetic fields act on is (qv), that is, a moving charge. And, by inference from the discussion above, magnetic fields themselves must be produced by moving charges. Hence there is no need to introduce any third agent equivalent to the mass or the charge (such as a magnetic pole); we already have the electrostatic charge and all we need is for it to be in motion But a moving charge is a current and it is well known that currents generate magnetic fields. But what about the field associated with a bar magnet? Good question. We will return to this later, but essentially the "currents" are on the very small scale and are associated with the electrons circulating the atoms. | 822 | 4,006 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.946731 |
http://www.programmingwithbasics.com/2017/04/geeksforgeeks-solution-for-sum-of_54.html | 1,516,390,345,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888113.39/warc/CC-MAIN-20180119184632-20180119204632-00436.warc.gz | 541,378,976 | 22,893 | # Geeksforgeeks Solution For " Sum of Middle Elements of two sorted arrays "
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 :- Sum of Middle Elements of two sorted arrays
Submit Your Solution :- Click Here
Solution :-
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main(){
int t,n,val;
cin>>t;
while(t--){
cin>>n;
int a[n],b[n];
vector<int>c(2*n);
for(int i=0;i<n;i++){
cin>>a[i];
}
for(int i=0;i<n;i++){
cin>>b[i];
}
merge(a,a+n,b,b+n,c.begin());
int s=c.size();
cout<<(c[(s/2)-1]+c[s/2])<<endl;
}
}
Output:-
#### Ghanendra Yadav
Hello, I Am Ghanendra Yadav Owner of This Blog, I am professional Blogger and Programmer. I Love Programming, Logo Making, And Banner Designing. My Highest Qualification is MCA From NIT Warangal. You Can Find Me On Social Media Through Below Link And If You Have Any Query Related To Programming And Other Subject Comment Below or You Can Mail Me I Will Try To Answer Within 24 Hours Email:- yghanendra@student.nitw.ac.in
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Facebook | Twitter | Google+ | RSS Feed | 343 | 1,274 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-05 | latest | en | 0.447824 |
http://www.onbarcode.com/tech/788/55/ | 1,603,454,596,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881369.4/warc/CC-MAIN-20201023102435-20201023132435-00449.warc.gz | 165,873,606 | 8,676 | # code 128 font vb.net NS SECTOR ALGEBRA in Java Generation QR-Code in Java NS SECTOR ALGEBRA
NS SECTOR ALGEBRA
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For the NS sector, the following relations are satis ed: [ Ln , Lm ] = ( n m ) Ln + m + c 3 ( n n ) n +m ,0 12 (7.42) (7.43) (7.44)
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1 [ Ln , Gr ] = ( n 2r )Gn +r 2 c {Gr , Gs } = 2 Lr +s + ( 4r 2 1) r +s ,0 12
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The central charge is related to the space-time dimension by c = D + D /2. Let be a physical state in the NS sector. The NS sector super-Virasoro constraints are ( L0 aNS ) = 0 Ln = 0 n > 0 Gr = 0 r > 0 (7.45) (7.46) (7.47)
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Here, following the quantization of the bosonic string, aNS is a normal-ordering constant. The open string mass formula is taken by setting L0 = aNS , which gives m2 = Where the number operator is N = n n +
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1 ( N aNS )
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(7.48)
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br
(7.49)
String Theory Demysti ed
R SECTOR ALGEBRA
In the R sector, the commutation and anticommutation relations are [ Lm , Ln ] = ( m n ) Lm + n + m [ Lm , Fn ] = n Fm +n 2 {Fm , Fn } = 2 Lm +n + The conditions on the physical states are ( L0 a R ) = 0 Ln = 0 n > 0 Fm = 0 m 0 Here, aR is the normal-ordering constant for the R sector. EXAMPLE 7.6 Deduce that aR = 0. SOLUTION We start with the anticommutation relation satis ed by the Fm : {Fm , Fn } = 2 Lm +n + Notice that if m = n = 0, we obtain {F0 , F0 } = F0 F0 + F0 F0 = 2 F02 = 2 L0 L0 = F02 The Fm annihilate physical states . Therefore, F0 = 0 D 2 m m +n ,0 2 (7.53) (7.54) (7.55) D 2 m m +n ,0 2 D 3 m m +n ,0 8 (7.50) (7.51) (7.52)
CHAPTER 7 RNS Superstrings
From this we obtain, by acting on the equation with F0 , the relation F0 F0
)= F
=0
L0 = 0 But we know that ( L0 aR ) = 0. Hence, 0 = ( L0 a R ) = L0 a R = a R aR = 0
The Open String Spectrum
Now let s examine the states of the string. We will look at states of the open string in this chapter. We must consider the NS and R sectors independently. Working in the NS sector rst, the ground state is 0,k NS and it is annihilated by the modes
i n 0, k NS
= bri 0, k
(7.56)
where n, r > 0. The zero mode 0 as discussed in the bosonic string case is a momentum operator: 0 0, k
= 2 0, k
(7.57)
It can be shown that the normal-ordering constant in the NS sector is aNS = 1 2 (7.58)
Using this we can nd the mass of the ground state, which is m2 = 1 2 (7.59)
Once again, we have a state with m 2 < 0, so the theory still contains a tachyon state. We will see later that we can get rid of the tachyon state in the superstring theory. The ground state in the NS sector is a unique spin-0 state. To nd massive states, we progressively act on the state with negative mode oscillators.
String Theory Demysti ed
Next we consider the R sector, which describes space-time fermions in the open string case. The ground state is annihilated by
m 0, k = dm 0, k
(7.60)
for m > 0. The zero mode d0 is actually a Dirac operator. That is,
d0 =
(7.61)
We will see below that the critical space-time dimension is 10, so the states in the R sector are 10-dimensional spinors. The ground state satis es the massless Dirac wave equation. In our notation, this is written in the following way, recalling that the momentum operator is 0 : | 1,594 | 5,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.675639 |
http://highschoolmath.blogspot.com/2006/06/number-8.html | 1,501,137,408,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549427749.61/warc/CC-MAIN-20170727062229-20170727082229-00590.warc.gz | 147,753,221 | 5,891 | # High School Math
MISSION STATEMENT: To encourage and promote a greater use of the internet and computer technology in the math classroom. For educators, students, parents and homeschoolers.
## Sunday, June 04, 2006
### Number 8
9 x 9 + 7 = 88
9 x 98 + 6 = 888
9 x 987 + 5 = 8888
9 x 9876 + 4 = 88888
9 x 98765 + 3 = 888888
9 x 987654 + 2 = 8888888
9 x 9876543 + 1 = 88888888
9 x 98765432 + 0 = 888888888.
More Patterns at www.TheMathWebSite.com. | 168 | 452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-30 | longest | en | 0.810202 |
https://www.educationindex.com/essay/Great-Lake-F38WWHEEY | 1,720,897,373,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00387.warc.gz | 665,205,140 | 10,975 | # Great Lake
2142 words 9 pages
GREAT LAKES PIPE & TUBE, INC. “If we do decide to produce the 10- and 12-inch pipe internally, it could solve our overstaffing problem,” Mark Rubin, owner of Great Lakes Pipe & Tube, Inc. (GLPT), remarked to Vinny Patricko, the plant manager. “I’m reluctant to lay anyone off or even cut back hours. It’s not good business and it’s not the right thing to do if it can be at all avoided.”
THE FIRM Mark Rubin had no intentions of starting his own firm in 1972. Since graduating from college in 1964, he had worked for ML Pipe, a company based in Youngstown, Ohio. In January 1972, the company decided to relocate to New Jersey, and Rubin went also. Rubin and his wife were quite unhappy in Virginia, mainly because they felt
For the purpose of analysis, Rubin will assume straight-line depreciation to zero salvage value over the eight-year life of the project. (Ideally, you should use MACRS depreciation). The market value of the equipment after eight years is expected to be \$180,000 before taxes.
THE ACCOUNTANT’S ESTIMATES
The firm’s accountant, Abe Komansky, has developed a set of numbers that, in his view, “strongly indicates” in-house production is a “losing proposition.” (See Exhibit 2.) Komansky estimates it will cost 54.3 cents per pound to produce the 10-inch and 12-inch pipe internally. He notes that GLPT can purchase the same pipe for 45 cents per pound from another manufacturer and incurs another 2 cents per pound to get the pipe to GLPT’s customers. Thus, Komansky argues, internal production results in an 7.3 cent per pound loss, or \$87,600 per year assuming 1.2 million pounds of pipe.
As Rubin scans these figures, he smiles as he notices that Komansky used Rubin’s sales estimates and annual sales probabilities. He wonders, though, how accurate the accountant’s numbers really are. For one thing, the estimates are based on the “most likely” sales figure and do not consider the other sales possibilities. In addition, Rubin questions the appropriateness of including depreciation, given that it is a non-cash item. For these and other reasons, he decides to rethink the figures the
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1592 words | 7 pages | 688 | 2,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.938918 |
https://www.jiskha.com/display.cgi?id=1353690213 | 1,503,538,474,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886126017.3/warc/CC-MAIN-20170824004740-20170824024740-00590.warc.gz | 943,008,430 | 3,578 | # math
posted by .
if 3 bananas cost £1.14 how much will 5 of them cost? and how do ypu work it out?
• math -
if 3 bananas cost £1.14 how much will 5 of them cost? and how do you work it out?
• hannah - math -
Hannah , you re-posted this same question at 2:15 for the third time, when you can clearly see that I answered your same question, posted just below this, at 1:04.
I don't understand, don't your read replies to your postings ??
## Similar Questions
1. ### Math
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2. ### Algebra
three apples and four bananas cost \$4.85,three apples and ten bananas cost \$8.75, find the cost of an apple
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if 3 bananas cost £1.14 how much will 5 of them cost?
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More Similar Questions | 455 | 1,637 | {"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} | 3.578125 | 4 | CC-MAIN-2017-34 | latest | en | 0.928496 |
https://discover.hubpages.com/education/The-New-Pythagoreans-Modern-Mathematicians | 1,669,791,987,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710733.87/warc/CC-MAIN-20221130060525-20221130090525-00106.warc.gz | 255,014,584 | 74,953 | # The New Pythagoreans: The Secret Society of Modern Math
## Donald Duck in Mathmagic Land
When I was a child, I remember seeing a Disney short about mathematics called Donald in Mathmagic Land. In it, I learned about the Pythagoreans, a secret math society from Ancient Greece who believed that all the universe consisted of countable numbers.
Pythagorus, the founder of the society, never published any mathematical work. Or at least, nothing survives that can be attributed to him. Pythagorus started the group after discovering that musical harmony at its root was a ratio of whole numbers. Pythagorus wondered if all orderings of the universe were in truth ratios of countable numbers (what today we would call rational numbers) and formed his society with the task of unlocking these secrets.
The story is that Pythagorus was passing a blacksmith when he heard what sounded like musical notes each time the blacksmith hammered on his anvil. Pythagorus stopped in to investigate and discovered that that the blacksmith had three anvils of different sizes. One anvil was 1/2 the size of the largest and the smallest was 2/3 the size of the largest.
## Modern Mathematics: A New Secret Society
On the surface, mathematics seems to be in the open. After all, it's required study in both high school and college. But, deeper down, if you think about it, it is a secret society.
On the surface, mathematics is about problem solving. You are given a formula, then a word problem, and then you must use the formula to solve the world problem. The goal of mathematics seems to be to challenge your problem solving skills and logic.
The way that mathematics is taught, it is perhaps very surprising to learn that the problem solving aspect has little or no interest to the professional mathematician. Indeed, the basic mathematics of school: algebra, geometry, trigonometry, and calculus, are for the most part, completely ignored. They are after all well established and not in need of expansion.
The technique of interest to the mathematician is the mathematical proof. While on the surface, math seems to be about intellectual development and problem solving, it is for the practitioner, a search for underlying patterns and harmonies. Mathematics, the way it is studied by mathematicians, is really about beauty and truth. While it may appear to be about engineering and accounting, mathematics really has more in common with art and science.
## Math Papers: A secret code
It is as if all math papers are encoded in unfathomable mathematical terminology. According to the Disney documentary, each Pythagoreans need to show the symbol of the pentagram in order to gain admittance to Pythagorean gathering. In today's, mathematics, to gauge the underyling logic requires far more substantial knowledge.
The reason that the Pythagoreans chose the pentagram is for me very interesting. The ratio of the sides of the pentragram represent the golden ratio. The details of this ratio can be found here.
Professional mathematicians today seem to feel little or no obligation to share their knowledge with the uninitiated. Indeed, it seems to me that the true wonders of mathematics, today, are only communicated to other mathematicians. If you are not a member of the Modern Pythagoreans, then you are not invited to share in their wisdom and their insights.
There is a coterie of brave souls who try to translate mathematics for the general reader. They try to write on topics that will appeal to the general reader and share some of the amazing ideas which has been part of the last two hundred years of mathematics. Unfortunately, because of the math drillings in school, they have a difficult market. Few people are open to hearing about math voluntarily.
## Making Math Interesting
This raises an important question: why can't a math problem be easy and then interesting? Science, when taught by a gifted teacher, can be about the fascinating experiments that challenge our intuitions and reveal the often surprising physical world. Why can't mathematics be presented in the same way?
What if mathematics was taught with the goal of making the underlying ideas easier to understand. Consider this example:
Is it possible to create a three-dimensional one-sided figure?
This is not meant as a trick question. It is really an effort to challenge our intuitions about three-dimensional space. The answer, which should be as obvious as cat is spelled c-a-t, is yes and it's called a Moebius Strip.
Here's another question that I would like to see as a fundamental part of mathematics eduation:
What's wrong with this proof that 1=2?
Scroll to Continue
1. Let a=b
2. a2 = ab
3. 2a2 = a2 + ab
4. 2a2 - 2ab = a2 - ab
5. 2(a2 - ab) = 1(a2 - ab)
6. 2 = 1
The answer is that we used division by zero. If a=b, then a2 - ab=0 and to get from step #5 to step #6, we did something which is not allowed in mathematics. In other words, all we proved was that 2*0 = 1*0.
Here's the last one:
If x2 = 2, can we represent x as the ratio of two whole numbers?
1. Let (a/b)2 = 2 where a,b are whole numbers
2. Let a,b be in the simplest terms so that a,b do not have any common factors (ie the simplest term for 2/6 is 1/3)
3. So a2/b2 = 2 which means a2 = 2b2
4. Since 2 divides a2, we know that a is even (an odd*odd = odd)
5. So, there exists c such that a=2c.
6. (2c)2 = 2b2
7. So, after dividing 2 from both sides, we get 2c2 = b2
8. And there is our contradiction. Do you see it?
9. If 2 divides b2, then b is also even so 2 divides both a,b but by step #2, we assumed that a,b were in the simplest terms.
For mathematicians, this is enough to prove that the square root of 2 is an irrational number (that is, it cannot be represented by a ratio of two whole numbers).
This last proof was especially significant to the Pythagoreans. Remember, they believed that all the world could be represented as the ratio of two whole numbers. In other words, they believed that all numbers are rational. The existence of irrational numbers disproves this hypothesis.
If we wished to, we could teach mathematics the same way as we teach science. We could focus on understanding the details and reasoning behind a proof rather than focusing solely on problem solving.
## A Mathematician's Challenge
Now, the secret society nature of mathematics is completely unintentional by mathematicians. They would love to share their learning and insights. My main point is that we need to have math appreciation classes in addition to the problem solving classes.
cute stuff 69! on March 07, 2013:
THIS HELPED ME A lot THANKS SOOOOO MUCH HUB PAGES!
Ireno Alcala from Bicol, Philippines on April 14, 2012:
It's like Functions during our fourth year in high school. Anything is possible in Mathematics. We can formulate our own equation, just like you did here.
I'm still learning, just like the others.
Manna in the wild from Australia on December 20, 2011:
Voted up and interesting. You explain things well and this generated some interesting comments.
Pathyy on December 19, 2011:
great this information
larryfreeman (author) from Fremont, CA on July 27, 2010:
Hi Simone,
I'm glad that you enjoyed this hub. I think that when mathematics is presented in its natural form, it becomes more like an architectural structure or a Bach symphony and less like an engineering problem.
Simone Haruko Smith from San Francisco on July 24, 2010:
Wow. As a complete math dunce, it's fascinating to read all this background! I feel like I've been missing out on something really amazing.
larryfreeman (author) from Fremont, CA on January 19, 2010:
Hi HabMath,
HABMATH on January 16, 2010:
I read the comic "Donald in Mathmagic Land" as a first grader. I loved it then ( and still do) and it did get me interested in math.I am presently writing a hub called: " how math really works" I would love your comments on it. I would like a little patience on your part because ( like many mathematicians) I am a terribly slow writer
larryfreeman (author) from Fremont, CA on August 27, 2009:
Hi Deborah,
My father was a computer engineer in the 70's so I got into computers through him.
I consider myself more a computer geek than a mathematician. For me, mathematics is more of a hobby.
I got a bit deeper into mathematics because I was bothered about not understanding why calculus works. In college, I learned the basic rules about derivatives and integration but I had skipped the Fundamental Theorem of Calculus.
Once I started getting to understand the classic mathematical proofs, I was surprised how interesting I found them.
I am very glad that you enjoyed my hubs.
Deborah-Lynn from Los Angeles, California on August 27, 2009:
Hi LarryFreeman,
I liked your Hub and the interesting comments it has generated, was it Disney that inspired you to become a mathmetician? I saw those same cartoons, but I was not inspired, in fact I learned only enough math to get me through college chemistry and organic chemistry, only because I was only motivated to get a high G.P.A. in college. I wish I saw and felt what others do regarding math, not having that love of numbers makes me very aware that I am not a part of the "secret society". Thank you for still including me by allowing me access to your mathematical insights through reading your wonderful Hubs.
i from Earth on June 07, 2009:
"If I have 10 apples and each day I give you x apples, how long before the apples are gone?" Gone where?....to the hell?......do I know you? :) In reality it depends on our choices: If you gave me 10 apples,- I can recycle them 100% or little less, but keep the seeds. If you didn't give me any apples (0 apples),- you don't have many choices. Only recycle them 100% or little less, or keep the seeds?
What do you chose? :)
Jmell from El Paso, Texas, USA on May 19, 2009:
hi Larry. I'm no mathematician, not even close, but I loved your hub and agree whole heartedly that Math should be fun when learning.
Peter from Australia on May 02, 2009:
G'day Larry I'm still working on that apple one.
I'm sure there is an answer hidden away somewhere ?
highway star from Mostly Seattle, Amsterdam and Milan on February 21, 2009:
A really great hub!
larryfreeman (author) from Fremont, CA on February 20, 2009:
Hi Packerpack,
Thanks very much for nice comments. I am glad that you enjoyed the hub.
-Larry
Om Prakash Singh from India, Calcutta on February 20, 2009:
This was a really great Hub. Though I was able to figure out both the problem you had suggested, no not because I am a good mathematician but because I had come across this problem earlier. :)
But more one the serious note, yes you are correct it is very important to have classes that teaches about the passion of mathematics. It should be made a subject that should be learned out to passion and then only students will be able to understand the real power of mathematics. I no math geek but I know how important it is. Thanks for the Hub! It is nice to have you around and nice to be your fan!
Shalini Kagal from India on September 26, 2008:
Great hub! If only we could guide our children through that magical, mystical door and instill in them a love for mathmatics!
larryfreeman (author) from Fremont, CA on September 25, 2008:
Great question. My oldest child is 5 and does not yet understand division but he's getting close. :-) It sounds like your child is very smart!
Here's what I plan to tell my son when he is old enough. In mathematics, there are plenty of mysteries. Most of these mysteries surround very tough math problems. Some, such as division by zero,are more difficult than they appear.
For example, in most division, you can undo it. if 5* 6 = 30, then 30/5 = 6 and 30/6=5. If we have 5x = 30, then we know that x =6.
But with zero, we have a serious ambiguity. 5*0=0 and 10*0=0. If I say that 0x=0 then x can be any number we want. There's no way to recover the value that is multiplied by zero.
In any division problem, there are conditions. If I have 10 apples and each day I give you x apples, how long before the apples are gone? I can't give you 11 apples since I only have 10 and likewise if I don't give you any apples (0 apples), then there will never be a time when the apples run out so in such a situation, there is no answer.
-Larry
Aya Katz from The Ozarks on September 25, 2008:
Okay, Larry, great hub and I've given you a thumbs up.
But now, supposing your elementary school child doesn't accept that dividing by zero is not allowed. Supposing the child asks: "If zero is just a number, like any other number, and we can add and subtract and even multiply by zero, why can't we divide?" What would you say then? | 3,019 | 12,886 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.969453 |
https://metanumbers.com/13522 | 1,624,087,739,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00620.warc.gz | 357,464,998 | 10,805 | ## 13522
13,522 (thirteen thousand five hundred twenty-two) is an even five-digits composite number following 13521 and preceding 13523. In scientific notation, it is written as 1.3522 × 104. The sum of its digits is 13. It has a total of 2 prime factors and 4 positive divisors. There are 6,760 positive integers (up to 13522) that are relatively prime to 13522.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 13
• Digital Root 4
## Name
Short name 13 thousand 522 thirteen thousand five hundred twenty-two
## Notation
Scientific notation 1.3522 × 104 13.522 × 103
## Prime Factorization of 13522
Prime Factorization 2 × 6761
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 13522 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 13,522 is 2 × 6761. Since it has a total of 2 prime factors, 13,522 is a composite number.
## Divisors of 13522
1, 2, 6761, 13522
4 divisors
Even divisors 2 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 20286 Sum of all the positive divisors of n s(n) 6764 Sum of the proper positive divisors of n A(n) 5071.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 116.284 Returns the nth root of the product of n divisors H(n) 2.66627 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 13,522 can be divided by 4 positive divisors (out of which 2 are even, and 2 are odd). The sum of these divisors (counting 13,522) is 20,286, the average is 507,1.5.
## Other Arithmetic Functions (n = 13522)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 6760 Total number of positive integers not greater than n that are coprime to n λ(n) 6760 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1604 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 6,760 positive integers (less than 13,522) that are coprime with 13,522. And there are approximately 1,604 prime numbers less than or equal to 13,522.
## Divisibility of 13522
m n mod m 2 3 4 5 6 7 8 9 0 1 2 2 4 5 2 4
The number 13,522 is divisible by 2.
• Semiprime
• Deficient
• Polite
• Square Free
## Base conversion (13522)
Base System Value
2 Binary 11010011010010
3 Ternary 200112211
4 Quaternary 3103102
5 Quinary 413042
6 Senary 142334
8 Octal 32322
10 Decimal 13522
12 Duodecimal 79aa
16 Hexadecimal 34d2
20 Vigesimal 1dg2
36 Base36 afm
## Basic calculations (n = 13522)
### Multiplication
n×i
n×2 27044 40566 54088 67610
### Division
ni
n⁄2 6761 4507.33 3380.5 2704.4
### Exponentiation
ni
n2 182844484 2472423112648 33432105329226256 452068928261797433632
### Nth Root
i√n
2√n 116.284 23.8239 10.7835 6.70206
## 13522 as geometric shapes
### Circle
Radius = n
Diameter 27044 84961.2 5.74423e+08
### Sphere
Radius = n
Volume 1.03565e+13 2.29769e+09 84961.2
### Square
Length = n
Perimeter 54088 1.82844e+08 19123
### Cube
Length = n
Surface area 1.09707e+09 2.47242e+12 23420.8
### Equilateral Triangle
Length = n
Perimeter 40566 7.9174e+07 11710.4
### Triangular Pyramid
Length = n
Surface area 3.16696e+08 2.91378e+11 11040.7
## Cryptographic Hash Functions
md5 468cbac056133a996283cca7e2976336 620fcbe5a07a0a95b66f1a54205f3c7399aeec93 085a39668d5ccfdbd15d6b67d0449006ac79daaff7e3b53526d331a1bcda6900 6bb9d9f857c9d43b7b8885ce95119c807dab115b7f2d23af8bcb6db019c6ff77b2b7d6fa7385b3f0c00f6b3bbba78d41546ca4bf610351f2a05c27bcd7e1e3aa 145f6d713916371eb037434d7086af24a131e484 | 1,450 | 4,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2021-25 | latest | en | 0.808054 |
www.mavenec.com | 1,506,060,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688671.43/warc/CC-MAIN-20170922055805-20170922075805-00678.warc.gz | 511,381,916 | 24,875 | Why Should You Invest In Conversion Rate Optimization?
When you think about Conversion Rate Optimization you probably think in terms of increases in your sales or the number of leads you get. That’s perfectly right, and it’s pretty simple math: if you have 10,000 visitors and a 1% conversion rate, you get 100 conversions. If every conversion is worth \$1,000, you will generate \$100,000. If you increase the conversion rate by 50% and get a 1.5% conversion rate – you get 150 conversions / clients / leads. With 50 conversions more you earn an additional \$50,000.
So you get 50% more conversions, and at the same time – a 50% increase in revenue from your website. Darn simple.
But in fact there is much more to it than that. I’ll show why the conversion rate is much more important and why it has the power to transform your online business.
Why conversion rate optimization is the best choice to increase your profits
To give you a better picture, let’s assume that you have a clear conversion goal. You would like to gain more profit from your website (of course, you would).
Every website generates revenue in the same way: the user count multiplied by the conversion rate and the customer lifetime value.
So if you want to achieve your goal and boost your revenue you can focus on one of these metrics:
• Get more users
• Get a conversion rate lift
• Increase the Customer Lifetime Value
You can improve every one of these 3 elements to get more revenue. Most people choose a strategy of getting more users (invest in traffic acquisition). In fact, this is not always the best way: usually websites are like leaky buckets. You can pour twice as much water into it but it will still be leaking.
If you are not made of money (you have a limited budget for your website), it’s better to focus strongly on conversion rate optimization. Now I will show you why.
Conversion rate optimization gives you ongoing results
Let’s go back to our example: we have got 10,000 visitors a month, a 1% conversion rate, and every conversion is worth \$1,000. You generate \$100,000 every month.
Now you want to grow by \$50,000. So, you need 50 more conversions to achieve your goal.
If you pay \$2,000 for 5,000 visitors more, to get 50 conversions more every month, you will have to spend 12 x \$2,000, so \$24,000 annually. You have to spend money on traffic every single month to get traffic. And what if you run out of money? You will be back to 100 conversions a month, so you won’t achieve your goal.
But if you invest “3 months” worth of money (\$6,000) into conversion rate optimization and increase your conversion rate from 1% to 1.5% just once, the results will be ongoing. The conversion rate won’t change for the next 12 and you will receive “earned” profit every single month (50 conversions more every month). So, once invested, you get your revenue rise up to \$50,000 a month consistently.
That means you have saved ¾ of your money for the same results! And that is money you can spend on other things, like promotion and boosting your company profits.
Conversion rate optimization pumps up your profit margins
When increasing the conversion rate you bear almost no additional monthly costs (apart from a one-time investment in lifting the conversion rate).
When you grow your revenues with conversion rate optimization, many of the costs will stay the same, such as sales and other general costs like electricity or telephones. The only costs that are going to go up are the costs of goods sold, as you won’t spend even one additional dime on advertising.
Thanks to this, more of your additional revenue can go straight to profit, lifting your profit margins.
Gain a competitive advantage
Of course, you could always keep on increasing traffic at all costs in a race with your competitors. But that will definitely end up costing you an arm and a leg.
Optimizing your conversion rates, however, can gain you a competitive advantage because you don’t need that much traffic to beat your competitors. If you have substantial conversion rates, your competition may get twice as many users, but you will still get more sales or leads.
Another competitive advantage is the “process of constant growth” you can make by combining conversion rate optimization with traffic acquisition.
1. Higher conversion rates give you higher profit margins
2. Which means you can spend more on getting individual users (for example on promotions such as SEO or PPC campaigns)
3. Which means you can beat your competitors by spending more on traffic
Then you can increase your conversion rates, and the whole process repeats itself.
Companies that are aware of the opportunities offered by conversion rate optimization beat all of their competitors on selected keywords (SEO and PPC).
But there is more. Is your marketing partly based on affiliation? If so, then conversion rate optimization is a great opportunity to get your partners to be more willing to put your link on their website.
If you pay your partners per conversion they will definitely get a better deal with you, than with someone who has a lower conversion rate. It’s simple: they can count on bigger profits with the same number of users they send to your website.
No redesign threat
Last but not least: if you are considering a redesign, conversion rate optimization can come to the rescue.
“Oh, we completely redesigned our website, and we’re so happy: the project was such an easy ride, we finished before the deadline, we were under budget, and from the first day we had conversion rate lifts”. Let’s guess. Is that how a redesign really goes?
No, of course not. It’s a fairytale.
Because if you’ve ever been through a website redesign, you know it almost always turns into a nightmare. Redesigns are usually over-budget, late, and end up costing you even more because they invariably lead to big conversion rate drops. But this will not happen, as long as you choose conversion rate optimization over redesign.
Conversion rate optimization is a process that involves implementing small changes to your site, which will systematically increase your conversion rates. It’s not about taking a gamble, for example, on whether a red button is better than one that is green. It’s based on statistics: because thanks to A/B testing you know whether your users will convert more after a change, or not. It’s pure science.
If you do A/B tests step by step you could notice that 3 changes gave you a lift in conversion rates and the next 4 – a drop. But if you just do a classic redesign you won’t spot these details, because what you get is only the final result: a drop. You could lose the opportunity to make 3 great changes. Source: http://www.widerfunnel.com/conversion-rate-optimization/evolutionary-site-redesign
So you don’t need to worry about losing money on a potentially hapless redesign – you can change your website as much as you want, but in ways which will produce more and more profit.
Ready to profit by boosting your conversion rate?
There you have it. Conversion Rate Optimization will not only grow your revenues, but will impact your online business in more dramatic ways. Firstly, you will get constant revenue growth but even more than that: a much bigger part of the money you make will become genuine profit. You will beat your competitors by spending more dollars on individual users. And last but not least – you don’t have to risk bad redesigns, which can result in conversion drops.
There is simply no other way to improve your website and get more sales, transactions or leads without spending a dime on traffic acquisition. It’s not like you shouldn’t invest in increasing traffic on your website or lift customer lifetime value. But the conversion rate is crucial for your website’s effectiveness. That’s why there is no time to waste: start optimizing immediately, and get more profit!
If you want to gain a competitive advantage by increasing your conversion rates and grow your revenues and profits without spending an additional dime on advertising, contact Mavenec today to learn more about our approach to conversion rate optimization.
Psst… if you need any help, check our Free Ultimate Conversion Rate Optimization Toolkit. We’ll show you step by step how to improve your website using best practices and tools. We have been working on conversion rates for our biggest clients for over 5 years and we love to share our knowledge. So why not make good use of it? And take the first steps to making your website insanely effective :). Check it out!
Author: Elżbieta Brzóz
At Mavenec I am responsible for PR and Marketing. I am in charge of our content strategy, blog, social media channels, external partnerships and events. Being so close to digital analytics and conversion rate optimization enables me to make data-driven decisions and single out the most effective marketing efforts.
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Grow Your Conversion Rate
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Page 1 of 2
• posted on May 27, 2010, 3:03 pm
I've had the moment to work on 440 volt roof top heating and AC.
Once in a while, need some power to run a 120 volt device. Is it 120 volts from one leg to ground?
(Yeah, I could just check it. But I was on the roof yesterday, and not sure how long until I get to another roof.)
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Christopher A. Young
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• posted on May 27, 2010, 3:23 pm
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• posted on May 27, 2010, 9:22 pm
Oh, thank you. I didn't know.
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Christopher A. Young
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• posted on May 28, 2010, 5:31 pm
On May 27, 5:22 pm, "Stormin Mormon"
It is entirely possible that the 440 volts is from a delta configured transformer that is not ground referenced or is corner grounded. Point is that one shouldn't assume that the transformer that is supplying the current is Y configured. -- Tom Horne
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• posted on May 27, 2010, 4:05 pm
On May 27, 10:03 am, "Stormin Mormon"
No.
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• posted on May 27, 2010, 9:22 pm
Oh, thank you. I didn't know.
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Christopher A. Young
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• posted on May 27, 2010, 4:33 pm
On Thu, 27 May 2010 11:03:40 -0400, "Stormin Mormon"
The nominal 480 wye to neutral is 277 volts. That is why you see 277v lighting in big buildings
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• posted on May 27, 2010, 9:21 pm
Thank you. Saves me building a device, and not have it work.
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Christopher A. Young
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• posted on May 27, 2010, 4:48 pm
Stormin Mormon wrote:
120 volts is needed there. Don't you have a meter?
--
LSMFT
I haven't spoken to my wife in 18 months.
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• posted on May 27, 2010, 9:23 pm
Where I was working, there was a GFCI convenience outlet on the side (120 VAC) but that got me to thinking. And, I'm learning.
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Christopher A. Young
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• posted on May 27, 2010, 8:03 pm
On May 27, 11:03 am, "Stormin Mormon"
How old is this roof top unit... Seriously 440 volts ? Are you sure that it isn't 480 volts ?
I asked the _age_ of the unit because there is supposed to be a service outlet for 120 volt powered tools for maintaining the roof top equipment installed somewhere near the unit...
And no, with a 3 phase 480 volt system it is typically 277 volts on each leg...
~~ Evan
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• posted on May 27, 2010, 9:25 pm
How old is this roof top unit... Seriously 440 volts ? Are you sure that it isn't 480 volts ?
CY: The unit looks maybe ten years old. It may well been 480, and I typed it wong.
I asked the _age_ of the unit because there is supposed to be a service outlet for 120 volt powered tools for maintaining the roof top equipment installed somewhere near the unit...
CY: Actually, this one did have a convenience outlet. But,t hat gets me thinking.
And no, with a 3 phase 480 volt system it is typically 277 volts on each leg...
CY: Thanks.
~~ Evan
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• posted on May 27, 2010, 9:29 pm
On 5/27/2010 1:03 PM Evan spake thus:
Right; we should all tighten up our terminology, nominal voltage-wise.
It's 120, 240 and 480. *Not* 110, 220 and 440. (Not to mention all those weird older numbers that were used, like 117 volts: WTF did that come from???)
--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.
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• posted on May 28, 2010, 12:36 pm
">>> I've had the moment to work on 440 volt roof top heating and AC.
*I have found that the actual voltage varies by power company. One company in NJ supplies 460/265 volts while another supplies 480/277. It is quite possible that Stormin is dealing with 440 volts.
When I worked for my father decades ago he did a lot of industrial work and we quite often had to add buck and boost transformers to equipment because the juice from the power company did not match the nameplate requirements for a machine.
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• posted on May 28, 2010, 12:55 pm
John Grabowski wrote: ...
More than likely it's owing to variations in distribution or even local voltage drops rather than a difference in system generation setpoints between utilities.
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• posted on May 28, 2010, 1:31 pm
*Many years ago I had an issue with the voltage at a hospital that I was working at. It seemed that there was low voltage everywhere. I had the power company come in and monitor the building for a week, but they assured me that they only supply 460/265 and that is what we were getting. The ballasts for the fluorescent lighting were rated for 277 volts. I had to go to every transformer in the building and change the taps so that we could get 120 volts out of an outlet instead of the 102-105. I also had to install a few buck and boost transformers on some A/C equipment.
Of course variation in distribution, voltage drop and distance from transformers does have an effect.
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• posted on May 28, 2010, 6:56 pm
On 5/28/2010 5:36 AM John Grabowski spake thus:
Don't doubt you at all; however, the *nominal* voltages are still 120/240/480. What you're describing are basically under-voltage situations.
--
The fashion in killing has an insouciant, flirty style this spring,
with the flaunting of well-defined muscle, wrapped in flags.
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• posted on May 28, 2010, 8:26 pm
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• posted on May 28, 2010, 11:00 pm
More likely, I forgot the number, and typed it wong. I don't deal with 480/3 very often.
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Christopher A. Young
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• posted on May 28, 2010, 5:02 am
Evan wrote:
Usually, equipment is rated for a range of voltages. A lot of HVAC equipment has different taps on the 24 volt control transformer to pick a different primary voltage. I've seen some air handlers that had a tap on the control transformer for 110 volts and another for 120 volts. Of course on larger package units there will be taps on the primary of the control transformer to choose between 208 or 240. I've seen a lot of blower motors that show a rating of 100-125 VAC 60 Hz on the nameplate.
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posted by .
With acceleration of 2500 m/s2, how long will it take to rreach speed of light
• physics -
acceleration is change in velocity over time, or:
a = (v - v0)/t
where v = final velocity in m/s
v0 = initial velocity in m/s
t = time in s
assuming that it is initially at rest, v0 = 0,, the final velocity, v, is the speed of light, which is approximately 3.0 x 10^8 m/s
substituting:
2500 = (3.0x10^8 - 0)/t
now, solve for t.
hope this helps.
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Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Apr 7th 2010, 03:52 AM
EinStone
Holomorphic extension of Dirichlet Series
Define $\displaystyle L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $\displaystyle L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.
Note that $\displaystyle L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $\displaystyle L : \mathbb{C} \rightarrow \mathbb{C}$
• Apr 7th 2010, 11:13 AM
chiph588@
Quote:
Originally Posted by EinStone
Define $\displaystyle L : \{s \in \mathbb{C}| Re(s) > 0\} \rightarrow \mathbb{C}$ by $\displaystyle L(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$.
Note that $\displaystyle L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $\displaystyle L : \mathbb{C} \rightarrow \mathbb{C}$
Tell me more i.e. what else do you know?
Are you familiar with the derivation of the analytic continuation of $\displaystyle \zeta(s)$? Because this series has a similar continuation.
If not, that's ok because I don't think you're asking for the functional equation itself, just that it exists.
Before I make any head way on this problem, what are your thoughts on how to approach it?
• Apr 8th 2010, 12:12 AM
EinStone
What I know is how to show that the Gamma Function is meromorphic on $\displaystyle \mathbb{C}$. Then by evaluating $\displaystyle \Gamma(s) * \zeta(s)$ in a specific way, one can show that $\displaystyle \Gamma(s)*\zeta(s)$ is also meromorphic on $\displaystyle \mathbb{C}$, and therefore $\displaystyle \zeta(s)$ has to be meromorphic as well.
I actually thought that $\displaystyle \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$ can somehow be extended holomorphically to $\displaystyle \mathbb{C}$ similarly to the Gamma function. But thats all I can think of.
• Apr 8th 2010, 01:12 AM
Laurent
Quote:
Originally Posted by EinStone
Note that $\displaystyle L(s) = \frac{1}{\Gamma(s)} \int_0^\infty \frac{t^{s-1}}{e^t+e^{-t}}dt$
Show that L can be extended to an entire function $\displaystyle L : \mathbb{C} \rightarrow \mathbb{C}$
I guess you know that $\displaystyle \frac{1}{\Gamma}$ is an entire function. Then all you have to prove is that the integral defines an entire function as well. There is a general theorem for proving that functions of the form $\displaystyle s\mapsto\int f(s,t)dt$ are analytic. If you know it (which is probably the case), then you know what left you have to do... and this shouldn't be too hard (very similar to Gamma). What did you already try this way?
• Apr 8th 2010, 02:24 AM
EinStone
Actually I don't remember the proof exactly, it was very strange. Also I don't know the theorem you are talking about :(. What can I do?
• Apr 8th 2010, 01:08 PM
EinStone
I still need help!
Meanwhile I asked myself why the integral converges in the first place for $\displaystyle Re(s) > 0$ ?
• Apr 8th 2010, 01:39 PM
Laurent
Quote:
Originally Posted by EinStone
I still need help!
Actually I was wrong: the integral is only defined when $\displaystyle {\rm Re}(s)>0$... It is not simple to extend it to $\displaystyle \mathbb{C}$. Like Chiph588@ said, you should repeat arguments used for zeta.
Quote:
Meanwhile I asked myself why the integral converges in the first place for $\displaystyle Re(s) > 0$ ?
We have $\displaystyle \frac{|t^{s-1}|}{e^t+e^{-t}}\leq |t^{s-1}e^{-t}|$, so if you know that the integral defining Gamma converges, then so does this one.
• Apr 8th 2010, 02:53 PM
EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\displaystyle \frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $\displaystyle B_k$ are the Bernoulli numbers.
So I think I need a similar expression for $\displaystyle \frac{t}{e^t+e^{-t}}$, but I don't see it.
• Apr 8th 2010, 03:08 PM
chiph588@
Quote:
Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\displaystyle \frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $\displaystyle B_k$ are the Bernoulli numbers.
So I think I need a similar expression for $\displaystyle \frac{t}{e^t+e^{-t}}$, but I don't see it.
$\displaystyle \frac{t}{e^t+e^{-t}} = \frac12 t \,\text{sech}(t) = \sum_{n=0}^\infty \frac{E_{2n}t^{2n+1}}{2(2n)!} \;\; |t|<\frac\pi2$, where $\displaystyle E_k$ are the Euler numbers.
I don't know if this will help at all though, since $\displaystyle |t|<\frac\pi2$...
• Apr 8th 2010, 04:35 PM
chiph588@
Quote:
Originally Posted by EinStone
Ok, So to prove that Zeta is holomorphic everywhere, we used $\displaystyle \frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k$ where $\displaystyle B_k$ are the Bernoulli numbers.
What all was said to show $\displaystyle \zeta(s)$ is meromorphic everywhere?
• Apr 9th 2010, 02:33 AM
EinStone
Ok here is the proof for Zeta being meromorphic everywhere.
$\displaystyle \frac{t}{e^t-1} = \sum_{k=0}^\infty \frac{B_k}{k!} t^k = 1 - \frac{t}{2} + \frac{t^2}{12} + 0t^3 - \frac{t^4}{720} + \cdots$ where $\displaystyle B_k$ are the Bernoulli numbers.
Fix $\displaystyle n > 0$, $\displaystyle f_n(t):= \sum_{k=0}^n (-1)^k \frac{B_k}{k!} t^k = 1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots + \frac{B_n}{n!}t^n$ (all terms even)
Now for Re(s) > 1:
$\displaystyle \Gamma(s) * \zeta(s) = \int_0^\infty \frac{t e^t}{e^t-1} e^{-t} t^{s-2}dt = \int_0^\infty (\frac{t e^t}{e^t-1} - f_n(t))e^{-t} t^{s-2}dt$ $\displaystyle + \int_0^\infty f_n(t) e^{-t} t^{s-2}dt = I_1(s) + I_2(s)$
$\displaystyle \frac{t e^t}{e^t-1}$ is holomorphic at t = 0, has Taylor expansion: $\displaystyle 1 + \frac{t}{2} + \frac{B_2}{2!}t^2 + \frac{B_4}{4!}t^4 + \cdots = \frac{t}{e^t-1} +t$
hence $\displaystyle \frac{t e^t}{e^t-1}-f_n(t) = O(t^{n+1})$ near t = 0.
$\displaystyle \rightsquigarrow$ Integrand of $\displaystyle I_1$ near t = 0 is $\displaystyle \approx O(t^{n+1}t^{s-2}) = O(t^{n+s-1})$. So $\displaystyle I_1$ converges if Re(n+s-1) > -1, i.e. if Re(s) > -n.
For $\displaystyle I_2$: $\displaystyle I_2(s) = \int_0^\infty (1 + \frac{t}{2} + \sum_{k=2}^n \frac{B_k}{k!}t^{k}) e^{-t} t^{s-2}dt = \Gamma(s-1) + \frac{1}{2}\Gamma(s) + \sum_{k=2}^n \frac{B_k}{k!} \Gamma(s+k-1)$ , a finite sum of meromorphic functions on $\displaystyle \mathbb{C}$.
$\displaystyle \rightsquigarrow \zeta$ is meromorphic for Re(s) > -n, n arbitrary, hence $\displaystyle \zeta$ is meromorphic on $\displaystyle \mathbb{C}$.
• Apr 9th 2010, 04:34 AM
Laurent
I didn't know this proof; it is rather simple, and adaptable to many situations.
Notice that you only use the fact that there exists an expansion $\displaystyle \frac{t e^t}{e^t-1}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})=f_n(t)+O(t^{n+1})$ when $\displaystyle t\to 0$, regardless of the value of $\displaystyle a_n$ (or of the convergence of the series expansion).
Since $\displaystyle t\mapsto\frac{t e^t}{e^t+e^{-t}}$ is $\displaystyle \mathcal{C}^\infty$ (even analytic), there is also such an expansion $\displaystyle \frac{t e^t}{e^t+e^{-t}}=a_0+a_1t+\cdots+a_n t^n+O(t^{n+1})$ when $\displaystyle t\to0$ (this is Taylor-Young's theorem). And you can transpose the proof seamlessly.
It will show you that $\displaystyle L$ is meromorphic on $\displaystyle \mathbb{C}$, so you also have to justify it has no poles... That may require the values of the coefficients $\displaystyle a_n$ ; I'll think about it.
Sidenote: there is a justification missing in your proof, but maybe it was an implicit reference to a proof you did for Gamma. An argument is indeed necessary to justify that $\displaystyle I_1$ is analytic; it is of the form $\displaystyle \int_0^\infty f(t,s)dt$ where, for all $\displaystyle t>0$, $\displaystyle s\mapsto f(t,s)$ is analytic, but this is not sufficient to conclude that the integral itself is analytic. A "domination" is needed, like for the continuity or differentiability theorems under the integral sign. But it is simple to apply here, so I guess you were told it is just like for Gamma.
----
Addendum: in fact, the same method proves the analyticity on $\displaystyle \mathbb{C}$.
Indeed, we get an expression like $\displaystyle L(s)=\frac{1}{\Gamma(s)}{I_1(s)}+\sum_{k=0}^n a_k\frac{\Gamma(s+k-1)}{\Gamma(s)}$, for $\displaystyle {\rm Re}(s)>-n$, where $\displaystyle I_1(s)$ is analytic. We know that $\displaystyle \Gamma(s)\neq 0$, hence the first term $\displaystyle \frac{I_1(s)}{\Gamma(s)}$ defines an analytic function.
Furthermore, for $\displaystyle k\geq 1$, $\displaystyle \Gamma(s+k-1)=(s+k-2)(s+k-3)\cdots s \Gamma(s)$ (usual functional equation of $\displaystyle \Gamma$ used several times) hence $\displaystyle \frac{\Gamma(s+k-1)}{\Gamma(s)}=(s+k-2)(s+k-3)\cdots s$ is just a polynomial, hence it is analytic.
The only piece that is not analytic for $\displaystyle \zeta$ is the term $\displaystyle \frac{\Gamma(s-1)}{\Gamma(s)}=\frac{1}{s-1}$, which comes from $\displaystyle k=0$. (This proves that 1 is the only pole for $\displaystyle \zeta$; it is simple with residue 1).
However, for $\displaystyle L(s)$, the term $\displaystyle a_0$ is 0 because $\displaystyle \frac{te^t}{e^t+e^{-t}}$ is 0 at 0. Therefore, all terms are analytic. Thus $\displaystyle L$ is analytic on $\displaystyle {\rm Re}(s)>-n$ for all n, hence on $\displaystyle \mathbb{C}$. qed.
• Apr 9th 2010, 05:26 AM
EinStone
So I want to apply the same proof, but how do I find $\displaystyle I_1$ for L? Also to show that $\displaystyle I_2$ is meromorphic, don't I need to know the $\displaystyle a_n$?
• Apr 9th 2010, 05:29 AM
Laurent
Quote:
Originally Posted by EinStone
So I want to apply the same proof, but how do I find $\displaystyle I_1$ for L? Also to show that $\displaystyle I_2$ is meromorphic, don't I need to know the $\displaystyle a_n$?
All you have to do is replace $\displaystyle \frac{t e^t}{e^t-1}$ by $\displaystyle \frac{t e^t}{e^t+e^{-t}}$, and of course $\displaystyle \zeta(s)$ by $\displaystyle L(s)$. Everything is the same, except that $\displaystyle a_0=0$ for $\displaystyle L$ (hence the analyticity), but you don't need the other values (neither for zeta of for L), just the fact that they exist. They are constants, so they matter in no way.
• Apr 9th 2010, 05:36 AM
EinStone
Now I get it, its just the existence (even for Zeta) that is needed, which follows from analyticity.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last | 3,558 | 10,461 | {"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} | 3.28125 | 3 | CC-MAIN-2018-26 | latest | en | 0.774968 |
https://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Weber_(unit).html | 1,653,601,860,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662625600.87/warc/CC-MAIN-20220526193923-20220526223923-00398.warc.gz | 1,189,853,570 | 7,435 | # Weber (unit)
The weber (symbol Wb) is the SI unit of magnetic flux. It is named for the German physicist Wilhelm Eduard Weber.
The ninth Conférence Générale des Poids et Mesures (General Conference on Weights and Measures) ratified in 1948 a definition of weber based on Faraday's law for magnetic induction. Faraday's law connects electromotive force $\mathcal{E}$ (in volt) to the rate of change of magnetic flux $\Phi\,$ (in Wb/s) through one loop of conducting wire (a "turn"):
$\mathcal{E} = -\frac{d\Phi}{dt}$
If the magnetic flux Φ is linear in time—Φ changes with uniform rate—and if Φ = 0 at t = 0, then it follows that
$\Phi(t) = - t\, \mathcal{E}.$
This equation forms the basis of the formal definition, which reads:
One weber is the magnetic flux which, linking a circuit of one turn, would produce in it an electromotive force of 1 volt if it were reduced to zero at a uniform rate in 1 second. (Resolution 2, International Committee for Weights and Measures. 1946).[1]
Dimension: Wb = V⋅s = N⋅m⋅A−1 = kg⋅A−1⋅s−2⋅m2 = T⋅m2, where A stands for ampere, T for tesla, V for volt, and N for newton.
## Reference
1. BIPM brochure about SI. PDF page 52; paper page 144. | 360 | 1,188 | {"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} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.861783 |
https://thinktank.pmq.com/t/375-00-sq-ft/8716 | 1,713,459,036,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817222.1/warc/CC-MAIN-20240418160034-20240418190034-00510.warc.gz | 530,213,008 | 7,039 | # \$375.00 sq ft?
going back in the post s way back. i remember a post when people were asking about gross sales. the number came up\$375 sq ft . is a number for gross sales . so my place is pizza/bar 3,000 sqft would that be \$1.1 million.not to be stupid is that the potential of the place i cant imagine your first year that can be done. what kind of net could you get out of that number. 5-10%.population is 13,000, plus neighbor towns another 12,000
There really is not a formula that will predict sales based on sq ft. If getting to 1 million was as simple as expanding my place and putting in a bar I would have done it a long time ago. Formulas are great and should be used as an aid but its no substitute for going out in the neighborhood and seeing what other businesses are doing – not just pizza or restaurants.
Make a spreadsheet based on possible p&l’s – a performa. Determine your break-even point and at what point the investment makes sense. Then go see if other similar businesses in the area are coming close to that, exceeding or are not even close. You can determine that from their customer count and ticket avg based on what is one their menus.
A performa will tell you what kind of net you can get at those sales figures.
Sales per foot is more closely related to how much sales you CAN do than to how much you WILL do. It is also an excellent benchmark when evaluating a lease. (most businesses can do pretty well when occupancy expense is less than 10% of sales) In other words, if your sales are \$375 per foot a lease rate of \$35 per foot including rent and triple net expenses can be acceptable.
When all is said and done, lease rates migrate to a number that relects the business potential in the area. As the potential rises, more tenants will be willing to pay higher rents and will succeed. When potential declines, businesses will not sign leases at higher rates or tend to fail when they do so. So… you could take your annual lease rate with all expenses rolled in and multiply by 10 to get a number to use for a minimum annual sales goal. Clearly you will do better if you can get to 12 times that number or higher and you will be in danger of failure if you can not maintain 8 times.
Additionally, the suggestion to come to some understanding of what sales per foot other busineses in the area are doing is a good one. That is a good benchmark for custommer traffic which is the prime driver for a sit-down establishment.
thanks alot for the info greatly appreciated. the one thing i will have a great advantage in is i wont have any rent to pay. we own the building so rent wont be a problem the other stores will pay the mortgage.
Talk to your accountant. There are some very good reasons to pay market rent anyway. Rental income is passive income and not taxed the same way as earned income. If the business is loosing money after paying market rent, you might be better off knowng that. You can also write losses off against other income of the same category. You can always loan the money back to the business without interest. If that creates a negative balance sheet and this a problem, you add the money back in as capital instead.
Also, if/when the time comes to sell either the business or the building, having a consistent arms length relationship between the two where the business pays market rent is beneficial. It helps to establish fair value for both entities.
Also, you have to take downside into account. If your business fails (unpleasant thought, but one you have to include in your considerations) and there is an accumulated loss you can write it off. If you never paid the rent, you will lose that portion of the writeoff. | 797 | 3,695 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-18 | latest | en | 0.968697 |
https://www.varsitytutors.com/act_math-help/how-to-use-scientific-notation?page=2 | 1,643,205,719,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00670.warc.gz | 1,067,913,187 | 51,440 | # ACT Math : How to use scientific notation
## Example Questions
2 Next →
### Example Question #2 : How To Use Scientific Notation
What is the result when ,, is rounded to the nearest thousand and then put in scientific notation?
Explanation:
First, when we round to the nearest thousand we get 5, 679, 000 since we round up when the next digit is greater than 5.
Then, to put it in scientific notation, we arrange the digits so that a decimal point creates a number between 1 and 10. We get 5.679.
Then, we want the exponent of the 10 to be the number of times the decimal needs to move to the right. This is 6 times.
Thus, we get our answer.
### Example Question #3 : How To Use Scientific Notation
What is in scientific notation?
Explanation:
In order to write a number in scientific notation, you must shift the number of decimal places to get a number in the ones place.
Since the original number is a decimal, the exponent will need to be negative. This eliminates three answer choices.
In order to get into scientific notation with '5' in the ones place, you must shift the decimal over seven places.
Therefore, the final answer in scientific notation is .
### Example Question #11 : How To Use Scientific Notation
Which of the following is equal to ?
Explanation:
For this, you can handle the s separately from the coefficients. The s easily cancel:
Now, handle the coefficients like they are a separate fraction:
Notice, though, that :
Now, you need to get this into scientific notation for your answer. It is easiest ot think of it like this:
### Example Question #12 : How To Use Scientific Notation
Simplify .
Explanation:
The easiest way to work with a large number like this is to combine the factors and then the other coefficients. After that, you will have to convert the number into scientific notation to match the answer:
Combine the powers of :
Next, combine the other coefficients:
Now, rewrite the leading number like this:
Thus, you get:
### Example Question #13 : How To Use Scientific Notation
is equal to which of the following? | 468 | 2,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2022-05 | latest | en | 0.832164 |
https://help.scilab.org/docs/5.5.1/fr_FR/taucs_chget.html | 1,620,416,390,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00443.warc.gz | 329,096,237 | 7,845 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
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See the recommended documentation of this function
Aide de Scilab >> Interface avec UMFPACK (sparse) > taucs_chget
# taucs_chget
retrieve the Cholesky factorization at the scilab level
### Calling Sequence
`[Ct,p] = taucs_chget(C_ptr)`
### Arguments
C_ptr
a pointer to the Cholesky factorization (C,p : A(p,p)=CC')
Ct
a scilab sparse matrix (you get the upper triangle i.e. Ct is equal to C')
p
column vector storing the permutation
### Description
This function may be used if you want to plot the sparse pattern of the Cholesky factorization (or if you code something which use the factors). Traditionnaly, the factorization is written :
`P A P' = C C'`
with P' the permutation matrix associated to the permutation p. As we get the upper triangle Ct (= C'), in scilab syntax we can write :
`A(p,p) = Ct' * Ct`
### Examples
```// Example #1 : a small linear test system
A = sparse( [ 2 -1 0 0 0;
-1 2 -1 0 0;
0 -1 2 -1 0;
0 0 -1 2 -1;
0 0 0 -1 2] );
Cp = taucs_chfact(A);
[Ct, p] = taucs_chget(Cp);
full(A(p,p) - Ct'*Ct) // this must be near the null matrix
taucs_chdel(Cp)```
```// Example #2 a real example
stacksize(3000000) // the last PlotSparse need memory
// first load a sparse matrix
// compute the factorization
Cptr = taucs_chfact(A);
// retrieve the factor at scilab level
[Ct, p] = taucs_chget(Cptr);
// plot the initial matrix
xset("window",0) ; clf()
PlotSparse(A) ; xtitle("Initial matrix A (bcsstk24.rsa)")
// plot the permuted matrix
B = A(p,p);
xset("window",1) ; clf()
PlotSparse(B) ; xtitle("Permuted matrix B = A(p,p)")
// plot the upper triangle Ct
xset("window",2) ; clf()
PlotSparse(Ct) ; xtitle("The pattern of Ct (A(p,p) = C*Ct)")
// retrieve cnz
[OK, n, cnz] = taucs_chinfo(Cptr)
// cnz is superior to the realnumber of non zeros elements of C :
cnz_exact = nnz(Ct)
// do not forget to clear memory
taucs_chdel(Cptr)```
• taucs_chfact — cholesky factorization of a sparse s.p.d. matrix
• taucs_chsolve — solve a linear sparse (s.p.d.) system given the Cholesky factors
• taucs_chdel — utility function used with taucs_chfact
• taucs_chinfo — get information on Cholesky factors
• taucs_chget — retrieve the Cholesky factorization at the scilab level
• cond2sp — computes an approximation of the 2-norm condition number of a s.p.d. sparse matrix | 765 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2021-21 | latest | en | 0.529315 |
https://www.online-accounting.net/present-value-of-1-annuity-table/ | 1,713,014,906,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816734.69/warc/CC-MAIN-20240413114018-20240413144018-00456.warc.gz | 884,428,058 | 22,501 | # Present Value of \$1 Annuity Table
## Present Value of \$1 Annuity Table
For perpetuities, however, there are an infinite number of periods, so we need a formula to find the PV. The formula for calculating the PV is the size of each payment divided by the interest rate. The PV for both annuities -due and ordinary annuities can be calculated using the size of the payments, the interest rate, and number of periods. For an ordinary annuity, however, the payments occur at the end of the period. This means the first payment is one period after the start of the annuity, and the last one occurs right at the end. Another way to think of it is how much an annuity due would be worth when payments are complete in the future, brought to the present. The initial payment earns interest at the periodic rate (r) over a number of payment periods (n). PVIFA is also used in the formula to calculate the present value of an annuity. An annuity due arises when each payment is due at the beginning of a period; it is an ordinary annuity when the payment is due at the end of a period. A common example of an annuity due is a rent payment that is scheduled to be paid at the beginning of a rental period. The Present Value of Annuity table Calculator is used to calculate the present value of an ordinary annuity, which is the current value of a stream of equal payments made at regular intervals over a specified period of time. It is adjusted for risk based on the duration of the annuity payments and the investment vehicle utilized. Higher interest rates result in lower net present value calculations. Calculate the present value interest factor of an annuity (PVIFA) and create a table of PVIFA values. Create a printable compound interest table for the present value of an ordinary annuity or present value of an annuity due for payments of \$1. The present value interest factor of an annuity is useful when determining whether to take a lump-sum payment now or accept an annuity payment in future periods. Using estimated rates of return, you can compare the value of the annuity payments to the lump sum. The present value interest factor may only be calculated if the annuity payments are for a predetermined amount spanning a predetermined range of time. There are different FV calculations for annuities due and ordinary annuities because of when the first and last payments occur. Valuation of life annuities may be performed by calculating the actuarial present value of the future life contingent payments. It uses a payment amount, number of payments, and rate of return to calculate the value of the payments in today’s dollars. They can arise in loans, retirement plans, leases, insurance settlements, tax-related calculations, and so forth. This is because the value of \$1 today is diminished if high returns are anticipated in the future. There are some formulas to make calculating the FV of an annuity easier. The present value of an annuity due (PVAD) is calculating the value at the end of the number of periods given, using the current value of money. Sometimes, one may be curious to learn how much a recurring stream of payments will grow to after a number of periods. An annuity due is an annuity where the payments are made at the beginning of each time period; for an ordinary annuity, payments are made at the end of the time period. The formula for the present value of an annuity due, sometimes referred to as an immediate annuity, is used to calculate a series of periodic payments, or cash flows, that start immediately. An annuity is a series of payments that occur over time at the same intervals and in the same amounts. The fact that the value of the annuity-due is greater makes sense because all the payments are being shifted back (closer to the start) by one period. Moving the payments back means there is an additional period available for compounding. Note the under the annuity due the first payment compounds for 3 periods while under the ordinary annuity it compounds for only 2 periods. Likewise for the second and third payments; they all have an additional compounding period under the annuity due. Note also that the above formula implies that both the PV and the FVof an annuity due will be greater than their comparable ordinary annuity values.
• Calculate the present value interest factor of an annuity (PVIFA) and create a table of PVIFA values.
• Create a printable compound interest table for the present value of an ordinary annuity or present value of an annuity due for payments of \$1.
If annuity payments are due at the beginning of the period, the payments are referred to as an annuity due. To calculate the present value interest factor of an annuity due, take the calculation of the present value interest factor and multiply it by (1+r), with the variable being the discount rate. The discount rate used in the present value interest factor calculation approximates the expected rate of return for future periods.
## Understanding the Present Value of an Annuity
The moral is to save early and save often (and live long!) to take advantage of the power of compound interest. Here we use the same values as the PV of an annuity problem above to calculate PVwhen the payments are made at the end of the period (ordinary annuity) and at the beginning of the period (annuity due). Both annuities-due and ordinary annuities have a finite number of payments, so it is possible, though cumbersome, to find the PV for each period.
## Present Value Annuity Due Tables
The first and last payments of an annuity due both occur one period before they would in an ordinary annuity, so they have different values in the future. The present value (PV) of an annuity due is the value today of a series of payments in the future.
## Calculating Present and Future Value of Annuities
These annuities are called ordinary annuities (also known as annuities in arrears). The next graphic portrays a 5-year, 10%, ordinary annuity involving level payments of \$5,000 each. Notice the similarity to the preceding graphic – except that each year’s payment is shifted to the end of the year. This means each payment will accumulate interest for one less year, and the final payment will accumulate no interest! Be sure to note the striking difference between the accumulated total under an annuity due versus and ordinary annuity (\$33,578 vs. \$30,526).
## Accounting Ratios
Once you have the PVIFA factor value, you can multiply it by the periodic payment amount to find the current present value of the annuity. The preceding annuity table is useful as a quick reference, but only provides values for discrete time periods and interest rates that may not exactly correspond to a real-world scenario. Accordingly, use the annuity formula in an electronic spreadsheet to more precisely calculate the correct amount of the present value of an annuity due. This problem calculates the difference between the present value (PV) of an ordinary annuity and an annuity due. The timing difference in the payments is illustrated in an Excel schedule. This is illustrated graphically in the section that follows, “Visual Comparison of Cash Flows.” It can also be clearly seen in the discount and accumulation schedules constructed in the “Excel” section. The Present Value (PV) of an annuity can be found by calculating the PV of each individual payment and then summing them up. As in the case of finding the Future Value (FV) of an annuity, it is important to note when each payment occurs. Annuities-due have payments at the beginning of each period, and ordinary annuities have them at the end. | 1,605 | 7,639 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-18 | latest | en | 0.942993 |
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# Add one to nine system.
Topic closed. 81 replies. Last post 1 year ago by jstewart26.
Page 2 of 6
NE PA
United States
Member #127843
May 10, 2012
799 Posts
Online
Posted: November 16, 2012, 8:10 pm - IP Logged
Draw Mirror Total draw is 1397 break it down to Enter draw 397 2 410 7 965 8 725 enter 1 3 107 8 652 8 250 4 074 9 529 8 505 5 743 0 298 8 058 6 430 1 985 8 583 7 307 2 852 8 838 8 076 3 521 8 381 9 764 4 219 8 817 0 641 5 196 8 172
NE PA
United States
Member #127843
May 10, 2012
799 Posts
Online
Posted: November 16, 2012, 8:20 pm - IP Logged
the lines moved, crap!
if you can do two enter draws
xxx enter
x enter
then for draw, mirror, total, would be 6 rows across but give you the 4 digits each
draw,draw mirror,mirror total,total
x xxx x xxx x xxx
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 16, 2012, 8:28 pm - IP Logged
Dianne I am with you as he does not explain it clearly enough especially ysing the word total. It makes no sense.
Retrexx you should know what a total is by now as long as you have been on this site.
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 16, 2012, 8:42 pm - IP Logged
CarliG,
Can we test for Pick 4? Thanks!
It was intended for p3 but try this. Using the number 1234.
1234 1234 1234
0123 4567 8901
1357 5791 9135
Then overlap^^
1357
5757
5791
9191,7919, 9135
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 16, 2012, 9:00 pm - IP Logged
The program is intended as the original example. I used the total for NY as an example but the procedure was the same. Its simple addition. Read the first entry in the post and its very simple math and easy to understand. I will send the file to Cg to change from the current one.
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
United States
Member #1987
August 5, 2003
8812 Posts
Online
Posted: November 16, 2012, 9:51 pm - IP Logged
Can you explain this system for me?
Hello diane1248 I do not know if you received a full explanation as to how Lakerben's system works. This is how I see the system. I will use this:
Fri, Nov 16, 2012 Georgia Cash 3 Evening 0-9-3
Adding 1 through 9. Use lottery math, meaning do not carry.
093 093 093
+ 123 +456 +789
_____ ___ ___
116 449 772
Using the total. Subtract the draw # from 999.
999
- 093
___
906
______________________________________________
906 906 906
+123 +456 +789
___ ___ ___
029 352 685
Florida
United States
Member #66575
October 30, 2008
3544 Posts
Offline
Posted: November 16, 2012, 10:19 pm - IP Logged
Laker ,
I apologize for not doing exactly as you specified .. I will redo per the info you sent and upload new file
CG
United States
Member #134499
October 29, 2012
75 Posts
Offline
Posted: November 16, 2012, 10:50 pm - IP Logged
Hello EveryOne How Are You Doing Today Let Me Know If I Did This Right For Texas Pick 3 Mid 672
Texas Pick 3 672
672 672 672
+ 123 +456 + 789
_____ ___ ___
795 028 351
Using the total. Subtract the draw # from 999.
999
- 672
___
326
______________________________________________
326 326 326
+123 +456 +789
___ ___ ___
449 772 005
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 16, 2012, 10:50 pm - IP Logged
Hello diane1248 I do not know if you received a full explanation as to how Lakerben's system works. This is how I see the system. I will use this:
Fri, Nov 16, 2012 Georgia Cash 3 Evening 0-9-3
Adding 1 through 9. Use lottery math, meaning do not carry.
093 093 093
+ 123 +456 +789
_____ ___ ___
116 449 772
Using the total. Subtract the draw # from 999.
999
- 093
___
906
______________________________________________
906 906 906
+123 +456 +789
___ ___ ___
029 352 685
I used the total of the ny draw as an input and then added the 1 to 9 to it. I didn't anticipate the confusion since I know there are alot of experienced members on the site who know what a total is.
Again using 123 as the input.
1 2 3 1 2 3 1 2 3
1 2 3 4 5 6 7 8 9 <<<<<<<<<<<<<< one through nine added to the input shown in blue.
2 4 6 5 7 9 8 0 2<<<<<<<<<<<<overlap these for the picks.
246
465
657
798
802
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
Florida
United States
Member #66575
October 30, 2008
3544 Posts
Offline
Posted: November 16, 2012, 11:10 pm - IP Logged
Revised per Lakers instructions
https://www.box.com/s/ouhqq8tr7y5xl7nqayol
CG
PS. Lakerben did not intend this for pick 4, therefore I will not do a pick 4 version .
Thanks
Carlig
United States
Member #1987
August 5, 2003
8812 Posts
Online
Posted: November 16, 2012, 11:16 pm - IP Logged
I used the total of the ny draw as an input and then added the 1 to 9 to it. I didn't anticipate the confusion since I know there are alot of experienced members on the site who know what a total is.
Again using 123 as the input.
1 2 3 1 2 3 1 2 3
1 2 3 4 5 6 7 8 9 <<<<<<<<<<<<<< one through nine added to the input shown in blue.
2 4 6 5 7 9 8 0 2<<<<<<<<<<<<overlap these for the picks.
246
465
657
798
802
lakerben, when you are in a forum as vast as this with people joining everyday you must write as if everyone is a beginner. When you write in that manner you will draw more people into reading your thread & the benefit is that your readers will have a greater understanding.
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 17, 2012, 12:01 am - IP Logged
lakerben, when you are in a forum as vast as this with people joining everyday you must write as if everyone is a beginner. When you write in that manner you will draw more people into reading your thread & the benefit is that your readers will have a greater understanding.
I made the post as basic as possible with simple addition. I even highlighted the inputs etc. I think one problem is that people skim over the reading and instantly say they don't understand.
I will make a suggestion to the forum to have a beginers corner with all the fundamentals so the beginners can get up to speed before they stat posting on a regular basis.
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 17, 2012, 12:02 am - IP Logged
Revised per Lakers instructions
https://www.box.com/s/ouhqq8tr7y5xl7nqayol
CG
PS. Lakerben did not intend this for pick 4, therefore I will not do a pick 4 version .
Thanks
Carlig
Thanks!
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
Zeta Reticuli Star System
United States
Member #30470
January 17, 2006
9057 Posts
Offline
Posted: November 17, 2012, 12:32 am - IP Logged
Laverne Maloney,
I didn't have any trouble following lakerben's example in the OP.
In your example where does subtract the draw total from 999 come from, or is that just part of thus system or a separate thing?
Those who run the lotteries love it when players look for consistency in something that's designed not to have any.
There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.
New Mexico
United States
Member #86100
January 29, 2010
8323 Posts
Online
Posted: November 17, 2012, 12:49 am - IP Logged
Laverne Maloney,
I didn't have any trouble following lakerben's example in the OP.
In your example where does subtract the draw total from 999 come from, or is that just part of thus system or a separate thing?
The system is explained in the first post of the thread and CarliG has posted the file. Its basic addition to the input draw. Disregard the total.
The only percentage in gambling that exsists is 50%. You either win or lose. Any other claim is just endless babble.
Page 2 of 6 | 2,818 | 9,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-41 | longest | en | 0.617778 |
https://www.physicsforums.com/threads/disk-speed-non-uniform-and-unifor-circular-motion.66328/ | 1,508,635,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824931.84/warc/CC-MAIN-20171022003552-20171022023552-00275.warc.gz | 957,547,331 | 16,115 | # Disk speed NON UNIFORM and UNIFOR CIRCULAR MOTION
1. Mar 7, 2005
### sevens
A computer disk is 8.0cm in diameter. a refrence dot on the edge of the disk is initially located at theta= 45 degrees.
The disk accelerates steadily for 1/2 second, reaching 2000 rpm, then coasts at steady angular velocity for 1/2 second.
what is the speed of the refrence dot at 1 second. answer is in m/s.
it seems to me that the first half second of this problem could be ignored, since after the first 1/2 second we stop accelerating and are in Uniform Circular Motion.
so I thought i would use v = ( 2(pi)r ) / T.
i knew my radius was .04m.
and to find T i did ( 2000 rev) * (1 min / 60 sec) = 33.33 rev/sec
my V ended being .0075 m/s this answer is wrong and im not sure if it is because i ignored the first half second of the problem, i just dont see how that ties into finding the velocity at 1 second. i figured that if it was coasting steadily at 2000 RPM it turns into a unifor circular motion problem
:yuck:
any help is apreciated, thanks.
2. Mar 7, 2005
### Staff: Mentor
Good.
Good.
T is the period, the time for one revolution (sec/rev). You calculated the inverse of the period (rev/sec).
3. Mar 7, 2005
### ZapperZ
Staff Emeritus
You have a bit of a problem with dimensional analysis here. First of all, you're working way too hard! Secondly, T is the PERIOD of revolution, i.e. it has the dimension of time. If you look above, you have T having units of rev/sec. This is a frequency, not a period.
$$v = r \omega$$
This is one of those problem in which you could solve it if you know what each of these symbols mean PHYSICALLY. Here, $$\omega$$ means angular velocity or angular frequency, and it has units of radians/second. If an object makes one completely rotation in 1 second, then it has an angular velocity of $$2 \pi/sec$$.
Now look at yourproblem. When it is coasting with constant angular velocity, it is rotating at 2000 rpm, or rotations per minute. It means that in one minute, it is making 2000*2pi radians of rotations. This is the angular velocity $$\omega$$! If you convert the minute into seconds, you have now the angular velocity in rad/sec, just what you need to solve this problem.
Notice that what I did above made very little use of "formulas". All I did was make use of the definitions of each of the quantities involved and extract the values I need from those definitions.
Zz.
Last edited: Mar 7, 2005
4. Mar 7, 2005
### sevens
Thanks for the help
what i did was take my answer of 33.33 rev/sec and did (1/33.33rev/sec) to find the amount of time per rotation or period T. giving me .03 sec. now that i knew the right T i could solve my equation
v= (2(pi)r)/T = (2(pi)(.04m))/.03 sec = 8.38 m/s
thanks for all the help. | 750 | 2,772 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-43 | longest | en | 0.943935 |
https://newhorizons.studysixsigma.com/box-cox-transformation-with-sigmaxl/ | 1,726,178,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651498.46/warc/CC-MAIN-20240912210501-20240913000501-00198.warc.gz | 390,778,137 | 15,867 | # Box Cox Transformation with SigmaXL
### Box Cox Transformation
[unordered_list style=”star”]
• Data transforms are usually applied so that the data appear to more closely meet assumptions of a statistical inference model to be applied or to improve the interpret-ability or appearance of graphs.
• Power transformation is a class of transformation functions that raise the response to some power. For example, a square root transformation converts X to X1/2
• Box Cox transformation is a popular power transformation method developed by George E. P. Box and David Cox.
[/unordered_list]
#### Box Cox Transformation Formula
The formula of the Box Cox transformation is:
Where:
[unordered_list style=”star”]
• y is the transformation result
• x is the variable under transformation
• λ is the transformation parameter
[/unordered_list]
### Use SigmaXL to Perform a Box-Cox Transformation
SigmaXL provides the best Box-Cox transformation with an optimal λ that minimizes the model SSE (sum of squared error). Here is an example of how we transform the non-normally distributed response to normal data using Box-Cox method.
Data File: “Box-Cox” tab in “Sample Data.xlsx”
Step 1: Test the normality of the original data set.
1. Select the entire range of “Y” in column H
2. Click SigmaXL -> Graphical Tool -> Histograms & Descriptive Statistics
3. A new window named “Histograms & Descriptive” pops up and the selected range automatically appears in the box below “Please select your data”.
4. Click “Next >>”
5. A new window named “Histograms & Descriptive Statistics” pops up.
6. Select “Y” as “Numeric Data Variables (Y)”
7. Click “OK>>”
8. The analysis results are shown automatically in the new spreadsheet “Hist Descript(1)”
Normality Test:
[unordered_list style=”star”]
• H0: The data are normally distributed.
• H1: The data are not normally distributed.
[/unordered_list]
If p-value > alpha level (0.05), we fail to reject the null hypothesis. Otherwise, we reject the null. In this example, p-value = 0.029 < alpha level (0.05). The data are not normally distributed.
Step 2: Run the Box-Cox Transformation:
1. Select the entire range of Y in column H
2. Click SigmaXL -> Process Capability -> Nonnormal -> Box-Cox Transformation
3. A new window named “Box-Cox Transformation” pops up and the selected range appears automatically in the box under “Please select your data”
4. Click “Next >>”
5. A new window also named “Box-Cox Transformation” pops up.
6. Select “Y” as “Numeric Data Variables (Y)”
7. Click “OK>>”
8. The analysis results are shown automatically in the new spreadsheet “Box-Cox (1)”
The software looks for the optimal value of lambda that minimizes the SSE (Sum of Squares of Error). In this case the minimum value is 0.12. The transformed Y can also be saved in another column. The transformed Y is also listed in Column G in the newly generated tab “Box-Cox (1)
Use the Anderson–Darling test to test the normality of the transformed data
[unordered_list style=”star”]
• H0: The data are normally distributed.
• H1: The data are not normally distributed.
[/unordered_list]
Model summary: If p-value > alpha level (0.05), we fail to reject the null. Otherwise, we reject the null. In this example, p-value = 0.327 > alpha level (0.05). The data are normally distributed. | 780 | 3,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.731804 |
http://avidloanstudio.com/exact/interest-only-calculator.php | 1,529,554,638,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864022.18/warc/CC-MAIN-20180621040124-20180621060124-00207.warc.gz | 28,213,084 | 7,059 | # Interest-Only Calculator
Input Information
Loan Information Amount : (\$) Interest-Only Rate : (%) Conventional Loan Rate : (%) Length : (Yrs) Let Me Print That Form in PDF! Send calculation results to email Send to Email Address : Name : Phone # : Show Schedule Table
Financial Analysis (Switch to Plain English)
Interest-Only Loan Payment : \$989.58 Total Interest-Only : \$356,250.00 Monthly Principal & Interests : \$1,342.05 Total Principal & Interest : \$483,139.46
Plain English Help (Switch to Financial Analysis)
When applying for a mortgage loan for your home, you can choose between a standard loan and an interest only loan. With an interest only loan, you will pay only on the interest when you make your monthly payments and you will eventually be called upon to pay the principal. It is a wise financial decision to compare the two types of loans before deciding which one is best for you.If you wanted to borrow \$250,000.00 for the purchase of your home, you might be offered a standard loan with a 5.000% interest rate or an interest only loan with a 4.750% interest rate, with both being 30 year loans. With an interest only loan, your monthly payment would be \$989.58, while a standard loan would be \$1,342.05. Under this plan, the total interest only cost would be \$356,250.00, while the total standard loan cost would be \$483,139.46. | 318 | 1,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.913065 |
http://www.fit.vutbr.cz/study/course-l.php.en?id=13232 | 1,553,491,204,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203755.18/warc/CC-MAIN-20190325051359-20190325073359-00057.warc.gz | 267,644,282 | 4,949 | Title:
# Applied Evolutionary Algorithms
Code:EVO
Ac.Year:2019/2020
Sem:Summer
Curriculums:
ProgrammeField/
Specialization
YearDuty
IT-MSC-2MBI-Compulsory-Elective - group I
IT-MSC-2MBS-Elective
IT-MSC-2MGM-Elective
IT-MSC-2MIN-Elective
IT-MSC-2MIS-Elective
IT-MSC-2MMI-Elective
IT-MSC-2MMM-Elective
IT-MSC-2MPV-Compulsory-Elective - group B
IT-MSC-2MSK-Elective
Language of Instruction:Czech
Public info:http://www.fit.vutbr.cz/study/courses/EVO/public/
Credits:5
Completion:examination (written)
Type of
instruction:
Hour/semLecturesSeminar
Exercises
Laboratory
Exercises
Computer
Exercises
Other
Hours:26001214
ExamsTestsExercisesLaboratoriesOther
Points:60001822
Guarantor:Bidlo Michal, Ing., Ph.D. (DCSY)
Lecturer:Bidlo Michal, Ing., Ph.D. (DCSY)
Instructor:Hyrš Martin, Ing. (DCSY)
Faculty:Faculty of Information Technology BUT
Department:Department of Computer Systems FIT BUT
Substitute for:
Applied Evolutionary Algorithms (EVA), DCSY
Learning objectives:
Survey about actual optimization techniques and evolutionary algorithms for solution of complex, NP complete problems. To learn how to solve typical complex tasks from engineering practice using evolutionary techniques.
Description:
Overview of principles of stochastic search techniques: Monte Carlo methods, evolutionary algorithms. Detailed explanation of selected algorithms: Metropolis algorithm, simulated annealing, problems in statistical physics. Overview of basic principles of evolutionary algorithms (EA): evolutionary programming (EP), evolution strategies (ES), genetic algorithms (GA), genetic programming (GP), differential evolution (DE). Advanced evolutionary techniques: estimation of distribution algorithms (EDA), multiobjective optimization, parallel and distributed EA. Social computing algoritmhs: particle swarm optimization (PSO), ant colony optimization (ACO). Applications in engineering problems and artificial intelligence.
Learning outcomes and competencies:
Ability of problem formulation for the solution on the base of evolutionary computation. Knowledge of analysis and design methods for evolutionary algorithms.
Syllabus of lectures:
1. Introduction, terminology, principles of stochastic search algorithms.
2. Monte Carlo method and variants (Metropolis algorithm, Simulated Annealing).
3. Basic evolutionary algorithms (Evolutionary Programming, Evolution Strategies).
4. Genetic algorithms (control parameters, genetic operators).
5. Genetic programming (representation, symbolic regression, applications).
6. Evolutionary development and grammatical evolution.
7. Numerical optimization (Particle Swarm Optimization, Differential Evolution).
8. Ant Colony Optimization (basic principles, Ant System, Ant Colony System).
9. Advanced evolutionary techniques (Estimation of Distribution Algorithms).
10. Basic statistics for evolutionary computation.
11. Evaluation of evolutionary experiments.
12. Multiobjective optimization (basic techniques, engineering case study).
13. Advanced multiobjective evolutionary algorithms. Summary of the course.
Syllabus of laboratory exercises:
• Basic concepts of evolutionary computing, typical problems, solution of a technical task using a variant of Metropolis algorithm.
• Evolutionary algorithms in engineering areas, optimization of electronic circuits using genetic algorithm.
• Evolutionary design using genetic programming.
• Differential evolution-based optimization of neural networks.
• Edge detection based on ant algorithms.
• Solution of selected task from statistical physics.
Syllabus - others, projects and individual work of students:
Conducting and evaluation of experiment regarding application of evolutionary algorithms to solve a selected task.
By agreement there is a possibility to include solution of the project from other course (e.g. BIN) to EVO if its topic belongs to evolutionary computation.
Fundamental literature:
• Brabazon, A., O'Neill, M., McGarraghy, S.: Natural Computing Algorithms. Springer-Verlag Berlin Heidelberg, 2015, ISBN 978-3-662-43630-1
• Eiben, A.E., Smith, J.E.: Introduction to Evolutionary Computing, 2nd ed. Springer-Verlag Berlin Heidelberg, 2015, ISBN 978-3-662-44873-1
• Jansen, T.: Analyzing Evolutionary Algorithms. Springer-Verlag, Berlin Heidelberg, 2013, ISBN 978-3-642-17338-7
• Talbi, E.-G.: Metaheuristics: From Design to Implementation. Wiley, Hoboken, New Jersey, 2009, ISBN 978-0-470-27858-1
• Bäck, T.: Evolutionary Algorithms in Theory and Practice. Oxford University Press, Oxford, 1996, ISBN 978-0195099713
Study literature:
• Luke, S.: Essentials of Metaheuristics. Lulu, 2015, ISBN 978-1-300-54962-8
• Jansen, T.: Analyzing Evolutionary Algorithms. Springer-Verlag, Berlin Heidelberg, 2013, ISBN 978-3-642-17338-7
• Kvasnička, V., Pospíchal, J., Tiňo, P.: Evolučné algoritmy. STU Bratislava, Bratislava, 2000, ISBN 80-227-1377-5
• Oplatková, Z., Ošmera, P., Šeda, M., Včelař, F., Zelinka, I.: Evoluční výpočetní techniky - principy a aplikace. BEN - technická literatura, Praha, 2008, ISBN 80-7300-218-3
Progress assessment:
Evaluated practices, project.
Exam prerequisites:
None. | 1,238 | 5,107 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-13 | latest | en | 0.708979 |
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